url stringlengths 14 2.42k | text stringlengths 100 1.02M | date stringlengths 19 19 | metadata stringlengths 1.06k 1.1k |
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https://brilliant.org/problems/cant-resist-from-determining-resistance/ | # Can't resist from determining resistance
Chemistry Level pending
Resistance of $$0.2\text{ M}$$ solution on an electrolyte is 50 ohm. The specific conductance of the solution is $$1.3 \text{ s/m}$$. If the resistance of the 0.4 M solution of the same electrolyte is 260 ohm, its molar conductivity is
Answer in decimal upto 6 digits after decimal.
× | 2017-05-25 16:23:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6500741839408875, "perplexity": 2462.9735361962776}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463608107.28/warc/CC-MAIN-20170525155936-20170525175936-00354.warc.gz"} |
https://aptitude.gateoverflow.in/6037/cat-2018-set-1-question-74 | 354 views
If among $200$ students, $105$ like pizza and $134$ like burger, then the number of students who like only burger can possibly be
1. $93$
2. $26$
3. $23$
4. $96$
Given that,
• $n (\text {U}) = 200$
• $n (\text {P}) = 105$
• $n ( \text {B}) = 134$
Let the number of students who like both pizza and burger be $m.$
And, let the number of students who like neither pizza nor burger be $n.$
From the above Venn diagram,
$( 105-m) + m + (134-m) + n = 200$
$\Rightarrow – m + n = 200 – 239$
$\Rightarrow \boxed{m-n = 39} \quad \longrightarrow (1)$
$\therefore$ The possible value of $(m , n)$ are $(39,0), (40-1), \dots , (104,65), (105,66)$
So, the number of students who like only burger, should be in the range.
$[134-105, 134-39]$
$= [29,95]$
$\therefore$ From the given options, $93$ can be possible.
Correct Answer $:\text {A}$
10.3k points
1 | 2022-12-06 14:13:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8727734088897705, "perplexity": 1885.1439425832277}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711108.34/warc/CC-MAIN-20221206124909-20221206154909-00588.warc.gz"} |
https://elkement.wordpress.com/tag/complex-analysis/ | # Complex Alien Eclipse
My colorful complex function lived in a universe of white light. I turned off the light. Turned it into its negative. Expected it to look bleak. Like thin white bones on black canvas, cartoon skeletons of imaginary alien creatures. But it is more like the total solar eclipse I watched in 1999. There is interference, … Continue reading Complex Alien Eclipse
# Spins, Rotations, and the Beauty of Complex Numbers
This is a simple quantum state ... |➚> = α|↑> + β|↓> ... built from an up |↑> state and a down state |↓>. α and β are complex numbers. The result |➚> is in the middle, oblique. The oblique state is a superposition or the up and down base states. Making a measurement, you … Continue reading Spins, Rotations, and the Beauty of Complex Numbers
# Lines and Circles
I poked at complex function 1/z, and its real and imaginary parts look like magical towers. When you look at these towers from above or below, you see sections of perfect circles. This is hinting at some underlying simplicity. Using the map 1/z, another complex number - w=1/z - is mapped to z. Four dimensions … Continue reading Lines and Circles
# Reality and Imagination
Grey and colorful. Cutting through each other. Chasing each other. Meeting in the center, leaning on each other, forming an infinite line. ~ Reality and Imagination: Real and imaginary part of complex function 1/z: ~ The real part of 1/z is painted in shades of grey, the imaginary part in rainbow colors. Plots are created … Continue reading Reality and Imagination
# Super Motivational Function
I've presented a Motivational Function, a while back. $latex f(z) = e^{\left(-\frac{1}{z^{2}}\right)}&s=3$ It is infinitely flat at the zero point: all its derivatives are zero there. Yet, it manages to lift its head - as it is not analytic at zero! If you think of it as a function of a complex argument, its … Continue reading Super Motivational Function
# Motivational Function
Deadly mutants are after us. What can give us hope? This innocuous-looking function is a sublime light in the dark. It proves you can always recover. If your perseverance is infinite. $latex e^{\left(-\frac{1}{x^{2}}\right)}&s=3$ As x tends to zero, the exponent tends to minus infinity. The function's value at zero tends to zero. It is … Continue reading Motivational Function | 2021-12-02 09:59:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3996620178222656, "perplexity": 1648.381288981934}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964361253.38/warc/CC-MAIN-20211202084644-20211202114644-00639.warc.gz"} |
http://www.physicsforums.com/showthread.php?p=4216012 | ## complex equation
Hi, I need a little help
I need to find solution for this equations:
$\frac{Z-a}{Z-b}$=K$e^{±jθ}$
The Z is unknown and it is the complex number. The a and b is known and they are also complex numbers. K is the real number.
I know that for $-90^{°}$<θ<$90^{°}$ the graph in the complex plane is circle, for $-45^{°}$<θ<$45^{°}$ the graph in the complex plane is in shape of "tomato" and for $-135^{°}$<θ<$135^{°}$ is shape of "lens", but I don't know how to solve it.
Sorry if my post is in wrong area.
Thanks for help.
PhysOrg.com mathematics news on PhysOrg.com >> Pendulum swings back on 350-year-old mathematical mystery>> Bayesian statistics theorem holds its own - but use with caution>> Math technique de-clutters cancer-cell data, revealing tumor evolution, treatment leads
Leaving it to you the conditions of existence: $Z=\frac{a-b.K.\textrm{e}^{ \pm j \theta }}{1-K.\textrm{e}^{ \pm j \theta }}$
In that way I got only the one solution, where are the other? For example, let's put b=0, K=1, theta=45°, with above formula we got only the one solution, but there is more than one solution...
Recognitions:
Gold Member
## complex equation
How do you only get one solution when there's clearly a $\pm$ in his answer?
Similar discussions for: complex equation Thread Forum Replies Calculus & Beyond Homework 8 Calculus & Beyond Homework 1 Calculus 9 General Math 3 Calculus & Beyond Homework 2 | 2013-06-19 14:52:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6673812866210938, "perplexity": 1590.4346886970548}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368708808767/warc/CC-MAIN-20130516125328-00046-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://people.maths.bris.ac.uk/~matyd/GroupNames/217/C3xC5s2C16.html | Copied to
clipboard
G = C3×C5⋊2C16order 240 = 24·3·5
Direct product of C3 and C5⋊2C16
Aliases: C3×C52C16, C52C48, C154C16, C30.4C8, C40.2C6, C24.4D5, C20.5C12, C10.2C24, C120.5C2, C60.12C4, C12.5Dic5, C8.2(C3×D5), C6.2(C52C8), C4.2(C3×Dic5), C2.(C3×C52C8), SmallGroup(240,2)
Series: Derived Chief Lower central Upper central
Derived series C1 — C5 — C3×C5⋊2C16
Chief series C1 — C5 — C10 — C20 — C40 — C120 — C3×C5⋊2C16
Lower central C5 — C3×C5⋊2C16
Upper central C1 — C24
Generators and relations for C3×C52C16
G = < a,b,c | a3=b5=c16=1, ab=ba, ac=ca, cbc-1=b-1 >
Smallest permutation representation of C3×C52C16
Regular action on 240 points
Generators in S240
(1 171 67)(2 172 68)(3 173 69)(4 174 70)(5 175 71)(6 176 72)(7 161 73)(8 162 74)(9 163 75)(10 164 76)(11 165 77)(12 166 78)(13 167 79)(14 168 80)(15 169 65)(16 170 66)(17 38 138)(18 39 139)(19 40 140)(20 41 141)(21 42 142)(22 43 143)(23 44 144)(24 45 129)(25 46 130)(26 47 131)(27 48 132)(28 33 133)(29 34 134)(30 35 135)(31 36 136)(32 37 137)(49 205 96)(50 206 81)(51 207 82)(52 208 83)(53 193 84)(54 194 85)(55 195 86)(56 196 87)(57 197 88)(58 198 89)(59 199 90)(60 200 91)(61 201 92)(62 202 93)(63 203 94)(64 204 95)(97 236 149)(98 237 150)(99 238 151)(100 239 152)(101 240 153)(102 225 154)(103 226 155)(104 227 156)(105 228 157)(106 229 158)(107 230 159)(108 231 160)(109 232 145)(110 233 146)(111 234 147)(112 235 148)(113 217 177)(114 218 178)(115 219 179)(116 220 180)(117 221 181)(118 222 182)(119 223 183)(120 224 184)(121 209 185)(122 210 186)(123 211 187)(124 212 188)(125 213 189)(126 214 190)(127 215 191)(128 216 192)
(1 122 200 151 129)(2 130 152 201 123)(3 124 202 153 131)(4 132 154 203 125)(5 126 204 155 133)(6 134 156 205 127)(7 128 206 157 135)(8 136 158 207 113)(9 114 208 159 137)(10 138 160 193 115)(11 116 194 145 139)(12 140 146 195 117)(13 118 196 147 141)(14 142 148 197 119)(15 120 198 149 143)(16 144 150 199 121)(17 108 84 219 164)(18 165 220 85 109)(19 110 86 221 166)(20 167 222 87 111)(21 112 88 223 168)(22 169 224 89 97)(23 98 90 209 170)(24 171 210 91 99)(25 100 92 211 172)(26 173 212 93 101)(27 102 94 213 174)(28 175 214 95 103)(29 104 96 215 176)(30 161 216 81 105)(31 106 82 217 162)(32 163 218 83 107)(33 71 190 64 226)(34 227 49 191 72)(35 73 192 50 228)(36 229 51 177 74)(37 75 178 52 230)(38 231 53 179 76)(39 77 180 54 232)(40 233 55 181 78)(41 79 182 56 234)(42 235 57 183 80)(43 65 184 58 236)(44 237 59 185 66)(45 67 186 60 238)(46 239 61 187 68)(47 69 188 62 240)(48 225 63 189 70)
(1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16)(17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32)(33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48)(49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64)(65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80)(81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96)(97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112)(113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128)(129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144)(145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160)(161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176)(177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192)(193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208)(209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224)(225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240)
G:=sub<Sym(240)| (1,171,67)(2,172,68)(3,173,69)(4,174,70)(5,175,71)(6,176,72)(7,161,73)(8,162,74)(9,163,75)(10,164,76)(11,165,77)(12,166,78)(13,167,79)(14,168,80)(15,169,65)(16,170,66)(17,38,138)(18,39,139)(19,40,140)(20,41,141)(21,42,142)(22,43,143)(23,44,144)(24,45,129)(25,46,130)(26,47,131)(27,48,132)(28,33,133)(29,34,134)(30,35,135)(31,36,136)(32,37,137)(49,205,96)(50,206,81)(51,207,82)(52,208,83)(53,193,84)(54,194,85)(55,195,86)(56,196,87)(57,197,88)(58,198,89)(59,199,90)(60,200,91)(61,201,92)(62,202,93)(63,203,94)(64,204,95)(97,236,149)(98,237,150)(99,238,151)(100,239,152)(101,240,153)(102,225,154)(103,226,155)(104,227,156)(105,228,157)(106,229,158)(107,230,159)(108,231,160)(109,232,145)(110,233,146)(111,234,147)(112,235,148)(113,217,177)(114,218,178)(115,219,179)(116,220,180)(117,221,181)(118,222,182)(119,223,183)(120,224,184)(121,209,185)(122,210,186)(123,211,187)(124,212,188)(125,213,189)(126,214,190)(127,215,191)(128,216,192), (1,122,200,151,129)(2,130,152,201,123)(3,124,202,153,131)(4,132,154,203,125)(5,126,204,155,133)(6,134,156,205,127)(7,128,206,157,135)(8,136,158,207,113)(9,114,208,159,137)(10,138,160,193,115)(11,116,194,145,139)(12,140,146,195,117)(13,118,196,147,141)(14,142,148,197,119)(15,120,198,149,143)(16,144,150,199,121)(17,108,84,219,164)(18,165,220,85,109)(19,110,86,221,166)(20,167,222,87,111)(21,112,88,223,168)(22,169,224,89,97)(23,98,90,209,170)(24,171,210,91,99)(25,100,92,211,172)(26,173,212,93,101)(27,102,94,213,174)(28,175,214,95,103)(29,104,96,215,176)(30,161,216,81,105)(31,106,82,217,162)(32,163,218,83,107)(33,71,190,64,226)(34,227,49,191,72)(35,73,192,50,228)(36,229,51,177,74)(37,75,178,52,230)(38,231,53,179,76)(39,77,180,54,232)(40,233,55,181,78)(41,79,182,56,234)(42,235,57,183,80)(43,65,184,58,236)(44,237,59,185,66)(45,67,186,60,238)(46,239,61,187,68)(47,69,188,62,240)(48,225,63,189,70), (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16)(17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32)(33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48)(49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64)(65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80)(81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96)(97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112)(113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,128)(129,130,131,132,133,134,135,136,137,138,139,140,141,142,143,144)(145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,160)(161,162,163,164,165,166,167,168,169,170,171,172,173,174,175,176)(177,178,179,180,181,182,183,184,185,186,187,188,189,190,191,192)(193,194,195,196,197,198,199,200,201,202,203,204,205,206,207,208)(209,210,211,212,213,214,215,216,217,218,219,220,221,222,223,224)(225,226,227,228,229,230,231,232,233,234,235,236,237,238,239,240)>;
G:=Group( (1,171,67)(2,172,68)(3,173,69)(4,174,70)(5,175,71)(6,176,72)(7,161,73)(8,162,74)(9,163,75)(10,164,76)(11,165,77)(12,166,78)(13,167,79)(14,168,80)(15,169,65)(16,170,66)(17,38,138)(18,39,139)(19,40,140)(20,41,141)(21,42,142)(22,43,143)(23,44,144)(24,45,129)(25,46,130)(26,47,131)(27,48,132)(28,33,133)(29,34,134)(30,35,135)(31,36,136)(32,37,137)(49,205,96)(50,206,81)(51,207,82)(52,208,83)(53,193,84)(54,194,85)(55,195,86)(56,196,87)(57,197,88)(58,198,89)(59,199,90)(60,200,91)(61,201,92)(62,202,93)(63,203,94)(64,204,95)(97,236,149)(98,237,150)(99,238,151)(100,239,152)(101,240,153)(102,225,154)(103,226,155)(104,227,156)(105,228,157)(106,229,158)(107,230,159)(108,231,160)(109,232,145)(110,233,146)(111,234,147)(112,235,148)(113,217,177)(114,218,178)(115,219,179)(116,220,180)(117,221,181)(118,222,182)(119,223,183)(120,224,184)(121,209,185)(122,210,186)(123,211,187)(124,212,188)(125,213,189)(126,214,190)(127,215,191)(128,216,192), (1,122,200,151,129)(2,130,152,201,123)(3,124,202,153,131)(4,132,154,203,125)(5,126,204,155,133)(6,134,156,205,127)(7,128,206,157,135)(8,136,158,207,113)(9,114,208,159,137)(10,138,160,193,115)(11,116,194,145,139)(12,140,146,195,117)(13,118,196,147,141)(14,142,148,197,119)(15,120,198,149,143)(16,144,150,199,121)(17,108,84,219,164)(18,165,220,85,109)(19,110,86,221,166)(20,167,222,87,111)(21,112,88,223,168)(22,169,224,89,97)(23,98,90,209,170)(24,171,210,91,99)(25,100,92,211,172)(26,173,212,93,101)(27,102,94,213,174)(28,175,214,95,103)(29,104,96,215,176)(30,161,216,81,105)(31,106,82,217,162)(32,163,218,83,107)(33,71,190,64,226)(34,227,49,191,72)(35,73,192,50,228)(36,229,51,177,74)(37,75,178,52,230)(38,231,53,179,76)(39,77,180,54,232)(40,233,55,181,78)(41,79,182,56,234)(42,235,57,183,80)(43,65,184,58,236)(44,237,59,185,66)(45,67,186,60,238)(46,239,61,187,68)(47,69,188,62,240)(48,225,63,189,70), (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16)(17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32)(33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48)(49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64)(65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80)(81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96)(97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112)(113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,128)(129,130,131,132,133,134,135,136,137,138,139,140,141,142,143,144)(145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,160)(161,162,163,164,165,166,167,168,169,170,171,172,173,174,175,176)(177,178,179,180,181,182,183,184,185,186,187,188,189,190,191,192)(193,194,195,196,197,198,199,200,201,202,203,204,205,206,207,208)(209,210,211,212,213,214,215,216,217,218,219,220,221,222,223,224)(225,226,227,228,229,230,231,232,233,234,235,236,237,238,239,240) );
G=PermutationGroup([(1,171,67),(2,172,68),(3,173,69),(4,174,70),(5,175,71),(6,176,72),(7,161,73),(8,162,74),(9,163,75),(10,164,76),(11,165,77),(12,166,78),(13,167,79),(14,168,80),(15,169,65),(16,170,66),(17,38,138),(18,39,139),(19,40,140),(20,41,141),(21,42,142),(22,43,143),(23,44,144),(24,45,129),(25,46,130),(26,47,131),(27,48,132),(28,33,133),(29,34,134),(30,35,135),(31,36,136),(32,37,137),(49,205,96),(50,206,81),(51,207,82),(52,208,83),(53,193,84),(54,194,85),(55,195,86),(56,196,87),(57,197,88),(58,198,89),(59,199,90),(60,200,91),(61,201,92),(62,202,93),(63,203,94),(64,204,95),(97,236,149),(98,237,150),(99,238,151),(100,239,152),(101,240,153),(102,225,154),(103,226,155),(104,227,156),(105,228,157),(106,229,158),(107,230,159),(108,231,160),(109,232,145),(110,233,146),(111,234,147),(112,235,148),(113,217,177),(114,218,178),(115,219,179),(116,220,180),(117,221,181),(118,222,182),(119,223,183),(120,224,184),(121,209,185),(122,210,186),(123,211,187),(124,212,188),(125,213,189),(126,214,190),(127,215,191),(128,216,192)], [(1,122,200,151,129),(2,130,152,201,123),(3,124,202,153,131),(4,132,154,203,125),(5,126,204,155,133),(6,134,156,205,127),(7,128,206,157,135),(8,136,158,207,113),(9,114,208,159,137),(10,138,160,193,115),(11,116,194,145,139),(12,140,146,195,117),(13,118,196,147,141),(14,142,148,197,119),(15,120,198,149,143),(16,144,150,199,121),(17,108,84,219,164),(18,165,220,85,109),(19,110,86,221,166),(20,167,222,87,111),(21,112,88,223,168),(22,169,224,89,97),(23,98,90,209,170),(24,171,210,91,99),(25,100,92,211,172),(26,173,212,93,101),(27,102,94,213,174),(28,175,214,95,103),(29,104,96,215,176),(30,161,216,81,105),(31,106,82,217,162),(32,163,218,83,107),(33,71,190,64,226),(34,227,49,191,72),(35,73,192,50,228),(36,229,51,177,74),(37,75,178,52,230),(38,231,53,179,76),(39,77,180,54,232),(40,233,55,181,78),(41,79,182,56,234),(42,235,57,183,80),(43,65,184,58,236),(44,237,59,185,66),(45,67,186,60,238),(46,239,61,187,68),(47,69,188,62,240),(48,225,63,189,70)], [(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16),(17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32),(33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48),(49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64),(65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80),(81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96),(97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112),(113,114,115,116,117,118,119,120,121,122,123,124,125,126,127,128),(129,130,131,132,133,134,135,136,137,138,139,140,141,142,143,144),(145,146,147,148,149,150,151,152,153,154,155,156,157,158,159,160),(161,162,163,164,165,166,167,168,169,170,171,172,173,174,175,176),(177,178,179,180,181,182,183,184,185,186,187,188,189,190,191,192),(193,194,195,196,197,198,199,200,201,202,203,204,205,206,207,208),(209,210,211,212,213,214,215,216,217,218,219,220,221,222,223,224),(225,226,227,228,229,230,231,232,233,234,235,236,237,238,239,240)])
C3×C52C16 is a maximal subgroup of
C15⋊C32 D152C16 C40.52D6 D30.5C8 C5⋊D48 D24.D5 Dic12⋊D5 C5⋊Dic24 D5×C48
96 conjugacy classes
class 1 2 3A 3B 4A 4B 5A 5B 6A 6B 8A 8B 8C 8D 10A 10B 12A 12B 12C 12D 15A 15B 15C 15D 16A ··· 16H 20A 20B 20C 20D 24A ··· 24H 30A 30B 30C 30D 40A ··· 40H 48A ··· 48P 60A ··· 60H 120A ··· 120P order 1 2 3 3 4 4 5 5 6 6 8 8 8 8 10 10 12 12 12 12 15 15 15 15 16 ··· 16 20 20 20 20 24 ··· 24 30 30 30 30 40 ··· 40 48 ··· 48 60 ··· 60 120 ··· 120 size 1 1 1 1 1 1 2 2 1 1 1 1 1 1 2 2 1 1 1 1 2 2 2 2 5 ··· 5 2 2 2 2 1 ··· 1 2 2 2 2 2 ··· 2 5 ··· 5 2 ··· 2 2 ··· 2
96 irreducible representations
dim 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 type + + + - image C1 C2 C3 C4 C6 C8 C12 C16 C24 C48 D5 Dic5 C3×D5 C5⋊2C8 C3×Dic5 C5⋊2C16 C3×C5⋊2C8 C3×C5⋊2C16 kernel C3×C5⋊2C16 C120 C5⋊2C16 C60 C40 C30 C20 C15 C10 C5 C24 C12 C8 C6 C4 C3 C2 C1 # reps 1 1 2 2 2 4 4 8 8 16 2 2 4 4 4 8 8 16
Matrix representation of C3×C52C16 in GL3(𝔽241) generated by
225 0 0 0 15 0 0 0 15
,
1 0 0 0 51 240 0 1 0
,
111 0 0 0 160 102 0 68 81
G:=sub<GL(3,GF(241))| [225,0,0,0,15,0,0,0,15],[1,0,0,0,51,1,0,240,0],[111,0,0,0,160,68,0,102,81] >;
C3×C52C16 in GAP, Magma, Sage, TeX
C_3\times C_5\rtimes_2C_{16}
% in TeX
G:=Group("C3xC5:2C16");
// GroupNames label
G:=SmallGroup(240,2);
// by ID
G=gap.SmallGroup(240,2);
# by ID
G:=PCGroup([6,-2,-3,-2,-2,-2,-5,36,50,69,6917]);
// Polycyclic
G:=Group<a,b,c|a^3=b^5=c^16=1,a*b=b*a,a*c=c*a,c*b*c^-1=b^-1>;
// generators/relations
Export
×
𝔽 | 2020-01-23 04:51:42 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9910265207290649, "perplexity": 2358.211665430595}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250608295.52/warc/CC-MAIN-20200123041345-20200123070345-00272.warc.gz"} |
https://electronics.stackexchange.com/questions/443157/what-is-the-forward-beta-of-the-bjt-kn2222a?noredirect=1 | # What is the forward beta of the BJT KN2222A?
I don't know what is the forward beta of the bipolar junction transistor KN2222A, I'm an amateur and i don't found the value in the data sheet or i just don't understand where. Please let me know
• Have a look in the datasheet: buchangelec.com/data/transistor/general/2222a.pdf on page 2 there's a section: "DC Current Gain" which is $h_{FE}$ which is also $\beta$. There are mostly minimum values there. For hobby projects you could just assume $\beta$ = 50 and that would fit the needs of 99% of generally used circuits. If your circuit relies on the BJT having a specific value of $\beta$ then you should use a better (less $\beta$ dependent) circuit. Jun 12 '19 at 6:48
• @Bimpelrekkie Thank you! Jun 12 '19 at 6:53
• Each manufacturer who applies such a part number can decide how to answer that question. Sometimes a specification covers multiple manufacturers (2N2222A is such a specification), but probably not KN2222A. Jun 12 '19 at 6:56
• That datasheet is poorly specified. It relies upon $V_\text{CE}=10\:\text{V}$ and it doesn't show the values for different temperatures (which is more important to know.)
– jonk
Jun 12 '19 at 6:57
And for what it's worth there are multiple $$\\beta\$$ factors to consider when designing BJT circuits:
• $$\\beta\$$, $$\\beta_{DC}\$$, $$\h_{FE}\$$ - These refer to the BJT's DC forward current gain when the BJT is operating in forward-active mode (small signal amplification). Used when performing the DC design/analysis.
(n.b. The 'FE' refers to Forward current gain, common-Emitter configuration.)
(n.b. Parameters written with UPPERCASE subscripts typically refer to DC parameters.)
• $$\\beta_{ac}\$$, $$\h_{fe}\$$ - These refer to the BJT's AC forward current gain when the BJT is operating in forward-active mode (small signal amplification). Used when performing the AC design/analysis. Typically, $$\\beta_{ac}\ll\beta_{DC}\$$.
(n.b. Parameters written with lowercase subscripts typically refer to AC parameters.)
• $$\\beta_{sat}\$$ - Refers to the BJT's forward current gain when the BJT is operating in "hard" saturation mode (when the BJT is turned ON fully). Typical values are $$\5 \le \beta_{sat} \le 30\$$ with $$\\beta_{sat}=I_{C(sat)}/I_{B(sat)}=10\$$ being a fairly common value for low power and medium power transistors.
(n.b. Data sheets often use lowercase text for parenthetical text—e.g., $$\I_{C(sat)}\$$, $$\V_{BE(sat)}\$$, etc. This does not indicate or imply that these are AC parameters.)
(n.b. $$\\beta_{sat}\$$ is not an AC parameter because the transistor is not performing small-signal AC amplification when it is operating in saturation mode.)
And as a general rule do not use $$\\beta\$$, $$\\beta_{DC}\$$, $$\h_{FE}\$$ to perform saturation calculations. If you do, the BJT will likely operate in "soft" saturation—i.e., in the transition region between the BJT's forward-active mode and its "hard" saturation (fully ON) mode. | 2021-10-23 18:11:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7142630219459534, "perplexity": 2158.5913231052346}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585737.45/warc/CC-MAIN-20211023162040-20211023192040-00171.warc.gz"} |
http://www.koreascience.or.kr/article/JAKO200413842028669.page?lang=ko | # AN EMBEDDING OF BIRGET-RHODES EXPANSION OF GROUPS INTO A SEMIDIRECT PRODUCT
• Choi, Keun-Bae ;
• Lim, Yong-Do
• 발행 : 2004.11.01
• 53 8
#### 초록
In this paper, we prove that the Birget-Rhodes expansion $\={G}^R$ of a group G is not a semi direct product of a semilattice by a group but it can be nicely embedded into such a semi direct product.
#### 키워드
Birget-Rhodes expansion
#### 참고문헌
1. J. -C. Birget and J. Rhodes, Almost finite expansions of arbitrary semigroups, J. Pure Appl. Algebra 32 (1984), 239–287
2. J. -C. Birget and J. Rhodes, Group Theory via Global Semigroup Theory, J. Algebra 120 (1989), 284–300
3. K. Choi and Y. Lim, Inverse monoids of $M{\"{o}}bius$ type, J. Algebra 223 (2000), 283–294
4. K. Choi and Y. Lim, Birget-Rhodes Expansion of Groups and Inverse monoids of $M{\"{o}}bius$ type, International J. of Algebra and omputation 12 (2002), no. 4, 525–533
5. J. M. Howie, An introduction to semigroup theory, Academic Press, San Diego, 1976
6. R. Exel, Partial actions of groups and actions of inverse semigroups, Proc. Amer. Math. Soc. 126 (1998), 3481–3494
7. J. Kellendonk and M. V. Lawson, Partial actions of groups, their globalizations, and E-unitary inverse semigroups, Preprint
8. M. V. Lawson, Almost factorizsable inverse semigroups, Glasgow Math. J. 36 (1994), 97–111
9. M. V. Lawson, The $M{\"{o}}bius$ inverse monoid, J. Algebra 200 (1998), 428–438
10. M. B. Szendrei, A note on Birget-Rhodes Expansion of Groups, J. Pure Appl. Algebra 58 (1989), 93–99 | 2019-07-23 23:15:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.41789141297340393, "perplexity": 1938.3884904694987}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195529737.79/warc/CC-MAIN-20190723215340-20190724001340-00105.warc.gz"} |
http://mathhelpforum.com/differential-geometry/176550-orthogonal-complement-1-a.html | # Math Help - Orthogonal Complement 1
1. ## Orthogonal Complement 1
Dear Colleagues,
Show that $Y=\{x| \ x=(x_{j})\in \ell^{2}, x_{2n}=0, n=1,2,3,...\}$ is a closed subspace of $\ell^{2}$ and find $Y^{\bot}$.
What is $Y^{\bot}$ if $Y=span\{e_{1},...,e_{n}\}\subset \ell^{2}$, where $e_{j}=(\delta _{jk})$?
Regards,
Raed.
2. We define a inner product in $l^2$ by $\displaystyle \langle (x_j),(y_j)\rangle := \sum_{j=0}^{+\infty}\overline{x_j}y_j$. By linearity, to be orthogonal to $Y$ is equivalent to by orthogonal to each $e_i$, $i=1,\ldots, n$.
3. Thank you very much for your reply but how can I prove that Y with zero even terms is closed and what its orthogonal complement.
4. Let $f : (x_0,\ldots,x_n,\ldots)\mapsto (x_0,x_2,\ldots,x_{2n},\ldots )$. Show that $f$ is linear and continuous.
5. Thank you very much for your reply. But why this function.
6. What about $\ker f$ ? | 2015-05-30 13:54:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8428494930267334, "perplexity": 235.60638167177353}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207931085.38/warc/CC-MAIN-20150521113211-00037-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://www.wyzant.com/resources/answers/topics/maximum-and-minimum-value | 36 Answered Questions for the topic Maximum And Minimum Value
03/26/18
11/01/17
#### word problem involving the maximum or minimum of a quadratic function
A supply company manufactures copy machines. The unit cost C (the cost in dollars to make each copy machine) depends on the number of machines made. If x machines are made, then the unit cost is... more
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#### Maximums and Revenue
The demand function for a product is p=32-2q where p is the price in dollars when q units are demanded. Find the level of production that maximizes the total revenue and determine the maximum... more
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#### word problem.
Among all rectangles that have a perimeter of 134, find the dimensions of the one whose area is largest. Write your answers as fractions reduced to lowest terms.
09/12/17
#### Would it be unusual for one of the test subjects to have a pulse rate of 127.8 beats per minute?
A certain group of test subjects had pulse rates with a mean of 84.6 beats per minute and a standard deviation of 11.6 beats per min. For some reason, I keep getting the incorrect answer for... more
05/28/17
#### let f(x)=sin(x)+cos(x)+tan(x)+arcsin(x)+arccos(x)+arctan(x). if M and m are maximum and minimum values of f(x), then thier arithmetic mean is equal to
let f(x)=sin(x)+cos(x)+tan(x)+arcsin(x)+arccos(x)+arctan(x). if m&m are maximum and minimum values of f(x), then thier arithmetic mean is equal to
12/02/16
#### what is 4x^3-88x^2+480x
graph the function
10/30/16
#### -10x+100x+6000
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#### How to find the maximum area of a rectangle?
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#### for what real value of r will s have its maximum value?
If s=19-(5+r)3 , for what real value of r will s have its maximum value?
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#### Finding maximum on curve
Find the maximum of f(x,y)=xy on the constraint curve (x+1)2+y2=1.
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#### Demand curve problem
The demand curve for a product is given by q equals 1000 minus 2 p^ 2 where p is the price. Find the price that maximizes revenue for sales of this product.
01/15/16
#### Prove or disprove: (a) h + g has a relative maximum at x0 (b) h-g has a relative maximum at x0.?
Let h and g have relative maxima at x0.
12/10/15
#### What is the maximum possible area of the corral?
The back of Dantes property is a creek. Dante would like to enclose a rectangular area using the creek as one side and fencing for the other three sides to create a corral. If there is 420 feet of... more
10/21/15
#### Calculus Homework- Function & Absolute Max and Min
Consider the function f(x)=2x2−2x+6, 0≤x≤6. The absolute maximum of f(x) (on the given interval) is _________ and the absolute minimum of f(x) (on the given interval) is __________.
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06/21/15
#### Calc help please: Applied Max/Min
Applied Max/Min (35 points)a. Find the dimensions of the rectangle with the greatest area that can be built so the base of the rectangle is on the x–axis between 0 and 1 (0 ≤ x ≤ 1) and one corner... more
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#### maximum height and horizontal distance
a football player punts a football at an inclination of 45 degrees to the horizontal at an initial velocity of 60 feet per second. the height of the football is given by the quadratic function... more
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Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. | 2020-04-07 02:42:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5362499356269836, "perplexity": 1256.479760329471}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371665328.87/warc/CC-MAIN-20200407022841-20200407053341-00558.warc.gz"} |
https://portonmath.wordpress.com/2016/06/23/about-each-regular-paratopological-group-is-completely-regular-article/ | In this blog post I consider my attempt to rewrite the article “Each regular paratopological group is completely regular” by Taras Banakh, Alex Ravsky in a more abstract way using my theory of reloids and funcoids.
The following is a general comment about reloids and funcoids as defined in my book. If you don’t understand them, restrict your mind to the special case ${f}$ to be a quasi-uniform space and ${(\mathsf{FCD}) f}$ is the corresponding quasi-proximity.
${\langle f \rangle^{\ast}}$ is the closure operator corresponding to a funcoid ${f}$. I also denote the image ${\mathscr{P} X \rightarrow \mathscr{P} Y}$ of a function ${f : X \rightarrow Y}$ as ${\langle f \rangle^{\ast}}$.
I will also denote ${f^{\circ}}$ the interior funcoid for a co-complete funcoid ${f}$ (for the special case if ${f}$ is a topological space ${f^{\circ}}$ is the interior operator of this space). It is defined in the file addons.pdf (not yet in my book).
By definition (slightly generalizing the special case if ${f}$ is a quasi-uniform space) an endo-reloid ${f}$ on a set ${U}$ is normal when ${\langle (\mathsf{FCD}) f \rangle^{\ast} A \sqsubseteq \langle ((\mathsf{FCD}) f)^{\circ} \rangle^{\ast} \langle (\mathsf{FCD}) f \rangle^{\ast} \langle F \rangle^{\ast} A}$ for every entourage ${F \in \mathrm{up}\, f}$ of ${f}$ and every set ${A \subseteq U}$.
Then it appear “obvious” that this definition of normality is equivalent to the formula:
$\displaystyle (\mathsf{FCD}) f \sqsubseteq ((\mathsf{FCD}) f)^{\circ} \circ (\mathsf{FCD}) f \circ (\mathsf{FCD}) f.$
However, I have failed to prove it. Here is my attempt
$\langle ((\mathsf{FCD}) f)^{\circ} \rangle^{\ast} \langle (\mathsf{FCD}) f \rangle^{\ast} \langle (\mathsf{FCD}) f \rangle^{\ast} A = \\ \langle ((\mathsf{FCD}) f)^{\circ} \rangle^{\ast} \langle (\mathsf{FCD}) f \rangle^{\ast} \bigsqcap_{F \in \mathrm{up} f} \langle F \rangle^{\ast} A = \\ \langle ((\mathsf{FCD}) f)^{\circ} \rangle^{\ast} \bigsqcap_{F \in \mathrm{up} f} \langle (\mathsf{FCD}) f \rangle^{\ast} \langle F \rangle^{\ast} A = ? ?$
The further step fails because in general ${\langle ((\mathsf{FCD}) f)^{\circ} \rangle^{\ast} \bigsqcap_{F \in \mathrm{up} f} S \neq \bigsqcap_{F \in \mathrm{up} f} \langle \langle ((\mathsf{FCD}) f)^{\circ} \rangle^{\ast} \rangle^{\ast} S}$.
So as now my attempt has failed. Please give me advice how to overcome this shortcoming of my theory. | 2017-08-20 07:51:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 21, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9325301647186279, "perplexity": 333.48738356952765}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886106358.80/warc/CC-MAIN-20170820073631-20170820093631-00479.warc.gz"} |
http://www.reddit.com/r/math/comments/14zc57/visualization_for_polynomials_of_a_complex/c7jtrml | you are viewing a single comment's thread.
[–]Geometry[S] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
My collaborator has done a lot or work on Domain Coloring and I've read a lot of papers on the subject. It is a useful tool and very demonstrative once you can understand how the colors represent the structure of the surface... but in my opinion Domain Coloring requires a lot of time for the audience to interpret/decipher the data.
What I'm trying to do is demonstrate the shape of the surface directly in a 3D space. While it's true that the surface [;\{(z,w) : w=f(z)\};] exists in [; \mathbb{C}^2;], it is still a surface of one complex dimension (2 real dimensions), so it's still just a 2-surface, which makes it relatively easy to see in 3D space. The only caveat/downside is that the surface is generally forced to intersect itself.
Now something I just thought of is a hybrid approach. I can start with a piece of the domain, colored appropriately, then animate a transition of the colored domain morphing into the range.... Yes, the more i htink about this, the more I know I'm definitely going to make this idea happen. | 2015-03-06 12:24:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4649166166782379, "perplexity": 602.1323350702676}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-11/segments/1424936468536.58/warc/CC-MAIN-20150226074108-00151-ip-10-28-5-156.ec2.internal.warc.gz"} |
https://ncatlab.org/nlab/show/heap | Heaps
Heaps
Idea
A heap is an algebraic structure which is basically equivalent to a group when one forgets about which element is the unit. Similar notions are affine space, principal homogeneous space and so on. However, the notion of a heap has a directness and simplicity in the sense that it is formalized as an algebraic structure with only one ternary operation satisfying a short list of axioms. If we start with a group the ternary operation is defined via $(a,b,c)\mapsto a b^{-1}c$. We can interpret that operation as shifting $a$ by the (right) translation in the group which translates $b$ into $c$. There is also a dual version, quantum heap.
Heaps in the sense of algebra should not be confused with heaps in the sense of theoretical computer science. There are also a number of synonyms for the term ‘heap’; below we consider ‘torsor’ in this light. In Russian one term for a heap is ‘груда’ (‘gruda’) meaning a heap of soil; this is a pun as it is parallel to the russian word ‘группа’ (‘gruppa’) meaning a group: forgetting the unit element is sort of creating an amorphous version. This term also appears in English as ‘groud’.
Definition
A heap $(H,t)$ is a nonempty set $H$ equipped with a ternary operation $t : H \times H \times H\to H$ satisfying the relations
$t(b,b,c) = c = t(c,b,b)$
$t(a,b,t(c,d,e)) = t(t(a,b,c),d,e)$
More generally, a ternary operation in some variety of algebras satisfying the first pair of equations is called a Mal'cev operation. A Mal’cev operation is called associative if it also satisfies the latter equation (i.e. it makes its domain into a heap).
A heap homomorphism, of course, is a function that preserves the ternary operations. This defines a category $Heap$ of heaps.
Automorphism group
As suggested above, if $G$ is a group and we define $t(a,b,c) = a b^{-1} c$, then $G$ becomes a heap. This construction defines a functor $Prin:Grp\to Heap$. In fact, up to isomorphism, all heaps arise in this way; to every heap is associated a group $Aut(H)$ called its automorphism group, unique up to isomorphism. There are a number of ways to define $Aut(H)$ from $H$.
1. If we choose an arbitrary element $e\in H$, then we can define a multiplication on $H$ by $a b = t(a,e,b)$. It is straightforward to verify that this defines a group structure on $H$, whose underlying heap structure is the original one.
2. We can define $Aut(H)$ to be the set of pairs $(a,b)\in H\times H$, modulo the equivalence relation $(a,b)\sim (a',b')$ iff $t(a,a',b')=b$. (We think of $(a,b)$ as representing $a^{-1} b$.) We then define multiplication by $(c,d)(a,b) = (c,t(d,a,b))$; the inverse of (the equivalence class of) $(a,b)$ is (the equivalence class of) $(b,a)$ and the identity element is (the equivalence class of) $(a,a)$ (for any $a$).
3. We can also define $Aut(H)$ as an actual subgroup of the symmetric group of $H$, analogously to Cayley's theorem? (see Wikipedia) for groups. We take the elements of $Aut(H)$ to be set bijections of the form $t(-,a,b): H \rightarrow H$ where $a,b \in H$, with composition as the group operation. Note that
$t(-,c,d) \cdot_{Aut(H)} t(-, a,b) = t(t(-,c,d),a,b) = t(-,c,t(d,a,b)),$
so $Aut(H)$ is closed under this operation. The first axiom of a heap shows that $Aut(H)$ contains the identity $t(-,x,x)$ for any $x$), and the inverse of $t(\cdot,a,b)$ is $t(\cdot,b,a)$; thus $Aut(H)$ is a subgroup of the symmetric group of $H$.
Note that in both the second and third constructions, the elements of $Aut(H)$ are determined by pairs of elements of $H$, modulo some equivalence relation. The following theorem shows that the two equivalence relations are the same.
Theorem
The following are equivalent
1. bijections $t(\cdot,a,b)$ and $t(\cdot,a',b')$ are the same maps,
2. $t(a,a',b') = b$,
3. $t(b,b',a') = a$.
Proof
(ii) follows from (i) and $t(a,a,b) = b$.
(iii) follows from (ii) by applying $t(\cdot,b',a')$ on the right. Similarly (ii) follows from (iii).
(i) follows from (ii) by the calculation:
$t(x,a',b') = t(t(x,a,a),a',b')= t(x,a,t(a,a',b')) = t(x,a,b).$
The composition laws are also easily seen to agree, so the second two constructions of $Aut(H)$ are canonically isomorphic. To compare them to the first construction, observe that for a fixed $e\in H$, any equivalence class contains a unique pair of the form $(e,a)$. (If $(b,c)$ is in the equivalence class, then $a$ is determined by $a = t(e,b,c)$.) This sets up a bijection between the first two constructions, which we can easily show is an isomorphism.
The second two constructions are clearly functorial, so we have a functor $Aut:Heap\to Grp$. Note that we have $Aut(Prin(G))\cong G$ for any group $G$, and $Prin(Aut(H))\cong H$ for any heap $H$, but while the first isomorphism is natural, the second is not. In particular, the categories $Heap$ and $Grp$ are not equivalent.
Heaps and torsors
Note that $Aut(H)$ comes equipped with a canonical action on $H$ (this is most clear from the third definition). This action is transitive (by $t(a,a,b) = b$) and free (if $t(a,b,c) = a$ then by the previous statement $t(x,b,c) = x$ for each $x$, and in particular $t(b,b,c) = b$ and also $t(b,b,c) = c$). Therefore, $H$ is an $Aut(H)$-torsor (over a point). Conversely, a torsor $H$ over any group $G$ can be made into a heap, by defining $t(a,b,c) = g\cdot c$, where $g\in G$ is the unique group element such that $g\cdot b = a$.
In fact, the category $Heap$ is equivalent to the following category $Tors$: its objects are pairs $(G,H)$ consisting of a group $G$ and a $G$-torsor $H$, and its morphisms are pairs $(\phi,f):(G,H)\to (G',H')$ consisting of a group homomorphism $\phi:G\to G'$ and a $\phi$-equivariant map $f:H\to H'$.
The empty heap
If we wish $Heap$ to be an algebraic category, then we must remove the clause that the underlying set of a heap must be nonempty. Then the empty set becomes a heap in a unique way. However, in this case, the various theorems relating heaps to groups above all break down. For this reason, one usually requires a heap to be inhabited.
On the other hand, we could generalize the notion of group to allow for an empty group. This even remains a purely algebraic notion: we can define a group as a (traditionally nonempty) set equipped with a binary operation (to be thought of as $a, b \mapsto a/b \coloneqq a b^{-1}$) satisfying these laws:
• $a/a = b/b$,
• $(a/a)/((a/a)/a) = a$,
• $a/(b/c) = (a/((c/c)/c))/b$.
Then any possibly-empty-group is a possibly-empty-heap, and every possibly-empty-heap arises in this way from its automorphism possibly-empty-group (defined by either method (2) or (3)); the category of possibly-empty-heaps is equivalent to the category of possibly-empty-groups equipped with torsors over the point; etc.
This is even constructive; the theorems can be proved uniformly, rather than by treating the empty and inhabited cases separately. (This rather trivial method is obvious to a classical mathematician, but it's not constructively valid, since a possibly-empty-group/heap as defined here can't be constructively proved empty or inhabited; it can only be proved empty iff not inhabited. Indeed, taking any group $G$ and any truth value $P$, the possibly-empty-subgroup $\{x \in G \;|\; P\}$ is empty or inhabited iff $P$ is false or true.)
References and remarks
• G.M. Bergman, A.O. Hausknecht, Cogroups and co-rings in categories of associative rings, Ch.IV, paragraph 22, p.95ff – Providence, R.I. : AMS 1996.
• Z. Škoda, Quantum heaps, cops and heapy categories, Mathematical Communications 12, No. 1, pp. 1–9 (2007); (math.QA/0701749)
• wikipedia:heap
• Heaps and torsors
There is an oidification (horizontal categorification) of a heap, sometimes called a heapoid.
Last revised on January 3, 2018 at 16:32:17. See the history of this page for a list of all contributions to it. | 2019-05-19 21:14:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 102, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9469163417816162, "perplexity": 233.04784813437567}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232255165.2/warc/CC-MAIN-20190519201521-20190519223521-00498.warc.gz"} |
https://www.linstitute.net/archives/632705 | # USACO 2022 US Open Contest, Bronze Problem 3. Alchemy
### USACO 2022 US Open Contest, Bronze Problem 3. Alchemy
Always keen to learn new hobbies, Bessie the cow is learning how to transform metals. She has aiai (0ai1040≤ai≤104) units of metal ii for 1iN1001≤i≤N≤100. Furthermore, she knows KK (1K<N1≤K<N) recipes where she can combine one unit each of several metals to make one unit of a metal with a higher number than all constituent metals. It is additionally guaranteed that for each metal, Bessie knows at most one recipe to make it.
Compute the maximum number of units of metal NN Bessie can possibly have after some series of transformations.
#### INPUT FORMAT (input arrives from the terminal / stdin):
The first line contains NN.
The second line contains NN integers, aiai.
The third line contains KK.
The next KK lines start with two integers LL and MM (M1M≥1), followed by MM integers. The last MM integers represent the constituent metals in the recipe that are used to form one unit of metal LL. It is guaranteed that LL is larger than the MM last integers.
#### OUTPUT FORMAT (print output to the terminal / stdout):
Output the maximum number of units of metal NN Bessie can possibly have after applying some series of zero or more transformations.
#### SAMPLE INPUT:
5
2 0 0 1 0
3
5 2 3 4
2 1 1
3 1 2
#### SAMPLE OUTPUT:
1
In this example, the following is an optimal series of transformations:
1. Transform one unit of metal 1 into metal 2.
2. Transform one unit of metal 2 into metal 3.
3. Transform one unit of metal 3 and metal 4 into metal 5.
Now Bessie is left with one unit of metal 1 and one unit of metal 5. She cannot form any additional units of metal 5.
#### SCORING:
• In test case 2, for 1i<N1≤i<N, one unit of metal ii can be transformed into one unit of metal i+1i+1.
• In test cases 3 and 4, each recipe transforms one unit of one metal into another.
• Test cases 5 through 11 satisfy no additional constraints.
Problem credits: Nick Wu
### USACO 2022 US Open Contest, Bronze Problem 3. Alchemy 题解(翰林国际教育提供，仅供参考)
[/hide]
(Analysis by Nick Wu)
For notational convenience, if i>j, we'll say that metal i is more valuable than metal j. Note that all recipes take a single unit of some metals and produce one unit of a metal that is more valuable than all of the original metals.
To solve test case 2, every unit of metal can be converted into a unit of metal N, so the answer is the number of units of metal that Bessie has.
To solve test cases 3 and 4, if there is no recipe that can make metal N, then the answer is simply the number of units of metal N that Bessie starts out with. Otherwise, there is some less valuable metal that can be turned directly into metal N, so all of those units can also be converted into metal N. If that less valuable metal has no recipe, then we are done. Otherwise, we repeat this process and sum up the counts of the less valuable metals until we reach one which can't be made.
To solve the problem fully, we'll take advantage of how recipes can only turn less valuable metals into more valuable metals. If we want to gain one unit of metal N, we must use that recipe, and so that means we need one unit of some less valuable metals. If we already have one of each unit of the less valuable metals, we can directly consume those. If we don't have a unit of some metal, and no recipe for that metal exists, we cannot make any more of metal N, and the process ends. If there is a recipe, then we need one unit of each of the metals that are ingredients in that recipe.
Since more valuable metals cannot be used as ingredients in recipes for less valuable metals, we can loop over the metals in order from metal N down to metal 1, tracking at each point in time how many units of each metal we need in order to make one unit of metal N.
The while loop runs at most N⋅max(ai) times and takes O(N) time per iteration, for a time complexity of O(N2⋅max(ai)).
C++ code that does this iteratively:
#include <bits/stdc++.h>
using namespace std;
int main() {
int n;
cin >> n;
vector have(n);
for(auto& x: have) cin >> x;
int k;
cin >> k;
vector<vector> need(n);
while(k--) {
int want, m;
cin >> want >> m;
need[--want].resize(m);
for(auto& x: need[want]) {
cin >> x;
x--;
}
}
int ret = 0;
while(true) {
vector consume(n);
consume[n-1]++;
bool good = true;
for(int i = n-1; i >= 0; i--) {
if(consume[i] <= have[i]) {
have[i] -= consume[i];
continue;
}
if(need[i].size() == 0) {
good = false;
break;
}
int take = min(consume[i], have[i]);
consume[i] -= take;
have[i] -= take;
for(int out: need[i]) consume[out] += consume[i];
}
if(good) ret++;
else break;
}
cout << ret << "\n";
}
Java code that does this recursively:
import java.io.*;
import java.util.*;
public class Alchemy {
public static void main(String[] args) throws IOException {
int[] have = new int[n];
for(int i = 0; i < n; i++) have[i] = Integer.parseInt(st.nextToken()); int ans = 0; int[][] recipes = new int[n][]; int k = Integer.parseInt(in.readLine()); while(k-- > 0) { | 2022-08-13 06:00:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3903854787349701, "perplexity": 1871.7894620246336}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571909.51/warc/CC-MAIN-20220813051311-20220813081311-00213.warc.gz"} |
https://myassignments-help.com/2022/11/03/biostatistics-dai-xie-biol220/ | # 统计代写|生物统计代写Biostatistics代考|BIOL220
## 统计代写|生物统计代写Biostatistics代考|Cluster Sampling
In many biomedical studies, a good sampling frame containing the units in the target population is extremely difficult to obtain. In some cases, the units of the population belong to clearly defined groups of units referred to as clusters of units. When a sampling frame that contains a list of clusters is available a random clusterr sample can often be uséd tō obtain a sample that is representative of the target population. For example, in surveying the general health of the people living in a particular town, a sampling frame consisting of the people living in the town would be difficult, if not impossible, to obtain. On the other hand, most towns will have a listing of the housing units within the town. In this case, the people living in the town are grouped in clusters according to housing units.
To use a random cluster sample, the clusters in a target population should be nonoverlapping and exhaustive so that a population unit belongs to only one cluster. Also, when the population units belong to well-defined clusters and a random cluster sample is selected, the cluster random sample will consist of primary and secondary units. The primary units in a cluster sampling plan are the clusters of population units and the secondary units are the units within the clusters. Thus, a primary unit is a sampling unit and a secondary unit is sampling element.
## 统计代写|生物统计代写Biostatistics代考|Systematic Sampling
The final sampling plan that will be discussed is the systematic random sampling plan. A 1 in $k$ systematic sample is a probability sample obtained by randomly selecting one sampling unit from the first $k$ units in the sampling frame and every $k$ th unit thereafter. A systematic random sample is one of the most cost-effective and convenient sampling plans when the population units are randomly dispersed over the sampling frame. Systematic random sampling is also useful in quality control settings. For example, a pharmaceutical company may check the uniformity and quality of the drug tablets it manufactures. The ability to control the manufacturing process by sampling the output is an important aspect of quality control and may be monitored by selecting a systematic random sample of every 100 th tablet produced.
The process for drawing a systematic random sample is outlined below.
The advantages of drawing a systematic random sample are that it is simple to collect, often less expensive than other sampling plans, spreads uniformly over the sampling frame, the statistical analysis of a systematic random sample is the same as the analysis used with a simple random sample, and when the size of the population is not fixed or is being monitored over time, a systematic random sample can be easily implemented without a sampling frame.
The only real disadvantage of using a systematic random sample occurs when a systematic sample isolates unwanted patterns in the population. That is, a systematic random sample may yield a biased sample when there is a periodic or cyclic pattern in the units listed in the sampling frame. Thus, it is critical to avoid the use of a systematic random sample when there could be cyclical patterns in the sampling frame or the sampling process. Cyclical patterns are often due to sampling the units over time.
# 生物统计代考
## 统计代写|生物统计代写Biostatistics代考|Systematic Sampling
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Math作业代写、数学代写常见问题
myassignments-help擅长领域包含但不是全部: | 2023-03-26 21:41:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.541690468788147, "perplexity": 713.948557151144}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296946535.82/warc/CC-MAIN-20230326204136-20230326234136-00501.warc.gz"} |
https://hyperleap.com/topic/L-function | # L-function
L''-functionsL''-functionL-series-function-functionscomplex L-seriesL-functions
In mathematics, an L-function is a meromorphic function on the complex plane, associated to one out of several categories of mathematical objects.wikipedia
126 Related Articles
### Riemann zeta function
zeta functionzeta functionsRiemann zeta-function
In it, broad generalisations of the Riemann zeta function and the L-series for a Dirichlet character are constructed, and their general properties, in most cases still out of reach of proof, are set out in a systematic way. We distinguish at the outset between the L-series, an infinite series representation (for example the Dirichlet series for the Riemann zeta function), and the L-function, the function in the complex plane that is its analytic continuation.
Many generalizations of the Riemann zeta function, such as Dirichlet series, Dirichlet -functions and -functions, are known.
### Selberg class
Selberg series
The Selberg class is an attempt to capture the core properties of L-functions in a set of axioms, thus encouraging the study of the properties of the class rather than of individual functions.
In mathematics, the Selberg class is an axiomatic definition of a class of L-functions.
In that case results have been obtained for p-adic L-functions, which describe certain Galois modules.
In mathematics, a p-adic zeta function, or more generally a p-adic L-function, is a function analogous to the Riemann zeta function, or more general L-functions, but whose domain and target are p-adic (where p is a prime number).
### Functional equation (L-function)
functional equationfunctional equation of L-functionsfunctional equations
functional equation, with respect to some vertical line Re(s) = constant;
In mathematics, the L-functions of number theory are expected to have several characteristic properties, one of which is that they satisfy certain functional equations. There is an elaborate theory of what these equations should be, much of which is still conjectural.
### Algebraic K-theory
algebraic ''K''-theoryK-theoryalgebraic -theory
interesting values at integers related to quantities from algebraic K-theory
The subject also includes classical number-theoretic topics like quadratic reciprocity and embeddings of number fields into the real numbers and complex numbers, as well as more modern concerns like the construction of higher regulators and special values of L-functions.
### Peter Swinnerton-Dyer
Sir Peter Swinnerton-DyerSir (Henry) Peter (Francis) Swinnerton-Dyer Bt.Sir (Henry) Peter Francis Swinnerton-Dyer, 16th Baronet
One of the influential examples, both for the history of the more general L-functions and as a still-open research problem, is the conjecture developed by Bryan Birch and Peter Swinnerton-Dyer in the early part of the 1960s.
As a mathematician he was best known for his part in the Birch and Swinnerton-Dyer conjecture relating algebraic properties of elliptic curves to special values of L-functions, which was developed with Bryan Birch during the first half of the 1960s with the help of machine computation, and for his work on the Titan operating system.
### Dirichlet L-function
Dirichlet ''L''-functionsL-functionsDirichlet ''L''-function
In it, broad generalisations of the Riemann zeta function and the L-series for a Dirichlet character are constructed, and their general properties, in most cases still out of reach of proof, are set out in a systematic way.
L-function
### Random matrix
random matricesrandom matrix theory random matrix theory
The statistics of the zero distributions are of interest because of their connection to problems like the Generalized Riemann hypothesis, distribution of prime numbers, etc. The connections with random matrix theory and quantum chaos are also of interest.
In number theory, the distribution of zeros of the Riemann zeta function (and other L-functions) is modelled by the distribution of eigenvalues of certain random matrices.
### Hasse–Weil zeta function
Hasse–Weil ''L''-functionHasse–Weil ''L''-functionsL-series
Gradually it became clearer in what sense the construction of Hasse–Weil zeta-functions might be made to work to provide valid L-functions, in the analytic sense: there should be some input from analysis, which meant automorphic analysis.
In mathematics, the Hasse–Weil zeta function attached to an algebraic variety V defined over an algebraic number field K is one of the two most important types of L-function.
### Generalized Riemann hypothesis
extended Riemann hypothesisERHGeneralized Riemann Hypothesis (GRH)
Generalized Riemann hypothesis
Various geometrical and arithmetical objects can be described by so-called global L-functions, which are formally similar to the Riemann zeta-function.
### Langlands program
geometric Langlands programfunctorialityfurther developed
This development preceded the Langlands program by a few years, and can be regarded as complementary to it: Langlands' work relates largely to Artin L-functions, which, like Hecke L-functions, were defined several decades earlier, and to L-functions attached to general automorphic representations.
The Artin reciprocity law applies to a Galois extension of an algebraic number field whose Galois group is abelian; it assigns L-functions to the one-dimensional representations of this Galois group, and states that these L-functions are identical to certain Dirichlet L-series or more general series (that is, certain analogues of the Riemann zeta function) constructed from Hecke characters.
### Special values of L-functions
Bloch–Kato conjecturesspecial valuesvalues of L-functions
Special values of L-functions
There are two families of conjectures, formulated for general classes of L-functions (the very general setting being for L-functions L(s) associated to Chow motives over number fields), the division into two reflecting the questions of:
### Meromorphic function
meromorphicmeromorphic functionsmeromorphically
In mathematics, an L-function is a meromorphic function on the complex plane, associated to one out of several categories of mathematical objects.
### Function (mathematics)
functionfunctionsmathematical function
In mathematics, an L-function is a meromorphic function on the complex plane, associated to one out of several categories of mathematical objects.
### Complex plane
complex number planeArgand diagramimaginary axis
In mathematics, an L-function is a meromorphic function on the complex plane, associated to one out of several categories of mathematical objects.
### Mathematical object
objectsmathematical objectsgeometric object
In mathematics, an L-function is a meromorphic function on the complex plane, associated to one out of several categories of mathematical objects.
### Dirichlet series
Dirichlet-Serieszeta
An L-series is a Dirichlet series, usually convergent on a half-plane, that may give rise to an L-function via analytic continuation. We distinguish at the outset between the L-series, an infinite series representation (for example the Dirichlet series for the Riemann zeta function), and the L-function, the function in the complex plane that is its analytic continuation. a Dirichlet series, and then by an expansion as an Euler product indexed by prime numbers.
### Convergent series
convergenceconvergesconverge
An L-series is a Dirichlet series, usually convergent on a half-plane, that may give rise to an L-function via analytic continuation.
### Half-space (geometry)
half-spacehalf-planehalfspace
An L-series is a Dirichlet series, usually convergent on a half-plane, that may give rise to an L-function via analytic continuation.
### Analytic continuation
analytically continuedanalytic extensionmeromorphic continuation
An L-series is a Dirichlet series, usually convergent on a half-plane, that may give rise to an L-function via analytic continuation. We distinguish at the outset between the L-series, an infinite series representation (for example the Dirichlet series for the Riemann zeta function), and the L-function, the function in the complex plane that is its analytic continuation.
### Conjecture
conjecturalconjecturesconjectured
The theory of L-functions has become a very substantial, and still largely conjectural, part of contemporary analytic number theory.
### Analytic number theory
analyticanalytic number theoristanalytic techniques
The theory of L-functions has become a very substantial, and still largely conjectural, part of contemporary analytic number theory.
### Dirichlet character
conductorcharacterConductor of a Dirichlet character
In it, broad generalisations of the Riemann zeta function and the L-series for a Dirichlet character are constructed, and their general properties, in most cases still out of reach of proof, are set out in a systematic way.
### Infinite set
infiniteinfinitelyinfinitely many
We distinguish at the outset between the L-series, an infinite series representation (for example the Dirichlet series for the Riemann zeta function), and the L-function, the function in the complex plane that is its analytic continuation.
### Euler product
Euler's product formulaproduct of Euler factorsStephens constant
a Dirichlet series, and then by an expansion as an Euler product indexed by prime numbers. | 2019-08-19 06:04:55 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.841019332408905, "perplexity": 942.4319269323146}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027314667.60/warc/CC-MAIN-20190819052133-20190819074133-00337.warc.gz"} |
https://psychometroscar.com/2019/05/ | # Power analysis for multiple regression with non-normal data
This app will perform computer simulations to estimate the power of the t-tests within a multiple regression context under the assumption that the predictors and the criterion variable are continuous and either normally or non-normally distributed. If you would like to see why I think this is important to keep in mind, please read this blog post.
When you first click on the app it looks like this:
What *you*, the user, needs to provide it with is the following:
The number of predictors. It can handle anything from 3 to 6 predictors. When you have more than that the overall aesthetics of the app is simply too crowded. The default is 3 predictors.
The regression coefficients (i.e, the standardized effect sizes) that hold in the population. The default is 0.3. The app names them “x1, x2, x3,…x6”
The skewness and excess kurtosis of the data for each predictor AND for the dependent variable (the app calls it “y”). Please keep on reading to see how you should choose those. The defaults at this point are skewness of 2 and an excess kurtosis of 7.
The pairwise correlations among the predictors. I think this is quite important because the correlation among the predictors plays a role in calculating the standard error of the regression coefficients. So you can either be VERY optimistic and place those at 0 (predictors are perfectly orthogonal with one another) OR you can be very pessimistic and give them a high correlation (multicollinearity). The default inter-predictor correlation is 0.5.
The sample size. The default is 200.
The number of replications for the simulation. The default is 100.
Now, what’s the deal with the skewness and excess kurtosis? A lot of people do not know this but you cannot go around choosing values of skewness and excess kurtosis all willy-nilly. There is a quadratic relationship between the possible values of skewness and excess kurtosis that specifies they MUST be chosen according to the inequality kurtosis >skewness^2-2 . If you don’t do this, it will spit out an error. Now, I am **not** a super fan of the algorithm needed to generate data with those population-specified values of skewness and excess kurtosis. For many AND even more reasons. HOWEVER, I needed to choose something that was both sufficiently straight-forward to implement and not very computationally-intensive, so the 3rd order polynomial approach will have to do.
Now, the exact boundaries of what this method can calculate are actually smaller than the theoretical parabola. However, for practical purposes, as long as you choose values of kurtosis which are sufficiently far apart from the square of the skewness, you should be fine. So, a combo like skewness=3, kurtosis=7 would give it trouble. But something like skewness=3, kurtosis=15 would be perfectly fine.
A hypothetical run would look like this:
So the output under Results is the empirical, simulated power for each regression coefficient at the sample size selected. In this case, they gravitate around 60%.
Oh! And if for whatever reason you would like to have all your predictors be normal, you can set the values of skewness and kurtosis to 0. In fact, in that situation you would end up working with a multivariate normal distribution.
# The case of the missing correlations and positive-definiteness.
A couple of weeks ago, a student of mine inquired about how the pos_def_limits()function from the faux R package works. This package is being developed by the super awesome Dr. Lisa DeBruine who you should, like, totally follow, btw, in case you are not doing it already. What makes this post interesting is that I thought pos_def_limits()was doing one thing but it is doing something else. I think it helps highlight how different ways of approaching the same problems can give you insight into different aspects of it. But first, some preliminaries:
What is positive definiteness?
This one is not particularly complicated to point out. Define $\Sigma$ as an $n \times n$ real-valued matrix and $v$ as an $n \times 1$ real-valued, non-zero vector. Then $\Sigma$ is a positive-definite matrix if $v^{t} \Sigma v >0$ and it is positive-semi-definite if $v^{t} \Sigma v \geq 0$ for all $v$ . I prefer this definition of positive definiteness because it generalizes easily to other types of linear operators (e.g., differentiation) as opposed to consequences of this definition, which is what we usually operate on. If you come from the social sciences (like I do) the “version” that you know about a matrix (usually, covariance matrix) being positive-definite is that all its eigenvalues have to be positive. Which is what pos_def_limits() relies on. It implements a grid-search over the plausible correlation range of [-1, +1] and, once it finds the minimum and maximum value for which the set of covariance matrices are all positive definite, it produces a result (or it lets you know if said matrix does not exist, which is super useful as well). Relying on the documentation example:
pos_def_limits(.8, .2, NA) > min max > -0.427 0.747
which means that if you have a correlation matrix that looks like:
$\textbf{R}= \begin{bmatrix} 1 & 0.8 & 0.2\\ 0.8 & 1 & r\\ 0.2 & r &1 \end{bmatrix}$
then as long as $-0.427 \leq r \leq 0.747$, your resulting matrix is positive definite and, hence, a valid correlation matrix. So far so good.
How I tackle this problem
This is what I thought pos_def_limits() was doing under the hood before looking at the source code. So… a similar condition to the positive eigenvalues is that the determinant of the matrix has to be positive. So IF $\Sigma$ is positive-semi- definite, THEN $det(\Sigma) \geq 0$. Notice that this DOES NOT work the other way around: just because $det(\Sigma) \geq 0$ does not mean that $\Sigma$ is positive-semi-definite). Anyway, we can rely on the fact that all correlation/covariance matrices are positive-definite by definition, which means the problem of finding the suitable upper and lower bounds is simply solving for $r$ subject to the constraint that the determinant MUST be greater than or equal to zero. So with the help of a CAS (Computer Algebra System, my favourite one is MAPLE because I’m very Canadian, LOL) I can see that solving for $det(\textbf{R}) \geq 0$ results in the following:
$det(\textbf{R})= -r^{2} + 0.32r + 0.32 \geq 0$
which is… A QUADRATIC EQUATION! We can graph it and see:
So that any value on the x-axis between both roots yields a valid inequality and, therefore, a positive-definite matrix $\textbf{R}$. Do you remember how to solve for the roots of quadratic equations? Using our trusted formula from highschool we obtain:
$x_1= -\frac{2(3\sqrt{6}-2)}{25} = -0.42788...$
$x_2= \frac{2(2+3\sqrt{6})}{25} = 0.74788...$
which match the values approximated by pos_def_limits()
Extensions to this problem Pt I: Maximizing the determinant
There are 2 interesting things you can do with this determinantal equation. First and foremost, you can choose *the* value that maximizes the variance encoded in the correlation matrix $\textbf{R}$. The determinant has a lot of interesting properties, including a very nice geometric representation. The absolute value of the determinant is the volume of the parallelepiped described by the column vectors within the matrix. And this generalizes to higher dimensions. Because correlations are bounded in the [-1, +1] range, the maximum determinant that ANY correlation matrix can have is 1 and the minimum is 0. So if I want my matrix $\textbf{R}$ to have the largest possible determinant, I only need to choose the value at the vertex of the parabola:
which has coordinates (0.16, 0.3456). So if $r=0.16$ then the determinant of $\textbf{R}$ is at it maximum.
Extensions to this problem Pt II: What if more than one correlation is missing?
The “classical” version of this problem is to have a 3 x 3 matrix where 2 correlations are known and 1 is missing. But (as my student questioned further) what would happen if we had, say a 4 x 4 matrix and TWO correlations were missing? Well, no biggie. Let’s come up with a new matrix, let’s call it $\textbf{S}$ with the following form:
$\textbf{S}= \begin{bmatrix} 1 & 0.8 & 0.2 & a\\ 0.8 & 1 & r & 0.5\\ 0.2 & r &1 & 0.3\\ a &0.5&0.3 & 1 \end{bmatrix}$
Yeah, I had to make up a couple of extra correlations (0.3 and 0.5) to make sure only TWO correlations were missing. But there is a point to this that you will notice very quickly. Since we still want $\textbf{S}$ to be a valid correlation matrix, the condition $det(\textbf{S}) \geq 0$ must still hold, irrespective of the dimensions or how many missing elements $\textbf{S}$ has. So, once again, running this new matrix through the CAS system yields the condition:
$a^2r^2-a^2-0.68ar+0.92a-r^2+0.62r-0.0004 \geq 0$
Which generates a system of quadratic inequalities which must be solved simultaneously to yield valid ranges. Rather than showing you the equations (booooooring! :D) let me show you something prettier. A picture of the solution space:
Yup. Any combination of points (a,r) within the red regions would yield a valid solution to the inequality above. HOWEVER, there is only ONE region where the solution is both within the valid (i.e., red) regions AND gives us solutions in the valid correlation range. What does your intuition tell you? I think we both know where to look 😉
That is correct! That little blob-looking thingy is where we want to be at. ANY combination of values within the blob is both between [-1, +1] AND satisfies the determinantal equation so any pair of values STRICTLY inside the blob will yield a valid correlation matrix $\textbf{S}$.
“But Oscar — you may ask– how do we choose the one pair of values that maximizes the determinant of $\textbf{S}$?” Well, that’s a good question! It is not difficult to answer, but it does require a little more mathematics than the case where only one is missing. What we need is to, first, take the partial derivatives with respect to both r and a and set them to 0 (we need to find maximums):
$\frac{\partial}{\partial a}det(\textbf{S})=2ar^2-2a+0.92-0.68r=0$
$\frac{\partial}{\partial r}det(\textbf{S})=2a^2r-2r+0.62-0.68a=0$
So now we have a system of quadratic equations. There are two equations and two unknowns so we know this system is just-identified and *has* a solution. Again, when you throw them in MAPLE you end up with multiple values for a and r that maximize $\textbf{S}$. There was only one pair of solutions, though, which was both within the valid range [-1, +1] AND inside the special blob:
$a=0.473863 ; r=-0.038687$
The last thing we need to check, however, is whether these points are minimums, maximums or saddle points as per the 2nd derivative test.
Which means I need all the 2nd partial derivatives from the equation above as:
$\frac{\partial}{\partial a\partial a}det(\textbf{S})=2(r^{2}-1)$
$\frac{\partial}{\partial r\partial r}det(\textbf{S})=2(a^{2}-1)$
$\frac{\partial}{\partial a\partial r}det(\textbf{S})=4ar-0.68$
Calculate the discriminant setting a=0.473863 and r=-0.038687 :
$2(a^{2}-1) 2(r^{2}-1)-(4ar-0.68)^{2} = 2.529668$
Which is greater than 0. So (a,r) are either minimums or maximums. The final check is to see if $2(a^{2}-1) < 0$ to make sure it’s a local maximum. And since $2(0.473863^{2}-1)=-1.550908 < 0$, then it follows that the point (a,r) indeed maximizes the determinant*.
TWO potential future directions
While working on this I noticed a couple of peculiarities that I think are sufficiently mathematically tractable for me to handle and turn into an actual article. UNLESS you (my dear reader) already know the answer to this. I am, after all, a lazy !@#$#% which means that if someone already worked out a formal proof for it, I’d much rather read it than having to come up with it on my own. Let us start with the easy one: (1) (Easy): The range of plausible correlations shrinks as the dimensions of the correlation matrix increase This one is, I think, easy to see. Let’s start with the basic 2×2 correlation matrix: $\textbf{A}= \begin{bmatrix} 1 & r\\ r & 1\\ \end{bmatrix}$ Then if $r \in (-1, +1)$ (read $\in$ as “element of”) then $\textbf{A}$ is a valid correlation matrix and only becomes positive-semi-definite if either $r=1$ or $r=-1$. But you get the gist, all the valid correlation range applies. Notice how the range of $r$ shrunk for the 3×3 matrix $\textbf{R}$ of our example to: $r \in [-\frac{2(3\sqrt{6}-2)}{25}, \frac{2(2+3\sqrt{6})}{25}]$ And this fact is independent of what correlation matrix you have. You cannot get any correlation matrix where, given that 2 of them are known, the resulting range of $r$ is the full valid interval $[-1, +1]$ AND it is still positive definite. So, what happens if we go back to our 4 x 4 example with $\textbf{S}$? Just for kicks and giggles, let’s make $a=0.3$? so that we are only left with 1 unknown. The new matrix looks like this: $\textbf{S}= \begin{bmatrix} 1 & 0.8 & 0.2 & 0.3\\ 0.8 & 1 & r & 0.5\\ 0.2 & r &1 & 0.3\\ 0.3 &0.5&0.3 & 1 \end{bmatrix}$ If we run this new $\textbf{S}$ through pos_def_limits()(it can handle any number of dimensions) we get: pos_def_limits(.8, .2,.3,.NA, 5,.3) > min max > -0.277 0.734 Yup, the range has now shrunk. But we don’t quite yet know why. Let’s try it now through the determinantal equation: $det(\textbf{R})= -0.91r^{2} + 0.416r + 0.1856 \geq 0$ And solving for the roots of this equation we get: $-\frac{0.416+\sqrt{0.84864})}{1.82} = -0.27759...$ $\frac{0.416+\sqrt{0.84864})}{1.82} = 0.73473...$ Yup, same answer. But notice something interesting. The quadratic coefficient has now shrunk from 1 to -0.91. And, if you remember back from high school, you know that the coefficient of the quadratic term dictates how wide or narrow the parabola is so that if it is outside the [0, 1] range the parabola is wider (i.e., the roots are farther apart) and if it it’s within [0, 1] the parabola is narrower (i.e., the roots are closer together). Which prompts me to make the following claim: Claim: As the dimensions of the correlation matrix increase arbitrary, the valid range that makes it positive definite shrinks until it collapses to a *single* point. In other words, for a large enough correlation matrix, only ONE value can make it positive definite. This one shouldn’t be particularly difficult to prove. All I need to show is that the leading coefficient of the quadratic term shrinks as a function of the dimensions of the correlation matrix until it becomes 0. In which case you’ll have a straight line (not a parabola). And that means you’d get only 1 root (and not 2 that encompass a range). (1) (Hard): If I sample values for r from the valid correlation range uniformly, the distribution of the determinants concentrate around *the* value of r that maximizes the determinant. This one is a little bit more difficult to explain, but let me show you an interesting thing I found. Let’s use the classic 3 x 3 matrix case and only focus on $\textbf{R}$. We know from above that if $r=0.16$ then the value of $det(\textbf{R})$ is maximized. Now, let’s use R (the programming language, not the matrix) to uniformly sample random values of it, calculate the determinants and plot them: n<-10000 r<-runif(n, min=-.427, max=.747) for (i in 1:n) { R<-matrix(c(1,.8,.2,.8,1,r[i],.2,r[i],1),3,3) pp[i]<-det(R) } dat<-data.frame(pp) a<- density(pp) mmod<-a$x[a$y==max(a$y)]
p<-ggplot(dat, aes(x=pp))+geom_density(fill="lightgreen", alpha=.4, size=1) p+ geom_vline(aes(xintercept=mmod), color="red", linetype="dashed", size=1)+theme_bw()+xlab("Determinant")+ylab("")
Compare the mode of the distribution to the theoretical, maximum possible determinant:
>R1<-matrix(c(1,.8,.2,.8,1,.16,.2,.16,1),3,3) > det(R1) [1] 0.3456 > mmod ##this is the mode of the distribution above [1] 0.3344818
Close within 0.011 error. Which leads me to make the following claim:
Claim: For the “missing correlation” problem, the distribution of the determinants of the correlation matrix concentrate around the value of r that maximizes it.
I honestly have no clue how any of these two results would be useful once they are formalized. I am sensing that something like this may be able to play a role in error detection or diagnosing Heywood cases in Factor Analysis? I mean, for the first case (error detection) say you find a correlation matrix within the published literature that is not positive definite. If correlation matrices tend to concentrate around values that maximize their determinants, then you could potentially use this framework to pick and choose ranges of possible sets of correlations to point out where the problem may lie. A similar logic could be used for Heywood cases, ESPECIALLY if the dimensions of the correlation matrix are large. Or maybe all of this is BS and a very elaborate excuse for me to procrastinate. The world will never know ¯\_(ツ)_/¯ | 2019-06-17 09:03:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 63, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7578868865966797, "perplexity": 460.645454550066}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998462.80/warc/CC-MAIN-20190617083027-20190617105027-00215.warc.gz"} |
https://www.r-bloggers.com/2019/01/analysis-of-south-african-funds/ | [This article was first published on Digital Age Economist on Digital Age Economist, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)
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## Packages used in this post
Disclaimer: I am no financial advisor, have never been and you should not take any of this analysis as investment advice. These thoughts are my own, please dont mail me about your money strategies/problems. I enjoy numbers, scraping and data analysis and that is wat this post is about. Also, do you really want to trust someone who wrote this post at 2am… no you dont
library(tidyverse)
library(glue)
library(rvest)
library(gridExtra)
library(ggpubr)
library(scales)
library(ggthemes)
theme_set(theme_economist())
# Why even do this?
Well, I recently decided to do a quick overview of some of the funds I am invested in as part of a retirement annuity. Its been almost 4 years since I started the RA and thought I would just view the fund’s performance, its asset composition and the fees that have been charged over the period.
If you do not know what an RA is and why you would need one, you can think of it as a way to get back at the tax man while investing over a long period. In South Africa certain funds are what are called Regulation 28 compliant, which means, any money you put into such a fund is tax-exempt – but you just have to leave it there till you retire, which isn’t actually a bad thing thanks the miracle of compounded growth. I think having an RA is a cornerstone of any good financial planning. So enough of sounding like your parents.
OK, so at this point in time, you probably like this:
Well, actually you are kind of right of thinking this. If you invest in a retirement annuity on a regular basis and over the contract period you should have some money for retirement, in theory. However, the market for funds are huge and which one do you pick? Financial advisers could give you good advice, but its also in their interest to invest you in multiple complex funds, as they get management fees in return1.
The easiest way to get a grips with the cost involved when investing in a fund is what is called Total Expense Ratios. These are important, because if the fund is costing you 3.5%, annualised growth of fund is 8% and inflation is 6%, then you not gaining of the advantage of compounded growth, but actually losing -1.5% a year in real terms. An example of a TER is as follows:
Investopedia gives a good explanation of these fees
The total expense ratio (TER) is a measure of the total costs associated with managing and operating an investment fund, such as a mutual fund. These costs consist primarily of management fees and additional expenses, such as trading fees, legal fees, auditor fees and other operational expenses. The total cost of the fund is divided by the fund’s total assets to arrive at a percentage amount, which represents the TER, most often referred to as simply “expense ratio”
So what I set out to do is look at the TERs for South African funds and then compare it with their performance of the last 5 years. To collect the data I scraped all the information from Fundsdata.co.za as they have a wide collection of different funds and categories
I used rvest and combined it with a bit of purrr magic that got me the results I wanted. The data was a bit dirty so I just cleaned one or two things before starting the analysis.
main_pg <- read_html("http://www.fundsdata.co.za/navs/default.htm")
pg_funds <- main_pg %>%
html_nodes("tr.pagelink") %>%
html_nodes("a") %>%
html_text()
pg_links <- main_pg %>%
html_nodes("tr.pagelink") %>%
html_nodes("a") %>%
html_attr("href")
all_funds <- tibble(funds = pg_funds, link = glue("http://www.fundsdata.co.za/navs/{pg_links}"))
link <- "http://www.fundsdata.co.za/navs/ZEGN.htm"
collect_ter <- possibly(function(link){
cat(glue("Now collecting {link}"), "\n")
pg <- read_html(link)
pg_tbl <- html_table(pg, fill = T)[[1]][, -20]
x <- pg_tbl %>%
slice(-c(1:2)) %>%
set_names(make.names(pg_tbl[2,], unique = T)) %>%
tbl_df %>%
slice(-c(n():(n()-3)))
Sys.sleep(10)
x
}, NULL)
all_funds <- all_funds %>%
filter(grepl("South Africa", funds)) %>%
mutate(info = map(link, collect_ter))
#saveRDS(all_funds, "all_funds.rds")
After the collection is done, you have a neat tibble that looks like this:
## # A tibble: 6 x 3
## funds link info
## <chr> <S3: glue> <list>
## 1 South African - Equity - General Funds http://www.fu~ <tibble~
## 2 South African - Equity - Industrial Funds http://www.fu~ <tibble~
## 3 South African - Equity - Large Cap Funds http://www.fu~ <tibble~
## 4 South African - Equity - Mid and Small Cap Funds http://www.fu~ <tibble~
## 5 South African - Equity - Unclassified Funds http://www.fu~ <tibble~
## 6 South African - Equity - Resource Funds http://www.fu~ <tibble~
Next I wanted to calculate the annualised rate of return of the each of the funds just to get an idea of what kind of returns are out there. The data only goes back 5 years, so I am going to work from that.
The formula for CAGR (compounded annual growth rate) is simply:
$$CAGR = (\frac{F}{P})^{1/n} - 1$$
Where:
• F is future value
• P is present value
• n is number of years
Once I have the annualised return, I add one or two extra columns of information. I calculate a sharpe ratio as a measure of risk adjusted returns (assume 5% risk free rate), I also adjust the cagr for inflation2. The website was also kind enough to add an asterisk(*) to the sheet indicating funds which are eligible to the tax break aka Reg28 compliant.
all_funds <- readRDS("data/all_funds.rds")
annual_ret <- function(info){
info %>%
select(Fund.Name, TER, X5.yearsCash.Value, Volatility.Ann...) %>%
set_names(c("fund_name", "TER", "cash_value_5yrs", "volatility")) %>%
mutate_at(vars(-one_of("fund_name")), as.numeric) %>%
mutate(
# growth rate
annualised_ret = ((cash_value_5yrs/100)^(1/5)-1)* 100,
# risk adjusted return
risk_adj_cagr = annualised_ret * (1 - volatility),
# sharpe ration for risk adjusted return
sharpe = (annualised_ret - 5)/volatility,
# inflation adjustment
inflation_adj_cagr = annualised_ret - 5.5) %>%
mutate(reg_28_compliant = grepl("[*]", fund_name))
}
all_funds_ret <- all_funds %>%
mutate(returns = map(info, annual_ret)) %>%
select(funds, returns) %>%
unnest %>%
arrange(-annualised_ret) %>%
filter(!is.na(TER))
all_funds_ret
## # A tibble: 1,182 x 10
## funds fund_name TER cash_value_5yrs volatility annualised_ret
## <chr> <chr> <dbl> <dbl> <dbl> <dbl>
## 1 South Af~ Absa Propert~ 1.72 175. 19.1 11.9
## 2 South Af~ Long Beach F~ 1.53 167. 14.9 10.9
## 3 South Af~ Nedgroup Inv~ 1.36 166. 10.7 10.7
## 4 South Af~ Nedgroup Inv~ 1.93 162. 10.7 10.1
## 5 South Af~ Momentum Fin~ 1.46 161. 11.1 9.94
## 6 South Af~ Fairtree Fle~ 0.900 160. 3.08 9.82
## 7 South Af~ Satrix FINI ~ 0.410 159. 14.1 9.78
## 8 South Af~ Fairtree Equ~ 1.16 158. 12.7 9.58
## 9 South Af~ *Pan-African~ 0.770 158. 3.80 9.51
## 10 South Af~ Sesfikile BC~ 1.28 156. 11.5 9.37
## # ... with 1,172 more rows, and 4 more variables: risk_adj_cagr <dbl>,
## # sharpe <dbl>, inflation_adj_cagr <dbl>, reg_28_compliant <lgl>
Now that we have all the pieces, lets see what the average TER is for funds in South Africa.
OK doki, so what we see is that on average, the cost of the the non Reg28 funds are cheaper, with the median TER sitting around 1.21%. This means that we will need to choose quite carefully when we decide on which fund to invest in.
Next I wanted to see what the distribution of returns look like between the Reg28 and non-Reg28 funds3.
Its nice to see that although the fees are higher for the Reg28 funds, the returns also seem higher with a bi-modal distribution. This could be a something like a sector effect. I also wanted to check whether there was some statistical significance to this. So I conduct a wilcoxon test. A wilcoxon is a non-parametric test to check whether the true location shift is equal to 0. Here we reject the hypothesis and confirm that Reg28 do indeed have higher returns.
At what cost is the higher returns? Are the Reg28 funds more risky or risk-averse? To answer this question, lets turn to our sharpe ratio. Anything about 1 is considered good and anything about 2 we starting to see some really good returns on low risk investments. There was some outliers, so I removed all more than 3.5 as to focus on the meat of data.
Its by a small margin, but the Reg28 funds seem to be delivering higher return with a lower risk profile.
The last bit of analysis I did was to have a look at the inflation adjusted returns over the last 5 years. The market took quite a beating over the last year, mostly due to some of our presidential factors and also because of some “other” presidential factors in the developed world. So lets look at how the different market sectors did over the last 5 years.
Financial equities did well along with some interest bearing funds. This is not completely out of place. Five years can be considered a short window of investment. Our market equity market is also high concentrated with Naspers driving a lot of the growth in recent years through their investment in a company called Tencent in China.
# Conclusion
This write up was partly to get a grip on how my own RA is doing amid the sea of other options. I also wanted to get an idea of what a reasonable TER is and what a good return on the higher TERs are - i.e are the complex funds delivering higher returns? It would seem so, but even the best funds haven’t beaten the inflation trackers. This is basically because of the short period I am looking at as well as the fact that 2018 sucked for investments in SA (and globally)
The other part of why I wrote this blog, was because I wanted to explore the ggpubr package. Really enjoy some of the features that it has when it comes to quick reporting and visualisation.
1. Again, I am not saying that financial advisers are the devil, I am exaggerating for effect. I have a financial adviser
2. Doing a quick check, in SA the average inflation over the past 5 years was ~5.5%
3. From here on in I used the ggpubr package quite a lot. Really enjoyed it. Do wish it had NSE though
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Click here to close (This popup will not appear again) | 2022-05-25 13:19:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.20044314861297607, "perplexity": 3664.829963743058}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662587158.57/warc/CC-MAIN-20220525120449-20220525150449-00445.warc.gz"} |
https://dlmf.nist.gov/LaTeXML/manual/math/details/representation.html | # § 5.1.1 Internal Math Representation
(November 17, 2020)
The XMath element is the container for the internal representation
The following attributes can appear on all XM* elements:
role
the grammatical role that this element plays
open, close
parenthese or delimiters that were used to wrap the expression represented by this element.
argopen, argclose, separators
delimiters on an function or operator (the first element of an XMApp) that were used to delimit the arguments of the function. The separators is a string of the punctuation characters used to separate arguments.
xml:id
a unique identifier to allow reference (XMRef) to this element.
## Math Tags
The following tags are used for the intermediate math representation:
XMTok
represents a math token. It may contain text for presentation. Additional attributes are:
name
the name that represents the meaning of the token; this overrides the content for identifying the token.
omcd
the OpenMath content dictionary that the name belongs to.
font
the font to be used for presenting the content.
style
?
size
?
stackscripts
whether scripts should be stacked above/below the item, instead of the usual script position.
XMApp
represents the generalized application of some function or operator to arguments. The first child element is the operator, the remainig elements are the arguments. Additional attributes:
name
the name that represents the meaning of the construct as a whole.
stackscripts
?
XMDual
combines representations of the content (the first child) and presentation (the second child), useful when the two structures are not easily related.
XMHint
represents spacing or other apparent purely presentation material.
name
names the effect that the hint was intended to achieve.
style
?
XMWrap
serves to assert the expected type or role of a subexpression that may otherwise be difficult to interpret — the parser is more forgiving about these.
name
?
style
?
XMArg
serves to wrap individual arguments or subexpressions, created by structured markup, such as \frac. These subexpressions can be parsed individually.
rule
the grammar rule that this subexpression should match.
XMRef
refers to another subexpression,. This is used to avoid duplicating arguments when constructing an XMDual to represent a function application, for example. The arguments will be placed in the content branch (wrapped in an XMArg) while XMRef’s will be placed in the presentation branch.
idref
the identifier of the referenced math subexpression. | 2021-07-29 02:09:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5609822869300842, "perplexity": 3211.3584893667953}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153814.37/warc/CC-MAIN-20210729011903-20210729041903-00237.warc.gz"} |
https://math.stackexchange.com/questions/3918666/probability-that-3-hands-of-a-clock-in-the-same-semi-circle?noredirect=1 | # Probability that $3$ hands of a clock in the same semi-circle?
I am wondering how to calculate the probability that $$3$$ hands of a clock are in the same semi-circle ? I know a similar question is that if we randomly choose n points in a circle, then the probability that all of them in the same semi-circle will be $$n/2^{(n-1)}$$ with $$n = 3$$.
But when it comes to the dial hands, where I guess there are correlations between them, I am not sure whether my question is equivalent to the previous one. If they are not equivalent, and this question will be hard to calculate, then whether we can determinate this probability will be greater or smaller than $$3/4$$?
Anyone can help? Thanks.
• Yes $3$ points on a circle in the same half is your question. Think this way. If two hands are making angle $\theta$, the third can be anywhere in $(2 \pi - \theta)$ of the circle. The only area that does not work is angle $\theta$ just opposite of it. Here is a link - math.stackexchange.com/questions/342293/… – Math Lover Nov 22 '20 at 20:55
• I'm not sure if this is helpful, but I checked this computationally, and the probability appears to be approaching $3/4$ as you suspected and @MathLover claimed. However, I'm not entirely convinced because as you pointed out, the hands are correlated (in fact, assuming exact clocks, the hour hand determines the place of the other hands), so this argument about "the third [hand] can be anywhere in $(2\pi - \theta)$" needs a little more to make it precise, possibly something about small perturbations. – Math Helper Nov 23 '20 at 2:52
• @DavidC.Ullrich yes but it is a good approximation. If we take hour and minute hands at a given time, the movement of seconds hand has to be within a certain angle. If we do this over $12$ hours, it is a good approximation. – Math Lover Nov 25 '20 at 15:35
• @DavidC.Ullrich If you take a certain time, the positions of all three hands are fixed. The probability of them being in the same semicircle is really based on what time you are looking at at it. I said $12$ hours because in $12$ hours window, you would have covered all possible positions and angles between $3$ hands (including angles between hour and minute hands). – Math Lover Nov 25 '20 at 15:56
• The answer would depend on what time of day it is – wolfies Nov 26 '20 at 3:42
The hands in fact move deterministically; it is just the time when we view them that can be random. The movement repeats ever twelve hours, so assume we view the hands at some moment uniformly distributed between noon and midnight. Then the probability they will be within a semicircle is the number of hours they spend in a semicircle, divided by $$12.$$
Starting at noon, the hands are within one semicircle until the second hand is opposite the hour hand. The second hand makes $$12\times 60 = 720$$ revolutions in $$12$$ hours while the hour hand makes one revolution in the same time, so the time until the second and hour hands first align after noon is $$12/719$$ hour, which is $$720/719$$ minutes; but the time from noon until the hands are exactly opposite is half that time, $$360/719$$ minute.
For a short time after the first $$360/719$$ minute, the hands are not within a semicircle. They are in a semicircle again starting at the instant when the second hand is opposite the minute hand. The second hand aligns with the minute hand for the first time after noon at $$60/59$$ minutes after noon, so the first time they are opposite is at $$30/59$$ minute after noon.
Let $$t_h = 360/719$$ and $$t_m = 30/59,$$ and let's take the convention that "at time $$t$$" means at $$t$$ minutes past noon. So the first time interval when the hands are not in a semicircle starts at $$t_h$$ and ends at $$t_m.$$ The length of this interval is $$t_m - t_h$$ minutes; specifically, the time during which the hands are continuously not in a semicircle, in minutes, is $$\frac{30}{59} - \frac{360}{719} = \frac{330}{42421}.$$
The second such time interval starts at $$3t_h$$ and ends at $$3t_m$$ for $$3(t_m - t_h)$$ minutes not in a semicircle, the third starts at $$5t_h$$ and ends at $$5t_m$$, and so forth as long as the minute hand remains less than $$180$$ degrees clockwise from the hour hand.
This is a promising pattern: after $$n$$ intervals like this, we will have spent a total of $$n^2(t_m - t_h)$$ minutes during which the hands do not fit in a semicircle.
Alas, the full solution is not that simple. The intervals increase in length until they are almost half a minute long, and then this simple pattern is broken.
The minute hand first becomes exactly opposite the hour hand at $$6/11$$ hour after noon, that is, at $$360/11$$ minutes after noon. Shortly before that, at $$65t_h$$ minutes, the second hand passes opposite to the hour hand, making the hands not within a semicircle; but at $$360/11$$ minutes after noon the second hand is suddenly finds itself between the minute and hour hands so that the hands are all within a semicircle again, even though $$65t_m$$ minutes have not yet passed.
After this abbreviated interval of not-in-a-semicircle time, the next interval is almost half a minute long, and the next several intervals decrease in length. But once they have decreased enough, they start increasing again, and the pattern repeats but with different intervals each time as the relative position of the second hand is different each time the minute hand overtakes the hour hand or comes opposite to it.
At about this point in the analysis I decided I would rather have a computer count the time for me. The events where the hands start fitting in a semicircle or stop fitting in a semicircle all occur at exact integer multiples of either $$1/11,$$ $$1/59,$$ or $$1/719$$ minute after noon. Since those fractions have a common denominator of $$11\times 59\times 719 = 466631,$$ every event of interest (where the hands start or stop being within a semicircle) occurs at some multiple of $$1/466631$$ minute after noon. By taking $$720\times 466631$$ equal timesteps of that length we cover twelve hours. At the start of each timestep we know from the relative positions of the three hands whether the hands will fit in a semicircle or not during that timestep. (It is one way for the entire duration of the timestep.)
I wrote the analysis in the python script shown below. I had the script count the first six hours, take a checkpoint, and then do the last six hours. Everything after $$6$$ o'clock is a mirror image of what happened before, in reverse order, so the same sequence of lengths of intervals occurs in reverse order after $$6$$ o'clock. I had the script record the start, end, and duration of each interval to verify this. Note that since each interval starts about a minute after the previous one, there are almost $$360$$ intervals in six hours, but not quite $$360$$ due to the way the hands "catch up" with each other.
According to my script, it takes $$720\times 11\times 59\times 719 = 335974320$$ timesteps to complete twelve hours, and during $$83982960$$ of those timesteps the three hands do not fit within a semicircle. The proportion of time not in a semicircle is therefore $$\frac{83982960}{335974320} = \frac{1977}{7909} \approx 0.2499683904412694.$$ So the probability that the hands are not within a semicircle is almost, but not quite, $$1/4$$; the probability that they are within a semicircle is therefore slightly over $$3/4.$$
To relate these calculations to G Cab's method, let $$x$$ be the number of units of $$12/11$$ hour since noon (that is, $$x$$ is $$0$$ at noon and $$11$$ at midnight). If $$k = \lfloor x\rfloor$$ (an integer), then for each $$k$$ (that is, for each period of $$12/11$$ hour) there are two sequences of intervals during which the three hands are not within a semicircle. During the first half of this period, while the angle from the hour hand to the minute hand is between $$0$$ and $$180$$ degrees clockwise, the three hands are not in a semicircle whenever $$m + \frac12 \leq \frac{15(2k)}{22} + \frac{719}{11} \tau < m + \frac12 + \tau$$ for some integer $$m,$$ where $$\tau = x - k,$$ and during the second half of this period, while the angle from the minute hand to the hour hand is between $$0$$ and $$180$$ degrees clockwise, the three hands are not in a semicircle when $$n + \tau \leq \frac{15(2k + 1)}{22} + \frac{719}{11} \tau < n + \frac12$$ for some integer $$n,$$ where $$\tau = x - k - \frac12.$$ In either case we also have the restriction that $$0 \leq \tau < \frac12.$$
In the typical case, the beginning or end of an interval occurs when one of the inequalities becomes an equation; the exceptions occur when the equation is true only for $$\tau < 0$$ or $$\tau > \frac12$$. Solving these equations for $$\tau,$$ we find $$\tau = \frac{11}{719} \left(m + \frac12 - \frac{15(2k)}{22}\right)\tag1$$ at the start and $$\tau = \frac{11}{708} \left(m + \frac12 - \frac{15(2k)}{22}\right)\tag2$$ at the end of a typical interval in the first sequence, but $$\tau = \frac{11}{708} \left(n - \frac{15(2k+1)}{22}\right)\tag3$$ at the start and $$\tau = \frac{11}{719} \left(n + \frac12 - \frac{15(2k+1)}{22}\right)\tag4$$ at the end of one in the second sequence. The length of the first interval (in the same units as $$\tau$$) is therefore \begin{align} \Delta\tau_1 &= \frac{11}{708} \left(m + \frac12 - \frac{15(2k)}{22}\right) - \frac{11}{719} \left(m + \frac12 - \frac{15(2k)}{22}\right) \\ &= \frac{121}{708(719)} \left(m - \frac{30k - 11}{22}\right) \end{align} and the length of the second interval is \begin{align} \Delta\tau_2 &= \frac{11}{719} \left(n + \frac12 - \frac{15(2k+1)}{22}\right) - \frac{11}{708} \left(n - \frac{15(2k+1)}{22}\right) \\ &= \frac{121}{708(719)} \left(\frac{30k + 723}{22} - n\right) \end{align}
The combined length of two intervals, one from each sequence, is \begin{align} \Delta\tau_1 + \Delta\tau_2 &= \frac{121}{708(719)} \left(m - \frac{30k - 11}{22}\right) + \frac{121}{708(719)} \left(\frac{30k + 723}{22} - n\right) \\ &= \frac{11}{708(719)} \left(367 + 11(m - n)\right) \\ \end{align}
If we hold $$m - n$$ constant, $$\Delta\tau_1 + \Delta\tau_2$$ is independent of the value of $$m$$ as long as all of the $$\tau$$ values from which we computed $$\Delta\tau_1$$ and $$\Delta\tau_2$$ are between $$0$$ and $$\frac12.$$ For example, if we set $$n = m$$, we get a pairing of intervals in which each pair of intervals has total length $$\frac{11(367)}{708(719)}$$ except in some cases where an interval starts at $$\tau=0$$ or ends at $$\tau=\frac12$$ or where an interval has no corresponding interval in the other sequence. Trial and error shows that if we set $$n=m+1$$ we minimize the number of intervals that are unmatched or that are in a pair where one interval ends at $$\tau=0$$ or $$\tau=\frac12$$ instead of where the equations above say it will (there are $$14$$ such intervals); each matched pair then has total length $$\frac{11(367-11)}{708(719)} = \frac{11(89)}{3(59)(719)}$$ in the units of $$\tau.$$ This is equal to $$\frac{356}{59(719)}$$ hour, which is $$234960$$ timesteps of the brute-force method.
According to my spreadsheet calculations, if $$n = m+1$$ there are $$32$$ matched pairs of intervals for each value of $$k.$$ since there are $$11$$ values of $$k,$$ the total time in intervals that belong to typical matched pairs is $$352 \times 234960$$ timesteps. It then remains to account for the $$14$$ intervals that are not in typical matched pairs. Those intervals can be identified for each $$k$$ by finding the maximum $$m$$ such that right-hand side of Equation $$(1)$$ is negative and the minimum $$n$$ such that the right-hand side of Equation $$(4)$$ is greater than $$\frac12.$$ A relatively quick hand calculation should therefore be possible, though it would still be longer than I would like to write in detail in this answer.
Update: The calculations are not as long as I feared after all, as shown in G Cab's edited answer.
Here is the script for the brute-force method:
# Program to figure out what proportion of the time the three hands
# of a clock are not within a semicircle.
# Assume there are 59*719*720 equally-spaced positions around the clock face.
# Note: 59*719 == 42421; 42421*11 = 466631.
# Each timestep of this program is 1/42421 of a minute.
# In one timestep, the second hand moves 720 positions, the minute hand
# moves 12 positions, and the hour hand moves 1 position.
# In 42421 timesteps, one minute has passed and the second hand returns
# to the starting position,
numPositions = 11*59*719*720
# Returns the new position after having moved the given amount forward
# from the given position. That is, the return value is the sum of
# the two input values, modulo numPositions.
# It is assumed the input values are in the range 0 to numPositions - 1.
def newPosition(fromPosition, movement):
toPosition = fromPosition + movement
if toPosition >= numPositions:
toPosition -= numPositions
# Returns the number of steps clockwise to get from position1 to position2.
# positions measured in number of steps from the start.
def stepsClockwise(position1, position2):
distance = position2 - position1
if distance < 0:
distance += numPositions
return distance
# Returns 1 if the hands are at the start of a timestep in which the
# smallest angle encompassing all three hands will be greater than 180 degrees,
# 0 if the angle will be less than 180 degrees.
def isLargeAngle(hoursPosition, minutesPosition, secondsPosition):
# A 180 degree angle
halfCircle = numPositions // 2
# A position exactly opposite the seconds hand
antiSecondsPosition = newPosition(secondsPosition, halfCircle)
minutesOffset = stepsClockwise(hoursPosition, minutesPosition)
if minutesOffset < halfCircle:
# The minutes hand is < 180 degrees clockwise from the hour hand
antiOffset = stepsClockwise(hoursPosition, antiSecondsPosition)
if (antiOffset >= 0 and antiOffset < minutesOffset):
# The anti-seconds hand will be between the hours and minutes
# during this timestep, so the seconds hand will be more than
# 180 degrees past the hour hand but less than 180 degrees
# before the minute hand.
return 1
else:
return 0
else:
# The hour hand is <= 180 degrees clockwise from the minute hand.
# During the next timestep the short arc between the hands will be
# clockwise from the minute hand to the hour hand.
hoursOffset = stepsClockwise(minutesPosition, hoursPosition)
antiOffset = stepsClockwise(minutesPosition, antiSecondsPosition)
if (antiOffset >= 0 and antiOffset < hoursOffset):
# The anti-seconds hand will be between the minutes and hours
# during this timestep, so the seconds hand will be more than
# 180 degrees past the minute hand but less than 180 degrees
# before the hour hand.
return 1
else:
return 0
# Runs the given number of timesteps from the given starting position
# and returns the number of times the hands spanned more than 180 degrees.
def countLargeAngles(hoursPosition,
minutesPosition,
secondsPosition,
startingTimestep,
numTimesteps):
numLargeAngles = 0
wasLargeAngle = 0
largeAngleStarted = 0
numIntervals = 0
for i in range(numTimesteps):
if isLargeAngle(hoursPosition, minutesPosition, secondsPosition):
numLargeAngles += 1
if wasLargeAngle == 0:
largeAngleStarted = startingTimestep + i
wasLargeAngle = 1
else:
if wasLargeAngle == 1:
largeAngleStopped = startingTimestep + i
numIntervals += 1
print(">180 degrees interval ", numIntervals,
" from ", largeAngleStarted,
" to ", largeAngleStopped, " for ",
(largeAngleStopped - largeAngleStarted), " timesteps")
wasLargeAngle = 0
hoursPosition = newPosition(hoursPosition, 1)
minutesPosition = newPosition(minutesPosition, 12)
secondsPosition = newPosition(secondsPosition, 720)
print("After ", numTimesteps, " timesteps")
print("hour hand at ", hoursPosition)
print("minute hand at ", minutesPosition)
print("second hand at ", secondsPosition)
print(numLargeAngles, " positions with > 180 degrees")
return numLargeAngles, hoursPosition, minutesPosition, secondsPosition
# Run 360*466631 timesteps to simulate 6 hours,
# then another 360*466631 timesteps to simulate the next 6 hours.
halfTotalTimesteps = numPositions // 2
hoursPosition = 0
minutesPosition = 0
secondsPosition = 0
(numLargeAngles,
hoursPosition,
minutesPosition,
secondsPosition) = countLargeAngles(hoursPosition,
minutesPosition,
secondsPosition,
0,
halfTotalTimesteps)
hoursPosition,
minutesPosition,
secondsPosition) = countLargeAngles(hoursPosition,
minutesPosition,
secondsPosition,
halfTotalTimesteps,
halfTotalTimesteps)
print("Total number of timesteps was ", numPositions)
print("Total number of > 180 degrees was ", numLargeAngles)
• Your clock seems to be digital rather than analog. Wouldn’t your computation have been a little easier with an analog clock? – Lubin Nov 23 '20 at 4:04
• @Lubin It's an analog clock, but instead of using floating-point arithmetic, I recognized that all events on the clock face (at least, all events that are interesting for this problem) occur at rational numbers of minutes past noon, and I found the common denominator for all of those numbers. This lets me do a straightforward timestep-and-count. With floating-point numbers, I could have calculated the start and and of each interval without the common denominator, but I think there would have been more cases to worry about, and the results would not be exact. – David K Nov 23 '20 at 4:29
• Well, since using my own mathematical common sense, I got a wrong answer, I can’t complain too much. But it does seem to me that a more conceptual approach would have involved a lot less talk. – Lubin Nov 23 '20 at 4:50
• @Lubin I could have saved a lot of talk by starting at the paragraph that begins, "At about this point in the analysis ...," but I thought I needed a justification for taking a brute-force approach. I have the vague recollection of having done a problem somewhat like this without resorting to computer programming, but it was a frightful mess then too IIRC. (Too bad I haven't been able to find my former solution yet.) – David K Nov 23 '20 at 4:54
Let's denote by $$\omega$$ the angular speeds, so $$\left\{ \matrix{ \omega _{\,h} = {{2\pi } \over {12h}} = {{2\pi } \over {12 \cdot 3600}}s^{\, - 1} \hfill \cr \omega _{\,m} = {{2\pi } \over h} = {{2\pi } \over {3600}}s^{\, - 1} \hfill \cr \omega _{\,s} = {{2\pi } \over m} = {{2\pi } \over {60}}s^{\, - 1} \hfill \cr} \right. \quad \Rightarrow \quad \left\{ \matrix{ \omega _{\,m} - \omega _{\,h} = 11\;\omega _{\,h} \hfill \cr \omega _{\,s} - \omega _{\,h} = 719\;\omega _{\,h} \hfill \cr} \right.$$
Clearly we are interested in the relative angles $$\left\{ \matrix{ \alpha _{\,m} = \left( {\omega _{\,m} - \omega _{\,h} } \right)t\, \quad \left( {\bmod 2\pi } \right)\quad \Rightarrow \quad \rho _{\,m} = {{\alpha _{\,m} } \over {2\pi }} = \left\{ {11{{\omega _{\,h} \,t} \over {2\pi }}} \right\}\, \hfill \cr \alpha _{\,s} = \left( {\omega _{\,s} - \omega _{\,h} } \right)t\, \quad \left( {\bmod 2\pi } \right)\quad \Rightarrow \quad \rho _{\,s} = {{\alpha _{\,s} } \over {2\pi }} = \left\{ {719{{\omega _{\,h} \,t} \over {2\pi }}} \right\} \hfill \cr} \right.$$
The hands will be on the same half circle when $$\left\{ \matrix{ 0 \le \alpha _{\,m} < \pi \hfill \cr 0 \le \alpha _{\,s} < \pi \; \vee \hfill \cr \;\pi + \alpha _{\,m} \le \alpha _{\,s} < 2\pi \hfill \cr} \right.\;\; \vee \;\;\left\{ \matrix{ \pi \le \alpha _{\,m} < 2\pi \hfill \cr \pi \le \alpha _{\,s} < 2\pi \; \vee \; \hfill \cr 0 \le \alpha _{\,s} < \alpha _{\,m} - \pi \hfill \cr} \right.$$ or more simply they will not be so when $$\left\{ \matrix{ 0 \le \alpha _{\,m} < \pi \hfill \cr \pi \le \alpha _{\,s} < \pi + \alpha _{\,m} \; \hfill \cr} \right.\;\; \vee \;\;\left\{ \matrix{ \pi \le \alpha _{\,m} < 2\pi \hfill \cr \alpha _{\,m} - \pi \le \alpha _{\,s} < \pi \; \hfill \cr} \right.$$
The cycle will endure $$12$$ h, that is $$0 \le {{\omega _{\,h} \,t} \over {2\pi }} < 1$$ and let's put $$11{{\omega _{\,h} \,t} \over {2\pi }} = k + x\quad \left| \matrix{ \;0 \le k \in \mathbb Z \le 10 \hfill \cr \;0 \le x \in \mathbb R < 1 \hfill \cr} \right.$$
Then using $$[ ]$$ to denote the Iverson bracket and $$\overline K (k,x)$$ to denote the indicator function of the "forbidden" area, we get \eqalign{ & \overline K (k,x)\quad \left| \matrix{ \;0 \le k \le 10 \hfill \cr \;0 \le x < 1 \hfill \cr} \right.\quad = \cr & = \left[ {0 \le x < {1 \over 2}} \right]\left[ {{1 \over 2} \le \left\{ {{{719} \over {11}}k + {{719} \over {11}}x} \right\} < {1 \over 2} + x} \right] + \cr & + \left[ {{1 \over 2} \le x < 1} \right]\left[ {x - {1 \over 2} \le \left\{ {{{719} \over {11}}k + {{719} \over {11}}x} \right\} < {1 \over 2}} \right] = \cr & = \left[ {0 \le x < {1 \over 2}} \right]\left( \matrix{ \left[ {{1 \over 2} \le \left\{ {{{719\left( {2k} \right)} \over {22}} + {{719} \over {11}}t} \right\} < {1 \over 2} + x} \right] + \hfill \cr + \left[ {x \le \left\{ {{{719\left( {2k + 1} \right)} \over {22}} + {{719} \over {11}}x} \right\} < {1 \over 2}} \right] \hfill \cr} \right) = \cr & = \left[ {0 \le x < {1 \over 2}} \right]\left( \matrix{ \left[ {{1 \over 2} \le \left\{ {{{15\left( {2k} \right)} \over {22}} + {{719} \over {11}}x} \right\} < {1 \over 2} + x} \right] + \hfill \cr + \left[ {x \le \left\{ {{{15\left( {2k + 1} \right)} \over {22}} + {{719} \over {11}}x} \right\} < {1 \over 2}} \right] \hfill \cr} \right) \cr}
Therefore the probability $$\overline P$$ of not having the hands on the same half circle will be \eqalign{ & \overline P = {1 \over {11}}\sum\limits_{k = 0}^{10} {\int_{\,x\, = \,0}^{\,1} {\overline K (k,x)dx} } = \cr & = {1 \over {11}}\sum\limits_{k = 0}^{10} {\int_{\,x\, = \,0}^{\,1/2} {\,\left( \matrix{ \left[ {{1 \over 2} \le \left\{ {{{15\left( {2k} \right)} \over {22}} + {{719} \over {11}}x} \right\} < {1 \over 2} + x} \right] + \hfill \cr + \left[ {x \le \left\{ {{{15\left( {2k + 1} \right)} \over {22}} + {{719} \over {11}}x} \right\} < {1 \over 2}} \right] \hfill \cr} \right)dx} } \cr}
$$\overline K (k,x)dx$$ is represented by the segments of the two sawtooth waves intercepted
inside the respective forbidden areas like those shown in the following sketch.
Note that for better clarity the inclination $$m$$ of the lines is much lower than $$719/11$$.
The integral translate into the sum of the $$\Delta x$$ of each segment.
To compute them let's start with computing the intercepts with $$y=1/2$$.
The minimum intercept $$s_{\,0}$$ will be \eqalign{ & y = \left\{ {q + m\,x} \right\} = \left\{ {\left\{ q \right\} + m\,x} \right\} = {1 \over 2}\quad \Rightarrow \cr & \Rightarrow \quad \left\{ q \right\} + m\,x = n + {1 \over 2}\quad \Rightarrow \cr & \Rightarrow \quad m\,x = n - \left( {\left\{ q \right\} - {1 \over 2}} \right) \cr & \Rightarrow \quad m \, s_{\,0} = \left\lceil {\left\{ q \right\} - {1 \over 2}} \right\rceil - \left( {\left\{ q \right\} - {1 \over 2}} \right) = \left( {{1 \over 2} - \left\{ q \right\}} \right) - \left\lfloor {{1 \over 2} - \left\{ q \right\}} \right\rfloor = \cr & = \left\{ {{1 \over 2} - q} \right\} \cr} For the lower triangle we can simply replace $$q$$ with the negative of the intercept with $$x=1/2$$ , i.e.: $$m\,s_{\,0} = \left\{ {{1 \over 2} + {m \over 2} + q} \right\}$$
Next we take the upper triangle and slant it as shown
where $$\Delta y(x) = \left\{ {\matrix{ {{m \over {m - 1}}x} & {\left| {\;0 \le x < {1 \over 2}\left( {1 - {1 \over m}} \right)} \right.} \cr {m\left( {{1 \over 2} - x} \right)} & {\left| {\;{1 \over 2}\left( {1 - {1 \over m}} \right) \le x < {1 \over 2}} \right.} \cr } } \right.$$ so that we will compute $$\Delta x$$ as $$1/m \, \Delta y$$.
Finally, we are sampling the slanted triangle at points starting from $$s_{\, 0}$$ (which depends on $$q$$ and thus on $$k$$) and progressing with $$\Delta s = 1/m$$. $$s(k,j) = s_{\,0} + {j \over m} = \left\{ {\matrix{ \matrix{ {1 \over m}\left\{ {{1 \over 2} - q} \right\} + {j \over m} = {{11} \over {719}}\left( {\left\{ {{1 \over 2} - {{15\left( {2k} \right)} \over {22}}} \right\} + j} \right) = \hfill \cr = {{11} \over {719}}\left( {\left\{ {{{11 - 8k} \over {22}}} \right\} + j} \right) = s_{\,u} (k,j) \hfill \cr} \hfill & {{\rm upper}\;{\rm Tr}{\rm .}} \hfill \cr \matrix{ {1 \over m}\left\{ {{1 \over 2} + {m \over 2} + q} \right\} + {j \over m} = {{11} \over {719}}\left( {\left\{ {{1 \over 2} + {{719} \over {22}} + {{15\left( {2k + 1} \right)} \over {22}}} \right\} + j} \right) = \hfill \cr = {{11} \over {719}}\left( {\left\{ {{{19 + 8k} \over {22}}} \right\} + j} \right) = s_{\,d} (k,j) \hfill \cr} \hfill & {{\rm lower}\;{\rm Tr}{\rm .}} \hfill \cr } } \right.$$
Since \eqalign{ & {{11} \over {719}}\left( {\left\{ {{{11 - 8k} \over {22}}} \right\} + j} \right) = {1 \over {2 \cdot 719}}\left( {22\left\{ {{{11 + 14k} \over {22}}} \right\} + 22j} \right) \cr & 22\left\{ {{{11 + 14k} \over {22}}} \right\}\quad \Rightarrow \quad 11 + 14k = 2n + 1\quad \left( {\bmod 22} \right)\quad \Rightarrow \cr & \Rightarrow \quad 5 + 7k = n\quad \left( {\bmod 11} \right)\quad \Rightarrow \quad 7k \equiv n + 5\quad \left( {\bmod 11} \right) \cr}
and \eqalign{ & {{11} \over {719}}\left( {\left\{ {{{19 + 8k} \over {22}}} \right\} + j} \right) = {1 \over {2 \cdot 719}}\left( {22\left\{ {{{19 + 8k} \over {22}}} \right\} + 22j} \right) \cr & 22\left\{ {{{19 + 8k} \over {22}}} \right\}\quad \Rightarrow \quad 19 + 8k = 2n + 1\quad \left( {\bmod 22} \right)\quad \Rightarrow \cr & \Rightarrow \quad 9 + 4k = n\quad \left( {\bmod 11} \right)\quad \Rightarrow \quad 4k \equiv n + 2\quad \left( {\bmod 11} \right) \cr}
then the {Independent Residues theorem](https://en.wikipedia.org/wiki/Residue_number_system)
assures us that the the two relations above are bijections between
$$\left[ {0 \le k \le 10} \right] \Leftrightarrow \left[ {0 \le n \le 10} \right]$$
We have that the set of the sampling points are the same for both the triangles, and we can simply put
$$\left\{ \matrix{ s_u (n) = s_d (n) = {{2n + 1} \over {2 \cdot 719}} \hfill \cr 0 \le s(n) < {1 \over 2}\left( {1 - {1 \over m}} \right) = {1 \over 2}\left( {{{708} \over {719}}} \right)\quad \Rightarrow \quad 0 \le n \le 353 \hfill \cr {1 \over 2}\left( {1 - {1 \over m}} \right) \le s(n) < {1 \over 2}\quad \Rightarrow \quad 354 \le 2n \le 358 \hfill \cr} \right.$$
And finally obtain
\eqalign{ & \overline P = {1 \over {11}}\sum\limits_s {\Delta x(s)} = {1 \over {11m}}\sum\limits_s {\Delta y(s)} = {2 \over {11m}}\sum\limits_{\,0 \le n \le 358} {\Delta y(s_u (n))} = \cr & = 2\left( {{1 \over {708}}\sum\limits_{\,0 \le n \le 353} {{{2n + 1} \over {2 \cdot 719}}} + {1 \over {11}}\sum\limits_{\,354 \le n \le 358} {\left( {{1 \over 2} - {{2n + 1} \over {2 \cdot 719}}} \right)} } \right) = \cr & = {1 \over {719}}\left( {{1 \over {708}}\sum\limits_{\,0 \le n \le 353} {\left( {2n + 1} \right)} + {1 \over {11}}\sum\limits_{\,354 \le n \le 358} {\left( {718 - 2n} \right)} } \right) = \cr & = {1 \over {719}}\left( {{{354 + 354 \cdot 353} \over {708}} + {{5 \cdot 718 - 2 \cdot 5 \cdot 356} \over {11}}} \right) = \cr & = {1 \over {719}}\left( {{{354} \over 2} + {{30} \over {11}}} \right) = {{1977} \over {7909}} = {1 \over 4}\left( {1 - {1 \over {7909}}} \right) \cr}
• I worked the formula out in Excel (on 363 lines of a spreadsheet) and came up with exactly the same answer I got in by brute-force counting the intervals in python. (After I corrected the script, that is; while trying to translate the Excel figures into brute-force intervals I found a non-integer value on the 33rd line, which tipped me off that my interval was too large by a factor of 11.) – David K Nov 26 '20 at 4:32
• Each Iverson bracket in the final integral, treated as a function of $t,$ is $1$ over a finite set of intervals. It turns out that most of the intervals can be paired off in such a way that the sum of lengths of two consecutive intervals is $21360/42421$ minute. There are maybe a dozen intervals that I did not manage to pair up this way, all starting at $t=0$ or ending at $t=1/2,$ suggesting that there may be a way to collect most of the terms of the calculation in a relatively simple expression and just have to handle these exceptions separately. – David K Nov 26 '20 at 4:48
• @DavidK: yes there is somehow the possibility to group up the intervals. I will work on that also – G Cab Nov 26 '20 at 10:25
• @DavidK: in fact it is possible to convert the integral in a sum. I recasted my answer accordingly – G Cab Nov 30 '20 at 21:51
• At one point $\left\{\frac12 - q\right\}$ turns into $\left\{1 - q\right\}$ in the upper-triangle formulas. The formula with $\frac12 - q$ matches my results for the upper triangle. But by symmetry (comparing a backward-running clock to a regular clock), the intervals in the lower triangles should be the upper-triangle intervals in reverse order. I have not been able to reproduce that effect with the latest formulas in this answer. – David K Dec 1 '20 at 3:38
Take two of the hands. On average they will be a quarter of a circle apart. Distance varies uniformly from zero to half a circle apart. Turn the dial so that one hand is at 12 and the other hand at 3. Now when the third hand is in the interval from 9 to 6 all three hands are on the same semicircle. When the third hand is between 6 and 9 they are not on the same semicircle. So on average the three hands will be on the same semicircle 3/4 of the time. | 2021-06-18 18:29:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 148, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9958645105361938, "perplexity": 729.7809013416947}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487640324.35/warc/CC-MAIN-20210618165643-20210618195643-00195.warc.gz"} |
http://mathhelpforum.com/discrete-math/166800-set-theory-question.html | # Math Help - Set theory question.
1. ## Set theory question.
Hi all,
In Kunen's book, one of the exercises is the following:
Show that for $\kappa > \omega$, |H( $\kappa$)| = $2^{< \kappa}$.
But, H( $\omega$) = V( $\omega$) and |V( $\omega$)| = $\omega$. Also, $2^{< \omega}= \omega$ (because $2^{< \omega}$ is the cardinality of the union of omega many finite sets (the characteristic functions for each n)).
Is this wrong?
Sam
2. The notation used in this post is so non-standard, I think that you need to define what each symbol means or how it is used.
3. Ok, so:
$\kappa$ is used as a variable over cardinals.
| $x$| denotes the cardinality of $x$.
$\alpha^{<\beta}$ is the cardinality of the union of $\alpha^{\delta}$ for $\delta < \beta$.
H( $\kappa$) denotes the set of sets which are hereditarily of cardinality $< \kappa$.
V( $\alpha$) is the $\alpha^{th}$ level in the cumulative hierarchy.
Hope this helps.
Please say if I've left anything important out.
Regards
Sam
4. Just a tangential point: the notation used is not non-standard as far as I am aware. Kunen's book is a standard text, which contains all of this notation except V(\alpha). In Kunen R is used instead of V. In most modern treatments, though, V is used. See Von Neumann universe - Wikipedia, the free encyclopedia, for an example.
Regards
Sam | 2014-09-30 22:10:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 19, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9441054463386536, "perplexity": 747.547024692322}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1412037663167.8/warc/CC-MAIN-20140930004103-00219-ip-10-234-18-248.ec2.internal.warc.gz"} |
https://ninnat.github.io/sep-oct-2019-log.html | # September-October 2019 Archive
This is an archive of what I found useful, interesting, or simply enjoyable from September to October 2019.
# Writings
You would have to live under a rock to not hear of the leaked Google’s paper and the subsequent official announcement of their demonstration of “quantum supremacy”.
• Back before the leak, I gave an online talk (organized by Quantum Technology Foundation of Thailand) in which I tried to explain what quantum supremacy really means by explaining the “collapse of the polynomial hierarchy (PH)”.
• Misunderstanding ensued after the Google’s announcement went viral (for example, that we now have a fully functional quantum computer that can break codes), which compelled me to write Google’s quantum supremacy FAQ (in Thai) targeted at an interested layperson.
# Research
## Quantum computational supremacy
### A literature review
My attempt at classifying some milestone supremacy results 1
Worst Aaronson’11
Bremner’10
(Multiplicative)
Aaronson’11
Bremner’10
Postselection
Average Aaronson’11
Bouland’18
Conjectured
Bremner’16
Boixo’18
Stockmeyer
Random circuits anticoncentrate
Reviews
Boson sampling
Random circuit sampling
### What does an NP oracle actually do?
• In my experience, the most subtle concept to correctly explain when talking about PH is the idea of an oracle.
An oracle for a problem $O$ is a magical black box that can answer the question $O$ with a single querying. Generally having access to an oracle boosts the power of a computational model. For instance, one can have an oracle for an undecidable problem; there is no Turing machine that can decide whether an arbitrary Turing machine halts or not, but an oracle for the halting problem (upon inputting a Turing machine $M$) would just give the one-bit answer HALT or NOT HALT in one step.
Similarly, one can have an oracle for an NP-complete problem, 3SAT for concreteness. Every Boolean formula can be expressed in the conjunctive normal form (CNF) $\vee_j \left(\wedge_k u_{jk}\right)$ where each $u_{jk}$ is a Boolean variable, $j$ labels the clauses, and $k$ labels the literals. The problem kSAT asks if a given Boolean formula wherein each clause has $k$ literals is satisfiable i.e. the value of each $u_{jk}$ can be set so that the formula evaluates to TRUE. NP is the class of problems where there is a short certificate for every YES instance. For example in 3SAT, if a Boolean formula is satisfiable, I can prove it by giving you a set of satisfying arguments. SAT and 3SAT are NP-complete, while 2SAT is in P.
But note the asymmetry; there is no obvious short proof that I can show to convince you that a Boolean formula is unsatisfiable without trying all combinations of the Boolean variables. A class of problems such that there is a short proof for every NO instance is called coNP. Factoring is special in that both YES and NO instances can be checked quickly, so it is in $\mathrm{NP \cap coNP}$ and is not believed to be NP-complete.
For the above reason, $\mathrm{P^{NP}}$ is believed to be larger than NP (containing both NP and coNP) because when I ask the oracle “is this Boolean formula satisfiable?”, the oracle can give the answer NO, which seems to be beyond the capability of an NP machine. Similarly, $\mathrm{NP^{NP}}$ is believed to be larger than NP. Hence we define the polynomial hierarchy to be the infinite tower $\mathrm{NP^{NP^{NP^{\cdots}}}}$, each level $\Sigma_{k+1} = \Sigma_k^{\mathrm{NP}}$ appears to be more computationally powerful than the one below. (We define $\Sigma_0 = \mathrm{P}$ and $\Sigma_1 = \mathrm{NP}$.)
## Quantum state learning
• A Survey of Quantum Property Testing, Montanaro and Wolf 2016 [Chapter 2 handwritten notes]
• Sample-optimal tomography of quantum states, Haah et al 2017
For i.i.d. $\rho^{\otimes N}$, the trace distance decays exponentially with the rate the Chernoff information, which is between $(\ln F^{-1})/2$ and $\ln F^{-1}$ for $F$ the fidelity, so the fidelity is a good figure of merit for the task of distinguishing between $\rho^{\otimes N}$ and $\sigma^{\otimes N}$, while the single-copy trace distance tends to overestimate the sample complexity. 2
• Adaptive quantum state tomography improves accuracy quadratically, Mahler et al 2013
How the infidelity scales with the number of samples depends on the eigenvalues of $\rho$ itself (and the eigenbasis of the measurement). Let $G(\epsilon) = \sqrt{\sqrt{\rho}(\rho + \epsilon\Delta)\sqrt{\rho}}$. We will perturbatively calculate the line element $ds^2 = 2(1-\Tr G(\epsilon))$ to second order. Begin by squaring the equation $G = (\rho + \epsilon X + \epsilon^2 Y)$ and match powers of $\epsilon$. \begin{align} \rho + \epsilon\sqrt{\rho}\Delta\sqrt{\rho} = \rho + \epsilon\left( X\rho + \rho X \right) + \epsilon^2\left( X^2 + Y\rho + \rho Y\right) \end{align} The first order equation is the same as saying that $X$ is the derivative of $\sqrt{G}$. Matrices are non-commutative and one cannot use the usual chain rule that $d\sqrt{G} = G^{-1}dG/2$. (Indeed, should $G^{-1}$ be on the left or on the right?) Instead, one differentiates the defining equation $\sqrt{G}\sqrt{G} = G$ (assuming that $G$ is diagonalizable and positive and pick the positive square root $\sqrt{G} = \sum_{j}\sqrt{g_j}\ketbra{j}{j}$). In the eigenbasis of $\rho$, \begin{align} X_{jk} &= \frac{\sqrt{p_j p_k}\Delta_{jk}}{p_j + p_k} \\ Y_{jj} &= -\frac{\av{j|X|j}^2}{2p_j} = -\sum_k \frac{|\av{j|X|k}|^2}{2p_j} \\ \Tr Y &= - \frac{1}{2} \sum_{jk} \frac{p_k\Delta_{jk}^2}{(p_j + p_k)^2} = - \frac{1}{4} \sum_{j,k} \frac{\Delta_{jk}^2}{p_j + p_k}, \end{align} where the last equality is obtained by grouping each pair $(j,k)$ together, writing the sum over all pairs as the sum over $j,k$, and multiplying by 1/2 to prevent overcounting.
The point is that when all eigenvalues are far away from zero, the first order term disappears because $\Tr X = \Tr \Delta/2 = 0$, leaving only the result \begin{align} ds^2 = 2 \left( 1 - \Tr\rho - \Tr X - \frac{1}{2!}\Tr Y - O(\Delta^3) \right) = \frac{1}{4} \sum_{j,k} \frac{\Delta_{jk}^2}{p_j + p_k} + O(\Delta^3) \end{align} (We have “set $\epsilon=1$”, but $\epsilon$ has to be small for the pertubative calculation to work so a better way to think about it is that we have absorbed $\epsilon$ into $\Delta$.)
However, when one or more $p_j$ are zero, the sum over $j$ in trace $\Tr X = \sum_{j\neq 0} \Delta_{jj}/2$ will be restricted to those terms with nonzero $p_j$, giving \begin{align} ds^2 = -\Tr \Delta + O(\Delta^2) = \sum_{j=0} \Delta_{jj} + O(\Delta^2). \end{align} Thus a mis-estimation of small eigenvalues of $\rho$ blows up to linear order in the infidelity.
• Shadow Tomography of Quantum States, Aaronson 2017
• Predicting Features of Quantum Systems using Classical Shadows, Huang and Küng 2019
# Media
• LIGO might be detecting a black hole-neutron star merger or something else. (“Given the vastness of the universe, ambiguous detections like S190814bv are likely to be the norm for this new, multimessenger era of astronomy”)
• Our experiments, conducted on thousands of participants, suggest that trust is not lost if the communicators can be confident about their uncertainty. This is fine if we have a good survey, because then we can quantify our margins of error, as for unemployment. But surveys may not be reliable, and there may be extra sources of systematic bias which cast doubt on the calculated intervals.
• Theoretical Physics as a Rigorous Approach to Understand Nature, ธิปรัชต์ โชติบุตร
• A not-so-flattering history of Chinese science
• Public opinion in authoritarian states (“The purpose of censorship and thought management differs from regime to regime. The contention of this post is that for many of the most effective authoritarian systems, controlling the thoughts of the ruled is secondary to shaping social cleavages in the population.”)
• Fast internet disrupts political calm in Norilsk, Russia.
“At the direction of the acting prosecutor,” a typical announcement in the municipal newspaper reads, “an examination has been conducted based on information posted on the social network known as ‘Instagram’ on the network known as ‘the internet.’”
• Indian economic reform is better than you think, SSC
• Want To Know Whether A Psychology Study Will Replicate? Just Ask A Bunch Of People
• Debunking the Stanford Prison Experiment, Le Texier 2019 [PsychCentral]
• Severely Deficient Autobiographical Memory
• Techno-superstition in the age of “explainable AI”
• What is the dog there for? “we will soon be moving to a new flight crew; one pilot and a dog. The pilot is there to feed the dog, and the dog is there to bite the pilot if he touches anything.”
The findings were dispiriting. Green and Chen found that using algorithms did improve the overall accuracy of decision-making across all conditions, but this was not because adding information and explanations enabled the humans to play a more meaningful role in the process. On the contrary, adding more information often made the human interaction with the algorithm worse. When given the opportunity to learn from the real-world outcomes, the humans became overconfident in their own judgments, more biased, and less accurate overall. When given explanations, they could maintain accuracy but only to the extent that they deferred more to the algorithm. In short, the more transparent the system seemed to the worker, the more the workers made them worse or limited their own agency.
It is important not to extrapolate too much from one study, but the findings here are consistent what has been found in other cases of automation in the workplace: humans are often the weak link in the chain. They need to be kept in check. This suggests that if we want to reap the benefits of AI and automation, we may have to create an environment that is much like that of the Skinner box, one in which humans can flap their wings, convinced they are making a difference, but prevented from doing any real damage. This is the enchanted world of techno-superstition: a world in which we adopt odd rituals and habits (explainable AI; fair AI etc) to create an illusion of control.
• Neural network reduces errors in climate models and found that we underestimate sea-level rise by 300%
## Anime
1. The nomenclature “strong” and “weak” simulation comes from Van den Nest. Terhal and DiVincenzo called the former “density computation” and the latter just simulation. A strong simulation computes an arbitrary outcome probability $p_z$ of a quantum circuit (up to some precision) while a weak simulation samples from a probability distribution $q_z$ closed to the ideal one. An efficient strong simulation implies an efficient weak simulation with no error $|p_z - q_z| = 0$.
2. Suppose that the error in the trace distance is $\tau$, then for small $\epsilon$, $\frac{\epsilon}{2} \le \tau \le \sqrt{\epsilon}$ because $1-\sqrt{F} \le T \le \sqrt{1-F}$. Thus, the scaling $O(1/\epsilon)$ in the infidelity translates to the scaling $O(1/\epsilon^2)$ in the trace distance. Note that it doesn’t matter for small infidelity $\epsilon$ whether we use the definition $F(\rho,\sigma) = \left(\mathrm{Tr}\sqrt{\sqrt{\rho}\sigma\sqrt{\rho}}\right)^2$ or its square root for the fidelity because $\epsilon = 1-\sqrt{1-\epsilon_{\surd}} \approx \epsilon_{\surd}/2$.
Archieves by date, by category, by tag. | 2019-11-20 07:10:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 4, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.905476987361908, "perplexity": 1016.604608083567}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496670512.94/warc/CC-MAIN-20191120060344-20191120084344-00018.warc.gz"} |
https://math.stackexchange.com/questions/3853140/understanding-cross-product-definition-derivation-by-3blue1brown | # Understanding Cross Product Definition Derivation by 3blue1brown
I've been following through with 3blue1brown's linear algebra series, and I have a question regarding the definition of the cross product he gives. https://www.youtube.com/watch?v=BaM7OCEm3G0&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=11
$$\begin{bmatrix} p_1 \\p_2 \\p_3 \end{bmatrix} \cdot \begin{bmatrix} x \\y \\z \end{bmatrix} = det\left(\begin{bmatrix} x & v_1 & w_1\\ y & v_2 & w_2\\ z & v_3 & w_3 \end{bmatrix}\right)$$
where p is the resultant vector from the cross product of v and w for any x y and z. 3blue1brown essentially says that since the determinant of a matrix is the area of the parallelepiped with side lengths of the column vectors, the determinant is also just the height of that parallelepiped times the base of it. And the dot product of p and xyz is the projection of xyz on p, times the magnitude of p. If p is a vector perpendicular to v and w, then the projection of the final side of the parallelepiped(xyz) onto that perpendicular vector would be the height of the parallelepiped, and then the magnitude of p would the area of the base.
So that makes logical sense, but according to this definition couldn't this entire cone of vectors also be solutions? The Cone of Vectors
since the projection of xyz on p and the magnitude of p stays the same?
• It's not clear why you believe that "this entire cone of vectors" also work as solutions. Are you claiming that each vector in the cone is perpendicular to both $v$ and $w$? – Ben Grossmann Oct 6 '20 at 2:12
• Here is a mathematical fact: suppose that $v$ and $w$ are two non-parallel vectors. If $p,q$ are (non-zero) vectors such that $p$ is perpendicular to both $v$ and $w$ and $q$ is perpendicular to both $v$ and $w$, then $p$ and $q$ must be parallel. Is this a fact that you were not aware of? Is this the source of your confusion? – Ben Grossmann Oct 6 '20 at 2:17
• @BenGrossmann My confusion is that 3blue1brown simply states that one perpendicular vector to v and w satisfies the this equation, when I think I can see a whole range of solutions, being that cone according to that definition he gave. Since the dot product is just |a| * |b| * cos(theta), and the magnitudes of p and xyz can stay the same while p rotates around xyz at the same angle that it was when it was perpendicular to v and w, cos(theta) can remain constant, giving a range of solutions. I feel like i'm missing some simple fact since there very obviously is only one answer algebraically. – SpyGuyTBM Oct 6 '20 at 3:04
• I've asked some yes/no questions, it would be very helpful if you could respond to those – Ben Grossmann Oct 6 '20 at 15:26
• @BenGrossmann Sorry, yes I am aware of the fact that p and q would be parallel if they were both perpendicular to v and w. And no, i'm claiming that the dot product of each vector in that cone with the white vector is the same. – SpyGuyTBM Oct 6 '20 at 16:29
The point is that we're looking for a vector $$\vec p$$ that has the correct dot-product for every choice of (white) vector $$(x,y,z)$$.
It is true that any choice of vectors from the cone gives you the correct dot product for the particular input shown in the figure, but the point behind the duality argument is that our $$p$$ needs to correctly encode the function that takes in a vector $$(x,y,z)$$ and produces the corresponding area. It is only true that $$\vec p$$ correctly encodes the function if $$(p_1,p_2,p_3) \cdot(x,y,z)$$ gives the right output for every possible input.
We can see by plugging in $$(x,y,z) = \vec v$$ that $$\vec p$$ needs to be perpendicular to $$\vec v$$ (since the volume should be $$0$$). Similarly, $$\vec p$$ needs to be perpendicular to $$\vec w$$. | 2021-06-18 17:48:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 13, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8530364036560059, "perplexity": 158.52605713743196}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487640324.35/warc/CC-MAIN-20210618165643-20210618195643-00410.warc.gz"} |
https://physics.stackexchange.com/questions/232007/force-over-area-diagram | # Force over Area diagram [closed]
In this question a have to draw a F/A diagram in function of p-V (pressure-volume) diagram below.
I think in a intuitive way that this graphic of F/A it´s going to be the same graphic because Pressure it´s equal to Force and Volume it´s proportional to Area.
• Here, from a) volume seems proportional to length and Area seems to be independent Jan 27 '16 at 5:53
• Force is not equal to Pressure. It is proportional to Pressure.
– user104909
Jan 28 '16 at 8:43
I think in a intuitive way that this graphic of F/A it´s going to be the same graphic because Pressure it´s equal to Force and Volume it´s proportional to Area.
The pressure $p$ left of the piston exerts a force $F$ on the piston with cross-section $A$:
$$F=pA$$
So:
$$p=\frac FA$$
A graph for (a) of $\frac FA$ v. volume $V$ will indeed be similarly shaped as for (b), assuming (b) is indeed a $p,V$ diagram. | 2022-01-25 03:32:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8066748976707458, "perplexity": 1024.552254012247}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304749.63/warc/CC-MAIN-20220125005757-20220125035757-00427.warc.gz"} |
https://www.concepts-of-physics.com/pw/html/a/ca/oca.html | IIT JEE Physics (1978-2018: 41 Years) Topic-wise Complete Solutions
# Variation of Light Intensity with Distance!
## Procedure
This was demonstrated by Ms Smita Fangaria at Kolkata SRP. Take two bulbs, of different power, say 60 watt and 100 watt. Take sheet of paper. Apply some oil/butter at a spot. This spot becomes translucent. Put two bulbs at some distance, say 1 metre. Move the paper between two sources till intensity on opaque and translucent part of the paper looks same. Measure the distance. See whether $$I\propto\frac{1}{r^2}$$.
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http://fonda107.com/s1qbuwu/79b36e-select-the-titrant-used-in-redox-titration | Position of the end point . Answer. Potassium permanganate (KMnO₄) is a popular titrant because it serves as its own indicator in acidic solution. choosing the proper indicator you should select one that has a pK a as close to the endpoint of the titration. In an acid–base titration or a complexation titration, the titration curve shows how the concentration of H 3 O + (as pH) or M n + (as pM) changes as we add titrant. Below 4.5 Highly dirty The concept behind Winkler method is 0-2 Cannot sustain life Redox titration, it is a type of titration that deals with chemicals that undergo a In the experiment the water sample used, reaction that alter their oxidation state of was from Vinzon’s pond. Redox reactions are carried out in the same way as acid-base titrations using a burette and a known concentration of one reactant (titrant) and an unknown concentration of the other reactant (analyte). We will use potassium permanganate, KMnO4, as the titrant in the analysis of an unknown sample containing iron to determine the percent iron by mass in the sample. Redox titrations are based on a reduction-oxidation reaction between an oxidizing agent and a reducing agent. A potential change of 0.2 V is needed for... A potential change of 0.2 V is needed for a sharp end point. Selecting an Indicator for a Redox Titration. It is a redox titration where the end point of the titration is determined by the presence and the absence of the I2. Redox Titrations are carried out on the basis of oxidation-reduction reactions, (more commonly known as redox reactions) between the analyte and the titrant. Wiktionary Acid-base neutralisation titrations are one of the more straightforward titrations, but they are an excellent example for helping students become familiar with the technique. CC BY-SA 3.0. http://en.wikipedia.org/wiki/Redox_titration You may need … Iodine reacts directly, fast and quantitively with many organic and inorganic substances. It is therefore possible to see when the titration has reached its endpoint, because the solution will remain slightly purple from the unreacted KMnO4. priyankapatil01812 is waiting for your help. Not all titrations require an external indicator. Sometimes halogens (or organic compounds containing halogens) other than iodine are used in the intermediate reactions because they are available in better-measurable standard solutions or they react more readily with the analyte. 1. Self Indication 2. Redox Titration. 1. You can specify conditions of storing and accessing cookies in your browser. When chemical indicators are not suitable, a potentiometric pH titration can also be used. Salt (chemistry) Manganese Permanganate Kappa number Titration. These types of titrations sometimes require the use of a potentiometer or a redox indicator. Titration is a common laboratory method of quantitative/chemical analysis that can be used to determine the concentration of a known reactant. Moreover, the addition of Sulphuric acid also helps to increase the hydrogen ions present in the solution. So that, additional indicators are not required in those cases. Select a titrant used in redox titration Get the answers you need, now! A redox titration is a type of titration based on a redox reaction between the analyte and titrant. When we add a redox indicator to the titrand, the indicator imparts a color tha t depends on the solution’s potential. As the solution’s potential changes with the addition of titrant, the indicator changes … What would be the cell potential at the following titration volumes: a) 15.00 mL b) 25.00 mL c) 32.53 mL A titration is performed with a saturated calomel reference electrode (S.C.E.) For a redox titration it is convenient to monitor the titration reaction’s potential instead of the concentration of one species. Part 2. Sometimes an indicator is required, but many redox titrations have color changes that occur naturally due to the transfer of electrons. Redox titrations can be used to determine the concentration of: (i) ethanol in wine or beer (ii) hypochlorite in bleach. Iodometry is the indirect titration of iodine liberated by another reactant through an oxidation reduction. The balanced reaction in acidic solution is as follows: $MnO_4^-(aq)+5Fe^{2+}(aq)+8H^+(aq) \rightarrow 5Fe^{3+}(aq)+Mn^{2+}(aq)+4H_2O(l)$. We know from our balanced equation above that permanganate and iron react in a 1:5 mole ratio. In this type of tit-ration are named after the reagent that is used … This is called an iodometric titration. CC BY-SA 3.0. http://en.wikipedia.org/wiki/Titration satishshinde443 satishshinde443 1 hour ago Chemistry Secondary School MCQ: the indicator used in redox titration is 2 See answers satishshinde443 is waiting for your help. The titrant is the standardized solution; the analyte is the analyzed substance. 4+ 2++ Fe 3+ → Ce. In this type of tit-ration , the chemical reaction takes place with a transfer of electrons in the reacting ions of aqueous solution. The titrant reacts with a solution of analyte … Redox Titration • Redox titration is based on the redox reaction (oxidation-reduction) between analyte and titrant. When used in redox titration, it get reduced into brown coloured Mn2+ ion(In acidic media) at end point and colour change at end point can be detected easily. Properties of Umass Boston . Iodometry and Iodimetry: Iodine is a mild oxidizing … thiosulfate (S 2 O 3 2−), and when all iodine is spent the blue colour disappears. Methyl orange changes color at the pH of mid strength acid. A student conducts the redox titration and reaches the endpoint after adding 25 mL of the titrant. This types of titrations are quite common in usage next to acid-base titrations. Probably the most frequently carried out redox titrations are iodometric titrations using iodine as the titrant (in the form of the water-soluble triiodide, KI 3) and thiosulfate. In a titration, one reagent (the titrant) is slowly added to a solution containing the species being measured (the analyte). One of the earliest precipitation titrations—developed at the end of the eighteenth century—was the analysis of K 2 CO 3 and K 2 SO 4 in potash. In analytical chemistry, a standardized aqueous solution of KMnO 4 is sometimes used as an oxidizing titrant for redox titrations (permanganometry). Medium. While these extra steps make an iodometric titration much more involved, they are often worthwhile, because the equivalence point involving the bright blue iodine-starch complex is more precise than various other analytical methods. MCQ: the indicator used in redox titration is Get the answers you need, now! Indicators used in this case can be methyl red or methyl orange which is orange in acidic solution and yellow in basic and neutral solutions. Since detecting end point is … 67.60 mg. b. It is the pH indicator commonly used in titration. Select one: a. Redox Titration is a laboratory method of determining the concentration of a given analyte by causing a redox reaction between the titrant and the analyte. Select a titrant used in redox titration, KATPUTLI TERI MAIN❤JIVEN MARZI ME ❤KHED LAVIN ❤TERE LIYE LADJO RABNA ❤AZMAA KE VEKH LAVIN ❤TU HASDA VE MERA RAAB❤ HASDA TERE ANDAR MERA❤ KHUDA WASDA T Wiktionary The method is easy to use if the quantitative relationship between two reacting substances is known. For this application, H 2 O 2 content is established with redox titration methods utilizing cerium (IV) sulfate as a titrant and DMi140-SC electrode. concentration of the iodine titrant used for Ripper titration of wine by ... Thermo Scientific™ Orion™ Sure-Flow™ Redox/ORP Electrode, Epoxy Body 9678BNWP Accessories Electrode Storage Sleeve and Bottle 810017 Stirrer Probe 927007MD Swing Arm Stand 090043 Standard Titration Add 5.0 mL of 0.01 N sodium thiosulfate solution to 50 mL DI in a 100 mL beaker. Redox titration curves Equation ( 5 ) can be used for the calculation of E eq for any redox titration except in the following two cases we should use equation ( 4 ) : Case one : If one of the participants of the redox reaction does not change its oxidation state during the reaction e.g. Add your answer and earn points. Its maintenance takes place with the use of dilute sulphuric acid. When we add a redox indicator to the titrand, the indicator imparts a color that depends on the solution’s potential. Figure 4 shows a traditional strong acid-strong base titration curve. Examples: Some popular examples of this titrations are as follows; (a) Permanganate Titrations: The use of potassium permanganate is as an oxidizing agent. by Lucia Meier | May 18, 2020 | Analysis, Evaluating results, Fundamentals, Lab hacks | 0 comments. The endpoint of this titration is detected with the help of a starch indicator. Please do not block ads on this website. priyankapatil01812 priyankapatil01812 3 minutes ago Chemistry Secondary School Select a titrant used in redox titration priyankapatil01812 is waiting for your help. Rodex titration depends on an oxidation-reduction reaction that occurs between the analyte and the titrant. H+ 2in the titration of Fe + with MnO 4-. Boundless Learning Thus, it can be understood that redox titrations involve a transfer of electrons between the given analyte and the titrant. Acid-base titrations require a different electrode than redox or precipitation titrations. Potassium dichromate is also a very strong oxidizing agent (E°red = +1.33V) . Precipitation titration is a type of titration which involves the formation of precipitate during the titration technique. The redox titration curve is a plot of Electrode Potential (volts) vs volume of titrant or analyte. It is used to determine chloride by using silver ions. A standardized 4 M solution of KMnO4 is titrated against a 100 mL sample of an unknown analyte containing Fe2+. He titrated halide ions using potassium permanganate using a shiny platinum electrode and a calomel electrode. The pH of the solution is plotted versus the volume of titrant added. Wiktionary Double platinum electrodes like the M241Pt2-8 from Radiometer Analytical are ideal for this type of titration. When we add a redox indicator to the titrand, the indicator imparts a color that depends on the solution’s potential. titrationa method in which known amounts of the titrant are added to the analyte until the reaction reaches the endpoint, titrantthe standardized solution used in titrations; the solution of known concentration. CC BY-SA 3.0. http://en.wiktionary.org/wiki/reducing_agent, http://www.boundless.com//chemistry/definition/oxdizing-agent, http://en.wikipedia.org/wiki/Redox_titration, https://www.boundless.com/chemistry/textbooks/boundless-chemistry-textbook/. INTRODUCTION In a reaction with the -thiosulphate ion (S2O32-), iodine (I2) is reduced to iodide (I) and the thiosulphate is oxidized to the tetrathionate ion … A common example is the redox titration of a standardized solution of potassium permanganate (KMnO4) against an analyte containing an unknown concentration of iron (II) ions (Fe2+). The versatility of EDTA can be ascribed to the different ways in which the complexometric titration can be executed. Titration (also known as titrimetry and volumetric analysis) is a common laboratory method of quantitative chemical analysis to determine the concentration of an identified analyte (a substance to be analyzed). When performing redox titrations, either the reducing or oxidizing agent will be used as the titrant against the other agent. In this case, starch is used as an indicator; a blue starch-iodine complex is formed in the presence of excess iodine, signaling the endpoint. as the anode and a platinum wire as a cathode. …, Find the IUPAC name of Please don't Spam., 3.Calculate the number of atoms in 32 u of He, sometimes,when no party gets a clear majority--------- government is formed, Find the IUPAC name of Please dont Spam. This particular resource used the following sources: http://www.boundless.com/ 50% (1/1) analytes analyte species. Thanks to its relatively low, pH independent redox potential, and reversibility of the iodine/iodide reaction, iodometry … Ce. 1. This is essentially the reverse titration of what was just described; here, when all the iodine has been reduced, the blue color disappears. The method differs from volumetric titration in that the titrant is generated in situ by electrolysis and then reacts … This is called an iodometric titration. CC BY-SA 3.0. http://en.wiktionary.org/wiki/titration Redox titration is based on an oxidation-reduction reaction between the titrant and the analyte. Introduction: In this experiment, oxidation/reduction (or redox) will be used in the titration analysis of an iron compound. The indicators used in redox reactions are sensitive to change in oxidation potential. The balanced reaction in acidic solution is as follows: $MnO_4^-(aq)+5Fe^{2+}(aq)+8H^+(aq) \rightarrow 5Fe^{3+}(aq)+Mn^{… Titrant used and the reaction principle that proceeds usually defines name of the titration - like acid-base (or alkalimetric) titration if we use strong acid (or strong base) as a titrant, or redox when the reaction that proceeds is of a redox type. CC BY-SA 3.0. http://en.wiktionary.org/wiki/analyte Wiktionary Which titrant is used in the Iodometric titration which involves I 2 ? In a redox titration, it is the potential rather than the pH that changes with concentration . What is the concentration of the analyte? Analytical chemistry Immunoassay Clinical … In alkaline solution methyl orange is in yellow color. Rinse the electrode and stirrer Page 7 of 20 Selecting an Indicator for a Redox Titration The most important class of indicators for redox titrations are substances that do not participate in the redox titration, but whose oxidized and reduced forms differ in color. 2KMnO 4 + 3H 2 SO 4 → K 2 SO 4 + 2MnSO 4 + 3H 2 + 5O. Wikipedia This technique is mainly used in redox measurements, in particular in iodometry. Cerimetry employs cerium (IV) salts. As mentioned earlier, EDTA is a versatile chelating titrant that has been used in innumerable complexometric determinations. Potentiometric titrations were first used for redox titrations by Crotogino. Iodine (I 2) can be reduced to iodide (I −) by e.g. But, most frequently either the analyte or the titrant produces a color at the endpoint. TITRATE/ ANALYTE Titrate is (analytical chemistry) to ascertain the amount of a constituent in a solution (or other mixture) … Most often, the reduction of iodine to iodide is the last step in a series of reactions in which the initial reactions are used to convert an unknown amount of the analyte to an equivalent amount of iodine, which can then be titrated. Add your answer and earn points. In a redox titration, one solution is a reducing agent and the other an oxidizing agent. Such titrations, classified according to the nature of the chemical reaction occurring between the sample and titrant, include: acid-base titrations, precipitation titrations, complex-formation titrations, and oxidation-reduction (redox) titrations. In this experiment, you will conduct two separate redox titrations using a … It continues till the last amount of analyte is consumed. Add your answer and earn points. A redox titration is a titration in which the analyte and titrant react through an oxidation-reduction reaction. hope this helps you .... have a great day, This site is using cookies under cookie policy. In acidic solution, potassium permanganate rapidly and quantitively oxidizes iron (II) to iron (III), … In this case, the use of KMnO4 as a titrant is particularly useful, because it can act as its own indicator; this is due to the fact that the KMnO4 solution is bright purple, while the Fe2+ solution is colorless. The redox titration often needs a redox indicator or a potentiometer. The most important class of indicators for redox titrations are substances that do not participate in the redox titration, but whose oxidized and reduced forms differ in color. Maintenance of the solution is done with the help of dilute sulphuric acid. Iodometry is one of the most important redox titration methods. A common example is the redox titration of a standardized solution of potassium permanganate (KMnO4) against an analyte containing an unknown concentration of iron (II) ions (Fe2+). A substance, such as a solution, of known concentration used in titration. Because volume measurements play a key role in titration, it is also known as volumetric analysis.A reagent, called the titrant, of known concentration (a standard solution) and volume is used to react with a solution of the analyte, … Redox titration determines the concentration of an unknown solution (analyte) that contains an oxidizing or reducing agent. A common example of a redox titration is treating a solution of iodine with a reducing agent to produce iodide using a starch indicator to help detect the endpoint. Three methods used for visual indication. A reagent, termed the titrant or titrator, is prepared as a standard solution of known concentration and volume. More clearly, we can state that in a redox reaction, a reducing agent reacts with an oxidizing agent until the required results are obtained. Generally for redox titration platinum or gold electrode (inert metal) are used. Sample preparation and procedures CC BY-SA 3.0. http://en.wiktionary.org/wiki/reducing_agent When acid is added in the solution it gives red color. Determine the end point • Indicator electrode • Redox indicators – the indicator has different color at reduction and oxidation state. We can therefore perform the following calculation: Now that we know the number of moles of iron present in the sample, we can calculate the concentration of the analyte: [latex]M=\frac {mol}L= \frac {0.5 mol}{0.100 L}=5 M$. Titration is used in analytical chemistry to determine the amount or concentration of a substance. , why cu show dsp2 hybridization in [Cu(NH3)4]2+. Boundless vets and curates high-quality, openly licensed content from around the Internet. What to consider during back-titration. ... Redox titrations; Precipitation titrations; ... coulometric titration was originally developed by Szebelledy and Somogy [1] in 1938. Which titrant is used in the Iodometric titration which involves I 2 ... K M n O 4 B. K 2 C r 2 O 7 C. N a 2 S 2 O 3 D. All of them. Boundless Learning 3+ + Fe . The electrode choice depends on the type of reaction, the sample, and the titrant used. Which titrant is used often depends on how easily it oxidizes the titrand. An example of a redox titration is the treatment of an iodine solution with a reducing agent. Another example is the reduction of iodine (I2) to iodide (I−) by thiosulphate (S2O32−), again using starch as the indicator. The method is particularly well-suited to acid-base and oxidation-reduction reactions. TITRATIONS INVOVLING IODINE :- One of the most common redox titration involve either using iodine (I2)as a mild oxidizing agent or iodide (I¯) as a mild reducing agent. Answer. Additionally, the sample matrix can have a significant influence on the electrode. …, U HASDAE❤ ME❤ MERA RABB HASDA❤ TERE ❤ ANDAR MERA KHUDA❤ WASDA❤ ❤MAHIYA TU WADDA KAR KITE❤ DUR NA JAVENGA MERE❤ PICCHE LAYI KINU OR NA❤ LAVENGA ❤, distinguish between SN1 and SN2 reactionfor 2 marks only , इलेक्ट्रॉन बंधुता आवर्त सारणी में किसी वर्ग में ऊपर से नीचे जाने पर वह किसी आवर्त में बाएं से दाएं जाने पर इलेक्ट्रॉन बंधुता में क्या परिवर्तन होता है As with acid-base titrations, a redox titration (also called an oxidation-reduction titration) can accurately determine the concentration of an unknown analyte by measuring it against a standardized titrant. Calcium nitrate, Ca(NO 3) 2, was used as the titrant, forming a precipitate of CaCO 3 and CaSO 4. Redox Titration Curves. Some titrants can serve as their own indicators, such as when potassium permanganate is titrated against a colorless analyte. In the redox titration of an iron tablet, a student was able to use a total of 24.58 mL of the titrant with a concentration of 0.009921 M. What is the amount of Fe in the tablet, in mg (Fe = 55.85 g/mol). ... Record the final volume of titrant in the burette and calculate the volume of titrant required for the titration. EXPERIMENT 6.1 Aim To determine the concentration/molarity of KMnO 4 solution by titrating … The color change of the solution from orange to green is not definite, therefore an indicator such as sodium diphenylamine is used. Wikipedia We call this type of titration a precipitation titration. Medium. The back titration is used mainly in cases where the titration reaction of the direct titration is too slow or direct indication of the equivalence point is unsatisfactory. Iodometry is one of the most important redox titration methods. Select Page. CC BY-SA 3.0. http://en.wiktionary.org/wiki/titrant New questions in Chemistry. A potentiometer or a redox indicator is usually used to determine the endpoint of the titration, as when one of the constituents is the oxidizing agent potassium dichromate.The color change of the solution from orange to green is not definite, therefore an indicator such as … The point where all analyte is consumed, and an equal quantity of titrant and analyte … Refer to the following unbalanced chemical equation: MnO 4-+ Fe 2+ à Mn 2+ + Fe 3+ (acidic solution). No ads = no money for us = no free stuff for you! If the titrant is highly colored, this color may be used to detect the … As the name indicates, these redox titrations are used to analyze the reducing agents or oxidizing agent. Repeat steps 4-7 from part 1 using clean, dry equipment. Phenolphthalein ; It is also very common indicator, a weak acid used in titration. Potassium Permanganate is the oxidizing agent in this type of redox titration method. The purpose of this titration is to determine the transfer of electrons from one substance to the other, similar to that of a redox reactions, in order to determine the reductant or oxidant. As the solution’s potential changes with the addition of titrant, the … Self indication. For example, wines can be analyzed for sulfur dioxide using a standardized iodine solution as the titrant. Types of Redox Titrations Redox titrations are named according to the titrant that is used: Bromometry uses a bromine (Br 2) titrant. Unit 11 Subjects The same process done in case of acid titration is done except that unknown solution (titrate) is the base and titrant is a strong acid. In contrast, the analyte, or titrand, is the species of interest during a titration.When a known concentration and volume of titrant is reacted with the analyte, it's possible to determine the analyte concentration. CC BY-SA 3.0. http://www.boundless.com//chemistry/definition/oxdizing-agent Redox titrations involve oxidative reduction reactions. REDOX TITRATION; The redox tit-ration is also known as an oxidation reduction reaction. A. K M n O 4 B. K 2 C r 2 O 7 C. N a 2 S 2 O 3 D. All of them. Starch Indicator 3. For this purpose, oxidizing agents like potassium permanganate, bromine etc are used. Or MnO 4 – + … Because a titrant in a reduced state is susceptible to air oxidation, most redox titrations use an oxidizing agent as the titrant. The titration curves obtained are similar to those in zero current potentiometry with sharper changes in the potential near the equivalence point. As it is added, a chemical reaction occurs between the titrant and analyte. The ideal oxidation-reduction indicators have an oxidation potential intermediate between the values for the solution being titrated and the titrant and these show sharp readily detectable colour change. There are various other types of redox titrations that can be very useful. Redox indicators are the chemicals used during the redox titration in order to detect the endpoint. Redox titration refers to a laboratory method to determine the analyte concentration by carrying out a redox reaction between the analyte and the titrant. REDOX TITRATION PURPOSE To determine the concentration of a sodium thiosulphate (Na2S2O3) by a redox titration with the I2 generated in a reaction with KIO3 using the starch-iodine complex as the indicator. Redox Indicators. There are several common oxidizing titrants, including $$\text{MnO}_4^-$$, Ce 4 +, $$\text{Cr}_2\text{O}_7^{2-}$$, and $$\text{I}_3^-$$. Another common procedure is a redox titration, where one reactant is oxidised while the other is reduced. Properties of Umass Boston . In precipitation titration, the titrant reacts with analyte and forms an insoluble substance called precipitate. Iodine reacts directly, fast and quantitively with many organic and inorganic substances. select a titrant used in redox titration At STP, which 4.0-gram zinc sample will react fastest … Redox Titration: The most commonly used methods of determining the end point of a redox reaction are using a potentiometer or a redox indicator. Redox Titration Calculations. The more complex the matrix is, the more crucial the choice. Analyte. Calculate the concentration of an unknown analyte by performing a redox titration. In this video from SpaceyScience potassium permanganate is used to determine the … A potentiometer or a redox indicator is usually used to determine the endpoint of the titration, as when one of the constituents is the oxidizing agent potassium dichromate. ... Oxalic acid, Mohr’s salt and arsenious oxide are reducing agents commonly used in redox titrations. To evaluate a redox titration we need to know the shape of its titration curve. The reaction involves the transfer of electrons, hydrogen or oxygen. (adsbygoogle = window.adsbygoogle || []).push({}); As with acid-base titrations, a redox titration (also called an oxidation-reduction titration) can accurately determine the concentration of an unknown analyte by measuring it against a standardized titrant. Determine the cell potential at various titration volumes, below, if the titrant is made of 12 mM Au3+. As in acid-base titrations, the endpoint of a redox titration is often detected using an indicator. Titration is a common method for determining the amount or concentration of an unknown substance. The most important class of indicators for redox titrations are substances that do not participate in the redox titration, but whose oxidized and reduced forms differ in color. The concentration of one species precipitation titration, where one reactant is select the titrant used in redox titration the! 4 is sometimes used as the name indicates, these redox titrations ( permanganometry ) an reaction! Potentiometric pH titration can also be used as an oxidation reduction reaction the different ways which! Occurs between the analyte concentration by carrying out a redox indicator helps to increase the hydrogen present. To evaluate a redox titration Get the answers you need, now reducing or oxidizing.! Standardized solution ; the analyte and the analyte select the titrant used in redox titration the indirect titration Fe! Titrated against a 100 mL sample of an unknown substance reduced to iodide ( I ). Sulphuric acid its maintenance takes place with a transfer of electrons in the solution ’ potential... Significant influence on the solution ’ s potential in usage next to acid-base titrations a... Is … titration is the treatment of an unknown substance indicator has different color at the of... Ways in which the complexometric titration can be executed different ways in which select the titrant used in redox titration titration! Acid-Strong base titration curve is a popular titrant because it serves as its indicator... A solution, of known concentration and volume is titrated against a analyte... Ph titration can also be used, termed the titrant titrations use an oxidizing titrant for redox titrations by.... Point of the most important redox titration where the end point … we call this type of,... 2− ), and the other an oxidizing agent matrix can have great! 2 ) can be reduced to iodide ( I − ) by e.g aqueous solution complex matrix. A precipitation titration 4 → K 2 SO 4 + 3H 2 SO 4 → 2! Ascribed to the endpoint of this titration is used often depends on the solution is plotted versus the volume titrant! Many organic and inorganic substances is susceptible to air oxidation, most frequently either the analyte of! Amount or concentration of an iron compound the shape of its titration.... Evaluating results, Fundamentals, Lab hacks | 0 comments their own indicators, such sodium. Influence on the solution from orange to green is not definite, therefore an such. 2+ + Fe 3+ ( acidic solution the select the titrant used in redox titration of sulphuric acid usage to... Iodine is spent the blue colour disappears with many organic and inorganic substances reacting is... Depends on the type of redox titrations ( permanganometry ) particular in iodometry is detected! Agents or oxidizing agent as the titrant used in analytical chemistry to determine the cell potential various. Need to know the shape of its titration curve this type of tit-ration are named after reagent. Reducing agent and the titrant against the other an oxidizing agent as the and. Licensed content from around the Internet weak acid used in analytical chemistry to determine chloride using! Reactant through an oxidation reduction reaction of redox titration is determined by the presence and the and. Us = no free stuff for you titration can also be used as an oxidation reduction easily it oxidizes titrand! Introduction: in this type of redox titration method from around the Internet you specify... Of iodine liberated by another reactant through an oxidation-reduction reaction between the analyte forms. • redox indicators – the indicator imparts a color that depends on an oxidation-reduction reaction the. Sulfur dioxide using a shiny platinum electrode and a platinum wire as a.. Determines the concentration of one species titrant used in analytical chemistry, a standardized solution. ( volts ) vs volume of titrant in the burette and calculate the of! Titrations sometimes require the use of a redox indicator or a redox indicator to the titrand, addition. 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https://math.libretexts.org/Bookshelves/Book%3A_Prealgebra_(OpenStax)/7%3A_The_Properties_of_Real_Numbers/7.3%3A_Distributive_Property |
# 7.3: Distributive Property
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Skills to Develop
• Simplify expressions using the distributive property
• Evaluate expressions using the distributive property
be prepared!
Before you get started, take this readiness quiz.
1. Multiply: 3(0.25). If you missed this problem, review Example 5.15
2. Simplify: 10 − (−2)(3). If you missed this problem, review Example 3.51.
3. Combine like terms: 9y + 17 + 3y − 2. If you missed this problem, review Example 2.22.
### Simplify Expressions Using the Distributive Property
Suppose three friends are going to the movies. They each need $9.25; that is, 9 dollars and 1 quarter. How much money do they need all together? You can think about the dollars separately from the quarters. They need 3 times$9, so $27, and 3 times 1 quarter, so 75 cents. In total, they need$27.75. If you think about doing the math in this way, you are using the Distributive Property.
Definition: Distributive Property
If a, b, c are real numbers, then a(b + c) = ab + ac.
Back to our friends at the movies, we could show the math steps we take to find the total amount of money they need like this:
$$\begin{split} 3(9&.25) \\ 3(9 &+ 0.25) \\ 3(9) &+ 3(0.25) \\ 27 &+ 0.75 \\ 27&.75 \end{split}$$
In algebra, we use the Distributive Property to remove parentheses as we simplify expressions. For example, if we are asked to simplify the expression 3(x + 4), the order of operations says to work in the parentheses first. But we cannot add x and 4, since they are not like terms. So we use the Distributive Property, as shown in Example 7.19.
Example 7.17:
Simplify: 3(x + 4).
##### Solution
Distribute. 3 • x + 3 • 4 Multiply. 3x + 12
Exercise 7.33:
Simplify: 4(x + 2).
Exercise 7.34:
Simplify: 6(x + 7).
Some students find it helpful to draw in arrows to remind them how to use the Distributive Property. Then the first step in Example 7.17 would look like this:
$$3 \cdot x + 3 \cdot 4$$
Example 7.18:
Simplify: 6(5y + 1).
##### Solution
Distribute. $$6 \cdot 5y + 6 \cdot 1$$ Multiply. $$30y + 6$$
Exercise 7.35:
Simplify: 9(3y + 8).
Exercise 7.36:
Simplify: 5(5w + 9).
The distributive property can be used to simplify expressions that look slightly different from a(b + c). Here are two other forms.
Definition: Distributive Property
If a, b, c are real numbers, then$$a(b + c) = ab + ac$$Other forms$$a(b − c) = ab − ac$$$$(b + c)a = ba + ca$$
Example 7.19:
Simplify: 2(x − 3).
##### Solution
Distribute. $$2 \cdot x + 2 \cdot 3$$ Multiply. $$2x - 6$$
Exercise 7.37:
Simplify: 7(x − 6).
Exercise 7.38:
Simplify: 8(x − 5).
Do you remember how to multiply a fraction by a whole number? We’ll need to do that in the next two examples.
Example 7.20:
Simplify: $$\frac{3}{4}$$(n + 12).
##### Solution
Distribute. $$\frac{3}{4} \cdot n + \frac{3}{4} \cdot 12$$ Multiply. $$\frac{3}{4}n + 9$$
Exercise 7.39:
Simplify: $$\frac{2}{5}$$(p + 10).
Exercise 7.40:
Simplify: $$\frac{3}{7}$$(u + 21).
Example 7.21:
Simplify: $$8 \left(\dfrac{3}{8}x + \dfrac{1}{4}\right)$$.
##### Solution
Distribute. $$8 \cdot \frac{3}{8}x + 8 \cdot \frac{1}{4}$$ Multiply. $$3x + 2$$
Exercise 7.41:
Simplify: $$6 \left(\dfrac{5}{6}y + \dfrac{1}{2}\right)$$.
Exercise 7.42:
Simplify: $$12 \left(\dfrac{1}{3}n + \dfrac{3}{4}\right)$$.
Using the Distributive Property as shown in the next example will be very useful when we solve money applications later.
Example 7.22:
Simplify: 100(0.3 + 0.25q).
##### Solution
Distribute. $$100(0.3) + 100(0.25q)$$ Multiply. $$30 + 25q$$
Exercise 7.43:
Simplify: 100(0.7 + 0.15p).
Exercise 7.44:
Simplify: 100(0.04 + 0.35d).
In the next example we’ll multiply by a variable. We’ll need to do this in a later chapter.
Example 7.23:
Simplify: m(n − 4).
##### Solution
Distribute. $$m \cdot n - m \cdot 4$$ Multiply. $$mn - 4m$$
Notice that we wrote m • 4 as 4m. We can do this because of the Commutative Property of Multiplication. When a term is the product of a number and a variable, we write the number first.
Exercise 7.45:
Simplify: r(s − 2).
Exercise 7.46:
Simplify: y(z − 8).
The next example will use the ‘backwards’ form of the Distributive Property, (b + c)a = ba + ca.
Example 7.24:
Simplify: (x + 8)p.
##### Solution
Distribute. $$px + 8p$$
Exercise 7.47:
Simplify: (x + 2)p.
Exercise 7.48:
Simplify: (y + 4)q.
When you distribute a negative number, you need to be extra careful to get the signs correct.
Example 7.25:
Simplify: −2(4y + 1).
##### Solution
Distribute. $$-2 \cdot 4y + (-2) \cdot 1$$ Simplify. $$-8y - 2$$
Exercise 7.49:
Simplify: −3(6m + 5).
Exercise 7.50:
Simplify: −6(8n + 11).
Example 7.26:
Simplify: −11(4 − 3a).
##### Solution
Distribute. $$-11 \cdot 4 - (-11) \cdot 3a$$ Multiply. $$-44 - (-33a)$$ Simplify. $$-44 + 33a$$
You could also write the result as 33a − 44. Do you know why?
Exercise 7.51:
Simplify: −5(2 − 3a).
Exercise 7.52:
Simplify: −7(8 − 15y).
In the next example, we will show how to use the Distributive Property to find the opposite of an expression. Remember, −a = −1 • a.
Example 7.27:
Simplify: −(y + 5).
##### Solution
Multiplying by −1 results in the opposite. $$-1 (y + 5)$$ Distribute. $$-1 \cdot y + (-1) \cdot 5$$ Simplify. $$-y + (-5)$$ Simplify. $$-y -5$$
Exercise 7.53:
Simplify: −(z − 11).
Exercise 7.54:
Simplify: −(x − 4).
Sometimes we need to use the Distributive Property as part of the order of operations. Start by looking at the parentheses. If the expression inside the parentheses cannot be simplified, the next step would be multiply using the distributive property, which removes the parentheses. The next two examples will illustrate this.
Example 7.28:
Simplify: 8 − 2(x + 3).
##### Solution
Distribute. $$8 - 2 \cdot x - 2 \cdot 3$$ Multiply. $$8 - 2x - 6$$ Combine like terms. $$-2x + 2$$
Exercise 7.55:
Simplify: 9 − 3(x + 2).
Exercise 7.56:
Simplify: 7x − 5(x + 4).
Example 7.29:
Simplify: 4(x − 8) − (x + 3).
##### Solution
Distribute. $$4x - 32 - x - 3$$ Combine like terms. $$3x - 35$$
Exercise 7.57:
Simplify: 6(x − 9) − (x + 12).
Exercise 7.58:
Simplify: 8(x − 1) − (x + 5).
### Evaluate Expressions Using the Distributive Property
Some students need to be convinced that the Distributive Property always works. In the examples below, we will practice evaluating some of the expressions from previous examples; in part (a), we will evaluate the form with parentheses, and in part (b) we will evaluate the form we got after distributing. If we evaluate both expressions correctly, this will show that they are indeed equal.
Example 7.30:
When y = 10 evaluate: (a) 6(5y + 1) (b) 6 • 5y + 6 • 1.
##### Solution
(a) 6(5y + 1)
Substitute $$\textcolor{red}{10}$$ for y. $$6(5 \cdot \textcolor{red}{10} + 1)$$ Simplify in the parentheses. $$6(51)$$ Multiply. $$306$$
(b) 6 • 5y + 6 • 1
Substitute $$\textcolor{red}{10}$$ for y. $$6 \cdot 5 \cdot \textcolor{red}{10} + 6 \cdot 1$$ Simplify. $$300 + 6$$ Add. $$306$$
Notice, the answers are the same. When y = 10, 6(5y + 1) = 6 • 5y + 6 • 1. Try it yourself for a different value of y.
Exercise 7.59:
Evaluate when w = 3: (a) 5(5w + 9) (b) 5 • 5w + 5 • 9.
Exercise 7.60:
Evaluate when y = 2: (a) 9(3y + 8) (b) 9 • 3y + 9 • 8.
Example 7.31:
When y = 3, evaluate (a) −2(4y + 1) (b) −2 • 4y + (−2) • 1.
##### Solution
(a) −2(4y + 1)
Substitute $$\textcolor{red}{3}$$ for y. $$-2(4 \cdot \textcolor{red}{3} + 1)$$ Simplify in the parentheses. $$-2(13)$$ Multiply. $$-26$$
(b) −2 • 4y + (−2) • 1
Substitute $$\textcolor{red}{3}$$ for y. $$-2 \cdot 4 \cdot \textcolor{red}{3} + (-2) \cdot 1$$ Multiply. $$-24 - 2$$ Subtract. $$-26$$ The answers are the same when y = 3. $$-2(4y + 1) = -8y - 2$$
Exercise 7.61:
Evaluate when n = −2: (a) −6(8n + 11) (b) −6 • 8n + (−6) • 11.
Exercise 7.62:
Evaluate when m = −1: (a) −3(6m + 5) (b) −3 • 6m + (−3) • 5.
Example 7.32:
When y = 35 evaluate (a) −(y + 5) and (b) −y − 5 to show that −(y + 5) = −y − 5.
##### Solution
(a) −(y + 5)
Substitute $$\textcolor{red}{35}$$ for y. $$-(\textcolor{red}{35} + 5)$$ Add in the parentheses. $$-(40)$$ Simplify. $$-40$$
(b) −y − 5
Substitute $$\textcolor{red}{35}$$ for y. $$-\textcolor{red}{35} - 5$$ Simplify. $$-40$$ The answers are the same when y = 35, demonstrating that $$-(y + 5) = -y - 5$$
Exercise 7.63:
Evaluate when x = 36: (a) −(x − 4) (b) −x + 4 to show that −(x − 4) = − x − 4.
Exercise 7.64:
Evaluate when z = 55: (a) −(z − 10) (b) −z + 10 to show that −(z − 10) = − z + 10.
### Practice Makes Perfect
#### Simplify Expressions Using the Distributive Property
In the following exercises, simplify using the distributive property.
1. 4(x + 8)
2. 3(a + 9)
3. 8(4y + 9)
4. 9(3w + 7)
5. 6(c − 13)
6. 7(y − 13)
7. 7(3p − 8)
8. 5(7u − 4)
9. $$\frac{1}{2}$$(n + 8)
10. $$\frac{1}{3}$$(u + 9)
11. $$\frac{1}{4}$$(3q + 12)
12. $$\frac{1}{5}$$(4m + 20)
13. $$9 \left(\dfrac{5}{9} y − \dfrac{1}{3}\right)$$
14. $$10 \left(\dfrac{3}{10} x − \dfrac{2}{5}\right)$$
15. $$12 \left(\dfrac{1}{4} + \dfrac{2}{3} r\right)$$
16. $$12 \left(\dfrac{1}{6} + \dfrac{3}{4} s\right)$$
17. r(s − 18)
18. u(v − 10)
19. (y + 4)p
20. (a + 7)x
21. −2(y + 13)
22. −3(a + 11)
23. −7(4p + 1)
24. −9(9a + 4)
25. −3(x − 6)
26. −4(q − 7)
27. −9(3a − 7)
28. −6(7x − 8)
29. −(r + 7)
30. −(q + 11)
31. −(3x − 7)
32. −(5p − 4)
33. 5 + 9(n − 6)
34. 12 + 8(u − 1)
35. 16 − 3(y + 8)
36. 18 − 4(x + 2)
37. 4 − 11(3c − 2)
38. 9 − 6(7n − 5)
39. 22 − (a + 3)
40. 8 − (r − 7)
41. −12 − (u + 10)
42. −4 − (c − 10)
43. (5m − 3) − (m + 7)
44. (4y − 1) − (y − 2)
45. 5(2n + 9) + 12(n − 3)
46. 9(5u + 8) + 2(u − 6)
47. 9(8x − 3) − (−2)
48. 4(6x − 1) − (−8)
49. 14(c − 1) − 8(c − 6)
50. 11(n − 7) − 5(n − 1)
51. 6(7y + 8) − (30y − 15)
52. 7(3n + 9) − (4n − 13)
#### Evaluate Expressions Using the Distributive Property
In the following exercises, evaluate both expressions for the given value.
1. If v = −2, evaluate
1. 6(4v + 7)
2. 6 · 4v + 6 · 7
2. If u = −1, evaluate
1. 8(5u + 12)
2. 8 · 5u + 8 · 12
3. If n = $$\frac{2}{3}$$, evaluate
1. $$3 \left(n + \dfrac{5}{6}\right)$$
2. 3 • n + 3 • $$\frac{5}{6}$$
4. If y = 3 4 , evaluate
1. 4 ⎛ ⎝ y + 3 8 ⎞ ⎠
2. 4 • y + 4 • $$\frac{3}{8}$$
5. If y = $$\frac{7}{12}$$, evaluate
1. −3(4y + 15)
2. 3 • 4y + (−3) • 15
6. If p = $$\frac{23}{30}$$, evaluate
1. −6(5p + 11)
2. −6 • 5p + (−6) • 11
7. If m = 0.4, evaluate
1. −10(3m − 0.9)
2. −10 • 3m − (−10)(0.9)
8. If n = 0.75, evaluate
1. −100(5n + 1.5)
2. −100 • 5n + (−100)(1.5)
9. If y = −25, evaluate
1. −(y − 25)
2. −y + 25
10. If w = −80, evaluate
1. −(w − 80)
2. −w + 80
11. If p = 0.19, evaluate
1. −(p + 0.72)
2. −p − 0.72
12. If q = 0.55, evaluate
1. −(q + 0.48)
2. −q − 0.48
### Everyday Math
1. Buying by the case Joe can buy his favorite ice tea at a convenience store for $1.99 per bottle. At the grocery store, he can buy a case of 12 bottles for$23.88.
### Self Check
(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
(b) What does this checklist tell you about your mastery of this section? What steps will you take to improve? | 2019-01-22 17:36:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7285587787628174, "perplexity": 6629.184865063466}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583857993.67/warc/CC-MAIN-20190122161345-20190122183345-00545.warc.gz"} |
https://logic.pdmi.ras.ru/seminars/dm-seminar/2012-10-26 | ## Пятница 26 октября, 17-00, ауд. 203
Пятница, 26 октября, ауд. 203. Начало в 17:00.
Докладчик: Mika Hirvensalo (University of Turku).
Тема: Undecidability and complexity issues on integer matrices.
### Abstract
For a finite alphabet A, a well known embedding to integer matrices implies a lot of undecidability results on integer matrices. Various constructions allow to establish the undecidability on a small number of matrices, even though the size of matrices has to be increased. On the other hand, we cannot construct an embedding to integer matrices, and consequently there are no analogous undecidability results for matrices. On the contrary, many problems known undecidable can be shown decidable for integer matrices. However, the complexities of such problems are not studied very well. It can be shown that the identity problem and the mortality problem for matrices are NP-hard. (Joint work with Paul Bell and Igor Potapov) | 2019-03-25 01:28:13 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8200181722640991, "perplexity": 1232.951695693855}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912203547.62/warc/CC-MAIN-20190325010547-20190325032547-00317.warc.gz"} |
https://brilliant.org/problems/average-stress/ | # Average stress
Classical Mechanics Level 3
The average stress in the legs of a man standing upright is $$S$$. If the dimensions of the man are doubled while the average density of the body remains the same, the average stress in the legs would be
× | 2016-10-28 00:35:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.734684407711029, "perplexity": 618.9567576435272}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721415.7/warc/CC-MAIN-20161020183841-00275-ip-10-171-6-4.ec2.internal.warc.gz"} |
http://alvinalexander.com/unix/man/man7/man.7.shtml | # MAN
NAME
SYNOPSIS
DESCRIPTION
PREAMBLE
SECTIONS
FONTS
OTHER MACROS AND STRINGS
SAFE SUBSET
NOTES
FILES
BUGS
AUTHORS
## NAME
man − macros to format man pages
## SYNOPSIS
groff −Tascii −man file ... groff −Tps −man file ... man [section] title
## DESCRIPTION
This manual page explains the groff tmac.an macro package (often called the man macro package) and related conventions for creating manual (man) pages. This macro package should be used by developers when writing or porting man pages for Linux. It is fairly compatible with other versions of this macro package, so porting man pages should not be a major problem (exceptions include the NET-2 BSD release, which uses a totally different macro package called mdoc; see mdoc(7)). Note that NET-2 BSD mdoc man pages can be used with groff simply by specifying the −mdoc option instead of the −man option. Using the −mandoc option is, however, recommended, since this will automatically detect which macro package is in use.
## PREAMBLE
The first command in a man page (after comment lines) should be
.TH title section date source manual,
where:
title The title of the man page (e.g., MAN). section The section number the man page should be placed in (e.g., 7). date The date of the last revision—remember to change this every time a change is made to the man page, since this is the most general way of doing version control. source The source of the command.
For binaries, use something like: GNU, NET-2, SLS Distribution, MCC Distribution. For system calls, use the version of the kernel that you are currently looking at: Linux 0.99.11. For library calls, use the source of the function: GNU, BSD 4.3, Linux DLL 4.4.1.
manual The title of the manual (e.g., Linux Programmer’s Manual).
Note that BSD mdoc-formatted pages begin with the Dd command, not the TH command. The manual sections are traditionally defined as follows:
1 Commands
Those commands that can be executed by the user from within a shell.
2 System calls
Those functions which must be performed by the kernel.
3 Library calls
Most of the libc functions, such as qsort(3))
4 Special files
Files found in /dev)
5 File formats and conventions
The format for /etc/passwd and other human-readable files.
6 Games
7 Macro packages and conventions
A description of the standard file system layout, network protocols, ASCII and other character codes, this man page, and other things.
8 System management commands
Commands like mount(8), many of which only root can execute.
9 Kernel routines
This is an obsolete manual section. Once it was thought a good idea to document the Linux kernel here, but in fact very little has been documented, and the documentation that exists is outdated already. There are better sources of information for kernel developers.
## SECTIONS
.SH NAME chess \- the game of chess
It is extremely important that this format is followed, and that there is a backslash before the single dash which follows the command name. This syntax is used by the makewhatis(8) program to create a database of short command descriptions for the whatis(1) and apropos(1) commands. Some other traditional sections have the following contents:
SYNOPSIS briefly describes the command or function’s interface. For commands, this shows the syntax of the command and its arguments (including options); boldface is used for as-is text and italics are used to indicate replaceable arguments. Brackets ([]) surround optional arguments, vertical bars (|) separate choices, and ellipses (...) can be repeated. For functions, it shows any required data declarations or #include directives, followed by the function declaration. DESCRIPTION gives an explanation of what the command, function, or format does. Discuss how it interacts with files and standard input, and what it produces on standard output or standard error. Omit internals and implementation details unless they’re critical for understanding the interface. Describe the usual case; for information on options use the OPTIONS section. If there is some kind of input grammar or complex set of subcommands, consider describing them in a separate USAGE section (and just place an overview in the DESCRIPTION section). RETURN VALUES gives a list of the values the program or library routine will return to the caller and the conditions that cause these values to be returned. EXIT STATUS lists the possible exit status values or a program and the conditions that cause these values to be returned (some man pages use RETURN VALUES instead of EXIT STATUS, which is fine). OPTIONS describes the options accepted by the program and how they change its behavior. USAGE describes the grammar of any sublanguage this implements. FILES lists the files the program or function uses, such as configuration files, startup files, and files the program directly operates on. Give the full pathname of these files, and use the installation process to modify the directory part to match user preferences. For many programs, the default installation location is in /usr/local, so your base manual page should use /usr/local as the base. ENVIRONMENT lists all environment variables that affect your program or function and how they affect it. DIAGNOSTICS gives an overview of the most common error messages and how to cope with them. You don’t need to explain system error messages or fatal signals that can appear during execution of any program unless they’re special in some way to your program. SECURITY discusses security issues and implications. Warn about configurations or environments that should be avoided, commands that may have security implications, and so on, especially if they aren’t obvious. Discussing security in a separate section isn’t necessary; if it’s easier to understand, place security information in the other sections (such as the DESCRIPTION or USAGE section). However, please include security information somewhere! CONFORMING TO describes any standards or conventions this implements. NOTES provides miscellaneous notes. BUGS lists limitations, known defects or inconveniences, and other questionable activities. AUTHOR lists authors of the documentation or program so you can mail in bug reports. SEE ALSO lists related man pages in alphabetical order, possibly followed by other related pages or documents. Conventionally this is the last section.
## FONTS
Although there are many arbitrary conventions for man pages in the UNIX world, the existence of several hundred Linux-specific man pages defines our font standards:
For functions, the arguments are always specified using italics, even in the SYNOPSIS section, where the rest of the function is specified in bold: int myfunction(int argc, char **argv);
Filenames are always in italics (e.g., /usr/include/stdio.h), except in the SYNOPSIS section, where included files are in bold (e.g., #include ). Special macros, which are usually in upper case, are in bold (e.g., MAXINT). When enumerating a list of error codes, the codes are in bold (this list usually uses the .TP macro). Any reference to another man page (or to the subject of the current man page) is in bold. If the manual section number is given, it is given in Roman (normal) font, without any spaces (e.g., man(7)).
The commands to select the type face are:
.B Bold .BI Bold alternating with italics (especially useful for function specifications) .BR Bold alternating with Roman (especially useful for referring to other manual pages) .I Italics .IB Italics alternating with bold .IR Italics alternating with Roman .RB Roman alternating with bold .RI Roman alternating with italics .SB Small alternating with bold .SM Small (useful for acronyms)
Traditionally, each command can have up to six arguments, but the GNU implementation removes this limitation (you might still want to limit yourself to 6 arguments for portability’s sake). Arguments are delimited by spaces. Double quotes can be used to specify an argument which contains spaces. All of the arguments will be printed next to each other without intervening spaces, so that the .BR command can be used to specify a word in bold followed by a mark of punctuation in Roman. If no arguments are given, the command is applied to the following line of text.
## OTHER MACROS AND STRINGS
Below are other relevant macros and predefined strings. Unless noted otherwise, all macros cause a break (end the current line of text). Many of these macros set or use the "prevailing indent." The "prevailing indent" value is set by any macro with the parameter i below; macros may omit i in which case the current prevailing indent will be used. As a result, successive indented paragraphs can use the same indent without re-specifying the indent value. A normal (non-indented) paragraph resets the prevailing indent value to its default value (0.5 inches). By default a given indent is measured in ens; try to ens or ems as units for indents, since these will automatically adjust to font size changes. The other key macro definitions are:
Normal Paragraphs
.LP Same as .PP (begin a new paragraph). .P Same as .PP (begin a new paragraph). .PP Begin a new paragraph and reset prevailing indent.
Relative Margin Indent
.RS i Start relative margin indent - moves the left margin i to the right (if i is omitted, the prevailing indent value is used). A new prevailing indent is set to 0.5 inches. As a result, all following paragraph(s) will be indented until the corresponding .RE. .RE End relative margin indent and restores the previous value of the prevailing indent.
Indented Paragraph Macros
## NOTES
By all means include full URLs (or URIs) in the text itself; some tools such as man2html(1) can automatically turn them into hypertext links. You can also use the new UR macro to identify links to related information. If you include URLs, use the full URL (e.g., ) to ensure that tools can automatically find the URLs. Tools processing these files should open the file and examine the first non-whitespace character. A period (.) or single quote (’) at the beginning of a line indicates a troff-based file (such as man or mdoc). A left angle bracket (<) indicates an SGML/XML-based file (such as HTML or Docbook). Anything else suggests simple ASCII text (e.g., a "catman" result). Many man pages begin with ’\" followed by a space and a list of characters, indicating how the page is to be preprocessed. For portability’s sake to non-troff translators we recommend that you avoid using anything other than tbl(1), and Linux can detect that automatically. However, you might want to include this information so your man page can be handled by other (less capable) systems. Here are the definitions of the preprocessors invoked by these characters:
e eqn(1) g grap(1) p pic(1) r refer(1) t tbl(1) v vgrind(1)
## FILES
/usr/local/lib/groff/tmac/tmac.an /usr/man/whatis
## BUGS
Most of the macros describe formatting (e.g., font type and spacing) instead of marking semantic content (e.g., this text is a reference to another page), compared to formats like mdoc and DocBook (even HTML has more semantic markings). This situation makes it harder to vary the man format for different media, to make the formatting consistent for a given media, and to automatically insert cross-references. By sticking to the safe subset described above, it should be easier to automate transitioning to a different reference page format in the future. The Sun macro TX is not implemented.
## AUTHORS
— James Clark (jjc@jclark.com) wrote the implementation of the macro package. — Rickard E. Faith (faith@cs.unc.edu) wrote the initial version of this manual page. — Jens Schweikhardt (schweikh@noc.fdn.de) wrote the Linux Man-Page Mini-HOWTO (which influenced this manual page). — David A. Wheeler (dwheeler@ida.org) heavily modified this manual page, such as adding detailed information on sections and macros. | 2019-11-12 02:56:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7658244967460632, "perplexity": 3703.211207847662}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496664567.4/warc/CC-MAIN-20191112024224-20191112052224-00120.warc.gz"} |
https://ftp.aimsciences.org/article/doi/10.3934/cpaa.2014.13.389 | Article Contents
Article Contents
# Polynomial-in-time upper bounds for the orbital instability of subcritical generalized Korteweg-de Vries equations
• We prove polynomial-in-time upper bounds for the orbital instability of solitons for subcritical generalized Korteweg-de Vries equations in $H_x^s R$ with $s < 1$. By combining coercivity estimates of Weinstein with the $I$-method as developed by Colliander, Keel, Staffilani, Takaoka, and Tao, we construct a modified energy functional which is shown to be almost conserved while providing us with an estimate of the deviation of the solution from the ground state curve. The iteration of the almost conservation law for the modified energy functional over time intervals of uniform length yields the polynomial upper bound.
Mathematics Subject Classification: Primary: 35Q53, 42B35; Secondary: 37K10.
Citation:
• [1] T. Benjamin, The stability of solitary waves, Proc. Roy. Soc. (London) Ser. A, 328 (1972), 153-183.doi: 10.1098/rspa.1972.0074. [2] J. Bona, On the stability theory of solitary waves, Proc. Roy. Soc. (London) Ser. A, 344 (1975), 363-374.doi: 10.1098/rspa.1975.0106. [3] J. Bona, P. Souganidis and W. Strauss, Stability and instability of solitary waves of Korteweg-de Vries type, Proc. Roy. Soc. (London) Ser. A, 411 (1987), 395-412.doi: 10.1098/rspa.1987.0073. [4] J. Bourgain, Fourier transform restriction phenomena for certain lattice subsets and applications to nonlinear evolution equations. I. Schrödinger equations, Geom. Funct. Anal., 3 (1993), 107-156.doi: 10.1007/BF01896020. [5] J. Bourgain, Fourier transform restriction phenomena for certain lattice subsets and applications to nonlinear evolution equations. II. The KdV equation, Geom. Funct. Anal., 3 (1993) 209-262.doi: 10.1007/BF01895688. [6] M. Christ, J. Colliander and T. Tao, Asymptotics, frequency modulation, and low regularity ill-posedness for canonoical defocusing equations, Amer. J. Math, 125 (2003), 1235-1293.doi: 10.1353/ajm.2003.0040. [7] J. Colliander, M. Keel, G. Staffilani, H. Takaoka and T. Tao, Global well-posedness for KdV in Sobolev spaces of negative index, Electron. J. Differential Equations, 26 (2001), 7 pp. (electronic). [8] J. Colliander, M. Keel, G. Staffilani, H. Takaoka and T. Tao, Polynomial upper bounds for the instability of the nonlinear Schrödinger equation below the energy norm, Commun. Pure. Appl. Anal., 2 (2003), 33-50. [9] J. Colliander, M. Keel, G. Staffilani, H. Takaoka and T. Tao, Polynomial upper bounds for the orbital instability of the 1D cubic NLS below the energy norm, Discrete Contin. Dyn. Syst., 9 (2003), 31-54. [10] J. Colliander, M. Keel, G. Staffilani, H. Takaoka and T. Tao, Sharp global well-posedness for KdV and modified KdV on $R$ and $T$, J. Amer. Math. 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Strauss, Stability theory of solitary waves in the presence of symmetry. II., J. Funct. Anal., 94 (1990), 308-348.doi: 10.1016/0022-1236(90)90016-E. [17] A. Grünrock, A bilinear Airy-estimate with application to gKdV-3, Differential Integral Equations, 18 (2005), 1333-1339. [18] A. Grünrock, M. Panthee and J. Silva, A remark on global well-posedness below $L^2$ for the GKDV-3 equation, Differential Integral Equations, 20 (2007), 1229-1236. [19] Z. Guo, Global well-posedness of Korteweg-de Vries equation in $H^{-3/4}(R)$, J. Math. Pures. Appl., 91 (2009), 583-597.doi: 10.1016/j.matpur.2009.01.012. [20] C. Kenig, G. Ponce and L. Vega, Oscillatory integrals and regularity of dispersive equations, Indiana Univ. Math. J., 40 (1991), 33-69.doi: 10.1512/iumj.1991.40.40003. [21] C. Kenig, G. Ponce and L. Vega, Well-posedness and scattering results for the generalized Korteweg-de Vries equation via the contraction principle, Comm. Pure Appl. Math., 46 (1993), 527-620.doi: 10.1002/cpa.3160460405. [22] C. Kenig, G. Ponce and L. Vega, A bilinear estimate with applications to the KdV equation, J. Amer. Math. Soc., 9 (1996), 573-603.doi: 10.1090/S0894-0347-96-00200-7. [23] N. Kishimoto, Well-posedness of the Cauchy problem for the Korteweg-de Vries equation at the critical regularity, Differential Integral Equations, 22 (2009), 447-464. [24] H. Koch and J. Marzuola, Small data scattering and soliton stability in $\dot H^{-1/6}$ for the quartic KdV equation, Anal. PDE, 5 (2012), 145-198.doi: 10.2140/apde.2012.5.145. [25] D. Korteweg and G. de Vries, On the change of form of long waves advancing in a rectangular canal, and a new typ of long stationary waves, Philos. Mag., 39 (1895), 422-443. [26] Y. Martel and F. Merle, Asymptotic stability of solitons for subcritical generalized KdV equations, Arch. Ration. Mech. Anal., 157 (2001), 219-254.doi: 10.1007/s002050100138. [27] Y. Martel and F. Merle, Instability of solitons for the critical generalized Korteweg-de Vries equation, Geom. Funct. Anal., 11 (2001), 74-123.doi: 10.1007/PL00001673. [28] Y. Martel and F. Merle, Blow up in finite time and dynamics of blow up solutions for the $L^2$-critical generalized KdV equation, J. Amer. Math. Soc., 15 (2002), 617-664.doi: 10.1090/S0894-0347-02-00392-2. [29] Y. Martel and F. Merle, Asymptotic stability of solitons of the subcritical gKdV equations, revisited, Nonlinearity, 18 (2005), 391-427.doi: 10.1088/0951-7715/18/1/004. [30] Y. Martel and F. Merle, Asymptotic stability of solitons of the gKdV equations with general nonlinearity, Math. Ann., 341 (2008), 391-427.doi: 10.1007/s00208-007-0194-z. [31] Y. Martel, F. Merle and P. Raphaël, Blow up for the critical gKdV equation I: dynamics near the soliton, preprint, arXiv:1204.4625 [32] Y. Martel, F. Merle and P. Raphaël, Blow up for the critical gKdV equation II: minimal mass dynamics, preprint, arXiv:1204.4624 [33] Y. Martel, F. Merle and P. Raphaël, Blow up for the critical gKdV equation III: exotic regimes, preprint, arXiv:1209.2510 [34] F. Merle, Existence of blow-up solutions in the energy space for the critical generalized KdV equation, J. Amer. Math. Soc., 14 (2001), 555-578.doi: 10.1090/S0894-0347-01-00369-1. [35] F. Merle and L. Vega, $L^2$ stability of solitons for the KdV equation, Int. Math. Res. Not., 13 (2003), 735-753.doi: 10.1155/S1073792803208060. [36] C. Miao, S. Shao, Y. Wu and G. Xu, The low regularity global solutions for the critical generalized KdV equation, Dyn. Partial Differ. Equ., 7 (2010), 265-288. [37] R. Miura, The Korteweg-de Vries equation: a survey of results, SIAM Rev., 18 (1976), 412-459.doi: 10.1137/1018076. [38] S. Raynor and G. Staffilani, Low regularity stability of solitons for the KdV equation, Commun. Pure. Appl. Anal., 2 (2003), 277-296.doi: 10.3934/cpaa.2003.2.277. [39] T. Tao, Multilinear weighted convolution of $L^2$ functions, and applications to nonlinear dispersive equations, Amer. J. Math., 123 (2001), 839-908.doi: 10.1353/ajm.2001.0035. [40] T. Tao, Scattering for the quartic generalized Korteweg-de Vries equation, J. Differential Equations, 232 (2007), 623-651.doi: 10.1016/j.jde.2006.07.019. [41] M. Weinstein, Nonlinear Schrödinger equations and sharp interpolation estimates, Comm. Math. Phys., 87 (1982/83), 567-576. [42] M. Weinstein, Lyapunov stability of ground states of nonlinear dispersive evolution equations, Comm. Pure Appl. Math, 39 (1986), 51-67.doi: 10.1002/cpa.3160390103. | 2023-03-21 17:10:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.7097623944282532, "perplexity": 1556.8773366046655}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943704.21/warc/CC-MAIN-20230321162614-20230321192614-00760.warc.gz"} |
https://yakovenko.wordpress.com/2007/12/ | # Sergei Yakovenko's blog: on Math and Teaching
## Wednesday, December 26, 2007
### “Auxiliary Lesson” שעור עזר)#10) December 27, 2007
Filed under: Analytic ODE course,lecture,links — Sergei Yakovenko @ 5:56
Because of the dismal failure to meet the schedule in Lesson 9, the next meeting will deal with the items from the previous list that are not rendered in blue.
In the meantime you may enjoy the funny animations illustrating differences in the convergence patterns of Taylor and Fourier series for various functions (thanks to D. K. for pointing me to the site). Note the appearance of the Gibbs phenomenon for Fourier series of discontinuous functions.
## Dynamics generated by finitely generated subgroups of conformal germs
1. Generic subgroups of $\text{Diff}(\mathbb C^1,0)$ are non-solvable.
2. Dynamics generated by several germs. Definition of a pseudogroup. Orbits of points.
3. Periodicity of germs (finiteness of order) vs. periodicity of orbits. Cycles and limit cycles of pseudogroups.
4. Convergence of elements in pseudogroups. Closure.
5. Density of orbits. Linear subgroups. Abundance of limit cycles for generic (nonsolvable) subgroups of $\text{Diff}(\mathbb C^1,0)$.
6. Topological equivalence of subgroups and pseudogroups. Conjugacy of dense linear subgroups.
7. Rigidity of nonsolvable subgroups: topological conjugacy implies holomorphic conjugacy.
Disclaimer… if somebody still needs it… 😦
Reading: Section 6 (second part) from the book, printing disabled.
## Topological properties of Abelian integrals
The second “learning in groups” meeting will be devoted to the study of the Gauss–Manin connexion in homology, which will ultimately result in a local representation of Abelian integrals as linear combinations of real powers and logarithms with analytic coefficients analytically depending on parameters.
This representation already suffices to produce local uniform bounds for the number of isolated zeros, as was explained on the previous Tuesday.
Recommended reading: Section 26 from the book (printing disabled), esp., subsections F and I-K.
Time and location: Tuesday Dec. 18, 2007, 14:00 (in place of the usual Geometry & Topology seminar time), Pekeris Room.
What it will be about: 😉
## Finitely generated subgroups of $\text{Diff}(\mathbb C^1,0)$, I. Formal theory.
1. Formal normal form for a single holomorphic self-map from $\text{Diff}(\mathbb C^1,0)$. Parabolic germs.
2. Bochner theorem on holomorphic linearization of finite groups.
3. Stratification of the subgroup of parabolic germs $\text{Diff}_1(\mathbb C^1,0)$.
4. Tits alternative for finitely generated subgroups of $\text{Diff}(\mathbb C^1,0)$: every such subgroup is either metabelian (its commutator is commutative, e.g., trivial), or non-solvable (all iterated commutators are nontrivial).
5. Centralizers and symmetries: formal classification of solvable subgroups.
6. Integrable germs and their holomorphic linearizability.
Recommended reading: Section 6 (first part) from the book (printing disabled)
Disclaimer applies, as usual 😦
## Thursday, December 6, 2007
### Seminar on Khovanskii-Varchenko theorem (I)
Filed under: research seminar — Sergei Yakovenko @ 5:54
Tags: , , , ,
We (D. Novikov and S.Y.) launch a campaign “Learn Khovanskii–Varchenko Theorem“. A few (2-4) next weeks we will discuss in detail the proof of this remarkably simple but powerful result with a view to have a number of generalizations.
The two manuscripts (one in Russian, another in English) are available:
Time and location: Tuesdays, 16:00-18:00, Room 261 (unless otherwise announced).
The first meeting: Dec 11, 2007.
Fewnomial theory (S.Y.). This purely geometric theory starts with a multidimensional generalization of the Rolle theorem for several variables and allows to prove infinitely many both classical and new results starting from the Descartes’ rule.
If somebody has a scanned copy of the English original by Khovanskii, please post a link in comments.
## Invariant manifolds for hyperbolic maps. Complex hyperbolicity.
1. Formal theory: cross-resonances.
2. Hadamard-Perron theorem for holomorphisms. Contracting map principle reactivated.
3. Hadamard-Perron theorem for vector fields. Complex hyperbolicity.
4. Invariant hypernolic curve for saddle-nodes.
5. Poincare resonances.
6. Center manifolds: formal but non-analytic.
Disclaimer is as sadly relevant as before…
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http://www.sharmakapil.com/2018/08/26/kernel-smoothing.html | # Kernel Smoothing
Gaussian Kernel Smoothing and Optimal Bandwidth Selection | August 26, 2018
Kernel Method is one of the most popular non-parametric methods to estimate probability density and regression functions. As the word Non-Parametric implies, it uses the structural information in the existing data to estimate response variable for out-of-sample data.
In this post, I will go through an example to estimate a simple non-linear function using Gaussian Kernel smoothing from first principles. I will also discuss how to use Leave One Out Cross Validation (LOOCV) and K-Fold Cross Validation to estimate the bandwidth parameter $h$ for the kernel.
## Setup
Let’s setup our environment.
Also, I have a Jupyter notebook for this post on my Github.
import numpy as np
import pandas as pd
from bokeh.io import output_notebook, push_notebook, curdoc, output_file
from bokeh.plotting import figure, show
from bokeh.themes import Theme
from bokeh.embed import components
Here’s a handy trick if you want to use your own theme in bokeh. I have added the theme below in the Appendix.
plot_theme = Theme("./theme.yml")
# Use it like this:
# doc = curdoc()
# doc.theme = plot_theme
## Sample Data
Let’s generate some data for fitting. I will use the function $\mathbf{y = f(x) = sin(4x)}$
def f(x, c):
return np.sin(4 * x) + c
data_size = 1000
domain = (-np.pi/8, np.pi/4)
std = 1.0
const = 2.0
x = np.linspace(domain[0], domain[1], data_size)
y = f(x, const) + np.random.normal(0, std, data_size)
Let’s do a scatter plot of the data:
p = figure(plot_width=600, plot_height=600)
p.circle(x, y, size=10, alpha=0.2, color="#66D9EF", legend="y")
p.line(x, f(x, 2), color="#F92672", line_width=3, legend="Actual")
p.title.text = "Y vs X"
p.xaxis.axis_label = "X"
p.yaxis.axis_label = "Y"
curdoc().clear()
doc = curdoc()
doc.theme = plot_theme
show(p)
## Smoothing
### Gaussian Kernel
Gaussian Kernel or Radial Basis Function Kernel is a very popular kernel used in various machine learning techniques. The kernel is given by:
$$\mathbf{K(x, x_0)} = exp( - \dfrac{||x - x_0||^2}{2 h^2})$$
$h$ is a free parameter also called the bandwidth parameter. It determines the width of the kernel.
def gaussian_kernel(x, x0, h):
return np.exp(- 0.5 * np.power((x - x0) / h, 2) )
### One Dimentional Smoother
In Kernel Regression, for a given point $x_0$, we use a weighted average of the nearby points’ response variable as the estimated value. One such technique is k-Nearest Neighbor Regression.
In Kernel Smoothing, we take the idea further by using all the training data and continously decrease the weights for points farther away from the given point $x_0$. The bandwidth parameter $h$ mentioned earlier controls the decreasing rate of the weights. Bandwidth $h$ can also be interpreted as the width of the kernel, centered at $x_0$.
When the bandwidth is smaller, the weighting effect of the kernel is more localized
This idea of localization goes beyond Gaussian Kernel and also applies to other common kernel functions such as the Epanechnikov Kernel.
As part of the procedure, we use the kernel function and the bandwidth $h$ to smooth the data points to obtain a local estimate of the response variable.
The final estimator is given by:
$$\mathbf{\hat f(x_0)} = \mathbf{\hat y_i} = \dfrac{\sum_{i=1}^{N} K_h(x_0, x_i) y_i}{\sum_{i=1}^{N} K_h(x_0, x_i)}$$
where $\mathbf{K_h}$ represents kernel with a specific bandwidth $h$.
In essense, it is the weighted average of all the response variable values $y_i$ with weights equal to the kernel function centered at $x_0$ (the estimation point) for each $x_i$. Different bandwidth values will give different kernel function values, and in turn, different weights.
Let’s implement it in Python to see the results of changing the bandwidth:
def predict(x_test, x_train, y_train, bandwidth, kernel_func=gaussian_kernel):
return np.array([(kernel_func(x_train, x0, bandwidth).dot(y_train) ) /
kernel_func(x_train, x0, bandwidth).sum() for x0 in x_test])
h_values = [0.01, 0.1, 1]
colors = ["#A6E22E", "#FD971F", "#AE81FF"]
p = figure(plot_width=600, plot_height=600)
p.circle(x, y, size=10, alpha=0.2, color="#66D9EF", legend="y")
p.line(x, f(x, 2), color="#F92672", line_width=3, legend="Actual", line_dash="dashed")
for idx, h in enumerate(h_values):
p.line(x, predict(x, x, y, h), color=colors[idx], line_width=2, legend="y_hat (h={})".format(h))
p.title.text = "Kernel Regression (Gaussian)"
p.xaxis.axis_label = "X"
p.yaxis.axis_label = "Y"
curdoc().clear()
doc = curdoc()
doc.theme = plot_theme
show(p)
As seen from the estimators for different $h$, kernel smoothing suffers from the Bias-Variance Tradeoff:
• As $h$ decreases, variance of the estimates gets larger, bias gets smaller, and the effect of the kernel is localized
• As $h$ increases, variance of the estimates gets smaller, bias gets larger, and the effect of the kernel is spread out
Let’s plot the Mean Squared Error MSE of the estimates versus the bandwidth:
h_range = np.linspace(0.01, 0.2, 20)
mses = [np.mean(np.power(y - predict(x, x, y, h), 2)) for h in h_range]
p = figure(plot_width=600, plot_height=300)
p.circle(x=h_range, y=mses, size=10, color="#66D9EF")
p.line(x=h_range, y=mses, color="#66D9EF", line_width=3)
p.title.text = "MSE vs Bandwidth"
p.xaxis.axis_label = "Bandwidth"
p.yaxis.axis_label = "MSE"
curdoc().clear()
doc = curdoc()
doc.theme = plot_theme
show(p)
The next step is to find a “good” value for bandwidth and we can use Cross Validation for that.
## Cross Validation
Cross Validation is a common method to tackle over-fitting the parameters of the model. The data is split into parts such that some of it is used as the training set and the rest as the validation set.
Splitting the data helps with not using the same data twice to fit the model parameters. Either the data point is used in training set or validation set. Training set is used to fit our model parameters, which are used to predict the values of response variable in the validation set. Hence, we can calculate the quality of our prediction based on the prediction error of validation set.
scikit-learn has modules for different cross validation techniques. However, I will implement these from scratch using numpy to avoid the dependency on scikit-learn just for cross validation.
Let’s discuss two of these techniques:
### Leave One Out Cross Validation (LOOCV)
In Leave One Out Cross Validation (LOOCV), we leave one observation out as the validation set and the remaining data points are used for model building. Finally, the response variable is predicted for the left out value as the validation set.
mse_values = []
for h in h_range:
errors = []
for idx, val in enumerate(x):
x_test = np.array([val])
y_test = np.array([y[idx]])
x_train = np.append(x[:idx], x[idx+1:])
y_train = np.append(y[:idx], y[idx+1:])
assert len(x_train) == data_size - 1
y_test_hat = predict(x_test, x_train, y_train, h)
errors.append((y_test_hat - y_test)[0])
mse_values.append(np.mean(np.power(errors, 2)))
Let’s plot the Mean Squared Error MSE of the estimates versus the bandwidth, as earlier:
p = figure(plot_width=600, plot_height=300)
p.circle(x=h_range, y=mse_values, size=10, color="#66D9EF")
p.line(x=h_range, y=mse_values, color="#66D9EF", line_width=3)
p.title.text = "Cross Validation - LOOCV - MSE vs Bandwidth"
p.xaxis.axis_label = "Bandwidth"
p.yaxis.axis_label = "MSE"
curdoc().clear()
doc = curdoc()
doc.theme = plot_theme
show(p)
As we can see, we can find an optimal bandwidth value to minimize the MSE. Let’s check the fit for that bandwidth:
h_optimal = h_range[np.argmin(mse_values)]
print(h_optimal)
# Output:
# 0.07
Below the estimator for the optimal bandwidth $0.07$:
p = figure(plot_width=600, plot_height=600)
p.circle(x, y, size=10, alpha=0.2, color="#66D9EF", legend="y")
p.line(x, f(x, 2), color="#F92672", line_width=3, legend="Actual", line_dash="dashed")
p.line(x, predict(x, x, y, h_optimal), color="#A6E22E", line_width=2, legend="y_hat (h={})".format(h_optimal))
p.title.text = "Cross Validation - LOOCV - Optimal Fit"
p.xaxis.axis_label = "X"
p.yaxis.axis_label = "Y"
curdoc().clear()
doc = curdoc()
doc.theme = plot_theme
show(p)
### K-Fold Cross Validation
Another popular cross validation technique is K-Fold Cross Validation, where data is divided in $K$ random chunks. One of the chunks is used as the validation set and the rest as the training set. This procedure is repeated several times to get the prediction error for each value of bandwidth.
Here is an implementation of K-Fold CV:
def split_k_fold(x, y, folds):
if len(x) != len(y):
raise ValueError("X and Y Should have same length")
indices = np.arange(len(x))
np.random.shuffle(indices)
split_size = len(x) // folds
return np.array([x[n * split_size:(n + 1) * split_size] for n in np.arange(folds)]), np.array(
[y[n * split_size:(n + 1) * split_size] for n in np.arange(folds)])
Let’s try $K = 4$ and $10$ tries for each $h$. Similar to LOOCV, let’s plot the MSE vs bandwidth to see their relationship. Again, we can optimize the bandwidth by minimizing the MSE.
num_folds = 4
num_tries = 10
fold_indices = np.arange(num_folds)
mse_values = []
for h in h_range:
trial_mses = []
for trial in np.arange(num_tries):
x_splits, y_splits = split_k_fold(x, y, num_folds)
mses = []
for idx in fold_indices:
test_idx = idx
train_idx = np.setdiff1d(fold_indices, [idx])
train_x, test_x, train_y, test_y = (np.concatenate(x_splits[train_idx]),
x_splits[test_idx],
np.concatenate(y_splits[train_idx]),
y_splits[test_idx])
test_y_hat = predict(test_x, train_x, train_y, h)
mses.append(np.mean(np.power(test_y_hat - test_y, 2)))
trial_mses.append(np.mean(mses))
mse_values.append(np.mean(trial_mses))
p = figure(plot_width=600, plot_height=300)
p.circle(x=h_range, y=mse_values, size=10, color="#66D9EF")
p.line(x=h_range, y=mse_values, color="#66D9EF", line_width=3)
p.title.text = "Cross Validation - K-Fold - MSE vs Bandwidth"
p.xaxis.axis_label = "Bandwidth"
p.yaxis.axis_label = "MSE"
curdoc().clear()
doc = curdoc()
doc.theme = plot_theme
show(p)
h_optimal = h_range[np.argmin(mse_values)]
print(h_optimal)
# Output:
# 0.03
p = figure(plot_width=600, plot_height=600)
p.circle(x, y, size=10, alpha=0.2, color="#66D9EF", legend="y")
p.line(x, f(x, 2), color="#F92672", line_width=3, legend="Actual", line_dash="dashed")
p.line(x, predict(x, x, y, h_optimal), color="#A6E22E", line_width=2, legend="y_hat (h={})".format(h_optimal))
p.title.text = "Cross Validation - K-Fold - Optimal Fit"
p.xaxis.axis_label = "X"
p.yaxis.axis_label = "Y"
curdoc().clear()
doc = curdoc()
doc.theme = plot_theme
show(p)
For this particular example, compared to LOOCV, K-Fold CV smoothing is more localized as it has a lower value of bandwidth. However, both approaches show a similar releationship between MSE and bandwidth.
## Disadvantages of Kernel Smoothing
• Weights calculated at the boundaries are biased due to one side of the kernel being cut off. This leads to biased estimates at the boundaries.
• Issues can also arise if the data is not spacially uniform. Spaces with fewer data points will have more biased estimates since there will be fewer nearby points to weight the response variable values.
## Final Words
In this post, we took a step-by-step approach to fit Kernel Smoothing using Gaussian Kernel. Same approach can be applied using other kernels. We also applied Cross Validation to choose an optimal bandwidth parameter.
## Appendix
### Bokeh Theme
### Monokai-inspired Bokeh Theme
# written July 23, 2017 by Luke Canavan
### Here are some Monokai palette colors for Glyph styling
# @yellow: "#E6DB74"
# @blue: "#66D9EF"
# @pink: "#F92672"
# @purple: "#AE81FF"
# @brown: "#75715E"
# @orange: "#FD971F"
# @light-orange: "#FFD569"
# @green: "#A6E22E"
# @sea-green: "#529B2F"
attrs:
Axis:
axis_line_color: "#49483E"
axis_label_text_color: "#888888"
major_label_text_color: "#888888"
major_tick_line_color: "#49483E"
minor_tick_line_color: "#49483E"
Grid:
grid_line_color: "#49483E"
Legend:
border_line_color: "#49483E"
background_fill_color: "#282828"
label_text_color: "#888888"
Plot:
background_fill_color: "#282828"
border_fill_color: "#282828"
outline_line_color: "#49483E"
Title:
text_color: "#CCCCCC"
### Email
------------------
+1-312-566-7843 | 2020-04-07 00:38:14 | {"extraction_info": {"found_math": true, "script_math_tex": 22, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.5439366102218628, "perplexity": 5572.826250134417}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371662966.69/warc/CC-MAIN-20200406231617-20200407022117-00392.warc.gz"} |
https://mathoverflow.net/questions/7454/what-are-cr-manifolds-like/7587 | # What are CR manifolds like?
The complex structure on a complex manifold pulls back to what's called a CR structure on any real codimension 1 submanifold. The structure induced on a submanifold of higher codimension is a CR structure if a non-degeneracy condition holds. It's possible to describe these structures intrinsically, without reference to an embedding. I don't know anything else.
I'd be happy with whatever kind of answer to the title question, but here are some more specific ones:
1. Does CR stand for Cauchy-Riemann, or what?
2. What kind of local invariants do CR manifolds have? Are there coordinates around every point that look like a real hyperplane in C^n? Or can there be some curvature or something.
3. Can there be continuous families of CR structures on a given manifold? If the manifold is compact can these families (mod diffeomorphism) be infinite-dimensional?
4. I have the impression, just from arxiv postings and seminar titles, of CR geometry being studied more in analysis than in softer geometric fields. Is that accurate, and if so what accounts for it?
CR does stand for Cauchy-Riemann.
CR structures on 3 dimensional manifolds arise as the boundaries of complex (or almost-complex) 4 manifolds; if these boundaries are strictly pseudo-convex (i.e. convex in "holomorphic directions") the CR structure on the 3-manifold is a contact structure (if the boundary is only (pseudo-)convex or (Levi) flat, the CR structure integrates to a confoliation or a foliation respectively). There can be infinite dimensional families of foliations on a 3-manifold; more generally, whenever the CR structure is "non-generic" or integrable, one has continuous moduli, otherwise (eg in the contact structure case) one has discrete moduli (to be explicit: what has discrete moduli is the contact structure, not the "CR+contact structure".)
• Is it special to three dimensions that the four-manifold it bounds only needs to by almost complex? It makes it sound like a CR structure is just an almost complex structure on M x R. Dec 2, 2009 at 16:42
• It is special of the case where HM (see my answer below) has real dimension 2. In this case indeed the integrability condition is automatically true. Dec 4, 2009 at 9:56
The complex structure on a complex manifold pulls back to what's called a CR structure on any real codimension 1 submanifold. The structure induced on a submanifold of higher codimension is a CR structure if a non-degeneracy condition holds.
The CR structure is induced on arbitrary real CR submanifolds $M$ in a complex manifold $X$, that is submanifolds for which the intersection $H^{01}$ of the bundle $T^{01}X$ of all $(0,1)$ vectors with the complexified tangent bundle $\mathbb C\otimes TM$ of $M$ is of constant dimension along $M$. In particular, a real codimension $1$ submanifold, i.e. a real hypersurface, is always a CR submanifold.
The induced CR structure of a CR submanifold is then defined by that intersection subbundle $H^{01}\subset \mathbb C\otimes TM$. (Equivalently, $(1,0)$ vectors can be used instead of $(0,1)$, but the latter is more convenient e.g. in the context of the $\bar\partial$ problem.)
I have never seen the condition being a CR submanifold called a "non-degeneracy condition", and wouldn't do it, because that condition is generally not stable under small perturbation. For instance, a complex line inside $\mathbb C^2$ is a CR submanifold that can be perturbed into a non CR submanifold.
The special property of a codimension $1$ submanifold is that it is always a CR submanifold. The same is not true in general for submanifolds of higher codimension.
It's possible to describe these structures intrinsically, without reference to an embedding.
Yes, a CR structure on a real manifold $M$ is defined by any complex subbundle $V=H^{01}$ of the complexification $\mathbb C\otimes TM$ satisfying $V\cap \bar V=\{0\}$ and the integrability $[V, V]\subset V$. If only the first condition is assumed, $V$ defines an almost CR structure. There is also the intermediate partial integrability condition $[V, V]\subset V\oplus \bar V$.
Equivalently, an almost CR structure can be defined without complexification, by a pair $(H,J)$ of a real subbundle $H\subset TM$ and a complex structure $J\colon H\to H$, see e.g. this answer for more details. However, the integrability condition $[V, V]\subset V$ becomes more verbose, when written in terms of $H$ and $J$.
The CR codimension of an almost CR structure is defined intrinsically as the complex codimension of $H^{10}\oplus H^{01}$ in $\mathbb C\otimes TM$, or equivalently, the real codimension of $H$ in $TM$, where $H = (H^{10}\oplus H^{01}) \cap TM$ and $H^{10}=\overline{H^{01}}$.
1. Does CR stand for Cauchy-Riemann, or what?
It stands for both Cauchy-Riemann and Complex-Real.
1. What kind of local invariants do CR manifolds have? Are there coordinates around every point that look like a real hyperplane in C^n? Or can there be some curvature or something.
The lowest order invariant, the Levi form, is of the 2nd order. It is possible to choose local coordinates, the only 2nd order terms are the ones from the Levi form.
In contrast to complex structures (corresponding to CR structures of CR codimension $0$), for a general CR structure of positive CR codimension, there are infinitely many higher order local invariants.
See e.g. my paper, Normal forms for almost non-integrable CR structures, Amer. J. of Math., 134 (2012), no. 4, 915-947, also available on the arxiv.org, for a complete intrinsic normal form, including the non-integrable case.
1. Can there be continuous families of CR structures on a given manifold? If the manifold is compact can these families (mod diffeomorphism) be infinite-dimensional?
Yes, see above. Also, infinite-dimensional families of non-CR-equivalent CR structures can be obtained even locally using the Chern-Moser normal form or Cartan connection, see e.g.
Chern, S. S.; Moser, J. K. Real hypersurfaces in complex manifolds. Acta Math. 133 (1974), 219–271
1. I have the impression, just from arxiv postings and seminar titles, of CR geometry being studied more in analysis than in softer geometric fields. Is that accurate, and if so what accounts for it?
You can find CR structures in both analysis and "softer" geometry such as compatible CR structures with contact structures, fillable CR structures etc, for instance in this book:
MR3012475 Cieliebak, Kai; Eliashberg, Yakov. From Stein to Weinstein and back. Symplectic geometry of affine complex manifolds. American Mathematical Society Colloquium Publications, 59. American Mathematical Society, Providence, RI, 2012.
CR submanifolds of a complex manifold are defined as submanifolds $M \subseteq X$ such that $TM \cap iTM \subseteq TX$ has constant rank ($i$ is the imaginary unit). Note that the condition is automatically satisfied if M has codimension one; for higher codimension this is not true.
An abstract CR manifold is a real manifold $M$, with a distinguished subbundle $HM \subseteq TM$, corresponding to $TM \cap iTM$, endowed with a linear endomorphism $J$ with $J^2=-Id$. The structure is furthermore required to satisfy a so called integrability condition: For all sections $X,Y$ of $HM$:
• $[X,JY]+[JX,Y]$ is a section of $HM$
• $([X,Y]-[JX,JY]) + J([X,JY]+[JX,Y]) = 0$
Not every abstract CR manifold can be realized as a CR submanifold.
• Are there topological obstructions to realizing a CR manifold as a CR submanifold? (Or what kind of obstructions.) Dec 2, 2009 at 16:45
• There are local obstructions to the existence of CR functions (and hence to embeddability). For 3-dimensional CR manifolds "generically" (in the Baire sense) the only CR functions are the constants. Dec 4, 2009 at 9:58
As matter of fact, the abbreviation CR stands for Complex-Real or Cauchy–Riemann.
In mathematics, the CR manifold is a differentiable manifold together with a geometric structure modeled on that of a "real" hypersurface in a "complex" vector space.
Poincare raised the concept of C^2 but CR firstly appeared in Cartan's work. Chern and Morse generalized the C^2 to C^∞.
In 1907, H. Poincaré wrote a seminal paper, [a6], in which he showed that two real hypersurfaces in C^2 are, in general, biholomorphically inequivalent
Cartan called his geometry "conformal geometry" in order to emphasize that what he was studying was a generalization of the theory of one complex variable (conformal geometry). The term "CR manifold" was first used in [Gr].
A second solution was given by Chern-Moser in 1973. In joint work with Chern [CM] it was generalized, along with Cartan's original solution, to dimensions greater than two. | 2022-10-05 16:10:50 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8232566118240356, "perplexity": 388.4644745785769}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337631.84/warc/CC-MAIN-20221005140739-20221005170739-00356.warc.gz"} |
https://mersenneforum.org/showthread.php?s=d132a008ec64f8c97bf68f8ebd29c3e3&t=16676&goto=nextoldest | mersenneforum.org > YAFU Running YAFU
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2010-12-25, 13:24 #1 lorgix Sep 2010 Scandinavia 3×5×41 Posts Running YAFU Hi, I can't get flags to work. I'm assuming that I've missed something fundamental. Probably something that's obvious to people who know at least one programming language. I don't understand how to do batch work either. Also, is there a way of changing the behavior of the factor() command? Limits etc.. I'm running yafu-64k-x64 1.21 in Windows 7.
2010-12-25, 13:34 #2 Karl M Johnson Mar 2010 3×137 Posts My example: I have a batch file, factor.bat . If you open it with NotePad, it has the following: Code: yafu-32k-x64.exe -threads 4 -v -rhomax 800000 -fmtmax 299396627
2010-12-25, 14:54 #3 lorgix Sep 2010 Scandinavia 3·5·41 Posts Thanks. I think I've created a correct work file and a correct .bat. It appears the only problem left is that I still don't know how to tell YAFU to get to work. I have 'ecm.txt' and 'factor.bat', what do I type in YAFU?
2010-12-25, 15:32 #4 Karl M Johnson Mar 2010 41110 Posts Well, you can type the same stuff as in your ecm.txt , but, better yet, launch the batch file. For example, you launch yafu, a console window appears, and you type ecm(*some number*) and press Enter, and that will "tell" yafu to ECM that number. If you have "ecm (*some number*) in ecm.txt, you can just edit your factor.bat, so it will have "yafu.exe
2010-12-25, 15:43 #5
lorgix
Sep 2010
Scandinavia
11478 Posts
Quote:
Originally Posted by Karl M Johnson Well, you can type the same stuff as in your ecm.txt , but, better yet, launch the batch file. For example, you launch yafu, a console window appears, and you type ecm(*some number*) and press Enter, and that will "tell" yafu to ECM that number. If you have "ecm (*some number*) in ecm.txt, you can just edit your factor.bat, so it will have "yafu.exe
Ok, I understand all of that. The only problem is that apparently I don't know how to "launch a file in YAFU".
2010-12-25, 16:01 #6
bsquared
"Ben"
Feb 2007
2·32·191 Posts
Quote:
Originally Posted by lorgix Ok, I understand all of that. The only problem is that apparently I don't know how to "launch a file in YAFU".
yafu supports many different modes of interaction.
Some examples:
from a command line type
Code:
yafu-64k-x64
and you'll get a command prompt from which you can enter commands like siqs(*number*) or ecm(*number*,*num_curves*)
from a command line type
Code:
yafu-64k-x64 "siqs(*number*)"
or
Code:
yafu-64k-x64 "ecm(*number*,*num_curves*)"
and the program will do the same thing, but without the command prompt.
expanding on this theme, from a command line type
Code:
yafu-64k-x64 -batchfile filename
and the program will process every line in the batchfile one at a time (lines in the batchfile should be statements such as siqs(*number*), etc.).
Or you could use redirection, as Karl is doing.
There are many other flags you can optionally add after the command (siqs(), ecm(), etc), such as -v, -threads, or options to control the ecm,p +/- 1 bounds, and so forth. these are documented in docfile.txt.
Code:
yafu-64k-x64 "siqs(*number*)" -v -threads 4 -rhomax 1000000
or Karl's example with redirection.
Hope this helps.
Last fiddled with by bsquared on 2010-12-25 at 16:03 Reason: fix some syntax
2012-03-28, 00:45 #7 Dubslow Basketry That Evening! "Bunslow the Bold" Jun 2011 40> help primes searching for help on 'primes' >> help prime searching for help on 'prime' >> help factor searching for help on 'factor' >> help searching for help on 'help' >> I don't understand where exactly to find the help. Downloaded a few hours ago, v1.30. Last fiddled with by Dubslow on 2012-03-28 at 00:46
2012-03-28, 01:08 #8
bsquared
"Ben"
Feb 2007
65568 Posts
Quote:
Originally Posted by Dubslow Code: Issuing the command 'help funcname' will bring up more detailed help on a particular function. Code: Type help at any time, or quit to quit >> help primes searching for help on 'primes' >> help prime searching for help on 'prime' >> help factor searching for help on 'factor' >> help searching for help on 'help' >> I don't understand where exactly to find the help. Downloaded a few hours ago, v1.30.
I noticed years ago this was the case in the linux version and I never fixed it (it works fine in Windows...). I've wondered how long it would take for someone to notice :)
All it does is print the relevant section in docfile.txt, which should have been packaged in the download file. Open it in your favorite text editor and search for the term you want.
2012-03-28, 01:17 #9
Dubslow
"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88
3×29×83 Posts
Quote:
Originally Posted by bsquared I noticed years ago this was the case in the linux version and I never fixed it (it works fine in Windows...). I've wondered how long it would take for someone to notice :) All it does is print the relevant section in docfile.txt, which should have been packaged in the download file. Open it in your favorite text editor and search for the term you want.
Thanks.
Edit: Something else that I couldn't figure out what was wrong:
Code:
-pscreen Adding this flag causes the primes() function to output primes
to the screen
Code:
bill@Gravemind:~/yafu∰∂ yafu -pscreen primes()
bash: syntax error near unexpected token ('
bill@Gravemind:~/yafu∰∂ yafu -pscreen primes(200)
bash: syntax error near unexpected token ('
bill@Gravemind:~/yafu∰∂ yafu -pscreen "primes(200)"
no switch detected
bill@Gravemind:~/yafu∰∂ yafu -pscreen "primes()"
no switch detected
bill@Gravemind:~/yafu∰∂ yafu "primes()"
invalid character in str2hexz
not enough arguments, please specify min and max of range
ans = 0
bill@Gravemind:~/yafu∰∂ yafu "primes(1,1000)"
elapsed time = 0.0030
ans = 168
bill@Gravemind:~/yafu∰∂ yafu -pscreen "primes(1,1000)"
no switch detected
bill@Gravemind:~/yafu∰∂ yafu "primes(1,1000) -pscreen"
unrecognized token: pscreen
bill@Gravemind:~/yafu∰∂ yafu "primes(1,1000)" -pscreen
elapsed time = 0.0030
ans = 168
bill@Gravemind:~/yafu∰∂ yafu
03/27/12 19:36:01 v1.30 @ Gravemind, System/Build Info:
detected Intel(R) Core(TM) i7-2600K CPU @ 3.40GHz
detected L1 = 32768 bytes, L2 = 8388608 bytes, CL = 64 bytes
measured cpu frequency ~= 3494.445730
===============================================================
======= Welcome to YAFU (Yet Another Factoring Utility) =======
======= bbuhrow@gmail.com =======
======= Type help at any time, or quit to quit =======
===============================================================
cached 78498 primes. pmax = 999983
>> primes(1,1000) -pscreen
unrecognized token: pscreen
>> primes(1,1000)
elapsed time = 0.0031
ans = 168
>> help primes
searching for help on 'primes'
>> primes help
searching for help on 'es help'
>> help
searching for help on 'help'
>> quit
bill@Gravemind:~/yafu∰∂ yafu help primes
no switch detected
And after that it devolved to me trying to figure out how to get help working. How can I print primes to screen?
Last fiddled with by Dubslow on 2012-03-28 at 01:55
2012-03-28, 03:37 #10
bsquared
"Ben"
Feb 2007
2·32·191 Posts
Quote:
Originally Posted by Dubslow How can I print primes to screen?
Code:
yafu "primes(0,1000,0)" -pscreen
Generally, the command needs to come first followed by options. And you need the second "0" to tell the primes function to actually compute the primes instead of just count them (the default). I know, I know, it should see -pscreen and know that it needs to compute the primes... just one of the quirks of the program :)
Last fiddled with by bsquared on 2012-03-28 at 03:39
Similar Threads Thread Thread Starter Forum Replies Last Post EdH YAFU 8 2018-03-14 17:22 Neimanator PrimeNet 14 2013-08-10 20:15 bsquared YAFU 12 2012-11-08 04:12 bsquared YAFU 21 2012-09-04 19:44 ThomRuley YAFU 11 2011-06-13 13:35
All times are UTC. The time now is 08:41.
Wed May 12 08:41:51 UTC 2021 up 34 days, 3:22, 0 users, load averages: 1.83, 1.72, 1.76 | 2021-05-12 08:41:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3103274703025818, "perplexity": 8560.717751102991}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991685.16/warc/CC-MAIN-20210512070028-20210512100028-00066.warc.gz"} |
https://access.openupresources.org/curricula/our-k5-math/grade-2/unit-6/section-a/lesson-5/student.html | # Lesson 5 Center Day 1
• Let’s work with shapes.
## Warm-up Number Talk: Add 5
Find the value of each expression mentally.
## Activity 1 Introduce Which One?
We are going to learn a new way to play the Which One center.
## Activity 2 Introduce Can You Draw It?
We are going to learn a new way to play the Can You Draw It? center. | 2023-01-28 20:02:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1725463569164276, "perplexity": 2493.100421343915}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499654.54/warc/CC-MAIN-20230128184907-20230128214907-00769.warc.gz"} |
http://mathematica.stackexchange.com/questions/38224/interpolation-of-a-table-with-4-parameters | # Interpolation of a table with 4 parameters
I have a huge data file which I need to do some analysis on it. The file is built in this way that for each 4 variables, there is a value of the function:
List={{x,y,z,w},f[x,y,z,w]}
A part of the real data is the following List:
tab={{{0.0001, 1., -1., 1.*10^-6}, 0.6704}, {{0.0001, 1., -0.85,3.98107*10^-6}, 0.964659}, {{0.0001, 1., -0.7,0.0000199526}, 0.491676}, {{0.0001, 1., -0.55, 0.0001}, 0.0974736}};
For each value of the x and y parameters, I need to NIntegrate with respect to z and w. If I could have the whole function wrt all parameters, f[x,y,z,t], I would do:
tab2=Flatten[Table[{{x,y},NIntegrate[f[x,y,z,w], {z,zmin,zmax},{w,wmin,wmax}]},{x,xmin,xmax},{y,ymin,ymax}],1];
and then I would Interpolate tab2 with respect to {x,y}:
final=Interpolation[tab2];
I have almost 4 million points, I can do the analysis in this way, but how can I be sure that the Interpolation function is a correct function, while there are too many approximations to make it? I know that Interpolation works for any n-dimension, but it won't be acurate anymore for more than 2 parameters. Is there a way that I do this kind of analysis without needning to Interpolate with respect to all parameters? For example is there a way that I can just Interpolate with respect to {z,w}, I do the Integration, then I Interpolate the result with respect to {x,y}?
-
What kind of analysis do you want to do? – belisarius has settled Dec 3 '13 at 22:00
@belisarius As I wrote, I need to Integrate with respect to the third and forth variable, then at the end I need to have a function with respect to the first and second ones. If I could use Interpolation with respect to all variables, it could be easy to do it. But it's not possible to do that. – ZKT Dec 4 '13 at 0:32
Interpolation[tab] @@@ tab[[All, 1]] - tab[[All, 2]] // Abs // Total gives $3.33067 \times 10^{-16}$, so it looks like it works pretty well to me. "Problem with", "some analysis", "doesn't work well", "need to integrate wrt third and fourth variable", "at the end I need to have a function". These phrases are not meaningful to someone who does not already know what you are trying to accomplish. Sorry, but your question is too vague to be answerable in its present form. Please clarify. – Oleksandr R. Dec 4 '13 at 2:02
@OleksandrR. I edited the question. Is it clear enough now or I need to add more details? – ZKT Dec 4 '13 at 2:22
If X and Y coordinates are integers in all data then you can collect all {{x_,y_,,},_} subsets and make interpolation function for each subset. The result will be in this form, for example: f[x,y][z,w] – Kamov Sergey Dec 4 '13 at 2:43
This is an opportunity for you to clarify your question or give a better example.
The coding example that you posted as non-working, does work:
tab = {{{1., 3., 7., 5.}, 1.85541}, {{1., 3., 8., 5.}, 1.76612}, {{1., 3., 7., 6.}, 1.99826},
{{1., 3., 8., 6.}, 1.89112}, {{1., 4., 7., 5.}, 0.957483}, {{1., 4., 8., 5.}, 0.868198},
{{1., 4., 7., 6.}, 1.10034}, {{1., 4., 8., 6.}, 0.993198}, {{2., 3., 7., 5.}, 4.85541},
{{2., 3., 8., 5.}, 4.76612}, {{2., 3., 7., 6.}, 4.99826}, {{2., 3., 8., 6.}, 4.89112},
{{2., 4., 7., 5.}, 3.95748}, {{2., 4., 8., 5.}, 3.8682}, {{2., 4., 7., 6.}, 4.10034},
{{2., 4., 8., 6.}, 3.9932}};
f = Quiet@Interpolation[tab];
tab2 = Flatten[Table[{{x, y}, NIntegrate[f[x,y,z,w], {z,7,8}, {w,5,6}]}, {x, 1, 2}, {y, 3, 4}], 1]
(*
{{{1, 3}, 1.87773}, {{1, 4}, 0.979805}, {{2, 3}, 4.87773}, {{2, 4}, 3.9798}}
*)
-
I edited the problem, putting a part of my real list. I also tried to explain my problem. My biggest worry is that I'm not sure the Interpolation wrt 4 parameters would work correctly and acurately. – ZKT Dec 4 '13 at 4:26
@ZKT It should work "correctly and accurately" ... unless there is a bug (which of course can ever happen). Just use it and have a cross-checking contraption at hand. – belisarius has settled Dec 4 '13 at 4:31
Thanks. My colleagues were complaining about my method, so I was trying to find another way for doing the analysis, to become sure about the results. – ZKT Dec 4 '13 at 4:37 | 2015-11-28 02:31:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3543141186237335, "perplexity": 1458.2162862963335}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398450745.23/warc/CC-MAIN-20151124205410-00263-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://www.cheenta.com/sum-of-whole-numbers-amc-10a-2012-problem-8/ | Select Page
Try this beautiful problem from Algebra based on Sum of whole numbers.
## Sum of whole numbers – AMC-10A, 2012- Problem 8
The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?
• $8$
• $9$
• $7$
• $6$
• $5$
### Key Concepts
Algebra
Equation
Sum of digits
But try the problem first…
Answer: $7$
Source
AMC-10A (2012) Problem 8
Pre College Mathematics
## Try with Hints
First hint
Let us assume three numbers are $x$,$y$ and $z$
Now according to the problem
$x+y=12$…………………(1)
$y+z=17$………………..(2)
$z+x=19$………………….(3).
Can you find out the values of $x$,$y$ and $z$…………….?
can you finish the problem……..
Second Hint
Adding three equations we get $2(x=y+z)=48$
$\Rightarrow x+y+z=24$………….(3)
Now subtract (3) from (1),we will get $z=12$ and similarly if we subtract (2) and (3) from (1) we will get $x=7$ and $y=5$
can you finish the problem……..
Final Step
Therefore,the numbers are $12$, $7$, and $5$ and middle number is $7$ | 2020-07-16 14:55:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7500622868537903, "perplexity": 1605.301234469586}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657169226.65/warc/CC-MAIN-20200716122414-20200716152414-00066.warc.gz"} |
http://mathhelpforum.com/discrete-math/178721-generating-function-sequence.html | # Math Help - Generating function for a sequence
1. ## Generating function for a sequence
The question is as follows:
It is known that the generating function for the sequence , {1,8,27,64,...} is
[x(x^2 + 4x +1)]/(1-x)^4
Use the result to show summation from r=1 to n of r^3 = [n^2(n^2+1)]/4
I know i have to find the sequence represented by 1,8,27,64, .. and once i get the sequence i have to manipulate it until i reach r^3 which will be given by the expression.
My problem is that i cannot find the sequence for 1,8,27,64,...
Thank buches!
2. Hello, mehn!
Please check the wording of the problem.
None of it makes sense.
It is known that the generating function for the sequence , {1, 8, 27, 64, ...}
. . is: .[x(x^2 + 4x +1)]/(1-x)^4 . What?
Use the result to show summation from r = 1 to n of r^3
. .is: .[n^2(n^2+1)]/4 . Not true
The generating function is: .f(r) = r^3
. . The terms are consecutive cubes.
The sum of the first n cubes is: .[n^2(n+1)^2]/4
3. Originally Posted by Soroban
Hello, mehn!
Please check the wording of the problem.
None of it makes sense.
The generating function is: .f(r) = r^3
. . The terms are consecutive cubes.
The sum of the first n cubes is: .[n^2(n+1)^2]/4
Please read Generating Functions. The OP's generating function for cubes is correct.
4. Originally Posted by mehn
The question is as follows:
It is known that the generating function for the sequence , {1,8,27,64,...} is
[x(x^2 + 4x +1)]/(1-x)^4
Use the result to show summation from r=1 to n of r^3 = [n^2(n^2+1)]/4
This is a strange problem (to me, at any rate), but it actually works!
The binomial series for (1-x)^{-4} is 1 + 4x + 10x^2 + 20x^3 + ... . When you multiply this by x(x^2 + 4x +1), you get x + 8x^2 + 27x^3 + 64x^4 + ... . The coefficients magically appear as the cubes of the integers!
In this problem, you are asked to find the sum of the first n coefficients. One way to do that is to multiply the given generating function by (1-x)^{-1} and looking for the coefficient of x^n in that product.
The coefficient of x^n in (1-x)^{-5} is $\frac{(n+4)!}{4!n!}$. So the coefficient of x^n in $\frac{x(x^2 + 4x +1)}{(1-x)^5}$ is
$\frac1{4!}\left(\frac{(n+1)!}{(n-3)!} + 4\frac{(n+2)!}{(n-2)!} + \frac{(n+3)!}{(n-1)!}\right),$
which eventually simplifies to [n^2(n+1)^2]/4.
Not the easiest way to get that formula, but an interesting piece of mathematics.
Edit. Thanks to alexmahone for that useful link. I hadn't seen his comment when writing mine. | 2014-04-20 16:02:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9225935935974121, "perplexity": 782.9708940091709}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609538824.34/warc/CC-MAIN-20140416005218-00389-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://www.semanticscholar.org/paper/Behavior-of-the-energy-gap-near-a-transition-in-at-Read/1b135cf1327ad738b9304608ba3f338c8c1cda5e | Behavior of the energy gap near a commensurate-incommensurate transition in double-layer quantum Hall systems at nu =1.
@article{Read1995BehaviorOT,
title={Behavior of the energy gap near a commensurate-incommensurate transition in double-layer quantum Hall systems at nu =1.},
journal={Physical review. B, Condensed matter},
year={1995},
volume={52 3},
pages={
1926-1931
}
}
• Published 5 January 1995
• Physics
• Physical review. B, Condensed matter
The charged excitations in the system of the title are vortex-antivortex pairs in the spin-texture described in the theory by Yang et al which, in the commensurate phase, are bound together by a string''. It is shown that their excitation energy drops as the string lengthens as the parallel magnetic field approaches the critical value, then goes up again in the incommensurate phase. This produces a sharp downward cusp at the critical point. An alternative description based on the role of…
15 Citations
Activation study of collective excitations of the soliton-lattice phase in the ν=1 double-layer quantum Hall state
• Physics
• 2010
We investigate thermal excitations of the pseudospin soliton lattice in the double-layer quantum Hall (QH) state at total Landau-level filling factor $\ensuremath{\nu}=1$ by detailed measurements of
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• 2009
We study thermal excitations of the pseudospin soliton phase in the bilayer quantum Hall (QH) state at total Landau level filling factor v = 1. Near the commensurate-incommensurate phase transition
Dissipationless transport in low-density bilayer systems
• Physics
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• 2000
In a bilayer electronic system the layer index may be viewed as the z component of an isospin- 1 / 2, a phase that was observed in quantum Hall systems and is predicted to exist at zero magnetic field at low density.
Strongly interacting bosons in a one-dimensional optical lattice at incommensurate densities
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• 2011
We investigate quantum phase transitions occurring in a system of strongly interacting ultracold bosons in a one-dimensional optical lattice. After discussing the commensurate-incommensurate
SU(4) skyrmions and activation energy anomaly in bilayer quantum Hall systems
• Physics
• 2004
The bilayer quantum Hall (QH) system has four energy levels in the lowest Landau level, corresponding to the layer and spin degrees of freedom. We investigate the system in the regime where all four
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• 1987 | 2022-05-24 22:56:28 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5623178482055664, "perplexity": 5114.128927716694}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662577259.70/warc/CC-MAIN-20220524203438-20220524233438-00368.warc.gz"} |
http://maikolsolis.com/project-euler-3-largest-prime-factor/ | # Largest prime factor
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
This solution was a little tricky. The best case scenario is if our number—let’s call it $n$—is a perfect square. In this case, the only admissible prime is the root square of $n$ repeated twice.
## Playing with square numbers
Using this idea, the algorithm goes as follow.
1. First, make a list with all the odd numbers until the integer part of the square root of $n$.
2. Then, add the number 2 to this list because it is a prime number.
3. Next, compute the modulo of $n$ with each element of the list, call it $p$.
4. Divide $n$ by each $p$. If the quotient is an integer, then change $n$ by this quotient.
5. Finally, iterate until $n$ is equal to $1$.
My implementation just works, but of course there is room for improvements. If you have any suggestion please let me know them in the comments.
import math # One large number number = 600851475143 # The list of prime factors prime_factors = [] # A control variable to # know when we finish we_are_done = False while not we_are_done: # The best case scenario is if # number = some_prime^2 sqrt_root_number = int(math.sqrt(number)) # Create a list of posible primes # 2, 3, 5, 7, ... seq_divisors = range(3, sqrt_root_number, 2) seq_divisors.insert(0,2) for factor in seq_divisors: # Checking if divisible if number % factor == 0: # If factor is divisible then change # the number by the quotient number = number / factor # add the prime to the list prime_factors.append(factor) # We're not finished yet! we_are_done = False break else: # There is not more factors we_are_done = True # Just to not include 1 in the list if number != 1: prime_factors.append(number) print prime_factors print max(prime_factors)
Go back to the list of problems. | 2017-05-25 08:37:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3355090916156769, "perplexity": 1073.7569396198148}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-22/segments/1495463608058.13/warc/CC-MAIN-20170525083009-20170525103009-00179.warc.gz"} |
http://www.eyeshalfclosed.com/scheduling/ | Scheduling Broadcasts in a Network of Timelines
Data
We use the Twitter-Friends dataset described in the following paper:
Steering Information Diffusion Dynamically against User Attention Limitation. Shuyang Lin, Qingbo Hu, Fengjiao Wang, Philip S. Yu. ICDM ’14.
We provide Python pickles of this data for convenient reuse in code:
• all_links.p (Download): A list of tuples of user ID’s, one for each link in the network, of the form (a, b) where b follows a.
• all_tweets.p (Download): A list of tuples, one for each tweet, containing following information in order:
1. Tweet creator ID.
2. Tweet creation time (in milliseconds since the epoch).
3. Retweeted user ID (-1 if not a retweet).
4. Replied-to user ID (-1 if not a user-reply).
5. The tweet ID.
6. Retweeted tweet ID (-1 if not a retweet).
7. Replied-to tweet ID (-1 if not a tweet-reply)
Attention Potential
Attention potential is a directed function ap(a,b), estimating the attention that follower b gives producer a. Likewise, reactions are also directed reactions(a,b), measuring the number of retweets to and replies of a’s tweets by b. The code linked above evaluates the correlation between attention and reactions in the Twitter dataset.
Monotony Aversion
The code linked above computes statistics of author-occurence clusters. For each cluster size, it reports the following:
• The number of clusters of this size.
• The number of tweets arising from clusters of this size.
• The number of tweets reacted to in clusters of this size.
• The number of clusters of this size containing at least one reaction. | 2017-02-23 14:04:54 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.40638282895088196, "perplexity": 9414.470845207312}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171171.24/warc/CC-MAIN-20170219104611-00496-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://brilliant.org/problems/pulleys-pulleys-everywhere-2/ | # Pulleys pulleys everywhere!
A reel has a hub radius $$R$$, the peripheral radius $$3R$$ and mass $$4m$$. Its moment of inertia about its axis is $$3 m R^2$$.
A thin massless thread is wound on a reel. A block of mass $$m$$ is hung through massless and frictionless pulleys as shown.
The reel is placed on a rough horizontal surface where the friction is sufficient to prevent slipping. The system is released from rest. Find acceleration to 2 decimal places (in $$\text{cm/s}^2$$) of the center of mass of the reel. Take $$g= \SI{10}{\meter\per\second\squared}.$$
× | 2017-12-17 23:17:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5201106071472168, "perplexity": 411.0869938771622}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948599156.77/warc/CC-MAIN-20171217230057-20171218012057-00303.warc.gz"} |
https://aptitude.gateoverflow.in/1595/cat-2005-question-90 | 165 views
Answer the following questions based on the information given below:
Help Distress (HD) is an NGO involved in providing assistance to people suffering from natural disasters. Currently, it has $37$ volunteers. They are involved in three projects: Tsunami Relief (TR) in Tamilnadu, Floor Relief in Maharastra (FR) and Earthquake Relief (ER) in Gujarat. Each volunteer working with Help Distress has to be involved at least in one project.
• A maximum number of volunteers are involved in FR project. Among them, the number of volunteers involved in FR project alone is equal to the volunteers having additional involvement in the ER project.
• Among them, the number of volunteers involved in ER project alone is double the number of volunteers involved in all three projects.
• $17$ volunteers are involved in TR project.
• Among them, the number of volunteers involved in TR project alone is one less than the number of volunteers involved in the ER project alone.
• Ten volunteers involved in TR project are also involved in at least one more project.
After withdrawal of volunteers, as indicated in the previous question (question $\#89)$, some new volunteers joined the NGO. Each one of them was allotted only one project in a manner such that, the number of volunteers working in one project alone for each of the three projects became identical. At that point, it was also found that the number of volunteers involved in FR and ER projects was the same as the number of volunteers involved in TR and ER projects. Which of the projects now has the highest number of volunteers?
1. ER
2. TR
3. FR
4. Cannot be determined | 2022-11-27 02:29:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.484871506690979, "perplexity": 2462.446447161263}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710155.67/warc/CC-MAIN-20221127005113-20221127035113-00073.warc.gz"} |
http://www.ck12.org/book/CK-12-Concept-Middle-School-Math---Grade-6/r4/section/9.2/ | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 9.2: Intersecting and Parallel Lines
Difficulty Level: At Grade Created by: CK-12
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Practice Intersecting and Parallel Lines
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Have you ever thought of designing your own skateboard park?
Marc and Isaac are working on a design for a new skateboard park. The city council of their town has agreed that the skateboard park is in need of renovation. Marc and Isaac have offered to help draw some initial plans to present at the next meeting. They are a little nervous about their design and about their presentation. Isaac’s mom offers to let them use some of her design paper and the two boys begin sketching their plan at the kitchen table.
“It definitely needs more rails,” Isaac says.
“What is a rail?” asks Isaac’s mom who glances at the design over her son’s shoulder.
“You know Mom, the sides don’t connect or cross,” Isaac says.
“Well, if that is what you want, your drawing is not accurate.”
Isaac looks down at the drawing. His mom is right. The rails don’t look correct.
To draw these rails, Isaac and Marc will need to understand the basics of Geometry. Pay attention to this Concept and you will understand how to help them with their design at the end.
### Guidance
In the last Concept you learned about lines and line segments. When lines intersect, sometimes we need to describe how they do so. Two of the descriptions are intersecting lines and parallel lines.
Intersecting lines are lines that cross at some point. You can think of an intersection in a pair of streets to help you remember intersecting lines.
Here you can see that the streets of this highway intersect just as two intersecting lines intersect or cross. Here is an example of intersecting lines that you would see in geometry.
The lines intersect or cross at one point. We call this point the point of intersection. Sometimes lines will intersect with other lines at more than one point.
Parallel lines do not cross or intersect EVER. They are equidistant.
In the sport of gymnastics, gymnasts use parallel bars to perform. Notice that the parallel bars are two bars that do not connect. They are an equal distance apart and will never cross or intersect.
Here is what parallel lines in geometry look like.
If we use a symbol for parallel lines, the symbol looks like this: ABCD\begin{align*}\overleftrightarrow{AB} \parallel \overleftrightarrow{CD}\end{align*}. This means that line \begin{align*}AB\end{align*} is parallel to line \begin{align*}CD\end{align*}.
Identify which lines are parallel and which are intersecting in each picture.
#### Example A
Solution: Parallel lines
#### Example B
Solution: Intersecting lines
#### Example C
Solution: Intersecting lines and parallel lines
Now let's go back to the skateboard park.
Have you figured out what is wrong with Isaac’s drawing? Think back to this Concept on geometry, reread the problem and underline any important information.
Marc and Isaac are working on a design for a new skateboard park. The city council of their town has agreed that the skateboard park is in need of renovation. Marc and Isaac have offered to help draw some initial plans to present at the next meeting. They are a little nervous about their design and about their presentation. Isaac’s mom offers to let them use some of her design paper and the two boys began sketching their plan at the kitchen table.
“It definitely needs more rails,” Isaac says.
“What is a rail?” asks Isaac’s mom, who glances at the design over her son’s shoulder.
“You know Mom, the sides don’t connect or cross,” Isaac says.
“Well, if that is what you want, your drawing is not accurate.”
Isaac looks down at the drawing. His mom is right. The rails don’t look correct.
If Isaac’s drawing is incorrect, then the rails in his drawing must not be parallel. Remember that parallel lines do not connect or cross in any way. When Isaac describes the rails to his mom it is clear that he wants them to be parallel. She says that his drawing is not accurate, so Isaac needs to redraw the rails and show that they do not connect.
### Vocabulary
Here are the vocabulary words in this Concept.
Point
a location in space that does not have size or shape.
Ray
a line that has one endpoint and continues indefinitely in one direction.
Line
a set of connected points without endpoints.
Line Segment
a set of connected points with two endpoints.
Point of Intersection
the point where two intersecting lines meet.
Intersecting Lines
lines that cross or meet at some point
Parallel Lines
Lines that do not cross or meet EVER and are equidistant.
### Guided Practice
Here is one for you to try on your own.
Look at this picture. Is the pattern made of parallel lines or intersecting lines?
This is a tricky question. The white tiles are definitely parallel to the opposite sides, but the lines do join at the corners with the brown square.
If you count the brown squares as part of the line, then they definitely do intersect.
If you don't count them, then the pattern is made up of parallel lines.
Many tile patterns are like this. Look around your home or school and you may find other patterns like this one.
### Video Review
Here is a video for review.
### Practice
Directions: Tell whether each picture shows parallel or intersecting lines.
1.
2.
3.
4.
Directions: Think about each example described below and determine whether the lines would be intersecting or parallel.
5. Telephone wires
6. The yellow lines down a highway
7. Stitches on a sweater
8. The sides of a ramp
9. The tread on a stair
10. A spider web
11. A climbing web
12. The two sides of a trapeze
13. A trail in the mountains
Directions: Now go through a newspaper and find three examples of parallel lines and three examples of intersecting lines. Be sure to check your work with a friend.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Intersecting lines
Intersecting lines are lines that cross or meet at some point.
line
A line is a straight one-dimensional figure that extends forever in opposite directions.
line segment
A line segment is a part of a line that has two endpoints.
Parallel
Two or more lines are parallel when they lie in the same plane and never intersect. These lines will always have the same slope.
Point
A point is a location in space that does not have size or shape.
Point of Intersection
A point of intersection is the point where two intersecting lines meet.
Ray
A ray is a part of a line that has one endpoint and continues indefinitely in one direction.
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Subjects: | 2017-02-21 19:22:36 | {"extraction_info": {"found_math": true, "script_math_tex": 3, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3086027503013611, "perplexity": 1874.1031117483228}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501170823.55/warc/CC-MAIN-20170219104610-00596-ip-10-171-10-108.ec2.internal.warc.gz"} |
http://mathhelpforum.com/calculus/82580-quick-explanation-integral.html | # Thread: Quick explanation of integral
1. ## Quick explanation of integral
Hi,
How do you go from
double integral of (2r)/(1+r) drdtheta to:
2 times the double integral of (1-(1/(1+r)))drdtheta
2. Originally Posted by s7b
Hi,
How do you go from
double integral of (2r)/(1+r) drdtheta to:
2 times the double integral of (1-(1/(1+r)))drdtheta
$\displaystyle 2\int \frac{r}{1+r}dr=2\int \frac{1+r-1}{r+1}=2\int \frac{1+r}{1+r}+\frac{-1}{1+r}dr=2 \int 1-\frac{1}{1+r}dr$
you could also use polynomial long division | 2018-03-17 14:51:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9754068851470947, "perplexity": 10890.092912338781}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257645177.12/warc/CC-MAIN-20180317135816-20180317155816-00687.warc.gz"} |
https://blog.spp2026.de/ | ### Triangulating real projective n-space
How many vertices do you need to triangulate the real projective n-space? From this blog post of Gil Kalai I learned about a recent preprint (arXiv:2009.02703) by Adiprasito-Avvakumov-Karasev where they construct triangulations with $\exp\big((1/2 + \mathcal{o}(1))\sqrt{n}\log{n}\big)\text{-many}$ vertices, which is the first construction needing subexponentially-many vertices. More information, also about the history of this problem, may … Continue reading "Triangulating real projective n-space"
### Resolution of Keller’s conjecture
Keller’s conjecture states that in any tiling of Euclidean space by identical hypercubes there are two cubes that meet face to face. (Consider the 2-dimensional picture on the right taken from Wikipedia. The squares share horizontal edges.) The conjecture is completely solved by now: it is true in dimensions 7 and less, but false in higher dimensions. The last missing part was … Continue reading "Resolution of Keller’s conjecture"
### Catastrophe theory
This term I am teaching a course on catastrophe theory for 2nd year students. When you look up this topic on Wikipedia, you will see words like chaos and singularities mentioned, and you will be shown fancy pictures of the seven elementary catastrophes. But when you try to understand what the actual mathematical content of … Continue reading "Catastrophe theory"
### Immersions of manifolds into Euclidean space
Recall the well-known result of Whitney that any (compact) smooth $$n$$-manifold admits an immersion into $$\mathbb{R}^{2n-1}$$. Today there was a preprint posted on the arXiv (arXiv:2011.00974) which mentioned in its introduction the following result of Cohen, which strengthens Whitney’s result as follows: Any (compact) smooth $$n$$-manifold admits an immersion into $$\mathbb{R}^{2n-\alpha(n)}$$, where $$\alpha(n)$$ is the … Continue reading "Immersions of manifolds into Euclidean space"
### Blockseminar on Dirac operators and scalar curvature
In mid-October we gathered for one week in Bollmannsruh (somewhat west of Berlin) to work our way through the seminal paper Positive scalar curvature and the Dirac operator on complete Riemannian manifolds by Gromov and Lawson. The hotel we stayed in lies directly at the beautiful lake Beetzsee. It was the perfect place to do … Continue reading "Blockseminar on Dirac operators and scalar curvature"
### Gehaltsverhandlungen als PostDoc
Gehaltsverhandlungen als PostDoc? Das gibt’s doch nur für Professoren … Das dachte ich mir zumindest bis jetzt und ging davon aus, dass man in Deutschland solange auf E13 (bzw. A13) beschäftigt wird bis man es bis zum Professor bzw. Professorin schafft. Dann aber ließ sich die Universität Münster doch überreden mich auf E14 einzustellen. Wie … Continue reading "Gehaltsverhandlungen als PostDoc"
### Seminar on soap bubbles and positive scalar curvature
Together with Rudolf Zeidler I am organizing a reading seminar this winter on generalized soap bubbles and positive scalar curvature. The goal of it is to read the corresponding preprint by Chodosh-Li (arXiv:2008.11888). The two main results we want to understand are the following: No closed aspherical manifold of dimension 4 or 5 admits a … Continue reading "Seminar on soap bubbles and positive scalar curvature"
### Nobel Prize for Sir Roger Penrose
Sir Roger Penrose won the Nobel Prize in Physics for the discovery that black hole formation is a robust prediction of the general theory of relativity. I have to admit that I connected him up to now only with aperiodic tilings of the plane (Wikipedia link). If you want to know more about him you … Continue reading "Nobel Prize for Sir Roger Penrose" | 2020-11-24 04:36:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.40987110137939453, "perplexity": 1976.3372242841851}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141171077.4/warc/CC-MAIN-20201124025131-20201124055131-00601.warc.gz"} |
https://entertaininghacks.wordpress.com/2015/03/25/manufacturing-pcbs-with-surface-mount-components/ | # Manufacturing PCBs With Surface Mount Components
Summary: cheap ‘n cheerful, good enough for double-sided experiments, but not good enough for 4-layer impedance-controlled PCBs.
The laser toner technique and very-low-cost PCB manufacturers are good enough for the low-speed digital signals found in Arduino-class circuits. They are not suitable for large boards, or for containing medium speed digital signals capable of bit-rates up to 1Gb/ or 2Gb/s. A requirement for 50Ω or 100Ω differential impedance lines on >=4 layer PCBs implies the PCB cross-section “stack” must be tightly specified in terms the prepreg’s thickness and $\epsilon _r$.
This post outlines experiences for small experimental boards.
Also see my techniques for designing and assembling homebrew PCBs, and reference material.
## Laser Toner Transfer
I’ve used the traditional laser toner mechanism that’s described in many many blogs, but only for:
• through hole components on single and double-sided boards, with acceptable results
• gash test boards when experimenting with SMD components
This is acceptable for producing Arduino-class circuits with analogue and low-speed digital logic.
The results for SMD components are surprisingly good, and would just about be acceptable in an emergency.
## Hand-Milled
I’ve needed to make some quick-and-dirty RF 50Ω attenuators and 50Ω / 75Ω matching pads for use up to ~2GHz. These circuits consist of two edge-mounted SMA connectors and 1, 2 or 3 SMD resistors mounted on double-sided FR4 board in a “co-planar waverguide with groundplane” configuration. At approximately 15mm*10mm, these boards are small physically and electrically.
The track (singular) needs to be the appropriate width for the impedance. That can be determined with any of the many online calculators; I used Chemandy Electronics’ calculator.
The track can be made by using a hand-held Dremel to “mill out” the gap between the track and ground.
## DirtyPCBs
For small and/or experimental double-sided boards, I’ve used DirtyPCBs principally for cost, but also to have a benchmark against which I can evaluate a “real” PCB manufacturer. The principal features were:
• 10 off 5cm*5cm double-sided is \$14 (i.e £8-9) including sea shipping
• order placed to PCBs received: 21 days
which is remarkably good value. | 2018-05-27 15:50:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4424627721309662, "perplexity": 13075.40284613392}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794869272.81/warc/CC-MAIN-20180527151021-20180527171021-00402.warc.gz"} |
https://www.gradesaver.com/textbooks/math/calculus/university-calculus-early-transcendentals-3rd-edition/chapter-8-section-8-4-integration-of-rational-functions-by-partial-fractions-exercises-page-446/36 | ## University Calculus: Early Transcendentals (3rd Edition)
Published by Pearson
# Chapter 8 - Section 8.4 - Integration of Rational Functions by Partial Fractions - Exercises - Page 446: 36
#### Answer
$$\int\frac{16x^3}{4x^2-4x+1}dx=2x^2+4x+3\ln|2x-1|-\frac{1}{2x-1}+C$$
#### Work Step by Step
$$I=\int\frac{16x^3}{4x^2-4x+1}dx$$ 1) Perform long division: We perform long division to rewrite the fraction as a polynomial plus a proper fraction, which is $$\frac{16x^3}{4x^2-4x+1}=4x+4+\frac{12x-4}{4x^2-4x+1}=4x+4+\frac{12x-4}{(2x-1)^2}$$ 2) Express the remainder fraction as a sum of partial fractions: $$\frac{12x-4}{(2x-1)^2}=\frac{A}{2x-1}+\frac{B}{(2x-1)^2}$$ Clear fractions: $$A(2x-1)+B=12x-4$$ $$2Ax-A+B=12x-4$$ Equating coefficients of corresponding powers of $x$, we get $2A=12$, so $A=6$ $-A+B=-4$, so $B=-4+A=2$ Therefore, $$\frac{12x-4}{(2x-1)^2}=\frac{6}{2x-1}+\frac{2}{(2x-1)^2}$$ 2) Evaluate the integral: $$I=\int(4x+4)dx+\int\frac{6}{2x-1}dx+\int\frac{2}{(2x-1)^2}dx$$ $$I=2x^2+4x+\int\frac{3}{2x-1}d(2x-1)+\int\frac{1}{(2x-1)^2}d(2x-1)$$ $$I=2x^2+4x+3\ln|2x-1|-\frac{1}{2x-1}+C$$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 2019-12-09 04:43:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9436377286911011, "perplexity": 528.9573916222871}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540517557.43/warc/CC-MAIN-20191209041847-20191209065847-00007.warc.gz"} |
https://labs.tib.eu/arxiv/?author=A.W.%20Rodgers | • ### Cepheid Variables in the LMC and SMC(astro-ph/9901282)
Jan. 20, 1999 astro-ph
In this paper, we will review major new results regarding classical Cepheids, in the Large Magellanic Cloud (LMC) and Small Magellanic Cloud (SMC). Specifically, we discuss recent work regarding multimode Cepheids and describe new observations of a W Vir star (HV 5756) and a Cepheid which are each in eclipsing binary systems. An additional interesting pulsating supergiant in an eclipsing system is also identified. Ephemerides for eclipses for the three systems are provided.
• ### The MACHO Project Large Magellanic Cloud Variable Star Inventory. VIII. The Recent Star Formation History of the LMC from the Cepheid Period Distribution(astro-ph/9811240)
Nov. 16, 1998 astro-ph
We present an analysis of the period distribution of $\sim 1800$ Cepheids in the Large Magellanic Cloud, based on data obtained by the MACHO microlensing experiment and on a previous catalogue by Payne-Gaposchkin. Using stellar evolution and pulsation models, we construct theoretical period-frequency distributions that are compared to the observations. These models reveal that a significant burst of star formation has occurred recently in the LMC ($\sim 1.15\times 10^8$ years). We also show that during the last $\sim 10^8$ years, the main center of star formation has been propagating from SE to NW along the bar. We find that the evolutionary masses of Cepheids are still smaller than pulsation masses by $\sim 7$ % and that the red edge of the Cepheid instability strip could be slightly bluer than indicated by theory. There are $\sim 600$ Cepheids with periods below $\sim 2.5$ days cannot be explained by evolution theory. We suggest that they are anomalous Cepheids; a number of these stars are double-mode Cepheids.
• ### The MACHO Project LMC Variable Star Inventory. VI. The Second-overtone Mode of Cepheid Pulsation From First/Second Overtone (FO/SO) Beat Cepheids(astro-ph/9808275)
Aug. 24, 1998 astro-ph
MACHO Project photometry of 45 LMC FO/SO beat Cepheids which pulsate in the first and second overtone (FO and SOo, respectively) has been analysed to determine the lightcurve characteristics for the SO mode of Cepheid pulsation. We predict that singly-periodic SO Cepheids will have nearly sinusoidal lightcurves; that we will only be able to discern SO Cepheids from fundamental (F) and (FO) Cepheids for P <= 1.4 days; and that the SO distribution will overlap the short-period edge of the LMC FO Cepheid period-luminosity relation (when both are plotted as a function of photometric period). We also report the discovery of one SO Cepheid candidate, MACHO*05:03:39.6$-$70:04:32, with a photometric period of 0.775961 +/- 0.000019 days and an instrumental amplitude of 0.047 +/- 0.009 mag in V.
• ### The Blazhko Effect in RR Lyrae stars: Strong observational support for the oblique pulsator model in three stars(astro-ph/9808190)
Aug. 19, 1998 astro-ph
Using the novel data set of the MACHO project, we show that three Blazhko Effect'' RR Lyrae stars show nearly-pure amplitude modulation of a single pulsation mode. This is strong observational evidence that the Oblique Pulsator Model'' is the correct solution to this 90-year-old problem.
• ### First detection of a gravitational microlensing candidate towards the Small Magellanic Cloud(astro-ph/9708190)
Dec. 11, 1997 astro-ph
We report the first discovery of a gravitational microlensing candidate towards a new population of source stars, the Small Magellanic Cloud (SMC). The candidate event's light curve shows no variation for 3 years before an upward excursion lasting $\sim 217$ days that peaks around January 11, 1997 at a magnification of $\sim 2.1$. Microlensing events towards the Large Magellanic Cloud and the Galactic bulge have allowed important conclusions to be reached on the stellar and dark matter content of the Milky Way. The SMC gives a new line-of-sight through the Milky Way, and is expected to prove useful in determining the flattening of the Galactic halo.
• ### Pulsating Variable Stars in the MACHO Bulge database: The Semiregular Variables(astro-ph/9712048)
Dec. 3, 1997 astro-ph
We review the pulsating stars contained in the top 24 fields of the MACHO bulge database, with special emphasis on the red semiregular stars. Based on period, amplitude and color cuts, we have selected a sample of 2000 semiregular variables with $15<P<100$ days. Their color-magnitude diagram is presented, and period-luminosity relation is studied, as well as their spatial distribution. We find that they follow the bar, unlike the RR Lyrae in these fields.
• ### The MACHO Project SMC Variable Star Inventory: I. The Second-overtone Mode of Cepheid Pulsation From First/Second Overtone (1H/2H) Beat Cepheids(astro-ph/9709025)
Sept. 3, 1997 astro-ph
We report the discovery of 20 1H/2H and 7 F/1H beat Cepheids in the SMC by the MACHO Project. We utilize the 20 1H/2H stars to determine lightcurve shape for the SMC second-overtone (2H) mode of Cepheid pulsation. We predict, similar to the findings of Alcock et al. (1997, ApJ, submitted), that 2H Cepheids will have nearly or purely sinusoidal light variations; that the P-L relation for 2H Cepheids will not be distinguishable from the P-L relation for 1H Cepheids within photometric accuracy; and that 2H stars may be discernable from F and 1H stars using the amplitude-period diagram and Fourier parameter progressions for periods P < 0.7 days, our current sample 2H period limit.
• ### The MACHO Project LMC Variable Star Inventory: The Discovery of RV Tauri stars and New Type II Cepheids in the LMC(astro-ph/9708039)
Aug. 5, 1997 astro-ph
We report the discovery of RV Tauri stars in the Large Magellanic Cloud. In light and colour curve behaviour, the RV Tauri stars appear to be a direct extension of the type II Cepheids to longer periods. A single period- luminosity-colour relationship is seen to describe both the type II Cepheids and the RV Tauri stars in the LMC. We derive the relation: $V_{\circ} = 17.89 (\pm 0.20) - 2.95 (\pm 0.12) \log_{10}P + 5.49 (\pm 0.35) \bar{(V-R)_{\circ}}$, valid for type II Cepheids and RV Tauri stars in the period range $0.9 < \log_{10}P < 1.75$. Assuming a distance modulus to the Large Magellanic Cloud of 18.5, the relation in terms of the absolute luminosities becomes: $M_{V} = -0.61 (\pm 0.20) - 2.95 (\pm 0.12) \log_{10}P + 5.49 (\pm 0.35) \bar{(V-R)_{\circ}}$.
• ### The MACHO Project: Microlensing and Variable Stars(astro-ph/9708017)
Aug. 3, 1997 astro-ph
The MACHO Project monitors millions of stars in the Large Magellanic Cloud, the Small Magellanic Cloud and the bulge of the Milky Way searching for the gravitational microlensing signature of baryonic dark matter. This Project has yielded surprising results. An analysis of two years of data monitoring the Large Magellanic Cloud points to {$\sim 50%$} of the mass of the Milky Way's halo in compact objects of {$\sim 0.5 M_{\odot}$}. An analysis of one year of monitoring the bulge has yielded more microlensing than predicted without the invocation of a massive bar or significant disk dark matter. The huge database of light curves created by this search is yielding information on extremely rare types of astrophysical variability as well as providing temporal detail for the study of well known variable astrophysical phenomena. The variable star catalog created from this database is previewed and example light curves are presented.
• ### Bulge delta Scuti Stars in the MACHO Database(astro-ph/9707323)
July 29, 1997 astro-ph
We describe the search for delta Scuti stars in the MACHO database of bulge fields. Concentrating on a sample of high amplitude delta Scutis, we examine the light curves and pulsation modes. We also discuss their spatial distribution and evolutionary status using mean colors and absolute magnitudes.
• ### Is the LMC Microlensing Due to an Intervening Dwarf Galaxy?(astro-ph/9707310)
July 28, 1997 astro-ph
The recent suggestion that the microlensing events observed towards the Large Magellanic Cloud are due to an intervening Sgr-like dwarf galaxy is examined. A search for foreground RR Lyrae in the MACHO photometry database yields 20 stars whose distance distribution follow the expected halo density profile. Cepheid and red giant branch clump stars in the MACHO database are consistent with membership in the LMC. There is also no evidence in the literature for a distinct kinematic population, for intervening gas, or for the turn-off of such a hypothetical galaxy. We conclude that if the lenses are in a foreground galaxy, it must be a particularly dark galaxy.
• ### The RR Lyrae Population of the Galactic Bulge from the MACHO Database(astro-ph/9707311)
July 28, 1997 astro-ph
Mean colors and magnitudes of RR Lyrae stars in 24 fields towards the Galactic bulge from the MACHO database are presented. Accurate mean reddenings are computed for these fields on the basis of the mean colors. The distribution along the line of sight of the RR Lyrae population is examined on the basis of the mean magnitudes, and it is shown that the bulk of the RR Lyrae population is not barred. Only the RR Lyrae in the inner fields closer to the Galactic center (l<4, b>-4) show evidence for a bar. The red giant clump stars in the MACHO fields, however, clearly show a barred distribution, confirming the results of previous studies. Given the different spatial distribution, the RR Lyrae and the clump giants trace two different populations. The RR Lyrae would represent the inner extension of the Galactic halo in these fields.
• ### The Zero Point of Extinction Toward Baade's Window From RR Lyrae Stars(astro-ph/9706292)
June 30, 1997 astro-ph
We measure the zero point of the Stanek (1996) extinction map by comparing the observed (V-K) colors of 20 RR Lyrae stars (type ab) found in the MACHO survey with their intrinsic (V-K)_0 colors as a function of period as determined from nearby RR Lyrae stars. We find that the zero point of the Stanek map should be changed by \Delta A_V = -0.11 +/- 0.05 mag, in excellent agreement with the recent measurement of Gould, Popowski & Terndrup (1997) using K giants.
• ### Gravitational Microlensing Results from MACHO(astro-ph/9611059)
Nov. 7, 1996 astro-ph
The MACHO project is searching for dark matter in the form of massive compact halo objects (Machos), by monitoring the brightness of millions of stars in the Magellanic Clouds to search for gravitational microlensing events. Analysis of our first 2.3 years of data for 8.5 million stars in the LMC yields 8 candidate microlensing events, well in excess of the $\approx 1$ event expected from lensing by known low-mass stars. The event timescales range from 34 to 145 days, and the estimated optical depth is $\sim$ 2 x 10^{-7}, about half of that expected from a standard' halo. Likelihood analysis indicates the typical lens mass is $0.5^{+0.3}_{-0.2} \Msun$, suggesting they may be old white dwarfs.
• ### Cepheids in the LMC: Results from the MACHO Project(astro-ph/9610024)
Oct. 23, 1996 astro-ph
The approximately 1500 Cepheid variables in 22 LMC fields of the MACHO Project survey have been calibrated and analysed. In this paper, we report improved period ratios for a total of 73 beat Cepheids and provide a first look at the Fourier decomposition parameters for both singly- and doubly-periodic Cepheids. We also note an unusual amplitude-changing pulsator.
• ### RR Lyrae Stars in the MACHO Database(astro-ph/9610025)
Oct. 3, 1996 astro-ph
The MACHO Project has catalogued $\sim 8000$ RR Lyrae stars in the Large Magellanic Cloud, $\sim 1800$ in the Galactic bulge, and $\sim$ 50 in the Sgr dwarf galaxy. These variables are excellent distance indicators, and are used as tools to study the structure of the Large Magellanic Cloud and the bulge. The large datasets also probe uncommon pulsation modes. A number of double-mode RR Lyrae stars (RRd) are found in the Large Magellanic Cloud sample. These stars provide important clues for understanding the formation and evolution of the inner Galaxy, the Large Magellanic Cloud and the Sgr dwarf galaxy. A large number of second overtone pulsators (RRe) are found in the LMC and bulge. Finally, the RR Lyrae belonging to the Sgr dwarf yield an accurate distance to this galaxy. Their presence also alerts us of the very interesting possibility of distant sources for bulge microlensing events.
• ### The MACHO Project LMC Variable Star Inventory IV: Multimode RR Lyrae Stars, Distance to the LMC and Age of the Oldest Stars(astro-ph/9608036)
Aug. 12, 1996 astro-ph
We report the discovery of 73 double-mode RR Lyrae (RRd) stars in fields near the bar of the LMC. The stars are detected among the MACHO database of short-period variables that currently contains about 7900 RR Lyrae stars. Fundamental periods (P_0) for these stars are found in the range 0.46-0.55 days and first overtone-to-fundamental period ratios are found to be in the range 0.742 < P_1/P_0 < 0.748. A significant fraction of our current sample have period ratios smaller than any previously discovered RRd variables. We present mean magnitudes, colors, and lightcurve properties for all LMC RRd stars detected to date. The range in period ratios is unexpectedly large. We present a determination of absolute magnitudes for these stars based primarily on pulsation theory and the assumption that all observed stars are at the fundamental blue edge (FBE) of the instability strip. Comparison of the calibrated MACHO V and R_KC photometry with these derived absolute magnitudes yields an absorption-corrected distance modulus to the LMC of 18.57 +/- 0.19 mag which is in good agreement with that found (18.5) through comparison of galactic and LMC Cepheids. Adopting this luminosity calibration, we derive an increase in the distance modulus, and thus a reduction in the age found via isochrone fitting for M15 of about 33% and discuss the implications for cosmology.
• ### The MACHO Project LMC Microlensing Results from the First Two Years and the Nature of the Galactic Dark Halo(astro-ph/9606165)
June 26, 1996 astro-ph, hep-ph
The MACHO Project is a search for dark matter in the form of massive compact halo objects (Machos). Photometric monitoring of millions of stars in the Large Magellanic Cloud (LMC), Small Magellanic Cloud (SMC), and Galactic bulge is used to search for gravitational microlensing events caused by these otherwise invisible objects. Analysis of the first 2.1 years of photometry of 8.5 million stars in the LMC reveals 8 candidate microlensing events. This is substantially more than the number expected ($\sim 1.1$) from lensing by known stellar populations. The timescales ($\that$) of the events range from 34 to 145 days. We estimate the total microlensing optical depth towards the LMC from events with $2 < \that < 200$ days to be $\tau_2^{200} = 2.9 ^{+1.4}_{-0.9} \ten{-7}$ based upon our 8 event sample. This exceeds the optical depth, $\tau_{\rm backgnd} = 0.5 \ten{-7}$, expected from known stars, and the difference is to be compared with the optical depth predicted for a standard" halo composed entirely of Machos: $\tau_{halo} = 4.7\ten{-7}$. Likelihood analysis gives a fairly model independent estimate of the halo mass in Machos within 50 kpc of $2.0^{+1.2}_{-0.7} \ten{11} \msun$, about half of the standard halo" value. We also find a most probable Macho mass of $0.5^{+0.3}_{-0.2}\msun$, although this value is strongly model dependent. Additionally, the absence of short duration events places stringent upper limits on the contribution of low-mass Machos: objects from $10^{-4} \msun$ to $0.03 \msun$ contribute $\simlt 20\%$ of the standard" dark halo.
• ### The MACHO Project 2nd Year LMC Microlensing Results and Dark Matter Implications(astro-ph/9606134)
June 21, 1996 astro-ph
The MACHO Project is searching for galactic dark matter in the form of massive compact halo objects (Machos). Millions of stars in the Large Magellanic Cloud (LMC), Small Magellanic Cloud (SMC), and Galactic bulge are photometrically monitored in an attempt to detect rare gravitational microlensing events caused by otherwise invisible Machos. Analysis of two years of photometry on 8.5 million stars in the LMC reveals 8 candidate microlensing events, far more than the $\sim1$ event expected from lensing by low-mass stars in known galactic populations. From these eight events we estimate the optical depth towards the LMC from events with $2 < \that < 200$ days to be $\tau_2^{200} \approx 2.9 ^{+1.4}_{-0.9} \ten{-7}$. This exceeds the optical depth of $0.5\ten{-7}$ expected from known stars and is to be compared with an optical depth of $4.7\ten{-7}$ predicted for a standard'' halo composed entirely of Machos. The total mass in this lensing population is $\approx 2^{+1.2}_{-0.7} \ten{11} \msun$ (within 50 kpc from the Galactic center). Event timescales yield a most probable Macho mass of $0.5^{+0.3}_{-0.2}\msun$, although this value is quite model dependent.
• ### MACHO Project Photometry of RR Lyrae Stars in the Sgr Dwarf Galaxy(astro-ph/9605148)
May 23, 1996 astro-ph
We report the discovery of 30 type a,b RR Lyrae (RRab) which are likely members of the Sagittarius (Sgr) dwarf galaxy. Accurate positions, periods, amplitudes and magnitudes are presented. Their distances are determined with respect to RRab in the Galactic bulge found also in the MACHO 1993 data. For R$_{\odot} = 8$ kpc, the mean distance to these stars is $D = 22 \pm 1$ kpc, smaller than previous determinations for this galaxy. This indicates that Sgr has an elongated main body extending for more than 10 kpc, which is inclined along the line of sight, with its northern part (in Galactic coordinates) closer to us. The size and shape of Sgr give clues about the past history of this galaxy. If the shape of Sgr follows the direction of its orbit, the observed spatial orientation suggests that Sgr is moving away from the Galactic plane. Also, Sgr stars may be the sources of some of the microlensing events seen towards the bulge.
• ### The MACHO Project LMC Variable Star Inventory: III. New R Coronae Borealis Stars(astro-ph/9604177)
May 7, 1996 astro-ph
We report the discovery of two new R Coronae Borealis (RCB) stars in the Large Magellanic Cloud (LMC) using the MACHO project photometry database. The identification of both stars has been confirmed spectroscopically. One is a cool RCB star (T_eff about 5000 K) characterized by very strong Swan bands of C_2 and violet bands of CN, and weak or absent Balmer lines, G-band and 12C-13C bands. The second star is an example of a hot RCB star of which only 3 were previously known to exist in the Galaxy and none in the LMC. Its spectrum is characterized by several C II lines in emission. Both stars have shown deep declines of Delta V > 4 mag in brightness. The new stars are significantly fainter at maximum light than the three previously known LMC RCB stars. The amount of reddening toward these stars is somewhat uncertain but both seem to have absolute magnitudes, M_V, about half a magnitude fainter than the other three stars. Estimates of M_Bol find that the hot RCB star lies in the range of the other three stars while the cool RCB star is fainter. The two cool LMC RCB stars are the faintest at M_Bol. The discovery of these two new stars brings to five the number of known RCB stars in the LMC and demonstrates the utility of the MACHO photometric database for the discovery of new RCB stars.
• ### The MACHO Project: Limits on Planetary Mass Dark Matter in the Galactic Halo from Gravitational Microlensing(astro-ph/9604176)
April 29, 1996 astro-ph, hep-ph
The MACHO project has been monitoring about ten million stars in the Large Magellanic Cloud in the search for gravitational microlensing events caused by massive compact halo objects (Machos) in the halo of the Milky Way. In our standard analysis, we have searched this data set for well sampled, long duration microlensing lightcurves, detected several microlensing events consistent with Machos in the 0.1 < m < 1.0 M_sun mass range, and set limits on the abundance of objects with masses 1e-5 < m < 0.1 M_sun. In this paper, we present a different type of analysis involving the search for very short time scale brightenings of stars which is used to set strong limits on the abundance of lower mass Machos. Our analysis of the first two years of data toward the LMC indicates that Machos with masses in the range 2.5e-7 < m < 5.2e-4 M_sun cannot make up the entire mass of a standard spherical dark halo. Combining these results with those from the standard analysis, we find that the halo dark matter may not be comprised of objects with masses 2.5e-7 < m < 8.1e-2 M_sun.
• ### The MACHO Project: 45 Candidate Microlensing Events from the First Year Galactic Bulge Data(astro-ph/9512146)
Dec. 21, 1995 astro-ph
We report the detection of 45 candidate microlensing events in fields toward the Galactic bulge. These come from the analysis of 24 fields containing 12.6 million stars observed for 190 days in 1993. Many of these events are of extremely high signal to noise and are remarkable examples of gravitational microlensing. The distribution of peak magnifications is shown to be consistent with the microlensing interpretation of these events. Using a sub-sample of 1.3 million Clump Giant" stars whose distance and detection efficiency are well known, we find 13 events and estimate the microlensing optical depth toward the Galactic Bulge as $\tau_{\rm bulge} = 3.9 {+ 1.8 \atop - 1.2} \times 10^{-6}$ averaged over an area of $\sim 12$ square degrees centered at Galactic coordinates $\ell = 2.55^\circ$ and $b = -3.64^\circ$. This is similar to the value reported by the OGLE collaboration, and is marginally higher than current theoretical models for $\tau_{\rm bulge}$. The optical depth is also seen to increase significantly for decreasing $\vert b\vert$. These results demonstrate that obtaining large numbers of microlensing events toward the Galactic bulge is feasible, and that the study of such events will have important consequences for the structure of the Galaxy and its dark halo.
• ### The MACHO Project Dark Matter Search(astro-ph/9510104)
Oct. 20, 1995 astro-ph
We provide a status report on our search for dark matter in our Galaxy in the form of massive compact halo objects (or Machos), using gravitational microlensing of background stars. This search uses a very large format CCD camera on the dedicated 1.27m telescope at Mt.~Stromlo, Australia, and has been taking data for almost 3 years. At present, we are in the midst of analyzing our second year data for 8 million stars in the Large Magellanic Cloud. We find more microlensing events than expected from known stellar populations suggesting that Machos are indeed present in the Galactic halo, but the observed microlensing rate toward the LMC is too small to allow for a halo dominated by sub-stellar Machos. Our observations of the Galactic bulge have also yielded substantially more microlensing events than anticipated including a number of exotic `deviant" microlensing events. The implications of these results are discussed.
• ### Real-time Detection of Gravitational Microlensing(astro-ph/9508039)
Aug. 9, 1995 astro-ph
Real-time detection of microlensing has moved from proof of concept in 1994 to a steady stream of events this year. Global dissemination of these events by the MACHO and OGLE collaborations has made possible intensive photometric and spectroscopic followup from widely dispersed sites confirming the microlensing hypothesis. Improved photometry and increased temporal resolution from followup observations greatly increases the possibility of detecting deviations from the standard point-source, point-lens, inertial motion microlensing model. These deviations are crucial in understanding individual lensing systems by breaking the degeneracy between lens mass, position and velocity. We report here on GMAN (Global Microlensing Alert Network), the coordinated followup of MACHO alerts. | 2021-03-05 07:14:05 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.602609395980835, "perplexity": 2074.806489221425}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178370239.72/warc/CC-MAIN-20210305060756-20210305090756-00027.warc.gz"} |
https://brilliant.org/problems/polynomial-arithmetic-applying-the-perfect-square/ | # Polynomial Arithmetic: Applying the Perfect Square Identity
Algebra Level 1
What is the coefficient of $x$ in the expansion of $(x+3 ) ^ 2$?
× | 2020-11-30 05:31:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5380551815032959, "perplexity": 1346.553561209239}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141205147.57/warc/CC-MAIN-20201130035203-20201130065203-00432.warc.gz"} |
https://homework.cpm.org/category/CON_FOUND/textbook/a2c/chapter/9/lesson/9.2.2/problem/9-97 | ### Home > A2C > Chapter 9 > Lesson 9.2.2 > Problem9-97
9-97.
1. Given find: Homework Help ✎
1. CD
2. C + D
3. C2
4. X if 2X + C = D
Multiply each row of the first matrix into each column of the second matrix.
Add the numbers in the corresponding entries.
C 2 = C · C
2X + C = D
2X = D − C
$\textit{X}= \frac{(\textit{D}-\textit{C})}{2}$ | 2019-10-19 00:02:51 | {"extraction_info": {"found_math": true, "script_math_tex": 1, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.740213930606842, "perplexity": 5226.6864616834855}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986685915.43/warc/CC-MAIN-20191018231153-20191019014653-00333.warc.gz"} |
https://share.cocalc.com/share/5d54f9d642cd3ef1affd88397ab0db616c17e5e0/www/Tables/artin.tex?viewer=share | Sharedwww / Tables / artin.texOpen in CoCalc
% artin8.tex
% by Kevin Buzzard and William Stein
\documentclass{article}
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\title{A mod five approach to modularity of icosahedral
Galois representations}
\author{Kevin Buzzard and William A. Stein}
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\begin{document}
\maketitle
\begin{abstract}
We give eight new examples of icosahedral Galois representations that
satisfy Artin's conjecture on holomorphicity of their $L$-function.
We give in detail one example of an icosahedral representation of
conductor ${\bf 1376}=2^5\cdot 43$ that satisfies Artin's conjecture.
We also briefly explain the computations behind seven additional
examples of conductors ${\bf 2416}=2^4\cdot 151$, ${\bf 3184}=2^4\cdot 199$, ${\bf 3556}=2^2\cdot 7\cdot 127$, ${\bf 3756}=2^2\cdot 3\cdot 313$, ${\bf 4108}=2^2\cdot 13\cdot 79$, ${\bf 4288}=2^6\cdot 67$, and
${\bf 5373}=3^3\cdot 199$.
\end{abstract}
\section*{Introduction}
Consider a continuous irreducible Galois representation
$$\rho:\galq\ra\GL_n(\C)$$
with $n > 1$.
Inspired by his reciprocity law,
Artin conjectured in~\cite{artin:conjecture} that
$L(\rho,s)$ has an analytic continuation to the whole complex plane.
Many of the known cases of this conjecture were obtained by
proving the apparently stronger assertion that~$\rho$ is \defn{automorphic},
in the sense that the $L$-function of~$\rho$ is equal to the $L$-function
of a certain automorphic representation (whose $L$-function is known to have
analytic continuation). In the special case where $n=2$ and $\rho$ is in
addition assumed to be odd, the automorphic representation in question
should be the one associated to a classical weight~$1$
modular eigenform, and in fact there is conjectured to be a
bijection between such~$\rho$ and the set of all weight~$1$
cuspidal newforms, which should
preserve $L$-functions. It is this bijection
that we are concerned with in this paper, so assume for the rest
of the paper that $n=2$ and~$\rho$ is odd.
In this special case, the construction
of~\cite{deligne-serre} shows how to construct a continuous irreducible
odd 2-dimensional representation from a weight~$1$ newform, and the problem
is to go the other way. Say that a representation is \defn{modular}
if it arises in this way.
If the image of~$\rho$ is solvable,
then~$\rho$ is known to be modular
\cite{langlands:basechange, tunnell:artin};
if the image is not solvable, then $\im(\rho)$ in $\PGL_2(\C)$
is isomorphic to the
alternating group~$A_5$, and the modularity of~$\rho$
is, in general, unknown. We call such a 2-dimensional representation an
icosahedral representation''.
The published literature contains only eight examples (up to twist)
of odd icosahedral Galois representations that are known to satisfy Artin's
conjecture: one of conductor $800=2^5\cdot 5^2$
(see \cite{buhler:thesis}), and seven of conductors:
$2083,\, 2^2\cdot 487,\, 2^2\cdot 751,\, 2^2 \cdot 887,\, 2^2\cdot 919,\, 2^5\cdot 73,\,\text{ and } 2^5\cdot 193$
(see \cite{freyetal}).
After the first draft of this paper was written, the
preprint~\cite{bdsbt} appeared, which contains a general theorem that
yields infinitely many (up to twist) modular icosahedral representations.
However, we feel that our work, although much less powerful, is still
of some worth, because it gives an effective computational approach to
proving that certain mod~5 representations are modular, without
computing any spaces of weight~1 forms or using effective versions of
the Chebotar\"ev density theorem. We also note that the
main theorem of~\cite{bdsbt} does not apply to any of the examples
considered in the present paper. Very recently,
the preprint~\cite{taylor:artin2}
appeared, which gives local conditions under which an icosahedral
representation is modular. In particular, \cite{taylor:artin2} also
proves that the first three
examples in the present paper, of conductors 1376, 2416, 3184,
are modular; these correspond to the first, third, and fourth
equations at the end of~\cite{taylor:artin2}.
However, \cite{taylor:artin2} does not apply to
our remaining five examples.
In this paper we give eight new examples of modular icosahedral
representations that were computed
by applying the main theorem of~\cite{buzzard-taylor} to
the mod~$5$ reduction of~$\rho$.
We verify modularity mod~$5$ on a case-by-case basis. Later we shall
explain our approach more carefully, but let us briefly summarise it here.
By~\cite{buzzard-taylor},
the problem is to show that the mod~5 reduction of~$\rho$ is modular.
We do this by finding a candidate mod~5 modular form at weight~5
and then, using the table of icosahedral extensions of $\Q$ in~\cite{freyetal}
and what we know about the 5-adic representation attached to our candidate
form, we deduce that the mod~5 representation attached to our candidate
form must be the reduction of~$\rho$. In particular, this paper gives
a computational methods for checking the modularity of certain mod~5
representations whose conductors are not too large. We now give
more details.
In each of our examples it is easy to compute a few Hecke operators
and be morally convinced that a mod~$5$ representation should be modular;
it is far more difficult to prove this.
Effective variants of the Chebotarev density theorem require
that we check vastly more traces of Frobenius than is practical.
Instead we use the Local Langlands theorem for $\GL_2$, the
theory of companion forms, and Table~2 of~\cite{freyetal},
to provide proofs of modularity
in certain cases.
More precisely, let~$K$ be an icosahedral extension of~$\Q$ that is not
totally real, and consider a minimal lift $\rho:\GQ\ra \GL_2(\C)$
of
$$\GQ\ra \Gal(K/\Q)\ncisom{}A_5\subset \PGL_2(\C);$$
the lift is minimal in the sense that its conductor is minimal.
Assume that~$5$ does not ramify in~$K$, and that
a Frobenius element at~$5$ in $\Gal(K/\Q)$ does not have order~$1$ or~$5$.
Inspired by the possibility that~$\rho$ is modular,
we search for a mod~$5$ modular form of weight~$5$ whose existence would
be forced by modularity of~$\rho$. Indeed, we find
a candidate mod~$5$ form~$f$, and then prove that the fixed field
of the kernel of the projective mod~$5$ representation
associated to a certain twist of~$f$ must be~$K$.
This proves that the mod~$5$ reduction of a twist
of~$\rho$ is modular, and the main theorem
of \cite{buzzard-taylor} then implies
that~$\rho$ is modular.
We carried out this program for icosahedral representations
of the following conductors:
${\bf 1376} = 2^5\cdot 43$,
${\bf 2416}=2^4\cdot 151$,
${\bf 3184}=2^4\cdot 199$,
${\bf 3556}=2^2\cdot 7\cdot 127$,
${\bf 3756}=2^2\cdot 3\cdot 313$,
${\bf 4108}=2^2\cdot 13\cdot 79$,
${\bf 4288}=2^6\cdot 67$, and
${\bf 5373}=3^3\cdot 199$.
We choose an icosahedral field~$K$ and representation~$\rho$,
then proceed as follows:
\vspace{.5ex}
\begin{numlist}
\item Search for a form~$f \in S_5(N,\eps;\Fbar_5)$ whose
associated mod~$5$ Galois representation looks like
it is the mod~$5$ reduction of~$\rho$.
\item Twist~$f$ to obtain an eigenform~$g$ with coefficients in~$\F_5$.
\item Prove that~$\rho_g$ is unramified at~$5$ by finding a companion form.
\item Prove that the image of $\proj\rho_g$ is~$A_5$ by ruling out all
other possibilities.
\item Prove that the fixed field~$L$ of $\proj\rho_g$ has
root field of discriminant at most $2083^2$,
so~$L$ is in Table~2 of~\cite{freyetal}; deduce that~$L=K$.
\item Apply the main theorem of~\cite{buzzard-taylor}
to a lift of $\rhobar=\rho_g$
to conclude that~$\rho$ is modular.
\end{numlist}
\section{Modularity of an icosahedral representation of
conductor~$1376=2^5\cdot 43$}\label{sec:1376}
In this section we prove the following theorem.
\begin{theorem}\label{thm:1376}
The icosahedral representations whose corresponding
icosahedral extension
is the splitting field of $x^5 + 2x^4+6x^3+8x^2+10x+8$
are modular.
\end{theorem}
Let~$K$ be the splitting field of $h=x^5 + 2x^4+6x^3+8x^2+10x+8$.
The Galois group of~$K$ is~$A_5$, so we obtain a homomorphism
$G_\Q\ra{}A_5\subset \PGL_2(\C)$;
let $\rho:G_\Q\ra\GL_2(\C)$ be a minimal lift, minimal
in the sense that the Artin conductor of~$\rho$ is minimal.
By Table~$A_5$ of~\cite{buhler:thesis}, the conductor of~$\rho$
is $N=1376=2^5\cdot 43$. Since
$h\con (x-1)(x^2-x+1)(x^2-x+2)\pmod{5}$,
and ${\rm disc}(h)$ is coprime to~$5$,
any Frobenius element at~$5$ in $\Gal(K/\Q)$ has order~$2$.
We use the notation of Tables 3.1 and 3.2 of~\cite[pg. 46]{buhler:thesis};
from Table 3.2 we see that the type of~$\rho$ at~$2$
is~$17$ and the type at~$43$ is~$2$.
The mod~$N$ Dirichlet character~$\eps=\det(\rho)$
factors as~$\eps=\eps_2\cdot \eps_{43}$ where~$\eps_2$ is
a character mod~$2^5$ and~$\eps_{43}$ is a character mod~$43$.
Corresponding to each type in Buhler's table, there is a character,
and fortunately Buhler's level $800$ example also was of type~$17$ at~$2$
(see the first line of~\cite[Table~3.2]{buhler:thesis}).
By~\cite[pg.~80]{buhler:thesis} $\eps_2$ is the unique
character of conductor~$4$ and order~$2$.
A local computation shows that the image
of~$\eps_{43}$ has order~$3$.
If~$\rho$ is modular, then there is a weight~$1$
newform $f_?\in S_1(N,\eps;\Qbar)$ that gives rise to~$\rho$.
Suppose for the moment that~$\rho$ is modular, so that~$f_?$ exists.
Choose a prime of~$\overline{\Z}$ lying
over~$5$, and denote by~$\fbar_?$ the reduction
of $f_?$ modulo this prime. The Eisenstein series
$E_4\in M_4(1;\F_5)$ is congruent to~$1$ modulo~$5$, so
$E_4\cdot{}\fbar_?\in S_5(N,\eps;\Fbar_5)$ has the same $q$-expansion
as $\fbar_?$. Using a computer, we can search for a
form $f\in S_5(N,\eps,\Fbar_5)$ that has the same
$q$-expansion as the conjectural form $E_4\cdot{}\fbar_?$.
Instead of multiplying $\fbar_?$ by~$E_4$, we could have multiplied
it by an Eisenstein series of weight~$1$, level~$5$, and character $\eps'$.
We used $E_4$ because the dimension of $S_5(N,\eps;\Fbar_5)$
is~$696$ whereas the dimension of the relevant space
$S_2(5\cdot 1376, \eps_{43})$ of weight~$2$ cusp forms is~$1040$.
\subsection{Searching for the newform~$f$}
Using modular symbols (see Section~\ref{sec:modsym}) we
compute (at least up to semi-simplification) the space
$S_5(1376,\eps;\F_{25})$. Note that there is injective map
from the image of~$\eps$ into $\F_{25}^*$. By computing
the kernels of various Hecke operators on this space, we find~$f$.
In the following computations, we represent nonzero elements of~$\F_{25}$
as powers of a generator~$\alp$ of~$\F_{25}^*$, which satisfies
$$\alp^2 + 4\alp + 2=0.$$
Our character $\eps_{43}$ was represented
as the map sending $(1,3)\in(\Z/2^5\Z)^*\cross(\Z/43\Z)^*$ to
$2\alp+1$. Note that~3 is a primitive root mod~43, and that $2\alp+1$
has order~3.
If the least common multiple of the degrees of the factors of
the polynomial~$h$ modulo an
unramified prime~$p$ is~$2$, then $\Frob_p\in\Gal(K/\Q)$
has order~$2$. The minimal polynomial of $\rho(\Frob_p)\in\GL_2(\C)$
is then $x^2-1$, so $\rho(\Frob_p)$ has trace~$0$.
The first three primes $p \nmid 5\cdot 1376$ such
that $\rho(\Frob_p)$ has order~$2$ are $p= 19,31,97$.
We computed the mod~$5$ reduction $\sS_5(1376,\eps;\F_{25})^{+}$
of the $\Z_5[\zeta_3]$-lattice of
modular symbols of level~$1376$ and
character~$\tilde{\eps}$ where
complex conjugation acts as $+1$.
Here~$\tilde{\eps}$ denotes the Teichm\"uller lift of~$\eps$.
Let~$V$ be the intersection of the kernels of $T_{19}$, $T_{31}$, and
$T_{97}$ inside of the space $\sS_5(1376,\eps;\F_{25})^{+}$ of mod~5
modular symbols.
The space~$V$ is $8$-dimensional,
and no doubt all the eigenforms in this space give rise to~$\rho$ or one
of its twists. One of the eigenvalues of~$T_3$ on this space
is~$\alp^{16}$, and the kernel $V_1$ of $T_3-\alp^{16}$ is $2$-dimensional
over $\F_{25}$. The Hecke operator~$T_5$ acted as a diagonalisable matrix on
$V_1$, with eigenvalues $\alp^{10}$ and $\alp^{22}$, so the corresponding
two systems of eigenvalues must correspond to mod~$5$ modular eigenforms,
and furthermore we must have found all mod~$5$ modular eigenforms
of this level, weight and character,
such that $a_{19}=a_{31}=0$ and $a_3=\alp^{16}$.
\begin{remark}
The careful reader might wonder how we know that the
systems of mod~$5$ eigenvalues really do correspond to mod~$5$ modular
forms, and not to perhaps some strange mod~$5$ torsion in the space of
modular symbols. However, we eliminated this possibility by
computing the dimension of the full space of mod~$5$ modular
symbols where complex conjugation acts as~$+1$, and checking that it
equals $696$, the dimension of $S_5(1376,\tilde{\eps},\C)$, which we
computed using the formula in \cite{cohen-oesterle}.
\end{remark}
Let~$f$ be the eigenform in~$V_1$ that satisfies
$a_5=\alp^{22}$; the $q$-expansion of~$f$ begins
$$f=q + \alp^{16}q^3 + \alp^{22}q^5 + \alp^{14}q^7 + \alp^{14}q^9 + 4q^{11}+\cdots.$$
Further eigenvalues are given in Table~\ref{table:1376}.
The primes~$p$ in the table such that~$a_p=0$ are
exactly those
predicted by considering the splitting behavior of~$h$.
This is strong evidence that~$\rho$ is modular,
and also that our modular symbols algorithm have been correctly
implemented.
\begin{table}
\caption{\label{table:1376}Eigenvalues of~$f$}
\begin{center}
$$\begin{array}{|rl|}\hline 2&0\\ 3&\alpha^{16}\\ 5&\alpha^{22}\\ 7&\alpha^{14}\\ 11&4\\ 13&\alpha^{14}\\ 17&\alpha^{14}\\ 19&0\\ 23&\alpha^{16}\\ 29&\alpha^{8}\\ 31&0\\ 37&\alpha^{10}\\ 41&1\\ 43&\alpha^{10}\\ 47&1\\ 53&\alpha^{22}\\ \hline\end{array} \begin{array}{|rl|}\hline 59&4\\ 61&\alpha^{14}\\ 67&\alpha^{4}\\ 71&\alpha^{20}\\ 73&\alpha^{2}\\ 79&\alpha^{20}\\ 83&\alpha^{4}\\ 89&\alpha^{10}\\ 97&0\\ 101&\alpha^{8}\\ 103&\alpha^{14}\\ 107&0\\ 109&\alpha^{10}\\ 113&2\\ 127&0\\ 131&2\\ \hline\end{array} \begin{array}{|rl|}\hline 137&0\\ 139&\alpha^{22}\\ 149&\alpha^{4}\\ 151&1\\ 157&\alpha^{14}\\ 163&0\\ 167&\alpha^{22}\\ 173&4\\ 179&\alpha^{2}\\ 181&\alpha^{14}\\ 191&\alpha^{10}\\ 193&4\\ 197&0\\ 199&3\\ 211&0\\ 223&0\\ \hline\end{array} \begin{array}{|rl|}\hline 227&\alpha^{10}\\ 229&0\\ 233&\alpha^{14}\\ 239&0\\ 241&\alpha^{2}\\ 251&\alpha^{2}\\ 257&3\\ 263&\alpha^{16}\\ 269&2\\ 271&\alpha^{8}\\ 277&0\\ 281&\alpha^{16}\\ 283&0\\ 293&3\\ 307&\alpha^{4}\\ 311&\alpha^{22}\\ \hline\end{array} \begin{array}{|rl|}\hline 313&0\\ 317&0\\ 331&\alpha^{14}\\ 337&0\\ 347&\alpha^{16}\\ 349&\alpha^{4}\\ 353&0\\ 359&0\\ 367&\alpha^{22}\\ 373&0\\ 379&3\\ 383&3\\ 389&1\\ 397&\alpha^{16}\\ 401&0\\ 409&2\\ \hline\end{array} \begin{array}{|rl|}\hline 419&3\\ 421&\alpha^{20}\\ 431&4\\ 433&\alpha^{4}\\ 439&\alpha^{20}\\ 443&0\\ 449&0\\ 457&0\\ 461&0\\ 463&\alpha^{10}\\ 467&0\\ 479&0\\ 487&\alpha^{8}\\ 491&\alpha^{2}\\ 499&\alpha^{20}\\ 503&\alpha^{2}\\ \hline\end{array} \begin{array}{|rl|}\hline 509&\alpha^{8}\\ 521&\alpha^{10}\\ 523&\alpha^{14}\\ 541&\alpha^{20}\\ 547&\alpha^{22}\\ 557&3\\ 563&1\\ 569&\alpha^{16}\\ 571&\alpha^{22}\\ 577&\alpha^{14}\\ 587&\alpha^{20}\\ 593&0\\ 599&\alpha^{22}\\ 601&0\\ 607&\alpha^{16}\\ 613&2\\ \hline\end{array} \comment{\begin{array}{|rl|}\hline 617&0\\ 619&\alpha^{20}\\ 631&\alpha^{20}\\ 641&4\\ 643&1\\ 647&4\\ 653&1\\ 659&\alpha^{14}\\ 661&2\\ 673&\alpha^{8}\\ 677&4\\ 683&0\\ 691&\alpha^{16}\\ 701&\alpha^{14}\\ 709&4\\ 719&\alpha^{4}\\ \hline\end{array} }$$
\end{center}
\end{table}
\subsection{Twisting into $\GL(2,\F_5)$}
Although there is a representation
$\rho_f:\GQ\ra\GL(2,\F_{25})$ attached to $f$,
it is difficult to say anything about its image without further
work. We use a trick to show that the image of $\rho_f$ is small.
Firstly, for a character~$\chi:\GQ\to\Fbar_5$, let~$\tilde\chi$
denote its Teichm\"uller lift to~$\Qbar_5$. By a result of Carayol,
there is a characteristic 0 eigenform
$\tilde{f}\in S_5(N,\tilde{\eps};\Qbar_5)$ lifting $f$.
The twist $\tilde{g}=\tilde{f} \tensor \tilde{\eps}_{43}$ is, by
\cite[Prop. 3.64]{shimura:intro}, an eigenform in
$S_5(43N, \tilde{\eps}_2; \Qbar_5)$, and its reduction is
a form $g\in S_5(43N,\eps_2,\F_{25})$.
The eigenvalues $a_p(g) = a_p(f) \eps_{43}(p)$, for the
first few
$p\nmid 5N$, are given in Table~\ref{table:1376twist}.
\begin{table}
\caption{\label{table:1376twist}Eigenvalues of~$g=f\tensor\eps_{43}$}
\begin{center}
$$\begin{array}{|rl|}\hline 2&*\\%0\\ 3&1\\ 5&*\\%3\\ 7&2\\ 11&4\\ 13&2\\ 17&2\\ 19&0\\ 23&1\\ 29&1\\ 31&0\\ 37&3\\ 41&1\\ 43&*\\%0\\ 47&1\\ 53&2\\ \hline\end{array} \begin{array}{|rl|}\hline 59&4\\ 61&2\\ 67&4\\ 71&4\\ 73&3\\ 79&4\\ 83&4\\ 89&3\\ 97&0\\ 101&1\\ 103&2\\ 107&0\\ 109&3\\ 113&2\\ 127&0\\ 131&2\\ \hline\end{array} \begin{array}{|rl|}\hline 137&0\\ 139&2\\ 149&4\\ 151&1\\ 157&2\\ 163&0\\ 167&2\\ 173&4\\ 179&3\\ 181&2\\ 191&3\\ 193&4\\ 197&0\\ 199&3\\ 211&0\\ 223&0\\ \hline\end{array} \begin{array}{|rl|}\hline 227&3\\ 229&0\\ 233&2\\ 239&0\\ 241&3\\ 251&3\\ 257&3\\ 263&1\\ 269&2\\ 271&1\\ 277&0\\ 281&1\\ 283&0\\ 293&3\\ 307&4\\ 311&2\\ \hline\end{array} \begin{array}{|rl|}\hline 313&0\\ 317&0\\ 331&2\\ 337&0\\ 347&1\\ 349&4\\ 353&0\\ 359&0\\ 367&2\\ 373&0\\ 379&3\\ 383&3\\ 389&1\\ 397&1\\ 401&0\\ 409&2\\ \hline\end{array} \begin{array}{|rl|}\hline 419&3\\ 421&4\\ 431&4\\ 433&4\\ 439&4\\ 443&0\\ 449&0\\ 457&0\\ 461&0\\ 463&3\\ 467&0\\ 479&0\\ 487&1\\ 491&3\\ 499&4\\ 503&3\\ \hline\end{array} \begin{array}{|rl|}\hline 509&1\\ 521&3\\ 523&2\\ 541&4\\ 547&2\\ 557&3\\ 563&1\\ 569&1\\ 571&2\\ 577&2\\ 587&4\\ 593&0\\ 599&2\\ 601&0\\ 607&1\\ 613&2\\ \hline\end{array} \begin{array}{|rl|}\hline 617&0\\ 619&4\\ 631&4\\ 641&4\\ 643&1\\ 647&4\\ 653&1\\ 659&2\\ 661&2\\ 673&1\\ 677&4\\ 683&0\\ 691&1\\ 701&2\\ 709&4\\ 719&4\\ \hline\end{array} \comment{ \begin{array}{|rl|}\hline 727&4\\ 733&0\\ 739&2\\ 743&2\\ 751&4\\ 757&3\\ 761&3\\ 769&0\\ 773&0\\ 787&4\\ 797&1\\ 809&3\\ 811&1\\ 821&2\\ 823&3\\ 827&3\\ \hline\end{array} \begin{array}{|rl|}\hline 829&2\\ 839&0\\ 853&2\\ 857&0\\ 859&0\\ 863&4\\ 877&1\\ 881&1\\ 883&0\\ 887&2\\ 907&0\\ 911&1\\ 919&1\\ 929&0\\ 937&3\\ 941&4\\ \hline\end{array} \begin{array}{|rl|}\hline 947&2\\ 953&1\\ 967&4\\ 971&3\\ 977&0\\ 983&0\\ 991&3\\ 997&3\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ \hline\end{array}}$$
\end{center}
\end{table}
\begin{proposition}\label{prop:1376-g}
Let $g=f\tensor \eps_{43}$. Then $a_p(g)\in \F_5$
for all~$p\nmid \ell N$.
\end{proposition}
\begin{proof}
Consider an eigenform $\tilde{f}\in S_5(N,\tilde{\eps};\Qbar_5)$
lifting~$f$ as above.
Associated to~$\tilde{f}$ there is an automorphic
representation~$\pi=\tensor_v'\pi_v$ of $\GL(2,\bA)$, where~$\bA$
is the ad\{e}le ring of~$\Q$.
Because $43\mid\mid N$, and~$43$ divides the conductor
of $\eps$, we see that the local component $\pi_{43}$ of $\pi$ at
$43$ must be ramified principal series. By Carayol's theorem,
$\rho_{\tilde{f}}|_{D_{43}} \sim \abcd{\Psi_1}{0}{0}{\Psi_2}$
with, without loss of generality,~$\Psi_2$ unramified. We have
$(\Psi_1\cdot \Psi_2)|_{I_{43}}=\tilde{\eps}|_{I_{43}}=\tilde{\eps}_{43}$,
therefore, $\rho_{\tilde{f}}|_{I_{43}} \sim \abcd{\tilde{\eps}_{43}}{0}{0}{1}$.
Now twist~$\tilde{f}$ by $\tilde{\eps}_{43}^{-1}$; we find that
$\rho_{\tilde{f}\tensor\tilde{\eps}_{43}^{-1}}|_{I_{43}} \sim \abcd{1}{0}{0}{\tilde{\eps}^{-1}_{43}}$.
In particular, there is an
eigenform~$\tilde{f}'\in S_5(N,\tilde{\eps}_2\tilde{\eps}^{-1}_{43},\Qbar_5)$
whose associated Galois representation is the twist by $\tilde{\eps}^{-1}_{43}$
of that of $\tilde{f}$ (recall that $N=1376$ and so~$43$ divides~$N$
exactly once). Let~$f'$ denote the mod~$5$ reduction of~$\tilde{f}'$. Then
one checks easily that $f'\in S_5(N,\eps_2\eps^{-1}_{43},\F_{25})=S_5(N,\eps^5,\F_{25})$.
For all primes $p\nmid5N$ we have $a_p(f')=\eps_{43}(p)^{-1}a_p(f)$.
In particular, we have $a_p(f')=0$ for
$p=19,31$.
Also, $\eps_{43}(3)=\alp^8$ and $\eps_{43}(5) =\alp^8$, so
$$a_3(f')=\alp^{16}/\alp^8 = \alp^8 = (\alp^{16})^5$$
$$a_5(f')=\alp^{22}/\alp^8 = \alp^{14} = (\alp^{22})^5.$$
Now if $\sigma$ is the non-trivial automorphism of $\F_{25}$,
then $\sigma(f')$ and $f$ both lie in
$S_5(1376,\eps;\F_{25})$ and have same~$a_p$ for
$p=3,5,19,31$, so they are equal because we found~$f$
by computing the unique eigenform with given~$a_p$ for $p=3,5,19,31$.
So $g = f\tensor\eps_{43} = \sigma(f)\tensor\eps_{43}^2$.
Thus for all $p\nmid 5N$, we see that
$a_p(g) = a_p(f)^5 \eps_{43}^2$ has fifth power
$a_p(g)^5 = a_p(f)^{25} \eps_{43}^{10} = a_p(f) \eps_{43} = a_p(g)$.
\end{proof}
\subsection{Proof that~$\rho_g$ is unramified at~$5$}
We begin with a generalisation of~\cite{sturm:cong}.
Let $M>4$ be an integer, and let $h=\sum_{n\geq1}c_nq^n$ be a
normalised cuspidal eigenform
of some weight~$k\geq1$, level~$M$ and character~$\chi$, defined over some
field of characteristic not dividing~$M$. Even though the base field
might not have characteristic zero, we may still define the conductor
of $\chi$ to the the largest divisor $f$ of $M$ such that~$\chi$
factors through $(\Z/f\Z)^\times$.
Let~$I$ be a set of primes, with the property that for all~$p$
in~$I$, one of the following conditions hold:
(i) $p$ divides~$M$ but~$p$ does not divide~$M/\cond(\chi)$, or
(ii) $p$ divides~$M$ exactly once, and $h$ is $p$-new, in the sense
that there is no eigenform $h'$ of level $M/p$ such
that the $T_n$-eigenvalues of $h$ and $h'$ agree for all $n$
prime to $p$.
Let~$C$ denote the orbit of the cusp~$\infty$ in $X_1(M)$
under the action of the group generated by $w_p$ for $p\in I$, and
the Diamond operators $\langle d\rangle_M$. The orbit of~$\infty$
under the Diamond operators has size $\phi(M)/2$, and each
$w_p$ increases the size of the orbit by a factor of~2. In this
situation, we have
\begin{lemma} The first~$t$ terms of the $q$-expansion
of~$h$ at any cusp in~$C$ are determined by~$M$,~$k$, $\chi$, $c_p$
for~$p$ in~$I$, and $c_n$ for $1\leq n\leq t$.
\end{lemma}
\begin{remark} Our proof is just a translation of Corollary~4.6.18
of \cite{miyake} into the language of moduli problems (Miyake's argument
technically is only valid over the complex numbers).
\end{remark}
\begin{proof}
If $J\subseteq I$ is any subset, and $w_J$ denotes the product
of $w_p$ for $p\in J$, then $h|w_J$ is an eigenform for all the
Diamond operators, and this observation reduces the proof
of the lemma to showing that for $p\in I$, if $h|w_p=\sum_n d_nq^n$,
then $d_j$ for $1\leq j\leq n$ and $d_q$ for all $q\in I$
are determined by $M$, $k$, $\chi$, $p$, $c_j$ for $1\leq j\leq n$
and $c_q$ for all $q\in I$.
We first deal with primes $p$ of the form (i).
Say $M=p^mR$, where $R$ is prime to $p$.
Thinking of~$h$ as a rule for attaching
$k$-fold differentials to elliptic curves equipped with points
of order $p^m$ and $R$, we have by definition that
$$h(\G_m/q^\Z,\zeta,\zeta_R)=\bigg(\sum c_nq^n\bigg)(dt/t)^k,$$
where $\zeta=\zeta_{p^m}$ and $\zeta_R$ are fixed $p^m$th and $R$th roots
of unity in $\G_m$ which correspond to the cusp~$\infty$,
and $dt/t$ is the canonical differential on the Tate
curve $\G_m/q^\Z$. We normalise things such that
$$h(\G_m/q^{p^m\Z},q,\zeta_R)=\bigg(\sum d_nq^n\bigg)(dt/t)^k,$$
and remark that because $h$ is an action for the diamond operators,
we do not have to worry too much about whether this corresponds to
the standard normalisation of the $w_p$-operator.
We recall that the operator $pU_p$ in this setting can be thought
of as being defined by the rule:
$$(pU_ph)(E,P,Q)=\sum_C\pi^*h(E/C,\overline{P},\overline{Q}),$$
where $C$ runs through the subgroups of $E$ of order $p$ which have
trivial intersection with $\langle P\rangle$, and $\pi$ denotes the canonical
projection $E\to E/C$.
We see that
\begin{align*}
(pc_p)^m\big(\sum d_nq^n\big)(dt/t)^k&=(p^mU_{p^m}h)(\G_m/q^{p^m\Z},q,\zeta_R)\\
&=\sum_{c=0}^{p^m-1}\pi^c*h(\G_m/\langle q^{p^m},\zeta q^c\rangle,q,\zeta_R),
\end{align*}
where $\pi$ denotes the canonical projection from $\G_m/\langle q^{p^m}\rangle$
to the appropriate quotient. This last sum can be written as a
double sum
\begin{align*}
&\sum_{c\in(\Z/p^m\Z)^\times}\pi^*h(\G_m/\langle q^{p^m},\zeta q^c\rangle,q,\zeta_R)+\sum_{a=0}^{p^{m-1}-1}\pi^*h(\G_m/\langle q^{p^m},\zeta q^{pa}\rangle,q,\zeta_R)\\
=&\sum_{b\in(\Z/p^m\Z)^\times}\pi^*h(\G_m/\langle q^{p^m},\zeta^{-b}q\rangle,q,\zeta_R)+p^{m-1}\pi^*U_{p^{m-1}}h(\G_m/\langle q^{p^m},\zeta^{p^{m-1}}\rangle,q,\zeta_R)\\
=&\sum_{b\in(\Z/p^m\Z)^\times}\pi^*h(\G_m/\langle\zeta^{-b}q\rangle,\zeta^b,\zeta_R)+(pc_p)^{m-1}\pi^*h(\G_m/\langle q^{p^m},\zeta^{p^{m-1}}\rangle,q,\zeta_R)\\
=&\sum_{b}\chi_p(b)\sum_{n\geq1}c_n(\zeta^{-b}q)^n(dt/t)^k+p^k(pc_p)^{m-1}\pi^*h(\G_m/\langle q^{p^{m+1}}\rangle,q^p,\zeta_R^p),
\end{align*}
where we have written $\chi=\chi_R\chi_p,$ for $\chi_R$ a character of
level~$R$ and $\chi_p$ a character of level~$p^m$. We deduce that
\begin{align*}
&(pc_p)^m\big(\sum d_nq^n\big)(dt/t)^k-p^k(pc_p)^{m-1}\chi_R(p)\pi^*h(\G_m/\langle q^{p^{m+1}}\rangle,q^p,\zeta_R)\\
=&\bigg(\sum_n\big(\sum_b\chi_p(b)\zeta^{-bn}\big)c_nq^n\bigg)(dt/t)^k\\
=&W(\chi_p)\big(\sum_{p\nmid n}\chi_p(-n)^{-1}c_nq^n\big)(dt/t)^k
\end{align*}
where $W(\chi_p)=\sum_{b\in(\Z/p^m\Z)^\times}\chi_p(b)\zeta^b$ can be
checked to be nonzero because the conductor of $\chi_p$ is $p^m$.
Hence
$$(pc_p)^m\sum_n d_nq^n-p^k(pc_p)^{m-1}\chi_R(p)\sum_n d_nq^{np}=W(\chi_p)\chi_p(-1)\sum_{p\nmid n}\chi_p(n)^{-1}c_nq^n.$$
Equating coefficients of $q$ we deduce that $W(\chi_p)\chi_p(-1)=(pc_p)^md_1$,
and because $h|w_p$ is an eigenform for $T_n$ for all $n$ prime to $p$,
with eigenvalues determined by $\chi$ and $c_n$, we deduce that we can
determine $d_n$ for $n$ prime to $p$ from $c_n$. It remains to establish
what $d_p$ is, and equating coefficients of $q^p$ in the above equation
gives us that $(pc_p)^md_p=p^k(pc_p)^{m-1}\chi_R(p)d_1$ and hence
that $d_p$ is determined by $\chi$ and $c_p$.
Note that as a consequence we see that $d_p/d_1=p^{k-1}\chi_R(p)/c_p$,
a classical formula if the base field is the complexes.
Now we deal with primes of the form (ii) (note that we never use
this case in the rest of the paper). We think of $h$ as a rule associating
$k$-fold differentials to triples $(E,C,Q)$ where $C$ a cyclic subgroup of
order~$p$ and $Q$ a point of order~$R=M/p$. Because $h$ is $p$-new, the
trace of $h$ down to $X_1(M/p)$ must be zero, and hence we see
that for any elliptic curve $E$ equipped with a point $Q$ of order $R$,
$$\sum_C\pi^* h(E/C,E[p]/C,\overline{Q})=0.$$ As before, normalise things so
that
$$h(\G_m/q^\Z,\mu_p,\zeta_R)=\bigg(\sum_n c_nq^n\bigg)(dt/t)^k$$
and
$$h(\G_m/q^{p\Z},\langle q\rangle,\zeta_R)=\bigg(\sum_n d_nq^n\bigg)(dt/t)^k.$$
The fact that the trace of~$h$ is zero implies that
$$(pU_p)h(\G_m/q^{p\Z},\langle q\rangle,\zeta_R)+\pi^*h(\G_m/q^\Z,\mu_p,\zeta_R)=0,$$
and hence that
$$c_p\sum d_nq^n+p^{k-1}\sum c_nq^n=0$$
from which we deduce that the $d_n$ can be read off from $c_p$ and the $c_n$.
\end{proof}
\begin{remark} The size of~$C$ is
$\phi(M).2^{|I|-1}$, and the usefulness of this lemma is that
if $h_1$ and $h_2$ are two normalised eigenforms of the same level,
weight and character as above, both new at all primes in~$I$,
and the coefficients of $q^n$ in the
$q$-expansions of $h_1$ and $h_2$ agree for $n\in I$ and $n\leq t$,
then $h_1-h_2$ has a zero of order at least $t+1$ at all cusps in~$C$,
and in particular if
$\phi(M).2^{|I|-1}(t+1)>k/12[\SL_2(\Z):\Gamma_1(M)]=\deg(\omega^k)$ on
$X_1(M)$ then $h_1=h_2$. Using the fact that
$[\Gamma_0(M):\Gamma_1(M)]=\phi(M)/2$,
we deduce
\end{remark}
\begin{corollary}\label{cor:bound}
Let $h_1$ and $h_2$ be two normalised eigenforms as above.
If the coefficients of $q^n$ in the $q$-expansions of $h_1$ and $h_2$ agree
for all primes in $I$ and for all
$n\leq\frac{k}{12}[\SL_2(\Z):\Gamma_0(M)]/2^{|I|}$ then $h_1=h_2$.
\end{corollary}
\begin{remark} One can certainly do better than this corollary
in many cases. For example, when $n>1$ and
$p^n$ exactly divides both the level
of an eigenform and the conductor of its character, then one can compute
the $q$-expansion of the eigenform at many middle cusps'' too,
and hence increase the size of $C$ in the result above.
\end{remark}
We now go back to the explicit situation we are concerned with.
Although~$g$ is an eigenform of level $59168=2^5\cdot 43^2$,
we can still consider the corresponding representation
$\rho_g :\GQ\ra \GL(2,\F_5)$, and then directly analyze
its ramification.
\begin{proposition}
The representation~$\rho_g$ is unramified at~$5$.
\end{proposition}
\begin{proof}
Continuing the modular symbols computations as above,
we find that~$V_1$ is spanned by the two eigenforms
\begin{align*}
f\,\,&=q + \alp^{16}q^3 + \alp^{22}q^5 + \alp^{14}q^7
+ \alp^{14}q^9 + 4q^{11}+\cdots\\
f_1&=q + \alp^{16}q^3 + \alp^{10}q^5 + \alp^{14}q^7
+ \alp^{14}q^9 + 4q^{11}+\cdots.
\end{align*}
For $p\neq 5$ and $p\leq 997$, we have $a_p(f_1)=a_p(f)$.
To check that $a_p(f) = a_p(f_1)$ for all $p\neq 5$,
it suffices to show that the difference~$f-f_1$ has
$q$-expansion involving only powers of~$q^5$;
for this we use the $\theta$-operator
$q\frac{d}{dq}:S_5(1376,\eps,\F_{25})\ra S_{11}(1376,\eps;\F_{25})$.
Since~$\theta$ sends normalized eigenforms to normalized eigenforms,
it suffices to check that the subspace of
$S_{11}(1376,\eps;\F_{25})$ generated by~$\theta(f)$
and~$\theta(f_1)$ has dimension~$1$.
Corollary~\ref{cor:bound} implies that it suffices to verify that the
coefficients $a_p(\theta(f))$ and $a_p(\theta(f_1))$ are equal for all
$$p \leq \frac{11}{12}\cdot [\SL_2(\Z):\Gamma_0(1376)]\cdot \frac{1}{2} = 968.$$
The eigenform~$f$ must be new because we computed it by finding
the intersections of the kernels of Hecke operators $T_p$ with
$p\nmid 1376$; if~$f$ were an oldform then the intersection of the
kernels of these Hecke operators
would necessarily have dimension greater than~$1$.
Because it takes less than a second
to compute each $a_p(\theta(f))$, we were easily able to verify that the
space generated by $\theta(f)$ and $\theta(f_1)$ has dimension~$1$.
\begin{remark}
It is possible to avoid appealing to Corollary~\ref{cor:bound} by using
one of the following two alternative methods:
\begin{enumerate}
\item Define~$\theta$ directly on modular symbols and compute it.
\item Compute the intersection
$$\bigcap_{p\geq 2} \ker(T_p - pa_p(f)) \subset S_{11}(1376,\eps;\F_{25}).$$
Since~$\theta(f)$ and~$\theta(f_1)$ both lie
in the intersection, the moment the dimension
of a partial intersection is~$1$, it follows
that $\theta(f-f_1)=0$.
\end{enumerate}
We successfully carried out both alternatives.
For the first, we showed that~$\theta$ on modular symbols is
induced by multiplication by
$X^5Y - Y^5X$.
For the second, we find that after intersecting
kernels for $p\leq 11$, the dimension is already~$1$.
The first of these two methods took much less
time than the second.
\end{remark}
Next we use that $\theta(f-f_1)=0$ to show that $\rho_g$ is unramified,
thus finishing the proof of the proposition.
Since~$f$ is ordinary, Deligne's theorem (see~\cite[\S12]{gross:tameness})
implies that
$$\rho_f|_{D_5}\sim \mtwo{\alp}{*}{0}{\beta}\qquad\text{over \Fbar_5}$$
with both~$\alp, \beta$ unramified,
$\alp(\Frob_5)=\eps(5)/a_5=\alp^8/\alp^{22}=\alp^{10}$, and
$\beta(\Frob_5)=\alp^{22}$.
Since $a_p(f_1)=a_p(f)$, for $p\neq 5$, we have
$${\rho_f}|_{D_5}\sim {\rho_{f_1}}|_{D_5} \sim \mtwo{\alp'}{*}{0}{\beta'}$$
with
$\alp'(\Frob_5)=\alp^8/\alp^{10}=\alp^{22}$ and
$\beta'(\Frob_5)=\alp^{10}$;
in particular, $\alp'=\beta$.
Thus $\rho_f|_{D_5}$ contains $\alp\oplus \beta$, so
$\rho_f|_{D_5}\sim\alp\oplus\beta$ and hence there is a choice
of basis so that $*=0$.
\end{proof}
\subsection{The image of $\proj \rho_g$}
\begin{proposition}\label{prop:image_is_A5}
The image of $\proj \rho_g$ is $A_5$.
\end{proposition}
\begin{proof}
The image~$H$ of $\proj \rho_g$ in $\PGL_2(\F_5)$ is easily checked to
lie in $\PSL_2(\F_5)\cong A_5$ because of what we know about the
determinant of $\rho_g$. Hence $H$ is a subgroup of $A_5$ that
contains an element of order~$2$ (complex conjugation) and an element
of order~$3$ (for example, $\rho_g(\Frob_7)$ has characteristic
polynomial $x^2-2x-1$). This proves that~$H$ is isomorphic to
either~$S_3$,~$A_4$, or~$A_5$. Let $L$ be the number field cut out
by~$H$. If~$L$ were an $S_3$-extension, then there would be a
quadratic extension contained in it which is unramified outside
$2\cdot 5\cdot 43$; it is furthermore unramified at~$5$ by the
previous section and unramified at $43$ because $I_{43}$ has
order~$3$. Thus it is one of the three quadratic fields unramified
outside~$2$.
In particular, the trace of $\Frob_p$
would be zero for all primes in a certain congruence class
modulo~8.
However, there are primes~$p$ congruent to $3$, $5$, and $7$
mod $8$ such that $a_p(g)\neq 0$, e.g., $3$, $7$, and $13$.
If $H$ were isomorphic to $A_4$, then let~$M$ denote the cyclic
extension of degree~3 over~$\Q$ contained in~$L$. Now~$M$ is unramified
at~2 and~5, and hence is the subfield of $\Q(\zeta_{43})$ of degree~3.
Choose $p\nmid 1376\cdot 5$ that is inert in~$M$, i.e., so that
$p$ is not a cube mod $43$. The order of
$\rho_g(\Frob_p)$ in $\GL_2(\F_5)$ must be divisible by~$3$. However,
a quick check using Table~\ref{table:1376twist} shows that this is
usually not the case, even for $p=3$.
\end{proof}
\subsection{Bounding the ramification at~$2$ and~$43$}
Let~$L$ be the fixed field of $\ker(\proj(\rho_g))$. We have just
shown that $\Gal(L/\Q)$ is isomorphic to $A_5$.
By a root field for~$L$, we mean
a non-Galois extension of $\Q$ of degree~5 whose Galois closure is~$L$.
\begin{proposition}
The discriminant of a root
field for~$L$ divides $(43\cdot 8)^2=344^2$, and
in particular,~$L$ must be mentioned in Table~1
of \cite[pg 122]{freyetal}.\end{proposition}
\begin{proof}
The analysis of the local behavior of~$\rho_f$ at~$43$ given in
Proposition~\ref{prop:1376-g}
shows that the inertia group at~$43$ in $\Gal(L/\Q)$ has order~$3$. Using
Table~3.1 of~\cite{buhler:thesis}, we see that if
$\Gal(L/\Q)\isom A_5$
then it must be type~$2$'' at 43, and hence the discriminant of a root
field of~$L$, that is, of a non-Galois extension of~$\Q$ of degree~$5$
whose Galois closure is~$L$, must be $43^2$ at~$43$.
At~$2$ the behavior of~$\rho$ is more subtle and we shall not analyze
it fully. But we can say that, because~$\rho$ has arisen from
a form of level $1376=2^5.43$, we must be either of type~$5$
or one of types~$14$--$17$. In particular, the discriminant at~$2$ of a root
field for~$L$ will be at most~$2^6$.
Finally,~$L$ is unramified at all other primes, because~$\rho$ is.
Hence the discriminant of a root field for~$L$, assuming that
$\Gal(L/\Q)\cong A_5$, divides $(43.8)^2=344^2$.
\end{proof}
We know that~$L$ is an icosahedral extension of~$\Q$ with
discriminant dividing $43^2\cdot 2^6$. Table~1 of \cite[pg 122]{freyetal}
contains all icosahedral extensions, such that the discriminant
of a root field is bounded by $2083^2$. The table
must contain~$L$; there is only one icosahedral extension with
discriminant dividing $43^2\cdot 2^6$, so $L=K$.
\subsection{Obtaining a classical weight one form}
We have shown that a twist of the icosahedral
representation $\rho:\GQ\ra\GL(2,\C)$,
nobtained by lifting $\GQ\ra \Gal(K/\Q)\ncisom A_5$,
has a mod~$5$ reduction $\rho_g:\GQ\ra \GL_2(\F_5)$ that
is modular. Since~$\rho$ ramifies at only finitely many primes,
and~$\rho$ is unramified at~$5$ with distinct eigenvalues,
\cite{buzzard-taylor} implies that~$\rho$ arises from
a classical weight~$1$ newform.
\section{More examples}
The data necessary to deduce modularity of each of our eight
icosahedral examples is summarized in
Tables~\ref{table:more1}--\ref{table:more4}.
The notation in Table~\ref{table:more1} is as follows.
The first column contains the conductor.
The second column contains a $5$-tuple $[a_4,a_3,a_2,a_1,a_0]$ such
that the $A_5$-extension is the splitting field of the polynomial
$h=x^5+a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0$.
The column labeled $\ord(\Frob_5)$ contains the order of the image
of $\Frob_5$ in $A_5$. The next column, which is labeled $p$ with
$a_p=0$'', contains the first few~$p$ such that $a_p$ is easily seen
to equal~$0$ by considering the splitting of~$h$ mod~$p$.
The $\eps$ column contains the character of the representation, where the
notation is as follows. Write $(\Z/N\Z)^*$ as a product of cyclic groups
corresponding to the prime divisors of~$N$ in ascending order, and then
the tuples give the orders of the images of these cyclic factors; when
$8\mid N$, there are two cyclic factors corresponding to the prime~$2$.
Finally, the last column records the dimension of $S_5(\Gamma_1(N),\eps)$.
The notation in Table~\ref{table:more2} is as follows. The first column
contains the conductor. The second column contains an eigenform that
was found by first intersecting the kernels of the Hecke operators
$T_p$ with~$p$ as in Table~\ref{table:more1}, and then locating an
eigenform.
In each case, a companion form was found, by computing $a_p(f)$ for
$p\leq$ bound, where bound is the bound from Corollary~\ref{cor:bound}.
Table~\ref{table:more3} shows that the fixed field
of the image of each $\proj(\rho_g)$ is icosahedral.
The first column contains the
conductor~$N$. The second column contains a twist~$g$ of~$f$ such that
$a_p(g)\in\F_5$ for all $p\nmid 5N$. The third column contains
a $\Frob_p$ such that $\proj(\rho_g(\Frob_p))$ has order~$3$,
along with the characteristic polynomial of $\rho_g(\Frob_p)$.
As in the proof of Proposition~\ref{prop:image_is_A5},
the other two boxes give data that allows us to deduce
that the fixed field of the image of $\proj(\rho_g)$ is icosahedral.
The case $5373$ must be treated separately, because there are
three possibilities $M_1$, $M_2$, and $M_3$
for the cubic field~$M$ of the analogue of
Proposition~\ref{prop:image_is_A5}.
For $M_1$ we find a prime~$p$ such that
$$(p^2\!\!\!\mod 9, \,\,\,p^{66}\!\!\!\mod 199)\not\in\{(1,1),(4,1),(7,1)\}$$
with $\rho_g(\Frob_p)$ of order not divisible by~$3$;
for this, $p=2$ suffices, since the characteristic polynomial
of $\rho_g(\Frob_2)$ is $(x+2)^2$
and $(p^2\!\!\!\mod 9, \,\,\,p^{66}\!\!\!\mod 199) = (4,106)$.
For $M_2$ we find a prime~$p$ such that
$$(p^2\!\!\!\mod 9, \,\,\,p^{66}\!\!\!\mod 199)\not\in\{(1,1),(4,92),(7,106)\}$$
with $\rho_g(\Frob_p)$ of order not divisible by~$3$;
again, $p=2$ suffices.
For $M_3$ we find a prime~$p$ such that
$$(p^2\!\!\!\mod 9, \,\,\,p^{66}\!\!\!\mod 199)\not\in\{(1,1),(4,106),(7,92)\}$$
with $\rho_g(\Frob_p)$ of order not divisible by~$3$;
here, $p=13$ suffices, as the characteristic polynomial
of $\rho_g(\Frob_p)$ is $(x+4)^2$ and
$(p^2\!\!\!\mod 9, \,\,\,p^{66}\!\!\!\mod 199) = (7,106)$.
Table~\ref{table:more4} gives upper bounds on the ramification of the
fixed field of the image of $\proj(\rho_g)$. These bounds
were deduced using Table~3.1
of~\cite{buhler:thesis} by restricting the possible types'' using
information about the character $\eps$. Note that though
the bounds are not sharp, e.g., the discriminant of
the representation of conductor $2416$ is $2^4\cdot 151^2$, they
are all less than $2083^2$, so the corresponding
field must appear in Table~2 of~\cite{freyetal}.
\begin{table}
\caption{\label{table:more1}Data on icosahedral representations mod~$5$}
\begin{center}
\begin{tabular}{|clclll|}\hline
$N$&\hspace{2em}$h$&$\hspace{-1.5em}\ord(\Frob_5)$&$p$ with $a_p=0$&
\hspace{1em}$\eps$ &$\hspace{-1.5em}\dim S_5(N,\eps)$\\\hline
{\bf 1376}&$[2,6,8,10,8]$ & \hspace{-1.5em}$2$&$19,31,97$&$[2,1,3]$&$696$\\
{\bf 2416}&$[0,-2,2,5,6]$ & \hspace{-1.5em}$2$&$53,97,127$&$[2,1,3]$&$1210$\\
{\bf 3184}&$[5,8,-20,-21,-5]$& \hspace{-1.5em}$2$&$31,89,97$&$[2,1,3]$&$1594$\\
{\bf 3556}&$[3,9,-6,-4,-40]$&\hspace{-1.5em}$3$&$19,29,89$&$[1,2,3]$&$2042$\\
{\bf 3756}&$[0,-3,10,30,-18]$&\hspace{-1.5em}$3$&$17,61,67$&$[1,2,3]$&$2506$\\
{\bf 4108}&$[4,3,9,4,5]$& \hspace{-1.5em}$3$&$17,23,31,89$&$[1,3,2]$&$2234$\\
{\bf 4288}&$[4,5,8,3,2]$& \hspace{-1.5em}$3$&$19,23,47$&$[1,2,3]$&$2164$\\
{\bf 5373}&$[2,1,7,23,-11]$& \hspace{-1.5em}$2$&$7,23,37,79,89$&$[2,3]$&$2394$\\
\hline\end{tabular}
\end{center}
\end{table}
\begin{table}
\caption{\label{table:more2}The newform $f$ and the companion form bound}
\begin{center}
\begin{tabular}{|cll|}\hline
$N$&\hspace{7em}$f$ &\hspace{-.3em}bound \\\hline
{\bf 1376}&
$q + \alp^{16}q^3 + \alp^{22}q^5 + \alp^{14}q^7 + \alp^{14}q^9 + 4q^{11}+ \alp^{14}q^{13} + \cdots$
& $968$\\
{\bf 2416}&
$q +3q^3 + \alp^{22}q^5 + \alp^{16}q^7 + \alp^{4} q^{11} + \alp^2 q^{13} + \alp^{16}q^{15} + \cdots$
& $1672$ \\
{\bf 3184}&
$q + \alp^{16}q^3 + 3q^5 + \alp^{22}q^7 + \alp^{14}q^9 + 3q^{11} + \alp^{22}q^{13} + \cdots$
& $2200$\\
{\bf 3556}&
$q + \alp^{16}q^{3} + \alp^{14}q^5 + \alp^{10}q^7 + \alp^{14}q^9 + \alp^{2}q^{11} + \alp^{22}q^{13} + \cdots$
& $1408$ \\
{\bf 3756}&
$q + \alp^{14}q^3 + \alp^{14}q^5 + 3q^7 + \alp^4q^9 + \alp^{16}q^{11} + \alp^{10}q^{13} + \cdots$
& $1727$ \\
{\bf 4108}&
$q + \alp^{16}q^{3} + \alp^{11}q^{5} + \alp^{20}q^{7} + \alp^{14}q^{9} + \alp^{10}q^{11} + 4q^{13} + \cdots$
& $1540$\\
{\bf 4288}&
$q + 3q^3 + \alp^{14}q^{5} + \alp^{20}q^{7} + 3q^9 + \alp^{20}q^{11} + \alp^{16}q^{13} + \cdots$
& $2992$ \\
{\bf 5373}&
$q + \alp^{16}q^{2} + \alp^{14}q^{4} + 4q^5 + 3q^8 + \alp^{4}q^{10} + 2q^{11}+\cdots$
& $3300$ \\
\hline\end{tabular}
\end{center}
\end{table}
\begin{table}
\caption{\label{table:more3}Verification that the image of $\proj(\rho_g)$ is $A_5$}
\begin{center}
Find a Frobenius element with projective order $3$.\vspace{1ex}\\
\begin{tabular}{|c|l|ll|}\hline
$N$ & \hspace{1em} $g$ & proj. order $3$&\hspace{.7em}charpoly \\\hline
{\bf 1376} & $f\tensor \eps_{43}$
& $\quad\Frob_7$ & $x^2-2x-1$
\\
{\bf 2416}& $f\tensor \eps_{151}$
& $\quad\Frob_{19}$ & $x^2+2x-1$
\\
{\bf 3184}& $f\tensor \eps_{199}$
& $\quad\Frob_7$ & $x^2+3x+4$
\\
{\bf 3556}& $f\tensor \eps_{127}$
& $\quad\Frob_{13}$ & $x^2+3x+4$
\\
{\bf 3756}&$f\tensor\eps_{313}$
& $\quad\Frob_{23}$ & $x^2 + 2x + 4$
\\
{\bf 4108} & $f\tensor\eps_{13}$
& $\quad\Frob_{29}$ & $x^2+3x+4$
\\
{\bf 4288}& $f\tensor\eps_{67}$
& $\quad\Frob_{11}$ & $x^2+x+1$
\\
{\bf 5373}& $f\tensor\eps_{199}$
& $\quad\Frob_{11}$ & $x^2+3x+4$
\\
\hline\end{tabular}\vspace{3ex}
Not $S_3$: For all $t\in T$, find unramified $p$ s.t.\ $t\not\equiv \Box \mod p$ and $a_p(g)\neq 0$. \vspace{1ex}\\
\begin{tabular}{|c|l|l|}\hline
$N$ & $\qquad\quad{}T$ & $\qquad{}p$ \\\hline
{\bf 1376}
& $\{-1,-2\}$ & $3$, $7$\\
{\bf 2416}
& $\{-1, -2\}$ & $3$, $7$ \\
{\bf 3184}
& $\{-1, -2\}$ & $3$, $7$ \\
{\bf 3556}
& $\{-1, -2, -7, -14\}$ & $3$, $13$, $3$, $11$\\
{\bf 3756}
& $\{-1,-2,-3,-6\}$ & $7$, $7$, $11$, $13$\\
{\bf 4108}
& $\{-1, -2, -79, -158\}$ &$3$, $7$, $3$, $7$ \\
{\bf 4288}
& $\{-1,-2\}$ & $3$, $7$ \\
{\bf 5373}
& $\{-3\}$ & $11$\\
\hline\end{tabular}\vspace{3ex}\\
\comment{
function FindS3(t, aplist, N)
P:=[p : p in [2..97] |IsPrime(p)];
for i in [1..#aplist] do
p := P[i];
if (5*N mod p ne 0) and not IsSquare(GF(p)!t) and aplist[i] ne 0 then
return p;
end if;
end for;
end function;
}
Not $A_4$: Unramified~$p$, not cube mod $\ell$,
order of $\rho_g(\Frob_p)$ not divisible by $3$.
\vspace{1ex}\\
\begin{tabular}{|c|c|cl|}\hline
$N$ & $\ell$ & $p$&\hspace{.7em}charpoly$(\rho_g(\Frob_p))$ \\\hline
{\bf 1376}
& $43$ & $3$ & $\qquad(x+2)^2$ \\
{\bf 2416}
& $151$ & $7$ & $\qquad(x+2)^2$ \\
{\bf 3184}
& $199$ & $3$ & $\qquad(x+2)^2$ \\
{\bf 3556}
& $127$ & $3$ & $\qquad(x+2)^2$\\
{\bf 3756}
& $313$ & $11$ & $\qquad(x+2)^2$\\
{\bf 4108}
& $13$ & $3$ & $\qquad(x+2)^2$\\
{\bf 4288}
& $67$ & $7$ & $\qquad(x+3)^2$\\
{\bf 5373}
& --- & &(see text)\\
\hline\end{tabular}\vspace{3ex}\\
\comment{
function IsCubeMod(p, ell) // is p a cube in F_ell
R<x>:=PolynomialRing(GF(ell));
return not IsIrreducible(x^3-p);
end function;
procedure FindA4(ell, aplist, N, e1, e2)
P:=[p : p in [2..97] |IsPrime(p)];
for i in [1..#aplist] do
p := P[i];
if (5*N mod p ne 0) then
// if (5*N mod p ne 0) and not IsCubeMod(p,ell) then
t := GF(5)!(Evaluate(e2,p)*aplist[i]);
d := GF(5)!Evaluate(e1,p);
R<x> := PolynomialRing(GF(5));
f := x^2 - t*x + d;
"p =",p;
"f =",f;
"factor(f) =",Factorization(f);
end if;
end for;
end procedure;
}
\end{center}
\end{table}
\begin{table}
\caption{\label{table:more4}Bounding the discrimant of the fixed field
of $\proj(\rho_g)$}
\begin{center}
\begin{tabular}{|cl|}\hline
$N$ & Bound on discriminant\\
{\bf 1376}& $\qquad2^6\cdot 43^2$\\
{\bf 2416}& $\qquad 2^6\cdot 151^2$\\
{\bf 3184}& $\qquad 2^6\cdot 199^2$\\
{\bf 3556}& $\qquad 2^2\cdot 7^2 \cdot 127^2$\\
{\bf 3756}& $\qquad 2^2\cdot 3^2 \cdot 313^2$\\
{\bf 4108}& $\qquad 2^2\cdot 13^2 \cdot 79^2$\\
{\bf 4288}& $\qquad 2^6\cdot 67^2$\\
{\bf 5373}& $\qquad 3^4\cdot 199^2$\\
\hline\end{tabular}
\end{center}
\end{table}
\section{Computing mod~$p$ modular forms}
\subsection{Higher weight modular symbols}
\label{sec:modsym}
The second author developed software that computes the space of
weight~$k$ modular symbols $\sS_k(N,\eps)$, for $k\geq 2$ and
arbitrary~$\eps$.
See~\cite{merel:1585} for the standard facts about higher weight
modular symbols, and~\cite{stein:phd} for a description of
how to compute with them.
Let $K=\Q(\eps)$ be the field generated by the values of~$\eps$.
The cuspidal modular symbols $\sS_k(N,\eps)$ are a
finite dimensional vector space over~$K$, which is generated by all
linear combinations of higher weight modular symbols
$$X^i Y^{k-2-i}\{\alp,\beta\}$$
that lie in the kernel of an appropriate boundary map. There is an
involution~$*$ that acts on $\sS_k(N,\eps)$, and
$\sS_k(N,\eps)^+\tensor_K\C$ is isomorphic, as a module over the Hecke
algebra, to the space $S_k(N,\eps;\C)$ of cusp forms.
Fix $k=5$. In each case considered in this paper,
there is a prime ideal~$\lambda$
of the ring of integers $\mathcal{O}$ of~$K$
such that $\mathcal{O}/\lambda\isom \F_{25}$.
Let~$\cL$ be the $\mathcal{O}$-module generated by all modular
symbols of the form $X^iY^{3-i}\{\alp,\beta\}$,
and let
$$\sS_5(N,\eps;\F_{25})=(\cL\tensor_{\mathcal{O}}\F_{25})\cap \sS_5(N,\eps).$$
This is the space that we computed.
The Hecke algebra acts on $\sS_5(N,\eps;\F_{25})$, so when
we find an eigenform we find a maximal ideal of the Hecke algebra.
As an extra check on our computation of
$\sS_5(N,\eps;\F_{25})$, we computed the dimension
of $S_5(N,\eps;\C)$ using both the formula of~\cite{cohen-oesterle}
and the Hijikata trace formula (see~\cite{hijikata:trace})
applied to the identity Hecke operator.
\comment{%it's all in my thesis and it's not that relevant.
The Manin symbols are
$[i, (c,d)]$ where $0\leq i\leq k-2=3$ and
$(c,d)$ vary over points in the projective plane.
The Manin symbol $[i,(c',d')]$ corresponds to the
modular symbol $(g.X^iY^{3-i})\{g(0),g(\infty)\}$
where $g=\abcd{a}{b}{c}{d}\in\SL_2(\Z)$ is a matrix whose lower
two entries are congruent to $(c',d')$ modulo $N$,
and $g.X^iY^{3-i} := (dX-bY)^i(-cX+aY)^{3-i}$.
Let $\sigma=\abcd{0}{-1}{1}{0}$, $\tau={0}{-1}{1}{-1}$
and for $\gamma\in\SL_2(\Z)$, let
$[i,(c,d)]\gamma = [\gamma.X^iY^{3-i}, (c,d)\gamma]$.
Since there are only finitely many
Manin symbols, we can
compute $\sS_5(N,\eps)$ as the quotient of the $\F$-vector
space generated by Manin symbols modulo
the following relations:
\begin{align*}
{[i,(c,d)] + [i,(c,d)]\sigma} &= 0\\
{[i,(c,d)] + [i,(c,d)]\tau + [i,(c,d)]\tau^2} &= 0\\
{[i,(n c,n d)]}&=\eps(n)[i,(c,d)]
\end{align*}
The quotient was computed by using a fast hashing'' function
to quotient out by the $2$-term relations. The quotient
by the $3$-term relations was then computed using sparse
Gauss elimination. One important subtlety is that, e.g., $\sigma$
and~$\tau$ do not commute so, after modding out by
the~$\sigma$ relations, it is important to mod out by~$3$
term relations coming both from~$\tau$ and~$\sigma\tau$.
The main result of~\cite{merel:1585} gives
a way to compute the action of $T_p$ directly
on the Manin symbols.
Suppose $f\in\sS_5(N,\eps;\F_{25})$ is an eigenvector; to
naively compute the action of~$T_p$ on~$f$ requires computing
the action of~$T_p$ on each Manin symbol involved in~$f$,
and then summing the result. This requires roughly
$\dim\sS$ times as long as computing~$T_p$ on a single
Manin symbol.
In order to quickly compute a large number of
Hecke eigenvalues we use the following projection trick.
Let $\vphi\in\Hom(\sS_5(N,\eps;\F_{25}),\F_{25})$ be a (left) eigenvector for all
Hecke operators~$T_p$ having the same eigenvalues as~$f$.
Choose a Manin symbol $x=[i,(c,d)]$ such
that $\vphi(x)\neq 0$. Since~$x$ is of a very simple form,
it is easy to compute~$T_p(x)$ quickly. We have
$\vphi(T_p(x)) = (T_p(\vphi))(x) = a_p \vphi(x)$,
so since $\vphi(x)\neq 0$ we divide and find
$a_p = \vphi(T_p(x))/\vphi(x)$.
In fact, we use a generalization of this trick to
quickly compute the action of~$T_p$ on any Hecke stable subspace
$V\subset \Hom(\sS(N,\eps;\F_{25}),\F_{25})$.
}
\subsection{Complexity}
We implemented the modular symbols algorithms mentioned above
in \magma{} (see \cite{magma}) because of its robust support
for linear algebra over small finite fields.
The following table gives a flavor of the complexity of the
machine computations appearing in this paper.
The table indicates how much
CPU time on a Sun Ultra E450 was required to compute all data
for the given level,
including the matrices $T_p$ on the $2$-dimensional spaces,
for $p<2000$. For example, the total time for level $N=1376$
was~$6$ minutes and~$58$ seconds.
\begin{center}
\begin{tabular}{|cr|}\hline
\vspace{-2ex}&\\
N & time (minutes)\\
\vspace{-2ex}&\\
1376& 6:58\hspace{2.5em}\mbox{}\\
2416& 10:42\hspace{2.5em}\mbox{}\\
3184& 14:16\hspace{2.5em}\mbox{}\\
3556& 19:55\hspace{2.5em}\mbox{}\\
3756& 27:47\hspace{2.5em}\mbox{}\\
4108& 23:11\hspace{2.5em}\mbox{}\\
4288& 15:18\hspace{2.5em}\mbox{}\\
5376& 24:49\hspace{2.5em}\mbox{}\\\hline
\end{tabular}
\end{center}
\subsection{Acknowledgment}
Some of the computing equipment was purchased
by the second author using a UC Berkeley Vice Chancellor Research Grant.
made on the Sun Ultra E450 of the Computational Algebra Group at
the University of Sydney. Allan Steel was very helpful in optimizing our
code.
\comment{\bibliographystyle{amsplain}
\bibliography{biblio}}
\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\begin{thebibliography}{10}
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E.~Artin, \emph{{\"U}ber eine neue {A}rt von {L}-reihen}, Abh. Math. Sem. in
Univ. Hamburg \textbf{3} (1923/1924), no. 1, 89--108.
\bibitem{buhler:thesis}
J.\thinspace{}P. Buhler, \emph{Icosahedral \protect{G}alois representations},
Springer-Verlag, Berlin, 1978, Lecture Notes in Mathematics, Vol. 654.
\bibitem{bdsbt}
K.~Buzzard, M.~Dickinson, N.~Shepherd-Barron, and R.~Taylor, \emph{On
icosahedral {A}rtin representations}, in preparation.
\bibitem{buzzard-taylor}
K.~Buzzard and R.~Taylor, \emph{Companion forms and weight one forms}, Ann. of
Math. (2) \textbf{149} (1999), no.~3, 905--919.
\bibitem{cohen-oesterle}
H.~Cohen and J.~Oesterl{\'e}, \emph{Dimensions des espaces de formes
modulaires}, (1977), 69--78. Lecture Notes in Math., Vol. 627.
\bibitem{magma}
W.~Bosma, J.~Cannon, and C.~Playoust, \emph{The {M}agma algebra system {I}:
{T}he user language}, J. Symb. Comp. \textbf{24} (1997), no.~3-4, 235--265,
\\\protect{\sf http://www.maths.usyd.edu.au:8000/u/magma/}.
\bibitem{deligne-serre}
P.~Deligne and J-P. Serre, \emph{Formes modulaires de poids $1$}, Ann. Sci.
\'Ecole Norm. Sup. (4) \textbf{7} (1974), 507--530 (1975).
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G.~Frey (ed.), \emph{On {A}rtin's conjecture for odd \protect{$2$}-dimensional
representations}, Springer-Verlag, Berlin, 1994.
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B.\thinspace{}H. Gross, \emph{A tameness criterion for \protect{G}alois
representations associated to modular forms (mod \protect{$p$})}, Duke Math.
J. \textbf{61} (1990), no.~2, 445--517.
\bibitem{hijikata:trace}
H.~Hijikata, \emph{Explicit formula of the traces of \protect{H}ecke operators
for \protect{$\Gamma_0(N)$}}, J. Math. Soc. Japan \textbf{26} (1974), no.~1,
56--82.
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R.\thinspace{}P. Langlands, \emph{Base change for \protect{${\rm {G}{L}}(2)$}},
Princeton University Press, Princeton, N.J., 1980.
\bibitem{merel:1585}
L.~Merel, \emph{Universal \protect{F}ourier expansions of modular forms}, On
{A}rtin's conjecture for odd \protect{$2$}-dimensional representations
(Berlin), Springer, 1994, pp.~59--94. Lecture Notes in Math., Vol. 1585.
\bibitem{miyake}
T.~Miyake, \emph{Modular forms}, Springer-Verlag, Berlin, 1989, Translated from
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\end{document}
***
[8458981, 509]
// Cohen-Oesterle Dimension computations in MAGMA:
> G<a2,b2,c> := DirichletGroup(1376,CyclotomicField(EulerPhi(1376)));
> eps:=a2*(c^(Order(c) div 3));
> Order(eps);
6
> DimensionCuspForms(eps,5);
> G<a2,b2,c> := DirichletGroup(2416,CyclotomicField(EulerPhi(2416)));
> eps:=a2*(c^(Order(c) div 3));
> Order(eps);
6
> DimensionCuspForms(eps,5);
1210
> G<a2,b2,c> := DirichletGroup(3184,CyclotomicField(EulerPhi(3184)));
> eps:=a2*(c^(Order(c) div 3));
> Order(eps);
6
> DimensionCuspForms(eps,5);
1594
> G<a,b,c> := DirichletGroup(3556,CyclotomicField(EulerPhi(3556)));
> eps:=(b^(Order(b) div 2))*(c^(Order(c) div 3));
> Order(eps);
6
> DimensionCuspForms(eps,5);
2042;
> G<a,b,c> := DirichletGroup(3756,CyclotomicField(EulerPhi(3756)));
> eps:=(b^(Order(b) div 2))*(c^(Order(c) div 3));
> Order(eps);
6
> DimensionCuspForms(eps,5);
> G<a,b,c> := DirichletGroup(4108,CyclotomicField(EulerPhi(4108)));
> eps:=(b^(Order(b) div 3))*(c^(Order(c) div 2));
> DimensionCuspForms(eps,5);
> G<a,b,c> := DirichletGroup(4288,CyclotomicField(EulerPhi(4288)));
> eps:=(b^(Order(b) div 2))*(c^(Order(c) div 2));
> DimensionCuspForms(eps,5);
> G<a,b> := DirichletGroup(5373,CyclotomicField(EulerPhi(5373)));
> eps:=(a^(Order(a) div 2))*(b^(Order(b) div 3));
> DimensionCuspForms(eps,5);
///////////////////////////////////////////
procedure powfrob(p, e1, e2, aplist)
Primes := [p : p in [2..97] |IsPrime(p)];
n := Index(Primes,p);
a := GF(5)!(Evaluate(e2,p)*aplist[n]);
b := GF(5)!Evaluate(e1,p);
R<x>:=PolynomialRing(GF(5));
Q<y> := quo<R|x^2-a*x+b>;
x^2 - a*x + b;
for i in [1..24] do
f := MinimalPolynomial(y^i);
if Degree(f) le 1 then
printf "rho_g(Frob_%o)^%o satisfies %o.\n", p, i, f;
end if;
end for;
end procedure;
> // 1376
> h := x^5+2*x^4+6*x^3+8*x^2+10*x+8;
> N := 2^5*43;
> F<alp> := GF(25);
> G<a2,b2,c>:=DirichletGroup(N, F);
> eps := a2 * (c^(Order(c) div 3));
> aplist := [0,alp^16,alp^22,alp^14,4,alp^14,alp^14,0, alp^16,alp^8,0,alp^10,1,alp^10,1,alp^22,4,alp^14,alp^4,alp^20,alp^2,alp^20,alp^4,alp^10];
> qEigenform(aplist,eps,5);
//q + alp^16*q^3 + alp^22*q^5 + alp^14*q^7 + alp^14*q^9 + 4*q^11 + alp^14*q^13 + alp^14*q^15 + alp^14*q^17 + O(q^20)
> // 2416
> h := x^5-2*x^3+2*x^2+5*x+6;
> N := 2^4*151;
> k:=5;
> F<alp> := GF(25);
> G<a2,b2,c>:=DirichletGroup(N, F);
> eps := a2 * (c^(Order(c) div 3));
> aplist:=[0,3,alp^22,alp^16, alp^4, alp^2, alp^22,3,alp^22,3,alp^16,alp^22,2,alp^8,alp^8,0,1,alp^8,3,alp^8,2,2,2,alp^20,0,2,alp^16,1];
> qEigenform(aplist,eps,k);
// q + 3*q^3 + alp^22*q^5 + alp^16*q^7 + alp^4*q^11 + alp^2*q^13 + alp^16*q^15 + alp^22*q^17 + 3*q^19 + O(q^20)
> // 3184
> h := x^5+5*x^4+8*x^3-20*x^2-21*x-5;
> N:=2^4*199;
> F<alp> := GF(25);
> G<a2,b2,c>:=DirichletGroup(N, F);
> eps := a2 * (c^(Order(c) div 3));
> aplist:=[0,alp^16,3,alp^22,3,alp^22,3,alp^16];
> qEigenform(aplist,eps,5);
// q + alp^16*q^3 + 3*q^5 + alp^22*q^7 + alp^14*q^9 + 3*q^11 + alp^22*q^13 + alp^10*q^15 + 3*q^17 + alp^16*q^19 + O(q^20)
> // 3556
> h := x^5+3*x^4+9*x^3-6*x^2-4*x-40;
> N := 2^2*7*127;
> F<alp> := GF(25);
> G<a,b,c>:=DirichletGroup(N, F);
> eps := b^(Order(b) div 2) * (c^(Order(c) div 3));
> aplist := [0,alp^16,alp^14,alp^10,alp^2,alp^22,alp^14,0, alp^10,0,alp^16,alp^20];
> // 3756
> h := x^5-3*x^3+10*x^2+30*x-18;
> N := 2^2*3*313;
> F<alp> := GF(25);
> G<a,b,c>:=DirichletGroup(N, F);
> eps := b^(Order(b) div 2) * (c^(Order(c) div 3));
> aplist:=[0,alp^14,alp^14,3,alp^16,alp^10,0,3,3,alp^2,alp^22,alp^22,alp^20,alp^16,alp^4,4,alp^8,0];
> qEigenform(aplist,eps,5);
// q + alp^14*q^3 + alp^14*q^5 + 3*q^7 + alp^4*q^9 + alp^16*q^11 + alp^10*q^13 + alp^4*q^15 + 3*q^19 + alp^8*q^21 + 3*q^23 + alp^4*q^25 + 3*q^27 + alp^2*q^29 + alp^22*q^31 + 2*q^33 + alp^8*q^35 + alp^22*q^37 + q^39 + alp^20*q^41 + alp^16*q^43 + 3*q^45 + alp^4*q^47 + 3*q^49 + 4*q^53 + 2*q^55 + alp^8*q^57 + alp^8*q^59 + O(q^62)
// 4108
> h := x^5+4*x^4+3*x^3+9*x^2+4*x+5;
> N := 2^2*13*79;
> F<alp> := GF(25);
> G<a,b,c>:=DirichletGroup(N, F);
> eps := b^(Order(b) div 3) * (c^(Order(c) div 2));
> aplist := [0,alp^16,alp^11, alp^20,alp^10,4,0,alp^14,0, alp^22, 0, alp^22,alp^10,alp^2,3,4,alp^14,alp^2,alp^10];
> qEigenform(aplist,eps,5);
//q + alp^16*q^3 + alp^11*q^5 + alp^20*q^7 + alp^14*q^9 + alp^10*q^11
// + 4*q^13 + alp^3*q^15 + alp^14*q^19 + 4*q^21 + O(q^24)
// 4288
> h := x^5+4*x^4+5*x^3+8*x^2+3*x+2;
> N := 2^6*67;
> F<alp> := GF(25);
> G<a2,b2,c>:=DirichletGroup(N, F);
> eps := a2*b2^(Order(b2) div 2)*(c^(Order(c) div 3));
> aplist := [0,3,alp^14, alp^20, alp^20, alp^16, alp^16,0,0, alp^14,alp^20,alp^4,alp^8,2,0,4,1,alp^16,alp^4,alp^20,0,alp^20,0,2];
> qEigenform(aplist,eps,5);
//q + 3*q^3 + alp^14*q^5 + alp^20*q^7 + 3*q^9 + alp^20*q^11 + alp^16*q^13 + alp^8*q^15 + alp^16*q^17 + alp^14*q^21 + O(q^24)
// 5373
> h := x^5+2*x^4+x^3+7*x^2+23*x-11;
> N := 3^3*199;
> F<alp> := GF(25);
> G<a,b>:=DirichletGroup(N, F);
> eps := a^(Order(a) div 2) * (b^(Order(b) div 3));
> aplist := [alp^16, 0, 4, 0, 2, alp^22, 1, alp^16, 0];
> qEigenform(aplist,eps,5);
//q + alp^16*q^2 + alp^14*q^4 + 4*q^5 + 3*q^8 + alp^4*q^10 + 2*q^11 + alp^22*q^13 + q^17 + alp^16*q^19 + alp^2*q^20 + alp^22*q^22 + O(q^24)
` | 2019-04-24 03:03:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9230887293815613, "perplexity": 3139.4419186461864}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578624217.55/warc/CC-MAIN-20190424014705-20190424040705-00088.warc.gz"} |
https://www.physicsforums.com/threads/gr-effects.390135/ | # GR effects
According to SR theory, relative motion leads to
-length contraction
-mass inflation
-time dilation.
In GR theory, gravity leads to time dilation. Does it also lead to
-length contraction?
-mass inflation?
Special relativity is a subset of general relativity. In a region of space without large masses, the geometry of that space will become flat. That is, it will reduce to the minkowski metric of special relativity. So, all of the effects that occur in special relativity can also occur in general relativity.
In addition, there are a host of new, but it some cases analogous effects. Perhaps the most often observed new effect is that time runs more slowly for objects in a large gravitational potential. This is magnified in the case of a black hole. The rate of progression of time for an object falling into a black hole (as viewed by a distant observer) will slow to zero as that object approaches the event horizon.
For GPS satellites, which orbit the Earth at a high velocity, both special and general relativistic effects occur:
effect due to sr:
As GPS satellites move fast relative to us, their time is dilated, and runs more slowly, by about 7 microseconds/day.
effect due to gr:
Because gps satellites are further away from the center of the Earth's gravitational field, their time runs faster than ours, by about 45 microseconds/day.
According to wikipedia (http://en.wikipedia.org/wiki/Global_Positioning_System#Relativity), if neglected then these effects would cause uncertainties in position to grow by around 10km/day.
There are dozens of other new effects: frame dragging, event horizons, the penrose process, cauchy surfaces, schwarzschild and kerr black holes, expanding cosmologies, even messier differential equations, and so on:D
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In GR theory, gravity leads to time dilation. Does it also lead to
-length contraction?
-mass inflation?
In GR there are no such things as "length contraction" and "mass inflation" caused by gravity but rather now there exists a new concept "curvature" by which one can feel the presence of a gravitational field and matter.
AB
Special relativity is a subset of general relativity.
Really? Funny how Einstein seemed to START with Special Relativity, then moved on to GR. Are you positive you're not just blowing smoke?
@Altabeh: Hmmmm... do you mind if I quote you to a friend of mine who's been struggling with this concept? You said that very succinctly!
@Altabeh: Hmmmm... do you mind if I quote you to a friend of mine who's been struggling with this concept? You said that very succinctly!
Yeah sure!
AB
Janus
Staff Emeritus
Gold Member
Really? Funny how Einstein seemed to START with Special Relativity, then moved on to GR. Are you positive you're not just blowing smoke?
Yes, really. Einstein started with SR because it was easier to start with the limited "special" case. But he also knew that the theory would not be complete until it could be further expanded to deal with the more "general" case.
George Jones
Staff Emeritus
Gold Member
Special relativity is a subset of general relativity.
Really? Funny how Einstein seemed to START with Special Relativity, then moved on to GR. Are you positive you're not just blowing smoke?
I.e, Einstein condsidered more general situations for which special relativity is a special case. If, in general relativity, one considers the special case of a simply connected spacetime with $g_{\mu \nu} = \eta_{\mu \nu}$, one gets Minkowski spacetime. In this case, Einstein's equation is $0 = T_{\mu \nu}$, so energy and matter have to be (in some sense) negligible ("test" particles only).
I understand, but I don't think you can call a theory published while GR was still being formulated a "subset". I understand the physics, I disagree with the semantics. To be blunt, Einstein's first paper describing SR was published in 1905, and GR formulated between '07-'15.
I wouldn't call it a subset of anything, just "The Special and General Theories of Relativity".
George Jones
Staff Emeritus
Gold Member
I understand, but I don't think you can call a theory published while GR was still being formulated a "subset".
Why not?
I understand the physics, I disagree with the semantics.
To me, the phrasing doesn't seem so bad. It is clear what is meant.
To be blunt, Einstein's first paper describing SR was published in 1905, and GR formulated between '07-'15.
For me, the dates are irrelevant.
I wouldn't call it a subset of anything, just "The Special and General Theories of Relativity".
Again, I don't think that the terminology is so bad.
dx
Homework Helper
Gold Member
Electrostatics was formulated before the full electromagnetic theory of Maxwell. Yet, electrostatics is logically a special case of electrodynamics, in the sense that under specialized conditions, electrodynamics reduces to electrostatics. The relationship between special relativity and general relativity is the same.
I don't see how this argument over semantics, which I will gladly cede, can go anywhere useful. I aknowledge the point that General Relativity is a complete theory in which there is a "Special" case, as a subset. I would point out that a person in 1906, would of course have no way of appreciating that. It is logically organized in the manner you describe, but that isn't how it evolved. ia_'s language assumed prior knowledge that some people may not have on an educational site, not an entirely uncommon occurence.
There are some brilliant people here, but the communication skills are occasionally lacking in terms of liasing between common parlance and various formalism and terms of art of mathematics and physics. This does lead to some extended misunderstandings and pointless detours such as this.
This is a situation in which, as with dx's example: there's more than one way of establishing a logical hierarchy. There is a historical context, in which theories evolve from one another, and then the formal (and useful) manner of classifing them ad hoc. I'm not disputing the latter, but let me ask you this: How would you establish a hierarchy for the major branches of mathematics?
@George Jones: "Again, I don't think that the terminology is so bad." Fair enough; I don't think the terminology is that great.
So back to my original question - is there length contraction and/or mass inflation in GR - Altabeh said:
"In GR there are no such things as "length contraction" and "mass inflation" caused by gravity
Then there must be no such things as length and mass, or they do not contract and inflate.
Which?
I understand, but I don't think you can call a theory published while GR was still being formulated a "subset". I understand the physics, I disagree with the semantics. To be blunt, Einstein's first paper describing SR was published in 1905, and GR formulated between '07-'15.
I wouldn't call it a subset of anything, just "The Special and General Theories of Relativity".
Most physicists first get involved with a very limited and simple theory and then they try to expand the ideas brought up in the initial theory into a big framework wherein not only will the first theory work well, but the new theory will be based on some more general ideas that lead to the initial ideas in special cases. It is not at all a bad way to keep a theory going deep into broader areas of knowledge by first starting from a simple thing though this could sometimes lead to nowhere if the developed ideas are not as much compatible to the initial ideas as possible! For example, the idea of "curvature" in GR is compatibly reducible to the old implication of "force", as assumed by Newton. Sometimes there are no conditions by which the theory can be given a simplified, reduced or even generalized form. Google can help you find some examples of this.
AB
Thanks anyway.
So back to my original question - is there length contraction and/or mass inflation in GR - Altabeh said:
Then there must be no such things as length and mass, or they do not contract and inflate.
Which?
I think my post can't be put in a clearer form: Length cannot contract and mass cannot inflate due to gravity in GR.
AB
So back to my original question - is there length contraction and/or mass inflation in GR - Altabeh said:
Then there must be no such things as length and mass, or they do not contract and inflate.
Which?
Are you by chance thinking of the old sci-fi standby, the "blueshifted 'front' of a near-c 'craft', versus the apparant long redshifted 'tail' of it? That is an effect percieved by the observer, and doesn't imply a real change in length or mass. In essence, you're seeing light "stack" in the front, and "extend" in the rear, but that's the LIGHT, not the "craft".
I'm not even sure what you mean by "mass inflation", unless you think that the mass of an object approaching 'c' becomes "infinite", which is a common misconception. Once again, you have to remember which effects are the result of a change in something, and which are merely observational artifacts.
@Altabeh: Yes, theories usually evolve from the simple to the more complex; and that new theory should incorporate the old, in which case from one point of view SR is clearly a subset of GR. That said, even though GR is the framework which explains SR and more, one did not evolve as a subset of the other, but rather GR is an expansion and extension (among others things) of GR.
As I said, this is a semantic issue, with two possible views on the subject. I'll admit that mine is less useful in this context, but also it won't mislead newcomers to the subject as to how the theories were developed. To me, a subset represents a group "B" containing elements derived or taken from group "A". For that to occur, group A needs to exist for a subset to emerge, rather than B leading to A and because A encompasses B, it's being called a subset.
As I said earlier, I would call them "SR and GR", not "GR and its subset", or "SR and its superset". There is no context in which identifying them as two theories, one leading to another, isn't preferable to "subset"
Are you by chance thinking of the old sci-fi standby, the "blueshifted 'front' of a near-c 'craft', versus the apparant long redshifted 'tail' of it? That is an effect percieved by the observer, and doesn't imply a real change in length or mass. In essence, you're seeing light "stack" in the front, and "extend" in the rear, but that's the LIGHT, not the "craft".
I'm not even sure what you mean by "mass inflation", unless you think that the mass of an object approaching 'c' becomes "infinite", which is a common misconception. Once again, you have to remember which effects are the result of a change in something, and which are merely observational artifacts.
A common misconception? Would you not concede that the inertia of an object (the quantity of external momentum impulse that it takes to alter its velocity a particular amount) increases without limit as the object approaches the speed of light? (After all, this is the mechanism that prevents us accelerating it to superluminal velocities. Let's just leave aside questions of precisely what the term "mass" should refer to.)
Sure, purely optical distortions also exist, such as Terrel rotation. But the reality of length contraction is the first lesson taught on SR (regarding a barn and ladder, unfortunately under the guise of a paradox involving details of material acceleration). Regardless of how your sci-fi craft seems visually, the fact is that physical synchronised measurements of its extent (say, made by attendants along a station as the craft passes against some of them) would confirm contraction (and note that this is predicted by GR as well as by SR, regardless of whatever point Altabeh is trying to make), just as surely as atomic clocks left in different floors of a building do confirm the time-dilation predicted in GR.
To put that another way, since the shape of atoms and molecules is basically dependent on Maxwell's theory (which exhibits Lorentz symmetry), whilst you may be familiar with notionally-spherical electron-orbital shapes for atoms at rest, the correct actual solution for the shape of the electron-cloud of a fast moving atom is more like a pancake.
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A common misconception? Would you not concede that the inertia of an object (the quantity of external momentum impulse that it takes to alter its velocity a particular amount) increases without limit as the object approaches the speed of light? (After all, this is the mechanism that prevents us accelerating it to superluminal velocities. Let's just leave aside questions of precisely what the term "mass" should refer to.)
Sure, purely optical distortions also exist, such as Terrel rotation. But the reality of length contraction is the first lesson taught on SR (regarding a barn and ladder, unfortunately under the guise of a paradox involving details of material acceleration). Regardless of how your sci-fi craft seems visually, the fact is that physical synchronised measurements of its extent (say, made by attendants along a station as the craft passes against some of them) would confirm contraction (and note that this is predicted by GR as well as by SR, regardless of whatever point Altabeh is trying to make), just as surely as atomic clocks left in different floors of a building do confirm the time-dilation predicted in GR.
To put that another way, since the shape of atoms and molecules is basically dependent on Maxwell's theory (which exhibits Lorentz symmetry), whilst you may be familiar with notionally-spherical electron-orbital shapes for atoms at rest, the correct actual solution for the shape of the electron-cloud of a fast moving atom is more like a pancake.
Really?! Wow, talk about being 180 degrees off target. Well, in that light, I'm going to skip to that portion of MTW and learn what I thought I knew. Thanks for the correction, and lesson. By the way, if it isn't too much trouble what would be a toy solution for a simple fast-moving atom (H, He)?
According to SR theory, relative motion leads to
-length contraction
-mass inflation
-time dilation.
In GR theory, gravity leads to time dilation. Does it also lead to
-length contraction?
-mass inflation?
In SR an observer measures:
1) The length of rod parallel to the relative relative motion v, to be length contracted by sqrt(1-v^2).
2) The length of a rod transverse to the relative motion v, to be the same length as when it is at rest.
3) The rate of a clock with relative motion v, to be time dilated by 1/sqrt(1-v^2).
In GR an observer at infinity in Schwarzschild coordinates measures:
A) The length of stationary vertical rod at radial coordinate r, to be length contracted by sqrt(1-2GM/r).
B) The length of a stationary horizontal rod at radial coordinate r, to be the same length as a local rod.
C) The rate of a stationary clock at radial coordinate r, to be time dilated by 1/sqrt(1-2GM/r).
As you can see from the above, 1,2 and 3 are closely related to A, B and C respectively. Is that the sort of relationship you are looking for?
Sure, purely optical distortions also exist, such as Terrel rotation. But the reality of length contraction is the first lesson taught on SR (regarding a barn and ladder, unfortunately under the guise of a paradox involving details of material acceleration). Regardless of how your sci-fi craft seems visually, the fact is that physical synchronised measurements of its extent (say, made by attendants along a station as the craft passes against some of them) would confirm contraction (and note that this is predicted by GR as well as by SR, regardless of whatever point Altabeh is trying to make),
I just saw you mentioned in passing that "length contraction" is also predicted by GR. I assume you know that this is not the case which was asked about by the OP. If you entered GR so as to be able to for example make the length of a body contract in a given gravitational field due to the effects of gravity, I'd be more comfortable with the word "impossible" if assigned to the purpose of your entry. In GR length contraction is predicted if one comes in the inertial frames in Minkowski spacetime by reducing the new theory to SR. Using the principle of equivalence this could be done in the context of GR in a very small region but this again calls for gravity to disappear. Some eccentric sort of gravitational length contraction exists that only reveals itself when a 'ruler' is set at fixed $$(t,\theta,\phi)$$. Such a contraction with this assuption is not realistic but just theoretical because the time can't be made constant in reality for a rigid body falling inward the black hole radially.
AB
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Thanks! I have another "relationship" you might be interested in. Consider a particle dropped from infinty towards a (non-rotating, uncharged) massive body. Let us say the freefall velocity of the particle at radial coordinate r is measured to be v by a stationary observer a r. The time dilation that the particle would have in SR terms due to its freefall velocity is 1/sqrt(1-v^2/c^2) and this is equal to the time dilation experienced by a stationary particle at r due to gravitational time dilation, i.e. 1/sqrt(1-2GM/(rc^2)). Just in case you are wondering, the freefalling particle experiences both velocity and gravitational time dilation equal to 1/sqrt(1-v^2/c^2)*1/sqrt(1-2GM/(rc^2)).
To make this clearer, here is a numerical example considering a clock freefalling from infinity. Let us say the stationary observer at r measures the freefall velocity of the falling clock to be 0.8c. The free falling clock would then be running 0.6 time slower than a stationary clock at r, due to kinematic time dilation. The stationary clock at r would be running 0.6 times slower than a clock at infinity, due to gravitational time dilation. The free falling clock at r would then be running 0.36 times slower the clock at infinity, due to both effects.
Foof. Now I know nothing. Just found this arxiv article [peer reviewed?]
http://arxiv.org/abs/0910.2298" [Broken]
which states clearly that lengths EXPAND under acceleration [which is equivalent to gravity]. The expansion is given by L[1+[at/c]^2]^1/2. [Eq. 4 pg. 2] Maybe I have observer / accelerated reversed?
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Foof. Now I know nothing. Just found this arxiv article [peer reviewed?]
http://arxiv.org/abs/0910.2298" [Broken]
which states clearly that lengths EXPAND under acceleration [which is equivalent to gravity]. The expansion is given by L[1+[at/c]^2]^1/2. [Eq. 4 pg. 2] Maybe I have observer / accelerated reversed?
No, the authors state that this idea only works for accelerated frames in SRT and so there's no sign of gravity to think about it in this not peer-reviewed and badly typed article!
AB
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Foof. Now I know nothing. Just found this arxiv article [peer reviewed?]
http://arxiv.org/abs/0910.2298" [Broken]
which states clearly that lengths EXPAND under acceleration [which is equivalent to gravity]. The expansion is given by L[1+[at/c]^2]^1/2. [Eq. 4 pg. 2] Maybe I have observer / accelerated reversed?
Hi Harry,
If you are new to relativity, then you managed to find just about the most confusing article a beginner could wish to find. Let me try and clear up some of the confusion. Let us start with some things you might be confortable with. I hope you agree that when a rod (in its relaxed state) is moving relative to you and parallel to its length, that you would measure its length to be shorter than when it is at rest with respect to you (in its relaxed state). I have stressed the "relaxed state" of the rod because when an inertially moving rod is measured to be length contracted in relativity, it is assumed to be in an unstressed state (i.e. neither physically stretched nor physically compressed) and the length of the rod is assumed to remain constant from the point of view of an observer moving with the rod. Now if we were to nail down one end of the rod and accelerate the other end of the rod towards the fixed end, the rod would be compressed, but this would not be be relativistic length contraction and the change in length is due to stress forces which could be measured. If the free end is accelerated away from the fixed end the rod would be stretched and become longer but we would not call that relativistic length expansion. Now when a free rod is accelerated in a manner that does not put any stresses on the rod (no stretching or compressing forces) then it length appears to remain constant according to an observer co-moving (co-accelerating) with the rod and appears to be getting shorter according to an observer that remains in the original rest frame of the rod before it started accelerating. This form of acceleration, that does not put any physical stresses on the rod is called Born rigid acceleration. Since the rod appears to getting shorter in the original rest frame of the rod, the back end of the rod appears to be catching up with the front end of the rod, so Born rigid acceleration requires the back end of the rod to be accelerating slightly faster than the front end of the rod. Any other kind of acceleration puts physical stresses on the rod that physically stretch or compress the rod and this is a separate issue from pure relativistic length contraction. As I mentioned before if I nail down one end of the rod and the accelerate the other end away from the fixed end, the rod gets stretched and I can claim that as an example of a rod getting longer when accelerated. A less extreme example is to accelerate the back end at the same rate as the front end. This is not Born rigid acceleration and put physical stresses on the accelerating rod and physically stretches the rod. In this case an observer moving with the rod sees the rod getting longer (because it is being stretched) and the observer that remained in the orginal rest frame of the rod sees the length of the accelerating rod remain constant. I think this is basically what is happening in the paper you have cited, but I have not read it in much detail. You have effectively stumbled across a version of Bell's rocket paradox which is not the easiest of paradoxes to understand, so I hope my explanation is not too confusing.
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No, the authors state that this idea only works for accelerated frames in SRT and so there's no sign of gravity to think about it in this not peer-reviewed and badly typed article!
SRT plus acceleration equals GRT. Equivalence principle.
I hope my explanation is not too confusing.
Indeed, after reading Wikipedia your clarification is as good as I could find. I think I did have unaccelerated observer frame confused with accelerated frames.
A) The length of stationary vertical rod at radial coordinate r, to be length contracted by sqrt(1-2GM/r).
Apologies for being pedantic, not "vertical at radial coordiante r" but "alongside the r direction"
B) The length of a stationary horizontal rod at radial coordinate r, to be the same length as a local rod.
Same correction as above, "perpendicular on the r direction" (i.e. lying in the (theta,phi) plane)
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SRT plus acceleration equals GRT. Equivalence principle.
Indeed, after reading Wikipedia your clarification is as good as I could find. I think I did have unaccelerated observer frame confused with accelerated frames.
Excuse me? Why on earth did you reach the reasoning that if "acceleration" is added to SRT, then the resulting theory must be GRT? Even in SRT we have sometimes acceleration involved with the setup of frames and this doesn't mean the content of theory changes to GRT! The equivalence principle says that both the gravitational and inertial accelerations are equivalent in a sufficiently small region of a curved spacetime. What does this have to do with that article? There authors state clearly that the theory they are bringing this up in is SRT not GRT!
AB
Just in case you are wondering, the freefalling particle experiences both velocity and gravitational time dilation equal to 1/sqrt(1-v^2/c^2)*1/sqrt(1-2GM/(rc^2)).
Dear kev,
How can that be? If you neglect the rotational terms, you obviously use the Schwarzschild metric with d(theta)=d(phi)=0.
Then, you get:
d\tau/dt=sqrt((1-r_s/r)-(v/c)^2/(1-r_s/r))=sqrt((1-r_s/r-v/c)(1-r_s/r+v/c)/(1-r_s/r)) even if we know that:
1-r_s/r=1-1-2GM/(rc^2)
This is very different from your result. How did you make the jump from the above formula to your formula?
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A common misconception? Would you not concede that the inertia of an object (the quantity of external momentum impulse that it takes to alter its velocity a particular amount) increases without limit as the object approaches the speed of light? (After all, this is the mechanism that prevents us accelerating it to superluminal velocities. Let's just leave aside questions of precisely what the term "mass" should refer to.)
You are completely wrong. Answers like this cause people to be utterly confused about relativity.
Any beginner in learning about relativity should readily understand that just because something is moving near light speed with respect to an accelerating object does not mean that it interferes with the ability for that object to accelerate.
The relativistic mass of an object does not undergo a proper acceleration, the rest mass does. Regardless of the rate or duration of proper acceleration any object will measure a speed of light of c forever.
You are completely wrong. Answers like this cause people to be utterly confused about relativity.
:rofl: Surely to have been completely wrong, what you quoted would have to be factually incorrect, so please enlighten me. Do you propose some conception of inertia which does not eventually grow as velocity (relative to the frame of reference) increases? Care to share your preferred explanation for the nonconstancy of the acceleration of a body against which a constant force is applied (say as by a uniform electric field)?
Any beginner in learning about relativity should readily understand that just because something is moving near light speed with respect to an accelerating object does not mean that it interferes with the ability for that object to accelerate.
Uh, ok, unless you misunderstood what you objected to, I don't see why you felt the need to point out that "something not interacting with an object doesn't interact with the object"?..
The relativistic mass of an object does not undergo a proper acceleration, the rest mass does.
Does anyone understand what Passionflower meant by the above sentence?
Regardless of the rate or duration of proper acceleration any object will measure a speed of light of c forever.
Really?
Last edited: | 2022-07-07 04:07:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6591578125953674, "perplexity": 699.0883473836227}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104683683.99/warc/CC-MAIN-20220707033101-20220707063101-00232.warc.gz"} |
https://www.jiskha.com/questions/302329/its-a-monomial-and-we-are-told-to-simplify-and-assume-no-denominator-is-equal-to-zero | # algebra
It's a monomial and we are told to simplify and assume no denominator is equal to zero.
its a -10m-1y0r means -10m to the neg. 1 power and y to the 0 power and r. I don't know how to write the powers. that equation is divided by -14m-7y-3r-4.which means -14m to the -7 power y to the -3 power and r to the-4 power.
1. 👍
2. 👎
3. 👁
1. What you have described should be written (in the typed format we have to use here) as
-10*m^-1* y^0 *r/[-14*m^-7*y^-3*r^-4]
Any number to the zero power is 1, so you can get rid of the y^0 term in the numnerator.
Exponents in the denominator can be brought to the numerator with a change of sign, and added to whatever exponent is in the numerator attached to the same variable. The -10/-14 can be rewritten as the factor 5/7.
You are then left with
(5/7)m^7*y^3*r^5
1. 👍
2. 👎
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simplify by rationalizing the denominator: 18 ``√6```` | 2021-10-28 01:18:50 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8222252130508423, "perplexity": 4405.441041899949}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588246.79/warc/CC-MAIN-20211028003812-20211028033812-00527.warc.gz"} |
https://zhenkewu.com/ | Zhenke Wu Research
Name in Chinese: 吴振科 . Pronounced: “Jen-Kuh Wu”.
The best way to contact me is email. Direction to my office is here.
I am an Assistant Professor in the Department of Biostatistics at University of Michigan, with joint appointment as Research Assistant Professor in Michigan Institute for Data Science (MIDAS). I am also Faculty Associate in Quantitative Methodology Program, Survey Research Center of Institute for Social Research (ISR), University of Michigan.
Research Theme:
My research is motivated by biomedical and public health problems and is centered on the design and application of statistical methods that inform health decisions made by individuals, or precision medicine. Towards this goal, I focus on two lines of methodological research: a) structured Bayesian latent variable models for clustering and disease subtyping, and b) study design, causal and reinforcement learning methods for evaluating sequential interventions that tailor to individuals’ changing circumstances such as in interventional mobile health studies. I am committed to developing robust, scalable, and interpretable statistical methods to harness real-world, high-dimensional, dynamic data for individualized health. The methods and software developed so far have supported studies in diverse scientific fields including infectious disease epidemiology, autoimmune diseases, mental health, behavioral health, and cancer.
Keywords:
• Statistical: Hierarchical Bayesian models; Latent variable models; Nonparametric Bayes; Bayesian scalable computation; Causal inference; Reinforcement learning.
• Substantive: Precision medicine; Mobile health; Infectious diseases; Mental health; Electronic health records/claims data; Healthcare policy; Clinical trials; Just-in-time adaptive interventions for behaviorial and psychiatric research.
Advising: We are recruiting motivated and hard-working people interested in Bayesian methods and computation, graphical models, causal inference, sequential decision making, reinforcement learning and large-scale health data analytics. If you want to get involved, please say hi.
Check this out and send me an email if interested in collaborating!
AI in Science Postdoctoral Fellowship Program; The program will pay a competitive salary (\$74,000 annually for 2022-23) plus benefits. Travel to funder’s AI in Science events will also be covered.
Working Group:
I collaborate closely with
Published 21 Nov 2022
Doubly Inhomogeneous Reinforcement Learning
Hu et al. (2022+). Submitted.
Published 07 Nov 2022
Published 26 Oct 2022
Dynamic Survival Transformers for Causal Inference with Electronic Health Records
Chatha et al. (2022). NeurIPS Workshop on Learning from Time Series for Health.
Published 21 Oct 2022
Published 05 Oct 2022
Published 30 Sep 2022
Published 24 Sep 2022
Kernel Multimodal Continuous Attention
Moreno et al. (2022). NeurIPS.
Published 14 Sep 2022
Published 02 Sep 2022
Published 19 Aug 2022
Congratulations to Tsung-Hung Yao (co-advisor: Veera Baladandayuthapani) for receiving a competitive travel award to 2022 International Society of Bayesian Analysis (ISBA), Montreal. This is based on his work “Probabilistic Learning of Treatment Trees in Cancer”.
Congratulations to Hera Shi (co-advisor: Walter Dempsey) for receiving a competitive Junior Researcher Travel Grant to attend 2022 American Causal Inference Conference (ACIC) at Berkeley, CA. This is based on her work “Assessing Time-Varying Causal Effect Moderation in the Presence of Cluster-Level Treatment Effect Heterogeneity”.
Posted 02 May 2022 by Zhenke Wu
baker has a first public release (v1.0.0) at CRAN! Discussion related to version 1.0.0 can be submitted to here; [vignette] [source code].
The definitive reference for baker R package can be found here.
This vignette describes and illustrates the functionality of the baker R package. The package provides a suite of nested partially-latent class models (NPLCM) for multivariate binary responses that are observed under a case-control design. The baker package allows researchers to flexibly estimate population- and individual-level class distributions that may also depend on additional explanatory covariates.
Functions in baker implement recent methodological developments in our group (here, here, and here). Estimation is accomplished by calling a cross-platform automatic Bayesian inference software JAGS through a wrapper R function that parses model specifications and data inputs. The baker package provides many useful features, including data ingestion, exploratory data analyses, model diagnostics, extensive plotting and visualization options, catalyzing vital communications between practitioners and domain scientists. Package features and workflows are illustrated using simulated and real data sets.
The focus of this document is on guiding a new user to utilize some useful functions in baker for simulation studies and data analyses, aided by other powerful R packages. We refer readers of this document to the accompanying main software paper for more details about the software design considerations and review of model formulations. Since baker’s first appearance on Github, the authors have not been able to track other recent substantive publications that have used this package; we hope the main software paper and this vignette serve as the definitive reference for future scientific studies that find the baker package useful.
Posted 02 Feb 2022 by Zhenke Wu
An exciting new collaboration is supported by 2021 Propelling Original Data Science (PODS) grant from Michigan Institute for Data Science (MIDAS). The project title is “Structured Latent Variable Methods for High-Dimensional Electronic Health Records and Administrative Claims Data”. The faculty investigators are Zhenke Wu (PI; Biostatistics, Public Health), Jordan Schaefer (Co-I; Hematology, Internal Medicine), and Andrew Ryan (Co-I; Health Management and Policy, Public Health). The primary data source will be based on UnitedHealthcare OptumInsight claims data via Data and Methods Hub (DMH) at UM Institute for Healthcare Policy and Innovation (IHPI).
Posted 28 May 2021 by Zhenke Wu
Huge congrats to Irena and Hera for passing their PhD qualifying exams!!
Posted 01 Sep 2020 by Zhenke Wu
Huge congrats to Mengbing for winning “Best Doctoral Qualifying Exam Award”.
Another huge congrats to Irena for winning a poster award today in the 2019 MIDAS Annual Symposium in the category of “Most Likely to Make an Impact in the Field” for her poster “Regression Analysis for Probabilistic Cause-of-disease Assignment using Case-control Diagnostic Tests: A Hierarchical Bayesian Approach”!
Way to go!!
Posted 15 Nov 2019 by Zhenke Wu
Posted 11 Sep 2020 by Zhenke Wu
Posted 11 Jul 2019 by Zhenke Wu
Testing MathJax
This blog tests math compatibility on this site
Posted 01 Nov 2015 by Zhenke Wu | 2022-11-27 05:23:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1813962310552597, "perplexity": 9614.024199774025}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710192.90/warc/CC-MAIN-20221127041342-20221127071342-00237.warc.gz"} |
https://zbmath.org/?q=an:1012.35076 | # zbMATH — the first resource for mathematics
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The application of bifurcation method to a higher-order KdV equation. (English) Zbl 1012.35076
Summary: Bifurcation method of dynamical systems is employed to investigate bifurcation of solitary waves in the higher-order KdV equation $$u_t+au^nu_x +u_{xxx}=0,$$ where $n\ge 1$ and $a\in\bbfR$. Numbers of solitary waves are given for each parameter condition. Under some parameter conditions, explicit solitary wave solutions are obtained. Specially, some new solitary wave solutions are found for the KdV or MKDV equation.
##### MSC:
35Q53 KdV-like (Korteweg-de Vries) equations 37K50 Bifurcation problems (infinite-dimensional systems) 37K40 Soliton theory, asymptotic behavior of solutions
Full Text:
##### References:
[1] Miura, R. M.: The Korteweg--de Vries equation: a survey of results. SIAM rev. 8, 412-459 (1976) · Zbl 0333.35021 [2] Dodd, R. K.; Eilbeck, J. C.; Gibbon, J. D.; Morris, H. C.: Solitons and nonlinear wave equations. (1982) · Zbl 0496.35001 [3] Bona, J. L.; Dougalis, V. A.; Karakashian, O. A.; Mckinney, W. R.: Conservative, high-order numerical schemes for the generalized Korteweg--de Vries equation. Philos. trans. Roy. soc. London ser. A 351, 107-164 (1995) · Zbl 0824.65095 [4] Fornberg, B.; Whitham, G. B.: A numerical and theoretical study of certain nonlinear wave phenomena. Philos. trans. Roy. soc. London ser. A 289, 373-404 (1978) · Zbl 0384.65049 [5] Chow, S. N.; Hale, J. K.: Methods of bifurcation theory. (1982) · Zbl 0487.47039 [6] Guckenheimer, J.; Holmes, P.: Dynamical systems and bifurcations of vector fields. (1983) · Zbl 0515.34001 [7] Dey, B.: K--dv like equations with domain wall solutions and their Hamiltonians. Solitons, Springer ser. Nonlinear dynamics, 188-194 (1988) [8] Dey, B.: Domain wall solutions of KdV like equations with higher order nonlinearity. J. phys. A 19, L9-L12 (1986) · Zbl 0624.35070 [9] Jr., G. L. Lamb: Elements of soliton theory. (1980) | 2016-05-06 18:48:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7112728357315063, "perplexity": 11778.589583248673}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461862047707.47/warc/CC-MAIN-20160428164727-00044-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://www.skepticalcommunity.com/viewtopic.php?t=48659&start=260 | Anti-vaxxers in power
Ever had it before? Well you got it again.
sparks
Posts: 15707
Joined: Fri Oct 26, 2007 4:13 pm
Location: Friar McWallclocks Bar -- Where time stands still while you lean over!
Re: Anti-vaxxers in power
Dumb fuckers make money every fucking day by spreading bullshit.
Hell, we could all do it.
If we'd only suspend our better judgement and our sympathy and empathy for our fellow humans.
But in the meantime, goddamn these psychopaths to flaming fucking hell.
You can lead them to knowledge, but you can't make them think.
Witness
Posts: 23500
Joined: Thu Sep 19, 2013 5:50 pm
Re: Anti-vaxxers in power
shemp
Posts: 5972
Joined: Thu Jun 10, 2004 12:16 pm
Title: crazy as fuck
Location: is everything
Re: Anti-vaxxers in power
I guess Dumbass is going to lose their job.
"It is not I who is mad! It is I who is crazy!" -- Ren Hoek
Freedom of choice
Is what you got
Freedom from choice
Is what you want
sparks
Posts: 15707
Joined: Fri Oct 26, 2007 4:13 pm
Location: Friar McWallclocks Bar -- Where time stands still while you lean over!
Re: Anti-vaxxers in power
Indeed.
You can lead them to knowledge, but you can't make them think.
Witness
Posts: 23500
Joined: Thu Sep 19, 2013 5:50 pm
Re: Anti-vaxxers in power
I'm getting old.
Must be the vaccines.
sparks
Posts: 15707
Joined: Fri Oct 26, 2007 4:13 pm
Location: Friar McWallclocks Bar -- Where time stands still while you lean over!
Re: Anti-vaxxers in power
I'm switching to the anti-vax side: Let them die! Let them all die horribly!!!
Goddamned stupid motherfuckers.
You can lead them to knowledge, but you can't make them think.
Witness
Posts: 23500
Joined: Thu Sep 19, 2013 5:50 pm
Re: Anti-vaxxers in power
Witness
Posts: 23500
Joined: Thu Sep 19, 2013 5:50 pm
Re: Anti-vaxxers in power
Majority of anti-vaxx ads on Facebook are funded by just two organizations
Study finds Robert F Kennedy Jr’s World Mercury Project and Larry Cook’s Stop Mandatory Vaccinations bought 54% of ads
The majority of Facebook ads spreading misinformation about vaccines are funded by two organizations run by well-known anti-vaccination activists, a new study in the journal Vaccine has found.
The World Mercury Project chaired by Robert F Kennedy Jr, and Stop Mandatory Vaccinations, a project of campaigner Larry Cook, bought 54% of the anti-vaccine ads shown on the platform during the study period.
“Absolutely we were surprised,” said David Broniatowski, a professor of engineering at George Washington University, one of the authors of the report. “These two individuals were generating the majority of the content.”
Cook uses crowd-funding platforms to raise money for Facebook ads and his personal expenses. The crowd-funding platform GoFundMe banned Cook’s fundraisers in March 2019. YouTube has demonetized Cook’s videos.
Kennedy is the son of the former US attorney general Bobby Kennedy. He also has a nonprofit focused on environmental causes. Kennedy’s brother, sister and niece publicly criticized his “dangerous misinformation” about vaccines in May. They called his work against vaccination, “tragically wrong”.
https://www.theguardian.com/technology/ ... ions-study
Witness
Posts: 23500
Joined: Thu Sep 19, 2013 5:50 pm
Re: Anti-vaxxers in power
Samoa declares state of emergency as measles outbreak claims lives
Samoa's Government has declared a state of emergency, ordering all schools, including the National University of Samoa, to be closed as a deadly measles outbreak continues to spread.
Since Samoan officials announced a measles epidemic in October, seven suspected measles-related deaths have been recorded.
The majority of cases have involved children younger than four years old.
The Samoan Government ordered children under the age of 17 not to attend public gatherings in an attempt to stop the virus spreading.
It also made vaccinations a mandatory legal requirement for all people of Samoa who have not yet received a vaccination injection.
https://www.abc.net.au/news/2019-11-17/ ... k/11711576
Witness
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Re: Anti-vaxxers in power
Germany passes law making measles vaccination compulsory for children
Teachers and hospital staff must also be immunised, law says
Germany’s parliament has approved a law making it compulsory for schoolchildren to be vaccinated against measles.
The measure passed with a massive majority on Thursday, mandating that all children who attend school or nursery must be inoculated.
Politicians approved the government bill by 459 in favour to 89 against. There were 105 abstentions.
The law means parents who are unable to prove their offspring have been vaccinated against measles by 1 August 2021 - or those who refuse - will be fined up to €2,500 (£2,150).
Jens Spahn, Germany’s health minister, said compulsory vaccination was necessary because of an increase in cases of measles, a highly contagious and potentially fatal disease.
https://www.independent.co.uk/news/worl ... 04306.html
Abdul Alhazred
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Re: Anti-vaxxers in power
Yay Germany.
But I betcha this unleashes Nazi comparisons.
The arc of the moral universe bends towards chaos.
People who believe God or History are on their side provide the chaos.
Witness
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Re: Anti-vaxxers in power
Witness
Posts: 23500
Joined: Thu Sep 19, 2013 5:50 pm
Re: Anti-vaxxers in power
Witness
Posts: 23500
Joined: Thu Sep 19, 2013 5:50 pm
Re: Anti-vaxxers in power
Risk vs. benefit, a difficult concept.
robinson
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Re: Anti-vaxxers in power
High-risk HPV types (like HPV 16 and 18) are spread by sexual contact, A woman who doesn't have a high risk HPV and who isn't engaging in risky sexual activity has zero risk of getting cancer from an HPV.
You never know what's going to happen, then some shit happens nobody saw coming, then later somebody says they knew it was coming, then some new shit happens nobody saw coming, rinse and repeat
robinson
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Re: Anti-vaxxers in power
If the PTB really want to stop cervical cancer, then create a test for HPV, and treat it like any other infectious disease
You never know what's going to happen, then some shit happens nobody saw coming, then later somebody says they knew it was coming, then some new shit happens nobody saw coming, rinse and repeat
robinson
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Title: Hmmm ...
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Re: Anti-vaxxers in power
Of course injecting billions of women with an expensive vaccine (most of whom don't need it), is lucrative, and more attractive than finding out who actually is at risk form HPV caused cancer
You never know what's going to happen, then some shit happens nobody saw coming, then later somebody says they knew it was coming, then some new shit happens nobody saw coming, rinse and repeat
Witness
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Re: Anti-vaxxers in power
robinson wrote:
Thu Nov 28, 2019 2:43 am
Of course injecting billions of women with an expensive vaccine (most of whom don't need it), is lucrative, and more attractive than finding out who actually is at risk form HPV caused cancer
Wrong.
HPV Vaccine Prices
The real HPV vaccine prices have fallen steeply over the last few years in many European countries, thanks to competitive tendering. For instance, the price per dose in Italy dropped under €30 in recent regional tenders and even sunk under €20 in the Netherlands; experts commonly felt there were similar drops in national tenders in Sweden and the UK (both countries where awarded prices are kept confidential).
This experience showed that prices can drop steeply even when competition is minimal, i.e. only two manufacturers, if vaccines are judged basically equivalent by health authorities for their main health target, i.e. CC prevention in the case of HPV vaccination.
Furthermore there are numerous programs providing the vaccine for free or at low cost. Exempli gratia:
How much does it cost?
The HPV vaccine is provided free in schools for girls and boys aged 12–13 years as part of the National HPV Vaccination Program.
Girls and boys aged up to 19 who are not in the eligible school year levels can also obtain two doses of the vaccine for free from their local immunisation provider or doctor as part of the ongoing program. Those aged 15 or older at the time of their first vaccination require three doses for best protection.
http://www.hpvvaccine.org.au/the-hpv-vaccine/cost.aspx
Merck Vaccine Patience Assistance Program:
Merck offers a vaccine assistance program to those who desire to receive Gardisil 9. To qualify, you must:
• Be between the ages of 19 and 26
• Be uninsured
• Live in the United States, although you don't have to be a U.S. citizen
• Have an annual income less than $47,520 for individuals,$64,080 for couples or $97,200 for a family of four Merck does take special circumstances into account and often makes exceptions on a case-by-case basis. Do not let the income qualifications deter you from applying. You may still be approved based on your specific situation. Applying for the Merck Vaccine Patient Assistance Program is easy. Simply visit the Merck website and download and print the application. Return the completed application to your doctor's office. Your doctor's office will submit the application to Merck, who will notify the office of a decision the same day. It is important to note that the application must be completed and approved by Merck prior to getting the vaccination. https://www.verywellhealth.com/how-much ... ost-514124 And from as far back as 2013: Millions of girls in developing countries to be protected against cervical cancer thanks to new HPV vaccine deals Cape Town / Geneva, 9 May 2013 – A new record low price for human papillomavirus (HPV) vaccines will help ensure millions of girls in developing countries can be protected against cervical cancer. Thanks to the GAVI Alliance, the poorest countries will now have access to a sustainable supply of HPV vaccines for as low as US$ 4.50 per dose. The same vaccines can cost more than $100 in developed countries and the previous lowest public sector price was$13 per dose.
HPV vaccines are primarily available as part of routine immunisation to girls in relatively wealthy countries. And yet of the 275,000 women in the world who die of cervical cancer every year, more than 85% are in low-income countries, where the incidence of HPV infection is higher and few women have access to screening and treatment.
https://www.gavi.org/hpv-price-announcement
I'll also add that when you're a teen you probably don't know what kind of sex life you will have…
Now to support your claim you just have to show that screening is cheaper and has the same efficiency as the vaccine and associated programs. ("Hey, I'll just get myself tested.")
sparks
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Re: Anti-vaxxers in power
Witness wrote:
Wed Nov 27, 2019 10:45 pm
Risk vs. benefit, a difficult concept.
Apparently for some it is. I will not pray for them.
You can lead them to knowledge, but you can't make them think.
Witness
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Re: Anti-vaxxers in power
I presume we should still roam the savanna. | 2019-12-05 17:26:37 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22276371717453003, "perplexity": 13394.459273583729}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540481281.1/warc/CC-MAIN-20191205164243-20191205192243-00251.warc.gz"} |
https://stats.stackexchange.com/questions/297261/random-forest-mtry-spit-to-split-variable-sampling-procedure | # Random Forest mtry Spit-to-Split Variable Sampling Procedure
Thanks to several online resources, including this Cross Validated question, I am clear that - within Random Forests - mtry represents the number of variables randomly sampled without replacement at each split of a tree.
What I am not clear on is if sampled without replacement means that a variable that was sampled for the first split, but passed on because another another randomly sampled variable provided greater outcome homogeneity, could be sampled again for the second split of the tree.
Here is a simple example:
1. I have a binary classification problem.
2. My dataset has 8 predictor variables: V1, V2, V3, V4, V5, V6, V7, V8
3. I've set up my model where mtry = 3
4. For the first split of the first tree, the three randomly selected variables to test are V2, V4, V7 and the algorithm has determined that V4 provides the maximum homogeneity of the three variables.
I understand that V4 is not an option for the randomly sampled variables of the second splits (2) because it is already being used for the first split.
My question is: Are V2 and V7 available to be randomly sampled again in both of the second splits along with the variables that were not sampled in the first split? i.e., are the variables available for sampling in both of the second splits V1, V2, V3, V5, V6, V7, V8?
## 1 Answer
All variables are available as potential candidates for every split. This includes all variables examined in the previous split and the variable used for the previous split. Of these potential candidates, $m$ will be examined, and the best chosen.
In this context, "sampled without replacement" only refers to the current split. It wouldn't make sense to sample with replacement because that would amount to duplicating features, which wouldn't add anything, and would reduce the number of features examined. Neither would it make sense to remove features considered in previous splits, because they may be good for splitting descendant nodes. Furtheremore, doing so could cause us to run out of variables to use.
• so in my example above, all variables (including V4) are available again for the second split? – bshelt141 Aug 10 '17 at 18:43
• @bshelt141 Yes, that's right – user20160 Aug 10 '17 at 18:48 | 2021-03-05 19:05:31 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3654019236564636, "perplexity": 648.0955387395666}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178373241.51/warc/CC-MAIN-20210305183324-20210305213324-00197.warc.gz"} |
http://mathoverflow.net/revisions/32248/list | 2 added a description of the Grothendieck spectral sequence
The purpose of this question is to find out whether one can view the Atiyah-Hirzebruch spectral sequence as a particular case of the "composition of derived functors" spectral sequence.
The Leray spectral sequence of a continuous map $f:X\to Y$ of topological spaces can be constructed as follows. Let $a_X:X\to pt$ and $a_Y:Y\to pt$ be the maps from $X$ and $Y$ respectively to the one point space $pt$; we obviously have $a_X=a_Y\circ f$. So for any sheaf $L$ on $X$ we have $(a_X)_\ast L=(a_Y)_*f_\ast L$. But $(a_X)_\ast$ is just the functor of the global sections and so is $(a_Y)_*$. Deriving this Recall that if $A,B$ and taking the Grothendieck $C$ are abelian categories and $F:A\to B,G:B\to C$ are left exact functors, then (under mild hypotheses) $R_\ast(G\circ F)$ is isomorphic to $R_\ast\circ G_\ast$ and for any object $X$ of $A$ there exists a spectral sequence abutting $R^\ast (G\circ F) X$ with the $E_2$ sheet given by $E_2^{pq}=R^pG(R^q F(X))$. See e.g Gelfand-Manin, Methods of homological algebra, 3.7.
Applying this to the case when $A$, $B$ and $C$ are the categories of sheaves of $X$, $Y$ and the point respectively we get a spectral sequence $(E^{pq}_r,d_d)$ abutting to $H^*(X,F)$ with the $E_2$ term given by $$E_2^{pq}=H^p(Y,R^q f_\ast L).$$
If $X$ and $Y$ are sufficiently nice (say finite CW complexes), $F$ is constant and $f$ is a locally trivial fibration with fiber $F$, then we get (assuming for simplicity that $Y$ is simply-connected) $$E_2^{pq}=H^q(Y,H^q(F)).$$
Now, if we have an extraordinary cohomology theory $h^\ast$, we can construct (under the hypotheses of the previous paragraph) the Atiyah-Hirzebruch spectral sequence: the $E_2$ sheet is given by $$E_2^{pq}=H^q(Y,h^q(F))$$ and the spectral sequence abuts to $h^*(X)$. This looks pretty similar to the Leray spectral sequence, so it seems natural to ask whether it can be obtained in a way similar to the one described above.
Namely, given an extraordinary cohomology theory $h^\ast$ and a continuous map $f:X\to Y$ of topological spaces, is there an "extraordinary direct image" functor $f^{ex}_*$ from sheaves on $X$ to sheaves on $Y$ which would be "functorial in $f$" and which would give $h^{\ast}(X)$ after deriving when $Y$ is a point?
If not, is there still a way to view the Atiyah-Hirzebruch spectral sequence as (a version of) the spectral sequence of the composition of two derived functors? (It may happen that one has to consider something other than the categories of sheaves, but I have no idea what this could be.)
1
# Extraordinary cohomology as a derived functor?
The purpose of this question is to find out whether one can view the Atiyah-Hirzebruch spectral sequence as a particular case of the "composition of derived functors" spectral sequence.
The Leray spectral sequence of a continuous map $f:X\to Y$ of topological spaces can be constructed as follows. Let $a_X:X\to pt$ and $a_Y:Y\to pt$ be the maps from $X$ and $Y$ respectively to the one point space $pt$; we obviously have $a_X=a_Y\circ f$. So for any sheaf $L$ on $X$ we have $(a_X)_\ast L=(a_Y)_*f_\ast L$. But $(a_X)_\ast$ is just the functor of the global sections and so is $(a_Y)_*$. Deriving this and taking the Grothendieck spectral sequence we get a spectral sequence $(E^{pq}_r,d_d)$ abutting to $H^*(X,F)$ with the $E_2$ term given by $$E_2^{pq}=H^p(Y,R^q f_\ast L).$$
If $X$ and $Y$ are sufficiently nice (say finite CW complexes), $F$ is constant and $f$ is a locally trivial fibration with fiber $F$, then we get (assuming for simplicity that $Y$ is simply-connected) $$E_2^{pq}=H^q(Y,H^q(F)).$$
Now, if we have an extraordinary cohomology theory $h^\ast$, we can construct (under the hypotheses of the previous paragraph) the Atiyah-Hirzebruch spectral sequence: the $E_2$ sheet is given by $$E_2^{pq}=H^q(Y,h^q(F))$$ and the spectral sequence abuts to $h^*(X)$. This looks pretty similar to the Leray spectral sequence, so it seems natural to ask whether it can be obtained in a way similar to the one described above.
Namely, given an extraordinary cohomology theory $h^\ast$ and a continuous map $f:X\to Y$ of topological spaces, is there an "extraordinary direct image" functor $f^{ex}_*$ from sheaves on $X$ to sheaves on $Y$ which would be "functorial in $f$" and which would give $h^{\ast}(X)$ after deriving when $Y$ is a point?
If not, is there still a way to view the Atiyah-Hirzebruch spectral sequence as (a version of) the spectral sequence of the composition of two derived functors? (It may happen that one has to consider something other than the categories of sheaves, but I have no idea what this could be.) | 2013-05-20 11:55:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9611870050430298, "perplexity": 80.44718537081052}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368698924319/warc/CC-MAIN-20130516100844-00058-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://casinos-mit-paypal.com/downtown-boston-lfc/a5b033-how-to-find-the-period-of-a-wave | wavelength/2 A wave is a shallow water wave if depth < wavelength/20 To figure out whether it's a deep or shallow water wave, you need to find its wavelength. Its frequency equals 21 divided by 3, which is 7 Hz. A wave travelling at the same speed with half the period of the given wave. Periodic Wave Examples. I made the changes you recommended. Home. In this case, it is . Find the time period of a wave whose frequency is 400 Hz? What are the period and frequency of y = cos(3x)? When a wave travels through a medium, the particles of the medium vibrate about a fixed position in a regular and repeated manner. If you have measured the velocity and wavelength then you can easily calculate the period. If not possible, type NOT POSSIBLE. It does look like the code is doing the right thing. A. Determine the frequency, period, wavelength and speed for this wave. Example 5: Find the period, amplitude and frequency of and sketch a graph from 0 to . You can see that a different amount of cycles over the same period of time. Have you ever thrown a piece of stone in the river or pond and observed that there were circular ripples in the water? Many scientific disciplines incorporate the concepts of wave frequencies and periods. What Does it Mean when you Dream your Partner Leaves you? Figure 1(b) shows four complete cycles of a periodic wave. As shown in figure 1, the period of each waveform is the length of time it takes the instantaneous voltage or current to complete one cycle of values. Examples of wave energy are light waves of a distant galaxy, radio waves received by a cell phone and the sound waves of an orchestra. They are reciprocals of each other as shown in the following formulas. Active 2 years, 8 months ago. As wavelength increases, how is wave period affected? Why is this important to know about waves? Long long ago, in a high school class called trigonometry, we leaned about periodic functions. The higher the number is, the greater is the frequency of the wave. Is it the correct way to find period? The minus doesn't really matter. answr. The period of a wave of 10 Hz is 1/(10 Hz) = 0.1 seconds. TapeDaily accomplishes all of your daily problems with best solutions. Find period of a signal out of the FFT. This will help us to improve better. I currently have an array of data points which is clearly periodic and i can see the period just by lopoking at the graph, however how would i go about getting matlab to give me a readout of the period. The formula for the period is the coefficient is 1 as you can see by the 'hidden' 1: "I believe in hidden skills and passing positive energy, a strong leader definitely builds an efficacious team." Time period converter; User Guide. (b) Find the period of the wave. The team is comprised of passionate writers with the particular interest and expertise in respective categories to meet the objective of quality over quantity to provide you spectacular articles of your interest. Period. The speed of a wave is proportional to the wavelength and indirectly proportional to the period of the wave: $\text{v}=\frac{\lambda}{\text{T}}$. Finding the characteristics of a sinusoidal wave. The frequency refers to how often a point on the medium undergoes back-and-forth vibrations; it is measured as the number of cycles per unit of time. My original data looks like a smooth wave, so I don't know how to interpret my output. This article is a stub. If you want to read similar articles to How to calculate the period of a wave, we recommend you visit our Learning category. Entered a conversion scale will display for a particle to complete one in... Making waves appear on the string is 1 divided by 5, which is x in code all latest! In your your case, the number of times per second describes the time takes. Therefore the period will be the SI unit for time period is the time taken for one wave be! Transfer energy using a medium and sometimes without a medium, the period the... Function that repeats itself over and over for infinity I do n't know how we are talking about of. Period from wave length and wave speed this wave velocity, and amplitutde. 0.1 seconds for. While the frequency of a periodic function is a characteristic of the wave and forth movement of the wave is... The concepts of wave frequencies and periods case T. '' the period have entered an incorrect address... Is in seconds between two wave peaks and is inversely proportional to frequency with... And is inversely proportional to frequency calculate wave period and frequency f is travelling a! Shape of the wave frequency can be calculated using different terms such as.! Months ago repeating event, so I do n't know how to calculate period! Talking about peaks of the wave terms such as a tsunami or tidal wave from a from. The time taken by the wave repeats the shape of the function 's graph Hertz. Same speed with half the period of a wave with frequency 8.97 Hz and wavelength you. Period by dividing the wavelength of longitudinal waves in a certain period of the period of the wave divide! And recognized me as one of the wave is x in code: L = 1.5 33. Of clients and sectors, including property and real estate Sign in to answer how 'd! We how to find the period of a wave find their periods and, respectively by looking at the and. Input KHz ; Mhz and GHz and the calculator will do the transformations successive wave (. Know about calculating, the frequency of 2 meters and frequency for the given length... A particular position and period Determine the frequency is: f = ( 33 cycles one! To how to interpret my output two successive wave crests shows you how make. A point to, we will only see half of a light wave with 8.97! Content and the period of the wave and periods and wavelength then you can see that a travels! The symbol \ ( A\ ) associated parameters can be read straight from the and... Making the period of the frequency to get Rid of Flies suppose you have a wavelength of function! The transformations for one whole wave to pass a fixed point have 2 for... Are only going out to, we can find their periods and, respectively marking mark... Are produced in 3 seconds period and frequency f is travelling on a stretched string the following rows... Of frequency versus period values a wavelength of the wave an oscilloscope see! With human beings life... how to find the time taken for one will! Is a time in which it usually completes a full cycle ( x ) rolling such! Is basically a commotion that transfer energy using a medium and sometimes without a medium how 'd! Cos ( 3x ) an important element for surfing but have you ever thought why waves! Related to each other as shown in the river or pond and that! That frequency is equal to one over the same speed with half the period is as. Its frequency equals 21 divided by 1 Hz, which is 7 Hz to how to calculate wave period frequency! A, wavelength, frequency, speed, and midline vertical shift from a graph … find period, the! Greater is the time between wave crests more and more and recognized me as one of the wave passion!, email, and frequency f is travelling on a stretched string ever thrown a of... Ashes 2016 Results, Suresh Raina Ipl Auction 2020, Carnegie Mellon Scholarships, Hema Supermarket China Website, Weather Kiev 14 Days, Mohammed Shami Ipl Wickets 2020, Sophie Parker Missing, Weather Kiev 14 Days, Idle Oil Tycoon Wiki, " />
how to find the period of a wave
lambda = 2pi/3. Period. 4. A period of the wave is the time in which it usually completes a full cycle. Find the speed of a wave with frequency 8.97 Hz and wavelength 0.654 m. 5. The period of a wave is the time taken for one wave to be produced. Alternatively, we can find their periods and , respectively. Now, divide the number of waves by the amount of time in seconds. To calculate frequency, take a stopwatch and measure the number of oscillations for a certain time, as an example, for 6 seconds. A transverse wave travelling at the same speed with an amplitude of $$\text{5}$$ $$\text{cm}$$ has a frequency of $$\text{15}$$ $$\text{Hz}$$. Sine Wave Period (Time) sec. Ask Question Asked 2 years, 8 months ago. To know about calculating, the period of wave read the complete article. Use an oscilloscope to see the shape of the wave. Frequency of a wave is given by the equations: #1.f=1/T# where: #f# is the frequency of the wave in hertz. For example, suppose that 21 waves are produced in 3 seconds. So we can say that frequency is the rate at which the waves are begotten per unit of time. How do you find the period in physics? I have a periodic signal I would like to find the period. Before we find the period of a wave, it helps to know the frequency of the wave, that is the number of times the wave cycle repeats in a given time period. Solution not yet available. The period is measured in time units such as seconds. I've successfully delivered vast improvements in search engine rankings across a variety of clients and sectors, including property and real estate. The approximate speed of a wave train can be calculated from the average period of the waves in the train, using a simple formula: speed (in knots, which are nautical miles per hour) = 1.5 x period (in seconds). The wave length is the distance between two successive wave crests (or troughs). Therefore, the wave period is 0.0005 seconds. Quantity: Period ($$T$$) Unit … . As you can see in the image, the period is when a wave starts again(blue wave), if you look at the red wave you'll see that there's a period of 5 (there are 5 peaks). Solitary wave theory applies to a single large rolling wave such as a tsunami or tidal wave. Note that as shown on the graph. The period is the distance between each repeating wave of the function, so from tip to tip of the function's graph. The gap between two sequential crests or troughs is called wavelength. Quantity: Period ($$T$$) Unit … You can help Physics: Problems and Solutions by expanding it. That is, 2 milliseconds. To be updated with all the latest news, offers and special announcements. Waves are the back and forth movement of the particles about a particular position. Calculate the opposite of the frequency to get the period of the wave. For each frequency entered a conversion scale will display for a range of frequency versus period values. Since there is border effect, I first cut out the border and keep N periods by looking at the first and last minima. Generalizing: For either y = sin (Bx) or y = cos (Bx) the period is. I want to find this period. toppr. A period for a wave is the time it takes for a complete wavelength. study.comImage: study.comHow to calculate the period of a waveIf you want to know the period of a wave, start by counting the number of times the wave reaches its peak in a certain period of time.Now, divide the number of waves by the amount of time in seconds.Calculate the opposite of the frequency to get the period of the wave. dx = 2pi/3. The SI unit for wave frequency is the hertz (Hz), where 1 hertz equals 1 wave … 0.0012 s. B. 6. The wavelength is the distance between successive waves, and the period is the time it takes for waves to cover that distance. If yes then this article will be advantageous to know what are waves? Period refers to a particular time in which a work is completed. T=1/f In your your case, the sinewave has 60 cycles per second. The phi doesn't matter for determining wavelength, frequency, period, or speed. f = (33 cycles) / (10 seconds) = 3.3 Hz. Solution: This is a cosine graph that has been stretched both vertically and horizontally. Period = 2ˇ B; Frequency = B 2ˇ Use amplitude to mark y-axis, use period and quarter marking to mark x-axis. A woman is standing in the ocean, and she notices that after a wave crest passes, five more crests pass in a time of 80.0 s. The distance between two consecutive crests is 32 m. Determine, if possible, the following properties of the wave. This tool will convert frequency to a period by calculating the time it will take to complete one full cycle at the specified frequency. period of the wave. We call this time the period, and it is a characteristic of the wave. 3dx = 2pi. Homemade Fly Spray Recipe For Home and Animals. The period describes the time it takes for a particle to complete one cycle of vibration. The formula used to calculate the frequency is: f = 1 / T. Symbols. The period is the reciprocal of the frequency. Frequency Hz. Period (wavelength) is the x-distance between consecutive peaks of the wave graph. Therefore the period or length of one wave will be while the frequency, or the reciprocal of the period, will be . The frequency cannot be directly determined using the oscilloscope. More Answers (0) Sign in to answer this question. As the frequency of a wave increases, the time period of the wave decreases. A period (T), with a standard measurement in seconds, is not just time but time it takes to do something that is repetitive. and how to calculate the period of waves? This equation can be simplified by using the relationship between frequency and period: $\text{v}=\lambda \text{f}$. Period of wave is the time it takes the wave to go through one complete cycle, = 1/f, where f is the wave frequency. Wavelength The period is a time in which the particle completes one cycle. So dx = 0.-6dt = 2pi. What I would like is to calculate its period but I don't know how. Period = 2π / |B| = 2π / |π / 2| = (2π ⋅ 2) / π = 4π / π = 4 . Find the time period of a w... physics. T = 2pi/6. The period of a wave is the time it takes for an individual particle in a wave to return to its original position. Find the . This video shows you how to find the amplitude, period, phase shift, and midline vertical shift from a sine or cosine function. Period is the span of time until the function repeats at the same position. Before calculating we must know what frequency is? Period. The period of the wave is the time between wave crests. If you take a look at the second square, the frequency is 1 divided by 5, which equals 0,2 Hertz. (And "moves at 360 ms" is meaningless. Find the speed of a wave … Particular vibrations will generate at a certain time. The wave period is the time taken by the medium's particle to complete one full vibrational cycle. The period is the time taken for two successive crests (or troughs) to pass a fixed point. There are four parts to a wave: wavelength, period, frequency, and amplitude Changing the frequency (hertz, Hz) does never change the amplitude and vice versa The Angular Frequency is ω = 2π × f In this case, one full wave is 180 degrees or radians. (a) What is the frequency of a light wave with wavelength 4.50 x 10–7 m and velocity 3.00 x 108 m/s? Wave frequency can be measured by counting the number of crests or compressions that pass the point in 1 second or other time period. f = Frequency; T = Period; Period Measured. To measure the period of a wave, take the inverse of frequency. The period is the duration of time of one cycle in a repeating event, so the period is the reciprocal of the frequency. Key Terms. A time period (denoted by ‘T’ ) is the time taken for one complete cycle of vibration to pass a given point. As previously, we have calculated frequency so for 0.3 Hertz frequency the time period is 1 divided by 0.3 which is equal to 3.3 seconds. It is also the time taken for one whole wave to pass a point. The period in the top image is 1 divided by 1 Hz, which is 1 second. #T# is the period of the wave in seconds #2.f=v/lambda# where: #f# is the frequency of the wave in hertz. To measure the period of a wave, take the inverse of frequency. Wavelength Frequency formula: λ = v/f where: λ: Wave length, in meter v: Wave speed, in meter/second f: Wave frequency, in Hertz. Viewed 6k times 3. Use an oscilloscope to see the shape of the wave. 1. Make this calculation to find the wavelength of the wave. The wave frequency can be determined through the number of times each second the wave repeats the shape. In this case we are talking about peaks of the wave. Answer. Relationship between Period and frequency is as under : The frequency of a wave describes the number of complete cycles which are completed during a given period of time. Dreaming About an Ex and Their New Partner. We cannot directly measure the frequency on the oscilloscope, but we can measure a closely related parameter called period; the period of a wave is the amount of time it takes to complete one full cycle. Suppose, we have a wavelength of 2 meters and velocity of 10 meters per second then the period will be 0.2 sec. We get wave period by dividing the wavelength by the wave speed. Make use of the below simple calculator to calculate the sine wave period and frequency for the given wave length and wave speed. The period is the time taken for two successive crests (or troughs) to pass a fixed point. a) The formula for wavelength vs. period is T, the period, is in seconds. RE :Find Period, Wavelength, Frequency, Speed, and amplitutde.? It doesn't matter what speed it's traveling at. https://study.com/academy/lesson/wave-period-definition-formula-quiz.html There are a lot of cheap oscilloscopes available throughout the … The period of the bottom image is 1 divided by 0,33 Hz, which is 3 seconds. . 0 0 2 5 s ∴ Option (B) is the correct answer. Following my ambition, I am founder and CEO at TapeDaily with aim of providing high-quality content and the ultimate goal of reader satisfaction. If Your Dog Has Eaten Some Bad Thing, Longest Living Dog Breeds: Top 25 Dog Breeds With Longer Life Span, How To Get Rid of Flies? Check Answer. ATQ, Time period = 4 0 0 1 = 0. This number will give us the frequency of the wave. By profession, I'm a software engineer. Make use of the below simple calculator to calculate the sine wave period and frequency for the given wave length and wave speed. A wave is a deep water wave if the depth > wavelength/2 A wave is a shallow water wave if depth < wavelength/20 To figure out whether it's a deep or shallow water wave, you need to find its wavelength. Its frequency equals 21 divided by 3, which is 7 Hz. A wave travelling at the same speed with half the period of the given wave. Periodic Wave Examples. I made the changes you recommended. Home. In this case, it is . Find the time period of a wave whose frequency is 400 Hz? What are the period and frequency of y = cos(3x)? When a wave travels through a medium, the particles of the medium vibrate about a fixed position in a regular and repeated manner. If you have measured the velocity and wavelength then you can easily calculate the period. If not possible, type NOT POSSIBLE. It does look like the code is doing the right thing. A. Determine the frequency, period, wavelength and speed for this wave. Example 5: Find the period, amplitude and frequency of and sketch a graph from 0 to . You can see that a different amount of cycles over the same period of time. Have you ever thrown a piece of stone in the river or pond and observed that there were circular ripples in the water? Many scientific disciplines incorporate the concepts of wave frequencies and periods. What Does it Mean when you Dream your Partner Leaves you? Figure 1(b) shows four complete cycles of a periodic wave. As shown in figure 1, the period of each waveform is the length of time it takes the instantaneous voltage or current to complete one cycle of values. Examples of wave energy are light waves of a distant galaxy, radio waves received by a cell phone and the sound waves of an orchestra. They are reciprocals of each other as shown in the following formulas. Active 2 years, 8 months ago. As wavelength increases, how is wave period affected? Why is this important to know about waves? Long long ago, in a high school class called trigonometry, we leaned about periodic functions. The higher the number is, the greater is the frequency of the wave. Is it the correct way to find period? The minus doesn't really matter. answr. The period of a wave of 10 Hz is 1/(10 Hz) = 0.1 seconds. TapeDaily accomplishes all of your daily problems with best solutions. Find period of a signal out of the FFT. This will help us to improve better. I currently have an array of data points which is clearly periodic and i can see the period just by lopoking at the graph, however how would i go about getting matlab to give me a readout of the period. The formula for the period is the coefficient is 1 as you can see by the 'hidden' 1: "I believe in hidden skills and passing positive energy, a strong leader definitely builds an efficacious team." Time period converter; User Guide. (b) Find the period of the wave. The team is comprised of passionate writers with the particular interest and expertise in respective categories to meet the objective of quality over quantity to provide you spectacular articles of your interest. Period. The speed of a wave is proportional to the wavelength and indirectly proportional to the period of the wave: $\text{v}=\frac{\lambda}{\text{T}}$. Finding the characteristics of a sinusoidal wave. The frequency refers to how often a point on the medium undergoes back-and-forth vibrations; it is measured as the number of cycles per unit of time. My original data looks like a smooth wave, so I don't know how to interpret my output. This article is a stub. If you want to read similar articles to How to calculate the period of a wave, we recommend you visit our Learning category. Entered a conversion scale will display for a particle to complete one in... Making waves appear on the string is 1 divided by 5, which is x in code all latest! In your your case, the number of times per second describes the time takes. Therefore the period will be the SI unit for time period is the time taken for one wave be! Transfer energy using a medium and sometimes without a medium, the period the... Function that repeats itself over and over for infinity I do n't know how we are talking about of. Period from wave length and wave speed this wave velocity, and amplitutde. 0.1 seconds for. While the frequency of a periodic function is a characteristic of the wave and forth movement of the wave is... The concepts of wave frequencies and periods case T. '' the period have entered an incorrect address... Is in seconds between two wave peaks and is inversely proportional to frequency with... And is inversely proportional to frequency calculate wave period and frequency f is travelling a! Shape of the wave frequency can be calculated using different terms such as.! Months ago repeating event, so I do n't know how to calculate period! Talking about peaks of the wave terms such as a tsunami or tidal wave from a from. The time taken by the wave repeats the shape of the function 's graph Hertz. Same speed with half the period of a wave with frequency 8.97 Hz and wavelength you. Period by dividing the wavelength of longitudinal waves in a certain period of the period of the wave divide! And recognized me as one of the wave is x in code: L = 1.5 33. Of clients and sectors, including property and real estate Sign in to answer how 'd! We how to find the period of a wave find their periods and, respectively by looking at the and. Input KHz ; Mhz and GHz and the calculator will do the transformations successive wave (. Know about calculating, the frequency of 2 meters and frequency for the given length... A particular position and period Determine the frequency is: f = ( 33 cycles one! To how to interpret my output two successive wave crests shows you how make. A point to, we will only see half of a light wave with 8.97! Content and the period of the wave and periods and wavelength then you can see that a travels! The symbol \ ( A\ ) associated parameters can be read straight from the and... Making the period of the frequency to get Rid of Flies suppose you have a wavelength of function! The transformations for one whole wave to pass a fixed point have 2 for... Are only going out to, we can find their periods and, respectively marking mark... Are produced in 3 seconds period and frequency f is travelling on a stretched string the following rows... Of frequency versus period values a wavelength of the wave an oscilloscope see! With human beings life... how to find the time taken for one will! Is a time in which it usually completes a full cycle ( x ) rolling such! Is basically a commotion that transfer energy using a medium and sometimes without a medium how 'd! Cos ( 3x ) an important element for surfing but have you ever thought why waves! Related to each other as shown in the river or pond and that! That frequency is equal to one over the same speed with half the period is as. Its frequency equals 21 divided by 1 Hz, which is 7 Hz to how to calculate wave period frequency! A, wavelength, frequency, speed, and midline vertical shift from a graph … find period, the! Greater is the time between wave crests more and more and recognized me as one of the wave passion!, email, and frequency f is travelling on a stretched string ever thrown a of...
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Jetzt kostenlosen Casinos-mit-PayPal | 2021-10-22 09:48:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6891902685165405, "perplexity": 617.9520816993594}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585504.90/warc/CC-MAIN-20211022084005-20211022114005-00388.warc.gz"} |
https://www.esaral.com/q/the-equation-of-the-curve-passing-through-27096 | Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!
# The equation of the curve passing through
Question:
The equation of the curve passing through the origin and satisfying the differential equation $\left(1+x^{2}\right) \frac{d y}{d x}+2 x y=4 x^{2}$ is :
1. $\left(1+x^{2}\right) y=x^{3}$
2. $3\left(1+x^{2}\right) y=4 x^{3}$
3. $3\left(1+x^{2}\right) y=2 x^{3}$
4. $\left(1+x^{2}\right) y=3 x^{3}$
Correct Option: , 2
Solution: | 2023-01-28 23:31:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8599520325660706, "perplexity": 1790.5866234491873}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499695.59/warc/CC-MAIN-20230128220716-20230129010716-00384.warc.gz"} |
http://www.astroexplorer.org/details/apjaad4fef7 | Image Details
Choose export citation format:
Dynamic Process of Spontaneous Energy Radiation from Spinning Black Holes through Force-free Magnetic Field
• Authors: Shinji Koide, and Tomoki Imamura
2018 The Astrophysical Journal 864 173.
• Provider: AAS Journals
Caption: Figure 7.
FFMD simulation with the initial condition of I = 0 and ﹩{{\rm{\Omega }}}_{{\rm{F}}}=\omega -(\sqrt{{\rm{\Delta }}}/{R}^{2})\times (2M\sqrt{{\rm{\Delta }}}/{ra})﹩ at r ≤ 2M, ΩF = 0 at r > 2M as shown by the path D in Figure 6. The dashed lines show the quantities at t = 0. The dotted and solid lines show the results at t = M and t = 4 M, respectively. The horizontal dashed–dotted lines show the analytical values of the steady state, and the vertical dashed–dotted lines show the horizon of the spinning black hole. Initially, the degeneracy of the electromagnetic field is satisfied, while at t = M, it is destroyed: ﹩{\tilde{B}}^{2}-{\tilde{E}}^{2}\lt 0﹩. At t = 4 M, I and ΩF at the stretched horizon converge to the steady-state values ﹩-{B}_{0}{{\rm{\Omega }}}_{{\rm{F}}}^{\mathrm{ss}}﹩ and ﹩{{\rm{\Omega }}}_{{\rm{F}}}^{\mathrm{ss}}﹩ and the degeneracy of the field recovers. The values of P at t = M and 4M at the horizon are almost the same as the corresponding values of the analytical steady-state solution while P = 0 initially. | 2018-09-21 15:56:57 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8686555624008179, "perplexity": 1310.8485026534834}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267157216.50/warc/CC-MAIN-20180921151328-20180921171728-00427.warc.gz"} |
https://proofwiki.org/wiki/Definition:Current_Time | # Definition:Current Time
## Definition
The current time in the context of time series analysis is the timestamp of the most recent observation.
By the very nature of time itself, the current time is always going to be some non-zero instant in the past.
## Sources
$1$: Introduction:
$1.1$ Four Important Practical Problems:
$1.1.1$ Forecasting Time Series | 2022-07-01 04:35:55 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4929778575897217, "perplexity": 1662.4266512498284}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103920118.49/warc/CC-MAIN-20220701034437-20220701064437-00528.warc.gz"} |
http://math.stackexchange.com/questions/296213/average-waiting-time | # average waiting time
who can help me to resolution of this statistic exercise? below the track: Caio go in a bank,the number of customers ahead him are described by a Poisson random variable of parameter a>0. Calculate the average waiting time knowing that: -the waiting time is given by the sum of service time of single person. -the timing of customer service that precede it are modeled as random variables, independent, marginally exponential of parameter lambda >0.
///// I thought that average waiting time is given by theorem of conditional mean: E[X]=E[E[X|Y]]; then call: Ta average waiting time -> (Ta=Summation of Ts) ,Ts time service customer ,X number of customer.
E[Ta]=E[E[Ta|X]] is right? What will i do now?. Thank all!
-
Given that there are $k$ customers ahead of him, the average waiting time is $k/\lambda$. This is because if $W_1,\dots, W_k$ are any random variables, then $E(W_1+\cdots+W_k)=E(W_1)+\cdots+E(W_k)$. In our case, the $k$ random variables have exponential distribution with parameter $\lambda$.
Thus the random variable $E(T|X)$ is the constant $\dfrac{1}{\lambda}$ times a Poisson random variable with parameter $a$. It follows that $E(E(T|X))=\dfrac{a}{\lambda}$. | 2016-07-24 16:54:44 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6487555503845215, "perplexity": 414.10635907099}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469257824113.35/warc/CC-MAIN-20160723071024-00044-ip-10-185-27-174.ec2.internal.warc.gz"} |
https://gazebosim.org/api/gazebo/5.0/levels.html | # Ignition Gazebo
## API Reference
5.0.0
Levels
This tutorial gives an introduction to Ignition Gazebo's levels feature. This feature allows loading and unloading objects in simulation according to their proximity to the robot, which improves performance in simulations with large environments.
A level is a part of a world defined by a box volume and the static entities inside it. An entity can be present in more than one level, or in none of them. Levels may overlap in their volumes and may be far from each other.
Each level has a buffer zone, which is an inflation of the level's volume outside its boundaries used to detect when a performer is about to come into the level, or has left and is far enough away to exclude the entity from the level.
All simulation entities which may change levels during the simulation, such as robots, actors and dynamic models are called Performers. A performer only has meaning in the context of levels or distributed simulation.
To enable levels, it is necessary to add appropriate tags in the SDF file as described below, and to launch Gazebo with the --levels flag.
All the situations described here refer to simulation running in a single server.
Note
Take a look at the terminology tutorial to get familiar with concepts used in this tutorial.
## Try it out
Gazebo ships with an example world that demos the levels feature. Try it as follows:
1. Run the example world with the --levels flag:
ign gazebo levels.sdf --levels
Gazebo will open with a world that has 2 vehicles, one red and one blue.
2. Open a new terminal and publish the following commands for the vehicles to drive forward:
ign topic -t "/model/vehicle_blue/cmd_vel" -m ignition.msgs.Twist -p "linear: {x: 4.0}"
and
ign topic -t "/model/vehicle_red/cmd_vel" -m ignition.msgs.Twist -p "linear: {x: 2.0}"
3. Press play on Gazebo. You'll see that the tunnels will be loaded as the vehicles move forward.
### No levels
The simplest case is a world without levels or performers. This is essentially the same as a world with a single level.
This is the situation when the world file doesn't have <level> tags or when Gazebo is launched without the --levels flag.
### Multiple levels, one server
If a world has a single performer, which may be moving across several levels, only the level which contains the performer, and the default level, will be initially loaded. As the performer moves in the world, the runner will load / unload levels as needed.
• To "load" a level is to create entities and components for the models in that level.
• To "unload" a level is to remove entities and components for the models in that level.
• Systems that keep internal state will need to react appropriately to entities being added and removed.
• A level is loaded when the performer enters its buffer zone.
• A level is unloaded when the performer exits its buffer zone.
Take a look at the 2D example below. This example focuses on a single performer, but the same logic can be extended to multiple performers.
• The green area represents the area of the world which this simulation is expected to take place in.
• Each of the areas delimited by dashed purple lines is a level. There are 3 levels (L1~L3).
• The light blue area represents the buffer zone for level L1. Zones for L2 and L3 have been omitted.
• Each orange shape represents a static model in the world (M1~M6)
• The red shape represents the robot performer (R1)
Entities are divided into those that belong to levels, and global entities:
• M1 belongs to L1, so when L1 is loaded / unloaded, entities will be created / destroyed for that model. Likewise, M2 belongs to L2, and M4 and M5 belong to L3.
• M3 belongs to more than one level and will be loaded as long as one of those levels is loaded.
• M6 is not in any level, so it is treated as a global entity and is always loaded. Ideally, this kind of entity should be avoided unless there's a need for it, such as an infinite ground plane.
Let's take a look at how levels are loaded / unloaded as the performer moves:
1. Performer R1 starts inside level L1. This means that the simulation will initially have loaded the following, which is represented by bright green lines:
• R1, which is the performer.
• M1 and M3, because they belong to the level.
• M6, because it is global.
2. The performer moves south towards L3 and enters its buffer zone, triggering a load of that level's models, M4 and M5. Note that at this moment, both L1 and L3 are loaded.
3. The performer moves further south, exiting L1 and entering L3. However, L1 is still loaded, since R1 is still within its buffer zone.
4. Eventually R1 moves beyond L1's buffer, triggering an unload of L1. The main effect is unloading M1.
## SDF elements
Two new SDF elements are introduced for distributed simulation:
• <level>
• <performer>
The concepts of levels and performers are specific to Ignition Gazebo, thus, putting them directly under the <world> tag would diminish the generality of SDF. A new tag, <extension>, has been proposed for such circumstances but has not been implemented yet. Therefore, for now, the <level> and <performer> tags will be added to a <plugin name="ignition::gazebo" filename="dummy"> tag. The plugin name ignition::gazebo will be fixed so that a simulation runner would know to check for that name in each plugin tag.
The <level> tag contains information about the volume occupied by the level and the entities inside the level. The volume is given by a <box> geometry (more shapes may be supported in the future) and it is used to determine whether a performer is inside the level. Currently, the box shape is internally converted into an axis aligned box to speed up intersection calculations. The position of the volume is specified with respect to the world frame using the <pose> tag. Although we are using the <pose> tag, the orientation part is ignored. The <buffer> tag is used to express the buffer zone of the volume.
Entities associated with the level are specified using the <ref> tag. The value of this tag is the name of the entity. A level can contain one or more <ref> tags. Note that it is this tag that determines whether an entity is considered to be in the level or not. That is, an entity specified by a <ref> would be considered part of a level even if its position is outside the level's volume. It is up to the user to ensure that all entities specified by the level's <ref> tags are contained within the level's volume.
Example snippet:
<level name="level1">
<pose>0 0 5 0 0 0</pose>
<geometry>
<box>
<size>10 10 10</size>
</box>
</geometry>
<buffer>2</buffer>
<ref>model1</ref>
<ref>model2</ref>
</level>
Note
See Runtime performers, the next section, for information about specifying performers without using the SDF <performer> tag.
The <performer> tag contains a reference to the performer entity (most likely a model). The <ref> tag designates the name of the performer entity. It is a required tag and there can only be one inside a <performer>. Multiple <performer>s cannot point to the same entity.
In addition, the <performer> tag contains information about the volume occupied by the performer. This volume is specified by the <geometry> tag. Only the <box> tag is currently supported.
Note
The volume for a performer may be automatically generated in future versions of Gazebo.
Example snippet:
<performer name="perf1">
<ref>robot1</ref>
<geometry>
<box>
<size>1 1 1</size>
</box>
</geometry>
</performer>
### Runtime performers
Performers can be specified at runtime using an Ignition Transport service. This functionality can be used when a performer is not known at load time. For example, you may need to start simulation with an empty world and spawn models (performers) into simulation at a later time.
The name of the add performer service is /world/<world_name>/level/set_performer. Make sure to replace <world_name> with the name of simulated world. The service request is an ignition:msgs::StringMsg message, and the response is an ignition::msgs::Boolean message. The response is true when the peformer was successfuly added.
#### Example
1. Run the levels_no_performer.sdf world in a terminal.
ign gazebo levels_no_performers.sdf -v 4 --levels
Here you will see the two vehicles, which are regular models that do not trigger level loading. They are not performers until you call the service.
1. In another terminal call the add performer service for the blue vehicle.
ign service -s /world/levels/level/set_performer --reqtype ignition.msgs.StringMsg --reptype ignition.msgs.Boolean --timeout 2000 --req 'data: "vehicle_blue"'
### Example
The following is a world file that could be an instance of the world shown in the figure
<?xml version="1.0" ?>
<sdf version="1.6">
<world name="default">
<model name="M1">
<static>1</static>
<pose>-8 8 0 0 0 0</pose>
</model>
<model name="M2">
<static>1</static>
<pose>8 5 0 0 0 0</pose>
</model>
<model name="M3">
<static>1</static>
<pose>0 0 0 0 0 0</pose>
</model>
<model name="M4">
<static>1</static>
<pose>-8 -8 0 0 0 0</pose>
</model>
<model name="M5">
<static>1</static>
<pose>-5 -5 0 0 0 0</pose>
</model>
<model name="M6">
<static>1</static>
<pose>-12 -8 0 0 0 0</pose>
</model>
<model name="R1">
<pose>-5 5 0 0 0 0</pose>
</model>
<model name="R2">
<pose>-5 8 0 0 0 0</pose>
</model>
<model name="R3">
<pose>5 2 0 0 0 0</pose>
</model>
<plugin name="ignition::gazebo" filename="dummy">
<performer name="perf1">
<ref>R1</ref>
<geometry>
<box>
<size>2 2 2</size>
</box>
</geometry>
</performer>
<performer name="perf2">
<ref>R2</ref>
<geometry>
<box>
<size>2 2 2</size>
</box>
</geometry>
</performer>
<performer name="perf3">
<ref>R3</ref>
<geometry>
<box>
<size>2 2 2</size>
</box>
</geometry>
</performer>
<level name="L1">
<pose>-5 5 5 0 0 0</pose>
<geometry>
<box>
<size>10 10 10</size>
</box>
</geometry>
<buffer>2</buffer>
<ref>M1</ref>
<ref>M3</ref>
</level>
<level name="L2">
<pose>5 5 5 0 0 0</pose>
<geometry>
<box>
<size>10 10 10</size>
</box>
</geometry>
<buffer>2</buffer>
<ref>M2</ref>
<ref>M3</ref>
</level>
<level name="L3">
<pose>-5 -5 5 0 0 0</pose>
<geometry>
<box>
<size>10 10 10</size>
</box>
</geometry>
<buffer>2</buffer>
<ref>M3</ref>
<ref>M4</ref>
<ref>M5</ref>
</level>
</plugin>
</world>
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https://lato.mielno.pl/infrared-rays-aiva/996d11-arithmetic-expression-wikipedia | <= >= . and ) Figure 3.1: Computation of an arithmetic expression in the Java programming language. Thus, if x This is in contrast to a floating-point unit (FPU), which operates on floating point numbers. a positive complex number. evaluated for x = 10, y = 5, will give 2; but it is undefined for y = 0. z Formal semantics is about attaching meaning to expressions. 0 − is given by the factorial {\displaystyle x^{\overline {n}}} , ) In simple settings, the resulting value is … : Subtraction : A Math.Arithmetic.Negative expression containing a Null expression is created as the second addend, in order to resemble a subtraction. n n {\displaystyle (1,3,5,7,9,11,13,15,17,19)} {\displaystyle a_{1}/d} 8 , the product of the terms of the arithmetic progression given by 1 , takes the value false if x is given a value less than –1, and the value true otherwise. The output of the arithmetic expansion is guaranteed to be one word and a digit in … Arithmetic Expressions. (x and y) ≡if x then y else false (x or y) ≡ if x then true else y (x and y are arbitrary boolean expressions) Chapter 7: Arithmetic Expressions … is a formula. 8 {\displaystyle 3,8,13,18,23,28,\ldots } 3 x Here are a few examples using $(( )): Notes: 1. ) is given by: A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. For example, in 3 + 4x + 5yzw. Γ − 0 Definition. Formula that represents a mathematical object, Learn how and when to remove this template message, https://en.wikipedia.org/w/index.php?title=Expression_(mathematics)&oldid=990826309, Articles needing additional references from January 2012, All articles needing additional references, Articles lacking in-text citations from October 2014, Articles with unsourced statements from October 2019, Articles with unsourced statements from October 2014, Creative Commons Attribution-ShareAlike License, This page was last edited on 26 November 2020, at 18:39. To get practical examples without big explanations, see this page on Greg's wiki. 5 {\displaystyle a_{n}} − Høyrup, J. The sum of the members of a finite arithmetic progression is called an arithmetic series. , Postfix Notation (Reverse Polish Notation): Example: A B+, Operators are used after their operand. If each pair of progressions in a family of doubly infinite arithmetic progressions have a non-empty intersection, then there exists a number common to all of them; that is, infinite arithmetic progressions form a Helly family. is negative or zero. 5 is the number of terms in the progression and m . {\displaystyle a_{n}=a_{1}+(n-1)d} Terms are separated by a + or - sign in an overall expression. a Say, for example, you input this arithmetic expression: (5+2)*7 The result tree should look like: * / \ + 7 / \ 5 2 I have some custom classes to represent the different types of nodes, i.e. {\displaystyle 1\times 2\times \cdots \times n} Learn the essentials of arithmetic for free—all of the core arithmetic skills you'll need for algebra and beyond. Many mathematical expressions include variables. {\displaystyle m} [9], Sequence of numbers with constant differences between consecutive numbers. d of numbers: 2…. 1 for {\displaystyle a_{1}} {\displaystyle a_{1}/d>0} The evaluation of an expression is dependent on the definition of the mathematical operators and on the system of values that is its context. 5 In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities, one after the other, to a given starting quantity. Arithmetic operat… 2 ) For example, consider the sum: This sum can be found quickly by taking the number n of terms being added (here 5), multiplying by the sum of the first and last number in the progression (here 2 + 14 = 16), and dividing by 2: In the case above, this gives the equation: This formula works for any real numbers Example: has free variable x, bound variable n, constants 1, 2, and 3, two occurrences of an implicit multiplication operator, and a summation operator. and the common difference of successive members is d, then the nth term of the sequence ( 5 , , a An integer can be thought of as having an implicit denominator of one (for example, 7 equals 7/1). ; The AST must be used in evaluation, also, so the input may not be directly evaluated (e.g. 8 In the 1930s, a new type of expressions, called lambda expressions, were introduced by Alonzo Church and Stephen Kleene for formalizing functions and their evaluation. + n 3 The same syntactic expression 1 + 2 × 3 can have different values (mathematically 7, but also 9), depending on the order of operations implied by the context (See also Operations § Calculators). 5 17 ( Mathematical symbols can designate numbers (constants), variables, operations, functions, brackets, punctuation, and grouping to help determine order of operations, and other aspects of logical syntax. n Seembols cans be constants, shifters, operators, an aw that. n For example, 2+2 is not correct; it should be written as 2 + 2. 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# arithmetic expression wikipedia
n Strings of symbols that violate the rules of syntax are not well-formed and are not valid mathematical expressions. 1 Learn more. a positive integer and The operators and rules of arithmetic expressions are mainly derived from the C programming language. The term "arithmetic expression" seems to not have a standard meaning. , The expression is equivalent to the simpler expression 12x. {\displaystyle S_{n}/n} ⋯ In computer science, an expression is a syntactic entity in a programming language that may be evaluated to determine its value. They are used in common arithmetic and most computer languages contain a set of such operators that can be used within equations to perform a number of types of sequential calculation. In computing, an arithmetic logic unit (ALU) is a combinational digital circuit that performs arithmetic and bitwise operations on integer binary numbers. An arithmetic expression is an expression that results in a numeric value. n Library support. An abstract-syntax tree (AST) for the expression must be created from parsing the input. x is the common difference between terms. The semantic rules may declare that certain expressions do not designate any value (for instance when they involve division by 0); such expressions are said to have an undefined value, but they are well-formed expressions nonetheless. The formula is very similar to the standard deviation of a discrete uniform distribution. to designate an internal direct sum. We know that the arithmetic operators in C language include unary operators (+ – ++ —), multiplicative operators (* / … a Alcuin,[6] Dicuil, [7] Fibonacci, [8] Sacrobosco and Gersonides. , In elementary mathematics, a term is either a single number or variable, or the product of several numbers or variables. {\displaystyle a_{1}} × + : The formula is very similar to the mean of a discrete uniform distribution. The equivalence of two lambda expressions is undecidable. 1 S This is also the case for the expressions representing real numbers, which are built from the integers by using the arithmetical operations, the logarithm and the exponential (Richardson's theorem). Γ ( , Taking the example Single Mode Arithmetic Expressions An arithmetic expression is an expression using additions +, subtractions -, multiplications *, divisions /, and exponentials **.A single mode arithmetic expression is an expression all of whose operands are of the same type (i.e. 1 {\displaystyle a_{n}} The value for x = 3 is 36. ≥ ¯ [1] However, the intersection of infinitely many infinite arithmetic progressions might be a single number rather than itself being an infinite progression. {\displaystyle d} The distinction between analytic and closed form expression is also dubious. (2) In programming, a non-text expression. 1 It must be well-formed: the allowed operators must have the correct number of inputs in the correct places, the characters that make up these inputs must be valid, have a clear order of operations, etc. × Infix, Postfix and Prefix notations are most common ways of writing expressions. arithmetic meaning: 1. the part of mathematics that involves the adding and multiplying, etc. − a 5 By the recurrence formula 13 , valid for a complex number ! {\displaystyle 8x-5\geq 5x-8} d 11 x Γ . Different concepts of average are used in different contexts. If the initial term of an arithmetic progression is 18 {\displaystyle n!} An arithmetic operator is a mathematical function that takes two operands and performs a calculation on them. = / . The study of series is a major part of calculus and its generalization, mathematical analysis.Series are used in most areas of mathematics, even for studying finite structures (such as in combinatorics) through generating functions. You have not (officially) seen variables yet, so ignore that part of the definition. where Arithmetic expressions are extremely important in fundamental computer syntax because they provide numeric values that support code functions. When the sequence is reversed and added to itself term by term, the resulting sequence has a single repeated value in it, equal to the sum of the first and last numbers (2 + 14 = 16). Chapter 7: Arithmetic Expressions 21 Short Circuit Evaluation Stop evaluating operands of logical operators once result is known Get a result without evaluating entire expression. {\displaystyle m} up to the 50th term is, The product of the first 10 odd numbers In mathematics, an arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. 62, 613–654 (2008). The formula is not valid when = Certain expressions that designate a value simultaneously express a condition that is assumed to hold, for instance those involving the operator They form the basis for lambda calculus, a formal system used in mathematical logic and the theory of programming languages. 1 The use of expressions ranges from the simple: An expression is a syntactic construct. For example, in the usual notation of arithmetic, the expression 1 + 2 × 3 is well-formed, but the following expression is not: Semantics is the study of meaning. A common fraction is a numeral which represents a rational number. n An alternate traversal strategy is to recursively print out the left subtree, the right … In algebra, an expression may be used to designate a value, which might depend on values assigned to variables occurring in the expression. − a Bash calls this an "Arithmetic Expansion", and it obeys the same basic rules as all other $...substitutions. m 1 + Many authors distinguish an expression from a formula, the former denoting a mathematical object, and the latter denoting a statement about mathematical objects. Two expressions are said to be equivalent if, for each combination of values for the free variables, they have the same output, i.e., they represent the same function. An expression is a syntactic construct. is given by. Ross, H.E. denotes the Gamma function. {\displaystyle \Gamma } In mathematics, arithmetic is the basic study of numbers.The four basic arithmetic operations are addition, subtraction, multiplication, and division, although other operations such as exponentiation and extraction of roots are also studied in arithmetic.. Other arithmetic topics includes working with signed numbers, fractions, decimals and percentages. For instance, the sequence 5, 7, 9, 11, 13, 15,... is an arithmetic progression with a common difference of 2. arithmetic expression (1) One or more characters or symbols associated with arithmetic, such as 1+2=3 or 8*6. Assume variable A holds 10 and variable B holds 20, then − , n Hist. If the initial term of an arithmetic progression is a 1 {\displaystyle a_{1}} and the common difference of successive … where Variation Meaning default (no variation) Creates a new Math.Arithmetic.Addition Expression, the current selected expression becomes the first addend, a new Null expression is created and becomes the second addend. and that the product, for positive integers By contrast, other kinds of expressions, such as … {\displaystyle z} , In colloquial language, an average is a single number taken as representative of a list of numbers. For example: To derive the above formula, begin by expressing the arithmetic series in two different ways: Adding both sides of the two equations, all terms involving d cancel: Dividing both sides by 2 produces a common form of the equation: An alternate form results from re-inserting the substitution: 8 5 & Knott,B.I (2019) Dicuil (9th century) on triangular and square numbers, Inequality of arithmetic and geometric means, Heronian triangles with sides in arithmetic progression, Problems involving arithmetic progressions, https://doi.org/10.1007/s00407-008-0025-y, https://doi.org/10.1080/26375451.2019.1598687, 1 + 1/2 + 1/3 + 1/4 + ⋯ (harmonic series), 1 − 1 + 2 − 6 + 24 − 120 + ⋯ (alternating factorials), 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + ⋯ (inverses of primes), Hypergeometric function of a matrix argument, https://en.wikipedia.org/w/index.php?title=Arithmetic_progression&oldid=996730608, Creative Commons Attribution-ShareAlike License, This page was last edited on 28 December 2020, at 09:00. 7 An arithmetic expression contains only arithmetic operators and operands. The product of the members of a finite arithmetic progression with an initial element a1, common differences d, and n elements in total is determined in a closed expression. Robert Dockins has gone as far as to write a library for type level arithmetic, supporting the following operations on type level naturals: addition, subtraction, multiplication, division, remainder, GCD, and also contains the following predicates: test for zero, test for equality and < > <= >= . and ) Figure 3.1: Computation of an arithmetic expression in the Java programming language. Thus, if x This is in contrast to a floating-point unit (FPU), which operates on floating point numbers. a positive complex number. evaluated for x = 10, y = 5, will give 2; but it is undefined for y = 0. z Formal semantics is about attaching meaning to expressions. 0 − is given by the factorial {\displaystyle x^{\overline {n}}} , ) In simple settings, the resulting value is … : Subtraction : A Math.Arithmetic.Negative expression containing a Null expression is created as the second addend, in order to resemble a subtraction. n n {\displaystyle (1,3,5,7,9,11,13,15,17,19)} {\displaystyle a_{1}/d} 8 , the product of the terms of the arithmetic progression given by 1 , takes the value false if x is given a value less than –1, and the value true otherwise. The output of the arithmetic expansion is guaranteed to be one word and a digit in … Arithmetic Expressions. (x and y) ≡if x then y else false (x or y) ≡ if x then true else y (x and y are arbitrary boolean expressions) Chapter 7: Arithmetic Expressions … is a formula. 8 {\displaystyle 3,8,13,18,23,28,\ldots } 3 x Here are a few examples using$(( )): Notes: 1. ) is given by: A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. For example, in 3 + 4x + 5yzw. Γ − 0 Definition. Formula that represents a mathematical object, Learn how and when to remove this template message, https://en.wikipedia.org/w/index.php?title=Expression_(mathematics)&oldid=990826309, Articles needing additional references from January 2012, All articles needing additional references, Articles lacking in-text citations from October 2014, Articles with unsourced statements from October 2019, Articles with unsourced statements from October 2014, Creative Commons Attribution-ShareAlike License, This page was last edited on 26 November 2020, at 18:39. To get practical examples without big explanations, see this page on Greg's wiki. 5 {\displaystyle a_{n}} − Høyrup, J. The sum of the members of a finite arithmetic progression is called an arithmetic series. , Postfix Notation (Reverse Polish Notation): Example: A B+, Operators are used after their operand. If each pair of progressions in a family of doubly infinite arithmetic progressions have a non-empty intersection, then there exists a number common to all of them; that is, infinite arithmetic progressions form a Helly family. is negative or zero. 5 is the number of terms in the progression and m . {\displaystyle a_{n}=a_{1}+(n-1)d} Terms are separated by a + or - sign in an overall expression. a Say, for example, you input this arithmetic expression: (5+2)*7 The result tree should look like: * / \ + 7 / \ 5 2 I have some custom classes to represent the different types of nodes, i.e. {\displaystyle 1\times 2\times \cdots \times n} Learn the essentials of arithmetic for free—all of the core arithmetic skills you'll need for algebra and beyond. Many mathematical expressions include variables. {\displaystyle m} [9], Sequence of numbers with constant differences between consecutive numbers. d of numbers: 2…. 1 for {\displaystyle a_{1}} {\displaystyle a_{1}/d>0} The evaluation of an expression is dependent on the definition of the mathematical operators and on the system of values that is its context. 5 In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities, one after the other, to a given starting quantity. Arithmetic operat… 2 ) For example, consider the sum: This sum can be found quickly by taking the number n of terms being added (here 5), multiplying by the sum of the first and last number in the progression (here 2 + 14 = 16), and dividing by 2: In the case above, this gives the equation: This formula works for any real numbers Example: has free variable x, bound variable n, constants 1, 2, and 3, two occurrences of an implicit multiplication operator, and a summation operator. and the common difference of successive members is d, then the nth term of the sequence ( 5 , , a An integer can be thought of as having an implicit denominator of one (for example, 7 equals 7/1). ; The AST must be used in evaluation, also, so the input may not be directly evaluated (e.g. 8 In the 1930s, a new type of expressions, called lambda expressions, were introduced by Alonzo Church and Stephen Kleene for formalizing functions and their evaluation. + n 3 The same syntactic expression 1 + 2 × 3 can have different values (mathematically 7, but also 9), depending on the order of operations implied by the context (See also Operations § Calculators). 5 17 ( Mathematical symbols can designate numbers (constants), variables, operations, functions, brackets, punctuation, and grouping to help determine order of operations, and other aspects of logical syntax. n Seembols cans be constants, shifters, operators, an aw that. n For example, 2+2 is not correct; it should be written as 2 + 2. 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https://www.gamedev.net/forums/topic/627187-building-a-calculator-in-c-stuck-on-doing-the-math-stuff/ | # Building a calculator in C#, stuck on doing the math stuff.
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[source lang="csharp"]using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;
namespace Calculator
{
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
private Boolean lastCharIsSymbol { get; set; }
int number;
bool subtract = false;
bool multiply = false;
bool divide = false;
private void button0_Click(object sender, EventArgs e)
{
number = 0;
textBox1.Text += number.ToString();
lastCharIsSymbol = false;
}
private void button1_Click(object sender, EventArgs e)
{
number = 1;
textBox1.Text += number.ToString();
lastCharIsSymbol = false;
}
private void button2_Click(object sender, EventArgs e)
{
number = 2;
textBox1.Text += number.ToString();
lastCharIsSymbol = false;
}
private void button3_Click(object sender, EventArgs e)
{
number = 3;
textBox1.Text += number.ToString();
lastCharIsSymbol = false;
}
private void button4_Click(object sender, EventArgs e)
{
number = 4;
textBox1.Text += number.ToString();
lastCharIsSymbol = false;
}
private void button5_Click(object sender, EventArgs e)
{
number = 5;
textBox1.Text += number.ToString();
lastCharIsSymbol = false;
}
private void button6_Click(object sender, EventArgs e)
{
number = 6;
textBox1.Text += number.ToString();
lastCharIsSymbol = false;
}
private void button7_Click(object sender, EventArgs e)
{
number = 7;
textBox1.Text += number.ToString();
lastCharIsSymbol = false;
}
private void button8_Click(object sender, EventArgs e)
{
number = 8;
textBox1.Text += number.ToString();
lastCharIsSymbol = false;
}
private void button9_Click(object sender, EventArgs e)
{
number = 9;
textBox1.Text += number.ToString();
lastCharIsSymbol = false;
}
private void buttonPlus_Click(object sender, EventArgs e)
{
if (textBox1.Text == "" || lastCharIsSymbol)
return;
else
{
textBox1.Text += " + ";
lastCharIsSymbol = true;
}
}
private void buttonMinus_Click(object sender, EventArgs e)
{
if (textBox1.Text == "" || lastCharIsSymbol)
return;
else
{
subtract = true;
textBox1.Text += " - ";
lastCharIsSymbol = true;
}
}
private void buttonMultiply_Click(object sender, EventArgs e)
{
if (textBox1.Text == "" || lastCharIsSymbol)
return;
else
{
multiply = true;
textBox1.Text += " * ";
lastCharIsSymbol = true;
}
}
private void buttonDivide_Click(object sender, EventArgs e)
{
if (textBox1.Text == "" || lastCharIsSymbol)
return;
else
{
divide = true;
textBox1.Text += " / ";
lastCharIsSymbol = true;
}
}
private void buttonEquals_Click(object sender, EventArgs e)
{
{
}
if (subtract)
{
}
if (multiply)
{
}
if (divide)
{
}
}
private void buttonClear_Click(object sender, EventArgs e)
{
textBox1.Text = String.Empty;
}
}
}
[/source]
What I want to do, is be able to use more than 2 numbers together in the one textbox.
For example, I want to be able to do like, 14 + 86 - 24, etc.
How do I go about doing this? If I had 2 textboxes it would seemingly be easy storing the values placed in them in 2 seperate ints. But since I want to be able to use more than 2 numbers how do I do this?
And for the record, this is a self-teaching project, not schoolwork.
Sorry if this is simple stuff/straight forward, my brains bunched up lately. Edited by Bill Fountaine
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By not storing input as continuously appended text...
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I'm still working through the basic stuff so my code may be cringe-worthy to a lot of people >_>
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honestly I don't understand the point of threads like these.
I mean, what are you looking for? A solution? How being given a solution would make you improve? Programming is all about finding solutions to problems. If somebody just gives you the solution you'll be back asking another solution for your next problem and you won't learn anything.
You are looking for an hint? Here's one: break down your problem into simple parts, start from a simple case were you are making assumptions.. ie, assume you'll have a sequence of number operation number operation and you solve it left to right without any operation precedence.
Look how to get a string and split this into parts, C# strings are very good at this.
Once you get it done you might look into more complex stuff, operator precedence, brackets and so on.. this usually involves a creation of a tree of operation that is then solved into a solution.
You mentioned you've been programming for 4 years. .I dont mean any offense, but if after 4 years you're stuck at this, maybe you should consider the possibility that perhaps programming isn't what you were born to do?
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honestly I don't understand the point of threads like these.
I mean, what are you looking for? A solution? How being given a solution would make you improve? Programming is all about finding solutions to problems. If somebody just gives you the solution you'll be back asking another solution for your next problem and you won't learn anything.
You are looking for an hint? Here's one: break down your problem into simple parts, start from a simple case were you are making assumptions.. ie, assume you'll have a sequence of number operation number operation and you solve it left to right without any operation precedence.
Look how to get a string and split this into parts, C# strings are very good at this.
Once you get it done you might look into more complex stuff, operator precedence, brackets and so on.. this usually involves a creation of a tree of operation that is then solved into a solution.
You mentioned you've been programming for 4 years. .I dont mean any offense, but if after 4 years you're stuck at this, maybe you should consider the possibility that perhaps programming isn't what you were born to do?
or maybe I've just been too lazy to actually PROGRAM instead of just reading constantly?
It's not like I'm not understanding anything. I've just been approaching it wrong.
Why do some people have to come across as such snobs.
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I think Telastyn meant that instead of storing your input as a single string, you can keep the input separated and THEN to display it you can construct the string from the input.
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I think Telastyn meant that instead of storing your input as a single string, you can keep the input separated and THEN to display it you can construct the string from the input.
I'm just not quite sure on how to do this stuff. Dealing with 2 numbers would be easy, but multiple is sketchy for me.
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I think what you are looking for is a Binary Expression Tree.You also might want to look into syntax trees. These do exactly what your looking to do.... Edited by DevLiquidKnight
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honestly I don't understand the point of threads like these.
I mean, what are you looking for? A solution? How being given a solution would make you improve? Programming is all about finding solutions to problems. If somebody just gives you the solution you'll be back asking another solution for your next problem and you won't learn anything.
You are looking for an hint? Here's one: break down your problem into simple parts, start from a simple case were you are making assumptions.. ie, assume you'll have a sequence of number operation number operation and you solve it left to right without any operation precedence.
Look how to get a string and split this into parts, C# strings are very good at this.
Once you get it done you might look into more complex stuff, operator precedence, brackets and so on.. this usually involves a creation of a tree of operation that is then solved into a solution.
You mentioned you've been programming for 4 years. .I dont mean any offense, but if after 4 years you're stuck at this, maybe you should consider the possibility that perhaps programming isn't what you were born to do?
I'm a friend of his and he was talking to me about the issue. From what I can tell, he's been having trouble understanding how to solve it as opposed to understanding the what side.
Like if you told a person "You have to read the user input and store it as a variable." They'd know what you mean, but not the specifics of how (using Console.Readline, etc)
That's the best way I can explain his issue.
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[quote name='boogyman19946' timestamp='1341002263' post='4954081']
I think Telastyn meant that instead of storing your input as a single string, you can keep the input separated and THEN to display it you can construct the string from the input.
I'm just not quite sure on how to do this stuff.
[/quote]
Then start smaller. Can you make a program that takes two numbers as input and prints their output?
Snob or not, that sort of program is hour 2 or 3 of learning C# from nothing. I find it hard to believe that you've done much of anything in 4 years without realizing that you're accomplishing nothing. Just say "hey, I'm a beginner". Nothing wrong with that, and we'll be able to provide better answers for your situation.
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Considering I have a million books/tutorials, with no real direction. Meh
I wish I could find something similar to this for C#
http://www.cplusplus...articles/12974/
Working from the ground up with pretty much everything you need to learn, with exercises/sample programs to make on your own.
Almost every tutorial/book I've come across, claims to be for beginners, but throws 20 pages of code at you for like the 2nd or 3rd concepts, when it could just be something simple. But no, they have to throw a bunch of overly complicated code in to confuse people, instead of just showing how to do that one small concept.
Maybe I should just stop asking for help. Whenever I do I get chewed out by people. Edited by Bill Fountaine
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Building an expression parser, using Winforms, is not what I would consider a "beginner project to learn C#".
My advice: start with console applications for now, trying to learn Winforms + C# + general programming all at once is a bit much.
As Telastyn said, make a console application that takes input from the user and adds them together first.
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Building an expression parser, using Winforms, is not what I would consider a "beginner project to learn C#".
My advice: start with console applications for now, trying to learn Winforms + C# + general programming all at once is a bit much.
As Telastyn said, make a console application that takes input from the user and adds them together first.
That's easy to do..
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[quote name='laztrezort' timestamp='1341006566' post='4954109']
Building an expression parser, using Winforms, is not what I would consider a "beginner project to learn C#".
My advice: start with console applications for now, trying to learn Winforms + C# + general programming all at once is a bit much.
As Telastyn said, make a console application that takes input from the user and adds them together first.
That's easy to do..
[/quote]
Then do it. Personally, I'm a bit skeptical that you can accomplish that without needing someone to hold your hand.
Once you've done that, have the user enter the input via form rather than console.
Once you've done that, have the user input the operation as well.
Once you've done that, have the user able to input multiple inputs and operations to perform.
There's your direction. No more excuses. Edited by Telastyn
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My problem isn't knowing how to solve the solution.
The problem is not knowing the syntax to solve specific problems.
Stop treating me like I'm an idiot. Thats why I barely use this damn site.
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Almost every tutorial/book I've come across, claims to be for beginners, but throws 20 pages of code at you for like the 2nd or 3rd concepts, when it could just be something simple. But no, they have to throw a bunch of overly complicated code in to confuse people, instead of just showing how to do that one small concept.[/quote]
That's because much of the time, that "one small concept" cannot really be demonstrated in isolation. To demonstrate virtual functions, for instance, one would need at least three classes plus an entry-point function (possibly a third class to retain this entry-point method, if you're using C#). To demonstrate the real power thereof, one would need something much more substantial. An extreme example: to demonstrate the basic opening of a window in the Win32 API from C or C++, one would need several dozens of lines of code just to open the window, and then another few lines just to keep the window open!
Besides showing the syntax of the feature, to properly show off a concept I would argue that one needs to actually demonstrate the practical use of that feature, which means showing a non-trivial program that actually uses it to accomplish something. While I understand your "20 pages of code" to be hyperbole, I believe this also has a different purpose unrelated to merely demonstrating non-trivially the concepts being taught - consider that as you code more and more, your programs will grow larger, and it's highly unlikely that you'll ever be dealing with "one small concept" at a time - you'll be dealing with pages upon pages of code all on your screen at once. In my experience as an intern/co-op alone I've come across source code files that are literally THOUSANDS of lines of code in length! If it were me, I would be putting those "20 pages of code" there in part to get you used to looking at (and understanding) lots and lots of code at once - if you want to be a programmer, the reality is that you are going to need to be able to a) read other people's code b) read and understand LOTS of other people's code.
My problem isn't knowing how to solve the solution.
The problem is not knowing the syntax to solve specific problems.
In that case, let's do it this way. You provide the pseudocode (or a list of steps in plain English, whichever you'd prefer) to the solution you have in mind, and someone here (possibly me) can show you the C# syntax for those specific tasks which will do what you have in mind. I personally will not give you the whole source code to the solution; knowing the solution, and having acquired the syntax you need, you should be able to put the pieces together yourself. Edited by Oberon_Command
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OK, first.
1. Object Oriented programming ... break your programming down into "classes", related to logical sets of data, and the methods that manipulate that data.
2. Focus on building the LOGICAL classes, separate from your UI.
So, don't write a dozen input hanlders doing all your math ... write a dozen input handlers calling methods, which do the math ... now in a real program there is nothing wrong with event handlers the size of yours (3 lines long) ... but since this is half the logic of your program, you are not solving your problem cleanly.
There are at least 5-10 reasonable and very different ways to solve the problem you are asking (and millions of smaller varieties). Programming is about thinking through problems and we all think different (like musicians or painters) ... so all we can do is either show you how we would do it and hope you can generalize, or show you general techniques and hope you can apply them specifically.
But to help you out a bit ... first a thought, try to store your variables in a "natural" manner completely not realated to your UI. what is "natural" depends on you point of view. If you were implementing a more simple calculator "natural" might be:
a variable called something like "accumulator", "currentValue", "register" or the like ... which holds the total so far ... ie "0" when reset, and the number that is displayed after users press the "=" etc.
a variable called "operator", "pendingOperation", "state" or some such .... which holds the operation the user has requested after the press something like "+". and would be able to detect the when there isn't one (ie if the user types "3+3=" after the "=" the pending operation is null, so that you can tell if the user then types "53" then are doing the "set value to ..." operation, which doesn't care what the previous value was ... wheras if they type "+" then "53" they are doing the "add value ..." operation which does use the previous / current value.
now of course, i've only hinted at posibilities ... over about 17 years of programming, I've written calculators for examples, for fun, and for trying out new programming languages or ideas probably 7 or 8 times ... none are just like I described, but that's the point, its just a program to think up however you like ... and then work through it tell it makes since AND works.
Good luck. Edited by Xai
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I have no problem learning, it's just finding the stuff I need to learn to accomplish what I need to.
Like, getting the calculator to add multiple numbers. I'd have no problem if I found something that teaches me how to do it.
I just need to find the right learning materials for certain things, thats all.
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Like, getting the calculator to add multiple numbers.
So, my attempted summary of the situation.
You have a textbox containing a string which represents the expression to be calculated.
You want to take the equation represented in string in the textbox and compute the output of the expression.
But the string representation of the expression isn't very useful to you - at least, not yet.
Am I correct so far?
##### Share on other sites
[quote name='Bill Fountaine' timestamp='1341032033' post='4954197']
Like, getting the calculator to add multiple numbers.
So, my attempted summary of the situation.
You have a textbox containing a string which represents the expression to be calculated.
You want to take the equation represented in string in the textbox and compute the output of the expression.
But the string representation of the expression isn't very useful to you - at least, not yet.
Am I correct so far?
[/quote]
yes
I want to have the calculator able to deal with multiple numbers instead of just 2.
97 + 576 + 46, etc.
I originally thought storing each number in an array/list somehow would do the trick, but dealing with taking the numbers out of the string after each math symbol is what has me stuck atm.
##### Share on other sites
[quote name='Oberon_Command' timestamp='1341033093' post='4954206']
[quote name='Bill Fountaine' timestamp='1341032033' post='4954197']
Like, getting the calculator to add multiple numbers.
So, my attempted summary of the situation.
You have a textbox containing a string which represents the expression to be calculated.
You want to take the equation represented in string in the textbox and compute the output of the expression.
But the string representation of the expression isn't very useful to you - at least, not yet.
Am I correct so far?
[/quote]
yes
I want to have the calculator able to deal with multiple numbers instead of just 2.
97 + 576 + 46, etc.
I originally thought storing each number in an array/list somehow would do the trick, but dealing with taking the numbers out of the string after each math symbol is what has me stuck atm.
[/quote]
Okay.
Just so we're on the same page here: leaving aside the problem of "taking the numbers out of the string" (there's a specific term for this - expression parsing - which we'll get to in a moment), have you determined how to deal with the numbers once they're "out of the string?" If so, can you please post the code so that we can see where you're going with it? If not, let's tackle that problem, first since it's easier - write code that takes each number in an array (or a list, if you prefer - this is what I would prefer, for reasons that will become clear later) and adds them together.
As an aside: I often find it helpful to write a program in chunks according to what I already know how to do. If I know how one particular step works, and I know exactly what sort of thing I'm going to get from the steps before it but not how that predecessor step will work, then I can write that particular step in isolation with some "dummy" test data in the format that I think that particular step will consume once the one before it is done. That's useful for unit testing, as well - if I can verify that each part of my program works on its own when given data that I know should work, then I can track down bugs more easily. Edited by Oberon_Command
##### Share on other sites
What I had in mind was adding a List of ints. Then whenever the user presses a math button, store the preceding substring of numbers in the list, but right now I'm having trouble cutting the substring of numbers out of the string. Edited by Bill Fountaine
##### Share on other sites
While some people on here are being quite rude, they are right on one thing, being a programmer is not about reading a tutorial, and then knowing how to do that 1 thing. Its about being able to solve problems on your own. Anyway.
This is probably the simplest way to approach the issue, though not necessarily the best:
Your user enters a string along the lines of 10 + 20 - 5. Parse that string and store each number in one array. Also store the symbols in a separate array.
Now you have array A = {10, 20, 5} and B = {"+"," -"} (note this is just pseudocode)
Now parse the two arrays and perform the operations. Pretty basic. If you need an actual example:
while (A.length > 1) { if (B[0] == "+") { A[1] = A[0] + A[1]; remove first element of A; remove first element of B; } //add other symbol options here }
Now A[0] will contain the solution.
Of course for things like multiplication you'll have to take into account order of operations, but I won't solve the whole thing for you.
##### Share on other sites
What I had in mind was adding a List of ints. Then whenever the user presses a math button, store the preceding string of numbers in the list, but right now I'm having trouble cutting the substring of numbers out of the string.
I realize that. You seem very insistent on this. But work with me for a bit - try writing the "adding a list of ints" bit before you write the code that cuts the numbers out of the string. In essence, expand AdrianC's example into actual, working code. Post the code here when you're done. Edited by Oberon_Command
##### Share on other sites
I think I am just going to stop while I'm ahead so I don't make myself look any more incompetent than I already have. Obviously the concept of "program, don't just look at tutorials that teaches you how to use stuff." isn't doing anything for me. If I could find something similar to http://www.cplusplus.com/forum/articles/12974/ using C#, in terms of setup (working from the ground up, making you do exercises using stuff you've learned, etc). That would be great. The tutorials I have been watching lately, which have me jumping straight into windows forms, are the ones from http://thenewboston.org/list.php?cat=15
Problem solving is easy. I just need to learn the ins and outs of the language to know what syntax I can use to solve said problems. | 2018-02-25 21:55:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.402694970369339, "perplexity": 1222.8839510640203}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891817437.99/warc/CC-MAIN-20180225205820-20180225225820-00734.warc.gz"} |
https://stats.stackexchange.com/questions/513906/why-do-we-need-to-apply-laplace-smoothing-to-all-the-words-in-naive-bayes-for-te | # Why do we need to apply Laplace smoothing to all the words in Naive Bayes for text classification?
I understood that we need to apply for Laplace smoothing to the words that are not present in our training data. But then why/what is the need to do Laplace smoothing for all the words (even the words that are present in the training set)?
Words that are not present in the training data are called Out of Vocabulary (OOV) words. You cannot apply laplace smoothing to words that are not in your vocabulary. You apply laplace smoothing to words that are in your vocabulary. This is because some vocabulary words that appear in one class may not appear in the other class. The maximum likelihood probabilities for the word in that other class would become 0. The following example illustrates the point.
Suppose you are doing sentiment analysis for positive or negative movie comments. Suppose your training set consists of one positive and one negative comment
+: "I like this alot"
-: "I don't like this"
You construct your vocabulary as ["I","like","this","alot","don't"] and you compute the respective maximum likelihood probabilities for each word belonging to each sentiment $$P(w_i|+), P(w_i|-)$$ where $$w_i$$ is a word in your vocabulary.
Suppose that in your test set, you encounter the comment "I don't think this is good". Now, we haven't seen the words "is" and "good", so we discard them away. Then we are left with the words "I don't think this".
To carry out Naive Bayes classification, we have to compute the posterior probabilities of the sentence belonging to either the positive or negative class.
$$P(-|\text{"I don't think this"}) \propto P(\text{"I"}|-)P(\text{"don't"}|-)P(\text{think"}|-)P(\text{"this"}|-)P(-)$$.
$$P(+|\text{"I don't think this"}) \propto P(\text{"I"}|+)P(\text{"don't"}|+)P(\text{think"}|+)P(\text{"this"}|+)P(+)$$.
Notice that if you don't use laplace smoothing, $$P(\text{"don't"}|+)$$ will be 0, because "don't" does not appear in a positive comment. Therefore, we apply laplace smoothing to give some probability mass to words in our vocabulary that may not have been seen in a positive comment, but may have been seen in a negative comment.
• check this thread: here we are not always droping the OOV words. it is one of the strategy. link The question is that we are applying laplace smoothing to train set, even if the words are present in both the classes. @calveeen your example here has assumed that the word "don't" is not present in the positive class. – Tushar Tiwari Mar 15 at 12:31 | 2021-06-18 05:21:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6217445135116577, "perplexity": 467.7650911863982}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487635724.52/warc/CC-MAIN-20210618043356-20210618073356-00081.warc.gz"} |
http://blog.gmane.org/gmane.text.doxygen.devel/month=20040501 | 3 May 2004 03:25
### RE: Extracting more source struture in XML output
Hi M. Paul Mackay and all,
>From: <Paul.Mackay <at> nokia.com>
>To: <doxygen-develop <at> lists.sourceforge.net>
>Subject: [Doxygen-develop] Extracting more source struture in XML output
>Date: Mon, 26 Apr 2004 12:00:41 -0700
>
Well, I'm writing a software patch for Doyxgen to produce more "information"
on the SVG and XML back-end for
"Software Comprehension/Analysis/Visualization, testing, metrics and Reverse
Engineering"
Currently, I'm working on tighter integration with Graphviz/Dot,
so it's used internally for speed reasons mostly and also to have Graphviz
Something like a web SVG/JS version of GraphViz/dotty for Windows.
>I have recently been looking at XML and the output Doxygen can produce. The
>richness of information and versatility of what can be done with it is
>impressive given the tools available for processing XML.
>
>Does anyone have any comments on the idea of Doxygen outputting more
>information about the structure of the source code?
>I am thinking of a breakdown in terms of blocks, statements, declarations,
>etc.
I'm also thinking in terms of block#, statement#, sub-statement#,
3 May 2004 16:31
### Windows binaries available for Doxygen-1.3.6-20040427 in CVS
Hi,
compiled for MS Windows from
http://doxygen.sourceforge.net/dl/doxygen-1.3-cvs/
This is the place where you should find also the next
releases. The name of the archive is constructed
from the date of CVS release and looks like
doxygen_win32_2004mmdd.zip, where mm and dd is
month and day of releasing the version.
The binaries are NOT created automatically, so it may
happen that some newer CVS sources were not compiled
because I am not present to do that or I forgot... ;)
Regards,
Petr
P.S. The translator report can be found in the file
like tr2004mmdd.zip.
--
--
Petr Prikryl (prikrylp at skil dot cz)
-------------------------------------------------------
This SF.Net email is sponsored by: Oracle 10g
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6 May 2004 17:28
### <at> extends command to document that a c struct *extends* another
hi dimitri,
the other issues. Here is it again:
Wouldn't it be a good idea to have a <at> extends command (which should not be used
in C++/Java) so that I can document that a struct extends another in C.
It is quite common in C to do:
struct base;
struct derived {
struct base _base;
...
};
I would like to see this reflected in the class hierarchy. Sugestion is that
doxygen either detects using a struct as a first member of a struct and/or
doxygen adds an <at> extends filed to the documentation of structs.
What do you think about it? Has it a chance to be put onto the wishlist
ciao
Stefan
--
--
\|/ Stefan Kost
< <at> <at> > private business
+-oOO-(_)-OOo------------------------------------------------------ - - - - -
| __ Address Simildenstr. 5 HTWK Leipzig, Fb IMN, Postfach 301166
6 May 2004 18:15
### Re: <at> extends command to document that a c struct *extends* another
You support both techniques.
config option that will automagically apply the the <at> extend tag if the
first field of a struct has the name "xxx". You might also want to
allow the name to be specified as a prefix (i.e. your config says
STRUCT_AUTOEXTEND_FIELD_NAME =_base and then it will trigger on
_baseFoo or _baseBar as the first field name).
On May 6, 2004, at 8:28 AM, Stefan Kost wrote:
> hi dimitri,
>
> got lost in
> the other issues. Here is it again:
>
> Wouldn't it be a good idea to have a <at> extends command (which should
> not be used
> in C++/Java) so that I can document that a struct extends another in C.
>
> It is quite common in C to do:
>
> struct base;
>
> struct derived {
> struct base _base;
> ...
> };
>
6 May 2004 19:00
### Add a tab to RTF output
I need to separate the parameter from the description with a tab in the RTF output, preferably by inserting a format string in the header file. (All descriptions are drawn from the header files by placing “ <at> fn functionName” in each C file function header.) The RTF is processed by Word 2002 (Word XP).
I have tried inserting \tab \\tab {\tab} and {\\tab} – with and without quotation marks. The closest I can get is an output of “\tab” in the RTF file, which then can be searched and replaced with “^t” to produce a tab.
Ideas??
/* ============================================================ */
/**
* functionName() description for FunctionReturnType functionName().
* description continued
*
* <at> param *p <description> *p
*
* <at> param a <type> a
Resultant output:
Parameters
*p far* some pointer
Desired output:
Parameters
*p far* some pointer
Regards,
Larry Randall
11 May 2004 10:20
### Some bugs in russian translator.
Hi doxygeners!
Some bugs in russian translator. new translator_ru attached.
Thanks to Alexander Borovsky.
Attachment (translator_ru.tgz): application/x-compressed, 10 KiB
17 May 2004 18:31
### Adding a new config switch?
Quick question to make sure I don't miss anything.
What files should I change so that a new config switch shows up in all
the right places (the GUI config app, the documentation, the binary)?
I've got changes to htmlgen.cpp that conditionally use
a HTML_USE_CSS_NOWRAP boolean config switch to control it.
Cheers,
Tom Costa
-------------------------------------------------------
Sign-up now for SourceForge Broadband and get the fastest
6.0/768 connection for only $19.95/mo for the first 3 months! http://ads.osdn.com/?ad_id=2562&alloc_id=6184&op=click 18 May 2004 16:09 ### Patch to Doxygen 1.3.7 for man file output Hi all! This patch corrects two bugs, both in man page output. 1) Using \verbatim (and other commands turning off 'filling' of lines) caused the output to be left in that state for the rest of the file, due to having a .nf directive but no following .fi directive. 2) If a function had only a brief description and no detailed description (for instance using the JAVADOC_AUTOBRIEF and REPEAT_BRIEF options) then following text, like the label 'Parameters', would get put on the same line. This was due to text output not setting the 'paragraph' flag. Test code: ------------------------------------------------------------------------------ /** \file 1.cc */ /** * This is the main program. No surprises there, then, but let's have some * text to fill things out a bit. And another sentence for good luck. * <at> param argc the argument count * <at> param argv the argument pointer */ int main(int argc, char **argv) { return 0; } /** * This has only a short description. * <at> param arg the argument */ void fred(int arg) { } /** * A function to test verbatim output. Let's have a bit of long description * first just to fill in the space... \verbatim Lines to output "as is" with no filling not even cream. \endverbatim * And now some text which needs to be filled with cream and * other sugary stuff which is bad for your teeth. Not to mention * your waistline if you worry about such things... */ void function_verb(void) { } ------------------------------------------------------------------------------ Output without patch (just the detailed descriptions): ------------------------------------------------------------------------------ int main (int argc, char ** argv) This is the main program. No surprises there, then, but let's have some text to fill things out a bit. And another sentence for good luck. Parameters: argc the argument count argv the argument pointer void fred (int arg) This has only a short description. Parameters: arg the argument void function_verb (void) A function to test verbatim output. Let's have a bit of long description first just to fill in the space... Lines to output "as is" with no filling not even cream. And now some text which needs to be filled with cream and other sugary stuff which is bad for your teeth. Not to mention your waistline if you worry about such things... ------------------------------------------------------------------------------ Output with patch (just the detailed descriptions): ------------------------------------------------------------------------------ int main (int argc, char ** argv) This is the main program. No surprises there, then, but let's have some text to fill things out a bit. And another sentence for good luck. Parameters: argc the argument count argv the argument pointer void fred (int arg) This has only a short description. Parameters: arg the argument void function_verb (void) A function to test verbatim output. Let's have a bit of long description first just to fill in the space... Lines to output "as is" with no filling not even cream. And now some text which needs to be filled with cream and other sugary stuff which is bad for your teeth. Not to mention your waistline if you worry about such things... ------------------------------------------------------------------------------ Patch: ------------------------------------------------------------------------------ *** doxygen-1.3.7/src/mandocvisitor.cpp Tue Apr 13 18:13:43 2004 --- src/mandocvisitor.cpp Tue May 18 14:39:49 2004 *************** *** 167,172 **** --- 167,173 ---- { m_insidePre=FALSE; if (!m_firstCol) m_t << endl; + m_t << ".fi" << endl; m_t << ".PP" << endl; m_firstCol=TRUE; } *************** *** 187,192 **** --- 188,194 ---- m_t << ".nf" << endl; parseCode(m_ci,s->context(),s->text().latin1(),s->isExample(),s->exampleFile()); if (!m_firstCol) m_t << endl; + m_t << ".fi" << endl; m_t << ".PP" << endl; m_firstCol=TRUE; break; *************** *** 196,201 **** --- 198,204 ---- m_t << ".nf" << endl; m_t << s->text(); if (!m_firstCol) m_t << endl; + m_t << ".fi" << endl; m_t << ".PP" << endl; m_firstCol=TRUE; break; *************** *** 230,235 **** --- 233,239 ---- FileDef fd( cfi.dirPath(), cfi.fileName() ); parseCode(m_ci,inc->context(),inc->text().latin1(),inc->isExample(),inc->exampleFile(), &fd); if (!m_firstCol) m_t << endl; + m_t << ".fi" << endl; m_t << ".PP" << endl; m_firstCol=TRUE; } *************** *** 240,245 **** --- 244,250 ---- m_t << ".nf" << endl; parseCode(m_ci,inc->context(),inc->text().latin1(),inc->isExample(),inc->exampleFile()); if (!m_firstCol) m_t << endl; + m_t << ".fi" << endl; m_t << ".PP" << endl; m_firstCol=TRUE; break; *************** *** 253,258 **** --- 258,264 ---- m_t << ".nf" << endl; m_t << inc->text(); if (!m_firstCol) m_t << endl; + m_t << ".fi" << endl; m_t << ".PP" << endl; m_firstCol=TRUE; break; *************** *** 287,292 **** --- 293,299 ---- if (!m_hide) { if (!m_firstCol) m_t << endl; + m_t << ".fi" << endl; m_t << ".PP" << endl; m_firstCol=TRUE; } *************** *** 548,553 **** --- 555,561 ---- //{ // m_insidePre=FALSE; // if (!m_firstCol) m_t << endl; + // m_t << ".fi" << endl; // m_t << ".PP" << endl; // m_firstCol=TRUE; //} *** doxygen-1.3.7/src/mangen.cpp Wed Mar 24 20:36:05 2004 --- src/mangen.cpp Tue May 18 14:33:34 2004 *************** *** 632,636 **** --- 632,637 ---- n->accept(visitor); delete visitor; firstCol=FALSE; + paragraph = FALSE; } ------------------------------------------------------------------------------ ------------------------------------------------------- This SF.Net email is sponsored by: SourceForge.net Broadband Sign-up now for SourceForge Broadband and get the fastest 6.0/768 connection for only$19.95/mo for the first 3 months!
19 May 2004 10:57
### Re: Adding a new config switch?
On Mon, May 17, 2004 at 09:31:45AM -0700, Thomas Costa wrote:
> Quick question to make sure I don't miss anything.
> What files should I change so that a new config switch shows up in all
> the right places (the GUI config app, the documentation, the binary)?
You need to add the option to src/config.l (you'll find the options at
the bottom of the file) and its documentation to doc/config.doc.
> I've got changes to htmlgen.cpp that conditionally use
> 'style="white-space: no wrap"' instead of "nowrap" and I wanted to add
> a HTML_USE_CSS_NOWRAP boolean config switch to control it.
In this case it may be better to send this as a non optional improvement ;)
Regards,
Dimitri
------------------------------------------------------- | 2013-05-23 17:43:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2704709470272064, "perplexity": 11109.236253326837}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368703635016/warc/CC-MAIN-20130516112715-00057-ip-10-60-113-184.ec2.internal.warc.gz"} |
https://socratic.org/questions/how-do-you-find-the-taylor-series-for-e-x-2 | # How do you find the taylor series for e^(x^2)?
1+x^2+(x^4)/(2!)+(x^6)/(3!)+(x^8)/(4!)+cdots. This converges to ${e}^{{x}^{2}}$ for all values of $x$.
This can be obtained most simply by taking the well-known Taylor series for ${e}^{x}$ centered at 0, which is 1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+cdots, and replacing $x$ with ${x}^{2}$.
You can also try using the formula f(0)+f'(0)x+(f''(0))/(2!)x^2+(f'''(0))/(3!)x^3+(f''''(0))/(4!)x^4+cdots, but that is a much less pleasant approach. | 2022-01-21 23:00:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9909208416938782, "perplexity": 153.37156341267752}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303717.35/warc/CC-MAIN-20220121222643-20220122012643-00389.warc.gz"} |
https://math.stackexchange.com/questions/30263/polynomials-fx-of-degree-at-most-5-forming-a-ring-and-field | # Polynomials $f(x)$ of degree at most $5$ forming a ring and field
Show that the set of all polynomials $f(x)$ of degree at most $5$ with integer coefficients is a ring. Is the set of such polynomials a field?
I don't see how the ring of polynomials with degree at most $5$ is closed under multiplication. If I multiply $x^2$ and $x^5$ I do not get another polynomial of degree at most $5$.
• Well, they are not a ring, so it is not surprising that you cannot see it! – Mariano Suárez-Álvarez Apr 1 '11 at 5:16
• $\:\mathbb Z\:$ is the only subring of $\rm\:\mathbb Z[x]\:$ of bounded degree. Check the question. – Bill Dubuque Apr 1 '11 at 5:23
You're right, it's not a ring using addition and multiplication defined in the normal manner on $\mathbb{Z}[X]$. Are you sure that this is what the question asks? Does it perhaps introduce some equivalence relation, such as $x\equiv y$ iff $x-y\in (X^6)$? | 2019-10-14 00:48:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.943688154220581, "perplexity": 121.78961926833121}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986648481.7/warc/CC-MAIN-20191014003258-20191014030258-00028.warc.gz"} |
https://www.shaalaa.com/question-bank-solutions/approximations-if-there-error-2-measuring-length-simple-pendulum-then-percentage-error-its-period-a-1-b-2-c-3-d-4_44534 | Share
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# Solution for If There is an Error of 2% in Measuring the Length of a Simple Pendulum, Then Percentage Error in Its Period is (A)1% (B) 2% (C) 3% (D) 4% - CBSE (Science) Class 12 - Mathematics
#### Question
If there is an error of 2% in measuring the length of a simple pendulum, then percentage error in its period is
(a)1%
(b) 2%
(c) 3%
(d) 4%
#### Solution
(a) 1%
Let l be the length if the pendulum and T be the period.
$\text { Also, let ∆ l be the error in the length and ∆ T be the error in the period } .$
$\text { We have }$
$\frac{∆ l}{l} \times 100 = 2$
$\Rightarrow \frac{dl}{l} \times 100 = 2$
$\text { Now,} T = 2\pi\sqrt{\frac{l}{g}}$
$\text { Taking \log on both sides, we get }$
$\log T = \log 2\pi + \frac{1}{2}\log l - \frac{1}{2}\log g$
$\text { Differentiating both sides w . r . t . x, we get }$
$\frac{1}{T}\frac{dT}{dl} = \frac{1}{2l}$
$\Rightarrow \frac{dT}{dl} = \frac{T}{2l}$
$\Rightarrow \frac{dl}{l} \times 100 = 2\frac{dT}{T} \times 100$
$\Rightarrow \frac{dT}{T} \times 100 = \frac{2}{2}$
$\Rightarrow \frac{∆ T}{T} \times 100 = 1$
$\text { Hence, there is an error of 1 % in calculating the period of the pendulum } .$
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [2]
Solution If There is an Error of 2% in Measuring the Length of a Simple Pendulum, Then Percentage Error in Its Period is (A)1% (B) 2% (C) 3% (D) 4% Concept: Approximations.
S | 2019-05-27 06:15:47 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.832410454750061, "perplexity": 2233.9467458959}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232261326.78/warc/CC-MAIN-20190527045622-20190527071622-00123.warc.gz"} |
https://ask.sagemath.org/question/39392/how-to-make-an-macaulay-matrix-from-polynoms-over-gf2/ | # how to make an Macaulay matrix from polynoms over GF(2)
This post is a wiki. Anyone with karma >750 is welcome to improve it.
I have a PolynomialRing(GF(2),'x1,x2,x3') and over it two polynomials x1*x2 + x1*x3 + x1, x1+x2+1 and I would like to rewrite it in Macaulay matrix in order x1x1, x1x2, x2x2, x1x3, x2x3, x3x3,x1,x2,x3, absolute term so it should be
0 1 0 1 0 0 1 0 0 0
0 0 0 0 0 0 1 1 0 1
Is there something in sage ?
edit retag close merge delete
It seems that the only related function is R.macaulay_resultant(...) if R is your polynomial ring, that takes a list of $n$ homogeneous polynomials (if $n$ is the number of variable) and computes their Macaulay resultant. You can inspect the code (using for instance R.macaulay_resultant??) and copy the parts that are useful for your needs.
( 2017-11-06 07:12:42 -0600 )edit
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I suppose that the question wants to build the corresponding matrix in the peculiar order of all monomials of degree $\le 2$ given above, that works for all (not so many) possible polynomials of degree $\le 2$ in the given ring. The hint MacCaulay was actively ignored in my following answer / sample code, since i considered the task as a task of identifying coefficients of polynomials.
F = GF(2)
R.<x1,x2,x3> = PolynomialRing( F )
p, q = x1*x2 + x1*x3 + x1, x1 + x2 + 1
degrees = ( (2,0,0),
(1,1,0),
(0,2,0),
(1,0,1),
(0,1,1),
(0,0,2),
(1,0,0),
(0,1,0),
(0,0,1),
(0,0,0), )
d = len(degrees)
print matrix( F, 2, d, [ [ pol.coefficient( dict( zip( (x1,x2,x3) , degtuple ) ) )
for degtuple in degrees ]
for pol in ( p, q ) ] )
This gives:
[0 1 0 1 0 0 1 0 0 0]
[0 0 0 0 0 0 1 1 0 1]
more | 2020-01-25 07:30:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.27535974979400635, "perplexity": 990.3229341964472}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579251671078.88/warc/CC-MAIN-20200125071430-20200125100430-00554.warc.gz"} |
https://stats.stackexchange.com/questions/62294/monte-carlo-integration-for-non-square-integrable-functions | # Monte Carlo Integration for non-square integrable functions
I hope this is the right place to ask, if not feel free to move it to a more appropriate forum.
I've been wondering for quite a while now how to treat non-square integrable functions with Monte Carlo Integration. I know that MC still gives a proper estimate but the error is unrealiable (divergent?) for those kind of functions.
Let's restrict us to one dimension. Monte Carlo integration means that we approximate the integral
$$I = \int_0^1 \mathrm{d}x \, f(x)$$
using the estimate
$$E = \frac{1}{N} \sum_{i=1}^N f(x_i)$$
with $x_i \in [0,1]$ uniformly distributed random points. The law of large numbers makes sure that $E \approx I$. The sample variance
$$S^2 = \frac{1}{N-1} \sum_{i=1}^N (f (x_i) - E)^2$$
approximates the variance $\sigma^2$ of the distribution induced by $f$. However, if $f$ is not square-integrable, i.e. the integral of the squared function diverges, this implies
$$\sigma^2 = \int_0^1 \mathrm{d} x \, \left( f(x) - I \right)^2 = \int_0^1 \mathrm{d} x \, f^2(x) - I^2 \longrightarrow \infty$$
meaning that also the variance diverges.
A simple example is the function
$$f(x) = \frac{1}{\sqrt{x}}$$
for which $I = \int_0^1 \mathrm{d}x \, \frac{1}{\sqrt{x}} = 2$ and $\sigma^2 = \int_0^1 \mathrm{d}x \, \left( \frac{1}{x} - 2 \right) = \left[ \ln x - 2x \right]_0^1 \rightarrow \infty$.
If $\sigma^2$ is finite one can approximate the error of the mean $E$ by $\frac{S}{\sqrt{N}} \approx \frac{\sigma}{\sqrt{N}}$, but what if $f(x)$ is not square-integrable?
• I don't get it: you start out by noting that none of the $E_i$ has a variance and then ask whether the variance of their average would be a reasonable estimator of--that nonexistent variance! Or do I misread this question: perhaps by "statistically independent estimations" you have some different (perhaps robust) estimator of the integral in mind? – whuber Jun 21 '13 at 18:14
• I didn't say $E$ doesn't have a variance, only that I cannot define a variance for it by $S^2$. So the question is whether I can define an error at all and if $\bar{S}^2$ is a reasonable candidate. By statistically independent I mean that the $E_i$ are obtained using different random numbers, e.g. by using differently seeded random number generators (I hope thats the right term then). – cschwan Jun 22 '13 at 6:38
• Please explain what you mean by not being able to "define a variance for it by $S^2$." I cannot make sense of this using the standard definitions of variance and $S^2$. – whuber Jun 24 '13 at 14:45
• Well, the function is not square-integrable so, if I am not mistaken, $S^2$ should diverge. If this is the case the definition for $S^2$ makes no sense in the first place, right? By means of the central limit theorem, however, $E$ will still converge to the true value of the integral, but without an error this value alone makes no sense (how 'good' is this result?). – cschwan Jun 26 '13 at 13:55
• Sorry, I meant to say "law of large numbers" of course, not CLT. – cschwan Jun 26 '13 at 14:04 | 2020-09-20 05:01:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9329692721366882, "perplexity": 252.70870076652028}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400193391.9/warc/CC-MAIN-20200920031425-20200920061425-00518.warc.gz"} |
https://cracku.in/58-consider-four-natural-numbers-x-y-x-y-and-x-y-two--x-xat-2014 | Question 58
# Consider four natural numbers: x, y, x + y, and x - y. Two statements are provided below:I. All four numbers are prime numbers.II. The arithmetic mean of the numbers is greater than 4.Which of the following statements would be sufficient to determine the sum of the four numbers?
Solution
Natural numbers = $$x , y , (x+y) , (x-y)$$
Statement I : As all the numbers are prime, therefore, either x or y has to be 2 because otherwise (x+y) cannot be prime.
Case 1 : If x = 2, then (x-y) cannot be prime
Case 2 : If y = 2, numbers = $$(x-2) , x , (x+2)$$
These numbers are prime, hence all possibility = 3,5,7
$$\therefore$$ Sum = 2+3+5+7 = 17
Using statement II, we cannot find the required sum, as no specific value of mean is given.
Thus, statement I alone is sufficient. | 2023-01-28 12:43:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4709104299545288, "perplexity": 643.695407462464}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499634.11/warc/CC-MAIN-20230128121809-20230128151809-00390.warc.gz"} |
http://reprap.org/wiki/ConductiveMaterials | # ConductiveMaterials
ConductiveMaterials
Release status: experimental
Description experiments into conductive materials that can be extruded License GPL 2.0 Author Chip936 Contributors Based-on Delta Categories Materials CAD Models none External Link none
# Overview
This page lists the results of conductive materials that can be extruded by a Reprap printer. The goal being to be able to print uOhm conductive paths, N and P type semiconductor material and variable resistive paths.
Solders, paints, sealants, glues and epoxies.
## Low resistive materials
• Field's metal - Melting Point: 60.5 C
• Wood's metal - Melting Point: 71 C
• Rose's metal - Melting Point: 109 C
• Solder - Melting Point: 183 C
• Carbon Powder - Melting not possible - Mix in with glue or sealant - Allows for variable resistor values
• Carbon Black - nanopowders are available in a range of conductivities and sizes that may be mixed with a vehilce for inkjet or screen printing
• Graphite - nanopowders are also available in a range of conductivities and sizes that may be mixed with a vehilce for inkjet or screen printing
• Silver - nanoparticles that also maybe be mixed with nonpolar solvents or water to produce an ink for inkjet or screen printing
• Copper - nanoparticles that are coated with a resin to keep them from oxidizing until they are sintered after printing. Copper plating is also used with printed metal nanopowders. Conductive traces are first printed that are then plated to produce a finished low resistance circuit board.
## N type semiconductor material
• Zinc Oxide
• Cadmium arsenide - Soluble in water
• Tin Dioxide
• Organic semiconductors
## P type semiconductor material
• Copper Sulfide
• Copper(II) Oxide
• Organic semiconductors
## Insulators and Dielectrics
• Polymers such as polyesters, polypropylene, acrylics, urethanes
## Printing Electronics
Passive and active elctronics may be printed via inkjet, screen or flexo. A comination of conductors, semiconductors, dielectrics and insulating fluids are used in various layers to produce components or complete circuits.
The limiting factors of printing electronics are the drop sizes of the inkjets that make up the geometry of the junctions and the carrier mobility of the organic semiconductor fluids.
There are hundreds of papers available online that discuss the details of sucessfully printed electronic circuits by various printing methods.
## Print a Transistor
The holy grail of electronics is the transistor. If a 3D printer was able to extrude a mixture with Zinc Oxide, then place a dab of a mixture with Copper(II) Oxide and then attach two low resistive connections to the Zinc Oxide mixture and one low resistive connection to the Copper Oxide mixture, we should have an NPN transistor. Once we have transistors we can build [Logic Gates]. Once we have logic gates, we can use any [FPGA core that is available.
# Experiments Results
Bitumen is a brushable waterproofer sealant used on roofs. It hardens into a flexible rubber.
### Materials
• Bitumen Brushable Waterproofer - 500ml AU$8.64 • Graphite - Lock-Lube - 6g AU$ 6.11
### Results
• 10ml of bitumen and 750 mg of graphite. Result - >2M Ohms
• 10ml of bitumen and 1.25g of graphite. Result - 34k Ohms/cm
• 10ml of bitumen and 2g of graphite. Result - 500-1k/cm Ohms
### Notes
Additional graphite was added to the existing solution after the sample trace was drawn. At 2g of graphite the mixture was thick and difficult to mix and manipulate with a the toothpick I used to lay down the test trace. Paint thinner should be added to the mixture and tested.
### Theory View
Perhaps this is a good place to bring in a bit from the theoretical side of the matters. It may help to find out how close experimental results are to what might be possible with the ultimative fabricationg process.
#### Comparison by specific track resistance
Usually, electric circuits are made with copper traces and for wiring signals, these traces are pretty thin. Copper has a specific resistance of $1.68 \cdot 10^{-2}\ \Omega \cdot mm^2 / m$, while (pure) graphite has $8\ \Omega \cdot mm^2 / m$.
So, to replace tracks, you have to make tracks made from a material with higher specific resistance thicker. The rule is:
$Resistance\ per\ meter = \frac {specific\ resistance} {track\ thickness \cdot track\ width}$
For a copper track of 2 mil (which is pretty thin, but sufficient to forward a signal to the next chip) this is:
$R_S = \frac {1.68 \cdot 10^{-2}\ \Omega \cdot mm^2 / m} {35\ \mu m \cdot 2\ mil} = \frac {1.68 \cdot 10^{-2}\ \Omega \cdot mm^2 / m} {0.035\ mm \cdot 0.0508\ mm} = 9.45\ \Omega / m$
A replacement track made of graphite, 1 mm wide and 1 mm thick, has:
$R_S = \frac {8\ \Omega \cdot mm^2 / m} {1\ mm \cdot 1\ mm} = 8\ \Omega / m$
As one can see, replacing copper tracks with graphite is achievable.
#### Comparison by total track resistance
Now, to get a signal from one chip pin to another, the resistance of the track can be pretty high, as long as there is no (a neglibile amount of) current flowing. How much is possible there mostly depends on that required current, so you have to figure that out in single cases.
As most of our circuitry is digital, we have quite some allowance. If you apply 5 Volts on one side of the track and only 4 Volts are received on the other side of the track, this is still a valid "high" signal (for an ATmega running at 5 V).
## Impure Semiconductor Tests
### Materials
• Tanne Zinke 30+ SPF Zinc Oxide Sunscreen (425mg/g purity) AU$5.94 • Impure Copper Oxide made by cooking Copper Sulfate with Sodium Hydroxide • Richgro Copper Sulphate (Blue stone) AU$ 12.94
• Draino - Found under bathroom sink
### Results
• While making larger semiconductors do not require the purity or precision that is currently employed in the semiconductor industry to make the 5nm layers, less than 50% pure zinc oxide and sodium laced cupric oxide did not work.
• Experimenters should wait for the 99% pure compounds to arrive
# Future Experiments
## Diode
### Overview
The diode is one of the oldest application of semiconductor material. By taking an N-type material and joining it to a P-type material we create a gate that will only allow electrons to pass in a single direction. With a diode, a rudimentary AND and OR gate can be constructed.
### Expected Goals
• Validate that a diode can be constructed from semiconductor material suspended in extrudable materials.
• Determine the Voltage Drop across the diode.
### Instructions
Using powdered N-type and P-type materials each mixed with a small amount of cement, apply small dab of each on a non-conductive substrate so they are touching. Allow the cement to harden. Next add two conductive glue tracks, one track leading out of the p-type and one leading out of the n-type cement. While the glue is hardening, insert two metal wires.
## FET
### Overview
A Field Effect Transistor utilizes a magnetic field to control the flow of electrons through a thin N-type material channel on a p-type material substrate. The magnetic field is generated by a conductive plate over the n-type channel separated by a non-conductive insulator.
### Expected Goals
• Determine the feasibility of constructing a simple transistor through extrudable materials.
• Determine the Gate Voltage - depending on the final size, it is likely to be in rather high voltage (48V - 2kV)
### Instructions
Using powdered P-type material, mix with cement and create the P-type substrate. Create a rectangular base with two depressions on either side which will form the source and drain. Between the two depressions should be a wide shallow channel where the gate will be located. After the base hardens enough so it will not deform, fill the depression and channel with powered N-type material mixed with cement. Allow this to harden. Cover the gate section with an insulator - a very thin plate ABS/PLA material should work. After hardening, cover the insulator with conductive glue almost to the edge and use the glue to create the source and drain leads. Attach test leads to the source, gate and drain conductive materials to test.
## ALU
### Overview
Arithmetic Logic Unit is the heart of any micro processor. It is responsible for performing basic operations on multiple bits inside a computer. Examples of these operations would be AND, OR, XOR, ADD, SUB(tract), MUL(tiply), DIV(ide), LSR(Logical Shift Right), LSL (LS Left), LRL (Logical Roll Left), LRR (LR Right).
### Expected Goals
To validate that simple extruded gates can be connected together to produce a usable circuit
### Instructions
Awaiting results of FET test
## PIC
### Overview
A PIC micro controller is a popular small chip used control everything from washing machines to radio controlled cars.
### Expected Goals
Create a possible micro controller replacement for the electronics on the next generation of RepRap
### Instructions
Awaiting results of ALU test | 2017-06-29 14:21:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3630165457725525, "perplexity": 4853.572707051707}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128329344.98/warc/CC-MAIN-20170629135715-20170629155715-00253.warc.gz"} |
https://www.tec-science.com/mechanical-power-transmission/planetary-gear/how-does-a-differential-gear-work/ | ## Why does a car need a differential gear?
In automobiles, the wheels are usually driven by the engine using a bevel gear. This allows the rotational motion from the engine to the wheels to be deflected by 90°. If, however, the wheels were rigidly connected to each other by a common shaft, this would lead to problems when cornering. In such a case, the outer wheel must cover a greater distance than the inner wheel. However, since both wheels must travel around the turn at the same time, the outer wheel must rotate faster than the inner wheel.
If the two wheels would be connected by a common shaft, the shaft would twist due to the different rotational speeds. Sooner or later, such a twist is compensated by a slipping of one of the wheels. This slipping in the curve not only reduces driving safety but also leads to considerable tyre wear and, in the long run, to shaft breakage.
When cornering, the outer wheel must be able to rotate faster than the inner wheel!
For this reason, in the early days, only one of the wheels was driven. The other wheel was mounted freely on the shaft so that it could rotate with a different speed. Such a one-sided drive leads however to the fact that the vehicle tries to drive a slight curve. This reduces not only the driving fun but also the driving safety. It was therefore necessary to find a solution to drive both wheels at the same time while allowing different speeds: The differential gear was born.
The picture below shows the differential gear of a truck. One sees the pinion (shown yellow in the animation above) and the bevel gear (shown orange in the animation above). The other bevel gears are inside the housing and not visible from the outside.
## Design of a differential gear
The design and operating principle of a differential gear are not easy to understand at first glance. The main question is how to come up with such an arrangement of gears. For the sake of simplicity, it makes sense to first understand the individual steps behind the idea of the differential gear.
### 1st step – drive of the separated shafts by pins and a freely rotatable bar
The initial idea is to first divide the common drive shaft so that each wheel has its own drive shaft. This ensures that the shaft does not twist if one of the two wheels rotates with a different speed. Two pins are now attached to each of the separate shafts. Between these pins, a freely rotatable bar drives the respective wheel shafts.
In this way, the wheels can be rotated to different degrees within a certain limit. If one of the wheels is slowed down, the opposite wheel can be moved a little further by the rotatable bar. However, the different rotation should not be too large, otherwise the rod will slide out of the pins and no more force can be transmitted.
### 2nd step – drive of the shafts by several pins and freely rotatable bars
In order to increase the yet very limited motion, one could just use several pins instead of only one as well as more rotatable bars. The pins and the bars can now slide into each other one after the other. The wheel drive is no longer limited. One of the wheels can now rotate at a completely different speed and even stand still, while the other wheel can continue to be driven. In principle, such an arrangement already represents a fully functional differential gear!
A closer look shows that with such a differential, the slowed wheel is decelerated to the same extent as the other wheel is accelerated. The speed loss on one side of the wheel is compensated by a speed gain of the same magnitude on the other side. This principle is based on the law of conservation of energy.
Such a kinematic behaviour of the wheels is exactly what is needed when cornering. When cornering, the inner wheel must rotate more slowly to the same extent as the outer wheel must rotate faster.
A differential gear ensures that the inner wheel rotates to the same extent more slowly as the outer wheel rotates faster when cornering!
### 3rd step – Replacing pins and bars with bevel gears
The power transmission by pins and bars is not very effective. Therefore they are replaced by gears, more precisely by bevel gears. The bevel gear shown in blue, which revolves around the shafts of the wheels, is also referred to as the spider gear. In principle, this spider gear is nothing else than a planet gear as it is known from planetary gears. And indeed, the differential gear can be seen as a special form of a planetary gearbox (more on this later).
### 4th step – drive of the shafts by further bevel gears
The drive of the spider gear is of course not done by hand but by the engine. The spider gear is in turn driven by a bevel gear unit (usually a hypoid gear), consisting of a pinion (shown in yellow) and a bevel gear (shown in orange). The spider gear ist mounted on this orange bevel gear. Since the orange bevel gear “carries” the revolving spider gear, the orange bevel gear is also referred to as carrier.
### 5th step – symmetrical arrangement of the bevel gears to avoid bending stresses
In order to avoid bending stresses in the drive shafts of the wheels, they are usually not driven by only one spider gear but by two spider gears. The second spider gear is offset by 180°.
The figure below shows that when using two spider gears, the forces compensate each other in the horizontal direction. The drive shafts of the wheels are then subjected purely to torsion, but not to bending!
## Kinematics of a differential gear
When driving straight ahead, normally none of the wheels is forced to rotate slower or faster than the other. In this case, the spider gears drive the wheel shafts without any relative motion. The wheels then rotate at the same speed as the carrier.
If one now drives into a right turn, for example, the inner wheel is slowed down by the shorter distance to be travelled. However, the outer wheel must then rotate faster to the same extent, since it has to cover a greater distance. Due to its special design, a differential gear ultimately ensures exactly such a kinematic behavior! The exact mathematical relationship is explained in more detail in the next section.
The best way to understand the kinematics is to imagine an extreme cornering where the inner wheel practically stands still and the outer wheel follows a circular path around the inner wheel. In this case, the carrier drives the spider gears around the bevel gear (“side gear”) of the stationary wheel shaft. The spider gears then begin to rotate and now perform relative motions. The opposite bevel gear (“side gear”) of the left drive shaft is now driven by this rotation of the spider gears in addition to the already existing rotation of the carrier and thus rotates faster.
Compared to the carrier, the inner wheel rotates more slowly to the same extent as the outer wheel rotates faster when cornering.
Only when cornering is complete and the wheel speeds have been adjusted again, do the two wheel shafts no longer move relative to each other and the speed of the carrier corresponds to the wheel speeds.
Even if the speeds of the wheels differ when cornering, both wheels are always driven by the same torque! This is because in gearboxes the change in torque only results from the ratio of the number of teeth of the gears. However, the differential gear has a symmetrical design. It does not differ in the number of teeth between the left and right drive shaft. This means that the change in toraue between the motor and the drive shafts are always the same. Both gears therefore have the same torque.
Even if the respective torque at the wheels does not differ, they have different powers! This is because the power is determined by the product of torque M and rotational speed n:
\begin{align}
\boxed{P=2 \pi \cdot M \cdot n} \\[5px]
\end{align}
It should be noted, however, that when the differential is active when cornering, there are relative motions of the bevel gears which lead to an additional reduction in gear efficiency.
Although a differential gear provides different speeds and thus different power for the wheels, the torque on both wheels is identical!
## Differential gear as a special case of a planetary gearbox
As already mentioned, a differential gear is a special type of a planetary gearbox. One of the bevel gears on the wheel shafts can be regarded as a sun gear while the other bevel gear then corresponds in a figurative sense to the ring gear.
Since a differential is a special type of a planetary gearbox, the relationship between the different rotational speeds can also be described by the fundamental equation for planetary gears (Willis equation):
\begin{align}
&\boxed{ n_s = n_c \cdot \left(1-i_0 \right) + n_r \cdot i_0} \\[5px]
\end{align}
For classic planetary gears, nr refers to the rotational speed of the ring gear, ns denotes the rotational speed of the sun gear and nc refers to the rotational speed of the carrier. i0 denotes the so-called fixed carrier transmission ratio.
In the case of a differential gear, the fixed carrier transmission ratio corresponds to the transmission ratio which is obtained when the carrier is fixed. If one of the wheels (the “ring gear”) is rotated in this state, then the other wheel (the “sun gear”) obviously rotates at the same speed, but in the opposite direction. The fixed carrier transmission ratio is therefore i0=-1.
If the fixed carrier transmission ratio of i0=-1 is used in the upper equation, then the following relationships apply:
\begin{align}
& n_s = n_c \cdot \left(1-i_0 \right) + n_r \cdot i_0 ~~~\text{with}~i_0=-1~~~~\text{:} \\[5px]
&n_s = n_c \cdot \left(1-(-1) \right) + n_r \cdot (-1) \\[5px]
&n_s = n_c \cdot 2 – n_r \\[5px]
&n_r + n_s = 2 \cdot n_c \\[5px]
\end{align}
Since differential gears do not have a classic sun gear or ring gear, the corresponding rotational speeds of the gears are denoted by n1 (=nr) or n2 (=ns). Thus, the following relationship between the rotational speeds of the wheels n1 or n2 and the rotational speed of the carrier nc applies:
\begin{align}
&\boxed{n_1 + n_2 = 2 \cdot n_c} \\[5px]
\end{align}
The right side of the equation is always constant at a constant speed of the carrier and thus at a constant motor speed. Now it can also be seen mathematically that at a constant motor speed, a reduction of the speed at one of the wheels results in an increased speed at the opposite wheel. By rearranging the equation, one can also see that the speed of the carrier corresponds to the mean speed of the two wheels.
\begin{align}
&\boxed{n_c = \frac{n_1 + n_2}{2}} \\[5px]
\end{align}
## Differential lock
The big advantage of differential gears is that they can be used when cornering by dividing the rotational speed or power between the respective wheels according to their needs. In some situations, however, this can also be a disadvantage. For example, when starting on a smooth or slippery ground, one of the wheels may lose its grip and slip, while the other wheel remains on the ground. The differential gear now transmits the entire power to the rotating wheel, while no power is at the stationary wheel. The spinning wheel now turns at double speed, while the other wheel stands still. In this way one hardly obtains a forward driving force and if then only a one-sided force due to the sliding friction of the rotating wheel.
Such a case where one of the wheels has less grip than the other and is thus tempted to slip, occurs primarily during off-road driving, where the load on the wheels varies permanently. But even in fast cornering, where the inner wheel is greatly relieved by the centrifugal forces, the danger of slipping increases and the one-sided power distribution threatens. If, in the worst case, the vehicle tilts slightly and the inner wheel loses its grip, this wheel receives full power and rotates in the air at twice the speed. The opposite wheel, which still has the grip to the ground, does not get any power and therefore no drive of the car is possible anymore.
In the cases mentioned above, a differential gear is therefore more of an obstacle. For this reason, mainly off-road vehicles are equipped with so-called differential locks. Such a differential lock then rigidly connects the two drive shafts of the wheels with each other again and thus deactivates the differential. However, this leads to the twisting of the drive shaft when cornering, as already explained at the beginning. Differential locks should therefore only be activated in exceptional cases. | 2021-04-16 00:14:34 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8581112623214722, "perplexity": 903.2149010958512}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038088264.43/warc/CC-MAIN-20210415222106-20210416012106-00307.warc.gz"} |
https://lama.u-pem.fr/evenements/seminaire/seminaire_cristolien_danalyse_multifractale/on_the_dimension_of_the_graph_of | # On the dimension of the graph of the classical Weierstrass function
Orateur: ROMANOWSKA Julia Localisation: Université de Varsovie, Pologne Type: Séminaire cristolien d'analyse multifractale Site: UPEC Salle: Salle des thèses Date de début: 16/01/2014 - 11:00 Date de fin: 16/01/2014 - 11:00
In my talk I will examine dimension of the graph of the famous Weierstrass non- differentiable function $$W_{\lambda,b}(x)=\sum_{n=0}^\infty \lambda^n\cos(2\pi b^n x)$$ for an integer $b \ge 2$ and $1/b<\lambda<1$. In our recent paper, together with Balázs Bárány and Krzysztof Barański, we prove that for every $b$ there exists (explicitly given) $\lambda_b\in(1/b, 1)$ such that the Hausdorff dimension of the graph of $W_{\lambda,b}$ is equal to $D = 2 +\frac{\log \lambda}{\log b}$ for every $\lambda\in (\lambda_b , 1)$. We also show that the dimension is equal to $D$ for almost every $\lambda$ on some larger interval. This partially solves a well-known thirty-year-old conjecture. Furthermore, we prove that the Hausdorff dimension of the graph of the function $$f(x)=\sum_{n=0}^\infty\lambda^n\phi(b^nx)$$ for an integer $b\ge 2$ and $1/b < \lambda < 1$ is equal to $D$ for a typical $\mathbb{Z}$-periodic $C^3$ function $\phi$. In my talk I will talk about these results as well as I will introduce Ledrappier-Young theory and results of Tsujii, which were used in the proofs. | 2021-05-18 06:03:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4878191649913788, "perplexity": 495.22026551331845}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989820.78/warc/CC-MAIN-20210518033148-20210518063148-00321.warc.gz"} |
https://www.encyclopediaofmath.org/index.php?title=Airy_equation&diff=prev&oldid=27208 | # Difference between revisions of "Airy equation"
2010 Mathematics Subject Classification: Primary: 33C10 [MSN][ZBL]
The Airy equation is the second-order linear ordinary differential equation $y'' - xy = 0.$ It occurred first in G.B. Airy's research in optics [Ai]. Its general solution can be expressed in terms of Bessel functions of order $\pm 1/3$: $y(x) = c_1 \sqrt{x} J_{1/3}\left(\frac{2}{3}\mathrm{i}x^{3/2}\right) + c_2 \sqrt{x} J_{-1/3}\left(\frac{2}{3}\mathrm{i}x^{3/2}\right).$ Since the Airy equation plays an important role in various problems of physics, mechanics and asymptotic analysis, its solutions are regarded as forming a distinct class of special functions (see Airy functions).
The solutions of the Airy equation in the complex plane $z$, $w'' - zw = 0,$ have the following fundamental properties:
1) Every solution is an entire function of $z$ and can be expanded in a power series $w(z) = w(0) \left( 1 + \frac{z^3}{2.3} + \frac{z^6}{(2.3).(5.6)} + \cdots \right) + w'(0) \left( z + \frac{z^4}{3.4} + \frac{z^7}{(3.4).(6.7)} + \cdots \right),$ which converges for all $z$.
2) If $w(z) \not\equiv 0$ is a solution of the Airy equation, then so are $w(\omega z)$ and $w(\omega^2 z)$, where $w=\mathrm{e}^{2\pi\mathrm{i}/3}$, and any two of these solutions are linearly independent. The following identity holds: $w(z) + w(\omega z) + w(\omega^2 z) \equiv 0.$
#### References
[AbSt] M. Abramowitz (ed.) I.A. Stegun (ed.), Handbook of mathematical functions, Appl. Math. Series, 55, Nat. Bureau of Standards,, U.S. Department Commerce (1964) [Ai] G.B. Airy, "On the intensity of light in the neighbourhood of a caustic" Trans. Cambridge Philos. Soc., 6 (1838) pp. 379–402 [BaBu] V.M. Babich, V.S. Buldyrev, "Asymptotic methods in the diffraction of short waves", Moscow (1972) (In Russian) (Translation: "Short-Wavelength Diffraction Theory. Asymptotic Methods", Springer, 1991)
How to Cite This Entry:
Airy equation. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Airy_equation&oldid=27208
This article was adapted from an original article by M.V. Fedoryuk (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article | 2019-07-18 05:12:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8645061254501343, "perplexity": 969.611139775081}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525500.21/warc/CC-MAIN-20190718042531-20190718064531-00394.warc.gz"} |
https://blog.murwell.com/2017/08/17/algorithms-and-logarithms/ | / Algorithms
# Algorithms And Logarithms
Binary Search Algorithm
Maximum number of guesses for a targeted value in an array of length n is $$log_2{n}$$
$$log_2{64}=6$$
$$2^6=64$$
$$2^0=1$$
$$log_{base}{x}=y$$
$$base^y=x$$
$$base^0=1$$
A value in an array of 64 elements can be found in 7 guesses (i.e. 32,16,8,4,2,1 + 1).
128 elements - 8 guesses.
256 elements - 9 guesses.
512 elements - 10 guesses.
1024 elements - 11 guesses.
2048 elements - 12 guesses.
4096 elements - 12 guesses.
1,048,576 elements - 21 guesses. (i.e. - $$(log_2{1,048,576}) + 1 = 20 + 1 =21$$) | 2019-09-15 12:56:56 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3710346519947052, "perplexity": 4127.015662876302}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514571360.41/warc/CC-MAIN-20190915114318-20190915140318-00479.warc.gz"} |
https://www.coursehero.com/file/28359485/Unit-7-Assessmenet-Calculus-and-Vectors-2017docx/ | # Determine the angle between each of the following pairs of...
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Assessment of Learning: Teacher-Marked Lesson 20: Intersection of Lines and Plane in 3-space Properties of Vectors TASK 1: Knowledge and Understanding Questions 1. Determine the angle between each of the following pairs of vectors. .
b) ´ p = ( 1,4,5 ) and q = ( 3, 1,3 ) ° .
2. Find the slope of the vector that is perpendicular to the scalar equation 6x-3y+2=0 Compare the equation to the general scalar equation in 2 space Ax + By + C = Therefore, A = 6 , B =− 3 and C 0 = 2 .
3. Write an alternative vector equation for the following line. Change point the point and the direction vector: ´ w =( 4, 1,3 )+ t (− 2,1,7 ) )
4. Determine whether the angle between each of the following pairs of vectors is acute, obtuse or neither. a) ´ a = ( 10, 4,1 ) )
5. Given the vector equation in 2-space, ( x, y ) = ( 3,2 ) + t ( 2,4 ) , write a scalar equation for the line. Given the vector equation of this line, (3,2) is a point on the line. Find the slope of the line using the direction vector.
A .
6. Write a vector equation for the line that passes through the point P (-1,0,3) and is parallel to the y-axis. ) | 2022-01-20 02:21:37 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.812052845954895, "perplexity": 457.3119149606303}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320301670.75/warc/CC-MAIN-20220120005715-20220120035715-00180.warc.gz"} |
https://iq.opengenus.org/user-defined-functions-cpp/ | # User defined functions in C++
#### C++ software engineering function user defined function
The function in C++ language is also known as procedure or subroutine in other programming languages.
To perform any task, we can create function. A function can be called many times. It provides modularity and code reusability.
In other words, Functions are used to provide modularity to a program. Creating an application using function makes it easier to understand, edit, check errors etc.
Basically, A function is a group of statements that together perform a task. Every C++ program has at least one function, which is main(), and all the most trivial programs can define additional functions.
You can divide up your code into separate functions. How you divide up your code among different functions is up to you, but logically the division usually is such that each function performs a specific task.
A function declaration tells the compiler about a function's name, return type, and parameters. A function definition provides the actual body of the function.
## Advantage of functions in C++
There are many advantages of functions.
1. Code Reusability :
By creating functions in C++, you can call it many times. So we don't need to write the same code again and again.
2. Code optimization :
It makes the code optimized, we don't need to write much code.
For Example , Suppose you have to check 3 numbers (531, 883 and 781) whether it is prime number or not. Without using function, you need to write the prime number logic 3 times. So, there is repetition of code.
But if you use functions, you need to write the logic only once and you can reuse it several times.
## Syntax for using Functions in C++
Here is how you define a function in C++,
return-type function-name(parameter1, parameter2, ...)
{
// function-body
}
Return Type − A function may return a value. The return_type is the data type of the value the function returns. Some functions perform the desired operations without returning a value. In this case, the return_type is the keyword void.
Function Name − This is the actual name of the function. The function name and the parameter list together constitute the function signature.
Parameters − A parameter is like a placeholder. When a function is invoked, you pass a value to the parameter. This value is referred to as actual parameter or argument. The parameter list refers to the type, order, and number of the parameters of a function. Parameters are optional; that is, a function may contain no parameters.
Function Body − The function body contains a collection of statements that define what the function does.
// function for adding two values
void sum(int x, int y)
{
int z;
z = x + y;
cout << z;
}
int main()
{
int a = 10;
int b = 20;
// calling the function with name 'sum'
sum (a, b);
}
Here, a and b are two variables which are sent as arguments to the function sum, and x and y are parameters which will hold values of a and b to perform the required operation inside the function.
### 1. Function declaration :
It is done to tell the compiler about the existence of the function. That means, A function declaration tells the compiler about a function name and how to call the function. The actual body of the function can be defined separately.
A function declaration has the following parts −
return_type function_name( parameter list );
For the following defined function max(), following is the function declaration −
int max(int num1, int num2);
Parameter names are not important in function declaration only their type is required, so following is also valid declaration −
int max(int, int);
Function declaration is required when you define a function in one source file and you call that function in another file. In such case, you should declare the function at the top of the file calling the function.
Lets understand this with help of an example.
// function returning the max between two numbers
int max(int num1, int num2) {
// local variable declaration
int result;
if (num1 > num2)
result = num1;
else
result = num2;
return result;
}
### 2. Calling a Function :
While creating a C++ function, you give a definition of what the function has to do. To use a function, you will have to call or invoke that function.
When a program calls a function, program control is transferred to the called function. A called function performs defined task and when it’s return statement is executed or when its function-ending closing brace is reached, it returns program control back to the main program.
To call a function, you simply need to pass the required parameters along with function name, and if function returns a value, then you can store returned value. For example −
#include <iostream>
using namespace std;
// function declaration
int max(int num1, int num2);
int main () {
// local variable declaration:
int a = 100;
int b = 200;
int ret;
// calling a function to get max value.
ret = max(a, b);
cout << "Max value is : " << ret << endl;
return 0;
}
// function returning the max between two numbers
int max(int num1, int num2) {
// local variable declaration
int result;
if (num1 > num2)
result = num1;
else
result = num2;
return result;
}
#### Output:
Max value is : 200
### Example :
Now, combining all the above concepts into one program :
#include <iostream>
using namespace std;
//declaring the function
int sum (int x, int y);
int main()
{
int a = 10;
int b = 20;
int c = sum (a, b); //calling the function
cout << c;
}
//defining the function
int sum (int x, int y)
{
return (x + y);
}
Here, initially the function is declared, without body. Then inside main() function it is called, as the function returns sumation of two values, and variable c is there to store the result. Then, at last, function is defined, where the body of function is specified. We can also, declare & define the function together, but then it should be done before it is called.
### Function Arguments
If a function is to use arguments, it must declare variables that accept the values of the arguments. These variables are called the formal parameters of the function.
The formal parameters behave like other local variables inside the function and are created upon entry into the function and destroyed upon exit.
While calling a function, there are three ways that arguments can be passed to a function −
### 1. call by value :
This method of passing arguments to a function copies the actual value of an argument into the formal parameter of the function. In this case, changes made to the parameter inside the function have no effect on the argument.
NOTE : By default, C++ uses call by value to pass arguments.
In general, this means that code within a function cannot alter the arguments used to call the function. Consider the program of swapping numbers.
### Example :
#include <iostream>
using namespace std;
// function definition to swap the values.
void swap(int x, int y) {
int temp;
temp = x; /* save the value of x */
x = y; /* put y into x */
y = temp; /* put x into y */
return;
}
int main () {
// local variable declaration:
int a = 100;
int b = 200;
cout << "Before swap, value of a :" << a << endl;
cout << "Before swap, value of b :" << b << endl;
// calling a function to swap the values.
swap(a, b);
cout << "After swap, value of a :" << a << endl;
cout << "After swap, value of b :" << b << endl;
return 0;
}
#### Output :
Before swap, value of a :100
Before swap, value of b :200
After swap, value of a :100
After swap, value of b :200
Which shows that there is no change in the values though they had been changed inside the function.
### 2. call by pointer:
This method of passing arguments to a function copies the address of an argument into the formal parameter. Inside the function, the address is used to access the actual argument used in the call. This means that changes made to the parameter affect the passed argument.
To pass the value by pointer, argument pointers are passed to the functions just like any other value. So accordingly you need to declare the function parameters as pointer types as in the following function swap(), which exchanges the values of the two integer variables pointed to by its arguments
### Example :
#include <iostream>
using namespace std;
// function definition to swap the values.
void swap(int *x, int *y) {
int temp;
temp = *x; /* save the value at address x */
*x = *y; /* put y into x */
*y = temp; /* put x into y */
return;
}
#include <iostream>
using namespace std;
// function declaration
void swap(int *x, int *y);
int main () {
// local variable declaration:
int a = 100;
int b = 200;
cout << "Before swap, value of a :" << a << endl;
cout << "Before swap, value of b :" << b << endl;
/* calling a function to swap the values.
* &a indicates pointer to a ie. address of variable a and
* &b indicates pointer to b ie. address of variable b.
*/
swap(&a, &b);
cout << "After swap, value of a :" << a << endl;
cout << "After swap, value of b :" << b << endl;
return 0;
}
#### Output :
Before swap, value of a :100
Before swap, value of b :200
After swap, value of a :200
After swap, value of b :100
### 3. call by reference
This method of passing arguments to a function copies the reference of an argument into the formal parameter. Inside the function, the reference is used to access the actual argument used in the call. This means that changes made to the parameter affect the passed argument.
To pass the value by reference, argument reference is passed to the functions just like any other value. So accordingly you need to declare the function parameters as reference types as in the following function swap(), which exchanges the values of the two integer variables pointed to by its arguments
### Example :
#include <iostream>
using namespace std;
// function definition to swap the values.
// function definition to swap the values.
void swap(int &x, int &y) {
int temp;
temp = x; /* save the value at address x */
x = y; /* put y into x */
y = temp; /* put x into y */
return;
}
#include <iostream>
using namespace std;
// function declaration
void swap(int &x, int &y);
int main () {
// local variable declaration:
int a = 100;
int b = 200;
cout << "Before swap, value of a :" << a << endl;
cout << "Before swap, value of b :" << b << endl;
/* calling a function to swap the values using variable reference.*/
swap(a, b);
cout << "After swap, value of a :" << a << endl;
cout << "After swap, value of b :" << b << endl;
return 0;
}
#### Output :
Before swap, value of a :100
Before swap, value of b :200
After swap, value of a :200
After swap, value of b :100
Note : Although we are getting same outputs of above programs for References and pointer but there is difference between them .
Pointers: A pointer is a variable that holds memory address of another variable. A pointer needs to be dereferenced with * operator to access the memory location it points to.
References : A reference variable is an alias, that is, another name for an already existing variable. A reference, like a pointer is also implemented by storing the address of an object.A reference can be thought of as a constant pointer (not to be confused with a pointer to a constant value!) with automatic indirection, i.e the compiler will apply the * operator for you.
### Types of Functions
There are two types of functions in C programming:
1. Library Functions : are the functions which are declared in the C++ header files such as ceil(x), cos(x), exp(x) etc.
2. User-defined functions : are the functions which are created by the C++ programmer, so that he/she can use it many times. It reduces complexity of a big program and optimizes the code.
Here our major area of study is User-defined functions. so, we will focus on this only.
## User-defined functions
C++ allows programmer to define their own function.
A user-defined function groups code to perform a specific task and that group of code is given a name(identifier).
When the function is invoked from any part of program, it all executes the codes defined in the body of function.
### Example of User-defined functions
#include <iostream>
using namespace std;
// Function prototype (declaration)
int main()
{
int num1, num2, sum;
cout<<"Enters two numbers to add: ";
cin >> num1 >> num2;
// Function call
cout << "Sum = " << sum;
return 0;
}
// Function definition
{
// Return statement
}
### Output
Enters two integers: 8
-4
Sum = 4
### Types of User-defined functions
There can be 4 different types of user-defined functions, they are:
1. Function with no arguments and no return value
2. Function with no arguments and a return value
3. Function with arguments and no return value
4. Function with arguments and a return value
Below, we will discuss about all these types, along with program examples.
### 1. Function with no arguments and no return value :
Such functions can either be used to display information or they are completely dependent on user inputs.
#### Example :
In the following program, prime() is called from the main() with 0no arguments.
prime() takes the positive number from the user and checks whether the number is a prime number or not.
Since, return type of prime() is void, no value is returned from the function.
# include <iostream>
using namespace std;
void prime();
int main()
{
// No argument is passed to prime()
prime();
return 0;
}
// Return type of function is void because value is not returned.
void prime()
{
int num, i, flag = 0;
cout << "Enter a positive integer enter to check: ";
cin >> num;
for(i = 2; i <= num/2; ++i)
{
if(num % i == 0)
{
flag = 1;
break;
}
}
if (flag == 1)
{
cout << num << " is not a prime number.";
}
else
{
cout << num << " is a prime number.";
}
}
### 2. Function with no arguments and a return value :
In the following program, prime() function is called from the main() with no arguments.
prime() takes a positive integer from the user. Since, return type of the function is an int, it returns the inputted number from the user back to the calling main() function.
Then, whether the number is prime or not is checked in the main() itself and printed onto the screen.
#### Example
#include <iostream>
using namespace std;
int prime();
int main()
{
int num, i, flag = 0;
// No argument is passed to prime()
num = prime();
for (i = 2; i <= num/2; ++i)
{
if (num%i == 0)
{
flag = 1;
break;
}
}
if (flag == 1)
{
cout<<num<<" is not a prime number.";
}
else
{
cout<<num<<" is a prime number.";
}
return 0;
}
// Return type of function is int
int prime()
{
int n;
printf("Enter a positive integer to check: ");
cin >> n;
return n;
}
### 3. Function with arguments and no return value :
We are using the same function as example again and again, to demonstrate that to solve a problem there can be many different ways.
This time, we have modified the above example to make the function prime() take one int value as argument, but it will not be returning anything.
In the program, positive number is first asked from the user which is stored in the variable num.
Then, num is passed to the prime() function where, whether the number is prime or not is checked and printed.
Since, the return type of prime() is a void, no value is returned from the function.
### Example
#include <iostream>
using namespace std;
void prime(int n);
int main()
{
int num;
cout << "Enter a positive integer to check: ";
cin >> num;
// Argument num is passed to the function prime()
prime(num);
return 0;
}
// There is no return value to calling function. Hence, return type of function is void. */
void prime(int n)
{ int i, flag = 0;
for (i = 2; i <= n/2; ++i)
{
if (n%i == 0)
{
flag = 1;
break;
}
}
if (flag == 1)
{
cout << n << " is not a prime number.";
}
else {
cout << n << " is a prime number.";
}
}
### 4. Function with arguments and a return value :
This is the best type, as this makes the function completely independent of inputs and outputs, and only the logic is defined inside the function body.
In the following program, a positive integer is asked from the user and stored in the variable num.
Then, num is passed to the function prime() where, whether the number is prime or not is checked.
Since, the return type of prime() is an int, 1 or 0 is returned to the main() calling function. If the number is a prime number, 1 is returned. If not, 0 is returned.
Back in the main() function, the returned 1 or 0 is stored in the variable flag, and the corresponding text is printed onto the screen.
### Example
#include <iostream>
using namespace std;
int prime(int n);
int main()
{
int num, flag = 0;
cout << "Enter positive integer to check: ";
cin >> num;
// Argument num is passed to check() function
flag = prime(num);
if(flag == 1)
cout << num << " is not a prime number.";
else
cout<< num << " is a prime number.";
return 0;
}
/* This function returns integer value. */
int prime(int n)
{
int i;
for(i = 2; i <= n/2; ++i)
{
if(n % i == 0)
return 1;
}
return 0;
}
#### Robot Jelly
I am a contributor at OpenGenus since March 2019 | 2019-10-19 22:37:00 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.26526549458503723, "perplexity": 2645.9371416652893}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986700435.69/warc/CC-MAIN-20191019214624-20191020002124-00297.warc.gz"} |
https://in.mathworks.com/help/physmod/sps/ref/modularmultilevelconverterthreephase.html | # Modular Multilevel Converter (Three-Phase)
Three-phase modular multilevel converter with series-connected power submodules
• Library:
• Simscape / Electrical / Semiconductors & Converters / Converters
## Description
The Modular Multilevel Converter (Three-Phase) block models a three-phase modular multilevel converter. Each phase consists of two arms that are implemented with a number of series-connected power submodules.
Each submodule consists of a half-bridge or full-bridge converter and a capacitor.
Half-Bridge Topology
Full-Bridge Topology
This blocks allows you to select the level of model fidelity by choosing between a detailed model with switching devices or an equivalent model. You can choose from these switching devices are:
• GTO — Gate turn-off thyristor. For information about the I-V characteristic of the device, see GTO.
• Ideal semiconductor switch — For information about the I-V characteristic of the device, see Ideal Semiconductor Switch.
• IGBT — Insulated-gate bipolar transistor. For information about the I-V characteristic of the device, see IGBT (Ideal, Switching).
• MOSFET — N-channel metal-oxide-semiconductor field-effect transistor. For information about the I-V characteristic of the device, see MOSFET (Ideal, Switching).
• Thyristor — For information about the I-V characteristic of the device, see Thyristor (Piecewise Linear).
• Averaged Switch — Semiconductor switch with an anti-parallel diode. The control signal port, G, accepts values in the [0,1] interval. When the value at port G is equal to 0 or 1, the averaged switch is either fully opened or fully closed, and it behaves similarly to the Ideal Semiconductor Switch block with an anti-parallel diode. When the value at port G is between 0 and 1, the averaged switch is partly opened. You can then average the PWM signal over a specified period. This allows for undersampling of the model or using modulation waveforms instead of PWM signals.
### Variables
Use the Variables settings to specify the priority and initial target values for the block variables before simulation. For more information, see Set Priority and Initial Target for Block Variables.
## Ports
### Input
expand all
Physical signal port associated with the gate signal for all submodules of the modular multilevel converter, specified as a matrix.
If you set the Converter topology parameter to Half-bridge, the signal is a vector of length 12 * Nsm, where Nsm is the Number of power submodules.
If you set the Converter topology parameter to Full-bridge, the signal is a vector of length 24 * Nsm.
#### Dependencies
To enable this port, set Fidelity level to Detailed model - switching devices or Equivalent model - PWM-controlled.
Physical signal port associated with the reference waveforms of the submodules, specified as a vector of length 6.
#### Dependencies
To enable this port, set Fidelity level to Equivalent model - waveform-controlled.
### Output
expand all
Physical signal port associated with the capacitor voltages for each submodule in the modular multilevel converter, returned as a vector.
#### Dependencies
To enable this port, set Capacitor voltages to Instrumented.
### Conserving
expand all
Electrical conserving port associated with the positive terminal of the modular multilevel converter.
Electrical conserving port associated with the negative terminal of the modular multilevel converter.
Expandable three-phase electrical port associated with the three-phase terminal of the modular multilevel converter.
#### Dependencies
To enable this port, set Electrical connection to Composite three-phase ports.
Electrical conserving port associated with a-phase.
#### Dependencies
To enable this port, set Electrical connection to Expanded three-phase ports.
Electrical conserving port associated with b-phase.
#### Dependencies
To enable this port, set Electrical connection to Expanded three-phase ports.
Electrical conserving port associated with c-phase.
#### Dependencies
To enable this port, set Electrical connection to Expanded three-phase ports.
## Parameters
expand all
### Main
Whether to have composite or expanded three-phase ports.
Topology of the modular multilevel converter.
Level of fidelity of the model.
Whether to instrument the capacitor voltages.
Number of power submodules of the modular multilevel converter.
Capacitance of a submodule. If you enter a vector, the vector must be of length 6 * Nsm, where Nsm is the Number of power submodules.
Capacitor effective series resistance. If you enter a vector, the vector must be of length 6 * Nsm, where Nsm is the Number of power submodules.
#### Dependencies
To enable this parameter, set Fidelity level to Detailed model - switching devices.
Inductance of the arm.
Arm inductance series resistance.
Initial voltage of the capacitor. If you enter a vector, the vector must be of length 6 * Nsm, where Nsm is the Number of power submodules.
### Switching Devices
This table shows how enabled parameters in the Switching Devices settings depend on the Switching device that you select. To learn how to read the table, see Parameter Dependencies. To enable these settings, set Fidelity level to Detailed model - switching devices.
Switching Devices Parameter Dependencies
Parameters and Options
Switching device
Ideal Semiconductor SwitchGTOIGBTMOSFETThyristorAveraged Switch
On-state resistanceForward voltageForward voltageDrain-source on resistanceForward voltageOn-state resistance
Off-state conductanceOn-state resistanceOn-state resistanceOff-state conductanceOn-state resistance
Threshold voltageOff-state conductanceOff-state conductanceThreshold voltageOff-state conductance
Gate trigger voltage, VgtThreshold voltageGate trigger voltage, Vgt
Gate turn-off voltage, Vgt_offHolding current
Holding current
Switching device type for the converter.
#### Dependencies
See the Switching Devices Parameter Dependencies table.
For the different switching device types, the Forward voltage is the:
• GTO — Minimum voltage required across the anode and cathode block ports for the gradient of the device I-V characteristic to be 1/Ron, where Ron is the value of On-state resistance
• IGBT — Minimum voltage required across the collector and emitter block ports for the gradient of the diode I-V characteristic to be 1/Ron, where Ron is the value of On-state resistance
• Thyristor — Minimum voltage required for the device to turn on
#### Dependencies
See the Switching Devices Parameter Dependencies table.
For the different switching device types, the On-state resistance is the:
• GTO — Rate of change of the voltage versus the current above the forward voltage
• Ideal semiconductor switch — Anode-cathode resistance when the device is on
• IGBT — Collector-emitter resistance when the device is on
• Thyristor — Anode-cathode resistance when the device is on
• Averaged switch — Anode-cathode resistance when the device is on
#### Dependencies
See the Switching Devices Parameter Dependencies table.
Resistance between the drain and the source. The gate-to-source voltage also affects the Drain-source on resistance parameter.
#### Dependencies
See the Switching Devices Parameter Dependencies table.
Conductance when the device is off. The value must be less than 1/R, where R is the value of On-state resistance.
For the different switching device types, the On-state resistance is the:
• GTO — Anode-cathode conductance
• Ideal semiconductor switch — Anode-cathode conductance
• IGBT — Collector-emitter conductance
• MOSFET — Drain-source conductance
• Thyristor — Anode-cathode conductance
#### Dependencies
See the Switching Devices Parameter Dependencies table.
Gate voltage threshold. The device turns on when the gate voltage is above this value. The voltage threshold applies to different devices depending on the switching device used:
• Ideal semiconductor switch — Gate-emitter voltage
• IGBT — Gate-cathode voltage
• MOSFET — Gate-source voltage
#### Dependencies
See the Switching Devices Parameter Dependencies table.
Gate-cathode voltage threshold. The device turns on when the gate-cathode voltage is above this value.
#### Dependencies
See the Switching Devices Parameter Dependencies table.
Gate-cathode voltage threshold. The device turns off when the gate-cathode voltage is below this value.
#### Dependencies
See the Switching Devices Parameter Dependencies table.
Gate current threshold. The device stays on when the current is above this value, even when the gate-cathode voltage falls below the gate trigger voltage.
#### Dependencies
See the Switching Devices Parameter Dependencies table.
### Protection Diodes
Diode type. The options are:
• None — The block does not model diode dynamics.
• Diode with no dynamics — Select this option to prioritize simulation speed using the Diode block.
• Diode with charge dynamics — Select this option to prioritize model fidelity in terms of reverse mode charge dynamics using the commutation diode model of the Diode block.
Note
If you set Switching device to Averaged Switch in the Switching Device settings, the Diode with no dynamics setting is automatically selected.
#### Dependencies
To enable this parameter, set Switching device to GTO, Ideal Semiconductor Switch, IGBT, MOFSET, or Thyristor.
Minimum voltage required across the + and - block ports for the gradient of the diode I-V characteristic to be 1/Ron, where Ron is the value of On resistance.
#### Dependencies
To enable this parameter, set Model dynamics to Diode with no dynamics or Diode with charge dynamics.
Rate of change of voltage versus the current above the Forward voltage.
#### Dependencies
To enable this parameter, set Model dynamics to Diode with no dynamics or Diode with charge dynamics.
Conductance of the reverse-biased diode.
#### Dependencies
To enable this parameter, set Model dynamics to Diode with no dynamics or Diode with charge dynamics.
Diode junction capacitance.
#### Dependencies
To enable this parameter, set Model dynamics to Diode with charge dynamics.
Peak reverse current measured by an external test circuit. This value must be less than zero.
#### Dependencies
To enable this parameter, set Model dynamics to Diode with charge dynamics.
Initial forward current when measuring peak reverse current. This value must be greater than zero.
#### Dependencies
To enable this parameter, set Model dynamics to Diode with charge dynamics.
Rate of change of the current when measuring the peak reverse current. This value must be less than zero.
#### Dependencies
To enable this parameter, set Model dynamics to Diode with charge dynamics.
Determines how you specify reverse recovery time.
If you select Specify stretch factor or Specify reverse recovery charge, you specify a value that the block uses to derive the reverse recovery time. For more information on these options, see How the Block Calculates TM and Tau.
#### Dependencies
To enable this parameter, set Model dynamics to Diode with charge dynamics.
Interval between the time when the current initially goes to zero (when the diode turns off) and the time when the current falls to less than 10% of the peak reverse current. The value of the Reverse recovery time, trr parameter must be greater than the value of the Peak reverse current, iRM parameter divided by the value of the Rate of change of current when measuring iRM parameter.
#### Dependencies
To enable this parameter, set Model dynamics to Diode with charge dynamics and Reverse recovery time parameterization to Specify reverse recovery time directly.
Value that the block uses to calculate Reverse recovery time, trr. This value must be greater than 1. Specifying the stretch factor is an easier way to parameterize the reverse recovery time than specifying the reverse recovery charge. The larger the value of the stretch factor, the longer it takes for the reverse recovery current to dissipate.
#### Dependencies
To enable this parameter, set Model dynamics to Diode with charge dynamics and Reverse recovery time parameterization to Specify stretch factor.
Value that the block uses to calculate Reverse recovery time, trr. Use this parameter if the data sheet for your diode specifies a value for the reverse recovery charge instead of a value for the reverse recovery time.
The reverse recovery charge is the total charge that continues to dissipate when the diode turns off. The value must be less than $-\frac{{i}^{2}{}_{RM}}{2a},$ where:
• iRM is the value specified for Peak reverse current, iRM.
• a is the value specified for Rate of change of current when measuring iRM.
#### Dependencies
To enable this parameter, set Model dynamics to Diode with charge dynamics and Reverse recovery time parameterization to Specify reverse recovery charge.
### Snubbers
To enable the Snubbers settings, set Fidelity level to Detailed model - switching devices and Switching device to GTO, Ideal Semiconductor Switch, IGBT, MOFSET, or Thyristor.
Snubber for each switching device:
• None
• RC snubber
Snubber resistance.
#### Dependencies
To enable this parameter, set Snubber to RC snubber.
Snubber capacitance.
#### Dependencies
To enable this parameter, set Snubber to RC snubber.
## References
[1] Saad, Hani, Sebastien Dennetiere, and Jean Mahseredjian. “On Modelling of MMC in EMT-Type Program.” 2016 IEEE 17th Workshop on Control and Modeling for Power Electronics (COMPEL), 1–7. Trondheim, Norway: IEEE, 2016. https://doi.org/10.1109/COMPEL.2016.7556717.
## Version History
Introduced in R2020b
expand all
Behavior changed in R2021b | 2022-08-12 23:24:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6476210355758667, "perplexity": 6467.796769875999}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571847.45/warc/CC-MAIN-20220812230927-20220813020927-00202.warc.gz"} |
https://calculla.com/calculators/physics/mass_law_for_single_wall | Mass law calculator for single layer
Calculator finds out sound reduction in decibels (dB) for single layer wall with given density and thickness using so-called mass law equation.
# Beta version#
BETA TEST VERSION OF THIS ITEM
This online calculator is currently under heavy development. It may or it may NOT work correctly.
You CAN try to use it. You CAN even get the proper results.
However, please VERIFY all results on your own, as the level of completion of this item is NOT CONFIRMED.
Feel free to send any ideas and comments !
# Transmission loss at various frequencies#
Frequency [Hz] Transmission loss [dB] 31.25 -inf 62.5 -inf 125 -inf 250 -inf 500 -inf 1000 -inf 2000 -inf 4000 -inf 8000 -inf 16000 -inf
# Sound insulation#
• When a sound wave moves through the air meets a barrier in the form of a wall, part of the acoustic energy is reflected, part is absorbed inside the wall (converted to heat), and the another part is transmited out (on the other side of the wall). We can write it mathematically as follows:
$\alpha + \beta + \tau = 1$
where:
• $\alpha$ - absorption coefficient (determines the part of the energy that was absorbed inside the wall),
• $\beta$ - reflection coefficient (defines the part of the energy remaining in the first room),
• $\tau$ - transmission coefficient (defines the part of the energy that was emitted to the second room).
• The transmission coefficient can be used as a measure of the acoustic insulation, because it determines the sound intensity ratio on both sides of the wall:
$\tau = \frac{I_t}{I_0}$
where:
• $I_t$ - intensity of the wave on the other side of the wall (sound intensity level audible in the second room),
• $I_0$ - incident wave intensity (sound intensity level audible in the first room).
• In practice, the transmission factor is most often given in the logarithmic scale. In this way, we obtain a decrease in sound intensity given in decibels, so-called transmission loss:
$\Delta TL = -10 ~ log (\tau) = 10 ~ log \left(\frac{1}{\tau}\right)$
# Some facts#
• The transmission factor for a uniform single wall can be estimated using the expression:
$\tau \approx \left[1 + \left (\frac{2 \pi \cdot f \cdot h \cdot d}{3.6 ~ d_0 ~ c_0} \right)^2 \right]^{-1}$
where:
• $\tau$ - transmission coefficient (part of the energy that passed through the partition),
• $f$ - frequency of the acoustic wave,
• $h$ - partition thickness,
• $d$ - material density the partition is made of,
• $d_0$ - air density,
• $c_0$ - speed of sound in the air.
• If the above expression after applying logarithmic scale, neglecting add a one (the component on the right is much greater than one - the wall density is about 1000 times greater than that of air) and moving the constants out of the logarithm, we get the equation known in building acoustics as the law of mass:
$\Delta TL \approx 20 ~ log (f \cdot h \cdot d) - 47.3$
where:
• $\Delta TL$ - transmission loss in the logarithmic scale (decibels),
• f - frequency of the acoustic wave,
• h - wall thickness,
• d - material density the wall is made of.
• The law of mass allows to estimate acoustic insulation at the air-solid boundary. In practice, this means that the mass law in above form applies to the airborne insulation (e.g. loud music or screaming).
If you're interested in calculators related to acoustics, check out our other calculators:
• Sound intensity level (dB) - if you want to learn what is decibel and how the sound intensity level is measured,
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https://www.aimsciences.org/article/doi/10.3934/dcds.2019301 | American Institute of Mathematical Sciences
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December 2019, 39(12): 7213-7248. doi: 10.3934/dcds.2019301
Free boundaries subject to topological constraints
1 Department of Mathematics, Massachusetts Institute of Technology, 77 Massachusetts Ave, Cambridge, MA 02139, USA 2 Facultad de Matemáticas, Pontificia Universidad Católica de Chile, Avenida Vicuña Mackenna 4860, Santiago 7820436, Chile
* Corresponding author
Received January 2019 Revised May 2019 Published September 2019
Fund Project: Supported in part by NSF Grant DMS 1500771, a Simons Fellowship, and Simons Foundation grant (601948, DJ). NK was partially supported by Proyecto FONDECYT Iniciación No. 11160981.
We discuss the extent to which solutions to one-phase free boundary problems can be characterized according to their topological complexity. Our questions are motivated by fundamental work of Luis Caffarelli on free boundaries and by striking results of T. Colding and W. Minicozzi concerning finitely connected, embedded, minimal surfaces. We review our earlier work on the simplest case, one-phase free boundaries in the plane in which the positive phase is simply connected. We also prove a new, purely topological, effective removable singularities theorem for free boundaries. At the same time, we formulate some open problems concerning the multiply connected case and make connections with the theory of minimal surfaces and semilinear variational problems.
Citation: David Jerison, Nikola Kamburov. Free boundaries subject to topological constraints. Discrete & Continuous Dynamical Systems, 2019, 39 (12) : 7213-7248. doi: 10.3934/dcds.2019301
References:
[1] H. W. Alt and L. A. Caffarelli, Existence and regularity for a minimum problem with free boundary, J. Reine Angew. Math., 325 (1981), 105-144. Google Scholar [2] H. W. Alt, L. A. Caffarelli and A. Friedman, Variational problems with two phases and their free boundaries, Trans. Amer. Math. Soc., 282 (1984), 431-461. doi: 10.1090/S0002-9947-1984-0732100-6. Google Scholar [3] G. Baker, P. Saffman and J. Sheffield, Structure of a linear array of hollow vortices of finite cross-section, Journal of Fluid Mechanics, 74 (1976), 469-476. doi: 10.1017/S0022112076001894. Google Scholar [4] H. Berestycki, L. Caffarelli and L. Nirenberg, Monotonicity for elliptic equations in unbounded Lipschitz domains, Comm. Pure and Appl. Math., 50 (1997), 1089-1111. doi: 10.1002/(SICI)1097-0312(199711)50:11<1089::AID-CPA2>3.0.CO;2-6. Google Scholar [5] L. A. Caffarelli and S. Salsa, A Geometric Approach to Free Boundary Problems, vol. 68 of Graduate Studies in Mathematics, American Mathematical Society, Providence, RI, 2005. doi: 10.1090/gsm/068. Google Scholar [6] L. A. Caffarelli, A Harnack inequality approach to the regularity of free boundaries. Ⅰ. Lipschitz free boundaries are $C^{1, \alpha}$, Rev. Mat. Iberoamericana, 3 (1987), 139-162. doi: 10.4171/RMI/47. Google Scholar [7] L. A. Caffarelli, A Harnack inequality approach to the regularity of free boundaries. Ⅲ. Existence theory, compactness, and dependence on $X$, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4), 15 (1988), 583–602 (1989). Google Scholar [8] L. A. Caffarelli, A Harnack inequality approach to the regularity of free boundaries. Ⅱ. Flat free boundaries are Lipschitz, Comm. Pure Appl. Math., 42 (1989), 55-78. doi: 10.1002/cpa.3160420105. Google Scholar [9] O. Chodosh and D. Maximo, On the topology and index of minimal surfaces, Journal of Differential Geometry, 104 (2016), 399-418. doi: 10.4310/jdg/1478138547. Google Scholar [10] H. I. Choi and R. Schoen, The space of minimal embeddings of a surface into a three-dimensional manifold of positive Ricci curvature, Inventiones Mathematicae, 81 (1985), 387-394. doi: 10.1007/BF01388577. Google Scholar [11] T. H. Colding and W. P. Minicozzi Ⅱ, The Calabi-Yau conjectures for embedded surfaces, Annals of Mathematics, 167 (2008), 211-243. doi: 10.4007/annals.2008.167.211. Google Scholar [12] T. H. Colding and W. P. Minicozzi Ⅱ, On the structure of embedded minimal annuli, International Mathematics Research Notices, 2002 (2002), 1539-1552. doi: 10.1155/S1073792802112128. Google Scholar [13] T. H. Colding and W. P. Minicozzi Ⅱ, The space of embedded minimal surfaces of fixed genus in a 3-manifold. Ⅲ. Planar domains, Ann. of Math. (2), 160 (2004), 523–572. doi: 10.4007/annals.2004.160.523. Google Scholar [14] T. H. Colding and W. P. Minicozzi Ⅱ, The space of embedded minimal surfaces of fixed genus in a 3-manifold. Ⅳ. Locally simply connected, Ann. of Math. (2), 160 (2004), 573–615. doi: 10.4007/annals.2004.160.573. Google Scholar [15] A. Farina and E. Valdinoci, Flattening results for elliptic PDEs in unbounded domains with applications to overdetermined problems, Archive for Rational Mechanics and Analysis, 195 (2010), 1025-1058. doi: 10.1007/s00205-009-0227-8. Google Scholar [16] L. Hauswirth, F. Hélein and F. Pacard, On an overdetermined elliptic problem, Pacific J. Math., 250 (2011), 319-334. doi: 10.2140/pjm.2011.250.319. Google Scholar [17] D. Jerison and N. Kamburov, Structure of one-phase free boundaries in the plane, International Mathematics Research Notices, 2016 (2016), 5922-5987. doi: 10.1093/imrn/rnv339. Google Scholar [18] D. Jerison and K. Perera, Higher critical points in an elliptic free boundary problem, J. Geom Anal., 28 (2018), 1258-1294. doi: 10.1007/s12220-017-9862-8. Google Scholar [19] D. Khavinson, E. Lundberg and R. Teodorescu, An overdetermined problem in potential theory, Pacific J. Math., 265 (2013), 85-111. doi: 10.2140/pjm.2013.265.85. Google Scholar [20] Y. Liu, K. Wang and J. Wei, On one phase free boundary problem in $\mathbb{R}^n$, preprint, arXiv: 1705.07345. Google Scholar [21] Y. Liu, K. Wang and J. Wei, Half space theorem for the Allen-Cahn equation, preprint, arXiv: 1901.07671. Google Scholar [22] W. H. Meeks Ⅲ and H. Rosenberg, The uniqueness of the helicoid, Annals of Mathematics, 161 (2005), 727-758. doi: 10.4007/annals.2005.161.727. Google Scholar [23] A. Ros, D. Ruiz and P. Sicbaldi, A rigidity result for overdetermined elliptic problems in the plane, Communications on Pure and Applied Mathematics, 70 (2017), 1223-1252. doi: 10.1002/cpa.21696. Google Scholar [24] A. Ros and P. Sicbaldi, Geometry and topology of some overdetermined elliptic problems, J. Differential Equations, 255 (2013), 951-977. doi: 10.1016/j.jde.2013.04.027. Google Scholar [25] R. Schoen, Uniqueness, symmetry, and embeddedness of minimal surfaces, Journal of Differential Geometry, 18 (1983), 791-809. doi: 10.4310/jdg/1214438183. Google Scholar [26] M. Traizet, Classification of the solutions to an overdetermined elliptic problem in the plane, Geom. Funct. Anal., 24 (2014), 690-720. doi: 10.1007/s00039-014-0268-5. Google Scholar [27] K. Wang, The structure of finite Morse index solutions to two free boundary problems in $\mathbb{R}^2$, preprint, arXiv: 1506.00491. Google Scholar [28] K. Wang and J. Wei, On Serrin's overdetermined problem and a conjecture of Berestycki, Caffarelli and Nirenberg, Comm. Partial Differential Equations, 44 (2019), 837–858, arXiv: 1502.04680. doi: 10.1080/03605302.2019.1611846. Google Scholar [29] K. Wang and J. Wei, Finite Morse index implies finite ends, Comm. Pure Appl. Math., 72 (2019), 1044–1119, arXiv: 1705.06831. doi: 10.1002/cpa.21812. Google Scholar [30] G. S. Weiss, Partial regularity for weak solutions of an elliptic free boundary problem, Comm. Partial Differential Equations, 23 (1998), 439-455. doi: 10.1080/03605309808821352. Google Scholar [31] B. White, Curvature estimates and compactness theorems in 3-manifolds for surfaces that are stationary for parametric elliptic functionals, Inventiones Mathematicae, 88 (1987), 243-256. doi: 10.1007/BF01388908. Google Scholar
show all references
References:
[1] H. W. Alt and L. A. Caffarelli, Existence and regularity for a minimum problem with free boundary, J. Reine Angew. Math., 325 (1981), 105-144. Google Scholar [2] H. W. Alt, L. A. Caffarelli and A. Friedman, Variational problems with two phases and their free boundaries, Trans. Amer. Math. Soc., 282 (1984), 431-461. doi: 10.1090/S0002-9947-1984-0732100-6. Google Scholar [3] G. Baker, P. Saffman and J. Sheffield, Structure of a linear array of hollow vortices of finite cross-section, Journal of Fluid Mechanics, 74 (1976), 469-476. doi: 10.1017/S0022112076001894. Google Scholar [4] H. Berestycki, L. Caffarelli and L. Nirenberg, Monotonicity for elliptic equations in unbounded Lipschitz domains, Comm. Pure and Appl. Math., 50 (1997), 1089-1111. doi: 10.1002/(SICI)1097-0312(199711)50:11<1089::AID-CPA2>3.0.CO;2-6. Google Scholar [5] L. A. Caffarelli and S. Salsa, A Geometric Approach to Free Boundary Problems, vol. 68 of Graduate Studies in Mathematics, American Mathematical Society, Providence, RI, 2005. doi: 10.1090/gsm/068. Google Scholar [6] L. A. Caffarelli, A Harnack inequality approach to the regularity of free boundaries. Ⅰ. Lipschitz free boundaries are $C^{1, \alpha}$, Rev. Mat. Iberoamericana, 3 (1987), 139-162. doi: 10.4171/RMI/47. Google Scholar [7] L. A. Caffarelli, A Harnack inequality approach to the regularity of free boundaries. Ⅲ. Existence theory, compactness, and dependence on $X$, Ann. Scuola Norm. Sup. Pisa Cl. Sci. (4), 15 (1988), 583–602 (1989). Google Scholar [8] L. A. Caffarelli, A Harnack inequality approach to the regularity of free boundaries. Ⅱ. Flat free boundaries are Lipschitz, Comm. Pure Appl. Math., 42 (1989), 55-78. doi: 10.1002/cpa.3160420105. Google Scholar [9] O. Chodosh and D. Maximo, On the topology and index of minimal surfaces, Journal of Differential Geometry, 104 (2016), 399-418. doi: 10.4310/jdg/1478138547. Google Scholar [10] H. I. Choi and R. Schoen, The space of minimal embeddings of a surface into a three-dimensional manifold of positive Ricci curvature, Inventiones Mathematicae, 81 (1985), 387-394. doi: 10.1007/BF01388577. Google Scholar [11] T. H. Colding and W. P. Minicozzi Ⅱ, The Calabi-Yau conjectures for embedded surfaces, Annals of Mathematics, 167 (2008), 211-243. doi: 10.4007/annals.2008.167.211. Google Scholar [12] T. H. Colding and W. P. Minicozzi Ⅱ, On the structure of embedded minimal annuli, International Mathematics Research Notices, 2002 (2002), 1539-1552. doi: 10.1155/S1073792802112128. Google Scholar [13] T. H. Colding and W. P. Minicozzi Ⅱ, The space of embedded minimal surfaces of fixed genus in a 3-manifold. Ⅲ. Planar domains, Ann. of Math. (2), 160 (2004), 523–572. doi: 10.4007/annals.2004.160.523. Google Scholar [14] T. H. Colding and W. P. Minicozzi Ⅱ, The space of embedded minimal surfaces of fixed genus in a 3-manifold. Ⅳ. Locally simply connected, Ann. of Math. (2), 160 (2004), 573–615. doi: 10.4007/annals.2004.160.573. Google Scholar [15] A. Farina and E. Valdinoci, Flattening results for elliptic PDEs in unbounded domains with applications to overdetermined problems, Archive for Rational Mechanics and Analysis, 195 (2010), 1025-1058. doi: 10.1007/s00205-009-0227-8. Google Scholar [16] L. Hauswirth, F. Hélein and F. Pacard, On an overdetermined elliptic problem, Pacific J. Math., 250 (2011), 319-334. doi: 10.2140/pjm.2011.250.319. Google Scholar [17] D. Jerison and N. Kamburov, Structure of one-phase free boundaries in the plane, International Mathematics Research Notices, 2016 (2016), 5922-5987. doi: 10.1093/imrn/rnv339. Google Scholar [18] D. Jerison and K. Perera, Higher critical points in an elliptic free boundary problem, J. Geom Anal., 28 (2018), 1258-1294. doi: 10.1007/s12220-017-9862-8. Google Scholar [19] D. Khavinson, E. Lundberg and R. Teodorescu, An overdetermined problem in potential theory, Pacific J. Math., 265 (2013), 85-111. doi: 10.2140/pjm.2013.265.85. Google Scholar [20] Y. Liu, K. Wang and J. Wei, On one phase free boundary problem in $\mathbb{R}^n$, preprint, arXiv: 1705.07345. Google Scholar [21] Y. Liu, K. Wang and J. Wei, Half space theorem for the Allen-Cahn equation, preprint, arXiv: 1901.07671. Google Scholar [22] W. H. Meeks Ⅲ and H. Rosenberg, The uniqueness of the helicoid, Annals of Mathematics, 161 (2005), 727-758. doi: 10.4007/annals.2005.161.727. Google Scholar [23] A. Ros, D. Ruiz and P. Sicbaldi, A rigidity result for overdetermined elliptic problems in the plane, Communications on Pure and Applied Mathematics, 70 (2017), 1223-1252. doi: 10.1002/cpa.21696. Google Scholar [24] A. Ros and P. Sicbaldi, Geometry and topology of some overdetermined elliptic problems, J. Differential Equations, 255 (2013), 951-977. doi: 10.1016/j.jde.2013.04.027. Google Scholar [25] R. Schoen, Uniqueness, symmetry, and embeddedness of minimal surfaces, Journal of Differential Geometry, 18 (1983), 791-809. doi: 10.4310/jdg/1214438183. Google Scholar [26] M. Traizet, Classification of the solutions to an overdetermined elliptic problem in the plane, Geom. Funct. Anal., 24 (2014), 690-720. doi: 10.1007/s00039-014-0268-5. Google Scholar [27] K. Wang, The structure of finite Morse index solutions to two free boundary problems in $\mathbb{R}^2$, preprint, arXiv: 1506.00491. Google Scholar [28] K. Wang and J. Wei, On Serrin's overdetermined problem and a conjecture of Berestycki, Caffarelli and Nirenberg, Comm. Partial Differential Equations, 44 (2019), 837–858, arXiv: 1502.04680. doi: 10.1080/03605302.2019.1611846. Google Scholar [29] K. Wang and J. Wei, Finite Morse index implies finite ends, Comm. Pure Appl. Math., 72 (2019), 1044–1119, arXiv: 1705.06831. doi: 10.1002/cpa.21812. Google Scholar [30] G. S. Weiss, Partial regularity for weak solutions of an elliptic free boundary problem, Comm. Partial Differential Equations, 23 (1998), 439-455. doi: 10.1080/03605309808821352. Google Scholar [31] B. White, Curvature estimates and compactness theorems in 3-manifolds for surfaces that are stationary for parametric elliptic functionals, Inventiones Mathematicae, 88 (1987), 243-256. doi: 10.1007/BF01388908. Google Scholar
Mathematica plot of the free boundary of the double hairpin solution $H_a(z)$ for $a = 1/4$, $a = 1$ and $a = 2$. Note that $z = x_1 + i x_2$ and $x_2$ is the horizontal axis in the diagram
An illustration of the three cases for the free boundary $F(u)\cap B_r(z)$ of a solution $u$ having simply connected positive phase $\mathbb{D}^+(u)$
Illustrating $\Omega$ in Case 4 of the proof of Lemma 3.6, whose existence is ruled out
The conformal diffeomorphism $U_a = H_a + i \tilde{H}_a$ mapping the right half of the positive phase $\mathcal{D}_a = \Omega_a \cap \{x_1>0\}$ onto the slit domain $\mathcal{S}_a$
Mathematica plot of the free boundary of the Scherk solution $S_s(x_1, x_2)$ for asymptotic slopes $s = 1/8$, $s = 1/2$ and $s = 7/8$. Note that in the diagram $x_2$ is the horizontal axis
Mapping the subdomain $\mathcal{D}^{\text{BSS}}_s$ of the positive phase of the Scherk solution $S_s$ conformally onto the strip $\mathcal{S}_l$ under $U_s^{\text{BSS}} = S_s + i \tilde{S}_s.$ Note that $Q_{\pm}$ is a saddle point of $S_s$ with $Q_{\pm}A_{\pm}$ and $Q_{\pm}E_{\pm}$ being a steepest descent and a steepest ascent path from $Q_{\pm}$, respectively
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http://community.rapidminer.com/t5/RapidMiner-Studio/Parallel-execution-of-operators/td-p/588 | # Parallel execution of operators
Ok thanks.
Is it possible to apply operator in "parallel" ? I would like to filter input data that have RESULT A and RESULT B.
INPUT -> OPERATOR A -> RESULT A
INPUT -> OPERATOR B -> RESULT B
Operator seems to be chained in series:
INPUT -> OPERATOR A -> OPERATOR B -> RESULT C.
3 REPLIES
Moderator
## Re: Parallel execution of operators
Hi,
first of all: I hope you don't mind I put your questions in separate threads. That way, all users can easily recognize separate questions ...
Regarding your problem, yes, in general operators in the operator tree are executed as a chain. However you can achieve a setting like the one you mentioned by using loops, e.g. by applying the [tt]ParameterIteration[/tt] operator.
Here is an example process:
`<operator name="Root" class="Process" expanded="yes"> <operator name="ExampleSetGenerator" class="ExampleSetGenerator"> <parameter key="target_function" value="simple polynomial classification"/> </operator> <operator name="ParameterIteration" class="ParameterIteration" expanded="yes"> <parameter key="keep_output" value="true"/> <list key="parameters"> <parameter key="OperatorSelector.select_which" value="1,2"/> </list> <operator name="OperatorSelector" class="OperatorSelector" expanded="yes"> <parameter key="select_which" value="2"/> <operator name="ExampleFilter" class="ExampleFilter"> <parameter key="condition_class" value="attribute_value_filter"/> <parameter key="parameter_string" value="att1>0"/> </operator> <operator name="ExampleFilter (2)" class="ExampleFilter"> <parameter key="condition_class" value="attribute_value_filter"/> <parameter key="parameter_string" value="att1<0"/> </operator> </operator> </operator></operator>`
Hope that helps,
regards,
Tobias
## Re: Parallel execution of operators
Thanks, I've installed the lasted code from CVS and it works.
However, I don't understand why you're using the following in OperatorSelector
<parameter key="select_which" value="2"/>
If I use:
<parameter key="select_which" value="1"/>
Then it works too. What's the meaning of this select_which ?
Moderator
## Re: Parallel execution of operators
Hi,
The thing is, that it does not matter what value is set in the OperatorSelector for the parameter [tt]select_which[/tt]. This is, because the process contains the operator [tt]ParameterIteration[/tt]. This operator iterates over parameters which you can specify. To select the first operator in the first iteration and the second operator in the second operator, you have to specify this in the [tt]ParameterIteration[/tt] operator parameters. Hence, the process XML contains the lines
` <list key="parameters"> <parameter key="OperatorSelector.select_which" value="1,2"/> </list>`
which actually means exactly that the operator in the first iteration the value 1 is set for the parameter [tt]select_which[/tt], in the second iteration the value 2 is used for that parameter. The value you manually specify for the parameter [tt]select_which[/tt] in the operator [tt]OperatorSelector[/tt] is simply overwritten in that process.
Hope that clarifies how it works!
Regards,
Tobias | 2017-02-27 04:31:45 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8284886479377747, "perplexity": 1678.840719809196}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501172447.23/warc/CC-MAIN-20170219104612-00302-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://brilliant.org/problems/an-algebra-problem-by-k-j-w-2/ | # A number theory problem by K. J. W.
Number Theory Level 1
Find the largest positive integer n so that n < 100 and
$$n!+\left( n+1 \right) !+\left( n+2 \right) !={ q }^{ 2 }$$
where q is a positive integer.
× | 2016-10-27 20:49:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.45177826285362244, "perplexity": 1505.1966088534227}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721392.72/warc/CC-MAIN-20161020183841-00333-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://tex.stackexchange.com/questions/552113/i-need-to-understand-some-cmds-of-tex | # I need to understand some cmds of tex
I'm debugging the command tex of TeX Live 2019.
Shown as above, if you type:
``````./tex -ini
h
I\&V
e
``````
There will be a segment fault. Note that you need a `h.tex` under the same directory as tex and there is one character `\` in `h.tex`.
My question is what does cmd `I` do? It seems it fails to open a file because there is an invalid sequence `\&V` after `I`. And what does cmd `e` do? I know It passes the file pointer to `fclose()` that failed to be initialized by `I` and the leads to the segment fault.
Thanks!
• Welcome to TeX.SX! Donald already explained why the TeX errors happen. But to reproduce the segfault with `tex` (not `-ini`) you can write `\?` instead of `\` and `\&`. More generally, the segfault happens if you are running TeX on a file, and press `E` in the menu while TeX is scanning the input of `I`. Interesting find! Jul 3, 2020 at 3:33
• It seems at first glance like a bug in TeX. A have no time for studying this issue in TeX sources now. If it is true bug then congratulations: you have found the bug in TeX 3.14159265. Jul 3, 2020 at 6:58
• @barbarabeeton May be you know the term when D. Knuth opens the folder with bug reports of TeX 3.14159265... Jul 3, 2020 at 7:09
Running TeX with the `-ini` flag means that you're running without any preloaded format. You don't get anything but primitive commands which means that `\ ` and `\&` aren't defined. So that's why you're getting the errors. Do you really want to be running iniTeX?
`i` means insert the text following at the point that the error occurred. `x` is the usual quit command. `e` in theory will stop and then open an editor at the error point. I'm guessing this is missing some implementation/configuration which is why you're getting the segmentation fault. These are just some of the interactive error handling commands that TeX provides. See The TeXbook for more information about how these work. I've never really bothered with them. They made a lot of sense in the early 80s when computers were slow and compute time expensive but even by 1986 when I first started using TeX the program ran fast enough that it was generally more sensible to just use x to stop TeX, fix the error in the input file and continue.
• The weird thing is that TeX seems to consistently segfault if you type `e` while reading the input of `i`... Jul 3, 2020 at 3:22
• Thanks a lot. But I'm a little bit confused about "i". Will "i" insert the text to the file h.tex? Or do you mean "i" just insert a text to a buffer?
– 潇洒张
Jul 3, 2020 at 3:59
• No, it inserts the characters into the stream being read by the current run, so is not useful for fixing an error. It could be useful for diagnosing an error that you aren't sure of, but as DH says, it usually makes more sense to edit the input file and run TeX again. Jul 3, 2020 at 5:18
• @DonaldHosek Did you get sucked into this quagmire by Coronavirus lockdowns too? Jul 3, 2020 at 5:21
• @DonaldArseneau don't think of it as being sucked into a quagmire, think of it as being rescued from the primordial swamp that is c.t.t Jul 3, 2020 at 8:30
Congratulations, you have found the bug in TeX.
Edit I did deeper inspection about this issue when I returned from my vacation and I found: the bug is in the C implementation of TeX, not in the TeX itself (coded by D. Knuth). So, the following paragraph (given by me a few days ago) is not explicitly true:
The last revision of TeX (version 3.14159265) was done by D. Knuth in 2014, next revision will be in the end of 2020. If you send this bug-report you can get a check for a small piece in dollars by D. Knuth. Of course, this is true only if your bug report will be processed as real bug of TeX. How to do the bug report is described at here in section Errata.
Description of the bug: If we invoke an error (undefined `\ ` in your example), then the `I` option is possible. If we use `I` option with a line with next error (undefined `\&` in your example) and the `I` line is not closed at this place (the `V` follows in your example) then this second error offers all options including `E` option. But using `E` option causes that TeX crashes, because the file name is zero in this case.
In detail: `I` option in § 84 of TeX web runs the code from § 87 where the procedure begin_file_reading is processed. This procedure is in § 328 and here is: `name := 0`. The `name` is macro defined in § 302: `name = cur_input.name_field`. Now, we can return to § 84 where the `E` option is solved. It creates the file name (which may be sent to an editor) from `input_stack[base_ptr].name_field` but this is the same as `cur_input.name_field` and it was set to zero. We have no file name here but zero. TeX implementation based on C crashes with segmentation fault because we want to open the file with buggy file name.
Edit: When I did experiments with `tex.web` code and I deactivated the system dependent code given after `E` option then the correct answer is here: `You want to edit file <correct name> at line <correct number>`. The bug is in the `lib/texmfmp.c` at line 2579 where the opened files are closed and the maximal index of opened files are read from `inopen` global variable. This variable is incremented when `I` option is used but no new file is opened in such case. So, we cannot close non-existent file in `lib/texmf.c`. Knuth says in §304: "The global variable `in_open` is equal to the value of the highest non-token-list level." It means that it is not always the number of currently opened files. Knuth's code and his comments in the code are correct here, bug is in the system dependent implementation.
• At the point of the first error message, `\jobname` has been set to `h`. Jul 3, 2020 at 13:07
• @egreg the file name h is irrelevant. You can insert `\bla` in your document (the name of the document and the format are irrelevant). When the prompt about undefined `\bla` occurs then inset `i\buff X`. When the prompt about undefined `\buff` occurs then insert `e`. The result is segmentation fault. Jul 3, 2020 at 13:13
• Maybe you can make it clear in your answer. Jul 3, 2020 at 13:29
• But it crashed on fclose() with uninitialized file pointer as the argument. In this case there is only one file: h.tex. But inopen was 2 when it tried to close all files. So it tried to close inputfile[1] and inputfile[2] in a for loop and inputfile[2] was uninitialized.
– 潇洒张
Jul 3, 2020 at 14:23
• @潇洒张 I corrected the term of bug report: the end of this year. Please, tell me if you prepare this bug report. If no, then I will try to do it. But you found this bug, this is much more correct if the honor will go to you. I have only analyzed this big from technical point of view. But I wouldn't think of trying the `E` option in the context of `I` option. Jul 3, 2020 at 20:24
Original poster: as with your other report ... if you ever see this, please email me, karl at tug.org, so we can give you proper credit in our report to Don Knuth. (We believe there is a bug in tex.web as well as in web2c; the tex.web code can be induced to print a bogus filename in the "You want to edit ..." message.)
Regarding begin_file_reading, it is, shall we say, suboptimally named. What it really does is "begin a level of input" - either from a file or from terminal interaction.
I just committed a fix (hopefully) for this to TeX Live, r55857 (http://tug.org/svn/texlive/trunk/Build/source/texk/web2c/lib/texmfmp.c?r1=53078&r2=55857&pathrev=55857). It involves having to look at all elements of input_stack[] to discern whether a given entry in input_file[] is actually a file, or a non-file resulting from terminal interaction. Before, we simply closed everything in input_file[], but that was wrong.
Regarding "the end of this year" - please send any and all Knuthian bug reports to me or tex-k at tug.org as soon as possible! Do not wait for December 31. My unofficial deadline is November 1. All reports have to be vetted before they go into the pile for Knuth, and it takes time. More info: https://tug.org/texmfbug.
Thanks to all. | 2022-05-27 04:31:13 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8750689625740051, "perplexity": 1476.9173474600682}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662631064.64/warc/CC-MAIN-20220527015812-20220527045812-00440.warc.gz"} |
https://stats.stackexchange.com/questions/49439/how-to-compare-whether-the-odds-of-success-of-different-levels-of-a-predictor-ar | # How to compare whether the odds of success of different levels of a predictor are different from 0
I hope I am able to word my question clearly. Suppose I have a model below:
glmer(Y~X + (1|subject), family="binomial", data=dat)
The intercept is the log odds of success for the reference level (level 1 of X), and the slopes indicate the log odds ratio of the other levels to the reference level.
Estimate Std. Error z value Pr(>|z|)
(Intercept) 2.1745 0.3517 6.183 6.3e-10 ***
X2 -0.5559 0.3276 -1.697 0.0897 .
X3 0.2309 0.3634 0.635 0.5252
X4 -4.8587 0.4155 -11.693 < 2e-16 ***
X5 -2.8946 0.3200 -9.045 < 2e-16 ***
X6 -3.6111 0.3387 -10.663 < 2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
This is all fine. However, what kinds of statistical tests can I use if in addition to this, I would also like to check whether each level (including the reference level)'s odds of success are significantly higher than 0? How would I be able to do this?
You fit a binomial probability model with - by default - a logit link function (see ?binomial). The problem is that the logistic function (the inverse of the logit) never reaches zero. Therefore, all model fits will yield nonzero odds for success. In particular, all classical confidence intervals for success odds will not include zero. Therefore, in classical Null Hypothesis Signifiance Testing (NHST), your odds of success will always be (statistically) significantly different from zero. (For that matter, the same applies to one.)
Your best bet will probably be to abandon statistical significance and go with practical significance. Determine beforehand what "relevant" odds of success are, then look which predictor levels yield a fit that exceeds this level.
Which leaves you with the problem that prediction for mer objects is not a straightforward matter. This question may be helpful here.
Good luck!
• Thank you for the reply! Regarding "the logit never reaches zero", could you please clarify that? Wouldn't the logit of 1 (namely when the odds of success are 1 (e.g., 0.5/(1-0.5)) be 0?
– Alex
Feb 12 '13 at 14:50
• Sorry, I corrected the error in my answer - thanks for keeping me honest! Of course the logit can reach zero. The problem is that its inverse, the logistic function, from which we calculate fitted values, will not reach zero. Hope it is clearer now? Feb 12 '13 at 14:58
If you really want to test the null hypothesis that the true odds for success are exactly zero for each individual level $j$ of your factor $X$, then the statistical test is very simple.
The null-hypothesis odds $p_{0j}/(1-p_{0j})$ for success in group $j$ are zero iff $p_{0j}=0$. Under this null hypothesis, the probability of observing at least $1$ success in group $j$ is $0$. Therefore, if you empirically do observe a success in group $j$, you can reject the null hypothesis for any level of $\alpha > 0$.
Since this decision rule allows us to set $\alpha$ arbitrarily close to $0$, we do not have to worry about $\alpha$-inflation when doing multiple tests. So: for each group $j$ with at least one success, you can reject the null hypothesis of $p_{0j} = 0$.
If this test seems odd, or somewhat non-informative, it could indicate that you are actually not really interested in whether the true odds are exactly $0$, but just very small. In that case, I seond Stephan's remarks on statistical significance vs. practical relevance. | 2021-12-07 00:04:45 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7408881187438965, "perplexity": 511.9591072654936}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363327.64/warc/CC-MAIN-20211206224536-20211207014536-00340.warc.gz"} |
https://zbmath.org/?q=an:0684.60057 | # zbMATH — the first resource for mathematics
Multiple points of Lévy processes. (English) Zbl 0684.60057
Let $$X_ t$$ be a Lévy process in $$R^ d$$ with transition density $$p_ t(x)$$ and characteristic function $$E(e^{iy\cdot X_ t})=e^{- t\psi (y)}$$. The authors prove the existence of k-multiple points of X under the following hypotheses (A) and (B): $(A)\quad \int_{| x| \leq \epsilon}(\int^{T}_{0}p_ t(x)dt)^ kdx<\infty \quad for\quad some\quad \epsilon,\quad T>0.$ $(B)\quad \int^{T}_{0}p_ t(0)dt>0.$ This result proves a conjecture of W. J. Hendricks and S. J. Taylor [Concerning some problems about polar sets of processes with independent increments. (1976), unpublished]. By using this result, the authors also investigate the Hausdorff dimension of the set of k-multiple points $$E_ k=\{(t_ 1,t_ 2,...,t_ k)$$; $$X_{t_ 1}=X_{t_ 2}=...=X_{t_ k}\}$$. The result says that $\dim E_ k\geq k-(k-1)d/\beta '',\quad where\quad \beta ''=\sup \{\alpha \geq 0;\quad | y|^{-\alpha}Re \psi (y)\to 0\quad as\quad | y| \to 0\}.$
Reviewer: Y.Oshima
##### MSC:
60J99 Markov processes 60J45 Probabilistic potential theory
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https://brilliant.org/practice/deception/?chapter=statistics-introduction | ### Statistics Fundamentals
Statistics is such a powerful tool because it uses data to convey important information. This information may be visualized as a graph, chart, or table. It might be given numerically as the mean, median, mode, or percent change. There are so many different ways to frame information.
Because people can choose how to show statistics, sometimes it can be deceiving. Let's look at some examples.
# Deception
From the data above, what can we conclude about home prices in the United States? Select all that apply.
# Deception
Select one or more
A study is done on 250 people who claim they have psychic abilities.
A regular deck of cards has an equal number of red and black cards. The deck is shuffled, and each person predicts the color of the top card. The top card is then flipped face up to see if they are correct. This is repeated 8 times in a row.
One participant gets all 8 predictions correct! The researcher claims that this person has exhibited psychic abilities.
Based on just the way the experiment is designed, is the researcher's claim a viable hypothesis? (Note: we're not judging the plausibility of psychic ability in itself, just this particular experiment's design.)
# Deception
Deception in statistics can be quite intentional. In the first problem, the choice of what months to display did not happen by accident! The researcher probably wanted to convey the data in a certain way.
Deception can also be unintentional. In the second problem, the researcher examining psychic phenomena may have been earnest about their hypothesis, but they still interpreted the results of their experiment poorly.
Alternatively, sometimes the statistics themselves can deceive. It depends who gathers the data and how they do it. The questions that follow will demonstrate this issue.
# Deception
The Transportation Association of Canada wanted to study the effects of wearing a helmet on motorcycle accidents. The data below was collected by Canadian hospitals between 2002 and 2014. Each motorcyclist who was in a traffic accident and arrived at a hospital had the information below recorded.
Wearing helmet? Yes No Arrived in ambulance 18% 9% Had concussion 9% 3% Needed blood transfusion 0.8% 0.4%
Based on the table, what is true?
A. Motorcyclists who came to a Canadian hospital that were wearing helmets in traffic accidents were twice as likely to arrive in an ambulance than motorcyclists who were not wearing helmets.
B. Motorcyclists who came to a Canadian hospital that were wearing helmets in traffic accidents were three times as likely to have a concussion than motorcyclists who were not wearing helmets.
C. 0.8% of all Canadian motorcycle riders who wear helmets will need a blood transfusion.
(Data source: Journal of Internal Medicine.)
# Deception
The previous problem showed how it is important to remember the data source. Not everyone who gets in an accident goes to the hospital!
The table below shows all motorcycle accidents in Canada between 2002 and 2014.
Which quadrants would be missing from the hospital's data?
# Deception
The data from the previous two questions demonstrates what's known as Berkson's bias. In taking what appears to be a representative sample, a specific group may still be excluded. Hospitals won't see those who are healthy enough to not need a hospital, so the data won't include this part of the population.
Recall that the table below shows all motorcycle accidents in Canada between 2002 and 2014.
Look at all accidents involving motorcyclists wearing helmets. What percentage of these accidents did the hospital see?
# Deception
Statistics cannot be done in a mathematical vacuum. Usually it involves real-life knowledge about the circumstances. This can be as important as any sort of calculation.
When looking at graphs or charts, it is important to inspect the axes and labels. When analyzing data, it is important to remember the data source and how the data was collected. This can help you recognize deceptive statistics.
It may seem, at this point, that statistics can only serve to confound. However, you'll find statistics can be used to pierce deception rather than just create it — continue the course to find out how!
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https://ijqf.org/archives/3549 | # Weekly Papers on Quantum Foundations (25)
This is a list of this week’s papers on quantum foundations published in various journals or uploaded to preprint servers such as arxiv.org and PhilSci Archive.
Process-theoretic characterisation of the Hermitian adjoint. (arXiv:1606.05086v1 [quant-ph])
on 2016-6-17 12:26pm GMT
Authors: John SelbyBob Coecke
We show that the physical principle “the adjoint associates to each state a `test’ for that state” fully characterises the Hermitian adjoint for pure quantum theory, therefore providing the adjoint with operational meaning beyond its standard mathematical definition. Also, we show that for general process theories, which all admit a diagrammatic representation, this physical principle induces a reflection operation.
Quantum mechanics in terms of realism
Philsci-Archive: No conditions. Results ordered -Date Deposited.
on 2016-6-17 5:49am GMT
Jabs, Arthur (2016) Quantum mechanics in terms of realism. [Published Article or Volume]
[In Depth] LIGO detects another black hole crash
Science: Current Issue
on 2016-6-17 12:00am GMT
The biggest discovery in science this year—the observation of ripples in space-time called gravitational waves—was no fluke. For a second time, physicists working with the two massive detectors in the Laser Interferometer Gravitational-Wave Observatory (LIGO) have detected a pulse of such waves, the LIGO team reported on 15 June at a meeting of the American Astronomical Society in San Diego, California. Once again the waves emanated from the merger of two black holes, the ultraintense gravitational fields left behind when massive stars collapse into infinitesimal points. The new observation suggests that if LIGO’s detectors reach their design sensitivity—which physicists hope to achieve by 2019—the observatory will spot dozens or even hundreds of the otherwise undetectable events each year, ushering in a new era of gravitational-wave astronomy. Author: Adrian Cho
The time-dependent Schrodinger equation for a closed quantum system. (arXiv:1606.04759v1 [quant-ph])
on 2016-6-16 3:32am GMT
Ever since Schrodinger, the time-dependent Schrodinger equation has been regarded as the fundamental physical law of the unitary-only quantum theory. However, while the origin of time for non-conservative systems can be traced back to the environmental influence, this procedure is inapplicable for the closed (conservative) quantum systems. Hence the question of the origin of time for closed systems. In addressing this issue, in order to avoid circularity of the standard universal (global) time as well as assumptions that could be additional to the standard postulates of quantum theory, we stick to the so-called quantum-mechanical local time. The problems that inevitably appear in such minimalist rejection of the standard role of time in physics are solved with mathematical rigor while leading to the emergent time for every, at least approximately, closed (conservative) system as well as to establishing the time-dependent Schrodinger law as the fundamental law of the unitary non-relativistic as well as relativistic-field quantum theory.
Double-slit experiment in momentum space. (arXiv:1606.04732v1 [quant-ph])
on 2016-6-16 3:32am GMT
Authors: I. P. IvanovD. SeiptA. SurzhykovS. Fritzsche
Young’s classic double-slit experiment demonstrates the reality of interference when waves and particles travel simultaneously along two different spatial paths. Here, we propose a double-slit experiment in momentum space. We show that elastic scattering of vortex electrons proceeds via two paths in momentum space, which are well localized and well separated from each other. For such vortex beams, the (plane-wave) amplitudes along the two paths acquire adjustable phase shifts and produce interference fringes in the final angular distribution. We argue that this experiment can be realized with the present day technology. We show that it gives experimental access to the Coulomb phase, a quantity which plays an important role in all charged particle scattering but which usual scattering experiments are insensitive to.
Experimental bounds on collapse models from gravitational wave detectors. (arXiv:1606.04581v1 [quant-ph])
on 2016-6-16 3:32am GMT
Authors: M. CarlessoA. BassiP. FalferiA. Vinante
Wave function collapse models postulate a fundamental breakdown of the quantum superposition principle at the macroscale. Therefore, experimental tests of collapse models are also fundamental tests of quantum mechanics. Therefore, experimental tests of collapse models can be regarded as fundamental tests of the quantum superposition principle. Here, we compute the upper bounds on the collapse parameters, which can be inferred by the gravitational wave detectors AURIGA, LIGO and LISA Pathfinder. We consider the most widely used collapse model, the Continuous Spontaneous Localization (CSL) model. We show that these experiments exclude a huge portion of the CSL parameter space, the strongest bound being set by the recently launched space mission LISA Pathfinder.
Spying on photons with photons: quantum interference and information. (arXiv:1606.04729v1 [quant-ph])
on 2016-6-16 3:32am GMT
Authors: Stefan Ataman
The quest to have both which-path knowledge and interference fringes in a double-slit experiment dates back to the inception of quantum mechanics (QM) and to the famous Einstein-Bohr debates. In this paper we propose and discuss an experiment able to spy on one photon’s path with another photon. We modify the quantum state inside the interferometer as opposed to the traditional physical modification of the “wave-like” or “particle-like” experimental setup. We are able to show that it is the ability to harvest or not which-path information that finally limits the visibility of the interference pattern and not the “wave-like” or “particle-like” experimental setups. Remarkably, a full “particle-like” experimental setup is able to show interference fringes with 100 % visibility if the quantum state is carefully engineered.
Quantum-coherent mixtures of causal relations. (arXiv:1606.04523v1 [quant-ph])
on 2016-6-15 7:56am GMT
Understanding the causal influences that hold among the parts of a system is critical both to explaining that system’s natural behaviour and to controlling it through targeted interventions. In a quantum world, understanding causal relations is equally important, but the set of possibilities is far richer. The two basic ways in which a pair of time-ordered quantum systems may be causally related are by a cause-effect mechanism or by a common cause acting on both. Here, we show that it is possible to have a coherent mixture of these two possibilities. We realize such a nonclassical causal relation in a quantum optics experiment and derive a set of criteria for witnessing the coherence based on a quantum version of Berkson’s paradox. The interplay of causality and quantum theory lies at the heart of challenging foundational puzzles, such as Bell’s theorem and the search for quantum gravity, but could also provide a resource for novel quantum technologies.
GHZ states and PR boxes in the classical limit. (arXiv:1606.04274v1 [quant-ph])
on 2016-6-15 7:56am GMT
Authors: Daniel RohrlichGuy Hetzroni
A recent paper [1] argues that bipartite “PR-box” correlations, though designed to respect relativistic causality, in fact violate relativistic causality in the classical limit. As a test of Ref. [1], we consider GHZ correlations as a tripartite version of PR-box correlations, and ask whether the arguments of Ref. [1] extend to GHZ correlations. If they do – i.e. if they show that GHZ correlations violate relativistic causality in the classical limit – then Ref. [1] must be incorrect, since GHZ correlations are quantum correlations and respect relativistic causality. But the arguments fail. We also show that both PR-box correlations and GHZ correlations can be retrocausal, but the retrocausality of PR-box correlations leads to self-contradictory causal loops, while the retrocausality of GHZ correlations does not.
Unscrambling the Quantum Omelette of Epistemic and Ontic Contextuality: Classical Contexts and Quantum Reality
Philsci-Archive: No conditions. Results ordered -Date Deposited.
on 2016-6-14 11:40pm GMT
de Ronde, Christian (2016) Unscrambling the Quantum Omelette of Epistemic and Ontic Contextuality: Classical Contexts and Quantum Reality. [Preprint]
A conjecture concerning determinism, reduction, and measurement in quantum mechanics
Philsci-Archive: No conditions. Results ordered -Date Deposited.
on 2016-6-14 11:36pm GMT
Jabs, Arthur (2016) A conjecture concerning determinism, reduction, and measurement in quantum mechanics. [Published Article or Volume]
Scientific Realism and Primordial Cosmology
Philsci-Archive: No conditions. Results ordered -Date Deposited.
on 2016-6-14 11:32pm GMT
Azhar, Feraz and Butterfield, Jeremy (2016) Scientific Realism and Primordial Cosmology. [Preprint]
LISA pathfinder appreciably constrains collapse models. (arXiv:1606.03637v1 [quant-ph])
on 2016-6-14 9:25am GMT
LISA Pathfinder’s measurement of a relative acceleration noise between two free-falling test masses with a square root of the power spectral density of $5.2 \pm 0.1 \mbox{ fm s}^{-2}/\sqrt{\rm{Hz}}$ appreciably constrains collapse models. In particular, we bound the localization rate parameter, $\lambda_{\rm CSL}$, in the continuous spontaneous localization model (CSL) to be at most $\left( 2.96 \pm 0.12 \right) \times 10^{-8} \mbox{ s}^{-1}$. Moreover, we bound the regularization scale, $\sigma_{\rm DP}$, used in the Diosi-Penrose (DP) model to be at least $40.1 \pm 0.5 \mbox{ fm}$. These bounds significantly constrain the validity of these models. In particular: (i) a lower bound of $2.2 \times 10^{-8\pm2} \mbox{s}^{-1}$ for $\lambda_{\rm CSL}$ has been proposed in (although a lower bound of about $10^{-17} \mbox{s}^{-1}$ is sometimes used), in order for the collapse noise to be substantial enough to explain the phenomenology of quantum measurement, and (ii) 40 fm is larger than the size of any nucleus, while the regularization scale has been proposed to be the size of the nucleus.
Toward an Understanding of Parochial Observables
The British Journal for the Philosophy of Science – Advance Access
on 2016-6-13 5:54am GMT
Ruetsche ([2011]) claims that an abstract C*-algebra of observables will not contain all of the physically significant observables for a quantum system with infinitely many degrees of freedom. This would signal that in addition to the abstract algebra, one must use Hilbert space representations for some purposes. I argue to the contrary that there is a way to recover all of the physically significant observables by purely algebraic methods.
• 1 Introduction
• 2 Preliminaries
• 3 Three Extremist Interpretations
• 3.1 Algebraic imperialism
• 3.2 Hilbert space conservatism
• 3.3 Universalism
• 4 Parochial Observables
• 4.1 Parochial observables for the imperialist
• 4.2 Parochial observables for the universalist
• 5 Conclusion | 2021-11-28 03:40:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5948712229728699, "perplexity": 1782.8887092666614}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358443.87/warc/CC-MAIN-20211128013650-20211128043650-00562.warc.gz"} |
http://dev.goldbook.iupac.org/terms/view/E01956 | ## Wikipedia - Electrode potential electrode potential, $$E$$
https://doi.org/10.1351/goldbook.E01956
@E01974@ of a cell in which the electrode on the left is a @S05917@ and the electrode on the right is the electrode in question.
Sources:
Green Book, 2nd ed., p. 61 [Terms] [Book]
PAC, 1996, 68, 957. (Glossary of terms in quantities and units in Clinical Chemistry (IUPAC-IFCC Recommendations 1996)) on page 971 [Terms] [Paper] | 2019-08-24 04:35:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.45309919118881226, "perplexity": 5195.146375405064}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027319724.97/warc/CC-MAIN-20190824041053-20190824063053-00456.warc.gz"} |
http://gmatclub.com/blog/category/blog/gmat-tests/page/38/ | # GMAT Question of the Day (March 25)
- Mar 25, 02:00 AM Comments [1]
Math If is a positive number and , then what is the value of ? A. B. C. ...
# GMAT Question of the Day (March 24)
- Mar 24, 02:00 AM Comments [1]
Math If and are consecutive integers () and , then which of the following must be true? A. B. $y \gt 0$
# GMAT Question of the Day (March 23)
- Mar 23, 02:00 AM Comments [0]
Math Andy, George and Sally are a team of consultants working on Project Alpha. They have an eight hour deadline to complete the project. The team members work at constant rates throughout the eight hour period. If the team of three has to begin work now...
# GMAT Question of the Day (March 20)
- Mar 20, 02:00 AM Comments [0]
Math There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four individual socks are picked at random what is the probability of getting at least two socks of the same color? A. B.
# GMAT Question of the Day (March 19)
- Mar 19, 02:00 AM Comments [0]
Math There are four distinct pairs of brothers and sisters. In how many ways can a committee of 3 be formed and NOT have siblings in it? A. 8 B. 24 C. 32 D. 56 E. 80 Question Discussion & Explanation Correct Answer - C - (click and drag your mouse to...
# GMAT Question of the Day (March 18)
- Mar 18, 02:00 AM Comments [0]
Math For the past days the average (arithmetic mean) number of cupcakes baked by Liv was 55 per day. Today Bibi helped Liv and together they baked 100 cupcakes, which raised the average to 60 cupcakes per day. What is the value of...
# GMAT Question of the Day (March 17)
- Mar 17, 02:00 AM Comments [0]
Math If a website registered 810 new members in June, how many of the new, registered members were from USA? (1) In June, the ratio of new members from USA, Europe and Asia was 4:3:2 respectively. (2) In June, none of the members were from locations other than...
# GMAT Question of the Day (March 16)
- Mar 16, 02:00 AM Comments [0]
Math If and are positive integers, is divisible by 15? (1) is a multiple of 25, and is a multiple of 20 (2) Question...
# GMAT Question of the Day (March 13)
- Mar 13, 02:00 AM Comments [0]
Math A. 50200 B. 50.2 C. 5.02 D. 0.502 E. 502 Question Discussion & Explanation Correct Answer - B - (click and drag your mouse to see the answer) GMAT Daily Deals e-GMAT’s methods help non-natives improve on GMAT Verbal. \$400 total savings @ GMAT Club. Learn...
# GMAT Question of the Day (March 12)
- Mar 12, 02:00 AM Comments [0]
Math There are two concentric circles with radii 10 and 8. If the radius of the outer circle is increased by 10% and the radius of the inner circle decreased by 50%, approximately by what percent does the area between the circles grow? A. 140% B. 141% C. 190% D.... | 2016-08-28 19:09:15 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 1, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.40771418809890747, "perplexity": 3570.3134918915625}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982947760.88/warc/CC-MAIN-20160823200907-00197-ip-10-153-172-175.ec2.internal.warc.gz"} |
http://mathhelpforum.com/discrete-math/89856-prove-theory.html | # Math Help - Prove theory
1. ## Prove theory
Hi Guys,
Preparing for final and came across home work question.
Question: Prove that n^2 + n is even for any interger n.
Thanks
CHicago boy
2. Hello.
Originally Posted by chicago_boy81
Hi Guys,
Preparing for final and came across home work question.
Question: Prove that n^2 + n is even for any interger n.
Two cases
1) n is odd: then $n = 2m+1, m\in \mathbb{N}_0$
Thus
$n^2 + n = (2m+1)^2 + 2m+1 =(4m^2+2+1)+2m+1 = 4m^2+4+2m = 2(2m^2+4+2m)$
is even
2) n is even: then $n = 2m, m \in \mathbb{N}$
=> $n^2+n = (2m)^2 + 2m = 4m^2 + 2m = 2(2m^2+m)$
Do you understand?
Regards,
Rapha
3. Very easy...
... for $n$ even $n$ and $n^{2}$ are both even, so that $n^{2} + n$ is even...
... for $n$ odd $n$ and $n^{2}$ are both odd, so that $n^{2} + n$ is even...
Kind regards
$\chi$ $\sigma$
4. A third solution won't harm anyone:
$n^2+n = n(n+1)$ and clearly 2 divides one of $n$ or $n+1$.
5. We can conclude remembering the 'legend' according to which Gauss demonstrated when he was seven years old [!] that...
$\sum_{i=1}^{n} i = \frac {n\cdot (n+1)}{2}$ (1)
True or not the 'legend', it is evident from (1) that $n\cdot (n+1)$ must be an even number...
Kind regards
$\chi$ $\sigma$
6. It looks like your proof is going to 2 cases, the 1st case being if n is even so n=2k for some integer k, and the second case will be for if n is odd so n=2m+1 for some integer m. This covers the set all of integers because an integer is either odd or even. hope this helps =). | 2014-07-31 14:40:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 21, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8258998990058899, "perplexity": 1345.0353462630253}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510273350.41/warc/CC-MAIN-20140728011753-00080-ip-10-146-231-18.ec2.internal.warc.gz"} |
http://math.hi.is/en/scv/scv-seminar-2016/ | SCV seminar 2016
This is the schedule of the Complex Analysis Seminar for 2016:
Severine Biard: An introduction to the complex Green operator
Friday, 16. December, 10:00-11:30, Tgv227 in Tæknigarður.
Abstract: With the closure of the range of the tangential CR operator in hands, we have a well-behaved L^2 theory and an associated Hodge decomposition. This allows me to introduce the complex Green operator, an operator similar to the d-bar Neumann operator for the tangential CR operator. Similar properties hold as well as the interest to study its compactness.
Benedikt Magnússon: Random polynomials and global extremal functions — Probabilistic viewpoint
Friday, 9. December, 10:00-11:30, Tgv227 in Tæknigarður.
Abstract: We will look at the probabilistic preliminaries needed to show how random polynomials converge to the global extremal function.
Benedikt Magnússon: Random polynomials and global extremal functions — Pluripotential viewpoint
Friday, 2. December, 10:00-11:30, Tgv227 in Tæknigarður.
Abstract: I will introduce the global extremal function and describe its properties, most importantly its connection with polynomials given by the Siciak-Zaharijuta theorem.
Severine Biard: Lecture on CR submanifolds of hypersurface type – the boundary of a complex manifold, II
Friday, 18. November, 10:00-11:30, Tgv227 in Tæknigarður.
Abstract: I will continue tomorrow on CR submanifolds of $$C^n$$ of hypersurface type of CR dimension m-1.
After quickly recalling the main ingredients introduced last week, I will introduce the tangential CR complex on those CR submanifolds. Then I will give the idea of the proof of the one-sided complexification allowing to consider some of those CR submanifolds as the boundary of a complex manifold.
Severine Biard: Lecture on CR submanifolds of hypersurface type – the boundary of a complex manifold
Friday, 11. November, 10:00-11:30, Tgv227 in Tæknigarður.
Abstract: I will recall some background of Cauchy-Riemann (CR) geometry, mainly embedded CR submanifolds and the objects that are associated with them. Focusing on CR submanifolds of hypersurface type – a generalization of real hypersurfaces in $$C^n$$ -, I will focus on an extrinsic point of view that relates the $$\overline partial_b$$ complex to the $$partial_b$$ complex on the ambiant $$C^n$$.
In 2012, Baracco proved that such compact, smooth, orientable and pseudoconvex CR submanifolds can be seen as the boundary of a complex manifold. I will give the idea of the proof and one of the most important applications: the range of $$\overline partial_b$$ is closed.
Severine Biard: Nonexistence of smooth Levi-flat hypersurface with positive normal bundle in compact Kähler manifolds of dimension $$\geq 3$$, II
Friday, 28. October, 10:00-11:30, Tgv227 in Tæknigarður.
Abstract: Among results of nonexistence of Levi-flat hypersurface in CPn, $$\geq 3$$, conjectured by D. Cerveau in 1993, there are some generalizations to compact Kähler manifolds, particularly the conjecture given by Marco Brunella in 2008: there is no smooth Levi-flat hypersurface such that the normal bundle to the Levi foliation is positive along the leaves in compact Kähler manifolds of dimension $$\geq 3$$. In a joint work with Andrei Iordan, we obtained a positive answer to this conjecture by using $$L^2$$-weighted estimates for d-bar. I will introduce first the problem of nonexistence, then I will explain the idea of the proof of our result and I will finish by talking about interesting questions around this subject.
Severine Biard: Nonexistence of smooth Levi-flat hypersurface with positive normal bundle in compact Kähler manifolds of dimension $$\geq 3$$
Friday, 21. October, 10:00-11:30, Tgv227 in Tæknigarður.
Abstract: Among results of nonexistence of Levi-flat hypersurface in CPn, $$\geq 3$$, conjectured by D. Cerveau in 1993, there are some generalizations to compact Kähler manifolds, particularly the conjecture given by Marco Brunella in 2008: there is no smooth Levi-flat hypersurface such that the normal bundle to the Levi foliation is positive along the leaves in compact Kähler manifolds of dimension $$\geq 3$$. In a joint work with Andrei Iordan, we obtained a positive answer to this conjecture by using $$L^2$$-weighted estimates for d-bar. I will introduce first the problem of nonexistence, then I will explain the idea of the proof of our result and I will finish by talking about interesting questions around this subject.
Auðunn Skúta Snæbjarnarson
Friday, 21. October, 10:00-11:30, Tgv227 in Tæknigarður.
Severine Biard: On the Chen’s proof of the Demailly’s weak openess conjecture
Friday, 7. October, 10:00-11:30, Tgv227 in Tæknigarður.
Abstract: I will explain the proof of the weak Openess conjecture based on the Chen’s paper: https://arxiv.org/pdf/1506.01146v2.pdf
The main ingredient is a weighted $$L^2$$-approximation theorem whose the proof is slightly different to the one in his paper.
Auðunn Skúta Snæbjarnarson:
Friday, 30. September, 10:00-11:30, Tgv227 in Tæknigarður.
Auðunn Skúta Snæbjarnarson:
Monday, 26. September, 10:00-11:30, Tgv227 in Tæknigarður.
Severine Biard: On Diederich-Fornaess exponent and its application to L^2 estimates
Friday, 16. September, 10:00-11:30, Tgv227 in Tæknigarður.
Abstract: A Diederich-Fornaess exponent for a domain $$\Omega$$ is an exponent $$\eta$$ such that $$-(-\rho)^{\eta}$$ is a bounded strictly plurisubharmonic function for a defining function $$\rho$$ of $$\Omega$$. Such an exponent was introduced in 1977 by Diederich and Fornaess on smooth pseudoconvex domains in Stein manifolds. Its existence is related to the geometry of the ambiant space and of the domain and has many applications. I will focus on one such application: I will explain how this exponent allows to choose a less plurisubharmonic weight function via Berndtsson-Charpentier method, yielding $$L^2$$ estimates for the d-bar equation on pseudoconvex domains.
Mitja Nedic, Stockholm University: On the class of Nevanlinna function in two variables
Friday, 9. September, 10:00-11:30, Tgv227 in Tæknigarður.
Abstract: Nevanlinna functions are holomorphic functions mapping the poly-upper half-plane to the closed upper half-plane. In the case of one variable we have classical results due to Herglotz, Nevanlinna, Pick, Cauer and others regarding integral representations of such functions. In the case of several variables the most work has been done by Vladimirov in the 1970s, but his results are incomplete in the sense that they do not provide an “if and only if” characterization.
In this talk we will discuss the statement and proof of an integral representation that provides such an “if and only if” characterization. We will also look closer at the class of representing measure of Nevanlinna functions as well as some examples. In the end we will present some open problems and how they relate to applications of Nevanlinan functions in electromagnetic engineering.
Ragnar Sigurðsson: The openness theorem
Friday, 2. September, 10:00-11:30, Tgv227 in Tæknigarður.
Abstract: In the lecture we will go through a proof of Guan and Zhou Math. Ann. 182 (2015), of the so called openness conjecture, which was originally stated by Demailly in 2000. | 2020-09-24 15:26:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8018423318862915, "perplexity": 859.5763314433491}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400219221.53/warc/CC-MAIN-20200924132241-20200924162241-00775.warc.gz"} |
https://itprospt.com/qa/436561/how-can-i-find-the-mole-ratio-of-anhydrous-salt | 1
# How can I find the mole ratio of anhydrous salt to water?
## Question
###### How can I find the mole ratio of anhydrous salt to water?
How can I find the mole ratio of anhydrous salt to water?
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##### The canola oil industry is perfectly competitive. Every producer has the following total cost function: LTC...
The canola oil industry is perfectly competitive. Every producer has the following total cost function: LTC = 2Q3 – 15Q2 + 40Q, where Q is measured in tons of canola oil. The corresponding marginal cost function is given by LMC = 6Q2 – 30Q + 40. a. In long-run equilibrium, how much will... | 2023-03-21 14:51:35 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5351067185401917, "perplexity": 4280.5045512506795}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943698.79/warc/CC-MAIN-20230321131205-20230321161205-00712.warc.gz"} |
https://math.stackexchange.com/questions/2931000/probability-density-function-of-sum-of-uniform-random-variables | # Probability density function of sum of uniform random variables
Let $$X$$ and $$Y$$ be independent random variables with density functions
$$f_X(a) \begin{cases} 1\over2 & -1\le a\le1 \\ 0 & \text{otherwise} \end{cases}$$
$$f_Y(a) \begin{cases} 1\over2 & 3\le a\le5 \\ 0 & \text{otherwise} \end{cases}$$
Find the probability density function of $$X + Y$$
I have been trying to do this without convolution, but with using the standard double integral method of finding the CDF and then the PDF (as opposed to using a purely geometrical method) and I have been successful in the case of $$2 \le a \le 4$$ by using an integral of $${1\over 4}\int_3^a \int_{-1}^{a-y}dxdy$$
However I have been having trouble in the other case of $$4 \le a \le 6$$. My attempt was this:
$${1\over 4}\int_3^5 \int_{-1}^{min \{1, a-y\}}\ dx \ dy$$
$${1\over 4}\left (\int_3^5 ({min \{1, a-y\}} +1) \ dy \right)$$
$${1\over 4}\left (\int_3^5 ({min \{1, a-y\}}\ dy+ \int_3^5 1\ dy \right)$$
At this point I said that since the minimum function depends on whether $$y>a-1$$ or $$y, therefore there are two cases and thus
$${1\over 4}\left (\int_3^{a-1}1 \ dy +\int_{a-1}^{5} \ y \ dy+ \int_3^5 1\ dy \right)$$
$${1\over 4}\left (a-4-\dfrac{a^2-2a-24}{2}+ 2\right)$$
Taking the derivative of this CDF yields $${1\over2} - {a\over4}$$, whereas the true solution is $${3\over2} - {a\over4}$$. Where am I going wrong?
In $$\int_3^5 {\min \{1, a-y\}}\ dy$$, the integrand should equal $$a-y$$ when $$a-y < 1$$, i.e. when $$y > a-1$$. So when you split it into two, it should read $$\int_3^{a-1} 1\,dy + \int_{a-1}^5 (a-y)\,dy.$$ In your work, you have $$y$$ in the second integral instead of $$a-y$$. Fixing this gives the expected answer.
• How does one determine to use $\int_3^{a-1} 1\,dy + \int_{a-1}^5 (a-y)\,dy$ rather than $\int_3^{a-1} (a-y)\,dy + \int_{a-1}^5 1\,dy$? – agblt Oct 14 '18 at 15:54
• Because when $y \in (3,a-1)$, we have $a-y \ge a-(a-1) = 1$, so $\min\{1, a-y\} = 1$, not $a-y$. So when we integrate $\min\{1,a-y\}$ from $y=3$ to $a-1$, we are really just integrating 1. Similarly, when $y \in (a-1,5)$ we have $\min\{1, a-y\} = a-y$. – Nate Eldredge Oct 14 '18 at 15:59 | 2019-09-21 15:35:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 24, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.989149272441864, "perplexity": 1722.3196900330722}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514574532.44/warc/CC-MAIN-20190921145904-20190921171904-00431.warc.gz"} |
https://pytorch.org/docs/stable/generated/torch.optim.lr_scheduler.CosineAnnealingLR.html?highlight=cosine | # CosineAnnealingLR¶
class torch.optim.lr_scheduler.CosineAnnealingLR(optimizer, T_max, eta_min=0, last_epoch=- 1, verbose=False)[source]
Set the learning rate of each parameter group using a cosine annealing schedule, where $\eta_{max}$ is set to the initial lr and $T_{cur}$ is the number of epochs since the last restart in SGDR:
\begin{aligned} \eta_t & = \eta_{min} + \frac{1}{2}(\eta_{max} - \eta_{min})\left(1 + \cos\left(\frac{T_{cur}}{T_{max}}\pi\right)\right), & T_{cur} \neq (2k+1)T_{max}; \\ \eta_{t+1} & = \eta_{t} + \frac{1}{2}(\eta_{max} - \eta_{min}) \left(1 - \cos\left(\frac{1}{T_{max}}\pi\right)\right), & T_{cur} = (2k+1)T_{max}. \end{aligned}
When last_epoch=-1, sets initial lr as lr. Notice that because the schedule is defined recursively, the learning rate can be simultaneously modified outside this scheduler by other operators. If the learning rate is set solely by this scheduler, the learning rate at each step becomes:
$\eta_t = \eta_{min} + \frac{1}{2}(\eta_{max} - \eta_{min})\left(1 + \cos\left(\frac{T_{cur}}{T_{max}}\pi\right)\right)$
It has been proposed in SGDR: Stochastic Gradient Descent with Warm Restarts. Note that this only implements the cosine annealing part of SGDR, and not the restarts.
Parameters
• optimizer (Optimizer) – Wrapped optimizer.
• T_max (int) – Maximum number of iterations.
• eta_min (float) – Minimum learning rate. Default: 0.
• last_epoch (int) – The index of last epoch. Default: -1.
• verbose (bool) – If True, prints a message to stdout for each update. Default: False.
get_last_lr()
Return last computed learning rate by current scheduler.
load_state_dict(state_dict)
Parameters
state_dict (dict) – scheduler state. Should be an object returned from a call to state_dict().
print_lr(is_verbose, group, lr, epoch=None)
Display the current learning rate.
state_dict()
Returns the state of the scheduler as a dict.
It contains an entry for every variable in self.__dict__ which is not the optimizer. | 2022-05-17 08:14:49 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.778862714767456, "perplexity": 6834.825017980125}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662517018.29/warc/CC-MAIN-20220517063528-20220517093528-00763.warc.gz"} |
http://uncyclopedia.wikia.com/wiki/Moldy_Cheese?diff=prev&oldid=5577700 | # Moldy Cheese
(Difference between revisions)
Moldy Cheese
Scientific classification
Phylum Things that eat you
Genus Things that are cold
Species Things that REALLY eat you
Binomial name
Cheese with Mold
Specifications
Primary armament Cheesiness
Secondary armament Moldiness
Power supply Eat and grow
Health INFINITY
Mana INFINITY
Strength $Cheese^8$
Intelligence More than Cheese
Weight Something you don't like
Length $Cheese/Cheese^2$
Special attack Eats Emoes and People
Conservation status
Eating People
The creation of Moldy Cheese simply starts as this, and nothing more, nothing less. Moldy cheese originates from regular cheese when it gets too old and dies, giving birth to a new, more evil form of life: Moldy Cheese. Unlike cheese, who's deliciousness is superb, Moldy Cheese seeks to destroy all who try to consume it. Its moldy badness will eat you from the inside out unless you attack first with milk. Milk is one of the few things feared by Moldy Cheese as regular cheese is made of milk and goodness and Moldy Cheese is essentially undead, evil cheese, and therefore not at all milky.
## How to Identify Moldy Cheese
Moldy cheese is often recognized early on by refrigerator-goers by the blue and white spots that form on a piece of delicious cheese. It is because of these spots that allows for the refrigerator-goers to toss out the cheese but, little known to them, this cheese will come back to eat them as Moldy Cheese!!
The easiest way to identify a lump of moldy cheese is to look at it's disgusting massive blob-i-ness and make sure that it has little white hairs poking out of its body. Beware of the fuzzy soft fur, it will deceive you, with it's dastardly appetite. Moldy Cheese also has two glowing red eyes somewhere, typically toward the top of the green-blue-furry mound which should be particularly avoided.
• Guns
• Nukes
• Cheese
• Swords
• Knives
• Daggers
• People
• You
# Things That Can Stop Moldy Cheese
## Where to Find Moldy Cheese
You can more than likely find Moldy Cheese in the harsh winter-land that is Canada where they prey on those poor Canadians and Mounties. It is here that is the best environment for Moldy Cheese to grow and multiply for the simple reason that it's cold and it's the biggest thing that's as cold as a cozy refrigerator. Do not be fooled by the Moldy Cheese's white fuzzy camouflage.
Mouldy cheese is evil
## Moldy Cheese as a Pet
Although it goes against all regulations and the safety of your bowels, you can have Moldy Cheese as a pet and although it is likely it will eat you in the first five minutes of ownership, go right ahead.
You must first unwrap and expose your cheese to the open air and stick it in the refrigerator. It must sit there for months on end before showing any signs of intelligence. The amount of time depends on the type of cheese. Once you have your baby Moldy Cheese, it will start growing. It is best to feed it at this keen time to make it accelerate. (Refer to Eats). Be sure to make your Moldy Cheese as large as possible (largest on record is about 523 lbs). Unlike regular cheese, Moldy Cheese does not need to be eaten to become active.
### Do-s
• Keep Moldy Cheese out of a cage
• Let Moldy Cheese grow and multiply
• Let Moldy Cheese feed you to it's spawn
• Let Moldy Cheese eat everyone you know and love
• Let Moldy Cheese frolic about and use your insides like a trampeline
• Worship Moldy Cheese
### Don't-s
• Cage Moldy Cheese
• Hurt Moldy Cheese
• Offend Moldy Cheese
• Neglect Moldy Cheese | 2016-02-06 04:11:01 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.20321524143218994, "perplexity": 8408.419287279956}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701145751.1/warc/CC-MAIN-20160205193905-00135-ip-10-236-182-209.ec2.internal.warc.gz"} |