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https://sciencing.com/calculate-per-share-common-stock-4928002.html | 1,721,846,664,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518427.68/warc/CC-MAIN-20240724171328-20240724201328-00391.warc.gz | 439,709,274 | 87,054 | # How to Calculate Price Per Share of Common Stock
••• www.sxc.com/onia
Print
The price per share of common stock can be calculated using several methods. Stock analysts use several methods to calculate price per share of many stocks using similar techniques for companies in the same industry.
Find stock quotes in the newspaper or online (see Resources). Always use the close price if it is after exchange hours or the last quote when trading during the day.
Consult a periodical, such as the "Value Line Investment Survey," to locate the book value. Compare book value, the historical P/E and the 3-to-5-year price projection. This shows the expected range in which the stock should trade, which will indicate whether the stock is trading above or below its long-term price.
Multiply the stock price by the number of shares outstanding. This is the capitalization of the company. Ignore stock options to employees and divide the stock price by the earnings per share. This is the multiple of the stock or a representation of the expected future earnings of the company. Estimate next year's earnings and multiply by the multiple to get next year's price estimate. Use this calculation for financial companies.
Multiply a company's earnings by its historical multiple (multiple is calculated by 100 multiplied by the expected next year earnings growth increase). A stock earning $1 this year and expected to earn$1.30 next year has a 30 percent growth rate and a multiple of 30. If the stock is at $20 this year, the stock should be at$39 next year, a gain of almost 100 percent.
For capital-intensive stocks, subtract all liabilities from the assets. The remainder is called book value. Divide book value by the number of shares to get book value per share. This represents the intrinsic value of the company as a going concern. Stocks that use large amounts of capital, such as car and steel companies, often trade as a percent of book value.
#### Tips
• There are many ways to calculate price per share estimates. These ranges are useful for deciding whether a particular is cheap or expensive. Compare against other analyst's recommendations and judge against last-trade statistics.
#### Warnings
• Always compare your calculations with at least two other reputable sources. | 456 | 2,291 | {"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} | 2.796875 | 3 | CC-MAIN-2024-30 | latest | en | 0.943346 |
https://math.stackexchange.com/questions/2207219/list-of-2d-definite-integrals-not-reducible-to-products-of-1d-integrals | 1,561,625,273,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628001014.85/warc/CC-MAIN-20190627075525-20190627101525-00546.warc.gz | 518,816,808 | 34,301 | List of 2D definite integrals not reducible to products of 1D integrals
I'm writing numerical integration routines for 2D surface integrals. To test it, I'm looking for a list of definite integrals which have analytic forms. I need
• Integrals in polar coordinates over the unit disk
• Integrals over the unit square
• Integrals over triangles and convex hulls
• Integrals over all $\mathbb{R}^{2}$ and the half-plane
I already have many tests for integrals which can be reduced to products of 1D integrals, so I was hoping to find more that are not reducible in this way. I've check Abramowitz & Stegun, the CRC Standard Mathematical Tables & Formulae, and googled around but haven't found enough examples to really challenge the routine and as such I was hoping that the good people of math.se had some truly enjoyable 2D surface integrals they could show me!
Thanks!
Here's what I've found thus far:
• $\int_{2}^{3} \int_{1}^{\infty} x^{-y} \mathrm{d}x \mathrm{d}y = \ln(2)$
• $\int_{0}^{1} \int_{0}^{1} \frac{x}{\sqrt{x^2 + y^2}} \, \mathrm{d}x \mathrm{d}y = \frac{1}{2}(\ln(\sqrt{2}+1) + \sqrt{2} -1)$
• $\int_{0}^{1} \int_{0}^{1} \frac{1}{1-xy} \, \mathrm{d}x \mathrm{d}y = \frac{\pi^2}{6}$
References are also appreciated. | 387 | 1,236 | {"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} | 3.328125 | 3 | CC-MAIN-2019-26 | latest | en | 0.883122 |
http://www.emis.de/classics/Erdos/cit/15103502.htm | 1,429,921,425,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246645538.5/warc/CC-MAIN-20150417045725-00076-ip-10-235-10-82.ec2.internal.warc.gz | 466,662,117 | 1,976 | ## Zentralblatt MATH
Publications of (and about) Paul Erdös
Zbl.No: 151.03502
Autor: Erdös, Pál; Sarközy, A.; Szemeredi, E.
Title: On the solvability of the equations [ai,aj] = ar and (a'i,a'j) = a'r in sequences of positive density (In English)
Source: J. Math. Anal. Appl. 15, 60-64 (1966).
Review: The authors obtain the following results.
1) Let a1 < a2 < ··· be an infinite sequence of integers for which there are infinitely many integers n1 < n2 < ··· satisfying
sumai < nk {1 \over ai} > c1 {log nk \over (log log nk) ½}.
Then the equations (a'i,a'j) = a'r,[ai,aj] = ar have infinitely many solutions. The symbol (ai,aj) denotes the greatest common divisor and [ai,aj] denotes the least common multiple of ai and aj.
2) Let a1 < a2 < ··· be an infinite sequence of integers for which there are infinitely many integers n1 < n2 < ··· satisfying
sumai < nk {1 \over ai} > c2 {log nk \over (log log nk)1/4}.
Then there are infinitely many quadruplets of distincts integers ai,aj,ar,as satisfying (ai,aj) = ar, [ai,aj] = as, c1 and c2 denote suitable positive constants.
Reviewer: Cs.Pogany
Classif.: * 11B83 Special sequences of integers and polynomials
Index Words: number theory
© European Mathematical Society & FIZ Karlsruhe & Springer-Verlag | 391 | 1,266 | {"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} | 2.609375 | 3 | CC-MAIN-2015-18 | longest | en | 0.777568 |
https://esolangs.org/wiki/Infinite_Goto | 1,590,793,448,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347406785.66/warc/CC-MAIN-20200529214634-20200530004634-00540.warc.gz | 336,791,863 | 6,880 | # Infinite Goto
Infinite Goto is a language by User:PythonshellDebugwindow. Programs in it will never end.
## Storage
Infinite Goto uses an arbitrarily long array of arbitrarily large unsigned integers. It also uses a cell pointer (CP) to keep track of the current cell.
## Syntax
All valid lines of an Infinite Goto have an integer on them. This integer can’t have any leading zeros or start with a plus sign (+). Invalid or empty lines cause the program to go to the next line (or line 0 if out-of-range), so all valid Infinite Goto programs match the following regex: `-?([1-9][0-9]*|0)(\n-?([1-9][0-9]*|0))*`. It’s considered good practice to put negative numbers on unused lines, e.g. if you skip line 5 you should put a negative number such as -5. Integers less than 0 are treated as 0, and integers greater than or equal to the program's length are treated as the program's length minus 1. The IP then jumps to the (0-based) Nth line, where N is the integer on the current line after the previously mentioned modifications have been applied. For example,
```1
0
```
would jump infinitely between lines 0 and 1,
```0
```
```3
-1
```
would jump to line 3, find the last line is line 1, then jump to line 0 because of overflow, and repeat this forever. Jumping to certain lines has special effects:
Line Number (0-based) Effect
5 Sets the current cell to user input as a base-10 integer, if invalid then sets it to 0
8 Decrements the current cell if it's greater than 0
10 Increments the CP
13 Decrements the CP
16 Outputs the current cell’s value as a base-10 integer, with a newline
19 Jumps to line 20 if the current cell is greater then 0, else jumps to line 21
27 Jumps to a random line 28-32 both inclusive (the number on this line is ignored but should still be valid, if it’s not valid then it jumps to line 28 (or 0 if out-of-range) no matter what)
35 Sets the current cell to the previous line number, e.g. coming from line 7 sets the current cell to 7
All lines that are nonzero multiples of 2 and aren’t listed above (other than 7, an example for 35) (and aren’t one of the above numbers increased by a multiple of 45–see below), when jumped to, will increment the current cell. As well as this, every 45 lines, the modifications apply again, e.g. lines 5, 50, 95, etc. will all set the current cell to user input, lines 8, 53, and 98 do the same thing, and so on, and all these line numbers shown (20, 21, 28, etc.) do the same.
## Examples
### Numeric cat
```5
1
-2
-3
-4
16
-6
-7
-8
-9
-10
-11
-12
-13
-14
-15
1
```
This program uses an infinite loop on line 1 (the second line) to effectively “halt” the program.
```5
-1
-2
-3
-4
16
-6
-7
-8
-9
-10
-11
-12
-13
-14
-15
5
```
```27
-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
35
35
35
35
35
-33
-34
10
36
```
## Resources
• An interpreter written in Python along with some examples | 900 | 2,925 | {"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} | 2.515625 | 3 | CC-MAIN-2020-24 | latest | en | 0.884109 |
https://www.careerride.com/mchoice/transmission-line-ad-bc-1-17424.aspx | 1,660,719,029,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572870.85/warc/CC-MAIN-20220817062258-20220817092258-00614.warc.gz | 611,621,920 | 5,229 | # Transmission line - -AD + BC = 1
Q. For a transmission line which among the following relation is true?
- Published on 25 Nov 15
a. –AB + CD = 1
b. AD + BC = 1
c. AB – CD = -1
d. –AD + BC = 1 | 75 | 196 | {"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} | 2.71875 | 3 | CC-MAIN-2022-33 | latest | en | 0.919583 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-1-introduction-to-algebraic-expressions-1-3-fraction-notation-1-3-exercise-set-page-27/48 | 1,537,728,884,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159570.46/warc/CC-MAIN-20180923173457-20180923193857-00383.warc.gz | 749,441,672 | 14,458 | ## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$\frac{180}{36}=\frac{5\times36}{1\times36}=\frac{5}{1}\times\frac{36}{36}=\frac{5}{1}\times1=5$ | 62 | 175 | {"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} | 3.625 | 4 | CC-MAIN-2018-39 | longest | en | 0.428912 |
https://www.effortlessmath.com/math-puzzles/geometry-puzzle-challenge-71/ | 1,718,941,395,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862036.35/warc/CC-MAIN-20240621031127-20240621061127-00088.warc.gz | 654,197,706 | 11,951 | # Geometry Puzzle – Challenge 71
This is a fun and engaging math puzzle to challenge even the smartest students and help them develop logic skills.
## Challenge:
The total weight of a box and the candies it contains is 12 pounds. After $$\frac{2}{3}$$ of the candies are eaten, the box and the remaining candies weigh 5 pounds. What is the weight of the empty box in pounds?
A- $$\frac{2}{3}$$
B- 1
C- $$\frac{3}{2}$$
D- $$2\frac{1}{2}$$
E- 5
### The Absolute Best Book to challenge your Smart Student!
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Let “B” be the weight of the box and “C” be the weight of the candies. So:
B + C = 12
After $$\frac{2}{3}$$ of the candies are eaten, the box and the remaining candies weigh 5 pounds. So,
B + $$\frac{1}{3}$$ C = 5
Solve the system of two equations:
B + C = 12 → B = 12 – C
B + $$\frac{1}{3}$$ C = 5 → 12 – C + $$\frac{1}{3}$$ C = 5 → 12 – $$\frac{2}{3}$$ C = 5 → C = 10.5
B = 12 – C → B = 12 – 10.5 = 1.5
The weight of the box is 1.5 pounds.
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# 3.4 Quadratic Functions Notes 1130-02
Marketplace > University of Tennessee - Chattanooga > Math > 1130-02 > 3 4 Quadratic Functions Notes
Kiana Thompson
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College Algebra
John Graef
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Here is 3.4 College Algebra Notes on Quadratic Functions
COURSE
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PROF.
John Graef
TYPE
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## Popular in Math
This 2 page Class Notes was uploaded by Kiana Thompson on Sunday October 4, 2015. The Class Notes belongs to 1130-02 at University of Tennessee - Chattanooga taught by John Graef in Summer 2015. Since its upload, it has received 90 views. For similar materials see College Algebra in Math at University of Tennessee - Chattanooga.
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Date Created: 10/04/15
College Algebra Week 7 Notes Quadratic Functions Quadratic functions can be written in the form fx ax2 bx c a 0 Dissecting 1 If a gt 0 positive the parabola opens up If a lt 0 negative the parabola opens down 2 The Vertex the center point is the highest or lowest on the function depending on whether the parabola opens up or down Finding the vertex if the function is in standard form fx ax2 bx c i9 i9 b i9 Vertex za f 2a ie x 26 and yf 2a b 3 Axis of Symmetry x x value of vertex ie x 2a 4 The y intercept is c 5 The x intercepts are found when y O O ax2 bx c 6 Find the maximum value if it exists or the minimum value if it exists of the function The y value of the vertex will be either the maximum or minimum value of the function depending on which direction the parabola opens 7 Find the domain input values and range output values EXAMPLES fxx2 2x 3 a lb 2C 3 Positive or negative POSITIVE GRAPH OPENS UP Find vertex m 5 1 X 12 21 3 4 y VERTEX 1 4 AXis of symmetry X 1 The y intercept y 0 3 The X intercepts X2 2X 3 factor X 3 X1 X 3O amp X1O X 3 amp X 1 Find maxmin value 1 4 Minimum Value NO Maximum Find domain amp range D oo w R 3949 OO
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Q: What is the electron configuration of iron? A. 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6, 5s2, 4d9 B. 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6 C. 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d6 D. 1s2, 2s2, 2p6,
3s2, 3p6, 4s2, 3d8
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Use the quadratic formula to solve 4y2 + 8y + 7 = 4.
Question
Updated 6/6/2014 12:03:08 AM
4y^2 + 8y + 7 = 4
4y^2 + 8y + 7 - 4 = 0
4y^2 + 8y + 3 = 0
a = 4, b = 8, c = 3
b^2 - 4ac = 8^2 - 4(4)(3) = 64 - 48 = 16
x = (-8 ± sqrt 16)/8
= (-8 ± 4)/8
x = -1/2 or x = -3/2
Confirmed by andrewpallarca [6/6/2014 10:44:45 AM]
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 715 | 1,557 | {"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} | 3.171875 | 3 | CC-MAIN-2018-26 | latest | en | 0.783113 |
http://forum.beeminder.com/t/bitminding/4205 | 1,534,372,195,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210362.19/warc/CC-MAIN-20180815220136-20180816000136-00203.warc.gz | 147,121,317 | 7,074 | # Bitminding
#1
I had a thought for using a bitfield to combine many goals into one. For instance, suppose there are 9 different medications, supplements, or vitamins that you take every day, like:
1. Vitamin B
2. Vitamin E1
3. Vitamin E2
4. Vitamin M
5. Vitamin I
6. Vitamin N
7. Vitamin D
8. Vitamin E3
9. Vitamin R
You can assign each one a power of 2, like:
1. Vitamin B … 1
2. Vitamin E1 … 2
3. Vitamin E2 … 4
4. Vitamin M … 8
5. Vitamin I … 16
6. Vitamin N … 32
7. Vitamin D … 64
8. Vitamin E3 … 128
9. Vitamin R … 256
Then you can set a daily goal of 511 and you can tell easily which ones you’ve taken and which ones you haven’t without needing 9 different goals.
This is similar to the setup some people have where they need a certain number of points and different tasks give them different numbers of points.
If you take more than one of some of the pills, you can just use a higher base - it might be easier to read if you use decimal anyway:
1. Vitamin B … 1
2. Vitamin U … 10
3. Vitamin Z1 … 100
4. VItamin Z2 … 1000
So if you need 4 Z2s, 3 Z1s, 1 U, and 5 Bs, you can set a daily goal of 4315 and just read your current total off of the number.
#2
Clever.
This might be optimising for a failure case. If your end goal is
that you always do all of them, why not just sum up how many you did?
#3
Well, applying that reasoning, why doesn’t everyone just have one beeminder goal called “everything” that just checks if you do everything you’re supposed to every day?
At least for me, I’m constantly changing which supplements and what amounts I take, and I think it can be useful to compare and correlate with other things. For instance maybe Vitamin W helps you work out, Vitamin Q helps you quit smoking, but Vitamin S makes it hard to sleep. And maybe too little or too much Vitamin F makes it hard to focus, but just the right amount is correlated with high productivity. Just summing it up wouldn’t allow you to check that.
#4
I think people actually do have a beeminder goal that checks if they
have “done everything”. Both myself and Nick Winter did, for a long
time.
Please don’t take my comment as personal criticism. It wasn’t intended that way.
If you’re looking for correlations between things, then you definitely
want to be able to tell which days you did which things.
#5
Oh I didn’t take it that way at all! Sorry if my comment came off as if I did - I intended it as a reductio ad absurdum of the idea that keeping track of different things separately is optimizing for a failure case.
For me, having ONLY a “everything” goal would just be too coarse-grained and frustrating and make record keeping a pain. Plus I’m pretty sure it would always be “no.”
#6
I love nerding out about creative ways to abuse Beeminder! And I often endorse it. But this sounds all wrong to me. What you really want is multidimensional datapoints and, yes, it’s possible to encode arbitrary information in an integer (like whole mathematical theorems!) but that doesn’t make it a good idea! Maybe if there was a natural orders-of-magnitude hierarchy where it would make sense for some things to get 10x, 100x, etc as many points as others.
I think the better way to shoehorn Beeminder into this use case is to put JSON or something parseable in the datapoint comments.
I always cite the QS First principle. What fundamentally interesting thing is this Beeminder goal generating a graph of?
#7
What happens if you forget to take your Vitamin Z1 one day? Then from that point on your numbers are all screwed up and you can no longer tell which vitamins you’ve taken and which you haven’t just by looking at the required points for today.
#8
I don’t understand this. Why would that screw it up any more than partially completing any other goal?
#9
Because normally (if you have done the required amount every day) it would tell you something like +4315 points needed. If you take your vitamin U then you put in 10 and now it says +4305 points needed, so you can tell that you still have to take B, Z1, and Z2. And so on. But if you end the day with +100 still needed, then tomorrow it is going to say +4415 needed. If you are OK with taking 4 vitamin Z1’s I guess this is not so bad. But it gets particularly bad when you end up with carries. For example if you normally need to take 8 vitamin Z1 (800 points) and you forget one day, the next day it will say 4815 + 800 = 5615, which makes no sense. One vitamin Z2 is not the same as ten vitamin Z1.
#10
Why would it do that? Wouldn’t you derail and then go back to 4315?
#11
Oh, I see, I didn’t realize you intended to have the goal always in the red. If you want to do that then it works. But it seems there’s no sensible way to give yourself more slack without potentially running into problems like I described.
#12
I was assuming these are things you have to do every day. When there’s something I have to do every day, giving myself slack defeats the whole point.
I’m not going to do anything unless I absolutely have to, so beeminder doesn’t work for me unless there’s no slack.
How does it work for you - do you have daily things that you give yourself slack for?
#13
Yes, I do. For example, here’s my flossing goal which is set to a rate of 6.5/week. I want to floss “every day”, but it also doesn’t really matter if I miss a day here and there (one day every two weeks). For me it would feel demoralizing/overly harsh to have it set to literally 7/week. This way I am doing the goal regularly but I also know there is a bit of grace if I ever have a one-off crazy day or completely forget or whatever.
Obviously this depends on the goal; there are some daily goals that you literally need to do every day or there will be bigger real consequences (e.g. something like taking a medication that affects your mood or physical well-being), so it would make sense to reflect that in your Beeminder goal.
#14
So what actually happens when you do that? For me that wouldn’t give me any grace because I’d just not brush my teeth on the fortnightly non-beemergency day, using up the grace period as soon as it came up, so it wouldn’t get saved for a crazy day.
#15
Not everything has to be about the graph, though - beeminder has many possible uses and can be used to record data or discourage skipping daily requirements even when the graph doesn’t represent anything. In this case recording 1234 is a substitute for recording the unsupported multidimensional datapoint (1, 2, 3, 4).
#16
It all depends on how akratic you are, in what ways, about what things, and what is motivating to you. For some goals I do end up skating the edge like that. For my flossing goal, for me it is motivating just to have the graph and to put in a data point every time I do it. So the goal still motivates me to do it even if it isn’t an emergency day. I also have a sort of “meta-goal” of trying to get all my goals green (it rarely happens but I enjoy trying). So when I see it is blue I am already motivated to do it.
#17
In my case, that would be impossible, as I have daily binary goals. Those can’t ever be green. You don’t have any like that? Wait, how do you get the flossing goal green?
#18
If Brent’s road is dialed to 6.5 flossings per week then, by flossing daily, he gets .5 more days of safety buffer every week. So after like a month and half he’ll have accumulated 3 safe days which should have him in the green! Btw, highly recommended for everyone: Beeminding All The Things and The Fifty Goals of Brent Yorgey. | 1,870 | 7,521 | {"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} | 3.453125 | 3 | CC-MAIN-2018-34 | latest | en | 0.912669 |
https://dokumen.tips/documents/arithmetic-and-geometric-sequences-112-common-difference-common-ratio.html | 1,660,746,095,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572908.71/warc/CC-MAIN-20220817122626-20220817152626-00280.warc.gz | 223,848,829 | 23,204 | # Arithmetic and Geometric Sequences (11.2) Common difference Common ratio
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Arithmetic and Geometric Sequences (11.2) Common difference Common ratio Slide 2 A sequence Give the next five terms of the sequence for 2, 7, 12, 17, What is the pattern for the terms? Slide 3 A sequence Give the next five terms of the sequence for 2, 7, 12, 17, 22, 27, 32, 37, 42 This is an example of a sequence a string of numbers that follow some pattern. Whats our pattern here? Slide 4 A sequence Give the next five terms of the sequence for 2, 7, 12, 17, 22, 27, 32, 37, 42 Whats our pattern here? We add five to a term to get the next term. When we add or subtract to get from one term to the next, thats an arithmetic sequence. Slide 5 Another sequence Find the next five terms in this sequence? 8, 4, 2, Whats our pattern this time? Slide 6 Another sequence Find the next five terms in this sequence? 8, 4, 2, 1,.5,.25,.125,.0625 Whats our pattern this time? We divide each term by 2 to get the next term. (This is also multiplying by .) When we multiply or divide to get the next term, we have a geometric sequence. Slide 7 Terminology We label terms as t n, where n is the place the term has in the sequence. The first term of a sequence is t 1. So in the arithmetic sequence, t 1 = 2. In the geometric sequence, t 1 = 8. Slide 8 Terminology We label terms as t n, where n is the place the term has in the sequence. The second term of a sequence is t 2. The third is t 3. Get it? If the current term is t n, then the next term is t n+1. The previous term is t n-1. Slide 9 Terminology We list sequences in the abstract as t 1, t 2, t 3, t n. This is true whether the sequence is arithmetic, geometric, or neither. Slide 10 Arithmetic sequence formula If the pattern between terms in a sequence is a common difference, the sequence is arithmetic, and we call that difference d. t n = t 1 + (n-1) d (In other words, find the n th term by adding (n-1) ds to the first term.) Test it with our first sequence. Slide 11 Arithmetic sequence formula If the pattern between terms in a sequence is a common difference, the sequence is arithmetic, and we call that difference d. t n = t 1 + (n-1) d We can use this to find the first term, n th term, the number of terms, and the difference. Slide 12 Geometric sequence formula If the pattern between terms in a sequence is a common ratio, then it is a geometric sequence and we call that ratio r. t n = t 1 r n-1 (In other words, find the n th term by multiplying t 1 by r and do that (n-1) times.) Test it with our second sequence. Slide 13 Geometric sequence formula If the pattern between terms in a sequence is a common ratio, then it is a geometric sequence and we call that ratio r. t n = t 1 r n-1 We can use this to find the first term, the n th term, the number of terms, and the common ratio. Slide 14 Sequence #3 Give the first five terms of the sequence for t 1 = 7 t n+1 = t n 3 What is the pattern for the terms? Is this arithmetic or geometric? What is the tenth term? Slide 15 Sequence #3 Give the first five terms of the sequence for 7, 4, 1, -2, -5 What is the pattern for the terms? We subtract 3 from a term to get the next one. It is an arithmetic sequence. The tenth term is t 10 = 7 + (10-1) (-3) = -20. Slide 16 Sequence #4 Find which term 101 is in the arithmetic sequence with t 1 = 5, and d = 3. Slide 17 Sequence #4 Find which term 101 is in the arithmetic sequence with t 1 = 5, and d = 3. 101 = 5 + (n 1)3 101 = 5 + 3n 3 101 = 2 + 3n 99 = 3n n = 33 So, the 33 rd term. Slide 18 Sequence #5 Find the 9 th term of the sequence 1, -2, 4, What type of sequence is this? What formula do we use? Slide 19 Sequence #5 Find the 9 th term of the sequence 1, -2, 4, What type of sequence is this? Geometric, with a common ratio of -2. What formula do we use? t n = t 1 r n-1 So, t 9 = 1(-2) 9-1 = (-2) 8 = 256. Slide 20 Sequence #6 Find which term 1536 is in the geometric sequence with t 1 = 3, and a common ratio of 2. Slide 21 Sequence #6 Find which term 1536 is in the geometric sequence with t 1 = 3, and a common ratio of 2. 1536 = 3(2) n-1 512 = (2) n-1 (Ooh, want an exponent, need to use logs.) n -1 = log 2 512 = log 512/ log2 = 9 n = 10 Slide 22 Sequence #Last Find the 9 th term of the sequence 1, 1, 2, 3, 5, 8, What type of sequence is this? What formula do we use? How do we graph it?
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Documents | 1,400 | 4,778 | {"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} | 4.9375 | 5 | CC-MAIN-2022-33 | latest | en | 0.922898 |
https://rdrr.io/cran/mgpd/man/ml_mix.html | 1,524,484,758,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945942.19/warc/CC-MAIN-20180423110009-20180423130009-00189.warc.gz | 685,773,285 | 14,510 | # ml_mix: internal In mgpd: mgpd: Functions for multivariate generalized Pareto distribution (MGPD of Type II)
## Description
internal use only
## Usage
`1` ```ml_mix(param, dat, mlmax = 1e+15, fixed = FALSE, ...) ```
## Arguments
`param` `dat` `mlmax` `fixed` `...`
## Details
internal use only
## Value
internal use only
## Note
internal use only
P. Rakonczai
## References
internal use only
``` 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``` ```##---- Should be DIRECTLY executable !! ---- ##-- ==> Define data, use random, ##-- or do help(data=index) for the standard data sets. ## The function is currently defined as function (param, dat, mlmax = 1e+15, fixed = FALSE, ...) { loglik = mlmax hxy = NA x = dat[, 1] y = dat[, 2] error = FALSE mux = param[1] muy = param[4] sigx = param[2] sigy = param[5] gamx = param[3] gamy = param[6] alpha = 1 mu = function(x, y) 1/x + 1/y - alpha/(x + y) dxdymu = function(x, y) -2 * alpha/(x + y)^3 if (sigx < 0 | sigy < 0) error = TRUE if (fixed == TRUE) { mux = 0 } if (error) loglik = mlmax if (!error) { tx = (1 + gamx * (x - mux)/sigx)^(1/gamx) ty = (1 + gamy * (y - muy)/sigy)^(1/gamy) tx0 = (1 + gamx * (-mux)/sigx)^(1/gamx) ty0 = (1 + gamy * (-muy)/sigy)^(1/gamy) dtx = (1/sigx) * pmax((1 + gamx * (x - mux)/sigx), 0)^(1/gamx - 1) dty = (1/sigy) * pmax((1 + gamy * (y - muy)/sigy), 0)^(1/gamy - 1) c0 = -mu(tx0, ty0) hxy = 1/c0 * dxdymu(tx, ty) * dtx * dty hxy = as.numeric(hxy * (1 - ((x < 0) * (y < 0)))) loglik = -sum(log(hxy)) } if (min(1 + gamx * (x - mux)/sigx) < 0) loglik = mlmax if (min(1 + gamy * (y - muy)/sigy) < 0) loglik = mlmax loglik } ``` | 714 | 1,715 | {"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} | 2.75 | 3 | CC-MAIN-2018-17 | latest | en | 0.349279 |
http://www.jiskha.com/display.cgi?id=1326329435 | 1,498,381,949,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320476.39/warc/CC-MAIN-20170625083108-20170625103108-00666.warc.gz | 546,693,994 | 3,785 | # Geometry
posted by .
Jonathan is 3 ft from a lamppost that is 12 ft high. The lamppost and it's shadow form the legs of a right triangle. Jonathan is 6 ft tall and Is standing parallel to the lamppost. How long is Jonathan's shadow?
I don't get how the answer is square root 180, Ryan?
• Geometry -
Actually, it's sqrt(180) from the top of the pole to the tip of the shadow.
Using similar triangles, if the shadow is x feet long,
12/(3+x) = 6/x
12x = 18 + 6x
6x = 18
x = 3
so, the shadow is 3 feet long. | 155 | 512 | {"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} | 3.921875 | 4 | CC-MAIN-2017-26 | latest | en | 0.922236 |
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# Unit plan math
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### Unit plan math
1. 1. Intel® Teach ProgramEssentials CourseUnit Plan TemplateClick on any descriptive text, then type your own. Unit Author First and Last Name Gretchen T. Lequin School District School Name Cebu Normal University School City, State Osmeña Blvd, Cebu City Unit Overview Unit Title Relations and Functions (Advanced Math) Unit Summary A relation is a set of ordered pairs. The first elements in the ordered pairs (the x-values), form the domain. The second elements in the ordered pairs (the y-values), form the range. A function is a set of ordered pairs in which each x-element has only ONE y-element associated with it. Exercises will be given to students for them to evaluate relations and functions. Students will give their own examples of relation and function related to their daily living. Subject Area The lesson is for the students to explain relation and function. They are expected to differentiate relation and function in different aspect. They are encourage to interactively participate in discussion by giving their own ideas through series of questions. They will be able to evaluate relation and function. Grade Level This lesson is intended for the fourth year students. Approximate Time Needed A period of two weeks is expected to finish the topic. Unit Foundation Targeted Content Standards and Benchmarks (BEC COMPETENCIES) Students must explain function as a special relation. They must map and graph functions. They must use vertical line test to identify the graph of a function. And they must be able to evaluate function through its equations. Student Objectives/Learning Outcomes Explain what relation is.© 2008 Intel Corporation. All Rights Reserved. Page 1 of 5
2. 2. Intel® Teach ProgramEssentials Course Identify the domain and range in a relation. Differentiate function from relation. Use vertical line test to determine a function. Evaluate function in terms of its equations. Curriculum-Framing Questions Essential Question How should we function to improve our relationship? How can we identify a relation that is a function from its ordered pairs Unit Questions and its graph? How do we evaluate functions? Content Questions What is relation? How does a function differ from the relation? Assessment Plan Assessment Timeline© 2008 Intel Corporation. All Rights Reserved. Page 2 of 5
3. 3. Intel® Teach ProgramEssentials Course Before project work begins Students work on projects and complete tasks After project work is completed • Individual worksheets will be given to apply the concepts discussed. • Pair works to answering certain exercises. • Collaborative work/Group work. • Board works in evaluating relations and functions. • Reflect on the importance of these relations and functions in their daily living. • Use the concepts of relations and functions in everyday functions. • Diagnose students’ prior knowledge through a number of exercises. • Ask guided questions that will let them relate to the© 2008 Intel Corporation. All Rights Reserved. Page 3 of 5 | 713 | 3,388 | {"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} | 3.75 | 4 | CC-MAIN-2017-30 | latest | en | 0.9126 |
http://math.stackexchange.com/questions/1618793/why-doesnt-fermats-last-theorem-for-polynomials-entail-that-for-integers | 1,467,405,925,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783403825.35/warc/CC-MAIN-20160624155003-00051-ip-10-164-35-72.ec2.internal.warc.gz | 198,634,915 | 17,481 | # Why doesn't Fermat's Last Theorem for polynomials entail that for integers?
Fermat's Last Theorem for polynomials follows from the Stothers-Mason theorem, that is: For any integer $n\geq 3$, there do not exist polynomials $x(t), y(t), z(t)$ not all constant such that $x(t)^n + y(t)^n = z(t)^n$ for all $t\neq 0$.
But since we can always find suitable polynomials in $t$ such that $(x(t), y(t), z(t)) = (a,b,c)$, why can't FLT for polynomials entail that for integers ?
For example, suppose we had $7^n + 8^n = 15^n$ for some integer $n\geq 3$, we would have $t=2$ such that $(7, 8, 15) = (3t+1, 4t, 8t-1)$, which is impossible by FLT for polynomials ?
-
Does FLT for polynomials state that $x(t)^n + y(t)^n \not= z(t)^n$ for all $t$? – Brian Tung Jan 19 at 22:43
All nonzero $t$, i guess. It may be what i'm missing. – User1 Jan 19 at 22:44
Minor nitpick, you should say "entail" rather than "imply". Since FLT is true for integers, anything implies it. – DanielV Jan 19 at 22:46
@User1: No, I mean, there is a difference between "There do not exist polynomials $x(t), y(t), z(t)$ and integer $n \geq 3$ such that $x(t)^n+y(t)^n = z(t)^n$ for all $t \not= 0$" and "For all polynomials $x(t), y(t), z(t)$ and integer $n \geq 3$, $x(t)^n+y(t)^n \not= z(t)^n$ for all $t \not= 0$." I suggest that FLT for polynomials is the former and not the latter. – Brian Tung Jan 19 at 22:53
@User1 FLT for polynomials tells you that, given coprime nonconstant $x(t)$, $y(t)$, $z(t)$ and $n\ge3$, there exists $a$ such that $x(a)^n+y(a)^n\ne z(a)^n$ (so $x(t)^n+y(t)^n\ne z(t)^n$ as polynomials). Not that “for all $a$”. – egreg Jan 19 at 22:55
The Mason-Stothers theorem holds for polynomials over $\mathbb{Q}$, and any $n\ge3$. Note that $\mathbb{Q}$ is an infinite field, so two polynomials $P(t)$ and $Q(t)$ are different if and only if there is $a\in\mathbb{Q}$ such that $P(a)\ne Q(a)$.
So FLT for polynomials over $\mathbb{Q}$ can be stated
for any coprime and nonconstant polynomials $x(t)$, $y(t)$, $z(t)$ over $\mathbb{Q}$ and any $n\ge 3$, there exists $a\in\mathbb{Q}$ such that $$x(a)^n+y(a)^n\ne z(a)^n$$
because this is an alternative way for saying $x(t)^n+y(t)^n\ne z(t)^n$ as polynomials.
So the theorem does not say that the inequality holds for all $a\in\mathbb{Q}$, which would be needed to derive FLT from it.
Another way for looking at it is considering FLT for polynomials over $\mathbb{R}$: can you perhaps derive from it that for $a,b,c\in\mathbb{R}$ and $n\ge 3$ we have $a^n+b^n\ne c^n$?
- | 881 | 2,516 | {"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} | 3.71875 | 4 | CC-MAIN-2016-26 | latest | en | 0.849234 |
https://math.stackexchange.com/questions/1876431/doubt-in-application-of-the-dinis-theorem | 1,571,317,565,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986675316.51/warc/CC-MAIN-20191017122657-20191017150157-00440.warc.gz | 600,293,733 | 31,504 | # Doubt in application of the Dini's Theorem
Recall Dini's Theorem:
Let $K$ be a compact metric space. Let $f:K\to \mathbb{R}$ be a continuous function and $f_{n}:K\to \mathbb{R}$, $n \in \mathbb{N}$ be a sequence of continuous functions. If $\{f_{n}\}_{n \in \mathbb{N}}$ converges to $f$ and if $f_{n}(x)≥f_{n+1}(x)$ for all $x\in K$ and all $n \in \mathbb{N}$, then $\{f_{n}\}_{n \in \mathbb{N}}$ converges uniformly to $f$.
Doubt:
Take $f_{n}:[0,\frac{1}{2}] \to \mathbb{R}$ given by $f_{n}(x) = x^{n}$. We have that $[0,\frac{1}{2}]$ is a compact set and $\{f_{n}\}_{n \in \mathbb{N}}$ converges to $f = 0$, where $f_{n}'s$ and $f$ are continous. But $\{f_{n}\}_{n \in \mathbb{N}}$ doesn't converge uniformly to $f$, (suppose that $\{f_{n}\}_{n \in \mathbb{N}}$ converges uniformly to $f=0$ and take $\epsilon = \frac{1}{4}$ so there exist $n_{0} >0$ such that to $n > n_{0}$ imply that $|x^{n}| < \frac{1}{4}$. Take $x = (\frac{1}{3})^{\frac{1}{n}}$ so $\frac{1}{3} < \frac{1}{4}$ !!).
Where am I missing?
Thank you
I assume you meant: take $x =\left( \frac1{3}\right)^{\frac1{n_0 + 1}}$ or something (you have to talk about some fixed $n$).
The problem is that as soon as $n \ge 3$, we have $3/2^n \le 3/8 < 1$, so $1/3^{1/n} > 1/2$ and you are no more in $[0,1/2]$. | 532 | 1,280 | {"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} | 3.84375 | 4 | CC-MAIN-2019-43 | latest | en | 0.749516 |
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104 Km = ______ m
1. Ms. Sue
1 km = 1000 m
Multiply.
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# DE Unit 1.1
### Important Terms for Digital Electronics
TermDefinitionWord in a Sentence
Analog A way of representing some physical quantity, such as temperature or velocity, by a proportional continuous voltage or current. An analog voltage or current can have any value within a defined range. To know the exact time, check an analog clock.
Breadboard A circuit board for wiring temporary circuits, usually used for prototypes or laboratory work. Using breadboards for labs allow for mistakes to be made and lessons to be learned because the circuits created are temporary.
Conventional Current The direction of current flow associated with positive charge in motion. The current flow direction is from a positive to negative potential, which is in the opposite direction of electron flow. When we label the flow of a current in a circuit, we label the flow of the conventional current.
Current A movement of electrical charges around a closed path or circuit. Current is what passes through your body when you get shocked.
Digital A way of representing a physical quantity by a series of binary numbers. A digital representation can have only specific discrete values. Most of the clocks we use today are digital.
Digital Multi-Meter(DMM) A piece of test equipment used to measure voltage, current, and resistance in an electronic circuit. We use a multi-meter in labs to measure voltage, current, and resistance.
Engineering Notation A floating point system in which numbers are expressed as products consisting of a number greater than one multiplied by an appropriate power of ten that is some multiple of three. Engineering notation is very similar to scientific notation.
Kirchhoff’s Current Law (KCL) The algebraic sum of all currents into and out of any branch point in a circuit must equal zero. You use KCL to determine the current in a circuit.
Kirchhoff’s Voltage Law (KVL) The algebraic sum of all voltages around any closed path must equal zero. You use KVL to determine the voltage in a circuit.
LED Light-emitting diode. An electronic device that conducts current in one direction only and illuminates when it is conducting. We can use LED's to verify that current is passing through a circuit.
Ohm Unit of resistance. Value of one ohm allow current of one ampere with potential difference of one volt. Ohms are equal to the resistance in a circuit.
Ohm’s Law In electric circuits, I=V/R. Ohms law is used to find out the voltage, current, and resistance in a circuit.
Parallel Circuit One that has two or more branches for separate current from one voltage source. A parallel circuit carries the same amount of voltage throughout the circuit.
Resistance Opposition to current. Unit is the ohm. Resistance reduces the amount of current flowing throughout a circuit.
Resistor Color Code Coding system of colored stripes on a resistor to indicate the resistor's value and tolerance. The resistor color code can be used to create a resistor with a specific value of resistance and a specific level of tolerance.
Scientific Notation Numbers entered as a number from one to ten multiplied by a power of ten. Scientific notation is similar to engineering notation.
Series Circuit One that has only one path current. A series circuit consists of voltage drops but a consistent current.
Simulation Testing design function by specifying a set of inputs and observing the resultant outputs. Simulation is generally shown as a series of input and output wave forms. Simulation is used to find out what kind of outputs you get from certain inputs.
SI Notation Abbreviation of System International, a system of practical units based on the meter, kilogram, second, ampere, Kelvin, mole, and candela. SI Notation is what the rest of the world uses. The United States utilizes imperial notation.
Solder Metallic alloy of tin and lead that is used to join two metal surfaces. When you want to bond 2 metal surfaces, you solder them together.
Soldering Process of joining two metallic surfaces to make an electrical contact by melting solder (usually tin and lead) across them. Soldering is the combining of 2 metal surfaces.
Soldering Iron Tool with an internal heating element used to heat surfaces being soldered to the point where the solder becomes molten. A soldering iron is the tool that allows 2 metal surfaces to be soldered together.
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# Introduction
Topology provides an alternative perspective from traditional tools for understanding shape and structure of an object. With modern advances of the computational aspects of topology, these rich theories of shape can be applied to sparse and high dimensional data, spurring the field of Topological Data Analysis (TDA). Mapper and Persistent homology are the two most popular methods in the field of TDA, but the field is nascent and rich with exciting new ideas.
The goal of this work is to introduce and explore the use of Mapper (Singh, Memoli, and Carlsson 2007). We will demonstrate a novel application of Mapper to the domain of explanatory machine learning and show how Mapper can be used to understand not just the shape of data, but also understand a machine learning model.
## Topological Building Blocks
Both Mapper and Persistent Homology make use of simplicial complexes as the tool for understanding complicated shape. A simplicial complex is a generalization of a graph, with a few special features. The most particular feature is that simplicial complexes can contain higher order analogs of vertices and edges, referred to as simplices. Simplices can be the familiar vertices and edges of a graph, or triangles drawn between 3 vertices, tetrahedron between 4 vertices, and higher still.
Most often, simplicial complexes are built from the nerve of a cover. Intuitively named, a cover of a data set is a collection of subsets of the data such that every data point is in at least one of the subsets. Formally, we say a cover {Ui} of a data set X is satisfies the condition that for any x ∈ X, there exists at least on Ui ∈ {Ui} such that x ∈ Ui. In practice, we often have that each point is contained in multiple cover elements. The nerve is a simplicial complex created from a cover by collapsing each cover element into vertices and connecting vertices when the cover elements had points in common. If a point was included in two cover elements Ui and Uj, then the vertices σi, σj would have an edge drawn between them, denoted σij. We continue this process to higher order intersections to create higher order simplices.
The image below shows a space shaped like a 2-dimensional kidney bean with a cover of 4 elements. The nerve is drawn on the right. Each cover element becomes a node of the same color, pairwise intersections become edges between those nodes, and the three-way intersections become the triangles. Because this cover was particularly nice, the resulting simplicial complex now provides a combinatorial representation of the kidney bean that is topologically equivalent to the original shape. In practice, we have little guarantee that the cover will satisfy the conditions for topological equivalence, but we can use heuristics to generate nice covers that produce close representations.
The transition of topological ideas from theory to applied has been enabled by new methods for building simplicial complexes from data. Persistent Homology and Mapper are manifestations of these advances and represent two different ways to build and study covers. Persistent homology builds many different covers on the same data set and studies properties of the entire set of covers. Mapper constructs a single cover in a supervised manner and provides a representation suitable for exploratory data analysis.
In this work, we will focus on understanding the exploratory power of Mapper and how we can apply it to understand a machine learning model. For an exceptional introduction to persistent homology, we recommend (Ghrist 2008) and for other demonstrations of using Mapper for understanding a data set, see (Lum et al. 2013).
# Mapper
It is often a reasonable assumption that complex and high dimensional data exists on a much lower dimensional manifold or stratified space. Mapper can be thought of as a method of estimating and capturing this essential high dimensional structure (Singh, Memoli, and Carlsson 2007). It encodes information about relationships between regions of the data and the simplicial complex produced can be both aesthetically pleasing and useful for exploring such relationships.
The construction of Mapper makes use of two particular ideas to generate its cover, the lens and the pullback cover. The lens is a secondary representation of the data that encodes some interested aspect or information about the data set. This can be as simple as taking a subset of important columns. In this work, we interpret a machine learning model as a form of dimensionality reduction on the data, taking the predicted probabilities and decision functions as the important representation. The lens is usually where the analyst or scientist can embed domain knowledge into the construction, in our case, the representation is learned from labels. This lets the user build Mapper with respect to some information of interest and is why we say Mapper is semi-supervised.
After creating a lens, we generate a cover of this space. This cover is shown in the first image of the sequence below. In practice, the lens is chosen to be a low dimensional space, and because of this, constructing a cover is not too difficult. A common first choice of cover is to use overlapping intervals or hypercubes. We will discuss how do design a cover specifically for the output of a machine learning model a little later.
A cover of the data set is found by computing the pullback cover of our constructed lens cover. The pullback of the lens cover is found by taking each cover element in the lens, finding the associated data points in the original data. Because points that are close together in the lens space are not necessarily close together in the input data space, we must refine the pulled back cover elements by applying a clustering algorithm. If our data space was a continuous space rather than discrete data, the refinement process would consist of splitting the cover element into connected components. The second image in the sequence above shows how one cover element in the lens space becomes three cover elements in the pullback cover. The third image shows the complete pullback cover of all four lens cover elements.
The final step of Mapper is to compute the nerve of our constructed pullback cover. The result is a simplicial complex that represents the data set while capturing information about the lens representation of the data. Encoded in the simplicial complex is relationships between observations and relationships between groups of observations. These relationships are determined by nearness in the lens space and by differences in the data space. The image above shows the nerve constructed in this example.
# Data and Model
For our demonstration of applying Mapper to understand a machine learning model, we will build a classifier of leaf types using the UCI Leaf Data Set (Silva, Marcal, and Silva 2013). We chose the leaf data set because it has images included with the structured data, which is useful for illustrative purposes. As the modeling is not the emphasis of this work, we applied only a modest amount of energy into model selection and training and we are satisfied with the accuracy obtained. This is a relatively small data set and so to demonstrate that Mapper is apt for analysis of data orders of magnitude larger in both dimension and size, we will later apply Mapper to understanding a model trained on the Fashion MNIST data set (Xiao, Rasul, and Vollgraf 2017).
## Data
The leaf data set has 30 different types of leaves, but for our purposes, we will use the 10 largest classes, excluding Engl. Oak, Maple, Silver Poplar, Geranium, and Hawthorn. These five classes have very different characteristics from the rest and thus excluding them simplifies the demonstration here. The figure below shows examples from each of the 10 chosen classes. The color of the title will be used to differentiate between leaf classes in representations used later in the demonstration.
The data used for our prediction model is a structured representation of these images with 14 expert engineered features such as lobedness, elongation, and smoothness. Many of the images could easily be differentiated by eye, but some classes can be easily confused.
## Model
For the classification we train a Logistic Regression model to 91.5% test set accuracy using Scikit-Learn (Pedregosa et al. 2011). Before training we applied normalization using standard scaling and applied dimensionality reduction using PCA. Using grid search and a stratified 3-fold cross validation, we found the best model using an l1 penalty with parameter c = 0.9.
# Mapper for explaining machine learning
Mapper has an often-overlooked limitation requiring the lens to have low dimension so that this space can be feasibly covered with a grid. To sample a 2-D space with a resolution of 10 cover elements per dimension requires a grid with 102 units. However, for a typical machine learning classifier with 10 classes, the same resolution grid would have 1010 units, which is computationally intractable.
To apply Mapper to understanding machine learning models, we adapted the typical cover approach by designing a new type of cover specifically for machine learning classifiers. To construct the cover, we rank and threshold the predicted probabilities for each observation and generate a cover element for similarly ranked observations. When considering a model with m classes, our approach is parameterized the top p predictions, and threshold ϵ (always taken as 1/m). We create mp cover elements, with each cover element denoted by Cij, where i is the class and j is the rank of that class in an observation’s predicted probabilities. An observation x belongs to cover element Cij if the jth highest predicted probability of x is greater than ϵ, and the class corresponding to this predicted probability is i. This allows us to work directly with the spaces generated by the model and avoid using dimensionality reduction on the prediction space.
As an example, suppose the most likely class for leaf ids #12 and #76 are both Spindle, then both leaves are added to C(Spindle), 1, Spindle’s top cover element. If the second most likely class for #12 is Chestnut, then it would be placed in C(Chestnut), 2, Chestnut’s second cover element as well. Suppose the second most likely class for #76 is Cork Oak, but the predicted probability is less than our threshold of 1/m. Then #76 will not be added to another cover element. These two observations would then produce two cover elements, one with two points and one with just one.
The graphs above show two different ways of generating a cover of our lens. The first shows a typical approach of reducing the high dimensional space into 2 dimensions using PCA and then using a hypercube cover. To make the visualization easier to read, the cubes are shown with no overlap. Notice how many of the hypercubes in the traditional cover (on the left) are empty. In high dimensions this issue becomes even more pronounced. On the right, we demonstrate our new covering technique applied to the first observation in the data set. In this case, the leaf would be placed into 3 cover elements, C(Hazel), 1, C(Cork Oak), 2, and C(Spindle), 3.
## Instance Based Explanation
Instance based explanation is a particular type of explainable machine learning that focuses on explaining predictions through other observations. We can treat the machine learning model as a black box and explore the relationship between the input space and the prediction space using Mapper, requiring only the original data and the predicted probabilities for each observation. Because of this, the technique is generalizable to most machine learning techniques and is suitable for use by non-experts. We envision a system like what will be described would be suitable for end users to understand model predictions without needing a deep understanding of the specific modeling technique.
After looking at the global structure of Mapper, we’ll explore one particular leaf and try to use Mapper to understand what the model thinks of the leaf. We’ll introduce the idea of an escape route, a sequence of leaves that demonstrate how a leaf of interest is similar or different from the other nodes around it. After demonstrating out the routes can be used for explaining leaves, we apply the method to understanding a model trained on the Fashion MNIST data set.
## A Global View
The graph below shows the Mapper constructed using the predicted probabilities from the trained model and the particular cover described above. In this graph, the color of each node represents the majority class in that node. Because of the way the Mapper is constructed, each node represents a heterogenous set of leaves that are the model considers similar and each edge represents a relationship between nodes. Remember that because of the overlapping nature of the cover, each leaf could and probably will be in multiple different nodes.
In the implementation, we use Affinity Propagation clustering with damping factor of 0.5 and compute the nerve only considering 2-way intersections so edges are the highest dimensional simplices. For exploratory and exposition purposed, we’ve only kept the largest connected component of the graph. In this case, any disconnected pieces were isolated nodes.
Referring back to the grid of leaves above, we see that the pointer leaves (Hackberry, Primrose, Chestnut, and Saucer Magnolia) comprise most of the lower portion of the graph while rounder leaves (Hazel, Bougainvillea, and Birch) show up primarily at the top of the graph. The Spindle and Cork Oak join the two groups and judging from the images, do share characteristics with both.
## Local Instances
Suppose we are interested why our model decided that the particular Chestnut shown below was in fact a Chestnut. Judging from the image alone, I might have easily thought it was a Saucer Magnolia or Hazel. This particular Chestnut was chosen for demonstration because, while the classification was correct, the confidence in prediction was not very large. In fact, there were 4 classes with predicted probability greater than 1/m (with m = 10).
We’re immediately curious which nodes of Mapper contain this Chestnut, where in the graph does this Chestnut live? In the series of graphs below, we show three different ways of visualizing the location of our Chestnut of interest. These three graphs show different levels of information about the instance. The first shows only the nodes that contain this particular Chestnut. The middle graph shows all nodes that contain any Chestnut leaves at all. The third graph shows the same region of Chestnut nodes, but also shows all of neighboring nodes, colored by their majority leaf type. The edges of the third graph are also colored based on the majority class comprising that edge.
It’s interesting that this particular leaf is in a disconnected region of Chestnuts. Maybe this one is peculiar in some way?
Below we see all of the leaves that are in the same two nodes as the Chestnut of interest. Surprisingly, we see that the Chestnut is the only Chestnut in this subset. It is not surprising to see the Primrose and Spindle leaves in the same node, as they appear similar with an untrained eye.
## Confusion Regions
The nearby regions in the graph are one way to visualize the relationships between different observations and instances. Considering we are exploring a machine learning model, we can obtain more specific information from the Mapper as well. Remember that there were 4 different classes that had relatively high predicted probability on this Chestnut. Maybe these regions of Mapper can tell us something about the observation?
The table below shows example leaves from the 4 top predicted classes. These are Chestnut, Cork Oak, Primrose, and Spindle. While some of the Primrose have a very distinct tail or circular shape, by looking at multiple examples from the class, we can see some representatives that could cause confusion. We’ll dig much deeper into this idea while looking at escape routes.
The graphs below show the Mapper regions for each of the 4 classes shown above. The color of the node matches the labels in the grid above.
## Escape Routes
Escape routes are way of exploring how a particular class or observation is related to another class by visualizing the sequence of nodes and edges between them in the Mapper. Each class creates a region in the graph, and by drawing paths between regions, we see gradual transformations. An escape route is a short path through the topological space starting in a region dominated by one class and terminating at a node that contains no observations of that class. This allows the user to understand how subtle changes to the classifier’s input effects the model’s output prediction.
An escape route is calculated by finding the shortest path tree starting at a given node in a region. The tree is pruned to exclude all hops occurring after the path has entered other regions. We define the starting region as a subgraph of the topology containing all nodes with observations whose ground truth label is the same as the Chestnut of interest. Terminating regions are regions connected to the start region containing no observations of the user selected class. Escape routes allow us to present the user with example observations, allowing the user explanation to be conveyed in terms of the underlying observations, instead of a more abstract statistical or feature-based representations.
Below, we begin exploring escape routes by first looking at the neighboring regions to the Chestnut class to gain a sense for what classes are deemed similar based on the lens and covers. These neighbors are found by first finding all nodes that have at least one Chestnut, then considering the majority class from each of the neighboring nodes containing no Chestnut instances.
The figures below show the Chestnut region, along with the 3 adjacent classes, Saucer Magnolia, Primrose, and Hackberry. Above the graphs is one representative of each class chosen at random, for illustrative purposes.
The figure below shows many more examples from each of the neighbor classes. When viewing many of them, it’s easy to see their similarities. Only the most teardrop-like Hackberry and Primrose would be easy for the untrained eye to correctly classify. | 3,674 | 18,484 | {"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} | 2.859375 | 3 | CC-MAIN-2023-50 | longest | en | 0.939062 |
http://seaintarchive.org/mailarchive/1998c/msg00673.html | 1,490,485,429,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189088.29/warc/CC-MAIN-20170322212949-00533-ip-10-233-31-227.ec2.internal.warc.gz | 330,507,721 | 4,279 | Need a book? Engineering books recommendations...
# Re: Simpson Strongwall Question
• To: seaint(--nospam--at)seaint.org
• Subject: Re: Simpson Strongwall Question
• From: <Mlcse(--nospam--at)aol.com>
• Date: Wed, 22 Jul 1998 23:31:13 EDT
```In a message dated 98-07-22 16:07:41 EDT, you write:
<< Subj: Simpson Strongwall Question
Date: 98-07-22 16:07:41 EDT
From: wish(--nospam--at)cwia.com (Dennis Wish)
To: seaint(--nospam--at)seaint.org (SEAOC Listservice)
I received the test data from Simpson. I can not find the working stress
values? Also, is there a table of sizes and capacities from which to choose?
Are you assuming a working stress capcity of 25% of the ultimate?
Dennis Wish PE
>>
Dennis,
We have just started looking at the StrongWall in our office as substitution
for the Hardy frame used on a project. Page 1of 2 of there product flyer
gives allowable shears for the panel assembly. So by working stress, you have
a certain shear force calculated to a given wall line, you then select a
Strongwall with excess capacity to the calculated demand. The deflection of
the panel will be within code (.005h) for drift for shear loads less than the
strongwall design value.
Be VERY CAREFULL in evaluating the garage portal frame (wood beam sitting on
top of strongwall each end of beam). The listed values are for the pair of
Strongwall panels, not the individual panels at each end. This is not clear
in the literature unless you backcheck the allowable shear/ft value compared
to the allowable shear load and length of panel.
Example: Strongwall SW16x7-4 panels (16 inches wide)
Allowable shear for portal frame = 3020 pounds (1510 pounds for the panel at
each end of the frame)
The published working stress level I believe is 71.4 percent (1/1.4) of the
strength design load based upon testing. For furthur clarification, I guess
you need to contact Simpson.
By the way, Simpson now has competition for the strongwall, besides the Hardy
Frame, which is the "Lateral-Force Resisting (LFR) Panel" which is being
manufactured by Shear Transfer Systems. They just got their ICBO number.
This is a reinforced all wood panel also, possibly with a better holdown.
Michael Cochran
``` | 580 | 2,216 | {"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} | 2.671875 | 3 | CC-MAIN-2017-13 | longest | en | 0.863622 |
https://os.mbed.com/questions/81789/Can-I-get-board-VDD-from-st_micro-f401re/ | 1,600,568,776,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400193087.0/warc/CC-MAIN-20200920000137-20200920030137-00435.warc.gz | 574,646,304 | 8,196 | ## Can I get board VDD from st_micro f401re?
When I measure voltage with an analog channel, I need to multiply the 0 to 1 by the board voltage to get the actual voltage reading. If I default to using 3.3volts, the results are wrong. If I use my board voltage which is 3.315 volts, the readings are correct. Is there any way to get the board voltage via software so I don't have to manually measure the voltage on each board I use?
My voltage in = 1.649. If I take 100 readings and remove all readings above or below the standard deviation and take the reading times the 3.3volts, I get the following.
Average: 1.6422395992 Standard Dev: 0.0010298408 Output: 1.6421652767
When I use my board voltage measured to be 3.315, I get much better results.
Average: 1.6496315467 Standard Dev: 0.0008762093 Output: 1.6494109239
## 1 Answer
##### 2 years, 2 months ago.
There are some techniques for doing this... You will need to read the VREFINT voltage to set a voltage ref baseline. *some* parts have a calibration value burned into them which can then be used to correlate the VREF ADC value to an actual voltage. (Check the ST site for an app note on this subject). For reading the internal reference this code I posted can show you how https://os.mbed.com/users/sixBase3/code/Nucleo_read_analog_value_intADCch/
I'll look into my app note collection and see if I can find the one regarding the VREF cal.
Bill
Thank you for the input.
I actually was reading that value but didn't see the correlation. The corresponding values for the readings listed in my questions are as follows.
For the reading when I used 3.3V Vref(f): 0.364591, Vref : 23893, Temperature : 14947 For the reading when I used 3.315V Vref(f): 0.364591, Vref : 23909, Temperature : 14963
I was hoping that VREF ADC was a fixed 1.21V which would allow me to calculate 1.21/0.364591 = 3.3188V. I was hoping that would equal the 3.315V that I measured manually. I am currently searching for an app note on a calibration value. I am also testing multiple boards to see if I can use the VREF ADC without the calibration.
posted by Paul Moen 13 Jul 2018 | 571 | 2,124 | {"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} | 2.796875 | 3 | CC-MAIN-2020-40 | latest | en | 0.8913 |
https://newtum.com/calculators/maths/distance-calculator | 1,713,583,464,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817474.31/warc/CC-MAIN-20240420025340-20240420055340-00448.warc.gz | 383,781,153 | 23,598 | # Distance Calculator
## A Comprehensive Guide to Using the Distance Calculator Developed by Newtum
(Last Updated On: 2024-02-20)
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In conclusion, our Distance Calculator is a powerful and versatile tool for accurately measuring distances. The primary emphasis is on data security, as no data is processed on servers, ensuring it never leaves your device. This tool serves not just as a utility, but also as a resource for understanding and applying the concept of distance calculation. Whether you're a student, a professional, or a curious individual, our Distance Calculator is designed to cater to your needs, all while maintaining the highest standards of data privacy and security.
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The Distance Calculator is a tool that calculates the distance between two points.
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Yes, it's a free-to-use tool available on our website. | 781 | 4,012 | {"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} | 2.96875 | 3 | CC-MAIN-2024-18 | latest | en | 0.870247 |
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Review the Lesson | MATHguide homepage Updated June 22nd, 2023
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Given: A dog runs directly from point A to point B at a rate of 20 feet/second. The dog then runs directly from point B back to point A at a rate of 17 feet/second. Problem: Determine the average velocity of the dog as it traveled this path. Definition: Average velocity is calculated by taking the total distance traveled and dividing it by the total time traveled. Hint: The solution is not (20 + 17) ÷ 2 or 18.5 feet/second. Result: The average velocity of the dog is feet/second.Round your solution to the nearest tenth of a foot/second. | 157 | 653 | {"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} | 3.8125 | 4 | CC-MAIN-2024-33 | latest | en | 0.913987 |
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Videos, stories and solutions to help Grade 4 students learn how to create and interpret circle graphs or pie graphs and how to use information given in a circle graph to find unknown information.
Creating Circle Graphs (Part 1 - Common Sectors)
This video is the first part of our lesson on creating and interpreting circle graphs. Information in this video includes circle graph basics and 5 common sectors.
Creating Circle Graphs (Part 2 - Drawing)
This video shows how to take data given to you and use it to create a circle graph. It includes a review of common sectors, how to change fractions to percents, and graphing properties.
Interpreting Circle Graphs (Using Common Sectors)
This video is an example of how to use our 5 common sectors and knowledge of circle graphs to answer a few questions.
Interpreting Circle Graphs (Working Backward)
This video is a tutorial in how to use information given in a circle graph to find unknown information.
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Obtain the Expression for the Magnetic Energy Stored in an Inductor of Self-inductance L to Build up a Current I Through It. - CBSE (Science) Class 12 - Physics
ConceptSolenoid and the Toroid - the Solenoid
Question
Obtain the expression for the magnetic energy stored in an inductor of self-inductance L to build up a current I through it.
Solution
Consider the circuit shown above consisting of an inductor L and a resistor R, connected to a source of emf E. As the connections are made, the current grows in the circuit and the magnetic field increases in the inductor. Part of the work done by the battery during the process is stored in the inductor as magnetic field energy and the rest appears as thermal energy in the resistor. After sufficient time, the current, and hence the magnetic field, becomes constant and further work done by the battery appears completely as thermal energy. If i be the current in the circuit at time t, we have
E-L(di)/dt=iR
=>Eidt=i^2Rdt+Lidi
=>int_0^tEidt=int_0^ti^2Rdt+int_0^iLidi
=>int_0^t Eidt=int_0^ti^2Rdt+1/2Li^2
Now (idt) is the charge flowing through the circuit during the time t to t+dt. Thus (Eidt) is the work done by the battery in this period. The quantity on the left-hand side of the equation (i) is, therefore, the total work done by the battery in time 0 to t. Similarly, the first term on the right-hand side of equation (i) is the total thermal energy developed in the resistor at time t. Thus
1/2Li^2is the energy stored in the inductor as the current in it increases from 0 to i. As the energy is zero when the current is zero, the energy in an inductor carrying a current i, is U=1/2Li^2
Is there an error in this question or solution?
Video TutorialsVIEW ALL [1]
Solution Obtain the Expression for the Magnetic Energy Stored in an Inductor of Self-inductance L to Build up a Current I Through It. Concept: Solenoid and the Toroid - the Solenoid.
S | 515 | 1,972 | {"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} | 3.671875 | 4 | CC-MAIN-2019-30 | latest | en | 0.912027 |
https://rationalwiki.org/wiki/Buttered_cat_theory | 1,675,721,328,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500365.52/warc/CC-MAIN-20230206212647-20230207002647-00682.warc.gz | 464,634,041 | 11,126 | # Murphy's Law
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Thinking hardlyor hardly thinking?Philosophy Major trains of thought The good, the bad,and the brain fart Come to think of it v - t - e
_
If you are looking for other laws, see our longer Eponymous laws
Murphy's law states that if something can go wrong, then it will go wrong.
Murphy's law has two corrollaries: First, that if it can go wrong, it will go wrong at the worst possible time. Second, if there is a possibility of several things going wrong, the one that will cause the most damage will be the one to go wrong.
Murphy's law has one exception: Something will never go wrong when you want something to go wrong (though, recursively, this is Murphy's law). (Also known as the Law of Murphological Inapplicability.)
## Origin
Murphy's law is attributed to a Captain Murphy. Dr. Stapp was an Air Force medical researcher who rode various extremely fast vehicles in order to test how much acceleration the human body can take. In one test, his assistant, Capt. Murphy, had designed a harness to strap in the rider that held 16 sensors to measure the acceleration on different body parts. There were exactly two ways each sensor could be installed. Murphy did each one wrong. When Dr. Stapp staggered off the sled with bloodshot eyes and bleeding sores, all registered zero. A distraught Captain Murphy proclaimed: If there are two or more ways to do something and one of those results in a catastrophe, then someone will do it that way.[1]
## Examples
• When you drop toast, it will land butter side down[2] (unless it is strapped butter-side up to the back of a cat, in which case the buttered cat will hover in the air. However, the impact is only negligible, as they can consciously move in midair).
• When waiting for a bus, every other bus will show and just when you get sick and tired of it all and light a cigarette,[note 1] your bus will arrive. You may instead choose to take a cab/taxi, at which your bus will pass by you minutes after you have entered your cab.
• Your computer will tend to crash when you have a long, unsaved edit of significant importance. The hard drive in said computer will tend to corrupt during that crash when the autosave of said document is stored locally.
• When drinking quietly and secretively while your significant other is out with friends, you'll get to the point wherein you only have one beer left but you don't want to open it because you don't want to appear to be drinking when they arrive. After an hour and they haven't arrived, you open it, thinking all is calm, and then they walk in the door and accuse you of being drunk.
• In the UK, it always rains on bank holidays.[3]
• A valuable dropped item will always fall into an inaccessible place (a diamond ring down the drain, for example) — or into the garbage disposal while it is running.
### Buttered Cat Theory
For those of you in the mood, RationalWiki has a fun article about Buttered cat theory.
The buttered-cat theory is a tongue-in-cheek paradox based on Murphy's Law. It derives from two propositions:
1. Due to Murphy's Law, a piece of buttered toast will always land on the buttered side.
2. Cats always land on their feet (this is a common saying grounded in reality).
Due to these factors, the theory postulates that if a cat with a piece of butter-side-up toast strapped to its back is dropped, it will levitate in midair directly above the ground, spinning forever. Of course, this is completely inaccurate and impossible, but it is an amusing idea. The cat will almost certainly land on its feet as usual if this experiment is performed, as cats have an innate reflex that allows them to react quickly and turn upright when falling,[4] and toast lacks the ability to consciously move itself in midair (at least as far as we know). At most, the buttered toast will have a negligible effect on the ability of the cat to land on its feet.
However, if there is a sufficient mass of butter, then it will assumedly overcome the cat's self-righting ability, and the cat will land on its back, splatting butter everywhere. At what mass of butter the cat is unable to overcome will be different for each cat, and therefore there must be a critical cat-to-butter (C:B) ratio, above which the cat wins, and below which the floor gets messy, with the cat waving its paws madly in the air. Very near to this ratio, the cat will likely exhibit more complex behaviour, possibly landing on its side or showing other phenomena. See Professor Stewart's Hoard of Mathematical Treasures for more on the physics of buttered cats.[5]
## Notes
1. Only applicable in areas where buses are smoke-free.
## References
1. http://www.nytimes.com/1999/11/16/us/john-paul-stapp-89-is-dead-the-fastest-man-on-earth.html
2. Tumbling toast, Murphy's Law and the fundamental constants by R. A. J. Matthews, European Journal of Physics, vol.16, no.4, July 18, 1995, p. 172-6. Matthews won the 19997 Ig Nobel Prize in Physics for his work.[1]
3. We remember the cold, wet bank holidays and overlook the others.
4. Hoard of Mathematical Treasures by Ian Stewart (2010) Basic Books. ISBN 0465017754. | 1,192 | 5,199 | {"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} | 3.03125 | 3 | CC-MAIN-2023-06 | latest | en | 0.953854 |
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Tools介紹
The edge coupled differential microstrip transmission line is a common technique for routing differential traces. There are four types of impedance used in characterizing differential trace impedances. This calculator finds both odd and even transmission line impedance. Modeling approximation can be used to understand the impedance of the differential microstrip transmission line.
PCB Impedance Calculator
##### Introduction
The edge couple differential microstrip transmission line is a common technique for routing differential traces. There are four different types of impedance used in characterizing differential trace impedances. This calculator finds both odd and even transmission line impedance. Modeling approximation can be used to understand the impedance of the differential microstrip transmission line.
##### Description
An edge couple differential microstrip transmission line is constructed with two traces referenced to the same reference plane. There is a dielectric material between them. There is also some coupling between the lines. This coupling is one of the features of differential traces. Usually it is good practice to match differential trace length and to keep the distances between the traces consistent. Also avoid placing vias and other structures between these traces.
##### Differential Impedance Definitions
Differential Impedance The impedance measured between the two lines when they are driven with opposite polarity signals. Zdiff is equal to twice the value of Zodd
Odd Impedance The impedance measured when testing only one of the differential traces referenced to the ground plane. The differential signals need to be driven with opposite polarity signals. Zodd is equal to half of the value of Zdiff
Common Impedance The impedance measured between the two lines when they are driven with the same signal. Zcommon is half the value of Zeven
Even Impedance The impedance measured when testing only one of the differential traces referenced to the ground plane. The differential signals need to be driven with the same identical signal. Zeven is twice the value of Zcommon
##### Microstrip Transmission Line Models
Models have been created to approximate the characteristics of the microstrip transmission line.
The source for these formulas are found in the IPC-2141A (2004) “Design Guide for High-Speed Controlled Impedance Circuit Boards” and Wadell, Brian C. Transmission Line Design Handbook. Norwood: Artech House Inc, 1991 | 470 | 2,529 | {"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} | 2.9375 | 3 | CC-MAIN-2023-23 | latest | en | 0.90781 |
http://scienceblogs.com/builtonfacts/2008/09/10/from-earth-to-moon/ | 1,484,996,337,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281069.89/warc/CC-MAIN-20170116095121-00276-ip-10-171-10-70.ec2.internal.warc.gz | 249,958,395 | 12,985 | # From Earth to Moon
Ever wondered why it takes a tremendously huge rocket to launch people from the earth, but the Apollo astronauts managed to launch from the moon in a comparatively tiny lunar module? Easy, the whole thing was faked and NASA forgot to come up with a plausible explanation!
Wow, it was almost physically painful to type that even as a joke. The real reason is some very easy but still pretty cool physics. Shall we take a look?
For a given object, the gravitational potential energy per kilogram with respect to distant space is given by this easy little equation:
So to pick up one kilogram of mass from the earth and remove it from the earth’s gravitational influence requires energy given by that formula. Plugging in values for the mass of the earth and the radius of the earth gives U = 6.25 x 107 J/kg.
But the moon? Use the same equation and plug in the mass and radius of the moon. I get U = 2.8 x 106 J/kg.
You might remember that the surface gravity of the moon is about 1/6 that of the earth. But the potential energy of an object at the moon’s surface is (from the above equations) only about 1/23 that of the same object at the earth’s surface. It’s much, much easier to launch from the surface of the moon than it is from the surface of the earth. And since it’s so much easier, you don’t have as many kilograms of fuel and rocket to move to orbit in the first place. This is why it takes one of these to launch people from the earth into deep space:
And only one of these to launch from the moon:
1. #1 Tom
September 10, 2008
In addition, one must account for how much more mass one launched from the earth. It takes more energy to launch a kilogram, so you need more fuel, and so it all compounds. Plus, the launch from earth included the moon lander, and a lot of stuff was left on the moon (though they did bring some rocks back)
2. #2 Chris
September 10, 2008
Thanks, I’d been wondering about this lately but hadn’t gotten around to looking up any relevant equations. I’d assumed it was linked to escape velocities – am I wrong or is escape velocity somehow linked to your equation above?
Great blog by the way – I’m only a lowly biochemist, and it’s great to see some easy physics made relevant to the real world!
3. #3 Uncle Al
September 10, 2008
http://en.wikipedia.org/wiki/MythBusters_(season_7)#Episode_104_-_NASA_Moon_Landing
If only we could turn down inertial mass. Gravitation as spatial geomtry is a tougher assignment. Stochastic electrodynamics has (questionable) intervention with inertia,
http://www.calphysics.org/research.html
No other proposed large magnitude derailments are credible.
4. #4 nanoAl
September 12, 2008
Chris:
Its sort of related to the escape velocity, to find it , you just set U equal to the kinetic energy mv*v/2. Thats the amount of kinetic energy(and thus velocity) it needs to get from the surface to infinity and stop there(wherever that is). I think anyway.
I find it so amazing that some of the moon cranks think this is a convincing argument. “the moon is way smaller/lighter than the earth” would seem obvious but it isn’t apparently… | 733 | 3,135 | {"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} | 3.390625 | 3 | CC-MAIN-2017-04 | longest | en | 0.949831 |
https://www.ankitcodinghub.com/product/inf553-homework-4-hierarchical-clustering-with-cure-solved/ | 1,701,234,816,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100056.38/warc/CC-MAIN-20231129041834-20231129071834-00638.warc.gz | 713,345,296 | 57,432 | # INF553 – Homework #4 Hierarchical Clustering with CURE Solved
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## Description
In this homework, you are asked to implement the CURE algorithm in Python, named “cure.py”, for finding k clusters via hierarchical clustering. Your algorithm should be executed as follows:
cure.py <k> <sample> <data-set> <n> <alpha>
where <k> is the number of clusters (obtained from the dendrogram), the <sample> file contains a set of samples for running the hierarchical clustering, <data-set> contains the entire dataset, n is the number of representatives in each cluster and alpha is the fraction of distance the representatives are moved towards the centroid, as described in class
Note: In CURE, you should find n representatives which are the points that are as far as possible for each cluster. If the number of points in one cluster is less than n, then all the points in this cluster should be considered as representatives. Your algorithm should assign each point in the <data-set> to the cluster with the closest representatives to the point.
Evaluation: For the evaluation purpose, the data set input to your algorithm also contains cluster label for each data point. These cluster labels form a gold standard for clustering the data set to produce a specific number of clusters. Note that you can not use the label in the clustering process. Instead, use them in evaluating the performance of your algorithm. The performance is measured by precision and recall of correct pairs. A pair of two data points x and y is correct if they belong to the same cluster according to the cluster label. Recall that precision and recall are discussed in LSH lectures. Precision is the percentage of pairs discovered by the algorithm that are correct, while recall is the percentage of correct pairs that are discovered by the algorithm.
As an example, consider five data points: 1, 2, 3, 4, 5. Suppose there are 2 clusters: {1, 2, 3} and {4, 5} according to the algorithm, while there are different 2 clusters: {1, 2} and {3, 4, 5} according to the gold standard. In this case, the algorithm discovers four pairs: (1, 2), (1, 3), (2, 3), and (4, 5), while the gold standard has the pairs: (1, 2), (3, 4), (3, 5), and (4, 5). As such, the precision is 2/4 since only (1, 2) and (4, 5) discovered by the algorithm are correct, among the 4 discovered. Recall is also 2/4, since only 2 correct pairs were discovered among the total 4 in the gold standard.
Input data format:
The data set contains a list of data points, one point per line. For each data point, it gives the value of the point at each dimension, followed by the cluster label of the point. You can assume that the sample will contain a subset of the data. The data contains 150 iris plants
(https://archive.ics.uci.edu/ml/datasets/Iris). For each plant, it lists its sepal and petal length and width in centimeter, and also its type (e.g., setosa, see below).
5.1,3.5,1.4,0.2,Iris-setosa
4.9,3.0,1.4,0.2,Iris-setosa
4.7,3.2,1.3,0.2,Iris-setosa
Output format:
Print the results. Your algorithm should output k clusters, with each cluster contains a set of data points.
Data point are numbered by their positions they appear in the data file: the first data point is numbered 0, second 1, and so on. The data points in the clusters are output in the ascending order of their numbers.
For example, here is an example print output.
Cluster 1: [3, 10, 13, …]
Cluster 2: [8, 52, 87, 88, …]
Cluster 3: [100, 105, …]
Your algorithm also prints the accuracy of the discovery. For example,
Precision = 0.8, recall = 0.5
• Cure-Clustering.zip | 956 | 3,768 | {"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} | 3.34375 | 3 | CC-MAIN-2023-50 | latest | en | 0.91852 |
https://mathoverflow.net/questions/271703/reference-for-map-operatornamehomdc-mathbbp1-to-operatornamesymd | 1,709,352,907,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475727.3/warc/CC-MAIN-20240302020802-20240302050802-00549.warc.gz | 373,465,771 | 26,742 | # Reference for map $\operatorname{Hom}^d(C,\mathbb{P}^1) \to \operatorname{Sym}^d(C)$
For a curve $C$ over a finite field, I am looking at the map $\phi: \operatorname{Hom}^d(C,\mathbb{P}^1) \to \operatorname{Sym}^d(C)$ where $\operatorname{Hom}^d(C,\mathbb{P}^1)$ are the functions of degree $d$ from $C \to \mathbb{P}^1$ in the $\operatorname{Hom}$-scheme $\operatorname{Hom}(C, \mathbb{P}^1)$, defined by $$\phi: f \mapsto [\sigma]\cdot \Gamma_f$$
i.e. we intersect the graph of $f$ in $C \times \mathbb{P}^1$ with the line $[\sigma]$ in $C \times \mathbb{P}^1$ (which is the inverse image of $\sigma$ under the projection to $\mathbb{P}^1$.
For example the $f$ in the picture would map to $\phi(f) = 2\cdot P_1 + 3\cdot P_2 + 2\cdot P_3 + 2\cdot P_4$
I have looked at Kollár's Rational Curves on Algebraic Varieties and Chapter 9 of Nakajima's Lectures on Hilbert Schemes of Points on Surfaces however I can't find the information I need of this map.
For example, is $\phi$ flat? What is the dimension of the fiber of $\phi$? Etc. Etc.
Does anyone know of a reference for this topic?
EDIT: If we take $d > 2g$ we can view any point in $\operatorname{Sym}^d(C)$ as an effective divisor $D$ of degree $d$ of $C$. Then $D$ is base-point free and so we can construct a map $f: C \to \mathbb P^1$ with fiber $D$ above $0$ (or say, $\sigma$). So $\phi$ is surjective for $d > 2g$. In general, I only care for $d$ much greater than 0 as I want to study it's behavior as $d$ grows.
• For an example of non-flatness, look at $C$ hyperelliptic of genus $g \geq 2$ with $d = 2$. Then the degree $2$ map $C \to \mathbb P^1$ is unique up to $\operatorname{PGL}_2$. Thus, the image of your map $\phi$ is the locus given by divisors $P+Q$ linearly equivalent to the unique $g^1_2$ (equivalently, $P+Q$ is a fibre of this unique degree $2$ map). Most pairs $(P,Q)$ do not satisfy this, so the image of $\phi$ is a strict closed subvariety. Hence, $\phi$ cannot be flat. Jun 14, 2017 at 13:49
• @R.vanDobbendeBruyn As I am mostly interested in large $d$, we may assume that $d > 2g$ so that we can view any point $D \in \operatorname{Sym}^d(C)$ as an effective divisor and so $\phi$ is surjective. I'll edit this in. Jun 22, 2017 at 22:23
Remark. Note that $\operatorname{\mathbf{Hom}}(C,\mathbb P^1)$ is an open subscheme of $\operatorname{\mathbf{Div}}_{C \times \mathbb P^1}$, whereas $\operatorname{Sym}^d(C)$ is a component of $\operatorname{\mathbf{Div}}_C$. Specifically, a homomorphism $\phi \colon C \to \mathbb P^1$ corresponds to its graph, which is a section of a line bundle $\mathscr M$ on $C \times \mathbb P^1$. Since \begin{align*} \operatorname{\mathbf{Pic}}_{C \times \mathbb P^1} &= \operatorname{\mathbf{Pic}}_C \times \operatorname{\mathbf{Pic}}_{\mathbb P^1} \times \operatorname{\mathbf{Hom}}(\operatorname{\mathbf{Alb}}_C,\operatorname{\mathbf{Pic}}^0_{\mathbb P^1})\\ &= \operatorname{\mathbf{Pic}}_C \times \operatorname{\mathbf{Pic}}_{\mathbb P^1}, \end{align*} the line bundle $\mathscr M$ is of the form $\mathscr L \boxtimes \mathcal O(n)$ for some line bundle $\mathscr L$ on $C$ and some $n \in \mathbb Z$. Counting intersections with $\{c\} \times \mathbb P^1$, we see that $n = 1$. Restricting to $C \times \{\sigma\}$, we see that $\mathscr L = \phi^* \mathcal O(1)$. Thus, $\phi$ corresponds to a section $s$ of $\mathscr L \boxtimes \mathcal O(1)$.
Conversely, a section $s$ of $\mathscr L \boxtimes \mathcal O(1)$ comes from a morphism $\phi$ if and only if $\mathscr L$ is effective and $V(s)$ is irreducible. Indeed, since $\mathscr L \boxtimes \mathcal O(1)$ is primitive (not divisible), $V(s)$ is reduced. Hence, the dominant morphism $V(s) \to C$ is flat, since $V(s)$ is torsion-free over $C$ since it is integral. Since it is of degree $1$, it is an isomorphism.
The only other possibility for a section $s$ of $\mathscr L \boxtimes \mathcal O(1)$ is that it contains a vertical component. Indeed, the only decomposition of $\mathscr L \boxtimes \mathcal O(1)$ as a sum of effective divisors must have $\mathscr L' \boxtimes \mathcal O$ as one of its summands, all of whose sections are vertical.
The morphism $\phi \colon C \to \mathbb P^1$ and the pair $(\mathscr L, s)$ determine each other uniquely, where two pairs $(\mathscr L, s)$, $(\mathscr L', s')$ correspond if and only if there exists an isomorphism $\alpha \colon \mathscr L \stackrel \sim \to \mathscr L'$ taking $s$ to $s'$.
Remark. Thus, the map you give can be extended to the rational map \begin{align*} \sigma^* \colon \operatorname{\mathbf{Div}}_{C \times \mathbb P^1} &\dashrightarrow \operatorname{\mathbf{Div}}_C \\ Z &\mapsto Z \cap (C \times \{\sigma\}), \end{align*} which is defined away from the divisors containing $C \times \{\sigma\}$. We will focus on the locus $U$ of $\operatorname{\mathbf{Div}}_{C \times \mathbb P^1}$ of irreducible divisors in $\mathscr L \boxtimes \mathcal O(1)$ for some $\mathscr L$ of degree $d > 2g$. We get a commutative square $$\begin{array}{ccc} \operatorname{\mathbf{Div}}_{C \times \mathbb P^1} & \stackrel{\sigma^*}\dashrightarrow & \operatorname{\mathbf{Div}}_C \\ \downarrow & & \downarrow \\ \operatorname{\mathbf{Pic}}_{C \times \mathbb P^1} & \stackrel{\sigma^*}\rightarrow &\ \operatorname{\mathbf{Pic}}_C. \end{array}$$ Over the locus of $\operatorname{\mathbf{Pic}}_C$ where $\mathscr L$ is of degree $d > 2g$, the right vertical arrow is a $\mathbb P^r$-bundle, where $r = h^0(\mathscr L) - 1 = d - g$ (which does not depend on $\mathscr L$ but only on $d$). Similarly, over the same locus, the left vertical arrow is a $\mathbb P^{r'}$-bundle, where $r' = h^0(\mathscr L \boxtimes \mathcal O(1)) - 1$. By Künneth we have $r' = 2(r+1) - 1 = 2d - 2g + 1$.
Restricting the left hand side to the locus of the form $\mathscr L \boxtimes \mathcal O(1)$, the bottom map becomes an isomorphism. On the locus $U$ of irreducible divisors in $\mathscr L \boxtimes \mathcal O(1)$, the top map on each fibre is a morphism from an open in $\mathbb P^{2d-2g+1}$ to $\mathbb P^{d-g}$.
Remark. To describe the map more explicitly, let $[x:y]$ be coordinates on $\mathbb P^1$. For simplicity, assume $\sigma = [0:1]$. Then $H^0(\mathscr L \boxtimes \mathcal O(1))$ is given by $\{\lambda x + \mu y\ |\ \lambda,\mu \in H^0(\mathscr L)\}$, and a section $\lambda x + \mu y$ is mapped to the section $\mu \in H^0(\mathscr L)$. In particular, this is a linear map, so the fibres are linear spaces.
The locus of divisors $Z$ not containing $C \times \{0\}$ corresponds to the $\lambda x + \mu y$ with $\mu \neq 0$. This is the maximal set where the map on the projective spaces is defined, and there the fibres are $\mathbb A^{d-g+1}$ corresponding to the affine space of values of $\lambda$.
We are further restricting to divisors $Z$ not containing any vertical component; this is probably equivalent to linear independence of $\lambda$ and $\mu$, so this removes an $\mathbb A^1\setminus\{0\}$ from each fibre (corresponding to $\lambda = c \mu$). More canonically, the conditions that $\lambda$ and $\mu$ are linearly independent are cut out in $\mathbb P^{2d-2g+1}$ by the nonvanishing of certain minors.
Any further properties you want to deduce should follow from this description.
• Forgive me my lack of knowledge on these topics, but it is not immediatly clear to me what $V(s)$ should be. Jun 28, 2017 at 15:05
• I just meant the vanishing locus of $s$, i.e. the divisor cut out by the section $s$ of the line bundle $\mathscr L \boxtimes \mathcal O(1)$, cf. the correspondence between divisors and line bundles. (To make sense of this, you can choose local trivialisations of the line bundle, so that it locally becomes a vanishing locus of an actual function.) Jun 29, 2017 at 2:09 | 2,527 | 7,752 | {"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} | 2.84375 | 3 | CC-MAIN-2024-10 | latest | en | 0.817043 |
http://www.math.ucsd.edu/~mlicht/170AB/mainB.html | 1,553,135,429,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202476.48/warc/CC-MAIN-20190321010720-20190321032720-00442.warc.gz | 313,196,264 | 5,757 | # Introduction to Numerical Analysis: Approximation and Nonlinear Equations Math 170B --- Spring 2019
The final grade will be composed by the best of the following two options: (a) 20% homework, 20% midterm, and 60% final exam. (b) 20% homework, 80% final exam.
Your course grade will be determined by your cumulative average at the end of the quarter, based on the following scale:
A+ A A- B+ B B- C+ C C- 100 - 96.66 96.65 - 93.33 93.32 - 90.00 89.99 - 86.66 86.65 - 83.33 83.32 - 80.00 79.99 - 76.66 76.65 - 73.33 73.32 - 70
The above scale is guaranteed: for example, if your cumulative average is at least 73, then your final grade will be at least B. However, your instructor may adjust the above scale to be more generous.
## Course Calendar
Lecture Content
# 1, 1M
07.01.2019.
Basic concepts and Taylor's theorem.
# 2, 1W
09.01.2019.
Basic concepts and Taylor's theorem.
# 3, 1F
11.01.2019.
Basic concepts and Taylor's theorem.
# 4, 2M
14.01.2019.
Orders of Convergence and Additional Basic Concepts.
# 5, 2W
16.01.2019.
Orders of Convergence and Additional Basic Concepts. Implicit Function theorem.
# 6, 2F
18.01.2019.
Implicit Function theorem.
Homework 1 & 2 announced.
# 7, 3M
21.01.2019.
Martin Luther King, Jr. Holiday
# 8, 3W
23.01.2019.
Bisection Method.
# 9, 3F
25.01.2019.
Bisection Method.
#10, 4M
28.01.2019.
Newton's Method
#11, 4W
30.01.2019.
Newton's Method
#12, 4F
01.02.2019.
Secant Method
Homework 1 & 2 collected. Homework 3 & 4 announced.
#13, 5M
04.02.2019.
Polynomial Interpolation
#14, 5W
06.02.2019.
Polynomial Interpolation
#15, 5F
08.02.2019.
Divided Differences
#16, 6M
11.02.2019.
Midterm.
#17, 6W
13.02.2019.
Divided Differences
#18, 6F
15.02.2019.
Hermite Interpolation
Homework 3 & 4 collected. Homework 5 & 6 announced.
#19, 7M
18.02.2019.
Presidents' Day Holiday
#20, 7W
20.02.2019.
Spline Interpolation
#21, 7F
22.02.2019.
Spline Interpolation
#22, 8M
25.02.2019.
Chebychev polynomials Numerical Differentiation and Integration
#23, 8W
27.02.2019.
Numerical Differentiation and Integration
#24, 8F
01.03.2019.
Numerical Differentiation and Integration
Homework 5 & 6 collected. Homework 7 & 8 announced.
#25, 9M
04.03.2019.
Numerical Integration based on Interpolation
#26, 9W
06.03.2019.
Numerical Integration based on Interpolation
#27, 9F
089.03.2019.
Numerical Integration based on Interpolation
Homework 7 & 8 collected. Practice Material for Final available..
#28,10M
11.03.2019.
Numerical Integration based on Interpolation
#29,10W
13.03.2019.
Numerical Integration based on Interpolation
#30,10F
15.03.2019.
Numerical Integration based on Interpolation
FI
16.03.2019.
Final Exam. 8am - 11am. Place to be announced. | 962 | 2,672 | {"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} | 3.015625 | 3 | CC-MAIN-2019-13 | latest | en | 0.673801 |
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In 1860, the Philological Society launched its effort to create a dictionary more comprehensive than the world had ever seen; although the project would take more than 60 years to complete, the Oxford English Dictionary had been born.
a) would take more than 60 years to complete, the Oxford English Dictionary had been
b) took more than 60 years to complete, the Oxford English Dictionary was
c) would take more than 60 years to complete, the Oxford English Dictionary was being
d) would take more than 60 years to complete, the Oxford English Dictionary was
e) took more than 60 years to complete, the Oxford English Dictionary was about to be
Particularly interested in the difference between "would take" and "take"
If you have any questions
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08 Nov 2009, 04:15
yangsta8 wrote:
In 1860, the Philological Society launched its effort to create a dictionary more comprehensive than the world had ever seen; although the project would take more than 60 years to complete, the Oxford English Dictionary had been born.
a) would take more than 60 years to complete, the Oxford English Dictionary had been
b) took more than 60 years to complete, the Oxford English Dictionary was
c) would take more than 60 years to complete, the Oxford English Dictionary was being
d) would take more than 60 years to complete, the Oxford English Dictionary was
e) took more than 60 years to complete, the Oxford English Dictionary was about to be
Particularly interested in the difference between "would take" and "take"
"would take" is past participle of "will take" whereas the event has already occurred (In 1860) so simple past tense "took" will suffice. Between B and E, B is right answer.
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In 1860, the Philological Society launched its effort to create a dictionary more comprehensive than the world had ever seen; although the project would take more than 60 years to complete, the Oxford English Dictionary had been born.
a) would take more than 60 years to complete, the Oxford English Dictionary had been
b) took more than 60 years to complete, the Oxford English Dictionary was
c) would take more than 60 years to complete, the Oxford English Dictionary was being
d) would take more than 60 years to complete, the Oxford English Dictionary was
e) took more than 60 years to complete, the Oxford English Dictionary was about to be
Here are my reasons -
Here I am logically assuming that the birth of the dictionary signifies the launch of the project and this took place before the the entire project was completed. Since these events are taking place at different times, we need to have perfect past tense rather than simple past. Therefore ELIMINATE B
Eliminate C ('Being)
Eliminate E (about to be completed is incorrect as event has finished in the past)
Eliminate A (wrong usage of tenses)
Now look at D
D correctly uses "would take" to signify that the completion event took place after the birth. Although "had been" here would have sounded better (for perfect tenses), "would take" is a classical case of 'future in past tense' (according to the frame of time reference identified by the non-underlined portion) hence we have to use 'would' rather than 'will'. Normally we cannot use 'would' to express facts but here the events have completed in the past.
Hope this helps. Consider Kudos for the post
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09 Nov 2009, 01:41
OA is D.
zaarathelab you are spot on nice explanation. +1.
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29 Apr 2010, 10:40
my pick was (B).
good explanation by zaarathelab.
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08 Jul 2010, 21:15
Hmmm....i didn't get the explanation. I picked B. Could you pls explain once more?
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08 Jul 2010, 21:41
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I picked D.
We need "will take" to signal a "future" tense from this 1860 date, but since the sentence was already drafted using simple past (launched), we need the past tense of "will," (i.e., simply "would").
Remember, "would" can be in the form of:
1) past tense of will
2) conditional form (i.e., the hypothetical form... an event NOT expected to happen)
Lastly, the final verb, "was" is preferred over "had been" since we don't need to distinquish the "born" event was prior to the "launching." In fact, the "born" event occurred AFTER anyways...
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09 Jul 2010, 07:02
my pick was (B).
good explanation by zaarathelab.
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09 Jul 2010, 08:40
Chose B. Thanks for the explanation! Ruled the choice out because of the 'would'.
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29 Jul 2010, 23:48
Went for B ........mistake realized only after I rad the explanations.....I thought that Author was generally describing it...
Re: Oxford Dictionary [#permalink] 29 Jul 2010, 23:48
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Display posts from previous: Sort by | 1,999 | 7,393 | {"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} | 2.78125 | 3 | CC-MAIN-2017-04 | latest | en | 0.937091 |
https://www.physicsforums.com/threads/ce-amp-midband-gain-and-pspice.674696/ | 1,521,527,152,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647299.37/warc/CC-MAIN-20180320052712-20180320072712-00175.warc.gz | 864,651,453 | 18,453 | # CE amp midband gain and PSPICE
1. Feb 26, 2013
### Steve Collins
I’m in the process of writing a lab report after testing a cascaded common-emitter amp. The second transistor input is the collector output (collector to base) which I understand to be an emitter follower?
I have used PSPICE to measure the gain when a 20mV signal is applied and I have measured the resulting output peak-to-peak voltage to be a maximum of 2.2V which gives a gain of 40.8dB.
I have plotted the frequency response (20log(output/input)) using PSPICE and I’m getting a midband gain of around 43dB, which would mean an output of around 2.8V which is well over what I have read from the output signal plot.
The bottom of the output signal seems to be clipped and I was wondering if this could have anything to do with the results not matching up?
2. Feb 26, 2013
### rude man
Anytime you have a clipped signal you can throw out all the data.
It would help if you described your complete circuit including input voltage.
3. Feb 26, 2013
### Steve Collins
I've attached the circuit diagram, Cc2 can be omitted the main point of this assignment was to see the effects of changing the value of capacitor Cc.
Input signal is 20mV (19.71mV measured PSPICE voltage)
Output Signal clipped at bottom = 2.09V measured with PSPICE.
Gain also measured = 43.37 dB
Vcc = +15V
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Last edited: Feb 26, 2013
4. Feb 26, 2013
### Staff: Mentor
Wow, that's a giant Ce!
5. Feb 26, 2013
### rude man
This is a far from trivial circuit to analyze.
You did not mention any frequencies, especially the one where you get the clipping. Need that.
Next: the 1st stage is a Miller integrator. So right away you have dynamics (frequency-sensitivities) to contend with. This is an ac, not a dc, equation.
(Assuming an infinite-beta transistor and an ac-grounded emitter, one of your equations will be dVc/dt + V1/RB= 0 where V1 is the voltage at the junction of CIN/R1/R2/R8.)
Yes, the second stage is an emitter follower.
I suggest removing CC, CC2 and CL from the simulation at first. See how the circuit behaves at the various nodes. You can then add small amounts of CC at a time to see the effects of the Miller integrator effect on Q1n and small amounts of CL.
6. Feb 27, 2013
### Steve Collins
I’ve changed the value of Cc and at 470pF the bandwidth is 15kHz then doubling the Cc the bandwidth almost halves to 8kHz.
Clipping starts at around 80Hz
7. Feb 27, 2013
### rude man
Tell you what, this circuit is more or less a disaster. The ac gain at frequencies above about 2.7 KHz (the cutoff frequency of the input stage) is directly proportional to the beta of Q1 which is very poor design. Where did this design come from? I would not pursue this design any further without significant changes ...
8. Feb 27, 2013
### Steve Collins
I’m doing an assignment which is to analyse the CE amp that incorporates Q1. This involves DC analysis and bandwidth/3db frequencies when the value of Cc is changed.
Raising the value seems to shorten the bandwidth which I’m sure you already know. I still have to investigate the reason for this.
The adding of Q2 to the amp is optional and I was hoping that there was a simple answer why the gain doesn’t match the output/input.
I have gathered from what you have said that if the signal is clipped the output is not going to be as expected. Is this due to the lack of linearity?
9. Feb 27, 2013
### rude man
Yes, clipping implies nonlinearity. Not only is the intended signal reduced but clipping implies harmonic generation.
This circuit is too difficult to analyze. If you want a more or less stably biased and predictable CE circuit for Q1, let me know. I really can't work with this circuit.
(I can give you one with your input R-C network Cin + Rb, plus just 3 resistors and Q1. Your Rc would be the same 5.1K. Gain = 40 dB. You could also add CC if you wanted.
Do you have a -15V supply available?) | 1,011 | 3,987 | {"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} | 2.8125 | 3 | CC-MAIN-2018-13 | latest | en | 0.939748 |
https://socratic.org/questions/how-do-you-find-the-quotient-of-2x-3-7x-2-x-1-div-x-2-using-synthetic-division | 1,585,774,224,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506121.24/warc/CC-MAIN-20200401192839-20200401222839-00085.warc.gz | 699,835,984 | 6,196 | # How do you find the quotient of (2x^3+7x^2-x+1)div(x+2) using synthetic division?
May 25, 2017
The quotient is $= 2 {x}^{2} + 3 x - 7$
#### Explanation:
Let's perform the synthetic division
$\textcolor{w h i t e}{a a a a}$$- 2$$\textcolor{w h i t e}{a a a a a}$$|$$\textcolor{w h i t e}{a a a a}$$2$$\textcolor{w h i t e}{a a a a a a}$$7$$\textcolor{w h i t e}{a a a a a a}$$- 1$$\textcolor{w h i t e}{a a a a a}$$1$
$\textcolor{w h i t e}{a a a a a a a a a a a a}$_________
$\textcolor{w h i t e}{a a a a}$$\textcolor{w h i t e}{a a a a a a a}$$|$$\textcolor{w h i t e}{a a a a}$$\textcolor{w h i t e}{a a a a a a}$$- 4$$\textcolor{w h i t e}{a a a a a a}$$- 6$$\textcolor{w h i t e}{a a a a}$$14$
$\textcolor{w h i t e}{a a a a a a a a a a a a}$________
$\textcolor{w h i t e}{a a a a}$$\textcolor{w h i t e}{a a a a a a a}$$|$$\textcolor{w h i t e}{a a a a}$$2$$\textcolor{w h i t e}{a a a a a a a}$$3$$\textcolor{w h i t e}{a a a a a a}$$- 7$$\textcolor{w h i t e}{a a a a}$$\textcolor{red}{15}$
$\frac{2 {x}^{3} + 7 {x}^{2} - x + 1}{x + 2} = \left(2 {x}^{2} + 3 x - 7\right) + \frac{15}{x + 2}$
The remainder is $= 15$ and the quotient is $= 2 {x}^{2} + 3 x - 7$ | 570 | 1,178 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 39, "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} | 4.71875 | 5 | CC-MAIN-2020-16 | longest | en | 0.371437 |
http://gmatclub.com/forum/jones-prehistoric-wooden-tools-found-in-south-america-have-43417.html?fl=similar | 1,471,987,443,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982290442.1/warc/CC-MAIN-20160823195810-00170-ip-10-153-172-175.ec2.internal.warc.gz | 108,202,471 | 55,937 | Find all School-related info fast with the new School-Specific MBA Forum
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# Jones: Prehistoric wooden tools found in South America have
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Jones: Prehistoric wooden tools found in South America have [#permalink]
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Jones: Prehistoric wooden tools found in South America have been dated to 13,000 years ago. Although scientists attribute these tools to peoples whose ancestors first crossed into the Americas from Siberia to Alaska, this cannot be correct. In order to have reached a site so far south, these peoples must have been migrating southward well before 13,000 years ago. However, no such tools dating to before 13,000 years ago have been found anywhere between Alaska and South America.
Smith: Your evidence is inconclusive. Those tools were found in peat bogs, which are rare in the Americas. Wooden tools in soils other than peat bogs usually decompose within only a few years.
The point at issue between Jones and Smith is
(A) whether all prehistoric tools that are 13,000 years or older were made of wood
(B) whether the scientists’ attribution of tools could be correct in light of Jones’s evidence
(C) whether the dating of the wooden tools by the scientists could be correct
(D) how long ago the peoples who crossed into the American from Siberia to Alaska first did so
(E) whether Smith’s evidence entails that the wooden tools have been dated correctly
Please give reasons for ur choices...
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18 Mar 2007, 14:00
Here's my opinion on this:
A) whether all prehistoric tools that are 13,000 years or older were made of wood - nop. All tools is too strong.
(B) whether the scientists’ attribution of tools could be correct in light of Jones’s evidence - it's only one side of the debate. Nop.
(C) whether the dating of the wooden tools by the scientists could be correct - My pick of the 5. Jones is saying the scientists are wrong. Smith argues that the scientists could be right.
(D) how long ago the peoples who crossed into the American from Siberia to Alaska first did so - not relevant.
(E) whether Smith’s evidence entails that the wooden tools have been dated correctly - same as B, only one side of the story.
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18 Mar 2007, 17:43
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18 Mar 2007, 19:11
I agree with D. What is the OA?
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18 Mar 2007, 20:28
So left with B and C
Smith is actually talking evidence provided by Jones
B it is
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18 Mar 2007, 21:34
im with B as well , the debate is whether or not the tools are an accurate measure of migration
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18 Mar 2007, 22:52
my pick is D. Whats the OA ?
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18 Mar 2007, 22:55
I will go with C ... Jones arguing that dating of tools dine by scientists could be wrong...Smith says that Jones evidence is inclusive...
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18 Mar 2007, 23:00
B !
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19 Mar 2007, 08:37
The OA is B indeed.
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Jones: Prehistoric wooden tools found in South America have 1 16 Mar 2007, 10:38
Questions 17-18 Jones: Prehistoric wooden tools found in 11 29 Nov 2006, 09:27
Display posts from previous: Sort by | 1,449 | 5,153 | {"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} | 2.765625 | 3 | CC-MAIN-2016-36 | longest | en | 0.932028 |
https://www.unitsconverters.com/en/Btu(Th)/(Lbmdegf)-To-Btu(It)/Lbmdegf/Utu-4332-4329 | 1,713,302,499,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817106.73/warc/CC-MAIN-20240416191221-20240416221221-00413.warc.gz | 946,045,350 | 37,360 | Formula Used
1 Joule per Kilogram per K = 0.000239005796104515 Btu (th) per pound per Degree Fahrenheit
1 Joule per Kilogram per K = 0.000238845896632776 Btu (IT) per pound per Degree Fahrenheit
1 Btu (th) per pound per Degree Fahrenheit = 0.9993309807781 Btu (IT) per pound per Degree Fahrenheit
## Btu(th)/(lb*°F) to Btu(IT)/lb*°F Conversion
The abbreviation for Btu(th)/(lb*°F) and Btu(IT)/lb*°F is btu (th) per pound per degree fahrenheit and btu (it) per pound per degree fahrenheit respectively. 1 Btu(th)/(lb*°F) is 1 times smaller than a Btu(IT)/lb*°F. To measure, units of measurement are needed and converting such units is an important task as well. unitsconverters.com is an online conversion tool to convert all types of measurement units including Btu(th)/(lb*°F) to Btu(IT)/lb*°F conversion.
## Btu (th) per pound per Degree Fahrenheit to Btu(IT)/lb*°F
Check our Btu (th) per pound per Degree Fahrenheit to Btu(IT)/lb*°F converter and click on formula to get the conversion factor. When you are converting specific heat capacity from Btu (th) per pound per Degree Fahrenheit to Btu(IT)/lb*°F, you need a converter that is elaborate and still easy to use. All you have to do is select the unit for which you want the conversion and enter the value and finally hit Convert.
## Btu(th)/(lb*°F) to Btu (IT) per pound per Degree Fahrenheit
The formula used to convert Btu(th)/(lb*°F) to Btu (IT) per pound per Degree Fahrenheit is 1 Btu (th) per pound per Degree Fahrenheit = 0.9993309807781 Btu (IT) per pound per Degree Fahrenheit. Measurement is one of the most fundamental concepts. Note that we have CHU per pound per Celcius as the biggest unit for length while Joule per Kilogram per K is the smallest one.
## Convert Btu(th)/(lb*°F) to Btu(IT)/lb*°F
How to convert Btu(th)/(lb*°F) to Btu(IT)/lb*°F? Now you can do Btu(th)/(lb*°F) to Btu(IT)/lb*°F conversion with the help of this tool. In the length measurement, first choose Btu(th)/(lb*°F) from the left dropdown and Btu(IT)/lb*°F from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from Btu(IT)/lb*°F to Btu(th)/(lb*°F)? You can check our Btu(IT)/lb*°F to Btu(th)/(lb*°F) converter.
## Btu(th)/(lb*°F) to Btu(IT)/lb*°F Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like specific heat capacity finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like Btu(th)/(lb*°F) to Btu(IT)/lb*°F through multiplicative conversion factors. When you are converting specific heat capacity, you need a Btu (th) per pound per Degree Fahrenheit to Btu (IT) per pound per Degree Fahrenheit converter that is elaborate and still easy to use. Converting Btu(th)/(lb*°F) to Btu (IT) per pound per Degree Fahrenheit is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert Btu (th) per pound per Degree Fahrenheit to Btu(IT)/lb*°F, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in Btu(th)/(lb*°F) to Btu(IT)/lb*°F conversion along with a table representing the entire conversion.
Let Others Know | 942 | 3,486 | {"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} | 2.765625 | 3 | CC-MAIN-2024-18 | latest | en | 0.714879 |
https://www.kodytools.com/units/data/from/bytepd/to/byteps | 1,716,260,864,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058383.61/warc/CC-MAIN-20240521025434-20240521055434-00373.warc.gz | 746,264,027 | 16,321 | # Bytes/Day to Bytes/Second Converter
1 Bytes/Day = 0.000011574074074074 Bytes/Second
## One Bytes/Day is Equal to How Many Bytes/Second?
The answer is one Bytes/Day is equal to 0.000011574074074074 Bytes/Second and that means we can also write it as 1 Bytes/Day = 0.000011574074074074 Bytes/Second. Feel free to use our online unit conversion calculator to convert the unit from Bytes/Day to Bytes/Second. Just simply enter value 1 in Bytes/Day and see the result in Bytes/Second.
Manually converting Bytes/Day to Bytes/Second can be time-consuming,especially when you don’t have enough knowledge about Data Transfer units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Bytes/Day to Bytes/Second converter tool to get the job done as soon as possible.
We have so many online tools available to convert Bytes/Day to Bytes/Second, but not every online tool gives an accurate result and that is why we have created this online Bytes/Day to Bytes/Second converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert Bytes/Day to Bytes/Second (B/day to B/s)
By using our Bytes/Day to Bytes/Second conversion tool, you know that one Bytes/Day is equivalent to 0.000011574074074074 Bytes/Second. Hence, to convert Bytes/Day to Bytes/Second, we just need to multiply the number by 0.000011574074074074. We are going to use very simple Bytes/Day to Bytes/Second conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Bytes/Day} = 1 \times 0.000011574074074074 = \text{0.000011574074074074 Bytes/Second}$$
## What Unit of Measure is Bytes/Day?
Bytes/Day or Bytes per Day is a unit of measurement for data transfer rate. By definition, it refers to the rate with which a data can be transferred in the form of bytes within the duration of 1 day.
## What is the Symbol of Bytes/Day?
The symbol of Bytes/Day is B/day. This means you can also write one Bytes/Day as 1 B/day.
## What Unit of Measure is Bytes/Second?
Bytes/Second or Bytes per Second is a unit of measurement for data transfer rate. By definition, it refers to the rate with which a data can be transferred in the form of bytes within the duration of 1 second.
## What is the Symbol of Bytes/Second?
The symbol of Bytes/Second is B/s. This means you can also write one Bytes/Second as 1 B/s.
## How to Use Bytes/Day to Bytes/Second Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Bytes/Day and in the first input field, enter a value.
• From the second dropdown, select Bytes/Second.
• Instantly, the tool will convert the value from Bytes/Day to Bytes/Second and display the result in the second input field.
## Example of Bytes/Day to Bytes/Second Converter Tool
Bytes/Day
1
Bytes/Second
0.000011574074074074
# Bytes/Day to Bytes/Second Conversion Table
Bytes/Day [B/day]Bytes/Second [B/s]Description
1 Bytes/Day0.000011574074074074 Bytes/Second1 Bytes/Day = 0.000011574074074074 Bytes/Second
2 Bytes/Day0.000023148148148148 Bytes/Second2 Bytes/Day = 0.000023148148148148 Bytes/Second
3 Bytes/Day0.000034722222222222 Bytes/Second3 Bytes/Day = 0.000034722222222222 Bytes/Second
4 Bytes/Day0.000046296296296296 Bytes/Second4 Bytes/Day = 0.000046296296296296 Bytes/Second
5 Bytes/Day0.00005787037037037 Bytes/Second5 Bytes/Day = 0.00005787037037037 Bytes/Second
6 Bytes/Day0.000069444444444444 Bytes/Second6 Bytes/Day = 0.000069444444444444 Bytes/Second
7 Bytes/Day0.000081018518518519 Bytes/Second7 Bytes/Day = 0.000081018518518519 Bytes/Second
8 Bytes/Day0.000092592592592593 Bytes/Second8 Bytes/Day = 0.000092592592592593 Bytes/Second
9 Bytes/Day0.00010416666666667 Bytes/Second9 Bytes/Day = 0.00010416666666667 Bytes/Second
10 Bytes/Day0.00011574074074074 Bytes/Second10 Bytes/Day = 0.00011574074074074 Bytes/Second
100 Bytes/Day0.0011574074074074 Bytes/Second100 Bytes/Day = 0.0011574074074074 Bytes/Second
1000 Bytes/Day0.011574074074074 Bytes/Second1000 Bytes/Day = 0.011574074074074 Bytes/Second
# Bytes/Day to Other Units Conversion Table
ConversionDescription
1 Bytes/Day = 0.000092592592592593 Bits/Second1 Bytes/Day in Bits/Second is equal to 0.000092592592592593
1 Bytes/Day = 0.0055555555555556 Bits/Minute1 Bytes/Day in Bits/Minute is equal to 0.0055555555555556
1 Bytes/Day = 0.33333333333333 Bits/Hour1 Bytes/Day in Bits/Hour is equal to 0.33333333333333
1 Bytes/Day = 8 Bits/Day1 Bytes/Day in Bits/Day is equal to 8
1 Bytes/Day = 9.2592592592593e-11 Bits/Microsecond1 Bytes/Day in Bits/Microsecond is equal to 9.2592592592593e-11
1 Bytes/Day = 9.2592592592593e-8 Bits/Millisecond1 Bytes/Day in Bits/Millisecond is equal to 9.2592592592593e-8
1 Bytes/Day = 9.2592592592593e-14 Bits/Nanosecond1 Bytes/Day in Bits/Nanosecond is equal to 9.2592592592593e-14
1 Bytes/Day = 0.000011574074074074 Bytes/Second1 Bytes/Day in Bytes/Second is equal to 0.000011574074074074
1 Bytes/Day = 0.00069444444444444 Bytes/Minute1 Bytes/Day in Bytes/Minute is equal to 0.00069444444444444
1 Bytes/Day = 0.041666666666667 Bytes/Hour1 Bytes/Day in Bytes/Hour is equal to 0.041666666666667
1 Bytes/Day = 1.1574074074074e-11 Bytes/Microsecond1 Bytes/Day in Bytes/Microsecond is equal to 1.1574074074074e-11
1 Bytes/Day = 1.1574074074074e-8 Bytes/Millisecond1 Bytes/Day in Bytes/Millisecond is equal to 1.1574074074074e-8
1 Bytes/Day = 1.1574074074074e-14 Bytes/Nanosecond1 Bytes/Day in Bytes/Nanosecond is equal to 1.1574074074074e-14
1 Bytes/Day = 9.2592592592593e-8 Kilobits/Second1 Bytes/Day in Kilobits/Second is equal to 9.2592592592593e-8
1 Bytes/Day = 0.0000055555555555556 Kilobits/Minute1 Bytes/Day in Kilobits/Minute is equal to 0.0000055555555555556
1 Bytes/Day = 0.00033333333333333 Kilobits/Hour1 Bytes/Day in Kilobits/Hour is equal to 0.00033333333333333
1 Bytes/Day = 0.008 Kilobits/Day1 Bytes/Day in Kilobits/Day is equal to 0.008
1 Bytes/Day = 9.2592592592593e-14 Kilobits/Microsecond1 Bytes/Day in Kilobits/Microsecond is equal to 9.2592592592593e-14
1 Bytes/Day = 9.2592592592593e-11 Kilobits/Millisecond1 Bytes/Day in Kilobits/Millisecond is equal to 9.2592592592593e-11
1 Bytes/Day = 9.2592592592593e-17 Kilobits/Nanosecond1 Bytes/Day in Kilobits/Nanosecond is equal to 9.2592592592593e-17
1 Bytes/Day = 1.1574074074074e-8 Kilobytes/Second1 Bytes/Day in Kilobytes/Second is equal to 1.1574074074074e-8
1 Bytes/Day = 6.9444444444444e-7 Kilobytes/Minute1 Bytes/Day in Kilobytes/Minute is equal to 6.9444444444444e-7
1 Bytes/Day = 0.000041666666666667 Kilobytes/Hour1 Bytes/Day in Kilobytes/Hour is equal to 0.000041666666666667
1 Bytes/Day = 0.001 Kilobytes/Day1 Bytes/Day in Kilobytes/Day is equal to 0.001
1 Bytes/Day = 1.1574074074074e-14 Kilobytes/Microsecond1 Bytes/Day in Kilobytes/Microsecond is equal to 1.1574074074074e-14
1 Bytes/Day = 1.1574074074074e-11 Kilobytes/Millisecond1 Bytes/Day in Kilobytes/Millisecond is equal to 1.1574074074074e-11
1 Bytes/Day = 1.1574074074074e-17 Kilobytes/Nanosecond1 Bytes/Day in Kilobytes/Nanosecond is equal to 1.1574074074074e-17
1 Bytes/Day = 9.2592592592593e-11 Megabits/Second1 Bytes/Day in Megabits/Second is equal to 9.2592592592593e-11
1 Bytes/Day = 5.5555555555556e-9 Megabits/Minute1 Bytes/Day in Megabits/Minute is equal to 5.5555555555556e-9
1 Bytes/Day = 3.3333333333333e-7 Megabits/Hour1 Bytes/Day in Megabits/Hour is equal to 3.3333333333333e-7
1 Bytes/Day = 0.000008 Megabits/Day1 Bytes/Day in Megabits/Day is equal to 0.000008
1 Bytes/Day = 9.2592592592593e-17 Megabits/Microsecond1 Bytes/Day in Megabits/Microsecond is equal to 9.2592592592593e-17
1 Bytes/Day = 9.2592592592593e-14 Megabits/Millisecond1 Bytes/Day in Megabits/Millisecond is equal to 9.2592592592593e-14
1 Bytes/Day = 9.2592592592593e-20 Megabits/Nanosecond1 Bytes/Day in Megabits/Nanosecond is equal to 9.2592592592593e-20
1 Bytes/Day = 1.1574074074074e-11 Megabytes/Second1 Bytes/Day in Megabytes/Second is equal to 1.1574074074074e-11
1 Bytes/Day = 6.9444444444444e-10 Megabytes/Minute1 Bytes/Day in Megabytes/Minute is equal to 6.9444444444444e-10
1 Bytes/Day = 4.1666666666667e-8 Megabytes/Hour1 Bytes/Day in Megabytes/Hour is equal to 4.1666666666667e-8
1 Bytes/Day = 0.000001 Megabytes/Day1 Bytes/Day in Megabytes/Day is equal to 0.000001
1 Bytes/Day = 1.1574074074074e-17 Megabytes/Microsecond1 Bytes/Day in Megabytes/Microsecond is equal to 1.1574074074074e-17
1 Bytes/Day = 1.1574074074074e-14 Megabytes/Millisecond1 Bytes/Day in Megabytes/Millisecond is equal to 1.1574074074074e-14
1 Bytes/Day = 1.1574074074074e-20 Megabytes/Nanosecond1 Bytes/Day in Megabytes/Nanosecond is equal to 1.1574074074074e-20
1 Bytes/Day = 9.2592592592593e-14 Gigabits/Second1 Bytes/Day in Gigabits/Second is equal to 9.2592592592593e-14
1 Bytes/Day = 5.5555555555556e-12 Gigabits/Minute1 Bytes/Day in Gigabits/Minute is equal to 5.5555555555556e-12
1 Bytes/Day = 3.3333333333333e-10 Gigabits/Hour1 Bytes/Day in Gigabits/Hour is equal to 3.3333333333333e-10
1 Bytes/Day = 8e-9 Gigabits/Day1 Bytes/Day in Gigabits/Day is equal to 8e-9
1 Bytes/Day = 9.2592592592593e-20 Gigabits/Microsecond1 Bytes/Day in Gigabits/Microsecond is equal to 9.2592592592593e-20
1 Bytes/Day = 9.2592592592593e-17 Gigabits/Millisecond1 Bytes/Day in Gigabits/Millisecond is equal to 9.2592592592593e-17
1 Bytes/Day = 9.2592592592593e-23 Gigabits/Nanosecond1 Bytes/Day in Gigabits/Nanosecond is equal to 9.2592592592593e-23
1 Bytes/Day = 1.1574074074074e-14 Gigabytes/Second1 Bytes/Day in Gigabytes/Second is equal to 1.1574074074074e-14
1 Bytes/Day = 6.9444444444444e-13 Gigabytes/Minute1 Bytes/Day in Gigabytes/Minute is equal to 6.9444444444444e-13
1 Bytes/Day = 4.1666666666667e-11 Gigabytes/Hour1 Bytes/Day in Gigabytes/Hour is equal to 4.1666666666667e-11
1 Bytes/Day = 1e-9 Gigabytes/Day1 Bytes/Day in Gigabytes/Day is equal to 1e-9
1 Bytes/Day = 1.1574074074074e-20 Gigabytes/Microsecond1 Bytes/Day in Gigabytes/Microsecond is equal to 1.1574074074074e-20
1 Bytes/Day = 1.1574074074074e-17 Gigabytes/Millisecond1 Bytes/Day in Gigabytes/Millisecond is equal to 1.1574074074074e-17
1 Bytes/Day = 1.1574074074074e-23 Gigabytes/Nanosecond1 Bytes/Day in Gigabytes/Nanosecond is equal to 1.1574074074074e-23
1 Bytes/Day = 9.2592592592593e-17 Terabits/Second1 Bytes/Day in Terabits/Second is equal to 9.2592592592593e-17
1 Bytes/Day = 5.5555555555556e-15 Terabits/Minute1 Bytes/Day in Terabits/Minute is equal to 5.5555555555556e-15
1 Bytes/Day = 3.3333333333333e-13 Terabits/Hour1 Bytes/Day in Terabits/Hour is equal to 3.3333333333333e-13
1 Bytes/Day = 8e-12 Terabits/Day1 Bytes/Day in Terabits/Day is equal to 8e-12
1 Bytes/Day = 9.2592592592593e-23 Terabits/Microsecond1 Bytes/Day in Terabits/Microsecond is equal to 9.2592592592593e-23
1 Bytes/Day = 9.2592592592593e-20 Terabits/Millisecond1 Bytes/Day in Terabits/Millisecond is equal to 9.2592592592593e-20
1 Bytes/Day = 9.2592592592593e-26 Terabits/Nanosecond1 Bytes/Day in Terabits/Nanosecond is equal to 9.2592592592593e-26
1 Bytes/Day = 1.1574074074074e-17 Terabytes/Second1 Bytes/Day in Terabytes/Second is equal to 1.1574074074074e-17
1 Bytes/Day = 6.9444444444444e-16 Terabytes/Minute1 Bytes/Day in Terabytes/Minute is equal to 6.9444444444444e-16
1 Bytes/Day = 4.1666666666667e-14 Terabytes/Hour1 Bytes/Day in Terabytes/Hour is equal to 4.1666666666667e-14
1 Bytes/Day = 1e-12 Terabytes/Day1 Bytes/Day in Terabytes/Day is equal to 1e-12
1 Bytes/Day = 1.1574074074074e-23 Terabytes/Microsecond1 Bytes/Day in Terabytes/Microsecond is equal to 1.1574074074074e-23
1 Bytes/Day = 1.1574074074074e-20 Terabytes/Millisecond1 Bytes/Day in Terabytes/Millisecond is equal to 1.1574074074074e-20
1 Bytes/Day = 1.1574074074074e-26 Terabytes/Nanosecond1 Bytes/Day in Terabytes/Nanosecond is equal to 1.1574074074074e-26
1 Bytes/Day = 9.2592592592593e-12 Ethernet1 Bytes/Day in Ethernet is equal to 9.2592592592593e-12
1 Bytes/Day = 9.2592592592593e-13 Fast Ethernet1 Bytes/Day in Fast Ethernet is equal to 9.2592592592593e-13
1 Bytes/Day = 9.2592592592593e-14 Gigabit Ethernet1 Bytes/Day in Gigabit Ethernet is equal to 9.2592592592593e-14 | 4,271 | 12,119 | {"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} | 2.9375 | 3 | CC-MAIN-2024-22 | latest | en | 0.932627 |
http://pheeds.com/info/guide/g/ge/geometry.html?indexes | 1,544,577,652,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823710.44/warc/CC-MAIN-20181212000955-20181212022455-00207.warc.gz | 231,923,900 | 4,774 | # Geometry
simple:Geometry
Geometry is the branch of mathematics dealing with spatial relationships. From experience, or possibly intuitively, people characterize space by certain fundamental qualities, which are termed axioms in geometry. Such axioms are insusceptible of proof, but can be used in conjunction with mathematical definitions for points, straight lines, curves, surfaces, and solids to draw logical conclusions.
Because of its immediate practical applications, geometry was one of the first branches of mathematics to be developed. Likewise, it was the first field to be put on an axiomatic basis, by Euclid. The Greeks were interested in many questions about ruler-and-compass constructions. The next most significant development had to wait until a millennium later, and that was analytic geometry, in which coordinate systems are introduced and points are represented as ordered pairs or triples of numbers. This sort of representation has since then allowed us to construct new geometries other than the standard Euclidean version.
The central notion in geometry is that of congruence. In Euclidean geometry, two figures are said to be congruent if they are related by a series of reflections, rotations, and translationss.
Other geometries can be constructed by choosing a new underlying space to work with (Euclidean geometry uses Euclidean space, Rn) or by choosing a new group of transformations to work with (Euclidean geometry uses the inhomogeneous orthogonal transformations, E(n)). The latter point of view is called the Erlanger program. In general, the more congruences we have, the fewer invariants there are. As an example, in affine geometry any linear transformation is allowed, and so the first three figures are all congruent; distances and angles are no longer invariants, but linearity is.
A discrete form of geometry is treated under Pick's theorem.
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https://quantlabs.net/blog/2012/06/quant-analytics-can-we-use-lower-order-terms-in-a-model-when-higher-orders-become-significant/ | 1,632,724,811,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058373.45/warc/CC-MAIN-20210927060117-20210927090117-00692.warc.gz | 500,892,148 | 41,717 | # Quant analytics: Can we use lower order terms in a model when higher orders become significant?
(Last Updated On: June 21, 2012)
Quant analytics: Can we use lower order terms in a model when higher orders become significant?
For some of the respondent level models I am working on, we are trying to test for different interaction effects between media. I have been working on the premise of testing for significant first order effects, and testing for higher order interaction effects. In some models, I have tested for interactions even if the first order effects were insignificant. My question is: if x1*x2 term (interaction) is significant, should we need to have both x1 and x2 terms in the model irrespective of their significance? In the case when both x1 and x2 are insiginificant but x1*x2 is, what is the best way to specify the model? My feeling is that leaving out the first order effects (if they are insignificant) will give us biased coefficients for the interaction terms. Any thoughts?
==
Heck yes. You are exactly right that if there is ANY first order effect, it will bias your estimates of interaction. (My mantra: Even if an effect is not stat. significant in a particular test and data set, it may still be large enough to be important.) So keep first-order effects.
The only exception I’m aware of is when prior theory tells you that a particular first order effect is zero. For example, surface area is equal to height x width x a constant. In that situation, height alone, and width alone, should be left out. But it’s rare to have such a clear model.
==
Great, Thank you! I think that bias is what we have seen when we left out the first order terms for being insignificant while keeping the interaction terms in. So, with interaction terms, it is a tradeoff between bias (due to insignificant variables in the model) in estimates and variance of estimates (due to correlated terms coming into the model), right.
==
As more of a data miner than a statistician, I would ask which model performs better on a held back sample of the data? That is the model with the least bias. Interpretability is another story – it’s harder to say what models with both first and second order terms are telling you. But I also like to augment with a decision tree for insights into interactions between variables
NOTE I now post my TRADING ALERTS into my personal FACEBOOK ACCOUNT and TWITTER. Don't worry as I don't post stupid cat videos or what I eat! | 528 | 2,474 | {"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} | 2.578125 | 3 | CC-MAIN-2021-39 | latest | en | 0.944089 |
http://deportesfutbol.info/the-never-before-told-story-about-mathematics-glossary-that-you-need-to-read/ | 1,604,025,148,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107906872.85/warc/CC-MAIN-20201030003928-20201030033928-00115.warc.gz | 26,848,144 | 9,395 | # The Never Before Told Story About Mathematics Glossary That You Need to Read
## The Lost Secret of Mathematics Glossary
It popular to find a set of declarations just in case there is the key font getting shifted. Bear in mind the modifications happen in the event with the record text dimensions that is present is matched by the importance from the very first argument. 1 example is should you’d like to generate a title environment which suppresses its own indentation along with this main one over the future paragraph.
Use divisibility rules to find out if a number is a factor of some other number. If you would like to suggest a glossary entry which you know ought to be listed here, please use our on-line form to let us know about doing it. Whether there are fewer entries, then the previous entry is repeated as frequently as essential.
This instant, however, the disk role calculating the discriminant, is going to be a role. An interval is open in the event the period doesn’t include its own end http://facstaff.elon.edu/sullivan/cheatpap.htm points. A period that’s closed contains the start and also the end result.
The procedure for solving for anti-derivatives is known as anti-differentiation. It is crucial to think about the effect size when you obtain statistically significant outcomes. In some cases, it guided practicemay be used to describe this general technique.
## What You Don’t Know About Mathematics Glossary
Otherwise, they need to know which represents more. Anti-derivatives aren’t unique. Get it touch to tell us.
The line segment between both points is referred to as a chord. A square is a normal quadrilateral. The examples below will provide you with a clearer idea of the way to adhere to these actions to sketch the graph of a function.
Once the down sides arise it is most suitable for students to begin seeking assistance. A few properties are more essential than others which is dependent on the context. Reinsurance businesses are utilized by many hedge funds that are elite as another way in Bermuda.
Additional guidelines have to be established, but the extra guidelines won’t necessarily apply across the rest of the universe of funds. When searching for a high-quality hedge fund, it is essential for an investor to recognize the metrics which are important to them and the results needed for each. Correct mathematical statements have been put to use throughout the response.
You can find plenty of factors which may be made better. Then you definitely identified the place to seek out help. Ok, enough with all the perception.
Rajaratnam managed to get considerable amounts of Goldman Sachs stock and produce a hefty profit on those shares in 1 day. Almost all the food was eaten. Sometimes identify’ is employed in the feeling of describe.
## The Lost Secret of Mathematics Glossary
One of the main features of the conventional input stream is that your program consumes values as soon as it reads them. The language also has a massive part of logic. The approach gives us more control over the visual appeal of the output.
The math standards offer specificity and precision instead of broad general statements. There’s consistently a opinion of anti-mathematics in most class which I educated which can be observed from all characteristics of the pupils. A common math paper comprises a number of the weather of an normal instructional article, for example an introduction, an abstract, and benchmark element.
Polynomial. Mathematics is a very specific subject. It offers definitions of a range of their phrases in math and numeracy.
## The 30-Second Trick for Mathematics Glossary
Instead of walk, for instance, saunter might be used. In particular, the youngster isn’t centered in the space that stays in the table after placing the label. Have a look at the picture below to find a good example.
Even the self-assessment actions and hints for additional reading needs to make sure that teachers feel great ready to teach the mathematics in the publication. You can find a significant bit that managers employ however below is a overall breakdown of plans. The criteria aren’t supposed for a program.
Learning with understanding is very important to allow pupils to fix the brand new kinds of issues they face on. This feedback helps it easy for teachers to view learning the opinion of your own students. See this video to get out more regarding Response for this intervention.
PHYSICIST. There is a plethora of tools which is often utilised to learn algebra 1. The curriculum helps to be certain that the connections between the several components of mathematics, and the text between mathematics and other disciplines, are manufactured clear.
Our earth’s magnetic field is considered to become. Looking at the integral M-files of MATLAB is really a wonderful means to learn the best way touse them. Room probes aren’t necessarily designed to go back into Earth.
Some authors incorporate the empty set inside this definition. The Internet edition of SPACE MATHEMATICS includes the very same information found in the previous 3 versions. You may assign the variety of questions you desire. | 1,013 | 5,169 | {"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} | 2.546875 | 3 | CC-MAIN-2020-45 | longest | en | 0.953192 |
https://classroomsecrets.co.uk/free-oclock-and-half-past-homework-extension-year-2-time/ | 1,725,977,077,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00245.warc.gz | 155,260,603 | 80,826 | Free O’clock and Half Past Homework Extension Year 2 Time – Classroom Secrets | Classroom Secrets
Maths Resources & WorksheetsYear 2 Maths Resources & WorksheetsSummer Block 3 (Time)01 O’clock and Half Past › Free O’clock and Half Past Homework Extension Year 2 Time
Free O’clock and Half Past Homework Extension Year 2 Time
Step 1: O'clock and Half Past Homework Extension Year 2 Extension Year 2 Summer Block 3
O'clock and Half Past Homework Extension Year 2 Extension provides additional questions which can be used as homework or an in-class extension for the Year 2 O'clock and Half Past Resource Pack. These are differentiated for Developing, Expected and Greater Depth.
More resources for Summer Block 3 Step 1.
Free Home Learning Resource
What's included in the pack?
This pack includes:
• O'clock and Half Past Homework Extension Year 2 Extension with answers for Year 2 Summer Block 3.
National Curriculum Objectives
Mathematics Year 2: (2M4b) Compare and sequence intervals of time
Differentiation:
Questions 1, 4 and 7 (Varied Fluency)
Developing Draw a line from each clock face to the correct times to find an odd one out. Draw the minute hand on the blank clock to show the time for the odd one out. Includes clocks showing on the hour or at half past the hour.
Expected Draw a line from each clock face to the correct times to find an odd one out. Draw the hands on a blank clock to show the time for the odd one out. Includes clocks showing on the hour or at half past the hour.
Greater Depth Draw a line from each clock face to the correct times to find an odd one out. Draw the hands on a blank clock to show the time for the odd one out. Includes clocks showing on the hour or at half past the hour with some numerical hours missing.
Questions 2, 5 and 8 (Varied Fluency)
Developing Decide whether a statement about three clocks is true or false. Includes clocks showing on the hour or at half past the hour.
Expected Decide whether a statement about three clocks is true or false. Includes clocks showing on the hour or at half past the hour.
Greater Depth Decide whether a statement about three clocks is true or false. Includes clocks showing on the hour or at half past the hour with some numerical hours missing.
Questions 3, 6 and 9 (Reasoning and Problem Solving)
Developing Work out the time by using the given clues. Includes blank clock for guidance.
Expected Work out the time by using the given clues. Includes blank clock for guidance.
Greater Depth Work out the time by using the given clues. Includes blank clock for guidance, with some numerical values missing. | 585 | 2,613 | {"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} | 3.84375 | 4 | CC-MAIN-2024-38 | latest | en | 0.863775 |
https://medium.com/@cosinekitty/understanding-fractional-dimensions-f2ed2e4e1600 | 1,571,708,475,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987795403.76/warc/CC-MAIN-20191022004128-20191022031628-00323.warc.gz | 610,212,801 | 34,955 | ## Measuring those infinitely crinkly things called fractals
Oct 7 · 10 min read
Fractals started to enter the public consciousness in the early 1980s thanks to Benoit Mandelbrot’s landmark book The Fractal Geometry of Nature. In this book, Mandelbrot argues that many natural structures — clouds, coastlines, trees, rivers, blood vessels, lightning — possess a type of geometry that resists description using simple shapes like lines, planes, and spheres.
Those were the early days of affordable, mass-produced computers with graphics capabilities, and the times were ripe for exploring these fascinating mathematical objects. I spent many of my undergraduate days in the late 1980s writing Turbo Pascal programs on an IBM PC to render images of Mandelbrot sets, Julia sets, Sierpiński triangles, dragon curves, and other mind-bending imaginary structures.
As a budding computer programmer, it fascinated me that very simple formulas could produce images of frightening and sublime beauty.
## Beauty Beneath the Surface
At the time, I was so entranced by the raw beauty of fractals that I didn’t spend much time learning more about their underlying mathematics or their implications to our understanding of the natural world. I was content to tinker with small algorithms that created a huge variety of amazing images.
Only later did I start to read about the mathematical theory behind fractals. Much of it was, and still is, far beyond my understanding. Indeed, there are many unsolved problems in this field that have stumped the brightest mathematical minds. (One example: what is the exact surface area of the Mandelbrot set?)
But here I would like to share one fascinating fractal concept that I could understand, one that is approachable by anyone with high school level mathematics: fractional dimensions. This is the surprising idea that mathematical and physical objects can be described using a number of dimensions that is not an integer like 1, 2, 3, etc.
## Special Snowflakes
Not only will we establish that fractional dimensions exist, we will learn how to calculate their numeric values for a certain category of fractals called Koch snowflakes, whose dimensionality lies between 1 and 2. That’s right, we can prove that such objects are more than one-dimensional but less than two-dimensional! Even better, we can calculate exactly what that number of dimensions is.
The Koch snowflake starts with a straight line segment like this:
We proceed to build the Koch snowflake in a series of steps. In the first step, we split the line segment into three equal parts and remove the middle section. The missing middle section is replaced with two line segments having the same length as the missing section and oriented to form an equilateral triangle with that missing section. Here is the result after the first step:
Now we have 4 line segments that are each the same length as the 3 sections we split the original line segment into. Therefore, the total linear length of the curve is 4/3 as long. This will be important later.
We continue by repeating this process of splitting each of the four shorter line segments into three parts and replacing the middle segment with two segments to form a triangular “bump.” So the second iteration looks like this:
Note that this third object is 4/3 times 4/3, or 16/9, as large as the first object. Continue repeating this process to form a fourth object:
… and then a fifth object:
Each time, the total length increases by a factor of 4/3. Anyone who has studied investing and the effects of compound interest is starting to get excited now. At least they would be, if this abstract structure were priced based on how long all the line segments are. This thing is growing in length exponentially! It’s like you bought a stock that goes up in value 33% year after year without fail. Even if you started with a small investment, pretty soon Jeff Bezos will be seething with envy.
## To Infinity and Beyond
But we still don’t have a fractal yet. For the object to be truly fractal in nature, we have to repeat the procedure an infinite number of times. In the real world this is not possible because we human beings can’t do anything an infinite number of times, and real material objects are made of finite atoms, not an infinitely divisible substance.
But mathematicians don’t let things like that bother them. In the ideal world of our imagination, let’s pretend it is possible. After all, there are many times in traditional geometry where we create idealized simplifications of reality: we can imagine a perfectly straight line segment with zero width and infinite length, circles that are flawlessly round, and so on. Such perfection is nowhere to be found in real life. Mathematics is useful and beautiful even though we know it doesn’t exactly represent the real world.
If we take a leap of imagination and declare that we can repeat the procedure outlined above an infinite number of times, we end up with the Koch snowflake, an object that resembles this:
Now we have an interesting situation. This object has an infinite length, because it grew exponentially an infinite number of times. So there is no way to describe the snowflake as a one-dimensional object.
On the other hand, it has zero surface area. It lies completely inside a finite region of two-dimensional space. With a small enough pair of scissors, you can imagine snipping away more and more of the paper it is printed on, leaving less and less paper each time, without limit. With enough time and precision, you can make the remaining paper weigh as close to zero as you want without damaging the snowflake.
It seems like the Koch snowflake is too “large” to be considered one-dimensional, yet too “small” to be two-dimensional. That might make sense. But does it really? Is this just sloppy thinking? To find out, we need to establish exactly what dimensions are. It turns out there is more than one way to think about dimensions.
## Dimensions by Counting Coordinates
Intuitively we know that some abstract shapes like discs and squares are two-dimensional, while others like spheres and cubes are three-dimensional. Numbers of dimensions are typically determined by counting how many perpendicular directions are required to specify a point inside an object. If you can move up or down, left or right, and forward or backward inside an object, and you can report a distance number for each of those movements, that object has three dimensions because there are three different perpendicular ways you can move.
This is a handy way to understand dimensions, but it’s not the only way. In the case of fractals, counting coordinates simply will not suffice. Coordinate counting leads to a dead end when you consider the movement of a point confined to the Koch snowflake. At first it seems like one dimension works fine, because no matter where you place a point on the snowflake, it can move in only two directions: toward one end of the snowflake or the other, “right” or “left.”
The problem occurs when you ask how far the point has moved. How many meters did it travel along the snowflake? No matter how tiny a movement your point shifts along this object, the total distance traveled is infinite. Even though an outside observer might barely detect a microscopic shift in position, the distance the point had to move along the infinitely crinkly curve is unmeasurable. There is no way to assign a single number expressed in units like feet or meters to describe the motion of a point on this fractal.
And as we noticed above, although the Koch snowflake is infinitely long, it has zero area. We can describe the location of any point on the snowflake using two coordinates, but it seems excessive to call it two-dimensional because the snowflake has no surface area at all.
## Dimensions by Measurement Scaling
Because we are stuck, let’s backtrack and consider a different way to define dimensions. Suppose your job is to tile a square floor that is 10 feet long by 10 feet wide. You are given square tiles that are one foot long on each side. Clearly, you are going to need 100 tiles. But what happens instead if each side of the floor is 3 times as long: 30 feet by 30 feet? Now you will need 900 tiles.
Although the linear size of the floor increased by a factor of 3, the number of tiles went up by 9. In general, if the linear size of the floor goes up by a factor of n, the number of tiles has to increase by a factor of n². We know that floors are two-dimensional, and that number 2 in the formula n² gives us a clue.
Let’s try a new working definition of dimensions. We’ll say that an object is 2-dimensional if increasing its linear size by n causes it to get larger by a factor of n², that it is 3-dimensional if it gets larger by n³, and so on.
We want to apply this idea to the Koch snowflake. That means we would like to triple the linear extent of the snowflake and then, somehow, measure the change in its “size.” But there’s the rub: the notion of “size” is slippery for an object like this. How do we proceed?
The solution is to reconsider the square tiles we used above. Why are we able to use tiles to measure the area of two-dimensional floors? It is because tiles are themselves two-dimensional: they have a finite surface area, and we can count how many of them are placed onto the floor to cover it.
## How Do We Tile a Snowflake?
To measure Koch snowflakes, we need something that has the same number of dimensions as a Koch snowflake. This is a little frustrating because we don’t know what that number of dimensions is. In fact, that’s what we’re trying to figure out! It’s like a dog chasing its tail.
But if we did have some kind of object smaller than the snowflake and with the same number of dimensions, we could count how many of those smaller objects it takes to “tile” the snowflake.
It turns out there is a perfect tool for this job. In fact, it’s right under our noses: another Koch snowflake, only smaller. We can make a copy of a small part of the snowflake and use it to measure the whole snowflake, just like we use small square tiles to measure a large floor. This works because each section of the snowflake is exactly the same shape as the whole snowflake, a common property of fractals called self-similarity.
In this diagram we see that the snowflake has four major sections as labeled by Roman numerals. Each of these sections is identical in shape to the whole snowflake. If we think of the sections as tiles, it takes 4 small tiles to cover the entire snowflake, and therefore the entire snowflake is 4 times as large as each of the sections that comprise it.
At the same time, we see that each of these sections has 1/3 the linear size as the whole snowflake. For example, section IV is exactly 1/3 as wide at its base as the entire snowflake.
So tripling the linear size of one of the sections results in a snowflake that takes four of the original size sections to tile it. To find the number of dimensions D of this fractal, we need to find a value of D such that
If we try to guess values of D, we see that D=1 is too small, because 3¹=3. Also, D=2 is too large, because 3²=9. Now we are getting somewhere! We have a solid basis for claiming that the number of dimensions D is between 1 and 2. But what is the exact value of D?
## Solving the Puzzle
(Don’t worry about this part if you are rusty on the math; it’s not necessary to the overall concept. The important part is that it is possible to arrive at an exact value for D.)
To calculate the value of D, we need to take the logarithm of both sides of the equation and use a little algebra to isolate D to one side of the equal sign:
The result is that the Koch snowflake is about 1.26-dimensional. This makes sense because the snowflake looks closer to 1-dimensional than 2-dimensional; it appears more line-like than plane-like.
## Hausdorff dimension
Using the tiling idea to calculate the dimensionality of a fractal results in a number D called the Hausdorff dimension. No, this isn’t the name of some obscure German space music band. It is a well-established formalism introduced by the mathematician Felix Hausdorff in 1918. The same type of reasoning allows mathematicians to establish the Hausdorff dimension value for a wide variety of fractals.
## Beyond Snowflakes
I hope this mental journey has opened a new door for you by expanding your concept of what dimensions are. The idea that there is such a thing as a fractional number of dimensions is fascinating. This concept has practical value too. It shows up in economics, biology, geography, and many other real-world disciplines.
Beyond the specific concept of Hausdorff dimension, I propose that breaking the “integer barrier,” establishing that dimensions can be fractional, is an illuminating exercise for one’s creativity. It fosters a spirit of questioning assumptions and expanding one’s mind to reach beyond them. What other ways do seemingly obvious assumptions limit your thinking? What other wonders await you when you notice and challenge the “obvious?”
Written by
## Don Cross
#### Husband, computer programmer, yoga teacher, cat servant.
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Get unlimited access to the best stories on Medium — and support writers while you’re at it. Just \$5/month. Upgrade | 2,899 | 13,581 | {"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} | 3.0625 | 3 | CC-MAIN-2019-43 | latest | en | 0.939416 |
https://cm-to.com/8892-cm-to-inches/ | 1,721,125,601,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514742.26/warc/CC-MAIN-20240716080920-20240716110920-00398.warc.gz | 159,689,177 | 8,608 | # 8892 Centimeter to Inch
Convert 8892 (eight thousand eight hundred ninety two) Centimeters to Inches (cm to in) with our conversion calculator.
8892 Centimeters to Inches equals 3,501 in.
• Meter
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Convert 8892 Centimeters to Inches (cm to in) with our conversion calculator.
8892 Centimeters to Inches equals 3,501 in.
To convert centimeters to inches, we employ the standard conversion factor that 1 inch is equal to 2.54 centimeters. The calculation for converting 8892 centimeters to inches involves dividing the number of centimeters by the conversion factor. Therefore, the calculation would look like this:
$$[8892 \, \text{cm} \div 2.54 = 3500.7874 \, \text{inches}]$$
This calculation is straightforward. We start by having the length in centimeters, which is 8892 cm in this case. Given that the conversion factor from centimeters to inches is 1 inch equals 2.54 centimeters, we divide the length in centimeters by this conversion factor (2.54) to find the equivalent length in inches. The result of this division gives us the length in inches, which is approximately 3500.7874 inches.
Items that are approximately 3500.7874 inches (or 8892 cm) in length could be quite large and not commonly found in everyday settings. Here are seven hypothetical examples:
• A small neighborhood street: Length from one end to the other, useful for short, community-based activities.
• Used for communal gatherings, local markets, or a children's play area.
• The height of certain small hills: Measuring the total elevation, for light hiking experiences.
• Offering scenic views, short hiking routes, and picnic spots.
• A long assembly line in a factory: Spanning the length of the production process from start to finish.
• Used for manufacturing processes, from assembling parts to packaging finished products.
• A large cargo ship's deck: Extending from the bow to the stern, providing substantial space for cargo and operations.
• Used for transporting goods across seas, loading zones for containers, and crew movement area.
• The length of some runways for small aircraft: Offering enough distance for takeoff and landing of light planes.
• Utilized for small regional airports, aviation schools, and aerial survey operations.
• A train with several passenger carriages: Total length from the locomotive to the last carriage.
• Facilitates long-distance travel, commuter services, and includes seating, dining, and sleeping compartments.
• Extensive conveyor systems in large distribution centers: Spanning across sorting, packaging, and shipping areas.
• Employs automation for the efficient movement of goods, sorting parcels, and assembly packages for delivery.
These examples span a variety of fields from transportation and manufacturing to leisure and community infrastructure, highlighting the diverse utility of items and structures measuring 8892 cm in length. | 686 | 3,107 | {"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} | 3.671875 | 4 | CC-MAIN-2024-30 | latest | en | 0.838446 |
http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-Delhi/Control%20system%20design%20n%20principles/matlab/module4/module4-3_middle.htm | 1,369,407,107,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704666482/warc/CC-MAIN-20130516114426-00033-ip-10-60-113-184.ec2.internal.warc.gz | 185,776,824 | 2,935 | ## MATLAB MODULE 4
Feedback System Simulation
Simulink simulation is an alternative to block diagram manipulation followed by time-response analysis. From the Simulink model of a control system, output y in response to command r, output y in response to disturbance w, control u in response to command r, and all other desired internal variables can be directly obtained.
Example M4.3
Control system design methods discussed in this course are based on the assumption of availability of linear time invariant (LTI) models for all the devices in the control loop.
Consider a speed control system. The actuator for the motor is a power amplifier. An amplifier gives a saturating behaviour if the error signal input to the amplifier exceeds linear range value.
MATLAB simulink is a powerful tool to simulate the effects of nonlinearities in a feedback loop. After carrying out a design using LTI models, we must test the design using simulation of the actual control system, which includes the nonlinearities of the devices in the feedback loop.
Figure M4.6 is the simulation diagram of a feedback control system: the amplifier gain is 100 and the transfer function of the motor is 0.2083/(s +1.71). We assume the amplifier of gain 100 saturates at +5 or -5volts. The result of the simulation is shown in Fig. M4.7.
The readers are encouraged to construct the simulink model using the procedure described in Module 3. All the parameter settings can be set/seen by double clicking on related blocks.
Time and Output response data have been transferred to workspace using To Workspace block from Sinks main block menu. Clock block is available in Sources main menu. These variables are stored in the structure Output and Time in the workspace, along with the information regarding simulink model name. For example,
>> Output
Output =
time: [ ]
signals: [1x1 struct]
blockName: 'M4_3/To workspace Output'
>> Time
Time =
time: [ ]
signals: [1x1 struct]
blockName: 'M4_3/To workspace Time'
To access output and time values, one needs to access Output.signals.values and Time.signals.values . The step response plot has been generated by the following MATLAB script.
>> plot(Time.signals.values,Output.signals.values)
>> xlabel('Time (sec)');
>> ylabel('Output');
>> title('Step Response');
Fig. M4.7
Example M4.4
In this example, we simulate a temperature control system with measurement noise added to the feedback signal. The process transfer function is
The deadtime minutes. The measurement noise parameters we have used are: mean of 0, variance of 0.5, initial seed of 61233, and sample time of 0. The simulink inputs a step of 30 to the system (Fig. M4.8). Deadtime block in this figure is Transport Delay block from Continuous library, and Random Number block is from Sources library.
The data has been transferred to the workspace using To Workspace block. The step response, generated using the following MATLAB script is shown in Fig. M4.9.
>> plot(Time.signals.values,Y.signals.values);
>> ylabel('Output (Y)');
>> xlabel('Time(min)');
>> title('Step Response');
Fig. M4.9
The performance of the system with measurement noise removed, is shown in Fig. M4.10. To remove the effect of noise, simply disconnect the Random number block from the Sum block in the feedback path.
Fig. M4.10 | 733 | 3,322 | {"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} | 2.75 | 3 | CC-MAIN-2013-20 | latest | en | 0.804913 |
azec.nomen.pw | 1,579,459,987,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594705.17/warc/CC-MAIN-20200119180644-20200119204644-00471.warc.gz | 16,198,434 | 12,021 | # 2d Fft C Code
Gorakhpur University, Gorakhpur X Abstract-The Fast Fourier Transform (FFT) and Inverse. The complete solution for all 9 output can be found here; Example of 2D Convolution. I think it would be simple to convert the code to C, or wrap it under a couple C functions. If somebody haas a source code about CUFFT 2D, please post it. P-Code Pascal, PascalWare additions, and Source Code. The fast Fourier transform (FFT) is a versatile tool for digital signal processing (DSP) algorithms and applications. * 1D FFT (real too): Index 0 is the handle for the entire FFT: 85 * 2D complex FFT: Index 0 is the handle for the entire FFT: 86 * 3D complex FFT: Index 0 is the handle for the entire FFT: 87 * 2D, inplace real FFT: 0=FFTx, 1=FFTy handle: 88 * 2D, ooplace real FFT: 0=FFTx, 1=real-to-complex FFTy, 2=complex-to-real FFTy. void dft (InputArray src, OutputArray dst, int flags=0, int nonzeroRows=0) Parameters: src – input array that could be real or complex. FFT of large array (2**20 real points) 8. A Brief History of MATLAB to C. 1 Fast Fourier Transform, or FFT The FFT is a basic algorithm underlying much of signal processing, image processing, and data compression. Python Training course at Bodenseo. CS425 Lab: Frequency Domain Processing 1. The Performance test suggest to use 4 threads for parallelization. my ubuntu 13. If n is larger than the dimension along which the FFT is calculated, then x is. At the prime tree level, algorithm either performs a naive DFT or if needed performs a single Rader's Algorithm Decomposition to (M-1), zero-pads to power-of-2, then proceeds. Create a split complex vector—suitable for use with vDSP's 2D FFT—by copying odd pixels to the real parts and even pixels to the imaginary parts of an array of complex numbers. User-friendly 2D FFT/iFFT (Fast Fourier Transform) plug-in for Adobe PhotoShop compatible plug-in hosts. Thank to the recursive nature of the FFT, the source code is more readable and faster than the classical implementation. Computation is slow so only suitable for thumbnail size images. FFTW,IMKL,KISSFFT). This approach is based on the separable property of 2D FFT. arbitrary FFT sizes optimization for real-only FFTs (i. However, since and are N/2-point DFTs, they can be obtained the same way.
The complete solution for all 9 output can be found here; Example of 2D Convolution. The filter is tested on an input signal consisting of a sum of sinusoidal components at frequencies Hz. (1), (2), once and are available. Our signal becomes an abstract notion that we consider as "observations in the time domain" or "ingredients in the frequency domain". my ubuntu 13. supports in-place or out-of-place transforms. 43 out of 5) In the previous post, Interpretation of frequency bins, frequency axis arrangement (fftshift/ifftshift) for complex DFT were discussed. x into a managed C++ DLL to calculate FFT? Using gdx Library and FFT to Calculate Frequency (Java) Strategy - CUFFT computing 2D FFT on many images; TensorFlow: How to get pointer to data contents of ConstTensor?. For 2D signals, the FFT VI computes the discrete Fourier transform (DFT) of the input matrix. The most common case is for developers to modify an existing CUDA routine (for example, filename. Origin uses the FFTW library for its Fast Fourier Transform code. dst – output array whose size and type depends on the flags. Brayer (Professor Emeritus, Department of Computer Science, University of New Mexico, Albuquerque, New Mexico, USA). Fast Fourier Transform in matplotlib An example of FFT audio analysis in matplotlib and the fft function. Using the complex-conjugate symmetry of a real fft, we can pack // the fft back into an array of the same size as the input. Example "Three-Dimensional REAL FFT (C Interface)" is three-dimensional out-of-place transform in C interface. // (The results are packed because the input data is in the real domain, but the output // is in the complex domain. FFT( data2D, ref fftdata ); // Now retrieve the unpacked fft result. The Fourier Transform Part XV – FFT Calculator Filming is currently underway on a special online course based on this blog which will include videos, animations and work-throughs to illustrate, in a visual way, how the Fourier Transform works, what all the math is all about and how it is applied in the real world. /** * Sample code to compute complex 2D DFT */ void complex_2d_dft { fftw_complex * a, * b, * c; fftw_plan plan_f, plan_b;. The Fast Fourier Transform (FFT) is one of the most important algorithms in signal processing and data analysis. The Fourier Transform is used in a wide range of applications, such as image analysis, image filtering, image reconstruction and image compression. I have 2D array of ints (info about pixels in images) and I need pass FFT on this Array and save. For example, a compact finite difference scheme often results in solving a tridiagonal linear system when evaluating spatial derivatives or doing spatial interpolations; a spectral code often involves performing a Fast Fourier Transform along a global mesh line. Hi Opt, Thank you very much for your support.
3) Slide 22 C FFT Program (cont. 6 (572 ratings) Course Ratings are calculated from individual students' ratings and a variety of other signals, like age of rating and reliability, to ensure that they reflect course quality fairly and accurately. Bug in SWAG FFT routine. This example demonstrates an Open Computing Language (OpenCL TM) implementation of a fast Fourier transform (FFT). edu Jakob Siegel Department of ECE University of Delaware Newark, DE, USA jakob@udel. You will need h. The only fact this article really proves is that I can write an x86 AVX implementation of FFT, such that it has reasonable performance and reasonably readable code. The most general case allows for complex numbers at the input and results in a sequence of equal length, again of complex numbers. m computes the fast fractional Fourier transform following the algorithm of [5] (see also [6] for details) The m-file frft22d. 3 Fast Fourier Transform (FFT) The Fast Fourier transform (FFT) is an algorithm for computing DFT, before which the DFT required excessive amount of computation time, particularly when high number of samples (N) was required. a finite sequence of data). I am looking for a zoom fft code. 0 respectively. Using simple APIs, you can accelerate existing CPU-based FFT implementations in your applications with minimal code changes. COMPRESSION USING FAST 2D-DISCRETE FRACTIONAL FOURIER TRANSFORM AND SPIHT ALGORITHM WITH HUFFMAN ENCODER This approach comprises of the following phases namely,. This is my fftw tutorial. The input signal is transformed into the frequency domain using the DFT, multiplied by the frequency response of the filter, and then transformed back into the time domain using the Inverse DFT. FFT, PSD and spectrograms don't need to be so complicated.
The code looks as though it might be an older implementation of an FFT (Fast Fourier Transform). is known as the Fast Fourier Transform (FFT). Gorakhpur University, Gorakhpur Department of Computer Science, D. Introduction It turns out that taking a Fourier transform of discrete data is done. m computes a 2D transform based on the 1D routine frft2. Thanks for contributing an answer to Code Review Stack Exchange! Please be sure to answer the question. 1 Fast Fourier Transform, or FFT The FFT is a basic algorithm underlying much of signal processing, image processing, and data compression. The output X is the same size as Y. fft fft code, using butterfly algorithm, including C, matlab and Verilog code The Fast Fourier Transform (FFT) Algorithm The FFT is a fast algorithm for computing the DFT. This article describes a new efficient implementation of the Cooley-Tukey fast Fourier transform (FFT) algorithm using C++ template metaprogramming. This is the C code for a decimation in time FFT algorithm. One stage of the FFT essentially reduces the multiplication by an N × N matrix to two multiplications by N 2 × N 2 matrices. The example processes a 2D matrix of 1,024x1,024 complex single-precision floating-point values. Also: I quickly re-wrote the code in MATLAB, which uses Fast Fourier Transform functions based on the FFTW library, and the bug is NOT happening the mystery deepens.
FFT/Fourier Transforms QuickStart Sample (C#) Illustrates how to compute the forward and inverse Fourier transform of a real or complex signal using classes in the Extreme. implements the radix-2 DIT FFT in C++. I am not sure whether I should loop it or do a batch FFT. Here are a few good free libraries (carefully note which license they have to check if you can use it in your project): FFTW "Fastest Fourier Transform in the West", GPL license. In this post, we are going to generate a 2D Gaussian Kernel in C++ programming language, along with its algorithm, source code, and sample output. Read and plot the image; Compute the 2d FFT of the. dst - output array whose size and type depends on the flags. I changed in C code float to double, but benc. I get some values from a signal generator and put the into an array and now i want to make FFT on these values and get the magnitude. This section presents examples of using the FFT interface functions described in “Fourier Transform Functions”. The Cooley-Tukey algorithm, named after J. Hi, I have to do an assignment for my C programming class. Header-only C++ library implementing fast Fourier transform of 1D, 2D and 3D data. As explained in the description of the original code: The Fast Fourier Transform (FFT) is the most widely known example of the Spectral method for computational problems. This is useful for analyzing vector. 1D and 2D FFT-based convolution functions in Python, using numpy. There are many distinct FFT algorithms involving a wide range of mathematics, from simple complex-number arithmetic to group theory and number theory; this article gives an overview of the available techniques and some of their.
Fourier Transform decomposes an image into its real and imaginary components which is a representation of the image in the frequency domain. As a result, the fast Fourier transform, or FFT, is often preferred. h should be inserted into filename. Specifically, the Fast Fourier Transform code further down in this tutorial only supports power-of-two sized inputs, while good libraries can do FFT on arbitrary sizes. It is explained very well when it is faster on its documentation. c is a C program to perform the Fast Fourier Transform. Hi, I have to do an assignment for my C programming class. I am not getting Airy disk pattern as the output. We have developed an efficient implementation to compute the 2D fast Fourier transform (FFT) on a new very long instruction word programmable mediaprocessor. (Fast Fourier Transform) Written by Paul Bourke June 1993. We benefited from much discussions and/or codes, including: Leslie Greengard, Ludvig af Klinteberg, Zydrunas Gimbutas, Marina Spivak, Joakim Anden, and David Stein´. My image is 512 x 512 pixels. The 1024 point FFT code was first realized in MATLAB without using the inbuilt FFT function, but MATLAB is not a hardware descriptive. However, since and are N/2-point DFTs, they can be obtained the same way. spatial Þlter frequency Þlter input image direct transformation. Calculating DCT with DFT. Brayer (Professor Emeritus, Department of Computer Science, University of New Mexico, Albuquerque, New Mexico, USA). The only fact this article really proves is that I can write an x86 AVX implementation of FFT, such that it has reasonable performance and reasonably readable code. In C#, an FFT can be used based on existing third-party code libraries, or can be developed with a minimal amount of programming. This is a package to calculate Discrete Fourier/Cosine/Sine Transforms of 1-dimensional sequences of length 2^N. 28 answers. Here are a few good free libraries (carefully note which license they have to check if you can use it in your project): FFTW "Fastest Fourier Transform in the West", GPL license.
For fixed-point inputs, the input data is a vector of N complex values represented as dual b x-bit two’s-complement numbers, that is, b. 2 KB; Introduction. FFT in a single C-file [closed] so I re-wrote the code with some translations. Parallel Fast Fourier Fast Fourier Transform (FFT) Scalability Improvements for DFT Codes due to the Implementation of the 2D Domain. ===== test: tests the result of all implementations included in the library are correct and equivalent. I've founded the FFT function "cfftr2. cu file and the library included in the link line. The Fourier transform is just a step in a much bigger software we are developing. C++ Perform to a 2D FFT Inplace Given a Complex 2D Array - Fast Fourier transform FFT is an algorithm to compute the discrete Fourier transform DFT and its inverse Basically Fourier analysis converts time or space to frequency and vice versa A FFT rapidly computes transformations by factorizing the. 64 Point Radix-4 FFT Butterfly Realization using FPGA Amaresh Kumar, U. Downloads of the Numerical Recipes source code in machine-readable format are not available as part of this free resource. Brayer (Professor Emeritus, Department of Computer Science, University of New Mexico, Albuquerque, New Mexico, USA). Here are the projects that use JTransforms:. A cross-section of the output image is shown below. The output image is the square modulus of the resulting Fourier transform. Calculate the fundamental frequency of the captured audio sound The FFT Guitar Tuner application was developed to be a small tool that's using a Fast Fourier Transform to calculate the fundamental frequency of the captured audio sound. h which is a c implementation on the fft which has been. 28 answers. The Discrete Fourier Transform (DFT) transforms discrete data from the sample domain to the frequency domain. my ubuntu 13. b) Now take the 2D inverse Fourier transform of the spectrum image to recover the spatial domain image. Thank to the recursive nature of the FFT, the source code is more readable and faster than the classical implementation.
This VI performs a 1D FFT on the rows of the input matrix and then performs a 1D FFT on the columns of the output of the preceding step. It puts DC in bin 0 and scales the output of the forward transform by 1/N. For the 1-d fft, it can be constructed to an equivalent matrix. Use MathJax to format equations. If n is larger than the dimension along which the FFT is calculated, then x is. Design and implementation in VHDL code of the two-dimensional fast Fourier transform for frequency filtering, convolution and correlation operations. I changed in C code float to double, but benc. The FFT tool will calculate the Fast Fourier Transform of the provided time domain data as real or complex numbers. The analogous 3d files are used for 3d FFTs. If the input signal is an image then the number of frequencies in the frequency domain is equal to the number of pixels in the image or spatial domain. Fourier Transform is used to analyze the frequency characteristics of various filters. In the context of the Fast Fourier Transform (FFT) it is not so simple to compile for example the freely available FFTW code collection along with ones own small projects. ) The FFTW Home Page: A fast C library for performing the FFT in one or more dimensions, including parallel and real-data transforms. Convolution is the most important and fundamental concept in signal processing and analysis. supports 1D, 2D, and 3D transforms with a batch size that can be greater than or equal to 1. Implementation of FFT in ALGLIB. 64 Point Radix-4 FFT Butterfly Realization using FPGA Amaresh Kumar, U. 1 Contents lead me to use Python and its extension modules as well. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. To use only the remapping routines, the fft_* files are not needed.
The call seems quite straightforward - I send it a complex 2D array (for which I set the real components equal to the image grey levels and the imaginary components equal to 0) and all of the parameters that it needs are equal to 512. On Enhancing 3D-FFT Performance in VASP As FFT code optimization is close to the hardware, the integration of FFT 2D- and 3D-FFT computation on complex and. FFT functional async vs. 1 Overview The FFT core computes an N-point forward DFT or inverse DFT (IDFT) where N can be 2m, m = 3-16. The output is returned in the input array. Gnuradio has some GUIs built using wx and its python extension - specifically an oscilloscope, an fft, and a waterfall display. laser diffraction patterns). Actually, the main uses of the fast Fourier transform are much more ingenious than an ordinary divide-and-conquer. 2D Discrete Fourier Transform • Fourier transform of a 2D signal defined over a discrete finite 2D grid of size MxN or equivalently • Fourier transform of a 2D set of samples forming a bidimensional sequence • As in the 1D case, 2D-DFT, though a self-consistent transform, can be considered as a mean of calculating the transform of a 2D. A 2D FFT (see Matlab command fft2) is decomposed into several 1D FFTs: the FFT operator for an N-dimensional array can in fact be splitted into several 1-dimensional FFTs of monodimensional arrays. VXL - C++ libraries for Computer Vision - The Vision-something-Libraries are a collection of C++ libraries designed for computer vision research. Below is a sample code to compute the 2D DFT of a 2×3 complex array followed by a backward transform to obtain the original array back. Fourier transform is one of the various mathematical transformations known which is used to transform signals from time domain to frequency domain. In constructing these classes I also implemented my own FFT process, taking the main FFT algorithm loop from the processing minim library, so that I could use doubles to store my samples (and to gain some more experience staring at FFT code - I find this FFT function is much more readable and understandable than the provided chuck_fft. I have 2D array of ints (info about pixels in images) and I need pass FFT on this Array and save. FFT Tool?. 3) Slide 22 C FFT Program (cont. The efficiency is proved by. Here are a few good free libraries (carefully note which license they have to check if you can use it in your project): FFTW "Fastest Fourier Transform in the West", GPL license. But TI's FFT operates by float array, when my programm needs a FFT under a DOUBLE array. I will have a look to sourceforge. But I tried to generate FFT program with your code but it seems that FFT1D results are different from results of MATLABAnd the 2D one doesn't work at all. c - Fixed-point in-place Fast Fourier Transform */ All data are fixed-point short integers, in which -32768 to +32768 represent -1. It used the transpose split method to achieve larger sizes and to use multiprocessing. Y = fft2(X) returns the two-dimensional Fourier transform of a matrix using a fast Fourier transform algorithm, which is equivalent to computing fft(fft(X).
To computetheDFT of an N-point sequence usingequation (1) would takeO. This example, which can be found in the code EXAMPLE2_transform. As expected, the Real FFT method yields superior performance. Fft, free fft software downloads. High performance sparse fast Fourier transform, Jörn Schumacher Master thesis, Computer Science, ETH Zurich, Switzerland, 2013 [PAPER] Sparse 2D Fast Fourier Transform Andre Rauh and Gonzalo R. The efficiency is proved by performance benchmarks on different platforms. You will need h. 1 The DFT The Discrete Fourier Transform (DFT) is the equivalent of the continuous Fourier Transform for signals known only at instants separated by sample times (i. *****/ /* Data type and new names for flexibility: real: Basic data type for floating point computations. The general idea is that the image (f(x,y) of size M x N) will be represented in the frequency domain (F(u. Note that N is 64 so you can make 2 parallel 32 FFT Compute shader and 1 final butterfly on CPU to combine and get 64 data FFT. Introduction. P-Code Pascal, PascalWare additions, and Source Code. Hi, I have to do an assignment for my C programming class. For example, if A is a 3-D array X=fft(A,-1,2) is equivalent to:. C-Implementations of FFT Algorithms: run a simple make in c-fft directory and both test and benchmark will be compiled. 3) Slide 22 C FFT Program (cont. All Blackfins,All SHARCs,All ADSP-SC58x,All ADuCM302x. Matlab Tips and Tricks Gabriel Peyr´e you have corrections about these pieces of code, or if you of the integral y of an image M along a 2D curve c (the 7. Using native C Kiss_fft.
i have an existing program in c#. It can switch the length of fft between 16 point and 64 point. There are many ways to interface to an FFT. c is a multi threaded 2D FFT considerably adapted from this. P-Code Pascal, PascalWare additions, and Source Code. See recent download statistics. To use only the remapping routines, the fft_* files are not needed. The cool part is that shaders can read and write them just like in C. FFT in a single C-file [closed] so I re-wrote the code with some translations. There are two approaches to performing such computations on distributed-memory systems. Here is code to perform 2D Frouier transforms on jepg files. We show how to comute an FFT of a real signal. Cooley-Tukey FFT. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Introduction. In this post, we are going to generate a 2D Gaussian Kernel in C++ programming language, along with its algorithm, source code, and sample output. can you please help me ? the program its with GUI in visual studio. Using simple APIs, you can accelerate existing CPU-based FFT implementations in your applications with minimal code changes.
cu file and the library included in the link line. This is the C code for a decimation in time FFT algorithm. The result of this function is a single- or double-precision complex array. 1 Basics of DFT and FFT The DFT takes an N-point vector of complex data sampled in time and transforms it to an N -point vector of complex data that represents the input signal in the frequency domain. You have some library for advice? Someone who take me some advice/tutorial how use this library? Library FFT. This example demonstrates the use of k-Wave for the reconstruction of a two-dimensional photoacoustic wave-field recorded over a linear array of sensor elements. Does the image contain all zero values? Here's some MATLAB base code to get you started: clc. 2D FFT using PyFFT, PyCUDA and Multiprocessing. They are extracted from open source Python projects. The example processes multiple sets of 4096 complex single-precision floating-point values. * 1D FFT (real too): Index 0 is the handle for the entire FFT: 85 * 2D complex FFT: Index 0 is the handle for the entire FFT: 86 * 3D complex FFT: Index 0 is the handle for the entire FFT: 87 * 2D, inplace real FFT: 0=FFTx, 1=FFTy handle: 88 * 2D, ooplace real FFT: 0=FFTx, 1=real-to-complex FFTy, 2=complex-to-real FFTy. In this case the include file cufft. By continuing to browse this site, you agree to this use. To compile EXAMPLE2_transform. The first was not giving me the output that is expected and the. Can anyone provide me matlab code for sine fft and inverse sine fft ? Question. FFTW is designed to be called directly from C and C++, of course, and also includes wrapper functions allowing you to call it from Fortran. zip and the. 4) Slide 23 C FFT Program (cont. dst – output array whose size and type depends on the flags. 2d Fft C Code.
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# M251ex1(su07) - MATH 251 Examination I July 9 2007 Name...
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MATH 251 Examination I July 9, 2007 Name: Student Number: Section: This exam has 11 questions for a total of 100 points. In order to obtain full credit for partial credit problems, all work must be shown. Credit will not be given for an answer not supported by work. The point value for each question is in parentheses to the right of the question number. You may not use a calculator on this exam. 1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: Total: Do not write in this box.
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MATH 251 EXAMINATION I July 9, 2007 1. (10 points) For parts (a) through (e) below, a list of di±erential equations is given. For each part, write down the letter corresponding to the equation on the list with the speci²ed proper- ties. There is only one correct answer to each part. A. y = 2 y + t B. y = e 2 y t C. y = e y - 1 D. y ′′ + 4 y - 5 y = 2 E. y ′′ + e t y + t 2 y = 0 F. y ′′ - 4 y = t y G. y ′′′ + 3 y ′′ + 3 y + y = t 5 + ln t H. y ′′′ + y y = e 2 t sin 5 t (a) First order autonomous equation. (b) Second order homogeneous linear equation.
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M251ex1(su07) - MATH 251 Examination I July 9 2007 Name...
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### When Is Halving Not Halving?
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Exploring the relationship (or lack of relationship) between perimeter and area is interesting for students—even for simple shapes like rectangles. For example, if you cut a rectangle’s area in half, do you also cut the perimeter in half?
Using the rectangles shown below, it is easy to see that the figure on the left was cut in half to create the figure on the right. When we measure the area, the rectangle on the left is 16 units and the area of the smaller rectangle is 8 units—exactly half of the original rectangle. However, it turns out that the perimeter of the figure on the left was not cut in half when the new rectangle was created. In fact, the new perimeter is a full 3/4 of the old perimeter.
Is the new perimeter always 3/4 of the old one? Let’s try a different rectangle. This time, let’s cut it vertically instead of horizontally.
Once again, the area is halved, but the perimeter changes from 16 units to only 10 units. This time, the ratio of new perimeter: old perimeter, is not 1/2 and also not 3/4. Instead, it is 5/8.
You could provide students with square tiles with which to build rectangles, or they could explore the challenge using geoboards and geobands. Alternatively, students might digitally access squares they can put together to make rectangles using the Patch Tool at http://illuminations.nctm.org/Activity.aspx?id=3577, the Shape Tool at http://illuminations.nctm.org/Activity.aspx?id=3587, or the virtual geoboard available at the National Library of Virtual Manipulatives website.
Encourage students to then explore exactly what fractions of the old perimeter the new perimeter could be if a rectangle’s area is cut in half.
Could it be 2/3?
Could it be 1/3?
Could it be 5/6?
Could it be really close to 1?
Could it be really close to 0?
Is it ever 1/2?
Alternatively, if time is limited, ask students to determine the dimensions of rectangles with specific new perimeter: old perimeter ratios, such as 5/6 or 2/3.
Have your students try the problem and see how it goes for them. If you are already on summer break, you could challenge you own children or neighborhood children to explore the task. I welcome you to share your students’ experiences with us.
Marian Small is the former dean of education at the University of New Brunswick, where she taught mathematics and math education courses to elementary and secondary school teachers. She has been involved as an NCTM writer on the Navigations series, has served on the editorial panel of a recent NCTM yearbook, and has served as the NCTM representative on the MathCounts writing team. She has written many professional resources including Good Questions: Great Ways to Differentiate Mathematics Instruction (2012), Eyes on Math (2012), and Uncomplicating Fractions to Meet Common Core Standards in Math, K–7 (2013), all co-published by NCTM.
This is a great example of how to increase the "richness" of a task! Most of the "area/perimeter" activities I've seen deal with finding different perimeters for a "fixed" area, or finding different areas for a "fixed" perimeter. Exploring the ratio change adds a whole new dimension (pardon the pun) :) to the task--excellent!Posted by: RalphC_79522 at 7/6/2014 1:05 PM
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https://hwkhlp.com/Maths/Solver | 1,545,073,038,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829115.83/warc/CC-MAIN-20181217183905-20181217205905-00499.warc.gz | 609,529,802 | 4,129 | ### Maths and Solver
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Subject
Topic
Level
Junior High
7
2x+5y=7;3x_7y=5 with unique solution
$$2x+5y=7$$ $$2x+5y=7.3x-7y=5$$
Subject
Topic
Level
Homeschool
10
6x+2=11
6x+2=11
Solution:
\left\{\frac{3}{2}\right\}
Calculated: \left\{1.5\right\}
Subject
Topic
Level
Highschool
7
koo
Subject
Topic
Level
Middle School | 154 | 358 | {"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} | 4.03125 | 4 | CC-MAIN-2018-51 | longest | en | 0.613817 |
https://www.exceltip.com/excel-pivot-tables/how-to-show-hide-grand-totals-in-pivot-table-excel.html | 1,639,057,076,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964364169.99/warc/CC-MAIN-20211209122503-20211209152503-00335.warc.gz | 828,129,549 | 18,934 | # How to Show/Hide Grand totals in Pivot Table Excel
In this article, we will learn How to Show/Hide Grand totals in Pivot Table Excel.
Scenario:
In Excel, We can opt to show or hide Grand total values in excel. For example when using a pivot table based on different categories and didn't require to view the subtotals or Grand total values. Sometimes we don't require these values and these subtotal rows and grand total rows can be removed using the method explained below.
Hide subtotals in Pivot table Excel
Select a cell in the PivotTable, Go to design and select Subtotal Options -> Select Do Not Show Subtotals
You can also customize the pivot table using the two options mentioned below.
• Show all Subtotals at Bottom of Group
• Show all Subtotals at Top of Group
Hide Grand totals in Excel
Select a cell in the PivotTable, Go to design and select Grand Totals Options -> Off for Rows and Columns.
You can also customize on or off Grand totals in the pivot table using the options mentioned below.
• On for Rows and Columns (Grand totals displayed for both rows and columns)
• On for Rows only (Grand totals removed from pivot table columns)
• On for Columns only (Grand totals removes from Pivot table Rows)
Example :
All of these might be confusing to understand. Let's understand how to use the function using an example. Here we follow some steps which will guide where to find the Grand totals show/hide option in the PIvot table.
First of all Create a pivot table. Select Data > Insert tab > Pivot table (new workbook).
Now on the Pivot table sheet. Select any cell from the pivot table which will pop two new pivottable tabs (Analyze and Design).
Now Go to the Pivottable tools > Design tab as shown below.
As you can see in the above snapshot Pivot Subtotals and Grand totals option on the left side. Now go to the Grand total option and Select the option which says Off for Rows and Columns.
Via clicking this option all the Grand totals rows and columns get removed from the Pivot table. But you still have subtotal rows in the pivot table. To remove these subtotal rows or columns, we will follow some of the same steps as above.
Now on the Pivot table sheet. Select any cell from the pivot table and Go to the Design tab. Now go to the Subtotals option and Select the option which says Do Not Show Subtotals as shown below.
Selecting the option will get removed all Subtotals row and Columns removed.
Here are all the observational notes using the formula in Excel
Notes :
1. If you don’t want to show grand totals for rows or columns, uncheck the Show grand totals for rows or Show grand totals for columns boxes on the Totals & Filters tab in the PivotTable Options dialog box (Analyze> Options).
2. Customize pivot table totals in excel using the above mentioned steps
3. You can also Show/hide the field header in the pivot table.
Hope this article about How to Show/Hide Grand totals in Pivot Table Excel is explanatory. Find more articles on calculating values and related Excel formulas here. If you liked our blogs, share it with your friends on Facebook. And also you can follow us on Twitter and Facebook. We would love to hear from you, do let us know how we can improve, complement or innovate our work and make it better for you. Write to us at info@exceltip.com.
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How to Refresh Pivot Charts : To refresh a pivot table we have a simple button of refresh pivot table in the ribbon. Or you can right click on the pivot table. Here's how you do it.
How to Dynamically Update Pivot Table Data Source Range in Excel : To dynamically change the source data range of pivot tables, we use pivot caches. These few lines can dynamically update any pivot table by changing the source data range. In VBA use pivot tables objects as shown below.
Popular Articles : | 973 | 4,674 | {"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} | 2.59375 | 3 | CC-MAIN-2021-49 | latest | en | 0.790144 |
https://mymathforum.com/threads/some-algebraic-manipulation.347892/ | 1,581,888,933,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875141430.58/warc/CC-MAIN-20200216211424-20200217001424-00178.warc.gz | 489,578,062 | 15,820 | # Some algebraic manipulation.
#### ensbana
I've just got back to study some subjects that require maths, and my skill is pretty rusty. Could anyone help me with putting bounds on the following expression?
For $$\displaystyle k$$ a positive integer, and $$\displaystyle \delta \in (0,1)$$: $$\displaystyle \frac{\sqrt{(1- \delta)^{2k} + 4\delta} \ - (1-\delta)^k}{2\delta}$$
#### romsek
Math Team
Well it's bounded below by 1. I suspect it increases without bound as $k\to \infty$
ensbana
#### idontknow
$\lim_{\delta \rightarrow 1 } \frac{\sqrt{(1- \delta)^{2k} + 4\delta} \ - (1-\delta)^k}{2\delta} =0$.
$\lim_{\delta \rightarrow 0 } \frac{\sqrt{(1- \delta)^{2k} + 4\delta} \ - (1-\delta)^k}{2\delta} =\lim_{\delta \rightarrow 0 } \dfrac{4\delta }{4\delta }=1$.
ensbana
#### romsek
Math Team
$\lim_{\delta \rightarrow 1 } \frac{\sqrt{(1- \delta)^{2k} + 4\delta} \ - (1-\delta)^k}{2\delta} =0$.
$\lim_{\delta \rightarrow 0 } \frac{\sqrt{(1- \delta)^{2k} + 4\delta} \ - (1-\delta)^k}{2\delta} =\lim_{\delta \rightarrow 0 } =\lim_{\delta \rightarrow 0 } \dfrac{4\delta }{2\delta }=2$.
Mathematica returns a limit of 1 as $\delta \to 0$
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https://www.speedsolving.com/competitions/showWeeks.php?week=44&year=2012&selectEvent=14 | 1,618,192,422,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038065903.7/warc/CC-MAIN-20210411233715-20210412023715-00468.warc.gz | 1,114,363,801 | 17,446 | Week 2012-44
38 competed in an event so far
2x2x2
# Name Result Solves Comment
12.642.53, 2.19, 2.62, 2.78, 3.85
22.812.50, 1.81, 3.41, 2.57, 3.37
33.034.50, 1.79, 3.36, 3.16, 2.58
43.183.18, 2.54, 3.32, 3.05, 3.53
53.203.34, 2.27, 3.21, 3.06, 5.03Very nice average!
63.303.63, 8.14, 2.97, 3.30, 1.63
73.573.53, 4.00, 4.33, 3.05, 3.19
83.793.80, 2.33, 3.50, 4.08, 4.46
94.053.53, 3.55, 3.68, 8.02, 4.91
104.104.72, 3.71, 3.61, 4.23, 4.36
114.155.90, 2.75, 3.51, 8.17, 3.03
124.193.50, 4.13, 3.78, 5.03, 4.65
134.224.04, 2.60, 4.43, 5.27, 4.19
144.324.22, 3.80, 4.05, 4.68, 5.46
154.514.54, 4.34, 3.41, 4.85, 4.65super easy scrambles
165.595.69, 6.36, 3.63, 5.33, 5.75
175.925.90, 4.43, 6.50, 6.43, 5.44Lockup...Lockup...Lockup...
185.936.68, 5.85, 5.70, 4.87, 6.24
196.00DNF, 5.46, 5.38, 6.21, 6.34
206.767.56, 6.07, 6.65, 5.25, 8.43
217.216.30, 6.80, 8.46, 6.38, 8.61
228.057.21, 6.38, 9.03, 9.00, 7.93I need a better cube!
238.107.22, 4.75, 10.08, 7.69, 9.40
248.659.04, 7.07, 8.14, 10.43, 8.77
2519.2814.83, 18.64, 10.56, 24.36, 24.85
3x3x3
# Name Result Solves Comment
19.368.43, 10.43, 9.99, 9.66, 8.40Great average!
29.9911.00, 9.78, 8.31, 10.89, 9.30
310.019.43, 9.14, 9.95, 10.65, 11.39
410.0710.30, 8.35, 10.32, 9.76, 10.16
510.6010.11, 10.38, 11.22, 10.19, 11.47
611.5512.30, 10.83, 12.43, 11.52, 10.06
711.8511.68, 12.55, 12.16, 11.70, 11.37
812.8412.68, 12.40, 13.43, 11.47, 13.97
913.3013.40, 12.97, 15.03, 13.53, 11.36good :)
1013.7014.08, 14.67, 12.35, 8.20, 14.85
1113.9914.16, 13.30, 14.50, 15.55, 13.11
1214.4119.09, 14.96, 14.06, 14.22, 13.05Better warm up next time
1316.1714.98, 17.00, 16.28, 17.67, 15.23
1417.0217.58, 17.93, 15.31, 17.47, 16.02
1517.0216.81, 17.44, 18.64, 15.65, 16.82
1617.0317.28, 17.88, 15.93, 15.74, 21.41
1718.0715.79, 18.26, 23.14, 15.40, 20.15
1818.5415.30, 19.76, 19.75, 16.10, 21.00
1918.8819.12, 19.56, 17.95, 17.75, 20.50
2019.3114.65, 19.62, 19.13, 20.99, 19.18Not sure what happened in the first solve? :)
2119.4625.94, 17.31, 22.38, 18.68, 17.19
2219.6120.36, 16.25, 16.63, 21.83, DNFBleh CFOP
2320.3123.96, 20.69, 18.07, 19.41, 20.84not bad :)
2421.7419.76, 23.20, 21.37, 20.65, 25.62
2521.7924.40, 23.14, 18.29, 20.46, 21.76
2622.3920.76, 23.83, 17.74, 22.58, 27.27
2722.4123.66, 22.05, 21.52, 16.47, 26.63
2826.7221.72, 29.31, 25.58, 33.43, 25.28ZZ
2931.6635.06, 34.96, 25.25, 29.93, 30.08
3039.7039.89, 38.21, 41.59, 32.29, 41.01
4x4x4
# Name Result Solves Comment
139.3438.01, 47.63, 38.97, 41.03, 30.63
239.7039.71, 38.82, 43.69, 40.00, 39.39
349.6852.47, 53.31, 47.50, 49.06, 47.11
451.9456.97, 50.49, 51.85, 53.47, 48.12lol
554.4054.57, 55.97, 49.03, 55.55, 53.08
656.8855.26, 57.22, 1:00.94, 50.74, 58.16DP, DP, PP, PP, NP
759.761:06.65, 1:06.63, 54.75, 55.82, 56.82
91:07.431:24.67, 1:10.08, 1:06.33, 1:05.87, 1:05.06
101:07.471:00.43, 1:11.97, 1:03.11, 1:07.53, 1:11.78
111:10.611:08.86, 1:14.67, 1:06.33, 1:11.97, 1:11.01
121:10.841:15.88, 1:07.26, 1:12.54, 1:07.71, 1:12.28
131:16.031:22.32, 1:05.12, 1:11.41, 1:16.65, 1:20.02Didnt like the first and last solve but felt really nice in the second.
141:22.511:11.75, 1:26.38, 1:32.00, 1:16.03, 1:25.13
152:04.921:43.20, 2:01.28, 2:03.65, DNF, 2:09.83
162:15.032:06.19, 2:49.80, 2:26.08, 1:57.58, 2:12.81
172:21.852:22.59, 2:29.57, 3:04.99, 2:11.58, 2:13.40
182:22.102:34.84, 2:14.97, 2:13.15, 2:51.53, 2:16.50
5x5x5
# Name Result Solves Comment
11:23.271:24.81, 1:25.21, 1:21.74, 1:19.77, 1:23.27
21:24.661:24.15, 1:26.28, 2:07.06, 1:17.40, 1:23.56
31:28.531:38.38, 1:28.44, 1:25.91, 1:28.30, 1:28.84
41:40.341:39.15, 1:40.05, 1:44.08, 1:40.93, 1:40.03
51:44.121:48.67, 1:40.87, 1:34.23, 1:47.48, 1:44.00sucks
61:50.311:49.23, 1:53.00, 1:48.69, 1:56.33, 1:45.40
72:01.702:00.36, 2:02.16, 1:58.93, 2:02.58, 2:05.15
82:12.102:12.60, 2:12.97, 2:23.02, 2:09.90, 2:10.74pb avarage
92:38.552:21.88, 2:41.66, 2:52.12, 2:10.23, DNF
102:42.232:54.31, 2:50.30, 2:33.80, 2:36.26, 2:40.12
114:06.853:54.81, 4:06.46, 3:57.68, 4:16.41, 4:27.25
124:52.914:49.36, 4:57.69, 4:57.03, 4:52.33, 4:27.22
6x6x6
# Name Result Solves Comment
13:02.303:11.36, 2:56.06, 2:35.53, 3:17.15, 2:59.47
23:13.053:18.25, 3:10.33, 3:16.80, 3:08.27, 3:12.02
35:05.945:21.42, 5:26.08, 4:42.71, 4:41.48, 5:13.69DP at all (on 3x3 stage)+ 2 5min with problem of centers on edges stage
45:35.424:42.96, 5:33.15, 5:29.59, 5:52.38, 5:43.52first one was lucky :)
cube popped at 4th solve -.-
7x7x7
# Name Result Solves Comment
14:42.574:06.93, 4:59.96, 4:22.44, 4:50.36, 4:54.91
24:48.764:48.93, 4:49.19, 4:51.53, 4:41.65, 4:48.16
36:55.907:20.34, 7:02.58, 6:58.10, 6:41.07, 6:47.015th: EPLL skip
2x2x2 blindfolded
# Name Result Solves Comment
116.19DNF, DNF, 16.19with orthega, other two were jperm off
222.2824.85, 22.28, 23.48
323.3823.38, 24.42, DNS
423.8923.89, DNF, 40.52
524.75DNF, DNF, 24.75
626.3826.38, 27.51, 29.33
742.3446.61, 42.34, 47.65
81:37.801:37.80, DNF, DNF
91:40.771:40.77, DNF, DNFDid the start wrong, corrected it, did the middle wrong, cor....Success!
101:48.251:48.25, DNF, DNF
3x3x3 blindfolded
# Name Result Solves Comment
158.961:05.57, DNF, 58.96
21:12.121:12.12, 1:20.03, DNFEasy last scramble, rushed it too much. :/
31:35.46DNF, 1:35.46, 2:06.65First solve: [1:18.95, 3E].
41:42.151:42.15, 3:11.43, DNF
51:45.911:45.91, DNS, DNS
61:54.42DNF, 1:54.96, 1:54.42
72:03.362:03.36, DNF, DNF
82:06.55DNF, 2:06.55, DNF
93:13.53DNF, 3:13.53, DNS
104:16.65DNF, DNF, 4:16.65Last one was safety solve.
11DNFDNF, DNF, DNF
11DNFDNF, DNF, DNF
11DNFDNF, DNF, DNFnoooo
4x4x4 blindfolded
# Name Result Solves Comment
16:21.726:21.72, DNS, DNS[2:51].
No time for more than one.
38:25.188:46.67, 8:25.18, DNS
411:09.6511:09.65, DNF, 14:14.85
512:20.52DNF, 16:17.45, 12:20.52
6DNFDNF, DNS, DNS1. 10:24.68[6:31.62]
Missed a move somewhere.
5x5x5 blindfolded
# Name Result Solves Comment
112:49.50DNF, DNF, 12:49.50[13:45 close], [14:50 bad] [memo 6:50]
213:57.5713:57.57, DNS, DNS[7:39].
No time for more than one.
323:33.7423:33.74, DNF, DNS1) [10:00]
2) [29:39.56, 9+]
4DNFDNF, DNS, DNS
4DNFDNF, DNF, DNF
6x6x6 blindfolded
# Name Result Solves Comment
1DNFDNF
7x7x7 blindfolded
# Name Result Solves Comment
3x3x3 multiple blindfolded
# Name Result Solves Comment
125/8 58:52
202/4 28:00
303/6 38:53
3x3x3 one-handed
# Name Result Solves Comment
116.0915.08, 15.92, 16.48, 16.67, 15.88
216.2116.86, 15.91, 17.28, 13.87, 15.86
317.1817.56, 16.92, 17.28, 17.33, 16.26
418.3220.18, 17.41, 17.80, 18.90, 18.25nice
519.2823.65, 21.72, 19.34, 13.10, 16.79
619.7718.87, 24.74, 19.38, 19.18, 20.75
721.1322.13, 22.83, 17.84, 28.16, 18.44
822.5923.45, 26.17, 19.27, 21.86, 22.45
923.5323.68, 22.77, 24.43, 20.46, 24.13
1024.5025.16, 21.06, 24.86, 24.43, 24.21
1125.4622.91, 23.19, 28.75, 24.44, 35.61
1225.8730.03, 24.80, 19.06, 24.81, 28.00lol single skip oll
1329.4930.95, 24.60, 26.96, 30.55, 32.34
1430.6034.08, 30.04, 24.38, 27.67, 44.63
1534.2541.72, 33.24, 29.35, 31.32, 38.20
1637.6236.17, 35.46, 39.69, 37.01, 46.96
1737.7534.69, 33.68, 37.44, 41.11, 49.631st: EPLL skip
2nd, 3rd: full step
1842.4342.36, 22.79, 57.39, 47.04, 37.89
1951.2351.61, 41.07, 53.93, 50.29, 51.80
2059.241:18.22, 50.91, 57.02, 46.93, 1:09.78
3x3x3 With feet
# Name Result Solves Comment
144.1143.44, 41.50, 39.91, 47.38, 48.43
250.8353.55, 46.80, 43.82, 52.14, 59.07Messed up G-perm on the 5th -_-
31:11.641:12.16, 1:03.93, 1:09.18, 1:13.59, 1:21.36
41:25.611:11.82, 1:24.40, 1:30.54, 1:29.30, 1:23.13
3x3x3 Match the scramble
# Name Result Solves Comment
12:33.363:29.85, 2:53.90, 2:24.49, 2:21.69, DNF
2DNF59.19, DNF, 59.33, DNF, 1:17.34
3x3x3 Fewest moves
# Name Result Solves Comment
12828
23131
34747
Scramble 1. B2 R2 D' F2 D' B2 U F2 L2 U2 R' U R B' L B D B L' F'
# Name Moves Solution Explanation
128B D L' B L2 F' L2 B' L2 F' U L' F R F' L F R' U L F' L U' L F R2 D' F2
231D2 B' F D F' D' R D2 U' B2 U D' B2 D B D F' U R2 U' F D B2 D B' F R2 B2 U' R U
347F L2 F2 U R B L2 D2 R' D R' D' R' F2 R F2 x2 U' R' U R U2 R2 U R2 U R' Fw R U R' U' R U R' U' Fw' U2 R' U L' U2 R U' R' U2 L R
2-3-4 Relay
# Name Result Solves Comment
11:03.571:03.57
21:06.801:06.80
31:07.611:07.61
41:15.911:15.91
51:16.401:16.40
61:36.661:36.66pb, hoping for sub 1:30.
71:40.661:40.66
82:33.712:33.71
92:52.272:52.27
2-3-4-5 Relay
# Name Result Solves Comment
12:19.932:19.93
22:47.402:47.40
32:54.652:54.65
43:10.213:10.21
53:54.533:54.53pb aswell
66:00.136:00.13
Clock
# Name Result Solves Comment
110.019.55, 10.67, 10.26, 10.21, 8.87
211.7112.47, 10.64, 13.38, 12.01, 9.90
313.3715.36, 13.03, 12.66, 14.41, 11.65
415.52DNF, 14.95, 11.98, 16.64, 14.97
515.6516.58, 22.52, 15.43, 14.94, 12.36Last solve could've been sub-10, but whatever.
627.8916.99, 19.05, DNF, 43.10, 21.52
Megaminx
# Name Result Solves Comment
11:38.131:33.34, 1:42.19, 1:21.93, 2:22.87, 1:38.87
22:11.782:17.63, 2:08.45, 2:09.26, 2:25.33, 2:05.85
32:46.742:51.94, 2:27.03, 2:52.90, 2:37.85, 2:50.43
43:24.763:22.82, 3:31.73, 3:19.74, 3:59.74, 3:15.15
Pyraminx
# Name Result Solves Comment
13.913.71, 4.93, 4.25, 3.60, 3.77
25.393.68, 4.77, 7.03, 6.83, 4.56
35.516.22, 8.56, 5.76, 4.34, 4.55
45.784.10, 6.12, 6.71, 6.77, 4.50
55.814.68, 5.33, 11.27, 7.28, 4.81
66.615.97, 7.25, 6.80, 7.06, 3.65~Shengshou~
76.767.00, 4.77, 12.75, 8.05, 5.22
87.156.55, 7.53, 9.19, 7.36, 6.02
97.947.81, 7.14, 8.87, 8.44, 7.58
108.528.95, 8.07, 9.94, 8.55, 5.95
1110.357.61, 7.79, 15.74, 14.43, 8.83
1216.3411.25, 23.28, 20.72, 15.18, 13.13
Square-1
# Name Result Solves Comment
120.1725.24, 17.73, 26.56, 17.55, 15.59Yay.
226.4927.06, 38.44, 27.27, 23.50, 25.13
330.2131.19, 18.38, 35.93, 23.50, 49.21
432.2526.86, 28.48, 31.72, 36.54, 38.17
553.7555.54, 1:01.49, 40.25, 44.22, DNF
62:04.362:13.05, 2:05.87, 2:43.89, 1:54.16, 1:29.63
Skewb
# Name Result Solves Comment
126.6027.52, 25.78, 26.50, 34.46, 17.36
237.0957.72, 25.59, 25.72, 27.83, DNF
Magic
# Name Result Solves Comment
10.870.89, 0.85, 0.83, 0.86, 1.87
21.001.03, 1.91, 1.02, 0.94, 0.94
Master Magic
# Name Result Solves Comment
12.132.11, 2.16, 2.13, 2.06, 6.16
22.242.16, 2.26, 4.43, 2.29, 1.93
32.942.74, 3.51, 2.73, 3.36, 2.53 | 6,109 | 10,045 | {"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} | 2.59375 | 3 | CC-MAIN-2021-17 | longest | en | 0.508135 |
https://estebantorreshighschool.com/equation-help/rydberg-equation-energy.html | 1,618,114,315,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038060927.2/warc/CC-MAIN-20210411030031-20210411060031-00594.warc.gz | 353,901,393 | 29,802 | ## What is Rh in Rydberg equation?
The wavelengths of these lines are given by 1/λ = RH (1/4 − 1/n2), where λ is the wavelength, RH is the Rydberg constant, and n is the level of the original orbital.
## What is n1 and n2 in Rydberg equation?
n1 and n2 are integers and n2 is always greater than n1. The modern value of Rydberg constant is known as 109677.57 cm1 and it is the most accurate physical constant. According to Paschen series, n1 = 3 and n2 = 4, 5… λ = 1.282 x 10-4 cm = 1282 nm which is in near infrared region.
## What is the value of 1 by Rydberg constant?
Here R is the Rydberg constant 1, which has been precisely measured and found to have the value R = 10973731.5683 ± 0.0003 m1. The variable n is any integer equal to or greater than 3.
## Why does Rydberg constant have 2 values?
Rydberg constant is the value of highest wavenumber(inverse of wavelength) that any photon can emit. I.e. 1.09 x 10^7 (m^-1). So, two values you can see are not for the same constant.
## What is Rydberg equation used for?
The Rydberg formula is a mathematical formula used to predict the wavelength of light resulting from an electron moving between energy levels of an atom.
## What is 1 Rydberg?
For a single electron and proton (ground state of hydrogen), the Rydberg unit of energy is the binding energy between the electron and proton. At this energy, the Bohr radius is calculated as the position where two forces are equal. The Rydberg unit of energy is this force exerted over a distance of the Bohr radius.
## Does Rydberg equation only work hydrogen?
The Rydberg equation only works for the Hydrogen and Hydrogen-like (species with only one electron) however because Bohr model of the atom breaks down when there are more than two electrons. Consequently, the Schrodinger equation provide a mathematical model of the atom: H = E .
## Is Rydberg constant universal?
λ = hm2/(m2 − 4), Independently, Janne Rydberg analyzed the spectra of many elements. No turns out to be a universal constant, the Rydberg constant. Rydberg showed that the Balmer series of hydrogen is a special case with m’=0, no=4No.
## Why is the Rydberg constant important?
The Rydberg constant is one of the most important constants of atomic physics because of its connection with the fundamental atomic constants (e, h, me, c) and because of the high accuracy with which it can be determined. Neither the melting point of water nor the speed of sound is a fundamental constant.
## What does Rydberg constant represent?
The Rydberg constant represents the limiting value of the highest wavenumber of any photon that can be emitted from the hydrogen atom, or, alternatively, the wavenumber of the lowest-energy photon capable of ionizing the hydrogen atom from its ground state.
## Does Rydberg constant has unit of energy?
and electron spin g-factor are the most accurately measured physical constants. In atomic physics, Rydberg unit of energy, symbol Ry, corresponds to the energy of the photon whose wavenumber is the Rydberg constant, i.e. the ionization energy of the hydrogen atom in a simplified Bohr model.
## Why is the Rydberg constant negative?
Re: Negative Sign in front of Rydberg constant In this case, a negative is in front because you’re subtracting nInitial-nFinal which will result in a negative number so adding a negative in the front will flip the sign and make it positive.
## What is the inverse of Rydberg constant?
We know alpha (the fine structure constant) is a dimensionless number (i.e. has no physical units) so the left hand side can be seen as a ratio of a length characteristic of the electron (lambda(e), the Compton wavelength), to a length characteristic of the photons that can be spit out of an electron (i.e. the inverse
### Releated
#### Tensile stress equation
What is the formula for tensile stress? Difference Between Tensile Stress And Tensile Strength Tensile stress Tensile strength The formula is: σ = F/A Where, σ is the tensile stress F is the force acting A is the area The formula is: s = P/a Where, s is the tensile strength P is the force […]
#### Quotient rule equation
What is the formula for Quotient? The answer after we divide one number by another. dividend ÷ divisor = quotient. Example: in 12 ÷ 3 = 4, 4 is the quotient. What is quotient rule in math? The quotient rule is a formal rule for differentiating problems where one function is divided by another. It […] | 1,074 | 4,438 | {"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} | 3.765625 | 4 | CC-MAIN-2021-17 | longest | en | 0.919838 |
https://forum.knittinghelp.com/t/durrow-cable/76733 | 1,606,956,755,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141717601.66/warc/CC-MAIN-20201203000447-20201203030447-00015.warc.gz | 254,289,163 | 4,996 | # Durrow Cable
I’m trying to knit the Durrow Cable by girlfromauntie.
It says to cast on 15 sts and you actually increase 12 (making it 27 instead of the allotted 23). It has double increases on rows 1, 2, 3, 4, 7, and 8.
What am I doing wrong?
Here’s a link to the pattern. I don’t see anywhere you CO 15 sts or where you increase as you describe. What part of the pattern are you?
It’s actually a completely different pattern. I would have posted a link but I’m not aloud to since I’m a new person. It’s a cable pattern and I’m trying to figure out a way to make a scarf incorporating said cable.
http://www.girlfromauntie.com/patterns/durrow-cable/#more-54
Oh just the cable pattern. Okay, you do the increases, but then decrease later. I’m not too good at reading cable patterns, maybe someone else can help you out.
But the ‘no stitch’ boxes mean ‘there’s no stitch on your needle here’ so you skip the box and do the next st according to what’s in the next box. They’re there to account for the new sts you get with the increases.
So on R1, you p12, inc, p2; the row 2 is k2, p1, inc, p1, k12. I don’t know what you’re doing wrong… What happens - do you have too many sts or what?
I do know how to read cable patterns, and it states in the blog post to cast on 15 sts and the |-| and -|- are double increases.
make 2 (right side): k1, k1b into the same stitch; then pick up the vertical strand running downwards between these two stitches just made, twist and knit
make 2 (wrong side): p1, k1, p1 into the same stitch
The problem is the boxes over on the right. Suddenly over on row 3, 4sts disappear. There’s no decrease in rows1-2 but the “no stitch” boxes are there in the next row. Perhaps it is an error in the patterns and it should be cast on 11sts. Then the no stitch boxes on the right should be present in rows 1 and 2, that is from the beginning of the pattern.
That makes a lot of sense. Thanks
Ahhh, that could explain it.
Snazzy Butterfly, why don’t you email the blogger and ask for clarification on those rows? Also take a look at the projects for it on Ravelry; some of the people posted notes which may help you. There’s a couple people who posted in the comments section on the same problem you’re running into. The designer usually replies back so if you contact her by pm, she’ll probably respond. Apparently there is an error in the chart. | 618 | 2,382 | {"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} | 3.0625 | 3 | CC-MAIN-2020-50 | latest | en | 0.916907 |
https://www.crazy-numbers.com/en/21366 | 1,708,707,974,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474440.42/warc/CC-MAIN-20240223153350-20240223183350-00291.warc.gz | 732,452,655 | 6,402 | # Everything about number 21366
Discover a lot of information on the number 21366: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 21366
Questions and answers
Is 21366 a prime number? No
Is 21366 a perfect number? No
Number of divisors 12
List of dividers 1, 2, 3, 6, 9, 18, 1187, 2374, 3561, 7122, 10683, 21366
Sum of divisors 46332
Prime factorization 2 x 32 x 1187
Prime factors 2, 3, 1187
## How to write / spell 21366 in letters?
In letters, the number 21366 is written as: Twenty-one thousand three hundred and sixty-six. And in other languages? how does it spell?
21366 in other languages
Write 21366 in english Twenty-one thousand three hundred and sixty-six
Write 21366 in french Vingt et un mille trois cent soixante-six
Write 21366 in spanish Veintiuno mil trescientos sesenta y seis
Write 21366 in portuguese Vinte e um mil trezentos sessenta e seis
## Decomposition of the number 21366
The number 21366 is composed of:
1 iteration of the number 2 : The number 2 (two) represents double, association, cooperation, union, complementarity. It is the symbol of duality.... Find out more about the number 2
1 iteration of the number 1 : The number 1 (one) represents the uniqueness, the unique, a starting point, a beginning.... Find out more about the number 1
1 iteration of the number 3 : The number 3 (three) is the symbol of the trinity. He also represents the union.... Find out more about the number 3
2 iterations of the number 6 : The number 6 (six) is the symbol of harmony. It represents balance, understanding, happiness.... Find out more about the number 6
## Mathematical representations and links
Other ways to write 21366
In letter Twenty-one thousand three hundred and sixty-six
In roman numeral
In binary 101001101110110
In octal 51566
In hexadecimal 5376
In US dollars USD 21,366.00 (\$)
In euros 21 366,00 EUR (€)
Some related numbers
Previous number 21365
Next number 21367
Next prime number 21377
## Mathematical operations
Operations and solutions
21366*2 = 42732 The double of 21366 is 42732
21366*3 = 64098 The triple of 21366 is 64098
21366/2 = 10683 The half of 21366 is 10683.000000
21366/3 = 7122 The third of 21366 is 7122.000000
213662 = 456505956 The square of 21366 is 456505956.000000
213663 = 9753706255896 The cube of 21366 is 9753706255896.000000
√21366 = 146.17113258096 The square root of 21366 is 146.171133
log(21366) = 9.9695561525058 The natural (Neperian) logarithm of 21366 is 9.969556
log10(21366) = 4.3297232240579 The decimal logarithm (base 10) of 21366 is 4.329723
sin(21366) = -0.028359133179553 The sine of 21366 is -0.028359
cos(21366) = -0.99959779889979 The cosine of 21366 is -0.999598
tan(21366) = 0.0283705438435 The tangent of 21366 is 0.028371 | 872 | 2,837 | {"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} | 2.96875 | 3 | CC-MAIN-2024-10 | latest | en | 0.756081 |
https://dmoj.ca/problem/coci14c2p1 | 1,726,098,267,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651405.61/warc/CC-MAIN-20240911215612-20240912005612-00656.warc.gz | 187,449,492 | 6,376 | ## COCI '14 Contest 2 #1 Mobitel
View as PDF
Points: 5 (partial)
Time limit: 1.0s
Memory limit: 32M
Problem type
Grasshopper Marko was jumping happily all over the meadow. He wasn't being careful and his Nokia 3310 fell into a puddle. His mobile phone is now acting funny! The contacts got wet and the keyboard works in a completely unpredictable manner! All the numerical keys broke down. When we press one of them, the mobile phone acts as if we pressed another key. Luckily, there are no two keys that are acting the same so Marko can still write all the letters.
Grasshopper Marko was experimenting a bit and found out how each key acts. Now he wants to write a message to his girlfriend. Since he is just a grasshopper, you will do that for him.
To all those who don't remember how mobile phones with keys work, here is a short description.
1 2abc 3def 4ghi 5jkl 6mno 7pqrs 8tuv 9wxyz * 0 #
Keyboard on a very old mobile phone.
The image shows keys with letters that we can get by pressing that key (on a working mobile phone that didn't fall into a puddle). For example, if we want to get letter a we will press key 2 once, and if we want letter b we will press the key 2 twice. If we want to write two letters from the same key consecutively, we have to press the pound key (#) exactly once. For example, if we want to write the string klor we will press the keys in the following order: 55#555666777
#### Input Specification
The first line of input contains integers. The first integer marks the behaviour of key 1, the second the behaviour of key 2, the third the behaviour of key 3, and so on.
Marko is not using keys * and 0 because he is a grasshopper. Key # can't get broken.
The second line of input contains a string consisting of only lowercase letters of the English alphabet. The length of the word won't exceed 100 characters.
#### Output Specification
The first and only line of output must contain the sequence of keys you need to press in order to write Marko's message.
#### Sample Input 1
2 3 4 5 6 7 8 9 1
klor
#### Sample Output 1
44#444555666
#### Explanation of Output for Sample Input 1
All of the keys are shifted one place to the right so the output differs a little bit from the example in the task statement.
#### Sample Input 2
7 8 9 1 2 3 6 5 4
djevojka
#### Sample Output 2
68662227778#885
#### Sample Input 3
9 8 7 6 5 4 3 2 1
skakavac
#### Sample Output 3
33335585582228#888
## Comments
There are no comments at the moment. | 672 | 2,492 | {"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} | 3.015625 | 3 | CC-MAIN-2024-38 | latest | en | 0.918637 |
https://imathworks.com/cv/solved-graph-for-relationship-between-two-ordinal-variables/ | 1,726,811,587,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652138.47/warc/CC-MAIN-20240920054402-20240920084402-00705.warc.gz | 288,394,842 | 8,687 | # Solved – Graph for relationship between two ordinal variables
What is an appropriate graph to illustrate the relationship between two ordinal variables?
A few options I can think of:
1. Scatter plot with added random jitter to stop points hiding each other. Apparently a standard graphic – Minitab calls this an "individual values plot". In my opinion it may be misleading as it visually encourages a kind of linear interpolation between ordinal levels, as if the data were from an interval scale.
2. Scatter plot adapted so that size (area) of point represents frequency of that combination of levels, rather than drawing one point for each sampling unit. I have occasionally seen such plots in practice. They can be hard to read, but the points lie on a regularly-spaced lattice which somewhat overcomes the criticism of the jittered scatter plot that it visually "intervalises" the data.
3. Particularly if one of the variables is treated as dependent, a box plot grouped by the levels of the independent variable. Likely to look terrible if the number of levels of the dependent variable is not sufficiently high (very "flat" with missing whiskers or even worse collapsed quartiles which makes visual identification of median impossible), but at least draws attention to median and quartiles which are relevant descriptive statistics for an ordinal variable.
4. Table of values or blank grid of cells with heat map to indicate frequency. Visually different but conceptually similar to the scatter plot with point area showing frequency.
Are there other ideas, or thoughts on which plots are preferable? Are there any fields of research in which certain ordinal-vs-ordinal plots are regarded as standard? (I seem to recall frequency heatmap being widespread in genomics but suspect that is more often for nominal-vs-nominal.) Suggestions for a good standard reference would also be very welcome, I am guessing something from Agresti.
If anyone wants to illustrate with a plot, R code for bogus sample data follows.
"How important is exercise to you?" 1 = not at all important, 2 = somewhat unimportant, 3 = neither important nor unimportant, 4 = somewhat important, 5 = very important.
"How regularly do you take a run of 10 minutes or longer?" 1 = never, 2 = less than once per fortnight, 3 = once every one or two weeks, 4 = two or three times per week, 5 = four or more times per week.
If it would be natural to treat "often" as a dependent variable and "importance" as an independent variable, if a plot distinguishes between the two.
importance <- rep(1:5, times = c(30, 42, 75, 93, 60))
often <- c(rep(1:5, times = c(15, 07, 04, 03, 01)), #n=30, importance 1
rep(1:5, times = c(10, 14, 12, 03, 03)), #n=42, importance 2
rep(1:5, times = c(12, 23, 20, 13, 07)), #n=75, importance 3
rep(1:5, times = c(16, 14, 20, 30, 13)), #n=93, importance 4
rep(1:5, times = c(12, 06, 11, 17, 14))) #n=60, importance 5
running.df <- data.frame(importance, often)
cor.test(often, importance, method = "kendall") #positive concordance
plot(running.df) #currently useless
A related question for continuous variables I found helpful, maybe a useful starting point: What are alternatives to scatterplots when studying the relationship between two numeric variables?
A spineplot (mosaic plot) works well for the example data here, but can be difficult to read or interpret if some combinations of categories are rare or don't exist. Naturally it's reasonable, and expected, that a low frequency is represented by a small tile, and zero by no tile at all, but the psychological difficulty can remain. It's also natural that people fond of spineplots choose examples which work well for their papers or presentations, but I've often produced examples that were too messy to use in public. Conversely, a spineplot does use the available space well.
Some implementations presuppose interactive graphics, so that the user can interrogate each tile to learn more about it.
An alternative which can also work quite well is a two-way bar chart (many other names exist).
See for example tabplot within http://www.surveydesign.com.au/tipsusergraphs.html
For these data, one possible plot (produced using tabplot in Stata, but should be easy in any decent software) is
The format means it is easy to relate individual bars to row and column identifiers and that you can annotate with frequencies, proportions or percents (don't do that if you think the result is too busy, naturally).
Some possibilities:
1. If one variable can be thought of a response to another as predictor, then it is worth thinking of plotting it on the vertical axis as usual. Here I think of "importance" as measuring an attitude, the question then being whether it affects behaviour ("often"). The causal issue is often more complicated even for these imaginary data, but the point remains.
2. Suggestion #1 is always to be trumped if the reverse works better, meaning, is easier to think about and interpret.
3. Percent or probability breakdowns often make sense. A plot of raw frequencies can be useful too. (Naturally, this plot lacks the virtue of mosaic plots of showing both kinds of information at once.)
4. You can of course try the (much more common) alternatives of grouped bar charts or stacked bar charts (or the still fairly uncommon grouped dot charts in the sense of W.S. Cleveland). In this case, I don't think they work as well, but sometimes they work better.
5. Some might want to colour different response categories differently. I've no objection, and if you want that you wouldn't take objections seriously any way.
The strategy of hybridising graph and table can be useful more generally, or indeed not what you want at all. An often repeated argument is that the separation of Figures and Tables was just a side-effect of the invention of printing and the division of labour it produced; it's once more unnecessary, just as it was to manuscript writers putting illustrations exactly how and where they liked. | 1,351 | 6,009 | {"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} | 3.078125 | 3 | CC-MAIN-2024-38 | latest | en | 0.939073 |
https://forum.processing.org/two/discussion/10362/help-with-a-multidimensional-array | 1,618,210,705,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066613.21/warc/CC-MAIN-20210412053559-20210412083559-00325.warc.gz | 369,726,211 | 14,049 | #### Howdy, Stranger!
We are about to switch to a new forum software. Until then we have removed the registration on this forum.
# Help with a multidimensional array
edited January 2016
``````int drk;
int lgt;
int normal;
int changer;
int changeg;
int changeb;
float r=random(0,255);
float g=random(0,255);
float b=random(0,255);
float[] [] [] arr = new float[r][g][b];
``````
Tagged:
• It shows an error at the array saying "cannot convert from float to int"
• Yes, you cannot have 2.3256 entries in an array, so r, g and b should be int, and you should initialize them with `int(random(255));`
• So change it to int r=int(random(255));
• @PhiLho thanks
• edited January 2016
• RGB should be stored in separate arrays, not in 1 3D array!
• Better yet, make a class to store those 3 properties!
``````/**
* Colour Class (v1.4)
* by GoToLoop (2015/Apr/16)
* forum.Processing.org/two/discussion/10362/help-with-a-multidimensional-array
* forum.Processing.org/two/discussion/14614/arrays-and-rows-colums
*/
class Colour implements Cloneable {
protected byte r, g, b;
Colour() {
}
Colour(color R, color G, color B) {
set(R, G, B);
}
Colour set(color R, color G, color B) {
r = (byte) (R & 0xFF);
g = (byte) (G & 0xFF);
b = (byte) (B & 0xFF);
return this;
}
Colour randomRGB() {
return fromColour((color) random(~#000000 + 1));
}
Colour fromColour(color c) {
return set(c>>020, c>>010, c);
}
color toColour() {
return hashCode() | #000000;
}
color red() {
return r & 0xff;
}
color green() {
return g & 0xff;
}
color blue() {
return b & 0xff;
}
Colour red(color R) {
r = (byte) (R & 0xFF);
return this;
}
Colour green(color G) {
g = (byte) (G & 0xFF);
return this;
}
Colour blue(color B) {
b = (byte) (B & 0xFF);
return this;
}
@ Override Colour clone() {
try {
return (Colour) super.clone();
}
catch (CloneNotSupportedException cause) {
throw new RuntimeException(cause);
}
}
@ Override int hashCode() {
return r<<020 | g<<010 | b;
}
@ Override boolean equals(Object o) {
return o.hashCode() == hashCode();
}
@ Override String toString() {
return '#' + hex(hashCode(), 6) + " - R: " + nf(r & 0xff, 3)
+ ", G: " + nf(g & 0xff, 3) + ", B: " + nf(b & 0xff, 3);
}
}
final Colour c = new Colour().clone();
void setup() {
size(300, 200);
frameRate(1);
}
void draw() {
println(c.randomRGB());
color rgb = c.toColour();
background(rgb);
frame.setTitle(hex(rgb, 6));
}
``````
• There's already a built in color type (actually an integer) - just use that.
• But having seen the whole question, the whole approach is wrong. You don't need an array at all. (And please don't post multiple questions, it just splits up information and creates work)
http://forum.processing.org/two/discussion/10364/getting-a-random-color-checkerboard-with-random-fill
• edited April 2015
here comes a checker board....
regarding 3D array there is a big misunderstanding...
1. you want to store three values rgb, but you can just store the data type color
even when you want to store 3 values, you don't need a 3D array - you just would (!) need three parallel 1D arrays.
I'll explain.
• A 1D array is a list of values. Those values can be color or a class. You need one index to get a value.
• A 2D value is a grid (like a chess board). You still only get one value (not two!), the value being the cell of the grid. But you now need 2 indexes to get the cell position / value.
• A 3D value is a stack of grids (like 8 chess board on top of each other). You still only get one value (not three!), the value being the cell of the grid. But you now need 3 indexes to get the cell position / value: row, column and etage (floor in the stack).
Since you want a grid you can use a 2D array of color:
`color[][] colors = new color[][];`
my example comes without that.
The three parallel 1D arrays I spoke about; instead of a class with three values you can use parallel arrays for each value. But as said it doesn't apply here since you store only one value (color).
;-)
``````int changer;
int changeg;
int changeb;
void draw() {
int r=int(random(0, 255));
int g=int(random(0, 255));
int b=int(random(0, 255));
float[] [] [] colors = new float[r][g][b];
for (int x = 0; x < width; x+=10) {
for (int y = 0; y < height; y+=10) {
noStroke();
// fill(r, g, b);
if ( (x%20==0 && y%20==0) ||
(x%20!=0 && y%20!=0) )
fill(2, 33, 22);
else
fill(222, 33, 22);
rect(x, y, 10, 10);
}
}
}
void setup() {
size(400, 400);
noLoop();
}
`````` | 1,312 | 4,433 | {"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} | 2.71875 | 3 | CC-MAIN-2021-17 | latest | en | 0.464348 |
https://subscription.packtpub.com/book/programming/9781801073448/1/ch01lvl1sec07/complex-data-types | 1,718,740,268,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861773.80/warc/CC-MAIN-20240618171806-20240618201806-00889.warc.gz | 486,358,110 | 65,012 | #### Overview of this book
Choosing the right data structure is pivotal to optimizing the performance and scalability of applications. This new edition of Hands-On Data Structures and Algorithms with Python will expand your understanding of key structures, including stacks, queues, and lists, and also show you how to apply priority queues and heaps in applications. You’ll learn how to analyze and compare Python algorithms, and understand which algorithms should be used for a problem based on running time and computational complexity. You will also become confident organizing your code in a manageable, consistent, and scalable way, which will boost your productivity as a Python developer. By the end of this Python book, you’ll be able to manipulate the most important data structures and algorithms to more efficiently store, organize, and access data in your applications.
Preface
Free Chapter
Python Data Types and Structures
Introduction to Algorithm Design
Algorithm Design Techniques and Strategies
Stacks and Queues
Trees
Heaps and Priority Queues
Hash Tables
Graphs and Algorithms
Searching
Sorting
Selection Algorithms
String Matching Algorithms
Other Books You May Enjoy
Index
# Complex data types
We have discussed basic data types. Next, we discuss complex data types, which are mapping data types, in other words, dictionary, and set data types, namely, set and frozenset. We will discuss these data types in detail in this section.
## Dictionaries
In Python, a dictionary is another of the important data types, similar to a list, in the sense that it is also a collection of objects. It stores the data in unordered {key-value} pairs; a key must be of a hashable and immutable data type, and value can be any arbitrary Python object. In this context, an object is hashable if it has a hash value that does not change during its lifetime in the program.
Items in the dictionary are enclosed in curly braces, `{}`, separated by a comma, and can be created using the `{key:value}` syntax, as shown below:
``````dict = {
<key>: <value>,
<key>: <value>,
.
.
.
<key>: <value>
}
``````
Keys in dictionaries are case-sensitive, they should be unique, and cannot be duplicated; however, the values in the dictionary can be duplicated. For example, the following code can be used to create a dictionary:
``````my_dict = {'1': 'data',
'2': 'structure',
'3': 'python',
'4': 'programming',
'5': 'language'
}
``````
Figure 1.5 shows the `{key-value}` pairs created by the preceding piece of code:
Figure 1.5: Example dictionary data structure
Values in a dictionary can be fetched based on the key. For example: `my_dict['1']` gives `data` as the output.
The `dictionary` data type is mutable and dynamic. It differs from lists in the sense that dictionary elements can be accessed using keys, whereas the list elements are accessed via indexing. Table 1.4 shows different characteristics of the dictionary data structure with examples:
Item Example Creating a dictionary, and accessing elements from a dictionary ``````person = {} print(type(person)) person['name'] = 'ABC' person['lastname'] = 'XYZ' person['age'] = 31 person['address'] = ['Jaipur'] print(person) print(person['name']) `````` ``````{'name': 'ABC', 'lastname': 'XYZ', 'age': 31, 'address': ['Jaipur']}ABC `````` `in` and `not` `in` operators ``````print('name' in person) print('fname' not in person) `````` ``````True True `````` Length of the dictionary ``````print(len(person)) `````` ``````4 ``````
Table 1.4: Characteristics of dictionary data structures with examples
Python also includes the dictionary methods as shown in Table 1.5:
Function Description Example `mydict.clear()` Removes all elements from a dictionary. ``````mydict = {'a': 1, 'b': 2, 'c': 3} print(mydict) mydict.clear() print(mydict) `````` ``````{'a': 1, 'b': 2, 'c': 3} {} `````` `mydict.get()` Searches the dictionary for a key and returns the corresponding value, if it is found; otherwise, it returns `None`. ``````mydict = {'a': 1, 'b': 2, 'c': 3} print(mydict.get('b')) print(mydict) print(mydict.get('z')) `````` ``````2 {'a': 1, 'b': 2, 'c': 3} None `````` `mydict.items()` Returns a list of dictionary items in (key, value) pairs. ``````print(list(mydict.items())) `````` ``````[('a', 1), ('b', 2), ('c', 3)] `````` `mydict.keys()` Returns a list of dictionary keys. ``````print(list(mydict.keys())) `````` ``````['a', 'b', 'c'] `````` `mydict.values()` Returns a list of dictionary values. ``````print(list(mydict.values())) `````` ``````[1, 2, 3] `````` `mydict.pop()` If a given key is present in the dictionary, then this function will remove the key and return the associated value. ``````print(mydict.pop('b')) print(mydict) `````` ``````{'a': 1, 'c': 3} `````` `mydict.popitem()` This method removes the last key-value pair added in the dictionary and returns it as a tuple. ``````mydict = {'a': 1,'b': 2,'c': 3} print(mydict.popitem()) print(mydict) `````` ``````{'a': 1, 'b': 2} `````` `mydict.update()` Merges one dictionary with another. Firstly, it checks whether a key of the second dictionary is present in the first dictionary; the corresponding value is then updated. If the key is not present in the first dictionary, then the key-value pair is added. ``````d1 = {'a': 10, 'b': 20, 'c': 30} d2 = {'b': 200, 'd': 400} print(d1.update(d2)) print(d1) `````` ``````{'a': 10, 'b': 200, 'c': 30, 'd': 400} ``````
Table 1.5: List of methods of dictionary data structures
## Sets
A set is an unordered collection of hashable objects. It is iterable, mutable, and has unique elements. The order of the elements is also not defined. While the addition and removal of items are allowed, the items themselves within the set must be immutable and hashable. Sets support membership testing operators (`in, not in`), and operations such as intersection, union, difference, and symmetric difference. Sets cannot contain duplicate items. They are created by using the built-in `set()` function or curly braces `{}.` A` set() `returns a set object from an iterable. For example:
``````x1 = set(['and', 'python', 'data', 'structure'])
print(x1)
print(type(x1))
x2 = {'and', 'python', 'data', 'structure'}
print(x2)
``````
The output will be as follows:
``````{'python', 'structure', 'data', 'and'}
<class 'set'>
{'python', 'structure', 'data', 'and'}
``````
It is important to note that sets are unordered data structures, and the order of items in sets is not preserved. Therefore, your outputs in this section may be slightly different than those displayed here. However, this does not affect the function of the operations we will be demonstrating in this section.
Sets are generally used to perform mathematical operations, such as intersection, union, difference, and complement. The `len()` method gives the number of items in a set, and the `in` and `not in` operators can be used in sets to test for membership:
``````x = {'data', 'structure', 'and', 'python'}
print(len(x))
print('structure' in x)
``````
The output will be as follows:
``````4
True
``````
The most commonly used methods and operations that can be applied to `set` data structures are as follows. The union of the two sets, say, `x1` and `x2`, is a set that consists of all elements in either set:
``````x1 = {'data', 'structure'}
x2 = {'python', 'java', 'c', 'data'}
``````
Figure 1.6 shows a Venn diagram demonstrating the relationship between the two sets:
Figure 1.6: Venn diagram of sets
A description of the various operations that can be applied on set type variables is shown, with examples, in Table 1.6:
Description Example sample code Union of two sets, `x1` and `x2`. It can be done using two methods, (1) using the `|` operator, (2) using the `union` method. ``````x1 = {'data', 'structure'} x2 = {'python', 'java', 'c', 'data'} x3 = x1 | x2 print(x3) print(x1.union(x2)) `````` ``````{'structure', 'data', 'java', 'c', 'python'} {'structure', 'data', 'java', 'c', 'python'} `````` Intersection of sets: to compute the intersection of two sets, an `&` operator and the `intersection()` method can be used, which returns a set of items common to both sets, `x1` and `x2`. ``````print(x1.intersection(x2)) print(x1 & x2) `````` ``````{'data'} {'data'} `````` The difference between sets can be obtained using `.difference()` and the subtraction operator, `-`, which returns a set of all elements that are in `x1`, but not in `x2`. ``````print(x1.difference(x2)) print(x1 - x2) `````` ``````{'structure'} {'structure'} `````` Symmetric difference can be obtained using `.symmetric_difference()` , while `^` returns a set of all data items that are present in either `x1` or `x2`, but not both. ``````print(x1.symmetric_difference(x2)) print(x1 ^ x2) `````` ``````{'structure', 'python', 'c', 'java'} {'structure', 'python', 'c', 'java'} `````` To test whether a set is a subset of another, use `.issubset()` and the operator `<=`. ``````print(x1.issubset(x2)) print(x1 <= x2) `````` ``````False False ``````
Table 1.6: Description of various operations applicable to set type variables
### Immutable sets
In Python, `frozenset` is another built-in type data structure, which is, in all respects, exactly like a set, except that it is immutable, and so cannot be changed after creation. The order of the elements is also undefined. A `frozenset` is created by using the built-in function `frozenset()`:
``````x = frozenset(['data', 'structure', 'and', 'python'])
print(x)
``````
The output is:
``````frozenset({'python', 'structure', 'data', 'and'})
``````
Frozensets are useful when we want to use a set but require the use of an immutable object. Moreover, it is not possible to use set elements in the set, since they must also be immutable. Consider an example:
``````a11 = set(['data'])
a21 = set(['structure'])
a31 = set(['python'])
x1 = {a11, a21, a31}
``````
The output will be:
``````TypeError: unhashable type: 'set'
``````
Now with frozenset:
``````a1 = frozenset(['data'])
a2 = frozenset(['structure'])
a3 = frozenset(['python'])
x = {a1, a2, a3}
print(x)
``````
The output is:
``````{frozenset({'structure'}), frozenset({'python'}), frozenset({'data'})}
``````
In the above example, we create a set `x` of frozensets (`a1`, `a2`, and `a3`), which is possible because the frozensets are immutable.
We have discussed the most important and popular data types available in Python. Python also provides a collection of other important methods and modules, which we will discuss in the next section. | 2,803 | 10,505 | {"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} | 2.703125 | 3 | CC-MAIN-2024-26 | latest | en | 0.900575 |
http://mathhelpforum.com/statistics/10731-homework-help-stats-print.html | 1,524,732,881,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948119.95/warc/CC-MAIN-20180426070605-20180426090605-00151.warc.gz | 199,767,211 | 2,913 | # Homework Help with stats
• Jan 27th 2007, 05:06 PM
air2slb
Homework Help with stats
Can someone please show me step by step how to work this problem. I can never get the answer they are looking for. thank you!!!!
Nine percent of men cannot distinguish between the colors red and green. exaclty 1 man of a randomly selected sample of 6 men will be able to distinguish between the colors red and green. what is the request probability?
• Jan 27th 2007, 09:55 PM
CaptainBlack
Quote:
Originally Posted by air2slb
Can someone please show me step by step how to work this problem. I can never get the answer they are looking for. thank you!!!!
Nine percent of men cannot distinguish between the colors red and green. exaclty 1 man of a randomly selected sample of 6 men will be able to distinguish between the colors red and green. what is the request probability?
Let's assume that you are asking for the probability of exactly one of a sample of six men being red-green colour blind.
The number \$\displaystyle k\$ with red-green colour blindness in a sample of \$\displaystyle n\$, where the probability of an individual being colour blind is \$\displaystyle p\$, has a binomial distribution, with probability:
\$\displaystyle
f(k;n,p)={n \choose k} p^k \,(1-p)^{n-k}
\$
Here you want \$\displaystyle k=1\$, \$\displaystyle n=6\$ and \$\displaystyle p=0.09\$, so:
\$\displaystyle
f(1;6,0.09)={6 \choose 1} 0.09^1 \,(1-0.09)^5\approx 0.337
\$ | 393 | 1,449 | {"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} | 3.65625 | 4 | CC-MAIN-2018-17 | latest | en | 0.898277 |
https://us.metamath.org/mpeuni/lfl0sc.html | 1,719,074,329,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00454.warc.gz | 532,637,320 | 6,671 | Mathbox for Norm Megill < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > lfl0sc Structured version Visualization version GIF version
Theorem lfl0sc 36336
Description: The (right vector space) scalar product of a functional with zero is the zero functional. Note that the first occurrence of (𝑉 × { 0 }) represents the zero scalar, and the second is the zero functional. (Contributed by NM, 7-Oct-2014.)
Hypotheses
Ref Expression
lfl0sc.v 𝑉 = (Base‘𝑊)
lfl0sc.d 𝐷 = (Scalar‘𝑊)
lfl0sc.f 𝐹 = (LFnl‘𝑊)
lfl0sc.k 𝐾 = (Base‘𝐷)
lfl0sc.t · = (.r𝐷)
lfl0sc.o 0 = (0g𝐷)
lfl0sc.w (𝜑𝑊 ∈ LMod)
lfl0sc.g (𝜑𝐺𝐹)
Assertion
Ref Expression
lfl0sc (𝜑 → (𝐺f · (𝑉 × { 0 })) = (𝑉 × { 0 }))
Proof of Theorem lfl0sc
Dummy variable 𝑘 is distinct from all other variables.
StepHypRef Expression
1 lfl0sc.v . . . 4 𝑉 = (Base‘𝑊)
21fvexi 6666 . . 3 𝑉 ∈ V
32a1i 11 . 2 (𝜑𝑉 ∈ V)
4 lfl0sc.w . . 3 (𝜑𝑊 ∈ LMod)
5 lfl0sc.g . . 3 (𝜑𝐺𝐹)
6 lfl0sc.d . . . 4 𝐷 = (Scalar‘𝑊)
7 lfl0sc.k . . . 4 𝐾 = (Base‘𝐷)
8 lfl0sc.f . . . 4 𝐹 = (LFnl‘𝑊)
96, 7, 1, 8lflf 36317 . . 3 ((𝑊 ∈ LMod ∧ 𝐺𝐹) → 𝐺:𝑉𝐾)
104, 5, 9syl2anc 587 . 2 (𝜑𝐺:𝑉𝐾)
116lmodring 19633 . . . 4 (𝑊 ∈ LMod → 𝐷 ∈ Ring)
124, 11syl 17 . . 3 (𝜑𝐷 ∈ Ring)
13 lfl0sc.o . . . 4 0 = (0g𝐷)
147, 13ring0cl 19313 . . 3 (𝐷 ∈ Ring → 0𝐾)
1512, 14syl 17 . 2 (𝜑0𝐾)
16 lfl0sc.t . . . 4 · = (.r𝐷)
177, 16, 13ringrz 19332 . . 3 ((𝐷 ∈ Ring ∧ 𝑘𝐾) → (𝑘 · 0 ) = 0 )
1812, 17sylan 583 . 2 ((𝜑𝑘𝐾) → (𝑘 · 0 ) = 0 )
193, 10, 15, 15, 18caofid1 7424 1 (𝜑 → (𝐺f · (𝑉 × { 0 })) = (𝑉 × { 0 }))
Colors of variables: wff setvar class Syntax hints: → wi 4 = wceq 1538 ∈ wcel 2114 Vcvv 3469 {csn 4539 × cxp 5530 ⟶wf 6330 ‘cfv 6334 (class class class)co 7140 ∘f cof 7392 Basecbs 16474 .rcmulr 16557 Scalarcsca 16559 0gc0g 16704 Ringcrg 19288 LModclmod 19625 LFnlclfn 36311 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2116 ax-9 2124 ax-10 2145 ax-11 2161 ax-12 2178 ax-ext 2794 ax-rep 5166 ax-sep 5179 ax-nul 5186 ax-pow 5243 ax-pr 5307 ax-un 7446 ax-cnex 10582 ax-resscn 10583 ax-1cn 10584 ax-icn 10585 ax-addcl 10586 ax-addrcl 10587 ax-mulcl 10588 ax-mulrcl 10589 ax-mulcom 10590 ax-addass 10591 ax-mulass 10592 ax-distr 10593 ax-i2m1 10594 ax-1ne0 10595 ax-1rid 10596 ax-rnegex 10597 ax-rrecex 10598 ax-cnre 10599 ax-pre-lttri 10600 ax-pre-lttrn 10601 ax-pre-ltadd 10602 ax-pre-mulgt0 10603 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1085 df-3an 1086 df-tru 1541 df-ex 1782 df-nf 1786 df-sb 2070 df-mo 2622 df-eu 2653 df-clab 2801 df-cleq 2815 df-clel 2894 df-nfc 2962 df-ne 3012 df-nel 3116 df-ral 3135 df-rex 3136 df-reu 3137 df-rmo 3138 df-rab 3139 df-v 3471 df-sbc 3748 df-csb 3856 df-dif 3911 df-un 3913 df-in 3915 df-ss 3925 df-pss 3927 df-nul 4266 df-if 4440 df-pw 4513 df-sn 4540 df-pr 4542 df-tp 4544 df-op 4546 df-uni 4814 df-iun 4896 df-br 5043 df-opab 5105 df-mpt 5123 df-tr 5149 df-id 5437 df-eprel 5442 df-po 5451 df-so 5452 df-fr 5491 df-we 5493 df-xp 5538 df-rel 5539 df-cnv 5540 df-co 5541 df-dm 5542 df-rn 5543 df-res 5544 df-ima 5545 df-pred 6126 df-ord 6172 df-on 6173 df-lim 6174 df-suc 6175 df-iota 6293 df-fun 6336 df-fn 6337 df-f 6338 df-f1 6339 df-fo 6340 df-f1o 6341 df-fv 6342 df-riota 7098 df-ov 7143 df-oprab 7144 df-mpo 7145 df-of 7394 df-om 7566 df-wrecs 7934 df-recs 7995 df-rdg 8033 df-er 8276 df-map 8395 df-en 8497 df-dom 8498 df-sdom 8499 df-pnf 10666 df-mnf 10667 df-xr 10668 df-ltxr 10669 df-le 10670 df-sub 10861 df-neg 10862 df-nn 11626 df-2 11688 df-ndx 16477 df-slot 16478 df-base 16480 df-sets 16481 df-plusg 16569 df-0g 16706 df-mgm 17843 df-sgrp 17892 df-mnd 17903 df-grp 18097 df-mgp 19231 df-ring 19290 df-lmod 19627 df-lfl 36312 This theorem is referenced by: lkrscss 36352 lfl1dim 36375 lfl1dim2N 36376
Copyright terms: Public domain W3C validator | 2,250 | 4,016 | {"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} | 3.234375 | 3 | CC-MAIN-2024-26 | latest | en | 0.299837 |
http://math.stackexchange.com/questions/tagged/eigenvectors+complex-numbers | 1,387,456,768,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1387345763533/warc/CC-MAIN-20131218054923-00021-ip-10-33-133-15.ec2.internal.warc.gz | 119,236,212 | 11,880 | # Tagged Questions
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I am reading Applied linear algebra: the decoupling principle by Lorenzo Adlai Sadun (btw very recommendable!) On page 69 it gives an example where a real, square matrix $A=[(a,-b),(b,a)]$ is raised ... | 498 | 1,736 | {"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} | 3.75 | 4 | CC-MAIN-2013-48 | latest | en | 0.750612 |
https://math.stackexchange.com/questions/2041989/olympiad-level-assignment-problem-having-a-use-of-geometry-and-trigonometry/2042032 | 1,623,906,647,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487629209.28/warc/CC-MAIN-20210617041347-20210617071347-00266.warc.gz | 358,238,355 | 37,481 | # Olympiad level assignment problem having a use of geometry and trigonometry.
I have got an assignment having olympiad problems. One entry of these questions went this way:
There are n line segments in the plane With the sum of lengths equal to one. Prove that there exist a straight line such that sum of length of projection of the segments on the line equal to $\frac{2}{\pi}$.
I was trying to use a bit of geometric construction and trigonometry but I don't think the data is sufficient for the method I want to use. I shall be highly thankful if you guys can suggest me how to approach for the problem. Thanks in advance.
• Show that $2/\pi$ is the average of the sum of projected lengths over all possible orientations of a projected-onto line. (By continuity, at least one orientation achieves the average.) To do this, express the "average of the sum of projected lengths" into "the sum of the average of projected lengths"; the average of projected lengths for any one segment happens to be a straightforward integral. – Blue Dec 3 '16 at 17:07
## 1 Answer
Let $l_i$ be the lengths of segments, and $\theta_i$ be the angles they make with +X-axis. $$\sum_{i=1}^{n}l_i=1$$ Sum of lengths of projections across line at angle $\theta$:-
$$f(\theta)=\sum_{i=1}^{n}l_i|\cos(\theta_i-\theta)|$$
Average value of $|\cos(\theta)|=\frac{2}{\pi}$. Hence average of $f(\theta)$:- $$avg(f(\theta))=\sum_{i=1}^{n}l_i\cdot avg(|\cos(\theta_i-\theta)|)=\frac{2}{\pi}\sum_{i=1}^{n}l_i=\frac{2}{\pi}$$
Since $f(\theta)$ is continuos, it must achieve its average value. | 426 | 1,570 | {"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} | 4.15625 | 4 | CC-MAIN-2021-25 | latest | en | 0.895654 |
https://www.physicsforums.com/threads/derviate-x-x-x.193310/ | 1,519,412,250,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814827.46/warc/CC-MAIN-20180223174348-20180223194348-00633.warc.gz | 905,140,028 | 15,742 | # Derviate x^x^x
1. Oct 23, 2007
### erjkism
1. The problem statement, all variables and given/known data
differentiate f(x)= x^x^x
2. Relevant equations
chain rule
product rule
3. The attempt at a solution
x^x (lnx)
i dont know what to do after this
2. Oct 23, 2007
### quasar987
A function raised to another function is an exponential:
In general, $$f(x)^{g(x)} = \exp(\ln(f(x)^{g(x)}))=\exp(g(x)\ln(f(x)))$$
And you know how to differentiate an exponential.
So, can you use what I wrote to write x^x^x as an exponential?
3. Oct 23, 2007
### arildno
Would that be:
1. $$x^{(x^{x})}=x^{x^{x}}$$
2. $$(x^{x})^{x}=x^{x^{2}}$$
Learn to use parentheses..
4. Oct 24, 2007
### HallsofIvy
Don't just leave x^x(ln x) by itself! If f= x^x^x, then ln(f)= x^x ln(x). Now DO IT AGAIN! ln(ln(f))= ln(x^x ln(x))= ln(x^x)+ ln(ln(x))= xln(x)+ ln(ln(x)).
Use the chain rule to differentiate both ln(ln(f(x)) and ln(ln(x)).
5. Oct 25, 2007
### Gib Z
I sometimes get annoyed with exponential notation for exactly that reason. My opinion is if the exponent is any larger than 1 term, write it is terms of exp(...).
6. Oct 26, 2007
### nizi
Neglecting the given attempt, I put
$$y = x^{x^{x}}$$
$$z = x^{x}$$
and develop as follows.
$$\ln y = \ln x^{x^{x}} = z \ln x$$
$$\frac{y'}{y} = z' \ln x + z \frac{1}{x}$$
here I calculate the differentiation of $$z$$
$$z = x^{x}$$
$$\ln z = x \ln x$$
$$\frac{z'}{z} = \ln x + x \frac{1}{x}$$
$$z' = z \left( { \ln x + 1 } \right) = x^{x} \left( { \ln x + 1 } \right)$$
Accordingly
$$y' = y \left( { x^{x} \left( { \ln x + 1 } \right) \ln x + x^{x} \frac{1}{x} } \right) = x^{x^{x}+x-1} \left( { x \left( { \ln x + 1 } \right) \ln x + 1 } \right)$$ | 668 | 1,697 | {"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} | 4.15625 | 4 | CC-MAIN-2018-09 | latest | en | 0.705249 |
https://fxsolver.com/browse/?like=2127&p=2 | 1,652,916,792,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522556.18/warc/CC-MAIN-20220518215138-20220519005138-00744.warc.gz | 327,665,595 | 49,985 | '
# Search results
Found 1825 matches
True anomaly - elliptic orbits
In celestial mechanics, true anomaly is an angular parameter that defines the position of a body moving along a Keplerian orbit. It is the angle between ... more
True anomaly - circular orbit
In celestial mechanics, true anomaly is an angular parameter that defines the position of a body moving along a Keplerian orbit. It is the angle between ... more
Normal segment for a given line
The normal segment for a given line is defined to be the line segment drawn from the origin perpendicular to the line. This segment joins the origin with ... more
True anomaly - circular orbit with zero inclination
In celestial mechanics, true anomaly is an angular parameter that defines the position of a body moving along a Keplerian orbit. It is the angle between ... more
True anomaly - as a function of eccentric anomaly, sin form
In celestial mechanics, true anomaly is an angular parameter that defines the position of a body moving along a Keplerian orbit. It is the angle between ... more
True anomaly - as a function of eccentric anomaly, Tan form
In celestial mechanics, true anomaly is an angular parameter that defines the position of a body moving along a Keplerian orbit. It is the angle between ... more
True anomaly - as a function of eccentric anomaly, cos form
In celestial mechanics, true anomaly is an angular parameter that defines the position of a body moving along a Keplerian orbit. It is the angle between ... more
In celestial mechanics, true anomaly is an angular parameter that defines the position of a body moving along a Keplerian orbit. It is the angle between ... more
Frequency (Doppler effect for a moving black body)
Black-body radiation is the thermal electromagnetic radiation within or surrounding a body in thermodynamic equilibrium with its environment, or emitted by ... more
Volume of a right prism (regular base)
A right prism is a prism in which the joining edges and faces are perpendicular to the base faces. The volume of a right prism whose base is a regular ... more
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Chapter2
# Chapter2 - Section 2.1 The Tangent and Velocity Problems...
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1 Section 2.1 The Tangent and Velocity Problems • Goals – Use two problems to explain the need for the notion of limit : The tangent problem The velocity problem What would it mean in general for a line to be tangent to a curve? Tangent Problem touching once touching twice Example Let’s use a specific example to see how tangent lines in general could be defined: We want to find an equation of the tangent line to the parabola y = x 2 at the point P (1, 1) . Since we know the point P , the only question is that of the slope m of the line. Example (cont’d) Our idea is to use a nearby point Q ( x , x 2 ) on the parabola and find the slope m PQ of the secant line PQ . We choose x 1 so that Q P . Then 2 1 1 PQ x m x = Example (cont’d) Now we allow x to approach the value 1 without ever being 1 . – Meanwhile we monitor m PQ . The tables on the next slide give the values of m PQ as x approaches 1 both from above and below. In both cases it seems that m PQ approaches 2 ! Example (cont’d)
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View Full Document | 325 | 1,290 | {"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} | 3.9375 | 4 | CC-MAIN-2018-13 | latest | en | 0.876086 |
https://puzzling.stackexchange.com/questions/77735/just-a-little-sum-thing | 1,561,510,904,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999964.8/warc/CC-MAIN-20190625233231-20190626015231-00326.warc.gz | 547,178,565 | 35,129 | # Just a little sum-thing
Select 5 numbers. Using exactly five copies of each, simultaneously create 15 (multi-)sets, one of each sum from 1 to 15 inclusive.
Thanks to Gordon Hamilton for the inspiration...
• Despite a successful solution (though perhaps not the only one possible!), it seems that some confusion remains. Could you please let me know what needs clarification, so that I can improve the puzzle? Thanks... – Zomulgustar Dec 24 '18 at 6:28
We must select 5 numbers with sum $$\frac{1}{5}(1+2+\dots+15) = \frac{1}{5} \cdot 120 = 24.$$
Select $$1, 3, 5, 7, 8.$$ Here is the list:
$$1$$
$$1, 1$$
$$3$$
$$1, 3$$
$$5$$
$$3, 3$$
$$7$$
$$8$$
$$8, 1$$
$$5, 5$$
$$8, 3$$
$$7, 5$$
$$8, 5$$
$$7, 7$$
$$8, 7$$
Motivation: $$1$$ is forced. I tried adding $$2$$ next, but that would've forced me to use a number above $$8$$ ($$\min \max = 9$$ achieved with $$\{1, 2, 4, 8, 9\}$$), so I added $$3$$ to the list. The only 5 element set with $$1, 3$$ and $$\max \le 8$$ was $$\{1, 3, 5, 7, 8\},$$ and it worked.
• Nicely done! The solution is unique if you restrict to natural numbers. – Zomulgustar Dec 23 '18 at 21:09 | 412 | 1,138 | {"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": 26, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-26 | latest | en | 0.889495 |
https://pixelatedworks.com/excel/formulae/trimmean-excel/ | 1,709,505,361,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476399.55/warc/CC-MAIN-20240303210414-20240304000414-00880.warc.gz | 439,076,993 | 31,495 | # Trimmean: Excel Formulae Explained
## Key Takeaway:
• TRIMMEAN is an Excel formula used to calculate the mean of a dataset, while excluding a certain percentage of top and bottom values. This is useful when dealing with skewed data or outliers, where the mean can be misleading.
• The syntax of TRIMMEAN involves specifying the data range and the percentage of values to exclude. For example, TRIMMEAN(A1:A10, 10%) will exclude the top and bottom 10% of values in the range A1:A10, and calculate the mean of the remaining values.
• Advanced ways to use TRIMMEAN include using wildcards to exclude values based on certain patterns, and using array formulas to apply TRIMMEAN to multiple ranges simultaneously. It is important to be aware of common errors that can occur when using TRIMMEAN, such as including non-numeric values in the data range.
Do you find it difficult to calculate the TRIMMEAN in Excel? Worry no more! In this article, you will learn how to easily calculate TRIMMEAN with step-by-step instructions. Eliminate the hassle of manually calculating TRIMMEAN and make your work easier.
### TRIMMEAN: Definition and Purpose
TRIMMEAN is an Excel formula that can be used for statistical analysis. It’s used when data has extreme values or outliers. TRIMMEAN calculates the average of a trimmed set of values after excluding a specified percentage of the highest and lowest values.
It’s used in place of the AVERAGE function when there are outliers present in the data. The formula takes three arguments: array, percent and two additional logical arguments. This returns the average value after excluding a specific percentage or number of data points from both ends.
Trimming Mean gives us a better idea as this excludes incorrect extreme numbers. It’s less sensitive to outliers than an arithmetic mean and can be more robust during statistical analysis. It also works well for skewed datasets, as it removes some extreme values to account for this skewness too.
Using TRIMMEAN is useful when analyzing large datasets with errors, missing data or inconsistent values. This automatically excludes outliers from its calculation, providing a more accurate reflection of the central tendency or average value.
For example, Thomas’s team analyzed sales data from multiple regions. They found outlier sales numbers were skewing the results. By using TRIMMEAN instead of AVERAGE, they removed these outlying numbers and produced more useful insights.
Next, we will discuss How TRIMEAN works and When to Use it in Excel.
### How TRIMMEAN Works and When to Use it in Excel
TRIMMEAN is an Excel function that helps you find the average of a trimmed data set. It discards some values from each end of the data set to stop skewed results due to outliers. Let’s look at how TRIMMEAN works and when to use it in Excel.
Here’s a 6-step guide on using TRIMMEAN:
1. Select a cell in your worksheet where you want to display the trimmed mean.
2. Type =TRIMMEAN( without quotes into the formula bar.
3. Highlight the cells in your range where the data exists.
4. Enter a percentage value for how much you want to trim from each end. For example, 0.2 to cut off 20% from each end.
5. Close with end parentheses and press Enter.
6. Format or modify your answer if necessary.
TRIMMEAN is useful when there are extreme or outlying values present. It eliminates outliers from each end of the dataset. It works well for non-normally distributed data sets containing extreme observations or outliers.
A Pro Tip – Use your judgement when selecting a trimming percentage to avoid over-generalizing conclusions.
In the next section, we shall discuss syntax types and examples for beginners to understand this Excel function.
## Syntax and Examples of TRIMMEAN Formulae: A Beginner’s Guide
Do you know the struggle of finding an Excel formula for specific data sets? As a data worker, I understand the annoyance of dealing with large quantities of info. In this guide, we’ll look at TRIMMEAN, a helpful Excel formula. We’ll explore the syntax and examples of how to apply it on a spreadsheet. We’ll finish this starter guide with a better idea of how to use TRIMMEAN formulae for data analysis.
### Understanding the Syntax of TRIMMEAN Formulae
TRIMMEAN requires two arguments: array and percent. Array is a range of cells or references with the data you want to calculate the TRIMMEAN for. Percent is the percentage of high and low values you remove to get the mean. For instance, 10% means you eliminate 10% of values from the high and low ends.
The syntax for TRIMMEAN in Excel is “=TRIMMEAN(array, percent)”. To display the result, input the formula in a cell.
Also, remember to separate arguments with a comma. Put array first, then a comma and then the percent value (0% to 100%).
Using TRIMMEAN properly can save time when working with large datasets. Not using it or not using it properly can mean manually calculating averages yourself, taking more time.
So don’t miss out on this useful Excel formula! Try TRIMMEAN now. In the next section, we’ll discuss some examples of how to use it.
### Examples of Using TRIMMEAN in Excel
The TRIMMEAN formula is great for Excel users! Examples of how to use it? Consider the data table with sample data. To find the trimmed mean, select all data cells or type them out separated by commas. For this case, select A2 to A6. Hit enter and you’ll get 12.67 as the trimmed mean.
You can also exclude data by typing a second argument. And for larger datasets, use array formulas with CTRL+SHIFT+ENTER. It’s a powerful tool for cleaning up data and getting an accurate picture.
My team used it for a survey. Some outliers skewed the results, so we used TRIMMEAN to exclude them. This made our analysis easier because we could focus on the normal age range.
Advanced ways to use TRIMMEAN? Wildcards and array formulas.
## Advanced Ways to Use TRIMMEAN: Wildcards and Array Formulas
Fed up with Excel formulas? There’s more to it than just basic calculations! Let’s learn some cool advanced ways to use the TRIMMEAN formula. We’ll check out its lesser-known features. Plus, we’ll explore two subsections. Firstly, how wildcards can broaden your search criteria. Then, using TRIMMEAN with array formulas to enhance data analysis skills. By the end, you’ll have useful techniques to take your Excel skills to a higher level!
### Unlocking TRIMMEAN’s Full Potential with Wildcards
Let’s look at an example with a table to show how to use wildcards with TRIMMEAN. The table has the average product price that starts with “A” in different regions.
Product Name Region Price
Apple North \$1
Ape South \$3
Banana East \$2
Use this formula: `=TRIMMEAN(IF(Region="North",IF(Product_Name="A*",Price)))` to get the average price of products that start with “A” in the North region.
Wildcards with TRIMMEAN is one way to use its full potential. You can also use array formulas to further your data analysis skills.
More ways to use TRIMMEAN:
1. Use multiple criteria, e.g. product name and region.
2. Incorporate TRIMMEAN into larger formulas, like nested IF statements or VLOOKUP functions.
3. Combine wildcards with logical operators, like AND or OR.
These techniques give you more insight into your data and help you find hidden patterns.
Using TRIMMEAN with Array Formulas for Enhanced Data Analysis is another useful technique. Array formulas let you do calculations across multiple cells at once, saving time and avoiding mistakes.
To get the most out of TRIMMEAN, you can use wildcards, combine with other formulas, or use array formulas. Experiment with these techniques and you’ll be able to take your data analysis to the next level.
### Using TRIMMEAN with Array Formulas for Enhanced Data Analysis
To boost your data analysis skills, use TRIMMEAN with array formulas. This advanced method lets you analyze multiple sets of data all at once, saving time and effort. Excel performs calculations automatically across a range of cells without manual input in each cell.
Let’s assume we have a table with four columns: Product Name, Sales Representative, Region and Sales Amount. To find the average sales for each rep in each region for every product, use TRIMMEAN with array formulas. Without it, we’d need to create multiple Pivot Tables or manually enter formulae into each cell.
Our table will have five rows:
1. Products (array)
2. Regions (array)
3. Representatives (array)
4. Result range 1 (non-array)
5. Total Result range 2 (non-array)
The first row will have arrays based on product type. For example, if there are three products, A, B and C, then the first row has three arrays – one for each product type.
The second row will have arrays based on different regions correlating to the products in row 1. For example, if product A is sold in regions X, Y and Z, then row two will have three arrays-one per region based on their relationship to product A.
In the third row, arrays are created for reps associated with the regions in row 2. Select an empty cell below, and enter an array formula to find the trimean sales value for each combination of product name, sales rep and region.
Pro Tip: Make entering an array formula easier and avoid errors – select a blank cell, type “=TRIMMEAN(“, select the range containing the sales data, add a comma, and then select the criteria range including all array formulas.
Troubleshooting TRIMMEAN Formulae: Common Errors & Solutions
To avoid errors while working with TRIMMEAN formulae, understand common issues and learn how to solve them. In our next section, we’ll explore some of these errors in detail and give solutions for each one.
## Troubleshooting TRIMMEAN Formulae: Common Errors and Solutions
When using TRIMMEAN formulae in Excel, errors can happen. Let’s take a look at some of these common errors. We’ll also discuss tips to prevent them, and how to get the most out of TRIMMEAN for data analysis. So, let’s begin!
### Common Errors to Look Out for When Using TRIMMEAN
When applying the TRIMMEAN formula in Excel, there are several errors to watch out for.
• Be careful when selecting the data range – it should not include empty cells or headings, or this can result in incorrect output and an error message.
• The alpha value (fraction argument) must be between 0 and 1; if it is greater than or equal to 1, a #NUM! error will occur.
• It is also important to remember that TRIMMEAN does not give you the mean of all values after trimming off a certain percentage from both ends. It only trims off the highest and lowest numbers; for example, if there are 10 values, trimming at 30% would trim the three highest and three lowest numbers.
• Using TRIMMEAN on large datasets may lead to errors like #REF!, #VALUE!, or #DIV/0!. This usually happens when there is insufficient memory available. To avoid this, try reducing the size of the dataset or using this formula on other Excel versions.
• TRIMMEAN is useful for getting accurate results in datasets with low-quality data points, but not for outliers.
• Microsoft Support also states that TRIMMEAN cannot take more than one set of numbers in its arguments.
By keeping these common errors in mind, most problems with this formula can be prevented.
### Tips on Avoiding Errors with TRIMMEAN in Excel
To use the TRIMMEAN Formula in Excel without errors, here are some tips to keep in mind:
• Check that your data range is accurate.
• Exclude any outliers or extreme values from your data range.
• Use a decimal separator (period or comma) that matches your regional settings.
• Check for typos or syntax errors, like missing commas or misspelled function names.
Pay attention to these factors. Even a small mistake can lead to wrong results. Additionally, you should understand how the formula works and what it does. This way, you can modify it to fit your needs.
For large data sets, it’s better to break them into smaller parts. This can help you find errors faster and make troubleshooting quicker.
To save time, use the built-in Excel error checking tools. These tools can detect and highlight errors in formulas and data ranges.
By following these tips and paying attention to details when using TRIMMEAN Formula in Excel, you’ll get accurate results in your data analysis.
### Recap of TRIMMEAN Formulae and Its Applications in Excel.
TRIMMEAN is a data-manipulation tool used in Excel. It trims the mean by excluding a certain percentage of outliers from the data set. This can be beneficial when managing large amounts of data with a lot of variance.
The formula is simple: “=TRIMMEAN (data range, percent to exclude).” Negative percentages can also be used to exclude values at both ends.
Investors use it to calculate an average return while disregarding outliers which may not reflect normal market trends.
In medical research, TRIMMEAN can help find correlations within a dataset. It was developed in 1959 by Tukey, one of the pioneers in modern statistics.
It’s clear that applying TRIMMEAN formulas can help better understand underlying data and draw meaningful insights.
## Some Facts About TRIMMEAN: Excel Formulae Explained:
• ✅ TRIMMEAN is an Excel function used for calculating the mean of a dataset while excluding a certain percentage of outliers from the top and bottom of the distribution. (Source: Microsoft)
• ✅ TRIMMEAN takes two arguments: the range of data and the percentage of outliers to be removed from the dataset. (Source: Excel Easy)
• ✅ TRIMMEAN is useful for datasets with extreme values or outliers, where the mean may not accurately represent the central tendency of the data. (Source: Investopedia)
• ✅ TRIMMEAN is widely used in finance, economics, and other fields where accurate data analysis is essential. (Source: WallStreetMojo)
• ✅ TRIMMEAN is only available in newer versions of Excel, such as Excel 2010 and later. (Source: Excel Jet)
## FAQs about Trimmean: Excel Formulae Explained
### What is TRIMMEAN in Excel formulae?
TRIMMEAN is a statistical function in Excel that calculates the mean of a range of values, excluding a percentage of the lowest and highest values in the range. It helps to exclude outliers and provide a more accurate representation of the central tendency of the data.
### How to use TRIMMEAN in Excel?
To use the TRIMMEAN function in Excel, enter “=TRIMMEAN(” followed by the data range in parentheses. Next, enter the percentage of trim required as a decimal value (e.g., 0.2 for 20%). Press enter to complete the formula, and Excel will return the trimmed mean value.
### What is the formula for TRIMMEAN in Excel?
The formula for TRIMMEAN in Excel is “=TRIMMEAN(range, percentage)”. The range argument specifies the data range that you want to include, and the percentage argument specifies the percentage of the highest and lowest values to exclude from the calculation.
### What are the advantages of using TRIMMEAN in Excel?
The TRIMMEAN function in Excel is useful when dealing with data that contains outliers or extreme values that could skew the results. By excluding a percentage of the highest and lowest values, TRIMMEAN provides a more representative average of the central tendency of the data.
### What are the limitations of TRIMMEAN in Excel?
TRIMMEAN can only exclude a fixed percentage of the highest and lowest values in the data range, which may not be appropriate for all types of data. Also, TRIMMEAN is affected by changes to the percentage of trim, which can significantly affect the calculated value.
### What is the difference between TRIMMEAN and AVERAGE in Excel?
The main difference between TRIMMEAN and AVERAGE in Excel is that TRIMMEAN excludes a percentage of the highest and lowest values in the range, while AVERAGE includes all the values in the range. As a result, TRIMMEAN provides a more accurate representation of the central tendency of the data when dealing with outliers. | 3,525 | 15,919 | {"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} | 3.75 | 4 | CC-MAIN-2024-10 | latest | en | 0.818884 |
https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_3&oldid=31801 | 1,597,349,982,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739073.12/warc/CC-MAIN-20200813191256-20200813221256-00319.warc.gz | 176,405,016 | 10,549 | # 1983 AIME Problems/Problem 3
## Problem
What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$?
## Solution
If we expand by squaring, we get a quartic polynomial, which obviously isn't very helpful.
Instead, we substitute $y$ for $x^2+18x+30$ and our equation becomes $y=2\sqrt{y+15}$.
Now we can square; solving for $y$, we get $y=10$ or $y=-6$. The second solution is extraneous since $2\sqrt{y+15}$ is positive. So, we have $y=10$ as the only solution for $y$. Substituting $x^2+18x+30$ back in for $y$,
$x^2+18x+30=10 \Longrightarrow x^2+18x+20=0.$
By Vieta's formulas, the product of our roots is therefore $\boxed{020}$.
## Solution 2
There is an error with solution one since the second solution for $y$ does NOT give us non-real roots, since the discriminant is positive. Another way to do this problem would be without substitution, and by just squaring both sides and making the quartic equation equal to 0. We have:
$x^4+36x^3+380x^2+1080x+900 = 0$
By Vieta's formula we get that the product is $\boxed{900}$. | 342 | 1,076 | {"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": 17, "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} | 4.375 | 4 | CC-MAIN-2020-34 | latest | en | 0.906283 |
https://www.coffeewithcodes.com/post/puzzle-10-measure-4-gallon-of-water-from-3-gallon-and-5-gallon-water-jar-amazon | 1,723,756,027,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641316011.99/warc/CC-MAIN-20240815204329-20240815234329-00304.warc.gz | 545,007,679 | 192,774 | top of page
Search
# Puzzle #10: Measure 4 Gallon of Water from 3 Gallon and 5 Gallon Water Jar | Amazon
### Puzzle Solution:
Step 1: Fill 3 gallon jar with water. ( 5p – 0, 3p – 3)
Step 2: Pour all its water into 5 gallon jar. (5p – 3, 3p – 0)
Step 3: Fill 3 gallon jar again. ( 5p – 3, 3p – 3)
Step 4: Pour 3 gallon jar water into 5 gallon jar until it is full. Now you will have exactly 1 gallon water remaining in 3 gallon jar. (5p – 5, 3p – 1)
Step 5: Empty 5 gallon jar, pour 1 gallon water from 3 gallon jar into 5 gallon jar. Now 5 gallon jar has exactly 1 gallon of water. (5p – 1, 3p – 0)
Step 6: Fill 3 gallon jar again and pour all its water into 5 gallon jar, thus 5 gallon jar will have exactly 4 gallon of water. (5p – 4, 3p – 0)
We are done !
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http://simont.dreamwidth.org/9143.html?thread=38071 | 1,503,201,135,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105970.61/warc/CC-MAIN-20170820034343-20170820054343-00422.warc.gz | 390,306,043 | 5,346 | Unsolved problem
simont
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Unsolved problem
Fri 2014-02-21 13:46
Indeed, you certainly can't get a longer smallest stick by small perturbations of that answer, because one of the 5-sticks is entirely made up of smallest-fragments and so if you lengthen them all by any ε > 0 then that stick overflows. What I was hoping to show was that the converse is also true – that the only way in which a dissection can be locally optimal is if some stick is made up entirely of smallest-fragments, by showing that in all other cases you can find a system of ε-adjustments that lengthens every smallest-fragment. No luck yet, though!
(In fact, has proved over on LJ that my dissection for 5-into-7 is globally as well as locally optimal, which is more than I had previously known.)
I've written a thing that looks for integer solutions
Ooh, I'd like to see whatever data it's generated.
Also it can't handle irrationals and so far can do fractions only where all stick-fragments share a denominator
In any rational dissection there must be some denominator shared by all fragments (just take the lcm of all denominators), so the latter isn't a problem. And I'm still convinced that irrationals can't appear in any solution unless there's an equally good or better one without them (I have a half-thought-out proof idea involving treating R as a vector space over Q), so I'm not worried about the former either.
Fri 2014-02-21 14:17
So far it is tediously slow and hasn't got anything harder than 5,7...
Fri 2014-02-21 16:50
Apparently I'm not exactly doing this the optimal way... because 7 x 10 in 1/3 steps is, er, ENOMEMORY. Fantastic.
5 x 8 sticks in half integer steps is best with 4 sticks cut into 2/3 and 4 uncut assembled into 4 lots of 3/5 and 1 of 2/2/2/2 which gives more than 3 stick-parts.
With m and n <= 10 and m<n in half integer steps 12 combinations have a smallest-stick-fragment larger than gcd(m,n); 4 of which are non-integer (5 x7 is the smallest). I have no idea how to prove whether any given answer could be bested by taking smaller steps other than by trying progressively smaller steps and seeing.
Fri 2014-02-21 17:18
5 x 8 sticks in half integer steps is best with 4 sticks cut into 2/3 and 4 uncut assembled into 4 lots of 3/5 and 1 of 2/2/2/2 which gives more than 3 stick-parts.
Borrowing 's proof technique: in a better solution, each of the five 8-sticks would have to be cut into at most three pieces (if you cut one in four then a piece must be <=2), which gives 15 pieces overall, and so at least one of the eight 5-sticks must end up as a single piece (15 isn't enough pieces for two each). The 8-stick including that whole 5-stick then has 3 units of length left over; if you divide that in two then you have a piece <=1.5 (no good) and OTOH if you leave it whole then that leaves 2 units on the 5-stick you cut it off.
So this dissection for 5 into 8 cannot be beaten even if you were to increase the denominator, and hence 's question is indeed answered.
(I do wonder how far that proof technique can be automated. It might give rise to a better search algorithm!) | 837 | 3,204 | {"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} | 3.15625 | 3 | CC-MAIN-2017-34 | latest | en | 0.969884 |
https://help.scilab.org/docs/5.5.2/fr_FR/sysconv.html | 1,702,077,026,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100779.51/warc/CC-MAIN-20231208212357-20231209002357-00067.warc.gz | 341,244,462 | 4,049 | Change language to:
English - 日本語 - Português - Русский
Please note that the recommended version of Scilab is 2024.0.0. This page might be outdated.
See the recommended documentation of this function
# sysconv
system conversion
### Calling Sequence
`[s1,s2]=sysconv(s1,s2)`
### Arguments
s1,s2
list (linear `syslin` systems)
### Description
Converts `s1` and `s2` into common representation in order that system interconnexion operations can be applied. Utility function for experts. The conversion rules in given in the following table.
"c"
continuous time system
"d"
discrete time system
n
sampled system with sampling period n
[]
system with undefined time domain For mixed systems `s1` and `s2` are put in state-space representation.
```s1\s2 | "c" | "d" | n2 | [] |
---------------------------------------------------------------
"c" | nothing |uncompatible | c2e(s1,n2) | c(s2) |
---------------------------------------------------------------
"d" |uncompatible| nothing | e(s1,n2) | d(s2) |
---------------------------------------------------------------
n1 | c2e(s2,n1) | e(s2,n1) | n1<>n2 uncomp | e(s2,n1) |
| | | n1=n2 nothing | |
---------------------------------------------------------------
[] | c(s1) | d(s1) | e(s1,n2) | nothing |
---------------------------------------------------------------```
With the following meaning:
n1,n2
sampling period
c2e(s,n)
the continuous-time system s is transformed into a sampled system with sampling period n.
c(s)
conversion to continuous (time domain is `"c"`)
d(s)
conversion to discrete (time domain is `"d"`)
e(s,n)
conversion to samples system with period `n`
### Examples
```s1=ssrand(1,1,2);
s2=ss2tf(s1);
[s1,s2]=sysconv(s1,s2);``` | 469 | 1,859 | {"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} | 2.59375 | 3 | CC-MAIN-2023-50 | latest | en | 0.679751 |
https://www.thedataschool.com.au/mipadmin/how-to-rank-in-tableau/ | 1,719,024,506,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862249.29/warc/CC-MAIN-20240622014659-20240622044659-00361.warc.gz | 890,294,825 | 39,731 | Select Page
Someone asked me earlier today, how can I rank a list of four items in Tableau if there are two items of the same value? I’ll talk about the answer below, but that prompted me to think about the common questions regarding ranks in Tableau and write this post.
I’ll go over the basic rank table calculation and the 4 other options that you have at your disposal when some values are of the same value.
### The basic — Rank()
The first option someone will turn to when investigating how to rank their data. Tableau defines it here:
Assigns a whole number rank starting with 1, in ascending or descending order to each row. If rows have the same value, they share the rank that is assigned to the first instance of the value. The number of rows with the same rank is added when calculating the rank for the next row, so you may not get consecutive rank values.
Assigns a whole number rank starting with 1, in ascending or descending order to each row. If rows have the same value, they share the rank that is assigned to the first instance of the value. The number of rows with the same rank is added when calculating the rank for the next row, so you may not get consecutive rank values.
In this case, when rows share the same value, they will receive the same rank. The following rank number will be skipped. I’ll use the Sample — Superstore data to demonstrate.
In the above image, we can see that once we get to row 4, multiple people have ordered 34 times. With Rank(), they share number just after the previous rank i.e 4.
Something that’s important to note, is that null values will be ignored by all ranking functions.
What if I didn’t want to skip any rankings?
### RANK_DENSE()
RANK_DENSE() returns a ranked list that gives identical values an identical rank, however no gaps are inserted into the ranks.
This will return the following:
We can see that this means in the case of tied values, the current rank increments by 1, and the next rank will also be an increment of 1.
### RANK_MODIFIED()
This rank function will return rank as a modified competition rank. This means that identical values will be given an identical rank, however the rank will be as if it were in a competition. E.g 1,3,3,4.
We can see an example below:
### RANK_UNIQUE()
Rank Unique, as the name suggests, will return a unique rank for every value. Identical values will also be assigned a different rank number. For the customer orders data set, the following result is displayed:
### RANK_PERCENTILE()
The final rank function that we’ll cover will be the rank percentile. What this rank function does, is normalise the values. This means that it will assign a rank between 0 and 1, and assigns the percentile rank of the value between the min and max values. See below the rank percentile function for the customer orders data.
Now you’ll have seen the five rank functions in Tableau, and will be able to pick and choose the appropriate rank function for your purpose. Happy visualising! | 656 | 3,005 | {"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} | 2.75 | 3 | CC-MAIN-2024-26 | latest | en | 0.875174 |
https://flylib.com/books/en/2.705.1/stdev_.html | 1,713,907,869,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818740.13/warc/CC-MAIN-20240423192952-20240423222952-00599.warc.gz | 217,858,905 | 10,795 | # StDev()
Category: Aggregate
Syntax: StDev ( field {; field...} )
Parameters:
fieldAny related field, repeating field, or set of non-repeating fields that represent a collection of numbers. Parameters in curly braces { } are optional.
Data type returned: Number
Description:
Returns the standard deviation of the non-blank values represented in the parameter list. Standard deviation is a statistical measurement of how spread out a collection of values is. In a normal distribution, about 68% of the values are within one standard deviation of the mean, and about 95% are within two standard deviations of the mean.
The difference between the StDevP() and StDev() functions is that StDev divides the sum of the squares by n-1 instead of by n.
StDev() can also be calculated as the square root of the Variance() of a set of numbers.
Examples:
There are several ways you can manually calculate the standard deviation of a set of numbers. One is to take the square root of the sum of the squares of each values distance from the mean, divided by n-1, where n is the number of values in the set.
For instance, given the set of numbers 8, 10, and 12, the mean of this set is 10. The distances of each value from the mean are therefore -2, 0, and 2. The squares of these are 4, 0, and 4. The sum of the squares is 8. The standard deviation of the population is Sqrt (8/(3-1)), which is 2.
``` StDev (8; 10; 12)
```
returns 2.
Given a portal that contains a field called Scores with the following values (64, 72, 75, 59, 67),
``` StDev (People::Scores)
```
returns 6.35.
FileMaker 8 Functions and Scripts Desk Reference
ISBN: 0789735113
EAN: 2147483647
Year: 2004
Pages: 352 | 431 | 1,688 | {"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} | 2.9375 | 3 | CC-MAIN-2024-18 | latest | en | 0.862618 |
https://statsthinking21.github.io/statsthinking21-figures-python/06-Probability.html | 1,695,902,576,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510387.77/warc/CC-MAIN-20230928095004-20230928125004-00635.warc.gz | 588,373,378 | 7,851 | # Chapter 6: Probability#
import pandas as pd
import sidetable
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
from scipy.stats import norm, binom
import rpy2.robjects as ro
from rpy2.robjects.packages import importr
from rpy2.robjects import pandas2ri
pandas2ri.activate()
from rpy2.robjects.conversion import localconverter
# import NHANES package
base = importr('NHANES')
with localconverter(ro.default_converter + pandas2ri.converter):
NHANES = ro.conversion.rpy2py(ro.r['NHANES'])
NHANES = NHANES.drop_duplicates(subset='ID')
NHANES['isChild'] = NHANES.Age < 18
rng = np.random.RandomState(123)
## Figure 6.1#
fig, ax = plt.subplots(1, 2, figsize=(12,6))
nsamples = 30000
pDf = pd.DataFrame({'trial_number': list(range(1, nsamples + 1))})
pDf['outcomes'] = rng.binomial(1, .5, nsamples)
pDf['mean_probability'] = np.cumsum(pDf.outcomes) / pDf.trial_number
pDf = pDf.query('trial_number >= 10')
sns.lineplot(data=pDf, x='trial_number', y='mean_probability', ax=ax[0])
ax[0].plot([10, nsamples], [0.5, 0.5])
ax[0].set_xlabel("Number of trials")
electionReturns = pd.melt(electionReturns, id_vars=['pctResp'],
ax[1].set_xlabel("Percentage of precincts reporting")
Text(0, 0.5, 'Percentage of votes')
## Figure 6.2#
imgmtx = np.zeros((6, 6))
imgmtx[:, -1] = 1
imgmtx[-1, :] = 1
plt.imshow(imgmtx, cmap='bwr')
plt.xticks(np.arange(0.5, 6))
plt.yticks(np.arange(0.5, 6))
plt.grid(color='white', linestyle='-', linewidth=1, which='major')
for i in range(0, 6):
for j in range(0, 6):
plt.annotate(f'{i+1},{j+1}', [i-0.2, j+0.1], color='white')
plt.setp(plt.gca().get_xticklabels(), visible=False)
_ = plt.setp(plt.gca().get_yticklabels(), visible=False)
## Table 6.1#
curry_df = pd.DataFrame({'numSuccesses': list(range(0, 5))})
curry_df['Probability'] = binom.pmf(curry_df.numSuccesses, 4, 0.91)
curry_df['CumulativeProbability'] = binom.cdf(curry_df.numSuccesses, 4, 0.91)
curry_df
numSuccesses Probability CumulativeProbability
0 0 0.000066 0.000066
1 1 0.002654 0.002719
2 2 0.040246 0.042965
3 3 0.271286 0.314250
4 4 0.685750 1.000000
## Table 6.2#
NHANES_diabetes_activity = NHANES_adult[['PhysActive', 'Diabetes']].dropna()
diabetes_summary = pd.DataFrame(NHANES_diabetes_activity.value_counts('Diabetes'), columns=['counts'])
diabetes_summary['prob'] = diabetes_summary.counts / NHANES_diabetes_activity.shape[0]
print(diabetes_summary)
activity_summary = pd.DataFrame(NHANES_diabetes_activity.value_counts('PhysActive'), columns=['counts'])
activity_summary['prob'] = activity_summary.counts / NHANES_diabetes_activity.shape[0]
print(activity_summary)
counts prob
Diabetes
No 4251 0.887659
Yes 538 0.112341
counts prob
PhysActive
Yes 2456 0.512842
No 2333 0.487158
## Table 6.3#
NHANES_diabetes_stats_by_activity = pd.DataFrame(
NHANES_diabetes_activity.value_counts(['Diabetes', 'PhysActive']), columns=['counts'])
NHANES_diabetes_stats_by_activity['prob'] = NHANES_diabetes_stats_by_activity / NHANES_diabetes_activity.shape[0]
NHANES_diabetes_stats_by_activity
counts prob
Diabetes PhysActive
No Yes 2261 0.472124
No 1990 0.415536
Yes No 343 0.071622
Yes 195 0.040718
## Table 6.4#
NHANES_mh = NHANES_adult.dropna(subset=['PhysActive', 'DaysMentHlthBad'])
'Bad Mental Health' if i > 7 else "Good Mental Health" for i in NHANES_mh.DaysMentHlthBad]
NHANES_mentalhealth_by_physactive_counts = pd.crosstab(
NHANES_mh.PhysActive,
margins = True, normalize='all')
NHANES_mentalhealth_by_physactive_counts
PhysActive
No 0.085177 0.402088 0.487265
Yes 0.060543 0.452192 0.512735
All 0.145720 0.854280 1.000000
## Table 6.5#
NHANES_mentalhealth_by_physactive_condp = pd.crosstab(
NHANES_mh.PhysActive, | 1,244 | 3,728 | {"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} | 2.84375 | 3 | CC-MAIN-2023-40 | latest | en | 0.402469 |
http://ecetutorials.com/question-answers/basic-electronics/be1/ | 1,516,338,849,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887746.35/warc/CC-MAIN-20180119045937-20180119065937-00348.warc.gz | 100,360,171 | 7,621 | # be1
## 1) What is an Electric Potential?
The potential of a particular point, is defined as the work done to move a positive charge particle to move to that point in opposite to electric field. It may be an external energy applied to move that positive charge or may be internal potential energy of the positive charge.
## 2) What is Potential Difference?
The differential in potentials at two points in an electric field is called as potential difference between two points. It is measure in Volts i.e. VAB is called the potential difference between points A & B. If the VAB is positive then it illustrates that the potential at point A is high compared to potential at point B. In this case there will be chance of movement of charged particle from A to B against to electrical field between the points.
VAB = VA – VB
## 3) What is a Transit time of an Electron?
The time taken by an electron to travel from one electrode to the other electrode is called electron transit time. It is denoted as ‘τ’.
${\color{DarkBlue}&space;\tau&space;=\sqrt{\frac{2*m}{e*V}}*d}$
Where V = Potential applied to electron
d= Distance between electrodes
e = Electron charge
m = mass of an electron
Note: The transit time of an electron decreases with the increase in potential (V).
## 4) What is an Electron Volt(eV)?
One Electron volt is the energy gained by the electron when it falls from one point to another point which has a potential difference of 1 volt. I.e. under potential difference of one volt, the electron moves from higher potential to lower potential, during this free fall in the electron field it gains an electron volt energy.
In other case it the energy required to move an electron in the potential difference of one volt. This is against to the electric field.
1 electron volt = 1.60217657 × 10-19 joules
#### 5) What is an Electric Field intensity(ε)?
The force experienced by an electric charge particle at a point by an electric field is called as Electric field intensity. It is denoted as ε and its unit is V/m (Volts/meter).
${\color{Blue}&space;\varepsilon&space;=\frac{f}{q}}$
${\color{Blue}&space;\varepsilon&space;=&space;\frac{V}{d}}$
## 6) What is an Ionization Potential?
The energy required to detach an loosely coupled electron from the influence of parent nucleus is called ionization potential.
## 7) What is an electron spin?
The electron generally orbits around the parent nucleus and in addition to this it also rotates around itself like an earth. This intrinsic angular momentum property of en electron is called electron spin. There are only two direction of spins when it is subjected to a magnetic field i.e. either parallel or anti-parallel to the field.
## 8) What is N type Semiconductor?
When extra valance electrons are introduced in a pure semi-conductor material (silicon) by putting or injecting dopants or impurities, an N-type material is produced. The dopants are used to create an N-Type material are Group-V elements in the materials table.
N-type semiconductor bond diagram
Group V elements: Arsenic, antimony, phosphorus.
In N-type semiconductor electrons are majority carriers and holes are minority carriers.
## 9) What is P- Type Semiconductor?
When a dopant from Group III elements from the elements table is introduced in to pure semiconductor, produces a P-Type semiconductor. The Group III elements have only 3 valance electrons in its outer orbit which there combines with four valance electrons semiconductor material (silicon). This results in missing of electron which creates a hole (P+) , or positive charge that can freely move around in the material.
P-type semiconductor bond diagram
Group III elements: aluminum, boron, and gallium.
In P-Type semiconductor, holes are the majority carriers and electrons are the minority carriers.
## 10) What are passive and active components?
Passive components: The components which are passive in nature i.e. which can attenuate the input applied signal to a desired level. These are energy consumers and cannot boost the input signal. Resistors, capacitors and inductors are the examples of passive components.
Active components: The components which plays an active role in the electronic circuit i.e. which can add energy to the input signal are called active components. These can amplify the signal to the desired level. Transistors, operational amplifiers are the examples of active components.
1 2 3 4 5 6 7 | 975 | 4,458 | {"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} | 3.796875 | 4 | CC-MAIN-2018-05 | latest | en | 0.939916 |
https://scribesoftimbuktu.com/combine-2-n26-n/ | 1,675,759,826,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500392.45/warc/CC-MAIN-20230207071302-20230207101302-00640.warc.gz | 529,352,482 | 14,525 | # Combine 2/(n^2)+6/n
2n2+6n
To write 6n as a fraction with a common denominator, multiply by nn.
2n2+6n⋅nn
Write each expression with a common denominator of n2, by multiplying each by an appropriate factor of 1.
Multiply 6n and nn.
2n2+6nn⋅n
Raise n to the power of 1.
2n2+6nn1n
Raise n to the power of 1.
2n2+6nn1n1
Use the power rule aman=am+n to combine exponents.
2n2+6nn1+1
2n2+6nn2
2n2+6nn2
Combine the numerators over the common denominator.
2+6nn2
Factor 2 out of 2+6n.
Factor 2 out of 2.
2(1)+6nn2
Factor 2 out of 6n.
2(1)+2(3n)n2
Factor 2 out of 2(1)+2(3n).
2(1+3n)n2
2(1+3n)n2
Combine 2/(n^2)+6/n
### Solving MATH problems
We can solve all math problems. Get help on the web or with our math app
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Absolute BS Crap Reasonable Nice Amazing
# Whats the big deal about pi?
A HOMELESS NOBODY
User ID: 21761661
United States
12/09/2012 02:33 AM
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Whats the big deal about pi?
Whats the big deal about pi?
Why do mathematicians go crazy about it, does pi show us we lived in a simulated reality?
Anonymous Coward
User ID: 29373288
Australia
12/09/2012 02:35 AM
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Re: Whats the big deal about pi?
There is nothing miraculous about it at all the same as Fibonacci sequences.
Anonymous Coward
User ID: 21300769
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12/09/2012 02:36 AM
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Re: Whats the big deal about pi?
Kirk
User ID: 25384388
United States
12/09/2012 02:36 AM
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Re: Whats the big deal about pi?
pi are round
Government is a body largely ungoverned.
A HOMELESS NOBODY (OP)
User ID: 21761661
United States
12/09/2012 02:44 AM
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Re: Whats the big deal about pi?
its said that pi can prove to us we live in a simulated reallity, anyone care to explain what pi is, how it reelects to the eye on horus?
Anonymous Coward (OP)
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12/09/2012 02:49 AM
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Re: Whats the big deal about pi?
Johann Martin Zacharias Dase calculated pi to 200 decimal places, with the first zero appearing at the 32nd decimal place ---meaning, possibly that the exercise should have ended there. It has not;
Anonymous Coward
User ID: 28117669
12/09/2012 02:53 AM
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Re: Whats the big deal about pi?
It represents a circle, which represents a 360 degree angle which is 180 x 2 and it also represents the sun and the moon.
It represents time, energy. Life.
It shows up in a number of patterns when your apply
a number of algorithms to a number of other algorithms
in conjunction with linear algebra.
Because nothing is more linear algebraically speaking
than a limit function and it's transforming derivative.
Like the sun or the moon rising and setting. And eventually
pi begins at 0 to +3.14 to 0 to - 3.14 and back to 0
again when it is subtracted from itself.
Anonymous Coward (OP)
User ID: 21761661
United States
12/09/2012 02:57 AM
Report Abusive Post
Re: Whats the big deal about pi?
It represents a circle, which represents a 360 degree angle which is 180 x 2 and it also represents the sun and the moon.
It represents time, energy. Life.
It shows up in a number of patterns when your apply
a number of algorithms to a number of other algorithms
in conjunction with linear algebra.
Because nothing is more linear algebraically speaking
than a limit function and it's transforming derivative.
Like the sun or the moon rising and setting. And eventually
pi begins at 0 to +3.14 to 0 to - 3.14 and back to 0
again when it is subtracted from itself.
Quoting: Anonymous Coward 28117669
why do they say pi never ends...
Anonymous Coward (OP)
User ID: 21761661
United States
12/09/2012 01:20 PM
Report Abusive Post | 853 | 3,056 | {"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} | 2.78125 | 3 | CC-MAIN-2016-40 | longest | en | 0.863385 |
https://www.physicsforums.com/threads/minima-maxima-of-a-circle.1050538/ | 1,679,556,423,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945030.59/warc/CC-MAIN-20230323065609-20230323095609-00781.warc.gz | 1,065,996,410 | 18,736 | # Minima/maxima of a circle
notgoodatphysics
We have the equation for a circle, and its derivative:
$$(y-a)^2 + (x-a)^2 = r^2$$
$$\frac{dy}{dx} = \frac{a-x}{(r^{2}-(x-a)^2)^{1/2}} = 0$$
So ##x = a## then he subs it into the original equation to get the max/min.
Why does ##x = a## give the points of minima/ maxima if we didn’t know it was a circle in this case? Is there a specific rule? It’s not really explained well in the book.
It’s from calculus made easy and he says “Since no value whatever of x will make the denominator infinite, the only condition to give zero is x = a”
I’m pretty confused what is meant by that statement.
Mentor
Isn't the slope at a max/min point 0? From there you can see that x must equal a to get a zero slope.
notgoodatphysics
Isn't the slope at a max/min point 0? From there you can see that x must equal a to get a zero slope.
Right, but you’d know this without having to find the derivative. I guess if you didn’t know it was a circle there’s an explanation to “Since no value whatever of x will make the denominator infinite, the only condition to give zero is x = a”?
Homework Helper
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Right, but you’d know this without having to find the derivative. I guess if you didn’t know it was a circle there’s an explanation to “Since no value whatever of x will make the denominator infinite, the only condition to give zero is x = a”?
If the centre of the circle is the point ##(a,b)##, then ##y## has its maximum and minimum values above and below the centre. The maximum and minimum ##y## are the points ##(a, b \pm r)##. Where ##r## is the radius.
You don't need calculus to see that.
dextercioby
notgoodatphysics
I understand that bit. But it’s presented in the book as an example, and he mentions specifically that you have no idea it’s circle, so suppose you were presented with the original equation and asked to find the minimum/ maximum.
You’d need find the derivative if I’m not mistaken? My confusion comes about when the derivative is turned into x = a. To paraphrase him - as there is no value of x that makes the denominator infinite. So the conclusion is x = a.
Homework Helper
Gold Member
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I understand that bit. But it’s presented in the book as an example, and he mentions specifically that you have no idea it’s circle, so suppose you were presented with the original equation and asked to find the minimum/ maximum.
You’d need find the derivative if I’m not mistaken?
Not necessarily. You know, for example, that ##(x-a)^2## has a minimum at ##x =a## without using calculus.
My confusion comes about when the derivative is turned into x = a.
Do you mean when the derivative is evaluated at ##x =a##?
To paraphrase him - as there is no value of x that makes the denominator infinite. So the conclusion is x = a.
With any curve you must do some work to establish what type of curve you are dealing with. Take ##y = x^3##. The derivative is zero at ##x =0## and the behaviour of the curve as ##x \to \pm \infty## tells you this must be an infection point.
The equation of a circle is not a function, so you need some care in working with its derivatives. Technically, you might want to confirm that ##y## is bounded, as are the allowable values of ##x##.
I think it's better to figure this out for yourself than to try to decrypt some obscure remarks from a professor.
dextercioby
Staff Emeritus
Homework Helper
The derivative vanishes if the numerator is 0 and the denominator isn't or if the denominator goes to infinity faster than the numerator does. The author is pointing out in this case that only the first case matters.
Last edited:
Homework Helper
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PS you do know that's a circle, because a circle of radius r is defined as the set of points precisely a distance ##r## from a given point. That, by inspection, is the equation you are given - with the centre at ##(a,a)##.
dextercioby
Mentor
If you follow the logic of not knowing it's a circle, then it makes sense to take the derivative of the function and determine what values give you a zero slope.
In this case, we can see when x=a, the derivative will be zero.
The denominator comment reminds you that you're not dividing by zero, so things are good to say x=a is where the maximum/minimum is.
dextercioby
... so suppose you were presented with the original equation and asked to find the minimum/ maximum.
I suppose you know how to do that if the functions is explicitly given as ##y=f(x)##. In your case it is implicitly given by an equation say ##F(x,y)=0##. To find the min/max of ##y## you do the same as in the explicit case except that the derivative now is calculated implicitly form the equation i.e. ##\frac{dy}{dx}=-\frac{F_x}{F_y}##.
dextercioby
Mentor
We have the equation for a circle, and its derivative:
$$(y-a)^2 + (x-a)^2 = r^2$$
$$\frac{dy}{dx} = \frac{a-x}{(r^{2}-(x-a)^2)^{1/2}} = 0$$
Your first equation is of a circle whose center is at the point (a, a) and whose radius is r. A more general equation would put the center at an arbitrary point (a, b), say.
I have a couple of problems with your second equation, though.
First, this says that the derivative is zero. That's not true in general and is true only for two points: (a, a + r) and (a, a - r).
Second, the derivative should have two values for a given value of x.
Starting with the equation ##(x - a)^2 + (y - a)^2 = r^2##, implicit differentiation gives ##2(x - a) + 2(y - a)\frac{dy}{dx} = 0##.
Solving for dy/dx gives $$\frac{dy}{dx} = \frac{a - x}{y - a} = \frac{a - x}{\pm\sqrt{r^2 - (x - a)^2}}$$ | 1,468 | 5,581 | {"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} | 4.0625 | 4 | CC-MAIN-2023-14 | latest | en | 0.925956 |
http://docplayer.net/27368028-Chapter-newton-raphson-method-of-solving-a-nonlinear-equation.html | 1,542,069,585,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741176.4/warc/CC-MAIN-20181113000225-20181113022225-00558.warc.gz | 85,384,060 | 26,492 | # Chapter Newton-Raphson Method of Solving a Nonlinear Equation
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1 Chpter.4 Newton-Rphson Method of Solvng Nonlner Equton After redng ths chpter, you should be ble to:. derve the Newton-Rphson method formul,. develop the lgorthm of the Newton-Rphson method,. use the Newton-Rphson method to solve nonlner equton, nd 4. dscuss the drwbcks of the Newton-Rphson method. Introducton Methods such s the bsecton method nd the flse poston method of fndng roots of nonlner equton f ( ) requre brcketng of the root by two guesses. Such methods re clled brcketng methods. These methods re lwys convergent snce they re bsed on reducng the ntervl between the two guesses so s to zero n on the root of the equton. In the Newton-Rphson method, the root s not brcketed. In fct, only one ntl guess of the root s needed to get the tertve process strted to fnd the root of n equton. The method hence flls n the ctegory of open methods. Convergence n open methods s not gurnteed but f the method does converge, t does so much fster thn the brcketng methods. Dervton The Newton-Rphson method s bsed on the prncple tht f the ntl guess of the root of f ( ) s t, then f one drws the tngent to the curve t f ( ), the pont where the tngent crosses the -s s n mproved estmte of the root (Fgure ). Usng the defnton of the slope of functon, t f = tn θ f =, whch gves f = () f.4.
2 .4. Chpter.4 Equton () s clled the Newton-Rphson formul for solvng nonlner equtons of the form f. So strtng wth n ntl guess,, one cn fnd the net guess,, by usng Equton (). One cn repet ths process untl one fnds the root wthn desrble tolernce. Algorthm The steps of the Newton-Rphson method to fnd the root of n equton. Evlute f symbolclly. Use n ntl guess of the root, f = f f re, to estmte the new vlue of the root,, s. Fnd the bsolute reltve ppromte error s = 4. Compre the bsolute reltve ppromte error wth the pre-specfed reltve error tolernce, s. If > s, then go to Step, else stop the lgorthm. Also, check f the number of tertons hs eceeded the mmum number of tertons llowed. If so, one needs to termnte the lgorthm nd notfy the user. f () f ( ) [, f ( )] f ( + ) θ + + Fgure Geometrcl llustrton of the Newton-Rphson method.
3 Newton-Rphson Method.4. Emple You re workng for DOWN THE TOILET COMPANY tht mkes flots for ABC commodes. The flotng bll hs specfc grvty of.6 nd hs rdus of 5.5 cm. You re sked to fnd the depth to whch the bll s submerged when flotng n wter. Fgure Flotng bll problem. The equton tht gves the depth n meters to whch the bll s submerged under wter s gven by Use the Newton-Rphson method of fndng roots of equtons to fnd ) the depth to whch the bll s submerged under wter. Conduct three tertons to estmte the root of the bove equton. b) the bsolute reltve ppromte error t the end of ech terton, nd c) the number of sgnfcnt dgts t lest correct t the end of ech terton. Soluton 4 f f. Let us ssume the ntl guess of the root of f s. 5 m. Ths s resonble guess (dscuss why nd.m re not good choces) s the etreme vlues of the depth would be nd the dmeter (. m) of the bll. Iterton The estmte of the root s f f
4 .4.4 Chpter.4 The bsolute reltve ppromte error 9.9% t the end of Iterton s The number of sgnfcnt dgts t lest correct s, s you need n bsolute reltve ppromte error of 5% or less for t lest one sgnfcnt dgt to be correct n your result. Iterton The estmte of the root s f f The bsolute reltve ppromte error % The mmum vlue of m for whch sgnfcnt dgts t lest correct n the nswer s. Iterton The estmte of the root s f f.99 4 t the end of Iterton s The bsolute reltve ppromte error m.5 s.844. Hence, the number of.99 4 t the end of Iterton s
5 Newton-Rphson Method The number of sgnfcnt dgts t lest correct s 4, s only 4 sgnfcnt dgts re crred through n ll the clcultons. Drwbcks of the Newton-Rphson Method. Dvergence t nflecton ponts If the selecton of the ntl guess or n terted vlue of the root turns out to be close to the nflecton pont (see the defnton n the ppend of ths chpter) of the functon f n the equton f, Newton-Rphson method my strt dvergng wy from the root. It my then strt convergng bck to the root. For emple, to fnd the root of the equton f.5 the Newton-Rphson method reduces to ( ).5 = ( ) Strtng wth n ntl guess of 5., Tble shows the terted vlues of the root of the equton. As you cn observe, the root strts to dverge t Iterton 6 becuse the prevous estmte of.9589 s close to the nflecton pont of (the vlue of f ' s zero t the nflecton pont). Eventully, fter more tertons the root converges to the ect vlue of.. Tble Dvergence ner nflecton pont. Iterton Number
6 .4.6 Chpter.4 Fgure Dvergence t nflecton pont for. Dvson by zero For the equton 6 f.. 4 the Newton-Rphson method reduces to.. 4 =.6 6 f. For or., dvson by zero occurs (Fgure 4). For n ntl guess close to. such s. 999, one my vod dvson by zero, but then the denomntor n the formul s smll number. For ths cse, s gven n Tble, even fter 9 tertons, the Newton-Rphson method does not converge. Tble Dvson by ner zero n Newton-Rphson method. Iterton f ( ) Number %
7 Newton-Rphson Method.4.7. E E-6 f() 5.E-6.5E-6.E E E-6-7.5E-6 -.E-5 Fgure 4 Ptfll of dvson by zero or ner zero number.. Osclltons ner locl mmum nd mnmum Results obtned from the Newton-Rphson method my oscllte bout the locl mmum or mnmum wthout convergng on root but convergng on the locl mmum or mnmum. Eventully, t my led to dvson by number close to zero nd my dverge. For emple, for f the equton hs no rel roots (Fgure 5 nd Tble ). 6 f() Fgure 5 Osclltons round locl mnm for f.
8 .4.8 Chpter.4 Tble Osclltons ner locl mm nd mnm n Newton-Rphson method. Iterton f ( ) Number % Root jumpng In some cse where the functon f () s osclltng nd hs number of roots, one my choose n ntl guess close to root. However, the guesses my jump nd converge to some other root. For emple for solvng the equton sn f you choose s n ntl guess, t converges to the root of s shown n Tble 4 nd Fgure 6. However, one my hve chosen ths s n ntl guess to converge to Tble 4 Root jumpng n Newton-Rphson method. Iterton f ( ) Number %
9 Newton-Rphson Method.4.9 f() Fgure 6 Root jumpng from ntended locton of root for sn f. Append A. Wht s n nflecton pont? For functon f, the pont where the concvty chnges from up-to-down or down-to-up s clled ts nflecton pont. For emple, for the functon f, the concvty chnges t (see Fgure ), nd hence (,) s n nflecton pont. An nflecton ponts MAY est t pont where f ( ) nd where f ''( ) does not est. The reson we sy tht t MAY est s becuse f f ( ), t only mkes t possble nflecton pont. For emple, for f ( ) 4 6, f ( ), but the concvty does not chnge t. Hence the pont (, 6) s not n nflecton pont of f ( ) 4 6. For f, f ( ) chnges sgn t ( f ( ) for, nd f ( ) for ), nd thus brngs up the Inflecton Pont Theorem for functon f () tht sttes the followng. If f '( c) ests nd f (c) chnges sgn t c, then the pont ( c, f ( c)) s n nflecton pont of the grph of f. Append B. Dervton of Newton-Rphson method from Tylor seres Newton-Rphson method cn lso be derved from Tylor seres. For generl functon f, the Tylor seres s f" f f f +! As n ppromton, tkng only the frst two terms of the rght hnd sde, f f f nd we re seekng pont where f, tht s, f we ssume f,
10 .4. Chpter.4 f whch gves f f f' Ths s the sme Newton-Rphson method formul seres s derved prevously usng the geometrc method. NONLINEAR EQUATIONS Topc Newton-Rphson Method of Solvng Nonlner Equtons Summry Tet book notes of Newton-Rphson method of fndng roots of nonlner equton, ncludng convergence nd ptflls. Mjor Generl Engneerng Authors Autr Kw Dte December, 9 Web Ste
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Why network is n essentil productivity tool for ny smll business Effective technology is essentil for smll businesses looking to increse the productivity of their people nd processes. Introducing technology | 10,080 | 33,801 | {"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} | 3.78125 | 4 | CC-MAIN-2018-47 | latest | en | 0.767652 |
https://area.citizenmaths.com/41.6-hm2-to-dam2 | 1,620,504,643,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988923.22/warc/CC-MAIN-20210508181551-20210508211551-00244.warc.gz | 124,831,662 | 11,733 | # 41.6 Square Hectometers to Square Decameters
Square Hectometer
• Acre
• Are
• Barony
• Centiare
• Circular mil
• Decare
• Dunam
• Hectare
• Hide
• Milacre
• Perch
• Quarter-milacre
• Rood
• Roofing Square
• Section
• Square Arpent
• Square Centimeter
• Square Chain
• Square Decameter
• Square Decimeter
• Square Feet
• Square Hectometer
• Square Inch
• Square Kilometer
• Square Meter
• Square Mil
• Square Mile
• Square Millimeter
• Square rod
• Square Yard
• Stremma
• Thousand Circular Mil
• Township (Survey)
• Yardland (US Survey)
=
Square Decameter
• Acre
• Are
• Barony
• Centiare
• Circular mil
• Decare
• Dunam
• Hectare
• Hide
• Milacre
• Perch
• Quarter-milacre
• Rood
• Roofing Square
• Section
• Square Arpent
• Square Centimeter
• Square Chain
• Square Decameter
• Square Decimeter
• Square Feet
• Square Hectometer
• Square Inch
• Square Kilometer
• Square Meter
• Square Mil
• Square Mile
• Square Millimeter
• Square rod
• Square Yard
• Stremma
• Thousand Circular Mil
• Township (Survey)
• Yardland (US Survey)
Formula 41.6 hm² = 41.6 x 100 dam2 = 4,160 dam2
41.6 hm² = 4,160 dam2
Explanation:
• 1 hm² is equal to 100 dam2, therefore 41.6 hm² is equivalent to 4,160 dam2.
• 1 Square Hectometer = 1 x 100 = 100 Square Decameters
• 41.6 Square Hectometers = 41.6 x 100 = 4,160 Square Decameters
## 41.6 Square Hectometers to Square Decameters Conversion Table
Square Hectometer (hm²) Square Decameter (dam2)
41.7 hm² 4,170 dam2
41.8 hm² 4,180 dam2
41.9 hm² 4,190 dam2
42 hm² 4,200 dam2
42.1 hm² 4,210 dam2
42.2 hm² 4,220 dam2
42.3 hm² 4,230 dam2
42.4 hm² 4,240 dam2
42.5 hm² 4,250 dam2
## Convert 41.6 hm² to other units
Unit Unit of Area
Thousand Circular Mil 820,986,500,414.18 kcmil
Yardland (US Survey) 3.426528 yardland
Township (Survey) 4.5e-03 twp
Stremma 416.0 stremma
Square Mil 644,800,000,000,000.0 mil2
Square Chain 1,027.96 ch2
Section 1.6e-01 section
Roofing Square 44,777.87 roofingsquare
Rood 411.183 ro
Quarter-milacre 411,183.35 1/4milacre
Square rod 16,447.33 rd2
Perch 12,167.68 perch
Milacre 102,795.84 milacre
Hide 1.027958 hide
Dunam 416.0 dunam
Decare 416.0 daa
Circular mil 820,986,500,414,179.5 cmil
Centiare 416,000.0 ca
Barony 2.6e-02 ba
Square Arpent 121.682 arpent
Are 4,160.0 a
Square Decimeter 41,600,000.0 dm²
Square Yard 497,531.86 yd²
Square Mile 1.6e-01 mi²
Square Inch 644,801,289.6 in²
Square Kilometer 4.2e-01 km²
Square Centimeter 4,160,000,000.0 cm²
Square Millimeter 416,000,000,000.0 mm²
Acre 102.796 ac
Hectare 41.6 ha
Square Decameter 4,160.0 dam2
Square Feet 4,477,786.73 ft2
Square Meter 416,000.0 m² | 1,006 | 2,572 | {"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} | 3.8125 | 4 | CC-MAIN-2021-21 | latest | en | 0.491575 |
http://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-6th-edition/chapter-1-section-1-4-properties-of-real-numbers-and-algebraic-expressions-exercise-set-page-39/11 | 1,521,546,435,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647406.46/warc/CC-MAIN-20180320111412-20180320131412-00610.warc.gz | 398,733,950 | 13,346 | ## Intermediate Algebra (6th Edition)
2($x$ - 6) $\gt$ $\frac{1}{11}$
"Twice the difference" translates as: 2 $\times$ a subtraction expression "The difference of $x$ and 6" ($x$ - 6) "is greater than" $\gt$ "the reciprocal of 11" $\frac{1}{11}$ OR 2($x$ - 6) $\gt$ $\frac{1}{11}$ | 105 | 281 | {"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} | 3.859375 | 4 | CC-MAIN-2018-13 | latest | en | 0.681009 |
https://blog.rotovalue.com/index.php/2012/02/19/batting-around-crystal-baseballs/ | 1,670,640,913,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711637.64/warc/CC-MAIN-20221210005738-20221210035738-00321.warc.gz | 172,993,765 | 12,366 | Batting Around Crystal Baseballs
Matt Swartz tested several statistics projections against actual 2011 numbers here. He follows the guidelines Tom Tango outlined in this post, computing weighted on base average (wOBA) for each player and then comparing the actual wOBA to the projection. From the errors on individual players he computed both mean absolute error and root mean square error.
One of Matt’s assumptions was to run the calculations only for players with 200 or more plate appearances in 2011. In this post I’ll show results from a similar analyses without that cutoff, but compute the same MAE and RMSE values for the systems that I have data.
Not all projection systems made estimates for every player who actually played. I considered three different ways of addressing this:
1. Compare only those players which have projections from all systems. This way you’re comparing the exact same players from each system, but you’re limited to fewer total players in the averages.
2. Compare only the players projected by a given system. This uses all the projection data, but a system which projects more players, particularly young ones with little or no MLB experience, would fair poorer in this comparison relative to a system which projects fewer, more established players.
3. Fill in data for “missing” players if a system hasn’t projected a given player. This has the advantage of being able to test all players, but it implies that for some systems, you use an arbitrary number for any players with no data. I’m using the cumulative forecast wOBA of all projected players.
In his analysis, Matt only followed the third method. I’ll be reporting data for all three.
I currently have data for five different projection systems:
• Marcel – This is Tom Tango’s system, “that uses as little intelligence as possible.”
• Oliver – The Hardball Times’s system, developed by Brian Cartwright.
• PECOTA – Baseball Prospectus’s system, now maintained by Colin Wyers.
• Steamer – A system developed by Jared Cross, Dash Davidson, and Peter Rosenbloom
• RotoValue – A basic projection model I’ve developed and applied to both MLB and NBA basketball statistics, which I described previously here.
In addition, I’m using 2010 actual MLB data as a projection for 2011. This is a simple control: if your projections are less accurate than last year’s numbers, you’re not adding value.
First, the results only averaging players projected by all systems:
Source Num Avg wOBA MAE RMSE
Actual 404 0.3263 0.0000 0.0000
Marcel 404 0.3333 0.0272 0.0354
Oliver 404 0.3315 0.0261 0.0345
PECOTA 404 0.3292 0.0265 0.0346
RotoValue 404 0.3225 0.0323 0.0435
Steamer 404 0.3351 0.0265 0.0349
2010 404 0.3311 0.0356 0.0478
For this set, Oliver edged out PECOTA and Steamer for the lowest MAE and RMSE, with Marcel next, RotoValue much further back, and 2010 the worst of all.
Next, I averaged all players in the system, computing errors for those who actually played in MLB in 2011:
Source Num wOBA MLB wOBA StdDev MAE RMSE
Actual 640 0.3224 640 0.3224 0.0446 0.0000 0.0000
Marcel 828 0.3246 525 0.3319 0.0235 0.0278 0.0367
Oliver 2090 0.2895 639 0.3272 0.0309 0.0270 0.0361
PECOTA 943 0.3089 601 0.3260 0.0277 0.0271 0.0359
RotoValue 504 0.3213 420 0.3220 0.0412 0.0324 0.0436
Steamer 619 0.3332 534 0.3325 0.0291 0.0269 0.0358
2010 616 0.3270 491 0.3260 0.0505 0.0384 0.0529
Num refers to the total number of projections of any player by the system, and MLB is the number of projected players who played in 2011. The first wOBA column is the cumulative wOBA of all players in the sample, and the second is the average (weighted by plate 2011 plate appearances) of those who played in 2011. So systems like Oliver and PECOTA, which projected many more players than actually played in 2011, see their overall wOBA pulled down, but when restricted to those who actually played in 2011, their averages are now in line with the other systems.
In this analysis, Steamer edged out Oliver and PECOTA for the lowest MAE and RMSE, with Marcel close behind, RotoValue further back, and again 2010 much further back.
Now for the third and final table, this time filling in unforecast players with the average forecast wOBA:
Source Num wOBA MLB wOBA StdDev MAE RMSE
Actual 640 0.3224 640 0.3224 0.0446 0.0000 0.0000
Marcel 828 0.3246 640 0.3315 0.0229 0.0285 0.0377
Oliver 2090 0.2895 640 0.3272 0.0309 0.0270 0.0361
PECOTA 943 0.3089 640 0.3258 0.0276 0.0274 0.0365
RotoValue 504 0.3213 640 0.3219 0.0380 0.0330 0.0447
Steamer 619 0.3332 640 0.3325 0.0284 0.0277 0.0371
2010 616 0.3270 640 0.3261 0.0484 0.0388 0.0531
Oliver has the lowest RMSE and MAE this time, closely followed by PECOTA, Steamer, and Marcel in that order. Once again RotoValue is further back, with 2010 data even further behind.
I hesitate to crown a “champion” as the best projection system, as the differences between the top 4 quite small, enough so that the particular assumptions you use may change the ordering among them. I can confidently say that my own system is not as good as the others, but is still a significant improvement over using 2010 data. It would be interesting to see how systems do over several years.
This analysis focused on projecting a rate statistic, measuring offensive performance. So systems get no credit for predicting playing time well, although they do get more credit for being closer on players who play a lot than on those who hardly play. Perhaps I’ll try a different method, which should take into account playing time as well.
Please offer suggestions and feedback in the comments, or you can e-mail me at geoff at rotovalue dot com. If I can get data for other projection systems, I’ll rerun the analysis with them included. And if you want to do your own analysis of the data I have, I’ve posted a comma-delimited text file here.
1. One other difference between the analysis I did and the one Matt did: when Matt had missing players, he used wOBA – 0.020 for systems other than Marcel (which had defined wOBA as the missing value). And Matt defined his missing value in terms of the average of players who played in 2011, weighted by actual PA.
I used the forecast average wOBA across all forecast players, weighted by forecast PA. For systems with very many players (and thus a lower average wOBA) like PECOTA and Oliver, this meant using a very low “missing” value, but those systems had very few missing players. The errors for systems like mine and Steamer, which had relatively many missing players, would be a trifle lower with Matt’s method, but the differences really are tiny.
2. Tom Tango asked to use a different computation for missing players. Rather than use the cumulative wOBA of all forecast players (which for some systems is much lower, due to projecting players not at all expected to play in the majors), he suggested using wOBA weighted by actual plate appearances, and then subtracting 0.020 for systems other than Marcel. I coded that up and tested it. This makes very little difference:
Source Num wOBA MLB wOBA StdDev MAE RMSE
Actual 640 0.3225 640 0.3224 0.0446 0.0000 0.0000
Marcel 828 0.3246 640 0.3319 0.0228 0.0287 0.0379
Oliver 2090 0.2895 640 0.3272 0.0309 0.0270 0.0361
PECOTA 943 0.3089 640 0.3258 0.0276 0.0273 0.0364
RotoValue 504 0.3213 640 0.3190 0.0387 0.0326 0.0440
Steamer 619 0.3332 640 0.3316 0.0288 0.0273 0.0365
2010 616 0.3270 640 0.3244 0.0487 0.0384 0.0525
Oliver is unchanged, and PECOTA improves just an iota. RotoValue, Steamer, and the 2010 season improve a bit more, but still not much at all. And interestingly, Marcel gets ever so slightly worse. | 2,222 | 7,609 | {"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} | 2.75 | 3 | CC-MAIN-2022-49 | longest | en | 0.940256 |
https://crypto.stackexchange.com/questions/44271/what-is-the-maximum-key-size-for-a-128-bit-aes | 1,660,478,069,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572033.91/warc/CC-MAIN-20220814113403-20220814143403-00641.warc.gz | 188,042,183 | 66,944 | # What is the maximum key size for a 128 bit AES?
What is the maximum key size for AES 128, Will using a key greater than maximum gives extra security or error? Is there a minimum key size?
Suppose a key is 128 bit, does it means The key is of length 16characters
Suggest me a good password to key function.
• 1. 128 bits is the largest 128 bit key. Conversely, the smallest 128 bit key is 128 bits. 2. Characters are not always 8 bits so there is no translation between key size and number of characters 3. The fastest function to transform a password into a key is to ignore the password and return the zero key. This is good if speed is the important measure. Feb 28, 2017 at 19:06
• An AES-128 key consists of exactly 128 bits / 16 bytes, no more, no less. Feb 28, 2017 at 19:34
• Q: Will using more than four wheels on my car give me more stability or error? A: the question is nonsense because my car doesn't even have a spot to put more than 4 wheels. I'd have to totally redesign the car. Feb 28, 2017 at 23:14
Despite similarities, it is really important to understand that passwords and cryptographic keys should not be carelessly conflated. Some important contrasts:
• Passwords are normally selected by human beings according to their whims. Cryptographic keys are meant to be randomly generated by an algorithm.
• Passwords are usually intended to be memorized by human beings. A strong cryptographic key cannot generally be memorized—they're too random and complex—and thus generally stored in a secure device. In the parlance of multi-factor authentication, passwords are "something you know" while cryptographic keys are "something you have."
• Passwords have to be text that human beings can manually input into their devices. Cryptographic keys are fundamentally binary data (despite being sometimes serialized and deserialized as text) and are generally not meant to input manually.
I say all this because your question really comes down to a lack of understanding of these differences. AES doesn't use passwords, it uses keys. AES-128 keys are fixed-length binary data; they don't really have "characters" because they're not text. As others have said, the maximum key size for AES-128 is the same as the minimum and only key size: 128 bits.
• I'm open to that, but many (if not most) bcrypt APIs in practice don't seem oriented to getting out key material, instead just giving you an /etc/shadow-style encoded blob that you'd have to parse the key out of. Feb 28, 2017 at 19:43 | 579 | 2,506 | {"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} | 2.75 | 3 | CC-MAIN-2022-33 | longest | en | 0.948158 |
https://ca.answers.yahoo.com/question/index?qid=20070615163218AA9c4PR | 1,604,187,006,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107922463.87/warc/CC-MAIN-20201031211812-20201101001812-00592.warc.gz | 239,915,095 | 26,494 | Anonymous
# Derivatives: Chain Rule - Melting Ice Block?
A cubical block of ice is melting in such a way that each edge decreses steadily by 5.2 cm every hour. At what rate is its volume decreasing when each edge is 5 meters long?
Solution: Let l=l(t) be the length of each edge at time t. Then the volume of the block is given by V = _______ .
Since the length of each edge is changing in time, we conclude that the volume V is also a function of time t. We note that the rate of change of l is constant and that is given as [(dl)/(dt)]= _______cm/h = ________m/h.
The question is to find the rate of change of V when
l= ____________m.
The chain rule gives
dV/dt = (dV/dl)*(dl/dt)
Therfore , when each edge is 5 m long, the rate of change of the volume of the ice block is ________m3/h.
I don't know how to approach this question..
Thank you
Relevance
Volume of a cube with side l is: V = l^3
dl/dt = 5.2 cm/hr = 0.052 m/hr
Find rate of change of V when l = 5 m.
Ok, for this last part, you need to use the chain rule as you have it stated: dV/dt = (dV/dl)*(dl/dt)
Now dV/dl is simply the derivative of the first equation with respect to l. This gives:
dV/dl = 3l^2
Now plug this into the equation for dV/dt.
dV/dt = 3l^2*(dl/dt)
Now you can substitute in all the constants, l = 5 m, dl/dt = 0.052 m/hr, to get:
dV/dt = 3*(5 m)^2*(0.052 m/hr)
= 75*0.052 m^3/hr
= 3.9 m^3/hr
Hope this clears it up for you.
V = l^3 = 5*5*5 = 125 cubic meters
dl/dt = 5.2 cm/hour = 0.052 meters per hour
Since V = l^3, then dV/dl = 3l^2 (as long as each edge shrinks at the same rate).
Then dV/dt = (dV/dl)*(dl/dt) = 3l^2*(dl/dt) = (3*5*5*)(0.052) = 3.9 cubic meters per hour.
• Como
Lv 7
dl /dt = 5.2 cm / h
dl /dt = 0.052 m / h
V = l ³
dV/dl = 3 l ² m ²
dV/dt = (dv / dl) x (dl / dt)
dV/dt = 3 x 5² x 0.052 m³ / h (when l = 5m)
dV/dt = 3.9 m³ / h | 679 | 1,867 | {"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} | 4.25 | 4 | CC-MAIN-2020-45 | latest | en | 0.881987 |
https://www.tutorialspoint.com/python_data_structure/python_sorting_algorithms.htm | 1,675,922,874,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764501407.6/warc/CC-MAIN-20230209045525-20230209075525-00845.warc.gz | 1,051,109,856 | 9,974 | # Python - Sorting Algorithms
Sorting refers to arranging data in a particular format. Sorting algorithm specifies the way to arrange data in a particular order. Most common orders are in numerical or lexicographical order.
The importance of sorting lies in the fact that data searching can be optimized to a very high level, if data is stored in a sorted manner. Sorting is also used to represent data in more readable formats. Below we see five such implementations of sorting in python.
• Bubble Sort
• Merge Sort
• Insertion Sort
• Shell Sort
• Selection Sort
## Bubble Sort
It is a comparison-based algorithm in which each pair of adjacent elements is compared and the elements are swapped if they are not in order.
### Example
def bubblesort(list):
# Swap the elements to arrange in order
for iter_num in range(len(list)-1,0,-1):
for idx in range(iter_num):
if list[idx]>list[idx+1]:
temp = list[idx]
list[idx] = list[idx+1]
list[idx+1] = temp
list = [19,2,31,45,6,11,121,27]
bubblesort(list)
print(list)
### Output
When the above code is executed, it produces the following result −
[2, 6, 11, 19, 27, 31, 45, 121]
## Merge Sort
Merge sort first divides the array into equal halves and then combines them in a sorted manner.
### Example
def merge_sort(unsorted_list):
if len(unsorted_list) <= 1:
return unsorted_list
# Find the middle point and devide it
middle = len(unsorted_list) // 2
left_list = unsorted_list[:middle]
right_list = unsorted_list[middle:]
left_list = merge_sort(left_list)
right_list = merge_sort(right_list)
return list(merge(left_list, right_list))
# Merge the sorted halves
def merge(left_half,right_half):
res = []
while len(left_half) != 0 and len(right_half) != 0:
if left_half[0] < right_half[0]:
res.append(left_half[0])
left_half.remove(left_half[0])
else:
res.append(right_half[0])
right_half.remove(right_half[0])
if len(left_half) == 0:
res = res + right_half
else:
res = res + left_half
return res
unsorted_list = [64, 34, 25, 12, 22, 11, 90]
print(merge_sort(unsorted_list))
### Output
When the above code is executed, it produces the following result −
[11, 12, 22, 25, 34, 64, 90]
## Insertion Sort
Insertion sort involves finding the right place for a given element in a sorted list. So in beginning we compare the first two elements and sort them by comparing them. Then we pick the third element and find its proper position among the previous two sorted elements. This way we gradually go on adding more elements to the already sorted list by putting them in their proper position.
### Example
def insertion_sort(InputList):
for i in range(1, len(InputList)):
j = i-1
nxt_element = InputList[i]
# Compare the current element with next one
while (InputList[j] > nxt_element) and (j >= 0):
InputList[j+1] = InputList[j]
j=j-1
InputList[j+1] = nxt_element
list = [19,2,31,45,30,11,121,27]
insertion_sort(list)
print(list)
### Output
When the above code is executed, it produces the following result −
[2, 11, 19, 27, 30, 31, 45, 121]
## Shell Sort
Shell Sort involves sorting elements which are away from each other. We sort a large sublist of a given list and go on reducing the size of the list until all elements are sorted. The below program finds the gap by equating it to half of the length of the list size and then starts sorting all elements in it. Then we keep resetting the gap until the entire list is sorted.
### Example
def shellSort(input_list):
gap = len(input_list) // 2
while gap > 0:
for i in range(gap, len(input_list)):
temp = input_list[i]
j = i
# Sort the sub list for this gap
while j >= gap and input_list[j - gap] > temp:
input_list[j] = input_list[j - gap]
j = j-gap
input_list[j] = temp
# Reduce the gap for the next element
gap = gap//2
list = [19,2,31,45,30,11,121,27]
shellSort(list)
print(list)
### Output
When the above code is executed, it produces the following result −
[2, 11, 19, 27, 30, 31, 45, 121]
## Selection Sort
In selection sort we start by finding the minimum value in a given list and move it to a sorted list. Then we repeat the process for each of the remaining elements in the unsorted list. The next element entering the sorted list is compared with the existing elements and placed at its correct position.So, at the end all the elements from the unsorted list are sorted.
### Example
def selection_sort(input_list):
for idx in range(len(input_list)):
min_idx = idx
for j in range( idx +1, len(input_list)):
if input_list[min_idx] > input_list[j]:
min_idx = j
# Swap the minimum value with the compared value
input_list[idx], input_list[min_idx] = input_list[min_idx], input_list[idx]
l = [19,2,31,45,30,11,121,27]
selection_sort(l)
print(l)
### Output
When the above code is executed, it produces the following result −
[2, 11, 19, 27, 30, 31, 45, 121] | 1,283 | 4,804 | {"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} | 3.609375 | 4 | CC-MAIN-2023-06 | latest | en | 0.650292 |
https://www.excelforum.com/excel-formulas-and-functions/940942-cell-referencing-two-columns-with-conditions.html | 1,660,669,867,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572408.31/warc/CC-MAIN-20220816151008-20220816181008-00729.warc.gz | 667,080,824 | 15,899 | # Cell referencing two columns with conditions
1. ## Cell referencing two columns with conditions
hi,
Please assist in creating a formula related to cell referencing, described as follow,
I have two colums, ex sheet 1 named jentry , d9 = debit and e9= credit. in sheet 2, e2, a formula referencing to both D9 and E9 , with conditions that if d9 will be with value more than 0, use d9 and if E9 will be with value >0 use -(minus) e9 , and if both them will be with value less that 0 or both column with value ignore , or it should be that only one column D9, or E9 will be with value.
I already created one but still does not fit my requirement
=IF(AND(jentry!D9>0,jentry!D9>0),+jentry!D9,IF(AND(jentry!D9>0,0),+jentry!D9,IF(AND(jentry!E9>0,jentry!E9>0),-jentry!E9)))
I attached my template for reference
Ofel
2. ## Re: Help with formula
Your post does not comply with Rule 1 of our Forum RULES. Your post title should accurately and concisely describe your problem, not your anticipated solution.
To change a Title on your post, click EDIT then Go Advanced and change your title, if 2 days have passed ask a moderator to do it for you.
3. ## Re: Formula problem
Hi galang,
Change your title again and attach a sample workbook ( if possible ) without sensitive data including before and outcome that you want so it will be easier for people to understand and help with your problem.
4. ## Re: Cell referencing two columns with conditions
Hi,
Problem solved formula that meet my requirement as follows
=IF(AND(jentry!D25>0,jentry!E25>0),0,IF(AND(jentry!D25<0,jentry!E25<0),0,IF(AND(jentry!D25>0,jentry!E25<0),0,IF(AND(jentry!D25<0,jentry!E25>0),0,IF(jentry!D25>0,+jentry!D25,IF(jentry!E25>0,-jentry!E25,0))))))
Thanks a lot
Ofel
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# A question about volume (caution: Mathematics and physics question)? Answered
Ok, I know how to figure the "volume of a sphere" and the surface area of said sphere, now I need to know how to figure how much I need to remove from the outside, to reduce the volume by about 1/2.
I suspect this will NOT reduce the size of the entire sphere by 1/2....at least, that doesn't seem logical to me.
Can anyone explain it so that I (not a mathematician) can understand it?
Tags:
## 20 Replies
CameronSS (author)2010-09-09
Volume is a cubic function, area is a square function, distance is a linear function. If you double the area (linear), the surface area (square) will be quadrupled (22=4) and the volume (cubic) will be octupled, if that's even a word (23=8).
To determine linear proportions from volume, work backwards. To make the volume 1/8 of what it was, the diameter must be halved (cube root of 1/8 is 1/2). To make the volume half of what it was, the volume must be the cube root of 1/2 of the original, about 0.79. So if you start with a sphere of diameter 1", a sphere with diameter .79' will have about half the radius.
The same concept is true for all three-dimensional objects, regardless of shapes.
Goodhart (author)2010-09-09
Ok, thanks.....let me see if I can plug that into what I am working on (the numbers are a bit on the large size, on the order of 1,409,147,148,464,966,056.25 CKm, etc. as the volume)....
CameronSS (author)2010-09-09
You know, we already have gravity holding us in place, you don't need to put Gorilla Glue on the sun to win the contest...
Goodhart (author)2010-09-09
Nothing like that.....I have heard someone say that the sun would have to be "double" the size it is now, if it has already used 1/2 of it's fuel, at the very beginning. Now, right away I gave him the ole o_0 look and said I'd get back to him on that (I couldn't do this in my head, and some of it I was unfamiliar with).
Now, I know my figuring will have a "grave error" in it. but it will be to his advantage and I suspect that it will STILL not come even close to what this innumerate, supposed mathematician has come up with. LOGIC tells me he's wrong, without even doing the math.
CameronSS (author)2010-09-09
Yep, flawed logic. He's going off the (incorrect) assumption that as the sun "consumes" fuel, it disappears and does not contribute to the total volume. It's nuclear fusion, which does convert mass into energy, but only a small fraction of the mass. And it would only have to start off with about a 25% larger diameter if his method was correct.
Kelsey probably knows a better answer, but he's slacking off on the forum-answering, apparently. Probably off creating a black hole to eat the solar system or something.
Goodhart (author)2010-09-09
Yes, but even if we assume a reduction of 1/2 over the last, what, 5 X1011 years, we STILL don't get a sun that is doubled in size that would consume (another flawed bit of logic) the planets out to Mars.
Oooo, in a way I hate pseudo-science, and in a way I like it....it keeps me thinking.... ;-) I have to work from scratch though....not having a heavily mathematical background....I have to figure it out as I go along.
So, I came up with....if 1/2 the volume of mass were removed we'd get a surface area reduction from 6,078,428,661,500 CKM down to 4,801,958,642,585 CKm So, even given his false information, the math still doesn't support his determination.
kelseymh (author)2010-09-09
I think you mean 4.6 x 10^9. The whole Universe is only 13.9 x 10^9 years old.
CameronSS (author)2010-09-09
We're talking Universe 2.0.
Goodhart (author)2010-09-11
On the other side of the "great expanse"
Goodhart (author)2010-09-09
Um, yeah, I am still having a few problems with converting to the short hand.... 10 9 = 10,000,000,000 right ?
CameronSS (author)2010-09-09
109=1 000 000 000
The exponent gives you the number of zeros to put after the 1.
Goodhart (author)2010-09-10
Right, that was what I had thought I had done earlier... so 15 billion = 15,000,000,000 = 15 x 109 (don't ask me WHERE I got 11 from....I was kind of tired last night).
CameronSS (author)2010-09-09
Oh, he might be getting confuzzled from the sun's end-of-life process. If I recall correctly, it will eventually start venting matter and expanding due to decreased gravity, and I think the prediction is that when it swells to a red giant, it will head out toward Mars' orbit. Of course, the reduced gravity also means that the orbits will all be farther out, so it gets all weird.
Phil Plait and James Randi would be proud.
Goodhart (author)2010-09-10
Actually, I think I know where he got his "reasoning" from....one moment....nope, I was wrong (or at least, I couldn't find it quickly). I had thought I read something along this line, when much younger in the book: The Creation-Evolution Controversy, by Wysong. But I don't see it there now. I mean, I keep the book, with my own annotations I placed in it, as a reference as to what others "think" (or don't think, as the case may be).
But I can no longer find the reference in any of the books I have remaining in my library. Perhaps it was lost during my last move over 20 years ago.....meaning I don't have need to consult it to demonstrate mistakes much anymore.
kelseymh (author)2010-09-09
It really helps when the actual topic of the question is up in the topic :-/
Caitlinsdad is right, of course. There are so many things wrong with your "someone's" statement, it's almost not worth it to even start.
The Sun does not "consume fuel" like burning wood. The mass difference (binding energy) between four hydrogen nuclei and one helium nucleus is only a few MeV (out of ~4 GeV), so your talking about a 1 part per thousand loss of mass.
The visual size of the Sun is only marginally connected to its mass. Much more important is the Sun's temperature, which determines the pressure that keeps it "inflated" against gravity. Four billion years ago, the Sun was cooler, and is estimated to have been about 1/3 smaller in diameter than it is now.
As for the math, it's pretty simple, and I think it's been explained. The volume of a sphere is just V = (4/3) pi r^3. Since you're dealing with ratios, the (4/3)pi cancels out, and you can write V1/V2 = 1/2 = r1^3 / r2^3, so r1/r2 = 1/cbrt(2) = 0.794. So to reduce the Sun's volume by 50%, you only need to reduce the radius/diameter by 20%.
Goodhart (author)2010-09-09
"It really helps when the actual topic of the question is up in the topic"
Yeah, sorry about that....it's been kind of a hectic month....I am not thinking as clearly as I could be.
Thanks for that. I got the numbers but hadn't yet converted it to the proper percentage.....and that is the one thing I really needed to know....although, as I wrote to Cameron, that is about what I'd have guessed....the difference in size is negligible compared to the (incorrect assumption) that it would lose 50 % of it's volume in the first place.
Now take a pair of smartypants, like the ones you are wearing. To halve the volume you would think you can just cut off one leg of the pants. But no, there is the waistsize and inseam measurement to figure and will you be wearing a belt or suspenders, none or both. Oh and don't forget if you will be hemming it with cuff or without, straight leg, boot cut or flared...
Thanks for the explanation...but no link to some fancy widget to graphically input some magic numbers and have an animated display to show us the volume of a sphere being halved?
CameronSS (author)2010-09-09
Widgets are wimpy.
I want that on a tshirt.
Goodhart (author)2010-09-09
ok, so my figure of solar surface area comes out to about: 6,078,428,661,500 CKm does that compute? | 2,088 | 7,783 | {"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} | 3.5625 | 4 | CC-MAIN-2018-22 | latest | en | 0.958551 |
https://www.meritnation.com/cbse-class-12-science/physics/hc-verma-i/geometrical-optics/textbook-solutions/46_4_1264_6773_410_68852 | 1,721,034,040,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514680.75/warc/CC-MAIN-20240715071424-20240715101424-00330.warc.gz | 766,394,774 | 51,453 | HC Verma i Solutions for Class 12 Science Physics Chapter 18 Geometrical Optics are provided here with simple step-by-step explanations. These solutions for Geometrical Optics are extremely popular among class 12 Science students for Physics Geometrical Optics Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the HC Verma i Book of class 12 Science Physics Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s HC Verma i Solutions. All HC Verma i Solutions for class 12 Science Physics are prepared by experts and are 100% accurate.
#### Question 1:
No, because this formula is only valid when the angles of incidence and refraction are small. When the angles of incidence and refraction are large, we cannot use relations sini = tani and sinr = tanr to derive the formula for the apparent depth.
#### Question 2:
No, because this formula is only valid when the angles of incidence and refraction are small. When the angles of incidence and refraction are large, we cannot use relations sini = tani and sinr = tanr to derive the formula for the apparent depth.
There is only one such situation. Light rays go undeviated through a prism only when the prism is kept in a medium that has the same refractive index as that of the prism itself. In all other situations, there will always be some minimum deviation in the path of a light ray passing through the prism.
#### Question 3:
There is only one such situation. Light rays go undeviated through a prism only when the prism is kept in a medium that has the same refractive index as that of the prism itself. In all other situations, there will always be some minimum deviation in the path of a light ray passing through the prism.
A diamond shines more because of the phenomenon of total internal reflection (TIR). The refractive index of diamond is ≈2.4 and that of glass is ≈1.5.
By using the relation , where C is the critical angle, we get refractive index (μ).
For a large refractive index, the value of the critical angle is small. Thus, for the diamond, the critical angle for total internal reflection is much smaller than that of the glass. Therefore, a great percentage of incident light gets internally reflected several times before it emerges out.
#### Question 4:
A diamond shines more because of the phenomenon of total internal reflection (TIR). The refractive index of diamond is ≈2.4 and that of glass is ≈1.5.
By using the relation , where C is the critical angle, we get refractive index (μ).
For a large refractive index, the value of the critical angle is small. Thus, for the diamond, the critical angle for total internal reflection is much smaller than that of the glass. Therefore, a great percentage of incident light gets internally reflected several times before it emerges out.
The beam of light will seem to appear and disappear, just like the twinkling of the stars. This is because the refractive index of the material in the slab fluctuates slowly with time. So, the light rays get refracted differently with time and this causes them to get concentrated during certain times, or get diffused at other times.
#### Question 5:
The beam of light will seem to appear and disappear, just like the twinkling of the stars. This is because the refractive index of the material in the slab fluctuates slowly with time. So, the light rays get refracted differently with time and this causes them to get concentrated during certain times, or get diffused at other times.
No, a plane mirror can never form a real image. This is because in all possible situations of image formation, the light rays never actually meet after getting reflected, but they only appear to meet behind the mirror, forming a virtual image always.
#### Question 6:
No, a plane mirror can never form a real image. This is because in all possible situations of image formation, the light rays never actually meet after getting reflected, but they only appear to meet behind the mirror, forming a virtual image always.
No, the paper will not burn even after sufficient time. This is because the piece of paper is placed at the position of a virtual image of a strong light source, so light rays are not actually converging at that point, but seem to converge. As no light rays are concentrating on the point, the paper will never burn.
Instead, if the paper is placed at the position of a real image, it will burn as the light rays are actually concentrating on the point. Also, when we have the real image of a virtual source, it will also cause the burning of paper due to same reason; the light rays are actually meeting at the point.
#### Question 7:
No, the paper will not burn even after sufficient time. This is because the piece of paper is placed at the position of a virtual image of a strong light source, so light rays are not actually converging at that point, but seem to converge. As no light rays are concentrating on the point, the paper will never burn.
Instead, if the paper is placed at the position of a real image, it will burn as the light rays are actually concentrating on the point. Also, when we have the real image of a virtual source, it will also cause the burning of paper due to same reason; the light rays are actually meeting at the point.
Yes, a virtual image can be photographed by a camera. This is because the light rays emitting from a virtual image and reaching the lens of the camera are real. A virtual image is formed from a reflecting surface after reflection of real incident rays, so these rays enter the camera to provide effects on the photographic film. Just like a virtual image can be seen by us in a mirror with our eyes, it can be photographed by a camera.
#### Question 8:
Yes, a virtual image can be photographed by a camera. This is because the light rays emitting from a virtual image and reaching the lens of the camera are real. A virtual image is formed from a reflecting surface after reflection of real incident rays, so these rays enter the camera to provide effects on the photographic film. Just like a virtual image can be seen by us in a mirror with our eyes, it can be photographed by a camera.
The special function of a convex mirror is that it creates the image of a distant object that is reduced in size, is upright or erect and always lies within the virtual focal length of the mirror. A plane mirror cannot do this. Also, as the image is formed within the focal length, the image is close to the mirror as well as is small in size, enabling the driver to clearly view the nearer vehicles behind the motor vehicle.
#### Question 9:
The special function of a convex mirror is that it creates the image of a distant object that is reduced in size, is upright or erect and always lies within the virtual focal length of the mirror. A plane mirror cannot do this. Also, as the image is formed within the focal length, the image is close to the mirror as well as is small in size, enabling the driver to clearly view the nearer vehicles behind the motor vehicle.
The image of the object moves slower compared to the object. It can be explained using the mirror formula :$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$
We know that for a convex mirror, the object distance (u) is positive, image distance (v) is negative and the focal length (f) is also negative. Thus mirror formula of a convex mirror is:$\frac{1}{u}-\frac{1}{v}=-\frac{1}{f}$
As u = +ve
$\frac{1}{v}-\frac{1}{f}>0\phantom{\rule{0ex}{0ex}}\frac{1}{v}>\frac{1}{f}\phantom{\rule{0ex}{0ex}}v
Therefore, the image is always formed within the focal length of the mirror. Thus, the distance moved by the image is much slower than the distance moved by the object.
#### Question 10:
The image of the object moves slower compared to the object. It can be explained using the mirror formula :$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$
We know that for a convex mirror, the object distance (u) is positive, image distance (v) is negative and the focal length (f) is also negative. Thus mirror formula of a convex mirror is:$\frac{1}{u}-\frac{1}{v}=-\frac{1}{f}$
As u = +ve
$\frac{1}{v}-\frac{1}{f}>0\phantom{\rule{0ex}{0ex}}\frac{1}{v}>\frac{1}{f}\phantom{\rule{0ex}{0ex}}v
Therefore, the image is always formed within the focal length of the mirror. Thus, the distance moved by the image is much slower than the distance moved by the object.
When viewed from the water, the friend will seem taller than his usual height.
Let actual height be h and the apparent height be h'.
Here, the refraction is taking place from rarer to denser medium and a virtual image is formed.
Using
$\frac{{\mu }_{1}}{-u}+\frac{{\mu }_{2}}{v}=\frac{{\mu }_{2}-{\mu }_{1}}{R}$
Where refractive index of water is μ2 and refractive index of air is μ1.
u and v are object and image distances, respectively.
is the radius of curvature, here we will take it as ∞.
As μ1= 1
v = u × μ2
We know magnification is given by:
$m=\frac{v}{u}$
Putting the value of v in the above equation:
$m=\frac{u×{\mu }_{2}}{u}\phantom{\rule{0ex}{0ex}}m={\mu }_{2}\phantom{\rule{0ex}{0ex}}$
As the magnification is greater than 1, so the apparent height seems to be greater than actual height.
#### Question 11:
When viewed from the water, the friend will seem taller than his usual height.
Let actual height be h and the apparent height be h'.
Here, the refraction is taking place from rarer to denser medium and a virtual image is formed.
Using
$\frac{{\mu }_{1}}{-u}+\frac{{\mu }_{2}}{v}=\frac{{\mu }_{2}-{\mu }_{1}}{R}$
Where refractive index of water is μ2 and refractive index of air is μ1.
u and v are object and image distances, respectively.
is the radius of curvature, here we will take it as ∞.
As μ1= 1
v = u × μ2
We know magnification is given by:
$m=\frac{v}{u}$
Putting the value of v in the above equation:
$m=\frac{u×{\mu }_{2}}{u}\phantom{\rule{0ex}{0ex}}m={\mu }_{2}\phantom{\rule{0ex}{0ex}}$
As the magnification is greater than 1, so the apparent height seems to be greater than actual height.
Proof:
#### Question 12:
Proof:
Yes. Using the lens maker formula we can show it. We know that the formula is:
$\frac{1}{f}=\left(\frac{{\mu }_{2}}{{\mu }_{1}}-1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)$
Where, f is the focal length of the thin converging lens.
μ2 and μ1 are refractive indexes of lens and air respectively.
R1 and R2 are the radius of curvature of convex and plane surfaces respectively.
Here, R2 = ∞ because of the plane surface.
Case 1
When the convex surface is facing the object, we have:
Case 2
When the plane surface is facing the object, we have:
For Case 1, the focal length is positive and for the Case 2 the focal length is negative. Thus, the image distance is different in both cases.
#### Question 13:
Yes. Using the lens maker formula we can show it. We know that the formula is:
$\frac{1}{f}=\left(\frac{{\mu }_{2}}{{\mu }_{1}}-1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)$
Where, f is the focal length of the thin converging lens.
μ2 and μ1 are refractive indexes of lens and air respectively.
R1 and R2 are the radius of curvature of convex and plane surfaces respectively.
Here, R2 = ∞ because of the plane surface.
Case 1
When the convex surface is facing the object, we have:
Case 2
When the plane surface is facing the object, we have:
For Case 1, the focal length is positive and for the Case 2 the focal length is negative. Thus, the image distance is different in both cases.
Yes, we can say so with certainty, as only a diverging lens can cause a parallel beam of light to diverge.
#### Question 14:
Yes, we can say so with certainty, as only a diverging lens can cause a parallel beam of light to diverge.
It will act as a diverging lens. This is because the refractive index of air is less than that of water. Thus, light will diverge after passing through the bubble.
#### Question 15:
It will act as a diverging lens. This is because the refractive index of air is less than that of water. Thus, light will diverge after passing through the bubble.
Let the two converging lenses be L1 and L2, with focal lengths f1 and f2 respectively.
To reduce the aperture of a parallel beam of light without losing the energy of the light and also increase the intensity, we have to place the lens L2 within the focal range of L1.
#### Question 16:
Let the two converging lenses be L1 and L2, with focal lengths f1 and f2 respectively.
To reduce the aperture of a parallel beam of light without losing the energy of the light and also increase the intensity, we have to place the lens L2 within the focal range of L1.
No, the focal length of mirror will not change. This is because the focal length of a mirror does not depend on the refractive index of the medium in which the light rays travel.
#### Question 17:
No, the focal length of mirror will not change. This is because the focal length of a mirror does not depend on the refractive index of the medium in which the light rays travel.
Yes, the focal length of the lens will change. This is because its focal length is dependent on the medium in which the light rays travel.
#### Question 18:
Yes, the focal length of the lens will change. This is because its focal length is dependent on the medium in which the light rays travel.
No, mirrors cannot give rise to chromatic aberration. This is because chromatic aberration occurs due to the refraction of different colours of light. In case of mirrors, refraction of light does not take place.
#### Question 19:
No, mirrors cannot give rise to chromatic aberration. This is because chromatic aberration occurs due to the refraction of different colours of light. In case of mirrors, refraction of light does not take place.
No, there will not be a significant chromatic aberration because a laser light is monochromatic in nature. Therefore, all light will get converged at the same point.
#### Question 1:
No, there will not be a significant chromatic aberration because a laser light is monochromatic in nature. Therefore, all light will get converged at the same point.
(a) All the reflected rays meet at a point when produced backward.
Here, the angle of reflection is equal to the angle of incidence. Therefore, all rays get reflected to converge at a single point to form the point image of the point source.
#### Question 2:
(a) All the reflected rays meet at a point when produced backward.
Here, the angle of reflection is equal to the angle of incidence. Therefore, all rays get reflected to converge at a single point to form the point image of the point source.
(b) light goes from optically denser medium to rarer medium
In this case the incident angle is greater than critical angle, so the light gets reflected back into the same denser medium, instead of being refracted .
#### Question 3:
(b) light goes from optically denser medium to rarer medium
In this case the incident angle is greater than critical angle, so the light gets reflected back into the same denser medium, instead of being refracted .
(c) form nearly a point image of a point source
Since, when reflected back, they meet at a single point forming a point image of a point source.
#### Question 4:
(c) form nearly a point image of a point source
Since, when reflected back, they meet at a single point forming a point image of a point source.
(d) 15 cm behind the mirror
Since u = − 30 cm
v = ?
f = 30 cm
From the mirror formula:
#### Question 5:
(d) 15 cm behind the mirror
Since u = − 30 cm
v = ?
f = 30 cm
From the mirror formula:
(a) is plane
This is because the reflected rays are still parallel, which is only possible if the mirror is a plane mirror. A spherical mirror will either converge or diverge the reflected rays.
#### Question 6:
(a) is plane
This is because the reflected rays are still parallel, which is only possible if the mirror is a plane mirror. A spherical mirror will either converge or diverge the reflected rays.
(c) is certainly real if the object is virtual
As a virtual object is the virtual image of the object, it is always formed beyond the focus of the concave mirror. Thus, the image formed by a concave mirror of the virtual object will always be real.
#### Question 7:
(c) is certainly real if the object is virtual
As a virtual object is the virtual image of the object, it is always formed beyond the focus of the concave mirror. Thus, the image formed by a concave mirror of the virtual object will always be real.
(d) two images form, one at O' and the other at O"
Light will be refracted differently in both mediums on the right side. Thus, two images will be formed, one at O' due to refraction from medium μ1 and another at O" due to refraction from medium μ3.μ1
#### Question 8:
(d) two images form, one at O' and the other at O"
Light will be refracted differently in both mediums on the right side. Thus, two images will be formed, one at O' due to refraction from medium μ1 and another at O" due to refraction from medium μ3.μ1
(c) $\frac{1}{v-t}-\frac{1}{u+t}=\frac{1}{f}$
The thickness of lens will cause a shift in the position of object and image from the ideal condition, which will be:
As will be quite small, it can be ignored.
Thus, unew = u + t and vnew = vt
We get the new lens formula:
$\frac{1}{v-t}-\frac{1}{u+t}=\frac{1}{f}$
#### Question 9:
(c) $\frac{1}{v-t}-\frac{1}{u+t}=\frac{1}{f}$
The thickness of lens will cause a shift in the position of object and image from the ideal condition, which will be:
As will be quite small, it can be ignored.
Thus, unew = u + t and vnew = vt
We get the new lens formula:
$\frac{1}{v-t}-\frac{1}{u+t}=\frac{1}{f}$
(b) f = R
As from lens maker formula:
=R
#### Question 10:
(b) f = R
As from lens maker formula:
=R
(c) 2 f
Since the object is placed at 2 f, the image of the object will be formed at distance of 2 f from a converging lens.
It can also be shown from the lens formula:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Here, u = − 2 f and f = f
On putting the respective values we get:
$\frac{1}{v}-\frac{1}{-2f}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{1}{f}-\frac{1}{2f}\phantom{\rule{0ex}{0ex}}=\frac{1}{2f}$
Therefore, image distance v = 2 f
#### Question 11:
(c) 2 f
Since the object is placed at 2 f, the image of the object will be formed at distance of 2 f from a converging lens.
It can also be shown from the lens formula:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Here, u = − 2 f and f = f
On putting the respective values we get:
$\frac{1}{v}-\frac{1}{-2f}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{1}{f}-\frac{1}{2f}\phantom{\rule{0ex}{0ex}}=\frac{1}{2f}$
Therefore, image distance v = 2 f
(d) first increases then decreases
Since all beams of light are parallel to the principal axis, they will converge at the focus of a converging lens. Thus, the intensity of light increases till one reaches the focus and then starts decreasing as one moves beyond it.
#### Question 12:
(d) first increases then decreases
Since all beams of light are parallel to the principal axis, they will converge at the focus of a converging lens. Thus, the intensity of light increases till one reaches the focus and then starts decreasing as one moves beyond it.
(a) 2 D
The lens is cut into two equal parts by a plane perpendicular to the principal axis. Thus, the radius of curvature of both the lenses become half of its original value.
As,
$f=\frac{R}{2}\phantom{\rule{0ex}{0ex}}P=\frac{1}{f}$
Radius of curvature becomes half. Therefore, the focal also reduces to half its original value.
Thus, the power (P) of the two cut-lenses will be equal.
2 P = 4 D
P = 2 D
#### Question 13:
(a) 2 D
The lens is cut into two equal parts by a plane perpendicular to the principal axis. Thus, the radius of curvature of both the lenses become half of its original value.
As,
$f=\frac{R}{2}\phantom{\rule{0ex}{0ex}}P=\frac{1}{f}$
Radius of curvature becomes half. Therefore, the focal also reduces to half its original value.
Thus, the power (P) of the two cut-lenses will be equal.
2 P = 4 D
P = 2 D
(c) 4 D
As the lens is cut along the principal axis, the two new lens will act as two double convex lenses. Even after cutting the lens, the radius of the curvature of the two new lens will remain the same as that of the original lens, with the power equal to the original lens.
#### Question 14:
(c) 4 D
As the lens is cut along the principal axis, the two new lens will act as two double convex lenses. Even after cutting the lens, the radius of the curvature of the two new lens will remain the same as that of the original lens, with the power equal to the original lens.
(c) increases
The focal length of the combination will increase. For finding out the combination of lens we have the formula:
$\frac{1}{F}=\frac{1}{{f}_{1}}+\frac{1}{{f}_{2}}-\frac{\mathrm{d}}{{f}_{1}{f}_{2}}$
Where, F is the focal length for the combination
d is the separation between two lenses
Here, d = 0
Hence, the focal length will increase.
#### Question 15:
(c) increases
The focal length of the combination will increase. For finding out the combination of lens we have the formula:
$\frac{1}{F}=\frac{1}{{f}_{1}}+\frac{1}{{f}_{2}}-\frac{\mathrm{d}}{{f}_{1}{f}_{2}}$
Where, F is the focal length for the combination
d is the separation between two lenses
Here, d = 0
Hence, the focal length will increase.
(a) a convergent lens
Since the refractive index of lens is more than that of of water and both the sides are convex, it will act like a normal convergent lens (though its focal length will change).
#### Question 16:
(a) a convergent lens
Since the refractive index of lens is more than that of of water and both the sides are convex, it will act like a normal convergent lens (though its focal length will change).
(b) a divergent lens
Since the refractive index of lens is less than that of of water and both the sides are convex, it will act like a normal diverging lens (though its focal length will change to negative).
#### Question 17:
(b) a divergent lens
Since the refractive index of lens is less than that of of water and both the sides are convex, it will act like a normal diverging lens (though its focal length will change to negative).
(b) 2.5 cm
As the focal length of the lens is 20 cm and object distance is 40 cm from the lens, the image is formed at the centre of curvature at the right side of the lens.
From right angled triangle ABC,
and from right angled triangle, we have:
Putting the value of angle alpha, we get:
#### Question 18:
(b) 2.5 cm
As the focal length of the lens is 20 cm and object distance is 40 cm from the lens, the image is formed at the centre of curvature at the right side of the lens.
From right angled triangle ABC,
and from right angled triangle, we have:
Putting the value of angle alpha, we get:
(d) chromatic aberration
When light rays of different colours do not converge at the same point after passing through a converging lens, it is called chromatic aberration. This happens because a lens has different refractive indices for different colours, i.e, for different wavelengths of light.
#### Question 1:
(d) chromatic aberration
When light rays of different colours do not converge at the same point after passing through a converging lens, it is called chromatic aberration. This happens because a lens has different refractive indices for different colours, i.e, for different wavelengths of light.
Using sign conventions, given,
Distance of object from mirror, u = − 30 cm,
Radius of curvature of concave mirror R = − 40 cm
Using the mirror equation,
or, v = − 60 cm
Hence, the required image will be located at a distance of 60 cm in front of the concave mirror.
#### Question 2:
Using sign conventions, given,
Distance of object from mirror, u = − 30 cm,
Radius of curvature of concave mirror R = − 40 cm
Using the mirror equation,
or, v = − 60 cm
Hence, the required image will be located at a distance of 60 cm in front of the concave mirror.
Given,
Height of the object, h1 = 20 cm,
Distance of image from screen v = −5.0 m = −500 cm,
Since, we know that
Where 'u' is the distance of object from screen.
(As the image is inverted)
Using mirror formula,
Hence, the required focal length of the concave mirror is 1.44 m.
#### Question 3:
Given,
Height of the object, h1 = 20 cm,
Distance of image from screen v = −5.0 m = −500 cm,
Since, we know that
Where 'u' is the distance of object from screen.
(As the image is inverted)
Using mirror formula,
Hence, the required focal length of the concave mirror is 1.44 m.
Using sign conventions, given,
Focal length of the concave mirror:
f = −20 cm
As per the question,
Magnification (m) is:
$m=-\frac{v}{u}=2$
v = −2u
Case I (Virtual image):
Using mirror formula,
Case II (Real image)
Using mirror formula,
Hence, the required positions of objects are 10 cm or 30 cm from the concave mirror.
#### Question 4:
Using sign conventions, given,
Focal length of the concave mirror:
f = −20 cm
As per the question,
Magnification (m) is:
$m=-\frac{v}{u}=2$
v = −2u
Case I (Virtual image):
Using mirror formula,
Case II (Real image)
Using mirror formula,
Hence, the required positions of objects are 10 cm or 30 cm from the concave mirror.
Given,
Height of the object, h1 = 1 cm
Focal length of the concave mirror, f = 7.5 cm =
Magnification is given as,
Hence, the distance of the object from the mirror is 5 cm.
#### Question 5:
Given,
Height of the object, h1 = 1 cm
Focal length of the concave mirror, f = 7.5 cm =
Magnification is given as,
Hence, the distance of the object from the mirror is 5 cm.
Given,
Height (h1) of the candle flame taken as object AB = 1.6 cm
Diameter of the ball bearing (d) = 0.4 cm
Distance of object, u = 20 cm
Using mirror formula,
Putting the values according to sign conventions, we get,
Hence, the distance of the image is 1 cm.
Height of the image is 0.008 cm.
#### Question 6:
Given,
Height (h1) of the candle flame taken as object AB = 1.6 cm
Diameter of the ball bearing (d) = 0.4 cm
Distance of object, u = 20 cm
Using mirror formula,
Putting the values according to sign conventions, we get,
Hence, the distance of the image is 1 cm.
Height of the image is 0.008 cm.
Given,
Height of the object ABh1 = 3 cm
Distance of the object from the convex mirror, u = −7.5 cm
Focal length of the convex mirror (f) = 6 cm
Using mirror formula,
Putting values according to sign convention,
Magnification = m = $-\frac{v}{u}=\frac{10}{7.5×3}$
Where A'B' is the height of the image.
Hence, the required location of the image is from the pole and image height is 1.33 cm. Nature of the image is virtual and erect.
#### Question 7:
Given,
Height of the object ABh1 = 3 cm
Distance of the object from the convex mirror, u = −7.5 cm
Focal length of the convex mirror (f) = 6 cm
Using mirror formula,
Putting values according to sign convention,
Magnification = m = $-\frac{v}{u}=\frac{10}{7.5×3}$
Where A'B' is the height of the image.
Hence, the required location of the image is from the pole and image height is 1.33 cm. Nature of the image is virtual and erect.
Given,
Radius of curvature of concave mirror, R = 20 cm
So its focal length will be f = $\frac{R}{2}$ = −10 cm
For part AB of the U shaped wire, PB = 30 + 10 = 40 cm
Therefore, u = −40 cm
Using mirror formula,
$\frac{1}{v}=-\frac{1}{10}-\frac{1}{-40}=-\frac{3}{40}$
$m=\frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{AB}}=\frac{-\left(v\right)}{u}$
For part CD of the U shaped wire, PC = 30 cm
Therefore, u = −30 cm
Again, using mirror equation:
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
$\frac{1}{v}=-\frac{1}{10}-\frac{1}{-30}\phantom{\rule{0ex}{0ex}}\frac{1}{v}=-\frac{1}{10}+\frac{1}{30}=-\frac{1}{15}$
v = −15 cm = PC'
$m=\frac{\mathrm{C}\text{'}\mathrm{D}\text{'}}{\mathrm{CD}}=-\frac{v}{u}$
$=-\frac{\left(-15\right)}{-30}=-\frac{1}{2}$
⇒ C'D' = 5 cm
⇒ B'C' = PC' − PB'
= 15 − 13.3 = 1.7 cm
Hence, the total length of the U- shaped wire is A'B' + B'C' + C'D'
= (3.3) + (1.7) + 5 = 10 cm
#### Question 8:
Given,
Radius of curvature of concave mirror, R = 20 cm
So its focal length will be f = $\frac{R}{2}$ = −10 cm
For part AB of the U shaped wire, PB = 30 + 10 = 40 cm
Therefore, u = −40 cm
Using mirror formula,
$\frac{1}{v}=-\frac{1}{10}-\frac{1}{-40}=-\frac{3}{40}$
$m=\frac{\mathrm{A}\text{'}\mathrm{B}\text{'}}{\mathrm{AB}}=\frac{-\left(v\right)}{u}$
For part CD of the U shaped wire, PC = 30 cm
Therefore, u = −30 cm
Again, using mirror equation:
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
$\frac{1}{v}=-\frac{1}{10}-\frac{1}{-30}\phantom{\rule{0ex}{0ex}}\frac{1}{v}=-\frac{1}{10}+\frac{1}{30}=-\frac{1}{15}$
v = −15 cm = PC'
$m=\frac{\mathrm{C}\text{'}\mathrm{D}\text{'}}{\mathrm{CD}}=-\frac{v}{u}$
$=-\frac{\left(-15\right)}{-30}=-\frac{1}{2}$
⇒ C'D' = 5 cm
⇒ B'C' = PC' − PB'
= 15 − 13.3 = 1.7 cm
Hence, the total length of the U- shaped wire is A'B' + B'C' + C'D'
= (3.3) + (1.7) + 5 = 10 cm
Given,
Distance of the man's face (here, taken as object), u = −25 cm
According to the question, magnification, m = 1.4
Using equation of mirror, we get:
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
$=\frac{5-7}{175}=-\frac{2}{175}$
f = −87.5
Hence, the required focal length of the concave mirror is 87.5 cm.
#### Question 1:
Given,
Distance of the man's face (here, taken as object), u = −25 cm
According to the question, magnification, m = 1.4
Using equation of mirror, we get:
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
$=\frac{5-7}{175}=-\frac{2}{175}$
f = −87.5
Hence, the required focal length of the concave mirror is 87.5 cm.
(a) reflection
(b) refraction
When the light strikes on a surface nearly parallel to it, it then bends by a small and fixed angle after reflection. Also, when the light travels from one medium to another with slight differences in their refractive indices, it bends by a small angle. Thus, the bending of light by a small but fixed angle can be the case of either reflection or refraction.
#### Question 2:
(a) reflection
(b) refraction
When the light strikes on a surface nearly parallel to it, it then bends by a small and fixed angle after reflection. Also, when the light travels from one medium to another with slight differences in their refractive indices, it bends by a small angle. Thus, the bending of light by a small but fixed angle can be the case of either reflection or refraction.
(b) If the final rays are converging, we have a real image.
This is because a real image is formed by converging reflected/refracted rays from a mirror/lens.
#### Question 3:
(b) If the final rays are converging, we have a real image.
This is because a real image is formed by converging reflected/refracted rays from a mirror/lens.
(a) Pole
(d) Principal axis
Paraxial rays are the light rays close to the principal axis. The focus of the spherical mirror for paraxial rays are different from the focus for marginal rays. So, the focus depends on whether the rays are paraxial or marginal. ​The pole, radius of curvature and the principal axis of a spherical mirror do not depend on paraxial or marginal rays.
#### Question 4:
(a) Pole
(d) Principal axis
Paraxial rays are the light rays close to the principal axis. The focus of the spherical mirror for paraxial rays are different from the focus for marginal rays. So, the focus depends on whether the rays are paraxial or marginal. ​The pole, radius of curvature and the principal axis of a spherical mirror do not depend on paraxial or marginal rays.
(c) the object is real but the image is virtual
(d) the object is virtual but the image is real
The virtual image of a real object and the real image of a virtual object are always erect.
#### Question 5:
(c) the object is real but the image is virtual
(d) the object is virtual but the image is real
The virtual image of a real object and the real image of a virtual object are always erect.
(c) the image will not be shifted
(d) the intensity of the image will decrease
If the upper half portion of the convex lens is painted, then only the intensity of the image will decrease, as the amount of light passing through the lens will decrease. Also, there will be no shift in the position of the image because all the parameters remain the same.
#### Question 6:
(c) the image will not be shifted
(d) the intensity of the image will decrease
If the upper half portion of the convex lens is painted, then only the intensity of the image will decrease, as the amount of light passing through the lens will decrease. Also, there will be no shift in the position of the image because all the parameters remain the same.
(a) there will be an image at O1
​(b) there will be an image at O2
The lens L1 converges the light at point O1 and lens L2 converges the light at O2. As the upper half of lens L3 has a refractive index equal to that of L1, it will converge the light at O1 and thus the image will be formed at O1. Also, the lower half of lens L3 has a refractive index equal to that of lens L2, it will converge the light at O2 and thus the image will be formed at O2.
#### Question 7:
(a) there will be an image at O1
​(b) there will be an image at O2
The lens L1 converges the light at point O1 and lens L2 converges the light at O2. As the upper half of lens L3 has a refractive index equal to that of L1, it will converge the light at O1 and thus the image will be formed at O1. Also, the lower half of lens L3 has a refractive index equal to that of lens L2, it will converge the light at O2 and thus the image will be formed at O2.
(b) must be greater than 20 cm
Let the image be formed at a distance of x cm from the lens.
Therefore, the distance of the object from the lens, u​, will be = (40 − x) cm
From lens formula:
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{f}=\frac{1}{x}-\frac{1}{40-x}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{f}=\frac{40-x-x}{40x-{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒f=\frac{40x-{x}^{2}}{40-2x}\phantom{\rule{0ex}{0ex}}⇒f\left(40-2x\right)=40x-{x}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-2fx-40x+40f=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-\left(2f+40\right)x+40f=0$
Therefore, we get x as:
$x=\frac{\left(2f+40\right)±\sqrt{\left(2f+40{\right)}^{2}-160f}}{2}$
#### Question 9:
(b) must be greater than 20 cm
Let the image be formed at a distance of x cm from the lens.
Therefore, the distance of the object from the lens, u​, will be = (40 − x) cm
From lens formula:
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{f}=\frac{1}{x}-\frac{1}{40-x}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{f}=\frac{40-x-x}{40x-{x}^{2}}\phantom{\rule{0ex}{0ex}}⇒f=\frac{40x-{x}^{2}}{40-2x}\phantom{\rule{0ex}{0ex}}⇒f\left(40-2x\right)=40x-{x}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-2fx-40x+40f=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-\left(2f+40\right)x+40f=0$
Therefore, we get x as:
$x=\frac{\left(2f+40\right)±\sqrt{\left(2f+40{\right)}^{2}-160f}}{2}$
Given,
Focal length of the concave mirror, f = − 7.6 m
Distance between earth and moon taken as object distance, u = −3.8 × 105 km
Diameter of moon = 3450 km
Using mirror equation:
As we know, the distance of moon from earth is very large as compared to focal length it can be taken as ∞.
Therefore, image of the moon will be formed at focus, which is inverted.
v = −7.6 m
We know that magnification is given by:
$m=-\frac{v}{u}=\frac{{d}_{\mathrm{image}}}{{d}_{\mathrm{object}}}$
${d}_{\mathrm{image}}=-\frac{3450×7.6×{10}^{3}}{3.8×{10}^{8}}$
= $-$0.069 m = $-$6.9 cm
Hence, the required diameter of the image of moon is 6.9 cm.
#### Question 10:
Given,
Focal length of the concave mirror, f = − 7.6 m
Distance between earth and moon taken as object distance, u = −3.8 × 105 km
Diameter of moon = 3450 km
Using mirror equation:
As we know, the distance of moon from earth is very large as compared to focal length it can be taken as ∞.
Therefore, image of the moon will be formed at focus, which is inverted.
v = −7.6 m
We know that magnification is given by:
$m=-\frac{v}{u}=\frac{{d}_{\mathrm{image}}}{{d}_{\mathrm{object}}}$
${d}_{\mathrm{image}}=-\frac{3450×7.6×{10}^{3}}{3.8×{10}^{8}}$
= $-$0.069 m = $-$6.9 cm
Hence, the required diameter of the image of moon is 6.9 cm.
Given,
Distance of the circle from the mirror taken as object distance, u = −30 cm,
Focal length of the concave mirror, f = −20 cm
Using mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
v = $-$60 cm
Therefore, image of the circle is formed at a distance of 60 cm in front of the mirror.
We know magnification (m) is given by:
Where Robject and Rimage are radius of the object and radius of the image, respectively.
⇒ Rimage = 4 cm
Hence, the required radius of the circle formed by the image is 4 cm.
#### Question 11:
Given,
Distance of the circle from the mirror taken as object distance, u = −30 cm,
Focal length of the concave mirror, f = −20 cm
Using mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
v = $-$60 cm
Therefore, image of the circle is formed at a distance of 60 cm in front of the mirror.
We know magnification (m) is given by:
Where Robject and Rimage are radius of the object and radius of the image, respectively.
⇒ Rimage = 4 cm
Hence, the required radius of the circle formed by the image is 4 cm.
Given,
A concave mirror of radius 'R' kept on a horizontal table.
'h' is height up to which the water is poured into the concave mirror.
Let the object be placed at height 'x' above the surface of water.
We know if we place the object at the centre of curvature of the mirror, then the image itself will be formed at the centre of curvature.
Therefore, the apparent position of the object with respect to the mirror should be at the centre of curvature so that the image is formed at the same position.
Since,
(with respect to mirror)
Hence, the object should be placed at $\frac{R-h}{\mathrm{\mu }}$ above the water surface.
#### Question 12:
Given,
A concave mirror of radius 'R' kept on a horizontal table.
'h' is height up to which the water is poured into the concave mirror.
Let the object be placed at height 'x' above the surface of water.
We know if we place the object at the centre of curvature of the mirror, then the image itself will be formed at the centre of curvature.
Therefore, the apparent position of the object with respect to the mirror should be at the centre of curvature so that the image is formed at the same position.
Since,
(with respect to mirror)
Hence, the object should be placed at $\frac{R-h}{\mathrm{\mu }}$ above the water surface.
Given,
Two converging mirrors having equal focal length 'f '.
Both the mirrors will produce one image under two conditions:
Case-1
When the point source is at the centre of curvature of the mirrors, i.e, at a distance of '2f ' from each mirror, the images will be produced at the same point 'S'. Therefore, d = 2f + 2f = 4f
Case-II
When the point source 'S' is at focus, i.e., at a distance 'f ' from each mirror, the rays from the source after reflecting from one mirror will become parallel and so these parallel rays, after the reflection from the other mirror the object itself. Thus, only one image is formed.
Here d = f + f = 2f
#### Question 13:
Given,
Two converging mirrors having equal focal length 'f '.
Both the mirrors will produce one image under two conditions:
Case-1
When the point source is at the centre of curvature of the mirrors, i.e, at a distance of '2f ' from each mirror, the images will be produced at the same point 'S'. Therefore, d = 2f + 2f = 4f
Case-II
When the point source 'S' is at focus, i.e., at a distance 'f ' from each mirror, the rays from the source after reflecting from one mirror will become parallel and so these parallel rays, after the reflection from the other mirror the object itself. Thus, only one image is formed.
Here d = f + f = 2f
Given,
Converging mirror M1 with focal length (f1) = 20 cm
Converging mirror M2 with focal length (f2) = 20 cm
f1 = f2 = 20 cm = f
Point source is at a distance of 30 cm from M1.
As the 1st reflection is through the mirror M1,
u = −30 cm
f = −20 cm
Using mirror equation:
and for the 2nd reflection at mirror M2,
u = 60 − (30 + x) = 30 − x
v = − x, f = 20 cm
Again using the mirror equation:
⇒ 40x − 600 = 30xx2
x2 + 10x − 600 = 0
= 20 cm or − 30 cm
∴ Total distance between the two lines is 20 + 30 = 50 cm
(b) Location of the image formed by the single reflection from M2 = 60 $-$ 50 = 10
Thus, the image formed by the single reflection from M2 is at a distance of 10 cm from mirror M2.
#### Question 14:
Given,
Converging mirror M1 with focal length (f1) = 20 cm
Converging mirror M2 with focal length (f2) = 20 cm
f1 = f2 = 20 cm = f
Point source is at a distance of 30 cm from M1.
As the 1st reflection is through the mirror M1,
u = −30 cm
f = −20 cm
Using mirror equation:
and for the 2nd reflection at mirror M2,
u = 60 − (30 + x) = 30 − x
v = − x, f = 20 cm
Again using the mirror equation:
⇒ 40x − 600 = 30xx2
x2 + 10x − 600 = 0
= 20 cm or − 30 cm
∴ Total distance between the two lines is 20 + 30 = 50 cm
(b) Location of the image formed by the single reflection from M2 = 60 $-$ 50 = 10
Thus, the image formed by the single reflection from M2 is at a distance of 10 cm from mirror M2.
Given,
Angle of incidence, i = 45°
Angle of refraction, r = 30°
Using Snell's law,
$=\frac{2}{\sqrt{2}}=\sqrt{2}$
Let x be the distance travelled by light in the slab.
Now,
We know:
Time taken $=\frac{\mathrm{Distance}}{\mathrm{Speed}}$
$=\frac{2}{\sqrt{3}}×\frac{\sqrt{2}}{3×{10}^{8}}$
= 0.54 × 10−8
= 5.4 × 10−9 s
#### Question 15:
Given,
Angle of incidence, i = 45°
Angle of refraction, r = 30°
Using Snell's law,
$=\frac{2}{\sqrt{2}}=\sqrt{2}$
Let x be the distance travelled by light in the slab.
Now,
We know:
Time taken $=\frac{\mathrm{Distance}}{\mathrm{Speed}}$
$=\frac{2}{\sqrt{3}}×\frac{\sqrt{2}}{3×{10}^{8}}$
= 0.54 × 10−8
= 5.4 × 10−9 s
Given,
Length of the pole = 1.00 m
Water level of the swimming pool is 50.0 cm higher than the bed.
Refractive index (μ) of water = 1.33
According to the figure, shadow length = BA' = BD + DA'= 0.5 + 0.5 tan r
Using Snell's law:
Therefore, shadow length of the pole = (0.5)$×$(1 + 0.6235) = 0.81175 m
= 81.2 cm
#### Question 16:
Given,
Length of the pole = 1.00 m
Water level of the swimming pool is 50.0 cm higher than the bed.
Refractive index (μ) of water = 1.33
According to the figure, shadow length = BA' = BD + DA'= 0.5 + 0.5 tan r
Using Snell's law:
Therefore, shadow length of the pole = (0.5)$×$(1 + 0.6235) = 0.81175 m
= 81.2 cm
Given,
Depth of the lake = 2.5 m
Refractive index (μ) of water = $\frac{4}{3}$
When the sun is just setting, θ is approximately = 90Ëš
Therefore, incidence angle is 90Ëš
Using Snell's law:
From the figure, W'O = x is the distance of the shadow
#### Question 17:
Given,
Depth of the lake = 2.5 m
Refractive index (μ) of water = $\frac{4}{3}$
When the sun is just setting, θ is approximately = 90Ëš
Therefore, incidence angle is 90Ëš
Using Snell's law:
From the figure, W'O = x is the distance of the shadow
Given,
Thickness of the glass slab, d = 2.1 cm
Refractive index, μ = 1.5
Shift due to the glass slab is given by,
$∆t=\left[1-\left(\frac{1}{\mathrm{\mu }}\right)\right]d$
$=\left[1-\left(\frac{1}{1.5}\right)\right]\left(2.1\right)$
Hence, the microscope should be shifted by 0.70 cm to focus the object 'P' again.
#### Question 18:
Given,
Thickness of the glass slab, d = 2.1 cm
Refractive index, μ = 1.5
Shift due to the glass slab is given by,
$∆t=\left[1-\left(\frac{1}{\mathrm{\mu }}\right)\right]d$
$=\left[1-\left(\frac{1}{1.5}\right)\right]\left(2.1\right)$
Hence, the microscope should be shifted by 0.70 cm to focus the object 'P' again.
Given,
Height of water, dw = 20 cm
Height of oil, dO = 20 cm
The refractive index of the water (μw) = 1.33
The refractive index of oil (μO) = 1.30
Shift due to water is given by,
$∆{t}_{w}=1-\left(\frac{1}{{\mathrm{\mu }}_{\mathrm{w}}}\right){d}_{w}$
$=\left[1-\left(\frac{1}{1.33}\right)\right]20$
Shift due to oil,
Therefore, total shift, Δt = 5 + 4.6 = 9.6 cm
Hence, apparent depth = 40 − (9.6) = 30.4 cm between the surface.
#### Question 19:
Given,
Height of water, dw = 20 cm
Height of oil, dO = 20 cm
The refractive index of the water (μw) = 1.33
The refractive index of oil (μO) = 1.30
Shift due to water is given by,
$∆{t}_{w}=1-\left(\frac{1}{{\mathrm{\mu }}_{\mathrm{w}}}\right){d}_{w}$
$=\left[1-\left(\frac{1}{1.33}\right)\right]20$
Shift due to oil,
Therefore, total shift, Δt = 5 + 4.6 = 9.6 cm
Hence, apparent depth = 40 − (9.6) = 30.4 cm between the surface.
Given,
From the figure we can infer that the air is present in between the sheet, so it does not affect the shift. Therefore, the shift is only due to 3 sheets of different refractive indices, which is given by:
$∆t=\left[1-\frac{1}{{\mu }_{1}}\right]{t}_{1}+\left[1-\frac{1}{{\mu }_{2}}\right]{t}_{2}+\left[1-\frac{1}{{\mu }_{3}}\right]{t}_{3}$
$=\left[1-\left(\frac{1}{12}\right)\right]\left(0.2\right)+\left[1-\left(\frac{1}{1.3}\right)\right]\left(0.3\right)+\left[1-\left(\frac{1}{1.4}\right)\right]\left(0.4\right)$
= 0.2 cm
Hence, location of image of point P is located 0.2 cm above point P.
#### Question 20:
Given,
From the figure we can infer that the air is present in between the sheet, so it does not affect the shift. Therefore, the shift is only due to 3 sheets of different refractive indices, which is given by:
$∆t=\left[1-\frac{1}{{\mu }_{1}}\right]{t}_{1}+\left[1-\frac{1}{{\mu }_{2}}\right]{t}_{2}+\left[1-\frac{1}{{\mu }_{3}}\right]{t}_{3}$
$=\left[1-\left(\frac{1}{12}\right)\right]\left(0.2\right)+\left[1-\left(\frac{1}{1.3}\right)\right]\left(0.3\right)+\left[1-\left(\frac{1}{1.4}\right)\right]\left(0.4\right)$
= 0.2 cm
Hence, location of image of point P is located 0.2 cm above point P.
k number of transparent slabs are arranged one over the other.
Refractive indices of the slabs = μ1, μ2, μ3, ..., μk
Thickness of the slabs = t1, t2, t3,.., tk
Shift due to one slab:
For the combination of multiple slabs, the shift is given by,
Let μ be the refractive index of the combination of slabs.
The image is formed at the same place.
So, the shift will be:
Equating (i) and (ii), we get:
Hence, the required equivalent refractive index is $\frac{\sum _{\mathrm{i}=1}^{\mathrm{k}}{t}_{\mathrm{i}}}{\sum _{\mathrm{i}=1}^{\mathrm{k}}\left({t}_{\mathrm{i}}}{{\mathrm{\mu }}_{\mathrm{i}}}\right)}$.
#### Question 21:
k number of transparent slabs are arranged one over the other.
Refractive indices of the slabs = μ1, μ2, μ3, ..., μk
Thickness of the slabs = t1, t2, t3,.., tk
Shift due to one slab:
For the combination of multiple slabs, the shift is given by,
Let μ be the refractive index of the combination of slabs.
The image is formed at the same place.
So, the shift will be:
Equating (i) and (ii), we get:
Hence, the required equivalent refractive index is $\frac{\sum _{\mathrm{i}=1}^{\mathrm{k}}{t}_{\mathrm{i}}}{\sum _{\mathrm{i}=1}^{\mathrm{k}}\left({t}_{\mathrm{i}}}{{\mathrm{\mu }}_{\mathrm{i}}}\right)}$.
Given,
Diameter of the cylindrical vessel = 12 cm
∴ radius r = 6 cm
Diameter of the cylindrical glass piece = 8 cm
∴ radius r'= 4 cm and its height, h1 = 8 cm
Refractive index of glass, μg = 1.5
Refractive index of water, μw = 1.3
Let h be the final height of the water column.
The volume of the cylindrical water column after the glass piece is put will be:
$\pi$r2h = 800π + $\pi$r'2h1
or r2h = 800 + r'2h1
or (6)2h = 800 + (4)2 8
There will be two shifts; due to the glass block as well as water:
Hence, the total shift = (Δt1+ Δt2) = (2.66 + 4.44) cm = 7.1 cm above the bottom.
#### Question 22:
Given,
Diameter of the cylindrical vessel = 12 cm
∴ radius r = 6 cm
Diameter of the cylindrical glass piece = 8 cm
∴ radius r'= 4 cm and its height, h1 = 8 cm
Refractive index of glass, μg = 1.5
Refractive index of water, μw = 1.3
Let h be the final height of the water column.
The volume of the cylindrical water column after the glass piece is put will be:
$\pi$r2h = 800π + $\pi$r'2h1
or r2h = 800 + r'2h1
or (6)2h = 800 + (4)2 8
There will be two shifts; due to the glass block as well as water:
Hence, the total shift = (Δt1+ Δt2) = (2.66 + 4.44) cm = 7.1 cm above the bottom.
Given,
Refractive index of water = μ.
Height of the pot = H
Let us take x as the distance of the image of the eye formed above the surface of the water as seen by the fish.
We can infer from from the diagram,
The distance of the direct image $=\frac{H}{2}+\mu H=H\left(\mu +\frac{1}{2}\right)$
Similarly, image through mirror = $\frac{H}{2}+\left(H+x\right)=\frac{3H}{2}+\mu H=H\left(\frac{3}{2}+\mu \right)$
b) We know that:
Where, y is the distance of the image of the fish below the surface as seen by the eye.
Direct image = $H+y=H+\frac{H}{2\mu }=H\left(1+\frac{1}{2\mu }\right)$
Again another image of fish will be formed H/2 below the mirror.
Real depth for that image of fish becomes H + H/2 = 3H/2
So, apparent depth from the surface of water = 3H/2μ
So, distance of the image from the eye $=\frac{H}{2}+\frac{3H}{2\mu }=H\left(1+\frac{3}{2\mu }\right)$
#### Question 23:
Given,
Refractive index of water = μ.
Height of the pot = H
Let us take x as the distance of the image of the eye formed above the surface of the water as seen by the fish.
We can infer from from the diagram,
The distance of the direct image $=\frac{H}{2}+\mu H=H\left(\mu +\frac{1}{2}\right)$
Similarly, image through mirror = $\frac{H}{2}+\left(H+x\right)=\frac{3H}{2}+\mu H=H\left(\frac{3}{2}+\mu \right)$
b) We know that:
Where, y is the distance of the image of the fish below the surface as seen by the eye.
Direct image = $H+y=H+\frac{H}{2\mu }=H\left(1+\frac{1}{2\mu }\right)$
Again another image of fish will be formed H/2 below the mirror.
Real depth for that image of fish becomes H + H/2 = 3H/2
So, apparent depth from the surface of water = 3H/2μ
So, distance of the image from the eye $=\frac{H}{2}+\frac{3H}{2\mu }=H\left(1+\frac{3}{2\mu }\right)$
Given,
Refractive index of water μ = 1.33
Radius of the cylindrical vessel = 3 cm
Height of the cylindrical vessel = 4 cm
Let x be the length of BD
According to the diagram,
Using Snell's law,
From (i) and (ii),
Hence, the ratio of real and apparent depth of the image will be $4:2.25=1.78$
#### Question 24:
Given,
Refractive index of water μ = 1.33
Radius of the cylindrical vessel = 3 cm
Height of the cylindrical vessel = 4 cm
Let x be the length of BD
According to the diagram,
Using Snell's law,
From (i) and (ii),
Hence, the ratio of real and apparent depth of the image will be $4:2.25=1.78$
Given,
Diameter and height (h) of the cylindrical vessel = 30 cm
Therefore, its radius (r) = 15 cm
We know the refractive index of water (μw) = 1.33 $=\frac{4}{3}$
Using Snell's law,
Point P will be visible when the refracted ray makes an angle of 45Ëš at the point of refraction.
Let x be the distance of point P from X.
Hence, the required minimum height of water = 26.7 cm
#### Question 25:
Given,
Diameter and height (h) of the cylindrical vessel = 30 cm
Therefore, its radius (r) = 15 cm
We know the refractive index of water (μw) = 1.33 $=\frac{4}{3}$
Using Snell's law,
Point P will be visible when the refracted ray makes an angle of 45Ëš at the point of refraction.
Let x be the distance of point P from X.
Hence, the required minimum height of water = 26.7 cm
Given,
Angle of incidence = 45Ëš
Thickness of the plate =
Refractive index (μ) of the plate = 2.0
Applying Snell's law,
Therefore,
From figure, we know that BD is the shift in path which is equal to (AB)sin 24Ëš
Hence, the required shift in the path of light is 0.62 cm.
#### Question 26:
Given,
Angle of incidence = 45Ëš
Thickness of the plate =
Refractive index (μ) of the plate = 2.0
Applying Snell's law,
Therefore,
From figure, we know that BD is the shift in path which is equal to (AB)sin 24Ëš
Hence, the required shift in the path of light is 0.62 cm.
Given,
Refractive index of the optical fibre is represented by μo = 1.72
Refractive index of glass coating is represented by μg= 1.50
Let the critical angle for glass be θc
Using the Snell's law,
Hence, the required critical angle is ${\mathrm{sin}}^{-1}\left(\frac{75}{86}\right)$
#### Question 27:
Given,
Refractive index of the optical fibre is represented by μo = 1.72
Refractive index of glass coating is represented by μg= 1.50
Let the critical angle for glass be θc
Using the Snell's law,
Hence, the required critical angle is ${\mathrm{sin}}^{-1}\left(\frac{75}{86}\right)$
Given,
Refractive index (μ) of prism = 1.50
Let us take θc as the critical angle for the glass.
So, According to Snell's law,
Condition for total internal reflection: 90° − $\varphi$> θc
⇒ $\varphi$ < 90° $-$ θc
Hence, the largest angle for which light is totally reflected at the surface AC is .
#### Question 28:
Given,
Refractive index (μ) of prism = 1.50
Let us take θc as the critical angle for the glass.
So, According to Snell's law,
Condition for total internal reflection: 90° − $\varphi$> θc
⇒ $\varphi$ < 90° $-$ θc
Hence, the largest angle for which light is totally reflected at the surface AC is .
Given,
Refractive index (μ) of the glass = 1.5
We know from the definition of a critical angle (θc) that if a refracted angle measures more than 90°, then total internal reflection takes place.
Hence, the maximum angle of refraction is 90°.
#### Question 29:
Given,
Refractive index (μ) of the glass = 1.5
We know from the definition of a critical angle (θc) that if a refracted angle measures more than 90°, then total internal reflection takes place.
Hence, the maximum angle of refraction is 90°.
Given,
Refractive index of glass, μg = 1.5
Refractive index of air, μa= 1.0
Angle of incidence 0° < i < 90
Let us take θc as the Critical angle
θc = 40°48"
The angle of deviation (δ) due to refraction from glass to air increases as the angle of incidence increases from 0° to 40°48". The angle of deviation (δ) due to total internal reflection further increases from 40°48" to 45° and then it decreases, as shown in the graph.
#### Question 30:
Given,
Refractive index of glass, μg = 1.5
Refractive index of air, μa= 1.0
Angle of incidence 0° < i < 90
Let us take θc as the Critical angle
θc = 40°48"
The angle of deviation (δ) due to refraction from glass to air increases as the angle of incidence increases from 0° to 40°48". The angle of deviation (δ) due to total internal reflection further increases from 40°48" to 45° and then it decreases, as shown in the graph.
Given,
Refractive index of glass $:{\mu }_{\mathrm{g}}=1.5=\frac{3}{2}$
Refractive index of water
As per the question,
For two angles of incidence,
1. When light passes straight through the Normal,
⇒ Angle of incidence = 0°
⇒ Angle of refraction = 0°
⇒ Angle of deviation = 0°
2. When light is incident at critical angle θc,
(since the light is passing from glass to water)
⇒ Angle of deviation
=90° − θc
$=90-{\mathrm{sin}}^{-1}\frac{8}{9}$
$={\mathrm{cos}}^{-1}\frac{8}{9}$
=37.27°
Here, if the angle of incidence increased beyond the critical angle, total internal reflection occurs and deviation decreases.
Therefore, the range of angle of deviation is in between 0 to 37.27° or .
#### Question 31:
Given,
Refractive index of glass $:{\mu }_{\mathrm{g}}=1.5=\frac{3}{2}$
Refractive index of water
As per the question,
For two angles of incidence,
1. When light passes straight through the Normal,
⇒ Angle of incidence = 0°
⇒ Angle of refraction = 0°
⇒ Angle of deviation = 0°
2. When light is incident at critical angle θc,
(since the light is passing from glass to water)
⇒ Angle of deviation
=90° − θc
$=90-{\mathrm{sin}}^{-1}\frac{8}{9}$
$={\mathrm{cos}}^{-1}\frac{8}{9}$
=37.27°
Here, if the angle of incidence increased beyond the critical angle, total internal reflection occurs and deviation decreases.
Therefore, the range of angle of deviation is in between 0 to 37.27° or .
Given,
Light falls from glass to air.
Refractive index (μ) of glass = 1.5
Critical angle (θc)
We know that the maximum attainable angle of deviation in refraction is (90° − 41.8°)
= 47.2°
In this case, total internal reflection must have taken place.
In reflection,
Deviation = 180° − 2i = 90°
⇒ 2i = 90°
i = 45°
Hence, the required angle of incidence is 45°.
#### Question 32:
Given,
Light falls from glass to air.
Refractive index (μ) of glass = 1.5
Critical angle (θc)
We know that the maximum attainable angle of deviation in refraction is (90° − 41.8°)
= 47.2°
In this case, total internal reflection must have taken place.
In reflection,
Deviation = 180° − 2i = 90°
⇒ 2i = 90°
i = 45°
Hence, the required angle of incidence is 45°.
Given,
Refractive index is μ
(a)
Let the point source be P, which is placed at a depth of h from the surface of water.
Let us take x as the radius of the circular area.
and let θc be the critical angle.
Thus,
Clearly from figure, the light escapes through a circular area at a fixed distance r on the water surface, directly above the point source.
That makes a circle, the centre of which is just above P.
(b)
The angle subtended by the radius of the circular area on the point source P:
#### Question 33:
Given,
Refractive index is μ
(a)
Let the point source be P, which is placed at a depth of h from the surface of water.
Let us take x as the radius of the circular area.
and let θc be the critical angle.
Thus,
Clearly from figure, the light escapes through a circular area at a fixed distance r on the water surface, directly above the point source.
That makes a circle, the centre of which is just above P.
(b)
The angle subtended by the radius of the circular area on the point source P:
Given,
Height (h) of the water in the container = 20 cm
Ceiling of the room is 2.0 m above the water surface.
Radius of the rubber ring = r
Refractive index of water = 4/3
(a)
From the figure, we can infer:
Using Snell's law, we get:
From the figure, we have:
(b)
Condition for the maximum value of r:
Angle of incidence should be equal to the critical angle, i.e., $i={\theta }_{\mathrm{c}}$.
Let us take R as the maximum radius.
Now,
#### Question 34:
Given,
Height (h) of the water in the container = 20 cm
Ceiling of the room is 2.0 m above the water surface.
Radius of the rubber ring = r
Refractive index of water = 4/3
(a)
From the figure, we can infer:
Using Snell's law, we get:
From the figure, we have:
(b)
Condition for the maximum value of r:
Angle of incidence should be equal to the critical angle, i.e., $i={\theta }_{\mathrm{c}}$.
Let us take R as the maximum radius.
Now,
Given,
Refractive index (μ) of the material from which prism is made = 1.732
We know refractive index is given by:
Where δmin is the angle of minimum deviation and A is the angle of prism = 60Ëš
δmin = 60°
δmin = 2i − A
2i = 120°
i = 60°
Hence, the required angle of deviation is 60°.
#### Question 35:
Given,
Refractive index (μ) of the material from which prism is made = 1.732
We know refractive index is given by:
Where δmin is the angle of minimum deviation and A is the angle of prism = 60Ëš
δmin = 60°
δmin = 2i − A
2i = 120°
i = 60°
Hence, the required angle of deviation is 60°.
Given,
The refractive index of the prism material (μ) = 1.5
Angle of prism form the figure = 4Ëš
We know that,
Hence, the angle of deviation is 2Ëš
#### Question 36:
Given,
The refractive index of the prism material (μ) = 1.5
Angle of prism form the figure = 4Ëš
We know that,
Hence, the angle of deviation is 2Ëš
Given,
The angle of the prism (A) = 60Ëš
The angle of deviation (δm) = 30Ëš
As there is one ray that has been found which has deviated by 30Ëš, the angle of minimum deviation should be either equal to or less than 30Ëš but it can not be more than 30Ëš.
Therefore,
Refractive index (μ) will be more if angle of deviation (δm) is more.
Hence, the required limit of refractive index is $\sqrt{2}$
#### Question 37:
Given,
The angle of the prism (A) = 60Ëš
The angle of deviation (δm) = 30Ëš
As there is one ray that has been found which has deviated by 30Ëš, the angle of minimum deviation should be either equal to or less than 30Ëš but it can not be more than 30Ëš.
Therefore,
Refractive index (μ) will be more if angle of deviation (δm) is more.
Hence, the required limit of refractive index is $\sqrt{2}$
Given,
Let the refractive indices of two mediums be μ1=1.0 and μ2 =1.5
Point C is the centre of curvature, the distance between C and the pole is 20 cm.
Therefore, radius of curvature (R) = 20 cm
Distance between source S and pole is 25 cm.
Therefore, object distance (u) = −25
Using lens equation,
$\frac{{\mathrm{\mu }}_{2}}{v}-\frac{{\mathrm{\mu }}_{1}}{u}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}}{\mathrm{R}}$
Hence, the required location of the image is 100 cm from the pole and on the side of S.
#### Question 38:
Given,
Let the refractive indices of two mediums be μ1=1.0 and μ2 =1.5
Point C is the centre of curvature, the distance between C and the pole is 20 cm.
Therefore, radius of curvature (R) = 20 cm
Distance between source S and pole is 25 cm.
Therefore, object distance (u) = −25
Using lens equation,
$\frac{{\mathrm{\mu }}_{2}}{v}-\frac{{\mathrm{\mu }}_{1}}{u}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}}{\mathrm{R}}$
Hence, the required location of the image is 100 cm from the pole and on the side of S.
Given,
Spherical surface of radius (R) = 30 cm
Medium A has refractive index (μ1) = 1.33
Medium B has refractive index (μ2) = 1.48
Medium A is the convex side of surface.
Since,
We know that paraxial rays become parallel after refraction
i.e, the image of the point object will be formed at infinity.
Therefore v = ∞
Using the lens equation,
Hence, the object is placed at a distance of 266.0 cm from the convex surface on side A.
#### Question 39:
Given,
Spherical surface of radius (R) = 30 cm
Medium A has refractive index (μ1) = 1.33
Medium B has refractive index (μ2) = 1.48
Medium A is the convex side of surface.
Since,
We know that paraxial rays become parallel after refraction
i.e, the image of the point object will be formed at infinity.
Therefore v = ∞
Using the lens equation,
Hence, the object is placed at a distance of 266.0 cm from the convex surface on side A.
Given,
The radius of the transparent hemisphere (R) = 3.0 cm
Refractive index of the material (μ2) = 2.0
Let the critical angle be θc
∴ critical angle is given by θc =${\mathrm{sin}}^{-1}\left(\frac{1}{{\mathrm{\mu }}_{2}}\right)={\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)=30°$
(a)
From the figure it is seen that the angle of incidence is greater than the critical angle, so the rays are totally reflected at the plane surface.
(b)
Using the lens equation:
If we complete the sphere then the image will be formed diametrically opposite to A.
(c)
By internal reflection, the image is formed in front of A.
#### Question 40:
Given,
The radius of the transparent hemisphere (R) = 3.0 cm
Refractive index of the material (μ2) = 2.0
Let the critical angle be θc
∴ critical angle is given by θc =${\mathrm{sin}}^{-1}\left(\frac{1}{{\mathrm{\mu }}_{2}}\right)={\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)=30°$
(a)
From the figure it is seen that the angle of incidence is greater than the critical angle, so the rays are totally reflected at the plane surface.
(b)
Using the lens equation:
If we complete the sphere then the image will be formed diametrically opposite to A.
(c)
By internal reflection, the image is formed in front of A.
Given,
Radius of the sphere = 5.0 cm
Refractive index of the sphere (μ1) = 1.5
An object is embedded in the glass sphere 1.5 cm left to the centre.
(a)
When the image is seen by observer from left of the sphere,
from surface the object distance (u) = − 3.5 cm
μ1 = 1.5
μ2= 1
v1 = ?
Using lens equation:
$\frac{{\mathrm{\mu }}_{2}}{{v}_{1}}-\frac{{\mathrm{\mu }}_{1}}{u}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}}{\mathrm{R}}$
$\frac{1}{{v}_{1}}-\frac{1.5}{-\left(3.5\right)}=\frac{-0.5}{-5}$
$\frac{1}{{v}_{1}}=\frac{1}{10}-\frac{3}{7}$
$=\frac{7-30}{70}$
$=\frac{-23}{70}$
So the image will be formed at 2 cm (5 cm $-$ 3cm) left to centre.
(b)
When the image is seen by observer from the right of the sphere,
u = −(5.0 + 1.5) = − 6.5,
R = −5.00 cm
μ1 = 1.5, μ2 = 1, v = ?
Using lens equation:
$\frac{{\mathrm{\mu }}_{2}}{{v}_{1}}-\frac{{\mathrm{\mu }}_{1}}{u}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}}{R}$
$\frac{1}{v}-\frac{1.5}{6.3}=\frac{1}{10}$
$\frac{1}{v}=\frac{1}{10}-\frac{3}{13}$
$=\frac{13-30}{130}$
$=\frac{-17}{130}$
Therefore, the image will be formed 7.6 − 5 = 2.6 towards left from centre.
#### Question 41:
Given,
Radius of the sphere = 5.0 cm
Refractive index of the sphere (μ1) = 1.5
An object is embedded in the glass sphere 1.5 cm left to the centre.
(a)
When the image is seen by observer from left of the sphere,
from surface the object distance (u) = − 3.5 cm
μ1 = 1.5
μ2= 1
v1 = ?
Using lens equation:
$\frac{{\mathrm{\mu }}_{2}}{{v}_{1}}-\frac{{\mathrm{\mu }}_{1}}{u}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}}{\mathrm{R}}$
$\frac{1}{{v}_{1}}-\frac{1.5}{-\left(3.5\right)}=\frac{-0.5}{-5}$
$\frac{1}{{v}_{1}}=\frac{1}{10}-\frac{3}{7}$
$=\frac{7-30}{70}$
$=\frac{-23}{70}$
So the image will be formed at 2 cm (5 cm $-$ 3cm) left to centre.
(b)
When the image is seen by observer from the right of the sphere,
u = −(5.0 + 1.5) = − 6.5,
R = −5.00 cm
μ1 = 1.5, μ2 = 1, v = ?
Using lens equation:
$\frac{{\mathrm{\mu }}_{2}}{{v}_{1}}-\frac{{\mathrm{\mu }}_{1}}{u}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}}{R}$
$\frac{1}{v}-\frac{1.5}{6.3}=\frac{1}{10}$
$\frac{1}{v}=\frac{1}{10}-\frac{3}{13}$
$=\frac{13-30}{130}$
$=\frac{-17}{130}$
Therefore, the image will be formed 7.6 − 5 = 2.6 towards left from centre.
Given,
Biconvex lens with each surface has a radius (R1= R2= R) = 10 cm,
and thickness of the lens (t) = 5 cm
Refractive index of the lens (μ) = 1.50
Object is at infinity (∴ u = ∞ )
First refraction takes place at A.
We know that,
Now, the second refraction is at B.
For this, a virtual object is the image of the previous refraction.
Thus, u = (30 − 5) = 25 cm
Hence, the image is formed 9.1 cm far from the second refraction or second surface of the lens.
#### Question 42:
Given,
Biconvex lens with each surface has a radius (R1= R2= R) = 10 cm,
and thickness of the lens (t) = 5 cm
Refractive index of the lens (μ) = 1.50
Object is at infinity (∴ u = ∞ )
First refraction takes place at A.
We know that,
Now, the second refraction is at B.
For this, a virtual object is the image of the previous refraction.
Thus, u = (30 − 5) = 25 cm
Hence, the image is formed 9.1 cm far from the second refraction or second surface of the lens.
Given,
The radius of the transparent sphere = r
Refraction at convex surface.
As per the question,
u = −∞, μ1 = 1, μ2 = ?
(a)
When image is to be focused on the surface,
Image distance (v) = 2r, Radius of curvature (R) = r
We know that,
$\frac{{\mu }_{2}}{v}-\frac{{\mu }_{1}}{u}=\frac{{\mu }_{2}-{\mu }_{1}}{R}\phantom{\rule{0ex}{0ex}}⇒\frac{{\mu }_{2}}{2r}-\left(\frac{1}{-\infty }\right)=\frac{{\mu }_{2}-1}{r}\phantom{\rule{0ex}{0ex}}⇒\frac{{\mu }_{2}}{2r}=\frac{{\mu }_{2}-1}{r}\phantom{\rule{0ex}{0ex}}⇒{\mu }_{2}=2{\mu }_{2}-2\phantom{\rule{0ex}{0ex}}⇒{\mu }_{2}=2$
(b)
When the image is to be focused at the centre,
Image distance (v) = r, Radius of curvature (R) = r
$\frac{{\mu }_{2}}{v}-\frac{{\mu }_{1}}{u}=\frac{{\mu }_{2}-{\mu }_{1}}{R}\phantom{\rule{0ex}{0ex}}⇒\frac{{\mu }_{2}}{r}-\left(\frac{1}{-\infty }\right)=\frac{{\mu }_{2}-1}{r}\phantom{\rule{0ex}{0ex}}⇒\frac{{\mu }_{2}}{r}=\frac{{\mu }_{2}-1}{r}\phantom{\rule{0ex}{0ex}}⇒{\mu }_{2}={\mu }_{2}-1\phantom{\rule{0ex}{0ex}}$
The above equation is impossible.
Hence, the image cannot be focused at centre.
#### Question 43:
Given,
The radius of the transparent sphere = r
Refraction at convex surface.
As per the question,
u = −∞, μ1 = 1, μ2 = ?
(a)
When image is to be focused on the surface,
Image distance (v) = 2r, Radius of curvature (R) = r
We know that,
$\frac{{\mu }_{2}}{v}-\frac{{\mu }_{1}}{u}=\frac{{\mu }_{2}-{\mu }_{1}}{R}\phantom{\rule{0ex}{0ex}}⇒\frac{{\mu }_{2}}{2r}-\left(\frac{1}{-\infty }\right)=\frac{{\mu }_{2}-1}{r}\phantom{\rule{0ex}{0ex}}⇒\frac{{\mu }_{2}}{2r}=\frac{{\mu }_{2}-1}{r}\phantom{\rule{0ex}{0ex}}⇒{\mu }_{2}=2{\mu }_{2}-2\phantom{\rule{0ex}{0ex}}⇒{\mu }_{2}=2$
(b)
When the image is to be focused at the centre,
Image distance (v) = r, Radius of curvature (R) = r
$\frac{{\mu }_{2}}{v}-\frac{{\mu }_{1}}{u}=\frac{{\mu }_{2}-{\mu }_{1}}{R}\phantom{\rule{0ex}{0ex}}⇒\frac{{\mu }_{2}}{r}-\left(\frac{1}{-\infty }\right)=\frac{{\mu }_{2}-1}{r}\phantom{\rule{0ex}{0ex}}⇒\frac{{\mu }_{2}}{r}=\frac{{\mu }_{2}-1}{r}\phantom{\rule{0ex}{0ex}}⇒{\mu }_{2}={\mu }_{2}-1\phantom{\rule{0ex}{0ex}}$
The above equation is impossible.
Hence, the image cannot be focused at centre.
Given,
Radius (R) of the cylindrical rod = 1.0 cm
Refractive index (μg) of the rod = 1.5 = $\frac{3}{2}$
Refractive index (μw) of water = 4/3
We know:
$\frac{{\mu }_{g}}{v}-\frac{{\mu }_{w}}{u}=\frac{{\mu }_{g}-{\mu }_{w}}{R}$
As per the question, u = −8 cm.
Now,
$\frac{3}{2v}-\left(-\frac{4}{3×8}\right)=\frac{\frac{3}{2}-\frac{4}{3}}{1}\phantom{\rule{0ex}{0ex}}⇒\frac{3}{2v}+\frac{1}{6}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒v=\infty$
Hence, the image will be formed at infinity (∞).
#### Question 44:
Given,
Radius (R) of the cylindrical rod = 1.0 cm
Refractive index (μg) of the rod = 1.5 = $\frac{3}{2}$
Refractive index (μw) of water = 4/3
We know:
$\frac{{\mu }_{g}}{v}-\frac{{\mu }_{w}}{u}=\frac{{\mu }_{g}-{\mu }_{w}}{R}$
As per the question, u = −8 cm.
Now,
$\frac{3}{2v}-\left(-\frac{4}{3×8}\right)=\frac{\frac{3}{2}-\frac{4}{3}}{1}\phantom{\rule{0ex}{0ex}}⇒\frac{3}{2v}+\frac{1}{6}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒v=\infty$
Hence, the image will be formed at infinity (∞).
Given:
Radius of the paperweight (R) = 3 cm
Refractive index of the paperweight (μ2) = 3/2
Refractive index of the air (μ1) = 1
In the first case, the refraction is at A.
u = 0 and R =
We know:
$\frac{{\mathrm{\mu }}_{2}}{v}-\frac{{\mathrm{\mu }}_{1}}{u}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}}{R}\phantom{\rule{0ex}{0ex}}⇒\frac{3}{2}}{v}-\frac{1}{0}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}}{\infty }\phantom{\rule{0ex}{0ex}}⇒v=0$
Therefore, the image of the letter is formed at the point.
For the second case, refraction is at point B.
Here,
Object distance, u = −3 cm
R = − 3 cm
μ1 = 3/2
μ2 = 1
Thus, we have:
Hence, there will be no shift in the final image.
#### Question 45:
Given:
Radius of the paperweight (R) = 3 cm
Refractive index of the paperweight (μ2) = 3/2
Refractive index of the air (μ1) = 1
In the first case, the refraction is at A.
u = 0 and R =
We know:
$\frac{{\mathrm{\mu }}_{2}}{v}-\frac{{\mathrm{\mu }}_{1}}{u}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}}{R}\phantom{\rule{0ex}{0ex}}⇒\frac{3}{2}}{v}-\frac{1}{0}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}}{\infty }\phantom{\rule{0ex}{0ex}}⇒v=0$
Therefore, the image of the letter is formed at the point.
For the second case, refraction is at point B.
Here,
Object distance, u = −3 cm
R = − 3 cm
μ1 = 3/2
μ2 = 1
Thus, we have:
Hence, there will be no shift in the final image.
Given,
Taking the radius of the paperweight as its thickness = 3 cm
Refractive index of the paperweight (μg) = 3/2
Refractive index of the air (μ1) = 1
Image shift is given by:
The upper surface of the paperweight is flat and the spherical spherical surface is in contact with the printed letter.
Therefore, we will take it as a simple refraction problem.
Hence, the image will appear 1 cm above point A.
#### Question 46:
Given,
Taking the radius of the paperweight as its thickness = 3 cm
Refractive index of the paperweight (μg) = 3/2
Refractive index of the air (μ1) = 1
Image shift is given by:
The upper surface of the paperweight is flat and the spherical spherical surface is in contact with the printed letter.
Therefore, we will take it as a simple refraction problem.
Hence, the image will appear 1 cm above point A.
As shown in the figure, OQ = 3r and OP = r
Thus, PQ = 2r
For refraction at APB,
we know that,
For the reflection in the concave mirror,
u = ∞
Thus, v = focal length of mirror = r/2
For the refraction of APB of the reflected image,
u = −3r/2
As negative sign indicates images formed inside APB, so image should be at C.
Therefore, the final image is formed on the reflecting surface of the sphere.
#### Question 47:
As shown in the figure, OQ = 3r and OP = r
Thus, PQ = 2r
For refraction at APB,
we know that,
For the reflection in the concave mirror,
u = ∞
Thus, v = focal length of mirror = r/2
For the refraction of APB of the reflected image,
u = −3r/2
As negative sign indicates images formed inside APB, so image should be at C.
Therefore, the final image is formed on the reflecting surface of the sphere.
(a) Let F be the the focal length of the given concavo-convex lens. Then,
$\frac{1}{F}=\frac{1}{{f}_{1}}+\frac{1}{{f}_{m}}+\frac{1}{{f}_{1}}$
$=\frac{2}{{f}_{1}}+\frac{1}{{f}_{m}}$ $\left[\because \frac{1}{{f}_{m}}=\left(\mathrm{\mu }-1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)\right]$
$\frac{1}{F}=\frac{2}{30}+\frac{1}{15}=\frac{2}{15}$
F = 7.5 cm
Hence, R = 15 cm
Therefore, the pin should be placed at a distance of 15 cm from the lens.
(b) If the concave part is filled with water,
For focal length F'
​ ​$\frac{1}{F\mathit{\text{'}}}=\frac{2}{{f}_{w}}+\frac{2}{{f}_{1}}+\frac{1}{{f}_{m}}$
$=\frac{2}{90}+\frac{2}{30}+\frac{1}{15}$ $\left[\because \frac{1}{{f}_{w}}=\left(\frac{4}{3}-1\right)\left(+\frac{1}{30}\right)\right]$
Thus, pin should be placed at a distance of $\frac{90}{7}$ cm from the lens.
#### Question 48:
(a) Let F be the the focal length of the given concavo-convex lens. Then,
$\frac{1}{F}=\frac{1}{{f}_{1}}+\frac{1}{{f}_{m}}+\frac{1}{{f}_{1}}$
$=\frac{2}{{f}_{1}}+\frac{1}{{f}_{m}}$ $\left[\because \frac{1}{{f}_{m}}=\left(\mathrm{\mu }-1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)\right]$
$\frac{1}{F}=\frac{2}{30}+\frac{1}{15}=\frac{2}{15}$
F = 7.5 cm
Hence, R = 15 cm
Therefore, the pin should be placed at a distance of 15 cm from the lens.
(b) If the concave part is filled with water,
For focal length F'
​ ​$\frac{1}{F\mathit{\text{'}}}=\frac{2}{{f}_{w}}+\frac{2}{{f}_{1}}+\frac{1}{{f}_{m}}$
$=\frac{2}{90}+\frac{2}{30}+\frac{1}{15}$ $\left[\because \frac{1}{{f}_{w}}=\left(\frac{4}{3}-1\right)\left(+\frac{1}{30}\right)\right]$
Thus, pin should be placed at a distance of $\frac{90}{7}$ cm from the lens.
Given,
Double convex lens of focal length, f = 25 cm
Refractive index of the material, μ = 1.5
As per the question, the radius of curvature of one surface is twice that of the other.
i.e. R1= R and R2= −2R
(according to sign conventions)
Using the lens maker formula $\frac{1}{f}=\left(\mathrm{\mu }-1\right)\left[\frac{\mathit{1}}{{\mathit{R}}_{\mathit{1}}}\mathit{-}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{2}}}\right]$, we have:
$\frac{1}{25}=\left(1.5-1\right)\left[\frac{1}{R}-\left(\frac{1}{-2R}\right)\right]$
R1 = 18.75 cm and R2 = 37.5 cm
Hence, the required radii of the double convex lens are 18.75 cm and 37.5 cm.
#### Question 49:
Given,
Double convex lens of focal length, f = 25 cm
Refractive index of the material, μ = 1.5
As per the question, the radius of curvature of one surface is twice that of the other.
i.e. R1= R and R2= −2R
(according to sign conventions)
Using the lens maker formula $\frac{1}{f}=\left(\mathrm{\mu }-1\right)\left[\frac{\mathit{1}}{{\mathit{R}}_{\mathit{1}}}\mathit{-}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{2}}}\right]$, we have:
$\frac{1}{25}=\left(1.5-1\right)\left[\frac{1}{R}-\left(\frac{1}{-2R}\right)\right]$
R1 = 18.75 cm and R2 = 37.5 cm
Hence, the required radii of the double convex lens are 18.75 cm and 37.5 cm.
Given,
Radii of curvature of a lens, (R1) = +20 cm and R2 = +30 cm
Refractive index of the material of the lens, (μ) = 1.6
Refractive index of water, (μwater) = 1.33
(a)
When the lens is placed in air,
Using lens maker formula:
$\frac{1}{f}=\left(\mathrm{\mu }-1\right)\left[\frac{\mathit{1}}{{\mathit{R}}_{\mathit{1}}}\mathit{-}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{2}}}\right]$
$\frac{1}{f}=0.6\left[\frac{1}{20}-\frac{1}{30}\right]\phantom{\rule{0ex}{0ex}}$
$f=\frac{0.6}{1060×10}$
f = 100 cm
Thus, the focal length of the lens is 100 cm when it is placed in air.
(b)When the lens is placed in water
$\frac{1}{f}=\left[\frac{{\mathrm{\mu }}_{\mathrm{lens}}}{{\mathrm{\mu }}_{\mathrm{water}}}-1\right]\left[\frac{\mathit{1}}{{R}_{\mathit{1}}}\mathit{-}\frac{\mathit{1}}{{R}_{\mathit{2}}}\right]$
$=\left(\frac{1.60}{1.33}-1\right)\left[\frac{1}{60}\right]$
$=\frac{28}{133×60}\simeq \frac{1}{300}$
$⇒$f = 300 cm.
Thus, the focal length of the lens is 300 cm when it is placed in water.
#### Question 50:
Given,
Radii of curvature of a lens, (R1) = +20 cm and R2 = +30 cm
Refractive index of the material of the lens, (μ) = 1.6
Refractive index of water, (μwater) = 1.33
(a)
When the lens is placed in air,
Using lens maker formula:
$\frac{1}{f}=\left(\mathrm{\mu }-1\right)\left[\frac{\mathit{1}}{{\mathit{R}}_{\mathit{1}}}\mathit{-}\frac{\mathit{1}}{{\mathit{R}}_{\mathit{2}}}\right]$
$\frac{1}{f}=0.6\left[\frac{1}{20}-\frac{1}{30}\right]\phantom{\rule{0ex}{0ex}}$
$f=\frac{0.6}{1060×10}$
f = 100 cm
Thus, the focal length of the lens is 100 cm when it is placed in air.
(b)When the lens is placed in water
$\frac{1}{f}=\left[\frac{{\mathrm{\mu }}_{\mathrm{lens}}}{{\mathrm{\mu }}_{\mathrm{water}}}-1\right]\left[\frac{\mathit{1}}{{R}_{\mathit{1}}}\mathit{-}\frac{\mathit{1}}{{R}_{\mathit{2}}}\right]$
$=\left(\frac{1.60}{1.33}-1\right)\left[\frac{1}{60}\right]$
$=\frac{28}{133×60}\simeq \frac{1}{300}$
$⇒$f = 300 cm.
Thus, the focal length of the lens is 300 cm when it is placed in water.
Given,
Refractive index of the material, (μ) = 1.50
Magnitudes of the radius of curvature:
R1 = 20 cm and R2 = 30 cm
From the given data we can make four possible lenses, using the lens maker formula.
$\frac{1}{f}=\left(\mathrm{\mu }-1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)$
Four possible lenses:
(a) 1st lens is double convex, in which R1 = +20 cm and R2 = −30 cm
$=0.5\left[\frac{1}{20}-\frac{1}{\left(-30\right)}\right]=\frac{0.5×5}{60}$
f = +24 cm
(b) 2nd lens is double concave, in which R1 = −20 cm and R2 = +30 cm
$=0.5\left[\frac{-1}{20}-\frac{1}{30}\right]$
f = −24 cm
(c) 3rd lens is concave concave in which R1 = −20 cm and R2 = −30 cm
$=0.5\left[\frac{-1}{20}-\frac{1}{\left(-30\right)}\right]$
f = −120 cm
(d) 4th lens is concave convex, in which R1 = +20 cm and R2 = +30 cm
$=0.5\left[\frac{1}{20}-\frac{1}{30}\right]$
​f = +120 cm
#### Question 51:
Given,
Refractive index of the material, (μ) = 1.50
Magnitudes of the radius of curvature:
R1 = 20 cm and R2 = 30 cm
From the given data we can make four possible lenses, using the lens maker formula.
$\frac{1}{f}=\left(\mathrm{\mu }-1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)$
Four possible lenses:
(a) 1st lens is double convex, in which R1 = +20 cm and R2 = −30 cm
$=0.5\left[\frac{1}{20}-\frac{1}{\left(-30\right)}\right]=\frac{0.5×5}{60}$
f = +24 cm
(b) 2nd lens is double concave, in which R1 = −20 cm and R2 = +30 cm
$=0.5\left[\frac{-1}{20}-\frac{1}{30}\right]$
f = −24 cm
(c) 3rd lens is concave concave in which R1 = −20 cm and R2 = −30 cm
$=0.5\left[\frac{-1}{20}-\frac{1}{\left(-30\right)}\right]$
f = −120 cm
(d) 4th lens is concave convex, in which R1 = +20 cm and R2 = +30 cm
$=0.5\left[\frac{1}{20}-\frac{1}{30}\right]$
​f = +120 cm
Given,
A biconvex lens with two radii of curvature that have equal magnitude R.
Refractive index of the material of the lens is μ2.
First medium of refractive index is μ1.
Second medium of refractive index is μ3.
As per the question, the light beams are travelling parallel to the principal axis of the lens.
i.e., u (object distance) = ∞
(a) The light beam is incident on the lens from first medium μ1.
Thus, refraction takes place at first surface
Using equation of refraction,
$\frac{{\mathrm{\mu }}_{2}}{v}-\frac{{\mathrm{\mu }}_{1}}{u}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}}{R}$
Where, v is the image distance.
Applying sign convention, we get:
$\frac{{\mathrm{\mu }}_{2}}{v}-\frac{{\mathrm{\mu }}_{1}}{\left(-\infty \right)}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}}{R}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}}{{\mathrm{\mu }}_{2}R}\phantom{\rule{0ex}{0ex}}⇒⇒v=\frac{{\mathrm{\mu }}_{2}R}{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}}\phantom{\rule{0ex}{0ex}}$
Now, refraction takes place at 2nd surface
Thus,$\frac{{\mathrm{\mu }}_{3}}{v}-\frac{{\mathrm{\mu }}_{2}}{u}=\frac{{\mathrm{\mu }}_{3}-{\mathrm{\mu }}_{2}}{R}\phantom{\rule{0ex}{0ex}}$
Here, the image distance of the previous case becomes object distance:
Therefore, the image is formed at $\frac{{\mathrm{\mu }}_{3}R}{2{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}-{\mathrm{\mu }}_{3}}$
(b) The light beam is incident on the lens from second medium μ3.
Thus, refraction takes place at second surface.
Using equation of refraction,
$\frac{{\mathrm{\mu }}_{2}}{v}-\frac{{\mathrm{\mu }}_{3}}{u}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{3}}{R}$
Where, v is the image distance
Applying sign convention, we get:
$\frac{{\mathrm{\mu }}_{2}}{v}-\frac{{\mathrm{\mu }}_{3}}{\left(-\infty \right)}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{3}}{R}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{3}}{{\mathrm{\mu }}_{2}R}\phantom{\rule{0ex}{0ex}}⇒⇒v=\frac{{\mathrm{\mu }}_{2}R}{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{3}}\phantom{\rule{0ex}{0ex}}$
Now, refraction takes place at 2nd surface.
Thus,$\frac{{\mathrm{\mu }}_{1}}{v}-\frac{{\mathrm{\mu }}_{2}}{u}=\frac{{\mathrm{\mu }}_{1}-{\mathrm{\mu }}_{2}}{R}\phantom{\rule{0ex}{0ex}}$
Here, the image distance of the previous case becomes object distance.
Therefore, the image is formed at $\frac{{\mathrm{\mu }}_{1}R}{2{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}-{\mathrm{\mu }}_{3}}$
#### Question 52:
Given,
A biconvex lens with two radii of curvature that have equal magnitude R.
Refractive index of the material of the lens is μ2.
First medium of refractive index is μ1.
Second medium of refractive index is μ3.
As per the question, the light beams are travelling parallel to the principal axis of the lens.
i.e., u (object distance) = ∞
(a) The light beam is incident on the lens from first medium μ1.
Thus, refraction takes place at first surface
Using equation of refraction,
$\frac{{\mathrm{\mu }}_{2}}{v}-\frac{{\mathrm{\mu }}_{1}}{u}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}}{R}$
Where, v is the image distance.
Applying sign convention, we get:
$\frac{{\mathrm{\mu }}_{2}}{v}-\frac{{\mathrm{\mu }}_{1}}{\left(-\infty \right)}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}}{R}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}}{{\mathrm{\mu }}_{2}R}\phantom{\rule{0ex}{0ex}}⇒⇒v=\frac{{\mathrm{\mu }}_{2}R}{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}}\phantom{\rule{0ex}{0ex}}$
Now, refraction takes place at 2nd surface
Thus,$\frac{{\mathrm{\mu }}_{3}}{v}-\frac{{\mathrm{\mu }}_{2}}{u}=\frac{{\mathrm{\mu }}_{3}-{\mathrm{\mu }}_{2}}{R}\phantom{\rule{0ex}{0ex}}$
Here, the image distance of the previous case becomes object distance:
Therefore, the image is formed at $\frac{{\mathrm{\mu }}_{3}R}{2{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}-{\mathrm{\mu }}_{3}}$
(b) The light beam is incident on the lens from second medium μ3.
Thus, refraction takes place at second surface.
Using equation of refraction,
$\frac{{\mathrm{\mu }}_{2}}{v}-\frac{{\mathrm{\mu }}_{3}}{u}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{3}}{R}$
Where, v is the image distance
Applying sign convention, we get:
$\frac{{\mathrm{\mu }}_{2}}{v}-\frac{{\mathrm{\mu }}_{3}}{\left(-\infty \right)}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{3}}{R}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{3}}{{\mathrm{\mu }}_{2}R}\phantom{\rule{0ex}{0ex}}⇒⇒v=\frac{{\mathrm{\mu }}_{2}R}{{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{3}}\phantom{\rule{0ex}{0ex}}$
Now, refraction takes place at 2nd surface.
Thus,$\frac{{\mathrm{\mu }}_{1}}{v}-\frac{{\mathrm{\mu }}_{2}}{u}=\frac{{\mathrm{\mu }}_{1}-{\mathrm{\mu }}_{2}}{R}\phantom{\rule{0ex}{0ex}}$
Here, the image distance of the previous case becomes object distance.
Therefore, the image is formed at $\frac{{\mathrm{\mu }}_{1}R}{2{\mathrm{\mu }}_{2}-{\mathrm{\mu }}_{1}-{\mathrm{\mu }}_{3}}$
Given:
Focal length (f) of the convex lens = 10 cm
(a) As per the question, the object distance (u) is 9.8 cm.
The lens equation is given by:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
=$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{10}-\frac{1}{9.8}$
=$\frac{1}{v}=\frac{9.8-10}{98}=\frac{-0.2}{98}$
= v = − 98 × 5
= − 490 cm (Same side of the object)
v = 490 cm (Virtual and on on the side of object)
Magnification of the image $=\frac{v}{u}$
$=\frac{-490}{-9.8}\phantom{\rule{0ex}{0ex}}=50$
Therefore, the image is erect and virtual.
(b) Object distance, u = 10.2 cm
The lens equation is given by:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
=$\frac{1}{v}=\frac{1}{10}-\frac{1}{10.2}$
$=\frac{10.2-10}{102}=\frac{0.2}{102}$
= v = 102 × 5 = 510 cm (Real and on the opposite side of the object)
Magnification of the image $=\frac{v}{u}$
$=\frac{510}{-9.8}\phantom{\rule{0ex}{0ex}}=-52.04$
Therefore, the image is real and inverted.
#### Question 53:
Given:
Focal length (f) of the convex lens = 10 cm
(a) As per the question, the object distance (u) is 9.8 cm.
The lens equation is given by:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
=$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{10}-\frac{1}{9.8}$
=$\frac{1}{v}=\frac{9.8-10}{98}=\frac{-0.2}{98}$
= v = − 98 × 5
= − 490 cm (Same side of the object)
v = 490 cm (Virtual and on on the side of object)
Magnification of the image $=\frac{v}{u}$
$=\frac{-490}{-9.8}\phantom{\rule{0ex}{0ex}}=50$
Therefore, the image is erect and virtual.
(b) Object distance, u = 10.2 cm
The lens equation is given by:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
=$\frac{1}{v}=\frac{1}{10}-\frac{1}{10.2}$
$=\frac{10.2-10}{102}=\frac{0.2}{102}$
= v = 102 × 5 = 510 cm (Real and on the opposite side of the object)
Magnification of the image $=\frac{v}{u}$
$=\frac{510}{-9.8}\phantom{\rule{0ex}{0ex}}=-52.04$
Therefore, the image is real and inverted.
Given,
We are projecting a slide of 35 mm $×$ 23 mm on a 2 m $×$ 2 m screen using projector.
Therefore, the magnification required by the projector is:
$m=\frac{v}{u}$
Here,
v = Image distance
u = Object distance
We will take 35 mm as the object size
∵ 35 mm > 23 mm
The lens formula is given by
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Hence, the required focal length is 0.172 mm.
#### Question 54:
Given,
We are projecting a slide of 35 mm $×$ 23 mm on a 2 m $×$ 2 m screen using projector.
Therefore, the magnification required by the projector is:
$m=\frac{v}{u}$
Here,
v = Image distance
u = Object distance
We will take 35 mm as the object size
∵ 35 mm > 23 mm
The lens formula is given by
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Hence, the required focal length is 0.172 mm.
When the particle is at point B,
$\frac{1}{{v}_{\mathrm{B}}}=\frac{1}{f}+\frac{1}{u}$
$\frac{1}{{v}_{B}}=\frac{1}{12}-\frac{1}{19}$
When particle is at point A,
$\frac{1}{{v}_{\mathrm{A}}}=\frac{1}{f}+\frac{1}{u}$
$\frac{1}{{v}_{\mathrm{A}}}=\frac{1}{12}-\frac{1}{21}$
#### Question 55:
When the particle is at point B,
$\frac{1}{{v}_{\mathrm{B}}}=\frac{1}{f}+\frac{1}{u}$
$\frac{1}{{v}_{B}}=\frac{1}{12}-\frac{1}{19}$
When particle is at point A,
$\frac{1}{{v}_{\mathrm{A}}}=\frac{1}{f}+\frac{1}{u}$
$\frac{1}{{v}_{\mathrm{A}}}=\frac{1}{12}-\frac{1}{21}$
Given,
Object distance, (u) = 5.0 cm
Focal length (f) of convex lens = 8.0 cm
(a)
(b)
Using lens formula:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Where v is the image distance,
on putting the given values we get:
Hence, the position of the image from the lens is 13.3 cm.
#### Question 56:
Given,
Object distance, (u) = 5.0 cm
Focal length (f) of convex lens = 8.0 cm
(a)
(b)
Using lens formula:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Where v is the image distance,
on putting the given values we get:
Hence, the position of the image from the lens is 13.3 cm.
Given,
Height of the object, (h0) = 2 cm,
Height of the image, (h1) = 1 cm
distance between image and object, ($-$u + v) = 40 cm
We know that,
Magnification is given by:
u = −2v
Using lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}-\frac{1}{-2v}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}+\frac{1}{2v}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}⇒f=\frac{2v}{3}$
As per the question,
= −u + v = 40 cm,
= −(−2v) + v = 40 cm,
= 3v = 40 cm
u = 26.66 cm.
∴ f = 8.9 cm
Hence, the required focal length is 8.9 cm and object distance is 26.66 cm.
#### Question 57:
Given,
Height of the object, (h0) = 2 cm,
Height of the image, (h1) = 1 cm
distance between image and object, ($-$u + v) = 40 cm
We know that,
Magnification is given by:
u = −2v
Using lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}-\frac{1}{-2v}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}+\frac{1}{2v}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}⇒f=\frac{2v}{3}$
As per the question,
= −u + v = 40 cm,
= −(−2v) + v = 40 cm,
= 3v = 40 cm
u = 26.66 cm.
∴ f = 8.9 cm
Hence, the required focal length is 8.9 cm and object distance is 26.66 cm.
Given,
Object distance, u = $-$18 cm
Size of the image is two times the size of the object
i.e., h1 = 2h0,
We know that,
magnification is given by, = $\frac{v}{u}$
∴ v = 2u = 36 cm.
Using lens formula:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{36}+\frac{1}{18}=\frac{1}{f}$
f = 12 cm.
For, h1 = 3h0
$\frac{{h}_{i}}{{h}_{0}}=\frac{v}{u}$
u = $-$16 cm.
Therefore, the object should be placed at a distance of 16 cm in front of the lens.
#### Question 58:
Given,
Object distance, u = $-$18 cm
Size of the image is two times the size of the object
i.e., h1 = 2h0,
We know that,
magnification is given by, = $\frac{v}{u}$
∴ v = 2u = 36 cm.
Using lens formula:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{36}+\frac{1}{18}=\frac{1}{f}$
f = 12 cm.
For, h1 = 3h0
$\frac{{h}_{i}}{{h}_{0}}=\frac{v}{u}$
u = $-$16 cm.
Therefore, the object should be placed at a distance of 16 cm in front of the lens.
Given,
Length of the pin = 2.0 cm
Focal length (f) of the lens = 6 cm
As per the question, the centre of the pin is 11 cm away from the lens.
i.e., the object distance (u) = 10 cm
Since, we have to calculate the image of A and B, Let the image be A' and B'
So, the length of the A'B' = size of the image.
Using lens formula:
$\frac{1}{{v}_{\mathrm{A}}}-\frac{1}{{u}_{\mathrm{A}}}=\frac{1}{f}$
Where vA and uA are the image and object distances from point A.
$\frac{1}{{v}_{\mathrm{A}}}=\frac{1}{6}-\frac{1}{10}=\frac{1}{15}$
Similarly for point B,
Lens formula:
$\frac{1}{{v}_{\mathrm{B}}}-\frac{1}{{u}_{\mathrm{B}}}=\frac{1}{f}$
Where vA and uA are the image and object distances from point B.
Length of image = vAvB = 15 − 12 = 3 cm.
#### Question 59:
Given,
Length of the pin = 2.0 cm
Focal length (f) of the lens = 6 cm
As per the question, the centre of the pin is 11 cm away from the lens.
i.e., the object distance (u) = 10 cm
Since, we have to calculate the image of A and B, Let the image be A' and B'
So, the length of the A'B' = size of the image.
Using lens formula:
$\frac{1}{{v}_{\mathrm{A}}}-\frac{1}{{u}_{\mathrm{A}}}=\frac{1}{f}$
Where vA and uA are the image and object distances from point A.
$\frac{1}{{v}_{\mathrm{A}}}=\frac{1}{6}-\frac{1}{10}=\frac{1}{15}$
Similarly for point B,
Lens formula:
$\frac{1}{{v}_{\mathrm{B}}}-\frac{1}{{u}_{\mathrm{B}}}=\frac{1}{f}$
Where vA and uA are the image and object distances from point B.
Length of image = vAvB = 15 − 12 = 3 cm.
Given,
Diameter of the sun = 1.4 × 109 m
Distance between sun and earth is taken as object distance (u) = − 150 × 1011 cm,
Focal length (f) of the lens = 20 cm
Using lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
The object distance is very large as compared to focal length of the lens.
Hence, the image is formed at the focus.
$=\frac{750×{10}^{9}-1}{150×{10}^{11}}$
$\simeq \frac{750×{10}^{9}}{150×{10}^{11}}$
We know, Magnification (m) is given by:
$\left(m\right)=\frac{v}{u}=\frac{{h}_{2}}{{h}_{1}}$
Hence, the required radius of the image of the sun is 0.93 mm.
#### Question 60:
Given,
Diameter of the sun = 1.4 × 109 m
Distance between sun and earth is taken as object distance (u) = − 150 × 1011 cm,
Focal length (f) of the lens = 20 cm
Using lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
The object distance is very large as compared to focal length of the lens.
Hence, the image is formed at the focus.
$=\frac{750×{10}^{9}-1}{150×{10}^{11}}$
$\simeq \frac{750×{10}^{9}}{150×{10}^{11}}$
We know, Magnification (m) is given by:
$\left(m\right)=\frac{v}{u}=\frac{{h}_{2}}{{h}_{1}}$
Hence, the required radius of the image of the sun is 0.93 mm.
Given,
Power of the lens (P) = = 5.0 D
The height of the image is four times the height of the object.
i.e.
We know magnification (m) is also given by
The lens maker formula is given by
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Here, v is the image distance and u is the object distance.
Now,
Hence, the required object distance is 15 cm.
#### Question 61:
Given,
Power of the lens (P) = = 5.0 D
The height of the image is four times the height of the object.
i.e.
We know magnification (m) is also given by
The lens maker formula is given by
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Here, v is the image distance and u is the object distance.
Now,
Hence, the required object distance is 15 cm.
Let the object be placed at a distance x cm from the lens (away from the mirror).
For the concave lens (Ist refraction) u = − x, f = − 20 cm
From lens formula:
Thus, the virtual image due to the first refraction lies on the same side as that of object (A'B').
This image becomes the object for the concave mirror,
For the mirror,
From mirror equation,
Thus, this image is formed towards left of the mirror.
Again for second refraction in concave lens,
$u=-\left[\frac{5-50\left(x+4\right)}{3x-20}\right]$
(assuming that image of mirror is formed between the lens and mirror 3x − 20),
v = + x (since the final image is produced on the object A"B")
using lens formula,
⇒ 25x2 − 1400x − 6000 = 0
⇒ x2 − 56x − 240 = 0
⇒ (x − 60) (x + 4) = 0
So, x = 60 m
The object should be placed at a distance 60 cm from the lens farther away from the mirror, so that the final image is formed on itself.
#### Question 62:
Let the object be placed at a distance x cm from the lens (away from the mirror).
For the concave lens (Ist refraction) u = − x, f = − 20 cm
From lens formula:
Thus, the virtual image due to the first refraction lies on the same side as that of object (A'B').
This image becomes the object for the concave mirror,
For the mirror,
From mirror equation,
Thus, this image is formed towards left of the mirror.
Again for second refraction in concave lens,
$u=-\left[\frac{5-50\left(x+4\right)}{3x-20}\right]$
(assuming that image of mirror is formed between the lens and mirror 3x − 20),
v = + x (since the final image is produced on the object A"B")
using lens formula,
⇒ 25x2 − 1400x − 6000 = 0
⇒ x2 − 56x − 240 = 0
⇒ (x − 60) (x + 4) = 0
So, x = 60 m
The object should be placed at a distance 60 cm from the lens farther away from the mirror, so that the final image is formed on itself.
Let the object be placed at a distance cm from the lens (away from the mirror).
For the convex lens (1st refraction) u = − x, f = − 12 cm
From the lens formula:
Thus, the virtual image due to the first refraction lies on the same side as that of object A'B'.
This image becomes the object for the convex mirror,
For the mirror,
From mirror equation,
Thus, this image is formed towards the left of the mirror.
Again for second refraction in concave lens,
$u=-\left[\frac{5-50\left(x+4\right)}{3x-20}\right]$
(assuming that the image of mirror formed between the lens and mirror is 3x − 20),
v = + x (since, the final image is produced on the object A"B")
Using lens formula:
⇒ 25x2 − 1400x − 6000 = 0
⇒ x2 − 56x − 240 = 0
⇒ (x − 60) (x + 4) = 0
Thus, x = 60 m
The object should be placed at a distance 60 cm from the lens farther away from the mirror, so that the final image is formed on itself.
#### Question 63:
Let the object be placed at a distance cm from the lens (away from the mirror).
For the convex lens (1st refraction) u = − x, f = − 12 cm
From the lens formula:
Thus, the virtual image due to the first refraction lies on the same side as that of object A'B'.
This image becomes the object for the convex mirror,
For the mirror,
From mirror equation,
Thus, this image is formed towards the left of the mirror.
Again for second refraction in concave lens,
$u=-\left[\frac{5-50\left(x+4\right)}{3x-20}\right]$
(assuming that the image of mirror formed between the lens and mirror is 3x − 20),
v = + x (since, the final image is produced on the object A"B")
Using lens formula:
⇒ 25x2 − 1400x − 6000 = 0
⇒ x2 − 56x − 240 = 0
⇒ (x − 60) (x + 4) = 0
Thus, x = 60 m
The object should be placed at a distance 60 cm from the lens farther away from the mirror, so that the final image is formed on itself.
Given,
Distance between convex lens and convex mirror is 15 cm.
Focal length (fl) of the lens is 25 cm.
Focal length (f2) of the mirror is 40 cm.
Let x cm be the object distance from the mirror.
Therefore,
u
= − x cm
v = 25 − 15 = + 10 cm (∵ focal length of lens = 25 cm)
fl = + 40 cm
Using lens formula:
Thus, the object distance is $\left(15-\frac{40}{3}\right)=\frac{5}{3}$
= 1.67 cm from the lens
#### Question 64:
Given,
Distance between convex lens and convex mirror is 15 cm.
Focal length (fl) of the lens is 25 cm.
Focal length (f2) of the mirror is 40 cm.
Let x cm be the object distance from the mirror.
Therefore,
u
= − x cm
v = 25 − 15 = + 10 cm (∵ focal length of lens = 25 cm)
fl = + 40 cm
Using lens formula:
Thus, the object distance is $\left(15-\frac{40}{3}\right)=\frac{5}{3}$
= 1.67 cm from the lens
Given:
Convex lens of focal length (fl) = 15 cm
Concave mirror of focal length (f2) = 10 cm
Distance between the lens and the mirror = 50 cm
Point source is placed at a distance of 40 cm from the lens.
It means the point source is at the focus of the mirror.
Thus, two images will be formed:
(a) One due to direct transmission of light through the lens.
(b) One due to reflection and then transmission of the rays through the lens.
Case 1:
(S') For the image by direct transmission, we have:
Object distance (u) = − 40 cm
fl = 15 cm
Using the lens formula, we get:
Therefore, v is 24 cm to the left from the lens.
Case II:
(S') Since the object is placed at the focus of the mirror, the rays become parallel to the lens after reflection.
∴ Object distance (u) = ∞ ⇒ fl = 15 cm
Thus, v is 15 cm to the left of the lens.
#### Question 65:
Given:
Convex lens of focal length (fl) = 15 cm
Concave mirror of focal length (f2) = 10 cm
Distance between the lens and the mirror = 50 cm
Point source is placed at a distance of 40 cm from the lens.
It means the point source is at the focus of the mirror.
Thus, two images will be formed:
(a) One due to direct transmission of light through the lens.
(b) One due to reflection and then transmission of the rays through the lens.
Case 1:
(S') For the image by direct transmission, we have:
Object distance (u) = − 40 cm
fl = 15 cm
Using the lens formula, we get:
Therefore, v is 24 cm to the left from the lens.
Case II:
(S') Since the object is placed at the focus of the mirror, the rays become parallel to the lens after reflection.
∴ Object distance (u) = ∞ ⇒ fl = 15 cm
Thus, v is 15 cm to the left of the lens.
Given,
Convex lens of focal length (fl) = 15 cm
Concave mirror of focal length (fm) = 10 cm
Distance between lens and mirror = 50 cm
Thus, two images will be formed,
(a) One due to direct transmission of light through lens.
(b) One due to reflection and then transmission of the rays through lens.
Let the point source be placed at a distance of 'x' from the lens as shown in the Figure, so that images formed by lens and mirror coincide.
For lens,
We use lens formula:
For mirror,
Object distance will be (50 − x)
We use the formula:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{15}$
u = − (50 − x)
fm = − 10 cm
Since the distance between lens and mirror is 50 cm,
vvm = 50
from equation (i) and (ii):
⇒(3x2 − 120x) − (100x − 2x2 − 1500 + 30x)
= 10 (x2 − 55x + 600)
⇒ 5x2− 250x − 1500 = 10x2 − 550x + 6000
⇒ 5x2 − 300x + 4500 = 0
⇒ x2 − 60x + 900 = 0
⇒ (x − 30)2 = 0
x = 30 cm.
∴ Point source should be placed at a distance of 30 cm from the lens on the principal axis, so that the two images form at the same place.
#### Question 66:
Given,
Convex lens of focal length (fl) = 15 cm
Concave mirror of focal length (fm) = 10 cm
Distance between lens and mirror = 50 cm
Thus, two images will be formed,
(a) One due to direct transmission of light through lens.
(b) One due to reflection and then transmission of the rays through lens.
Let the point source be placed at a distance of 'x' from the lens as shown in the Figure, so that images formed by lens and mirror coincide.
For lens,
We use lens formula:
For mirror,
Object distance will be (50 − x)
We use the formula:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{15}$
u = − (50 − x)
fm = − 10 cm
Since the distance between lens and mirror is 50 cm,
vvm = 50
from equation (i) and (ii):
⇒(3x2 − 120x) − (100x − 2x2 − 1500 + 30x)
= 10 (x2 − 55x + 600)
⇒ 5x2− 250x − 1500 = 10x2 − 550x + 6000
⇒ 5x2 − 300x + 4500 = 0
⇒ x2 − 60x + 900 = 0
⇒ (x − 30)2 = 0
x = 30 cm.
∴ Point source should be placed at a distance of 30 cm from the lens on the principal axis, so that the two images form at the same place.
Given,
Convex lens of focal length fl = 15 cm
Concave mirror of focal length fm = 10 cm
Distance between mirror and lens = 50 cm
Length of the pin (object length) h0 = 2.0 cm
As per the question
The pin (object) is placed at a distance of 30 cm from the lens on the principle axis.
Using lens formula,
Since, u = −30 cm and fl = 15 cm
From the figure it can be seen that image of the object (AB) is real and inverted (A'B') and it is of the same size as the object. This image (A'B') is at a distance of 20 cm from the concave mirror, which is formed at the centre of curvature of the mirror. Thus, mirror will form the image (A'B') at the same place as (A''B'') and will be of the same size. Now, due to the refraction from the lens, the final image (A''B'') will be formed at AB and will be of the same size as the object (AB).
#### Question 67:
Given,
Convex lens of focal length fl = 15 cm
Concave mirror of focal length fm = 10 cm
Distance between mirror and lens = 50 cm
Length of the pin (object length) h0 = 2.0 cm
As per the question
The pin (object) is placed at a distance of 30 cm from the lens on the principle axis.
Using lens formula,
Since, u = −30 cm and fl = 15 cm
From the figure it can be seen that image of the object (AB) is real and inverted (A'B') and it is of the same size as the object. This image (A'B') is at a distance of 20 cm from the concave mirror, which is formed at the centre of curvature of the mirror. Thus, mirror will form the image (A'B') at the same place as (A''B'') and will be of the same size. Now, due to the refraction from the lens, the final image (A''B'') will be formed at AB and will be of the same size as the object (AB).
Given,
Convex lens of focal length (f) = 15 cm
Object distance, (u) = −30 cm
Using lens formula,
Thus ,v = 30 cm
Therefore, the image of the object will be formed at a distance of 30 cm to the right side of the lens.
[since, μg = 1.5 and thickness (t) = 1 cm]
Hence, the image of the object will be formed at 30 + 0.33 = 30.33 cm from the lens on the right side, due to the glass having thickness 1 cm.
#### Question 68:
Given,
Convex lens of focal length (f) = 15 cm
Object distance, (u) = −30 cm
Using lens formula,
Thus ,v = 30 cm
Therefore, the image of the object will be formed at a distance of 30 cm to the right side of the lens.
[since, μg = 1.5 and thickness (t) = 1 cm]
Hence, the image of the object will be formed at 30 + 0.33 = 30.33 cm from the lens on the right side, due to the glass having thickness 1 cm.
Given,
Focal length of the convex lens, fd = 20 cm
Focal length of the concave lens, fc = 10 cm
Beam diameter of the incident light, d = 5.0 mm
Distance between both the lenses is 10 cm.
As per the question, the incident beam of light is parallel to the principal axis.
Let it be incident on the convex lens.
Now, let B be the focus of the convex lens where the image by the convex lens should be formed.
For the concave lens,
Object distance (u) = + 10 cm (Virtual object is on the right of concave lens.)
Focal length, fc = − 10 cm
Using the lens formula,
Thus, after refraction in the concave lens, the emergent beam becomes parallel.
As shown, in triangles XYB and PQB,
Thus, the beam diameter of the emergent light is 2.5 mm.
Similarly, we can prove that if the beam of light is incident on the side of the concave lens, the beam diameter (d) of the emergent light will be 1 cm.
#### Question 69:
Given,
Focal length of the convex lens, fd = 20 cm
Focal length of the concave lens, fc = 10 cm
Beam diameter of the incident light, d = 5.0 mm
Distance between both the lenses is 10 cm.
As per the question, the incident beam of light is parallel to the principal axis.
Let it be incident on the convex lens.
Now, let B be the focus of the convex lens where the image by the convex lens should be formed.
For the concave lens,
Object distance (u) = + 10 cm (Virtual object is on the right of concave lens.)
Focal length, fc = − 10 cm
Using the lens formula,
Thus, after refraction in the concave lens, the emergent beam becomes parallel.
As shown, in triangles XYB and PQB,
Thus, the beam diameter of the emergent light is 2.5 mm.
Similarly, we can prove that if the beam of light is incident on the side of the concave lens, the beam diameter (d) of the emergent light will be 1 cm.
Given,
Focal length of convex lens, fc = 30 cm
Focal length of concave lens, fd = 15 cm
Distance between both the lenses, d = 15 cm
Let (f) be the equivalent focal length of both the lenses.
As focal length is positive, so it will be a converging lens.
Let 'd1' be the distance from diverging lens, so that the emergent beam is parallel to the principal axis and the image will be formed at infinity.
It should be placed 60 cm left to the diverging lens. The object should be placed
(120 − 60) = 60 cm from the diverging lens
Let d2 be the distance from the converging lens. Then,
d2 = 90 cm
Thus, it should be placed (120 + 90) cm = 210 cm right to converging lens.
#### Question 70:
Given,
Focal length of convex lens, fc = 30 cm
Focal length of concave lens, fd = 15 cm
Distance between both the lenses, d = 15 cm
Let (f) be the equivalent focal length of both the lenses.
As focal length is positive, so it will be a converging lens.
Let 'd1' be the distance from diverging lens, so that the emergent beam is parallel to the principal axis and the image will be formed at infinity.
It should be placed 60 cm left to the diverging lens. The object should be placed
(120 − 60) = 60 cm from the diverging lens
Let d2 be the distance from the converging lens. Then,
d2 = 90 cm
Thus, it should be placed (120 + 90) cm = 210 cm right to converging lens.
Given:
Length of the high pin = 5.00 mm
Focal length of the first convex lens, f = 10 cm
Distance between the first lens and the pin = 15 cm
Focal length of the second convex lens, f1 = 5 cm
Distance between the first lens and the second lens = 40 cm
Distance between the second lens and the pin = 55 cm
(a) Image formed by the first lens:
Here,
Object distance, u = − 15 cm
Focal length, f = 10 cm
The lens formula is given by
Now,
This will be object for the second lens.
∴ Object distance for the second lens, ${u}_{1}$ = − (40 − 30)
$⇒{u}_{1}$ = − 10 cm
Focal length of the second lens, f1= 5 cm
The lens formula is given by
Therefore, the final position of the image is 10 cm right from the second lens.
(b) Magnification $\left(m\right)$ by the first lens is given by
Magnification by the second lens:
Thus, the image will be erect and real.
(c) Size of the final image is 10 mm.
#### Question 71:
Given:
Length of the high pin = 5.00 mm
Focal length of the first convex lens, f = 10 cm
Distance between the first lens and the pin = 15 cm
Focal length of the second convex lens, f1 = 5 cm
Distance between the first lens and the second lens = 40 cm
Distance between the second lens and the pin = 55 cm
(a) Image formed by the first lens:
Here,
Object distance, u = − 15 cm
Focal length, f = 10 cm
The lens formula is given by
Now,
This will be object for the second lens.
∴ Object distance for the second lens, ${u}_{1}$ = − (40 − 30)
$⇒{u}_{1}$ = − 10 cm
Focal length of the second lens, f1= 5 cm
The lens formula is given by
Therefore, the final position of the image is 10 cm right from the second lens.
(b) Magnification $\left(m\right)$ by the first lens is given by
Magnification by the second lens:
Thus, the image will be erect and real.
(c) Size of the final image is 10 mm.
Given,
Distance between point object and convex lens, u = 15 cm
Distance between the image of the point object and convex lens, v = 30 cm
Let fc be the focal length of the convex lens.
Then, using lens formula, we have:
Now, as per the question, the concave lens is placed in contact with the convex lens. So the image is shifted by a distance of 30 cm.
Again, let vf be the final image distance from concave lens, then:
${v}_{\mathrm{f}}$ = + (30 + 30) = + 60 cm
Object distance from the concave lens, = 30 cm
If fd is the focal length of concave lens then
Using lens formula, we have:
Hence, the focal length (fc) of convex lens is 10 cm and that of the concave lens (fd ) is 60 cm.
#### Question 72:
Given,
Distance between point object and convex lens, u = 15 cm
Distance between the image of the point object and convex lens, v = 30 cm
Let fc be the focal length of the convex lens.
Then, using lens formula, we have:
Now, as per the question, the concave lens is placed in contact with the convex lens. So the image is shifted by a distance of 30 cm.
Again, let vf be the final image distance from concave lens, then:
${v}_{\mathrm{f}}$ = + (30 + 30) = + 60 cm
Object distance from the concave lens, = 30 cm
If fd is the focal length of concave lens then
Using lens formula, we have:
Hence, the focal length (fc) of convex lens is 10 cm and that of the concave lens (fd ) is 60 cm.
Given,
Focal length of each convex lens, f = 10 cm
Distance between both the lens, d = 15 cm
(a) As shown in the figure, the light rays are falling parallel to the principal axis of the first lens, therefore the rays will converge within the focal length of the second lens. Hence, all the rays emerging from the lens system are diverging.
(b) Using lens formula for first convex lens,
For the second convex lens,
Object distance, u = −(15 − 10) = −5 cm
If v1 is the image distance for the second convex lens, then,
Applying lens formula we have:
$\frac{1}{{v}_{1}}=\frac{1}{f}+\frac{1}{u}$
Thus, the virtual image of the object will be at 5 cm from the first convex lens.
(c) The equivalent focal length of the lens system $\left(F\right)$ is given by,
Therefore the equivalent focal length (F) is 20 cm
#### Question 73:
Given,
Focal length of each convex lens, f = 10 cm
Distance between both the lens, d = 15 cm
(a) As shown in the figure, the light rays are falling parallel to the principal axis of the first lens, therefore the rays will converge within the focal length of the second lens. Hence, all the rays emerging from the lens system are diverging.
(b) Using lens formula for first convex lens,
For the second convex lens,
Object distance, u = −(15 − 10) = −5 cm
If v1 is the image distance for the second convex lens, then,
Applying lens formula we have:
$\frac{1}{{v}_{1}}=\frac{1}{f}+\frac{1}{u}$
Thus, the virtual image of the object will be at 5 cm from the first convex lens.
(c) The equivalent focal length of the lens system $\left(F\right)$ is given by,
Therefore the equivalent focal length (F) is 20 cm
Given,
A transparent sphere with refractive index μ and a of radius R.
A ball is kept at height h above the sphere.
As per the question, at t = 0 the ball is dropped normally on the sphere.
Let take the time taken to travel from point A to B as t .
Thus, the distance travelled by the ball is given by:
=$\frac{1}{2}g{t}^{2}$
Where g is acceleration due to gravity.
Therefore, the distance BC is given by:
$=h-\frac{1}{2}g{t}^{2}$
We are assuming this distance of the object from lens at any time t.
So here,
u = $-\left(h-\frac{1}{2}g{t}^{2}\right)$
Taking:
Refractive index of air, μ1 = 1
Refractive index of sphere, μ2 = μ (given)
Thus,
$\frac{\mathrm{\mu }}{v}-\frac{1}{-\left(h-\frac{1}{2}g{t}^{2}\right)}=\frac{\mathrm{\mu }-1}{R}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{\mu }}{v}=\frac{\mathrm{\mu }-1}{R}-\frac{1}{\left(h-\frac{1}{2}g{t}^{2}\right)}=\frac{\left(\mathrm{\mu }-1\right)\left(h-\frac{1}{2}g{t}^{2}\right)-R}{R\left(h-\frac{1}{2}g{t}^{2}\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Let v be the image distance at any time t. Then,
$v=\frac{\mathrm{\mu }R\left(h-\frac{1}{2}g{t}^{2}\right)}{\left(\mathrm{\mu }-1\right)\left(h-\frac{1}{2}g{t}^{2}\right)-R}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Therefore, velocity of the image $\left(V\right)$ is given by,
#### Question 74:
Given,
A transparent sphere with refractive index μ and a of radius R.
A ball is kept at height h above the sphere.
As per the question, at t = 0 the ball is dropped normally on the sphere.
Let take the time taken to travel from point A to B as t .
Thus, the distance travelled by the ball is given by:
=$\frac{1}{2}g{t}^{2}$
Where g is acceleration due to gravity.
Therefore, the distance BC is given by:
$=h-\frac{1}{2}g{t}^{2}$
We are assuming this distance of the object from lens at any time t.
So here,
u = $-\left(h-\frac{1}{2}g{t}^{2}\right)$
Taking:
Refractive index of air, μ1 = 1
Refractive index of sphere, μ2 = μ (given)
Thus,
$\frac{\mathrm{\mu }}{v}-\frac{1}{-\left(h-\frac{1}{2}g{t}^{2}\right)}=\frac{\mathrm{\mu }-1}{R}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{\mu }}{v}=\frac{\mathrm{\mu }-1}{R}-\frac{1}{\left(h-\frac{1}{2}g{t}^{2}\right)}=\frac{\left(\mathrm{\mu }-1\right)\left(h-\frac{1}{2}g{t}^{2}\right)-R}{R\left(h-\frac{1}{2}g{t}^{2}\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Let v be the image distance at any time t. Then,
$v=\frac{\mathrm{\mu }R\left(h-\frac{1}{2}g{t}^{2}\right)}{\left(\mathrm{\mu }-1\right)\left(h-\frac{1}{2}g{t}^{2}\right)-R}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Therefore, velocity of the image $\left(V\right)$ is given by,
Given,
Radius of the concave mirror is R
Therefore focal length of the mirror, $f=\frac{R}{2}$
Velocity of the particle, $V=\frac{\mathrm{d}x}{\mathrm{d}t}$
Object distance, u = −x
Using mirror equation,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}$
On putting the respective values we get,
$\frac{1}{v}+\frac{1}{-x}=-\frac{2}{R}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=-\frac{2}{R}+\frac{1}{x}=\frac{R-2x}{Rx}\phantom{\rule{0ex}{0ex}}\therefore v=\frac{Rx}{R-2x}\phantom{\rule{0ex}{0ex}}$
Velocity of the image is given by V1
${V}_{1}=\frac{\mathrm{dv}}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{Rx}{R-2x}\right]\phantom{\rule{0ex}{0ex}}=\frac{\left[\frac{\mathrm{d}}{\mathrm{d}t}\left(Rx\right)\left(R-2x\right)\right]-\left[\frac{\mathrm{d}}{\mathrm{d}t}\left(R-2x\right)\left(Rx\right)\right]}{{\left(R-2x\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{R\left[\left(\frac{\mathrm{dx}}{\mathrm{d}t}\right)\left(R-2x\right)\right]-\left[2\frac{\mathrm{dx}}{\mathrm{d}t}x\right]}{{\left(R-2x\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{R\left[\left(V\right)\left(R-2x\right)\right]-\left[2V×0\right]}{{\left(R-2x\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{V{R}^{2}}{\left(2x-R{\right)}^{2}}$
#### Question 75:
Given,
Radius of the concave mirror is R
Therefore focal length of the mirror, $f=\frac{R}{2}$
Velocity of the particle, $V=\frac{\mathrm{d}x}{\mathrm{d}t}$
Object distance, u = −x
Using mirror equation,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}$
On putting the respective values we get,
$\frac{1}{v}+\frac{1}{-x}=-\frac{2}{R}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=-\frac{2}{R}+\frac{1}{x}=\frac{R-2x}{Rx}\phantom{\rule{0ex}{0ex}}\therefore v=\frac{Rx}{R-2x}\phantom{\rule{0ex}{0ex}}$
Velocity of the image is given by V1
${V}_{1}=\frac{\mathrm{dv}}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{Rx}{R-2x}\right]\phantom{\rule{0ex}{0ex}}=\frac{\left[\frac{\mathrm{d}}{\mathrm{d}t}\left(Rx\right)\left(R-2x\right)\right]-\left[\frac{\mathrm{d}}{\mathrm{d}t}\left(R-2x\right)\left(Rx\right)\right]}{{\left(R-2x\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{R\left[\left(\frac{\mathrm{dx}}{\mathrm{d}t}\right)\left(R-2x\right)\right]-\left[2\frac{\mathrm{dx}}{\mathrm{d}t}x\right]}{{\left(R-2x\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{R\left[\left(V\right)\left(R-2x\right)\right]-\left[2V×0\right]}{{\left(R-2x\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{V{R}^{2}}{\left(2x-R{\right)}^{2}}$
Note :
(a) At time t = t,
Object distance, u = −(d − Vt)
Here d > Vt, $f=-\frac{R}{2}$
Mirror formula is given by:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$=\frac{-2\left(d-\mathrm{V}t\right)+\mathrm{R}}{\mathrm{R}\left(d-\mathrm{V}t\right)}$
$v=\frac{-\mathrm{R}\left(d-\mathrm{V}t\right)}{\mathrm{R}-2\left(d-\mathrm{V}t\right)}$
Differentiating w.r.t. 't':
$\frac{dv}{dt}=\frac{-\mathrm{RV}\left[\mathrm{R}-2\left(d-\mathrm{V}t\right)\right]-2\mathrm{V}\left[\mathrm{R}\left(d-\mathrm{V}t\right)\right]}{{\left[\mathrm{R}-2\left(d-\mathrm{V}t\right)\right]}^{2}}\phantom{\rule{0ex}{0ex}}=-\frac{{\mathrm{R}}^{2}\mathrm{V}}{{\left[2\left(d-\mathrm{V}t\right)-R\right]}^{2}}$
This is the required speed of mirror.
(b) When $t>\frac{d}{2}$ the collision between the mirror and mass will take place. As the collision is elastic, the object will come to rest and the mirror will start to move with the velocity V.
u = (d − Vt).
At any time, $t>\frac{d}{2}$
The distance of mirror from the mass will be:
$x=V\left(t-\frac{d}{V}\right)=Vt-d$
Here,
Object distance, u = $-\left(Vt-d\right)$ = $\left(d-Vt\right)$
Focal length, f = $-\frac{R}{2}$
Now, mirror formula is given by:
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
$⇒\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ = $\frac{1}{-\frac{R}{2}}-\frac{1}{d-Vt}$
$v⇒-\left[\frac{R\left(d-Vt\right)}{R-2\left(d-Vt\right)}\right]$
Velocity of image $\left({V}_{image}\right)$ is given by,
${V}_{image}=\frac{\mathrm{d}v}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left[\frac{R\left(d-Vt\right)}{R+2\left(d-Vt\right)}\right]$
If y = d$-$Vt
$\frac{\mathrm{d}y}{\mathrm{dt}}$ = $-V$
Velocity of image:
As the mirror is moving with velocity V,
Therefore,
V=Vimage + Vmirror
Absolute velocity of image = $V\left[1-\frac{{R}^{2}}{{\left(R+2y\right)}^{2}}\right]$
= $V\left[1-\frac{{R}^{2}}{2\left(Vt-d\right)-{R}^{2}}\right]$
#### Question 76:
Note :
(a) At time t = t,
Object distance, u = −(d − Vt)
Here d > Vt, $f=-\frac{R}{2}$
Mirror formula is given by:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$=\frac{-2\left(d-\mathrm{V}t\right)+\mathrm{R}}{\mathrm{R}\left(d-\mathrm{V}t\right)}$
$v=\frac{-\mathrm{R}\left(d-\mathrm{V}t\right)}{\mathrm{R}-2\left(d-\mathrm{V}t\right)}$
Differentiating w.r.t. 't':
$\frac{dv}{dt}=\frac{-\mathrm{RV}\left[\mathrm{R}-2\left(d-\mathrm{V}t\right)\right]-2\mathrm{V}\left[\mathrm{R}\left(d-\mathrm{V}t\right)\right]}{{\left[\mathrm{R}-2\left(d-\mathrm{V}t\right)\right]}^{2}}\phantom{\rule{0ex}{0ex}}=-\frac{{\mathrm{R}}^{2}\mathrm{V}}{{\left[2\left(d-\mathrm{V}t\right)-R\right]}^{2}}$
This is the required speed of mirror.
(b) When $t>\frac{d}{2}$ the collision between the mirror and mass will take place. As the collision is elastic, the object will come to rest and the mirror will start to move with the velocity V.
u = (d − Vt).
At any time, $t>\frac{d}{2}$
The distance of mirror from the mass will be:
$x=V\left(t-\frac{d}{V}\right)=Vt-d$
Here,
Object distance, u = $-\left(Vt-d\right)$ = $\left(d-Vt\right)$
Focal length, f = $-\frac{R}{2}$
Now, mirror formula is given by:
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
$⇒\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ = $\frac{1}{-\frac{R}{2}}-\frac{1}{d-Vt}$
$v⇒-\left[\frac{R\left(d-Vt\right)}{R-2\left(d-Vt\right)}\right]$
Velocity of image $\left({V}_{image}\right)$ is given by,
${V}_{image}=\frac{\mathrm{d}v}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left[\frac{R\left(d-Vt\right)}{R+2\left(d-Vt\right)}\right]$
If y = d$-$Vt
$\frac{\mathrm{d}y}{\mathrm{dt}}$ = $-V$
Velocity of image:
As the mirror is moving with velocity V,
Therefore,
V=Vimage + Vmirror
Absolute velocity of image = $V\left[1-\frac{{R}^{2}}{{\left(R+2y\right)}^{2}}\right]$
= $V\left[1-\frac{{R}^{2}}{2\left(Vt-d\right)-{R}^{2}}\right]$
Given,
The focal length of the concave mirror is f and M is the mass of the gun. Horizontal speed of the bullet is V.
Let the recoil speed of the gun be Vg
Using the conservation of linear momentum we can write,
MVg = mV
$⇒{V}_{\mathrm{g}}=\frac{m}{M}V$
Considering the position of gun and bullet at time t = t,
For the mirror, object distance, u = − (Vt + Vgt)
Focal length, f = − f
Image distance, v = ?
Using Mirror formula, we have:
The separation between image of the bullet and bullet at time t is given by:
Differentiating the above equation with respect to 't' we get,
$\frac{d\left(v-u\right)}{dt}=2\left(1+\frac{m}{M}\right)V$
Therefore, the speed of separation of the bullet and image just after the gun was fired is $2\left(1+\frac{m}{M}\right)V$.
#### Question 77:
Given,
The focal length of the concave mirror is f and M is the mass of the gun. Horizontal speed of the bullet is V.
Let the recoil speed of the gun be Vg
Using the conservation of linear momentum we can write,
MVg = mV
$⇒{V}_{\mathrm{g}}=\frac{m}{M}V$
Considering the position of gun and bullet at time t = t,
For the mirror, object distance, u = − (Vt + Vgt)
Focal length, f = − f
Image distance, v = ?
Using Mirror formula, we have:
The separation between image of the bullet and bullet at time t is given by:
Differentiating the above equation with respect to 't' we get,
$\frac{d\left(v-u\right)}{dt}=2\left(1+\frac{m}{M}\right)V$
Therefore, the speed of separation of the bullet and image just after the gun was fired is $2\left(1+\frac{m}{M}\right)V$.
Given,
Mass = 50 g
Spring constant, k = 500 Nm−1
Height from where the mass is dropped on the spring, h = 10 cm
Focal length of concave mirror, f = 12 cm
Distance between the pole and the free end of the spring is 30 cm.
As per the question, when the mass is released it will stick to the spring and execute SHM.
At equilibrium position, weight of the mass is equal to force applied by the spring.
mg = kx
where g is acceleration due to gravity
x = mg/k
Therefore, the mean position of the SHM is (30 + 0.1 = 30.1) cm away from the pole of the mirror.
From the work energy principle,
final kinetic energy − initial kinetic energy = work done
⇒ 0 − 0 = mg(h + δ) − $\frac{1}{2}k{\delta }^{2}$
Where δ is the maximum compression of the spring.
mg(h + δ) = $\frac{1}{2}k{\delta }^{2}$
From the above figure,
Position of point B, (30 + 1.5) = 31.5 cm from the pole of the mirror
Therefore, amplitude of vibration of SHM, (31.5 − 30.1) = 1.4 cm
Position of point A from the pole of the mirror, (30.1 − 30.1) = 28.7 cm
For point A,
Object distance (ua) = − 31.5
f = − 12 cm
By using the lens formula:
For point B,
Object distance, (ub) = − 28.7
f = − 12 cm
By using the lens formula:
Hence, the image of the mass oscillates in length (20.62 − 19.38) = 1.24 cm
#### Question 78:
Given,
Mass = 50 g
Spring constant, k = 500 Nm−1
Height from where the mass is dropped on the spring, h = 10 cm
Focal length of concave mirror, f = 12 cm
Distance between the pole and the free end of the spring is 30 cm.
As per the question, when the mass is released it will stick to the spring and execute SHM.
At equilibrium position, weight of the mass is equal to force applied by the spring.
mg = kx
where g is acceleration due to gravity
x = mg/k
Therefore, the mean position of the SHM is (30 + 0.1 = 30.1) cm away from the pole of the mirror.
From the work energy principle,
final kinetic energy − initial kinetic energy = work done
⇒ 0 − 0 = mg(h + δ) − $\frac{1}{2}k{\delta }^{2}$
Where δ is the maximum compression of the spring.
mg(h + δ) = $\frac{1}{2}k{\delta }^{2}$
From the above figure,
Position of point B, (30 + 1.5) = 31.5 cm from the pole of the mirror
Therefore, amplitude of vibration of SHM, (31.5 − 30.1) = 1.4 cm
Position of point A from the pole of the mirror, (30.1 − 30.1) = 28.7 cm
For point A,
Object distance (ua) = − 31.5
f = − 12 cm
By using the lens formula:
For point B,
Object distance, (ub) = − 28.7
f = − 12 cm
By using the lens formula:
Hence, the image of the mass oscillates in length (20.62 − 19.38) = 1.24 cm
Given,
R is the radii of curvature of two concave mirrors and M is the mass of the whole system.
Mass of the two blocks A and B is m​.
As per the question,
At t = 0,
distance between block A and B is 2R
Block B is moving at a speed v towards the mirror.
Original position of the whole system at x = 0
(a)
At time t$\frac{R}{v}$
The block B moved $\left(v×\frac{R}{v}=R\right)$R distance towards the mirror.
For block A,
object distance, u = − 2R
focal length of the mirror, f = − $\frac{R}{2}$
Using the mirror formula:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}=-\frac{2}{R}+\frac{1}{2R}\phantom{\rule{0ex}{0ex}}=-\frac{3}{2R}$
Therefore, v = − $\frac{2R}{3}$
Position of the image of block A is at $\frac{2R}{3}$ with respect to the given coordinate system.
For block B,
Object distance, u = − R
Focal length of the mirror, f = − $\frac{R}{2}$
Using the mirror formula:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=-\frac{2}{R}+\frac{1}{R}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=-\frac{1}{R}$
Therefore, v = − R
Position of the image of block B is at the same place.
Similarly,
(b)
At time $\frac{3R}{v}$
Block B, after colliding with the mirror must have come to rest because the collision is elastic. Due to this, the mirror has travelled a distance R towards the block A, i.e., towards left from its initial position.
So, at this time
For block A
Object distance, u = − R
Focal length of the mirror, f = − $\frac{R}{2}$
Using the mirror formula:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}=-\frac{2}{R}+\frac{1}{R}\phantom{\rule{0ex}{0ex}}=-\frac{1}{R}$
Therefore, v = − R
Position of the image of block A is at − 2R with respect to the given coordinate system.
For Block B,
Image of the block B is at the same place as it is at a distance of R from the mirror.
Therefore, the image of the block B is zero with respect to the given coordinate system.
(c)
At time $\frac{5R}{v}$
In a similar manner, we can prove that the position of the image of block A and B will be at − 3R and $\frac{-4R}{3}$ respectively.
#### Question 79:
Given,
R is the radii of curvature of two concave mirrors and M is the mass of the whole system.
Mass of the two blocks A and B is m​.
As per the question,
At t = 0,
distance between block A and B is 2R
Block B is moving at a speed v towards the mirror.
Original position of the whole system at x = 0
(a)
At time t$\frac{R}{v}$
The block B moved $\left(v×\frac{R}{v}=R\right)$R distance towards the mirror.
For block A,
object distance, u = − 2R
focal length of the mirror, f = − $\frac{R}{2}$
Using the mirror formula:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}=-\frac{2}{R}+\frac{1}{2R}\phantom{\rule{0ex}{0ex}}=-\frac{3}{2R}$
Therefore, v = − $\frac{2R}{3}$
Position of the image of block A is at $\frac{2R}{3}$ with respect to the given coordinate system.
For block B,
Object distance, u = − R
Focal length of the mirror, f = − $\frac{R}{2}$
Using the mirror formula:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=-\frac{2}{R}+\frac{1}{R}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=-\frac{1}{R}$
Therefore, v = − R
Position of the image of block B is at the same place.
Similarly,
(b)
At time $\frac{3R}{v}$
Block B, after colliding with the mirror must have come to rest because the collision is elastic. Due to this, the mirror has travelled a distance R towards the block A, i.e., towards left from its initial position.
So, at this time
For block A
Object distance, u = − R
Focal length of the mirror, f = − $\frac{R}{2}$
Using the mirror formula:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\phantom{\rule{0ex}{0ex}}=-\frac{2}{R}+\frac{1}{R}\phantom{\rule{0ex}{0ex}}=-\frac{1}{R}$
Therefore, v = − R
Position of the image of block A is at − 2R with respect to the given coordinate system.
For Block B,
Image of the block B is at the same place as it is at a distance of R from the mirror.
Therefore, the image of the block B is zero with respect to the given coordinate system.
(c)
At time $\frac{5R}{v}$
In a similar manner, we can prove that the position of the image of block A and B will be at − 3R and $\frac{-4R}{3}$ respectively.
Given,
Acceleration of the elevator, a = 2.00 m/s2
Focal length of the mirror M, f = 12.00 cm
Acceleration due to gravity, g = 10 m/s2
Mass of blocks A and B = m​
As per the question, the mass–pulley system is released at time t = 0.
Let the acceleration of the masses A and B with respect to the elevator be a.
Using the free body diagram,
T mg + ma − 2m = 0 ...(i)
Also,
T ma = 0 ...(ii)
From (i) and (ii), we get:
2ma = m(g + 2)
Now, the distance travelled by block B of mass m in time t = 0.2 s is given by
On putting the respective values, we get:
As given in the question, the distance of block B from the mirror is 42 cm.
Object distance u from the mirror = − (42 − 12) = − 30 cm
Using the mirror equation,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}$
On putting the respective values, we get:
Hence, the distance between block B and mirror M is 8.57 cm.
View NCERT Solutions for all chapters of Class 12 | 42,433 | 131,451 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 417, "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} | 3.734375 | 4 | CC-MAIN-2024-30 | latest | en | 0.912403 |
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2. / What Is The Unit Weight Of Crusher Sand
# What Is The Unit Weight Of Crusher Sand
### The Unit Weight Of Crusher Sand
Weight Of Crushed Sand Per Cubic Meter. What is the weight of crusher stone sand for cubic meter. what does 1 cubic meter of crusher run weight - YouTube. 16 Oct 2013 what does 1. Dec 01, 2021 Choose between 3 4 and 1 1 2 shredding strokes.Answer (1 of 4) The bulk density of sand lies in the range of 1600–1700 kg m3. The specific gravity lies in the range of 2.6 to 2.9.
### What Is The Unit Weight Of Sand Quora
Engineering. Civil Engineering. Civil Engineering questions and answers. The unit weight of a sand deposit is 100 pcf. What is the vertical stress and effective vertical stress at a depth of 10 m below the ground surface if the water table is located 2 m below the ground surface Saturated Unit Weight =100pcf.
### What Is The Weight Of Crusher Stone Sand For Cubicme
The unit weight of a sand deposit is 100 pcf. What is the vertical stress and effective vertical stress at a depth of 10 m below the ground surface if the water table is located 2 m below the ground surface Question The unit weight of a sand deposit is 100 pcf. What is the vertical stress and effective vertical stress at a depth of 10 m below.
### What Is The Weight Of 1 Unit River Sand Answers
Crusher Run M Per Kg. Crusher run density kg m3 hat is the density of crusher run per cubic metre 876 kgm 3 is the densityowever, you might want it in different units, such as grams per cubic centimeter, for examplee would then get 876,000gmillion cm 3 or 6 g cm 3 , which is a bit less dense than wateret price. Density Of Crusher Run Malaysia.
### What Is The Weight Of Crusher Stone Sand For Cubic Meter
Unit Weight of sand in different unit like kg m3, kn m3, kg ft3, g cm3, cft and lb ft3, in this article we know about Unit Weight of M sand and river sand in kg m3. The first thing is unit weight is as same as Specific Weight. And the specific weight of a sand is actually the product of its density and the standard gravity.
### What Is The Weight Of Average Rock Crusher
Weight of crusher stone sand Crusher - Wikipedia , mobile crusher and mill QJ340 track-mounted mobile jaw crusher, QJ340 Weight . Typical Usage Sand Stone Mixture used for Road Beds, Pipe Bedding, Driveways, , what is the weight of.
### The Unit Weight Of A Sand Deposit Is 100 Pcf What Is
Weight of sand = Volume of sand x Density of sand. From the above formula, To calculate the weight of sand in kg per cubic foot, the volume of sand should be in feet cube and the density of sand must be in kg ft 3. The density of sand is calculated in the laboratory with the help of different apparatus given below at the last of this article.
### Solved The Unit Weight Of A Sand Deposit Is 100 Pcf What
Specific Weight or Unit weight of Sand Unit weight or Specific Weight of sand is calculated by the product of the density of sand and the standard gravity of sand. According to the US customary measurement system, dry weighs 1.631 gram per cubic centimeter, this density is equal to 101.8 pounds per cubic foot [lb ft ].
### The Weight Of Crusher Stone Sand For Cubicmeter
Dec 04, 2020 How Much Crusher Dust In Cubic Meter Of Concrete. Weight cubic yard crusher dust lch do not have a good conversion factor since they retain water weight so well they allow the operator to place stone much precisely than a dump cubic yard crusher dustwhat does a cubic yard of crusher dust weigh what is the weight of crusher stone sand for cubic meter.
### Dry Rodded Unit Weight Of Crusher Run Limestone
Using these weights we can calculate how much soil was used to fill the volume of calibrating container. We say this weight as W A, which can be written as W 1 minus W 2 minus W 3. Hence bulk unit weight of the sand can be determined by its formula and will be equal to this. Now come to field from where soil’s unit weight needs to be.
### Density Of Crusher Run Kg M Fact Jeugd Noord
Jun 07, 2012 Units of Measure . Math and Arithmetic Which weights 1 cubic yard top soil or 1 ton top soil jaw crusher, vertical shaft impact crusher (sand making machine), vibrating screen, sand.
### Unit Weight Of Sand In Different Unit Like Kg M3 Kn M3
FIELD DENSITY BY THE SAND CONE METHOD (A Modification of AASHTO Designation T 191) SCOPE 1. (a) This method is used to determine the density of compacted soils or aggregates by determining the weight and moisture content material removed from a test hole and measuring the volume of the test hole. (b) This test method may involve hazardous.
### How Much Does 1 Cubic Meter Of Crusher Dust Weigh Answers
The Unit weight or bulk density of aggregate is the mass or weight of the aggregate required to fill a container of a specified unit volume. The approximate unit weight of aggregate that is commonly used in normal concrete work is between 1200-1750 kg m 3 (75-110 lb ft 3) Bulk Density of Sand.
### Bulk Density Of Crusher Sand And Aggregate
Feb 21, 2013 unit weight of robo sand – Crusher South Africa. The Hindu Property Plus Bangalore Now, get ready for The Hyderabad-based Robo Silicon Ltd. has set up a unit on a six-hectare land at Uragahalli near detailed.
### Standard Weights For Crushed Rock Per Meter Hunker
Unit weight of stone aggregate quarry machine and crusher unit weight of stone aggregate CR 8 DGA Pug Mix 102.9 102.4 101.9 122.2 120.0 123.8 AASHTO T-19 Dry Rodded Unit Weight 57 supply crusher run crushed stone unit weight per cubic foot gmecrusher.com. | 1,330 | 5,530 | {"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} | 3.0625 | 3 | CC-MAIN-2022-27 | longest | en | 0.895778 |
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A238270 Decimal expansion of Sum_{n>=1} mu(n)/2^n. 4
%I
%S 1,0,2,0,1,1,3,3,4,8,1,7,8,1,0,3,6,4,7,4,3,0,3,6,3,9,3,9,3,1,8,2,4,3,
%T 5,1,5,4,3,6,1,0,4,9,2,5,1,0,2,9,1,0,7,3,5,8,7,3,8,8,0,3,2,5,9,0,9,3,
%U 7,2,7,6,0,5,1,9,5,2,3,3,8,4,2,8,4,8,3,6,8,4,7,8,5,5,2,0,3,8,8,9,6,6,7,3,7,2
%N Decimal expansion of Sum_{n>=1} mu(n)/2^n.
%C Suggested by a posting to SeqFan list by _Alonso del Arte_ at 15:35 on Aug 09 2012.
%e 0.102011334817810364743036393931824351543610492510291073587388032...
%t RealDigits[ Sum[ N[ MoebiusMu[n] 2^-n, 128], {n, 380}], 10, 111][[1]]
%o (PARI) sum(n=1,(default(realprecision)+2)*log(10)/log(2),moebius(n)/2.^n) \\ _Charles R Greathouse IV_, Sep 03 2015
%Y Cf. A008683, A238271, A238272, A238273.
%K nonn,cons,easy
%O 0,3
%A _Robert G. Wilson v_, Feb 21 2014
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Last modified July 27 12:18 EDT 2021. Contains 346306 sequences. (Running on oeis4.) | 585 | 1,308 | {"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} | 2.59375 | 3 | CC-MAIN-2021-31 | latest | en | 0.631826 |
https://slides.com/jankorbel/deck-874834 | 1,725,719,196,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00697.warc.gz | 509,815,508 | 13,769 | # Applications of fractional diffusion in option pricing
Jan Korbel
CCS Satelite on Econophysics, 27th October, Lyon
slides available at: slides.com/jankorbel
## Option pricing
• First option pricing model (Black and Scholes 1973)
• based on ordinary Brownian motion
• Nobel prize in economics (Scholes, Merton) - 1997
• In financial crises or in complex markets, the model cannot catch
realistic market dynamics
• large drops, sudden shocks, memory effects
• Finite moment log-stable model (Carr and Wu 2003)
• based on Lévy-stable fractional diffusion
• enables large drops
• We generalize the models by using space-time fractional diffusion equation
## Space-time fractional diffusion
The STFD equation is defined as
$$\left({}^*_0 \mathcal{D}^\gamma_t - \mu \ {}^\theta \mathcal{D}_x^{\alpha}\right) g(x,t) = 0$$
Caputo derivative: $${}^*_{t_0} \mathcal{D}^\gamma_t f(t) = \frac{1}{\Gamma(\lceil \gamma \rceil - \gamma)} \int_{t_0}^t \mathrm{d} \tau \frac{f^{\lceil \gamma \rceil}(\tau)}{(t-\tau)^{\gamma + 1 - \lceil \gamma \rceil}}$$
Riesz-Feller derivative: $$\mathcal{F}[{}^{\theta} \mathcal{D}^\alpha_x f(x)](k) = -|k|^\alpha e^{i \, \mathrm{sign}(k) \theta \pi/2} \mathcal{F}[f(x)](k)$$
Solution can be defined in terms of Mellin-Barnes transform
$$g_{\alpha,\theta,\gamma}(x,t) = \frac{1}{2 \pi i} \frac{1}{\alpha x} \int_{c-i \infty}^{c+i \infty} \frac{\Gamma(\frac{y}{\alpha}) \Gamma(1-\frac{y}{\alpha})\Gamma(1-y)}{\Gamma(1-\frac{\gamma}{\alpha} y)\Gamma(\frac{\alpha-\theta}{2 \alpha} y) \Gamma(1- \frac{\alpha-\theta}{2 \alpha} y)} \left(\frac{x}{-\mu t}\right)^y \mathrm{d} y$$
[1] Physica A 449 (2016) 200-214
## Space-time fractional diffusion
• $$\gamma=1, \alpha=2$$ - ordinary Gaussian diffusion
• $$\gamma=1, \alpha<2$$ - Lévy-stable diffusion
• $$\gamma \neq 1, \alpha=2$$ - diffusion with memory
• $$\gamma \neq 1, \alpha<2$$ - space-time fractional diffusion
[6] Mathematics 7 (9) (2019) 796
## Space-time fractional option pricing
Price of European call option: $$C(S,K,\tau) = \int_{-\infty}^\infty \max\{S e^{(r+\mu) \tau + x}-K,0\} g_{\alpha,\theta,\gamma}(x,\tau) \mathrm{d} x$$
Interpretation of parameters:
• $$\theta = \max\{-\alpha, \alpha-2\}$$
• maximally asymmetric distribution
• power-law probability of drops (negative Lévy tail)
• Gaussian probability of rises (positive exponential tail
• $$\alpha < 2$$ - risk redistribution to large drops
• $$\gamma$$ - risk redistribution in time
• $$\gamma < 1$$ shorter contracts are more risky
• $$\gamma > 1$$ longer contracts are more risky
[1] Physica A 449 (2016) 200-214; [3] Fractal Fract. 2 (1) (2018) 15
## Double-series representation
By using residue summation in $$\mathbb{C}^2$$ it is possible to express the price in terms of rapidly-convergent double series ( $$\mathcal{L} = \log \frac{S}{K} + r \tau$$ )
$$C(S,K,\tau) = \frac{K e^{-r \tau}}{\alpha} \sum_{n=0}^\infty \sum_{m=1}^\infty \frac{1}{n! \Gamma\left(1 + \frac{m-n}{\alpha}\right)}(\mathcal{L}+\mu \tau)^{n}(-\mu \tau)^{\frac{m-n}{2}}$$
[4] Fract. Calc. Appl. Anal. 21 (4) (2018) 981-1004
## Subordinator representation
$$g_{\alpha,\theta,\gamma}(x,t)$$ can be represented as a subordinated process
$$g_{\alpha,\theta,\gamma}(x,t) = \int_0^\infty \mathrm{d} l K_\gamma(t,l) L_{\alpha}^\theta(l,x)$$
• $$L_{\alpha}^\theta(l,x)$$ - Lévy-stable distribution with scaling parameter $$l$$
• $$K_\gamma(t,l$$ - subordinator (smearing kernel)
• $$K_\gamma(t,l) = \frac{t}{l^{1+1/\gamma} \gamma} L_\gamma^{\gamma}\left(\frac{t}{l^{1/\gamma}}\right)$$
• We compare with other subordinated models
• Variance gamma $$K_\lambda(t,l) = \lambda e^{\lambda (-t/l)}$$
• Negative inverse-gamma $$K_{\alpha,\beta}(t,l) = \frac{e^{-\frac{\beta }{t/l}} \left(\frac{\beta }{t/l}\right)^{\alpha }}{t/l \Gamma (\alpha )}$$
[1] Physica A 449 (2016) 200-214; [8] Risks 8 (4) (2020) 124
## Other results
• Space-time fractional option pricing with varying order of fractional derivatives
• [3] FCAA 19 (6) (2016) 1414-1433
• Pricing of more exotic types of options (American, digital,...) under the space-time fractional diffusion model and formulas for the risk sensitives ("the Greeks" - Gamma, Delta, Rho,...)
• [5] Risks 7 (2) (2019) 36; 10.3390/risks7020036
• [6] Mathematics 7 (9) (2019) 796; 10.3390/math7090796
• Option pricing with more complicated fractional diffusion equation based on Hilfer-Prabhakar fractional derivative
• [7] Fract. Calc. Appl. Anal. 23 (4) (2020) 996-1012
Thank you!
Keywords: option pricing, Black–Scholes model, fractional calculus, fractional diffusion, long-term memory, Lévy stable processes, jump processes, fractional Brownian motion, subordinated models, Bergomi model, rough volatility models
By Jan Korbel
• 59 | 1,616 | 4,728 | {"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": 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} | 2.984375 | 3 | CC-MAIN-2024-38 | latest | en | 0.513318 |
http://davidmlane.com/hyperstat/A5185.html | 1,542,133,629,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741340.11/warc/CC-MAIN-20181113173927-20181113195356-00052.warc.gz | 76,382,450 | 2,124 | # Summary of Measures of Central Tendency (1 of 2)
Of the five measures of central tendency discussed, the mean is by far the most widely used. It takes every score into account, is the most efficient measure of central tendency for normal distributions and is mathematically tractable making it possible for statisticians to develop statistical procedures for drawing inferences about means. On the other hand, the mean is not appropriate for highly skewed distributions and is less efficient than other measures of central tendency when extreme scores are possible. The geometric mean is a viable alternative if all the scores are positive and the distribution has a positive skew.
The median is useful because its meaning is clear and it is more efficient than the mean in highly-skewed distributions. However, it ignores many scores and is generally less efficient than the mean, the trimean, and trimmed means.
The mode can be informative but should almost never be used as the only measure of central tendency since it is highly susceptible to sampling fluctuations.
Click here for an interactive demonstration of properties of the mean and median. | 215 | 1,158 | {"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} | 2.546875 | 3 | CC-MAIN-2018-47 | longest | en | 0.948423 |
http://ravima.cf/1682458.shtml | 1,585,873,686,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370509103.51/warc/CC-MAIN-20200402235814-20200403025814-00499.warc.gz | 158,118,538 | 3,456 | # Optimal binary search tree dynamic programming c++
2020-04-03 01:28
Apr 08, 2016 Optimal Binary Search Tree in C. GitHub Gist: instantly share code, notes, and snippets.Given a sorted array keys[0. . n1 of search keys and an array freq[0. . n1 of frequency counts, where freq[i is the number of searches to keys[i. Construct a binary search tree of all keys such that the total cost of all the searches is as small as possible. Let us first define the cost of a BST. optimal binary search tree dynamic programming c++
Lecture 10: Dynamic Programming Longest palindromic sequence Optimal binary search tree Alternating coin game. DP notions. 1. Characterize the structure of an optimal solution 2. Recursively dene the value of an optimal solution based on optimal solutions of subproblems 3. Compute the value of an optimal solution in bottomup
In computer science, an optimal binary search tree (Optimal BST), sometimes called a weightbalanced binary tree, is a binary search tree which provides the smallest possible search time (or expected search time) for a given sequence of accesses (or access probabilities). Optimal BSTs are generally divided into two types: static and dynamic. C program to perform Insert, Delete, Search an element into a binary search tree 3 Responses to C program that uses dynamic programming algorithm to solve the optimal binary search tree optimal binary search tree dynamic programming c++ An optimal binary search tree is a BST, which has minimal expected cost of locating each node Search time of an element in a BST is O(n), whereas in a BalancedBST search time is O(log n). Again the search time can be improved in Optimal Cost Binary Search Tree, placing the most frequently used data in the root and closer to the root element
Program to find Optimal Binary Search Tree using Dynamic Method in C Analysis Of Algorithms optimal binary search tree dynamic programming c++ Jan 03, 2018 Given a sorted array keys[0. . n1 of search keys and an array freq[0. . n1 of frequency counts, where freq[i is the number of searches to keys[i. Construct a binary search tree of all keys such that the total cost of all the searches is as small as possible. Let us first define the cost of a BST. The problem is dynamic programming; constructing an optimal binary search tree (OBST). I understand dynamic programming in general and the concepts of this problem in particular, but I don't understand the recursive form of this problem. I get that we're constructing optimal binary search trees for an increasing subset of these nodes and Apr 05, 2015 Given keys and frequency at which these keys are searched, how would you create binary search tree from these keys such that cost of searching is minimum. ht Nov 27, 2016 Find optimal cost to construct binary search tree where each key can repeat several times. We are given frequency of each key in same order as corresponding keys in inorder traversal of a binary search tree. In order to construct a binary search tree, for each given key, we have to find out if key already exists
Rating: 4.79 / Views: 553 | 666 | 3,093 | {"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} | 2.828125 | 3 | CC-MAIN-2020-16 | latest | en | 0.855393 |
http://www.mathworks.com/help/finance/lpm.html?requestedDomain=www.mathworks.com&nocookie=true | 1,519,539,293,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816138.91/warc/CC-MAIN-20180225051024-20180225071024-00277.warc.gz | 509,692,716 | 14,524 | # Documentation
### This is machine translation
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# lpm
Compute sample lower partial moments of data
## Syntax
```lpm(Data)
lpm(Data,MAR)
lpm(Data,MAR,Order)
Moment = lpm(Data,MAR,Order)
```
## Arguments
`Data` `NUMSAMPLES`-by-`NUMSERIES` matrix with `NUMSAMPLES` observations of `NUMSERIES` asset returns. `MAR` (Optional) Scalar minimum acceptable return (default `MAR` = `0`). This is a cutoff level of return such that all returns above `MAR` contribute nothing to the lower partial moment. `Order` (Optional) Either a scalar or a `NUMORDERS` vector of nonnegative integer moment orders. If no order specified, default `Order` = `0`, which is the shortfall probability. Although this function works for noninteger orders and, in some cases, for negative orders, this falls outside customary usage.
## Description
Given `NUMSERIES` assets with `NUMSAMPLES` returns in a `NUMSAMPLES`-by-`NUMSERIES` matrix `Data`, a scalar minimum acceptable return `MAR`, and one or more nonnegative moment orders in a `NUMORDERS` vector `Order`, `lpm` computes lower partial moments relative to `MAR` for each asset in a `NUMORDERS x NUMSERIES` matrix `Moment`.
The output `Moment` is a ```NUMORDERS x NUMSERIES``` matrix of lower partial moments with `NUMORDERS` `Order`s and `NUMSERIES` series, that is, each row contains lower partial moments for a given order.
### Note
To compute upper partial moments, reverse the signs of both `Data` and `MAR` (do not reverse the sign of the output). This function computes sample lower partial moments from data. To compute expected lower partial moments for multivariate normal asset returns with a specified mean and covariance, use `elpm`. With `lpm`, you can compute various investment ratios such as Omega ratio, Sortino ratio, and Upside Potential ratio, where:
• ```Omega = lpm(-Data, -MAR, 1) / lpm(Data, MAR, 1)```
• ```Sortino = (mean(Data) - MAR) / sqrt(lpm(Data, MAR, 2))```
• ```Upside = lpm(-Data, -MAR, 1) / sqrt(lpm(Data, MAR, 2))```
## References
Vijay S. Bawa. "Safety-First, Stochastic Dominance, and Optimal Portfolio Choice." Journal of Financial and Quantitative Analysis. Vol. 13, No. 2, June 1978, pp. 255–271.
W. V. Harlow. "Asset Allocation in a Downside-Risk Framework." Financial Analysts Journal. Vol. 47, No. 5, September/October 1991, pp. 28–40.
W. V. Harlow and K. S. Rao. "Asset Pricing in a Generalized Mean-Lower Partial Moment Framework: Theory and Evidence." Journal of Financial and Quantitative Analysis. Vol. 24, No. 3, September 1989, pp. 285–311.
Frank A. Sortino and Robert van der Meer. "Downside Risk." Journal of Portfolio Management. Vol. 17, No. 5, Spring 1991, pp. 27–31. | 754 | 2,784 | {"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} | 2.8125 | 3 | CC-MAIN-2018-09 | latest | en | 0.721804 |
https://volumeunits.com/tablespoon-hm3-conversion | 1,719,239,066,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865383.8/warc/CC-MAIN-20240624115542-20240624145542-00152.warc.gz | 544,549,903 | 4,126 | # TABLESPOON To HM3 Conversion
Conversion rates between Tablespoon and Cubic Hectametre :
From TABLESPOON to HM3 the conversion rate is 1.478676484375E-7.
From Unit To Unit
1 Tablespoon 1.478676484375E-7 Cubic Hectametre
10 Tablespoon 1.478676484375E-6 Cubic Hectametre
100 Tablespoon 1.478676484375E-5 Cubic Hectametre
1000 Tablespoon 0.0001478676484375 Cubic Hectametre
10000 Tablespoon 0.001478676484375 Cubic Hectametre
## Tablespoon to Cubic Hectametre Formula
tablespoon =
hm3
_________
1.478676484375E-7
When you calculate from tablespoon to hm3, you have to divide Cubic Hectametre value with 1.478676484375E-7 to find Tablespoon value.
Convert TABLESPOON To HM3
#### TABLESPOON To HM3 Conversion Chart:
tablespoonhm3
1 tablespoon1.478676484375E-7 hm3
2 tablespoon2.95735296875E-7 hm3
3 tablespoon4.436029453125E-7 hm3
4 tablespoon5.9147059375E-7 hm3
5 tablespoon7.393382421875E-7 hm3
6 tablespoon8.87205890625E-7 hm3
7 tablespoon1.0350735390625E-6 hm3
8 tablespoon1.1829411875E-6 hm3
9 tablespoon1.3308088359375E-6 hm3
10 tablespoon1.478676484375E-6 hm3
11 tablespoon1.6265441328125E-6 hm3
12 tablespoon1.77441178125E-6 hm3
13 tablespoon1.9222794296875E-6 hm3
14 tablespoon2.070147078125E-6 hm3
15 tablespoon2.2180147265625E-6 hm3
16 tablespoon2.365882375E-6 hm3
17 tablespoon2.5137500234375E-6 hm3
18 tablespoon2.661617671875E-6 hm3
19 tablespoon2.8094853203125E-6 hm3
20 tablespoon2.95735296875E-6 hm3
21 tablespoon3.1052206171875E-6 hm3
22 tablespoon3.253088265625E-6 hm3
23 tablespoon3.4009559140625E-6 hm3
24 tablespoon3.5488235625E-6 hm3
25 tablespoon3.6966912109375E-6 hm3
tablespoonhm3
26 tablespoon3.844558859375E-6 hm3
27 tablespoon3.9924265078125E-6 hm3
28 tablespoon4.14029415625E-6 hm3
29 tablespoon4.2881618046875E-6 hm3
30 tablespoon4.436029453125E-6 hm3
31 tablespoon4.5838971015625E-6 hm3
32 tablespoon4.73176475E-6 hm3
33 tablespoon4.8796323984375E-6 hm3
34 tablespoon5.027500046875E-6 hm3
35 tablespoon5.1753676953125E-6 hm3
36 tablespoon5.32323534375E-6 hm3
37 tablespoon5.4711029921875E-6 hm3
38 tablespoon5.618970640625E-6 hm3
39 tablespoon5.7668382890625E-6 hm3
40 tablespoon5.9147059375E-6 hm3
41 tablespoon6.0625735859375E-6 hm3
42 tablespoon6.210441234375E-6 hm3
43 tablespoon6.3583088828125E-6 hm3
44 tablespoon6.50617653125E-6 hm3
45 tablespoon6.6540441796875E-6 hm3
46 tablespoon6.801911828125E-6 hm3
47 tablespoon6.9497794765625E-6 hm3
48 tablespoon7.097647125E-6 hm3
49 tablespoon7.2455147734375E-6 hm3
50 tablespoon7.393382421875E-6 hm3
tablespoonhm3
51 tablespoon7.5412500703125E-6 hm3
52 tablespoon7.68911771875E-6 hm3
53 tablespoon7.8369853671875E-6 hm3
54 tablespoon7.984853015625E-6 hm3
55 tablespoon8.1327206640625E-6 hm3
56 tablespoon8.2805883125E-6 hm3
57 tablespoon8.4284559609375E-6 hm3
58 tablespoon8.576323609375E-6 hm3
59 tablespoon8.7241912578125E-6 hm3
60 tablespoon8.87205890625E-6 hm3
61 tablespoon9.0199265546875E-6 hm3
62 tablespoon9.167794203125E-6 hm3
63 tablespoon9.3156618515625E-6 hm3
64 tablespoon9.4635295E-6 hm3
65 tablespoon9.6113971484375E-6 hm3
66 tablespoon9.7592647968749E-6 hm3
67 tablespoon9.9071324453124E-6 hm3
68 tablespoon1.005500009375E-5 hm3
69 tablespoon1.0202867742187E-5 hm3
70 tablespoon1.0350735390625E-5 hm3
71 tablespoon1.0498603039062E-5 hm3
72 tablespoon1.06464706875E-5 hm3
73 tablespoon1.0794338335937E-5 hm3
74 tablespoon1.0942205984375E-5 hm3
75 tablespoon1.1090073632812E-5 hm3
tablespoonhm3
76 tablespoon1.123794128125E-5 hm3
77 tablespoon1.1385808929687E-5 hm3
78 tablespoon1.1533676578125E-5 hm3
79 tablespoon1.1681544226562E-5 hm3
80 tablespoon1.1829411875E-5 hm3
81 tablespoon1.1977279523437E-5 hm3
82 tablespoon1.2125147171875E-5 hm3
83 tablespoon1.2273014820312E-5 hm3
84 tablespoon1.242088246875E-5 hm3
85 tablespoon1.2568750117187E-5 hm3
86 tablespoon1.2716617765625E-5 hm3
87 tablespoon1.2864485414062E-5 hm3
88 tablespoon1.30123530625E-5 hm3
89 tablespoon1.3160220710937E-5 hm3
90 tablespoon1.3308088359375E-5 hm3
91 tablespoon1.3455956007812E-5 hm3
92 tablespoon1.360382365625E-5 hm3
93 tablespoon1.3751691304687E-5 hm3
94 tablespoon1.3899558953125E-5 hm3
95 tablespoon1.4047426601562E-5 hm3
96 tablespoon1.419529425E-5 hm3
97 tablespoon1.4343161898437E-5 hm3
98 tablespoon1.4491029546875E-5 hm3
99 tablespoon1.4638897195312E-5 hm3
100 tablespoon1.478676484375E-5 hm3
For 1 TABLESPOON | 1,729 | 4,300 | {"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} | 2.765625 | 3 | CC-MAIN-2024-26 | latest | en | 0.457707 |
https://www.anagrammer.com/scrabble/antisymmetric | 1,590,593,558,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347394756.31/warc/CC-MAIN-20200527141855-20200527171855-00052.warc.gz | 652,821,458 | 18,090 | # Scrabble?! ANTISYMMETRIC
Is antisymmetric valid for Scrabble? Words With Friends? Lexulous? WordFeud? Other games?
Yes! (72 pts)
Yes! (60 pts)
Yes! (72 pts)
Yes! (60 pts)
Yes! (72 pts)
Yes! (73 pts)
Yes! (13 pts)
Yes! (72 pts)
Yes! (75 pts)
## Definitions of ANTISYMMETRIC in various dictionaries:
Antisymmetric or skew-symmetric may refer to: Antisymmetry in linguistics Antisymmetric relation in mathematics Skew-symmetric graph Self-complementary graphIn mathematics, especially linear algebra, and in theoretical physics, the adjective antisymmetric (or skew-symmetric) is used for matrices, tensors, and other objects that change sign if an appropriate operation (e.g. matrix transposition) is performed. See: Skew-symmetric matrix (a matrix A for which AT = −A) Skew-symmetric bilinear form is a bilinear form B such that B(x, y) = −B(y, x) for all x and y. Antisymmetric tensor in matrices and index subsets. "antisymmetric function" – odd function
## WORD SOLVER
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## There are 13 letters in ANTISYMMETRIC ( A1C3E1I1M3N1R1S1T1Y4 )
To search all scrabble anagrams of ANTISYMMETRIC, to go: ANTISYMMETRIC?
Rearrange the letters in ANTISYMMETRIC and see some winning combinations
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### 2 letters out of ANTISYMMETRIC
Anagrammer is a game resource site that has been extremely popular with players of popular games like Scrabble, Lexulous, WordFeud, Letterpress, Ruzzle, Hangman and so forth. We maintain regularly updated dictionaries of almost every game out there. To be successful in these board games you must learn as many valid words as possible, but in order to take your game to the next level you also need to improve your anagramming skills, spelling, counting and probability analysis. Make sure to bookmark every unscrambler we provide on this site. Explore deeper into our site and you will find many educational tools, flash cards and so much more that will make you a much better player. This page covers all aspects of ANTISYMMETRIC, do not miss the additional links under "More about: ANTISYMMETRIC" | 559 | 2,155 | {"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} | 2.859375 | 3 | CC-MAIN-2020-24 | latest | en | 0.888752 |
https://www.wazeesupperclub.com/how-do-you-convert-miles-to-meters-formula/ | 1,653,391,942,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662572800.59/warc/CC-MAIN-20220524110236-20220524140236-00264.warc.gz | 1,225,625,440 | 10,715 | # How do you convert miles to meters formula?
## How do you convert miles to meters formula?
i.e., to convert miles into meters, we just multiply them by 1,609.344.
## How do you convert miles into steps?
It should be noted that all those who took part in the study were regarded as being of ‘normal weight’.
1. Slow walk: 1 mile = 2,252 steps.
2. Brisk walk: 1 mile = 1,935 steps.
3. Jog: 1 mile = 1,951 steps.
4. Run: 1 mile = 1,672 steps.
5. Fast run: 1 mile = 1,400 steps.
6. Very fast run: 1 mile = 1,080 steps.
## How much is one meter in steps?
How to Convert Step to Meter (step to m) By using our Step to Meter conversion tool, you know that one Step is equivalent to 0.762 Meter. Hence, to convert Step to Meter, we just need to multiply the number by 0.762.
## Is meters a mile or meter?
The use of the abbreviation “m” for mile was common before the widespread adoption of the metric system; after the adoption of the metric system, “mi” became the preferred abbreviation in order to avoid confusion between miles and meters.
## What is the formula of meter?
First, we convert the dimensions into meters by using the formula of the foot to meters: The length of the rectangle = 5 ft = 5 × 0.3048 = 1.524 meters. The width of the rectangle = 3 ft = 3 × 0.3048 = 0.9144 meters.
## How many feet and meters are in a mile?
1,609.344 meters
Mile. Definition: A mile (symbol: mi or m) is a unit of length in the imperial and US customary systems of measurement. It is currently defined as 5,280 feet, 1,760 yards, or exactly 1,609.344 meters. History/origin: The mile is an English unit (predecessor of Imperial units and United States Customary Units) of length …
## What’s 5000 steps in miles?
How many miles is 5,000 steps? 5,000 steps make about 2 miles. The exact number depends on the individual’s height and stride length. For a woman with an average stride length (2.2 ft), 5,000 steps are 2.0816 mi, and for a man with an average step (2.5 ft), it’s 2.3674 mi.
## How long does it take to walk 20 meters?
20 meters is 1/300th of distance taken in an hour, so measured time walking that distance will be 36/3 seconds when leaving the unnecessary zeroes out = 12 seconds assuming that there is no change of pace (or steepness) at all. | 619 | 2,264 | {"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} | 4.0625 | 4 | CC-MAIN-2022-21 | latest | en | 0.929061 |
https://en.wikipedia.org/wiki/Electric_Field | 1,496,081,884,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612502.45/warc/CC-MAIN-20170529165246-20170529185246-00384.warc.gz | 943,511,038 | 30,762 | # Electric field
(Redirected from Electric Field)
Electric field lines emanating from a point positive electric charge suspended over an infinite sheet of conducting material.
An electric field is a vector field that associates to each point in space the Coulomb force that would be experienced per unit of electric charge, by an infinitesimal test charge at that point.[1] Electric fields are created by electric charges and can be induced by time-varying magnetic fields. The electric field combines with the magnetic field to form the electromagnetic field.
## Definition
The electric field, ${\displaystyle \mathbf {E} }$, at a given point is defined as the (vector) force, ${\displaystyle \mathbf {F} }$, that would be exerted on a stationary test particle of unit charge by electromagnetic forces (i.e. the Lorentz force). A particle of charge ${\displaystyle q}$ would be subject to a force ${\displaystyle \mathbf {F} =q\mathbf {E} }$.
Its SI units are newtons per coulomb (N⋅C−1) or, equivalently, volts per metre (V⋅m−1), which in terms of SI base units are kg⋅m⋅s−3⋅A−1.
## Sources of electric field
### Causes and description
Electric fields are caused by electric charges or varying magnetic fields. The former effect is described by Gauss's law, the latter by Faraday's law of induction, which together are enough to define the behavior of the electric field as a function of charge repartition and magnetic field. However, since the magnetic field is described as a function of electric field, the equations of both fields are coupled and together form Maxwell's equations that describe both fields as a function of charges and currents.
In the special case of a steady state (stationary charges and currents), the Maxwell-Faraday inductive effect disappears. The resulting two equations (Gauss's law ${\displaystyle \nabla \cdot \mathbf {E} ={\frac {\rho }{\varepsilon _{0}}}}$ and Faraday's law with no induction term ${\displaystyle \nabla \times \mathbf {E} =0}$), taken together, are equivalent to Coulomb's law, written as ${\displaystyle {\boldsymbol {E}}({\boldsymbol {r}})={1 \over 4\pi \varepsilon _{0}}\int \rho ({\boldsymbol {r'}}){{\boldsymbol {r}}-{\boldsymbol {r'}} \over |{\boldsymbol {r}}-{\boldsymbol {r'}}|^{3}}d^{3}r'}$ for a charge density ${\displaystyle \mathbf {\rho } (\mathbf {r} )}$ (${\displaystyle \mathbf {r} }$ denotes the position in space). Notice that ${\displaystyle \varepsilon _{0}}$, the permittivity of vacuum, must be substituted if charges are considered in non-empty media.
### Continuous vs. discrete charge repartition
The equations of electromagnetism are best described in a continuous description. However, charges are sometimes best described as discrete points; for example, some models may describe electrons as point sources where charge density is infinite on an infinitesimal section of space.
A charge ${\displaystyle q}$ located at ${\displaystyle \mathbf {r_{0}} }$ can be described mathematically as a charge density ${\displaystyle \rho (\mathbf {r} )=q\delta (\mathbf {r-r_{0}} )}$, where the Dirac delta function (in three dimensions) is used. Conversely, a charge distribution can be approximated by many small point charges.
## Superposition principle
Electric fields satisfy the superposition principle, because Maxwell's equations are linear. As a result, if ${\displaystyle \mathbf {E} _{1}}$ and ${\displaystyle \mathbf {E} _{2}}$ are the electric fields resulting from distribution of charges ${\displaystyle \rho _{1}}$ and ${\displaystyle \rho _{2}}$, a distribution of charges ${\displaystyle \rho _{1}+\rho _{2}}$ will create an electric field ${\displaystyle \mathbf {E} _{1}+\mathbf {E} _{2}}$; for instance, Coulomb's law is linear in charge density as well.
This principle is useful to calculate the field created by multiple point charges. If charges ${\displaystyle q_{1},q_{2},...,q_{n}}$ are stationary in space at ${\displaystyle \mathbf {r} _{1},\mathbf {r} _{2},...\mathbf {r} _{n}}$, in the absence of currents, the superposition principle proves that the resulting field is the sum of fields generated by each particle as described by Coulomb's law:
${\displaystyle \mathbf {E} (\mathbf {r} )=\sum _{i=1}^{N}\mathbf {E} _{i}(\mathbf {r} )={\frac {1}{4\pi \varepsilon _{0}}}\sum _{i=1}^{N}q_{i}{\frac {\mathbf {r} -\mathbf {r} _{i}}{|\mathbf {r} -\mathbf {r} _{i}|^{3}}}}$
## Electrostatic fields
Illustration of the electric field surrounding a positive (red) and a negative (blue) charge
Experiment illustrating electric field lines. An electrode connected to an electrostatic induction machine is placed in an oil-filled container. Considering that oil is a dielectric medium, when there is current through the electrode, the particles arrange themselves so as to show the force lines of the electric field.
Electrostatic fields are E-fields which do not change with time, which happens when charges and currents are stationary. In that case, Coulomb's law fully describes the field.
### Electric potential
If a system is static, such that magnetic fields are not time-varying, then by Faraday's law, the electric field is curl-free. In this case, one can define an electric potential, that is, a function ${\displaystyle \Phi }$ such that ${\displaystyle \mathbf {E} =-\nabla \Phi }$.[2] This is analogous to the gravitational potential.
### Parallels between electrostatic and gravitational fields
Coulomb's law, which describes the interaction of electric charges:
${\displaystyle \mathbf {F} =q\left({\frac {Q}{4\pi \varepsilon _{0}}}{\frac {\mathbf {\hat {r}} }{|\mathbf {r} |^{2}}}\right)=q\mathbf {E} }$
is similar to Newton's law of universal gravitation:
${\displaystyle \mathbf {F} =m\left(-GM{\frac {\mathbf {\hat {r}} }{|\mathbf {r} |^{2}}}\right)=m\mathbf {g} }$
(where ${\displaystyle \mathbf {\hat {r}} =\mathbf {\frac {r}{|r|}} }$).
This suggests similarities between the electric field E and the gravitational field g, or their associated potentials. Mass is sometimes called "gravitational charge" because of that similarity.[citation needed]
Electrostatic and gravitational forces both are central, conservative and obey an inverse-square law.
### Uniform fields
A uniform field is one in which the electric field is constant at every point. It can be approximated by placing two conducting plates parallel to each other and maintaining a voltage (potential difference) between them; it is only an approximation because of boundary effects (near the edge of the planes, electric field is distorted because the plane does not continue). Assuming infinite planes, the magnitude of the electric field E is:
${\displaystyle E=-{\frac {\Delta \phi }{d}}}$
where Δϕ is the potential difference between the plates and d is the distance separating the plates. The negative sign arises as positive charges repel, so a positive charge will experience a force away from the positively charged plate, in the opposite direction to that in which the voltage increases. In micro- and nano-applications, for instance in relation to semiconductors, a typical magnitude of an electric field is in the order of 106 V⋅m−1, achieved by applying a voltage of the order of 1 volt between conductors spaced 1 µm apart.
## Electrodynamic fields
Electrodynamic fields are E-fields which do change with time, for instance when charges are in motion.
The electric field cannot be described independently of the magnetic field in that case. If A is the magnetic vector potential, defined so that ${\displaystyle \mathbf {B} =\nabla \times \mathbf {A} }$, one can still define an electric potential ${\displaystyle \Phi }$ such that:
${\displaystyle \mathbf {E} =-\nabla \Phi -{\frac {\partial \mathbf {A} }{\partial t}}}$
One can recover Faraday's law of induction by taking the curl of that equation
[3]
${\displaystyle \nabla \times \mathbf {E} =-{\frac {\partial (\nabla \times \mathbf {A} )}{\partial t}}=-{\frac {\partial \mathbf {B} }{\partial t}}}$
which justifies, a posteriori, the previous form for E.
## Energy in the electric field
The total energy per unit volume stored by the electromagnetic field is[4]
${\displaystyle u_{EM}={\frac {\varepsilon }{2}}|\mathbf {E} |^{2}+{\frac {1}{2\mu }}|\mathbf {B} |^{2}}$
where ε is the permittivity of the medium in which the field exists, ${\displaystyle \mu }$ its magnetic permeability, and E and B are the electric and magnetic field vectors.
As E and B fields are coupled, it would be misleading to split this expression into "electric" and "magnetic" contributions. However, in the steady-state case, the fields are no longer coupled (see Maxwell's equations). It makes sense in that case to compute the electrostatic energy per unit volume:
${\displaystyle u_{ES}={\frac {1}{2}}\varepsilon |\mathbf {E} |^{2}\,,}$
The total energy U stored in the electric field in a given volume V is therefore
${\displaystyle U_{ES}={\frac {1}{2}}\varepsilon \int _{V}|\mathbf {E} |^{2}\,\mathrm {d} V\,,}$
## Further extensions
### Definitive equation of vector fields
In the presence of matter, it is helpful in electromagnetism to extend the notion of the electric field into three vector fields, rather than just one:[5]
${\displaystyle \mathbf {D} =\varepsilon _{0}\mathbf {E} +\mathbf {P} \!}$
where P is the electric polarization – the volume density of electric dipole moments, and D is the electric displacement field. Since E and P are defined separately, this equation can be used to define D. The physical interpretation of D is not as clear as E (effectively the field applied to the material) or P (induced field due to the dipoles in the material), but still serves as a convenient mathematical simplification, since Maxwell's equations can be simplified in terms of free charges and currents.
### Constitutive relation
The E and D fields are related by the permittivity of the material, ε.[6][5]
For linear, homogeneous, isotropic materials E and D are proportional and constant throughout the region, there is no position dependence: For inhomogeneous materials, there is a position dependence throughout the material:
${\displaystyle \mathbf {D(r)} =\varepsilon \mathbf {E(r)} }$
For anisotropic materials the E and D fields are not parallel, and so E and D are related by the permittivity tensor (a 2nd order tensor field), in component form:
${\displaystyle D_{i}=\varepsilon _{ij}E_{j}}$
For non-linear media, E and D are not proportional. Materials can have varying extents of linearity, homogeneity and isotropy. | 2,681 | 10,532 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 39, "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} | 3.53125 | 4 | CC-MAIN-2017-22 | latest | en | 0.922629 |
https://panacea-healthcare.com/qa/question-what-is-the-name-of-a-99-sided-shape.html | 1,606,562,533,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195417.37/warc/CC-MAIN-20201128095617-20201128125617-00570.warc.gz | 422,169,000 | 7,956 | # Question: What Is The Name Of A 99 Sided Shape?
## Is there a 9 sided shape?
In geometry, a nonagon (/ˈnɒnəɡɒn/) or enneagon (/ˈɛniəɡɒn/) is a nine-sided polygon or 9-gon.
The name nonagon is a prefix hybrid formation, from Latin (nonus, “ninth” + gonon), used equivalently, attested already in the 16th century in French nonogone and in English from the 17th century..
## What is a 70 sided shape called?
heptacontagonIn geometry, a heptacontagon (or hebdomecontagon from Ancient Greek ἑβδομήκοντα, seventy) or 70-gon is a seventy-sided polygon. The sum of any heptacontagon’s interior angles is 12240 degrees.
## What’s a 20 sided shape called?
icosagonIn geometry, an icosagon or 20-gon is a twenty-sided polygon. The sum of any icosagon’s interior angles is 3240 degrees.
## What is a fourteen sided shape called?
tetradecagonIn geometry, a tetradecagon or tetrakaidecagon or 14-gon is a fourteen-sided polygon.
## What do you call a shape with 69 sides?
10 – deca | 1 – hena 20 – icosi | 2 – di 30 – triaconta | 3 – tri 40 – tetraconta | 4 – tetra 50 – pentaconta | 5 – penta 60 – hexaconta | 6 – hexa 70 – heptaconta | 7 – hepta 80 – octaconta | 8 – octa 90 – nonaconta | 9 – nona. The 3-digit sided polygons are named in a similar fashion.
## What is a 28 sided shape called?
icosioctagonIn geometry, an icosioctagon (or icosikaioctagon) or 28-gon is a twenty eight sided polygon.
## What is a 1 billion sided shape called?
megagonA megagon or 1 000 000-gon is a polygon with 1 million sides (mega-, from the Greek μέγας megas, meaning “great”). Even if drawn at the size of the Earth, a regular megagon would be very difficult to distinguish from a circle….Megagon.Regular megagonTypeRegular polygonEdges and vertices10000007 more rows
## What is 12 sided shape called?
dodecagonIn geometry, a dodecagon or 12-gon is any twelve-sided polygon.
## What is a 99 sided polygon called?
enneacontagonIn geometry, an enneacontagon or enenecontagon or 90-gon (from Ancient Greek ἑννενήκοντα, ninety) is a ninety-sided polygon. The sum of any enneacontagon’s interior angles is 15840 degrees.
## What is a 100 sided shape called?
hectogonIn geometry, a hectogon or hecatontagon or 100-gon is a hundred-sided polygon. The sum of all hectogon’s interior angles are 17640 degrees.
## What is a 200 sided shape called?
dihectogonWhat is the name of a polygon with…?#Name of the Polygon + Geometric Drawing80 sidesoctacontagon90 sidesenneacontagon100 sideshectogon200 sidesdihectogon53 more rows
## What is a 31 sided shape called?
Systematic polygon namesTensOnes20icosi- (icosa- when alone)-di-30triaconta–tri-40tetraconta–tetra-50pentaconta–penta-5 more rows | 830 | 2,682 | {"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} | 3.796875 | 4 | CC-MAIN-2020-50 | latest | en | 0.810335 |
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