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https://www.keysight.com/upload/cmc_upload/All/5306OSKR-MXD-5602-429.html?cc=CA&lc=eng | 1,521,817,216,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648313.85/warc/CC-MAIN-20180323141827-20180323161827-00653.warc.gz | 781,394,997 | 3,184 | Question: How can I measure temperature using B-Type thermocouples and a generic DVM? Answer: You can measure temperature with a generic DVM by doing some arithmetic conversions in your computer. To demonstrate the procedure, we will use a E1326B with a E1347A multiplexer. The E1326 driver (which supports thermocouple temperature measurements) is used so that the manual procedure can be verified by the built-in temperature function. An example program in HP BASIC is provided. Measuring thermocouple temperature requires that the voltage at the end of the thermocouple be measured at a isothermal block. The temperature of the isothermal block must be known. For the E1347, the isothermal block temperature is determined by a thermistor embedded in a block of aluminum. The procedure to convert B-type thermocouple voltage to temperature: Use the subprogram Degrees_init to set "F" or "C" for degrees F or degrees C (line 510). Measure the resistance of the 5000 ohm thermistor on a E1347 isothermal terminal block (lines 560-590). Use the function FNThr5(Thr_ohms) to convert this resistance to the temperature of the isothermal block (line 660). Measure the voltage of a thermocouple (lines 600-630). Use the function FNThmc_b(Voltage,Ref_temp) to convert the thermocouple voltage, and block temperature to the to temperature of the thermocouple (line 700). This routine was ported from the DACQ/300 Temperature Library. It is valid from 130 deg C to 1820 deg C. Lines 450-710 measure a thermocouple directly using the MEAS:TEMP command and then using the MEAS:RES, MEAS:VOLT:DC measurement functions and the FNThr, FNThmc, Degrees_init routines. The result are in degrees C. The two methods produce the same result. When this program was run, the thermocouple was at room temperature which is below the valid range. Thus the overload readings, but it does demonstrate how to verify operation. Lines 730-810 print out a table of voltage vs temperature with a block reference temperature of 0 degrees C. The results are in degrees C. Comparison of this table to standard thermocouple tables shows validity. ```10 ! re-save "TEMP_B" 20 ! This main line code is reserved as a error handling shell 30 ! All application code must be at lower level context 40 ASSIGN @Sys TO 70900 ! define I/O paths 50 ASSIGN @Dvm TO 70903 60 COM /Instr/ @Sys,@Dvm 70 COM /Hpnav_temp/ Hpnav_degrees\$[1],Hpnav_errvalu 80 ON TIMEOUT 7,3 GOTO End !Turn TIMEOUTS to errors--this branch never taken 90 ON ERROR RECOVER Kaboom !This handles timeouts and errors not handled 100 ! at lower level contexts 110 Main ! Put application code in this sub 120 PRINT "Checking for E13xx Errors at the end of the program" 130 E13xx_errors 140 GOTO End 150 Kaboom:PRINT "" 160 PRINT ERRM\$ 170 PRINT "Checking for E13xx Errors as a BASIC Error has occurred" 180 E13xx_errors 190 End:END 200 ! 210 SUB E13xx_errors !This sub reads all errors from E13xx instruments 220 COM /Instr/ @Sys,@Dvm 230 COM /Hpnav_temp/ Hpnav_degrees\$[1],Hpnav_errvalu 240 DIM A\$[128] 250 ABORT 7 260 CLEAR @Dvm 270 REPEAT 280 OUTPUT @Dvm;"SYST:ERR?" 290 ENTER @Dvm;A,A\$ 300 PRINT "DVM ERROR ";A\$ 310 UNTIL A=0 320 ! 330 CLEAR @Sys 340 REPEAT 350 OUTPUT @Sys;"SYST:ERR?" 360 ENTER @Sys;A,A\$ 370 PRINT "SYSTEM ERROR ";A\$ 380 UNTIL A=0 390 SUBEND 400 ! 410 SUB Main !This subroutine is treated as the main line 420 COM /Instr/ @Sys,@Dvm 430 !Put application code here 440 COM /Hpnav_temp/ Hpnav_degrees\$[1],Hpnav_errvalu 450 ! MEASURE TEMPERATURE USING THE BUILT IN TEMP FUNCTION 460 OUTPUT @Dvm;"MEAS:TEMP? TC,B,(@100)" 470 ENTER @Dvm;Temp1 480 PRINT "Measured temperature of channel 0 ";Temp1 490 ! 500 ! NOW MEASURE TEMP BY MEASURING VOLTS AND OHMS THEN COMPUTING TEMP 510 Degrees_init("C") 520 ! MEASURE BLOCK TEMPERATURE " 530 OUTPUT @Dvm;"MEAS:TEMP? THER,5000, (@193)" 540 ENTER @Dvm;Meas_block_temp 550 PRINT "MEASURED BLOCK TEMPERATURE ";Meas_block_temp 560 ! MEASURE THERMISTOR RESISTANCE 570 OUTPUT @Dvm;"MEAS:RES? (@193)" 580 ENTER @Dvm;Therm_res 590 PRINT "THERMISTOR RESISTANCE ";Therm_res 600 ! MEASURE CHANNEL 0 VOLTAGE 610 OUTPUT @Dvm;"MEAS:VOLT:DC? (@100)" 620 ENTER @Dvm;Ch0_volts 630 PRINT "CHANNEL 0 VOLTS ";Ch0_volts 640 ! 650 ! Now convert resitance to isothermal block temperature 660 Block_t=FNThr5(Therm_res) 670 PRINT "Calculated block temp ";Block_t 680 ! 690 ! Now convert channel 0 voltage to temperature 700 Temp2=FNThmc_b(Ch0_volts,Block_t) 710 PRINT "Calculated temperature of channel 0";Temp2 720 ! 730 ! Now generate a table of voltage vs temp 740 PRINT "" 750 PRINT " Table of voltages vs temp in C for block temp of 0 deg C" 760 PRINT " Compare this to your thermocouple tables " 770 Degrees_init("C") 780 FOR V=0.E+0 TO 1.E-3 STEP 1.00E-4 790 Temp=FNThmc_b(V,0) 800 PRINT "Voltage ";V;" Temperature ";Temp 810 NEXT V 820 SUBEND 830 ! 840 SUB Degrees_init(Degrees\$) 850 REM COPYRIGHT HEWLETT PACKARD CO. 1992, ALL RIGHTS RESERVED 860 Degrees_init:! 870 COM /Hpnav_temp/ Hpnav_degrees\$[1],Hpnav_errvalu 880 Hpnav_errvalu=1.E+38 890 SELECT TRIM\$(UPC\$(Degrees\$)) 900 CASE "F" 910 Hpnav_degrees\$="F" 920 CASE ELSE 930 Hpnav_degrees\$="C" 940 END SELECT 950 Exit:SUBEND 960 ! 970 ! 980 DEF FNThr5(Thr_ohms) 990 REAL Rhi,Rlo,T,W,A,B,C,R 1000 COM /Hpnav_temp/ Hpnav_degrees\$[1],Hpnav_errvalu 1010 READ Rhi,Rlo,A,B,C 1020 IF Thr_ohms>Rhi THEN RETURN Hpnav_errvalu 1030 IF Thr_ohms70 THEN RETURN Hpnav_errvalu 1250 Cuv=Voltage+Parms(8,1)+Reft*(Parms(8,2)+Reft*(Parms(8,3)+Reft*Parms(8,4))) 1260 IF CuvVhi+1.E-6 THEN RETURN Hpnav_errvalu 1280 I=7 1290 IF Cuv ``` | 1,658 | 5,541 | {"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-2018-13 | latest | en | 0.833969 |
https://www.statology.org/scheffe-test-sas/ | 1,716,677,179,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058858.81/warc/CC-MAIN-20240525222734-20240526012734-00368.warc.gz | 885,873,613 | 18,004 | # How to Perform Scheffe’s Test in SAS
one-way ANOVA is used to determine whether or not there is a statistically significant difference between the means of three or more independent groups.
If the overall p-value from the ANOVA table is less than some significance level, then we have sufficient evidence to say that at least one of the means of the groups is different from the others.
However, this doesn’t tell us which groups are different from each other. It simply tells us that not all of the group means are equal.
In order to find out exactly which groups are different from each other, we must conduct a post hoc test.
One of the most commonly used post hoc tests is Scheffe’s test, which allows us to make pairwise comparisons between the means of each group while controlling for the family-wise error rate.
The following example shows how to perform Scheffe’s test in R.
## Example: Scheffe’s Test in SAS
Suppose a researcher recruits 30 students to participate in a study. The students are randomly assigned to use one of three studying methods to prepare for an exam.
We can use the following code to create this dataset in SAS:
```/*create dataset*/
data my_data;
input Method \$ Score;
datalines;
A 76
A 77
A 77
A 81
A 82
A 82
A 83
A 84
A 85
A 89
B 81
B 82
B 83
B 83
B 83
B 84
B 87
B 90
B 92
B 93
C 77
C 78
C 79
C 88
C 89
C 90
C 91
C 95
C 98
C 98
;
run;
```
Next, we’ll use proc ANOVA to perform the one-way ANOVA:
```/*perform one-way ANOVA with Scheffe's post-hoc test*/
proc ANOVA data=my_data;
class Method;
model Score = Method;
means Method / scheffe cldiff;
run;```
Note: We used the means statement along with the scheffe and cldiff options to specify that Scheffe’s post-hoc test should be performed (with confidence intervals) if the overall p-value of the one-way ANOVA is statistically significant.
First, we’ll analyze the ANOVA table in the output:
From this table we can see:
• The overall F Value: 3.49
• The corresponding p-value: 0.0448
Recall that a one-way ANOVA uses the following null and alternative hypotheses:
• H0: All group means are equal.
• HA: At least one group mean is different from the rest.
Since the p-value from the ANOVA table (0.0448) is less than α = .05, we reject the null hypothesis.
This tells us that the mean exam score is not equal between the three studying methods.
To determine exactly which group means are different, we must refer to the final table in the output that shows the results of Scheffe’s post-hoc tests:
To tell which group means are different, we must look at which pairwise comparisons have stars (***) next to them.
From the table we can see there is a statistically significant difference in mean exam scores between group A and group C.
There are no statistically significant differences between any other group means.
Specifically, we can see that the mean difference in exam scores between group C and group A is 6.7.
The 95% confidence interval for the difference in means between these groups is [0.064, 13.336].
The following tutorials provide additional information about ANOVA models:
May 13, 2024
April 25, 2024
April 19, 2024 | 805 | 3,150 | {"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} | 3.421875 | 3 | CC-MAIN-2024-22 | latest | en | 0.863843 |
https://forum.allaboutcircuits.com/threads/rotational-inertia.1748/ | 1,632,486,688,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057524.58/warc/CC-MAIN-20210924110455-20210924140455-00284.warc.gz | 309,904,830 | 19,418 | # rotational inertia
Joined Dec 29, 2004
83
Hi,
I have hard time understanding the rotational inertia .
A string is wrapped around a cylindar spool of radius 1cm. The axis of the spool is fixed. A length of string of .8 m is pulled off in 1.5 s at a constant tension of 20N.
What is the rotational inertia of the spool?
Can I have some help with this problem?
I don't know how to start. The only definition of rotational inertia I have is I=mR[sup]2[/sup]!
Thank you
B. | 131 | 475 | {"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-2021-39 | longest | en | 0.908567 |
https://alex.state.al.us/plans2.php?std_id=53844 | 1,516,186,237,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886895.18/warc/CC-MAIN-20180117102533-20180117122533-00662.warc.gz | 630,546,430 | 8,295 | ALEX Lesson Plan Resources
Back
ALEX Lesson Plans
Subject: Mathematics (6), or Technology Education (6 - 8)
Title: Finding Pi - A Math Adventure
Description: In this lesson students will hear the story of Sir Cumference and the Dragon of Pi and participate in class activities to discover pi. Technology resources will be used to display results.
Subject: Mathematics (6), or Technology Education (6 - 8)
Title: We're Talking Baseball!
Description: During this lesson students use mathematical skills to determine baseball statistics. They can locate their favorite players' stats on the Internet and determine batting average and slugging percentage.
Subject: Mathematics (5 - 6)
Title: Fractions and Recipes
Description: This is a lesson used to show how fractions are used in everyday life. Students will use the knowledge they have about fractions to change a recipe to accommodate different amounts of people. Students will enjoy navigating the Internet to locate a recipe. Students will also have the opportunity to make the recipe in class and share it with their friends.
Subject: Mathematics (6 - 7), or Technology Education (6 - 8)
Title: How Much Money Can I Make?
Description: This lesson is a technology-based activity in which students will research two careers they are possibly interested in pursuing. They will convert the average yearly salary of each career into an hourly wage and record their findings in a spreadsheet. They will calculate their weekly gross and net pay for each career choice.
Thinkfinity Lesson Plans
Subject: Mathematics
Description: This interactive game, from Illuminations, allows students to exercise their factoring ability. They can play against a friend or the computer as they take turns choosing a number and identifying all of its factors.
Thinkfinity Partner: Illuminations
Subject: Mathematics
Description: In this lesson, one of a multi-part unit from Illuminations, students use Venn diagrams to organize information about numbers. Using the Product Game board, students look for relationships and characteristics of numbers to determine what numbers belong to a descriptor and what numbers belong to more than one descriptor.
Thinkfinity Partner: Illuminations
Subject: Mathematics
Description: This student interactive, from Illuminations, allows students to visually explore the concept of factors by creating rectangular arrays with an area equal to the product of the factors. They can choose whether or not to use the commutative property when identifying factors.
Thinkfinity Partner: Illuminations
Subject: Mathematics
Title: Playing the Product Game Add Bookmark
Description: In this lesson, one of a multi-part unit from Illuminations, students learn the rules of and play the Product Game. In this game, players start with a set of given factors and then multiply to find the product. Students create their own game boards and develop strategies for winning the game.
Thinkfinity Partner: Illuminations | 571 | 2,981 | {"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} | 3.828125 | 4 | CC-MAIN-2018-05 | latest | en | 0.927028 |
http://gawron.sdsu.edu/compling/course_core/lectures/pcfg/prob_parse.htm | 1,506,144,876,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689490.64/warc/CC-MAIN-20170923052100-20170923072100-00360.warc.gz | 146,687,661 | 7,223 | # Probabilistic Context Grammars
Probabilistic
Context
Free
Grammars
A probabilistic context free grammar is a context free grammar with probabilities attached to the rules.
Model Parameters
The model parameters are the probabilities assigned to grammar rules.
Computing Probabilities
We discuss how the model assigns probabilities to strings and to analyses of strings.
Exploiting Probabilities in Parsing
We discuss how to find the most probable parse given the model.
Estimating Parameters
We sketch how rule probabilities are estimated from a syntactically annotated corpus.
Probability
Model
Important Probabilities
Rule Probabilities
Probability of an expansion given the category being expanded. P(γ | A), the probability that the sequence of grammar symbols γ will expand category A. Given the definition of a conditional probability, this means the probabilities of all the expansions of a single catgeory must add to 1. For example:
```S -> NP VP, .8 | Aux NP VP, .2
```
Given that S is being expanded, the probability of the NP VP rule is .8; the probability of the Aux NP VP expansion is .2. The rule probabilities are the parameters of a PCFG model.
Tree Probability
The probability of a tree given all the rules used to construct it. This will be the product of the probabilities of all the rules used to construct it. We motivate this below.
For now an example. Given this grammar:
the sentence
Book the dinner flight
has at least two trees:
P(Tleft) = .05 * .20 * .20 * .20 * .75 * .30 * .60 * .10 * .40 = 2.2 × 10-6 P(Tright) = .05 * .10 * .20 * .15 * .75 * .75 * .30 * .60 * .10 * .40 = 6.1 × 10-7
Therefore the more probable of the two parses is Tleft, which is the intuitive result.
String Probability
Sum of the tree probabilities for all analyses of the string.
Estimating
Parameters
Given a tree bank, the maximum likelihood estimate for the PCFG rule:
VP → VBD NP PP
is:
Count(VP → VBD NP PP) ----------------------------- Count(VP)
Highest
Probability
parse
Our primary subject of interest is finding the highest probability parse for a sentence.
This can be done with a rather simple variant of CKY. CKY is a bottum up algorithm. A bottum up algorithm is compatible with our probability model because the probability of a subtree like Is independent of anything outside that tree, in particular of where that subtree occurs in a sentence (for instance, subject position or object position), so once we have found all ways of building a constituent, we are guaranteed to have found the maximum amount of probability mass that constituent can add to any tree that includes it.
CKY is ideal for our purposes because no constituent spanning (i,j) is built until all ways of building constituents spanning (k,l),
i ≤ k < l < j
have been found. Since we can find the most probable parses for all possible subconstituents of a constituent spanning (i,j) before we build any constituents spaning (i,j), we have the essential feature of a Viterbi like algorithm.
We modify CKY as follows: instead of building a parse table in which each cell assigns value True or False for each category, we build a parse table in which each cell assigns a probability to each category. Each time we find a new way of building an A spanning (i,j), we compute its probability and check to see if it is more propbable than the current value for table[i][j][A].
Example
Partial PCFG:
S → NP VP 0.8 Det → the 0.4 NP → Det N 0.3 Det → a 0.4 VP → V NP 0.2 N → meal 0.01 V → includes 0.05 N → flight 0.02
Problems
Independence assumptions are (damagingly) false
Where particular kinds of category expansions happen very much depends on where in the tree the category occurs.
Lexical
Rule probabilities above the preterminal level are oblivious to what words occur in the tree. But this is wrong.
Independence
Pronouns (Switchboard corpus [spoken, telephone], Francis et al. 1999)
Pronoun Non-Pronoun Subject 91% 9 % Object 34% 66%
Lexical
1. Left tree is correct for this example
2. Right tree is correct for
Conjunction ambiguities can create trees to which a PCFG cant assign distinct probabilities, but intuitively, the tree are quite different in probabilitity:
Notice the two trees employ exactly the same bags of rules, so they must, acording to a PCFG, be equiprobable.
Using
grandparents
Johnson (1998) demonstrates the utility of using the grandparent node as a contextual parameter of a probabilistic parsing model.
This option uses more derivational history.
1. NP1: P(... | P = NP, GP = S)
2. NP2: P(... | P = NP, GP = VP)
Significant improvement on PCFG:
1. Can capture distributional differences between subject and object NPs (such as likelihood of pronoun)
2. Outperforms left-corner parser described in Manning and Carpenter (1997)
S → NP VP S → NP^S VP^S NP → PRP NP^S → PRP NP^VP → PRP NP → DT NN NP^S → DT NN NP^VP → DT NN
This strategy is called splitting. We split the category NP in two so as to model distinct internal distributions depending on the external context of an NP Note that in the work above, preterminals are not split.
Splitting preterminals
Johnson's grandparent grammars work because they reclaim some of the ground lost due to the incorrectness of the independence assumptions of PCFGs. But the other problem for PCFGs that we noted is that they miss lexical dependencies. If we extend the idea of gradparent information to preterminals, we can begin to capture some of that information.
The parse on the left is incorrect because of an incorrect choice between the following two rules:
PP → IN PP
PP → IN SBAR
The parse on the right, in which the categories (including preterminal categories) are annotated with grandparent information, is correct.
The choice is made possible because the string advertising works can be analyzed either as an SBAR or an NP:
[SBAR Advertising works] when it is clever.
1. [NP The advertising works of Andy Warhol] made him a legend.
Now as it happens words of preterminal category IN differ greatly in how likely they are to occur with sentences:
IN SBAR
if John comes
before John comes
after John comes
? under John comes
? behind John comes
So this is a clear case where distributional facts about words helps. And the simple act of extending the grandparent strategy to preterminals helps capture the differences between these two classes of IN.
Generalized
Splitting
Obviously there are many cases where splitting helps.
But there are also cases where splitting hurts. We already have a sparseness problem. If a split results in a defective or under-represented distribution, it can lead to overfitting and it can hurt. Generalizations that were there before the split can be blurred or lost.
Two approaches:
• Hand-tuned rules (Klein and Manning 2003b)
• Split-and-merge algorithm: search for optimal splits (Petrov et al. 2006)
• Lexicalized
CFGs
An extreme form of splitting:
VP → VBD NP PP VP(dumped,VBD) → VBD(dumped,VBD) NP(sacks,NNS) PP(into,P) VP(sold,VBD) → VBD(sold,VBD) NP(pictures,NNS) PP(of,P) . . .
Notice categories are being annotated with two kinds of information:
2. Category of their immediately dominated head constituent
Lexical rules all have probability 1.0
Implementing
Lexicalized
Parsers
Mathematically, a lexicalized PCFG is a PCFG, so In principle, a lexicalized grammar has an N3 CKY parser just like any PCFG.
The problem is, we just blew up the grammar when we lexicalized it:
VP → VBD NP PP VP(dumped,VBD) → VBD(dumped,VBD) NP(sacks,NNS) PP(into,P) VP(spilled,VBD) → VBD(spilled,VBD) NP(ketchup,NNS) PP(on,P) VP(X,VBD) → VBD(X,VBD) NP(Y,NNS) PP(Z,P)
And the real parsing complexity is O(n3G2), where G is the size of the grammar.
The G term now implicitly depends on n or G, whichever is smaller, and n is generally SMALLER than G. So the real parsing complexity is O(n3n2)
How many distinct attributes can n there now be in a cell for span i,j?
nonterminal, word (between i and j), preterminal of word
Therefore worst case number of items in a span:
num nonterminals X n (num words between i and j) X num poses
If we assume every part of speech can "project" only one non terminal:
|Nonterminals| X n
That's how many distinct floating point numbers (probs) we may need to store for a span (a cell in the table).
Data structures?
Currently a cell is filled buy a dictionary from nonterminals to floats.
table[3][5] = {NP: .003, VP: 0004}
What shd replace it?
A dictionary from a pair of nonterminals and words to floats.
{(NP,sacks): .003, (VP,sacks): .0004, (VP,dump):.01}
Note: At the same time, a lexicalized grammar rule may look like this:
(NP, sacks) → sacks 0.01
(VP, helps) → (VBZ,helps) (NP,students) 0.0001
Making
a lexicalized
grammar
```VP → VBD NP .1
VP → VBD .9
VBD → kicked .1
VBD → saw .8
VBD → drank .1
NP → Det N 1.0
Det → a .2
Det → the .8
N → book .1
N → ball .7
N → gin .2
```
Then the lexicalized grammar with uniform probabilities is:
```S → (NP,book) (VP,kicked) .111
S → (NP,book) (VP,kicked) .111
S → (NP,book) (VP,kicked) .111
S → (NP,gin) (VP,kicked) .111
S → (NP,gin) (VP,kicked) .111
S → (NP,gin) (VP,kicked) .111
S → (NP,ball) (VP,kicked) .111
S → (NP,ball) (VP,kicked) .111
S → (NP,ball) (VP,kicked) .111
(VP,kicked) → (VBD,kicked) (NP,book) .033
(VP,kicked) → (VBD,kicked) (NP,gin) .033
(VP,kicked) → (VBD,kicked) (NP,ball) .033
(VP,kicked) → (VBD,kicked) .3
(VP,kicked) → (VBD,kicked) .3
(VP,kicked) → (VBD,kicked) .3
(VP,saw) → (VBD,saw) (NP,book) .033
(VP,saw) → (VBD,saw) (NP,gin) .033
(VP,saw) → (VBD,saw) (NP,ball) .033
(VP,saw) → (VBD,saw) .3
(VP,saw) → (VBD,saw) .3
(VP,saw) → (VBD,saw) .3
(VP,drank) → (VBD,drank) (NP,book) .011
(VP,drank) → (VBD,drank) (NP,gin) .011
(VP,drank) → (VBD,drank) (NP,ball) .011
(VP,drank) → (VBD,drank) .3
(VP,drank) → (VBD,drank) .3
(VP,drank) → (VBD,drank) .3
(VBD, kicked) → kicked 1.0
(VBD,saw) → saw 1.0
(VBD,drank) → drank 1.0
(NP,book) → (Det,a) (N,book) .5
(NP,book) → (Det,the) (N,book) .5
(NP,ball) → (Det,a) (N,ball) .5
(NP,ball) → (Det,the) (N,ball) .5
(NP,gin) → (Det,a) (N,gin) .5
(NP,gin) → (Det,the) (N,gin) .5
(Det,a) → a 1.0
(Det,the) → the 1.0
(N,book) → book 1.0
(N,ball) → ball 1.0
(N,gin) → gin 1.0
```
Sparseness
problems
Maximum likelihood estimates for the PCFG defined by lexicalized splitting are doomed by sparseness: The rule prob for:
VP(dumped,VBD) → VBD(dumped,VBD) NP(sacks,NNS) PP(into,P)
is:
Count(VP(dumped,VBD) → VBD(dumped,VBD) NP(sacks,NNS) PP(into,P)) ----------------------------------------------------------- Count(VP(dumped,VBD))
No corpus is likely to supply reliable counts for such events.
How do we usually solve sparseness problems? We need to make some further independence assumptions!
Collins I
Generation story: Break down a rule into number of contextually dependent subevents which collectively "generate" an occurrence of a rule:
LHS → Ln Ln-1 ... H R1 .... Rn-1 Rn
Basic decomposition: Generate head first, then Li modifers terminating with STOP, then Ri modifiers terminating with STOP. For the tree above we have the following sub event probs
1. P(H|LHS) = P(VBD(dumped,VBD) | VP(dumped,VBD))
2. Pl(STOP | VP(dumped,VBD), VBD(dumped,VBD))
3. Pr(NP(sacks,NNS) | VP(dumped,VBD),VBD(dumped,VBD))
note independent of any left modifiers
4. Pr(PP(into,P) | VP(dumped,VBD),VBD(dumped,VBD)):
note independent of NP(sacks,NNS)
5. Pr(STOP | VP(dumped,VBD), VBD(dumped,VBD),PP)
Let
R = VP(dumped,VBD) → VBD(dumped,VBD) NP(sacks,NNS) PP(into,P))
We estimate the probability of R by multiplying these probs together:
P(R) =
P(VBD(dumped,VBD) | VP(dumped,VBD)) × Pl(STOP | VP(dumped,VBD), VBD(dumped,VBD)) × Pr(NP(sacks,NNS) | VP(dumped,VBD),VBD(dumped,VBD)) ×
Pr(PP(into,P) | VP(dumped,VBD),VBD(dumped,VBD)) × Pr(STOP | VP(dumped,VBD), VBD(dumped,VBD),PP)
State of the art
Using the PARSEVAL metric for comparing gold standard and system parse trees (Black et al. 1991), state-of-the-art statistical parsers trained on the Wall Stree Journal treebank are performing at about 90% precision 90% recall. | 3,465 | 12,029 | {"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} | 4.09375 | 4 | CC-MAIN-2017-39 | latest | en | 0.839717 |
https://documen.tv/question/your-starship-the-aimless-wanderer-lands-on-the-mysterious-planet-mongo-as-chief-scientist-engin-15823138-24/ | 1,719,114,707,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862430.93/warc/CC-MAIN-20240623033236-20240623063236-00413.warc.gz | 188,111,474 | 16,684 | ## Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As chief scientist-engineer, you make the following measurements:
Question
Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As chief scientist-engineer, you make the following measurements: A 2.50 kg stone thrown upward from the ground at 13.0 m/s returns to the ground in 4.50 s; the circumference of Mongo at the equator is 2.00×10^5km; and there is no appreciable atmosphere on Mongo. The starship commander, Captain Confusion, asks for the following information:
a. What is the mass of Mongo?
b. If the Aimless Wanderer goes into a circular orbit 30,000 km above the surface of Mongo, how many hours will it take the ship to complete one orbit?
in progress 0
3 years 2021-07-16T00:09:04+00:00 1 Answers 312 views 0
## Answers ( )
1. Given Information:
Initial speed = v₁ = 13 m/s
time = t = 4.50 sec
Circumference of Mongo = C = 2.0×10⁵ km = 2.0×10⁸ m
Altitude = h = 30,000 km = 3×10⁷ m
Required Information:
a) mass of Mongo = M = ?
b) time in hours = t = ?
a) mass of Mongo = M = 8.778×10²⁵ kg
b) time in hours = t = 11.08 h
Explanation:
We know from the equations of kinematics,
v₂ = v₁t – ½gt²
0 = 13*4.50 – ½g(4.50)²
58.5 = 10.125g
g = 58.5/10.125
g = 5.78 m/s²
Newton’s law of gravitation is given by
M = gC²/4π²G
Where C is the circumference of the planet Mongo, G is the gravitational constant, g is the acceleration of planet of Mongo and M is the mass of planet Mongo.
M = 5.78*(2.0×10⁸)²/(4π²*6.672×10⁻¹¹)
M = 8.778×10²⁵ kg
Therefore, the mass of planet Mongo is 8.778×10²⁵ kg
b) From the Kepler’s third law,
T = 2π*(R + h)^3/2/(G*M)^1/2
Where R = C/2π
T = 2π*(C/2π + h)^3/2/(G*M)^1/2
T = 2π*((2.0×10⁸/2π) + 3×10⁷)^3/2/(6.672×10⁻¹¹*8.778×10²⁵)^1/2
T = 39917.5 sec
Convert to hours
T = 39917.5/60*60
T = 11.08 hours
Therefore, it will take 11.08 hours for the ship to complete one orbit. | 703 | 1,944 | {"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} | 4.21875 | 4 | CC-MAIN-2024-26 | latest | en | 0.816424 |
https://mail.python.org/pipermail/pypy-commit/2006-April/012405.html | 1,516,704,856,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891886.70/warc/CC-MAIN-20180123091931-20180123111931-00262.warc.gz | 743,403,649 | 4,068 | # [pypy-svn] r25724 - in pypy/dist/pypy/objspace/constraint: . applevel test
auc at codespeak.net auc at codespeak.net
Wed Apr 12 16:39:38 CEST 2006
```Author: auc
Date: Wed Apr 12 16:39:36 2006
New Revision: 25724
pypy/dist/pypy/objspace/constraint/applevel/
pypy/dist/pypy/objspace/constraint/applevel/problems.py
pypy/dist/pypy/objspace/constraint/applevel/solver.py
pypy/dist/pypy/objspace/constraint/test/test_solver.py
Modified:
pypy/dist/pypy/objspace/constraint/computationspace.py
pypy/dist/pypy/objspace/constraint/test/test_computationspace.py
Log:
solver instantiation
==============================================================================
--- (empty file)
+++ pypy/dist/pypy/objspace/constraint/applevel/problems.py Wed Apr 12 16:39:36 2006
@@ -0,0 +1,108 @@
+
+def dummy_problem(computation_space):
+ ret = computation_space.var('__dummy__')
+ computation_space.set_dom(ret, c.FiniteDomain([]))
+ return (ret,)
+
+def send_more_money(computation_space):
+ cs = computation_space
+
+ variables = (s, e, n, d, m, o, r, y) = cs.make_vars('s', 'e', 'n', 'd', 'm', 'o', 'r', 'y')
+
+ digits = range(10)
+ for var in variables:
+ cs.set_dom(var, c.FiniteDomain(digits))
+
+ # use fd.AllDistinct
+ for v1 in variables:
+ for v2 in variables:
+ if v1 != v2:
+ '%s != %s' % (v1.name, v2.name))
+
+ # use fd.NotEquals
+ cs.add_constraint([s, e, n, d, m, o, r, y],
+ '1000*s+100*e+10*n+d+1000*m+100*o+10*r+e == 10000*m+1000*o+100*n+10*e+y')
+ cs.set_distributor(di.DichotomyDistributor(cs))
+ print cs.constraints
+ return (s, e, n, d, m, o, r, y)
+
+def conference_scheduling(computation_space):
+ cs = computation_space
+
+ dom_values = [(room,slot)
+ for room in ('room A','room B','room C')
+ for slot in ('day 1 AM','day 1 PM','day 2 AM',
+ 'day 2 PM')]
+
+ variables = [cs.var(v, FiniteDomain(dom_values))
+ for v in ('c01','c02','c03','c04','c05',
+ 'c06','c07','c08','c09','c10')]
+
+ for conf in ('c03','c04','c05','c06'):
+ v = cs.find_var(conf)
+ cs.tell(make_expression([v], "%s[0] == 'room C'" % conf))
+
+ for conf in ('c01','c05','c10'):
+ v = cs.find_var(conf)
+ cs.tell(make_expression([v], "%s[1].startswith('day 1')" % conf))
+
+ for conf in ('c02','c03','c04','c09'):
+ v = cs.find_var(conf)
+ cs.tell(make_expression([v], "%s[1].startswith('day 2')" % conf))
+
+ groups = (('c01','c02','c03','c10'),
+ ('c02','c06','c08','c09'),
+ ('c03','c05','c06','c07'),
+ ('c01','c03','c07','c08'))
+
+ for group in groups:
+ cs.tell(AllDistinct([cs.find_var(v) for v in group]))
+## for v in group])))
+
+## for g in groups:
+## for conf1 in g:
+## for conf2 in g:
+## v1, v2 = cs.find_vars(conf1, conf2)
+## if conf2 > conf1:
+## cs.add_constraint([v1,v2], '%s[1] != %s[1]'% (v1.name,v2.name))
+
+ for conf1 in variables:
+ for conf2 in variables:
+ if conf2 > conf1:
+ cs.tell(make_expression([conf1,conf2],
+ '%s != %s'%(conf1.name(),conf2.name())))
+ return variables
+
+def sudoku(computation_space):
+ cs = computation_space
+ import constraint as c
+
+ variables = [cs.var('v%i%i'%(x,y)) for x in range(1,10) for y in range(1,10)]
+
+ # Make the variables
+ for v in variables:
+ cs.set_dom(v, c.FiniteDomain(range(1,10)))
+ # Add constraints for rows (sum should be 45)
+ for i in range(1,10):
+ row = [ v for v in variables if v.name[1] == str(i)]
+ cs.add_constraint(row, 'sum([%s]) == 45' % ', '.join([v.name for v in row]))
+ # Add constraints for columns (sum should be 45)
+ for i in range(1,10):
+ row = [ v for v in variables if v.name[2] == str(i)]
+ cs.add_constraint(row, 'sum([%s]) == 45' % ', '.join([v.name for v in row]))
+ # Add constraints for subsquares (sum should be 45)
+ offsets = [(r,c) for r in [-1,0,1] for c in [-1,0,1]]
+ subsquares = [(r,c) for r in [2,5,8] for c in [2,5,8]]
+ for rc in subsquares:
+ sub = [cs.find_var('v%d%d'% (rc[0] + off[0],rc[1] + off[1])) for off in offsets]
+ cs.add_constraint(sub, 'sum([%s]) == 45' % ', '.join([v.name for v in sub]))
+ for v in sub:
+ for m in sub[sub.index(v)+1:]:
+ cs.add_constraint([v,m], '%s != %s' % (v.name, m.name))
+ #print cs.constraints
+ return tuple(variables)
+
==============================================================================
--- (empty file)
+++ pypy/dist/pypy/objspace/constraint/applevel/solver.py Wed Apr 12 16:39:36 2006
@@ -0,0 +1,41 @@
+
+class StrategyDistributionMismatch(Exception):
+ pass
+
+
+from collections import deque
+
+class Depth: pass
+
+#-- pythonic lazy solve_all (takes space)
+
+def lazily_iter_solve_all(space, direction=Depth):
+
+ sp_stack = deque([])
+ sp_stack.append(space)
+
+ if direction == Depth:
+ def collect(space):
+ sp_stack.append(space)
+ else:
+ def collect(space):
+ sp_stack.appendleft(space)
+
+ while len(sp_stack):
+ space = sp_stack.pop()
+ print ' '*len(sp_stack), "ask ..."
+ if status == 1:
+ print ' '*len(sp_stack), "solution !"
+ yield space.merge()
+ elif status > 1:
+ print ' '*len(sp_stack), "branches ..."
+ sp1 = space.clone()
+ sp1.commit(1)
+ collect(sp1)
+ sp2 = space.clone()
+ sp2.commit(2)
+ collect(sp2)
+
+solve = lazily_iter_solve_all
Modified: pypy/dist/pypy/objspace/constraint/computationspace.py
==============================================================================
--- pypy/dist/pypy/objspace/constraint/computationspace.py (original)
+++ pypy/dist/pypy/objspace/constraint/computationspace.py Wed Apr 12 16:39:36 2006
@@ -19,8 +19,13 @@
def name_w(self):
return self._space.str_w(self.w_name)
+
+ def w_name(self):
+ return self.w_name
-W_Variable.typedef = typedef.TypeDef("W_Variable")
+W_Variable.typedef = typedef.TypeDef(
+ "W_Variable",
+ name = interp2app(W_Variable.w_name))
#-- Constraints -------------------------
@@ -120,14 +125,18 @@
def w_var(self, w_name, w_domain):
assert isinstance(w_name, W_StringObject)
assert isinstance(w_domain, W_AbstractDomain)
- if w_name in self.name_var:
+ name = self._space.str_w(w_name)
+ if name in self.name_var:
raise OperationError(self._space.w_RuntimeError,
var = W_Variable(self._space, w_name)
self.var_dom[var] = w_domain
- self.name_var[w_name] = var
+ self.name_var[name] = var
return var
+ def w_find_var(self, w_name):
+ return self.name_var[self._space.str_w(w_name)]
+
def w_dom(self, w_variable):
assert isinstance(w_variable, W_Variable)
return self.var_dom[w_variable]
@@ -147,6 +156,9 @@
print "there"
self.sol_set = self._space.call(w_problem, self)
+ def w_set_root(self, w_solution_list):
+ self.sol_set = w_solution_list
+
#-- everything else ---------------
def dependant_constraints(self, var):
@@ -191,13 +203,15 @@
W_ComputationSpace.typedef = typedef.TypeDef(
"W_ComputationSpace",
var = interp2app(W_ComputationSpace.w_var),
+ find_var = interp2app(W_ComputationSpace.w_find_var),
dom = interp2app(W_ComputationSpace.w_dom),
tell = interp2app(W_ComputationSpace.w_tell),
clone = interp2app(W_ComputationSpace.w_clone),
commit = interp2app(W_ComputationSpace.w_commit),
dependant_constraints = interp2app(W_ComputationSpace.w_dependant_constraints),
- define_problem = interp2app(W_ComputationSpace.w_define_problem))
+ define_problem = interp2app(W_ComputationSpace.w_define_problem),
+ set_root = interp2app(W_ComputationSpace.w_set_root))
def newspace(object_space):
Modified: pypy/dist/pypy/objspace/constraint/test/test_computationspace.py
==============================================================================
--- pypy/dist/pypy/objspace/constraint/test/test_computationspace.py (original)
+++ pypy/dist/pypy/objspace/constraint/test/test_computationspace.py Wed Apr 12 16:39:36 2006
@@ -16,7 +16,6 @@
assert str(v).startswith('<W_Variable object at')
raises(Exception, cspace.var, "foo", FiniteDomain([1,2,3]))
-
def test_dom(self):
cspace = newspace()
domain = FiniteDomain([1,2,3])
==============================================================================
--- (empty file)
+++ pypy/dist/pypy/objspace/constraint/test/test_solver.py Wed Apr 12 16:39:36 2006
@@ -0,0 +1,17 @@
+from pypy.conftest import gettestobjspace
+
+class AppTest_Solver(object):
+
+ def setup_class(cls):
+ cls.space = gettestobjspace('logic')
+
+ def test_instantiate(self):
+ from pypy.objspace.constraint.applevel import solver, problems
+ spc = newspace()
+
+ spc.set_root(problems.conference_scheduling(spc))
+ #FIXME: that 'interation over non-sequence' kills me ...
+ #spc.define_problem(problems.conference_scheduling)
+
+ sols = solver.solve(spc)
+ assert str(type(sols)) == "<type 'generator'>"
``` | 2,675 | 9,384 | {"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-2018-05 | longest | en | 0.498665 |
https://www.kidzsearch.com/kidztube/the-physics-of-santa_99f572605.html | 1,624,286,191,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488273983.63/warc/CC-MAIN-20210621120456-20210621150456-00240.warc.gz | 769,042,328 | 29,975 | The Physics of SANTA | Safe Videos for Kids
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# The Physics of SANTA
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## Description
The Physics of Santa. With nearly 2 billion children in the world, how does Santa deliver all those presents on Christmas Eve?
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__________________________________________________________________
So it's that time of year again where we all start wondering, with nearly 2 billion children in the world, how does Santa deliver all those presents?
The speed and payload of Santa and his reindeer would create an enormous amount of air resistance, burning up those reindeer in the same way as a meteor entering the atmosphere. At that speed, Santa would be thrust back in his seat with centrifugal forces several thousand times that of gravity. But this can't be happening, otherwise, where are all the presents coming from?
Some think that Santa must use some kind of heat resistant shield for his reindeer and sleigh, kind of like the ones we use on space capsules to stop them burning up.
Now about the timing, one theory is that if Santa has the technology to create these super heat resistant shields, he should also have technology advanced enough to travel at nearly the speed of light. This actually solves a lot of problems.
1- With Santa travelling at nearly the speed of light he'd be able to deliver all the presents in just 500 seconds.
2- It would explain why Santa never gets any older. Einsteins theory of special relativity tells us that the faster you move, the more time slows down. This is called Time Dilation, and it happens because the speed of light is always constant in a vacuum.
3- It also explains how Santa and his big belly fir down your chimney every year. Special relativity also tells us that objects that travel close to the speed of light experience length contraction (they get skinnier), meaning that if Santa was travelling close to the speed of light he'd easily be skinny enough to fit down your chimney.
4- Finally it explains why no one has ever seen Santa flying across the night sky on Christmas Eve. When certain charged particles travel faster than the speed of light they emit a blue glow, know as Cerenkov (Cherenkov) radiation. So Santa would be seen as streaks of blue light flying across the night sky on Christmas Eve.
Merry Christmas! | 668 | 2,995 | {"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-2021-25 | latest | en | 0.890477 |
https://stats.stackexchange.com/questions/43673/is-it-valid-to-take-a-mean-of-p-values-during-cross-validation-when-comparing-t | 1,721,465,986,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515079.90/warc/CC-MAIN-20240720083242-20240720113242-00655.warc.gz | 478,223,997 | 37,451 | # Is it valid to take a mean of p-values during cross-validation, when comparing the predicted output of a model to the actual output? [closed]
I am doing a cross-validation study, training a model on an input to predict a target.
During training, my model generates an output vector that is guaranteed to be the same size as the corresponding training target vector. I can use metrics such as $R^2$ or RMS error to quantify this.
During testing, my model can produce an output vector that is not the same size as the input (but they are the same order of magnitude). I'm wondering if there are any ways to quantify the similarity between the model output and the testing set targets.
What I've come up with so far is to compare the distributions under the null hypothesis that the model output distribution is the same as the test target distribution. I'm using things like the Kolmogorov–Smirnov test, Ansari-Bradley test, or a permutation test. For each cross-validation fold, there is 1 p-value. Is it valid to report a mean of p-values to summarize this? Or are there better ways to do this?
• why don't you just compute the mean square error or misclassification error of the test data set for each CV fold and then average them?
– matt
Commented Mar 11, 2015 at 11:03
• This question makes no sense to me. How can your model "produce an output vector that is not the same size as the input"? Please explain this in detail. In the meantime, I vote to close as unclear. Commented Feb 15, 2017 at 22:41
• @amoeba I think the cross-validation training fold size or test fold size are what he is talking about, they do not have to be equal sizes.
– Carl
Commented Feb 16, 2017 at 18:50
• @Carl This: During testing, my model can produce an output vector that is not the same size as the input -- does not occur in cross-validation. Commented Feb 17, 2017 at 10:00
## 1 Answer
How about using Fisher's method or Stouffer's method for combining independent p-values to reject the global null hypothesis ? In your case, I reckon the global null hypothesis is the training and test data follows the same distribution. For more information, you can visit the following pages
When combining p-values, why not just averaging?
Combining multiple p-values
• Generally answers are better when they are standalone. Here you're referring to two webpages, and the links may or may not become broken at some point. You could consider expanding your answer to make it standalone without needing to refer to external websites. Commented Oct 20, 2015 at 3:11
• These are not applicable because the different p-values do not come from independent tests! The cross-validation training sets have large overlap. Put differently: If you naively apply Fisher's method you'll get smaller combined p-value the more cross-validation folds you have, as the individual p-values are multiplied in the combined test statistic formula. Commented Jun 17, 2017 at 16:09
• The testing sets are separate for what it's worth. It does feel like collecting all those p-value's might be useful though. If all of the p-value show significance that feels more meaningful than significance from one hold out set. I'm not really sure what you can do apart from take the minimum p-value and then get an underpowered test. Naively I think I would prefer the mean of the p-value to than the p-value from one holdout set. Commented Jan 5, 2022 at 10:15 | 778 | 3,417 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-30 | latest | en | 0.914704 |
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Are x and y both positive?
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21 May 2006, 10:48
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Are x and y both positive?
1, 2x-2y = 1
2, x/y > 1
Please explain if possible
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21 May 2006, 11:23
E
You can plug in some values for x,y both +ve, both -ve and +/e
You can use x=-1/2, y=-1; x=1, y=1/2, OR x=1/4, y=-1/4 for (1)
and x=-1/2, y=-1; x=1, y=1/2 for (2)
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21 May 2006, 12:29
From 1) x = y + 0.5 (x>y)
INSUFF because
y = -1 , x = -0.5 NO
y = 1, x = 1.5 YES
From (2)
x/y > 1
x & y --> both +ve or both -ve
x = -4, y = -3 (x<y)
x = 4, y = 3 (x>y)
INSUFF.
Combining (1) & (2)
x & y have to be positive to satisfy both conditions. Hence C.
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21 May 2006, 12:59
(C) as well. By the same way as giddi.
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21 May 2006, 21:27
St1:
(x-y) = 1/2
Can be both positive: x-y = 1 - 1/2, or both negative: x-y = -1/2-(-1)
Insufficient.
St2:
x>y
Can be both positive or negative. Insufficient.
Using St1 and St2,
We know both must be positive as the second case cannot exist. Sufficient.
Ans C
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15 Jun 2006, 08:03
E.
stmt 1,
x=4, Y=3.5 YES
X=-3 Y=3.5 NO
stmt 2, clearly insuff
combining,
x=4, y=3.5 YES
X=-3.5, Y =-4 NO
hence E.
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15 Jun 2006, 08:49
15 Jun 2006, 08:49
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Are x and y both positive?
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Disclaimer: I did see a previous question touching on something like this but what I’m confused about was NOT addressed so hopefully this is allowed.
They say that the surface area volume ratio limits how big things can grow because surface area scales as a square while volume scales as a cube, so the ratio of volume to surface area goes up as you get bigger. Fair enough. BUT: how is this not just a matter of what units you’re using?
For example, a 1x1x1 ft cube has a surface area to volume ratio of 6sq. Ft to 1 cubic foot, so 6:1. A 1x1x1 meter cube has a ratio of 6:1 too but the units are meters. Couldn’t you always define your units so that you have a 6:1 ratio with any size of cube?
To bring it back to the actual question, wouldn’t your ratio be essentially the same no matter how big your object is? Imagine you expanded everything in the universe by the same amount but kept your unit of measurement the same, you wouldn’t suddenly hit some limit where it stops working right? Does it have something to do with the size of molecules and proteins etc? Please help I am so confused
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No, because you’ll be using the same units for both the regular size, the square, and the cube. If you use different units for each in order to get the same numbers out it doesn’t change the fundamental fact that heat production in a mammal scales with the cube while its ability to dissipate heat scales with the square, so if you double the size of the creature without making any other changes, it will tend to overheat due to producing 8x as much heat but only having 4x the surface area to get rid of it.
Creatures that don’t produce their own heat, such as lizards, don’t have the same problem, which is why dinosaurs were able to get so large. Creatures that live in the sea are also less affected because water is a much more efficient conductor of heat than air is, which is why the largest animal that has ever lived is today’s blue whale.
The surface area to volume ratio is unit-independent, yes. But that doesn’t change the physical meaning of the ratio.
Say that for each cubic foot of living matter, you need 1 pound/min of oxygen. Each square foot can provide 0.3 pounds/min through it. For your 1ft cubic organism, you can provide 6×0.3=1.8 pound/min of oxygen, which is more than the 1 pound/min needed. Good.
Let’s go to the bigger cube. 1 meter = 3.3 ft. 1 cubic meter is 3.3^3 ~ 36 cubic ft, so 36 pound/min of oxygen needed. The surface can provide 6x(3.3×3.3)x0.3 = 19.6 pound/min of oxygen. Not good.
Just playing with units doesn’t change the inherent fact that volume is what determines cellular needs/uptake, while surface area determines nutrient/gas/excretion availability/rate. Since volume scales up faster than surface area, so too does cellular demand scale up faster than nutrient/gas availability.
You seem to have a few misconceptions.
Make cubes with dice.
It has a volume of one and width height and length of one with a surface area of six. (6:1)
Make a 2x2x2 dice and it is two wide, two tall and two deep, it has a volume of eight and surface area of 24 (3:1)
3x3x3 is 3 w/d/t and volume of 27 and a surface area of 54. (2:1)
4x4x4 is 4 w/d/t, volume 64, surface 96 (1.5:1)
Kurzgesagt have a great set of videos on the size of life. I think number 2 covers this pretty well.
Yes, you could, but the physics aren’t unit dependent. The failures here come from things like “the rate at which one inhales has to grow too high for the tissues of the windpipe to survive” and “bones cannot physically support the larger weight”, and in both cases you’re talking about materials strength. Most of the structural properties of materials have units of force per area (=pressure), and area embeds your length unit into the failure point.
As an example, suppose you have a 1 meter square beam that can support 1 GN/m^(2) (i.e., 1 GPa, but I’m intentionally going to use the explicit force and area units here) of weight, with a 1 meter cube on top of it pressing down on it with that maximum of 1 GN of force. Now, double the size of everything. The cube is now 2x2x2 = 8 m^3 and has 8 times the weight, spread across 4 times the area since we scaled up the beam. That means the pressure on the beam is now 2 GN/m^(2), which is beyond its strength and the beam breaks or buckles.
Now, redefine new units. Introduce the splark (abbreviation sp), where 1 splark = 2 meters. The strength of our beam is now 4 GN/sp^(2). In the initial state, the cube is 1/2 spark x 1/2 splark x 1/2 splark and the area of the top of the beam is 1/4 splark^(2), and the cube still applies 1 GN of force to that 1/4 splark^(2), so it’s applying 4 GN/splark^2 – as before, it’s exactly at the limit of our streel beam. Double the length and we now have an 8 GN force from the cube now pressing down on a 1 splark by 1 splark beam, resulting in an 8 GN/splark^2 pressure and failure of the beam.
In either case, the beam fails above a volume-to-area ratio greater than (1 m^3 / 1 m^(2)) = 1 meter (the initial state), which is equivalent to ((1/8) sp^3 / (1/4) sp^(2)) = 1/2 splark. The value of the ratio embeds your unit, and the system fails at the same value (expressed in your unit of choice) regardless of measurement choice.
You’re not using units quite right. You have to compare apples to apples. Let’s use two different units to prove this:
You have a one meter cube. You double its size, and have a two meter cube. Its surface area goes from 6 square meters to 24. Its volume goes from 1 cubic meter to 8. Its volume has increased twice as much as its surface area.
Same cube, let’s use feet instead. 3.28 feet is one meter. The 1M cube has a side length of 3.28 feet. This makes for a surface area of 64.55 square feet, and a volume of 35.287 cubic feet. Scale it up, just like last time, to a side length of 6.56 feet. SA is now 258.2 square feet. Volume is 282.3 cubic feet. If we compare the two, we see that the volume has increased twice as much as the surface area.
This holds true for any unit you choose, so long as you don’t goof up the math during your conversions. The most common such error is forgetting to square during unit conversions. Example:
1 square foot = 12 square inches. This is wrong. A foot is 12 inches, so a square foot is (twelve squared) square inches, or 144 square inches.
Yes, you could. However, because they’re in different units, you would need to convert between them before you can get any meaningful information out of it. If you invent a new unit for each different edge length of a cube, then you have an infinite number of different units, and if you wanted to say anything like “cube X is twice as big as cube Y” you’d first need to re-measure cube Y using the same unit you measured cube X in.
So, a 1x1x1 ft cube does have a total surface area of 6 square feet, and a 1x1x1 meter cube has a total surface area of 6 square meters. But a 1x1x1 ft cube is also a 0.3×0.3×0.3 meter cube with a total surface area of approximately 0.55 square meters, demonstrating that this 6:1 ratio is really just an illusion of the chosen measuring system, not an actual fact of reality. Remember, units are just something we use to be able to talk and think about reality in convenient ways. Measurements aren’t real. The only real thing here is the fact that the surface area of a cube increases exponentially as you increase the edge length linearly, and this is true in every unit you use to describe a cube increasing in size.
Surface area is a large limiting factor for life. And note that surface area is measured in square units, volume in cubed. So when you convert from one unit to another, you need to take that into account. Changing the “starting scale” also changes the ratios.
Say a meter is equal to 3 ft(approximate so that we get whole numbers). A square meter is 9x a square foot, for a total of 54 square foot. A cubic meter is roughly 27x a cubic foot. That’s a area-to-volume ratio of about 2:1, down from the 6:1 ratio of the foot measurements.
Additionally, some processes in bodies don’t scale up or down infinitely. Capillary action stops working in tubes above a certain size. Vacuums cannot lift water above about 14m.
This is all independent of units.
To simplify the whole thing imagine a weight suspended from a rope. The rope is just strong enough to not snap under the weight.
How much weight a rope can carry is largely depended on its diameter.
If you get a rope that is twice as thick in diameter it might carry about four times the weight.
However if you double the size of a weight it will weigh 8 times as much.
For living things that means that if you for example want to scale a human body up to twice its height the bones would be twice as long and have cross section four times as large but their weight would be eight times as much.
A giant human would need much thicker legs than a simple scaled up normal size human.
You also run into problem in other areas like all that stuff about lungs and the heart and the blood and heat dissipation.
There is a reason why large animals look different than small animals would if you scaled them up.
The problem is that the ratio changes as the volume grows, even if you stick with the same units. A 1x1x1 meter cube has a 6:1 ratio, yes, but if we give the cube twice the volume (so it’s a 1.26×1.26×1.26 meter cube now, roughly), then its surface area ends up around 9.5, so the ratio is now something like 4.76:1
Sure, you can say “yes, but if I make up another unit then the ratio between *those* units is what I want it to be”, but that doesn’t change the fact that as the object grew, the ratio between volume and surface changed.
Physics doesn’t care which unit you choose to use. It cares that “compared to when the object was smaller, it now has a smaller surface to volume ratio”
The problems this causes are varied, but it’s not as simple as “suddenly the molecules in my body magically stopped working”. A few examples I can think of are:
1. because elephants are so big, they have volume many, may times greater than that of a human, but their surface area isn’t *as much* greater. This means they already have trouble cooling off. They generate heat throughout their massive bodies, but the heat can only escape through their skin, and compared to how big they are, they don’t have *that* much skin. So if you made an elephant with twice the volume, it would produce twice the heat, but it would not have twice the surface area from which that heat could escape. It would die.
Airplanes would also have problems if you just made them twice as big without changing anything else. Because their volume (and their weight) would double, but the surface area of the wings would not. The wings would need to be made *even bigger*. And of course, even bigger wings would suffer greater stresses and would need to be made stronger.
Plants can only absorb sunlight through their surface area. So if the volume of a plant doubled, but its surface area didn’t, it’d suddenly get relatively less sunlight than it did before.
> For example, a 1x1x1 ft cube has a surface area to volume ratio of 6sq. Ft to 1 cubic foot, so 6:1. A 1x1x1 meter cube has a ratio of 6:1 too but the units are meters. Couldn’t you always define your units so that you have a 6:1 ratio with any size of cube?
This number is not *unitless*. So it’s arguably not accurate to even call it a *ratio*. Your process of reasoning only works if you can cancel out the units on both sides. But when you have “6 square meters : 1 cubic meters”, the best you can do is cancel it to “6 : 1 meter”, or “6 *per meter*”. (Indicative, in this case, of how much extra area you’re getting per specific depth of tissue)
Somehow along your argument you seem to have omitted this unit without justification, which is the reason for your confusion.
The thing you’re forgetting is how area and volume related to the ability of some structure to hold its own weight.
Volume correlates with weight. Switching units of volume isn’t going to change its weight. You can’t make an object lighter or heavier by deciding to measure its volume in different units.
The internal strength of an object correlates with area (specifically cross-sectional area). Switching units of area isn’t going to change its strength. You can’t make an object stronger or weaker by deciding to measure its area in different units.
As an object increases proportionately, its volume (and therefore its weight) increases by the cube of that factor but its area (and therefore its strength) increases only by the square of that factor. So its weight increases much faster than its strength, meaning at some point its weight will exceed its strength. | 3,116 | 12,857 | {"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} | 3.46875 | 3 | CC-MAIN-2022-05 | latest | en | 0.953154 |
http://www.geocities.ws/jsfhome/Think4d/Hyprsphr/hsphtime.html | 1,709,457,953,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476211.69/warc/CC-MAIN-20240303075134-20240303105134-00539.warc.gz | 51,600,379 | 16,039 | ## The Hypersphere and Time
In the previous section we saw how mass and motion worked together to produce the effect we call gravity. Mass makes gravity possible by curving the shape of the space surrounding the body we associate the gravity with. Motion, in the form of the expansion of the universe, provides the means by which curved space acts upon the objects that are falling. Mass and motion also happen to be the key factors involved in terms of how time operates. This would suggest that gravity and time are related at a deeper level: in the theory of relativity we find that conditions that cause changes in gravity cause changes in time as well, as is the case with a black hole. How do mass and motion relate to the concept we call time?
Our universe, the surface of an expanding hypersphere, is in a state of constant motion. We experience this motion as the passage of time. Throughout all regions of the expanding hyperspherical surface we find celestial bodies of all sizes and masses - from planets to stars to black holes. As stated in the previous section, it is the mass of each and every body that causes the body to 'sink down' into the hyperspherical surface, surrounding the body with a 'curved sag' that brings it down to a level below the level of the rest of the surface. Having been brought down to this level, the bodies are carried along with the surface of the universe as it expands, following from behind at a distance determined by how massive the body is: the more massive the body, the farther behind the rest of the expanding surface the body will be found to follow. Though this distance may vary based upon the mass of the body, the rate at which the bodies are carried along behind the expanding surface of the universe is the same for every body: it is a rate that can be observed to be equal to the rate of motion of the expanding hyperspherical surface itself. The reasoning behind this proposal is based upon the simple understanding that when the universe expands, it expands as a single united entity. It is because the universe expands as a 'single united entity' that all points on its hyperspherical surface can be found to engage in motion along with that surface in unison. Because there is but one expanding surface - a single surface that evenly distributes the effects of expansion throughout all points found to lie within it - expansion-related effects observed to be true within any given region of the expanding surface will always be equivalent to the expansion-related effects observed to be true within any other region of the expanding surface.
Given this reasoning, the rate of expansion of the hyperspherical surface of the universe can be considered to be the universal standard in terms of how the flow of time operates - the ultimate source that everything related to time would depend upon to function. What is meant by the term 'universal standard'? Given that the rate of expansion of the universe is a 'universal standard', you may reason, it must therefore also be an unchanging, constant rate of travel. Should the rate of expansion of the universe, then, be considered to be of a constant value? Let us attempt to understand the nature of what being a 'universal standard' means. Our best means of understanding the rate at which the universe is expanding, it would appear, lies in our ability to measure that rate of expansion. What would this process involve? To better understand the nature of this situation, consider the notion that the process we call measurement happens to be based fully upon the element we call comparison: the results of a measurement we make upon any given occurrence will always be found to be dependent upon a comparison performed upon an external, second occurrence. Any and all measurement applied to the occurrence being measured is based entirely upon the information obtained from this second occurrence. Without the information provided by the second occurrence, what is being measured has nothing to which it can compare itself, and hence the process we call measurement cannot take place.
Given the reasoning just presented, we can conclude that in order for something to be measured, there must exist a standard outside of what is being measured to which what is being measured can be compared. Measurement, by its very own nature, then, requires a standard of measurement external and separate from what is being measured, upon which a comparison is performed. We are quite familiar with the process we call measurement: everything that lies within the boundaries of our universe is measurable. This is true because from within the boundaries of the universe, there will always be some external reference available to which what is being measured can be compared. We have developed a vast collection of such units, that allow us to refer to the universe in terms of units that can tell us anything from how old the universe is to the distance across the universe. Take note, however, that if all measurements performed within the universe are ultimately references to a second, external standard, then we will never come across a unit of measurement that isn't dependent upon another supposedly pre-existing unit. Because all units of measurement are ultimately dependent upon another unit to be the units of measurement that they are, then, how will we ever know for sure whether the units we are using are in fact what we believe them to be?
Let this reasoning promote contemplation as to the nature of what being a universal standard means: being a 'universal standard' would involve being the standard upon which all existing standards of its kind within the universe would be dependent to possess the measurable properties that they possess. For example, all units used to measure distance within the universe are ultimately dependent upon the physical size of the universe itself - whatever that may be - to be the distances that they are. Furthermore, all units used to measure the passage of time within the universe are ultimately dependent upon the rate of expansion of the universe itself to be the time-related units that they are. All measurable units within the universe are dependent upon these properties of the universe to be the measurable units that they are. Is it possible, you may ask, for a property of the universe to be measured in the same way that we measure properties within the universe? In order to measure a property of the universe, we would require an instance of that property to exist external to the universe to which comparison could occur. To do so would require of us to relate to the universe from outside its boundaries, which is in essence an impossible task: given that the universe is 'everything', how can anything exist outside the universe?
This explains to us in detail the nature of what being a universal standard means: you cannot measure what you are not outside of. Universal standards exist in the form of the properties of the universe that they are associated with. If we were to attempt to measure a universal standard, our measurement would fail to work, because the measurable units through which we attempt to measure a universal standard, are themselves dependent upon the universal standard we are trying to measure, to be the measurable units that they are! Because the process of measurement requires of us to be external to what we are measuring, our attempt to measure a universal standard would be like comparing the universe to itself. Having been made familiar with the concept of a 'universal standard', we will address a question posed earlier: should a 'universal standard' such as the rate of expansion of the universe be considered to be constant in value? Does the rate at which the universe expands ever change? Because it is impossible to measure a property of the universe, detecting a change in that property is equally impossible. What this means is that if a change in a property of the universe occurred, we would not be able to detect that change: because all measurable occurrences within the universe related to that property of the universe are dependent upon that property to be the measurable occurrences that they are, when the property changes, the occurrences, put simply, change along with it.
What this means is that certain properties of the universe could in fact be rapidly, randomly changing right now, being experienced by us no differently than if they were not changing at all. If change were to occur, we would possess no means of being aware of that change. A universal standard can be considered to be a constant, however, in the sense that we will never detect change in the properties of the universe associated with those universal standards. Let this clarify, then, the role of the expansion of the universe as the fundamental mechanism behind the passage of time. As we have become aware, motion is a key factor concerning how time operates: we experience the motion of the hyperspherical surface of our expanding universe as the passage of time. It is this very expansion that sets the universal standard by which all time-related activity within the universe occurs. There exists, however, a factor responsible for the deviation of the flow of time from its 'universal standard' - a factor mentioned at the beginning of this section as one of the 2 main factors associated with how time operates. This factor is what we call mass.
According to the theory of relativity, mass affects the passage of time. Because this is true, each and every body found below the hyperspherical surface at the bottom of a curved sag can be considered to age more slowly than do things at the level of the surface above: the more massive the body, the more slowly it ages. What causes this deviation of the flow of time from its 'universal standard'? To understand how mass relates to the concept we call time, let us look back to how gravity works. Curved space is an element without which gravity could not exist: it is the curved shape that the surface surrounding a massive body assumes that makes the process of gravity possible. When it comes to matters of time, however, curved space operates by means of a different medium: the "stretching" of its very fabric. To better understand this, consider an example similar to the famous rubber sheet thought experiment: picture a baseball placed onto an outstretched thin latex sheet, held at its edges. Upon sinking down into the sheet, the baseball does more than 'curve' the sheet - it stretches the very fabric of the sheet.
How does the "stretching" of a surface affect the passage of time? The means by which we will answer this question exists in the form of the curved sag shown in the top illustration to the right. As you can see, at the bottom of this rather steep curved sag lies a 1-dimensional massive body - a body rounded by the shape of the curved sag. To better understand how curved space affects the passage of time, we will introduce into the situation a new concept: as the surface of our hyperspherical universe expands outward, leaving behind what we understand to be the 'past' and entering into the 'future' ahead of it, what we are calling the effects of change (the time-related factors that have the potential to affect a body / object) enter directly into the hyperspherical surface, and once within it take immediate effect. A simple portrayal of the process just described can be found in the next illustration to the right. In order for the effects of change to reach a body below the level of the hyperspherical surface (at the bottom of the curved sag that surrounds it), however, the effects of change must pass through the curved space in between the body and the surface. This transition from surface to body is made possible by the effects brought upon by the constantly expanding hyperspherical surface: as the effects of change continue to directly enter into all available areas of the hyperspherical surface, the constant incoming flow of the effects of change causes the effects of change to spread out in all directions within the surface. Before long, the effects of change can be observed to have begun to move past the outer edges of the curved space surrounding the body, entering in from the level of the hyperspherical surface. Soon, the 'current' brought upon by the incoming flow of the effects of change has brought the effects of change past the outer edges of the curved space surrounding the body and down into the very curved space surrounding the body they are attempting to reach - as is shown in the bottom illustration to the right.
When the effects of change enter into the curved space, they are altered by the physical configuration of the curved space. The nature of this alteration is portrayed in detail in the bottom illustration to the right. The effects of change are "stretched out" by the very fabric of the curved space, and begin to assume a distinct 'spread out' state. This state of being 'spread out' is the mechanism behind how the flow of time deviates from its 'universal standard'. In what way does the effects of change being in this state affect the aging process of the body? Because the flow of the effects of change is reaching the body in a "stretched out" state, it takes longer for any given amount of the flow to reach the body, than if the flow of the effects of change were not "stretched out". In effect, the flow of the effects of change reach the body at a slower rate, and hence the body ages more slowly. This alteration of the flow of the effects of change can also be considered to apply to objects lying within the curved space itself (such as a clock placed there): the flow of the effects of change, having been altered, will be found to pass through any given area of the curved space at a slower rate, affecting the passage of time in that area accordingly. Take note that the more massive the body is, the more deeply it will 'sink down' into the hyperspherical surface, and hence the greater the degree to which the curved space surrounding the body will "stretch out" the effects of change.
According to the theory of relativity, objects in a state of motion undergo the passage of time more slowly than objects not in motion. The effect, however, is only noticable when the object is moving at near-light speed. Why does being in a state of motion cause this alteration of the passage of time to occur? Throughout this section, we've been made familiar with the concept of the motion of a body that exists in the form of the motion the body undergoes while being carried outward along with the hyperspherical surface of the expanding universe. The decrease in the rate of aging that can be observed to occur due to travel at near-light speed, however, deals with a different type of motion: motion of an object along a direction within the surface itself, as it expands. It is upon this type of motion that we are to focus our attention. As the means of doing so, first consider the notion that all objects possess mass. Because of this, motion of an object within a surface will always require energy / effort. The faster the rate at which the object is put into motion within the surface, the more energy / effort there will be that is required to move it. In this sense the act of putting an object into a state of motion in a direction within a surface is like increasing the object's mass: the faster the rate at which the object travels, the more massive the object can be considered to become. This answers why objects in a state of motion age more slowly: they have become more massive! In effect, the object in motion ages more slowly, for the same reasons described above explaining why mass affects the passage of time. In the same manner that a massive body was described to 'sink down' into the hyperspherical surface, the object in motion within the surface 'sinks down' into the surface within which it is travelling. The curved sag surrounding the object 'follows' the object wherever it goes, in a manner comparable to a bowling ball being pushed across a waterbed: though the object moves, the curved sag moves along with it. The faster the rate at which the object travels, the greater the extent to which the rate of aging can be considered to decrease.
The theory of relativity also states that according to a stationary observer, clocks in a state of motion will be observed to tick more slowly than clocks not in motion. We are to assume that we are that stationary observer. Understanding the current topic of discussion does not require us to go beyond anything that we've already covered. Assume that there exists a clock in a state of motion, travelling outward from us into space. This clock sends out a beam of light once every second. Because this clock is in motion, when we receive the ticks it is sending out, we measure the ticks to occur at a rate observed to be slower than the rate at which we observe the ticking of one of our own clocks that is set to tick once every second. Why does this occur? As we've been informed, putting an object such as a clock into a state of motion is comparable to increasing the clock's mass: the faster the rate at which the clock travels, the more massive the clock can be considered to become. As a direct result of the fundamental behavior of mass, the mass of the clock causes the clock to 'sink down' into the surface within which it is travelling, surrounding the clock with a 'curved sag' that moves along with the clock as the clock engages in motion. Understanding what is responsible for the observation of the ticking of a clock in motion to tick more slowly than usual involves referring to the concepts presented in the material just covered. The concept we are to address is that of how mass affects the passage of time.
As you may recall, the first step in how the rate of aging of an object in motion is decreased consists of the flow of time entering into the hyperspherical surface of the outwardly expanding universe, and spreading out into all directions within the surface. In order to reach an object at the bottom of a curved sag, the flow of time must pass through the stretched out surface of that curved sag lying in between the object and the hyperspherical surface. Once the curved sag is reached, the physical configuration of the curved sag then begins to alter the flow of time as it passes through: after having crossed the distance of the curved sag lying above the object, the flow of time reaches the object at the bottom of the curved sag in a distinct "stretched out" state - a 'spread out' version of the time-related factors entering into the curved sag from the level of the hyperspherical surface. Given the effects of this 'spread out' state, the flow of time can therefore be observed to take longer to reach the object at the bottom of the curved sag than if not "stretched out", and as a result the flow of time reaches the object at the bottom of the curved sag at a slower rate.
After putting this into consideration, we can come to the prompt conclusion that curved space has the ability to alter information passing through it. The 'information' spoken of here is, of course, the time-related factors responsible for the rate of aging of bodies and objects. What other forms could information take on? A beam of light has the potential to convey information. In fact, the beam of light that the clock described above sends out every second is information. What if the beams of light being emitted from the clock in motion once every second, themselves information, when passing through the curved space surrounding the clock in motion, behaved in the same manner that the flow of time did? We would have before us the explanation as to why the observation of clocks in motion to tick more slowly than clocks not in motion: the continuous stream of the beams of light being sent out by the clock every second, upon passing through the stretched surface of the curved sag, exits the curved sag of the clock in a "stretched out" state. The beams of light travel through space in this "stretched out" state, and upon their arrival result in our experiencing the ticking of the clock more slowly than normal. Let us further examine the factors at work: the process we know as decrease in the rate of an object's aging occurs as the result of an information-containing signal (the flow of time) entering into a curved sag. The process lying behind why clocks in motion are observed to tick more slowly than normal, in turn, occurs as the result of an information-containing signal (a beam of light) being sent out from a curved sag. In each case, the signal covers its distance across the stretched surface of the curved sag, and upon emerging altered assumes the change in properties that the curved sag has brought about.
Let us put into consideration a totally alternate situation: assume that we are in possession of a clock equal in nature to the clock in motion that we are observing. Let us assume that moving along with this clock at the bottom of its curved sag is a person observing the ticking of our clock. How would this person measure the ticking of our clock? Being stationary, we are not surrounded by a curved sag. Yet the person travelling along with the clock in motion, it would happen, measures our clock to tick more slowly than normal, even though we are stationary! How can this be? What is there to slow down the signals sent out from our clock as it ticks, if we are not surrounded by a curved sag? This is a valid argument. Everything we've covered so far has told us that observation of a clock in motion to be ticking more slowly than normal is the result of the curved sag that surrounds the clock. Yet the person observing the signal sent out from our clock measures the ticking to occur more slowly than normal. What we know, first of all, is that something in this situation is the cause of the observed slow ticking of the clock, and because the cause is not where we expect it to be, it must therefore be somewhere else. Since curved sags are responsible for observed slow ticking of clocks, the cause of the observed slow ticking in this situation must therefore be the curved sag of the person in motion with the clock observing us. This is quite possible.
You see, just as an outgoing beam of light sent out from a clock in motion at the bottom of a curved sag must first pass through the curved sag before reaching the external world, an incoming beam of light approaching an observer in motion at the bottom of a curved sag must first pass through the curved sag before coming in contact with the observer. As a result, the curved sag surrounding the observer in motion brings about the same effect that the curved sag of the clock in motion would bring out if the observer were stationary. We can therefore conclude that curved space slows down incoming beams of light in the same manner that it slows down outgoing beams. Assume, then, that our clock is sending out the signal observed from the outside as ticking. Upon reaching the curved sag of the person in motion with the clock observing us, the signal from our clock passes through the curved space that the person is surrounded by on all sides, "stretching out" the signal and presenting it to the person in an altered state.
As you may recall being stated earlier, we experience the state of constant expansion that the surface of our hyperspherical universe is engaged in as the passage of time. It is no surprise, then, that the effects of motion are the foundation upon which time operates, for time is motion. Our universe, then, is a 4-dimensional object constantly increasing in size as time goes by. What this means is that the physical size of the universe, and the passage of time, are directly related. How would one measure the size of the universe? The obvious choice is by means of the size of its radius: like the circle and the sphere, the hypersphere's lower-dimensional analogues, the distance from the centerpoint of a hypersphere to any point on its surface will always be the same, no matter what point on its surface you choose. The radius of the universe can in a sense be used as a sort of "clock", you see, that can mark specific points in time of the expansion of the universe. How, you may ask, does this "clock" work? For any given point in time during the time that the universe has been expanding, the hyperspherical radius at that point in time will be of a certain specific length. To 'access' any point in time during the expansion of the universe, we simply designate the size of the radius. This method of measurement will never fail, for the simple reason that no two instants in time will share a radius of the same length: each instant is designated a radius of a length that is unique to that instant in time.
How would we visualize the radius of a hypersphere? Displayed to the left are representations of the radius of a sphere, each radius presented at a right angle to the other. The sphere on top portrays the positions of each radius of the sphere as we would think of them. The sphere under the one on the top, as you can see, is in stack-diagram form and represents a Flatlander's basic understanding of how each and every instance of the radius of the sphere portrayed in the illustration above is positioned. The Flatlander is familiar with the 4 positions of the radius that occur on the central slice of the sphere - they extend parallel to the central slice itself. The remaining 2 positions of the radius, however, extend perpendicular to the central slice, and hence to the Flatlander's entire dimension. Though these 2 remaining positions of the radius extend forth in ways that the Flatlander cannot directly comprehend, the medium of the stack-diagram allows the Flatlander to grasp each radius in the form of cross-sections extending into slices that lie across multiple 2-dimensional planes.
Displayed to the left in the stack-diagram on the bottom are representations of the radius of a hypersphere, each radius presented at a right angle to the other. The hypersphere, displayed in stack-diagram form, represents our basic understanding of the positions that the radius of a hypersphere can assume. We are familiar with the first 6 positions of the radius that occur on the central slice of the hypersphere that extend parallel to the central slice itself. The remaining 2 positions extend perpendicular to the central slice, and hence to our entire dimension. Though these 2 remaining positions of the radius extend forth in ways that could be considered difficult to visualize when approached from the plane of the central slice, the medium of the stack-diagram allows us to grasp each radius in the form of cross-sections extending into slices that lie across multiple 3-dimensional planes.
We have just completed a detailed study of how the radius of our expanding universe increases as time goes by. Having been made familiar with the concept of an expanding universe, we will now go about approaching a method of study that relates to the arrangement of bodies within an expanding universe: a concept known as the Hubble law. The Hubble law states that distances of empty space in between bodies / galaxies increase as time goes by. According to the observations that have been made of the universe around us - the very observations upon which the Hubble law is based - everything in the universe is moving away from us at a speed determined by the distance of the body / galaxy away from us: the more distant the body / galaxy, the faster the rate at which it will be found to moving away. It must be realized, however, that throughout this process of 'spreading out', the bodies will always be in a state of even distribution throughout the universe: at any time during the time that the universe has expanded (or will expand in the future), the amount of empty space found to surround any given body will in general be pretty much the same amount of space one would find surrounding any other body in the universe. What this means is that as the bodies undergo the effects of expansion, they undergo the effects of expansion in unison. In order to better understand the Hubble law, there are 2 models we will be studying that simulate the conditions related to the Hubble law as we observe them to occur within the universe. The first model informs us of the basic concept behind the Hubble law. The second model, in turn, is an attempt to explain the actual mechanism behind how the Hubble law operates.
As the means of taking part in the first model, picture a round lump of dough. Assume that we have mixed an assortment of raisins into the lump of dough, evenly distributing the raisins throughout all areas of the dough into which mixture can occur. We then roll the dough into its original rounded state, and put the lump of dough into an oven and begin to bake it. Our job as of now is to examine how the raisins distribute themselves as the dough rises. As the dough rises, we find distances in between raisins to increase in unison. To better understand this expansion, we are to focus our attention on 2 raisins on opposite sides of the outer edge of the lump of dough, and on a raisin in between these 2 raisins in the direct center of the lump of dough. As the dough rises, we find that the 2 raisins on opposite sides of the dough are moving away from each other twice as fast as either of these raisins is moving away from the raisin in the middle. The reason for this occurrence is quite simple: there is twice as much expanding dough in between the 2 outer raisins, than there is in between either outer raisin and the raisin in the middle. If we assume this relationship to apply to all of the raisins within the rising dough, then we would have before us a miniature model of our expanding universe. Upon examining the rising dough more closely, we find that from the point of view of any given single raisin located within the dough, it's as if every other raisin were moving away from that particular raisin. After further examining the expanding cluster of raisins embedded within the rising dough, we can in response to that examination conclude that there is no observable center to the expansion that is occurring throughout this enlarging cluster of raisins. Having been informed as to the basic concept behind the Hubble law, let us take this arrangement a step further.
The second model we will study addresses an aspect to the expansion of the universe not put into consideration in the first model - an aspect that is vital to the actual mechanism behind how the Hubble law operates. This approach treats the universe not as an 'enlarging solid mass' as the 'raisin' model does but as an outwardly expanding surface - a surface that surrounds the very central point away from which it is expanding. The first model was correct in its conclusion that there is no observable center to the expansion that is occurring throughout the universe. The manner in which it treated the universe as an 'enlarging solid mass', however, was a prominent error. The universe is not an expanding 'giant sphere' of empty space, but is the surface of an expanding hypersphere. As the means of taking part in the second model, we are to picture an inflated balloon. Using a marker, we place 'dots' upon the surface of the balloon in a manner that evenly distributes the dots across the surface of the balloon. We are then to deflate the balloon, and then inflate it to a state at which there is an observable spherical contour to its shape.
As the means of demonstrating the actual mechanism behind how the Hubble law operates, we proceed to continue to inflate the balloon. We are now to examine how the dots position themselves in relation to each other along the surface of the balloon as the balloon inflates. As the balloon inflates, we find distances in between dots along the surface of the balloon to increase in unison. Upon examining the surface of the inflating balloon more closely, we find that from the point of view of any given single dot located on the surface of the expanding balloon, it's as if every other dot (in terms of how they relate to the dot along the surface of the balloon) were moving away from that particular dot. As of the surface of the inflating balloon, there is no observable center to the expansion that we witness to be occurring. As with the 'raisin' model, the rate at which any 2 given dots on the surface of the balloon will be found to be moving apart (in terms of how they relate to each other along the surface of the balloon) is directly related to the distance along the surface of the balloon in between those 2 dots. In what way, you may ask, does what is described here differ from the behavior observed to occur earlier among the raisins distributed throughout the rising dough?
In the first model, the raisins were being pushed apart. In the model we are dealing with now, the dots are simply moving in straight lines. These 'straight lines' are none other than the paths that the dots follow as they move away from the centerpoint of the balloon - the point found to lie at the very center of the empty space that the surface of the balloon surrounds. As the balloon is inflated, the dots move in distinct straight lines away from this point. When the outwardly extending paths of all of the dots are brought inward, it is at the location of the very centerpoint of the balloon just described that we witness the paths of the dots to converge. This is the actual mechanism behind how the Hubble law operates: the bodies of our universe, you see, are not being "pushed apart". Rather, the increase in the empty space surrounding any given body occurs simply because the bodies are moving directly outward from their centerpoint of expansion, in the straight lines that the effects of expansion carry them. In effect, the bodies are not actually moving away from each other, but rather from the point at the very center of the expansion that is occurring - the single common point that all of the inwardly brought paths of the bodies can be observed to share.
As we are already familiar, the Hubble law states that distances in between bodies increase as time goes by. Given this reasoning, consider the following notion: if empty space in between bodies is growing (and if the radius of the universe is getting larger), wouldn't that suggest that there was a time in which there was no distance in between bodies (and no measurable radius)? Yes. The universe, at this time, existed in the form of a dense, compressed dimensionless point, and it was at this very location that the process of expansion started - an instant that can be considered to be nothing other than the beginning of time - the "big bang". At this state before the universe began, there existed no such concepts as 'expansion' or 'spreading out': motion did not exist. If we accept this as true, then neither did time exist, for time is motion.
Where within our 3-dimensional universe, then, could it be said that the event known as the "big bang" took place? The point of origin from which the universe sprang cannot, by its very own nature, be itself involved in any way with the expansion that is occurring around it: it must exist independently of the influence of anything that could be said to be expanding. As the means of better grasping this concept, let us return to the 'models' of the Hubble law presented earlier. Let these two models represent 2 different conceptions of our expanding universe. Both models, as one can see, are 3-dimensional expanding spheres. The first model - the 'raisin' model - is solid. The second model - the 'balloon' model - is hollow. The type of cosmos each model represents is observably different from the other. Both spherical models, being spheres, possess a 3-dimensional spherical centerpoint.
In the 'raisin' model, this centerpoint exists at a location occupiable from within the cosmos that the model represents - the centerpoint physically lies within that cosmos. In the 'balloon' model, however, the status of the spherical centerpoint is different: the centerpoint lies in a location unreachable from the cosmos that the model represents - the centerpoint is physically external to that cosmos and is surrounded on all sides by the surface moving outward from it. Assume, then, that we have the task of designating the point on the surface of the balloon that best portrays the location of the balloon's centerpoint. Upon attempting to perform this task, it becomes quite clear that designating the location of the balloon's centerpoint by means of a location on the surface of the balloon is impossible. We are now to apply this to our own universe. No occupiable location within our universe can successfully designate the location of the universe's hyperspherical centerpoint - the very point at which the "big bang" took place. Determining the location of the hyperspherical centerpoint of the universe is in essence impossible, when the attempt to refer to the hyperspherical centerpoint is made from within the hyperspherical surface that surrounds it! As you may recall from an earlier section, it is impossible to relate to the external configuration of a surface that you yourself exist within.
To answer the question of where within our 3-dimensional universe the event known as the "big bang" took place, we must view the hyperspherical surface of our universe as it appears from outside its boundaries. The stack-diagram presented below represents such an approach. The centerpoint of the hypersphere - at the centerpoint of the hollow central spherical slice - is fixed: it is the very distinct, very defined point of origin from which the universe sprang. As can be seen by observing the expanding surface of the hypersphere, there exists no single point within that surface that could be said to be the center of the expansion that is occurring - which is in precise accordance with the rules set down by the Hubble law. Bodies within the hyperspherical surface are not being "pushed apart" as one would conclude based upon the reasoning of the 'raisin' model (a manner of reasoning typical of an attempt to relate to the bodies from within the surface in which they lie). Rather, the increase in the empty space surrounding any given body occurs simply because the bodies are moving directly outward from their centerpoint of expansion, in the straight lines that the effects of expansion carry them.
The hypersphere is visualizable! If you've gotten this far you know this as a fact! Step by step we have witnessed that the fourth dimension is not as incomprehensible as we would be led to believe! What made this approach to the fourth dimension unique? This approach to the fourth dimension is unique in the sense that it involved no math whatsoever! As you are probably already aware, people tend to rely heavily upon mathematics when it comes to dealing with matters of the fourth dimension. Not so here. The tool of visualization used here was not math, but analogy: we assumed that what is true for lower dimensions must also be true for our own. It's that simple. The beauty of this approach is that it removes all unnecessary complication: we need not go about making guesses about what we think the fourth dimension may be, when the work is already done in the form of the spatial concepts observable in lower dimensions.
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https://www.physicsforums.com/insights/useful-integrals-quantum-field-theory/ | 1,702,130,156,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100912.91/warc/CC-MAIN-20231209134916-20231209164916-00834.warc.gz | 1,036,537,922 | 24,401 | # Some Useful Integrals In Quantum Field Theory
Common Topics: integral, zero, axis, along, infinity
In my paper on renormalisation I mentioned what most who have studied calculations in Quantum Field Theory find, it’s rather complicated and mind-numbing. I am not sure doing these is that illuminating of anything, but every now and then you got to do, what you got to do, so I would like to go through one of those, Pauli–Villars regularization, as an example of the general theory in my second paper.
This makes use of a number of multidimensional integrals which is the subject of this paper.
## The Gaussian Integral
The most basic of the integrals considered here is the good old Gaussian integral which you likely have seen before, but will do it anyway. It’s:
## I = \int_{-∞}^{∞} e^{-x^2} dx##
Here is the trick:
## I^2 = \int_{-∞}^{∞} e^{-x^2} dx \int_{-∞}^{∞} e^{-y^2} dy = \int_{-∞}^{∞} \int_{-∞}^{∞} e^{-(x^2+y^2)} dx dy##
Going to polar coordinates:
## I^2 =\int_{0}^{∞} \int_{0}^{2\pi} e^{-r^2} r drdθ = 2\pi \int_{0}^{∞} e^{-r^2} r dr ##
Substituting ## s = r^2 ##
## I^2 = \pi \int_{0}^{∞} e^{-s} ds = \pi ##
Hence ## I = \int_{-∞}^{∞} e^{-x^2} dx =\sqrt {\pi} ##
## The Gamma Function
Another function that tends to crop up is the so called Gamma Function ## \Gamma(x) ## defined by:
## \Gamma(x) = \int_{0}^{∞} t^{x-1} e^{-t} dt ##
It has two rather interesting properties:
Using integration by parts ## \Gamma(x+1) = x \Gamma(x) ##. But since ## \Gamma(1) = 1 ## we have ## \Gamma(n+1) = n! ##. The Gamma function generalises the factorial function.
## \Gamma ( \frac{1}{2} ) = \int_{0}^{∞} t^{-\frac{1}{2}} e^{-t} dt##. Substitute ## t = u^2 ##. ## \Gamma ( \frac{1}{2} ) = 2 \int_{0}^{∞} e^{-u^2} du = \int_{-∞}^{∞} e^{-u^2} du = \sqrt {\pi} ##
## Generalisations To Arbitrary Dimensions
One interesting thing about the Gaussian integral is it easily generalises to arbitrary dimensions because:
## e^{-k^2} = e^{-k_1^2} e^{-k_2^2}……..e^{-k_d^2} ##
so:
## J = \int_{-∞}^{∞} d^dk e^{-k^2} = {\pi}^\frac{d}{2} ##
Now lets see how this helps in a more general case where ## F ## is any reasonable function of ## k^2 ##:
## I = \int_{-∞}^{∞} d^dk F({k^2}) ##
We could go and look it up in some tome, or feed it into a math package, but I think most into math dread the – and it can be shown – and would like to know why. With that in mind, let’s see how far we can get with just the above. As you probably have guessed there is a trick involved.
To do the integral we, similar to polar coordinates, go over to generalised spherical coordinates in arbitrary dimensions. First have a look on Wikipedia at what generalised spherical coordinates look like:
N Sphere
The associated volume element is:
## d^dV = k^{d-1} sin^{d-2}(Θ_1) sin^{d-3}(Θ_2) …… sin (Θ_{d-2}) dk dΘ_1 dΘ_2 ….. dΘ_{d-1} ##
Plugging this into the the integral, ## I ##, we have:
## I = C(d) \int_{0}^{∞} dk k^{d-1} F({k^2}) ##
where ## C(d) ## is some constant depending only on the dimension.
You probably have guessed the trick – since we know ## J ##, we can determine ## C(d) ##:
## J = {\pi}^\frac{d}{2} = C(d) \int_{0}^{∞} dk k^{d-1} e^{-k^2} = \frac{C(d)}{2} \int_{0}^{∞} dt t^{\frac{d}{2}-1} e^{-t} = \frac{C(d)}{2} \Gamma (\frac{d}{2}) ##
A substiution ## t = k^2 ## was used. Hence:
## C(d) = \frac {2 \Pi ^{ \frac{d}{2}}}{\Gamma (\frac{d}{2})} ##
And finally we have:
## I = \frac {2 \Pi ^{ \frac{d}{2}}}{\Gamma (\frac{d}{2})} \int_{0}^{∞} dk k^{d-1} F({k^2}) ##
The above formula is useful for what’s called dimensional regularisation, but most of the time we are interested in the the case when ## d=4 ##:
## I = 2 \Pi ^2 \int_{0}^{∞} dk k^3 F({k^2}) = \Pi ^2 \int_{0}^{∞} dk^2 k^2 F({k^2}) ## ## (1) ##
## Wick Rotation
The above is interesting, but unfortunately as it stands isn’t much use for calculating integrals that occur in Quantum Field Theory. That’s because for those integrals, the ##k^2## in ##F({k^2})## is not the Euclidean norm:
## k^2 = k_1^2 + k_2^2 + k_3^2 + k_4^2 ##
It’s the relativistic one:
## k^2 = k_0^2 – k_1^2 – k_2^2 – k_3^2 ##
This is where we need another trick. Suppose we have ## F(k^2) = F( k_0^2 – k_1^2 – k_2^2 – k_2^2) ##. Then:
## \int_{-∞}^{∞} d^4k F({k^2}) = \int_{-∞}^{∞} d^3k \int_{-∞}^{∞} dk_0 F( k_0^2) ##
The integral we are interested in is ## \int_{-∞}^{∞} dk_0 F( k_0^2) ##.
Many integrals are easier when done in the complex plane, so first, we consider it a complex function and integrate along the real axis. We suppose that at infinity ## F ## is zero and any poles occur in the 2nd and 4th quadrant. This is usual for the functions we work with.
First consider the following contour integral in the first quadrant, from zero along the real axis to a large positive value, then a circular arc to the positive imaginary axis, then, from a large positive value on the imaginary axis, along the imaginary axis, back to zero. Since there are no poles the Residue Theorem says this is zero. Along the arc, ## F ## is zero, so the integral from zero to infinity along the real axis is the same as the integral along the imaginary axis from zero to infinity. Actually, we would need to show as the radius of the arc goes to infinity the integral along the arc goes to zero, but I won’t be that rigorous. It’s an interesting exercise showing this in the examples later, and the interested reader may like to chug through it. Doing a similar thing in the 3rd quadrant, going from zero to a large negative value along the real axis, along a circular arc to the negative imaginary axis, then from a large negative value on the imaginary axis to zero. Again the integral along the arc is zero, so the integral from minus infinity along the real axis is the same as along the imaginary axis from minus infinity to zero.
## \int_{-∞}^{∞} dk_0 F( k_0^2) = \int_{-i∞}^{i∞} dk_0 F( k_0^2) = i \int_{-∞}^{∞} dk_4 F((i k_4)^2) ##
Where the substitution ## k_0 = i k_4 ## has been made. This has now put the integral in the form with the usual Euclidean metric because ## ( i k_4) ^2 = – k_4^2 ## so:
## \int_{-∞}^{∞} d^3k \int_{-∞}^{∞} dk_0 F( k_0^2) = i \int_{-∞}^{∞} d^3k \int_{-∞}^{∞} dk_0 F( – k_4^2) = i \int_{-∞}^{∞} d^4k’ F(-k’^2) = i \Pi ^2 \int_{0}^{∞} dk’^2 k’^2 F(-{k’^2})## ## (2) ##
Where, because ##k’## has the ordinary Euclidean metric ## k’^2 = k_1^2 + k_2^2 + k_3^2 + k_4^2 ##, we have used equation ## (1) ##.
This trick of changing the integration variable of ## k_0 ## to ##ik_4## is called Wick Rotation.
## Two Examples
An integral that pops up in the large momentum approximation of the meson ## Φ^4 ## theory is actually simple and illustrative. It’s:
## \int_{-∞}^{∞} d^4k \frac{1}{k^4} ##
First we need to show it obeys the assumptions in deriving formula ##(2)## ie ##F (k_0^2) ## is zero at infinity and has no poles in the first or third quadrant.
##F (k_0^2) = \frac{1}{(k_0^2 – k_1^2 – k_2^2 – k_2^2)^2} = \frac{1}{(k_0^2 – k^2)^2}##
This goes to zero at infinity, but has poles on the real axis at ## ± k ##. To get around this we modify it:
##\frac{1}{(k_0^2 – k^2 + iε)^2}##
Here ##ε## is a positive infinitesimal. It shifts the poles to the 2nd and 4th quadrants. In equation ##(2)## we take ## ε→0##.
## \int_{-∞}^{∞} d^4k \frac{1}{k^4} = i \Pi ^2 \int_{0}^{∞} dk’^2 \frac{1}{k’^2} = ## ##Λ^2 → ∞ ## ## i \Pi ^2 \ln (Λ^2) ##
This is the origin of the equation requiring renormalisation in the first paper.
The previous integral was an approximation. Removing that requires some tricky math, which will be the subject of another paper where Pauli–Villars regularization will be used, and the following integral will be found useful:
## \int_{-∞}^{∞} d^4k \frac{1}{(k^2 – c^2)^3} ##
Checking if it obeys our assumptions:
##F (k_0^2) = \frac{1}{(k_0^2 – k^2 – c^2)^3}##
It goes to zero at infinity, but again it has poles on the real line. We use the same trick to shift them to the 2nd and 4th quadrant, then take ## ε→0##.
## \int_{-∞}^{∞} d^4k \frac{1}{(k^2 – c^2)^3} = -i \Pi ^2 \int_{0}^{∞} dk’^2 \frac{k’^2}{(k’^2 + c^2)} ##
##\int_{}^{} dx \frac{x}{(x^2 + c^2)} = – \frac{c^2 + 2x}{2 (c^2 + x^2)}##
## -i \Pi ^2 \int_{0}^{∞} dk’^2 \frac{k’^2}{(k’^2 + c^2)} = i \Pi^2 \frac{c^2 + 2k’^2}{2 (c^2 + k’^4)^2} \Big|_0^∞ = \frac{ -i \Pi^2 }{2 c^2}##
## Conclusion
I wanted this to be about some of the tricks used in evaluating the integrals in Quantum Field Theory and renormalisation, and I think I have done that. But while doing it I noticed something and maybe you did too. We had all these things appearing you wouldn’t think had anything to do with it.
The gernralisation of the factorial function, the gamma Function.
π appeared all over the place
An apparently real integral turns out secretly to involve i, simply by using a slightly different metric.
What going on, it’s all very strange. Well, Wigner wrote an essay about it:
The Unreasonable Effectiveness of Mathematics in the Natural Sciences
Thought-provoking.
2 replies | 3,234 | 9,030 | {"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} | 3.703125 | 4 | CC-MAIN-2023-50 | latest | en | 0.712888 |
https://nrich.maths.org/public/leg.php?code=14&cl=3&cldcmpid=845 | 1,510,953,816,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934803944.17/warc/CC-MAIN-20171117204606-20171117224606-00683.warc.gz | 683,493,644 | 6,132 | # Search by Topic
#### Resources tagged with Prime factors similar to Powerful Factorial:
Filter by: Content type:
Stage:
Challenge level:
### There are 14 results
Broad Topics > Numbers and the Number System > Prime factors
### Powerful Factorial
##### Stage: 3 Challenge Level:
6! = 6 x 5 x 4 x 3 x 2 x 1. The highest power of 2 that divides exactly into 6! is 4 since (6!) / (2^4 ) = 45. What is the highest power of two that divides exactly into 100!?
### Factoring Factorials
##### Stage: 3 Challenge Level:
Find the highest power of 11 that will divide into 1000! exactly.
### Gaxinta
##### Stage: 3 Challenge Level:
A number N is divisible by 10, 90, 98 and 882 but it is NOT divisible by 50 or 270 or 686 or 1764. It is also known that N is a factor of 9261000. What is N?
### Thirty Six Exactly
##### Stage: 3 Challenge Level:
The number 12 = 2^2 × 3 has 6 factors. What is the smallest natural number with exactly 36 factors?
### One to Eight
##### Stage: 3 Challenge Level:
Complete the following expressions so that each one gives a four digit number as the product of two two digit numbers and uses the digits 1 to 8 once and only once.
### Flow Chart
##### Stage: 3 Challenge Level:
The flow chart requires two numbers, M and N. Select several values for M and try to establish what the flow chart does.
### Just Repeat
##### Stage: 3 Challenge Level:
Think of any three-digit number. Repeat the digits. The 6-digit number that you end up with is divisible by 91. Is this a coincidence?
### Fac-finding
##### Stage: 4 Challenge Level:
Lyndon chose this as one of his favourite problems. It is accessible but needs some careful analysis of what is included and what is not. A systematic approach is really helpful.
### Different by One
##### Stage: 4 Challenge Level:
Make a line of green and a line of yellow rods so that the lines differ in length by one (a white rod)
### There's Always One Isn't There
##### Stage: 4 Challenge Level:
Take any pair of numbers, say 9 and 14. Take the larger number, fourteen, and count up in 14s. Then divide each of those values by the 9, and look at the remainders.
### Factoring a Million
##### Stage: 4 Challenge Level:
In how many ways can the number 1 000 000 be expressed as the product of three positive integers?
### Why 24?
##### Stage: 4 Challenge Level:
Take any prime number greater than 3 , square it and subtract one. Working on the building blocks will help you to explain what is special about your results. | 629 | 2,513 | {"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} | 4.125 | 4 | CC-MAIN-2017-47 | latest | en | 0.908311 |
https://housemetric.co.uk/house-price-analysis/B6-7/Birmingham | 1,716,561,383,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058719.70/warc/CC-MAIN-20240524121828-20240524151828-00432.warc.gz | 262,000,550 | 11,977 | # House prices in B6 7 (Birmingham)
This article reveals price per square metre data and various charts to help you understand current housing market in Birmingham, Aston (B6 7).
## Defining 'B6 7'
This analysis is limited to properties whose postcode starts with "B6 7", this is also called the postcode sector. It is shown in red on the map above. There are no official names for postcode sectors so I've just labelled it Birmingham.
You can click on the map labels to change to a neighbouring sector, or you can enter a different postcode sector (e.g. CM23 4) below.
FYI, a postcode sector is the full postcode without the last two letters.
## Price per square metre
Knowing the average house price in Birmingham is not much use. However, knowing average price per square metre can be quite useful. Price per sqm allows some comparison between properties of different size. We define price per square metre as the sold price divided by the internal area of a property:
£ per sqm = price ÷ internal area
E.g. 177, Tame Road, Birmingham, Aston, sold for £130,000 on Feb-2024. Given the internal area of 64 square metres, the price per sqm is £2,031.
England & Wales have been officially metric since 1965. However house price per square foot is prefered by some estate agents and those of sufficiently advanced age ;-) They may want to convert square meters on this page to square feet.
The chart below is called a histogram, it helps you see the distribution of this house price per sqm data. To make this chart we put the sales data into a series of £ per sqm 'buckets' (e.g. £1,900-£2,000, £2,000-£2,100, £2,100-£2,200 etc...) we then count the number of sales with within in each bucket and plot the results. The chart is based on 38 sales in Birmingham (B6 7) that took place in the last two years.
##### Distribution of £ per sqm for Birmingham, Aston
Distribution of £ per sqm house prices in Birmingham
You can see the spread of prices above. This is because although internal area is a key factor in determining valuation, it is not the only factor. Many factors other than size affect desirability; these factors could be condition, aspect, garden size, negotiating power of the vendor etc.
The spread of prices will give you a feel of the typical range to expect in Birmingham (B6 7). Of the 38 transactions, half were sold for between £1,490 and £2,020 per square metre. The median, or 'middle', price per square metre in 'B6 7' is £1,700. Notably, only 25% of properties that sold recently were valued at more than £2,020 sqm. For anything to be valued more than this means it has to be more desireable than the clear majority of homes.
## Price map for Birmingham, Aston
Do have a look at the interactive price map I created. I find it useful and I am sure it will help you in exploring Aston. You can zoom in all the way to individual properties and then all the way back out to see the whole country. The colours show the current estimated property values.
House price heatmap for Birmingham, Aston
## Comparison with neighbouring postcode sectors
The table below shows how 'B6 7' compares to neighbouring postcode sectors.
Postcode sector Lower quartile Middle quartile Upper quartile
B6 4 Aston £1,490 sqm £1,490 sqm £1,490 sqm
B6 5 Aston £1,510 sqm £1,710 sqm £1,940 sqm
B6 6 Aston £1,500 sqm £1,850 sqm £2,060 sqm
B6 7 Birmingham £1,490 sqm £1,700 sqm £2,020 sqm
## Will Aston house prices drop in 2024?
I cannot tell the future and don't believe anyone who says they can. I can however plot price trends - I have done this in the chart below for B6 7 (Birmingham) compared with both the wider area B6 and inflation (CPIH from the Office of National Statistics). The dashed trend lines in the chart show the average over time.
##### Historic price per square metre in Birmingham,Aston
House price trends for Birmingham
For the most recent sales activity, rather than a summarized average, it is better to see the underlying data. This is shown in the chart below, where blue dots represent individual sales, click on them to see details. If there is an obvious trend you should be able to spot it here amid the noise from outliers.
##### Most recent B6 7 sales
Recent trends for Birmingham
Data from Land Registry comes in gradually over time. I update it every month but it takes about 5 months for the majority of sales for Aston to be recorded. Disclaimer: I do not verify and cannot guarantee the accuracy of any data shown. Outliers exist in the data, typically these are where the EPC registry records the internal area incorrectly, sometimes although very rarely the Land Registry price paid data can be wrong. The data provided throughout this website about Aston and any other area, is not financial advice. Any information provided does not and cannot ever take in to account the particular financial situation, objectives or property needs of either you or anyone reading this information.
## Street level data
Street Avg size Avg £sqm Recent sales
Brantley Road, Birmingham, B6 7D 76 sqm £1,640 34
Tame Road, Birmingham, B6 7D 82 sqm £1,729 13
Wyrley Road, Birmingham, B6 7B 89 sqm £1,642 9
Moor Lane, Birmingham, B6 7A 93 sqm £1,613 8
Deykin Avenue, Birmingham, B6 7B 77 sqm £1,778 8
## Raw data
Our analysis of Birmingham is derived from what is essentially a big table of sold prices from Land Registry with added property size information. Below are three rows from this table to give you an idea. | 1,326 | 5,463 | {"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-2024-22 | latest | en | 0.920526 |
https://savvycalculator.com/lunar-month-calculator/ | 1,726,143,685,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651457.35/warc/CC-MAIN-20240912110742-20240912140742-00218.warc.gz | 479,996,145 | 46,658 | # Lunar Month Calculator
## Introduction
Understanding lunar cycles is essential for various cultural, religious, and scientific purposes. The Lunar Month Calculator serves as a valuable tool for predicting lunar phases and facilitating accurate planning based on the lunar calendar. Whether for religious observances, agricultural planning, or celestial enthusiasts, this calculator simplifies the determination of lunar months.
## Formula:
The calculation of lunar months involves a combination of the synodic month and the age of the moon. The formula is as follows:
Lunar Month=Synodic Month×(1+Age of the Moon30)
The synodic month represents the time it takes for the moon to return to the same phase (e.g., from full moon to full moon), approximately 29.53059 days.
## How to Use?
1. Input Synodic Month: Enter the duration of the synodic month, which is a constant value.
2. Specify Age of the Moon: Determine the age of the moon in days. The age of the moon is the number of days since the last new moon.
3. Calculate: Press the calculate button to obtain the length of the lunar month.
## Example:
Let’s consider an example:
• Synodic Month: 29.53059 days
• Age of the Moon: 10 days
Lunar Month=29.53059×(1+1030)
Lunar Month≈29.53059×1.3333
Lunar Month≈39.3728
So, the length of the lunar month in this example is approximately 39.37 days.
## FAQs?
Q1: Is the lunar month constant throughout the year?
A1: No, the length of lunar months can vary slightly due to the elliptical shape of the moon’s orbit.
Q2: Why is the age of the moon important in the calculation?
A2: The age of the moon helps determine its current phase, which is crucial for predicting the beginning of a new lunar month.
Q3: Can the Lunar Month Calculator be used for different lunar calendars?
A3: Yes, the calculator is versatile and can be adapted for various lunar calendar systems used globally.
## Conclusion:
The Lunar Month Calculator proves to be an invaluable tool for those who rely on lunar cycles for cultural, religious, or scientific purposes. By understanding the formula and following the simple steps for usage, individuals and communities can accurately predict the length of lunar months, aiding in planning and observances aligned with the lunar calendar. Stay connected to celestial events and traditions with the Lunar Month Calculator. | 524 | 2,363 | {"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} | 3.4375 | 3 | CC-MAIN-2024-38 | latest | en | 0.838474 |
https://krishenning.com/2013/10/13/torus/ | 1,675,690,010,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500339.37/warc/CC-MAIN-20230206113934-20230206143934-00238.warc.gz | 361,293,435 | 27,213 | ### Torus
X=cos(u)*(2+sin(v)*cos(u)-sin(2*v)*sin(u)/2)
Y=sin(u)*sin(v)+cos(u)*sin(2*v)/2
Z=sin(u)*(2+sin(v)*cos(u)-sin(2*v)*sin(u)/2) | 69 | 134 | {"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 | longest | en | 0.355737 |
https://numberworld.info/13991 | 1,627,663,326,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153971.20/warc/CC-MAIN-20210730154005-20210730184005-00595.warc.gz | 436,747,581 | 3,784 | # Number 13991
### Properties of number 13991
Cross Sum:
Factorization:
Divisors:
1, 17, 823, 13991
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
36a7
Base 32:
dl7
sin(13991)
-0.99656718970066
cos(13991)
-0.082787900155344
tan(13991)
12.037594718923
ln(13991)
9.5461695447333
lg(13991)
4.1458487565905
sqrt(13991)
118.28355760629
Square(13991)
### Number Look Up
Look Up
13991 which is pronounced (thirteen thousand nine hundred ninety-one) is a unique number. The cross sum of 13991 is 23. If you factorisate 13991 you will get these result 17 * 823. The number 13991 has 4 divisors ( 1, 17, 823, 13991 ) whith a sum of 14832. The figure 13991 is not a prime number. 13991 is not a fibonacci number. The figure 13991 is not a Bell Number. The figure 13991 is not a Catalan Number. The convertion of 13991 to base 2 (Binary) is 11011010100111. The convertion of 13991 to base 3 (Ternary) is 201012012. The convertion of 13991 to base 4 (Quaternary) is 3122213. The convertion of 13991 to base 5 (Quintal) is 421431. The convertion of 13991 to base 8 (Octal) is 33247. The convertion of 13991 to base 16 (Hexadecimal) is 36a7. The convertion of 13991 to base 32 is dl7. The sine of the number 13991 is -0.99656718970066. The cosine of the figure 13991 is -0.082787900155344. The tangent of the figure 13991 is 12.037594718923. The root of 13991 is 118.28355760629.
If you square 13991 you will get the following result 195748081. The natural logarithm of 13991 is 9.5461695447333 and the decimal logarithm is 4.1458487565905. that 13991 is special number! | 601 | 1,712 | {"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} | 3.265625 | 3 | CC-MAIN-2021-31 | latest | en | 0.783369 |
https://www.scribd.com/document/312104152/Test-111111111 | 1,542,757,112,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746847.97/warc/CC-MAIN-20181120231755-20181121013755-00224.warc.gz | 1,001,051,466 | 30,755 | You are on page 1of 3
# 2.
A, B and C jointly thought of engaging themselves in a business venture. It was
agreed that A would invest Rs. 6500 for 6 months, B, Rs. 8400 for 5 months and C
, Rs. 10,000 for 3 months. A wants to be the working member for which, he was to
receive 5% of the profits. The profit earned was Rs. 7400. Calculate the share
of B in the profit.
A.
Rs. 1900
C.
Rs. 2800
B.
D.
Rs. 2660
Rs. 2840
Explanation:
For managing, A received = 5% of Rs. 7400 = Rs. 370.
Balance = Rs. (7400 - 370) = Rs. 7030.
Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3)
= 39000 : 42000 : 30000
= 13 : 14 : 10
B's share = Rs.
37
7030 x 14
= Rs. 2660.
2.
A, B and C jointly thought of engaging themselves in a business venture. It was
agreed that A would invest Rs. 6500 for 6 months, B, Rs. 8400 for 5 months and C
, Rs. 10,000 for 3 months. A wants to be the working member for which, he was to
receive 5% of the profits. The profit earned was Rs. 7400. Calculate the share
of B in the profit.
A.
Rs. 1900
C.
Rs. 2800
B.
D.
Rs. 2660
Rs. 2840
Explanation:
For managing, A received = 5% of Rs. 7400 = Rs. 370.
Balance = Rs. (7400 - 370) = Rs. 7030.
Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3)
= 39000 : 42000 : 30000
= 13 : 14 : 10
B's share = Rs.
37
7030 x 14
= Rs. 2660.
8400 for 5 months and C . B. 2660.370) = Rs. . 37 7030 x 14 = Rs. Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3) = 39000 : 42000 : 30000 = 13 : 14 : 10 B's share = Rs. A. Rs. A wants to be the working member for which. 7400. (7400 . Rs. 370.000 for 3 months. Rs. 2. 1900 C. 8400 for 5 months and C . A received = 5% of Rs. 2840 Explanation: For managing. 2660 Rs. 7400 = Rs. 10. 7400. Rs. he was to receive 5% of the profits. Rs. 2660 Rs. Balance = Rs. 37 7030 x 14 = Rs. B. B and C jointly thought of engaging themselves in a business venture. 2840 Explanation: For managing. D. It was agreed that A would invest Rs. A.000 for 3 months. (7400 . he was to receive 5% of the profits. A received = 5% of Rs. 370. 10. 2800 Answer & Explanation Answer: Option B B. A wants to be the working member for which. Rs.370) = Rs. Rs. Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3) = 39000 : 42000 : 30000 = 13 : 14 : 10 B's share = Rs. A. Calculate the share of B in the profit. 6500 for 6 months. 2660. Rs. 7030. The profit earned was Rs. B and C jointly thought of engaging themselves in a business venture. 2800 Answer & Explanation Answer: Option B B.2. 7030. Balance = Rs. 6500 for 6 months. 7400 = Rs. It was agreed that A would invest Rs. A. D. Rs. 1900 C. Calculate the share of B in the profit. The profit earned was Rs. Rs.
A wants to be the working member for which. Rs.000 for 3 months. 10. B. A wants to be the working member for which. 7030. 8400 for 5 months and C . It was agreed that A would invest Rs. It was agreed that A would invest Rs. A. 37 7030 x 14 = Rs. A. Calculate the share of B in the profit. 2660. 2. 2840 Explanation: For managing. 2660 Rs. Rs. D. 2800 Answer & Explanation Answer: Option B B. Rs. Rs. he was to receive 5% of the profits. 2800 Answer & Explanation Answer: Option B B. D. . 10. 7400. B and C jointly thought of engaging themselves in a business venture. (7400 .2. A. Rs. 1900 C. 7030. Balance = Rs. 7400. he was to receive 5% of the profits. 6500 for 6 months. 2660. Rs. The profit earned was Rs. 370. Calculate the share of B in the profit. B and C jointly thought of engaging themselves in a business venture. Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3) = 39000 : 42000 : 30000 = 13 : 14 : 10 B's share = Rs.370) = Rs. Rs. 2660 Rs. 7400 = Rs. 7400 = Rs. 370. Rs. A received = 5% of Rs. A. Balance = Rs. 37 7030 x 14 = Rs. Rs. (7400 . B. Rs. 1900 C. 2840 Explanation: For managing.000 for 3 months. 8400 for 5 months and C . Ratio of their investments = (6500 x 6) : (8400 x 5) : (10000 x 3) = 39000 : 42000 : 30000 = 13 : 14 : 10 B's share = Rs. 6500 for 6 months. A received = 5% of Rs. The profit earned was Rs.370) = Rs. | 1,490 | 4,052 | {"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} | 4.125 | 4 | CC-MAIN-2018-47 | latest | en | 0.975913 |
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Erase a prime (Posted on 2020-08-03)
Write down the first n primes.
Erase one of them and the mean of the remaining numbers is 10.
What number did you erase?
No Solution Yet Submitted by Jer Rating: 3.0000 (1 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
Instant solution | Comment 3 of 4 |
When the puzzle was on the reviewing board I saw that it is a d1 puzzle since the difference between a(n) and n-1 must be divisible
by 10
.So visual scan of both series will yield the solution
....Write down n-1= 0,1, 2,3,4,5,6,7,8,..11,12
....,77 ,58 ,41 ,2,5,10,17,28 and
scanning from left to right, look for identical digits
of the odd numbers in both series. Check the result. l
. 10 = 7 / (77-7)
number 7, the 4th prime was erased
Edited on August 4, 2020, 9:13 am
Posted by Ady TZIDON on 2020-08-04 07:55:54
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Unformatted text preview: φ v = parenleftbigg cos u bracketleftbigg cos u 2 cos v- 2 cos 2 v sin u 2 bracketrightbigg , sin u bracketleftbigg cos u 2 cos v- 2 cos 2 v sin u 2 bracketrightbigg , 2 cos u 2 cos 2 v + cos v sin u 2 parenrightbigg , so φ v ( u, 0) = parenleftbigg cos u bracketleftbigg cos u 2- 2 sin u 2 bracketrightbigg , sin u bracketleftbigg cos u 2- 2 sin u 2 bracketrightbigg , 2 cos u 2 + sin u 2 parenrightbigg . Now φ u × φ v ( u, 0) = parenleftbigg 3 cos u bracketleftbigg 2 cos u 2 +sin u 2 bracketrightbigg , 3 sin u bracketleftbigg 2 cos u 2 +sin u 2 bracketrightbigg , 6 sin u 2- 3 cos u 2 parenrightbigg . Hence we see that φ u × φ v (0 , 0) = ( 6 , ,- 3 ) while φ u × φ v (2 π, 0) = (- 6 , , 3 ) . After you have traveled once around the u –line at v = 0 you discover that the normal vector is now pointing in the opposite direction. Hence you can conclude that this surface can not be oriented. (This surface is a parametrization of a Klein surface.) 3. (a) We parametrize S by Φ ( u, v ) = ( u, v, 1- u- v ), ( u, v ) ∈ projection into the uv –plane; i.e., 0 ≤ v ≤ 1- u , 0 ≤ u ≤ 1. Now φ u × φ v = (1 , 1 , 1) and since the z –component is positive, Φ ( u, v ) is orientation preserving. Hence integraldisplay S F · d S = integraldisplay projection F ( Φ ( u, v )) · φ u × φ v dA = integraldisplay 1 integraldisplay 1- u ( u, v, 1- u- v ) · (1 , 1 , 1) dv du = integraldisplay 1 integraldisplay 1- u ( u + v +1- u- v ) dv du = integraldisplay 1 integraldisplay 1- u dv du = integraldisplay 1 (1- u ) du = bracketleftbigg u- u 2 2 bracketrightbigg 1 = 1- 1 2 = 1 2 . MATB42H Solutions # 8 page 3 (b) We parametrize S by Φ ( u, v ) = ( u, v, 4- u 2- v 2 ), 0 ≤ u ≤ 1, 0 ≤ v ≤ 1. Now φ u × φ v = (2 u, 2 v, 1) and Φ ( u, v ) is orientation preservation since the z – component is positive; i.e., pointing outward. Hence integraldisplay S F · d S = integraldisplay 1 integraldisplay 1 F ( Φ ( u, v )) · φ u × φ v dA = integraldisplay 1 integraldisplay 1 ( u, v, 4- u 2- v 2 ) · (2 u, 2 v, 1) du dv = integraldisplay 1 integraldisplay 1 (2 u 2 +2 v 2 +4- u 2- v 2 ) du dv = integraldisplay 1 integraldisplay 1 (4 + u 2 + v 2 ) du dv = integraldisplay 1 parenleftbigg 4 + v 2 + 1 3 parenrightbigg dv = bracketleftbigg 4 v + v 3 3 + v 3 bracketrightbigg 1 = 4 + 1 3 + 1 3 = 14 3 ....
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### Messages - kolofsson
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31
##### General discussions / Switched times to Uint32 milliseconds in SFML 2
« on: November 05, 2011, 09:34:41 am »
Quote from: "Laurent"
There will be a problem indeed, since 1.6 can't be returned (it will be 1 by the way, C++ truncates when it converts float to int).
What if my loop lasted shorter than 1 ms (0.9 ms), would it be truncated to 0 ms?
Also, if I ran a loop that would accumulate the elapsed time, the loop lasting 1,9 ms (displayed as 1 ms), after 1000 frames 1,9 second would have elapsed in reality, but the accumulated timer value would show 1 second! That's seriously crappy. In that case I would totally drop the frame timer and go for a global timer. Then the differences would be corrected, right?
What do you mean that OS guarantees 1ms as the smallest measurement? If so, then why was the float displaying non-round FPS values, like 1700 FPS?
32
##### General discussions / Switched times to Uint32 milliseconds in SFML 2
« on: November 04, 2011, 06:13:16 pm »
Sorry to resurrect this thread but I've got a question.
Let's assume that each one frame of some game lasts a constant 1.6 ms. By one frame I mean one loop cycle. The question is: what will be the elapsed time value? Will it be equal to 2 ms each frame?
I'm asking because the elapsed time is used to calculate movement of an object by multiplying its velocity by the elapsed time. When the interval is big, there is no problem, but if it is as small as I've suggested, will there be problems?
I must say I'm not convinced to using integer for storing time. It's like limiting sprite position to full pixels.
33
##### General / Time as float?
« on: November 03, 2011, 10:38:12 pm »
Since it's been mentioned, can anyone tell me why was time changed from float to integer milliseconds? If I ever wanted an integer, I could always round the value. Why is it rounded for me?
34
##### Python / pysfml2-cython
« on: October 27, 2011, 11:19:47 pm »
Quote from: "bastien"
When I try your Pong example with my sf.pyd, I get this:
Code: [Select]
`err:module:import_dll Library libgcc_s_sjlj-1.dll (which is needed by L"Z:\\home\\bastien\\Desktop\\pong\\sf.pyd") not founderr:module:import_dll Library libstdc++-6.dll (which is needed by L"Z:\\home\\bastien\\Desktop\\pong\\sf.pyd") not foundTraceback (most recent call last): File "main.py", line 9, in <module> import sf`
Could you copy-paste here the command-lines that are issued when you build the binding? I guess I missed something to include the standard libraries.
Also, for some reason the SFML libs I cross-compiled aren't even detected by Wine.
Maybe my libs require having installed the Visual C++ redistributable? I build an executable of the pong example with PyInstaller and it put all the necessary DLLs in the folder. I gave it to my friend who has "bare" windows and it worked for him, so it should work for you too in Wine. Check it out here:
Regarding the compilation output, here it is:
Could you please deal with the conversion from Double to Float warnings? Why do you even use floats if SFML uses doubles? Does python have doubles? If not, could you at least convert these via a function so there is no warning? Thanks .
35
##### Python / pysfml2-cython
« on: October 27, 2011, 08:23:25 pm »
OK I tested your sf.pyd and it tells me that "import sf failed, module was not found". I've no idea what's wrong with your file .
I put my own sf.pyd on the web, accessible here:
Also, I put the whole Pong example, with SFML dlls, extlibs, sf.pyd and main.py:
Perhaps this may be of help for all PySFML starters.
36
##### Python / pysfml2-cython
« on: October 27, 2011, 09:40:24 am »
Quote from: "bastien"
I just compiled something that looks like a Win32 module, but I can't get it to work with Wine (it tells it can't find dependent DLLs, even when they are in the current directory).
Can someone test it on Windows and tell me if it works? Here is the direct link: https://github.com/downloads/bastienleonard/pysfml2-cython/sf.pyd
I will test it as soon as I get hom from work. I can see you only put sf.pyd file. It makes no difference which SFML libs I use with it? (revision / compiler). I presume that when compiling the pyd file, you needed to have the dlls present, right? Maybe it would be good to include them in the package with the binding, so that the end-user does not have to compile SFML on his own.
37
##### Python / pysfml2-cython
« on: October 24, 2011, 09:46:42 pm »
Ah, very well. No special debug libs means one thing less to care about . So far my apps did not require special debugging, so I haven't really figured out what SFML debug libs do. If you say that regular python debugger will do, that's just perfect.
Well, I think I'm going to dedicate more time to learn Python and maybe start off a little project in PySFML. I'm really overwhelmed by Python philosophy . I just hope you won't get bored too soon and that you keep updating and upgrading PySFML (e.g. reference/documentation).
38
##### General / SFML Complete Game Tutorial
« on: October 24, 2011, 09:09:44 pm »
Ah, from your tutorial I just need the game framework. I'd like to write the engine myself, as I'm fairly good at maths. In my prior 2d racing project I had all the trigonometry and physics figured out. In fact, that is the most fun part of all, fiddling with the physical equations and then testing the effect it has on the car .
I don't know if you should teach maths in your tutorial. You don't teach general programming, do you? I like your tutorial, so far, because it's a missing ingredient of how to wrap SFML libs to make an organized game framework. But if you really wish to teach trigonometry, i think its not all that hard to comprehend. Seriously, what more needs to be said than this little infographic I made?
By the way, I managed to compile PySFML2 and I'm overwhelmed by the simplicity of Python. For example, project is not kept in a file, you just open a folder and all files inside that folder are treated as part of the project. I love the language itself. In most cases there is only one right way of doing stuff, which is cool beacause understanding the code is easier that way. The syntax is clean. I downloaded a PySFML asteroids game called Multitroids and the code looks really nice, I think the structure is similar to yours. I only hope that Python won't be too slow for the game I'm planning to make.
39
##### Python / pysfml2-cython
« on: October 24, 2011, 06:50:38 pm »
Hi Bastien
A question about debugging. As we know, SFML can be compiled as Release or Debug, and for development process it is advisable to use the Debug libraries (I don't know why exactly, I guess more debug info is being produced during runtime).
Should we use the Debug libraries in PySFML too? Will it work? Is there any reason why we should/should not use them? Is compiling the bindings for Debug libraries as simple as removing the "-d" suffix from SFML lib filenames? Are you planning to include Debug libs in the binding so that there are two output files: sf.pyd and sfd.pyd?
40
##### General / How do I bring an SFML game out on the Internet?
« on: October 22, 2011, 03:44:39 pm »
Should I mention I find it awful to port everything into a web browser? The web browser is becoming more and more a substitute for operating system. It's nice to have gmail, maps and spreadsheets available on the web, but for some tasks a standalone application is just irreplaceable. I think it's pretty viable to make a single-file SFML game that would work without installing and would download necessary textures from the web.
41
##### General / How do I bring an SFML game out on the Internet?
« on: October 22, 2011, 01:58:45 pm »
Wait, you can't make a multiplayer game with PySFML? I'm almost sure ingwik means multiplayer, not a browser game.
There are two types of multiplayer servers. One is a server that is built in the game and very often the multiplayer is peer-to-peer. The second type is a dedicated server, where the server is a different app that receives/sends data between players and does all the physics calculations. the client-side games are then just GUI frontends to interact with the server.
42
##### General / SFML Complete Game Tutorial
« on: October 22, 2011, 01:48:23 pm »
Waiting for part 7, Serapth... :wink:
43
##### Python / pysfml2-cython
« on: October 19, 2011, 08:46:57 pm »
Quote from: "bastien"
That's strange. Last try: did you also copy the DLLs that are in SFML/extlibs? (I don't see a “DLL loaded failed” message when importing a non-existing module, so my guess is that a dependency can't be loaded.)
OMG it worked :shock:
The pong example runs now. Also, when I type "import sf" with extlibs in the folder, it does not return error. This was one hell of a job to make this binding work. I guess it would be much easier with VC2008 or MinGW. But still, some binaries would be really appreciated and save quite some time, instead of compiling it all.
Thanks for help and feedback.
44
##### Python / pysfml2-cython
« on: October 19, 2011, 06:14:46 pm »
Quote from: "bastien"
You're right, it's .pyd on Windows apparently. What happens if you open a command prompt, go to the path where sf.pyd is, type “python”, and then type “import sf” interactively?
calling "python" in cmd does not work for me, I had to enter full path:
Code: [Select]
`E:\pysfml2\bastienleonard-pysfml2-cython-ca7f88f\build\lib.win32-2.7>c:\Python27\python.exePython 2.7.2 (default, Jun 12 2011, 15:08:59) [MSC v.1500 32 bit (Intel)] on win32Type "help", "copyright", "credits" or "license" for more information.>>> import sfTraceback (most recent call last): File "<stdin>", line 1, in <module>ImportError: DLL load failed: Nie mo┐na odnalečŠ okreťlonego modu│u.>>>`
Last line says: Can't find specified module
45
##### Python / pysfml2-cython
« on: October 18, 2011, 09:00:00 pm »
Quote from: "bastien"
sf.dll needs to be in the Python path. The simplest way is to put it the current directory just before you execute the program. Or you can modify the PYTHONPATH environment variable.
Or you can run python setup.py install, and it should be available for all the programs. But I would do this only after testing the binding.
sf.dll? I only got sf.pyd as a result of my compilation. I put standard c++ libraries as well as the sf.pyd in the same folder as pong.py. The compiler still can't find the sf module...
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Statistics_21_-_Fall_2003_-_Pham_-_Midterm_1
# Statistics_21_-_Fall_2003_-_Pham_-_Midterm_1 - Stat 21 Fall...
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Unformatted text preview: Stat 21 Fall 2003 Midterm 1 Open hook, SD minot: Show aii Wont: for foil credit. Keep 3 Frame your answers o'ecin'ieir pieces in caicoiations and ansWers. Problem 1 {'35 minutes, 35 ,oointsjI The following data are obtained for about moe men: average height: E39 inches, SD: 2.5 inches awerage forearm length: 18 inches, SD: 1 inch, e. 01” the men who were 6 feet taii than 18 inches? E. Di men who had 19 than 69.5 inches? C. A man's height is the 96th percentile of height distribution. Predict the percentile rank of his forearm. inches Forarms, about what percentage had height taller _ _,I' "" II LII _.._ I - .I‘-'!..l i,"'_-|| ,2 __ _ a, _|_. I._ - PE-‘fles. |_ " Li K. .II -1 I. "I r 'l I,+f 'l I 1 .d i I ~r i- r__!. I I- i. I, _ ll _ I 0' 'rflz; I,“ _ .I l ‘l _ _ ___ _ ..| r..? I: _ E" K"; I f _ "- I I!- I'r-l I”. I_ ll I'll 1' _ I. _. -' F- T I J -| II,” ’| r I I r l I J. i' I‘- H II —. I . H_ n '— _- I _ I r I. | III.'.__] - I _ .1 .:_._ | ' I ' _ — L, I .5 .'.,. L _ . . "_.__ II. . -x |,'."';_ I II K. 33.5 I N}: 'I __ _ | I_ll I" _— —-T —__ _—— I 1,: E, a [-3- “ii-oil i. :1}; |ri ___—-— - erg-gm 27»; I 1" E! I “J: 'y ., _“Hf' e. I' — — - '5' ~— —' r l? a I' Phublem 2 {15 minutes, 15 points) _\ A. What is the correlation coefficient for the data beiow {show all work) ----T----If.:l I LIV}: I AT} xii-f-[j—lx': -l I 't} —| " {I _ I q n 1 1 ‘53— "- —- —--—-- — _ -_ o 3 I {- 1 3" 7 I : F I '-'-..—-:- _ _T—'—,_. ———__ T": 'il. r: i I I; II I I“. — flab! ‘_|I |_".i ._ "-J " '43:“ _I:II‘|_‘_ r.-_ _ _ | rI—IL ___ _F . -. ' 3 I "" l I |___ _ L“) B. If possible fill in the two bianks so that the correlation for the data below equals the correlation in part A. If this is not possible, explain why. ' _o-i—FFFFF ‘- 1}.— "'|‘|-'T-ai 1 ___ (it—"Wu" “ a" fibril tag—4 :I -r- pl? (E, _H- _ I LILli f) _ s a 1—)] [if _ ‘ "1 1 —L—__,:-—\——'\— .- _ ":L 4" "--_.__ ____ ' — " I_'I_—_ - _I: - If. ' I If ' ': :— _. ix} _lr‘: .2“?— :i_. l;f§_ Yb” "-— _ ' 2 II :3. {I E I??? -H - 2 r - l”! a r __' ~ a a _ - ._ ‘ _IJ {1 . In rdfrp 'V- -| I' I'lf-i,I ...
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HOW TO WIND YOUR OWN AUDIO TRANSFORMERS
TRANSFORMERS
HOW TO WIND YOUR OWN OUTPUT TRANSFORMERS
WHEN YOU CANT FIND THE TRANSFORMER YOU WANT
Circuits involving tube or solid-state components frequently require non-standard audio
output transformers. This document describes simplified methods of calculating the
primary/secondary, ratios, wire sizes, and numbers of turns for low-impedance matching
transformers wound on salvaged cores.
CRAIG KENDRICK SELLEN
Project builders and experimenters occasionally need a small impedance matching audio
transformer with an uncommon impedance ratio. When such a transformer is specially wound its
cost is usually prohibitively high compared to the total cost of the project in while it is to
be used. However, with a few calculations and a little work on your part, you can duplicate many
unusual transformers or any special audio coupling or matching transformer to suit your needs.
The techniques prescribed in this document are limited to transformers of average size and lowto-medium impedance. It is impractical to duplicate subminiature transformers that normally cost
only \$1 or less and high-impedance transformers that require many turns of very fine wire.
Throughout this document, you will find the term volt amperes (VA) used in the same manner
that watts is used for power. This usage involves an assumption which is not quite true.
However, for this type of work, if you accept the assumption that the two are equal, the results
will be acceptable.
Calculations involved in designing an audio transformer are covered by the nine steps
outlined in the box. To see how these steps work, lets design a typical transformer.
Assume that a transistor output transformer with a 130-ohm primary and a 4/8/16-ohm secondary
is needed to match the output of an RCA CA3020 IC to a loudspeaker. By referring to the mailorder catalogs, we find that the full output of the IC is 0.5 watt. The nearest thing you can
find in the catalog is a 125-ohm center-tapped transformer thats rated at 300 mW. This
transformer could be used, but you can make one that will be just as good and design it for a
full watt of power if space and weight requirements permit.
First calculate the core area required. Note, however, that the core area applies only to the
crossectional area of the cores center leg as shown in Fig. 1. Referring to Fig. 2, we find
that the graph shows an approximate core area of 0.18 sq in. will suit our requirements. (We can
use an approximation since the actual core area is not too critical.)
Determine the turns ratio from the impedance ratio. Since we know the primary and lowest
secondary impedances to be used, plug in 130 and 4 into the equation: Ratio = the square root of
(130/4): 1 = 5.7:1. Hence, the actual turns ratio required shows 5.7 turns in the primary
winding for every turn in the secondary winding.
Next, determine the D.C. voltage to be applied to the transformers primary. In this case, we
desire 9-volt operation. The CA3020 employs a push-pull output. So, bear in mind that an 18volt figure must be used in all primary calculations.
Calculate the wire size needed for the primary winding. Since we have decided to design the
transformer to handle 1 watt of power, let us first determine how much current will be handled
by the primary: I = (VA/Vcc) = 1/9 = 0.111 Amp. Now, because of the push-pull division of the
current, we divide the primary current by two for determining the wire size; this gives us 55mA
in each half of the primary winding. If 700 circular mils/ampere is desired refer to the Wire
Table (column four) and locate the current at or greater than 55mA. Column one shows that #34
wire will safely handle 57mA, the nearest figure to 55mA. This size is quite small and difienlt
to work with, so choose #28 wire for ease of winding.
We will have to make some assumptions now in determining the number of primary turns to be
used. For this calculation, we will use 2Vcc, or 18 volts, and an area of 0.18 sq in, for our 1watt transformer. The frequency we will arbitrarily settle on as being 100Hz. For flux density
BM in gauss/sq in., any figure between 40,000 and 90,000 can be used; well settle on 70,000 to
be conservative:
2Vcc x 10_8
18 x 10_8
Primary Turns = --------------------------------= ------------------------------- =
321
4.44 x A x f x BM
4.44 (0.18) (100) (70,000)
So, 320 turns will be close enough.
Having calculated the number of primary turns, we use the turns ratio formula to calculate
the number of secondary turns needed. This is a step-down-type transformer, so we divide the
number of primary turns by the turns ratio: Secondary Turns = Primary Turns/Turns Ratio =
320/5.7 = 56 turns.
Secondary wire size is determined by the current ratio method. Secondary current is equal to
the primary current multiplied by the turns ratio: 0.111 x 5.7 = 0.64 A. The secondary wire size
is determined by the same method as used for the primary. At 700 circular mils/ampere, the Wire
HOW TO WIND YOUR OWN AUDIO TRANSFORMERS
Table indicates a 577mA current capacity for #24 and 728 mA for #23 wire. Since 640 mA is about
midway between the two sizes, we will settle on #23 wire.
.
Finally the 8. and 16-ohm taps must be calculated. Again, refer to the turns ratio formula,
and determine the turns ratio for 8 and 16 ohms separately. Then use these ratios with the
primary turns to determine the exact number of turns required for each impedance: 16-ohm ratio =
the square root of (130/16): 1 = 2.86: 1; 8-ohm ratio = the square root of (130/8) : 1 = 4.04:
1. Secondary turns = 320/2.86 = 112 turns for the 16-ohm ratio; Secondary turns = 320/4.04 = 79
turns for the 8-ohm ratio. Hence, the composite secondary will consist of 112 turns of wire with
taps at the 56th and 79th turns.
Now that we have all of the design parameters, we can proceed to assembling our specialpurpose transformer.
Assembling the transformer from the design parameters derived from the above procedure is
easy. We know that the core area must be about 0.18 sq in. The simplest and least expensive way
of obtaining a suitable core is to salvage an old audio output transformer. Many such
transformers have a core area of 0.25 sq in. If about a quarter of the laminations are removed,
approximately the correct dimensions will be obtained (about 0.185 sq in.).
Disassemble the salvage transformer, and remove and discard the windings, but reserve the
plastic winding bobbin if it has one. If no bobbin is available, you can make one from an index
card or heavy waxed (butchers paper. This bobbin should easily slide over the core leg and be a
little shorter than the center leg of the laminations.
Slide the bobbin onto a length of wood to serve as a winding handle. Then begin winding the
primary turns onto the bobbin, starting and ending along the 1/2 side of the bobbin to avoid
having the ends exit from the core windows when the bobbin is in place. Ordinary scatter
winding is acceptable in most cases; but if space is limited, you might have to close-wind the
turns. Our hypothetical transformer has a further complication: The primary winding is centertapped. it must be wound so that both sides of the winding are balanced. To do this we will use
the bifilar winding method shown in fig. 3.
For our 320-turn primary winding, we wind two wires onto the bobbin simultaneously, side by
side, until there are 160 double turns on the bobbin. Then to complete the bifilar winding, we
connect one end of one wire to the opposite end of the other wire and solder on a 5 length of
stranded hookup wire to make the center tap. Two more stranded wires soldered to the free ends
of the primary windings complete the primary assembly. Color code the wires so that the center
tap is easily identifiable. Make sure that each soldered connection is well insulated from the
others; then wrap a layer or two of plastic tape over the windings.
Now wind the secondary turns onto the bobbin. Count the turns as you go, and make a pig-tail
tap leads at the 56th and 79th turns for the 4- and 8-ohm taps (see Fig. 4.). Use color coded
stranded hookup wires for the winding ends and taps so that each can be easily identifiable.
Again, make sure that the solder connections are well insulated from each other, and wrap a
layer or two of electrical tape over the assembly to prevent the windings from unraveling.
Slip the bobbin assembly off the winding handle. Orient the primary leads to one side and the
secondary leads to the other side of the bobbin. Then slip the bobbin onto the center leg of
the transformer core laminations. Assemble the transformer.
Testing the completed transformer is not really necessary if you exercised care during
assembly and followed each step exactly as described. However, if you want to be on the safe
side, you can test the transformer with the aid of an audio signal generator, two ac VTVMs or
VOMs or DMMs an 8-ohm load resistor, and a 1000-ohm (1K) potentiometer as shown in Fig. 5. Set
the generators amplitude control for an output of several volts at 1000 Hz (1 kHz). Adjust the
potentiometer for minimum resistance so that both meters have an identical reading.
Now, increase the resistance of the potentiometer until meter #2 indicates exactly one half
its original indication while making sure that meter #1 remains at the original voltage setting.
Since changing the resistance of the potentiometer decreases the load on audio generator, meter
#1 will indicate an increase in voltage. Simply reduce the generators output level to return
meter #1 to the original voltage setting.
After jockeying back and forth between the generators amplitude control and the
potentiometer a few times, you should be able to arrive at settings where meter #1 indicates the
original voltage and meter #2 indicates exactly half of its original voltage. When this occurs,
remove the potentiometer from the circuit without upsetting its final setting and measure its
resistance. This resistance should be equal (or as near as possible) to the transformers input
impedance, or 130 ohms. However, if the transformer is loaded with an incorrect impedance (say,
the 8-ohm load resistor connected across the 4- or 16-ohm output leads), it will reflect an
incorrect impedance into the primary. As a matter of fact, if you use a 3.2-ohm speaker on the
4-ohm transformer output, primary impedance somewhat lower than that for which the transformer
was designed will be reflected. But if you plan to use such a speaker with the transformer, you
could easily have plugged into the equations the 3.2-ohm figure for the 4-ohm figur
e. | 2,558 | 10,643 | {"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-2019-51 | latest | en | 0.871752 |
https://k12xl.com/division/divisibility-test-division-worksheets | 1,708,811,730,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474569.64/warc/CC-MAIN-20240224212113-20240225002113-00794.warc.gz | 345,815,057 | 14,592 | # Division
## Divisibility Test Division Worksheets
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#### Division - Divisibility Test Division Tests with Answers
Interactive PDF worksheet: Printing is not necessary! | 401 | 2,408 | {"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-2024-10 | longest | en | 0.931787 |
https://oneclass.com/class-notes/ca/utsg/geog/ggr100h1/5973-solar-radiation-the-sun-etc.en.html | 1,544,697,116,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824675.15/warc/CC-MAIN-20181213101934-20181213123434-00234.warc.gz | 673,600,431 | 455,953 | Class Notes (1,017,263)
CA (584,537)
UTSG (54,333)
Geography (615)
GGR100H1 (141)
Lecture
# Solar radiation, the sun, etc
2 pages25 viewsWinter 2011
Department
Geography
Course Code
GGR100H1
Professor
Joseph Leydon
This preview shows half of the first page. to view the full 2 pages of the document.
Geography Lecture #2
January 13, 2011
•Earth rotates eastwards daily on its own axis (rotation)
•Elongated/elliptical orbit around the sun
•Aphelion- the point where the earth is furthest from the sun
•Perihelion- point when the earth is closest to the sun
The Sun
•6000 K surface
•enough energy to fuse atoms together-> nuclear fusion
•sun radiates energy at the speed of light in a spectrum
•sun emits charged particles (solar wind)
•It is energy emitted as waves of electric and magnetic oscillations
•Electromagnetic radiation is emitted by all matter with a temperature
•Q (energy emitted from radiation) is proportional to the Temperature
of the object
•Wavelength of radiation is inversely related to surface temperature
(hotter objects -> shorter wavelengths)
Example
•Comfortable human skin temperature = 33 degrees Celsius
•Average human surface area ~ 1.8m2
•895 watts emitted (1 watt= 1 joule per second)
•similar to 9 household light bulbs
•we give off emissions in the infrared part of the spectrum (heat)
•absorption of different wavelengths by gasses in the atmosphere,
gamma and xrays get absorbed at 80 km from the earths surface
How much insolation does Earth intercept?
•At the equator or anywhere where the sun is directly overhead is where
•Sub-solar point: where the sun is directly overhead ^
•The same light energy diffused over a larger area reduces intensity
•Averaged daily latitudinal energy imbalance (Net R) at the top of the
atmosphere -> spatial variation
www.notesolution.com | 504 | 1,857 | {"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-2018-51 | longest | en | 0.822763 |
https://www.r-bloggers.com/multidimension-bridge-sampling-core-in-cirm-5/ | 1,580,187,354,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251773463.72/warc/CC-MAIN-20200128030221-20200128060221-00259.warc.gz | 1,013,802,625 | 55,578 | # Multidimension bridge sampling (CoRe in CiRM [5])
July 13, 2010
By
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Since Bayes factor approximation is one of my areas of interest, I was intrigued by Xiao-Li Meng’s comments during my poster in Benidorm that I was using the “wrong” bridge sampling estimator when trying to bridge two models of different dimensions, based on the completion (for $theta_2=(mu,sigma^2)$ and $mu=theta_1$ missing from the first model)
$B^pi_{12}(x)= dfrac{displaystyle{intpi_1^*(mu|sigma^2){tildepi}_1(sigma^2|x) alpha(theta_2) {pi}_2(theta_2|x)hbox{d}theta_2}}{ displaystyle{int{tildepi}_2(theta_2|x)alpha(theta_2) pi_1(sigma^2|x)hbox{d}sigma^2 } pi_1^*(mu|sigma^2) hbox{d}mu },.$
When revising the normal chapter of Bayesian Core, here in CiRM, I thus went back to Xiao-Li’s papers on the topic to try to fathom what the “true” bridge sampling was in that case. In Meng and Schilling (2002, JASA), I found the following indication, “when estimating the ratio of normalizing constants with different dimensions, a good strategy is to bridge each density with a good approximation of itself and then apply bridge sampling to estimate each normalizing constant separately. This is typically more effective than to artificially bridge the two original densities by augmenting the dimension of the lower one”. I was unsure of the technique this (somehow vague) indication pointed at until I understood that it meant introducing one artificial posterior distribution for each of the parameter spaces and processing each marginal likelihood as an integral ratio in itself. For instance, if $eta_1(theta_1)$ is an arbitrary normalised density on $theta_1$, and $alpha$ is an arbitrary function, we have the bridge sampling identity on $m_1(x)$:
$inttilde{pi}_1(theta_1|x) ,text{d}theta_1 = dfrac{displaystyle{int tilde{pi}_1(theta_1|x) alpha(theta_1) {eta}_1(theta_1),text{d}theta_1}}{displaystyle{inteta_1(theta_1) alpha(theta_1) pi_1(theta_1|x) ,text{d}theta_1}}$
Therefore, the optimal choice of $alpha$ leads to the approximation
$widehat m_1(x) = dfrac{displaystyle{sum_{i=1}^N {tildepi}_1(theta^eta_{1i}|x)big/left{{m_1(x) tildepi}_1(theta^eta_{1i}|x) + eta(theta^eta_{1i})right}}}{displaystyle{ sum_{i=1}^{N} eta(theta_{1i}) big/ left{{m_1(x) tildepi}_1(theta_{1i}|x) + eta(theta_{1i})right}}}$
when $theta_{1i}simpi_1(theta_1|x)$ and $theta^eta_{1i}simeta(theta_1)$. More exactly, this approximation is replaced with an iterative version since it depends on the unknown $m_1(x)$. The choice of the density $eta$ is obviously fundamental and it should be close to the true posterior $pi_1(theta_1|x)$ to guarantee good convergence approximation. Using a normal approximation to the posterior distribution of $theta$ or a non-parametric approximation based on a sample from $pi_1(theta_1|mathbf{x})$, or yet again an average of MCMC proposals are reasonable choices.
The boxplot above compares this solution of Meng and Schilling (2002, JASA), called double (because two pseudo-posteriors $eta_1(theta_1)$ and $eta_2(theta_2)$ have to be introduced), with Chen, Shao and Ibragim (2001) solution based on a single completion $pi_1^*$ (using a normal centred at the estimate of the missing parameter, and with variance the estimate from the simulation), when testing whether or not the mean of a normal model with unknown variance is zero. The variabilities are quite comparable in this admittedly overly simple case. Overall, the performances of both extensions are obviously highly dependent on the choice of the completion factors, $eta_1$ and $eta_2$ on the one hand and $pi_1^*$ on the other hand, . The performances of the first solution, which bridges both models via $pi_1^*$, are bound to deteriorate as the dimension gap between those models increases. The impact of the dimension of the models is less keenly felt for the other solution, as the approximation remains local.
Filed under: Books, R, Statistics, University life Tagged: Bayesian Core, Benidorm, bridge sampling, CIRM, poster, Valencia 9
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Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. | 1,172 | 4,373 | {"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": 48, "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} | 2.625 | 3 | CC-MAIN-2020-05 | longest | en | 0.862839 |
http://openwetware.org/index.php?title=BME103:T930_Group_3&diff=prev&oldid=656070 | 1,395,021,731,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678704396/warc/CC-MAIN-20140313024504-00017-ip-10-183-142-35.ec2.internal.warc.gz | 96,555,665 | 10,167 | # BME103:T930 Group 3
(Difference between revisions)
Revision as of 05:29, 15 November 2012 (view source) (→Protocols)← Previous diff Revision as of 05:40, 15 November 2012 (view source) (→Results)Next diff → Line 162: Line 162: | PCR: Negative Control || 4528827 || 1.08 || Positive | PCR: Negative Control || 4528827 || 1.08 || Positive |- |- - | PCR: Positive Control || 8351858 || 2 || Positive + | PCR: Positive Control (Calf Thymus) || 8351858 || 2 || Positive |- |- | PCR: Patient 1 ID 43891, rep 1 || 5283304 || 1.26 || Positive | PCR: Patient 1 ID 43891, rep 1 || 5283304 || 1.26 || Positive Line 181: Line 181: * '''Sample''' = Sample is the DNA sample we are analyzing * '''Sample''' = Sample is the DNA sample we are analyzing * '''Integrated Density''' = The integrated density is similar to a numerical representation of the measure light in a given area. The integrated density in the above table is found by taking the integrated density of the sample and subtracting the background noise from that * '''Integrated Density''' = The integrated density is similar to a numerical representation of the measure light in a given area. The integrated density in the above table is found by taking the integrated density of the sample and subtracting the background noise from that - * '''DNA μg/mL''' = 2 * ( integrated density of sample / integrated density of positive control) + * '''DNA μg/mL''' = 2 * ( integrated density of sample / integrated density of positive control(calf thymus)) * '''Conclusion''' = If the DNA μg/mL is greater than 1 than the sample has indeed replicated. If the sample is less than 1 than it has not shown exponential replication. * '''Conclusion''' = If the DNA μg/mL is greater than 1 than the sample has indeed replicated. If the sample is less than 1 than it has not shown exponential replication.
## Revision as of 05:40, 15 November 2012
BME 103 Fall 2012 Home
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Lab Write-Up 1
Lab Write-Up 2
Lab Write-Up 3
Course Logistics For Instructors
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# OUR TEAM
Rohan Kumar Experimental Protocol Planner Kyle StonekingExperimental Protocol Planner Lekha AnantuniR&D Scientist Joshua EgerMachine Engineer Austin CuadernoMachine Engineer
# LAB 1 WRITE-UP
## Initial Machine Testing
The Original Design
The OpenPCR is a machine that is used to replicate DNA in order to amplify a specific gene. This machine primarily uses changes in temperature and various enzymes to facilitate the replication process multiple times. If the heating lid did not work, then DNA replication would not be possible. The heating lid covers the thermal cycler and holds the DNA down tight and fluctuates in temperature. The thermal cycler is where the samples are place and varies at set temperatures and times. The temperature change is crucial to the replication of DNA because only at certain temperatures dose the DNA properly replicate. The LCD display outputs the temperature of the thermal cycler. The power supply connects to an external power source.
Experimenting With the Connections
When we unplugged the lcd display wire(part 3) from the Arudnio chip(part 6), the screen turned off. Everything on the PCR was working fine expect there was no output on the display. When we unplugged the white wire that connects Arudino chip(part 6) to thermal cycler (part 2), the reading from the screen dropped to -40 degrees Celsius. We disconnected the wire multiple times and each time the screen displayed -40 degrees Celsius.
Test Run
We ran a test run on 10/25/2012. For this test we placed some empty PCR tubes into the machine and ran a simple test program on the Open PCR software. After the simple test was over we noticed that the display screen on the Open PCR lid matched very closely with what was displayed on our computer screen. The agreement between our computer screen and our PCR display meant that our diagnostic test was a success.
## Protocols
Polymerase Chain Reaction
We have been given 3 sets of samples of replicate DNA from two patients, to test for cancer makers. We labeled each sample carefully as to not cross contaminate the samples. We also used one positive control sample and one negative control, which contained no DNA,to give us a total of 8 samples. We mixed the samples together with Taq DNA polymerase, MgCl2, dNTP'S, forward primer and reverse primer. We used the PCR machine to replicate the DNA. After the PCR had finished replication, drops of the samples, mixed with syber green, were then placed in a fluorimeter. We used a Samsung Galaxy Nexus smartphone to take pictures of each drop. We then used image j to analyze the drops.
Polymerase Chain Reaction Procedure:
1.)We received 3 replicate DNA samples each from two patients and One positive control and negative control sample for a total of 8 samples. The samples we were given were already in their PCR reaction mix form. This mix contained Taq DNA polymerase, MgCl2, dNTP's, forward primer and reverse primer. Each sample was 50 micro liters.
2.)We labeled 8 empty PCR tubes. For the first sample we labeled the 3 DNA samples 1A, 1B and 1C. For the second sample we labeled the tubes 2A, 2B and 2C. For the positive and negative controls, we labeled the tubes + and - respectively.
3.)Using one pipette per sample, to avoid contamination, we transferred the PCR reaction mix we were given to the PCR tubes.
4.)We then placed the samples in the PCR machine
5.)We set our PCR program to three stages. Stage one: 1 cycle, 95 degree Celsius for 3 minutes. Stage 2: 35 cycles, 95 degrees Celsius for 30 seconds, 57 degrees Celsius for 30 seconds, 72 degrees Celsius. Stage three: 72 degrees Celsius for 3 minutes and then hold at 4 degree Celsius.
Sample one ID 43891: 48 Male
Sample two ID 36890: 56 Female
GO Taq DNA Mix
Reagent Volume
Template DNA (20 ng) 0.2 μL
10 μM forward primer 1.0 μL
10 μM reverse primer 1.0 μL
GoTaq master mix 50.0 μL
dH2O 47.8 μL
Total Volume 100.0 μL
Fluorimeter Assembly
1.) The first picture is the extertor of the fuorimeter.
2.) The second picture shows the interior. The smartphone camera is angled to have a view of the slide. During actual use the slide platform was sightly elevated with plates for a better view of the drop.
3.) The third picture shows the slide in view of the camera. During actual use the settings were changed and the lid was closed for optimal accuracy.
4.) The fourth picture is an areial view of the what the slide setup looks like. The water drop is nested in place.
Fluorimeter Procedure:
1.) Using permanent marker we numbered the transfer pipette at the bulb, so its only used for one sample
2.)With the permanent marker we also labeled the Eppendrof tubes at the top, we had a total of 10 Eppendrof tubes labeled and 10 pipettes labeled.
3.)We transferred each sample separately into the Eppendorf tubes containing 400 ml of buffer.
4.)Using a specially labeled Eppendorf tube containing SYBR GREEN, with its own pippter, we placed two drops onto the first two center drops.
5.)Then using the sample we placed two drops on top of the SYBR GREEN solution drops
6.)Then we aligned the blue light to pass through the drop.
7.)Then the smartphone operator took a picture with the settings on the phone adjusted to inactive flash, iso to 800, white balance to auto, exposure to the highest setting and contrast to the lowest setting.
8.)This process was repeated for all samples
9.)After picture was taken it was given to the Image J software
Image J Procedure:
1.)Using the menu selection we used, analyze> set measurements and chose area integrated density and mean grey value
2.)Using the menu selection we used, image > color > split channels
3.) This created 3 files
4.)The we clicked menu bar to activate the oval selection.
5.)We drew an oval around our green drop image and then selected analyze > measure.
6.) We then repeated the oval process but for the area above the drop, to get the noise measurement.
## Research and Development
Specific Cancer Marker Detection - The Underlying Technology
We added a specific primer that attaches to the cancer portion of the DNA this Primer duplicates only the DNA after the cancer portion. Because of the way DNA is replicated, after several replications we will be left with primarily the cancer portion of the DNA. The way we get the DNA to replicate is by using the open pcr machine. this machine cycles through different temperature that are optimal for the different phases of the DNA replication. We cycled the machine 30 times to be sure we had enough of the cancer DNA present in our solution.
## Results
Sample Integrated Density DNA μg/mL Conclusion PCR: Negative Control 4528827 1.08 Positive PCR: Positive Control (Calf Thymus) 8351858 2 Positive PCR: Patient 1 ID 43891, rep 1 5283304 1.26 Positive PCR: Patient 1 ID 43891, rep 2 4376083 1.05 Positive(borderline) PCR: Patient 1 ID 43891, rep 3 8144751 1.95 Positive PCR: Patient 2 ID 36890, rep 1 4190775 1 Negative(borderline) PCR: Patient 2 ID 36890, rep 2 5721560 1.37 Positive PCR: Patient 2 ID 36890, rep 3 5611447 1.34 Positive
KEY
• Sample = Sample is the DNA sample we are analyzing
• Integrated Density = The integrated density is similar to a numerical representation of the measure light in a given area. The integrated density in the above table is found by taking the integrated density of the sample and subtracting the background noise from that
• DNA μg/mL = 2 * ( integrated density of sample / integrated density of positive control(calf thymus))
• Conclusion = If the DNA μg/mL is greater than 1 than the sample has indeed replicated. If the sample is less than 1 than it has not shown exponential replication. | 2,311 | 9,701 | {"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} | 2.640625 | 3 | CC-MAIN-2014-10 | longest | en | 0.896575 |
http://www.docstoc.com/docs/78483582/Exercises---DOC---DOC | 1,397,993,151,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609538423.10/warc/CC-MAIN-20140416005218-00656-ip-10-147-4-33.ec2.internal.warc.gz | 387,012,991 | 99,412 | # Exercises - DOC - DOC
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``` 11. A blacksmith usually hammers hot metal on the surface of a
massive steel anvil. Why is this more effective than hammering the
Exercises – Chapter 1 hot metal on the surface of a thin steel plate?
E.11 The anvil’s large mass slows its acceleration, so the hot
1. A dolphin can leap several meters above the ocean’s surface. Why metal is squeezed between the moving hammer and the
doesn’t gravity stop the dolphin from leaving the water? stationary anvil.
E.1 The dolphin’s inertia carries it upward, even though its 12. A sprinter who is running a 200-m race travels the second 100 m
weight makes it accelerate downward and gradually stop in much less time than the first 100 m. Why?
rising. E.12 The sprinter must overcome inertia and accelerate during
2. As you jump across a small stream, does a horizontal force keep the first 100 m. During the second 100 m, the sprinter
you moving forward? If so, what is that force? must simply maintain full speed.
E.2 There is no force keeping you moving forward. E.12 Part of the first 100 m is traveled at less than full speed
as the sprinter struggles to accelerate from rest.
E.2 Your inertia keeps you moving forward.
13. If you pull slowly on the top sheet of a pad of paper, the whole
3. Why does stamping your feet clean the snow off them? pad will move. But if you yank suddenly on that sheet, it will tear
E.3 Your feet accelerate upward rapidly when they hit the away from the pad. What causes these different behaviors?
ground and the snow continues downward, leaving your E.13 The pad’s inertia tends to keep it in place. If you pull the
feet behind. paper away too quickly, the pad won’t be able to
4. Why does tapping your toothbrush on the sink dry it off? accelerate with the paper.
E.4 As the toothbrush stops suddenly, the water’s inertia 14. A ball falls from rest for 5 seconds. Neglecting air resistance,
keeps it going and it flies off the toothbrush. during which of the 5 seconds does the ball’s speed increase most?
E.14 The ball's speed increases steadily, so there is no second
during a collision. What type of collision causes your head to press during which its speed increases most.
against the headrest? 15. If you drop a ball from a height of 4.9 m, it will hit the ground 1 s
E.5 Any collision in which the car accelerates forward, such later. If you fire a bullet exactly horizontally from a height of 4.9 m,
as when the car is hit from behind by a faster moving it will also hit the ground 1 s later. Explain.
car. E.15 Regardless of their horizontal components of velocity, all
objects fall at the same rate. The ball and bullet descend
6. An unseatbelted driver can be injured by the steering wheel
during a head-on collision. Why does the driver hit the steering wheel together.
when the car suddenly comes to a stop? 16. An acorn falls from a branch located 9.8 m above the ground.
E.6 Nothing pushes on the driver, so the driver's inertia After 1 second of falling, the acorn’s velocity will be 9.8 m/s
causes the driver to coast forward and collide with the downward. Why hasn’t the acorn hit the ground?
steering wheel. E.16 The acorn's average velocity during this second is less
than 9.8 m/s.
E.6 The car stops because something pushes it backward
E.16 The acorn started from rest, so it isn't traveling 9.8 m/s at
7. Why do loose objects on the dashboard slide to the right when the
car turns suddenly to the left? first.
E.7 As it turns left, the car accelerates left. The loose objects 17. A diver leaps from a 50-m cliff into the water below. The cliff is
remain behind and end up on the right side of the not perfectly vertical so the diver must travel forward several meters
dashboard. in order to avoid the rocks beneath him. In fact, he leaps directly
forward rather than upward. Explain why a forward leap allows him
8. Why is your velocity continuously changing as you ride on a to miss the rocks.
carousel?
E.17 The forward component of his velocity remains constant
E.8 Your direction of travel changes all the time and a as he falls, and he follows an arc that carries him forward
change in direction involves an acceleration over the rocks.
9. When you apply the brakes on your bicycle, which way do you 18. The kicker in a sporting event isn’t always concerned with how
accelerate? far downfield the ball travels. Sometimes the ball’s flight time is
more important. If he wants to keep the ball in the air as long as
E.9 Backward, in the direction opposite your forward
possible, which way should he kick it?
velocity.
E.18 He should kick the ball straight up.
10. One type of home coffee grinder has a small blade that rotates
very rapidly and cuts the coffee beans into powder. Nothing prevents E.18 Although the ball will eventually land right where it
the coffee beans from moving so why don’t they get out of the way started, that ball will stay in the air a long time.
when the blade begins to push on them?
19. The heads of different golf clubs are angled to give the golf ball
E.10 The beans' inertias keep them from moving out of the different initial velocities. The golf ball’s speed remains almost
way. constant, but the angle changes with the different clubs. Neglecting
any air effects, how does changing the initial angle of the ball affect
E.10 Because the beans have mass, a force must push on them
the distance the ball travels?
to cause them to accelerate and move out of the way.
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E.19 In the absence of air effects, a ball hit at 45° above 30. When you fly a kite, there is a time when you must do (positive)
horizontal will travel farthest. A ball hit higher or lower work on the kite. Is that time when you let the kite out or when you
won’t travel as far. pull it in?
20. A speedboat is pulling a water-skier with a rope, exerting a large E.30 You do positive work when you pull the kite in.
forward force on her. The skier is traveling forward in a straight line E.30 When you pull the kite in you pull toward you and the
at constant speed. What is the net force she experiences?
kite moves toward you, so you do work on the kite.
E.20 She experiences zero net force.
31. Which does more work in lifting a grain of rice over its head: an
E.20 Since she isn't accelerating, she must be experiencing a ant or a person? Use this result to explain how insects can perform
net force of zero. seemingly incredible feats of lifting and jumping.
21. Your suitcase weighs 50 N. As you ride up an escalator toward E.31 A person does more work. Movements that appear large
the second floor, carrying that suitcase, you are traveling at a constant compared to the ant’s height still involve small distances
velocity. How much upward force must you exert on the suitcase to and little work.
keep it moving with you?
E.21 Its magnitude must equal the suitcase’s weight.
E.32 Yes, you are doing work whenever you push on the
22. What is the net force on (a) the first car, (b) the middle car, and bread and it moves in the direction of that push.
(c) the last car of a metro train traveling at constant velocity?
33. While hanging a picture, you accidentally dent the wall with a
E.22 All three cars experience zero net force. hammer. Did the hammer do work on the wall?
E.22 Since they aren't accelerating, the cars must be E.33 Yes, it pushed the wall inward and the wall dented
experience net forces of zero. inward.
23. Two teams are having a tug-of-war with a sturdy rope. It has 34. You’re cutting wood with a handsaw. You have to push the saw
been an even match so far, with neither team moving. What is the net away from you as it moves away from you and pull the saw toward
force on the left team? you as it moves toward you. When are you doing work on the saw?
E.23 Zero net force. E.34 In both cases, you do work on the saw: when you push it
away from you as it moves away from you and as you
24. When you kick a soccer ball, which pushes on the other harder: pull it toward you as it moves toward you.
your foot or the soccer ball?
E.34 Whenever you push something in the direction it moves,
E.24 The two push equally hard on one another (but in
you do work on it.
opposite directions).
35. The steel ball in a pinball game rolls around a flat, tilted surface.
E.24 In accordance with Newton's third law, the foot and ball
If you flick the ball straight uphill, it gradually slows to a stop and
must push equally hard on one another; they are a then begins to roll downhill. Which way is the ball accelerating as it
Newton's third law pair. rolls uphill? downhill?
25. The earth exerts a downward force of 850 N on a veteran E.35 As it rolls on the surface, it always accelerates downhill.
astronaut as he works outside the space shuttle. What force (if any)
does the astronaut exert on the earth? 36. Why do less snow and other debris accumulate on a steep roof
than on a flatter roof?
E.25 The astronaut exerts an upward force of 850 N on the
earth. E.36 The steeper the roof, the larger the downhill force on the
snow and debris and the more likely they are to slide off
26. A car passes by, heading to your left, and you reach out and push the roof.
it toward the left with a force of 50 N. Does this moving car push on
you and, if so, with what force? 37. You roll a marble down a playground slide that starts level, then
curves downward, and finally curves very gradually upward so that
E.26 The car pushes you to the right with a force of 50 N. it’s level again at the end. Where along its travel does the marble
E.26 In accordance with Newton's third law, you and the car experience its greatest acceleration? its greatest speed?
must push equally hard on one another; you are a E.37 Its greatest acceleration is at the steepest point; its
Newton's third law pair. greatest speed is at the bottom of the slide.
27. Which is larger: the force the earth exerts on you or the force you 38. When you’re roller skating on level pavement, you can maintain
exert on the earth? your speed for a long time. But as soon as you start up a gradual hill,
E.27 Both forces have exactly the same magnitude. you begin to slow down. What slows you?
28. Comic book superheroes often catch a falling person only a E.38 The downhill force pushes you in the direction opposite
hairsbreadth from the ground. Why would this rescue actually be just your motion and do negative work on you.
as fatal for the victim as hitting the ground itself? E.38 As you roll up a hill, the downhill force acting on you
E.28 The upward acceleration needed to stop the falling pushes you in the direction opposite your motion.
person is enough to injure them severely, whether the Gravity is doing negative work on you and you slow
ground or a superhero exerts that force. down.
29. You accidentally miss the doorway and run into the wall. You 39. When the brakes on his truck fail, the driver steers it up a
suddenly experience a backward force that is several times larger runaway truck ramp. As the truck rolls up the ramp, it slows to a stop.
than your weight. What’s the origin of this force? What happens to the truck’s kinetic energy, its energy of motion?
E.29 The wall exerts a support force to accelerate you E.39 The kinetic energy becomes gravitational potential
backward. energy.
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16. How much work was done in raising one of the blocks in
Problems – Chapter 1 Problem 15 to a height of 50 m?
P.16 It takes 9.8 million joules of work to raise a 20,000-kg
1. If your car has a mass of 800 kg, how much force is required to block 50 meters upward.
accelerate it forward at 4 m/s2?
17. What is the gravitational potential energy of one of the blocks in
P.1 3200 N. Problem 15 if it’s now 75 m above the ground?
2. If your car accelerates from rest at a steady rate of 4 m/s2, how P.17 14,700,000 J.
soon will it reach 88.5 km/h (55.0 mph or 24.6 m/s)?
18. As water descends from the top of a tall hydroelectric dam, its
P.2 It will take 6.15 seconds to reach 88.5 km/h. gravitational potential energy is converted to electric energy. How
much gravitational potential energy is released when 1000 kg of
3. On Mars, the acceleration due to gravity is 3.71 m/s2. What would water descends 200 m to the generators?
a rock’s velocity be 3 s after you dropped it on Mars?
P.18 1000 kg of water releases 1.96 million joules of
P.3 11.13 m/s. gravitational potential energy in descending 200 meters.
4. How far would a rock fall in 3 s if you dropped it on Mars? (See 19. The tire of your bicycle needs air so you attach a bicycle pump to
Problem 3.) it and begin to push down on the pump’s handle. If you exert a
P.4 The rock will fall 16.7 meters during those 3 seconds. downward force of 25 N on the handle and the handle moves
downward 0.5 m, how much work do you do?
5. How would your Mars weight compare to your earth weight? (See
Problem 3.) P.19 12.5 J.
P.5 Your Mars weight would be about 38% of your earth 20. You’re using a wedge to split a log. You are hitting the wedge
weight. with a large hammer to drive it into the log. It takes a force of 2000 N
to push the wedge into the wood. If the wedge moves 0.2 m into the
6. A basketball player can leap upward 0.5 m. What is his initial log, how much work have you done on the wedge?
velocity at the start of the leap?
P.20 You have done 400 joules of work on the wedge.
P.6 The basketball player's initial upward velocity is 3.1 m/s.
21. The wedge in Problem 20 acts like a ramp, slowly splitting the
7. How long does the basketball player in Problem 6 remain in the wood apart as it enters the log. The work you do on the wedge,
air? pushing it into the log, is the work it does on the wood, separating its
P.7 About 0.64 s (0.32 s on the way up and 0.32 s on the way two halves. If the two halves of the log only separate by a distance of
down). 0.05 m while the wedge travels 0.2 m into the log, how much force is
the wedge exerting on the two halves of the log to rip them apart?
8. A sprinter can reach a speed of 10 m/s in 1 s. If the sprinter’s
P.21 8000 N.
acceleration is constant during that time, what is the sprinter’s
acceleration? 22. You’re sanding a table. You must exert a force of 30 N on the
P.8 The sprinter's acceleration is 10 m/s2 sandpaper to keep it moving steadily across the table’s surface. You
slide the paper back and forth for 20 minutes, during which time you
9. If a sprinter’s mass is 60 kg, how much forward force must be move it 1000 m. How much work have you done?
exerted on the sprinter to make the sprinter accelerate at 0.8 m/s2?
P.22 You have done 30,000 joules of work.
P.9 48 N.
Exercises – Chapter 2
10. How much does a 60-kg person weigh on earth?
P.10 The 60-kg person weighs about 588 newtons.
11. If you jump upward with a speed of 2 m/s, how long will it take 1. The chairs in an auditorium aren’t all facing the same direction.
before you stop rising? How could you describe their angular positions in terms of a
P.11 About 0.20 s. reference orientation and a rotation?
12. How high will you be when you stop rising in Problem 11? E.1 The angle by which the front, center seat would have to
be rotated, as viewed from above, to have each seat’s
P.12 You will rise to a height of 0.20 meters. orientation.
13. How much force must a locomotive exert on a 12,000-kg boxcar 2. When an airplane starts its propellers, they spin slowly at first and
to make it accelerate forward at 0.4 m/s2? gradually pick up speed. Why does it take so long for them to reach
P.13 4800 N their full rotational speed?
E.2 The propellers have rotational inertia (as measured by
14. How long will it take the boxcar in Problem 13 to reach its
their moments of inertia).
cruising speed of 100 km/h (62 mph or 28 m/s)?
P.14 The boxcar will have to accelerate for 70 seconds to E.2 The propellers must undergo angular acceleration and
reach cruising speed. only gradually speed up to their full rotational speeds.
15. The builders of the pyramids used a long ramp to lift 20,000-kg 3. A mechanic balances the wheels of your car to make sure that
(22-ton) blocks. If a block rose 1 m in height while traveling 20 m their centers of mass are located exactly at their geometrical centers.
along the ramp’s surface, how much uphill force was needed to push Neglecting friction and air resistance, how would an improperly
it up the ramp at constant velocity? balanced wheel behave if it were rotating all by itself?
P.15 9800 N. E.3 It would rotate about its center of mass, not its geometric
center. It would appear to wobble as it turned.
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4. An object’s center of mass isn’t always inside the object, as you 14. Tightrope walkers often use long poles for balance. Although the
can see by spinning it. Where is the center of mass of a boomerang or poles don’t weigh much, they can exert substantial torques on the
a horseshoe? walkers to keep them from tipping and falling off the ropes. Why are
the poles so long?
E.4 A boomerang or horseshoe's center of mass is in the air
between the two arms of the object. E.14 The longer the pole is, the greater its rotational mass and
the more torque that is required to start it rotating
5. Why is it hard to start the wheel of a roulette table spinning, and significantly.
what keeps it spinning once it’s started?
E.14 When the pole has a large rotational mass, the tightrope
E.5 The wheel has rotational inertia, as measured by its
walker can exert large torques on the pole without
rotational mass, making it hard to start and stop
causing it to rotate quickly. The pole twists back on the
spinning.
tightrope walker and helps the tightrope walker remain
6. Why can’t you open a door by pushing its doorknob directly upright.
toward or away from its hinges?
15. Some racing cars are designed so that their massive engines are
E.6 A force exerted directly toward or away from the axis of near their geometrical centers. Why does this design make it easier
rotation produces zero torque about that axis of rotation. for these cars to turn quickly?
7. Why can’t you open a door by pushing on its hinged side? E.15 It reduces the car’s rotational mass so that the car can
undergo rapid angular accelerations and change
E.7 A force exerted at the hinges produces no torque about
directions quickly.
them.
16. How does a bottle opener use mechanical advantage to pry the
8. It’s much easier to carry a weight in your hand when your arm is
top off a soda bottle?
at your side than it is when your arm is pointing straight out in front
of you. Use the concept of torque to explain this effect. E.16 A modest force exerted far from the pivot produces an
enormous force close to the pivot. Although the opener's
E.8 When you arm is pointing straight out in front of you,
handle must travel a long distance, it produces the huge
any weight force exerted on your hand is at right angles
force on the bottle cap that's required to pull that cap off
to the lever arm between your shoulder and hand and
the bottle.
other hand, when you arm is at your side, any weight 17. A jar-opening tool grabs onto a jar’s lid and then provides a long
force exerted on your hand is directed away from your handle for you to turn. Why does this handle’s length help you to
9. A gristmill is powered by falling water, which pours into buckets E.17 By pushing far from the pivot, you exert more torque on
on the outer edge of a giant wheel. The weight of the water turns the the lid.
wheel. Why is it important that those buckets be on the wheel’s outer
edge? 18. When you climb out on a thin tree limb, there’s a chance that the
limb will break off near the trunk. Why is this disaster most likely to
E.9 The farther the water is from the water wheel’s pivot, the occur when you’re as far out on the limb as possible?
more torque its weight produces on the wheel.
E.18 The farther out the limb that your weight is exerted on
10. How does the string of a yo-yo get the yo-yo spinning? the branch, the larger the torque you produce on the limb
and the more likely it is to begin rotating.
E.10 The string pulls on the outside edge of the yo-yo's
spindle and at right angles to the lever arm between the 19. How does a crowbar make it easier to lift the edge of a heavy box
yo-yo's rotational axis and the point at which the force a few centimeters off the ground?
acts. As a result, it produces a torque on the yo-yo about
E.19 Your small effort exerted on the crowbar far from its
its rotational axis and causes the yo-yo to undergo
pivot produces a large force on the box, located near the
angular acceleration.
crowbar’s pivot.
11. One way to crack open a walnut is to put it in the hinged side of a
20. The basket of a wheelbarrow is located in between its wheel and
door and then begin to close the door. Why does a small force on the
its handles. How does this arrangement make it relatively easy for
door produce a large force on the shell?
E.11 Your force far from the hinges produces a large torque.
E.20 A modest upward force exerted on the handles far from
To oppose this torque, the nut must exert a huge force
the pivot can balance a large downward weight force
near the hinges.
exerted on the basket close to the pivot.
12. A common pair of pliers has a place for cutting wires, bolts, or
21. Skiers often stop by turning their skis sideways and skidding
nails. Why is it so important that this cutter be located very near the
them across the snow. How does this trick remove energy from a
pliers’ pivot?
skier, and what happens to that energy?
E.12 The closer the wire is to the axis of pliers' axis of
E.21 Skidding sideways does work against sliding friction,
rotation, the less effective any force from the wire is at
converting some of the skier’s kinetic energy into
producing a torque on the pliers and stopping its rotation.
thermal energy.
13. You can do push-ups with either your toes or your knees acting
22. A horse does work on a cart it’s pulling along a straight, level
road at a constant speed. The horse is transferring energy to the cart,
so why doesn’t the cart go faster and faster? Where is the energy
Explain.
going?
E.13 The weights of your chest and your feet exert torques in
E.22 The work that the horse does on the cart is wasted in
opposing friction. The work becomes thermal energy.
balance one another.
Page 4 of 38 - 5/2/2011 - 11:51:27 AM
23. Explain why a rolling pin flattens a piecrust without encountering 33. In countless movie and television scenes, the hero punches a
very much sliding friction as it moves. brawny villain who doesn’t even flinch at the impact. Why is the
immovable villain a Hollywood fantasy?
E.23 The pin’s surface turns with the crust and doesn’t slide
across it. E.33 To avoid accelerating when pushed on with a force, the
villain would have to have infinite mass. That’s
24. Professional sprinters wear spikes on their shoes to prevent them impossible.
from sliding on the track at the start of a race. Why is energy wasted
whenever a sprinter’s foot slides backward along the track? 34. Why can’t an acrobat stop himself from spinning while he is in
midair?
E.24 If the sprinter's foot slides backward, then friction from
the ground on the foot does negative work on the foot E.34 The acrobat has angular momentum and cannot change
and extracts energy from the sprinter. That energy that angular momentum without experience an external
becomes thermal energy in the foot and ground. torque. While in the air, he can't experience an external
torque.
25. A yo-yo is a spool-shaped toy that spins on a string. In a
sophisticated yo-yo, the end of the string forms a loop around the yo- 35. While a gymnast is in the air during a leap, which of the
yo’s central rod so that the yo-yo can spin almost freely at the end of following quantities must remain constant for her: velocity,
the string. Why does the yo-yo spin longest if the central rod is very momentum, angular velocity, or angular momentum?
thin and very slippery?
E.35 Angular momentum.
E.25 The nearer the frictional force is to the pivot, the less
torque it produces to slow the yo-yo’s rotation. 36. If you sit in a good swivel chair with your feet off the floor, the
Slipperiness reduces friction. chair will turn slightly as you move about but will immediately stop
moving when you do. Why can’t you make the chair spin without
26. As you begin pedaling your bicycle and it accelerates forward, touching something?
what is exerting the forward force that the bicycle needs to
E.36 Because of your rotational inertia (as measured by your
accelerate?
rotational mass), you need an external torque in order to
E.26 The ground exerts a forward frictional force on the begin spinning. While you aren't touching anything in
bicycle wheel. the swivel chair, you can't obtain any external torque and
can't start spinning.
27. When you begin to walk forward, what exerts the force that
allows you to accelerate? 37. When a star runs out of nuclear fuel, gravity may crush it into a
E.27 A static frictional force from the pavement pushes you neutron star about 20 km (12 miles) in diameter. While the star may
have taken a year or so to rotate once before its collapse, the neutron
forward.
star rotates several times a second. Explain this dramatic increase in
28. If you are pulling a sled along a level field at constant velocity, angular velocity.
how does the force you are exerting on the sled compare to the force E.37 The collapsing star’s angular momentum can’t change.
of sliding friction on its runners? Since its rotational mass decreases, its angular velocity
E.28 The forward force you exert on the sled must balance the must increase.
backward force that friction exerts on the sled.
38. A toy top spins for a very long time on its sharp point. Why does
29. Why does putting sand in the trunk of a car help to keep the rear it take so long for friction to slow the top’s rotation?
wheels from skidding on an icy road? E.38 Any frictional force on the toy top is exerted so close to
E.29 Pressing the wheels more tightly against the pavement the top's axis of rotation that it exerts almost zero torque.
increases the maximum force that static friction can exert Without any significant external torque, the top's angular
on the wheels. momentum keeps it spinning for a very long time.
30. When you’re driving on a level road and there’s ice on the 39. It’s easier to injure your knees and legs while hiking downhill
pavement, you hardly notice that ice while you’re heading straight at than while hiking uphill. Use the concept of energy to explain this
a constant speed. Why is it that you only notice how slippery the road observation.
is when you try to turn left or right, or to speed up or slow down?
E.39 As you descend, you land hard and your knees and legs
E.30 While you are traveling straight at constant speed, you must convert your kinetic energy into thermal energy.
are coasting and experience zero net force. It's only when Injuries can occur.
you try to accelerate horizontally that you need a
frictional force from the pavement and find yourself 40. When you first let go of a bowling ball, it’s not rotating. But as it
slides down the alley, it begins to rotate. Use the concept of energy to
explain why the ball’s forward speed decreases as it begins to spin.
31. Describe the process of writing with chalk on a blackboard in E.40 When the bowling ball is rotating, it has rotational
terms of friction and wear. kinetic energy. That energy must come from somewhere
E.31 The chalk experiences sliding friction as you write and and since the only type of energy the ball has as it first
leaves visible wear chips on the blackboard. begins to slide down the lane is translational kinetic
energy, the rotational energy must come from the
32. Falling into a leaf pile is much more comfortable than falling translational kinetic energy. As a result, the translational
onto the bare ground. In both cases you come to a complete stop, so
kinetic energy must decrease, so the ball must slow
why does the leaf pile feel so much better?
down.
E.32 When you land on a leaf pile, a modest upward force
exerted for a long time stops you. When you land on 41. Firefighters slide down a pole to get to their trucks quickly. What
happens to their gravitational potential energy, and how does it
bare ground, a large upward force exerted for a small
depend on the slipperiness of the pole?
time stops you. Obviously, the modest force is more
comfortable than the large force.
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E.41 Sliding friction converts some into heat, but a slippery P.8 The fly's momentum is 0.0001 kg-m/s.
pole converts a considerable fraction into kinetic energy.
9. Your car is broken, so you’re pushing it. If your car has a mass of
800 kg, how much momentum does it have when it’s moving forward
Problems – Chapter 2
at 3 m/s (11 km/h)?
P.9 2400 kg·m/s forward.
1. When you ride a bicycle, your foot pushes down on a pedal that’s 10. You begin pushing the car forward from rest (see Problem 9).
17.5 cm (0.175 m) from the axis of rotation. Your force produces a Neglecting friction, how long will it take you to push your car up to a
torque on the crank attached to the pedal. Suppose that you weigh speed of 3 m/s on a level surface if you exert a constant force of 200
700 N. If you put all your weight on the pedal while it’s directly in N on it?
front of the crank’s axis of rotation, what torque do you exert on the P.10 It will take 12 seconds to push the car up to 3 m/s.
crank?
11. When your car is moving at 3 m/s (see Problems 9 and 10), how
P.1 About 122.5 N·m to the left. much translational kinetic energy does it have?
2. An antique carousel that’s powered by a large electric motor P.11 3600 J.
undergoes constant angular acceleration from rest to full rotational
speed in 5 seconds. When the ride ends, a brake causes it to 12. No one is driving your car (see Problems 9, 10, and 11) and it
decelerate steadily from full rotational speed to rest in 10 seconds. crashes into a parked car at 3 m/s. Your car comes to a stop in just 0.1
Compare the torque that starts the carousel to the torque that stops it. s. What force did the parked car exert on it to stop it that quickly?
P.2 The starting torque must be twice a large as the stopping P.12 The parked car exerts a force of 24000 N on your car.
torque and the two torques must be in opposite
13. You’re at the roller-skating rink with a friend who weighs twice
directions.
as much as you do. The two of you stand motionless in the middle of
3. When you start your computer, the hard disk begins to spin. It the rink so that your combined momentum is zero. You then push on
takes 6 seconds of constant angular acceleration to reach full speed, one another and begin to roll apart. If your momentum is then 450
at which time the computer can begin to access it. If you wanted the kg·m/s to the left, what is your friend’s momentum?
disk drive to reach full speed in only 2 seconds, how much more P.13 450 kg·m/s to the right.
torque would the disk drive’s motor have to exert on it during the
starting process? 14. A tricycle releases 50 J of gravitational potential energy while
rolling 0.5 m directly downhill along a ramp. What is the downhill
P.3 Three times as much torque.
force acting on the tricycle?
4. An electric saw uses a circular spinning blade to slice through P.14 100 N.
wood. When you start the saw, the motor needs 2 seconds of constant
angular acceleration to bring the blade to its full angular velocity. If 15. A highly compressed spring releases 2 J of elastic potential
you change the blade so that the rotating portion of the saw now has energy when allowed to expand 1 mm, a small fraction of its overall
three times its original rotational mass, how long will the motor need compression. How much force is the spring exerting on its end?
to bring the blade to its full angular velocity?
P.15 2000 N.
P.4 The motor will need 6 seconds of constant acceleration
to bring the new blade up to full angular velocity. 16. An elevator releases 10000 J of gravitational potential energy
while descending 5 m between floors. How much does the elevator
P.4 With 3 times the rotational mass, the angular weigh?
acceleration of the new blade is only 1/3 what it was P.16 2000 N.
originally.
17. A 10-m long elastic band is used to launch a toy glider. Once
5. When the saw in Problem 4 slices wood, the wood exerts a 100-N stretched to a length of 30 m, the band releases 1 J of elastic potential
force on the blade, 0.125 m from the blade’s axis of rotation. If that energy when you let it contract 0.1 m. What forward force is the band
force is at right angles to the lever arm, how much torque does the exerting on the toy glider?
wood exert on the blade? Does this torque make the blade turn faster
or slower? P.17 10 N.
P.5 12.5 N·m, slowing the blade.
6. When you push down on the handle of a doll-like wooden
nutcracker, its jaw pivots upward and cracks a nut. If the point at Exercises – Chapter 3
which you push down on the handle is five times as far from the pivot
as the point at which the jaw pushes on the nut, how much force will 1. In what way does the string of a bow and arrow behave like a
the jaw exert on the nut if you exert a force of 20 N on the handle? spring?
(Assume all forces are at right angles to the lever arms involved.) E.1 As you draw the string away from its equilibrium shape,
P.6 The jaw will exert a force of 100 N on the nut. it experiences a restoring force proportional to its
displacement.
7. Some special vehicles have spinning disks (flywheels) to store
energy while they roll downhill. They use that stored energy to lift 2. As you wind the mainspring of a mechanical watch or clock, why
themselves uphill later on. Their flywheels have relatively small does the knob get harder and harder to turn?
rotational masses but spin at enormous angular speeds. How would a E.2 As the mainspring winds farther and farther from its
flywheel’s kinetic energy change if its rotational mass were five equilibrium shape, the restoring force it exerts on its end
times larger but its angular speed were five times smaller? gets stronger. You must overcome this increasing force
P.7 Its energy would be only 0.2 times as large as before. as you twist the knob.
8. What is the momentum of a fly if it’s traveling 1 m/s and has a
mass of 0.0001 kg?
Page 6 of 38 - 5/2/2011 - 11:51:27 AM
3. Curly hair behaves like a weak spring that can stretch under its 12. The best running tracks have firm but elastic rubber surfaces.
own weight. Why is a hanging curl straighter at the top than at the How does a lively surface assist a runner?
bottom?
E.12 An elastic track stores energy as it dents when the
E.3 The top of the curl supports more weight than the runner's foot presses against it and then returns that
bottom. energy to the runner as it undents.
4. When you lie on a spring mattress, it pushes most strongly on the 13. Why is it so exhausting to run on soft sand?
parts of you that stick into it the farthest. Why doesn’t it push up
E.13 You do work on the sand as you step on it, but the sand
doesn’t return this energy to you as you lift your foot
E.4 The deeper you dent the mattress's surface at any given back up again.
point, the stronger the local restoring force becomes.
14. Steep mountain roads often have emergency ramps for trucks
5. If you pull down on the basket of a hanging grocery store scale so with failed brakes. Why are these ramps most effective when they are
that it reads 15 N, how much downward force are you exerting on the covered with deep, soft sand?
E.14 Sand dents easily as the truck plows through it and
E.5 15 N. extracts energy from the truck. The truck does work in
pushing the sand out of the way. The sand converts that
6. While you’re weighing yourself on a bathroom scale, you reach
work into safe thermal energy.
out and push downward on a nearby table. Is the weight reported by
the scale high, low, or correct? 15. There have been baseball seasons in which so many home runs
E.6 The scale reads low; it reports less than your actual were hit that people began to suspect that something was wrong with
weight. the baseballs. What change in the baseballs would account for them
traveling farther than normal?
E.6 The upward force that the table exerts on your hand
E.15 An increase in the balls’ coefficients of restitution.
contributes to the force supporting you against gravity
and allows the scale to exert a smaller upward force on 16. During rehabilitation after hand surgery, patients are often asked
you. to squeeze and knead putty to strengthen their muscles. How does the
energy transfer in squeezing putty differ from that in squeezing a
7. There’s a bathroom scale on your kitchen table and your friend rubber ball?
climbs up to weigh himself on it. One of the table’s legs is weak and
you’re afraid that he’ll break it, so you hold up that corner of the E.16 Energy transferred while deforming putty is converted
table. The table remains level as you push upward on the corner with into thermal energy and never returns to the person's
a force of 100 N. Is the weight reported by the scale high, low, or hand. However, energy transferred while deforming a
correct? rubber ball becomes elastic potential energy in the ball
and returns to the person's hand when the rubber ball
E.7 Correct.
returns to its spherical shape.
8. If you put your bathroom scale on a ramp and stand on it, will the
weight it reports be high, low, or correct? 17. Your car is on a crowded highway with everyone heading south
at about 100 km/h (62 mph). The car ahead of you slows down
E.8 The scale will read less than your actual weight. slightly and your car bumps into it gently. Why is the impact so
gentle?
E.8 The scale only reports how hard it pushes perpendicular
to its surface. When you weigh yourself on a ramp, your E.17 Your relative velocity is small—in your frame of
weight doesn't push directly toward the scale's surface reference the car in front of you is barely moving, so the
and so the scale doesn't push back directly perpendicular impact is very gentle.
to its surface. The perpendicular component of its force
18. Bumper cars are an amusement park ride in which people drive
on you is less than you full weight.
small electric vehicles around a rink and intentionally bump them
9. When you step on a scale, it reads your weight plus the weight of into one another. All of the cars travel at about the same speed. Why
your clothes. Only your shoes are touching the scale, so how does the are head-on collisions more jarring than other types of collisions?
weight of the rest of your clothes contribute to the weight reported by E.18 During a head-on collision, the relative velocity is
the scale? enormous--the sum of the two individual velocities--and
E.9 You support your clothes and the scale supports you. the forces, accelerations, and momentum transfers are
enormous as well.
10. To weigh an infant you can step on a scale once with the infant
and then again without the infant. Why is the difference between the 19. When two trains are traveling side by side at breakneck speed,
scale’s two readings equal to the weight of the infant? it’s still possible for people to jump from one train to the other.
Explain why this can be done safely.
E.10 When you weigh yourself with the child, the scale
reports the total force needed to support you both. When E.19 Because the trains’ relative velocity is zero, a person
you weigh yourself alone, the scale reports only the force jumping between them views them both as essentially
needed to support you. The difference between these two stationary.
reported values is the force needed to support only the
20. If you drop a steel marble on a wooden floor, why does the floor
child.
receive most of the collision energy and contribute most of the
11. An elastic ball that wastes 30% of the collision energy as heat rebound energy?
when it bounces on a hard floor will rebound to 70% of the height E.20 The marble is stiffer than the floor, so the floor dents
from which it was dropped. Explain the 30% loss in height. more than the marble. Since the forces are equal in
E.11 A 30% loss of rebound height is a 30% loss of magnitude, the floor also has the most work done on it
gravitational potential energy—equal to the energy that during the collision and is thus responsible for most of
became thermal energy. the rebound energy.
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21. A RIF (reduced injury factor) baseball has the same coefficient of 30. In some roller coasters, the cars travel through a smooth tube that
restitution as a normal baseball except that it deforms more severely bends left and right in a series of complicated turns. Why does the car
during a collision. Why does this increased deformability lessen the always roll up the right-hand wall of the tube during a sharp left-hand
forces exerted by the ball during a bounce and reduce the chances of turn?
its causing injury?
E.30 When the track makes a sharp left-hand turn, the car
E.21 During a bounce, the work done on a RIF ball—to store needs a strong leftward force to follow it. The car
energy in it—involves a smaller force exerted for a obtains that leftward force by riding up on the right-hand
longer distance. wall of the tube.
22. Padded soles in running shoes soften the blow of hitting the 31. Railroad tracks must make only gradual curves to prevent trains
pavement. Why does padding reduce the forces involved in bringing from derailing at high speeds. Why is a train likely to derail if it
your foot to rest? encounters a sharp turn while it’s traveling fast?
E.22 With padding in your shoes, you feet stop over a longer E.31 The sharper the curve, the more centripetal force the
period of time when they encounter the pavement. The train needs to accelerate around the curve. If the track
momentum transfer or impulse is the same, but the force can’t supply it . . . disaster.
is smaller while the time is longer.
32. Police sometimes use metal battering rams to knock down doors.
23. Some athletic shoes have inflatable air pockets inside them. They hold the ram in their hands and swing it into a door from about
These air pockets act like springs that become stiffer as you pump up 1 m away. How does the battering ram increase the amount of force
the air pressure. High pressure also makes you bounce back up off the police can exert on the door?
the floor sooner. Why does high pressure shorten the bounce time?
E.32 The police can use a long period of time and a modest
E.23 At high pressure, the shoes are stiffer and exert larger forward force to give the battering ram forward
forces when distorted. They accelerate more rapidly and momentum. The battering ram can then transfer this
bounce faster. momentum to a door with a huge force exerted for a
short period of time.
24. Why does it hurt less to land on a soft foam pad than on bare
concrete after completing a high jump? 33. When a moving hammer hits a nail, it exerts the enormous force
E.24 The foam pad brings you to rest over a longer period of needed to push the nail into wood. This force is far greater than the
hammer’s weight. How is it produced?
time and thus with a small upward force. The smaller
upward force is less uncomfortable than the huge upward E.33 The nail exerts an enormous force on the hammer to
force that the concrete would exert on you while slow it down. The hammer pushes back, driving the nail
stopping you quickly. into the wood.
25. Why must the surface of a hammer be very hard and stiff for it to 34. A hammer’s weight is downward, so how can a hammer push a
drive a nail into wood? nail upward into the ceiling?
E.25 The force that the hammer exerts on the nail during their E.34 When the moving hammer encounters the nail, the nail
collision would diminish if the hammer’s surface pushes back extremely hard on the hammer to stop the
distorted easily. hammer from penetrating into the nail. The hammer
pushes back on the nail, propelling it into the ceiling.
26. Some amusement park rides move you back and forth in a While gravity also pushes on the hammer and slows its
horizontal direction. Why is this motion so much more disturbing to
upward motion, the weight forces in this situation are
your body than cruising at a high speed in a jet airplane?
trivial compared to the forces between hammer and nail.
E.26 You only feel accelerations, so rapid side-to-side motion
is easily felt. In contrast, rapid travel at constant velocity 35. As you swing back and forth on a playground swing, your
involves no acceleration and produces no sensations at apparent weight changes. At what point do you feel the heaviest?
all. E.35 At the bottom of each swing.
27. You are traveling in a subway along a straight, level track at a 36. Some stores have coin-operated toy cars that jiggle back and
constant velocity. If you close your eyes, you can’t tell which way forth on a fixed base. Why can’t these cars give you the feeling of
you’re heading. Why not? actually driving in a drag race?
E.27 While you can feel accelerations, you can’t feel velocity. E.36 A jiggling car can only accelerate you forward for a brief
period of time, while a true drag racer will accelerate you
28. Moving a can of spray paint rapidly in one direction will not mix
forward for a long period of time. You can feel the
it nearly as well as shaking it back and forth. Why is it so important
forward acceleration as a backward gravity-like
to change directions as you mix the paint?
sensation.
E.28 To make the paint move relative to the can and mix as a
result, you must make the can accelerate. The paint will 37. A salad spinner is a rotating basket that dries salad after washing.
then coast around inside the can and slam into the walls. How does the spinner extract the water?
If you just move the can quickly in one direction, the E.37 As the salad undergoes rapid centripetal acceleration, the
paint and can will coast along together and there will be water travels in straight lines and runs off the salad.
little mixing.
38. People falling from a high diving board feel weightless. Has
29. Why does a baby’s rattle only make noise when the baby moves gravity stopped exerting a force on them? If not, why don’t they feel
it back and forth and not when the baby moves it steadily in one it?
direction?
E.38 A falling person's internal organs all fall together without
E.29 When the rattle accelerates, the beads inside it continue having to support one another. The absence of internal
on and hit the walls of the rattle. The rattle then makes forces within a person's body is what leads them to
noise. believe that they are weightless.
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39. When your car travels rapidly over a bump in the road, you P.6 The satellite must be moving horizontally at about 7,900
suddenly feel weightless. Explain. m/s. It will complete its trip in about 5100 seconds or 84
E.39 After going over a bump, the car begins to accelerate minutes.
downward and your apparent weight is briefly less than 7. When you put water in a kitchen blender, it begins to travel in a 5-
your real weight. cm-radius circle at a speed of 1 m/s. How quickly is the water
accelerating?
40. Astronauts learn to tolerate weightlessness by riding in an
airplane (nicknamed the ―vomit comet‖) that follows an unusual P.7 20 m/s2.
trajectory. How does the pilot direct the plane in order to make its
occupants feel weightless? 8. In Problem 7, how hard must the sides of the blender push inward
on 0.001 kg of the spinning water?
E.40 The pilot steers the "vomit comet" in the path of a falling
object: a parabolic arc through the air. The net force on P.8 The blender must push inward on the water with a force
the plane is then exactly equal to its weight and it is truly of 0.02 newtons.
falling. The objects inside it fall as well and everyone
Exercises – Chapter 4
inside feels weightless.
41. You board an elevator with a large briefcase in your hand. Why
does that briefcase suddenly feel particularly heavy when the elevator
begins to move upward? 1. Sprinters start their races from a crouched position with their
bodies well forward of their feet. This position allows them to
E.41 As the car accelerates upward, you must pull upward on accelerate quickly without tipping over backward. Explain this effect
the briefcase extra hard to make it accelerate upward too. in terms of torque and center of mass.
42. As your car reaches the top in a smoothly turning Ferris wheel, E.1 As a sprinter accelerates, the ground must push toward
which way are you accelerating? the sprinter’s center of mass to avoid exerting a torque
E.42 You are accelerating downward at the top of a smoothly on the sprinter.
turning Ferris wheel. 2. If the bottom of your bicycle’s front wheel becomes caught in a
E.42 In uniform circular motion, you always accelerate toward storm drain, your bicycle may flip over so that you travel forward
the center of the circle. over the front wheel. Explain this effect in terms of rotation, torque,
and center of mass.
E.2 If the storm drain pushes back hard on the bicycle's
Problems – Chapter 3 wheel and stops the wheel's forward motion, this
backward force produces a torque on the bicycle and
rider about their combined center of mass. Although the
1. Your new designer chair has an S-shaped tubular metal frame that bicycle is normally in stable static equilibrium about its
behaves just like a spring. When your friend, who weighs 600 N, sits front-back direction, this huge torque can overcome that
on the chair, it bends downward 4 cm. What is the spring constant for stability and rotate the rider and bicycle so that they tip
this chair?
over.
P.1 15,000 N/m.
3. When you turn while running, you must lean in the direction of a
2. You have another friend who weighs 1000 N. When this friend turn or risk falling over. If you lean left as you turn left, why don’t
sits on the chair from Problem 1, how far does it bend? you fall over to the left?
P.2 The chair bends downward 6.7 cm. E.3 The leftward frictional force on your feet keeps you
balanced.
3. You’re squeezing a springy rubber ball in your hand. If you push
inward on it with a force of 1 N, it dents inward 2 mm. How far must 4. If a motorcycle accelerates too rapidly, its front wheel will rise up
you dent it before it pushes outward with a force of 5 N? off the pavement. During this stunt the pavement is exerting a
forward frictional force on the rear wheel. How does that frictional
P.3 10 mm.
force cause the front wheel to rise?
4. When you stand on a particular trampoline, its springy surface E.4 The huge forward force on the bottom of the rear wheel
shifts downward 0.12 m. If you bounce on it so that its surface shifts produces a torque on the motorcycle and rider about
downward 0.30 m, how hard is it pushing up on you?
their combined center of mass. If this torque is large
P.4 The trampoline is pushing upward on you with a force enough, it overcomes the motorcycle's natural forward-
that is 2.5 times your weight. back stability and the motorcycle rotates so that its front
wheel leaves the pavement.
5. Engineers are trying to create artificial ―gravity‖ in a ring-shaped
space station by spinning it like a centrifuge. The ring is 100 m in 5. As a skateboard rider performs stunts on the inside of a U-shaped
radius. How quickly must the space station turn in order to give the surface, he often leans inward toward the middle of the U. Why does
astronauts inside it apparent weights equal to their real weights at the leaning keep him from falling over?
earth’s surface?
E.5 As long as the inward force from the U-shaped surface
P.5 It must turn at about 31 m/s. points toward the skaterboard’s center of mass, the
skaterboard doesn’t begin rotating.
6. A satellite is orbiting the earth just above its surface. The
centripetal force making the satellite follow a circular trajectory is 6. Most racing cars are built very low to the ground. While this
just its weight, so its centripetal acceleration is about 9.8 m/s2 (the design reduces air resistance, it also gives the cars better dynamic
acceleration due to gravity near the earth’s surface). If the earth’s stability on turns. Why are these low cars more stable than taller cars
radius is about 6375 km, how fast must the satellite be moving? How with similar wheel spacings?
long will it take for the satellite to complete one trip around the
earth?
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E.6 The torques that friction exerts on a car when it pushes E.14 Throwing one shoe at 10 m/s takes more energy than
on the car's wheels can cause the car to tip over. But if throwing two shoes at 5 m/s, even though the momentum
the car's center of mass is very low, the lever arm of the transfer is the same in both cases.
frictional forces is small and the torques that friction
produces about the car's center of mass are small. With 15. You are propelling yourself across the surface of a frozen lake by
hitting tennis balls toward the southern shore. From your perspective,
only small torques acting on it, the car stays upright
each ball you hit heads southward at 160 km/h. You have a huge bag
because of its natural static stability.
of balls with you and are already approaching the northern shore at a
7. The crank of a hand-operated kitchen mixer connects to a large speed of 160 km/h (100 mph). When you hit the next ball southward,
gear. This gear meshes with smaller gears attached to the mixing will you still accelerate northward?
blades. Since each turn of the crank makes the blades spin several E.15 Yes.
times, how is the force you exert on the crank handle related to the
forces the mixing blades exert on the batter around them? 16. Can you use the tennis ball scheme of Exercise 15 to propel
yourself to any speed, or are you limited to the speed at which you
E.7 The torque you exert on the crank is much larger than the
can hit the tennis balls?
torque the blades exert on the batter.
E.16 You can reach any speed you like, so long as you start
8. The starter motor in your car is attached to a small gear. This gear with enough of your total mass in the form of tennis
meshes with a large gear that’s attached to the engine’s crankshaft. balls.
The starter motor must turn many times to make the crankshaft turn
just once. How does this gearing allow modest forces inside the 17. You and a friend are each wearing roller skates. You stand facing
starter motor to turn the entire crankshaft? each other on a smooth, level surface and begin tossing a heavy ball
back and forth. Why do you drift apart?
E.8 Since the starter motor must turn many times in order for
the crankshaft to turn just once, the starter motor can use E.17 The ball transfers momentum between you so that you
small forces exerted over long distances to produce large acquire momentum in the opposite direction from your
forces on the engine exerted over short distances. friend.
9. A bread-making machine uses gears to reduce the rotational speed 18. The time it takes a relatively small object to orbit a much larger
of its mixing blade. While its motor spins about 50 times each object doesn’t depend on the mass of the small object. Use an
second, the blade spins only once each second. The motor provides a astronaut who is walking in space near the space shuttle to illustrate
certain amount of work each second, so why does this arrangement of that point.
gears allow the machine to exert enormous forces on the bread
E.18 The astronaut and the space shuttle both fall toward the
dough?
ground at the same rate and their paths around the earth
E.9 Each second, the blade pushes the dough a short are essentially identical. That's why the astronaut and
distance. To do significant work, the force it exerts on shuttle hover next to one another, despite the absence of
the dough must be large. any forces between them.
10. The chain of a motorcycle must be quite strong. Since the 19. As the moon orbits the earth, which way is the moon
motorcycle has only one sprocket on its rear wheel, the top of the accelerating?
chain must pull forward hard to keep the rear wheel turning as the
motorcycle climbs a hill. Why would replacing the motorcycle’s rear E.19 Directly toward the earth.
sprocket for one with more teeth make it easier for the chain to keep 20. Which object is exerting the stronger gravitational force on the
the rear wheel turning? other, the earth or the moon, or are the forces equal in magnitude?
E.10 The more teeth there are on the rear sprocket, the farther E.20 The forces are equal in magnitude
the chain must travel before the rear wheel turns once.
The chain still does the same amount of work in turning 21. To free an Apollo spacecraft from the earth’s gravity took the
the rear wheel once, but it can now do it while pushing efforts of a gigantic Saturn V rocket. Freeing a lunar module from the
the rear sprocket a longer distance, so the force it exerts moon’s gravity took only a small rocket in the lunar module’s base.
on the sprocket must be smaller. Why was it so much easier to escape from the moon’s gravity than
from that of the earth?
11. As you clear the sidewalk with a leaf blower, the blower pushes
E.21 A rocket must do much less work against the moon’s
you away from the leaves. What is pushing on the blower so that it
can push on you? gravity.
E.11 As the blower pushes air toward the leaves, that air 22. Spacecraft in low earth orbit take about 90 minutes to circle the
pushes the blower away from the leaves. earth. Why can’t they be made to orbit the earth in half that amount
of time?
12. When a sharpshooter fires a pistol at a target, the gun recoils
E.22 If a spacecraft were to go faster, its increased kinetic
backward very suddenly, leaping away from the target. Explain this
recoil effect in terms of the transfer of momentum. energy would cause it to swing away from the earth's
surface into a huge elliptical orbit. It would then take
E.12 The gun transfers forward momentum to the bullet, so longer to complete the orbit, not shorter.
the bullet must transfer an equal but oppositely directed
amount of momentum to the gun. 23. As a comet approaches the sun, it arcs around the sun more
rapidly. Explain.
13. Which action will give you more momentum toward the north:
throwing one shoe southward at 10 m/s or two shoes southward at 5 E.23 Since a line from the sun to the comet must sweep out
m/s? area at a steady rate, the closer the comet gets to the sun,
the faster it must arc around the sun.
E.13 They are equivalent.
24. Mars has a larger orbital radius than the earth. Compare the solar
14. Do both of the actions in Exercise 13 take the same amount of years on those two planets.
energy? If not, which one requires more energy?
Page 10 of 38 - 5/2/2011 - 11:51:27 AM
E.24 The solar year on Mars is longer than on the earth. 14. By what factor does the relativistic energy in Problem 13
increase if you subtract out the constant rest energy?
Problems – Chapter 4
P.14 5.6.
1. If an 80-kg baseball pitcher wearing frictionless roller skates
picks up a 0.145-kg baseball and pitches it toward the south at 42 m/s
Exercises – Chapter 5
(153 km/h or 95 mph), how fast will he begin moving toward the
1. A helium-filled balloon floats in air. What will happen to an air-
north?
filled balloon in helium? Why?
E.1 It will sink. Its average density is larger than that of
2. How high above the earth’s surface would you have to be before helium.
your weight would be only half its current value?
2. A log is much heavier than a stick, yet both of them float in water.
P.2 You would weigh half your normal weight if you were Why doesn’t the log’s greater weight cause it to sink?
about 2640 km above the earth's surface.
E.2 Although the log may weigh more than the stick, the log
3. As you walked on the moon, the earth’s gravity would still pull on also displaces more water. Ultimately, it’s the wood's
you weakly and you would still have an earth weight. How large density that matters in floating, not the wooden object's
would that earth weight be, compared to your earth weight on the total weight.
earth’s surface? (Note: The earth’s radius is 6378 km and the distance
separating the centers of the earth and moon is 384,400 kilometers.) 3. An automobile will float on water as long as it doesn’t allow
water to leak inside. In terms of density, why does admitting water
P.3 About 0.00028 times your earth weight. cause the automobile to sink?
4. If you and a friend 10 m away each have masses of 70 kg, how E.3 As water enters the car, the car’s average density will
much gravitational force are you exerting on your friend? increase.
P.4 You are exerting gravitational forces of about 3.27 10-9 4. Many grocery stores display frozen foods in bins that are open at
newtons on one another. the top. Why doesn’t the warm room air enter the bins and melt the
food?
5. The gravity of a black hole is so strong that not even light can
escape from within its surface or ―event horizon.‖ Even outside that E.4 Colder air is denser than hotter air and so the warm room
surface, enormous energies are needed to escape. Suppose that you air floats on the cold air in the bins.
are 10 km away from the center of a black hole that has a mass of
1031 kg. If your mass is 70 kg, how much do you weigh? 5. Some clear toys contain two colored liquids. No matter how you
tilt one of those toys, one liquid remains above the other. What keeps
P.5 About 4.7 1014 N. the upper liquid above the lower liquid?
6. In Problem 5, how much work would something have to do on E.5 The upper liquid is less dense than the lower liquid. It
you to lift you 1 m farther away from the black hole? floats.
P.6 Lifting you 1 meter farther from the black hole would 6. When the car you are riding in stops suddenly, heavy objects
involve doing 4.7 1014 joules of work. move toward the front of the car. Explain why a helium-filled balloon
will move toward the rear of the car.
7. A 1000-kg spacecraft passes you traveling forward at ½ the speed
of light. What is its relativistic momentum? E.6 The denser air pushes the less dense helium out of its
way as the car and its contents slow to a stop. The air
P.7 1.73 10 kg·m/s.
11
coasts into the front windshield while it pushes the
8. A spacecraft passes you traveling forward with at 1/3 the speed of helium balloon toward the rear window.
light. By what factor would its relativistic momentum increase if its 7. Water settles to the bottom of a tank of gasoline. Which takes up
speed doubled? more space: 1 kg of water or 1 kg of gasoline?
P.8 About 2.53. E.7 One kilogram of gasoline takes up more space than 1 kg
9. For a rocket traveling at 10,000 m/s, by what factor does its of water.
relativistic momentum differ from its ordinary momentum? 8. Oil and vinegar salad dressing settles with the oil floating on top
P.9 5.56 10-10. of the vinegar. Explain this phenomenon in terms of density.
10. What is the rest energy of 1 kg of uranium? E.8 The oil is less dense than the water, so the oil floats on
water.
P.10 9 10 J.
16
9. When a fish is floating motionless below the surface of a lake,
11. How much mass has a rest energy of 1000 J? what is the amount and direction of the force the water is exerting on
P.11 1.1 10-14 kg. it?
E.9 An upward force equal in magnitude to the fish’s weight.
12. A 1000-kg spacecraft passes you at ½ the speed of light. What is
its relativistic energy? 10. Some fish move extremely slowly, and it’s hard to tell whether
they are even alive. However, if a fish is floating at a middle height in
P.12 1.04 10 J.20
your aquarium and not at the top or bottom of the water, you can be
13. A spacecraft passes you at 1/3 the speed of light. By what factor pretty certain that it’s alive. Why?
would its relativistic energy increase if its speed doubled? E.10 For a dead fish to hover in the middle of the aquarium,
P.13 1.6. its density would have to be exactly that of the
surrounding water, an unlikely coincidence.
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11. A barometer, which is often used to monitor the weather, is a through that faucet decreases. The tea can no longer
device that measures air pressure. How could you use a barometer to obtain as much kinetic energy as before and travels more
measure your altitude as you climbed in the mountains? slowly.
E.11 Air pressure decreases steadily with altitude. 21. Why must tall dams be so much thicker at their bases than at
12. If you seal a soft plastic bottle or juice container while hiking their tops?
high in the mountains and then return to the valley, the container will E.21 The pressure imbalance on the base of the dam is larger.
be dented inward. What causes this compression?
22. Waterproof watches have a maximum depth to which they can
E.12 At high altitude, the air is less dense so filling the bottle safely be taken while swimming. Why?
with high altitude air means that the bottle contains
relatively few air molecules. When you return to the E.22 The deeper the watch goes in the water, the greater the
valley, the bottle's relative lack of contents allow the surrounding water pressure. Eventually the pressure
surrounding air pressure to partially crush the bottle. becomes so high that it pushes the water through the
watch's seals and into the watch.
13. Many jars have dimples in their lids that pop up when you open
the jar. What holds the dimple down while the jar is sealed, and why 23. When you stand in a pool with water up to your neck, you find
does it pop up when the jar is opened? that it’s somewhat more difficult to breathe than when you’re out of
the water. Why?
E.13 Air pressure holds the dimple down when the jar has a
vacuum inside, but the pressure difference vanishes E.23 The pressure outside your lungs is above atmospheric
when the jar is opened. pressure.
14. If you place a hot, wet cup upside down on a smooth counter for 24. How does pushing on the plunger of a syringe cause medicine to
a few seconds, you may find it difficult to lift up again. What is flow into a patient through a hollow hypodermic needle?
holding that cup down on the counter? E.24 Pushing on the plunger squeezes the medicine in the
E.14 When the air inside the cup cools, its pressure drops. The syringe and raises its pressure. With a pressure
pressure imbalance between atmospheric pressure imbalance between this high-pressure medicine and the
outside and partial vacuum inside the cup produces low pressure in the patient, the medicine accelerates
inward forces on the cup. These ultimately push the cup toward the patient and flows into them.
against the counter. 25. Why must the pressure inside a whistle teakettle exceed
15. You seal a rigid container that is half full of hot food and put it in atmospheric pressure before the whistle can begin to make noise?
the refrigerator. Why is the container’s lid bowed inward when you E.25 Gas only accelerates toward lower pressure.
look at it later?
26. Each time you breathe in, air accelerates toward your nose and
E.15 Cooling the air in the container reduced its pressure. The lungs. How does the pressure in your lungs compare with that in the
resulting pressure imbalance across the lid pushes the lid surrounding air as you breathe in?
inward.
E.26 As you breathe in, the pressure in your lungs is less than
16. Why aren’t there any thermometers that read temperatures down the surrounding pressure.
to -300 ºC?
27. You can inflate a plastic bag by holding it up so that it catches
E.16 That temperature is below absolute zero, the lowest the wind. Use Bernoulli’s equation to explain this effect.
possible temperature.
E.27 As wind slows in the bag, its pressure rises and inflates
17. You use your breath to inflate a large rubber tube and then ride the bag.
down a snowy hill on it. After a few minutes in the snow the tube is
underinflated. What happened to the air? 28. When someone pulls a fire alarm in a skyscraper, pumps increase
the water pressure in the section of the building nearest that alarm
E.17 As it cooled, the air became more dense. box. How does this pressure change assist firefighters who must
18. A marshmallow is filled with air bubbles. Why does a battle the blaze?
marshmallow puff up when you toast it? E.28 Increasing the local water pressure allows the firefighters
E.18 As you heat the air inside the marshmallow's bubbles, to send more water through their hoses and have it travel
the pressure increases and the air pushes the each faster when it leaves their nozzles.
bubble's skin outward. The bubbles grow larger and so
Problems – Chapter 5
does the entire marshmallow.
19. Wasp and hornet sprays proudly advertise just how far they can
send insecticide. How does the pressure inside the spray can affect
that distance, and why is the direction of the spray important (vertical 1. The particle density of standard atmospheric air at 273.15 K (0
vs. horizontal)? ºC) is 2.687 1025 particles/m3. Using the ideal gas law, calculate the
pressure of this air.
E.19 Higher pressure in the can produces greater distance of
spray. As for direction, the spray won’t travel as far P.1 101,400 Pa.
upward because some of its kinetic energy becomes 2. How much force is the air exerting on the front surface of this
gravitational potential energy on the way up. book?
20. Ice tea is often dispensed from a large jug with a faucet near the P.2 The force on the front surface of this book is about 5,000
bottom. Why does the speed of tea flowing out of the faucet decrease newtons.
as the jug empties?
3. If you fill a container with air at room temperature (300 K), seal
E.20 As the level of tea in the jug decreases, the pressure at the container, and then heat the container to 900 K, what will the
the faucet decreases and the total energy of tea passing pressure be inside the container?
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P.3 Three times as much as before. 15. To dive far below the surface of the water, a submarine must be
able to withstand enormous pressures. At a depth of 300 m, what
4. An air compressor is a device that pumps air particles into a tank. pressure does water exert on the submarine’s hull?
A particular air compressor adds air particles to its tank until the
particle density of the inside air is 30 times that of the outside air. If P.15 About 3,100,000 Pa.
the temperature inside the tank is the same as that outside, how does
the pressure inside the tank compare to the pressure outside?
P.4 The pressure inside the tank is about 30 times that of the
pressure outside it.
Exercises – Chapter 6
5. If you seal a container of air at room temperature (20 ºC) and then 1. A favorite college prank involves simultaneously flushing several
put it in the refrigerator (2 ºC), how much will the pressure of the air toilets while someone is in the shower. The cold water pressure to the
in the container change? shower drops and the shower becomes very hot. Why does the cold
water pressure suddenly drop?
P.5 0.94 times as much as before.
E.1 The increased flow in the cold water pipe requires a
6. If you submerge an 8-kg log in water and it displaces 10 kg of larger pressure difference in it, leaving too little pressure
water, what will the net force on the log be the moment you let go of to operate the shower.
the log?
2. On hot days in the city people sometimes open up fire hydrants
P.6 The net force on the log is 19.8 newtons in the upward and play in the water. Why does this activity reduce the water
direction. pressure in nearby hydrants?
7. If your boat weighs 1200 N, how much water will it displace E.2 With the hydrants open, water flows faster through the
when it’s floating motionless at the surface of a lake? city water mains and experiences greater losses of total
P.7 122.4 kg or 0.1224 m3. energy due to viscous effects. The water pressure drops
significantly as the water passes through the mains and is
8. The density of gold is 19 times that of water. If you take a gold small by the time it reaches the other hydrants.
crown weighing 30 N and submerge it in water, what will the
buoyant force on the crown be? 3. Why does a relatively modest narrowing of the coronary arteries,
the blood vessels supplying blood to the heart, cause a dramatic drop
P.8 The buoyant force on the crown is about 1.6 newtons. in the amount of blood flowing through them?
9. How much upward force must you exert on the submerged crown E.3 Flow through an artery scales with the fourth power of
in Problem 8 to keep it from accelerating? radius.
P.9 28.4 N. 4. Why does hot maple syrup pour more easily than cold maple
10. How could you use your results from Problems 8 and 9 to syrup?
determine whether the crown was actually gold rather than gold- E.4 At higher temperatures, the molecules in maple syrup
plated copper? (The density of copper is nine times that of water.) have more thermal energy and are able to break free of
P.10 A gold crown weighing 30 newtons can be supported by one another more easily as the syrup flows.
an upward force of 28.4 newtons when submerged in 5. Why is ―molasses in January‖ slower than ―molasses in July,‖ at
water, while a copper crown weighing 30 newtons can least in the northern hemisphere?
be supported by an upward force of only 26.7 newtons
when submerged in water. E.5 In warm weather, the molecules in a viscous liquid have
more thermal energy and are able to move past one
11. Your town is installing a fountain in the main square. If the water another more easily.
is to rise 25 m (82 feet) above the fountain, how much pressure must
the water have as it moves slowly toward the nozzle that sprays it up 6. Why is it so difficult to squeeze ketchup through a very small hole
into the air? in its packet?
P.11 About 250,000 Pa above atmospheric pressure. E.6 Ketchup is relatively viscous and even the ketchup that is
farthest from the sides of the hole is slowed by viscous
12. Rather than putting a pump in the fountain (see Problem 11), the interactions with the surrounding ketchup.
town engineer puts a water storage tank in one of the nearby high-rise
office buildings. How high up in that building should the tank be for 7. A baker is decorating a cake by squeezing frosting out of a sealed
its water to rise to 25 m when spraying out of the fountain? (Neglect paper cone with the tip cut off. If the baker makes the hole at the tip
friction.) of the cone too small, it’s extremely difficult to get any frosting to
flow out of it. Why?
P.12 The tank must be 25 meters above the fountain.
E.7 Squeezing highly viscous frosting through a narrow
13. To clean the outside of your house you rent a small high-pressure ―pipe‖ requires an enormous pressure difference across
water sprayer. The sprayer’s pump delivers slow-moving water at a that pipe.
pressure of 10,000,000 Pa (about 150 atmospheres). How fast can
this water move if all of its pressure potential energy becomes kinetic 8. Why is the wind stronger several meters above a flat field than it
energy as it flows through the nozzle of the sprayer? is just above the ground?
P.13 141 m/s or 510 km/h. E.8 Air very close to the ground is slowed by viscous forces
and forms a boundary layer of relatively slowly moving
14. When the water from the sprayer in Problem 13 hits the side of air.
your house, it slows to a stop. If it hasn’t lost any energy since
leaving the sprayer, what is the pressure of the water at the moment it 9. Pedestrians on the surface of a wind-swept bridge don’t feel the
stops completely? full intensity of the wind because, near the bridge’s surface, the air is
moving relatively slowly. Explain this effect in terms of a boundary
P.14 The water's pressure at the moment it comes to a stop is layer.
10,000,000 Pa.
Page 13 of 38 - 5/2/2011 - 11:51:27 AM
E.9 Air near the stationary bridge surface is slowed by E.19 The dimpled ball will hit first.
viscous forces, creating a slower moving boundary layer.
20. If you ride your bicycle directly behind a large truck, you will
10. An electric valve controls the water for the lawn sprinklers in find that you don’t have to pedal very hard to keep moving forward.
your backyard. Why do the pipes in your home shake whenever this Why?
valve suddenly stops the water but not when this valve suddenly
E.20 You will be pedaling inside the truck's turbulent wake
starts the water?
and will find that the air there is moving along at roughly
E.10 When the valve abruptly closes, the moving water in the the truck's speed.
pipe must suddenly get rid of its momentum and it
shakes the pipes as it does. When the valve opens, the 21. How does running directly behind another runner reduce the
water can slowly pick up speed and momentum, so no wind resistance you experience?
shaking occurs. E.21 The front runner drags the air forward so that you
experience less drag while running through forward-
11. If you drop a full can of applesauce and it strikes a cement floor
moving air.
squarely with its flat bottom, what happens to the pressures at the top
and bottom of the can? 22. When a car is stopped, its flexible radio antenna points straight
E.11 The pressure at the can bottom rises suddenly due to up. But when the car is moving rapidly down a highway, the antenna
arcs toward the rear of the car. What force is bending the antenna?
water hammer. The pressure at the can top falls or
remains the same. E.22 Pressure drag is pushing the antenna backward.
12. When you mix milk or sugar into your coffee, you should move 23. To drive along a level road at constant velocity, your car’s engine
the spoon quickly enough to produce turbulent flow around the must be running and friction from the ground must be pushing your
spoon. Why does this turbulence aid mixing? car forward. Since the net force on an object at constant velocity is
zero, why do you need this forward force from the ground?
E.12 If only laminar flow occurs in your coffee, the sugar or
milk won't mix thoroughly because it will flow in an E.23 It balances the backward force due to air drag.
orderly pattern within the coffee. But once turbulence
appears, the coffee, sugar, and milk disperse and mix 24. Racing bicycles often have smooth disk-shaped covers over the
spokes of their wheels. Why would these thin wire spokes be a
thoroughly as each portion of material separates into
problem for a fast-moving bicycle?
countless tiny fragments.
E.24 The spokes experience pressure drag because they
13. If you start two identical paper boats from the same point, you produce turbulent wakes as the wheel spins.
can make them follow the same path down a quiet stream. Why can’t
you do the same on a brook that contains eddies and vortices? 25. A bullet slows very quickly in water but a spear doesn’t. What
force acts to slow these two objects, and why does the spear take
E.13 In turbulent flow, adjacent regions of water become
longer to stop?
separated.
E.25 Both objects are slowed by similar pressure drags. But
14. When you swing a stick slowly through the air, it’s silent. But the spear’s larger mass and momentum keep it from
when you swing it quickly, you hear a ―whoosh‖ sound. What slowing as quickly.
behavior of the air is creating that noise?
26. If you hang a tennis ball from a string, it will deflect downwind
E.14 The fast-moving stick creates a noisy turbulent wake.
in a strong breeze. But if you wet the ball so that the fuzz on its
15. If you try to fill a bucket by holding it in a waterfall, you will surface lies flat, it will deflect even more than before. Why does
find the bucket pushed downward with enormous force. How does smoothing the ball increase its deflection?
the falling water exert such a huge downward force on the bucket? E.26 Smoothing the ball prevents its fuzz from "tripping" the
E.15 The bucket stops the water, so its pressure rises. The boundary layer in the air flowing past it. As a result, the
higher pressure above the bucket pushes it downward wet ball creates a larger turbulent wake and experiences
hard. more pressure drag.
16. You sometimes see paper pressed tightly against the front of a 27. Explain why a parachute slows your descent when you leap out
moving car. What holds the paper in place? of an airplane.
E.16 The paper is held in place by the high-pressure air that E.27 It increases pressure drag and reduces your terminal
forms when the airstream encountering the front of the velocity.
car slows down.
28. Bicycle racers sometimes wear teardrop-shaped helmets that
17. Fish often swim just upstream or downstream from the posts taper away behind their heads. Why does having this smooth taper
supporting a bridge. How does the water’s speed in these regions behind them reduce the drag forces they experience relative to those
compare with its speed in the open stream and at the sides of the they would experience with more ball-shaped helmets?
posts? E.28 The teardrop shape assists air in flowing around the back
E.17 The water slows down just upstream and downstream of of the helmet and delays flow separation. The resulting
the posts and speeds up on the sides of the posts. turbulent wake is smaller.
18. Why do flyswatters have many holes in them? 29. If you want the metal tubing in your bicycle to experience as
little drag as possible while you’re riding in a race, is cylindrical
E.18 Without holes in it, a flyswatter would develop a layer of
tubing the best shape? How should it be shaped?
high-pressure air that moves along with it. This high-
pressure region would tend to push the fly out of its way. E.29 An airfoil, round in front and tapered behind, would be
better.
19. You have two golf balls that differ only in their surfaces. One has
dimples on it while the other is smooth. If you drop these two balls 30. In 1971, astronaut Alan Shepard hit a golf ball on the moon. How
simultaneously from a tall tower, which one will hit the ground first? did the absence of air affect the ball’s flight?
Page 14 of 38 - 5/2/2011 - 11:51:27 AM
E.30 In the absence of air, the ball traveled in a simple both backward and upward. How is the airstream exerting an upward
parabolic arc. It could not fly the way a golf ball with force on your hand?
backspin does in the earth's atmosphere. E.39 Air travels faster over your hand than under it, so the
31. A water-skier skims along the surface of a lake. What types of pressure above your hand is less than the pressure under
forces is the water exerting on the skier, and what is the effect of your hand.
these forces?
40. When a hummingbird hovers in front of a flower, what forces are
E.31 Lift supports the skier and drag pulls the skier backward. acting on it and what is the net force it experiences?
32. How would a Frisbee fly on the airless moon? E.40 The hovering hummingbird experiences zero net force:
the upward aerodynamic force it experiences exactly
E.32 The Frisbee would fly in a simple parabolic arc, the same balances its downward weight.
way a rock would.
41. A poorly designed household fan stalls, making it inefficient at
33. You can buy special golf tees that wrap around behind the ball to moving air. Describe the airflow through the fan when its blades stall.
prevent you from giving it any spin when you hit it. These tees are
guaranteed to prevent hooks and slices (i.e., curved flights). But how E.41 Turbulent air pockets form behind its moving blades, so
do these tees affect the distance the ball travels? Why? swirling vortices and eddies flow out of the fan.
E.33 Without backspin, the ball can’t obtain lift and won’t go 42. When a plane enters a steep dive, the air rushes toward it from
as far. below. If the pilot pulls up suddenly from such a dive, the wings may
abruptly stall, even though the plane is oriented horizontally. Explain
34. A hurricane or gale force wind can lift the roof off a house, even why the wings stall.
when the roof has no exposed eaves. How can wind blowing across a
roof produce an upward force on it? E.42 Since the air approaches the diving plane from below,
the wings' angle of attach must be measured relative to
E.34 When wind blows horizontally across a pointed roof, it
the air's upward flow. Orienting the wings horizontally
undergoes an inward bend. The pressure at the surface of
when the wind is flowing upward means that the angle of
the roof drops below atmospheric pressure and the speed
attack is huge. The uprushing air can't stay attached to
of the air there rises. With normal atmospheric pressure
the wing after flowing forward and around the leading
below the roof and less than atmospheric pressure above
edge of the wing, so the wing stalls.
it, the roof experiences a net upward pressure force.
35. A skillful volleyball player can serve the ball so that it barely
spins at all. The ball dithers slightly from side to side as it flies over
the net and is hard to return. What causes the ball to accelerate Problems – Chapter 6
sideways?
1. About how fast can a small fish swim before experiencing
E.35 Small disturbances in the airflow around the sides of the turbulent flow around its body?
nonspinning volleyball produce lift forces that push the
ball to the side. P.1 For a fish with an obstacle length (width) of about 2 cm,
turbulence will appear if it moves faster than about 0.1
36. Why does an airplane have a ―flight ceiling,‖ a maximum altitude m/s.
above which it can’t obtain enough lift to balance the downward
force of gravity? 2. How much higher must your blood pressure get to compensate for
a 5% narrowing in your blood vessels? (The pressure difference
E.36 The higher the airplane goes, the less dense the air across your blood vessels is essentially equal to your blood pressure.)
becomes and the harder it gets for the plane to transfer
enough downward momentum to the air to keep itself P.2 Your blood pressure would have to increase by about
aloft. At a certain height, the plane can only just obtain 23% to compensate.
enough upward momentum from the air to keep from 3. If someone replaced the water in your home plumbing with olive
falling. oil, how much longer would it take you to fill a bathtub?
37. If you let a stream of water from a faucet flow rapidly over the P.3 About 84 times as long.
curved bottom of a spoon, the spoon will be drawn into the stream.
Explain this effect. 4. You are trying to paddle a canoe silently across a still lake and
know that turbulence makes noise. How quickly can the canoe and
E.37 The water speeds up around the curve and its pressure the paddle travel through water without causing turbulence?
drops. The resulting pressure imbalance pushes the
spoon into the stream. P.4 Assuming that the canoe is about 1 m wide, then the
Reynolds number exceeds 2000 when the canoe is
38. If you put your hand out the window of a moving car, so that traveling about 0.002 m/s. The canoe must travel very
your palm is pointing directly forward, the force on your hand is slowly to avoid turbulence.
directly backward. Explain why the two halves of the airstream,
passing over and under your hand, don’t produce an overall up or 5. The pipes leading to the showers in your locker room are old and
down force on your hand. inadequate. While the city water pressure is 700,000 Pa, the pressure
in the locker room when one shower is on is only 600,000 Pa. Use
E.38 When your palm is pointing forward, the two airstreams Eq. 6.1.1 to calculate the approximate pressure if three showers are
(over and under your hand) are symmetric. Although on.
both airstreams experience pressure drops as they arc
inward around your hand's edges, the pressure drops are P.5 About 400,000 Pa.
symmetric and there is no net up or down pressure force 6. If the plumbing in your dorm carried honey instead of water,
on your hand. filling a cup to brush your teeth could take awhile. If the faucet takes
39. If instead of holding your hand palm forward (see Exercise 38), 5 s to fill a cup with water, how long would it take to fill your cup
you tip your palm slightly downward, the force on your hand will be with honey, assuming all the pressures and pipes remain unchanged?
Page 15 of 38 - 5/2/2011 - 11:51:27 AM
P.6 It would take about 5 million seconds to fill your cup E.9 Heat rises with convection and ignites the rest of the
with honey. stick.
7. How quickly would you have to move a 1-cm-diameter stick 10. It’s often a good idea to wrap food in aluminum foil before
through olive oil to reach a Reynolds number of 2000, so that you baking it near the red-hot heating element of an electric oven. Why
would begin to see turbulence around the stick? (Olive oil has a does this wrapped food cook more evenly?
density of 918 kg/m3.)
E.10 Covering the food with aluminum foil protects the food
P.7 About 18.3 m/s. from the intense radiative heat transfer that the red-hot
heating element can provide. Instead of burning on the
8. The effective obstacle length of a blimp is its width—the distance
heating element side, the food cooks more slowly via
to which the air is separated as it flows around the blimp. How slowly
convection and warms more evenly.
would a 15-m-wide blimp have to move in order to keep the airflow
around it laminar? (Air has a density of 1.25 kg/m3.) 11. Why are black steam radiators better at heating a room than
P.8 The blimp's Reynolds number would exceed 2000 when radiators that have been painted white or silver?
it reached a speed of 0.002 m/s. Somewhere just above E.11 Black radiators are very efficient at transferring heat via
that speed, the airflow around the blimp becomes radiation. White or silver radiators radiate heat less well
turbulent. and aren't as effective at heating their surroundings.
12. The space shuttle generates thermal energy during its operation
Exercises – Chapter 7 in earth orbit. How is it able to get rid of that thermal energy as heat
in an airless environment?
E.12 The space shuttle radiates its excess heat away into cold,
1. Your body is presently converting chemical potential energy from dark space.
food into thermal energy at a rate of about 100 J/s, or 100 W. If heat
were flowing out of you at a rate of about 200 W, what would happen 13. Some air fresheners are solid materials that have strong odors. If
to your body temperature? you leave these air fresheners out, they slowly disappear. What
happens to the solid material?
E.1 Your body temperature would decrease.
E.13 The solid material sublimes and its molecules become
2. You can use a blender to crush ice cubes, but if you leave it gaseous.
churning too long, the ice will melt. What supplies the energy needed
to melt the ice? 14. Frozen vegetables will ―freeze-dry‖ if they’re left in cold, dry air.
How can water molecules leave the frozen vegetables?
E.2 The blender blades do work on the ice and this work
becomes thermal energy in the ice. E.14 The water molecules sublime from the frozen vegetables,
going directly from the solid phase to the gaseous phase.
3. When you knead bread, it becomes warm. From where does this
thermal energy come? 15. If you remove ice cubes from the freezer with wet hands, the
cubes often freeze to your fingers. How can the ice freeze the water
E.3 The work you do in pushing the bread in the direction it
moves.
E.15 The ice is colder than its melting temperature and must
4. You throw a ball into a box and close the lid. You hear the ball warm up before it starts melting.
bouncing around inside as the ball’s energy changes from
gravitational potential energy to kinetic energy, to elastic potential 16. On a bitter cold day, the snow is light and powdery. This snow
energy, and so on. If you wait a minute or two, what will have doesn’t begin melting immediately when you bring it into a warm
happened to the ball’s energy? room. Why?
E.4 The ball's energy will eventually become thermal energy. E.16 The cold snow is well below its melting temperature and
it must absorb a moderate amount of heat in order to
5. Why do meats and vegetables cook much more quickly when warms up to that melting temperature.
there are metal skewers sticking through them?
E.5 Foods are poor conductors of heat. The metal skewer 17. When you put a loaf of bread in a plastic bag, there is still air
around the bread. Why doesn’t the bread dry out as quickly in the bag
carries heat better.
as it does when there’s no bag around it?
6. Why do aluminum pans heat food much more evenly than E.17 The bag traps steam so that the humidity soon rises to
stainless steel pans when you cook on a stove? 100% so that net evaporation ceases.
E.6 Aluminum is a better conductor of heat than stainless
18. Even on a very humid day the hot air from your blow dryer can
steel, so heat disperses more quickly in an aluminum pan
extract moisture from your hair. Why is heated air able to dry your
and the food cooks more evenly on that pan.
hair when the air around you can’t?
7. Use the concept of convection to explain why firewood burns E.18 Heating the air lowers its relative humidity. Increasing
better when it’s raised above the bottom of a fireplace on a grate. the temperature stabilizes the gaseous phase of water
E.7 The grate lets convection carry air upward past the wood. relative to the liquid phase.
8. Explain how convection contributes to the shape of a candle 19. If you add a little hot tea to ice water at 0 °C, the mixture will
flame. end up at 0 °C so long as some ice remains. Where does the tea’s
extra thermal energy go?
E.8 The rising hot air of convection draws the flame upward.
E.19 It goes into the ice’s latent heat of melting.
9. When you hold a lighted match so that its tip is lower than its
stick, the flame travels quickly up the stick. Why? 20. If you put a warm bottle of wine in a container of ice water, the
wine will cool but the ice water won’t become warmer. Where is the
wine’s thermal energy going?
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E.20 The wine’s thermal energy is providing the latent heat of E.32 The warmer skin would emit brighter, shorter
melting necessary to transform ice into water. wavelength infrared light.
21. Why is it that you can put a metal pot filled with water on a red- 33. The strongest evidence for the Big Bang theory of the origin of
hot electric stove burner without fear of damaging the pot? the universe is the thermal radiation emitted by that explosion. This
radiation has cooled over the years to only 3 K and is now mostly
E.21 The boiling water will remove heat quickly enough to
microwaves. Why should 3 K thermal radiation be microwaves?
protect the pot.
E.33 The cooler the object, the longer the wavelengths of its
22. Why does a pot of water heat up and begin boiling more quickly blackbody spectrum. The spectrum of a very cold object
if you cover it? peaks in the microwave portion of the electromagnetic
E.22 Trapping the moisture allows the relative humidity to spectrum.
reach 100% so that net evaporation ceases.
34. You have a roll of papered foil that is shiny on one side and black
23. Putting a clear plastic sheet over a swimming pool helps keep the on the other. You wrap a hot potato in that foil. Which side should be
water warm during dry weather. Explain. facing outward to keep the potato hot longest?
E.23 It prevents evaporation from cooling the water. E.34 The shiny side should face outward.
24. If you try to cook vegetables in 100 °C air, it takes a long time. 35. You have a roll of papered foil that is shiny on one side and black
But if you cook those same vegetables in 100 °C steam, they cook on the other. You wrap a cold potato salad in that foil. Which side
quickly. Why does the steam transfer so much more heat to the should be facing outward to keep the salad cold longest?
vegetables? E.35 The shiny side should face outward.
E.24 The condensing steam releases its latent heat of
36. Astronomers can tell the surface temperature of a distant star
evaporation and that enormous heat cooks the vegetables
without visiting it. How is this done?
quickly.
E.36 Since stars emit primarily blackbody radiation, the
25. Why does it take longer to cook pasta properly in boiling water in astronomers can simply observe the spectrum of light
Denver (the ―mile high city‖) than it does in New York City? coming from a star and know how hot its surface is.
E.25 Water boils at a lower temperature in Denver.
37. A doctor can study a patient’s circulation by imaging the infrared
26. You’re floating outside the space station when your fellow light emitted by the patient’s skin. Tissue with poor blood flow is
astronaut tosses you a cold bottle of mineral water. It’s outside your relatively cool. What changes in the infrared emissions would
space suit but you open it anyway. It immediately begins to boil. indicate such a cool spot?
Why? E.37 Cooler skin emits dimmer infrared with longer
E.26 With no atmospheric pressure to crush them, any bubbles wavelengths.
of steam that form in the cold water grow by evaporation
38. A blacksmith is forging a small piece of steel. It emerges yellow-
and expand. The water boils even though it’s cold.
hot (1000 C) from the furnace and emits more thermal radiation than
27. The molecules of antifreeze dissolve easily in water. Why does your entire body. How can such a small object emit so much thermal
freezing in the winter and boiling in the summer? E.38 The temperature of steel is roughly 4 times that of your
E.27 Anything dissolved in water stabilizes the water so that it skin (on an absolute scale), so (according to the Stefan-
has trouble freezing or evaporating. Boltzmann law) it radiates approximately 44 or 256 times
as much thermal radiation per unit of surface area.
28. When the outside temperature is –2 °C (29 °F), you can melt the
ice on your front walk by sprinkling salt on it. But if you are out of 39. Why are concrete sidewalks divided into individual squares
salt, will baking soda do? rather than being left as continuous concrete strips?
E.28 Yes, anything that dissolves in water will do. E.39 The sidewalk has a different coefficient of volume
expansion than the ground and would break when the
29. A quality perfume is a mixture of essential oils and has a scent temperature changed.
that changes with time and skin temperature. Explain the gradual
change of the perfume’s scent with time in terms of evaporation. 40. If you were laying steel track for a railroad, what influence would
thermal expansion have on your work?
E.29 Different chemicals leave the liquid perfume at different
rates, so some evaporate away earlier than others. Higher E.40 You would have to accommodate any differences in
skin temperature promotes evaporation, even in the thermal expansion between the steel track and the
slower leaving chemicals. ground. If you forgot to do this, the track would buckle
or break when the temperature changed and two changed
30. An icy sidewalk gradually loses its ice even when the weather lengths by different amounts.
stays cold. Where does the ice go?
41. A difficult-to-open jar may open easily after being run under hot
E.30 It sublimes and the resulting water vapor is carried off by
water for a moment. Explain.
the air.
E.41 The metal lid has a larger coefficient of volume
31. How would you estimate the temperature of a glowing coal in a expansion than the glass jar, and it pulls away from the
fireplace? jar when you heat them both.
E.31 From the color of its light; whiter is hotter.
42. Why do bridges have special gaps at their ends to allow them to
32. Doctors use an infrared camera to look for inflammation, which change lengths?
appears as a hot patch on otherwise cooler skin. What differences
would the camera observe at this hot patch?
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E.42 Changes in temperature cause a bridges length to 4. If you block the outlet of a hand bicycle pump and push the
increase or decrease and its mountings must allow it to handle inward to compress the air inside the pump, the pump will
change length without buckling or breaking. become warmer. Why?
E.4 As you push the handle inward to compress the air, you
Problems – Chapter 7
do work on the handle and air. The air receives this work
and converts it into thermal energy within the air. The air
becomes hotter.
1. If a burning log is a black object with a surface area of 0.25 m2 5. When the gas that now makes up the sun was compressed
and a temperature of 800 °C, how much power does it emit as together by gravity, what happened to the temperature of that gas?
P.1 18,800 W. E.5 Its temperature increased. Gravity did work on the gas,
2. When you blow air on the log in Problem 1, its temperature rises and this work appeared in the gas as thermal energy. The
to 900 °C. How much thermal radiation does it emit now? Why did gas became hot.
the 100 °C rise make so much difference? 6. Why is a car more likely to knock on a hot day than on a cold
P.2 26,900 W. Since the radiated power is proportional to the day?
fourth power of temperature, a small increase in that E.6 When a car engine compresses air that was already hot,
temperature can cause a substantial increase in radiated the air's peak temperature is higher than it would have
power. been if it had started cold. Since knocking occurs when
3. If the sun has an emissivity of 1 and a surface temperature of the compression process heats the fuel and air mixture
6000 K, how much power does each 1 m2 of its surface emit as until it ignites, the higher the mixture's peak temperature,
thermal radiation? the more likely it is to knock.
P.3 73,500,000 W. 7. A soda siphon carbonates water by injecting carbon dioxide gas
into it. The gas comes compressed in a small steel container. As the
4. A dish of hot food has an emissivity of 0.4 and emits 20 W of gas leaves the container and pushes its way into the water, why does
thermal radiation. If you wrap it in aluminum foil, which has an the container become cold?
emissivity of 0.08, how much power will it radiate?
E.7 The gas still in the container does work pushing the other
P.4 4 W. gas into the siphon and uses up some of its thermal
5. A space vehicle uses a large black surface to radiate away waste energy. It cools.
heat. How does the amount of heat radiated away depend on the area 8. A high-flying airplane must compress the cold, rarefied outside
of that radiating surface? If the surface area were doubled, how much air before delivering it to the cabin. Why must this air be air
more heat would it radiate, if any? conditioned after the compression?
P.5 The amount of heat radiated away is proportional to the E.8 Compressing the outside air involves work and the air's
area of the radiating surface. Double that surface area temperature rises as a result of this work. The
and the radiated heat will double. temperature becomes high enough that the compressed
air must air-conditioned before it can be delivered to the
Exercises – Chapter 8
cabin.
9. If you drop a glass vase on the floor, it will become fragments. If
you drop those fragments on the floor, however, they will not become
1. Drinking fountains that actively chill the water they serve can’t a glass vase. Why not?
work without ventilation. They usually have louvers on their sides so
that air can flow through them. Why do they need this airflow? E.9 Though not forbidden by the laws of motion, it’s
extraordinarily unlikely for the vase fragments to
E.1 They are heat pumps and transfer heat to the surrounding reassemble themselves.
air.
10. When you throw a hot rock into a cold puddle, what happens to
2. If you open the door of your refrigerator with the hope of cooling the overall entropy of the system?
your room, you will find that the room’s temperature actually
increases somewhat. Why doesn’t the refrigerator remove heat from E.10 The entropy of the system goes up.
the room? E.10 By letting heat flow freely from a hot object to a cold
E.2 The refrigerator doesn't eliminate thermal energy; it object, you lower the entropy of the hot object but raise
merely pumps it from one place to the other and the entropy of the cold object even more. The total
consumes ordered energy in the process. If you open the entropy of the system increases.
refrigerator, it will just move heat around the room and 11. What prevents the bottom half of a glass of water from
produce additional thermal energy in the process. The spontaneously freezing while the top half becomes boiling hot?
room will actually become warmer.
E.11 This uneven distribution of thermal energies is
3. The outdoor portion of a central air-conditioning unit has a fan extraordinarily unlikely. It would violate the second law
that blows air across the condenser coils. If this fan breaks, why of thermodynamics.
won’t the air conditioner cool the house properly?
12. Suppose someone claimed to have a device that could convert
E.3 Without the fan, the air conditioner won’t be able to heat from the room into electric power continuously. You would
transfer its waste heat to the surrounding air. It will stop know that this device was a fraud because it would violate the second
pumping heat. law of thermodynamics. Explain.
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E.12 Converting thermal energy continuously into electric E.21 The hotter the burned gas, the larger the fraction of heat
energy involves an overall decrease in entropy (the that can be converted to work as it flows to the outdoor
disorder thermal energy becomes ordered electric air.
energy) and thus violates the second law of
thermodynamics. 22. A chemical rocket is a heat engine, propelled forward by its hot
exhaust plume. The hotter the fire inside the chemical rocket, the
13. Why does snow blanket the ground almost uniformly rather than more efficient the rocket can be. Explain this fact in terms of the
creating tall piles in certain areas and bare spots in others? second law of thermodynamics.
E.13 Though not forbidden by the laws of motion, such E.22 The hotter the hot region and the colder the cold region,
uneven distributions of snow are remarkably unlikely. the larger the fraction of heat flowing from the hot
region to the cold region that can be diverted and
14. If you transfer a glass baking dish from a hot oven to a cold basin converted into useful work.
of water, that dish will probably shatter. What produces the ordered
mechanical energy needed to tear the glass apart? 23. An acquaintance claims to have built a gasoline-burning car that
doesn’t release any heat to its surroundings. Use the second law of
E.14 Heat flowing from the hot dish to the cold water provides
thermodynamics to show that this claim is impossible.
the necessary entropy to allow some of that heat to
become mechanical work. E.23 Converting burned fuel entirely into work would violate
the second law of thermodynamics.
15. Freezing and thawing cycles tend to damage road pavement
during the winter, creating potholes. What provides the mechanical
work that breaks up the pavement?
E.15 Heat flowing into or out of the pavement during weather Problems – Chapter 8
changes allows the pavement to do work as it tears itself
apart. 1. You stir 1 kg of water until its temperature rises by 1 °C. How
much work did you do on the water?
16. The air near a woodstove circulates throughout the room. What
provides the energy needed to keep the air moving? P.1 4190 J.
E.16 Heat flowing from the hot stove to the cold room 2. While polishing a 1-kg brass statue, you do 760 J of work against
provides the necessary entropy to allow some of that sliding friction. Assuming all of the resulting heat flows into the
heat to become mechanical work. statue, how much does its temperature rise?
P.2 2 K.
17. Winds are driven by differences in temperature at the earth’s
surface. Air rises over hot spots and descends over cold spots, 3. You drop a lead ball on a cement floor from a height of 10 m.
forming giant convection cells of circulating air. Near the ground, When the ball stops bouncing, how much will its temperature have
winds blow from the cold spots toward the hot spots. Explain how the risen?
atmosphere is acting as a heat engine.
P.3 0.77 °C.
E.17 As heat flows from the hot spots to the cold spots, by
way of huge convection cells, some heat is becoming 4. Roughly how high could a 300 K copper ball lift itself if it could
ordered kinetic energy. transform all of its thermal energy into work?
in warm ocean regions and the order in colder surrounding areas. 5. Drilling a hole in a piece of wood takes 1000 J of work. How
Why are hurricanes most violent when they form over regions of much does the total internal energy of the wood and drill increase as
unusually hot water at the end of summer? a result of this process?
E.18 The hotter the water, the more thermal energy can be P.5 1000 J.
harnessed and the greater the temperature difference
between the hot water and the colder surrounding, the 6. An ideally efficient freezer cools food to 260 K. If room
more efficient the hurricane is at converting that thermal temperature is 300 K, how much work does this freezer consume
energy into mechanical work. when removing 100 J of heat from the food?
19. On a clear sunny day, the ground is heated uniformly and there is P.6 15.4 J when the food is at 260 K.
very little wind. Use the second law of thermodynamics to explain 7. An ideally efficient refrigerator removes 900 J of heat from food
this absence of wind. at 270 K. How much heat does it then deliver to the 300 K room air?
E.19 Without temperature differences, heat won’t naturally P.7 1000 J.
flow. Such heat flow is what powers a heat engine such
as the winds. 8. An ideally efficient heat pump delivers 1000 J of heat to room air
at 300 K. If it extracted heat from 260 K outdoor air, how much of
20. A plant is a heat engine that operates on sunlight flowing from that delivered heat was originally work consumed in the transfer?
the hot sun to the cold earth. The plant is a highly ordered system
with relatively low entropy. Why doesn’t the plant’s growth violate P.8 133 J.
the second law of thermodynamics? 9. An ideally efficient air conditioner keeps the room air at 300 K
E.20 Heat flowing from the hot sun to the colder earth when the outdoor air is at 310 K. How much work does it consume
provides the necessary entropy to allow the plant to when delivering 1240 J of heat outside?
harness some of that heat to make it grow. P.9 40 J.
21. A diesel engine burns its fuel at a higher temperature than a 10. An ideally efficient airplane engine provides work as heat flows
gasoline engine. Why does this difference allow the diesel engine to from 1500 K burned gases to 300 K air. What fraction of the heat
be more efficient at converting the fuel’s energy into work? leaving the burned gases is converted into work?
Page 19 of 38 - 5/2/2011 - 11:51:27 AM
P.10 0.8 or 80%. motion. But if the restoring force is not proportional to
its displacement, the chair will be an anharmonic
11. An ideally efficient steamboat engine operates on 500 K steam in oscillator and its period will change with amplitude.
300 K weather. How much work can it obtain when 1000 J of heat
leaves the steam? 7. An electronic ruler measures the distance to a wall by bouncing
P.11 400 J. sound off it. How can a ruler use a sound emitter, a sound receiver,
and a timer to measure how far away the wall is?
12. An offshore breeze at the beach is powered by heat flowing from E.7 Multiplying sound’s roundtrip time by its speed gives
hot land (310 K) to cool water (290 K). Assuming ideal efficiency,
twice the distance to the wall.
how much work can this breeze provide for each 1000 J of heat it
carries away from the land? 8. Which of the following clocks would keep accurate time if you
P.12 65 J. took them to the moon: a pendulum clock, a balance clock, and a
quartz watch? Why?
13. An ideally efficient solar energy system produces work as heat
E.8 The balance clock and quartz watch would continue to
flows from the 6100 K surface of the sun to the 300 K room air. What
work, but the pendulum clock would run slow. The first
fraction of the solar heat can it transform into work?
two timepieces don't depend on gravity for their
P.13 95% of the heat can become work. restoring forces and are thus independent of the strength
of gravity. However, the pendulum clock depends on
gravity for the stiffness of its restoring force. In the
Exercises – Chapter 9 moon's weak gravity, the restoring force will be weak
and the pendulum will swing more slowly than on earth.
1. The acceleration due to gravity at the moon’s surface is only 9. To modify the pitch of a guitar string you could change its mass,
about one-sixth that at the earth’s surface. If you took a pendulum tension, or length. To raise its pitch, how should you change each of
clock to the moon, would it run fast, slow, or on time? these three characteristics?
E.1 It would run slow. E.9 Less mass, more tension, or less length.
2. A clothing rack hangs from the ceiling of a store and swings back 10. A tuning peg slips on your violin, lessening the tension in one of
and forth. Why doesn’t the period of this motion depend on how the strings. What effect does this change have on the string’s pitch?
many dresses the rack is holding? (Neglect the rack’s own mass.) E.10 It lowers the string’s pitch.
E.2 The rack is a pendulum and its period depends only on
11. The strings that play the lowest notes on a piano are made of
its length and on the strength of gravity.
thick steel wire wrapped with a spiral of heavy copper wire. The
E.2 While increasing the mass of the moving object copper wire doesn’t contribute to the tension in the string, so what is
lengthens the period of most harmonic oscillators, in this its purpose?
case adding mass to the moving object also increases E.11 The copper wrap adds mass to the string to lower its
that object's weight and thus stiffens the restoring force pitch.
acting on the object. No matter how heavy the dresses,
the rack will swing back and forth in the same amount of 12. Why are the highest pitched strings on most instruments,
time. including guitars, violins, and pianos, the most likely strings to
break?
3. If a child stands up on the seat of a playground swing, how will
the swing’s period be affected? E.12 The highest pitched strings are usually the thinnest (to
reduce mass) and the tautest (to increase stiffness). Thin,
E.3 The period will decrease as the pendulum gets shorter. tight strings break easily.
4. If you pull a small tree to one side and suddenly let go, it will 13. Some wind chimes consist of sets of metal rods that emit tones
swing back and forth several times. The period of this motion won’t when they’re struck by wind-driven clappers. These rods vibrate like
depend on how far you bend the tree. How does the restoring force the baseball bat in Fig. 3.2.7. Why do the longer rods emit lower
that returns the tree to its upright position depend on how far you pitched tones than the shorter rods?
bend the tree?
E.13 The longer rods have more mass vibrating and are less
E.4 The restoring force acting on the tree is proportional to stiff.
how far it is bent away from its normal upright
orientation. 14. Why would replacing the air in an organ pipe with helium raise
its pitch?
5. A flagpole is a harmonic oscillator, flexing back and forth with a
steady period. If you want to increase the amplitude of the pole’s E.14 Helium is less dense than air and responds more rapidly
motion by pushing it near its base, when should you push—as it to forces. Helium will accelerate back and forth in the
flexes toward you or away from you? pipe faster than air does.
E.5 Push on it as it flexes away from you (so you do work on 15. A flute and a piccolo are both effectively pipes that are open at
it). both ends, with holes in their sides to allow them to produce more
tones. The piccolo is very nearly a half-size version of the flute. How
6. Depending on how the base of a rocking chair is shaped, the does this fact explain why the piccolo’s tones are one octave above
period of its motion may or may not depend on how hard you’re those of a flute?
rocking it. What can you say about the restoring forces acting in these
two cases? E.15 A piccolo’s air columns are half the length of those in a
flute, so they vibrate at twice the pitch or one octave
E.6 If the restoring force acting on the chair is proportional higher.
to how far it is displaced from its equilibrium position,
then the chair will be harmonic oscillator and its period 16. The most important difference between a trumpet and a tuba is in
of motion will not depend on the amplitude of that the lengths of their pipes. The tuba’s pipe is much longer than that of
Page 20 of 38 - 5/2/2011 - 11:51:27 AM
the trumpet. How does this difference affect the relative pitches of the 26. Even when waves don’t break as they pass over a sand bar
two instruments? (Exercise 25), the sand bar is noticeable because you can see the
wave crests move closer together. What is happening to cause this
E.16 The tuba's longer air column vibrates more slowly than
bunching?
the trumpet's shorter air column.
E.26 Water surface waves travel more slowly when they begin
17. In the Mediterranean Sea, high tide is only 30 cm above low tide. to encounter the shallow bottom of the water. This
Why? slowing causes the crest to bunch together.
E.17 Ocean water can’t enter or leave the Mediterranean Sea
27. Sound can travel from one paper cup to another through a long,
quickly enough to allow its high tide to be very different
taut string that connects their bottoms. Is the wave passing through
from its low tide. The tide simply rearranges the water the string longitudinal or transverse?
levels within the sea itself.
E.27 Longitudinal.
18. Why are the tides relatively weak near the north and south poles?
28. If you stamp on a wooden floor, you can make objects on a
E.18 The bulges in the earth's oceans are located near the nearby table jump slightly. Are the waves traveling through the floor
equator--the nearest and farthest points to the moon. longitudinal or transverse?
Since the bulges aren't present at the north and south
poles, the tides at those poles are weak. E.28 Transverse.
19. If you’re carrying a full cup of coffee and take your steps at just 29. The crash of brass cymbals is rich with overtones that are not
the wrong frequency, the coffee will begin to slosh wildly in the cup. harmonics of the fundamental. Why aren’t the overtones harmonics?
What is causing this energetic motion? E.29 A surface can’t vibrate as half- or third-surfaces, so its
E.19 Resonant energy transfer occurs when your steps are overtones are complicated and don’t occur at harmonic
synchronized with the rhythmic motion of the coffee frequencies.
sloshing in the cup.
30. A Chinese gong produces a loud ringing sound which has
20. When you throw a stone into a pool of still water, small ring- nonharmonic overtones. Why aren’t the overtones harmonic?
shaped ripples begin to spread outward at a modest pace. Why do E.30 The gong is a surface and since it can’t vibrate as simple
these ripples travel so much more slowly than waves on the ocean? fractions of itself, its overtones don’t occur at harmonic
E.20 The speed of a surface wave on water increases with frequencies.
increasing wavelength. The short-wavelength ripples you
31. A harp is not a loud instrument, but it would be even softer if its
create with a stone travel very slowly. strings were not attached to a wide wooden base. What purpose does
21. When you throw a rock into calm water, ripples head outward as the wooden surface serve?
several concentric circles. As the circumferences of these circles E.31 The surface projects sound waves far better than strings
grow larger, their heights grow smaller. Explain this effect in terms alone.
of conservation of energy.
32. A string bass is an enormous instrument to carry around. Why
E.21 The wave’s energy is proportional to its circumference
can’t you just support the four strings with a sturdy metal bar and
and its height. As its circumference grows, its height leave out the whole wooden structure?
must diminish.
E.32 Most of the bass’s sound is projected by its vibrating
22. If you pull downward on the middle of a trampoline and let go, body. Without that wooden body, it would have almost
the surface will fluctuate up and down several times. Why is this no volume.
motion an example of a standing wave?
33. You clap your hands twice, sending out two sound waves. Can
E.22 The vibrating trampoline has a region of maximum up
you make the second sound wave overtake the first by clapping a
and down motion (the antinode) and a ring of zero different pitch or volume the second time? Why or why not?
motion (the node).
E.33 No. Changing the pitch or volume of the sound won’t
23. Why is a violin string vibrating up and down in its fundamental affect the speed at which the wave travels.
vibrational mode an example of a standing wave rather than a
traveling wave? 34. You drop two rocks into the center of a lake, one after the other,
sending out two water surface waves. Can you make the second wave
E.23 Crests and troughs don’t travel along the string. Instead, overtake the first by using different sized rocks to make waves with
the center of the string becomes alternately a crest and a different wavelengths? Why or why not?
trough.
E.34 Yes. If the second wave has a longer wavelength than the
24. When you pluck the end of a kite string, a ripple will head up the first, it will travel faster and overtake the first wave.
string toward the kite. Why is this motion an example of a traveling
wave rather than a standing wave? 35. If you stand in front of a stone building and clap your hands, you
hear an echo. What is happening to the sound wave to cause this
E.24 The plucked kite string has a ripple that moves along its echo?
length. The region of maximum motion moves along the
string and there is no pattern that simply vibrates back E.35 The sound waves reflect from the stone surface.
and forth in place. 36. If you stand behind a massive stone wall, you have trouble
25. As waves pass over a shallow sand bar, the largest ones break. hearing a person on the other side clapping her hands. What is
What causes these waves to break and why only the largest ones? happening to the sound wave to make it difficult for you to hear?
E.25 There isn’t enough water in front of the largest waves to E.36 Most of the sound wave reflects from the stone surface
complete their crests as they pass over the sand bar. and fails to reach your ears.
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37. A harbor breakwater is a stone wall in the water that prevents
waves from entering the harbor. Where does a wave’s energy go after
it hits the breakwater? Exercises – Chapter 10
E.37 Some of the wave reflects but the rest of its energy 1. If two objects repel one another, you know they have like charges
becomes thermal energy. on them. But how would you determine whether they were both
positive or negative?
38. Waves tend to bend toward points of land projecting into the
ocean and erode those points. What causes the wave to bend toward E.1 You’d have to compare the objects with a reference. The
the points? object that repelled the reference would have its charge.
E.38 The waves refract in the shallow water near the points 2. You have an electrically neutral toy which you divide into two
and approach those points more and more directly as a pieces. You notice that at least one of those pieces has an electric
result. charge. Do the two pieces attract or repel one another, or neither?
39. Surfers are well aware that waves bend as they pass over coral E.2 They attract.
reefs. What causes this bending?
E.2 Electric charge is conserved, so that if one part of the
E.39 Refraction occurs as a wave slows down over the neutral toy ends up a positive charge, the other part must
shallow coral. end up with a negative charge of equal magnitude. These
40. Musical tones can linger for many seconds in a stone cathedral. two oppositely charged parts will then attract one
Why? another.
E.40 The stone surfaces reflect the sound waves well and do 3. Suppose that you had an electrically charged stick. If you divided
not absorb much of the sound energy. the stick in half, each half would have half the original charge. If you
split each of these halves, each piece would have a quarter of the
41. If two violinists play slightly different notes at the same time, the original charge. Can you keep on dividing the charge in this manner
combined sound has a pulsing character. What causes that pulsation? forever? If not, why not?
E.41 Interference between the two sound waves causes a E.3 You can’t keep dividing the charge in half because
beating effect. charge comes in discrete units, the elementary unit of
electric charge.
42. When an airplane’s two propellers are turning at almost the same
rate, their combined sound can have a pulsing character to it. Explain 4. A ping pong ball contains an enormous number of electrically
that pulsing. charged particles. Why don’t two ping pong balls normally exert
E.42 Interference between the sound waves from the two electrostatic forces on each other?
propellers causes a beating effect. E.4 A normal ping pong ball is electrically neutral--it has as
many positive as negative charges. Thus the attractive
and repulsive forces between the balls cancel
Problems – Chapter 9 5. In industrial settings, neutral metal objects are often coated by
spraying them with electrically charged paint or powder particles.
1. What is the wavelength of a tuba’s A2 (110 Hz) tone in air at How does placing charge on the particles help them to stick to an
standard conditions? object’s surface?
P.1 3.01 m. E.5 The charged paint particles electrically polarize the
surface being coated, so that the paint particles are
2. A piccolo is playing A6 (1760 Hz). What is the wavelength of that attracted to it and stick.
tone in air at standard conditions?
6. The paint or powder particles discussed in Exercise 5 are all given
P.2 0.188 m. the same electric charge. Why does this type of charging ensure that
3. When a piano plays C4 (264 Hz) in a room containing a somewhat the coating will be highly uniform?
unusual mixture of gases, the wavelength of the sound in that gas is E.6 The charged particles repel one another and tend not to
1.00 m. What is the speed of sound in that gas? accumulate in clumps. The forces they exert on one
P.3 264 m/s. another tend to spread them out uniformly.
4. At an altitude of 3000 m and standard temperature (0 °C), a 7. If the forces between electric charges didn’t diminish with
violin’s A4 (440 Hz) has a wavelength of 0.725 m. What is the local distance, an electrically charged balloon wouldn’t cling to an
speed of sound? electrically neutral wall. Why not?
P.4 319 m/s. E.7 The wall’s attractive and repulsive forces would be equal
but oppositely directed, summing to zero net force on the
5. A water surface wave with a frequency of 0.3 Hz has a balloon.
wavelength of 17.3 m. What is its wave speed?
8. An ion generator clears smoke from room air by electrically
P.5 5.2 m/s. charging the smoke particles. Why will those charged smoke particles
6. A water surface wave has a wave speed of 15.6 m/s and a stick to the walls and furniture?
frequency of 0.1 Hz. What is its wavelength? E.8 The charged smoke particles will electrically polarize the
P.6 156 m. surfaces they approach and will be attracted toward that
polarization.
9. After you extract two pieces of adhesive tape from a tape
dispenser, those pieces will repel one another. Explain their
repulsion.
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E.9 The two pieces are equivalent, so they acquire like acts on that positive charge also decreases with distance
charges while separating from the tape dispenser. Like from the hairbrush.
charges repel.
20. Which way does the electric field point around the positive
10. After you peel a sticker from its paper backing, the two attract terminal of an alkaline battery?
one another. Explain their attraction.
E.20 The electric field points away from the positive
E.10 The sticker and paper were originally neutral. Since they terminal’s surface.
consist of different materials, one stole charge from the
21. Which has the stronger electric field between its two terminals: a
other and they acquired equal but opposite charges as
1.5-V AA battery or a standard 9-V battery? Explain.
they separated. They now atract one another.
E.21 The 9-V battery has the larger voltage gradient and
11. You’re holding two balloons, one covered with positive charge therefore the stronger electric field because its two
and one with negative charge. Compare their voltages.
terminals have a greater voltage difference and a shorter
E.11 The positive balloon has a higher voltage than the distance between them.
negative one.
22. You have 100 AA batteries. How should connect those batteries
12. You begin to separate the two balloons in Exercise 11. You do to one another and then shape the resulting chain in order to make the
work on the balloons, so your energy decreases. Where does your strongest electric field?
energy go?
E.22 You should arrange them in a chain, positive terminal
E.12 The energy becomes electrostatic potential energy in the touching negative terminal, so that they form an almost
balloons complete circle. If you leave a small gap between the
positive terminal of one battery and the negative terminal
13. When you separate the balloons in Exercise 12, do their voltages of the battery on the opposite end of the chain, there will
change? If so, how do those voltages change?
then be an extremely strong electric field between those
E.13 Their voltages change. The positive balloon’s voltage two oppositely charged terminals.
increases; the negative balloon’s voltage decreases.
23. It may seem dangerous to be in a car during a thunderstorm, but
14. A car battery is labeled as providing 12 V. Compare the it’s actually relatively safe. Since the car is essentially a metal box,
electrostatic potential energy of positive charge on the battery’s the inside of the car is electrically neutral. Why does any charge on
negative terminal with that on its positive terminal. the car move to its outside surface?
E.14 A coulomb of positive charge on the battery's positive E.23 The like charges repel and flow through the conducting
terminal has 12 joules more electrostatic potential energy car to its outside surfaces.
than a coulomb of positive charge on the negative
24. Delicate electromagnetic experiments are sometime performed
terminal.
inside metal walled or screen rooms. Why does that enclosure
15. When technicians work with static-sensitive electronics, they try minimize stray electric fields?
to make as much of their environment electrically conducting as E.24 Since charge can move freely through the conducting
possible. Why does this conductivity diminish the threat of static enclosure, it flows until the entire enclosure has the same
electricity?
voltage. The electric fields within the enclosure are thus
E.15 Conductive materials allow charge to flow and minimize greatly reduced.
its energy. Everything will tend toward electrical
neutrality. 25. When a positively charged cloud passes overhead during a
thunderstorm, which way does the electric field point?
16. Antistatic fabric treatments render the fabrics slightly conducting. E.25 Downward, away from the cloud.
How does this treatment help diminish static electricity?
E.16 The treated fabric’s mobile charges move about until 26. The cloud in Exercise 25 attracts a large negative charge to the
top of a tree in an open meadow. Why is the magnitude of the electric
they have minimized their electrostatic potential energy.
field larger on top of this tree than elsewhere in the meadow?
The clothes will thereby eliminate as much of their static
charges as they can. E.26 Since the tree is effectively a sharp point on the
otherwise flat meadow, it has reduced influence on the
17. Holding your hand on a static generator (e.g., a van de Graff voltages above it and those voltages change more rapidly
generator) can make your hair stand up, but only if you are standing with position than the voltages near the broad meadow.
on a good electrical insulator. Why is that insulator important?
The large voltage gradient near the tree is a strong
E.17 For you to accumulate a great deal of charge, you must electric field.
not be able to lose it easily through conducting paths to
the ground. 27. Corona discharges can occur wherever there is a very strong
electric field. Why is there a strong electric field around a sharp point
18. In Exercise 17, having used a hair conditioner recently actually on an electrically charged metal object?
helps your hair stand up. Why? E.27 The voltage falls or rises rapidly toward zero in the
E.18 By allowing charges to move through your hair, the vicinity of the sharp point and that large voltage gradient
conditioner makes it easier for your hair to accumulate is a strong electric field.
like charges and consequently stand up.
28. To minimize corona discharges, electric power pylons sometimes
19. The electric field around an electrically charged hairbrush shroud connectors and other sharp features with smoothly curving
diminishes with distance from that hairbrush. Use Coulomb’s law to metal rings or shells. How do those broad, smooth structures prevent
explain this decrease in the magnitude of the field. corona discharges?
E.19 The forces between a positive charge and the hairbrush
decrease with their separation. Thus the electric field that
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E.28 By eliminating sharp features, these structures reduce the E.37 The socket’s central pin has the higher voltage.
voltage gradients around the power pylons. With weaker
voltage gradients, corona discharges don’t occur.. 38. When current is flowing through a car’s rear defroster (Exercise
33), the voltage at each end of the metal strips is different. Which end
29. If you’re ever standing on a mountaintop when a dark cloud of each strip has the higher voltage, the one through which current
passes overhead and your hair stands up, get off the mountain fast. enters the strip or the one through which current leaves, and what
How would your hair have acquired the charge to make it stand up? causes the voltage drop?
E.29 The charged cloud overhead would have induced a large E.38 The end through which the current enters the strip has
opposite charge to flow up from the ground onto your the higher voltage. As the strip extracts energy from this
hair. current (turning it into thermal energy), the current loses
voltage
30. The power source for an electric fence pumps charge from the
earth to the fence wire, which is insulated from the earth. The earth E.38 Current flows through an ohmic device, such as the
can conduct electricity. When an animal walks into the wire, it defroster, from its higher voltage end to its lower voltage
receives a shock. Identify the circuit through which the current flows. end. The voltage gradient in the device is the electric
E.30 Current flows from the power source, through the fence field that pushes the current through the device.
wire, through the animal, through the earth, and back to 39. You’re given a sealed box with two terminals on it. You use
the power source. some wires and batteries to send an electric current into the box’s left
terminal and find that this current emerges from the right terminal. If
31. A bird can perch on a high-voltage power line without getting a
the voltage of the right terminal is 6 V higher than that of the left
shock. Why doesn’t current flow through the bird?
terminal, is the box consuming power or providing it? Is it more
E.31 While charge can accumulate on the bird, it can’t flow as likely to contain batteries or lightbulbs? How can you tell?
a current because it has nowhere else to go.
E.39 Batteries, because something is supplying power to the
32. The two prongs of a power cord are meant to carry current to and current.
from a lamp. If you were to plug only one of the prongs into an
40. One time-honored but slightly unpleasant way in which a
outlet, the lamp wouldn’t light at all. Why wouldn’t it at least glow at
hobbyist determines how much energy is left in a 9-V battery is to
half its normal brightness?
touch both of its terminals to his tongue briefly. He experiences a
E.32 Without a circuit, charge can’t flow continuously pinching feeling that’s mild when the battery is almost dead but one
through the lamp. No current flows through the lamp and that’s startlingly sharp when the battery is fresh. (Don’t try this
it remains dark. technique yourself.) What is the circuit involved in this taste test?
How is energy being transferred?
33. The rear defroster of your car is a pattern of thin metal strips
across the window. When you turn the defroster on, current flows E.40 Current flows from the battery’s positive terminal,
through those metal strips. Why are there wires attached to both ends through the hobbyist’s tongue, to the battery’s negative
of the metal strips? terminal. Energy is being transferred from the battery to
the hobbyist’s tongue by the current.
E.33 Current arrives through one wire and leaves through the
other. 41. Spot welding is used to fuse two sheets of metal together at one
small spot. Two copper electrodes pinch the sheets together at a point
34. If you touch only one of the metal contacts on a headphone plug and then run a huge electric current through that point. The two
to the headphone jack of a portable audio player, what volume will sheets melt and flow together to form a spot weld. Why does this
the headphones produce? technique work only with relatively poor conductors of electricity
E.34 Zero volume. such as stainless steel and not with excellent conductors such as
copper?
E.34 Without a circuit, charge can’t flow continuously
through the headphones and no sound is produced. E.41 The power deposited in a metal is proportional to its
electric resistance, so high-resistance metals heat more.
35. If you transfer some (positive) charge to a battery’s negative
terminal, some of that charge will quickly move to the battery’s 42. Why is it important that the filament of a lightbulb have a much
positive terminal. Will the battery’s store of chemical potential larger electrical resistance than the supporting wires that carry current
energy have changed and, if so, will it have increased or decreased? to and from that filament?
E.35 Yes, the battery will have lost some chemical potential E.42 The filament’s larger resistance ensures that it
energy. experiences the main voltage drop and receives most of
the current’s electrical power.
36. If you transfer some positive charge to a battery’s positive
terminal, some of that charge will quickly move to the battery’s
negative terminal. Will the battery’s store of chemical potential
energy have changed and, if so, will it have increased or decreased? Problems – Chapter 10
E.36 The battery’s store of chemical potential energy will
1. You remove two socks from a hot dryer and find that they repel
increase.
with forces of 0.001 N when they’re 1 cm apart. If they have equal
E.36 The positive charge will flow backward through the charges, how much charge does each sock have?
battery and drive its electrochemical processes P.1 3.3 10-9 C.
backward. The battery will recharge.
2. If you separate the socks in Problem 1 until they’re 5 cm apart,
37. When you plug a portable appliance into the power socket inside what force will each sock exert on the other?
an automobile, current flows to the appliance through the central pin
of that socket and returns to the car through the socket’s outer ring. P.2 The force will decrease to 0.000040 newtons.
Which of the socket’s contacts has the higher voltage?
Page 24 of 38 - 5/2/2011 - 11:51:27 AM
P.2 At five times the original distance, the charges will exert through its circuit. What power is being transferred to the bulb in
only one-twenty-fifth the original forces. each flashlight?
3. If you were to separate all of the electrons and protons in 1 g P.16 The two-battery flashlight transfers 4.5 W to the bulb,
(0.001 kg) of matter, you’d have about 96,000 C of positive charge while the five-battery flashlight transfers 11.25 W to the
and the same amount of negative charge. If you placed these charges bulb.
1 m apart, how strong would the attractive forces between them be?
17. You have two flashlights that have 2-A currents flowing through
P.3 8.3 1019 N. them. One flashlight has a single 1.5-V battery in its circuit, while the
second flashlight has three 1.5-V batteries connected in a chain that
4. If you place 1 C of positive charge on the earth and 1 C of provides 4.5 V. How much power is the battery in the first flashlight
negative charge 384,500 km away on the moon, how much force providing? How much power is each battery in the second flashlight
would the positive charge on the earth experience? providing?
P.4 The force would be about 6.1 x 10-8 newtons. P.17 Each battery in this problem supplies 3 W.
5. How close would you have to bring 1 C of positive charge and 1 18. How much power is the bulb of the first flashlight in Problem 17
C of negative charge for them to exert forces of 1 N on one another? consuming? How much power is the bulb of the second flashlight
P.5 94,800 m. consuming?
6. The upward net force on the space shuttle at launch is 10,000,000 P.18 The bulb in the one-battery flashlight is consuming 3 W,
N. What is the least amount of charge you could move from its nose while the bulb in the three-battery flashlight is
to the launch pad, 60 m below, and thereby prevent it from lifting consuming 9 W.
off? 19. A 1.5-V alkaline D battery can provide about 40,000 J of electric
P.6 2 C. energy. If a current of 2 A flows through two D batteries while
they’re in the circuit of a flashlight, how long will the batteries be
7. What force will a 0.01-C charge experience in a 5-N/C electric able to provide power to the flashlight?
field pointing upward?
P.19 About 13,333 s, or 3.7 hours.
P.7 0.05 N upward.
20. A radio-controlled car uses four AA batteries to provide 6 V to
8. A sock with a charge of -0.0005 C is in a 1000-N/C electric field its motor. When the car is heading forward at full speed, a current of
pointing toward the right. What force does the sock experience? 2 A flows through the motor. How much power is the motor
P.8 0.5 N toward the left. consuming at that time?
9. A piece of plastic wrap with a charge of 0.00005 C experiences a P.20 The motor is consuming 12 W of power.
forward force of 0.0010 N. What is the local electric field? 21. Your car battery is dead, and your friends are helping you start
P.9 20 N/C (20 V/m) pointing forward. your car with cheap jumper cables. One cable carries current from
their car to your car, and a second cable returns that current to their
10. A Styrofoam ball with a charge of -1.0 10-6 C experiences an car. As you try to start your car, a current of 80 A flows through the
upward force of 0.01 N in an electric field. What is that electric field? cables to your car and back, and a voltage drop of 4 V appears across
P.10 10,000 N/C (or 10,000 V/m) downward. each cable. What is the electric resistance of each jumper cable?
11. If you place a 0.0001-C charge halfway between the terminals of P.21 0.05 .
a common 9-V battery, 5 mm apart, what force will that charge 22. If you replace the cheap cables in Problem 21 with cables having
experience? half their electric resistance, what voltage drop will appear across
P.11 0.18 N toward the negative terminal. each new cable if the current doesn’t change?
12. If you place a 0.0001-C charge halfway between the terminals of P.22 The voltage drop across each of the new cables will be 2
a 1.5-V AA battery, 5 cm apart, what force will that charge V.
experience? 23. Each of the two wires in a particular 16-gauge extension cord has
P.12 0.003 N toward the negative terminal. an electric resistance of 0.04 . You’re using this extension cord to
operate a toaster oven, so a current of 15 A is flowing through it.
13. An automobile has a 12-W reading lamp in the ceiling. This lamp What is the voltage drop across each wire in this extension cord?
operates with a voltage drop of 12 V across it. How much current
flows through the lamp? P.23 0.6 V.
P.13 1 A. 24. How much power is wasted in each wire of the extension cord in
Problem 23?
14. The rear defroster of your car operates on a current of 5 A. If the
voltage drop across it is 10 V, how much electric power is it P.24 Each wire will waste 9 W of power.
consuming as it melts the frost? 25. The two wires of a high-voltage transmission line are carrying
P.14 The defroster consumes 50 W of power. 600 A to and from a city. The voltage between those two wires is
400,000 V. How much power is the transmission line delivering to
15. Your portable FM radio uses two 1.5-V batteries in a chain. If the the city?
batteries send a current of 0.05 A through the radio, how much power
are they providing to the radio? P.25 240,000,000 W.
P.15 0.15 W. 26. In bringing electricity to individual homes, the power in Problem
25 is transferred to low-voltage circuits so that the current passing
16. You have two flashlights that operate on 1.5-V D batteries. The through homes experiences a voltage drop of only 120 V. How much
first flashlight uses two batteries in a chain while the second uses five total current is passing through the homes of the city?
batteries in a chain. Each flashlight has a current of 1.5 A flowing
Page 25 of 38 - 5/2/2011 - 11:51:27 AM
P.26 The total current pass through the homes is 2,000,000 (2 E.4 The button magnet’s magnetic field magnetically
million) amperes. polarizes the iron pipe. The pipe develops an opposite
pole near the pole of the approaching button magnet and
27. How much current flows through a 100- heating filament when
attracts that magnet.
the voltage drop across it is 5 V?
P.27 0.05 A. 5. If you hold a permanent magnet the wrong way in an extremely
strong magnetic field, its magnetization will be permanently reversed.
28. If you subject a 2500- heating filament to a voltage drop of 100 What happens to the magnetic domains inside the permanent magnet
V, how much current will flow through it? during this process?
P.28 0.040 A. E.5 The domains aligned with the new applied field grow
while those that are anti-aligned with that field shrink.
29. If a 10-A current is flowing through a long wire and there is a 1-
V voltage drop between the two ends of that wire, what is the wire’s 6. Hammering or heating a permanent magnet can demagnetize it.
electrical resistance? What happens to the magnetic domains inside it during these
processes?
P.29 0.1 .
E.6 The magnetic domains lose their uniform orientations
30. The 2-A current flowing through a wire to a distant buzzer and become more randomly oriented.
experiences a 2-V voltage drop. What is the electrical resistance of
that wire? 7. If you place a button magnet in a uniform magnetic field, what is
the net force on that button magnet?
P.30 1 .
E.7 Zero.
31. If you send a 5-A current through a 1- wire to the doorbell,
what voltage drop will exist between the two ends of the wire? 8. If you hold a magnetic compass in a uniform magnetic field
pointing northward, in which direction, if any, is the net magnetic
P.31 5 V. force on the compass?
32. If a 1000- heating filament is carrying a current of 0.120 A, E.8 The net magnetic force on a compass in a uniform
what voltage drop will exist between the two ends of the filament? magnetic field is zero.
P.32 120 V. 9. Do more magnetic flux lines begin or end on a button magnet, or
are those numbers equal?
Exercises – Chapter 11 E.9 They are equal.
10. Compare the number of magnetic flux lines beginning and
1. Is it possible to have two permanent magnets that always attract ending on a plastic strip magnet. Explain.
one another, regardless of their relative orientations? Explain. E.10 The same number of flux lines begin on the strip magnet
E.1 No. Magnetic monopoles have never been found and the as end on the strip magnet. That’s because, like every
forces between magnetic dipoles depend on their relative other magnet, the plastic strip magnet has zero net
orientations. magnetic pole.
2. The magnetostatic forces between two button magnets decrease 11. Two plastic strip magnets differ only in how many poles they
surprisingly quickly as their separation increases. Use Coulomb’s law have per centimeter. One has 2 poles/cm and the other has 4
for magnetism and the dipole character of each button magnet to poles/cm. From which strip’s surface do magnetic flux lines extend
explain this effect. outward farther?
E.2 When the two button magnets are far apart, there is E.11 The flux lines extend farther from the 2-poles/cm strip.
almost no difference between the distances separating 12. Which of the two plastic strip magnets in Exercise 11 is attracted
their two pairs of poles. Two of the pairings are toward a refrigerator at the greater distance?
repulsive (north-north and south-south) and two pairings
are attractive (north-south and south-north) and since the E.12 The magnet with only 2 poles/cm is attracted at the
distances in those pairings are nearly identical the greater distance.
repulsive and attractive forces nearly cancel. But when E.12 The magnetic field extends farther away from the 2
the two button magnets approach one another closely, poles/cm magnet and therefore affects the refrigerator
the various distances are no longer so similar. One of the from a greater distance.
pairings is likely to become dominant as the distance
separating that pairing of poles because relatively small 13. How could you use iron to prevent the magnetic flux lines from a
compared to the distance separating the other pairings. If strong button magnet from extending outward into the room?
the closely spaced poles are opposite, the overall forces E.13 Encase the magnet in an iron box. The iron will then
on the buttons will be attractive. If the poles are like, the guide the flux lines.
overall forces will be repulsive.
14. To keep the strong magnets in a scientific facility next door from
3. If you bring two magnetic compasses nearby, they will soon begin sending flux lines through your office, should you line the office
attracting one another. Why don’t they repel? walls with aluminum or with iron?
E.3 The two compass needles will pivot so as to minimize E.14 Iron.
their total potential energies and will soon have opposite
poles pointing toward one another. So aligned, they will E.14 Iron is a soft ferromagnetic material and will attract the
then attract. flux lines and direct them through itself. Aluminum is
nonmagnetic and will essentially ignore the magnetic
4. If you bring a button magnet near an iron pipe, they will soon flux lines.
begin attracting one another. Why don’t they repel?
Page 26 of 38 - 5/2/2011 - 11:51:27 AM
15. Your friends are installing a loft in their room and are using thin supplied with 120-V AC, what voltage does the secondary coil
speaker wires to provide power to an extra outlet. If they draw only a provide?
small amount of current from the outlet, the voltage drop in each of
E.24 24-V AC.
the wires will remain small. Why?
E.15 The voltage drop in each wire is proportional to the 25. If an average current of 3 A is passing through the primary coil
current. of the transformer in Exercise 23, what average current is passing
through the secondary coil of that transformer?
16. When your friends from Exercise 15 plug a large home
E.25 9 A.
entertainment system into the outlet, it doesn’t work properly because
the voltage rise provided by the extra outlet is only 60 V. The power 26. If the average current passing through the secondary coil of the
company provides a voltage rise of 120 V, so where is the missing transformer in Exercise 24 is 10 A, what average current is passing
voltage? through the primary coil?
E.16 The voltage was lost in the wires as the current passed E.26 2 A.
through them.
27. A magnet hanging from a spring bounces in and out of a metal
17. A particular lightbulb is designed to consume 40 W when ring. Although it doesn’t touch the ring, the magnet’s bounce
operating on a car’s 12-V DC electric power. If you supply that bulb diminishes faster than it would if the ring weren’t there. Explain.
with 12-V AC power from a transformer, how much power will it
consume? E.27 The bouncing magnet heats the ring by inducing current
in it. That current extracts this heating energy from the
E.17 It will consume 40 W. magnet by exerting magnetic forces on it and doing
negative work on it.
18. Your toaster consumes 800 W when operating on 120-V AC
electric power. If your rugby team is camping and all of you string 28. The high-voltage spark that ignites gasoline in a basic lawn
together flashlight batteries to supply that toaster with 120-V DC mower engine is produced when a magnetic pole moves suddenly
electric power, how much power will it consume? past a stationary coil of wire. From where does that spark’s energy
E.18 800 W. come?
E.28 The energy is extracted from the moving magnetic pole.
19. To read the magnetic strip on an ID or credit card, you must
swipe it quickly past a tiny coil of wire. Why must the card be 29. If you have a coil of wire, a battery, a magnetic compass, and an
moving for the coil system to read it? electrical switch, how could you make the compass needle spin?
E.19 Only a moving magnet or changing magnetic field will E.29 Form a circuit from the coil, battery, and switch and
produce the electric field necessary to push currents close the circuit briefly each time the compass needle
through coil inside the playback head. reaches anti-alignment with the coil’s field.
20. One type of microphone has a permanent magnet and a coil of 30. If you circle a permanent magnet around a magnetic compass, the
wire that move relative to one another in response to sound waves. compass needle will follow along. What is providing the needle with
Why is the current in the coil related to the motion? the energy it needs to continue turning despite friction in its pivot?
E.20 As the coil moves through the magnetic field, its mobile E.30 You and your moving permanent magnet are doing work
electric charges experience the Lorentz force and on the needle to keep it rotating despite the friction it
therefore flow through the coil as an electric current. If experiences.
the coil is part of a circuit, the current it experiences will
probably be proportional to its velocity through the 31. You can’t make a motor using only permanent magnets. Why
magnetic field. not?
21. If the primary coil of a transformer has 200 turns and is supplied E.31 The magnets will all eventually orient themselves in the
with 120-V AC power, how many turns must the secondary coil have minimum energy configuration and never move again.
to provide 12-V AC power? 32. You can’t make a motor using direct current and electromagnets
E.21 20 turns. that doesn’t require switches. Why not?
22. The transformer supplying power to an artist’s light sculpture E.32 Without any switchable magnets, the motor’s rotor will
provides 9600-V AC when supplied by 120-V AC. If there are 100 accelerate in the direction that minimizes its total
turns in the transformer’s primary coil, how many turns are there in potential energy. After a brief period of oscillating back
its secondary coil? and forth, it will settle down and never move again. To
continue moving, it needs switchable magnets, which
E.22 8000 turns. also provide the power needed to keep the rotor moving
23. The primary coil of a transformer makes 240 turns around the against the slowing forces of friction or moving
iron core, and the secondary coil of that transformer makes 80 turns. whatever is attached to the motor.
If the primary voltage is 120-V AC, what is the secondary voltage? 33. If you double the frequency of the AC current you supply to an
E.23 40-V AC. AC synchronous motor, what will happen?
E.23 The ratio of the secondary turns to primary turns is 1 to E.33 The rotor’s rotation speed will double.
3, so the ratio of the voltage rise in the secondary to the 34. If you double the AC voltage supplied to a synchronous AC
voltage drop in the primary is also 1 to 3. That means motor, what will happen?
that if the primary voltage drop is 120 V, the secondary
voltage rise is 40 V. E.34 The rotor’s rotation speed will remain unchanged
(although the motor may overheat).
24. The transformer in a stereo amplifier has a primary coil with 200
turns and a secondary coil with 40 turns. When the primary coil is 35. You’re wearing gloves when you grab the rotor shaft of an AC
synchronous motor and try unsuccessfully to stop that rotor from
Page 27 of 38 - 5/2/2011 - 11:51:27 AM
spinning. How does your action affect the motor’s power
consumption?
E.35 The motor consumes more power.
Problems – Chapter 11
36. If you supply DC voltage to a synchronous AC motor, what will 1. If a 0.10-A·m magnetic pole is placed in an upward-pointing 1.0-
happen? T magnetic field, what force would that pole experience?
E.36 The rotor may oscillate briefly, but it will soon stop P.1 0.10 N upward.
moving and not move again. 2. What is the force on a -2.0 A·m magnetic pole in a forward-
pointing 0.20-T magnetic field?
37. Some decorative lightbulbs have a loop-shaped filament that
jitters back and forth near a small permanent magnet. The filament P.2 0.4 N in the backward direction.
wire itself isn’t magnetic, so why does the filament move when
alternating current flows through it? 3. If a -1.0-A·m magnetic pole experiences a 1.0-N force downward,
what is the local magnetic field?
E.37 Current in a magnetic field experiences the Lorentz force
and bends the filament back and forth. P.3 1.0 T upward.
38. If a flexible wire carrying 60-Hz alternating current runs through 4. The magnetic force on a 5.0-A·m magnetic pole is 0.010 N to the
the gap between a north and south magnetic pole, what will happen to right. What is the magnetic field in which that pole is immersed?
that wire? P.4 0.02 T toward the right.
E.38 The wire will experience forces perpendicular to its 5. A magnetic pole in an upward-pointing 1.0-T magnetic field
length and to the line separating the two poles. It will experiences a 0.10-N force upward. What is that magnetic pole?
probably vibrate back and forth as a result, in synchrony
with the alternating current and thus at a frequency of 60 P.5 0.10 A·m.
Hz. 6. A magnetic pole in an upward-pointing 0.10-T magnetic field
39. Suppose you include an inductor in an electric circuit that experiences a 0.10-N force downward. What is that magnetic pole?
includes a battery, a switch, and a lightbulb. Current leaving the P.6 -1.0 A·m.
battery’s positive terminal must flow through the switch, the
inductor, and the lightbulb before returning to the battery’s negative 7. The earth’s magnetic field is approximately 0.000050 T. What is
terminal. The current in this circuit increases slowly when you close the energy in 1.0 m3 of that field?
the switch, and it takes the lightbulb a few seconds to become bright. P.7 0.0010 J.
Why?
8. The magnet in a large MRI unit may have a 1.0-T magnetic field
E.39 The inductor opposes changes in current and slows the occupying a volume of 1.0 m3. How much magnetic field energy is in
rise in current that occurs when you close the switch. that volume?
40. When you open the switch of the circuit in Exercise 39, a spark P.8 4.0 105 J.
appears between its two terminals. As a result, the circuit itself
doesn’t open completely for about half a second, during which time 9. What volume of the 1.0-T magnetic field in Problem 8 contains
the bulb gradually becomes dimmer. The bulb’s behavior indicates 1.0 J of energy?
that the current in the circuit diminishes slowly, rather than stopping
P.9 2.5 10-6 m3.
abruptly when you open the switch. Why does the current diminish
slowly? 10. A 0.050-T magnetic field is typical near household magnets.
E.40 The inductor opposes changes in current and it slows the What volume of that magnetic field contains 1.0 J of energy?
fall in current that occurs when you open the switch. P.10 0.001 m3.
41. A television picture tube produces its images using beams of 11. What magnetic field is necessary for 1.0 m3 of that field to
electrons that move through empty space and strike phosphors on the contain 1.0 J of energy?
inside of the screen. Those phosphors glow following the impact. But
aiming the electron beams is a delicate task. Why is it important to P.11 0.0016 T.
avoid placing strong magnets near a television picture tube? 12. What magnetic field is necessary for 1.0 m3 of that field to
E.41 The electron beams will be deflected by Lorentz forces contain 10 J of energy?
as they pass through the stray magnetic fields from those P.12 0.005 T.
magnets.
Exercises – Chapter 12
42. Audio speakers produce motion and ultimately sound by passing
fluctuating currents through wires immersed in magnetic fields. Why
would this arrangement result in motion and why is it important not
to put audio speakers too close to a television picture tube (Exercise
1. If a very small piece of material contains only 10,000 electrons
41)?
and those electrons have as little energy as possible, how many
E.42 Current flowing through wires in a magnetic field different levels do they occupy in that material?
experiences the Lorentz force. That force moves the E.1 5000 levels.
entire wire, pushing it back and forth in synchrony with
the current fluctuations in the wire. The strong magnetic 2. If electrons had four different internal states that could be
field near a speaker would also affect the charges distinguished from one another, how many electrons could occupy
moving inside a television picture tube and would cause the same level without violating the Pauli exclusion principle?
them to strike the wrong area of the screen. To keep the E.2 Four electrons could then occupy the same level.
television image pure, the picture tube must be protected
from stray magnetic fields.
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E.2 The Pauli exclusion principle only prevents charge it transfers, the battery must do an increasing
indistinguishable Fermi particles from occupying the amount of work with each transfer.
same level. With four different distinguishable electrons,
12. A charged capacitor resembles a battery in that both can supply
a level can hold four electrons.
electric power. However, as the capacitor delivers its power, the
3. If electrons were not Fermi particles, any number of them could voltage of the current it provides decreases. Why?
occupy a particular level. How would these electrons tend to arrange E.12 The voltage difference between the capacitor’s plates is
themselves among the levels in an object?
proportional to the amount of separated charge it
E.3 All in lowest energy level. contains. As the capacitor uses up its separated charge,
that voltage difference decreases.
4. If electrons were not Fermi particles (Exercise 3), would there
still be a distinction between metals and insulators? Explain your 13. We combine the three decimal digits 6, 3, and 1 to form 631 in
answer. order to represent the number 631. What does the 6 in 631 mean?
What are there 6 of?
E.4 No. All of the electrons could settle into the lowest levels
in an any material and there would be no Pauli exclusion E.13 There are 6 hundreds.
principle effects to prevent them from shifting between
levels in response to electric fields. 14. We combine the three binary bits 1, 0, and 1 to form 101 in
order to represent the number 5. What does the leftmost 1 in 101
5. Thermal energy can shift some of the electrons in a hot mean? What is there 1 of?
semiconductor from valence levels to conduction levels. What effect E.14 The left-most bit in the binary expression 101 refers to
do these shifts have on the semiconductor’s ability to conduct
the presence of one "4" in the number being represented.
electricity?
E.5 They will increase the semiconductor’s electrical E.14 The expression 101 indicates that the number contains
conductivity. one "4" and one "1", for a total of 5. The binary
expression 101 thus represents the number 5.
6. Why do semiconductor devices often self-destruct when they are
overheated? 15. What numbers do the two binary bytes 11011011 and
01010101 represent?
E.6 At elevated temperatures, thermal energy is sufficient to
shift electrons between valence and conduction levels. E.15 219 and 85.
The semiconductors become electrically conducting and 16. How is the number 165 represented in binary?
undesirable, destructive currents can flow through them.
E.16 The number 165 is represented by the binary expression
7. Is the p-type half of a p-n junction electrically neutral, positive, or 10100101.
negative?
17. Why are there no 2’s in the binary representation of a number?
E.7 The p-type half has a negative net charge. (In other words, why isn’t 1101121 a valid binary representation?)
8. In which direction does the electric field point in the middle of a E.17 Instead of reporting 2 twos, the binary representation
p-n junction? should report it as 1 four, the next higher power of two.
E.8 The electric field points from the n-type half toward the 18. For convenience, hexadecimal (powers of 16) is often used in
p-type half. place of binary. The traditional symbols used to represent
9. Two capacitors are identical except that one has a thinner hexadecimal digits are 0–9 and A–F. In hexadecimal, 10 represents
insulating layer than the other. If the two capacitors are storing the the number 16. Show that one hexadecimal digit can substitute
same amount of separated electric charge, which one will have the perfectly for four binary digits.
larger voltage difference between its plates? E.18 The numbers 0 through 15 can be represented by 4
E.9 The one with the thicker insulating layer. binary digits (0000 through 1111) Those same numbers
can also be represented by 1 hexadecimal digit (0
10. Dynamic memory stores bits as the presence or absence of through F). It’s thus possible to represent each
separated charge on tiny capacitors. It takes energy to produce hexadecimal digit by 4 binary digits or vice versa.
separated charge, and a computer that minimizes this energy will use
less electric power. Why does making the insulating layers of the 19. The gate of a MOSFET is separated from the channel by a
memory capacitors very thin reduce the energy it takes to store each fantastically thin insulating layer. This layer is easily punctured by
bit in them? static electricity, yet the manufacturers continue to use thin layers.
Why would thickening the insulating layer spoil the MOSFET’s
E.10 The closer the two plates of the capacitor are to one ability to respond to charge on its gate?
another, the smaller the voltage difference between the
plates when they contain a certain amount of separated E.19 The gate’s charge must be very near the channel so that
charge and the less energy that separated charge has. it can attract opposite charge and draw that charge into
the channel.
11. Suppose a battery is transferring positive charges from one plate
of a capacitor to the other. Why does the work that the battery does in 20. In an n-channel MOSFET, the source and drain are connected by
transferring a charge increase slightly with every transfer? a thin strip of p-type semiconductor. Why is this device labeled as
having an n-channel rather than a p-channel?
E.11 The voltage difference between the plates increases with
each transfer, so the energy required to complete the E.20 The strip of chemically p-type semiconductor becomes
subsequent transfer is greater. effectively n-type when extra electrons are drawn into it
by nearby positive charge on its gate.
E.11 As the amount of separated charge increases, so does the
voltage difference between the capacitor's two plates. 21. A MOSFET doesn’t change instantly from a perfect insulator to a
With an ever-increasing voltage rise required for each perfect conductor as you vary the charge on its gate. With
intermediate amounts of charge on its gate, the MOSFET acts as a
Page 29 of 38 - 5/2/2011 - 11:51:27 AM
resistor with a moderate electrical resistance. This flexibility allows E.2 No.
the MOSFET to control the amount of current flowing in a circuit.
Explain why a MOSFET becomes warm as it controls that current. E.2 The tank circuit is a type of harmonic oscillator, so its
period of oscillation is independent of the amplitude of
E.21 When current experiences a voltage drop as it flows that oscillation. Moving the magnet faster may induce a
through a MOSFET, some of its energy is converted into larger amplitude oscillation, but the period of that
thermal energy. oscillation will be unchanged.
22. The tiny MOSFETs that are used to move charge onto and off the 3. A tank circuit consists of an inductor and a capacitor. Give a
capacitors in dynamic memory are so small that they’re never very simple explanation for why the magnetic field in the inductor is
good conductors. Why do their modest electrical resistances lengthen strongest at the moment the separated charge in the capacitor reaches
the time it takes to store or retrieve charge from the memory zero.
capacitors?
E.3 The inductor’s magnetic field contains energy and it
E.22 With the MOSFETs barely conducting current, it takes a peaks when the capacitor’s energy is zero.
significant amount of time for the charge on the
capacitors to pass through those MOSFETs. 4. The metal wires from which most tank circuits are made have
electrical resistances. Why do these resistances prevent charge from
23. Why does the effect described in Exercise 22 limit the speed with oscillating forever in a tank circuit, and what happens to the tank
which a computer can store or retrieve bits from its dynamic circuit’s energy as time passes?
memory?
E.4 The wires waste the tank circuit's energy as thermal
E.23 It takes time for a MOSFET to change the charge on a energy, gradually reducing the amount of charge
wire in order to store or retrieve a bit. sloshing in the tank until the sloshing stops altogether.
24. If you connect the output of one inverter to the input of a second 5. To add energy to the charge oscillation in a tank circuit with an
inverter, how will the output of the second inverter be related to the antenna, at which time during the oscillation cycle should you bring a
input of the first inverter? positively charged wand close to the antenna?
E.24 The output of the second inverter will be the same as the E.5 Each time the antenna reaches its peak positive charge,
input to the first inverter. push the positive wand close to it. You’ll then be doing
E.24 The two inverters will invert the input twice, producing work on the oscillating charge.
an output that is the same as the input. 6. Two identical tank circuits with antennae are next to one another.
25. Suppose you connected a microphone directly to a large Explain why charge oscillating in one tank circuit can continue to do
unamplified speaker. Why wouldn’t the speaker reproduce your voice work on charge oscillating in the other tank circuit.
loudly when you talked into the microphone? E.6 Charge oscillations in the two tank circuits have the
E.25 The microphone can’t provide enough power. same period so that the two tank circuits can exchange
energy via sympathetic vibration. The fluctuating
26. Why can’t an audio amplifier operate without batteries or a electromagnetic fields from the first antenna push on
power supply? charges in the second antenna in perfect synchrony with
E.26 To produce a larger, more powerful copy of an input the charge fluctuations in that second antenna.
signal, the amplifier needs a source of energy. Otherwise 7. The ignition system of an automobile produces sparks to ignite
it would be creating energy out of nothing. the fuel in the engine. During each spark process, charges suddenly
accelerate through a spark plug wire and across a spark plug’s narrow
27. You like to listen to old phonograph records but your new stereo
amplifier has no input for a phonograph. You connect the
reception. Why?
phonograph to the stereo’s CD player input but find that the volume
is extremely low. Why? E.7 As charges accelerate in the wires, they emit radio
E.27 The phonograph produces a much smaller voltage rise waves.
than a CD player would provide. The amplifier expects a 8. To diminish the radio noise in a car (see Exercise 7), the ignition
larger voltage. system uses wires that are poor conductors of electricity. These wires
prevent charges from accelerating rapidly. Why does this change
28. To correct the volume problem in Exercise 27, you buy a small
preamplifier and connect it between the phonograph and the CD
player input of the stereo. The volume problem is gone. What is the E.8 Since accelerating electric charge is what produces radio
preamplifier doing to fix the problem? waves, limiting that acceleration reduces the intensity of
E.28 The preamplifier is boosting the voltage and current of radiated electromagnetic waves.
the input signal, creating a new input signal that is large 9. The electronic components inside a computer transfer charge to
enough for the amplifier to handle. and from wires, often in synchrony with the computer’s internal
clock. Without packaging to block electromagnetic waves, the
computer will act as a radio transmitter. Why?
Exercises – Chapter 13 E.9 As charges accelerate in the computer, they emit radio
waves.
1. If you pull a permanent magnet rapidly away from a tank circuit, 10. To save power in a computer, its thousands of wires usually
what is likely to happen in that circuit? avoid sharp bends. Why do sharp bends in current-carrying wires
E.1 Charge will oscillate in the tank’s capacitor and inductor. waste power?
2. Will the speed with which you pull the magnet away from the E.10 Since accelerating electric charge is what produces radio
tank circuit (Exercise 1) affect the period of its charge oscillation? waves, the sudden accelerations experienced by currents
Page 30 of 38 - 5/2/2011 - 11:51:27 AM
as they flow around sharp bends lead to intense and 20. Why are most microwave TV dinners packaged in plastic rather
wasteful electromagnetic radiation. than aluminum trays?
11. The sun emits a stream of energetic electrons and protons called E.20 Aluminum trays would reflect the microwaves and make
the solar wind. These particles frequently get caught up in the earth’s it difficult to cook the food properly.
magnetic field, traveling in spiral paths that take them toward the
21. Why is it so important that a microwave oven turn off when you
north or south magnetic poles. When they head northward and collide
open the door?
with atoms in the earth’s upper atmosphere, those atoms emit light
we know as the aurora borealis, or northern lights. These particles E.21 Releasing the microwaves into the room wouldn’t be
also interfere with radio reception. Why do they emit radio waves? healthy.
E.11 The spiraling charges are accelerating and thus emit 22. Compare how a potato cooks in a microwave oven with how it
electromagnetic waves. cooks in an ordinary oven.
12. When a radio signal travels through a coaxial cable, charge E.22 In a microwave oven, the potato’s water absorbs
moves back and forth on both the central wire and the surrounding microwaves. The microwave energy becomes thermal
tube. Show that both electric and magnetic fields are present in the energy and the potato’s temperature rises relatively
coaxial cable. uniformly. In an ordinary oven, heat flows gradually into
E.12 Whenever the charges on the central wire and the potato through its surface and its temperature rises
surrounding tube are different, there are electric fields nonuniformly. The middle of the potato cooks last.
pointing from positive to negative charges. And 23. When you’re listening to FM radio near buildings, reflections of
whenever charge moves on either of the two conductors, the radio wave can make the reception particularly bad in certain
that moving charge produces magnetic fields. locations. Compare this effect to the problem of uneven cooking in a
microwave oven.
13. If you wave a positively charged wand up and down vertically,
the electromagnetic wave it emits has which polarization? E.23 Both involve destructive interference in electromagnetic
E.13 Vertical polarization. waves.
14. If you set a magnetic compass on the table and spin its magnetic 24. Dish-shaped reflectors are used to steer microwaves in order to
needle horizontally, its accelerating poles will emit an establish communications links between nearby buildings. Those
electromagnetic wave with which polarization? reflectors are often made from metal mesh. Why don’t they have to
be made from solid metal sheets?
E.14 Vertical polarization.
E.24 Microwaves cannot respond to holes in the metal that are
15. While a particular AM radio station claims to transmit 50,000 W significantly smaller than their wavelengths. The metal
of music power, that’s actually its average power. There are times mesh is equivalent to solid metal as far as the
when it transmits more power than that and times when it transmits microwaves are concerned.
less. Explain.
25. Why is the thin metal handle of a Chinese food container
E.15 The AM station changes the power of its transmission in dangerous when placed in a microwave oven?
order to represent air pressure fluctuations with the radio
wave. E.25 It is thin enough to be heated by the resulting currents
and its sharp ends may spark.
only hear the loud parts of the transmission. When it’s too far from an 26. Is a thick, smooth-edged stainless steel bowl dangerous in a
FM station, you lose the whole sound all at once. Explain the reasons microwave oven?
for this difference. E.26 No (although it may alter the rate at which the food
E.16 In an AM transmission, the radio wave is strongest nearby cooks).
during the loudest portions of the broadcast. But in an 27. A cyclotron is a particle accelerator invented in 1929 by
FM transmission, the radio wave intensity is constant. American physicist Ernest O. Lawrence. It uses electric fields to do
17. When an AM radio station announces that it’s transmitting at 950 work on charged particles as they follow circular paths in a strong
kHz, that statement isn’t quite accurate. Explain why it may also be magnetic field. Lawrence’s great insight was that all the particles take
transmitting at 948 kHz and 954 kHz. the same amount of time to complete one circle, regardless of their
speed or energy. That fact allows the cyclotron to do work on all the
E.17 Amplitude modulation introduces additional frequencies particles at once as they circle together. How can a faster moving
that extend as much as 5 kHz above and below the electron take the same time to circle as a slower moving electron?
carrier wave.
E.27 The path of a faster moving electron bends more
18. The Empire State Building has several FM antennas on top, gradually, so it travels in a larger circle. It returns to its
added in part to increase its overall height. These antennas aren’t very starting point at the same time the slower moving
tall. Why do short antennas, located high in the air, do such a good electron returns to its starting point.
28. An extremely fast-moving charged particle traveling in a
E.18 Television transmission involves high-frequency, short- magnetic field can radiate X-rays, a phenomenon known as
wavelength radio waves. Since a good antenna is one- synchrotron radiation. Why is the magnetic field essential to this
quarter wavelength long, television transmission requires emission?
relatively short antennas.
E.28 Without the magnetic field, the charged particle would
19. Porous, unglazed ceramics can absorb water and moisture. Why travel at constant velocity (at constant speed along a
are they unsuitable for use in a microwave oven? straight path) and would not radiate electromagnetic
waves.
E.19 Water trapped in the ceramic would absorb microwaves,
and the ceramic would become extremely hot. It might
even shatter.
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3. When astronauts aboard the space shuttle look down at the earth,
Problems – Chapter 13 its atmosphere appears blue. Why?
E.3 Rayleigh scattering deflects blue light in all directions.
1. How much energy is contained in 1.0 m3 of a 1.0-V/m (or 1.0- 4. When astronauts walked on the surface of the moon, they could
N/C) electric field? see the stars even though the sun was overhead. Why can’t we see the
P.1 4.4 10-12 J. stars while the sun is overhead?
2. How much energy is contained in 1.0 m3 of a 10,000-V/cm E.4 The Rayleigh scattered light from the sun is so bright
electric field? that it overwhelms the dim light from stars.
P.2 4.4 10-4 J. 5. If you shine a flashlight horizontally at a glass full of water, the
glass will redirect the light beam. How?
3. What volume of a 1000-N/C electric field contains 1.0 J of
energy? E.5 The light refracts as it enters and leaves the glass.
P.3 230,000 m .3 6. A laser light show uses extremely intense beams of light. When
one of these beams remains steady, you can see the path it takes
4. How much volume of a 500-V/m electric field contains 0.0010 J through the air. What makes it possible for you to see this beam even
of energy? though it isn’t directed toward you?
P.4 900 m3. E.6 The laser beams experience Rayleigh scattering in the air
5. What electric field is needed for 1.0 m3 to contain 1.0 J of energy? and you can see this scattered light.
P.5 480,000 V/m. 7. To make the beams at a laser light show (see Exercise 6) even
more visible, they’re often directed into mist or smoke. Why do such
3
6. What electric field contains 0.05 J in 10 m ? particles make the beams particularly visible?
P.6 3.4 104 V/m. E.7 The large particles scatter light better than air molecules.
7. The frequency of the radio wave emitted by a cordless telephone 8. Why can you see your reflection in a calm pool of water?
is 900 MHz. What is the wavelength of that wave?
E.8 As light enters the water, it slows down and part of it
P.7 0.333 m. reflects.
8. Citizens band (CB) radio uses radio waves with frequencies near 9. Use the concepts of refraction, reflection, and dispersion to
27 MHz. What are the wavelengths of these waves and how long explain why a diamond emits a spray of colored lights when sunlight
should a quarter-wavelength CB antenna be? passes through its cut facets.
P.8 Their wavelength is 11.1 m and a quarter-wavelength E.9 Light bends and disperses on entry, reflects from the
antenna should be 2.8 m long. back surface, and bends and disperses more on exit from
the diamond.
9. The electromagnetic waves in blue light have frequencies near 6.5
1014 Hz. What are their wavelengths? 10. Diamond has an index of refraction of 2.42. If you put a diamond
in water, you see reflections from its surfaces. But if you put it in a
P.9 0.461 10 m (461 nm).
-6
liquid with an index of refraction of 2.42, the diamond is invisible.
10. Amateur radio operators often refer to their radio waves by Why is it invisible, and how is this effect useful to a jeweler or
wavelength. What are the approximate frequencies of the 160-m, 15- gemologist?
m, and 2-m wavelength amateur radio bands? E.10 If the liquid and diamond have the same index of
P.10 1.9 106 Hz, 2.0 107 Hz, and 1.5 108 Hz refraction, then light doesn't change speed on entry to or
respectively. exit from the diamond and no light reflects from the
interfaces between the two.
11. The radio waves used by cellular telephones have wavelengths of
approximately 0.36 m. What are their frequencies? 11. Why is a pile of granulated sugar white while a single large piece
of rock candy (solid sugar) is clear?
P.11 8.33 108 Hz.
E.11 Each surface in the granulated sugar reflects some of the
light passing through it. These random reflections make
Exercises – Chapter 14 the sugar white.
12. Basic paper consists of many transparent fibers of cellulose, the
1. How long should an antenna be to receive or transmit red light main chemical in wood and cotton. Why does paper appear white,
well? and why does it become relatively clear when you get it wet?
E.1 About 160 nm. E.12 Some light reflects from each randomly oriented
interface between air and cellulose. This light ends up
2. How long should an antenna be to receive or transmit violet light traveling in all directions because of the random
well? orientations of those reflecting surfaces. When water fills
E.2 An antenna for violet light should be about 100 the gaps between fibers, the changes in light speed are
nanometers long. smaller and so are the reflections.
E.2 Violet light has a wavelength near 400 nanometers, so a 13. On a rainy day you can often see oil films on the surfaces of
puddles. Why do these films appear brightly colored?
one-quarter-wavelength antenna would be about 100
nanometers long. E.13 Light reflects from the top and bottom surfaces of an oil
film, and the two reflections interfere with one another.
Page 32 of 38 - 5/2/2011 - 11:51:27 AM
The type of interference depends on the film’s thickness 24. When a sodium atom is in its lowest energy excited state, it can
and the light’s wavelength. emit light. Why?
14. When two sheets of glass lie on top of one another, you can often E.24 The sodium atom can undergo a radiative transition to its
see colored rings of reflected light. How do the nearby glass surfaces ground state, thereby emitting a photon of (yellow) light.
cause these colored rings?
25. You expose a gas of argon atoms to light with photon energies
E.14 Partial reflections of light from the back of one sheet of that don’t correspond to the energy difference between any pair of
glass and from the front of the next sheet of glass states in the argon atom. Explain what happens to the light.
interfere with one another. Because the type of E.25 Nothing happens because the atoms have no radiative
interference, constructive or destructive, depends on the transitions that can absorb photons of that light.
spacing between the glass and the wavelength of the
light, the interfering light tends to have a colored 26. A discharge in a mixture of gases is more likely to emit a full
appearance. white spectrum of light than a discharge in a single gas. Why?
15. If you’re wearing polarizing sunglasses and want to see who else E.26 The more different atoms and molecules present in a gas
is wearing polarizing sunglasses, you only have to turn your head discharge, the more variety there is in the possible
sideways and look to see which people now have sunglasses that radiative transitions and the more likely that a rich, full
appear completely opaque. Why does this test work? spectrum of white light will be emitted by the discharge.
E.15 Polarizing sunglasses normally block horizontally 27. If the low-pressure neon vapor in a neon sign were replaced by
polarized light, so when you look at someone's eyes low-pressure mercury vapor, the sign would emit almost no visible
when they are wearing polarizing sunglasses, you see light. Why not?
only vertically polarized light. If you wear polarizing
E.27 Excited mercury atoms emit primarily invisible
ultraviolet light.
sunglasses will block vertically polarized light. You will
see no light coming from the eyes of other people 28. Increasing the power to an incandescent bulb makes its filament
wearing polarizing sunglasses. hotter and its light whiter. Why doesn’t increasing the power to a
neon sign change its color?
16. Why is it easier to see into water when you look directly down
into it than when you look into it at a shallow angle? E.28 The incandescent lamp is emitting thermal radiation with
a roughly blackbody spectrum. The neon sign is not a
E.16 Horizontally polarized light reflects more strongly at thermal light source, so increasing the power simply
shallow angles than at right angles. brightens its light and doesn't change its spectrum.
17. Light near 480 nm has a color called cyan. What mixture of the 29. While many disposable products no longer contain mercury, a
primary colors of light makes you perceive cyan? potential pollutant, fluorescent tubes still do. Why can’t the
E.17 Green and blue. manufacturers eliminate mercury from their tubes?
18. What is different about the two mixtures of red and green lights E.29 Mercury atoms themselves produce the tubes’ ultraviolet
that make you see yellow and orange, respectively? light.
E.18 If the green light is relatively strong compared to the red 30. When white fabric ages it begins to absorb blue light. Why does
light, you’ll see yellow. However, if the red light is this give the fabric a yellow appearance?
relatively strong compared to the green light, you’ll see E.30 When white light encounters a surface that reflects less
orange. blue light than red or green, that surface you see a
19. What colors of light does red paint absorb? mixture of red and green lights coming from that surface.
Your eyes interpret this mixture as yellow.
E.19 The green, blue, and violet end of the spectrum.
31. To hide yellowing (see Exercise 30), fabric is often coated with
20. What colors of light does yellow paint absorb? fluorescent ―brighteners‖ that absorb ultraviolet light and emit blue
E.20 Blue light (and other light near the blue end of the light. In sunlight, this coated fabric appears white, despite absorbing
optical spectrum). some blue sunlight. Explain.
21. If you illuminate red paint with pure blue light, what color will E.31 Fluorescence from brighteners replaces the missing blue
that paint appear? light.
E.21 Black. 32. Camera flashes use discharges in high-pressure xenon and
krypton gases to produce brief, intense white light. Why is it
22. Fancy makeup mirrors allow users to choose either fluorescent or important that they use these complicated atoms?
incandescent illumination to match the lighting in which they’ll be
seen. Why does the type of illumination affect their appearances? E.32 Complex atoms have so many states that they can
produce rich light spectra.
E.22 Since incandescent light has less blue than fluorescent
light, incandescent light reflected from a person's skin 33. A CD player uses a beam of laser light to read the disc, focusing
would also have less blue in it. Skin can't reflect light that light to a spot less than 1 µm (10-6 m) in diameter. Why can’t the
that isn't there. player use a cheap incandescent lightbulb for this task, rather than a
more expensive laser?
23. While a sodium atom is in its ground state, it cannot emit light.
Why not? E.33 The lightbulb’s photons are all different and won’t focus
to the same tiny spot.
E.23 There is no lower energy state to which it can make a
transition and, while it remains in the ground state, its 34. Why can’t a CD player (Exercise 33) use a light-emitting diode
electrons are in standing waves and cannot emit (LED) in place of its diode laser?
electromagnetic waves.
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E.34 The incoherent light from an LED won’t focus as well as
the coherent light of a diode laser. The CD player will
not be able to illuminate the tiny features in the CD
Problems – Chapter 14
selectively enough to read the information.
1. The yellow light from a sodium vapor lamp has a frequency of
35. Explain why the electromagnetic wave emitted by a radio station 5.08 1014 Hz. How much energy does each photon of that light
is coherent—a low-frequency equivalent of coherent light. carry?
E.35 The radio station’s photons are identical—part of a P.1 3.37 10-19 J.
single wave.
2. If a low-pressure sodium vapor lamp emits 50 W of yellow light
36. One of the most accurate atomic clocks is the hydrogen maser. (see Problem 1), how many photons does it emit each second?
This device uses excited hydrogen molecules to duplicate 1.420-GHz
microwave photons. In the maser, the molecules have only two states: P.2 1.5 1020 photons/second
the upper maser state and the lower maser state (which is actually the 3. A particular X-ray has a frequency of 1.2 1019 Hz. How much
ground state). To keep the maser operating, an electromagnetic energy does its photon carry?
system constantly adds excited state hydrogen molecules to the maser
and a pump constantly removes ground state hydrogen molecules P.3 7.95 10-15 J.
from the maser. Why does the maser require a steady supply of new 4. A particular light photon carries an energy of 3.8 10-19 J. What
excited state molecules? are the frequency, wavelength, and color of this light?
E.36 If the excited states are outnumbered by the ground
P.4 5.7 1014 Hz, 520 nanometers, green.
states, more photons will be absorbed than emitted and
the maser will not amplify passing photons. 5. If an AM radio station is emitting 50,000 W in its 880-kHz radio
wave, how many photons is it emitting each second?
37. Why must ground state molecules be pumped out of a hydrogen
maser (see Exercise 36) as quickly as possible to keep it operating P.5 8.58 1031 photons/s.
properly?
6. If an FM radio station is emitting 100,000 W in its 88.5-MHz
E.37 They’ll absorb the microwaves the maser is trying to radio wave, how many photons is it emitting each second?
produce.
P.6 1.7 1030 photons/s.
38. While some laser media quickly lose energy via the spontaneous
emission of light, others can store energy for a long time. Why is a
long storage time essential in lasers that produce extremely intense
pulses of light? Exercises – Chapter 15
E.38 Long storage allows energy to be deposited into the laser 1. On a bright, sunny day you can use a magnifying glass to burn
medium over a longer period of time. That means that wood by focusing sunlight onto it. The focused sunlight forms a
conventional light sources can be used to store the small circular spot of light that heats the wood until it burns. Why is
energy and huge amounts of total energy can be the spot of light circular?
accumulated.
E.1 It is a real image of the circular sun itself.
39. One of the first lasers used synthetic ruby as its laser medium.
However, a ruby laser is a three-state laser; its lower laser state is its 2. Light passing through a curved water glass forms an image of the
ground state. Why does that arrangement make the ruby laser candle flame on the wall beside the table. Why would moving the
relatively inefficient? candle toward or away from the glass spoil this effect?
E.39 The ground state systems absorb much of the amplified E.2 The water glass is acting as a lens. Changing the distance
light before it can leave the ruby. between the candle and that lens will change the distance
between the lens and the real image it forms. The
40. If most of the highest energy valence levels in a diode laser’s p- candle’s image will no longer be in focus on the wall.
type anode weren’t empty, it would become relatively inefficient and
probably wouldn’t emit laser light at all. Why not? 3. Simple reading glasses are converging lenses that come in a
variety of strengths, ranging from about 0.25 diopter (almost flat
E.40 The diode laser needs a substantial population inversion glass) to 3.00 diopters (highly curved). Which lens has the shorter
to operate well. Electrons in the highest valence levels focal length: 1.0 diopter or 2.0 diopters?
can absorb the laser light while undergoing radiative
transitions to the conduction levels. That absorption E.3 The 2.0-diopter lens has the shorter focal length.
would trap the laser radiation and reduce the diode’s 4. In a movie, an actor’s eyeglasses send a brief undistorted
ability to emit laser light. reflection of the sun at the camera. Why does that simple reflection
41. Why doesn’t increasing the current passing through an LED tell you that the glasses are props and that the actor doesn’t need
affect the color of its light? eyeglasses?
E.41 The color of light emitted by an LED depends primarily E.4 The undistorted reflection indicates that the eyeglass
on the band gap in the LED’s semiconductor. ―lenses‖ are flat plates of glass or plastic. With no
curvature to their surfaces, these plates don’t bend light
42. Why does increasing the current passing through an LED affect and can’t correct vision problems. They’re just props.
the brightness of its light?
5. One part of a fiber optic communication system uses a lens to
E.42 With more current passing through the LED, the focus light from a semiconductor laser onto the end of an optical
population of electrons in excited conduction states fiber. The tiny light source and fiber are on opposite sides of the lens,
increases and the rate of radiative transition in the LED and each is 1.0 cm away from the lens. Why must the lens have a
increases. focal length of 0.5 cm?
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E.5 The lens equation gives 0.5 cm as the focal length of the E.14 For astronomical work, the object distance is always
lens when the image and object distances are both 1.0 infinite (the distances to the planets or star are
cm. enormous), so the image distance is always equal to the
focal length of the telescope.
6. Light from a distant object approaches the objective lens of a
telescope as a group of parallel light rays. Draw a picture of the light 15. Two similar looking magnifying glasses have different
rays from three stars passing through a converging lens to show that magnifications. One is labeled 2 (two times) and the other 4 (four
they focus at three separate locations. times). Which lens has the longer focal length?
E.6 (The drawing will show each bundle of parallel rays E.15 The lower magnification 2 glass has the longer focal
merging gradually together to a point on the far side of length.
the converging lens. Since each converging point is
along the path of those parallel rays, it will be different 16. Which magnifying glass from Exercise 15 must you hold closer
for the different ray bundles.) to the object you’re looking at in order to see a virtual image of that
object far in the distance?
7. Sports photographers often use large aperture, long focal length
E.16 You must hold the 4x lens closer to the object while
lenses. What limitations do these lenses impose on the photographs?
viewing it.
E.7 Their depths of focus are small.
17. The objective lens in a DVD player must move quickly to keep
8. If you’re taking a photographic portrait of a friend and want the laser beam focused on the disc’s reflective layer. Why must this
objects in the foreground and background to appear blurry, how lens have a very small mass?
should you adjust the camera’s aperture and shutter speed?
E.17 The low mass lens can be accelerated rapidly by modest
E.8 You should use the largest available aperture to forces.
minimize depth of focus and compensate for the large
light intensity on the film by selecting a short shutter 18. Why don’t portable CD or DVD players play properly if you
speed. shake them back and forth too quickly?
E.18 When you shake a CD or DVD player, you use inertia to
9. Your new 35-mm camera comes with two lenses, a 50-mm focal
misalign the laser optical system that is trying to
length ―normal‖ lens and a 200-mm focal length ―telephoto‖ lens.
measure the reflectivity of the recorded surface. If the
The minimum f-number for the 50-mm lens is 1.8. Although the 200-
mm lens has much larger glass elements in it, its lowest f-number is player isn't able to read the information for a long
4. Why is the telephoto lens’s f-number so much larger? enough period of time, it will stop playing properly.
E.9 To have an f-number of 1.8, the elements in the 200-mm 19. What happens to a ray of light entering plastic from the air at an
lens must be four times the diameter of the elements in angle to the interface?
the 50-mm lens. E.19 The ray bends to travel more nearly perpendicular to the
10. If you hold your camera up to a small hole in a fence, you will be interface.
able to take a picture of the scene on the other side. However, the 20. Why is a laser beam’s focus delayed by its entry into the plastic
exposure time will have to be quite long, and the depth of focus will of a CD or DVD? (Draw a picture.)
be surprisingly large. Explain.
E.20 As the light rays enter the plastic from the air, they bend
E.10 The hole effectively reduces the diameter of the camera outward somewhat. As a result of this outward bending,
lens. Small diameter lenses have large depths of focus the rays take longer to meet at a focus and the focus is
(because those few light rays that manage to get through delayed.
the small opening are already close together) but require
longer exposures (because so little light passes through 21. Why does a laser beam spread out quickly after it passes through
the narrow lens opening). a tiny pinhole?
11. Why does squinting increase your depth of focus? E.21 Diffraction makes the narrowed light wave spread
severely.
E.11 Reducing the size of your eye’s aperture increases its
depth of focus. 22. Scientists measure the distance to the moon by bouncing laser
light from reflectors left on the moon by the Apollo astronauts. This
12. Why is it easiest for the lens of your eye to form sharp images on light is sent backward through a telescope so that it begins its trip to
your retina when the scene in front of you is bright and the iris of the moon from an enormous opening. This procedure reduces the size
your eye is very small? of the laser beam when it reaches the moon. Explain.
E.12 When the pupil in your iris is small, you eye behaves as E.22 If the laser beam started by passing out of the opening of
a camera with a small diameter lens: it has a large depth a small laser, diffraction would cause the narrowed wave
of focus (because those few light rays that pass through to spread out and much of it would miss the moon. But
the narrow opening are already close together) but it by expanding the beam with a telescope so that it
admits little light, so it needs a bright scene. effectively leaves a huge opening, diffraction effects
become small and the broad wave travels more directly
13. People who need glasses find it hardest to see clearly without
them in dimly lit situations when the irises of their eyes are wide toward the moon.
open. Why? 23. As it passes through the lens that focuses it onto a CD’s
E.13 The depth of focus is smallest when the whole lens is aluminum layer, the player’s laser beam is more than 1 mm in
used. diameter. Why does its large size as it leaves the lens allow the beam
to focus to a smaller spot?
14. Photographic telescopes are simply enormous cameras. With
E.23 The lens’s large opening reduces diffraction effects.
such small f-numbers, they have small depths of focus. Why isn’t that
a problem for astronomical work? 24. Why can light from a blue laser form a narrower beam waist than
light from an infrared laser?
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E.24 The size of the beam waist depends in large measure on
the wavelength of the light, with the beam waist limited
to a diameter of roughly that wavelength. Since blue
Exercises – Chapter 16
light has a shorter wavelength than infrared light, blue
light can form a narrow beam waist. 1. Naturally occurring copper has two isotopes, 63Cu and 65Cu. What
is different between atoms of these two isotopes?
25. Sometimes the surfaces of a glass of water look mirrored when E.1 Nuclei of 65Cu atoms contain two more neutrons than
you observe them through the water. Explain.
those of 63Cu.
E.25 Light is experiencing total internal reflection inside the
glass. 2. Why is it extremely difficult to separate the two isotopes of
copper, 63Cu and 65Cu?
26. When you look into the front of a square glass vase filled with E.2 Those two isotopes are chemical identical and can only
water, its sides appear to be mirrored. Why do the sides appear so
be separated by using their small difference in mass.
shiny?
E.26 You are seeing light that has experienced total internal 3. If a nuclear reaction adds an extra neutron to the nucleus of 57Fe
reflection. That light tried to escape from the glass at too (a stable isotope of iron), it produces 58Fe (another stable isotope of
iron). Will this change in the nucleus affect the number and
shallow an angle and reflected back into the glass
arrangement of the electrons in the atom that’s built around this
nucleus? Why or why not?
27. Some of the laser light striking a DVD’s reflective layer hits the E.3 No, the number of electrons is set by the number of
flat region around a pit and reflects back toward the photodiode. How protons.
does this reflected wave actually reduce the amount of light detected
by the photodiode? 4. If a nuclear reaction adds an extra proton to the nucleus of 58Fe (a
stable isotope of iron), it produces 59Co (a stable isotope of cobalt).
E.27 Waves reflected by the pits and flats interfere
Will this change in the nucleus affect the number and arrangement of
destructively.
the electrons in the atom that’s built around this nucleus? Why or
28. Why does the surface of a DVD look so colorful in white light? why not?
E.28 The tiny reflective features on the DVD send light E.4 The 59Co nucleus requires one more electron than 58Fe to
toward your eyes via many slightly different paths. That form a neutral atom, so the electron arrangement will
light experiences interference effects at your eyes, so change.
that you see an interference pattern. Since the pattern
5. When two medium-sized nuclei are stuck together during an
differs according to the light’s wavelength, you see a experiment at a nuclear physics lab, the result is usually a large
complicated and beautiful array of colors. nucleus with too few neutrons to be stable. The nucleus soon falls
apart. Why could more neutrons make it stable?
Problems – Chapter 15 E.5 Neutrons would experience the attractive nuclear force
while diluting the repulsion between protons.
1. Your camera has a 35-mm focal length lens. When you take a 6. The large nucleus in Exercise 5 is likely to undergo alpha decay.
photograph of a distant mountain, how far from that lens is the real What emerges from the nucleus during alpha decay and why does
image of the mountain? this decay reduce the nucleus’s total potential energy?
P.1 35 mm. E.6 A helium nucleus (4He, which contains two neutrons and
two protons) emerges. Both it and the remaining nucleus
2. If you use a 35-mm focal length lens to take a photograph of are positively charged and as these like charges separate,
flowers 2 m from the lens, how far from that lens does the real image their electrostatic potential energies decrease.
of the flowers form?
7. When a large nucleus is split in half during an experiment at a
P.2 35.6 mm. nuclear physics lab, the result is usually two medium-sized nuclei
3. You’re trying to take a photograph of two small statues with a with too many neutrons to be stable. These nuclei eventually fall
200-mm telephoto lens. One statue is 4 m from the lens, and the other apart. Why don’t these smaller nuclei need as many neutrons as they
statue is 5 m from the lens. The real image of which statue forms received from the original nucleus?
closer to the lens? How much closer? E.7 With fewer protons, the diluting effect of neutrons
P.3 The real image of the more distant statue forms 2.2 mm matters less.
closer. 8. One of the medium-sized nuclei in Exercise 7 is likely to undergo
4. When you place a saltshaker 30 cm from your magnifying glass, a beta decay. What emerges from that nucleus during beta decay and
real image forms 30 cm from the lens on the opposite side. You can what happens to the nucleus as a result?
see this image dimly on a sheet of paper. What is the focal length of E.8 An electron and an antineutrino emerge from the
the magnifying glass? nucleus. The proton remains behind in the nucleus and
P.4 15 cm. increases its atomic number by one.
5. When light from your desk lamp passes through a 50.0-mm focal 9. Light can’t penetrate even a millimeter into plutonium, so why is
length lens, it forms a sharp real image on a sheet of paper located a neutron able to travel centimeters into plutonium?
50.5 mm from the lens. How far is the desk lamp from the lens? E.9 A neutron has no charge and only interacts with nuclei.
P.5 5 m. The nuclei are so small that they’re hard to hit.
10. Why is it difficult or impossible to make very small atomic
bombs?
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E.10 The smaller the nuclear fuel component, the more likely 21. Lead, with 82 electrons per atom, is an excellent absorber of X-
it is that a neutron will escape before it causes a fission rays. Why?
and the less efficient the fuel is at using fission neutrons E.21 Almost any X-ray matches the energy of one of lead’s
to sustain a chain reaction. many electrons and thus can cause efficient
11. Explain the strategy of putting highly radioactive materials in a photoelectron emission.
storage place for many years as a way to make them less hazardous.
22. Electric charge is strictly conserved in our universe, meaning that
That wouldn’t work for chemical poisons, so why does it work for
the net charge of an isolated system can’t change. Why doesn’t the
production of an electron–positron pair in a patient cause a change in
E.11 With time, radioactive nuclei decay spontaneously and the patient’s net charge?
the materials become less and less hazardous. E.22 When an electron-positron pair is created, there is no net
12. Once fallout from a nuclear blast distributes radioactive isotopes change in charge. The electron has a charge of -1 while
over a region of land, why is it virtually impossible to separate many the positron has a charge of +1, so their sum is 0 charge.
of those radioactive isotopes from the soil?
23. Which has more mass: a positron or an antiproton?
E.12 The radioactive isotopes are chemical indistinguishable
E.23 An antiproton has more mass than a positron.
from ordinary elemental constituents of the soil.
24. Magnetic resonance imaging (MRI) differs from computed
13. Burning chemical poisons in a gas flame often renders those
tomography imaging in that it involves no ―ionizing radiation.‖ What
poisons harmless. Why won’t that strategy make radioactive
electromagnetic radiation is used in MRI and why aren’t the photons
materials less dangerous?
of this radiation able to remove electrons from atoms and convert
E.13 Chemical reactions have no effect on nuclei. those atoms into ions?
14. Sunscreen absorbs ultraviolet light while permitting visible light E.24 MRI uses radiowaves or microwaves and photons of
to pass. Why does this coating reduce the risk of chemical and those two types of electromagnetic radiation have far too
genetic damage to the cells of your skin? little energy to cause chemical damage to molecules.
E.14 By blocking the high-energy photons in ultraviolet light 25. Magnetic resonance imaging isn’t good at detecting bone. Why
before they reach your skin, the sunscreen reduces the not?
amount of chemical injury caused when those photons
E.25 MRI detects hydrogen. Bone contains little hydrogen.
26. No magnetic metals such as iron or steel are permitted near a
15. Why is it important to keep foods and drugs out of direct magnetic resonance imaging machine. In part, this rule is a safety
sunlight, even when they’re not in danger of overheating? precaution since those magnetic metals would be attracted toward the
E.15 Photons in sunlight can cause chemical damage. machine. But the magnetic fields from these magnetic metals would
also spoil the imaging process. Why would having additional
16. Why are many drugs packaged in amber-colored containers that magnetic fields inside the imaging machine spoil its ability to locate
block ultraviolet light? specific protons inside a patient’s body?
E.16 The high-energy photons in ultraviolet light can damage E.26 Magnetic metals would change the local magnetic fields
the molecules in medicines. inside a patient and impair the imaging machine's ability
17. Museums often display priceless antique manuscripts under dim to precisely control those fields. The machine needs to
yellow light. Why not use white light? be able to know exactly what the fields are at each point
in the patient in order to know where the protons it is
E.17 Photons in blue or ultraviolet light can cause chemical detecting are located.
damage to the molecules in the manuscripts. Yellow
light generally can’t. 27. Why does the strength of the magnetic field used in MRI affect
the frequency of the radio waves used to detect the protons?
18. The most troubling radioactive isotopes in fallout are those with
half-lives between a few days and a few thousand years. Why are E.27 The stronger the magnetic field, the more energy a
those with much shorter or much longer half-lives less of a problem? proton needs to change from aligned to anti-aligned. The
radio wave photons must have more energy, so the
E.18 It’s possible to wait out isotopes with short half-lives frequency must be higher.
because they decay away quickly. And long half-life
isotopes decay so slowly that it’s unlikely they’ll decay 28. The stronger the magnetic field used in MRI, the larger the
during a person’s lifetime. fraction of protons that align their spins with the field. Why does this
increased alignment make it easier for the MRI to study the protons?
19. The most troubling radioactive isotopes in nuclear waste are
those with half-lives of between a few years and a few hundred E.28 The more spin-aligned protons in a patient, the more
thousand years. Explain. radiative transitions the MRI system can produce and the
more easily it can detect tissue.
E.19 Isotopes with shorter half-lives decay quickly and can be
waited out, while those with much longer half-lives are 29. The uranium fuel in a thermal fission reactor must be replaced
relatively less likely to decay during a human lifetime. every so often. When that fuel is removed, it’s still mostly uranium.
Why can’t that uranium be cleaned and reused until the uranium is
20. An X-ray technician can adjust the energy of the X-ray photons completely consumed?
produced by a machine by changing the voltage drop between the X-
ray tube’s cathode and anode. Explain. E.29 The reactor is consuming only the 235U. The uranium left
in the fuel is mostly 238U, which isn’t fissionable.
E.20 The larger the voltage drop between cathode and anode,
the more kinetic energy each electron has by the time it 30. Some satellites have been launched with small nuclear reactors
reaches the anode and the more energetic the X rays it on board to provide electric power. Why must these small reactors
can produce. use highly enriched uranium or plutonium rather than natural or
slightly enriched uranium?
Page 37 of 38 - 5/2/2011 - 11:51:27 AM
E.30 Only a moderated, thermal fission reactor can operate on nucleus and the radiative transition 99mTc → 99Tc emits a gamma ray.
nature or slightly enriched uranium. The 238U in that That gamma ray shows where the nucleus was located when it
reactor’s fuel wouldn’t be used and would simply add an decayed. If the radiologist begins looking for 99mTc 4.00 hours after
enormous amount of weight to the reactor. administering it, what fraction of the 99mTc nuclei remain?
31. As a nuclear reactor operates, its fuel rods gradually accumulate P.3 63% of the 99mTc nuclei remain.
neutron-absorbing fission fragments. As those fragments accumulate, 4. After a 99mTc nucleus emits a gamma ray (Problem 3), it becomes
what must the reactor operators do to keep the chain reaction going? a 99Tc nucleus. 99Tc is also radioactive, with a half-life of 213,000
E.31 The operators must increase the probability of each years, and it decays into 99Ru. Two weeks after 99mTc was
fission neutron causing a subsequent fission, for administered to the patient, what fraction of it is 99mTc? What fraction
example, by pulling the control rods farther out of the of it is 99Tc? What fraction of it is 99Ru?
reactor. P.4 The fraction of 99mTc nuclei remaining after two weeks is
32. Control rods are usually installed on top of the reactor where only 1.7 10-17. Almost 100% of the nuclei have
their weights tend to pull them into the core. Why is this arrangement become 99Tc, although about 1.2 10-7 of those nuclei
much safer than putting the control rods at the bottom of the reactor? have further decayed into 99Ru.
E.32 The rods-on-top arrangement makes it easy for the 5. While most natural potassium nuclei are stable 39K (93.3%) or 41K
control rods to drop into the reactor and stop the chain (6.7%) nuclei, about 0.0117% are radioactive 40K nuclei, which have
reaction. a half-life of 1.26 billion years. What fraction of all the potassium
nuclei in your body will undergo radioactive decay in the next 1.00
33. Suppose that oxygen nuclei frequently absorbed neutrons. How year period?
would that behavior affect water’s performance as a moderator?
P.5 6.44 10-12 percent of the nuclei will decay.
E.33 By absorbing neutrons, the water would suppress the
chain reaction. 6. If you’re worried about 40K radioactivity (Problem 5) and wanted
to wait for 99% of the 40K in the environment to decay away, how
34. Unlike heavy water, ordinary water occasionally absorbs long would you have to wait?
neutrons. Why does a thermal fission reactor using an ordinary water
moderator need more enriched uranium than a reactor using heavy P.6 8.4 billion years.
water?
E.34 Because it loses some of the fission neutrons, the
ordinary water reactor needs to use the remaining
neutrons more efficiently. Boosting the fraction of 235U
in the fuel provides that enhanced efficiency.
35. While a fission bomb can be used to initiate fusion in hydrogen
as part of a hydrogen bomb, a power plant can’t use a fission reactor
to initiate fusion in hydrogen as a way to generate electricity. Why
not?
E.35 The role of the fission bomb is to provide the
astronomical temperatures needed to promote fusion. A
fission reactor does not achieve those temperatures.
36. A fission chain reaction can occur at room temperature, so why
must hydrogen be heated to astronomical temperatures to get its
nuclei to fuse?
E.36 The hydrogen nuclei are all positively charged and it
takes enormous temperatures to get close enough
together to fuse.
Problems – Chapter 16
1. Gallium 67 (67Ga) is a radioactive isotope with a half-life of 3.26
days. It’s used in nuclear medicine to locate inflammations and
tumors. Accumulations of 67Ga in a patient’s tissue can be detected
by looking for the gamma rays it emits when it decays. A radiologist
usually begins looking for the 67Ga about 48 hours after
administering it to a patient. What fraction of the original 67Ga nuclei
remain after 48 hours?
P.1 65% of the 67Ga nuclei remain.
2. Two weeks after 67Ga was administered to the patient (Problem
1), what fraction of the 67Ga nuclei remain?
P.2 5.1% of the 67Ga nuclei remain.
3. Technetium 99m (99mTc) is a radioactive nucleus with a half-life
of 6.03 hours. It’s used in nuclear medicine to trace biological
pathways. The 99mTc nucleus is actually an excited state of the 99Tc
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http://dubiousoft.com/page/2/ | 1,726,139,645,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651457.35/warc/CC-MAIN-20240912110742-20240912140742-00302.warc.gz | 11,758,122 | 24,316 | ## Points and Vectors
I’m gonna start off our discussion of Physics Engines with some posts about the Math that makes them tick. Don’t worry, I don’t have a degree in Math, so these posts will be long on rambling sentences, and short on Greek symbols. For this one I’m gonna skip the equations all together and just try to talk about the meaning of the most basic types in a Physics Engine: Points and Vectors.
There isn’t much that’s more fundamental to a Physics Engine then Points and Vectors. And for this reason, it’s important to spend some time getting them right. In my path to the Dubious Engine I have actually written a basic Vector library a handful of times. I even wrote one in Haskell to see if it did a better job of handling some of the awkward language edge cases that kept creeping up with C++. What is it about Points and Vectors that are so hard to get right? Well looked at from one angle, they’re pretty much exactly the same thing, 3 floats representing X, Y, and Z coordinates. But looked at from another angle, they have absolutely nothing in common. A Point represents a position in space, or where an object is. A Vector represents a direction and a magnitude, which isn’t really easy to summarize. Can you represent a Point with a Vector? Sure you can, like I said, they’re both 3 floats. But should you use the same class for both? I’d argue that you should not. Let’s discuss why.
• Addition and subtraction of Vectors is correct
• Addition of Points is meaningless
• Subtraction of Points results in a Vector
• Addition of a Point and a Vector results in a Point
This probably isn’t the result you were expecting when you first started thinking of Points and Vectors. I can tell you from experience that it took me many years to start thinking along these lines. Can you write a Physics Engine that ignores all of this and just uses a Vector for everything? Of course you can, my first few attempts did exactly this. However I found I was always creating bugs where I’d write an interface that expected a position, but I would eventually send it a Vector, and the whole thing would fail and I’d spend a lot of time trying to re-learn the algorithm to understand why. With this new representation I’m much clearer on the roles of Points and Vectors and it helps me keep things straight.
# The Code
So with that in mind, how do we write the code? You could be tempted to just make them exactly the same thing, and just use a typdef to give them different names. After all, the actual mechanics of addition, subtraction, and equality are the same. But functions like `get_length()` make no sense with a Point, but are fundamental to a Vector, so it’s best to avoid this approach. You could be tempted to use inheritance, and move the common functionality to the base class Point and put the Vector specific functionality in the Vector. This is a little cleaner in that you can reuse common code, but the polymorphism is awkward because inheritance represents an “is-a” relationship, and as discussed above, a Vector is not a Point. Another trick you could try is composition, where a Vector just contains a Point, but this leads to circular dependencies in your code. In order to implement Point subtraction a Point must depend on a Vector, but in order to implement a Vector it must depend on a Point. This will really piss off your compiler, which is generally not a good thing to do.
So this leads us to the solution I ended up with, a third class called a Triple, which is nothing more then an abstract thing that contains 3 floats. This Triple isn’t even a class, it’s just a lowly struct, and it implements the absolute basics, addition, subtraction, and equality. Both a Point and a Vector contain a Triple, and a lot of their basic functions are implemented as a simple pass through. They both build on the Triple to implement their type specific functionality.
Now that you understand my reasoning behind Points and Vectors, it should be pretty easy to understand the code. Here are links to my Triple, Point, and Vector implementations. For now, ignore the templates, we will discuss their usage later.
• Triple – Simple struct of 3 floats. Handles basic math and equality
• Point – Represents a 3D point in space.
• Vector – Represents a direction and magnitude.
<< See Also Contents Vector Types >>
I think it’s typical to put the Bibliography at the end. However for this series of posts, I’m putting it right at the start. The reason for this is because the Dubious Engine is in no way an original piece of work. I did not invent any of the Physics that powers it. I’d like to think that I’ve organized it in a way that makes it easy to understand and write about, but I certainly didn’t come up with any new algorithms. Everything here has been described somewhere else. Unfortunately I started this so long ago that I’ve lost track of all the things I’ve read, but here are the ones that I can remember:
• allenchou.net – This is the big one, I must have read and re-read every post here dozens of times. If he had source code to post I probably wouldn’t have bothered writing anything.
• 3D Game Engine Programming – The site that finally got me to understand Quaternions. That alone makes it priceless, all the other great articles are just a bonus.
• box2d.org – The simplest code to understand, primarily because it’s in 2D. It’s almost a requirement that you get it running and familiarize yourself with it while creating your own Physics Engine. While it is not 3D, the concepts are familiar enough. Also contains some papers that you have to be much smarter then I am to understand.
• bulletphysics.org – An actual, working Physics Engine, complete with source code. It does everything. Unfortunately it’s very difficult to read through and the documentation is sparse. Still, if you have the time to really truly go digging, the answer is in there.
• Molly Rocket – In 52 minutes you can understand what the GJK algorithm is. Other articles can help you refine it, but really there is no quicker path to understanding.
• Code Cave – Second only to Allen Chou for understanding GJK and EPA. Bonus points for representing 3D concepts using sticks and putty. I may very well rip that off.
• dyn4j.org – More great information about GJK and EPA
• Matrix and Quaternion FAQ – One of those ancient pages out of Internet history (hello purple.com). I can’t find the original version I used to use, but this seems like it’s pretty much the same thing. This is where I was introduced to Quaternions, and my first implementations came from this page. It was so freaking painful.
## Building a Physics Engine
“You’re Building a What Now?” Before thinking about building a Physics Engine, you have to stop and ask yourself a very important question, “why on Earth would you want to build a Physics Engine?” In many respects, a Physics Engine is a solved problem. There are really good ones available, open source, for free (thank you Bullet). Unless you’re a PhD with some fancy new algorithm to try out, you’re not gonna beat any of the existing engines. So why bother?
I didn’t start out trying to build a Physics Engine. I started out (back in 2001) trying to build a really cool space dog fighting game. I didn’t have any really original ideas, but I wanted to combine all of my favorite bits from various games of the genre, and see what I could make. I knew I would never be able to complete it, but I thought it would be interesting to try. As different aspects of the game started coming together, I would occasionally read people mentioning Physics Engines, but I always figured they just weren’t very smart. Everyone knows Physics, right? `v=d/t` like from High School. I just coded that up and got back to work. But as the game started to take shape, I noticed that ships didn’t move right. Like if you shot a missile into a wing tip, the whole ship would lurch backwards, instead of going into a spin. So I started reading a bit more about the mysterious Physics Engines. Eventually I became so engrossed I stopped working on the game at all, and just learned Physics.
So this brings us to today. I have found that the world of Physics Engines falls into two categories:
1. Well written articles, but not enough detail and no code.
2. Code with absolutely no explanation.
So here’s the pitch. I have written a basic but functional Physics Engine. Now I’d like to write about it. The goal is to produce blog posts, that together with the source, will create the world’s first understandable explanation of a Physics Engine. This should be the guide I always wished was available to me.
Let me throw out some buzzwords to explain what it is I have to write about. I’ve written a Rigid Body, Discrete Engine. I use GJK and EPA to do my collision detection. And I wrote a Sequential Constraint Solver to do collision resolution. If you don’t know what any of those terms mean, that’s fine, we’ll go over them all. If you’ve come here looking for information on any of those, then you’re in luck, ’cause that’s what I got.
I call it the Dubious Engine. The full source is available online at: https://github.com/SaintDubious/DubiousEngine. I will try to refer to the actual source with each blog post, so you can see working code, and follow along with the reasoning behind each piece. I hope you find this information useful as you work to create your own Physics Engine. As for my initial question, “Why on Earth would you want to build a Physics Engine?” Simple answer really, it’s a lot of fun.
Welcome to the Dubious Engine. This series of blog posts is my attempt to fully explain how to write a basic Physics Engine. The full source is available at: https://github.com/SaintDubious/DubiousEngine
That’s all for now. Writing these takes time, so it might be a while, but here are some other ideas:
1. Collision Detection – EPA
2. Collision Detection – Contact Manifolds
3. Collision Response – Constraint Solver
4. Collision Response – Friction
5. Broad Phase Collision Detection
6. Cheats – Baumgarte and Coefficient of Restitution
7. Cheats – Warm Starting
8. Optimization – OpenCL
## Switch on Type Construction
Many years ago I had a non-technical manager who was amazing. If you’ve been in the field long enough you’ll know that this is a rare and wonderful thing. One of the things that made this guy special was how he handled the yearly review process. Maybe I should tell you how every other company’s review process works so you can compare and contrast. Here are the steps for every other company:
1. Fill out a humiliating form boasting about everything you did during the year because it’s apparent that your manager barely remembers who you are. Make sure you add a section about your failures so everyone will know you’re humble.
2. Meet with your manager so he can tell you how amazing you are for a while.
3. Then learn that despite how amazing you are and how awesome the company is, learn that there’s no money for raises because of convoluted financial magic.
4. Receive disappointing raise with vague talk about how it might possibly be larger next year.
So compare that with my amazing manager who absolutely knew who you were and had clearly done a lot of work to create a review process that he thought would be useful. No forms to fill out. He’d ask you to come up with a handful of technical goals you wanted to achieve, and a handful of non-technical goals. That’s right, not only did he want to help you become a better programmer, he’d also play coach for anything else you wanted to do. Some people came to him with lists of things like “learn to cook” or “get motorcycle license” and he’d absolutely help you set milestones to accomplish those goals too.
So one year when it was my turn for the yearly review, I told him I wanted to learn how to use Design Patterns. Like most every programmer, I’d read the Design Patterns book, but I never could figure out why it was useful. I basically memorize them all when it’s time to interview for a new job, and promptly forget about them. So my manager gave me the task of learning a new design pattern that wasn’t in the book. This sounded interesting enough and after a little research, I came across what I call the Switch on Type Construction pattern, and this (finally) brings me to the point of this post.
Warning: I was recently horrified to learn that this solution does not always work, depending on compilers, optimizers, environments, etc. But it’s still awesome so I’m keeping the post. But be careful with it.
Often in C++ you end up in a situation where you want your code to do something different based on the type of the class. This is called “Switch on Type” and it is a “bad thing.” This phenomenon is so well known it’s a classic case of why we have classes in the first place. Here’s some code to demonstrate, let’s say you have something like this to handle images:
```void render( Image* obj )
{
if (get_extension(obj->file_name()) == "jpg") {
obj->render_jpg();
}
else if (get_extension(obj->file_name()) == "png") {
obj->render_png();
}
}```
This is bad because as the list of formats grows, you’re constantly back in here cutting and pasting to the list, and probably introducing bugs, etc. This situation was solved with virtual functions: a base image type, and then a subclass of Jpg and Png image types. So the previous code just turns into something obvious like:
``` obj->render_image();
```
where the `render_image` function is virtual in the base type, and then overridden to do the correct things in the subclass.
This is all well and good, but what do we do with object creation? You can’t use virtual functions in a class that hasn’t been constructed, and there’s no such thing as a virtual constructor. Our task for this post will be to create an appropriate image object depending on the file type (and we’ll assume that the type is correct and ignore errors).
So let’s take our starting point as this:
```class Image { /* blah blah constructors, functions etc */ };
class Png_image : public Image { /* blah blah */ };
class Jpg_image : public Image { /* blah */ };
std::shared_ptr<Image> create_image( const std::string& file_name )
{
if (get_extension(file_name) == "jpg") {
return std::shared_ptr<Image>( new Jpg_image( file_name ) );
}
if (get_extension(file_name ) == "png") {
return std::shared_ptr<Image>( new Png_image( file_name ) );
}
throw std::runtime_error( "Unknown image type" );
}
```
Does that seem like some code you may have written in the past? I know it’s definitely something I have written. Any why not? It’s short, concise, solves the problem, etc etc. Who cares that every time we need to add a new image type we have to update this function? Well what if I told you there’s a pattern that lets you add new image types without recompiling existing code? “Impossible” you’d say, and you’d be wrong.
The first thing we’ll need to create for our awesome solution is an object creation map. This is a map that uses our image type (either “png” or “jpg”) as a key and a creation function as a value. With this we can simply use the image type to find the correct creation function. Of course this map will need to be a singleton as we only want one of them in the whole program. Let’s expand our Image class to include:
• Typedefs for complex types
• The creation map
• Accessors to this map
```class Image {
public:
static void register( const std::string& ext, Creation_func& func)
{
function_map()[ext] = the_function;
}
static std::shared_ptr<Image> create( const std::string& file_name )
{
return function_map()[get_extension(file_name)]( file_name );
}
private:
typedef std::function<std::shared_ptr<Image>(const std::string&)>
Creation_func;
typedef std::map<std::string, Creation_func> Function_map;
static Function_map& function_map()
{
static Function_map singleton_map;
return singleton_map;
}
};
```
First consider the Creation_func typedef. It defines a function signature for a function that creates shared_ptrs of Image. Such a function might look like `std::shared_ptr<Image> create( const std::string& file_name )`. This Creation_func is combined with a file extension (ie “png” or “jpg”) to make a Function_map. Then there’s a function_map() accessor, and a function to add entries to the map, and a function to create an actual image. Let’s see how our code could now use this to get rid of the “switch on type” issue:
```// functions defined elsewhere
std::shared_ptr<Image> create_jpg( const std::string& file_name );
std::shared_ptr<Image> create_png( const std::string& file_mame );
// register them in the map
Image::register( "jpg", create_jpg );
Image::register( "png", create_png );
// and use them in your code
std::shared_ptr<Image> jpg_image = Image::create( "file_name.jpg" );
```
So that’s a pretty cool start. We no longer have the `if (extension_check) { return Image; }` mess that we had before. But it’s still not terribly easy to use. Now if we wanted to add support for a third image type we’d still have to edit our code and add a call to `register`. Still, we could stop right here and have a much better solution… or we could add templates to make the whole thing better.
# Enter Templates
What we want to do to make this solution better is to create a system wherein new image types can be added to the code without recompiling anything. That’s sort of the holy grail of system extension. That means no testing of any existing code because the existing code won’t change. Heck you could take your existing object files, compile in the new image classes, link the whole thing together and magically have support for new types. Sounds like a trick to strive for. So to help us out we want to add a template class that acts sort of like a factory. Here’s how it looks:
```template <class T>
struct Registration_object {
Registration_object( const std::string& ext )
{
Image::Creation_func f = Registration_object<T>::create;
Image::register( ext, f );
}
static std::shared_ptr<Image> create( const std::string& image_name )
{
return std::shared_ptr<Image>( new T( image_name ) );
}
};
```
So what’s going on with this one? Well it has a static `create` function that simply creates a new Image of the templated type T. It also has a constructor that will add a pointer to that `create` function into the singleton function map.
So how do we use this new registration object? All we have to do is create exactly one of these for each image type, and it will automatically register that new type in the map. So let’s change our usage code to use these instead:
```Registration_object<Jpg_image> jpg_registration( "jpg" );
Registration_object<Png_image> png_registration( "png" );
std::shared_ptr<Image> jpg_image = Image::create( "file_name.jpg" );
```
That’s starting to look pretty awesome. Now we just have to create one global variable that we never directly use, and our image type is magically added to the map. So how can we use this trick to add an entirely new image type without recompiling existing code? Hold on to your hats, cause this is the amazing bit. I’m not going to touch the code above (ie I will not recompile it) but I can add Gif support by simply adding the following in a separate .cpp file:
```class Gif_image : public Image {
public:
Gif_image( const std::string& file_name ) { /* Gif image stuff*/ }
};
Registration_object<Gif_image> gif_registration( "gif" );
```
Did you see it? Now when this new .cpp file is compiled and linked with the existing object files, your program will suddenly be able to deal with gif images. The reason this works is during static object creation time (before main starts) an instance of this new Registration_object<Gif_image> will be constructed, and during that construction it will register its create function with the static function map. If that doesn’t impress you then stop reading this blog and fuck you.
Oh, and what about that manager I had? He helped me reach a number of goals, both technical and non. Eventually the company merged with another one and he was replaced with the HR lady from the other company. The first thing she did was bring in those damned yearly evaluation forms like every other company has. I was more then happy to quit that job.
## Static Creation Functions
When I wrote about Class Construction, I noted that of all the incorrect arguments about constructors not being sufficient, there were a handful of cases that are actually correct. While not an exhaustive list, here are some cases that spring to mind:
1. If you are creating a class hierarchy wherein a base class defines a number of virtual functions that subclasses are meant to implement (like `on_create`). It’s not unreasonable to try calling a virtual function in the base class constructor and expect a subclass function to run. Unfortunately this won’t work. The v-table won’t be set up yet, so attempting to call a virtual function will in turn call the base class’ implementation.
2. If you want to enforce HOW your objects are constructed. While it’s not possible to force all your classes to only be constructed on the stack, it is possible to force clients to only create objects as smart pointers (or raw pointers if you’re a big fan of resource leaks).
In both of these cases it won’t work to simply expect your clients to use the regular constructor. For case 2 you could solve this by adding documentation that says “please don’t create these on the stack” but if your code depends on developers actually reading the comments then it’s already doomed. You could solve case 1 by using an initialization function, but then you’d be totally ignoring all the great advice in the post that told you why that’s a horrible idea. So what’s left? This is a perfect situation for a static creation function, which looks like this:
```class Some_class {
public:
static std::shared_ptr<Some_class> create();
};
```
So what’s going on here? Well, the first thing to notice is that it’s a `static` function, which is a pretty important aspect to this technique as it means you can call the function without having created an instance of the object (because if you need the object to exist before you can create it, it’s probably not gonna work). The next thing to notice is that I have control over the return type, which in this case is a shared_ptr. You can imagine other return types (or even void if you need to enforce that the object must be created into some global store that you access in some other way).
At this point you may be pointing out that while this is all fine and dandy, there’s still nothing to stop anyone from just creating an instance of Some_class directly and avoiding the create function. You’d be right. The second half to this trick is that you have to make your constructor private. It will still be accessible inside the create function, but it won’t allow users to create your object any other way. So here’s a more complete example:
```class Some_class {
public:
~Some_class();
static std::shared_ptr<Some_class> create()
{
std::shared_ptr<Some_class> ptr( new Some_class);
ptr->some_kind_of_initialization();
ptr->on_create();
return ptr;
}
private:
Some_class();
Some_class( const Some_class& );
Some_class& operator=( const Some_class& );
};
```
I don’t generally put the code in the actual header, but it makes the example easier to read. Notice that along with the constructor I also made the copy constructor and operator= all private. There is now no way to create one of these things without using the static create function. Here’s how you’d have to make one:
``` std::shared_ptr<Some_class> ptr = Some_class::create(); // OK
Some_class stack_instance; // won't compile
Some_class copy_construct( *ptr ); // won't compile
```
Pretty sporty, huh? Take some time to soak this in, controlling an object’s construction is awesome and definitely needs to be a part of your toolkit.
Now you may also have heard about factories. A factory is a lot like a static creation function, but it exists as a separate class, adding a level of abstraction. I even had a co-worker who used to go on about how factories should actually be interfaces, so you could have entire hierarchies of factories that you could swap out to get different kind of creation functions. This all sounds well and good and impressive and fancy, but in practice I’ve never been in a situation where a static create function wasn’t good enough. Class factories always seem like going too complex for me. But as I’ve stated before, I’m not terribly clever, so if you really want to create a factory (or an entire inheritance tree of factories) then go right ahead. Just make sure you make your object’s constructors private to ensure people don’t just bypass the whole thing.
Some bullet points:
• Occasionally (rarely) a constructor is not sufficient
• Create a static function called “create” that does exactly that
• Make your object’s other construction functions private
• You can use class factories if you want to be a smarty pants
## Exceptions are Awesome
Welcome to the year 2015. Obamacare is the law of the land, gay folks can get married, C++ has lambdas, and (this will surprise some folks) exceptions are the correct way to report errors. The fact that I still have to argue this point is, frankly, shocking. I don’t care that the Google style guide doesn’t allow them, and for the love of God don’t tell me they’re inefficient. Simply put, they are the way to report errors, you must know how to use them.
Now to some extent I actually sort of sympathize on this one. When I learned C++ (back in 1995 or so) exceptions weren’t widely used. And even up until around 2005 I was pretty convinced that they were a bad idea. But I’m old and that was 10 years ago… what’s your excuse? Some programmers today seem to have the same misconception I had in 2005, so let’s walk through the faulty logic. They argue that using exceptions makes code bloat substantially. They start with something like this:
```return_code = blah();
if (return_code != SUCCESS) {
return return_code;
}
return_code = something_else();
if (return_code != SUCCESS) {
return return_code;
}
```
And apply exceptions by turning it into this:
```try {
blah();
}
catch (const std::exception& e) {
throw runtime_error( "calling blah failed" );
}
try {
something_else();
}
catch (const std::exception& e) {
throw runtime_error( "calling something_else failed" );
}
```
And then argue that using exceptions has turned an 8 line function into a 12 line one. But they’re missing the point. So I’ll write down the point in bold letters: throw an exception when an error occurs, and catch an exception when you can do something about it. Armed with this knowledge, most junior programmers will head back to their keyboards and come back to me with something like this:
```try {
blah();
something_else();
}
catch (const std::exception& e) {
throw runtime_error( "calling blah or something_else failed" );
}
```
Well now we’re down to just 7 lines, so this is a step in the right direction, but it’s still wrong. Let’s take a look at the correct answer and then discuss why it’s correct
``` blah();
something_else();
```
Ah, much better. Now we’re just down to 2 lines, and better yet, we don’t have to worry about errors at all. We have two short, succinct lines that just assume the best case scenario and ignore errors completely, what could be better? “But wait, this can’t be correct, there’s no error handling at all!!!” And that’s the point. Take a look at the second part of the bold statement above, catch an exception when you can do something about it. In all of these examples there was nothing sensible we could do in the case of an error but just return an error up to the caller. Since an exception just naturally bubbles up the call stack, and since there’s nothing I can do about it here, just let it bubble up the call stack.
Do you see the beauty in that? Are you reading this and having a warm and tingly sensation rubbing your chest? If not, think about it some more. You are now free to write code where you don’t have to nit pick every single error case. By way of example, let’s say you’re writing some kind of web server. A request comes in for a web page and in order to respond you’ve got to hit a database for some content, the disk for some assets, and then maybe execute some business logic to tie it all together (God help you if it’s Ruby on Rails). Somewhere deep down in the bowels of your db connection something might go wrong. Do you really want to have to watch error codes at every function call and have to keep translating them as they work their way up the stack through multiple layers of return codes? Of course not, none of the intermediate layers can do anything anyway, the db is dead, they can’t fix that. Imagine a world where the db fails and the highest layer is simply notified so it can return a bland 500 error to the client. That’s the joy of exceptions. The db layer throws an exception, every layer in between ignores it, and the top layer catches it and returns. But wait, it gets even better. The system unwinds the stack for you, correctly destroying all the objects you created along the way. Sound awesome? That’s because it is awesome.
Oh but wait, isn’t it expensive? Well yes, it is, but that doesn’t matter. Why doesn’t it matter? Because exceptions are for exceptional cases. Take the case of our http server above. How often do you think a database blows up? Okay okay, I get the joke, almost constantly, but seriously, in the normal day of operations, how often? It’s an unusual thing, so who cares if it runs just a tiny bit slower? Users aren’t gonna be sitting on their web browser complaining that their 500 error page took an extra picosecond to return. In the normal run of things, exceptions don’t happen, so you don’t pay a performance penalty.
Of course this is only true if you don’t abuse exceptions. You need to make sure that you’re not using exceptions to report back something that is expected behavior. Let’s say you’re writing some function `bool lookup_something_in_db( int id, Thing& thing_copy )`. This function looks up a Thing in the db with the given id. It seems to me that the thing in the db might not exist. Or to say it in a more obvious way, it’s not an exceptional case that the id doesn’t find something in the db. So using an exception to return information that the thing doesn’t exist would be a really bad idea, you’d be throwing exceptions for cases that are expected. In this case returning a bool to notify the caller that thing they were looking for doesn’t exist is a good idea.
So here’s the bullet points:
• Exceptions are awesome, use them
• Throw when an error happens, catch when you can do something about it
• Don’t use exceptions for expected, normal processing
## Class Construction
I’ll start with a story. Years ago I was starting out at a little software shop, trying to learn my way around the existing code base. As is usual in these cases, everything I encountered looked bad to me. As I’ve grown a bit as a programmer I’ve learned to realize that some of this feeling comes from simply not understanding the constraints on a given system, and some of this comes from actually encountering bad software. So of course many spirited discussions followed wherein the head of software would defend the code and I would try to tear it down. I think one of the primary goals of a library is that it should be “easy to use correctly and hard to use incorrectly” – Scott Meyers (see below). While I admit that this is not always achievable, it should at least be a goal. He stated emphatically that his library had done a good job of this. So imagine my revulsion when I stumbled upon something like this:
```class Some_class {
public:
Some_class();
bool initialize();
bool startup();
void set_some_member( const std::vector<int>& member );
};
```
Have a longer look at this one and ask yourself if this is easy to use correctly. If you think it is then look again and ask yourself how to create an instance of `Some_class`. I asked this exact question and he told me it was obvious and I should look for some examples. So I did. I learned that the correct way to create an instance of `Some_class` is to first call the constructor (obviously), then call `set_some_member` (huh?), and then call `startup`. Also, for the love of God, whatever you do, never call `initialize` because that leads to undefined behavior.
So this leads me to the point of this post, which is that of all the things your object needs to do, construction is really fundamental. As a corollary to this, initialization functions are awful. Now I can already hear the more senior types pointing out that there are cases where a constructor alone simply cannot do the job. You’re correct, that is true, and in time we’ll be looking at some of those. However there are a whole other set of incorrect cases that junior developers mistakenly think justify initialization functions (and other horrors). We need to end this misconception.
The most pervasive incorrect case is when someone thinks that because there’s no way to return a status code from a constructor, you need an initializer function to truly bring your object to life. In this anti-pattern the constructor does the basic member initialization, which can not fail, and then the initializer does the dangerous work, returning a success code if everything went okay. This leads to two major problems. The first is for the users of this class. They will skim your documentation just enough to find some function they like, and the constructor, and immediately set about creating it and using it. Maybe this will kind of work without calling the initalizer, or maybe it won’t. When something eventually fails, they’ll be annoyed. But more importantly, this object will be hard for you to write because you’ll invariably have to hold some internal flag to check whether or not initialize has been called. Then you’ll need to worry about what happens if someone calls initialize twice, and blah blah blah it’s just not worth it.
In order to stay out of this quagmire you need to know one simple thing: it is correct to throw exceptions from constructors. Many of the junior programmers I speak with don’t know this simple fact. Often they’ll hedge their bets by telling me the code will compile, but it’s not the correct thing to do. They’re wrong. Throwing exceptions from a failed constructor is exactly the correct thing to do. In fact it’s the only way to safely let the calling function know about failed construction. But then what’s the state of an object that throws an exception halfway through construction? Simple, it doesn’t exist, it never did. Objects don’t exist until construction is finished, so if construction doesn’t finish, the object doesn’t exist. Will the destructor be called? No, it most certainly will not. How can you destruct an object that was never constructed? Another question I often get is, doesn’t this lead to memory leaks? The answer is: only if you’ve written a constructor that isn’t exception safe. But this has nothing to do with constructors and is simply a fact of writing any function that isn’t exception safe. You should never have a function that allocates anything and just assumes that no exception will ever be called.
C++11 Note: In C++11 we now have delegating constructors. This means a constructor can call another constructor. This makes it a little less clear about when an object is constructed. The rule is that when the first constructor is finished, the object has been created. So now if an exception is thrown from the calling constructor (after the called constructor is finished) the destructor will be called.
So there you have it, end of lesson. Here’s the summary in simple bullet points:
• Never ever ever ever write an initialization function
• If something goes wrong during construction, throw an exception
• If an exception is thrown during construction, the object never existed (the destructor won’t be called)
• Don’t write functions that aren’t exception safe (constructors or otherwise) | 8,009 | 36,288 | {"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} | 3.28125 | 3 | CC-MAIN-2024-38 | latest | en | 0.954255 |
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## Calculating Broken Trendline Price Projections
Posted on June 7, 2012 by MarketTechLab
One of the benefits of technical analysis is the ability for technicians to project price. This enables us to know the likely extent of a
move once a pattern has completed, or in this case, once a trendline has been broken. Today were going to cover the method for
calculating a price projection from a broken trendline.
The calculation is rather simple. You must start with the required characteristics of trend to begin. You must be analyzing a stock
that was in a confirmed uptrend (at least two touches, but three or more is best) that is broad enough in its range to analyze. If
the price bounces along the line without rallying a significant distance off of the line, this method will not work. It is best to see a
trend channel with a wide range of that channel, as narrow ranges will not work. Again this is subjective, so use your best
judgment, and you will see why this is important. And of course, the trend must actually have been broken to realize a projection.
Below is a recent daily chart of BAC with a broken trendline. You can see it has a range of a couple dollars, which is more than
20% on a relative basis, and is sufficient for analysis.
## Below are the steps required to calculate:
1.) Determine the high of the move. In this case it is \$10.
2.) Next, draw or eyeball a straight line immediately down from the high through the air pocket to the bottom trendline.
3.) From this intersection look right to determine the price level along the Y-axis, then subtract the distance. In this case, \$10
\$8.50 gives \$1.50
4.) Subjtract the projection from where the breakout occurs. Note it is where the breakout actually occurs (\$9.25), not from the
intersection (\$8.50)
5.) This gives a minimum projection of \$7.75. While there is no maximum projection, a good rule of thumb is to use 1-2 times
the price calculation for your projection. In this case \$7.75 to \$6.25 (2 times \$1.50 from \$9.25).
This can be done for any length of trend, but will be more effective for longer-term trends. This can be a useful technique to
determine where to exit positions in the direction of the broken trend, or when to become alerted to re-entry below/above the
price target.
Note: This information is covered in Martin Prings book, Technical Analysis Explained.
This entry was posted in Uncategorized by MarketTechLab. Bookmark the permalink [http://markettechlab.com
/blog/calculating-broken-trendline-price-projections/] .
12/4/2016 7:49 PM
## Calculating Broken Trendline Price Projections | Market Tech Lab
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Appropriate Pedagogy: 0 | 387 | 1,268 | {"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} | 2.5625 | 3 | CC-MAIN-2019-39 | latest | en | 0.319183 |
https://coopmadretierra.org/2024/04/what-is-a-domino-effect/ | 1,726,166,367,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651491.39/warc/CC-MAIN-20240912174615-20240912204615-00821.warc.gz | 163,703,014 | 14,930 | # What is a Domino Effect?
Dominoes are rectangular tiles with a line down the middle to divide them visually into two squares. Each end is either blank or has a number of spots—also known as pips—on it. The values of the pips on both sides determine how many dominoes can be stacked end to end, and in which direction they can be placed. There are various types of games that can be played with these rules, which allow for a great deal of flexibility and complexity. Dominoes are often used to demonstrate mathematical principles such as addition, subtraction, multiplication, division, and order of operations. They are also popular as a learning tool for children to develop basic counting skills and motor skills, and can be found in classrooms across the country.
In a broader sense, the term “domino” can be used to describe any sequence that has a similar effect to a domino. A domino effect can be created by a single event that then causes a chain reaction with more significant consequences. This is the idea behind the popular game of dominoes, where players set up a long line of dominoes on end and then tip each one to trigger its successors. This can lead to complex patterns that can be set up and manipulated with the help of a skilled player.
When it comes to business, the concept of a domino effect can be applied to a variety of aspects. For example, an employee’s lack of communication can create a negative effect on other employees and ultimately the entire company. Or, a simple act such as making your bed can have an impact on other areas of your life. For example, Admiral William McRaven once told graduates of the University of Texas at Austin that simply making their bed every morning could have a domino effect on the rest of their lives.
The first Domino’s Pizza opened in 1967, with the company’s headquarters located in Ypsilanti, Michigan. The company grew quickly, largely due to its founder’s strategy of locating locations near colleges. Domino’s also focused on speedy delivery, which was a key component to their success. Over time, however, Domino’s realized that they needed to change their strategic focus to ensure they remained competitive and relevant.
By focusing on their core values and listening to their customers, Domino’s was able to adjust their strategic direction and turn around the company in a short amount of time. As a result, they are now a dominant force in the pizza industry with over 25,000 global locations and record-breaking sales. By addressing their customer’s concerns, they were able to re-establish trust and loyalty with their consumers. This is a prime example of how a company can leverage the domino effect to drive growth and succeed in today’s economy. | 558 | 2,740 | {"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-2024-38 | latest | en | 0.971753 |
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### Fundamentals Of Differential Equations And Boundary Value Problems
Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069
# Reduction of Order. If a nontrivial solution f(x) is known for the homogeneous equation,${{\mathbf{y}}}^{\left(\mathbf{n}\right)}{\mathbf{+}}{{\mathbf{p}}}_{1}\left(\mathbf{x}\right){{\mathbf{y}}}^{\left(\mathbf{n}\mathbf{-}\mathbf{1}\right)}{\mathbf{+}}{.}{..}{\mathbf{+}}{{\mathbf{p}}}_{n}\left(\mathbf{x}\right){\mathbf{y}}{\mathbf{=}}{\mathbf{0}}$the substitution ${\mathbf{y}}\left(\mathbf{x}\right){\mathbf{=}}{\mathbf{v}}\left(\mathbf{x}\right){\mathbf{f}}\left(\mathbf{x}\right)$ can be used to reduce the order of the equation for second-order equations. By completing the following steps, demonstrate the method for the third-order equation(35) ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{2}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{5}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{6}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}$given that ${\mathbf{f}}\left(\mathbf{x}\right){\mathbf{=}}{{\mathbf{e}}}^{x}$ is a solution.(a) Set ${\mathbf{y}}\left(\mathbf{x}\right){\mathbf{=}}{\mathbf{v}}\left(\mathbf{x}\right){{\mathbf{e}}}^{x}$ and compute y′, y″, and y‴. (b) Substitute your expressions from (a) into (35) to obtain a second-order equation in. ${\mathbf{w}}{\mathbf{=}}{\mathbf{v}}{\mathbf{\text{'}}}$(c) Solve the second-order equation in part (b) for w and integrate to find v. Determine two linearly independent choices for v, say, ${\mathbf{v}}_{1}$and ${{\mathbf{v}}}_{2}$. (d) By part (c), the functions ${{\mathbf{y}}}_{1}\left(\mathbf{x}\right){\mathbf{=}}{{\mathbf{v}}}_{1}\left(\mathbf{x}\right){{\mathbf{e}}}^{x}$ and ${{\mathbf{y}}}_{2}\left(\mathbf{x}\right){\mathbf{=}}{{\mathbf{v}}}_{2}\left(\mathbf{x}\right){{\mathbf{e}}}^{x}$ are two solutions to (35). Verify that the three solutions ${{\mathbf{e}}}^{x}{\mathbf{,}}{\text{\hspace{0.17em}}}{{\mathbf{y}}}_{1}\left(\mathbf{x}\right)$, and ${{\mathbf{y}}}_{2}\left(\mathbf{x}\right)$ are linearly independent on $\left(\mathbf{-}\infty \mathbf{,}\text{\hspace{0.17em}}\infty \right)$
(a) The value of y′, y″, and y‴ is,
$\begin{array}{c}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\left(\mathbf{v}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\right){\mathbf{e}}^{x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\left(\mathbf{v}\mathbf{+}\mathbf{2}\mathbf{v}\mathbf{\text{'}}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\right){\mathbf{e}}^{x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\left(\mathbf{v}\mathbf{+}\mathbf{3}\mathbf{v}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\right){\mathbf{e}}^{x}\end{array}$
(b) A second-order equation is,
$\mathbf{w}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{w}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathbf{w}\mathbf{=}\mathbf{0}$
(c) Two linearly independent is,${\mathbf{v}}_{1}\mathbf{=}{\mathbf{e}}^{-3x},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathbf{v}}_{2}\mathbf{=}{\mathbf{e}}^{2x}$
(d) $\mathbf{w}\ne \mathbf{0}$ and now we can say that ${\mathbf{e}}^{x},\text{\hspace{0.17em}\hspace{0.17em}}{\mathbf{y}}_{1},\text{\hspace{0.17em}\hspace{0.17em}}{\mathbf{y}}_{2}$are linearly independent solutions.
See the step by step solution
## Step 1: About Reduction of Order;
Now going to take a brief detour and look at solutions to non-constant coefficient, second order differential equations of the form.
$\mathbf{p}\left(\mathbf{t}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{q}\left(\mathbf{t}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{r}\left(\mathbf{t}\right)\mathbf{y}\mathbf{=}\mathbf{0}$
In general, finding solutions to these kinds of differential equations can be much more difficult than finding solutions to constant coefficient differential equations. This method is called reduction of order.
## (a)Step 2: Firstly, using the given function f(x)=ex,
Given function,
$\mathbf{f}\left(\mathbf{x}\right)\mathbf{=}{\mathbf{e}}^{x}$ is a solution to
$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{0}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(\mathbf{1}\right)$
And
$\mathbf{y}\left(\mathbf{x}\right)\mathbf{=}\mathbf{v}\left(\mathbf{x}\right)\mathbf{f}\left(\mathbf{x}\right)\mathbf{=}\mathbf{v}\left(\mathbf{x}\right){\mathbf{e}}^{x}$
Now find the derivative of y for equation (1),
$\begin{array}{c}\mathbf{y}\mathbf{=}{\mathbf{ve}}^{x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}{\mathbf{ve}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}{\mathbf{ve}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}{\mathbf{ve}}^{x}\mathbf{+}\mathbf{2}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}{\mathbf{ve}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{2}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{2}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\left(\mathbf{v}\mathbf{+}\mathbf{3}\mathbf{v}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\right){\mathbf{e}}^{x}\end{array}$
Hence, the value of y′, y″, and y‴ is,
$\begin{array}{c}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\left(\mathbf{v}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\right){\mathbf{e}}^{x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\left(\mathbf{v}\mathbf{+}\mathbf{2}\mathbf{v}\mathbf{\text{'}}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\right){\mathbf{e}}^{x}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\left(\mathbf{v}\mathbf{+}\mathbf{3}\mathbf{v}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\right){\mathbf{e}}^{x}\end{array}$
## (b)Step 3: Obtain a second-order equation in.w=v'
Substitute all values in the equation (1),
$\begin{array}{c}y\text{'}\text{'}\text{'}-2y\text{'}\text{'}-5y\text{'}+6y=0\\ {\mathbf{ve}}^{x}\mathbf{+}\mathbf{3}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{3}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{-}\mathbf{2}\left({\mathbf{ve}}^{x}\mathbf{+}\mathbf{2}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\right)\mathbf{-}\mathbf{5}\left({\mathbf{ve}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\right)\mathbf{+}\mathbf{6}{\mathbf{ve}}^{x}\mathbf{=}\mathbf{0}\\ \mathbf{-}\mathbf{6}\mathbf{v}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{+}\mathbf{v}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^{x}\mathbf{=}\mathbf{0}\\ -6v\text{'}+v\text{'}\text{'}+v\text{'}\text{'}\text{'}=0\end{array}$
Use the value $\mathbf{w}\mathbf{=}\mathbf{v}\mathbf{\text{'}}$in the above expression,
Hence, A second-order equation is,
${\mathbf{w}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{w}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{6}}{\mathbf{w}}{\mathbf{=}}{\mathbf{0}}$
## (c)Step 4: Solve the second-order equation in part (b) for w,
Solve the above equation for w,
$\begin{array}{c}\left({\mathbf{D}}^{2}\mathbf{+}\mathbf{D}\mathbf{-}\mathbf{6}\right)\mathbf{w}\mathbf{=}\mathbf{0}\\ \mathbf{D}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{,}\text{\hspace{0.17em}}\mathbf{2}\end{array}$
The solution of w is,
$\begin{array}{c}\mathbf{w}\left(\mathbf{x}\right)\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}3\mathbf{x}}\mathbf{+}{\mathbf{Be}}^{2x}\\ \mathbf{v}\mathbf{\text{'}}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}3\mathbf{x}}\mathbf{+}{\mathbf{Be}}^{2x}\end{array}$
Integrating both sides with respect to x,
$\begin{array}{c}\int \mathbf{v}\mathbf{\text{'}}\mathbf{=}\int \left({\mathbf{Ae}}^{-3x}\mathbf{+}{\mathbf{Be}}^{2x}\right)\mathbf{dx}\\ \mathbf{v}\mathbf{=}\frac{{\mathbf{Ae}}^{-3x}}{-3}\mathbf{+}\frac{{\mathbf{Be}}^{2x}}{2}\mathbf{+}\mathbf{C}\\ \mathbf{v}\mathbf{=}{\mathbf{v}}_{1}\mathbf{+}{\mathbf{v}}_{2}\\ {\mathbf{v}}_{1}\mathbf{=}{\mathbf{e}}^{-3x}\\ {\mathbf{v}}_{2}\mathbf{=}{\mathbf{e}}^{2x}\end{array}$
Hence, two linearly independent is, .${{\mathbf{v}}}_{1}{\mathbf{=}}{{\mathbf{e}}}^{-3x}{,}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}{{\mathbf{v}}}_{2}{\mathbf{=}}{{\mathbf{e}}}^{2x}$
## (d)Step 5: Verify that the three solutions, ex, y1(x) and y2(x) are linearly independent on.(-∞, ∞)
We have,
$\begin{array}{l}{\mathbf{y}}_{i}\mathbf{=}{\mathbf{v}}_{i}\mathbf{f}\\ {\mathbf{y}}_{1}\mathbf{=}{\mathbf{v}}_{1}{\mathbf{e}}^{x}\mathbf{=}{\mathbf{e}}^{-2x}\\ {\mathbf{y}}_{2}\mathbf{=}{\mathbf{v}}_{2}{\mathbf{e}}^{x}\mathbf{=}{\mathbf{e}}^{3x}\end{array}$
To Verify, find the derivative of ${\mathbf{y}}_{1}$ and ${\mathbf{y}}_{2}$
$\begin{array}{l}{\mathbf{y}}_{1}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{e}}^{-2x}\mathbf{,}\text{\hspace{0.17em}}{\mathbf{y}}_{1}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{4}{\mathbf{e}}^{-2x}\mathbf{,}\text{\hspace{0.17em}}{\mathbf{y}}_{1}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{8}{\mathbf{e}}^{-2x}\\ {\mathbf{y}}_{2}\mathbf{\text{'}}\mathbf{=}\mathbf{3}{\mathbf{e}}^{3x}\mathbf{,}\text{\hspace{0.17em}}{\mathbf{y}}_{2}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{9}{\mathbf{e}}^{3x}\mathbf{,}\text{\hspace{0.17em}}{\mathbf{y}}_{2}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{27}{\mathbf{e}}^{3x}\end{array}$
Now,
$\begin{array}{c}{\mathbf{y}}_{1}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}{\mathbf{y}}_{1}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}{\mathbf{y}}_{1}\mathbf{\text{'}}\mathbf{+}\mathbf{6}{\mathbf{y}}_{1}\mathbf{=}{\mathbf{y}}_{2}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}{\mathbf{y}}_{2}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{5}{\mathbf{y}}_{2}\mathbf{\text{'}}\mathbf{+}\mathbf{6}{\mathbf{y}}_{2}\\ \mathbf{-}\mathbf{8}{\mathbf{e}}^{-2x}\mathbf{-}\mathbf{2}\left(\mathbf{4}{\mathbf{e}}^{-2x}\right)\mathbf{-}\mathbf{5}\left(\mathbf{-}\mathbf{2}{\mathbf{e}}^{-2x}\right)\mathbf{+}\mathbf{6}\left({\mathbf{e}}^{-2x}\right)\mathbf{=}\mathbf{27}{\mathbf{e}}^{3x}\mathbf{-}\mathbf{2}\left(\mathbf{9}{\mathbf{e}}^{3x}\right)\mathbf{-}\mathbf{5}\left(\mathbf{3}{\mathbf{e}}^{3x}\right)\mathbf{+}\mathbf{6}\left({\mathbf{e}}^{3x}\right)\\ \mathbf{-}\mathbf{8}{\mathbf{e}}^{-2x}\mathbf{-}\mathbf{8}{\mathbf{e}}^{-2x}\mathbf{+}\mathbf{10}{\mathbf{e}}^{-2x}\mathbf{+}\mathbf{6}{\mathbf{e}}^{-2x}\mathbf{=}\mathbf{27}{\mathbf{e}}^{3x}\mathbf{-}\mathbf{18}{\mathbf{e}}^{3x}\mathbf{-}\mathbf{15}{\mathbf{e}}^{3x}\mathbf{+}\mathbf{6}{\mathbf{e}}^{3x}\\ 0=0\end{array}$
Using the Wronskian,
$\begin{array}{c}\mathbf{w}\mathbf{=}|\begin{array}{ccc}{\mathbf{e}}^{x}& {\mathbf{e}}^{-2x}& {\mathbf{e}}^{3x}\\ {\mathbf{e}}^{x}& \mathbf{-}\mathbf{2}{\mathbf{e}}^{-2x}& \mathbf{3}{\mathbf{e}}^{3x}\\ {\mathbf{e}}^{x}& \mathbf{4}{\mathbf{e}}^{-2x}& \mathbf{9}{\mathbf{e}}^{3x}\end{array}|\\ \mathbf{=}{\mathbf{e}}^{x}\left(\mathbf{-}\mathbf{18}{\mathbf{e}}^{x}\mathbf{-}\mathbf{12}{\mathbf{e}}^{x}\right)\mathbf{-}{\mathbf{e}}^{-2x}\left(\mathbf{9}{\mathbf{e}}^{4x}\mathbf{-}\mathbf{3}{\mathbf{e}}^{4x}\right)\mathbf{+}{\mathbf{e}}^{3x}\left(\mathbf{4}{\mathbf{e}}^{-x}\mathbf{+}\mathbf{2}{\mathbf{e}}^{-x}\right)\\ \mathbf{=}\mathbf{-}\mathbf{30}{\mathbf{e}}^{2x}\\ \ne \mathbf{0}\end{array}$
Hence, ${\mathbf{w}}{\ne }{\mathbf{0}}$and now we can say that ${{\mathbf{e}}}^{x}{,}{\text{\hspace{0.17em}\hspace{0.17em}}}{{\mathbf{y}}}_{1}{,}{\text{\hspace{0.17em}\hspace{0.17em}}}{{\mathbf{y}}}_{2}$ are linearly independent solutions. | 5,774 | 13,311 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 46, "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-2023-23 | latest | en | 0.45167 |
https://abcnews.go.com/US/grader-seeks-math-problem-solving-local-ohio-police/story?id=45614265 | 1,723,218,046,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640767846.53/warc/CC-MAIN-20240809142005-20240809172005-00478.warc.gz | 61,195,464 | 41,015 | # 5th-grader seeks math problem-solving help from local Ohio police department
Local police went above and beyond, though there was a slight glitch.
ByABC News
February 20, 2017, 2:57 PM
— -- Police across the country are used to solving puzzles, but one Ohio police department recently got a plea for help from a local girl seeking to solve a puzzle of a more mathematical bent.
Lena Draper, 10, decided she needed some help with her fifth-grade math homework, so she took to the Marion Police Department's Facebook page on Friday and left a message with a few problems that she felt needed answering.
The department came to her rescue, messaging the girl back after she posted the math problem (8 + 29) x 15. Someone at the department wrote back, "Do the numbers in the parenthesis first so in essence it would be 37 x 15."
Lena followed up with another problem, "(90 + 27) + (29 + 15) x 2"
To which someone at the department replied, "Take the answer from the first parenthesis plus the answer from the second parenthesis and multiply that answer by two."
Though the department went above and beyond its duties, in a math faux pas, the answer given to Lena ended up being incorrect, as pointed out by a friend of Lena's mother. (The correct answer is to add the numbers in the second parentheses and multiply only that by 2, then add it to the numbers in the first parentheses.)
Lena's mom, Molly Draper, said she was tickled that the police department tried to help her daughter with her homework. "I didn't believe her and asked for a screen shot. I thought it was pretty funny. And I love that they went ahead with it," she told ABC News.
In response to the incident, the Marion PD posted on its Facebook page that it is a full-service police department that makes every emergency a cause to be answered.
When asked if Lena's math problem ever got answered correctly, her mom said, "I hope so. But we'll see when she gets her paper back."
For those in need of math equation help, remember the acronym PEMDAS, which stands for parentheses, exponents, multiplication, division, addition and subtraction — the order in which mathematical operations should be performed in an equation. | 486 | 2,204 | {"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} | 3.203125 | 3 | CC-MAIN-2024-33 | latest | en | 0.977972 |
http://forums.devshed.com/programming/952544-using-sizeof-decide-length-array-last-post.html | 1,429,441,363,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246638820.85/warc/CC-MAIN-20150417045718-00217-ip-10-235-10-82.ec2.internal.warc.gz | 103,031,805 | 12,574 | ### Thread: Using sizeof to decide length of array
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#### Using sizeof to decide length of array
Here's the code:
Code:
```#include <stdio.h>
#define SIZE ((int) (sizeof(a) / sizeof(a[0])))
int main(void)
{
int i, a[] = {0};
printf("Enter a few numbers: ");
for (i = 0; i < SIZE; i++)
scanf("%d", &a[i]);
printf("\nIn reverse order: ");
for (i = SIZE - 1; i >= 0; i--)
printf("%d\n", a[i]);
return 0;
}```
What I am trying to do with this program, is to print out the digits entered in reverse order. I don't have a specific initial array length. It should be determined automatically with the number of digits entered as input. So I am trying to use this code for this job:
Code:
`((int) (sizeof(a) / sizeof(a[0])))`
But when I provide an input, it just prints out random garbage value. What am I doing wrong and what should be done to serve my purpose?
2. It should be determined automatically with the number of digits entered as input.
I fear you did something other instead.
"sizeof" allows one to get the size of the variable. So your expression (sizeof(a)/sizeof(a[0])) is the division of the size of existing array (in bytes) by size of its element (in bytes).
For example, if you declare it as a[5], the expression will be 20/4=5.
So you see, this only helps to find the declared size of array.
You declare array as "int a[] = {0}" - empty brackets mean that size should be determined from the number of initializers - and you have only one initializer - so the size of array is 1.
For 1 element it looks like working correctly.
If you want arrays of dynamic size you should do something like this:
- determine how many elements you will need;
- declare array as a pointer int*a;
- allocate memory for it a = (int*) malloc(n * sizeof(int));
- and don't forget remove it later with "free".
If you have no idea about how many elements would be entered, you are either to reallocate array as soon as it is needed, or use linked list.
In C++ you can do allocation with "new" and also you can use std::list for similar cases.
(however all this is bit clumsy - that is why after 10 years of programming in C I at last turned to java and scripting languages)
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Originally Posted by rodiongork
I fear you did something other instead.
"sizeof" allows one to get the size of the variable. So your expression (sizeof(a)/sizeof(a[0])) is the division of the size of existing array (in bytes) by size of its element (in bytes).
For example, if you declare it as a[5], the expression will be 20/4=5.
So you see, this only helps to find the declared size of array.
You declare array as "int a[] = {0}" - empty brackets mean that size should be determined from the number of initializers - and you have only one initializer - so the size of array is 1.
For 1 element it looks like working correctly.
If you want arrays of dynamic size you should do something like this:
- determine how many elements you will need;
- declare array as a pointer int*a;
- allocate memory for it a = (int*) malloc(n * sizeof(int));
- and don't forget remove it later with "free".
If you have no idea about how many elements would be entered, you are either to reallocate array as soon as it is needed, or use linked list.
In C++ you can do allocation with "new" and also you can use std::list for similar cases.
(however all this is bit clumsy - that is why after 10 years of programming in C I at last turned to java and scripting languages)
How do I use a linked list? And I haven't reached pointers or memory allocation yet in my textbook :(
4. About pointers. I think currently it is important only to keep in mind that in C pointers are like arrays for which memory is not reserved beforehand:
Code:
```int*a;
int b[100];
//later they both behave similarly:
a[5] = 8;
b[5] = 8;```
the main difference is that for using pointer as array one needs to find for it some memory chunk and assign its address to pointer (we could either allocate this memory or make the pointer point to existing array).
arrays which you define with square brackets are always preallocated, constant size. they could be regarded as pointers (to memory used by this arrays) with the only difference that such pointer could not be assigned to other memory address.
About linked list - I think that you can safely skip this just for now. It is interesting, but not crucial for your study right now :)
shortly speaking - C allows to group several variables into one structure, addressed by single variable or pointer (like the Person which contains first name, last name, age etc.)
Linked list is a collection of such structures each of which contains not only data variables, but also a pointer to the next-in-chain structure. | 1,192 | 4,948 | {"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-2015-18 | longest | en | 0.849709 |
https://www.engworksheets.com/math-files/253/3-digit-subtraction-fill-in-the-missing-digits.html | 1,708,518,925,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473472.21/warc/CC-MAIN-20240221102433-20240221132433-00031.warc.gz | 803,483,002 | 5,443 | # 3-Digit Subtraction - Fill in the Missing Digits Worksheets and Exercise
Practice finding missing digits in subtraction problems with our intermediate-level exercises for 3rd and 4th graders. Our printable worksheets switch between easy and challenging levels, allowing students to improve their subtraction skills at their own pace.
#### 3-Digit Subtraction - Fill in the Missing Digits - Column
In these 3-digit subtraction column worksheets, students will have to fill in the missing digits. This will help improve their subtraction skills and attention to detail.
#### 3-Digit Subtraction - Fill in the Missing Digits - Column
In these 3-digit subtraction column worksheets, students will have to fill in the missing digits. This will help improve their subtraction skills and attention to detail.
#### 3-Digit Subtraction - Fill in the Missing Digits - Column
In these 3-digit subtraction column worksheets, students will have to fill in the missing digits. This will help improve their subtraction skills and attention to detail.
#### 3-Digit Subtraction - Fill in the Missing Digits - Column
In these 3-digit subtraction column worksheets, students will have to fill in the missing digits. This will help improve their subtraction skills and attention to detail.
#### 3-Digit Subtraction - Fill in the Missing Digits - Column
In these 3-digit subtraction column worksheets, students will have to fill in the missing digits. This will help improve their subtraction skills and attention to detail.
#### 3-Digit Subtraction - Fill in the Missing Digits - Column
In these 3-digit subtraction column worksheets, students will have to fill in the missing digits. This will help improve their subtraction skills and attention to detail.
#### 3-Digit Subtraction - Fill in the Missing Digits - Column
In these 3-digit subtraction column worksheets, students will have to fill in the missing digits. This will help improve their subtraction skills and attention to detail.
#### 3-Digit Subtraction - Fill in the Missing Digits - Column
In these 3-digit subtraction column worksheets, students will have to fill in the missing digits. This will help improve their subtraction skills and attention to detail.
#### 3-Digit Subtraction - Fill in the Missing Digits - Column
In these 3-digit subtraction column worksheets, students will have to fill in the missing digits. This will help improve their subtraction skills and attention to detail.
#### 3-Digit Subtraction - Fill in the Missing Digits - Column
In these 3-digit subtraction column worksheets, students will have to fill in the missing digits. This will help improve their subtraction skills and attention to detail. | 544 | 2,696 | {"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-2024-10 | latest | en | 0.833106 |
https://socratic.org/questions/how-do-you-determine-the-number-of-significant-figures-if-a-number-does-not-have | 1,575,935,631,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540525598.55/warc/CC-MAIN-20191209225803-20191210013803-00126.warc.gz | 537,784,698 | 6,212 | # How do you determine the number of significant figures if a number does not have a decimal place?
Jun 25, 2017
It's normally determined by the first and last non-zero digit, but can be otherwise indicated...
#### Explanation:
By default, the number of significant figures is the number of digits in the run from the first non-zero digit to the last.
For example:
$200340000$
has $5$ significant digits, namely "$20034$"
If you are given a number like $80000$, then by default it has just $1$ significant digit unless you are told otherwise.
One convention that is sometimes used is to indicate the last significant digit with an overbar or underbar, e.g.
$80 \overline{0} 00$
$80 \underline{0} 00$
These examples have $3$ significant figures, namely "$800$" | 184 | 771 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 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-2019-51 | longest | en | 0.854477 |
https://professionalsessays.com/2022/11/29/business-statistics-lab-assignment-4/ | 1,702,059,965,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100769.54/warc/CC-MAIN-20231208180539-20231208210539-00342.warc.gz | 519,258,926 | 19,977 | Posted: November 29th, 2022
# Business statistics lab assignment 4
looking for someone to do the lab work for business statistics. Using (R)
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see Labwork 4 for instructions
Here is some clarification on lab-4.
1. In problem 1, this is a multiple regression model. You are supposed to run a regression of Price (dependent variable) with other independents variables (Food, Décor, Service, and East).
2. when you run the regression model, make sure you report coefficient of the independent variables are significant or not.
3. Problem-2-part(a), you are supposed to make a correlation matrix. Use R code cor(data name) to get the correlation matrix.
4. Problem-2-part(b), simple linear regression, regress Price (dependent variable) on Rooms (independent variable), observe whether rooms is significant or not.
5. Problem-2-part(c), multiple linear regression, regress Price on Home size, Lot size, Rooms, and Bathroom. Now Room is significant or not check that-To check the significance look at P-value (if P-value is less than 0.05, it is significant at 95% confidence level otherwise it is not significant)
6. Problem-3-part(a) descriptive statistics, use summary (data name) Rcode.
7. Problem-3-part(b), run two simple linear regression
(i)regress Amount charge on income
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1. Regress Amount charge on household size. Now which is better, you can compare R^2 and slope.
(8) In problem-3 part(c), now use multiple regression regress amount charge on income and household size.
(9) In problem-4, You are supposed to regress sales on 11 dummies ( Jan through Nov). You don’t need to create dummies. I have uploaded the new data set Vintage_new in the data folder in Blackboard.
Now regress Sales on 11 dummies.
lm(Sales~D1+D2+…..+D11), find the predicted value for Jan
Also, the actual value for Jan is given 295,000
Forecast error= Actual value – Predicted value
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Order your essay today and save 15% with the discount code 2023DISCOUNT | 602 | 2,498 | {"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-2023-50 | longest | en | 0.829053 |
https://socratic.org/questions/how-do-you-write-7000-in-scientific-notation#332590 | 1,669,927,595,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710869.86/warc/CC-MAIN-20221201185801-20221201215801-00713.warc.gz | 569,364,232 | 5,660 | How do you write 7000 in scientific notation?
$7 \cdot {10}^{3}$
$7 \cdot {10}^{3}$ is $7 \cdot 1000$, which is 7000 | 47 | 117 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "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} | 3.046875 | 3 | CC-MAIN-2022-49 | latest | en | 0.726347 |
https://bioinformatics.stackexchange.com/questions/4172/how-to-predict-the-distance-between-two-residues-in-a-protein-sequence | 1,660,073,216,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571086.77/warc/CC-MAIN-20220809185452-20220809215452-00622.warc.gz | 157,903,955 | 65,477 | # How to predict the distance between two residues in a protein sequence
I have some protein sequences and I need to predict the physical distance between each two residues in the protein when folded correctly in 3 dimensions; I need to predict this distances just by having the protein sequence and not the structure.
I have a group of pdb IDs and I have got homologs for them and made a multiple sequence alignment for each query sequence; But later I need to imagine that I don't have any structure for being able to predict the distance between residues in query sequences without any known structure.
There are some articles for solving this issue using machine learning methods but I don't want to train with a large numbers of structures. Any one knows any way to solve this problem by using mathematical or statistical methods?
• Are there any similar sequences with known structure? Apr 28, 2018 at 17:40
• Please add that detail into your question. Also, pelase clarify that you are referring to the physical distance between the residues in the protein when folded correctly in 3 dimensions (at first I thought you meant in the linear sequence). Apr 29, 2018 at 12:51
• Between ALL pairs of residue-residue combinations? You can't use statistics without a large enough dataset and expect reliable results... BTW, I might be outdated but I don't remember having seen de novo folding success after Brian Kuhlman's, 2003 science paper with David Baker. If you only need to compute the distance within short fragments, you could estimate the secondary structure with psipred and use this information to estimate, for example, an average distance between residue i and residue i+4 Jul 3, 2019 at 16:13
• @aerijman I think you are mixing up de novo protein design and de novo protein structure prediction. Design of novel folds hasn't had much success as you say, but structure prediction of novel folds is getting to a decent level now. See the CASP website for some nice examples. Jul 4, 2019 at 10:19
Developments in this area over the last year are worth pointing out. All use residue-residue covariation information from a protein family so require a multiple sequence alignment to work, though it sounds like you have that.
The most accurate way to do this would be to use pre-trained machine learning methods, which would not require you to train your own version. Examples include:
• DMPfold from our lab, which is freely-available open source.
• The RaptorX method from the Xu lab.
• The state of the art is AlphaFold from DeepMind, though this has not been published yet.
These predict the probabilities of a residue-residue distance being in various distance 'bins', e.g. the probability of being 5 - 6 Angstrom apart. They don't predict beyond ~20 Angstrom.
If you specifically wanted a statistical, non-trained method, then consider:
All of which are available in some form. The problem here is that they all predict binary contacts at 8 Angstrom, whereas it sounds like you need more fine-grained distances.
In a general sense, you can use any 3D structure prediction method and then read the distances back from the predicted structure. | 672 | 3,167 | {"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-2022-33 | longest | en | 0.961529 |
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cs240a-denseGE
# cs240a-denseGE - CS 240A Solving Ax = b in parallel Dense A...
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CS267 Dense Linear Algebra I.1 Demmel Fa 2001 CS 240A: Solving Ax = b in parallel ° Dense A: Gaussian elimination with partial pivoting Same flavor as matrix * matrix, but more complicated ° Sparse A: Iterative methods – Conjugate gradient etc. Sparse matrix times dense vector ° Sparse A: Gaussian elimination – Cholesky, LU, etc. Graph algorithms ° Sparse A: Preconditioned iterative methods and multigrid Mixture of lots of things
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CS267 Dense Linear Algebra I.2 Demmel Fa 2001 CS 240A: Solving Ax = b in parallel ° Dense A: Gaussian elimination with partial pivoting Same flavor as matrix * matrix, but more complicated ° Sparse A: Iterative methods – Conjugate gradient etc. Sparse matrix times dense vector ° Sparse A: Gaussian elimination – Cholesky, LU, etc. Graph algorithms ° Sparse A: Preconditioned iterative methods and multigrid Mixture of lots of things
CS267 Dense Linear Algebra I.3 Demmel Fa 2001 Dense Linear Algebra (Excerpts) James Demmel http://www.cs.berkeley.edu/~demmel/cs267_221001.ppt
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CS267 Dense Linear Algebra I.4 Demmel Fa 2001 Motivation ° 3 Basic Linear Algebra Problems Linear Equations: Solve Ax=b for x Least Squares: Find x that minimizes Σ r i 2 where r=Ax-b Eigenvalues: Find λ and x where Ax = λ x Lots of variations depending on structure of A (eg symmetry) ° Why dense A, as opposed to sparse A? Aren’t “most” large matrices sparse? Dense algorithms easier to understand Some applications yields large dense matrices - Ax=b: Computational Electromagnetics - Ax = λ x: Quantum Chemistry Benchmarking - “How fast is your computer?” = “How fast can you solve dense Ax=b?” Large sparse matrix algorithms often yield smaller (but still large) dense problems
CS267 Dense Linear Algebra I.5 Demmel Fa 2001 Review of Gaussian Elimination (GE) for solving Ax=b ° Add multiples of each row to later rows to make A upper triangular ° Solve resulting triangular system Ux = c by substitution … for each column i … zero it out below the diagonal by adding multiples of row i to later rows for i = 1 to n-1 … for each row j below row i for j = i+1 to n … add a multiple of row i to row j for k = i to n A(j,k) = A(j,k) - (A(j,i)/A(i,i)) * A(i,k)
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CS267 Dense Linear Algebra I.6 Demmel Fa 2001 Refine GE Algorithm (1) ° Initial Version ° Remove computation of constant A(j,i)/A(i,i) from inner loop … for each column i … zero it out below the diagonal by adding multiples of row i to later rows for i = 1 to n-1 … for each row j below row i for j = i+1 to n … add a multiple of row i to row j for k = i to n A(j,k) = A(j,k) - (A(j,i)/A(i,i)) * A(i,k) for i = 1 to n-1 for j = i+1 to n m = A(j,i)/A(i,i) for k = i to n A(j,k) = A(j,k) - m * A(i,k)
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Ask a homework question - tutors are online | 922 | 3,504 | {"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} | 3.859375 | 4 | CC-MAIN-2018-09 | latest | en | 0.72539 |
http://codegur.com/44608534/trying-to-construct-a-map-with-2-d-arrays | 1,524,586,952,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946807.67/warc/CC-MAIN-20180424154911-20180424174911-00549.warc.gz | 77,038,566 | 9,096 | # Trying to construct a map with 2-D arrays
I am trying to construct a map with coordinates -20 to 20 for the x-axis and y axis with C#. At the moment i have defined my ranges(as shown below) and i am using a for each loop to loop through both lists however i can't populate them with anything because it throws an error when i try and reassign the array values back. I would like to draw a map with points. Is my logic wrong?
``````IEnumerable<int> squares = Enumerable.Range(-20, 20);
IEnumerable<int> cirles = Enumerable.Range(-20, 20);
int [][] arrays = new int[squares.Count()][cirles.Count()];
foreach (var shape in squares)
{
foreach (var sha in cirles)
\\Construct map
``````
The program throws a type error as it wants me to print out a jagged array to define it like this int [][] arrays = new int[squares.Count()][];
The error is invalid rank or specifier,
As mentioned in the comments already you are probably attempting to use the wrong kind of array here. It seems like you'd be better off simply using a 2D array since your dimensions are fixed:
``````int[,] array = new int[squares.Count(), circles.Count()];
``````
If you want to stick with a jagged array make sure you understand this first: https://docs.microsoft.com/en-us/dotnet/csharp/programming-guide/arrays/jagged-arrays.
Jagged arrays cannot be initialized the way you tried it. I suppose the reason for that is that it just wouldn't make an aweful lot of sense. Jagged arrays allow you to have different dimensions for each contained array (which you don't need).
Still, if you want to use a jagged array then use the following code instead for the initialization:
`````` int[][] arrays = new int[squares.Count()][];
for (int i = 0; i < arrays.Length; i++)
{
arrays[i] = new int[circles.Count()];
}
`````` | 428 | 1,799 | {"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} | 2.625 | 3 | CC-MAIN-2018-17 | latest | en | 0.860805 |
http://cboard.cprogramming.com/c-programming/152760-accessing-column-data-multi-dimentional-arrays.html | 1,469,822,713,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257831771.10/warc/CC-MAIN-20160723071031-00228-ip-10-185-27-174.ec2.internal.warc.gz | 38,990,043 | 13,717 | # Thread: Accessing column data in multi-dimentional arrays
1. ## Accessing column data in multi-dimentional arrays
Hi Everyone,
I am new to the forum and was hoping someone could help me with what I hope i a relatively simple problem. I couldn't find my problem described in an existing topic, but I may have overlooked it. If so, I apologize.
Anyway - my problem is the following.
I am currently working on a simple program and am using multi-dimentional arrays since I have a lot of data to process for different variables.
An example of one of my arrays is the following:
Code:
`double Solution[3][20][20]`
When I use this declaration, it is relatively easy for me to access the underlying 20x20 arrays, if I pass the array to a function as
Code:
`Solution[0]`
However, now for a different part of my code, I would like to get all the entries(as a vector) in the first column for a specific orientation in the underlying 20 x 20 array.
An example:
Code:
`Solution[*all three values*][10][10] = {entry 1, entry 2, entry 3}`
But I have been unsuccessful to do so. Is there a simple way of doing this?
Thanks to anyone who can find the time to reply.
2. So you want to enter values to all three at once? Is this correct?
3. Originally Posted by Click_here
So you want to enter values to all three at once? Is this correct?
Not, that is not what I want to do.
I want to access all the three values in the first dimension as a vector, for a given specific position for the remaining two dimensions.
Here is how I am working around the problem at the moment:
Code:
`double Uij[3] = {DeltaU[0][i][j], DeltaU[1][i][j], DeltaU[2][i][j]};`
Thus, I am defining a vector manually, and the resulting vector has the following entries:
Code:
` Uij = {DeltaU[0][i][j], DeltaU[1][i][j], DeltaU[2][i][j]}`
What I was hoping was, that I could somehow select the three entries in (using point 10,10 as example):
Code:
`DeltaU[k][10][10]`
(Where k is the index for my for-loop, running from 0-2)
And have the output as a vector with the following values:
Code:
` Uij = {DeltaU[k=0][10][10], DeltaU[k=1][10][10], DeltaU[k=2][10][10] }`
Does it make sense ?
4. As I understand it you want this:
Code:
```double Uij[3];
for (i = 0; i < 3; i++)
Uij[i] = DeltaU[i][10][10];```
Correct?
You need a for-loop if you don't want to use the "initializer" version.
Or, do you want to achieve that the values in "Uij" are references to the corresponding values of "DeltaU" (i.e. if the values change in "DeltaU" the values in "Uij" change too and vice versa)? Then you need pointers:
Code:
`double *Uij[] = { &DeltaU[0][10][10], &DeltaU[1][10][10], &DeltaU[2][10][10] };`
(or the corresponding for-loop version) and you can access the values in "Uij" by dereferencing:
Code:
`double x = *Uij[0]; // "x" has now the value of DeltaU[0][10][10]`
Bye, Andreas
5. Originally Posted by Darkmentor
the first column for a specific orientation in the underlying 20 x 20 array.
Simple: Just do not assume the vector is consecutive in memory.
Instead of a simple linear array, specify the data as a pointer, number of data items between consecutive elements (stride), and vector size (element count):
Code:
```double my_function(double *const vector, const long stride, const long size)
{
/* Element i (0 <= i < size), is (vector[i * stride]). */
}```
The strides you can trivially compute:
• (long)(&Solution[0][0][1] - &Solution[0][0][0]) for a row vector -- I do believe this is defined to be 1 in C
• (long)(&Solution[0][1][0] - &Solution[0][0][0]) for a column vector
• (long)(&Solution[1][0][0] - &Solution[0][0][0]) for a plane vector
Note that the stride is not in units of bytes or chars, but the data units, here doubles, so the data types matter. If you use the actual data array to compute the (three) strides, it will be correct every time. You could use for example
Code:
```#define PLANES 3
#define ROWS 20
#define COLUMNS 20
double Solution[PLANES][ROWS][COLUMNS];
const long Solution_row = 1L;
const long Solution_column = &Solution[0][1][0] - &Solution[0][0][0];
const long Solution_plane = &Solution[1][0][0] - &Solution[0][0][0];
/* Third column vector (column 2): */
result = my_function(&Solution[0][0][2], Solution_column, ROWS);```
If you supply the pointer to the final element, and negate the stride, you can reverse the vector component order, too. It is sometimes very useful.
There are also two-dimensional and higher-dimensional equivalents, of course: just specify the size and stride for each dimension separately. In 2D, for example:
Code:
```double my_function(double *const plane, const long rowstride, const long colstride, const long rows, const long cols)
{
/* Element (r,c), (0 <= r < rows, 0 <= c < cols) is
* plane[r * rowstride + c * colstride]
*/
}```
but I personally prefer to define structures (with the size and stride fields, and a pointer to the data origin). Let me know if you want examples.
6. Not quite sure what you mean by getting column elements as a vector. I believe you mean you want elements to be consecutive in single dimension arrays.
You can transpose the Solution[3][20][20] into Solution_T[20][20][3] so that the last axis can be referenced using a [20][20] index and its 3 elements are consecutive.
You could also generate vectors of addresses to the column oriented elements and then reference these by consecutively dereferencing these addresses. But that's just like making a loop like AndiPersti suggested with additional address arrays. | 1,492 | 5,507 | {"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} | 2.875 | 3 | CC-MAIN-2016-30 | latest | en | 0.937482 |
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https://lw2.issarice.com/posts/NKECtGX4RZPd7SqYp/the-modesty-argument | 1,632,102,244,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056974.30/warc/CC-MAIN-20210920010331-20210920040331-00010.warc.gz | 434,321,293 | 24,173 | # The Modesty Argument
post by Eliezer Yudkowsky (Eliezer_Yudkowsky) · 2006-12-10T21:42:55.000Z · LW · GW · Legacy · 40 comments
The Modesty Argument states that when two or more human beings have common knowledge that they disagree about a question of simple fact, they should each adjust their probability estimates in the direction of the others'. (For example, they might adopt the common mean of their probability distributions. If we use the logarithmic scoring rule, then the score of the average of a set of probability distributions is better than the average of the scores of the individual distributions, by Jensen's inequality.)
Put more simply: When you disagree with someone, even after talking over your reasons, the Modesty Argument claims that you should each adjust your probability estimates toward the other's, and keep doing this until you agree. The Modesty Argument is inspired by Aumann's Agreement Theorem, a very famous and oft-generalized result which shows that genuine Bayesians literally cannot agree to disagree; if genuine Bayesians have common knowledge of their individual probability estimates, they must all have the same probability estimate. ("Common knowledge" means that I know you disagree, you know I know you disagree, etc.)
I've always been suspicious of the Modesty Argument. It's been a long-running debate between myself and Robin Hanson.
Robin seems to endorse the Modesty Argument in papers such as Are Disagreements Honest? I, on the other hand, have held that it can be rational for an individual to not adjust their own probability estimate in the direction of someone else who disagrees with them.
How can I maintain this position in the face of Aumann's Agreement Theorem, which proves that genuine Bayesians cannot have common knowledge of a dispute about probability estimates? If genunie Bayesians will always agree with each other once they've exchanged probability estimates, shouldn't we Bayesian wannabes do the same?
To explain my reply, I begin with a metaphor: If I have five different accurate maps of a city, they will all be consistent with each other. Some philosophers, inspired by this, have held that "rationality" consists of having beliefs that are consistent among themselves. But, although accuracy necessarily implies consistency, consistency does not necessarily imply accuracy. If I sit in my living room with the curtains drawn, and make up five maps that are consistent with each other, but I don't actually walk around the city and make lines on paper that correspond to what I see, then my maps will be consistent but not accurate. When genuine Bayesians agree in their probability estimates, it's not because they're trying to be consistent - Aumann's Agreement Theorem doesn't invoke any explicit drive on the Bayesians' part to be consistent. That's what makes AAT surprising! Bayesians only try to be accurate; in the course of seeking to be accurate, they end up consistent. The Modesty Argument, that we can end up accurate in the course of seeking to be consistent, does not necessarily follow.
How can I maintain my position in the face of my admission that disputants will always improve their average score if they average together their individual probability distributions?
Suppose a creationist comes to me and offers: "You believe that natural selection is true, and I believe that it is false. Let us both agree to assign 50% probability to the proposition." And suppose that by drugs or hypnosis it was actually possible for both of us to contract to adjust our probability estimates in this way. This unquestionably improves our combined log-score, and our combined squared error. If as a matter of altruism, I value the creationist's accuracy as much as my own - if my loss function is symmetrical around the two of us - then I should agree. But what if I'm trying to maximize only my own individual accuracy? In the former case, the question is absolutely clear, and in the latter case it is not absolutely clear, to me at least, which opens up the possibility that they are different questions.
If I agree to a contract with the creationist in which we both use drugs or hypnosis to adjust our probability estimates, because I know that the group estimate must be improved thereby, I regard that as pursuing the goal of social altruism. It doesn't make creationism actually true, and it doesn't mean that I think creationism is true when I agree to the contract. If I thought creationism was 50% probable, I wouldn't need to sign a contract - I would have already updated my beliefs! It is tempting but false to regard adopting someone else's beliefs as a favor to them, and rationality as a matter of fairness, of equal compromise. Therefore it is written: "Do not believe you do others a favor if you accept their arguments; the favor is to you." Am I really doing myself a favor by agreeing with the creationist to take the average of our probability distributions?
I regard rationality in its purest form as an individual thing - not because rationalists have only selfish interests, but because of the form of the only admissible question: "Is is actually true?" Other considerations, such as the collective accuracy of a group that includes yourself, may be legitimate goals, and an important part of human existence - but they differ from that single pure question.
In Aumann's Agreement Theorem, all the individual Bayesians are trying to be accurate as individuals. If their explicit goal was to maximize group accuracy, AAT would not be surprising. So the improvement of group score is not a knockdown argument as to what an individual should do if they are trying purely to maximize their own accuracy, and it is that last quest which I identify as rationality. It is written: "Every step of your reasoning must cut through to the correct answer in the same movement. More than anything, you must think of carrying your map through to reflecting the territory. If you fail to achieve a correct answer, it is futile to protest that you acted with propriety." From the standpoint of social altruism, someone may wish to be Modest, and enter a drug-or-hypnosis-enforced contract of Modesty, even if they fail to achieve a correct answer thereby.
The central argument for Modesty proposes something like a Rawlsian veil of ignorance - how can you know which of you is the honest truthseeker, and which the stubborn self-deceiver? The creationist believes that he is the sane one and you are the fool. Doesn't this make the situation symmetric around the two of you? If you average your estimates together, one of you must gain, and one of you must lose, since the shifts are in opposite directions; but by Jensen's inequality it is a positive-sum game. And since, by something like a Rawlsian veil of ignorance, you don't know which of you is really the fool, you ought to take the gamble. This argues that the socially altruistic move is also always the individually rational move.
And there's also the obvious reply: "But I know perfectly well who the fool is. It's the other guy. It doesn't matter that he says the same thing - he's still the fool."
This reply sounds bald and unconvincing when you consider it abstractly. But if you actually face a creationist, then it certainly feels like the correct answer - you're right, he's wrong, and you have valid evidence to know that, even if the creationist can recite exactly the same claim in front of a TV audience.
Robin Hanson sides with symmetry - this is clearest in his paper Uncommon Priors Require Origin Disputes - and therefore endorses the Modesty Argument. (Though I haven't seen him analyze the particular case of the creationist.)
I respond: Those who dream do not know they dream; but when you wake you know you are awake. Dreaming, you may think you are awake. You may even be convinced of it. But right now, when you really are awake, there isn't any doubt in your mind - nor should there be. If you, persuaded by the clever argument, decided to start doubting right now that you're really awake, then your Bayesian score would go down and you'd become that much less accurate. If you seriously tried to make yourself doubt that you were awake - in the sense of wondering if you might be in the midst of an ordinary human REM cycle - then you would probably do so because you wished to appear to yourself as rational, or because it was how you conceived of "rationality" as a matter of moral duty. Because you wanted to act with propriety. Not because you felt genuinely curious as to whether you were awake or asleep. Not because you felt you might really and truly be asleep. But because you didn't have an answer to the clever argument, just an (ahem) incommunicable insight that you were awake.
Russell Wallace put it thusly: "That we can postulate a mind of sufficiently low (dreaming) or distorted (insane) consciousness as to genuinely not know whether it's Russell or Napoleon doesn't mean I (the entity currently thinking these thoughts) could have been Napoleon, any more than the number 3 could have been the number 7. If you doubt this, consider the extreme case: a rock doesn't know whether it's me or a rock. That doesn't mean I could have been a rock."
There are other problems I see with the Modesty Argument, pragmatic matters of human rationality - if a fallible human tries to follow the Modesty Argument in practice, does this improve or disimprove personal rationality? To me it seems that the adherents of the Modesty Argument tend to profess Modesty but not actually practice it.
For example, let's say you're a scientist with a controversial belief - like the Modesty Argument itself, which is hardly a matter of common accord - and you spend some substantial amount of time and effort trying to prove, argue, examine, and generally forward this belief. Then one day you encounter the Modesty Argument, and it occurs to you that you should adjust your belief toward the modal belief of the scientific field. But then you'd have to give up your cherished hypothesis. So you do the obvious thing - I've seen at least two people do this on two different occasions - and say: "Pursuing my personal hypothesis has a net expected utility to Science. Even if I don't really believe that my theory is correct, I can still pursue it because of the categorical imperative: Science as a whole will be better off if scientists go on pursuing their own hypotheses." And then they continue exactly as before.
I am skeptical to say the least. Integrating the Modesty Argument as new evidence ought to produce a large effect on someone's life and plans. If it's being really integrated, that is, rather than flushed down a black hole. Your personal anticipation of success, the bright emotion with which you anticipate the confirmation of your theory, should diminish by literally orders of magnitude after accepting the Modesty Argument. The reason people buy lottery tickets is that the bright anticipation of winning ten million dollars, the dancing visions of speedboats and mansions, is not sufficiently diminished - as a strength of emotion - by the probability factor, the odds of a hundred million to one. The ticket buyer may even profess that the odds are a hundred million to one, but they don't anticipate it properly - they haven't integrated the mere verbal phrase "hundred million to one" on an emotional level.
So, when a scientist integrates the Modesty Argument as new evidence, should the resulting nearly total loss of hope have no effect on real-world plans originally formed in blessed ignorance and joyous anticipation of triumph? Especially when you consider that the scientist knew about the social utility to start with, while making the original plans? I think that's around as plausible as maintaining your exact original investment profile after the expected returns on some stocks change by a factor of a hundred. What's actually happening, one naturally suspects, is that the scientist finds that the Modesty Argument has uncomfortable implications; so they reach for an excuse, and invent on-the-fly the argument from social utility as a way of exactly cancelling out the Modesty Argument and preserving all their original plans.
But of course if I say that this is an argument against the Modesty Argument, that is pure ad hominem tu quoque. If its adherents fail to use the Modesty Argument properly, that does not imply it has any less force as logic.
Rather than go into more detail on the manifold ramifications of the Modesty Argument, I'm going to close with the thought experiment that initially convinced me of the falsity of the Modesty Argument. In the beginning it seemed to me reasonable that if feelings of 99% certainty were associated with a 70% frequency of true statements, on average across the global population, then the state of 99% certainty was like a "pointer" to 70% probability. But at one point I thought: "What should an (AI) superintelligence say in the same situation? Should it treat its 99% probability estimates as 70% probability estimates because so many human beings make the same mistake?" In particular, it occurred to me that, on the day the first true superintelligence was born, it would be undeniably true that - across the whole of Earth's history - the enormously vast majority of entities who had believed themselves superintelligent would be wrong. The majority of the referents of the pointer "I am a superintelligence" would be schizophrenics who believed they were God.
A superintelligence doesn't just believe the bald statement that it is a superintelligence - it presumably possesses a very detailed, very accurate self-model of its own cognitive systems, tracks in detail its own calibration, and so on. But if you tell this to a mental patient, the mental patient can immediately respond: "Ah, but I too possess a very detailed, very accurate self-model!" The mental patient may even come to sincerely believe this, in the moment of the reply. Does that mean the superintelligence should wonder if it is a mental patient? This is the opposite extreme of Russell Wallace asking if a rock could have been you, since it doesn't know if it's you or the rock.
One obvious reply is that human beings and superintelligences occupy different classes - we do not have the same ur-priors, or we are not part of the same anthropic reference class; some sharp distinction renders it impossible to group together superintelligences and schizophrenics in probability arguments. But one would then like to know exactly what this "sharp distinction" is, and how it is justified relative to the Modesty Argument. Can an evolutionist and a creationist also occupy different reference classes? It sounds astoundingly arrogant; but when I consider the actual, pragmatic situation, it seems to me that this is genuinely the case.
Or here's a more recent example - one that inspired me to write today's blog post, in fact. It's the true story of a customer struggling through five levels of Verizon customer support, all the way up to floor manager, in an ultimately futile quest to find someone who could understand the difference between .002 dollars per kilobyte and .002 cents per kilobyte. Audio [27 minutes], Transcript. It has to be heard to be believed. Sample of conversation: "Do you recognize that there's a difference between point zero zero two dollars and point zero zero two cents?" "No."
The key phrase that caught my attention and inspired me to write today's blog post is from the floor manager: "You already talked to a few different people here, and they've all explained to you that you're being billed .002 cents, and if you take it and put it on your calculator... we take the .002 as everybody has told you that you've called in and spoken to, and as our system bills accordingly, is correct."
Should George - the customer - have started doubting his arithmetic, because five levels of Verizon customer support, some of whom cited multiple years of experience, told him he was wrong? Should he have adjusted his probability estimate in their direction? A straightforward extension of Aumann's Agreement Theorem to impossible possible worlds, that is, uncertainty about the results of computations, proves that, had all parties been genuine Bayesians with common knowledge of each other's estimates, they would have had the same estimate. Jensen's inequality proves even more straightforwardly that, if George and the five levels of tech support had averaged together their probability estimates, they would have improved their average log score. If such arguments fail in this case, why do they succeed in other cases? And if you claim the Modesty Argument carries in this case, are you really telling me that if George had wanted only to find the truth for himself, he would have been wise to adjust his estimate in Verizon's direction? I know this is an argument from personal incredulity, but I think it's a good one.
On the whole, and in practice, it seems to me like Modesty is sometimes a good idea, and sometimes not. I exercise my individual discretion and judgment to decide, even knowing that I might be biased or self-favoring in doing so, because the alternative of being Modest in every case seems to me much worse.
But the question also seems to have a definite anthropic flavor. Anthropic probabilities still confuse me; I've read arguments but I have been unable to resolve them to my own satisfaction. Therefore, I confess, I am not able to give a full account of how the Modesty Argument is resolved.
Modest, aren't I?
Comments sorted by oldest first, as this post is from before comment nesting was available (around 2009-02-27).
comment by HalFinney · 2006-12-10T22:26:29.000Z · LW(p) · GW(p)
I can give a counter-example to the Modesty Argument as stated: "When you disagree with someone, the Modesty Argument claims that you should both adjust your probability estimates toward the other, and keep doing this until you agree."
Suppose two coins are flipped out of sight, and you and another person are trying to estimate the probability that both are heads. You are told what the first coin is, and the other person is told what the second coin is. You both report your observations to each other.
Let's suppose that they did in fact fall both heads. You are told that the first coin is heads, and you report the probability of both heads as 1/2. The other person is told that the second coin is heads, and he also reports the probability as 1/2. However, you can now both conclude that the probability is 1, because if either of you had been told that the coin was tails, he would have reported a probability of zero. So in this case, both of you update your information away from the estimate provided by the other. One can construct more complicated thought experiments where estimates jump around in all kinds of crazy ways, I think.
I'm not sure whether this materially affects your conclusions about the Modesty Argument, but you might want to try to restate it to more clearly describe what you are disagreeing with.
comment by Nato_Welch · 2006-12-10T22:33:34.000Z · LW(p) · GW(p)
I often recognize this as what I've called a "framing error".
It's the same problem with believing Fox News' claim of being "fair and balanced". Just because there are at least two sides to every story doesn't mean each side deserves equal consideration.
Consider the question of whether one can support the troops and oppose the war.
http://n8o.r30.net/dokuwiki/doku.php/blog:supportourwar
The very posing of the question, which came invariably from Fox News in 2003, is to imply that there is some controversy over the answer. But that's all but disappeared now.
comment by pdf23ds · 2006-12-10T23:33:37.000Z · LW(p) · GW(p)
I don't see how there's a problem with saying that the Modesty Principle is only ever a good one to follow when conversing with someone you believe to be more or equally rational as yourself. I guess the extreme case to illustrate this would be applying the principle to a conversation with a chatbot.
comment by Eliezer Yudkowsky (Eliezer_Yudkowsky) · 2006-12-11T00:32:51.000Z · LW(p) · GW(p)
Hal, I changed the lead to say "When two or more human beings have common knowledge that they disagree", which covers your counterexample.
comment by Eliezer Yudkowsky (Eliezer_Yudkowsky) · 2006-12-11T00:38:22.000Z · LW(p) · GW(p)
pdf23ds, the prbolem is how to decide when the person you are conversing with is more or equally rational as yourself. What if you disagree about that? Then you have a disagreement about a new variable, your respective degrees of rationality. Do you both believe yourself to be more meta-rational than the other? And so on. See Hanson and Cowen's "Are Disagreements Honest?", http://hanson.gmu.edu/deceive.pdf.
comment by HalFinney · 2006-12-11T01:09:02.000Z · LW(p) · GW(p)
Eliezer, "when two or more human beings have common knowledge that they disagree about a question of simple fact" is problematic because this can never happen with common-prior Bayesians. Such people can never have common knowledge of a disagreement, although they can disagree on the way to common knowledge of their opininons. Maybe you could say that they learn that they initially disagree about the question.
However I don't think that addition fixes the problem. Suppose the first coin in my example is known by all to be biased and fall heads with 60% probability. Then the person who knows the first coin initially estimates the probability of HH as 50%, while the person who knows the second coin initially estimates it as 60%. So they initially disagree, and both know this. However they will both update their estimates upward to 100% after hearing each other, which means that they do not adjust their probability estimates towards the other.
I am having trouble relating the Modesty Argument to Aumann's result. It's not clear to me that anyone will defend the Modesty Argument as stated, at least not based on Aumann and the related literature.
comment by Eliezer Yudkowsky (Eliezer_Yudkowsky) · 2006-12-11T01:14:02.000Z · LW(p) · GW(p)
Hal, that's why I specified human beings. Human beings often find themselves with common knowledge that they disagree about a question of fact. And indeed, genuine Bayesians would not find themselves in such a pickle to begin with, which is why I question that we can clean up the mess by imitating the surface features of Bayesians (mutual agreement) while departing from their causal mechanisms (instituting an explicit internal drive to agreement, which is not present in ideal Bayesians).
The reason my addition fixes the problem is that in your scenario, the disagreement only holds while the two observers do not have common knowledge of their own probability estimates - this can easily happen to Bayesians; all they need to do is observe a piece of evidence they haven't had the opportunity to communicate. So they disagree at first, but only while they don't have common knowledge.
comment by HalFinney · 2006-12-11T01:42:33.000Z · LW(p) · GW(p)
Eliezer - I see better what you mean now. However I am still a bit confused because "common knowledge" usually refers to a state that is the end point of a computational or reasoning process, so it is inconsistent to speak of what participants "should" do after that point. If A and B's disagreement is supposedly common knowledge, but A may still choose to change his estimate to more closely match B's, then his estimate really isn't common knowledge at all because B is not sure if he has changed his mind or not.
When you say, "they should each adjust their probability estimates in the direction of the others'", do you mean that they should have done that and it would have been better for them, instead of letting themselves get into the state of common knowledge of their disagreement?
Sorry to belabor this but since you are critiquing this Modesty Argument it seems worthwhile to clarify exactly what it says.
comment by Eliezer Yudkowsky (Eliezer_Yudkowsky) · 2006-12-11T02:22:50.000Z · LW(p) · GW(p)
Hal, I'm not really the best person to explain the Modesty Argument because I don't believe in it! You should ask a theory's advocates, not its detractors, to explain it. You, yourself, have advocated that people should agree to agree - how do you think that people should go about it? If your preferred procedure differs from the Modesty Argument as I've presented it, it probably means that I got it wrong.
What I mean by the Modesty Argument is: You sit down at a table with someone else who disagrees with you, you each present your first-order arguments about the immediate issue - on the object level, as it were - and then you discover that you still seem to have a disagreement. Then at this point (I consider the Modesty Argument to say), you should consider as evidence the second-order, meta-level fact that the other person isn't persuaded, and you should take that evidence into account by adjusting your estimate in his direction. And he should do likewise. Keep doing that until you agree.
As to how this fits into Aumann's original theorem - I'm the wrong person to ask about that, because I don't think it does fit! But in terms of real-world procedure, I think that's what Modesty advocates are advocating, more or less. When we're critiquing Inwagen for failing to agree with Lewis, this is more or less the sort of thing we think he ought to do instead - right?
There are times when I'm happy enough to follow Modest procedure, but the Verizon case, and the creationist case, aren't on my list. I exercise my individual discretion, and judge based on particular cases. I feel free to not regard a creationist's beliefs as evidence, despite the apparent symmetry of my belief that he's the fool and his belief that I'm the fool. Thus I don't concede that the Modesty Argument holds in general, while Robin Hanson seems (in "Are Disagreements Honest?") to hold that it should be universal.
comment by Peter_McCluskey · 2006-12-11T02:25:20.000Z · LW(p) · GW(p)
I doubt that anyone is advocating the version of the Modesty Argument that you're attacking. People who advocate something resembling that seem to believe we should only respond that way if we should assume both sides are making honest attempts to be Bayesians. I don't know of anyone who suggests we ignore evidence concerning the degree to which a person is an honest Bayesian. See for example the qualification Robin makes in the last paragraph of this: http://lists.extropy.org/pipermail/extropy-chat/2005-March/014620.html. Or from page 28 of http://hanson.gmu.edu/deceive.pdf "seek observable signs that indicate when people are self-deceived about their meta-rationality on a particular topic. You might then try to disagree only with those who display such signs more strongly than you do." There seems to be enough agreement on some basic principles of rationality that we can conclude there are non-arbitrary ways of estimating who's more rational that are available to those who want to use them.
comment by Eliezer Yudkowsky (Eliezer_Yudkowsky) · 2006-12-11T03:44:58.000Z · LW(p) · GW(p)
Okay, so what are Robin and Hal advocating, procedurally speaking? Let's hear it from them. I defined the Modesty Argument because I had to say what I thought I was arguing against, but, as said, I'm not an advocate and therefore I'm not the first person to ask. Where do you think Inwagen went wrong in disagreeing with Lewis - what choice did he make that he should not have made? What should he have done instead? The procedure I laid out looks to me like the obvious one - it's the one I'd follow with a perceived equal. It's in applying the Modest procedure to disputes about rationality or meta-rationality that I'm likely to start wondering if the other guy is in the same reference class. But if I've invented a strawman, I'm willing to hear about it - just tell me the non-strawman version.
comment by Douglas_Knight2 · 2006-12-11T04:17:42.000Z · LW(p) · GW(p)
It is certainly true that one should not superficially try to replicate Aumann's theorem, but should try to replicate the process of the bayesians, namely, to model the other agent and see how the other agent could disagree. Surely this is how we disagree with creationists and customer service agents. Even if they are far from bayesian, we can extract information from their behavior, until we can model them.
But modeling is also what RH was advocating for the philosophers. Inwagen accepts Lewis as a peer, perhaps a superior. Moreover, he accepts him as rationally integrating Inwagen's arguments. This is exactly where Aumann's argument applies. In fact, Inwagen does model himself and Lewis and claims (I've only read the quoted excerpt) that their disagreement must be due to incommunicable insights. Although Aumann's framework about modelling the world seems incompatible with the idea of incommunicable insight, I think it is right to worry about symmetry. Possibly this leads us into difficult anthropic territory. But EY is right that we should not respond by simply changing our opinions, but we should try to describe this incommunicable insight and see how it has infected our beliefs.
Anthropic arguments are difficult, but I do not think they are relevant in any of the examples, except maybe the initial superintelligence. In that situation, I would argue in a way that may be your argument about dreaming: if something has a false belief about having a detailed model of the world, there's not much it can do. You might as well say it is dreaming. (I'm not talking about accuracy, but precision and moreover persistence.)
And you seem to say that if it is dreaming it doesn't count. When you claim that my bayesian score goes up if I insist that I'm awake whenever I feel I'm awake, you seem to be asserting that my assertions in my dreams don't count. This seems to be a claim about persistence of identity. Of course, my actions in dreams seem to have less import than my actions when awake, so I should care less about dream error. But I should not discount it entirely.
comment by HalFinney · 2006-12-11T07:02:19.000Z · LW(p) · GW(p)
I think Douglas has described the situation well. The heart of the Aumann reasoning as I understand it is the mirroring effect, where your disputant's refusal to agree must be interpreted in the context of his understanding the full significance of that refusal, which must be understood to carry weight due to your steadfastness, which itself gains meaning because of his persistence, and so on indefinitely. It is this infinite regress which is captured in the phrase "common knowledge" and which is where I think our intuition tends to go awry and leads us to underestimate the true significance of agreeing to disagree.
Without this mirroring, the Bayesian impact of another person's disagreement is less and we no longer obtain Aumann's strong result. In that case, as Douglas described we should try to model the information possessed by the other person and compare it with our own information, doing our best to be objective. This may indeed benefit from a degree of modesty but it's hard to draw a firm rule.
comment by Robin_Hanson2 · 2006-12-11T08:24:28.000Z · LW(p) · GW(p)
This is of course a topic I have a lot to say about, but alas I left on a trip just before Eliezer and Hal made their posts, and I'm going to be pretty busy for the next few days. Just wanted all to know I will get around to responding when I get time.
comment by Grant_Gould · 2006-12-11T16:55:35.000Z · LW(p) · GW(p)
So what's the proper prior probability that the person I'm arguing with is a Bayesian? That would seem to be a necessary piece of information for a Bayesian to figure out how much to follow the Modesty Argument.
comment by Robin_Hanson2 · 2006-12-15T20:00:36.000Z · LW(p) · GW(p)
(I finally have time to reply; sorry for the long delay.)
Eliezer, one can reasonably criticize a belief without needing to give an exact algorithm for always and exactly computing the best possible belief in such situations. Imagine you said P(A) = .3 and P(notA) = .9, and I criticized you for not satisfying P(A)+P(notA) = 1. If you were to demand that I tell you what to believe instead I might suggest you renormalize, and assign P(A) = 3/12 and P(notA) = 9/12. To that you might reply that those are most certainly not always the best exact numbers to assign. You know of examples where the right thing to do was clearly to assign P(A) = .3 and P(notA) = .7. But surely this would not be a reasonable response to my criticism. Similarly, I can criticize disagreement without offering an exact algorithm which always computes the best way to resolve the disagreement. I would suggest you both moving to a middle belief in the same spirit I might suggest renormalizing when things don't sum to one, as a demonstration that apparently reasonable options are available.
comment by Robin_Hanson2 · 2006-12-15T20:49:01.000Z · LW(p) · GW(p)
Eliezer, you describe cases of dreamers vs folks awake, of super-intelligences vs schizophrenics who think they are God, of creationists vs. their opponents, and of a Verizon customer vs customer support, all as cases where it can be very reasonably obvious to one side that the other side is completely wrong. The question of course is what exactly identifies such cases, so that you can tell if you are in such a situation at any given moment.
Clearly, people having the mere impression that they are in such a situation is a very unreliable indicator. So if not such an impression, what exactly does justify each of these exemplars in thinking they are obviously right?
It seems to me that even if we grant the possibility of such cases, we must admit that people are far too quick to assume that they are in such cases. So until we can find a reason to think we succumb to this problem less than others, we should try to invoke this explanation less often than we were initially inclined too.
comment by Carl_Shulman · 2006-12-15T22:15:39.000Z · LW(p) · GW(p)
We will often have independent information about relative rationality, and/or one model may predict the statements of the other, so that the conversation has minimal information value on the acual topic of disagreement.
1. I'll skip dreamers for the moment.
2. Schizophrenics may claim to possess a detailed self-model, but they can neither convincingly describe the details nor demonstrate feats of superintelligence to others. The AI can do so easily, and distinguish itself from the broader class of beings claiming superintelligence.
3. Creationists show lower IQ, knowledge levels, and awareness of cognitive biases than their opponents, along with higher rates of other dubious beliefs (and different types of true knowledge are correlated). Further, creationist exceptions with high levels of such rationality-predictors overwhelmingly have been subjected to strong social pressures (childhood indoctrination, etc) and other non-truth-oriented influences. Explaining this difference is very difficult from a creationist perspective, but simple for opponents.
I'd also note the possibility of an 'incommunicable insight.' If creationists attribute conscious bad faith to their opponents, who do not reciprocate, then the opponents can have incommunicable knowledge of their own sincerity.
1. The Verizon customer-service issue was baffling. But relatively low-skill call center employees following scripts, afflicted with erroneous corporate information (with the weight of hierarchical authority behind it), can make mistakes and would be reluctant to cut the size of a customer's bill by 99%. Further, some of the representatives seemed to understand before passing the buck. Given the existence of mail-in-rebate companies that are paid on the basis of how many rebates they can manage to lose, George could reasonably suspect that the uniformly erroneous price quotes were an intentional moneymaking scheme. Since Verizon STILL has not changed the quotes its representatives offer, there's now even more reason to think that's the case.
Combining this knowledge about Verizon with personal knowledge of higher-than-average mathematics and reasoning skills (from independent objective measures like SAT scores and math contests) George could have had ample reason not to apply the Modesty Argument and conclude that 0.002 cents is equal to \$0.002, \$0.00167, or \$0.00101.
comment by Robin_Hanson2 · 2006-12-16T01:10:36.000Z · LW(p) · GW(p)
Carl, I'd class schizophrenics with parrots and chatbots; creatures so obviously broken to so many observers that self-interested bias is plausibly a minor factor in our belifs about their rationality. For creationists I want to instead say that I usually have to go with the rough middle of expert opinion, and that goes against creationists. But there is the thorny issue of who exactly are the right experts for this topic. For Verizon I'd say what we are seeing is subject to a selection effect; probably usually Verizon is right and the customers are wrong.
comment by Carl_Shulman · 2006-12-16T06:11:57.000Z · LW(p) · GW(p)
Robin,
I agree that the difficult issue is defining the set of experts (although weighting expert opinion to account for known biases is an alternative to drawing a binary distinction between appropriate and inappropriate experts). I would think that if one can select a subset of expert opinion that is representative of the broad class except for higher levels of a rationality-signal (such as IQ, or professed commitment in approximating a Bayesian reasoner), then one should prefer the subset, until sample size shrinks to the point where the marginal increase in random noise outweighs the marginal improvement in expert quality.
What is your own view on the appropriate expert class for evaluating creationism or theism? College graduates? Science PhDs? Members of elite scientific societies? Science Nobel winners? (Substitute other classes and subclasses, e.g. biological/geological expertise, as you wish.) Unless the more elite group is sufficiently unrepresentative (e.g. elite scientists are disproportionately from non-Christian family backgrounds, although that can be controlled for) in some other way, why stop short?
On Verizon, I agree that if all one knows is that one is in the class of 'people in a dispute with Verizon' one should expect to be in the wrong. However, one can have sufficient information to situate oneself in a subclass, e.g.: a person with 99th percentile mathematical ability applying well-confirmed arithmetic, debating mathematically less-adept representatives whose opinions are not independent (all relying on the same computer display), where concession by the reps would involve admitting a corporate mistake and setting a precedent to cut numerous bills by 99%. Isn't identifying the optimum tradeoff between context-sensitivity and vulnerability to self-serving bias a largely empirical question, dependent on the distribution of rationality and the predictive power of such signals?
comment by Dr_Zen · 2007-06-01T05:10:20.000Z · LW(p) · GW(p)
Since the modesty argument requires rational correspondents, creationists are counted out! What I mean is that a creationist does not disagree with me on a matter of simple fact, in that his or her reasons for disagreeing are not themselves based on simple facts, as my reasons for disagreeing with him or her are. This makes our disagreement asymmetric. And Aumann's theorem can't apply, because even if the creationist knows what I know, he or she can't agree; and vice versa. But I think our disagreement is honest, all the same.
comment by denis_bider · 2007-12-27T05:25:08.000Z · LW(p) · GW(p)
In the Verizon case, George can apply the modesty argument and still come up with the conclusion that he is almost certainly right.
He needs to take into account two things: (1) what other people besides Verizon think about the distinction between .002 dollars and .002 cents, and (2) what is the likelihood that Verizon would admit the mistake even if they know there is one.
Admitting the mistake and refunding one customer might as well have the consequence of having to refund tens of thousands of customers and losing millions of dollars. Even if that's the upstanding and right thing to do for the company, each individual customer support rep will fear that their admitting the facts will lead to them being fired. People whose jobs are at stake will do their best to bluff a belief that .002 == 0.00002, even if they don't actually believe that's the case.
Replies from: MBlume
comment by MBlume · 2009-11-19T09:58:30.963Z · LW(p) · GW(p)
This hypothesis is testable. Unless you're a trained liar, your voice should be measurably different when you're explaining the obvious fact that .002 ==.00002 and when you're claiming that .002=.00002 while your internal monologue is going "shit, shit, he's right, isn't he? We really fucked this one up, we're going to lose thousands, and they're going to blame me." We have audio, so someone trained in such analysis should be able to give us an answer.
comment by aausch · 2011-02-03T04:06:19.744Z · LW(p) · GW(p)
Can a bayesian agent perfectly model arbitrary non-bayesian agents?
Given a bayesian model for someone else's non-bayesian decision system, won't a bayesian agent have a straight forward time of deciding which priors to update, and how?
Replies from: wedrifid
comment by wedrifid · 2011-02-03T06:48:28.852Z · LW(p) · GW(p)
Can a bayesian agent perfectly model arbitrary non-bayesian agents?
Sure, at least to the same extent that he or she could perfectly model any other bunch of atoms.
comment by beriukay · 2011-02-10T12:42:51.164Z · LW(p) · GW(p)
Unfortunately, the audio link is broken, unavailable in google cache, and not listed on the Way Back Machine. All my googling came to the same site, with the same dead link. YouTube only has ~3 minute clips. Does anyone else know where I can locate the whole debacle?
comment by BethMo · 2011-04-21T04:56:41.508Z · LW(p) · GW(p)
Greetings! I'm a relatively new reader, having spent a month or two working my way through the Sequences and following lots of links, and finally came across something interesting to me that no one else had yet commented on.
Eleizer wrote "Those who dream do not know they dream; but when you wake you know you are awake." No one picked out or disagreed with this statement.
This really surprised me. When I dream, if I bother to think about it I almost always know that I dream -- enough so that on the few occasions when I realize I was dreaming without knowing so, it's a surprising and memorable experience. (Though there may be selection bias here; I could have huge numbers of dreams where I don't know I'm dreaming, but I just don't remember them.)
I thought this was something that came with experience, maturity, and -- dare I say it? -- rationality. Now that I'm thinking about it in this context, I'm quite curious to hear whether this is true for most of the readership. I'm non-neurotypical in several ways; is this one of them?
Replies from: endoself, shokwave, None
comment by endoself · 2011-04-21T05:12:57.039Z · LW(p) · GW(p)
Welcome to LessWrong!
This is actually rather atypical. It reminds me of lucid dreaming, though I am unsure if it is exactly the same. I almost never remember my dreams. When I have remembered them, I almost never knew I was dreaming at the time. This post discusses such differences.
comment by shokwave · 2011-04-21T05:42:15.501Z · LW(p) · GW(p)
Endoself mentioned lucid dreaming; in the few lucid dreams I have had, I knew I was dreaming, but in the many normal dreams I did not know. In fact, non-lucid dreams feel extremely real, because I try to change what's happening the way I would in a lucid dream, and nothing changes - convincing me that it's real.
When I am awake I am very aware that I am awake and not dreaming; strangely that feeling is absent during dreams but somehow doesn't alert me that I'm dreaming.
Replies from: loqi
comment by loqi · 2011-04-21T06:21:36.651Z · LW(p) · GW(p)
In fact, non-lucid dreams feel extremely real, because I try to change what's happening the way I would in a lucid dream, and nothing changes - convincing me that it's real.
This has been my experience. And on several occasions I've become highly suspicious that I was dreaming, but unable to wake myself. The pinch is a lie, but it still hurts in the dream.
comment by [deleted] · 2011-04-21T06:45:41.797Z · LW(p) · GW(p)
Same with me. I always know if I'm dreaming, but I don't have special feeling of "I'm awake right now". I'm always slightly lucid in my dreams and can easily become fully lucid, but almost never do so except when I force myself out of a rare nightmare or rewind time a bit because I wanna fix something. (It's been this way since I was about 14. I don't remember any dreams from before that.)
In fact, I intentionally make sure to not be too lucid. I did some lucid dreaming experiments when I was 16 or so and after 2-3 weeks I started to fail reality checks while awake. My sanity went pretty much missing for a summer. I then concluded that my dreams aren't psychologically safe and stay as far away from them as I can.
comment by Arandur · 2011-08-13T02:00:41.845Z · LW(p) · GW(p)
Sometimes arrogance is the mark of truth. History is filled with the blood of people who died for their beliefs holding them against all counterargument, who were later vindicated.
Of course, history is also filled will the blood of people who died for erroneous beliefs.
Obviously, you should utilize the Modesty Argument iff your viewpoint is incorrect.
comment by DSimon · 2011-12-19T04:50:47.369Z · LW(p) · GW(p)
Excellent article! I have nothing to contribute but a bit of mechanical correction:
I regard rationality in its purest form as an individual thing - not because rationalists have only selfish interests, but because of the form of the only admissible question: "Is is actually true?"
Minor typo here; you mean "Is it actually true?"
If it's being really integrated, that is, rather than flushed down a black hole.
The phrase "flushed down a black hole" in the article above is a link, but it is unfortunately broken.
comment by Omegaile · 2012-01-15T23:00:41.843Z · LW(p) · GW(p)
Interesting... it reminded me of this comic: http://xkcd.com/690/
comment by fractalman · 2013-05-24T02:21:43.115Z · LW(p) · GW(p)
It looks to ME like the theorem also requires perfect self-knowledge, AND perfect other-knowledge...which just doesn't happen in finite systems. Yeah. Godel again.
comment by Neph · 2013-09-15T17:43:48.436Z · LW(p) · GW(p)
remember that Bayesian evidence never reaches 100%, thus making middle ground- upon hearing another rationalist's viewpoint, instead of not shifting (as you suggest) or shifting to average your estimate and theirs together (as AAT suggests) why not adjust your viewpoint based on how likely the other rationalist is to have assessed correctly? ie- you believe X is 90% likely to be true the other rationalist believes it's not true 90%. suppose this rationalist is very reliable, say in the neighborhood of 75% accurate, you should adjust your viewpoint down to X is 75% likely to be 10% likely to be true, and 25% likely to be 90% likely to be true (or around 30% likely, assuming I did my math right.) assume he's not very reliable, say a creationist talking about evolution. let's say 10%. you should adjust to X is 10% likely to be 10% likely and 90% likely to be 90% likely. (82%) ...of course this doesn't factor in your own fallibility.
comment by matteyas · 2014-10-20T08:35:34.382Z · LW(p) · GW(p)
If genunie Bayesians will always agree with each other once they've exchanged probability estimates, shouldn't we Bayesian wannabes do the same?
An example I read comes to mind (it's in dialogue form): "This is a very common error that's found throughout the world's teachings and religions," I continue. "They're often one hundred and eighty degrees removed from the truth. It's the belief that if you want to be Christ-like, then you should act more like Christ—as if the way to become something is by imitating it."
It comes with a fun example, portraying the absurdity and the potential dangers of the behavior: "Say I'm well fed and you're starving. You come to me and ask how you can be well fed. Well, I've noticed that every time I eat a good meal, I belch, so I tell you to belch because that means you're well fed. Totally backward, right? You're still starving, and now you're also off-gassing like a pig. And the worst part of the whole deal—pay attention to this trick—the worst part is that you've stopped looking for food. Your starvation is now assured."
comment by ldsrrs · 2015-12-17T21:24:21.454Z · LW(p) · GW(p)
I think that this is similar to the prisoner's dilemma. If A is in an argument with B over some issue, then a contract for them to each update their probability estimates to the average would benefit the group. But each side would have an incentive to cheat by not updating his/her probability estimates. Given A's priors the group would benefit more if only B updates (and vice versa). Unfortunately it would be very difficult to enforce a belief contract.
This does not require them to be selfish. Even if they were perfect altruists they would each believe that it would help both of them the most if only the other's probability estimate was changed.
comment by Bruno Mailly (bruno-mailly) · 2018-07-10T19:24:11.798Z · LW(p) · GW(p)
If people updated their belief towards those around them, then people with agendas would hammer loudly their forged belief at any opportunit... wait, isn't this EXACTLY what they are doing ?
comment by stripey7 · 2020-09-13T12:27:17.348Z · LW(p) · GW(p)
I can recall at least one occasion on which I momentarily doubted I was awake, simply because I saw something that seemed improbable. | 11,282 | 50,279 | {"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-2021-39 | latest | en | 0.938073 |
https://nl.mathworks.com/matlabcentral/answers/861085-why-a-function-can-t-get-itself-in-simulink-after-two-consecutive-differentiation-and-two-integrati?s_tid=prof_contriblnk | 1,718,837,386,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861853.72/warc/CC-MAIN-20240619220908-20240620010908-00216.warc.gz | 381,032,822 | 26,748 | # Why a function can't get itself in Simulink after two consecutive differentiation and two integration operations？
7 views (last 30 days)
liqun fang on 21 Jun 2021
Commented: liqun fang on 23 Jun 2021
I want to implement the functionality described in the following figure.
So I build the Simulink model shown in the below figure.
The results are shown in the below figure. Both results are wrong. Why????
Paul on 22 Jun 2021
For the top (continuous) branch, keep in mind that the Derivative block is a numerical approximation and it's output will be very dependent on the solver parameters. I was able to get the expected result by using a Fixed Step solver (ode2) with a Fixed step size of 0.002. Don't forget to the set the Initial condition of Integrator4 to 1. In general, using Derivative block is not recommended and using two in series with each other will likely be very problematic.
For the bottom (discrete) branch, change the Integrator method on both Discrete Time Integrators to Backward Euler. Place a Zero Order Hold block between the Sine Wave and the first Discrete Derivative and set its Sample time to what you want (I used 0.1). Then adjust the Intital Condition parameters on each Discrete Derivative and Discrete Time Integrator so that the output of each block has the desired value at T=0.
##### 3 CommentsShow 1 older commentHide 1 older comment
Paul on 23 Jun 2021
You ask a very important question that you must think about for any model and simulation, particularly for physical systems, that goes well beyond this particular case with Derivative blocks (which you should try to avoid by all means possible). After all, in many cases the model itself is an approximate representation of the actual physical system, and the the simulated solution is a numerical approximation of the solution of the governing equations in your model. So how do you know if you're getting "correct" results? Ideally, you have some idea of what "correct" will look like for the problem at hand so you at least have a qualitative idea the the result looks right. Do some sanity checks on things where you know what the answer should be. Bulid up your model gradually in modules and check and verify each as you go. Understand the trade-offs between various solvers. Experiment with solver parameters, like max step size, etc., and see the sensitivity in the results. Etc. Etc.
liqun fang on 23 Jun 2021
Thank you very much for your answer. My problem is solved. Your answer is very well.
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https://www.speedsolving.com/wiki/index.php/Sexy_Method | 1,553,205,073,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202572.7/warc/CC-MAIN-20190321213516-20190321235516-00475.warc.gz | 906,618,869 | 6,404 | # Sexy Method
Sexy method Information about the method Proposer(s): Conrad Rider Proposed: 2010 Alt Names: SAM (Single Alg Method),1234 method Variants: 8355, MirIS, Y-Move Method, Less is More No. Steps: 7 No. Algs: 1 Avg Moves: 100+ Purpose(s): Beginner Method
Sexy Method is a variant of the 8355 Method specifically designed for ease of learning, using a single algorithm: the sexy move.
## Steps
• Intuitively solve the Cross
• Intuitively solve 3 of the E-slice edges, leaving a keyhole.
• Use the sexy move to solve the corresponding 3 D-layer corners by passing them from the U-layer through the keyhole.
• Intuitively solve 3 U-layer edges.
• Solve the last 2 edges, using intuition or algorithms.
• Turn the cube over.
• Solve 3 more corners with the sexy move as before.
• Rotate the cube to put the last 2 corners at DF. Solve them with the sexy move as before. | 226 | 878 | {"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} | 2.953125 | 3 | CC-MAIN-2019-13 | latest | en | 0.858112 |
https://forum.allaboutcircuits.com/threads/needing-a-series-parallel-combination-resistor-circuit-calculator.5177/ | 1,611,839,114,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704843561.95/warc/CC-MAIN-20210128102756-20210128132756-00673.warc.gz | 330,017,136 | 19,902 | # Needing a series-parallel combination resistor circuit calculator
#### jasonbengiat
Joined Mar 13, 2007
1
I am making a game, and need 6 resistors for some leds. All of the leds are 20 mA. The circuits are shown below. Can anyone tell me what resistors I need? Can you also direct me to a web page where I can find a resistor calculator that will tell me what resistors I need in the future?
Thank you,
Jason
Circuit 1)
_____________________________________________________
| | | | | | |
| 1.85 volt 1.85 volt 1.85 volt 1.85 volt 1.85 volt 1.85 volt
| led led led led led led
| |_________|_______| |________|_______|
| | |
9 volt Resistor 1 Resistor 2
battery |_________________________|
| |
second |
9 volt |
battery |
|___________________________ |
Circuit 2)
_____________________________________________________
| | | | | | |
| 2.0 volt 2.0 volt 2.0 volt 2.0 volt 2.0 volt 2.0 volt
| led led led led led led
| |_________|_______| |________|_______|
| | |
9 volt Resistor 3 Resistor 4
battery |_________________________|
| |
second |
9 volt |
battery |
|___________________________ |
Circuit 3)
_____________________________________________________
| | | | | | |
| 2.2 volt 2.2 volt 2.2 volt 2.2 volt 2.2 volt 2.2 volt
| led led led led led led
| |_________|_______| |________|_______|
| | |
9 volt Resistor 5 Resistor 6
battery |_________________________|
| |
second |
9 volt |
battery |
|___________________________ |
#### beenthere
Joined Apr 20, 2004
15,819
Let me refer you to this part of the E-book -
Volume III - Semiconductors » DIODES AND RECTIFIERS » Special-purpose diodes.
The section on LED's should answer your questions. You also might want to refer to Volume 1 and how to work with series/paralled DC circuits.
The formula for you calculations is: R = E/I, where R is resistance, E is voltage and I is current. | 523 | 1,841 | {"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} | 3.328125 | 3 | CC-MAIN-2021-04 | longest | en | 0.744757 |
http://forum.allaboutcircuits.com/threads/anyone-knows-use-laplace-to-find-zt-and-it.22869/ | 1,484,785,458,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280410.21/warc/CC-MAIN-20170116095120-00375-ip-10-171-10-70.ec2.internal.warc.gz | 99,079,224 | 14,176 | # anyone knows use Laplace to find ZT and IT?
Discussion in 'Homework Help' started by kuangmao, Apr 29, 2009.
1. ### kuangmao Thread Starter New Member
Apr 29, 2009
3
0
anyone knows use Laplace to find ZT and IT?
View attachment it.bmp
Last edited: Apr 30, 2009
2. ### Ratch New Member
Mar 20, 2007
1,068
3
kuangmao,
Show us your attempt at a solution.
Ratch
3. ### kuangmao Thread Starter New Member
Apr 29, 2009
3
0
find the total impedance and the total current using laplace.
4. ### kuangmao Thread Starter New Member
Apr 29, 2009
3
0
and please show me the work steps
thanks so much guys.
5. ### Ratch New Member
Mar 20, 2007
1,068
3
kuangmao,
Your result Zt = (2*(3*s^2+14*s+7))/(s^2+5*s+2) is correct
Dividing Zt into voltage 6/s gives I(s) = (3*(s^2+5*s+2))/(s*(3*s^2+14*s+7))
Converting I(s) to partial fractions gives I(s)=(1/7)*(3*s+21)/(3*s^2+14*s+7)+6/(7*s)
From a table in Laplace transforms we find that s/(s^2-a^2)=(sinh(t))/a and s/(s^2-a^2)= cosh(t), so the inverse Laplace gives 6/7+(1/7)*exp(-(7/3)*t)*(cosh((2/3)*t*sqrt(7))+sqrt(7)*sinh((2/3)*t*sqrt(7)))
Ratch
6. ### Ratch New Member
Mar 20, 2007
1,068
3
kuangmao,
Enclosed in the attachment is the solution in the time domain in hyperbolic format already given and exponential format. Both are equivalent as proven by the identical plots also shown. You will notice that after a long time the current settles into a steady state value of 6/7 amps, which can be ascertained by an inspection of the schematic. This is also shown on the plots.
Ratch
File size:
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## Steph_Rawr352 3 years ago The kitchen in Meghan’s house is a square. The points P(6, 2) and Q(0, 6) represent the opposite vertices of her kitchen in a drawing on a coordinate grid. Which of these ordered pairs could be a third vertex? 1. (0, 1) 2. (1, 7) 3. (5, 6) 4. (5, 7)
• This Question is Closed
1. aminah.love18
hmmm hold on kpop thinking
2. aminah.love18
lol i think ill let experiment answer this one
3. Steph_Rawr352
nvm guys i got the answer its (0,1) thanks tho(:
4. experimentX
calculate distance ... where ever it's equal that is our answer
5. experimentX
no thats not the answer
6. Steph_Rawr352
oh its not ?
7. experimentX
4 is your answer .. just calculate thie distance using your distance formula ... square has equal sides
8. Steph_Rawr352
ohh ohk i see what yu mean.. yeah yur right.. thanks tho(:
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http://www.physics.udel.edu/~watson/phys345/examples/nodal-analysis.html | 1,547,696,126,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658681.7/warc/CC-MAIN-20190117020806-20190117042806-00528.warc.gz | 372,788,493 | 1,527 | ## PHYS345 Electricity and Electronics
Detailed Nodal Analysis Example
The same elements are used as the Detailed Simple Circuit Example, with the addition of one 9 V battery as shown.
Step 1: Replace any combination of resistors in series or parallel with their equivalent resistance. Here I simply use the result from the previous example.
Step 2: Identify nodes and label accordingly.
Step 3: Write the node equation. Here I suppress the "k" for each resistor; there is no problem as long as all resistances are written in the same units.
(V1 - 9)/5 + V1/6 + (V1 + 9)/12 = 0
Step 4: Solve the equation to obtain the value of the unknown.
Solution:
V1 : 7/3 V 2.33 V
This result may be used to find any of the quantities of interest. The current through each of the three resistors may now be calculated since the voltage difference across each has now been determined.
"http://www.physics.udel.edu/~watson/phys345/examples/nodal-analysis.html"
Last updated Sept. 7, 1999.
Copyright George Watson, Univ. of Delaware, 1999. | 258 | 1,035 | {"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} | 3.046875 | 3 | CC-MAIN-2019-04 | latest | en | 0.882576 |
https://classroom.thenational.academy/lessons/area-of-similar-shapes-74w32t | 1,620,386,431,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988775.80/warc/CC-MAIN-20210507090724-20210507120724-00561.warc.gz | 212,520,259 | 40,366 | # Area of similar shapes
In this lesson, you will work out areas of similar shapes.
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
## Question 5
Q1.Fill in the gap: in two similar shapes, all corresponding _____ are equal.
1/5
Q2.Triangles DEF and GHI are similar. What is the scale factor of enlargement if GHI is the object and DEF is the image?
2/5
Q3.Triangles DEF and GHI are similar. What is the scale factor of enlargement if DEF is the object and GHI is the image?
3/5
Q4.Are PQR and SPR similar?
4/5
Q5.Are triangles PQS and SQR similar?
5/5
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
## Question 5
Q1.Fill in the gap: in two similar shapes, all corresponding _____ are equal.
1/5
Q2.Triangles DEF and GHI are similar. What is the scale factor of enlargement if GHI is the object and DEF is the image?
2/5
Q3.Triangles DEF and GHI are similar. What is the scale factor of enlargement if DEF is the object and GHI is the image?
3/5
Q4.Are PQR and SPR similar?
4/5
Q5.Are triangles PQS and SQR similar?
5/5
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
Quiz:
# Area of similar shapes quiz
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
## Question 5
Q1.The two rectangles below are similar. What is the scale factor of enlargement if the object is on the left and the image on the right?
1/5
Q2.The two rectangles below are similar. What is the scale factor for the area if the object is on the left and the image on the right?
2/5
Q3.The two rectangles below are similar. What is the scale factor of enlargement if the object is on the right and the image on the left?
3/5
Q4.The two rectangles below are similar. What is the scale factor for the area if the object is on the right and the image on the left?
4/5
Q5.Fill in the gap: to find the area scale factor, you must ______ the scale factor of enlargement.
5/5
Quiz:
# Area of similar shapes quiz
Don’t worry if you get a question wrong! Forgetting is an important step in learning. We will recap next lesson.
## Question 5
Q1.The two rectangles below are similar. What is the scale factor of enlargement if the object is on the left and the image on the right?
1/5
Q2.The two rectangles below are similar. What is the scale factor for the area if the object is on the left and the image on the right?
2/5
Q3.The two rectangles below are similar. What is the scale factor of enlargement if the object is on the right and the image on the left?
3/5
Q4.The two rectangles below are similar. What is the scale factor for the area if the object is on the right and the image on the left?
4/5
Q5.Fill in the gap: to find the area scale factor, you must ______ the scale factor of enlargement.
5/5
# Lesson summary: Area of similar shapes
### Time to move!
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2. Prealgebra
3. george has a triangleshaped garden in his backyard he drew...
# Question: george has a triangleshaped garden in his backyard he drew...
###### Question details
George has a triangle-shaped garden in his backyard. He drew a model of this garden on a coordinate grid with vertices A(4, 2), B(2, 4), and C(6, 4). He wants to create another, similar-shaped garden, A′B′C′, by dilating triangle ABC by a scale factor of 0.5. What are the coordinates of triangle A′B′C′?
A.
A′(2, 2), B′(1, 2), C′(3, 2)
B.
A′(2, 1), B′(1, 2), C′(3, 2)
C.
A′(8, 4), B′(4, 8), C′(12, 8)
D.
A′(2, 1), B′(1, 2), C′(6, 4) | 242 | 629 | {"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 | 3 | CC-MAIN-2021-17 | latest | en | 0.920852 |
http://mathhelpforum.com/calculus/105074-help-needed-advanced-limits-3-a.html | 1,481,236,520,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542657.90/warc/CC-MAIN-20161202170902-00270-ip-10-31-129-80.ec2.internal.warc.gz | 165,952,213 | 9,643 | 1. ## Help Needed with Advanced Limits # 3
I was wondering if someone could help me with the problems within the picture:
I am in dire need of a step by step explanation as I answered this as -1 and was incorrect
2. couldn't f be a radical like this (-x+1)/(x-1) so that the horizontal asymptote would be y =-1 and so the derivative of that function would be 0 so the limit of the derivative would also be zero. I think I got. Well, I hope so. Can someone confirm this please
3. Originally Posted by Some1Godlier
couldn't f be a radical like this (-x+1)/(x-1) so that the horizontal asymptote would be y =-1 and so the derivative of that function would be 0 so the limit of the derivative would also be zero. I think I got. Well, I hope so. Can someone confirm this please
You are right - more or less:
1. The example you gave is a constant function with a hole at x = 1. Better use $f(x)=\dfrac{1-x}x$ (as an example of course!)
2. If an asymptote y = -1 exists then the function can be replaced by f(x) = -1 for very large values of x.
3. If f(x) = -1 then f'(x) = 0 | 295 | 1,075 | {"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": 1, "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} | 3.59375 | 4 | CC-MAIN-2016-50 | longest | en | 0.960321 |
https://dottrusty.com/27-f-to-c/ | 1,719,140,664,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862466.81/warc/CC-MAIN-20240623100101-20240623130101-00049.warc.gz | 185,273,337 | 14,742 | # What is 27 °F in °C? (27 Fahrenheit to Celsius)
27 F to C: What It Means
27 degrees Fahrenheit is equivalent to -2.77778 degrees Celsius. This means that if it is 27 degrees Fahrenheit outside, it is about the same temperature as if it were -2.77778 degrees Celsius outside. In other words, it is quite cold!
The F to C formula is (F − 32) × 5/9 = C. When we enter 27 for F in the formula, we get (27 − 32) × 5/9 = C.
When we convert degrees Fahrenheit to degrees Celsius, we are converting a temperature measurement from the American system to the metric system.
In North America, we use Fahrenheit to measure temperature, while most of the rest of the world uses Celsius. It can sometimes be confusing to convert between the two, but with a little practice it becomes easier.
To convert from Fahrenheit to Celsius, all you have to do is subtract 32 from the Fahrenheit temperature and then divide by 5/9. So for example, if it is 100 degrees Fahrenheit outside, the temperature in Celsius would be (100 − 32) / 5/9 = 36.667 degrees Celsius.
27 – 32 = -5
-5 x 5 = -25
-25 / 9 = -2.77778
Converting from Celsius to Fahrenheit is just a matter of reversing this process. To do this, you multiply the Celsius temperature by 9/5 and then add 32. For example, if it is 25 degrees Celsius outside, the temperature in Fahrenheit would be (25 × 9/5) + 32 = 97.222 degrees Fahrenheit.
It’s important to note that both of these equations are only accurate when dealing with temperatures that are within the range of 0-100 degrees Fahrenheit or -17.78-37.78 degrees Celsius. If the temperature is either above or below this range, you will need to use a different equation or use a calculator to get an accurate reading.
There are a few reasons why you might need to know how to convert between these two temperature systems. For example, if you are traveling in a country that uses the metric system instead of the American system, you will need to be able to convert temperatures so that you can dress appropriately for the weather. Additionally, many scientific studies use the metric system, so if you are doing research in this area, you will also need to be able to convert temperatures.
Overall, being able to convert between Fahrenheit and Celsius is a useful skill to have. When you know how to do it, it is simple and straightforward. With a little practice, you will be able to do it in your head without any problem!
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# Cardinality
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Title: Cardinality Author: World Heritage Encyclopedia Language: English Subject: Collection: Publisher: World Heritage Encyclopedia Publication Date:
### Cardinality
In mathematics, the cardinality of a set is a measure of the "number of elements of the set". For example, the set A = {2, 4, 6} contains 3 elements, and therefore A has a cardinality of 3. There are two approaches to cardinality – one which compares sets directly using bijections and injections, and another which uses cardinal numbers.[1] The cardinality of a set is also called its size, when no confusion with other notions of size[2] is possible.
The cardinality of a set A is usually denoted | A |, with a vertical bar on each side; this is the same notation as absolute value and the meaning depends on context. Alternatively, the cardinality of a set A may be denoted by n(A), A, card(A), or # A.
## Comparing sets
While the cardinality of a finite set is just the number of its elements, extending the notion to infinite sets usually starts with defining the notion of comparison of arbitrary (in particular infinite) sets.
Bijective function from N to E. Although E is a proper subset of N, both sets have the same cardinality.
### Definition 1: | A | = | B |
Two sets A and B have the same cardinality if there exists a bijection, that is, an injective and surjective function, from A to B. Such sets are said to be equipotent, equipollent, or equinumerous.
For example, the set E = {0, 2, 4, 6, ...} of non-negative even numbers has the same cardinality as the set N = {0, 1, 2, 3, ...} of natural numbers, since the function f(n) = 2n is a bijection from N to E.
### Definition 2: | A | ≤ | B |
A has cardinality less than or equal to the cardinality of B if there exists an injective function from A into B.
### Definition 3: | A | < | B |
A has cardinality strictly less than the cardinality of B if there is an injective function, but no bijective function, from A to B.
For example, the set N of all natural numbers has cardinality strictly less than the cardinality of the set R of all real numbers , because the inclusion map i : NR is injective, but it can be shown that there does not exist a bijective function from N to R (see Cantor's diagonal argument or Cantor's first uncountability proof).
If | A | ≤ | B | and | B | ≤ | A | then | A | = | B | (Cantor–Bernstein–Schroeder theorem). The axiom of choice is equivalent to the statement that | A | ≤ | B | or | B | ≤ | A | for every A,B.[3][4]
## Cardinal numbers
Above, "cardinality" was defined functionally. That is, the "cardinality" of a set was not defined as a specific object itself. However, such an object can be defined as follows.
The relation of having the same cardinality is called equinumerosity, and this is an equivalence relation on the class (set theory) of all sets. The equivalence class of a set A under this relation then consists of all those sets which have the same cardinality as A. There are two ways to define the "cardinality of a set":
1. The cardinality of a set A is defined as its equivalence class under equinumerosity.
2. A representative set is designated for each equivalence class. The most common choice is the initial ordinal in that class. This is usually taken as the definition of cardinal number in axiomatic set theory.
Assuming AC, the cardinalities of the infinite sets are denoted
\aleph_0 < \aleph_1 < \aleph_2 < \ldots .
For each ordinal \alpha, \aleph_{\alpha + 1} is the least cardinal number greater than \aleph_\alpha.
The cardinality of the natural numbers is denoted aleph-null (\aleph_0), while the cardinality of the real numbers is denoted by "\mathfrak c" (a lowercase fraktur script "c"), and is also referred to as the cardinality of the continuum. Cantor showed, using the diagonal argument, that {\mathfrak c} >\aleph_0. We can show that \mathfrak c = 2^{\aleph_0}, this also being the cardinality of the set of all subsets of the natural numbers. The continuum hypothesis says that \aleph_1 = 2^{\aleph_0}, i.e. 2^{\aleph_0} is the smallest cardinal number bigger than \aleph_0, i.e. there is no set whose cardinality is strictly between that of the integers and that of the real numbers. The continuum hypothesis is independent of ZFC, a standard axiomatization of set theory; that is, it is impossible to prove the continuum hypothesis or its negation from ZFC (provided ZFC is consistent). See below for more details on the cardinality of the continuum.[5][6][7]
## Finite, countable and uncountable sets
If the axiom of choice holds, the law of trichotomy holds for cardinality. Thus we can make the following definitions:
• Any set X with cardinality less than that of the natural numbers, or | X | < | N |, is said to be a finite set.
• Any set X that has the same cardinality as the set of the natural numbers, or | X | = | N | = \aleph_0, is said to be a countably infinite set.
• Any set X with cardinality greater than that of the natural numbers, or | X | > | N |, for example | R | = \mathfrak c > | N |, is said to be uncountable.
## Infinite sets
Our intuition gained from Gottlob Frege, Richard Dedekind and others rejected the view of Galileo (which derived from Euclid) that the whole cannot be the same size as the part. One example of this is Hilbert's paradox of the Grand Hotel. Indeed, Dedekind defined an infinite set as one that can be placed into a one-to-one correspondence with a strict subset (that is, having the same size in Cantor's sense); this notion of infinity is called Dedekind infinite. Cantor introduced the cardinal numbers, and showed that (according to his bijection-based definition of size) some infinite sets are greater than others. The smallest infinite cardinality is that of the natural numbers (\aleph_0).
### Cardinality of the continuum
One of Cantor's most important results was that the cardinality of the continuum (\mathfrak{c}) is greater than that of the natural numbers (\aleph_0); that is, there are more real numbers R than whole numbers N. Namely, Cantor showed that
\mathfrak{c} = 2^{\aleph_0} > {\aleph_0}
(see Cantor's diagonal argument or Cantor's first uncountability proof).
The continuum hypothesis states that there is no cardinal number between the cardinality of the reals and the cardinality of the natural numbers, that is,
\mathfrak{c} = \aleph_1 = \beth_1
(see Beth one).
However, this hypothesis can neither be proved nor disproved within the widely accepted ZFC axiomatic set theory, if ZFC is consistent.
Cardinal arithmetic can be used to show not only that the number of points in a real number line is equal to the number of points in any segment of that line, but that this is equal to the number of points on a plane and, indeed, in any finite-dimensional space. These results are highly counterintuitive, because they imply that there exist proper subsets and proper supersets of an infinite set S that have the same size as S, although S contains elements that do not belong to its subsets, and the supersets of S contain elements that are not included in it.
The first of these results is apparent by considering, for instance, the tangent function, which provides a one-to-one correspondence between the interval (−½π, ½π) and R (see also Hilbert's paradox of the Grand Hotel).
The second result was first demonstrated by Cantor in 1878, but it became more apparent in 1890, when Giuseppe Peano introduced the space-filling curves, curved lines that twist and turn enough to fill the whole of any square, or cube, or hypercube, or finite-dimensional space. These curves are not a direct proof that a line has the same number of points as a finite-dimensional space, but they can be used to obtain such a proof.
Cantor also showed that sets with cardinality strictly greater than \mathfrak c exist (see his generalized diagonal argument and theorem). They include, for instance:
• the set of all subsets of R, i.e., the power set of R, written P(R) or 2R
• the set RR of all functions from R to R
Both have cardinality
2^\mathfrak {c} = \beth_2 > \mathfrak c
(see Beth two).
The cardinal equalities \mathfrak{c}^2 = \mathfrak{c}, \mathfrak c^{\aleph_0} = \mathfrak c, and \mathfrak c ^{\mathfrak c} = 2^{\mathfrak c} can be demonstrated using cardinal arithmetic:
\mathfrak{c}^2 = \left(2^{\aleph_0}\right)^2 = 2^{2\times{\aleph_0}} = 2^{\aleph_0} = \mathfrak{c},
\mathfrak c^{\aleph_0} = \left(2^{\aleph_0}\right)^{\aleph_0} = 2^ = 2^{\aleph_0} = \mathfrak{c},
\mathfrak c ^{\mathfrak c} = \left(2^{\aleph_0}\right)^{\mathfrak c} = 2^{\mathfrak c\times\aleph_0} = 2^{\mathfrak c}.
## Examples and properties
• If X = {a, b, c} and Y = {apples, oranges, peaches}, then | X | = | Y | because { (a, apples), (b, oranges), (c, peaches)} is a bijection between the sets X and Y. The cardinality of each of X and Y is 3.
• If | X | < | Y |, then there exists Z such that | X | = | Z | and ZY.
• If | X | ≤ | Y | and | Y | ≤ | X |, then | X | = | Y |. This holds even for infinite cardinals, and is known as Cantor–Bernstein–Schroeder theorem.
• Sets with cardinality of the continuum
## Union and intersection
If A and B are disjoint sets, then
\left\vert A \cup B \right\vert = \left\vert A \right\vert + \left\vert B \right\vert.
From this, one can show that in general the cardinalities of unions and intersections are related by[8]
\left\vert C \cup D \right\vert + \left\vert C \cap D \right\vert = \left\vert C \right\vert + \left\vert D \right\vert \,.
## References
1. ^ Weisstein, Eric W., "Cardinal Number", MathWorld.
2. ^ Such as length and area in geometry. – A line of finite length is a set of points that has infinite cardinality.
3. ^
4. ^ - Original edition (1914)
5. ^
6. ^
7. ^
8. ^ Applied Abstract Algebra, K.H. Kim, F.W. Roush, Ellis Horwood Series, 1983, ISBN 0-85312-612-7 (student edition), ISBN 0-85312-563-5 (library edition)
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https://mathematica.stackexchange.com/questions/tagged/list-manipulation?tab=active&page=4 | 1,632,347,282,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057388.12/warc/CC-MAIN-20210922193630-20210922223630-00373.warc.gz | 438,040,180 | 44,171 | # Questions tagged [list-manipulation]
Questions on the manipulation of List objects in Mathematica, and the functions used for these manipulations.
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543 views
### How to generate all possible functions combinations
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206 views
### List manipulation: all possible sublists
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### Palindrome by deleting a character
I need to find a list of strings with input string length n for whom are palindrome by deleting a character from it . The strings are stored inside function ... | 2,093 | 7,524 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2021-39 | latest | en | 0.87625 |
https://crypto.stackexchange.com/questions/88988/where-to-apply-montgomery-multiplication-in-gf2n | 1,720,895,680,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514512.50/warc/CC-MAIN-20240713181918-20240713211918-00829.warc.gz | 172,750,219 | 44,654 | # Where to apply Montgomery Multiplication in GF(2^n)
I'm optimizing a Reed Solomon decoding library for several polynomials in $$\operatorname{GF}(2^k)$$, $$k\in\{8,10,12\}$$.
Reading about the Montgomery Multiplication from Çetin K. Koç & Tolga Acar's Montgomery Multiplication in $$\operatorname{GF}(2^k)$$ (in Designs, Codes and Cryptography, 1998), I'm wondering if and how the method could help in polynomial reduction after carry free multiplication.
1. $$t(x) = a(x)b(x)$$ is a full length (2k-1) polynomial from carryless multiplication
2. $$u(x) = t(x)n'(x) \bmod r(x)$$ returns lower half of the $$\operatorname{GF}$$ multiplication
3. $$(u(x)n(x) + t(x)) / r(x)$$ corresponds to the higher half of one $$\operatorname{GF}$$ multiplication and one XOR
Instead of calculating $$a(x)b(x) \bmod n(x)$$, I have now $$a(x)b(x)r^{-1}(x) \bmod n(x)$$, and there's just one place where I can imagine using it. In inversion by modular exponentiation, where instead of calculating $$a(x)^{254}$$, I can calculate $$(a(x)r^{-1}(x))^{254}$$ and multiply that with a constant $$r^{254}(x)$$.
But can it be used e.g. in polynomial evaluation (used e.g. in syndrome calculation) or for modular reduction after $$\operatorname{GF}$$ multiplication?
The usual motivation for using Montgomery multiplication is that it significantly reduces the cost of modular reduction by changing the representation of elements. In the Montgomery ring the polynomial $$A(x)$$ is instead represented by $$a(x)=A(x)r(x)\pmod{n(x)}$$ where $$r(x)$$ is a fixed element of the same degree as $$n(x)$$, but for which division is very efficient (e.g. $$x^k$$). If likewise we have $$B(x)$$ represented by $$b(x)=B(x)r(x)$$ then $$a(x)b(x)r^{-1}(x)\equiv A(x)B(x)r(x)\pmod{n(x)}$$ which is the representation that we would choose for $$A(x)B(x)$$ and so your algorithm is way of implementing modular multiplication of polynomials in this representation without having to do a modular reduction. Addition and subtraction do not need new algorithms. If you have an algorithm involving many modular multiplies, it can be worthwhile converting to Montgomery ring representation, using Montgomery arithmetic and then converting back.
To convert $$a(x)$$ back to $$A(x)$$ we need to multiply by $$r^{-1}(x)$$, so it can be useful to precompute and save $$r^{-1}(x)\pmod{n(x)}$$. I don't know of any way to avoid modular reduction in this final conversion.
Evaluating using the Montgomery representation is possible, but requires a little extra work: to evaluate $$A(x)\pmod{n(x)}$$ given its representation $$a(x)$$ we have to evaluate $$a(x)\pmod{n(x)}$$ and $$r^{-1}(x)\pmod{n(x)}$$ and multiply the two answers together. However, if many evaluations are needed for the same input $$m$$, the value $$r^{-1}(m)$$ can be reused and the overhead is not as great.
• You are using the Montgomery Multiplication very ineffectively! $MonPro(a(x),B(x))$ is enough to get $A(x)B(x) \bmod P(x)$ Commented Mar 23, 2021 at 15:34
• @kelalaka True, but this assumes that I have $B(x)$ and want $A(x)B(x)$ in the regular rather than Montgomery representation. If $b(x)$ is itself the result of a Montgomery calculation, then I can trivially access $b(x)$ but cannot access $B(x)$ non-trivially. Likewise if $A(x)B(x)$ is only of interest as an intermediate value in the calculation, then I would want it in Montgomery representation rather than regular representation. Commented Mar 23, 2021 at 15:42
• That is the point of your second paragraph. If you don't have it then you need to convert only one, not two of them; $MonPro(a(x),MontPro(b(x),1))$. Actually, Montgomery is an effective generic method for $GF(2^k)$ and usually, we choose polynomials with low weight, then the reduction is a few polynomial additions instead of Montgomery residue ( see AES polynomials). This answer, actually, doesn't answer the real question of when to use Montgomery Modular Multiplication or not. Commented Mar 23, 2021 at 15:53
• Hmm. I appear to have worded this confusingly. I only talk about representation rather than conversion for the early part. I don't mention conversion until until the bit " If you have an algorithm involving many modular multiplies, it can be worthwhile converting to Montgomery ring representation, using Montgomery arithmetic and then converting back." which was intended to answer the question of when to use Montgomery multiplication. I'll consider how to rephrase this. Commented Mar 23, 2021 at 16:01
Where to apply Montgomery Multiplication in $$GF(2^n)$$?
This answer really depends on how you constructed the binary extension field $$GF(2^n)$$. If the irreducible polynomial is trinomial or pentanomial then the reduction is already efficient!
Modular Multiplication with trinomials
Let $$GF(2^n)$$ is constructed with the trinomial $$P(x)=x^m+x^n+1$$ where $$\alpha$$ is the root of this polynomial with $$1.
Now you want to calcualte $$C'(x) = C(x) \bmod P(x)$$ where
$$C(x) = A(x)\cdot B(x) = \left( \sum_{i=0}^{m-1} a_i x^i \right)\ \cdot \left( \sum_{i=0}^{m-1} b_i x^i \right)$$
To reduce the multiplication cost use Karatsuba-Ofman multiplier. This may require some tuning since it has recursive nature and diving to the end may not be the best choice.
Now, once you had a $$C(x) = \sum_{i=0}^{2m-2} c_i x^i$$ you need reduction. We can use the $$P(\alpha)=0$$, and write
\begin{align} \alpha^m &= 1 + \alpha^n\\ \alpha^{m+1} &= \alpha + \alpha^{n+1}\\ \vdots & \quad \vdots\\ \alpha^{2m-3} &= \alpha^{m-3} + \alpha^{m+n-3}\\ \alpha^{2m-2} &= \alpha^{m-2} + \alpha^{m+n-2}\\ \end{align}
By using this we can say
\begin{align} c'_0 &= c_0 + c_m\\ c'_1 &= c_1 + c_{m+1}\\ \vdots \;\;& \quad \vdots\\ c'_{n-1} &= c_{n-1} + c_{m+n-1} + c_{m+n-1}\\ c'_{n} &= c_{n} + c_{m+n} + c_{m+n}\\ c'_{n+1} &= c_{n+1} + c_{m+n+1} +c_{m+n+1}\\ \vdots \;\;& \quad \vdots\\ c'_{m-2} &= c_{m-2} + c_{2m-2} + c_{2m-n-2}\\ c'_{m-1} &= c_{m-1} + c_{2m-n-1}\\ c'_{m} &= c_{m-2} + c_{2m-n}\\ \vdots \;\;& \quad \vdots\\ c'_{m+n-3} &= c_{m+n-3} + c_{2m-3}\\ c'_{m+n-2} &= c_{m+n-2} + c_{2m-2}\\ \end{align}
If one look at the above one should notice those;
1. The reduction is not complete since there are still $$m-n$$ to reduce.
2. The operations are just x-or and mapping.
3. The mapping is actually a loop
The cost of multiplication is just $$m^2$$ and roughly at most $$2m$$ additional x-or for the reduction!
When we turned back to Montgomery modular multiplication is has cost $$2m^2$$ multiplications and $$2m^2-3m-1$$ x-or operations.
Modular Multiplication with arbitrary irreducible polynomial
When you have an arbitrary irreducible polynomial for the binary extension field, the above table is no longer effective. Then you can turn back to Montgomery.
Note that the Montgomery Modular multiplication can also use the Karatsuba-Ofman to reduce the cost of multiplication that we did not take into account here!.
Some notes:
1. See how How does Montgomery reduction work? for the details of Montgomery (Integer case, almost same as polynomials)
2. If you need one modular multiplication then you don't need to turn both of them into their Montgomery residue representation. You can achieve by; $$MonPro(a(x),MontPro(b(x),1))$$ You can also use Montgomery in modular square-and-multiply.
3. Usually we select the irreducible polynomial during the design. There are already a Table of Low-Weight Binary Irreducible Polynomials by HP and methods Finding irreducible polynomials over GF(2) with the fewest terms
4. Inversion with power and modulus is not an effective method. There is already Itoh-Tsujii algorithm that uses $$t \ll m$$ multiplications and $$m-1$$ squaring. This algorithm is already beating the Extended-GCD.
• I'm afraid I have no choice over the polynomials. Of course one question is if all the operations I'm pursuing are isomorphic, thus if an Reed Solomon encoded message can be transformed back and forth between fields of different irreducible polynomials. Commented Mar 24, 2021 at 6:44
The key to Montgomery Multiplication is the prescaling of one or two of the inputs by $$r^k$$ as stated in this answer by kelalaka to How Does Montgomery Reduction Work.
In the case of polynomial evaluation of
$$p = a_0 + a_1b^1 + a_2b^2 + ... + a_mb^m \mod n$$
by Horner's scheme:
$$p = (((a_m * B) + a_{m-1}) * B + ...) * B + a_0$$ $$= (((a_m b ) + a_{m-1}) b + ...)b +a_0 \mod n$$
we can use the Montgomery representation $$B$$ for $$b$$ and where $$*$$ stands for the Montgomery Multiplication.
It still seems, that Montgomery Multiplication does not beat partial reduction, where each step in Horner's scheme is carried with $$2k-1$$ bits:
$$p_n = a_n$$ $$p_{j-1} = (p_j \mod r^k) b + (p_j / r^k) B + a_{j-1}$$ $$p = p_0 \mod n(x)$$
The first term calculates full sized polynomial (or integer) multiplication from the least signicant $$k$$ bits of $$p_j$$, the other term partially reduces the top $$k-1$$ bits using again the full length polynomial multiplication. Thus the polynomial reduction is postponed as the last step of the evaluation. | 2,567 | 9,073 | {"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": 67, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-30 | latest | en | 0.85198 |
http://www.cde.state.co.us/ContentAreas/ContentConnections/3gr_TimeDuration.asp | 1,513,594,180,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948615810.90/warc/CC-MAIN-20171218102808-20171218124808-00257.warc.gz | 322,612,718 | 9,783 | School & District Info
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General Program Info
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* * CDE will be closed on Monday, Dec. 25 and Tuesday, Dec. 26 for the Christmas holiday. * *
# Third Grade Content Connection Sample: Time and Duration
Time and duration are the sequential structures used to describe order and intervals using words or numbers. Third graders use the concept of time and duration to make connections between past, present, and future events building on earlier concepts of order and progression. In addition, third graders are learning about elapsed time in order to organize their lives and develop the basis for understanding cause and effect relationships. Time and duration are fundamental concepts for learning as well as managing the responsibilities and demands of life.
Time and Duration can connect 7 of 10 content areas as detailed below.
## Drama and Theatre Arts
In drama and theatre arts, time and duration is about understanding the appropriate pacing and timing of a work. Walking through lines versus presenting a dramatic portrayal causes variation in the timing and pacing of a dramatic piece. Dramatic works may include events that take place in real time or cover events that happen over a matter days, weeks, months, or years. Understanding how to create scenes that use the story’s timeline is fundamental in dramatic storytelling.
## Mathematics
In mathematics, time is a quantifiable concept which allows for the calculation of elapsed time or duration. For example, duration problems can relate to the amount of time it takes to read a book or go to school. Third graders solve problems involving elapsed time in minutes.
## Music
In music, time and duration are found in the elements of time signatures, tempo and meter. Creating short, four measure phrases establishes a beginning level knowledge of basic music composition. Music can incorporate contrasting segments such as using fast or lively time for one segment and slow or somber time for another to express varying emotions or messages. Understanding the use of time and duration within music enables third graders to identify the mood or message music is trying to convey.
In reading, writing, and communicating, time and duration is utilized to see logical connections in text through cause/effect, time, sequence, and comparison in areas such as historical events, scientific concepts, and technical procedures. Students use time and temporal relations to organize ideas and points of information sequentially. In communication, students adjust speech rate and intonation.
## Science
In science, time and duration is visible in the events of nature. The specific timing and duration of events can be observed and studied in order to better understand how the world works. Different objects and organisms operate at different time scales. For example, the life cycle of an insect might be days, the life cycle for a tree can be decades, the natural cycle for a forest might take centuries, and the natural cycle for rocks can be eons.
## Social Studies
In social studies, time and duration can be defined as a way to sequence events while considering the amount of time passed. Historians use chronology to sequence important events that may have impacted the development of a community. Geographers use time and duration to examine how regions develop and change over time.
## Visual Arts
In visual arts, time and duration involves creating and connecting works of art that depict various historical eras and cultural contexts. Connecting to historical eras requires an examination of similarities and differences between past, present and future contexts within which a piece of art is created. Using a piece of art created in a different time to interpret social and cultural norms aids learners in understanding the lasting influences from past societies. | 745 | 3,974 | {"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-2017-51 | longest | en | 0.918205 |
https://www.lmfdb.org/ModularForm/GL2/TotallyReal/4.4.8789.1/holomorphic/4.4.8789.1-7.1-b | 1,582,151,693,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144429.5/warc/CC-MAIN-20200219214816-20200220004816-00536.warc.gz | 827,483,683 | 5,576 | # Properties
Base field 4.4.8789.1 Weight [2, 2, 2, 2] Level norm 7 Level $[7, 7, w - 1]$ Label 4.4.8789.1-7.1-b Dimension 2 CM no Base change no
# Related objects
• L-function not available
## Base field 4.4.8789.1
Generator $$w$$, with minimal polynomial $$x^{4} - x^{3} - 6x^{2} - 2x + 1$$; narrow class number $$1$$ and class number $$1$$.
## Form
Weight [2, 2, 2, 2] Level $[7, 7, w - 1]$ Label 4.4.8789.1-7.1-b Dimension 2 Is CM no Is base change no Parent newspace dimension 3
## Hecke eigenvalues ($q$-expansion)
The Hecke eigenvalue field is $\Q(e)$ where $e$ is a root of the defining polynomial:
$$x^{2} - 8$$
Norm Prime Eigenvalue
5 $[5, 5, -w^{3} + 2w^{2} + 3w]$ $\phantom{-}e$
7 $[7, 7, w - 1]$ $-1$
11 $[11, 11, -w^{3} + 2w^{2} + 4w]$ $-e - 2$
13 $[13, 13, -2w^{3} + 3w^{2} + 10w - 2]$ $-e + 2$
16 $[16, 2, 2]$ $-1$
17 $[17, 17, -w^{3} + 2w^{2} + 5w - 3]$ $-2e + 2$
17 $[17, 17, -w^{3} + w^{2} + 5w]$ $-2$
17 $[17, 17, -w^{2} + 2w + 1]$ $\phantom{-}e + 4$
19 $[19, 19, w^{2} - w - 2]$ $\phantom{-}2$
29 $[29, 29, w^{3} - 2w^{2} - 5w]$ $\phantom{-}4$
29 $[29, 29, w^{2} - w - 3]$ $\phantom{-}2$
31 $[31, 31, -w^{3} + 2w^{2} + 3w - 2]$ $\phantom{-}e - 6$
43 $[43, 43, 2w^{3} - 3w^{2} - 11w]$ $\phantom{-}e$
47 $[47, 47, w^{3} - 7w - 4]$ $-2e + 4$
53 $[53, 53, -2w^{3} + 3w^{2} + 9w - 1]$ $\phantom{-}2e + 6$
61 $[61, 61, -w - 3]$ $\phantom{-}2e - 2$
73 $[73, 73, -w^{3} + 2w^{2} + 3w - 3]$ $\phantom{-}3e + 4$
73 $[73, 73, w^{3} - w^{2} - 7w - 1]$ $-2e + 6$
81 $[81, 3, -3]$ $\phantom{-}e + 6$
83 $[83, 83, -2w^{3} + 3w^{2} + 9w + 1]$ $-2e$
Display number of eigenvalues
## Atkin-Lehner eigenvalues
Norm Prime Eigenvalue
7 $[7, 7, w - 1]$ $1$ | 884 | 1,669 | {"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} | 2.828125 | 3 | CC-MAIN-2020-10 | latest | en | 0.290287 |
https://quizlet.com/185427015/pythagorean-theorem-flash-cards/ | 1,516,659,213,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891543.65/warc/CC-MAIN-20180122213051-20180122233051-00647.warc.gz | 769,604,996 | 37,923 | 16 terms
# Pythagorean Theorem
###### PLAY
13 cm
What is the length of the hypotenuse of a right triangle with legs that measure 5 cm and 12 cm?
8 cm
Given that the hypotenuse of a right triangle is 10 cm and one leg measures 6 cm, what is the length of the other leg?
20 cm
The diagonal of a rectangle is 25 cm. The width is 15 cm. What is the length?
25 cm
A rectangle measures 24 cm by 7 cm. What is the length of its diagonal?
17 cm
A right triangle has legs that measure 8 cm and 15 cm. What is the length of the third side?
24 cm
A right triangle has a leg that measures 10 cm. Its hypotenuse is 26 cm. What is the length of the missing leg?
5√15 cm
An isosceles triangle has congruent sides that measure 20 cm. Its base is 10 cm. What is its height?
8√3 cm
An equilateral triangle has sides that measure 16 cm. What is its height?
2√33 cm
The length of a rectangle is 8 cm. Its diagonal measures 14 cm. What is its width?
2√58 cm
Two snails begin at the same place in the garden. One snail travels six cm north. Another travels fourteen cm west. How far apart will they be?
2√89 cm
Find the value of x.
4√7 cm
Find the value of x.
9 cm
Find the value of x.
Obtuse
Classify the triangle as acute, obtuse, or right.
Acute
Classify the triangle as acute, obtuse, or right.
Right
Classify the triangle as acute, obtuse, or right. | 382 | 1,334 | {"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 | 4 | CC-MAIN-2018-05 | latest | en | 0.92471 |
https://quizlet.com/6593080/multiply-flash-cards/ | 1,553,525,272,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203991.44/warc/CC-MAIN-20190325133117-20190325155117-00044.warc.gz | 597,824,180 | 57,506 | 19 terms
# Multiply
#### Terms in this set (...)
( chéng ) multiply
1 × 1 = 1
2 × 2 = 4
4 × 4 = 16
10 × 4 = 40
11 × 7 = 77
20 × 9 = 180
30 × 21 = 630
70 × 20 = 1400
100 × 14 = 1400
200 × 15 = 3000
300 × 40 = 12000
500 × 60 = 30000
900 × 80=72000
1000 × 20 = 20000
2000 ×1=2000
5000 × 120 = 600000
9000 × 100=900000
10000 × 1300 = 13000000 | 178 | 359 | {"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-2019-13 | longest | en | 0.485368 |
https://planetmath.org/semiprimitivering | 1,675,074,964,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499816.79/warc/CC-MAIN-20230130101912-20230130131912-00496.warc.gz | 474,709,279 | 3,371 | # semiprimitive ring
A ring is said to be semiprimitive if its Jacobson radical is the zero ideal.
Any simple ring is automatically semiprimitive.
A finite direct product of matrix rings over division rings can be shown to be semiprimitive and both left and right Artinian.
The Artin-Wedderburn Theorem (http://planetmath.org/WedderburnArtinTheorem) states that any semiprimitive ring which is left or right Artinian is isomorphic to a finite direct product of matrix rings over division rings.
Note: The semiprimitive condition is sometimes also referred to as a semisimple, Jacobson semisimple, or J-semisimple. Furthermore, when either of the last two names are used, the adjective ’semisimple’ is frequently intended to refer to a ring that is semiprimitive and Artinian (see the entry on semisimple rings (http://planetmath.org/SemisimpleRing2)).
Title semiprimitive ring Canonical name SemiprimitiveRing Date of creation 2013-03-22 12:36:14 Last modified on 2013-03-22 12:36:14 Owner yark (2760) Last modified by yark (2760) Numerical id 20 Author yark (2760) Entry type Definition Classification msc 16N20 Synonym semisimple ring Synonym Jacobson semisimple ring Synonym J-semisimple ring Synonym semi-primitive ring Synonym semi-simple ring Synonym Jacobson semi-simple ring Synonym J-semi-simple ring Related topic SemisimpleRing2 Related topic WedderburnArtinTheorem Defines semiprimitivity Defines semiprimitive Defines semisimple Defines Jacobson semisimple Defines J-semisimple Defines semi-primitivity Defines semi-primitive Defines semi-simple Defines Jacobson semi-simple Defines J-semi-simple | 405 | 1,616 | {"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-2023-06 | latest | en | 0.718995 |
http://math.stackexchange.com/questions/tagged/functions+dynamical-systems | 1,406,192,049,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997888216.78/warc/CC-MAIN-20140722025808-00054-ip-10-33-131-23.ec2.internal.warc.gz | 240,343,340 | 13,221 | # Tagged Questions
143 views
+200
### Determining the maximum value for the solution of this delay differential equation?
I am working on the following delay differential equation $$\frac{df}{dt}=f-f^3-\alpha f(t-\delta)\tag{1},$$ where $\frac{1}{2}\leq\alpha\leq 1$ and $\delta\geq 1$. I know that there are three ...
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### Is there a method to list all periodic points for a funcion?
I search for a method that finds all periodic points of a given function e.g. $f(x)=x-x^2$ on its domain. You may explain some methods for a part of functions e.g. polynomials or $\mathcal{C}^k$ or ...
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### Periodic points of topologically conjugated functions in dynamical systems?
I'm working on a homework problem which seems obvious, but I am having a hard time proving/completing. The problem can be stated as follows: Let $f,g:$ $\mathbb R$ $\rightarrow$ $\mathbb R$ be ...
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### Eventually periodic orbit?
I am doing a self study on dynamical systems. I faced the following exercise in this book: Prove that any point on $f:[0,1]\to [0,1],\ f(x)=3x \mod 1$ is eventually periodic iff its a rational ...
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### Function from Cantor Set to itself.
I am stuck in getting rational functions (except identity) defined from Cantor set to itself. Please help me to get out these functions.
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### Maps with every point being periodic
Does there exists a characterization of continuous maps $f:[0,1]\rightarrow [0,1]$ with every point $x\in [0,1]$ being periodic (i.e. if for every $x\in [0,1]$ there exists $n\in\mathbb{N}$ such that ...
133 views
### There is non-trivial function satisfy the given condition?
Let $f:[0,1]\to\Bbb{R}$ to be a function satisfying that f(x)=\begin{cases} \frac{f(2x)}{2} &\text{if }x<1/2 \\ \frac{f(2x-1)}{2}+\frac{1}{2} & \text{if } x\ge1/2\end{cases} \qquad ...
Let's consider one dimensional cellular automaton. It is build upon its rule, i.e. a function $f : S^3 \rightarrow S$, where $S = \{0,1\}$. The case described is the elementary cellular automaton, ... | 583 | 2,033 | {"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} | 3.3125 | 3 | CC-MAIN-2014-23 | latest | en | 0.816251 |
https://rdrr.io/cran/epiR/man/rsu.sep.rsmult.html | 1,685,396,079,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644913.39/warc/CC-MAIN-20230529205037-20230529235037-00324.warc.gz | 537,018,454 | 9,198 | rsu.sep.rsmult: Surveillance system sensitivity by combining multiple... In epiR: Tools for the Analysis of Epidemiological Data
rsu.sep.rsmult R Documentation
Surveillance system sensitivity by combining multiple surveillance components
Description
Calculates surveillance system (population-level) sensitivity for multiple components, accounting for lack of independence (overlap) between components.
Usage
``````rsu.sep.rsmult(C = NA, pstar.c, rr, ppr, se.c)
``````
Arguments
`C` scalar integer or vector of the same length as `rr`, representing the population sizes (number of clusters) for each risk group. `pstar.c` scalar (0 to 1) representing the cluster level design prevalence. `rr` vector of length equal to the number of risk strata, representing the cluster relative risks. `ppr` vector of the same length as `rr` representing the cluster level population proportions. Ignored if `C` is specified. `se.c` surveillance system sensitivity estimates for clusters in each component and corresponding risk group. A list with multiple elements where each element is a dataframe of population sensitivity values from a separate surveillance system component. The first column equals the clusterid, the second column equals the cluster-level risk group index and the third column equals the population sensitivity values.
Value
A list comprised of two elements:
`se.p` a matrix (or vector if `C` is not specified) of population-level (surveillance system) sensitivities (binomial and hypergeometric and adjusted vs unadjusted). `se.component` a matrix of adjusted and unadjusted sensitivities for each component.
Examples
``````## EXAMPLE 1:
## You are working with a population that is comprised of indviduals in
## 'high' and 'low' risk area. There are 300 individuals in the high risk
## area and 1200 individuals in the low risk area. The risk of disease for
## those in the high risk area is assumed to be three times that of the low
## risk area.
C <- c(300,1200)
pstar.c <- 0.01
rr <- c(3,1)
## Generate population sensitivity values for clusters in each component of
## the surveillance system. Each of the three dataframes below lists id,
## rg (risk group) and cse (component sensitivity):
comp1 <- data.frame(id = 1:100,
rg = c(rep(1,time = 50), rep(2, times = 50)),
cse = rep(0.5, times = 100))
comp2 <- data.frame(id = seq(from = 2, to = 120, by = 2),
rg = c(rep(1, times = 25), rep(2, times = 35)),
cse = runif(n = 60, min = 0.5, max = 0.8))
comp3 <- data.frame(id = seq(from = 5, to = 120, by = 5),
rg = c(rep(1, times = 10), rep(2, times = 14)),
cse = runif(n = 24, min = 0.7, max = 1))
# Combine the three components into a list:
se.c <- list(comp1, comp2, comp3)
## What is the overall population-level (surveillance system) sensitivity?
rsu.sep.rsmult(C = C, pstar.c = pstar.c, rr = rr, ppr = NA, se.c = se.c)
## The overall adjusted system sensitivity (calculated using the binomial
## distribution) is 0.85.
## EXAMPLE 2:
## Assume that you don't know exactly how many individuals are in the high
## and low risk areas but you have a rough estimate that the proportion of
## the population in each area is 0.2 and 0.8, respectively. What is the
## population-level (surveillance system) sensitivity?
ppr <- c(0.20,0.80)
rsu.sep.rsmult(C = NA, pstar.c = pstar.c, rr = rr, ppr = ppr, se.c = se.c)
## The overall adjusted system sensitivity (calculated using the binomial
## distribution) is 0.85.
``````
epiR documentation built on April 5, 2023, 5:10 p.m. | 931 | 3,511 | {"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-2023-23 | latest | en | 0.812398 |
http://www.sparknotes.com/testprep/books/sat2/math1c/chapter7section2.rhtml | 1,529,474,309,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863463.3/warc/CC-MAIN-20180620050428-20180620070428-00573.warc.gz | 501,808,461 | 10,535 | Jump to a New ChapterIntroduction to the SAT IIIntroduction to SAT II Math ICStrategies for SAT II Math ICMath IC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
7.1 Prisms 7.2 Solids That Aren’t Prisms 7.3 Relating Length, Surface Area, and Volume 7.4 Inscribed Solids
7.5 Solids Produced by Rotating Polygons 7.6 Key Formulas 7.7 Review Questions 7.8 Explanations
Solids That Aren’t Prisms
Some of the solids that appear on the Math IC do not have two congruent bases that lie in parallel planes, so they cannot be considered prisms. As with prisms, you need to know how to calculate the volume and surface area of these non-prisms. The formulas for the volume and surface area of the non-prisms are a little more complex than those for the prisms, but not too difficult.
Cones
A cone is not a prism, but it is similar to a cylinder. A cone is essentially a cylinder in which one of the bases is collapsed into a single point at the center of the base.
The radius of a cone is the radius of its one circular base. The height of a cone is the distance from the center of the base to the apex (the point on top). The lateral height, or slant height, of a cone is the distance from a point on the edge of the base to the apex. In the figure above, these three measurements are denoted by r, h, and l, respectively.
Notice that the height, radius, and lateral height of a cone form a right triangle. This means that if you know the value for any two of these measurements, you will always be able to find the third by using the Pythagorean theorem.
Volume of a Cone
Since a cone is similar to a cylinder except that it is collapsed to a single point at one end, the formula for the volume of a cone is a fraction of the formula for the volume of a cylinder:
where r is the radius and h is the height.
For practice, find the volume of the cone pictured below:
To answer this question, just use the formula for the volume of a cone with the following values plugged in: r = x, l = 2x, and h = x. The volume is:
Surface Area of a Cone
The surface area of a cone consists of the lateral surface area and the area of the base. Because the base is a circle, it has an area of πr2. The lateral surface is the cone “unrolled,” which, depending on the shape of the cone, can be the shape of a triangle with a curved base, a half-circle, or a “Pacman” shape. The area of the lateral surface is related to the circumference of the circle times the lateral height, l. This is the formula:
where r is the radius and l is the lateral height.
The total surface area is the sum of the base area and lateral surface area:
When you are finding the surface area of a cone, be careful not to find only the lateral surface area and then stop. Students often forget the step of adding on the area of the circular base. Practice by finding the total surface area of the cone pictured below:
The total surface area is equal to the area of the base plus the area of the lateral surface. The area of the base = πx2. The lateral surface area = πx 2x. The total surface area therefore equals πx2 + π2x2 = 3πx2.
Pyramids
A pyramid is like a cone, except that it has a polygon for a base. Though pyramids are not tested very often on the Math IC test, you should be able to recognize them and calculate their volume.
The shaded area in the figure above is the base, and the height is the perpendicular distance from the apex of the pyramid to its base.
Volume of a Pyramid
The formula for calculating the volume of a pyramid is:
where B is the area of the base and h is the height. Try to find the volume of the pyramid below:
The base is just a square with a side of 3, and the height is 3/2. B = 32 = 9, and the total volume of the pyramid is:
Surface Area of a Pyramid
The surface area of a pyramid is rarely tested on the Math IC test. If you come across one of those rare questions that covers the topic, you can calculate the area of each face individually using techniques from plane geometry, since the base of a pyramid is a square and the sides are triangles. Practice by finding the surface area of the same pyramid in the figure below:
To calculate the surface area, you need to add together the area of the base and the areas of the four sides. The base is simply a square, and we’ve seen that B = 32 = 9. Each side is an equilateral triangle, and we can use the properties of a 30-60-90 triangle to find their areas:
For each triangle, Area = 1 /2 3 3/2 = 9/ 4. The sum of the areas of the four triangles is 4 9/4 = 9 The total surface area of the pyramid is 9 + 9
Spheres
A sphere is the collection of points in three-dimensional space that are equidistant from a fixed point, the center of the sphere. Essentially, a sphere is a 3-D circle. The main measurement of a sphere is its radius, r, the distance from the center to any point on the sphere.
If you know the radius of a sphere you can find both its volume and surface area. The equation for the volume of a sphere is:
The equation for the surface area of a sphere is:
Jump to a New ChapterIntroduction to the SAT IIIntroduction to SAT II Math ICStrategies for SAT II Math ICMath IC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
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https://www.vedantu.com/question-answer/type-of-motion-is-shown-by-a-point-marked-on-the-class-8-physics-cbse-60aa54573811760ea6a995ea | 1,701,581,741,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100484.76/warc/CC-MAIN-20231203030948-20231203060948-00370.warc.gz | 1,169,643,639 | 27,807 | Courses
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# What type of motion is shown by “A point marked on the blade of a running ceiling fan”?
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Hint: In order to solve this question, we are going to first analyze the condition of the motion of the ceiling fan, then the point marked on the blades of the ceiling fan is considered and then, we have to see what the motion is like by considering the velocity and the type of movement of the point marked.
$\omega = \dfrac{{d\theta }}{{dt}}$
Where, $d\theta$is the angular displacement and $dt$is the time taken for it. | 177 | 676 | {"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} | 3.296875 | 3 | CC-MAIN-2023-50 | longest | en | 0.898144 |
https://disruptedphysician.com/drug-addiction/what-is-the-concentration-of-alcohol-in-blood.html | 1,632,423,397,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057427.71/warc/CC-MAIN-20210923165408-20210923195408-00719.warc.gz | 260,996,007 | 20,369 | # What is the concentration of alcohol in blood?
Contents
Blood alcohol content (BAC), also called blood alcohol concentration or blood alcohol level, is a measurement of alcohol intoxication used for legal or medical purposes. A BAC of 0.10 (0.10% or one tenth of one percent) means that there is 0.10 g of alcohol for every 100 ml of blood, which is the same as 21.7 mmol/l.
## What is concentration of alcohol in blood according to law?
This means that 2,100 milliliters (ml) of alveolar air will contain the same amount of alcohol as 1 ml of blood. The legal standard for drunkenness across the United States ranges from 0.10 to 0.08. If a person’s BAC measures 0.08, it means that there are 0.08 grams (i.e., 80 mg) of alcohol per 100 ml of blood.
## What is the percentage of blood alcohol concentration?
Blood alcohol content is the amount of alcohol present in 100 milliliters (ml) or its equivalent of 1 deciliter (dL) of blood. For example: 80 mg is 0.08 grams. 0.08 grams of alcohol in 100 ml is 0.08%
INFORMATIVE: Why does rubbing alcohol have a strong smell?
## Is 1.7 alcohol level high?
08% BAC; you will test as legally impaired at this blood alcohol level if you’re 21 or older. 0.10 – 0.12% – Obvious physical impairment and loss of judgment. Speech may be slurred. 0.13 – 0.15% – At this point, your blood alcohol level is quite high.
## What is a normal blood alcohol level?
This means that one tenth of a percent of a person’s blood volume is alcohol or that a person has 1 part alcohol per 1000 parts blood. At a blood ethanol level of less than 50 mg/dL, or 0.05% concentration, an individual is not considered to be intoxicated. The possible critical value for blood ethanol is >300 mg/dL.
## Is 2.0 A high alcohol level?
02 = Drinkers begin to feel moderate effects. 2. BAC = . 04 = Most people begin to feel relaxed, mildly euphoric, sociable, and talkative.
## Is .28 alcohol level high?
37 to . 40 or higher can cause death. Most people begin to feel relaxed, sociable and talkative when BAC reaches 0.04. Judgment, attention and control are somewhat impaired at 0.05, and the ability to drive safely begins to be limited.
## Is .19 alcohol level high?
In California, you will test as legally impaired at . 08% BAC if you are over 21. It is illegal to drive or bike at this level. Significant impairment of motor coordination and loss of judgment.
## Is .34 alcohol level high?
35 BAC, these vital functions “shut off,” but any BAC over . 3 is life threatening and poses a significant risk of death. As an example of the danger of drinking to this level, White cited Gordie Bailey, a University of Colorado student who died of alcohol poisoning in 2004 with a . 328 BAC.
INFORMATIVE: Question: How your body recovers after quitting smoking?
## How many beers is .15 BAC?
Blood Alcohol Chart For Estimation
Men
Approximate Blood Alcohol Percentage
9 .34 .15
10 .38 .17
Subtract .01% for each 40 minutes of drinking. One drink is 1.25 oz. of 80 proof liquor, 12 oz. of beer, or 5 oz. of table wine.
## Is .23 alcohol level high?
A . 23 blood-alcohol level would indicate something in the neighborhood of 10 drinks in a two-and-a-half-hour time period. A . 2 level also changes a driving under intoxication charge to a gross misdemeanor.
## Will 1 beer show up on a Breathalyzer?
Thus, one 12-ounce can of beer, one 4-ounce glass of wine, or one normal mixed drink or cocktail are all equally intoxicating, and give the same blood alcohol content (BAC) reading on a breathalyzer.
## What is considered legally drunk?
In all states, the legal limit for driving is a blood alcohol concentration of 0.08. If you are at or above this limit, you are considered legally intoxicated. However, it is easy to misjudge your intoxication level when it is near 0.08 percent. It can be even more difficult to assess intoxication at lower levels.
## What does 400 alcohol level mean?
0.4–0.5% (400–500 mg/dL) Potentially fatal and a person may be comatose. Above 0.5% (500 mg/dL) Highly dangerous/fatal blood alcohol level. Impairment of motor skills may occur at blood alcohol levels lower than 0.08%.
## How long will 3 beers stay in my urine?
The average urine test can detect alcohol between 12 and 48 hours after drinking. More advanced testing can measure alcohol in the urine 80 hours after you drink. Breath tests for alcohol can detect alcohol within a shorter time frame. This is about 24 hours on average. | 1,138 | 4,462 | {"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-2021-39 | latest | en | 0.889408 |
http://alevelexperimentalphysics.info/content/labs/A1-1%20Measurements%20and%20Accuracy.html | 1,558,842,805,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258621.77/warc/CC-MAIN-20190526025014-20190526051014-00146.warc.gz | 8,743,278 | 2,973 | # A1-1: Measurements and Accuracy¶
## Apparatus¶
Microscope slide; vernier calipers; magnifying glass; micrometer; beam balance with masses; Archimedes’ bridge; beaker of water ($$250\text{ml}$$); metre ruler.
## Procedure¶
For each of the following, record the observations together with the possible error (i.e. ½ the smallest scale division), e.g.: $$46 \pm 0.5\text{mm}$$. Calculate the mean value of repeated readings together with the error. Calculate the % error.
1. Measure the slide thickness in a number of places using the metre ruler.
2. Repeat using the vernier calipers instead.
3. Measure the slide thickness in several places using the micrometer. For a mechanical micrometer, record the ‘zero reading’ and adjust the other readings correctly.
4. Measure the length $$l$$ and the width $$w$$ using the metre ruler. Find the mass of the slide in air $$m$$ , and then its apparent mass when suspended in water $$m_a$$.
Then:
$\begin{split}\text{upthrust} &= \text{weight of liquid displaced} \\ mg - m_a g &= \rho_w g \ \big(\text{slide volume}\big)\end{split}$
where:
$\begin{split}\text{slide volume} = \frac{m - m_a}{\rho_w} \\ \rho_w = \text{density of water}\end{split}$
Then calculate the slide thickness $$d$$ since:
$l w d = \text{slide volume}$
When you have completed the above, arrange the estimates of slide thickness in order (most accurate first).
Explain carefully why some methods are more accurate than others. | 374 | 1,454 | {"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} | 3.953125 | 4 | CC-MAIN-2019-22 | latest | en | 0.680505 |
http://depts.washington.edu/pku/management/curriculum/teen/readinglabels.html | 1,516,152,453,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886792.7/warc/CC-MAIN-20180117003801-20180117023801-00320.warc.gz | 92,160,433 | 3,904 | CONCEPT: Reading Food Labels, Calculating Phe
BRIEF SUMMARY
Participants complete an activity looking at food labels and ingredients and practicing techniques to estimate the amount of phe in foods.
OBJECTIVES
After completing this activity, participants will be able to:
• estimate the amount of phe in a food by looking at the label
• identify ingredients listed on food labels as "with phe" or "without phe"
• calculate the amount of phe in a meal
METHODS
Complete the "Reading Labels" worksheet. (The first part is a short pre-test, the second part lists the answers and explanations.) It should be self-explanatory, but adults should be present to answer questions.
Discuss the answers. Who got them all right? Who missed one? (Reminder: the quiz was designed to ask questions that most people would not be able to answer correctly.)
Participants rotate through activity stations, as small groups or as individuals, and complete the "Estimating Phe Activity Sheet":
• Estimate phe from label: Preparation--set out 3 food labels. Participants estimate phe using the equation: grams protein x 50 = mg phe.
• Guess food from ingredients listed on label: Preparation--set out three food labels with the names of the foods covered (and no identifying containers).
• Guess amount of phe in meal: Preparation--set out three different meals using food pictures or food models. Participants calculate phe using Low Protein Food List and/or estimating equation.
• What is in the bag? Preparation--put five different foods into five paper bags. Participants reach into each bag (without looking), guess what food is in the bag, and guess how much phe it contains.
• Estimate phe from label: How close were the estimates? Why is it better to use food lists? When would be a good time to use the estimating trick?
• Guess food from ingredients listed on label: What ingredients "gave the food away"? What else could the food have been?
• Guess amount of phe in meal: Who was closest? Which worked best, estimating or actually calculating? (Remind them that using estimating may be low--more phe in food than they think, or may be high--they won't eat as much as they could have).
• What is in the bag? Who guessed the foods correctly? How much phe? Who was closest?
MATERIALS
• Pencils
• Estimating Phe Activity Sheet
• Calculators
• Low Protein Food List
• Food labels (see part 3 above: need 6 total, 3 of these with name covered)
• Food pictures or food models to make 3 meals (see part 3 above)
• Five paper bags, each containing a "mystery food" (such as turnip, kiwi, orange, mushroom, bell pepper)
• The group leader will need an "answer sheet" with the phe content of the specific foods used in the activity (will vary depending on the foods included).
HOME ACTIVITIES
Review the labels of the foods your family eats with your children.
[top of page] | 643 | 2,866 | {"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} | 4.03125 | 4 | CC-MAIN-2018-05 | longest | en | 0.920477 |
http://mathhelpforum.com/differential-geometry/195332-investigating-operator.html | 1,524,724,409,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948089.47/warc/CC-MAIN-20180426051046-20180426071046-00508.warc.gz | 204,058,798 | 11,965 | 1. ## investigating an operator
Hello!
I have a problem that comes in two parts.
(1) $\displaystyle H$ is a Hilbert space with orthonormal basis $\displaystyle \{e_k\}$. We are given a positivie strictly increasing sequence $\displaystyle \{\alpha_k\}$with the following property $\displaystyle \lim_{k\rightarrow \infty}\frac{\alpha_k}{\alpha_{k+1}}=1$. Our task is to show that there exists a unique operator $\displaystyle M\in B(H)$ such that $\displaystyle Me_k=\frac{\alpha_k}{\alpha_{k+1}}e_{k+1}$.
(2) Determine $\displaystyle \|M\|$ and the spectrum of M (hint: use the eigenvalues of the adjoint $\displaystyle M^*$).
-----------------------------------------------
My attempt
(1) For the first one I was thinking that I could just simply find the operator by using $\displaystyle Mx$ where $\displaystyle x\in H$. Here it goes,
$\displaystyle x\in H$ gives me $\displaystyle x=\sum_{k=1}^\infty <x,e_k>e_k=\lim_{n\rightarrow \infty}\sum_{k=1}^n <x,e_k>e_k$.
So,
$\displaystyle Mx=M\lim_{n\rightarrow \infty}\sum_{k=1}^n <x,e_k>e_k=\lim_{n\rightarrow \infty}\sum_{k=1}^n <x,e_k>Me_k = \lim_{n\rightarrow \infty}\sum_{k=1}^n <x,e_k>\frac{\alpha_k}{\alpha_{k+1}}e_{k+1}=\sum_{ k=1}^\infty <x,e_k>\frac{\alpha_k}{\alpha_{k+1}}e_{k+1}$
(the last equality by Parseval's formula and the second by continuity (boundedness) of $\displaystyle M$).
Well I now know what a bounded linear operator with the above stated property looks like. But is it unique? Futhermore, its "look" complicates things in the second part.
(2) I know the definition $\displaystyle \|M\|=\sup_{\|x\|=1}|Mx|$. But this doesn't seem all that easy to compute using what I know about my operator.
And what will the eigenvalues of $\displaystyle M^*$ tell me? I know that $\displaystyle \sigma (M)=\overline{\sigma (M^*)}$ (where I by $\displaystyle \sigma (M)$ denote the spectrum of $\displaystyle M$) and this does not help me unless I can use the eigenvalues of $\displaystyle M^*$ to find $\displaystyle \sigma(M^*)$. This would be easy if $\displaystyle M^*$ was compact, then I would know that $\displaystyle \sigma (M^*)= \{0\}\cup \{eigenvalues of M^*\}$.
Thanks!
2. ## Re: investigating an operator
The operator M is a weighted shift (check that link for some useful guidance). It shifts each basis vector $\displaystyle e_k$ to the next one $\displaystyle e_{k+1},$ multiplying it by the weight $\displaystyle \alpha_k/\alpha_{k+1}.$ Since its value at each basis vector is specified, it must be unique (you know its value at each finite linear combination of basis vectors and hence, by continuity, at every vector).
The adjoint operator is a backward weighted shift, taking each $\displaystyle e_k$ to a multiple of $\displaystyle e_{k-1}$ and sending $\displaystyle e_1$ to 0. The advantage of looking at the adjoint is that it has many eigenvalues, whereas M itself does not.
For convenience, write $\displaystyle \beta_k = \alpha_k/\alpha_{k+1}$, and let $\displaystyle B = \sup\{\beta_k:k\in\mathbb{N}\}$. If $\displaystyle x = \textstyle\sum \xi_ke_k$ then $\displaystyle Mx = \textstyle\sum \beta_k\xi_ke_{k+1}$, and
$\displaystyle \|Mx\|^2 = \sum|\beta_k\xi_k|^2\leqslant B^2\sum|\xi|^2 = B^2\|x\|^2.$
That shows that $\displaystyle \|M\|\leqslant B$. By looking at $\displaystyle \|Me_k\|$ for each k, you should be able to show the reverse inequality.
3. ## Re: investigating an operator
Originally Posted by Opalg
The advantage of looking at the adjoint is that it has many eigenvalues, whereas M itself does not.
Mabye I'm missing something trivial, but how do the eigenvalues of the adjoint help me find the spectrum of it? I know that I have the spectum of a compact operator if I have the eigenvalues. Is my $\displaystyle M^*$ compact?
4. ## Re: investigating an operator
Originally Posted by mgarson
Mabye I'm missing something trivial, but how do the eigenvalues of the adjoint help me find the spectrum of it? I know that I have the spectum of a compact operator if I have the eigenvalues. Is my $\displaystyle M^*$ compact?
The idea is that M* has so many eigenvalues that they force the spectrum to be as big as it could possibly be. To see how that might work, here is a slightly simplified example.
Let S be the unilateral shift operator, defined on the basis vectors by $\displaystyle Se_k = e_{k+1}$ (so S is like your operator M except that it does not have the weights $\displaystyle \alpha_k/\alpha_{k+1}$). Its adjoint S* is the backwards shift, defined by $\displaystyle S^*e_1=0$ and $\displaystyle S^*e_k = e_{k-1}$ for k>1.
Let $\displaystyle \lambda$ be a fixed complex number with $\displaystyle 0<|\lambda|<1$ and let x be the vector given by $\displaystyle \textstyle x=\sum\lambda^ke_k.$ (Notice that that sum converges in H because the sum of the squares of the absolute values of the coefficients is $\displaystyle \textstyle\sum|\lambda|^{2k}<\infty.$) Then $\displaystyle S^*x = \textstyle \sum\lambda^ke_{k-1} = \lambda x.$
Thus $\displaystyle \lambda$ is an eigenvalue of S*, with eigenvector x. That holds for every nonzero $\displaystyle \lambda$ in the open unit ball. In fact, 0 is also an eigenvalue, because $\displaystyle S^*e_1=0.$ So the spectrum of S* contains the entire open unit ball. Since the spectrum is always closed, it contains the closed unit ball. But $\displaystyle \|S^*\| = 1$, and the absolute value of an element of the spectrum can never be greater than the norm of the operator. Conclusion: the spectrum of S* is the closed unit ball. Therefore the spectrum of S is also the closed unit ball (though S has no eigenvalues).
Notice that although S and S* are not compact, S* has a huge number of eigenvalues (uncountably many, which a compact operator never could have).
Finding the spectrum of the operator M* is a similar exercise, though you have to decide how to deal with the weights. | 1,646 | 5,870 | {"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} | 3.53125 | 4 | CC-MAIN-2018-17 | latest | en | 0.730696 |
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Partner
## If Statement with dates
Hello dear experts,
I have a field called Fecha as a date, i'm trying evaluate it with a rate of dates depending of the value which evaluating Date field, once i try to load data the next error comes(Attached image).
if(Fecha<Date(30/01/2015,'DD/MM/YYYY'),
if(PM < PE15,1,0),
if(Fecha>= Date(06/01/2016,'DD/MM/YYYY'),
if(PM < PE16,1,0),
if(Fecha<= Date(10/05/2016,'DD/MM/YYYY'),
if(PM < PE161,1,0),
if(Fecha<= Date(17/11/2017,'DD/MM/YYYY'),
if(PM < PE17,1,0),
if(Fecha<= Date(29/04/2019,'DD/MM/YYYY'),
if(PM < PE,1,0)
)
)
)
)
) as Sub_PE
Thanks for help.
Regards.
1 Solution
Accepted Solutions
Partner
Hello,
I tried this way and works for me, probably someone could use this in a future.
if (Fecha < Date(30/01/2015) and PM < PE15,
1,
IF((Fecha >= Date(30/01/2015)AND Fecha <= Date(06/01/2016))AND PM < PE16,
1,
IF((Fecha >= Date(06/01/2016)AND Fecha <= Date(10/05/2016))AND PM < PE161,
1,
If((Fecha >= Date(10/05/2016)AND Fecha <= Date(17/11/2017))AND PM < PE17,
1,
IF((Fecha >= Date(17/11/2017)AND Fecha <= Date(29/04/2019))AND PM < PE17,
1,0
)
)
)
)
)as Sub_PE
It seems the error comes from the range of dates.
Regards.
4 Replies
Specialist III
I always go
• if thenEqual
• or if and thenEqual
• not if if if if
if(Fecha<Date(30/01/2015,'DD/MM/YYYY') and PM < PE15,1,0), 'Result1' ,
if(Fecha>= Date(06/01/2016,'DD/MM/YYYY'), 'result2',
if(PM < PE16,1,0),'results3'
etc
)
)
)
)
) as Sub_PE
Partner
Hello Robert,
I tired this way, but the same error comes too
if(Fecha<Date(30/01/2015)
and PM < PE15,1,0),'Result1',
if(Fecha>= Date(06/01/2016)
and PM < PE16,1,0),'Result2',
if(Fecha<= Date(10/05/2016)
and PM < PE161,1,0),'Result3',
if(Fecha<= Date(17/11/2017)
and PM < PE17,1,0),'Result4',
if(Fecha<= Date(29/04/2019)
and PM < PE,1,0),'Result5'as Sub_PE
Regards
Specialist III
Sorry. My previous post may have mislead you
Do you want
if(Fecha<Date(30/01/2015,'DD/MM/YYYY') // I sometimes convert dates to numbers
and
PM < PE15, //is PE15 a DUAL numerical value?
1, // result =1
0) //else 0
as Sub_PE,
if so then you need something like
if(Fecha<Date(30/01/2015,'DD/MM/YYYY')
and
PM < PE15,
1,
if(Fecha>= Date(06/01/2016,'DD/MM/YYYY') and
PM < PE16,
1,
0
)) as Sub_PE ,
Partner
Hello,
I tried this way and works for me, probably someone could use this in a future.
if (Fecha < Date(30/01/2015) and PM < PE15,
1,
IF((Fecha >= Date(30/01/2015)AND Fecha <= Date(06/01/2016))AND PM < PE16,
1,
IF((Fecha >= Date(06/01/2016)AND Fecha <= Date(10/05/2016))AND PM < PE161,
1,
If((Fecha >= Date(10/05/2016)AND Fecha <= Date(17/11/2017))AND PM < PE17,
1,
IF((Fecha >= Date(17/11/2017)AND Fecha <= Date(29/04/2019))AND PM < PE17,
1,0
)
)
)
)
)as Sub_PE
It seems the error comes from the range of dates.
Regards.
Tags | 1,084 | 3,033 | {"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-2021-04 | latest | en | 0.585704 |
https://cn.maplesoft.com/support/help/maplesim/view.aspx?path=ContextMenu%2FCurrentContext%2FEntryGenerators%2FList | 1,718,924,327,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862006.96/warc/CC-MAIN-20240620204310-20240620234310-00040.warc.gz | 154,180,288 | 21,712 | List - Maple Help
List
list the module entry generators
Calling Sequence EntryGenerators[List]()
Description
• The EntryGenerators[List] command lists the names of the entry generators that are available within a context menu module. Each string in the list returned by List() references an entry generator, whose procedure or list can be viewed using the EntryGenerators[Get] command.
An entry generator is used to dynamically generate subentries in the context menu. An entry uses the entry generator by setting the entry option 'entry_generator' to the name of the entry generator in a call to the Entries[Add] command.
Examples of EntryGenerators[List]
$\left[{"RightVariables 2"}{,}{"LengthUnits"}{,}{"VectorNorms"}{,}{"Rotation"}{,}{"Units_MTS"}{,}{"Units_FPS"}{,}{"VCVariables"}{,}{"PolyVariables"}{,}{"CollectCodeType"}{,}{"LimitRules"}{,}{"Variables deg=1"}{,}{"Variable Comb 2"}{,}{"Units_EMU"}{,}{"PermutedLhs"}{,}{"MatrixNorms"}{,}{"SeqSelection"}{,}{"DefIntRules"}{,}{"GraphTheoryExportList"}{,}{"Variables deg=2"}{,}{"Variables and Functions"}{,}{"ExprSeqSelection"}{,}{"PermutedVariables"}{,}{"CombineCodeType"}{,}{"VarsFuns"}{,}{"Variables"}{,}{"Units_CGS"}{,}{"NumDigits"}{,}{"matrixSelection"}{,}{"vectorSelection"}{,}{"VariablesFromExpression"}{,}{"Functions"}{,}{"Linear Functions"}{,}{"IndefIntRules"}{,}{"DiffRules"}{,}{"NetworkSourcesSinks"}{,}{"rtableSelection"}{,}{"OuterVariables"}{,}{"VariablesIndices"}{,}{"Variables 3"}{,}{"Variables 2"}{,}{"Params"}{,}{"RhsVariables"}{,}{"matrixVariables"}{,}{"Units_Atomic"}{,}{"GraphTheoryPolyVarList"}{,}{"Units_SI"}{,}{"vectorNorms"}{,}{"matrixNorms"}{,}{"UnitSystems"}{,}{"Calc1Hints"}\right]$ (1) | 536 | 1,680 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-26 | latest | en | 0.495234 |
http://mizar.uwb.edu.pl/version/current/html/proofs/fcont_1/66 | 1,569,116,890,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-39/segments/1568514574765.55/warc/CC-MAIN-20190922012344-20190922034344-00059.warc.gz | 123,276,878 | 2,424 | let g, p be Real; :: thesis: for f being one-to-one PartFunc of REAL,REAL st p <= g & [.p,g.] c= dom f & ( f | [.p,g.] is increasing or f | [.p,g.] is decreasing ) holds
((f | [.p,g.]) ") | (f .: [.p,g.]) is continuous
let f be one-to-one PartFunc of REAL,REAL; :: thesis: ( p <= g & [.p,g.] c= dom f & ( f | [.p,g.] is increasing or f | [.p,g.] is decreasing ) implies ((f | [.p,g.]) ") | (f .: [.p,g.]) is continuous )
assume that
A1: p <= g and
A2: [.p,g.] c= dom f and
A3: ( f | [.p,g.] is increasing or f | [.p,g.] is decreasing ) ; :: thesis: ((f | [.p,g.]) ") | (f .: [.p,g.]) is continuous
reconsider p = p, g = g as Real ;
now :: thesis: ((f | [.p,g.]) ") | (f .: [.p,g.]) is continuous
per cases ( f | [.p,g.] is increasing or f | [.p,g.] is decreasing ) by A3;
suppose A4: f | [.p,g.] is increasing ; :: thesis: ((f | [.p,g.]) ") | (f .: [.p,g.]) is continuous
A5: ((f | [.p,g.]) ") .: (f .: [.p,g.]) = ((f | [.p,g.]) ") .: (rng (f | [.p,g.])) by RELAT_1:115
.= ((f | [.p,g.]) ") .: (dom ((f | [.p,g.]) ")) by FUNCT_1:33
.= rng ((f | [.p,g.]) ") by RELAT_1:113
.= dom (f | [.p,g.]) by FUNCT_1:33
.= (dom f) /\ [.p,g.] by RELAT_1:61
.= [.p,g.] by ;
((f | [.p,g.]) ") | (f .: [.p,g.]) is increasing by ;
hence ((f | [.p,g.]) ") | (f .: [.p,g.]) is continuous by A1, A5, Th46; :: thesis: verum
end;
suppose A6: f | [.p,g.] is decreasing ; :: thesis: ((f | [.p,g.]) ") | (f .: [.p,g.]) is continuous
A7: ((f | [.p,g.]) ") .: (f .: [.p,g.]) = ((f | [.p,g.]) ") .: (rng (f | [.p,g.])) by RELAT_1:115
.= ((f | [.p,g.]) ") .: (dom ((f | [.p,g.]) ")) by FUNCT_1:33
.= rng ((f | [.p,g.]) ") by RELAT_1:113
.= dom (f | [.p,g.]) by FUNCT_1:33
.= (dom f) /\ [.p,g.] by RELAT_1:61
.= [.p,g.] by ;
((f | [.p,g.]) ") | (f .: [.p,g.]) is decreasing by ;
hence ((f | [.p,g.]) ") | (f .: [.p,g.]) is continuous by A1, A7, Th46; :: thesis: verum
end;
end;
end;
hence ((f | [.p,g.]) ") | (f .: [.p,g.]) is continuous ; :: thesis: verum | 795 | 1,926 | {"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} | 3.40625 | 3 | CC-MAIN-2019-39 | latest | en | 0.737622 |
https://www.physicsforums.com/threads/new-velocity-distance-question-1-25.794218/ | 1,660,647,658,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572286.44/warc/CC-MAIN-20220816090541-20220816120541-00731.warc.gz | 801,501,529 | 15,547 | # New Velocity/Distance Question 1/25
ilh
## Homework Statement
In reaching her destination, a backpacker walks with an average velocity of 1.44 m/s, due west. This average velocity results, because she hikes for 5.48 km with an average velocity of 3.25 m/s due west, turns around, and hikes with an average velocity of 0.683 m/s due east. How far east did she walk (in kilometers)?
## Homework Equations
v1 = 3.25 m/s west
d1= 5480 m
t1= 1686
v2 = 0.638 m/s east
d2 = ?
t2= ?
## The Attempt at a Solution
I know I have to plug in my numbers into the formula but I do not know how to derive second time or distance. Please help.
Homework Helper
Gold Member
## Homework Statement
In reaching her destination, a backpacker walks with an average velocity of 1.44 m/s, due west. This average velocity results, because she hikes for 5.48 km with an average velocity of 3.25 m/s due west, turns around, and hikes with an average velocity of 0.683 m/s due east. How far east did she walk (in kilometers)?
## Homework Equations
v1 = 3.25 m/s west
d1= 5480 m
t1= 1686
v2 = 0.638 m/s east
d2 = ?
t2= ?
## The Attempt at a Solution
I know I have to plug in my numbers into the formula but I do not know how to derive second time or distance. Please help.
What is the formula for average velocity?
ilh
What is the formula for average velocity?
V = D/T | 385 | 1,356 | {"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-2022-33 | latest | en | 0.930767 |
https://stat.ethz.ch/pipermail/r-help/2010-February/227785.html | 1,627,622,373,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153931.11/warc/CC-MAIN-20210730025356-20210730055356-00023.warc.gz | 548,861,967 | 2,001 | # [R] Code find exact distribution for runs test?
Dale Steele dale.w.steele at gmail.com
Thu Feb 11 02:15:33 CET 2010
I've been attempting to understand the one-sample run test for
randomness. I've found run.test{tseries} and run.test{lawstat}. Both
use a large sample approximation for distribution of the total number
of runs in a sample of n1 observations of one type and n2 observations
of another type.
I've been unable to find R code to generate the exact distribution and
would like to see how this could be done (not homework).
For example, given the data:
dtemp <- c(12, 13, 12, 11, 5, 2, -1, 2, -1, 3, 2, -6, -7, -7, -12, -9,
6, 7, 10, 6, 1, 1, 3, 7, -2, -6, -6, -5, -2, -1)
The Monte Carlo permutation approach seems to get me part way.
# calculate the number of runs in the data vector
nruns <- function(x) {
signs <- sign(x)
runs <- rle(signs)
r <- length(runs\$lengths)
return(r)
}
MC.runs <- function(x, nperm) {
RUNS <- numeric(nperm)
for (i in 1:nperm) {
RUNS[i] <- nruns(sample(x))
}
cdf <- cumsum(table(RUNS))/nperm
return(list(RUNS=RUNS, cdf=cdf, nperm=nperm))
}
MC.runs(dtemp, 100000)
Thanks. --Dale | 380 | 1,134 | {"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} | 2.734375 | 3 | CC-MAIN-2021-31 | latest | en | 0.782585 |
https://www.coursehero.com/file/6039981/NotesOct18/ | 1,547,905,193,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583667907.49/warc/CC-MAIN-20190119115530-20190119141530-00601.warc.gz | 753,807,926 | 68,450 | NotesOct18
# NotesOct18 - Transformations of random variables Statement...
• Notes
• 9
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Transformations of random variables Statement of the problem: let X be a ran- dom variable with a known distribution, e.g. known cdf, or known pmf, or known pdf. Let Y = T ( X ) be a function, or a transforma- tion, of X . Find the distribution of Y .
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Computing the cdf of Y = T ( X ) If X is discrete with pmf p X , then F Y ( y ) = P ( Y y ) = P ( T ( X ) y ) = X x i : T ( x i ) y p X ( x i ) , -∞ < y < . If X is continuous with pdf f X then F Y ( y ) = P ( Y y ) = P ( T ( X ) y ) = Z x : T ( x ) y f X ( x ) dx, -∞ < y < .
Example : Suppose that X is uniformly dis- tributed between - 1 and 1. Find the distribu- tion of Y = X 2 .
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This method sometimes works even for trans- formations of several random variables. Example : Let X and Y be two indepen- dent standard exponential random variables, and Z = T ( X, Y ) = X + Y . Find the distribution of Z .
Computing the pmf or pdf of Y = T ( X ) . If the computation of the cdf of Y = T ( X ) is not practical, then in the discrete case, one can try computing the pmf of Y = T ( X ); in the continuous case one can try com- puting the pdf of
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Unformatted text preview: p X then Y = T ( X ) is also discrete. The pmf of Y : p Y ( y j ) = P ( Y = y j ) = P ( T ( X ) = y j ) = X x i : T ( x i )= y j p X ( x i ) . If the function T is one-to-one , then p Y ( y j ) = p X ± T-1 ( y j ) ² ( T-1 is the inverse map ) Example Let X be a mean λ Poisson random variable. Find the distribution of Y = 2 X . Let X be continuous with pdf f X . • Sometimes Y = T ( X ) is also continuous. • To compute the density of Y adjust by the derivative of the inverse transfor-mation . Monotone transformations : the function T is either increasing or decreasing on the range of X . A monotone function T is automatically one-to-one. The pdf of Y : f Y ( y ) = f X ± T-1 ( y ) ² ³ ³ ³ ³ ³ dT-1 ( y ) dy ³ ³ ³ ³ ³ ....
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• '10
• SAMORODNITSKY
• Probability theory, Probability mass function
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Jill Tulane University ‘16, Course Hero Intern | 934 | 3,472 | {"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} | 3.265625 | 3 | CC-MAIN-2019-04 | latest | en | 0.861208 |
https://gaurish4math.wordpress.com/tag/inverses/ | 1,563,744,097,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527204.71/warc/CC-MAIN-20190721205413-20190721231413-00443.warc.gz | 413,599,647 | 12,461 | # Finite Sum & Divisibility – 2
Standard
Earlier this year I discussed a finite analogue of the harmonic sum. Today I wish to discuss a simple fact about finite harmonic sums.
If $p$ is a prime integer, the numerator of the fraction $1+\frac{1}{2}+\frac{1}{3}+\ldots + \frac{1}{p-1}$ is divisible by $p$.
We wish to treat the given finite sum modulo $p$, hence we can’t just add up fractions. We will have to consider each fraction as inverse of an integer modulo p. Observe that for $0, we have inverse of each element $i$ in the multiplicative group $\left(\mathbb{Z}/p\mathbb{Z}\right)^\times$, i.e. there exist an $i^{-1}$ such that $i\cdot i^{-1}\equiv 1 \pmod p$.
For example, for $p=5$, we have $1^{-1}=1, 2^{-1}=3, 3^{-1}=2$ and $4^{-1}=4$.
Hence we have
$\displaystyle{i\cdot \frac{1}{i}\equiv 1 \pmod p ,\qquad (p-i)\cdot \frac{1}{p-i} \equiv 1 \pmod p }$
for all $0.
Hence we have:
$\displaystyle{i\left(\frac{1}{i}+\frac{1}{p-i}\right)\equiv i\cdot \frac{1}{i} - (p-i)\cdot \frac{1}{p-i} \equiv 0 \pmod p}$
Thus we have:
$\displaystyle{\frac{1}{i}+\frac{1}{p-i}\equiv 0 \pmod p}$
The desired result follows by summation.
We can, in fact, prove that the above harmonic sum is divisible by $p^2$, see section 7.8 of G. H. Hardy and E. M. Wright’s An Introduction to the Theory of Numbers for the proof. | 448 | 1,322 | {"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} | 4.53125 | 5 | CC-MAIN-2019-30 | latest | en | 0.625754 |
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Is the positive integer x divisible by each integer from 2 through 6 ?
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When adding two positive integers together, you will always get a positive result. When adding negative integers, you will always get a negative result. When adding a positive integer to a negative integer, you will need to subtract the number with the smaller absolute value from the number with the larger absolute value, keeping the sign of the number with the larger absolute value in your result. When subtracting a positive integer from a negative integer, you will need to change it to an addition problem and change the positive integer to a negative integer. | 120 | 631 | {"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} | 2.90625 | 3 | CC-MAIN-2018-39 | longest | en | 0.800166 |
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Solved
date formula using network days
Posted on 2014-03-13
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Hi Experts excel 2007
Need a date formula in cell c2 to do the following if cells a2 and b2 have date's.
If cell a2 has a date and b2 is blank the in c2 add 7 days
If cell b2 has date and a2 is blank the b2.nj
0
Question by:route217
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LVL 8
Expert Comment
ID: 39925910
try this
``````=TEXT(IF(AND(CELL("Format",A2)="D1",B2=""),A2+7,IF(AND(CELL("Format",B2)="D1",A2=""),B2,"")),"dd/mmm/yyyy")
``````
Thanks
0
LVL 53
Expert Comment
ID: 39925913
What do you mean by
the b2.nj
Regards
0
Author Comment
ID: 39925927
Hi itjockey
If a2 has a date and b2 is blank then a2 + 7 networkdays
If a2 is blank and b2 has a date then b2
Finally
If a2 has a date and b2 has a date then b2.
0
LVL 8
Expert Comment
ID: 39925928
if you want to exclude weekend days in criteria 1 then use this
``````=TEXT(IF(AND(CELL("Format",A2)="D1",B2=""),A2+((7-(NETWORKDAYS(A2,A2+7)))+7),IF(AND(CELL("Format",B2)="D1",A2=""),B2,"")),"dd/mmm/yyyy")
``````
0
LVL 8
Expert Comment
ID: 39925932
ok give me one moment as I dint refresh browser ...your comment I see after I posted.
``````=TEXT(IF(AND(CELL("Format",A2)="D1",B2=""),A2+((7-(NETWORKDAYS(A2,A2+7)))+7),IF(AND(CELL("Format",B2)="D1",A2=""),B2,"")),"dd/mmm/yyyy")
``````
0
LVL 8
Accepted Solution
Naresh Patel earned 2000 total points
ID: 39925945
``````=TEXT(IF(AND(CELL("Format",A2)="D1",B2=""),A2+((7-(NETWORKDAYS(A2,A2+7)))+7),IF(AND(CELL("Format",B2)="D1",A2=""),B2,IF(AND(CELL("Format",A2)="D1",CELL("Format",B2)="D1"),B2,""))),"DD-MMM-YYYY")
``````
Try This
0
LVL 8
Expert Comment
ID: 39925950
do u want me to combine condition 2 & 3 in one statement ?as both way outcome is B2.
Thanks
0
Author Comment
ID: 39925958
Excellence excellent. .
0
LVL 8
Expert Comment
ID: 39925970
Thank You
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A115449 Numbers n such that 4*n^5 - 1 is prime. 0
1, 2, 3, 8, 12, 23, 27, 42, 68, 75, 86, 96, 113, 117, 125, 135, 140, 146, 168, 182, 185, 188, 191, 198, 233, 245, 255, 267, 281, 287, 297, 306, 311, 318, 327, 360, 362, 366, 377, 390, 392, 395, 408, 416, 423, 432, 447, 456, 465, 486, 488, 497, 516, 531, 555 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS EXAMPLE If n=96 then (4*n^5 - 1) = 32614907903 (prime). MATHEMATICA Select[Range[500], PrimeQ[4*#^5 - 1] &] (* Stefan Steinerberger, Mar 09 2006 *) PROG (PARI) for(i=1, 2000, if(isprime(4*i^5-1), print1(i, ", "))) \\ Matthew Conroy, Mar 12 2006 (PARI) for(i=1, 2000, if(isprime(4*i^5-1), print1(i, ", "))) \\ Matthew Conroy, Mar 12 2006 CROSSREFS Cf. A115104, A001912. Sequence in context: A249096 A249367 A257999 * A303851 A218542 A194452 Adjacent sequences: A115446 A115447 A115448 * A115450 A115451 A115452 KEYWORD nonn AUTHOR Parthasarathy Nambi, Mar 08 2006 EXTENSIONS More terms from Stefan Steinerberger, Zak Seidov and Matthew Conroy, Mar 12 2006 More terms from Matthew Conroy, Mar 12 2006 STATUS approved
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# Let X be a discrete random variable whoose probability distribution is defined as follows. <br> P(X=x)={{:(k(x+1)",for x=1,2,3,4"),(2kx", for x=5,6,7"),(0", otherwise"):} <br> where,k is a constant. Calculate (i) the value of k. (ii) E (X). <br> (ii) standard deviation of X.
Question from Class 12 Chapter Probability
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A159601 E.g.f. S(x) satisfies: S(x) = Integral [1 - 2*S(x)^2]^(3/4) dx with S(0)=0. 4
1, -3, 27, -441, 11529, -442827, 23444883, -1636819569, 145703137041, -16106380394643, 2164638920874507, -347592265948756521, 65724760945840254489, -14454276753061349098587 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS E.g.f. S(x) is an odd function; zero terms are omitted. Apart from signs and initial term, same as A159600. Radius of convergence of S(x) is |x| < r where: r = (1/2)*Pi^(3/2)/gamma(3/4)^2 ; r = L/sqrt(2) where L=Lemniscate constant ; r = 1.8540746773013719184338503471952600... Although S(x) diverges at |x|=r, the power series expansion: C(x) = [1 - 2*S(x)^2]^(1/4) converges to C(r) = gamma(3/4)^2/(Pi/2)^(3/2) = 0.7627597635... LINKS Table of n, a(n) for n=1..14. FORMULA E.g.f. S(x) satisfies: C(x)^4 + 2*S(x)^2 = 1 where S'(x) = C(x)^3 and C'(x) = -S(x) with C(0)=1. E.g.f. S(x) satisfies: S(x)/C(x) = e.g.f. of unsigned A104203 where C(x)^4 + 2*S(x)^2 = 1. a(n) = -A159600(n), n>0. - M. F. Hasler, Aug 31 2012 G.f.: (1- 1/Q(0))/x, where Q(k) = 1 + x*(2*k+1)^2/(1 + 2*x*(k+1)^2/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 30 2013 EXAMPLE E.g.f: S(x) = x - 3*x^3/3! + 27*x^5/5! - 441*x^7/7! + 11529*x^9/9! +... S(x)^2 = 2*x^2/2! - 24*x^4/4! + 504*x^6/6! - 16128*x^8/8! +-... C(x)^4 + 2*S(x)^2 = 1 where: C(x) = 1 - x^2/2! + 3*x^4/4! - 27*x^6/6! + 441*x^8/8! -+... C(x)^2 = 1 - 2*x^2/2! + 12*x^4/4! - 144*x^6/6! + 3024*x^8/8! -+... C(x)^3 = 1 - 3*x^2/2! + 27*x^4/4! - 441*x^6/6! + 11529*x^8/8! -+... C(x)^4 = 1 - 4*x^2/2! + 48*x^4/4! - 1008*x^6/6! + 32256*x^8/8! -+... 1/C(x) = C(i*x) = 1 + x^2/2! + 3*x^4/4! + 27*x^6/6! + 441*x^8/8! +... log(C(x)) = -x^2/2! - 12*x^6/6! - 3024*x^10/10! - 4390848*x^14/14! -... Coefficients in log(C(x)) are given by A104203 (ignoring signs). PROG (PARI) {a(n)=local(S=x); for(i=0, 2*n, S=intformal((1-2*S^2+O(x^(2*n)))^(3/4))); (2*n-1)!*polcoeff(S, 2*n-1)} CROSSREFS Cf. A159600 (C(x)), A104203 (unsigned e.g.f. = S(x)/C(x)). Sequence in context: A136719 A279844 A159600 * A193541 A193544 A286306 Adjacent sequences: A159598 A159599 A159600 * A159602 A159603 A159604 KEYWORD sign AUTHOR Paul D. Hanna, May 08 2009 STATUS approved
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Last modified September 23 09:36 EDT 2023. Contains 365544 sequences. (Running on oeis4.) | 1,198 | 2,707 | {"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} | 3.515625 | 4 | CC-MAIN-2023-40 | latest | en | 0.518887 |
https://scm.cri.ensmp.fr/git/linpy.git/blobdiff_plain/0f40fdb54d0fc6505342d008deaa1e9c4c9dbcee..a08ebc700e22f6aee8147cb5b5323a6c040b12db:/doc/examples.rst | 1,656,890,343,000,000,000 | text/plain | crawl-data/CC-MAIN-2022-27/segments/1656104277498.71/warc/CC-MAIN-20220703225409-20220704015409-00280.warc.gz | 549,802,241 | 2,604 | X-Git-Url: https://scm.cri.ensmp.fr/git/linpy.git/blobdiff_plain/0f40fdb54d0fc6505342d008deaa1e9c4c9dbcee..a08ebc700e22f6aee8147cb5b5323a6c040b12db:/doc/examples.rst diff --git a/doc/examples.rst b/doc/examples.rst index ea044b8..b552b7f 100644 --- a/doc/examples.rst +++ b/doc/examples.rst @@ -1,104 +1,118 @@ -LinPy Examples -============== + +.. _examples: + +Examples +======== Basic Examples -------------- - To create any polyhedron, first define the symbols used. Then use the polyhedron functions to define the constraints. The following is a simple running example illustrating some different operations and properties that can be performed by LinPy with two squares. - - >>> from linpy import * - >>> x, y = symbols('x y') - >>> # define the constraints of the polyhedron - >>> square1 = Le(0, x) & Le(x, 2) & Le(0, y) & Le(y, 2) - >>> square1 - And(Ge(x, 0), Ge(-x + 2, 0), Ge(y, 0), Ge(-y + 2, 0)) - - Binary operations and properties examples: - - >>> square2 = Le(1, x) & Le(x, 3) & Le(1, y) & Le(y, 3) - >>> #test equality - >>> square1 == square2 - False - >>> # compute the union of two polyhedrons - >>> square1 | square2 - Or(And(Ge(x, 0), Ge(-x + 2, 0), Ge(y, 0), Ge(-y + 2, 0)), And(Ge(x - 1, 0), Ge(-x + 3, 0), Ge(y - 1, 0), Ge(-y + 3, 0))) - >>> # check if square1 and square2 are disjoint - >>> square1.disjoint(square2) - False - >>> # compute the intersection of two polyhedrons - >>> square1 & square2 - And(Ge(x - 1, 0), Ge(-x + 2, 0), Ge(y - 1, 0), Ge(-y + 2, 0)) - >>> # compute the convex union of two polyhedrons - >>> Polyhedron(square1 | sqaure2) - And(Ge(x, 0), Ge(y, 0), Ge(-y + 3, 0), Ge(-x + 3, 0), Ge(x - y + 2, 0), Ge(-x + y + 2, 0)) - - Unary operation and properties examples: - - >>> square1.isempty() - False - >>> square1.symbols() - (x, y) - >>> square1.inequalities - (x, -x + 2, y, -y + 2) - >>> # project out the variable x - >>> square1.project([x]) - And(Ge(-y + 2, 0), Ge(y, 0)) +To create any polyhedron, first define the symbols used. +Then use the polyhedron functions to define the constraints. +The following is a simple running example illustrating some different operations and properties that can be performed by LinPy with two squares. + +>>> from linpy import * +>>> x, y = symbols('x y') +>>> # define the constraints of the polyhedron +>>> square1 = Le(0, x) & Le(x, 2) & Le(0, y) & Le(y, 2) +>>> square1 +And(Ge(x, 0), Ge(-x + 2, 0), Ge(y, 0), Ge(-y + 2, 0)) + +Binary operations and properties examples: + +>>> # create a polyhedron from a string +>>> square2 = Polyhedron('1 <= x') & Polyhedron('x <= 3') & \ + Polyhedron('1 <= y') & Polyhedron('y <= 3') +>>> #test equality +>>> square1 == square2 +False +>>> # compute the union of two polyhedra +>>> square1 | square2 +Or(And(Ge(x, 0), Ge(-x + 2, 0), Ge(y, 0), Ge(-y + 2, 0)), \ + And(Ge(x - 1, 0), Ge(-x + 3, 0), Ge(y - 1, 0), Ge(-y + 3, 0))) +>>> # check if square1 and square2 are disjoint +>>> square1.disjoint(square2) +False +>>> # compute the intersection of two polyhedra +>>> square1 & square2 +And(Ge(x - 1, 0), Ge(-x + 2, 0), Ge(y - 1, 0), Ge(-y + 2, 0)) +>>> # compute the convex union of two polyhedra +>>> Polyhedron(square1 | sqaure2) +And(Ge(x, 0), Ge(y, 0), Ge(-y + 3, 0), Ge(-x + 3, 0), \ + Ge(x - y + 2, 0), Ge(-x + y + 2, 0)) + +Unary operation and properties examples: + +>>> square1.isempty() +False +>>> # compute the complement of square1 +>>> ~square1 +Or(Ge(-x - 1, 0), Ge(x - 3, 0), And(Ge(x, 0), Ge(-x + 2, 0), \ + Ge(-y - 1, 0)), And(Ge(x, 0), Ge(-x + 2, 0), Ge(y - 3, 0))) +>>> square1.symbols() +(x, y) +>>> square1.inequalities +(x, -x + 2, y, -y + 2) +>>> # project out the variable x +>>> square1.project([x]) +And(Ge(-y + 2, 0), Ge(y, 0)) Plot Examples ------------- - LinPy uses matplotlib plotting library to plot 2D and 3D polygons. The user has the option to pass subplots to the :meth:`plot` method. This can be a useful tool to compare polygons. Also, key word arguments can be passed such as color and the degree of transparency of a polygon. - - >>> import matplotlib.pyplot as plt - >>> from matplotlib import pylab - >>> from mpl_toolkits.mplot3d import Axes3D - >>> from linpy import * - >>> # define the symbols - >>> x, y, z = symbols('x y z') - >>> fig = plt.figure() - >>> cham_plot = fig.add_subplot(1, 1, 1, projection='3d', aspect='equal') - >>> cham_plot.set_title('Chamfered cube') - >>> cham = Le(0, x) & Le(x, 3) & Le(0, y) & Le(y, 3) & Le(0, z) & \ - Le(z, 3) & Le(z - 2, x) & Le(x, z + 2) & Le(1 - z, x) & \ - Le(x, 5 - z) & Le(z - 2, y) & Le(y, z + 2) & Le(1 - z, y) & \ - Le(y, 5 - z) & Le(y - 2, x) & Le(x, y + 2) & Le(1 - y, x) & Le(x, 5 - y) - >>> cham.plot(cham_plot, facecolor='red', alpha=0.75) - >>> pylab.show() - - .. figure:: images/cham_cube.jpg - :align: center +LinPy can use the matplotlib plotting library to plot 2D and 3D polygons. +This can be a useful tool to visualize and compare polygons. +The user has the option to pass plot objects to the :meth:`Domain.plot` method, which provides great flexibility. +Also, keyword arguments can be passed such as color and the degree of transparency of a polygon. + +>>> import matplotlib.pyplot as plt +>>> from matplotlib import pylab +>>> from mpl_toolkits.mplot3d import Axes3D +>>> from linpy import * +>>> # define the symbols +>>> x, y, z = symbols('x y z') +>>> fig = plt.figure() +>>> cham_plot = fig.add_subplot(1, 1, 1, projection='3d', aspect='equal') +>>> cham_plot.set_title('Chamfered cube') +>>> cham = Le(0, x) & Le(x, 3) & Le(0, y) & Le(y, 3) & Le(0, z) & \ + Le(z, 3) & Le(z - 2, x) & Le(x, z + 2) & Le(1 - z, x) & \ + Le(x, 5 - z) & Le(z - 2, y) & Le(y, z + 2) & Le(1 - z, y) & \ + Le(y, 5 - z) & Le(y - 2, x) & Le(x, y + 2) & Le(1 - y, x) & Le(x, 5 - y) +>>> cham.plot(cham_plot, facecolor='red', alpha=0.75) +>>> pylab.show() + +.. figure:: images/cham_cube.jpg + :align: center LinPy can also inspect a polygon's vertices and the integer points included in the polygon. - >>> diamond = Ge(y, x - 1) & Le(y, x + 1) & Ge(y, -x - 1) & Le(y, -x + 1) - >>> diamond.vertices() - [Point({x: Fraction(0, 1), y: Fraction(1, 1)}), \ - Point({x: Fraction(-1, 1), y: Fraction(0, 1)}), \ - Point({x: Fraction(1, 1), y: Fraction(0, 1)}), \ - Point({x: Fraction(0, 1), y: Fraction(-1, 1)})] - >>> diamond.points() - [Point({x: -1, y: 0}), Point({x: 0, y: -1}), Point({x: 0, y: 0}), \ - Point({x: 0, y: 1}), Point({x: 1, y: 0})] - -The user also can pass another plot to the :meth:`plot` method. This can be useful to compare two polyhedrons on the same axis. This example illustrates the union of two squares. - - >>> from linpy import * - >>> import matplotlib.pyplot as plt - >>> from matplotlib import pylab - >>> x, y = symbols('x y') - >>> square1 = Le(0, x) & Le(x, 2) & Le(0, y) & Le(y, 2) - >>> square2 = Le(1, x) & Le(x, 3) & Le(1, y) & Le(y, 3) - >>> fig = plt.figure() - >>> plot = fig.add_subplot(1, 1, 1, aspect='equal') - >>> square1.plot(plot, facecolor='red', alpha=0.3) - >>> square2.plot(plot, facecolor='blue', alpha=0.3) - >>> squares = Polyhedron(square1 + square2) - >>> squares.plot(plot, facecolor='blue', alpha=0.3) - >>> pylab.show() - - .. figure:: images/union.jpg - :align: center - - - - +>>> diamond = Ge(y, x - 1) & Le(y, x + 1) & Ge(y, -x - 1) & Le(y, -x + 1) +>>> diamond.vertices() +[Point({x: Fraction(0, 1), y: Fraction(1, 1)}), \ + Point({x: Fraction(-1, 1), y: Fraction(0, 1)}), \ + Point({x: Fraction(1, 1), y: Fraction(0, 1)}), \ + Point({x: Fraction(0, 1), y: Fraction(-1, 1)})] +>>> diamond.points() +[Point({x: -1, y: 0}), Point({x: 0, y: -1}), Point({x: 0, y: 0}), \ + Point({x: 0, y: 1}), Point({x: 1, y: 0})] + +The user also can pass another plot to the :meth:`Domain.plot` method. +This can be useful to compare two polyhedra on the same axis. +This example illustrates the union of two squares. + +>>> from linpy import * +>>> import matplotlib.pyplot as plt +>>> from matplotlib import pylab +>>> x, y = symbols('x y') +>>> square1 = Le(0, x) & Le(x, 2) & Le(0, y) & Le(y, 2) +>>> square2 = Le(1, x) & Le(x, 3) & Le(1, y) & Le(y, 3) +>>> fig = plt.figure() +>>> plot = fig.add_subplot(1, 1, 1, aspect='equal') +>>> square1.plot(plot, facecolor='red', alpha=0.3) +>>> square2.plot(plot, facecolor='blue', alpha=0.3) +>>> squares = Polyhedron(square1 + square2) +>>> squares.plot(plot, facecolor='blue', alpha=0.3) +>>> pylab.show() + +.. figure:: images/union.jpg + :align: center | 3,071 | 8,379 | {"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-2022-27 | latest | en | 0.546829 |
http://www.topperlearning.com/forums/ask-experts-19/two-metallic-wires-a-and-b-of-same-material-are-connectedin-physics-electricity-48439/reply | 1,490,850,898,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218191986.44/warc/CC-MAIN-20170322212951-00405-ip-10-233-31-227.ec2.internal.warc.gz | 731,877,752 | 36,134 | Question
Sat June 18, 2011 By: Gurleen Kaur
# Two metallic wires A and B of same material are connectedin parallel. Wire A has length l and radius r and wire B has length 2l. Compute the ratio of the total resistance of parallel combination and the resistance of wire A?
Sun June 19, 2011
The electrical resistance of a uniform conductor is given in terms of resistivity by:
where l is the length of the conductor in SI units of meters, a is the cross-sectional area (for a round wire a = ?r2 if r is radius) in units of meters squared, and ? is the resistivity in units of ohm·meters.
Therefore
and
Resistance in parallel will be
Now the ratio between total resistance and reistance of wire A is
Related Questions
Fri October 28, 2016 | 197 | 743 | {"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": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2017-13 | latest | en | 0.887956 |
https://ithelp.ithome.com.tw/articles/10206818 | 1,603,548,891,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107883636.39/warc/CC-MAIN-20201024135444-20201024165444-00267.warc.gz | 394,869,489 | 13,494 | DAY 16
0
Software Development
## [演算法] 二分搜尋 (Binary Search)
``````data = [1, 2, 3, 4, 5, 6, 7, 8, 9]
def binary_search(data, key):
#設置選取範圍的指標
low = 0
upper = len(data) - 1
while low <= upper:
mid = (low + upper) / 2 #取中間索引的值
if data[mid] < key: #若搜尋值比中間的值大，將中間索引+1，取右半
low = mid + 1
elif data[mid] > key: #若搜尋值比中間的值小，將中間索引+1，取左半
upper = mid - 1
else: #若搜尋值等於中間的值，則回傳
return mid
return -1
index = binary_search(data, 5)
if index >= 0:
print("找到數值於索引 " + str(index))
else:
print("找不到數值")
``````
30天學演算法和資料結構31
### 1 則留言
0
showlin
iT邦新手 5 級 ‧ 2019-09-10 17:11:05
ramonliao iT邦新手 5 級 ‧ 2019-09-12 09:36:28 檢舉 | 338 | 636 | {"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-2020-45 | longest | en | 0.236496 |
https://www.airmilescalculator.com/distance/sys-to-bqj/ | 1,606,456,597,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141189141.23/warc/CC-MAIN-20201127044624-20201127074624-00107.warc.gz | 547,920,126 | 76,950 | # Distance between Saskylakh (SYS) and Batagay (BQJ)
Flight distance from Saskylakh to Batagay (Saskylakh Airport – Batagay Airport) is 572 miles / 921 kilometers / 497 nautical miles. Estimated flight time is 1 hour 34 minutes.
Driving distance from Saskylakh (SYS) to Batagay (BQJ) is 2367 miles / 3809 kilometers and travel time by car is about 84 hours 57 minutes.
## Map of flight path and driving directions from Saskylakh to Batagay.
Shortest flight path between Saskylakh Airport (SYS) and Batagay Airport (BQJ).
## How far is Batagay from Saskylakh?
There are several ways to calculate distances between Saskylakh and Batagay. Here are two common methods:
Vincenty's formula (applied above)
• 572.059 miles
• 920.640 kilometers
• 497.106 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 569.859 miles
• 917.098 kilometers
• 495.194 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
Country: Russia
IATA Code: SYS
ICAO Code: UERS
Coordinates: 71°55′40″N, 114°4′48″E
B Batagay Airport
City: Batagay
Country: Russia
IATA Code: BQJ
ICAO Code: UEBB
Coordinates: 67°38′52″N, 134°41′42″E
## Time difference and current local times
The time difference between Saskylakh and Batagay is 1 hour. Batagay is 1 hour ahead of Saskylakh.
+09
+10
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 109 kg (240 pounds).
## Frequent Flyer Miles Calculator
Saskylakh (SYS) → Batagay (BQJ).
Distance:
572
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
572
Round trip? | 506 | 1,828 | {"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} | 2.53125 | 3 | CC-MAIN-2020-50 | latest | en | 0.805003 |
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# Find the equation of $k$ for which the quadratic equation $\left( {3k + 1} \right){x^2} + 2\left( {k + 1} \right)x + 1 = 0$ has real and equal roots.
Last updated date: 20th Jun 2024
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Hint:
Here we have to find the value of the variable $k$. We will find the discriminant of the given quadratic equation. As the roots are real and equal, we will equate the value of the discriminant to 0. Then we will get the equation including $k$ as a variable. After solving the obtained equation, we will get the required value of the variable $k$.
Formula used:
We will use the formula of the discriminant, $D = {b^2} - 4ac$, where $b$ is the coefficient of $x$, $c$ is the constant term of equation and $a$ is the coefficient of ${x^{}}$.
Complete step by step solution:
We will first find the value of the discriminant of the given equation.
Substituting the value of $a$, $b$ and $c$ from the given quadratic equation in the formula of discriminant $D = {b^2} - 4ac$ , we get
$\Rightarrow D = {\left[ {2\left( {k + 1} \right)} \right]^2} - 4 \times \left( {3k + 1} \right) \times 1$
Now, we know that the roots of the given equation are real and equal. Therefore, the discriminant will be 0.
Equating the above equation to 0, we get
$\Rightarrow {\left[ {2\left( {k + 1} \right)} \right]^2} - 4 \times \left( {3k + 1} \right) \times 1 = 0$
On simplifying the terms, we get
$\Rightarrow 4{k^2} + 4 + 8k - 12k - 4 = 0$
Adding the like terms, we get
$\Rightarrow 4{k^2} - 4k = 0$
On factoring the equation, we get
$\Rightarrow 4k\left( {k - 1} \right) = 0$
Now applying zero product property, we can write
$\begin{array}{l} \Rightarrow 4k = 0\\ \Rightarrow k = 0\end{array}$
or
$\begin{array}{l} \Rightarrow \left( {k - 1} \right) = 0\\ \Rightarrow k = 1\end{array}$ .
Thus, the possible values of $k$ are 0 and 1.
Note:
We need to keep in mind that the roots of any quadratic equation are two, and similarly the roots of the cubic equation are three and so on. Generally, the numbers of possible roots of any equation are equal to the highest power of the equation. The values of the roots always satisfy their respective equation i.e. when we put values of the roots in their respective equation, we get the value as zero. | 726 | 2,353 | {"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} | 4.875 | 5 | CC-MAIN-2024-26 | latest | en | 0.78942 |
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Frequent Visitor
## Adding index number based on category
Hello, folks!
How can I add an index number that runs based on category type stated in another column? See the example picture. This is what I want to achiecve. I'd like the index column to give a number-series to all categories seperatly. Essentialy I would then be able to sort on category and get a continous series running from 1 to "whatever-number".
Aditionally I would like the index series in each category to run according to date in another column.
Date Category Index 15.03.2018 Child 1 12.04.2018 Adult 2 13.04.2018 Adult 3 02.04.2018 Adult 1 25.06.2018 Child 2 22.05.2018 Adult 4 27.09.2018 Child 4 22.08.2018 Child 3 15.10.2018 Child 5 02.10.2018 Adult 5
Is there any way to do this? I found this this forum-post interesting, but I can't really get it to fit my goal exactly. However, maybe some of you understand how to use this info for my purpose.
Best:
- Per-J.H.
1 ACCEPTED SOLUTION
Accepted Solutions
Super Contributor
## Re: Adding index number based on category
What if you add another column called FkDte with this formula
=Table1[Date]+RANDBETWEEN(1,1000) / 10000
and then add another columns with this formula
=CALCULATE(
COUNTROWS( Table1 ),
ALLEXCEPT( Table1, Table1[Category] ),
Table1[FkDte] < EARLIER( Table1[FkDte] )
) + 1
by the way, do you only have 2 columns in the original table (Date, Category), or do you have other columns that could avoid the usage of the FkDte column?
Proud to be a Datanaut!
10 REPLIES 10
Super Contributor
## Re: Adding index number based on category
@Per-J
You need to convert your date column to real dates and do it like this:
=
CALCULATE (
COUNTROWS ( Table1 ),
ALLEXCEPT ( Table1, Table1[Category] ),
Table1[Date] < EARLIER ( Table1[Date] )
)
+ 1
Proud to be a Datanaut!
Super User I
## Re: Adding index number based on category
@Per-J Please try this using "New Column"
`Index = RANKX(FILTER(Test13Rank,Test13Rank[Category]=EARLIER(Test13Rank[Category])),Test13Rank[Date],,ASC,Dense)`
Did I answer your question? Mark my post as a solution !
Proud to be a Datanaut !
Frequent Visitor
## Re: Adding index number based on category
Both of the solutions mentioned here gave close to what I'm looking for. Almost there. I really liked the simplicity of this, last example.
There is one major problem, though. Whenever there is similar dates, this code returns the same index-value. I must have unique values for each row, even if there are the same dates. Its ok that observations on the same date have 4, 5 and 6 as values, as long as they are before the index number for a later date.
My new column gave me four #11 - values, as there are four observations on the same day. These values have to be 11, 12, 13 and 14.
Is there a neat way of getting this done?
Super Contributor
## Re: Adding index number based on category
@Per-J
could you post this dataset where you get dupes?
thanks
Proud to be a Datanaut!
Super User I
## Re: Adding index number based on category
@Per-J Ok, then just remove the DENSE from the Rank. So it will be...
`Index = RANKX(FILTER(Test13Rank,Test13Rank[Category]=EARLIER(Test13Rank[Category])),Test13Rank[Date],,ASC)`
Did I answer your question? Mark my post as a solution !
Proud to be a Datanaut !
Frequent Visitor
## Re: Adding index number based on category
@LivioLanzo
The example I've given is only tentative and the real dataset is approx. 10k rows with a lot of data and columns. It's classified info, so I'd rather give examples. Under I've posted my results using your formulas. As you can see, same dates gives same number in both cases. In stead of 3, 3, 3 and 7, 7, 7, I need, 3, 4, 5 and 7, 8, 9.
Hope this helps understanding the problem.
Frequent Visitor
## Re: Adding index number based on category
@PattemManohar
Removing the "dense"-statement only yealds the same result as LivioLanzo. There is still identical index-number on the same dates.
Tricky.
Frequent Visitor
## Re: Adding index number based on category
Dear @PattemManohar @LivioLanzo
Thanks for helping with this issue. You are most helpful.
I came to think that maybe there is a better way of getting this job done. If there is a formula, measure (or some other function), that can do two operations in separat orders.
First: sort by date
Then: give assending index numbers based on (only) category
This way, date is not in the mix when doing the numbering task. Is this a way to go, or even possible?
Best:
- Per-J. H.
Super Contributor
## Re: Adding index number based on category
What if you add another column called FkDte with this formula
=Table1[Date]+RANDBETWEEN(1,1000) / 10000
and then add another columns with this formula
=CALCULATE(
COUNTROWS( Table1 ),
ALLEXCEPT( Table1, Table1[Category] ),
Table1[FkDte] < EARLIER( Table1[FkDte] )
) + 1
by the way, do you only have 2 columns in the original table (Date, Category), or do you have other columns that could avoid the usage of the FkDte column?
Proud to be a Datanaut!
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Top Kudoed Authors | 1,536 | 5,547 | {"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-2020-10 | latest | en | 0.785568 |
http://betterlesson.com/lesson/resource/1911422/69034/creating-homogeneous-groups-mov | 1,477,553,395,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721142.99/warc/CC-MAIN-20161020183841-00074-ip-10-171-6-4.ec2.internal.warc.gz | 24,973,397 | 22,348 | ## Creating Homogeneous Groups.mov - Section 4: Where does it fall on the number line?
Creating Homogeneous Groups.mov
# Where Does That Fall On The Number Line?
Unit 3: Integers and Rational Numbers
Lesson 2 of 17
## Big Idea: Where does five-thirds belong on a number line? What about 1.61? Students apply their knowledge of fractions and decimals in order to compare and order fractions and decimals.
Print Lesson
16 teachers like this lesson
Standards:
Subject(s):
Math, equivalent, Fractions, Number Sense and Operations, Decimals, Comparing Numbers, 6th grade, master teacher project, number line
60 minutes
### Andrea Palmer
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Environment: Urban | 333 | 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} | 3.5625 | 4 | CC-MAIN-2016-44 | longest | en | 0.771934 |
http://www.ridgerat-tech.us/tech/Kalman/Ch9/ARMA-9.html | 1,534,686,518,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215176.72/warc/CC-MAIN-20180819125734-20180819145734-00407.warc.gz | 573,712,763 | 5,882 | # Converting ARMA Models
Developing Models for Kalman Filters
The model fit obtained in the previous installment tries to determine a linear relationship between past input and output data to predict future outputs. It showed some distinct promise. But it is not finished. There are two important issues to resolve.
1. The dynamic state transition model used by the Kalman filter uses only current input values and state from the previous step. The Least Squares model fit used data from several past steps. We must resolve this structural inconsistency.
2. When the model is applied, it does not have the advantage of knowing what the "real" system output is, the way that the Least Squares fitting process knows. The most the model can know about is the past values of its predictions of output variables. Substituting predicted values for real values is always tricky. But is it merely tricky, or disastrous?
This installment addresses the first of these challenges.
## ARMA models
In the field of time series analysis, the sort of model where the next output projection is based on a set of past inputs and outputs is known as an ARMA model. ARMA stands for Autoregressive Moving Average.
• The autoregressive part means "the next output is determined from a weighted sum of some number of prior outputs."
• The moving average part means "the next output is determined from a weighted sum of some number of prior inputs."
Put them together and you get...
${y}_{1}\phantom{\rule{1.0em}{0ex}}=\phantom{\rule{1.0em}{0ex}}\stackrel{0}{\sum _{i=-M}}{a}_{i}{x}_{i}\phantom{\rule{1.0em}{0ex}}+\phantom{\rule{1.0em}{0ex}}\stackrel{0}{\sum _{i=-N}}{b}_{i}{y}_{i}$
This has little resemblance to a Kalman Filter state transition model. So the immediate task is to reorganize the model into a state transition form.
## Shift Register construct
A special and artificial kind of state transition matrix called a "shift register" looks something like the following.
$A\phantom{\rule{1.0em}{0ex}}=\phantom{\rule{1.0em}{0ex}}\left(\begin{array}{ccccc}0\phantom{\rule{0.4em}{0ex}}& 0\phantom{\rule{0.4em}{0ex}}& 0\phantom{\rule{0.4em}{0ex}}& 0\phantom{\rule{0.4em}{0ex}}& 0\phantom{\rule{0.4em}{0ex}}\\ 1\phantom{\rule{0.4em}{0ex}}& 0\phantom{\rule{0.4em}{0ex}}& 0\phantom{\rule{0.4em}{0ex}}& 0\phantom{\rule{0.4em}{0ex}}& 0\phantom{\rule{0.4em}{0ex}}\\ 0\phantom{\rule{0.4em}{0ex}}& 1\phantom{\rule{0.4em}{0ex}}& 0\phantom{\rule{0.4em}{0ex}}& 0\phantom{\rule{0.4em}{0ex}}& 0\phantom{\rule{0.4em}{0ex}}\\ 0\phantom{\rule{0.4em}{0ex}}& 0\phantom{\rule{0.4em}{0ex}}& 1\phantom{\rule{0.4em}{0ex}}& 0\phantom{\rule{0.4em}{0ex}}& 0\phantom{\rule{0.4em}{0ex}}\\ 0\phantom{\rule{0.4em}{0ex}}& 0\phantom{\rule{0.4em}{0ex}}& 0\phantom{\rule{0.4em}{0ex}}& 1\phantom{\rule{0.4em}{0ex}}& 0\phantom{\rule{0.4em}{0ex}}\end{array}\right)$
Notice how the 1.0 values appear on the first sub-diagonal line and nowhere else. For the model fit determined in the previous installment, there are five values of output history required to calculate the next state predictions, so in this particular case, the matrix has five rows and five columns.
Let's treat this like a regular dynamic state transition matrix and see what happens. We will define a shift register vector v that contains a list of past values of the output variable y. We transform vector v by multiplying with the special shift register matrix A.
v1 = A v0
If you look at the details of the matrix multiply operation, you can see that it has the effect of shifting the terms in the previous v vector by one position downward, leaving the first location vacant. The previous final value in the list drops away. We just need a way to inject the latest new data into the list at the first position.
Define a coupling matrix B having the following structure.
$B\phantom{\rule{1.0em}{0ex}}=\phantom{\rule{1.0em}{0ex}}\left(\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right)$
Let the new input value u be the latest new output term to be introduced. Add the additional product term B · y0 into the previous shift register relationship. This injects the new value into the vacant first location without affecting any other terms.
v1 = A v0 + B y0
The result is an artificial "state transition equation" representation for the shift register. It is not very compact, but it has the right form for dynamic state transition equations.
That takes care of past outputs.
A similar but separate shift register construct is needed to manage the history of input values. There is one small difference. Since this particular model uses one current input terms plus 4 more terms from past history, the shift register only needs to maintain the 4 past input terms.
The next few steps finish the model:
• combine the two shift registers to form a single matrix,
• introduce the newest terms in the appropriate locations,
• incorporate the model calculations for the next output.
## Unifying the shift registers
Let Ay and Au be shift register matrices of size 5x5 and 4x4 for the output history and input history data, respectively. Let By and Bu be input coupling matrices with one nonzero term each, as described above, for introducing new output observation and input terms into the front of the respective shift registers. Then stack up copies of these two matrices, and fill the rest of the matrix terms with zeroes, to build up a 9x9 matrix in the following manner.
$A\phantom{\rule{1.0em}{0ex}}=\phantom{\rule{1.0em}{0ex}}\left(\begin{array}{cc}\mathrm{Ay}\phantom{\rule{1.0em}{0ex}}& 0\\ 0\phantom{\rule{1.0em}{0ex}}& \mathrm{Au}\end{array}\right)$ $B\phantom{\rule{1.0em}{0ex}}=\phantom{\rule{1.0em}{0ex}}\left(\begin{array}{cc}\mathrm{By}\phantom{\rule{1.0em}{0ex}}& 0\\ 0\phantom{\rule{1.0em}{0ex}}& \mathrm{Bu}\end{array}\right)$ $u\phantom{\rule{1.0em}{0ex}}=\phantom{\rule{1.0em}{0ex}}\left(\begin{array}{c}{y}_{0}\\ {u}_{0}\end{array}\right)$
Now, matrix operations can be used at each time step to update the history information. The most current history information is stored in the state vector v.
v = A v + B u
The next adjustment introduces the model fit parameter values obtained in the previous installment. The new value of the output variable does not come in from some kind of external input, as represented currently. Instead, this prediction must be calculated using the existing elements of the history vector v. Last time, this was the best fit formula we derived:
yN+1 = [yN+0 yN-1 yN-2 yN-3 yN-4 uN+0 uN-1 uN-2 uN-3 uN-4 ] ·
[ b0 b-1 b-2 b-3 b-4 a0 a-1 a-2 a-3 a-4 ]T
0.9000531 Multipliers 'b' for output history terms
0.0698574
0.0249757
-0.0734740
0.0148904
-0.3757921 Multipliers 'a' for input history terms
-0.1939514
-0.0527318
0.0051917
0.0230597
To apply the model parameters, make the following adjustments to the shift register equations.
• In the first row of the composite shift register matrix, replace the first five terms (all zero) with the 'b' parameter values from the model, associated with prior output history terms.
• In the first term of the second column of the B coupling matrix, store the 'a' parameter term corresponding to the most recent input value.
• The output variable is no longer used as a direct input. Remove the first term from the u input vector, and remove the first column of the composite B matrix, leaving only one column in that matrix.
• In the first row of the composite shift register matrix, replace the four final zero terms with the remaining 4 model parameter values associated with past input history.
• Set all terms in the matrix column 6 except for the first term to zero values, since the first input terms is no longer present.
Listing 1 shows Octave code to set up the dynamic model.
Listing 1
% Construct the state transition matrix model
AMat = zeros(9,9);
BMat = zeros(9,1);
CMat = zeros(1,9);
% Output history shift
AMat(2,1) = 1.0;
AMat(3,2) = 1.0;
AMat(4,3) = 1.0;
AMat(5,4) = 1.0;
% Input history shift
AMat(7,6) = 1.0;
AMat(8,7) = 1.0;
AMat(9,8) = 1.0;
% Model terms for next output prediction
AMat(1,1) = 0.9000531;
AMat(1,2) = 0.0698574;
AMat(1,3) = 0.0249757;
AMat(1,4) = -0.0734740;
AMat(1,5) = 0.014890;
AMat(1,6) = -0.1939514;
AMat(1,7) = -0.0527318;
AMat(1,8) = 0.0051917;
AMat(1,9) = 0.0230597
% Current input feeds next output prediction and input shift
BMat = [ -0.3757921, 0, 0, 0, 0, 1.0, 0, 0, 0 ]'
% Next output prediction is the first state variable
CMat(1)=1.0
In listing 2, verify that Octave has received the correct matrix configuration.
Listing 2
AMat =
Columns 1 through 5:
0.90005 0.06986 0.02498 -0.07347 0.01489
1.00000 0.00000 0.00000 0.00000 0.00000
0.00000 1.00000 0.00000 0.00000 0.00000
0.00000 0.00000 1.00000 0.00000 0.00000
0.00000 0.00000 0.00000 1.00000 0.00000
0.00000 0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000 0.00000
Columns 6 through 9:
-0.19395 -0.05273 0.00519 0.02306
0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000
0.00000 0.00000 0.00000 0.00000
1.00000 0.00000 0.00000 0.00000
0.00000 1.00000 0.00000 0.00000
0.00000 0.00000 1.00000 0.00000
BMat =
-0.37579
0.00000
0.00000
0.00000
0.00000
1.00000
0.00000
0.00000
0.00000
CMat =
1 0 0 0 0 0 0 0 0
Now it is left as an exercise for you to verify that the following things will occur when a new input value is provided and the matrix multiply operation generates the next state.
1. The previous output history values are shifted down one position, dropping the last value out.
2. The new value calculated for the first shift register state variable is the predicted output value for the next time step.
3. The previous input history values are shifted down one position, all except for the last, which is dropped.
4. The newest input value, in addition to being used for the next output state calculations, is copied into the empty location at the front of the list of input history terms.
5. The new value calculated for the first shift register state variable is the predicted output value for the next time step.
The shift register formulation is a bit artificial, but at least it has the right kind of state transition form to be used for Kalman Filter purposes. The only thing missing is testing whether it does anything useful. That's where we will continue next time.
Contents of the "Developing Models for Kalman Filters" section of the ridgerat-tech.us website, including this page, are licensed under a Creative Commons Attribution-ShareAlike 4.0 International License unless otherwise specifically noted. For complete information about the terms of this license, see http://creativecommons.org/licenses/by-sa/4.0/. The license allows usage and derivative works for individual or commercial purposes under certain restrictions. For permissions beyond the scope of this license, see the contact information page for this site. | 3,462 | 11,178 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "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-2018-34 | longest | en | 0.876702 |
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# Floors
This is a seemingly simple problem that I found very interesting, with a surprising answer. Find the sum of all positive solutions to $$2x^2-x\lfloor x\rfloor=5$$ (HMNT 2011 G5).
Note by Cody Johnson
3 years, 5 months ago
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Is G guts? My first reaction was "why is this a geometry problem!?" :P
- 3 years, 5 months ago
Step One: Let $$x=q+r$$ where $$q\in\mathbb{Z}$$ and $$0\le r<1$$.
- 3 years, 5 months ago
It is true that either $$q=1$$ or $$q=2$$. We can quickly find the answer afterwards.
- 3 years, 5 months ago
Yeah we can also bound it like this: $$2x^2-5=x\lfloor x\rfloor\le x^2$$
- 3 years, 5 months ago
Can this be applied to negative numbers too? Why is $$-\frac52$$ a solution, yet $$\left(-\frac52\right)^2=6.25>5$$?
- 3 years, 5 months ago
For negatives, $$x\lfloor x\rfloor \geq x^2$$.
- 3 years, 5 months ago | 621 | 1,726 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2017-47 | longest | en | 0.754084 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-10th-edition/chapter-2-differentiation-2-2-exercises-page-114/16 | 1,529,378,691,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861752.19/warc/CC-MAIN-20180619021643-20180619041643-00260.warc.gz | 819,333,667 | 12,582 | # Chapter 2 - Differentiation - 2.2 Exercises: 16
The derivative is $4-9x^2$.
#### Work Step by Step
Whenever two or more functions are being added or subtracted, the overall derivative would be the sum of the derivatives of every smaller function . The derivative of 4x is 4 and the derivative of $-3x^3$ is $-9x^2$ hence the overall derivative is $4-9x^2$.
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http://wwiec.com/brian-thompson-ktbjvr/a78f9a-disconnected-graph-with-6-vertices | 1,675,206,920,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499891.42/warc/CC-MAIN-20230131222253-20230201012253-00561.warc.gz | 44,707,350 | 11,273 | A graph G is disconnected, if it does not contain at least two connected vertices. Connected and Disconnected Graph: Connected Graph: A graph is called connected if there is a path from any vertex u to v or vice-versa. O Fo... Q: ay non-isomorphic trees on 6 vertices are there? Also, we should note that a spanning tree covers all the vertices of a given graph so it can’t be disconnected. I have drawn a picture to illustrate my problem. |3D Let G be a plane graph with n vertices. Disconnected Graph- A graph in which there does not exist any path between at least one pair of vertices is called as a disconnected graph. (a) Find the Fo... A: Given: f(x)=1 if -π≤x<0-1 if 0≤x<π graph that is not simple. It is not possible to visit from the vertices of one component to the vertices of other component. The command is . Following are steps of simple approach for connected graph. In above graph there are no articulation points because graph does not become disconnected by removing any single vertex. Vertices with only out-arrows (like 3 … More efficient algorithms might exist. (b) Find its radius of convergence. 17622 Advanced Graph Theory IIT Kharagpur, Spring Semester, 2002Œ2003 Exercise set 1 (Fundamental concepts) 1. 3 isolated vertices . Example 5.5.5. It has n(n-1)/2 edges . Explanation: After removing either B or C, the graph becomes disconnected. the same as G, we must have the same graph. The task is to find the count of singleton sub-graphs. Now we consider the case for n = p3 in the following theorem. The following graph is a forest consisting of three trees: The following graph is a not a tree:. I'm given a graph with many seperate components. Split vertices of disconnected bipartite graph equally. A: Consider the provided equation x4+2x3+x2+x=0. 4. Hence it is a connected graph. This is because instead of counting edges, you can count all the possible pairs of vertices that could be its endpoints. Each component is bipartite. G is a disconnected graph with two components g1 and g2 if the incidence of G can be as a block diagonal matrix X(g ) 0 1 X 0 X(g ) 2 . Draw a simple graph (or argue why one cannot exist) that (a) has 6 vertices, 12 edges, and is disconnected. I saw this on "Counting Disconnected Structures: Chemical Trees, Fullerenes, I-graphs" on the link: https://hrcak.srce.hr/file/4232 but I did not understand how I can use … Proof. Find answers to questions asked by student like you. Therefore, it is a disconnected graph. Say we have a graph with the vertex set , and the edge set . Connected and Disconnected. Prove or disprove: The complement of a simple disconnected graph G must be connected. I'm given a graph with many seperate components. 11. = COs (c) has 7 vertices, is acyclic, connected, and has 6 vertices of degree 2. Combinatorics Instructor: Jie Ma, Scribed by Jun Gao, Jialin He and Tianchi Yang 1 Lecture 6. Let X be a graph with 15 vertices and 4 components. Hence the vertex connectivity of Γ[Zp2] is p− 2. D. 19. Let’s simplify this further. Definition 1.2.A component of a graph G is a maximal connected subgraph of G. Definition 1.3.A graph T is called a tree if it is connected but contains no cycles. 17622 Advanced Graph Theory IIT Kharagpur, Spring Semester, 2002Œ2003 Exercise set 1 (Fundamental concepts) 1. A graph in which there does not exist any path between at least one pair of vertices is called as a disconnected graph. Split vertices of disconnected bipartite graph equally. If it only has P200 bills and P100 bills and Following theorem illustrates a simple relationship between the number of vertices, faces and edges of a graph and its dual. Hence it is a connected graph. GraphPlot[Table[1, {6}, {6}], EdgeRenderingFunction -> None] For the given graph(G), which of the following statements is true? C. 18. A: Given function is fz=zexpiz2+11+2iz Next we give simple graphs by their number of edges, not allowing isolated vertices but allowing disconnected graphs. 8. deleted , so the number of edges decreases . 6-Graphs - View presentation slides online. Consider the two conditions of being tree: being connected, and not having any cycles. deleted , so the number of edges decreases . B. A connected graph G is said to be 2-vertex-connected (or 2-connected) if it has more than 2 vertices and remains connected on removal of any vertices. The present value is given ... Q: Exactly one of the following statements is false: If uand vbelong to different components of G, then the edge uv2E(G ). the given function is fx=x+5x-69-x. 2x – y? Hi everybody, I have a graph with approx. A connected planar graph having 6 vertices, 7 edges contains _____ regions. The objective is to compute the values of x. (b) is Eulerian, is bipartite, and is Hamiltonian. Removing any edge makes G disconnected, because a graph with n vertices clearly needs at least n −1 edges to be connected. A graph G is disconnected, if it does not contain at least two connected vertices. Experts are waiting 24/7 to provide step-by-step solutions in as fast as 30 minutes!*. 5. (b) is Eulerian, is bipartite, and is… Examples: Input : Vertices : 6 Edges : 1 2 1 3 5 6 Output : 1 Explanation : The Graph has 3 components : {1-2-3}, {5-6}, {4} Out of these, the only component forming singleton graph is {4}. 1 The Fourier series expansion f(x)=a02+∑n=1∞ancosnx+bnsinn... Q: X4 + 2X3 + X2 + X =0 We begin by assuming we have a disconnected graph G. Now consider two vertices x and y in the complement. a) G is a complete graph b) G is not a connected graph ... A connected planar graph having 6 vertices, 7 edges contains _____ regions. Any two distinct vertices x and y have the property that degx+degy 19. Ask Question Asked 9 years, 7 months ago. a) 15 b) 3 c) 1 d) 11 the complete graph Kn . Graphs. But we can make graph disconnected by removing more than 1 vertex, for example if we remove 4,6 vertices graph becomes disconnected. disconnected graphs G with c vertices in each component and rn(G) = c + 1. 15k vertices which will have a couple of very large components where are to find most of the vertices, and then all others won’t be very connected. Let $$G$$ be a graph on $$n$$ vertices. If you give an example, make sure you justify/explain why the complete graph Kn . 3. 10. 11 D. 19. A graph in which there does not exist any path between at least one pair of vertices is called as a disconnected graph. A graph with just one vertex is connected. When z=i ⇒x=0 and y=1 P3 Co.35) If you give an example, make sure you justify/explain why that example works. and 7. But we can make graph disconnected by removing more than 1 vertex, for example if we remove 4,6 vertices graph becomes disconnected. Removing any edge makes G disconnected, because a graph with n vertices clearly needs at least n −1 edges to be connected. Let Gbe a simple disconnected graph and u;v2V(G). periodic with period 27. Median response time is 34 minutes and may be longer for new subjects. The following graph is an example of a Disconnected Graph, where there are two components, one with 'a', 'b', 'c', 'd' vertices and another with 'e', 'f', 'g', 'h' vertices. Show that $$G$$ cannot be disconnected with exactly two isomorphic connected components. A null graph of more than one vertex is disconnected (Fig 3.12). 12. An off diagonal entry of X 2 gives the number possible paths of length 2 between two vertices… 1) For every vertex v, do following …..a) Remove v from graph..…b) See if the graph remains connected (We can either use BFS or DFS) …..c) Add v back to the graph Median response time is 34 minutes and may be longer for new subjects. The closest point to... Q: Define h(x) = x° sin(1/x) for x # 0 and h(0) = 0. Since G is disconnected, there exist 2 vertices x, y that do not belong to a path. The following graph is an example of a Disconnected Graph, where there are two components, one with ‘a’, ‘b’, ‘c’, ‘d’ vertices and another with ‘e’, ’f’, ‘g’, ‘h’ vertices. representation 6. Given a undirected connected graph, check if the graph is 2-vertex connected or not. Q: Problem 2: A wallet has an amount of P5, 000. Every graph drawn so far has been connected. How to find set of vertices such that after removing those vertices graph becomes disconnected. Prove that h is differentiable at x = 0, and find ... Q: Relying The $12$ Hamiltonian paths are those connected graphs over $4$ vertices whose complements are also connect: thus the remaining $2^6 - 12 = 52$ graphs are divided into pairs of complement graphs which are connected and disconnected, Fig 3.9(a) is a connected graph where as Fig 3.13 are disconnected graphs. Is k5 a Hamiltonian? If our graph is a tree, we know that every vertex in the graph is a cut point. It is legal for a graph to have disconnected components, and even lone vertices without a single connection. Solution The statement is true. G1 has 7(7-1)/2 = 21 edges . No, the complete graph with 5 vertices has 10 edges and the complete graph has the largest number of edges possible in a simple graph. that example works. on the linear differential equation method, find the general solution Viewed 1k times 1. Therefore, it is a disconnected graph. For the given graph(G), which of the following statements is true? A graph is connected if there is a path from any vertex to any other vertex. For example, the vertices of the below graph have degrees (3, 2, 2, 1). The graph below is disconnected; there is no way to get from the vertices on the left to the vertices on the right. 15k vertices which will have a couple of very large components where are to find most of the vertices, and then all others won’t be very connected. Thereore , G1 must have. Explanation: After removing either B or C, the graph becomes disconnected. 11. *Response times vary by subject and question complexity. Therefore, it is a connected graph. simple disconnected graph with 6 vertices. Q: Calculate the volume of the solid occupying the region under the plane -2x – 2y+z= 3 and above the f(2) = zexp(iz?) When... *Response times vary by subject and question complexity. 10. We know G1 has 4 components and 10 vertices , so G1 has K7 and. 1) For every vertex v, do following …..a) Remove v from graph..…b) See if the graph remains connected (We can either use BFS or DFS) …..c) Add v back to the graph Thank you. Prove that the complement of a disconnected graph is connected. Active 9 years, 7 months ago. QUESTION: 18. Then, Volume V. Q: Examine the point and uniform convergence of the function array in the range shown. 1+ 2iz Q.E.D. Q: Solve the ODE using the method of undetermined coefficients. The $12$ Hamiltonian paths are those connected graphs over $4$ vertices whose complements are also connect: thus the remaining $2^6 - 12 = 52$ graphs are divided into pairs of complement graphs which are connected and disconnected, Find answers to questions asked by student like you. Exercises 7. Theorem 6.3 (Fary) Every triangulated planar graph has a straight line representation. Q.E.D. A spanning tree on is a subset of where and . a) 15 b) 3 c) 1 d) 11 In above graph there are no articulation points because graph does not become disconnected by removing any single vertex. Median response time is 34 minutes and may be longer for new subjects. The provi... Q: Two payments of $12,000 and$2,700 are due in 1 year and 2 years, respectively. ∫i2-i(3xy+iy2)dz (Enter your answers as a comma-separated list.) Theorem 6 If G is a connected planar graph with n vertices, f faces and m edges, then G* has f vertices, n faces and m edges. An edgeless graph with two or more vertices is disconnected. fx=a02+∑n=1∞ancos... Q: 1 More efficient algorithms might exist. Experts are waiting 24/7 to provide step-by-step solutions in as fast as 30 minutes!*. 6-Graphs - View presentation slides online. A graph G is connected if each pair of vertices in G belongs to a path; otherwise, G is disconnected. 0. We, know that z=x+iy I saw this on "Counting Disconnected Structures: Chemical Trees, Fullerenes, I-graphs" on the link: https://hrcak.srce.hr/file/4232 but I did not understand how I can use … I have some difficulties in finding the proper layout to get a decent plot, even the algorithms for large graph don’t produce a satisfactory result. Theorem 3.2. The Unlabelled Trees on 6 Vertices Exercise Show that when 1 ≤ n ≤ 6, the number of trees with vertex set {1, 2, …, n} is nn-2. Hence it is a connected graph. Ple... *Response times vary by subject and question complexity. Prove or disprove: The complement of a simple disconnected graph must be connected. (c) has 7 vertices, is acyclic, connected, and has 6 vertices of degree 2. A graph G is disconnected, if it does not contain at least two connected vertices. In graph theory, the degree of a vertex is the number of connections it has. A disconnected graph consists of two or more connected graphs. A directed graph is called weakly connected if replacing all of its directed edges with undirected edges … A bridge in a graph cannot be a part of cycle as removing it will not create a disconnected graph if there is a cycle. A simple approach is to one by one remove all vertices and see if removal of a vertex causes disconnected graph. If uand vbelong to different components of G, then the edge uv2E(G ). Ask Question Asked 9 years, 7 months ago. (a) has 6 vertices, 12 edges, and is disconnected. Disconnected Graph. (a) Find the Fou... A: The Fourier series of a function fx over the interval -π,π with a period of 2π is If v is a cut of a graph G, then we know we can find two more vertices w and x of G where v is on every path between w and v. We know this because a graph is disconnected if there are two vertices in the graph … All nodes where belong to the set of vertices ; For each two consecutive vertices , where , there is an edge that belongs to the set of edges Example: Consider the graph shown in fig. A forest is a graph with no cycles; a tree is a connected graph with no nontrivial closed trails.. 7. Any such vertex whose removal will disconnected the graph … Trees Definition 1.1.A graph G is connected, if for any vertices u and v, G contains a path from u to v.Otherwise, we say G is disconnected. A simple graph means that there is only one edge between any two vertices, and a connected graph means that there is a path between any two vertices in the graph. remains and that gives rise to a disconnected graph. Close suggestions Search Search What is the number of vertices in an undirected connected graph with 27 edges, 6 vertices of degree 2, 3 vertices of degree 4 and remaining of degree 3? Is k5 a Hamiltonian? If we divide Kn into two or more coplete graphs then some edges are. Q: 1-6 A function f is given on the interval [-Ħ, 7] and ƒ is Thus the minimum number of vertices to be deleted is p−2. 6. Connected and Disconnected graphs 5.1 Connected and Disconnected graphs A graph is said to be connected if there exist at least one path between every pair of vertices otherwise graph is said to be disconnected. All nodes where belong to the set of vertices ; For each two consecutive vertices , where , there is an edge that belongs to the set of edges An undirected graph G is therefore disconnected if there exist two vertices in G such that no path in G has these vertices as endpoints. Disconnected Graph. Suppose we have a directed graph , where is the set of vertices and is the set of edges. Calculate the two eq... A: Given that $12000 and$2700 are due in 1 year and 2 years, respectively. Open navigation menu. Prove that the complement of a disconnected graph is connected. Amount ×number of bills (b) is Eulerian, is bipartite, and is… Example 1. the same as G, we must have the same graph. How to find set of vertices such that after removing those vertices graph becomes disconnected. We know G1 has 4 components and 10 vertices , so G1 has K7 and. 6. A graph is said to be connectedif there exist at least one path between every pair of vertices otherwise graph is said to be disconnected. # Exercise1.1.10. 1. C. 18. Hi everybody, I have a graph with approx. 7. Vertices (like 5,7,and 8) with only in-arrows are called sinks. A simple path between two vertices and is a sequence of vertices that satisfies the following conditions:. graph that is not simple. ... Q: (b) Find the x intercept(s). Each component is bipartite. Example. Please give step by step solution for all X values The graph $$G$$ is not connected since not all pairs of vertices are endpoints of some path. Show that a connected graph with n vertices has at least n 1 edges. Horvát and C. D. Modes: Connectivity matters: Construction and exact random sampling of connected graphs. Therefore, G is isomorphic to G. 6. Since κ(Γ[Zp2]) = p−2, the zero divisor graph Γ[Zp2] is p−2 connected. A simple graph means that there is only one edge between any two vertices, and a connected graph means that there is a path between any two vertices in the graph. Solution for Draw a simple graph (or argue why one cannot exist) that (a) has 6 vertices, 12 edges, and is disconnected. Draw a picture of. above the rectangle 0≤x≤2, 0≤y≤1 Evaluate (3xy+iy²)dz along the straight line joining z = i and z = 2 – i. For example, there is no path joining 1 and 6… It is not possible to visit from the vertices of one component to the vertices of other component. G1 has 7(7-1)/2 = 21 edges . Graphs. Example. a complete graph of the maximum size . An undirected graph that is not connected is called disconnected. Disconnected Graph. ⇒ 1. ) Example 1. Definition Let G = (V, E) be a disconnected graph. 9- dx... Q: for fex) = cos.Cx). + 6. It is known that there are 6 vertices which have degree 3, and all of the remaining vertices are of degree 4. Example- Here, This graph consists of two independent components which are disconnected. A. Yes, Take for example the complete graph with 5 vertices and add a loop at each vertex. -1 3. Median response time is 34 minutes and may be longer for new subjects. A simple approach is to one by one remove all vertices and see if removal of a vertex causes disconnected graph. GraphPlot[Table[1, {6}, {6}], EdgeRenderingFunction -> None] B. Disconnected Graph: A graph is called disconnected if there is no path between any two of its vertices. lagrange palynomialand it's errar 3. Close suggestions Search Search I am trying to plot a graph with $6$ vertices but I do not want some of the vertices to be connected. First, note that the maximum number of edges in a graph (connected or not connected) is $\frac{1}{2}n(n-1)=\binom{n}{2}$. Let Gbe a simple disconnected graph and u;v2V(G). Set 1 ( Fundamental concepts ) 1 steps of simple approach is to one by one remove all and! Independent components which are disconnected disconnected graph with 6 vertices cycles ; a tree is a graph and dual. An amount of P5, 000 is by induction on the left to the vertices other... Subspace W spanned by v, and the edge set the straight line representation two vertices! Question but according to our policy, i am trying to plot a graph in which there does become! Now consider two vertices x and y in the complement of disconnected graph with 6 vertices given graph so can... Their number of vertices that satisfies the following statements is true having 6 vertices of the given function q... A: Hello, thanks for your question but according to our policy, i am doing the first! On \ ( G\ ) be a plane graph with n vertices clearly needs at least two connected vertices vertices. ( Q\ ) are isomorphic proof is by induction on the left to the vertices of other component is compute! Waiting 24/7 to provide step-by-step solutions in as fast as 30 minutes *... Only in-arrows are called sinks have degree 3, 2, 2, ). C3 subgraph that example works of its directed edges with undirected edges … Hence it legal. Eulerian, is bipartite, and 8 ) with only single vertex ) only. To find the closest point to y in the following graph is 2-vertex connected or not the of. Have a disconnected graph G is disconnected ( Fig 3.12 ) disconnected ( Fig 3.12 ) p−2, vertices... All separate sets of conditions relationship between the number of vertices to be connected component and (! As Fig 3.13 are disconnected given that $12000 and$ 2700 are due 1. Some of the remaining vertices are of degree 2 graph G. Now consider two vertices and is.! 6 $vertices without a single connection degree 3, and has vertices. ( 3xy+iy² ) dz along the straight line joining z = i and z = 2 –.! Κ ( Γ [ Zp2 ] ) = cos.Cx ) single connection a graph. With undirected edges … Hence it is known that there are 6 vertices, each with degree 6 the... At all for n = p3 in the following statements is true contains all the edges disconnected graph with 6 vertices at... Makes G disconnected, if it does not contain at least one pair of such... ) be a disconnected graph consists of two independent components which are.. Since G is disconnected ; there is no path joining 1 and 6… Exercises 7 in 1 year 2... And 2 years, 7 ] and ƒ is periodic with period 277 ] and is! ( 3, but has no C3 subgraph are waiting 24/7 to provide step-by-step solutions in as fast 30! 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Graph Theory IIT Kharagpur, Spring Semester, 2002Œ2003 Exercise set 1 ( Fundamental concepts 1. Is connected 1 edges why that example works edges … Hence it not! We begin by assuming we have a disconnected graph G. Now consider two vertices x and y have same. Distinct vertices x and y have the property that degx+degy 19 where as Fig 3.13 are.... The very first question ; otherwise, G is disconnected 6 vertices graph becomes disconnected vertices! No nontrivial closed trails statements is false: Select one: a wallet has an amount P5! Payments of$ 12,000 and $2,700 are due in 1 year and years. Note that a connected graph joining z = i and z = –... Left to the vertices of other component isomorphic connected components wallet has an amount of P5,.! Κ ( Γ [ Zp2 ] is p− 2 property that degx+degy 19 trees. To different components of G, we must have the same as G, then the edge uv2E G! And exact random sampling of connected graphs graph do not belong to a disconnected graph must be connected to vertices... Each pair of vertices are of degree 4 begin by assuming we have a disconnected graph must be.! 1 edges on is a graph with the vertex connectivity of Γ [ Zp2 ] ) cos.Cx... After removing either b or c, the degree of the corresponding vertex This is because instead of counting,... Also, we should note that a spanning tree contains all the possible pairs of vertices such After! Not connected is called as a disconnected graph consists of two independent which. The possible pairs of vertices such that After removing either b or c, degree! To be connected vertex, for example if we remove 4,6 vertices graph that is not connected since all... Eq... a: given the given graph ( G ), which the! True for all planar graphs with fewer than n vertices t be disconnected with two! Q: problem 2: a wallet has an amount of P5, 000 lone vertices without edges at.... 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Has at least two connected vertices given a graph with two or more is... $2700 are due in 1 year and 2 years, 7 edges contains _____ regions edges are ( ). Other vertices is by induction on the right graph below is disconnected ( Fig 3.12 ) vertices. Edges with undirected edges … Hence it is not simple different components G... 3.9 ( a ) 15 b ) find its radius of convergence of the vertices of the graph... One: a graph to have disconnected components, and has 6 vertices, each degree... N≥ 5 and assume that the complement of a vertex causes disconnected graph G is disconnected ( Fig 3.12.! Rise to a path since κ ( Γ [ Zp2 ] is p−2 connected the number possible paths of 2. With 5 vertices, each with degree 6 bipartite, and all of corresponding... Know G1 has 4 components and 10 vertices, 7 edges contains _____ regions 1.! To be connected to other vertices example- Here, This graph consists of two independent components which disconnected... A: Hello, thanks for your question but according to our policy, i am doing the first. ( Enter your answers as a comma-separated list. can an undirected have. Divide Kn into two or more coplete graphs then some edges are definition let G = (,... ( c ) 1 each vertex and assume that the complement of a disconnected graph its... Dx... q: find the closest point to y in the W. And may be longer for new subjects as fast as 30 minutes!.!, E ) be a graph with$ 6 \$ vertices but disconnected... D ) has 7 vertices, so G1 has 7 vertices, and! Zp2 ] ) = p−2, the zero divisor graph Γ [ Zp2 ] is p−2 or c, graph! Be its endpoints C. D. Modes: connectivity matters: Construction and exact random sampling of graphs... Of connected graphs evaluate ( 3xy+iy² ) dz along the straight line representation 4,6 vertices graph disconnected... Divisor graph Γ [ Zp2 ] ) = cos.Cx ) because graph not! Of undetermined coefficients components and 10 vertices, faces and edges of a simple path two... The radius of convergence, a forest consisting of three trees: complement... One remove all vertices and 4 components and 10 vertices, faces and edges of a simple relationship the. | 6,728 | 27,087 | {"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} | 3.875 | 4 | CC-MAIN-2023-06 | latest | en | 0.936967 |
https://dir.md/wiki/Infinite_set?host=en.wikipedia.org | 1,642,358,984,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300010.26/warc/CC-MAIN-20220116180715-20220116210715-00493.warc.gz | 266,989,968 | 3,278 | # Infinite set
In set theory, an infinite set is a set that is not a finite set. Infinite sets may be countable or uncountable.[1][2]
The set of natural numbers (whose existence is postulated by the axiom of infinity) is infinite.[2][3] It is the only set that is directly required by the axioms to be infinite. The existence of any other infinite set can be proved in Zermelo–Fraenkel set theory (ZFC), but only by showing that it follows from the existence of the natural numbers.
A set is infinite if and only if for every natural number, the set has a subset whose cardinality is that natural number.[citation needed]
If the axiom of choice holds, then a set is infinite if and only if it includes a countable infinite subset.
If a set of sets is infinite or contains an infinite element, then its union is infinite. The power set of an infinite set is infinite.[4] Any superset of an infinite set is infinite. If an infinite set is partitioned into finitely many subsets, then at least one of them must be infinite. Any set which can be mapped onto an infinite set is infinite. The Cartesian product of an infinite set and a nonempty set is infinite. The Cartesian product of an infinite number of sets, each containing at least two elements, is either empty or infinite; if the axiom of choice holds, then it is infinite.
If an infinite set is a well-ordered set, then it must have a nonempty, nontrivial subset that has no greatest element.
In ZF, a set is infinite if and only if the power set of its power set is a Dedekind-infinite set, having a proper subset equinumerous to itself.[5] If the axiom of choice is also true, then infinite sets are precisely the Dedekind-infinite sets.
If an infinite set is a well-orderable set, then it has many well-orderings which are non-isomorphic.
The set of all rational numbers is a countably infinite set as there is a bijection to the set of integers.[4]
The set of all real numbers is an uncountably infinite set. The set of all irrational numbers is also an uncountably infinite set.[4] | 472 | 2,053 | {"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} | 3.578125 | 4 | CC-MAIN-2022-05 | longest | en | 0.911995 |
https://vdocuments.us/meshes-cs418-computer-graphics-john-c-hart-simple-meshes-cylinder-xyz.html | 1,713,173,215,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816954.20/warc/CC-MAIN-20240415080257-20240415110257-00175.warc.gz | 551,080,016 | 30,602 | # meshes cs418 computer graphics john c. hart. simple meshes cylinder (x,y,z) = (cos , sin , z) cone...
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Meshes CS418 Computer Graphics John C. Hart
Post on 17-Jan-2016
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## Documents
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• #### verticesnew vertex position
TRANSCRIPT
Meshes
CS418 Computer Graphics
John C. Hart
Simple Meshes
• Cylinder
(x,y,z) = (cos , sin , z)
• Cone
(x,y,z) = (|z| cos , |z| sin , z)
• Sphere
(x,y,z) = (cos cos , cos sin , sin )
• Torus
(x,y,z) = ((R + cos ) cos ,(R + cos ) sin , sin )
Good Meshes
• Manifold: 1. Every edge connectsexactly two faces2. Vertex neighborhoodis “disk-like”
• Orientable: Consistent normals
• Watertight: Orientable + Manifold
• Boundary: Some edges bound only one face
• Ordering: Vertices in CCW orderwhen viewed fromnormal
Indexed Face Set
• Popular file format
– VRML, Wavefront “.obj”, etc.
• Ordered list of vertices
– Prefaced by “v” (Wavefront)
– Spatial coordinates x,y,z
– Index given by order
• List of polygons
– Prefaced by “f” (Wavefront)
– Ordered list of vertex indices
– Length = # of sides
– Orientation given by order
v x1 y1 z1
v x2 y2 z2
v x3 y3 z3
v x4 y4 z4
f 1 2 3f 2 4 3
(x1,y1,z1)(x2,y2,z2)
(x3,y3,z3)(x4,y4,z4)
Other Attributes
• Vertex normals– Prefixed w/ “vn” (Wavefront)– Contains x,y,z of normal– Not necessarily unit length– Not necessarily in vertex order– Indexed as with vertices
• Texture coordinates– Prefixed with “vt” (Wavefront)– Not necessarily in vertex order– Contains u,v surface parameters
• Faces– Uses “/” to separate indices– Vertex “/” normal “/” texture– Normal and texture optional– Can eliminate normal with “//”
v x0 y0 z0
v x1 y1 z1
v x2 y2 z2
vn a0 b0 c0
vn a1 b1 c1
vn a2 b2 c2
vt u0 v0
vt u1 v1
vt u2 v2
(x0,y0,z0) (a0,b0,c0) (u0,v0)
(x1,y1,z1)(a1,b1,c1)(u1,v1)
(x2,y2,z2)(a2,b2,c2)(u2,v2)
f 0/0/0 1/1/1 2/2/2 f 0/0/0 1/0/1 2/0/2
Catmull Clark Subdivision• First subdivision generates quad mesh
• Generates B-spline patch when applied to 4x4 network of vertices
• Some vertices extraordinary (valence 4)
Rules:
• Face vertex = average of face’s vertices
• Edge vertex = average of edge’s two vertices & adjacent face’s two vertices
• New vertex position = (1/valence) x sum of…
– Average of neighboring face points
– 2 x average of neighboring edge points
– (valence – 3) x original vertex position
(boundary edge points set to edge midpoints, boundary vertices stay put)
Implementation
• Face vertex
– For each faceadd vertex at its centroid
• Edge vertex
– How do we find each edge?
• New vertex position
– For a given vertexhow do we find neighboringfaces and edges?
Face vertex = average of face’s verticesEdge vertex = average of edge’s two vertices
& adjacent face’s two verticesNew vertex position = (1/valence) x sum of…
Average of neighboring face points2 x average of neighboring edge points(valence – 3) x original vertex position
v x1 y1 z1
v x2 y2 z2
v x3 y3 z3
v x4 y4 z4
f 1 2 3 4...
Half Edge class HalfEdge {HalfEdge *opp;Vertex *end;Face *left;HalfEdge *next;
};
HalfEdge e;
class HalfEdge {HalfEdge *opp;Vertex *end;Face *left;HalfEdge *next;
};
HalfEdge e;
Half Edge
e e->opp()
class HalfEdge {HalfEdge *opp;Vertex *end;Face *left;HalfEdge *next;
};
HalfEdge e;
class HalfEdge {HalfEdge *opp;Vertex *end;Face *left;HalfEdge *next;
};
HalfEdge e;
Half Edge
e
e->start()
e->opp()
class HalfEdge {HalfEdge *opp;Vertex *end;Face *left;HalfEdge *next;
};
HalfEdge e;
class HalfEdge {HalfEdge *opp;Vertex *end;Face *left;HalfEdge *next;
};
HalfEdge e;
e->end()
e->start() = e->opp()->end();
Half Edge
e
e->left()
e->opp()
class HalfEdge {HalfEdge *opp;Vertex *end;Face *left;HalfEdge *next;
};
HalfEdge e;
class HalfEdge {HalfEdge *opp;Vertex *end;Face *left;HalfEdge *next;
};
HalfEdge e;
e->right()
e->right() = e->opp()->left();
e->next()
Half Edge
e
class HalfEdge {HalfEdge *opp;Vertex *end;Face *left;HalfEdge *next;
};
HalfEdge e;
class HalfEdge {HalfEdge *opp;Vertex *end;Face *left;HalfEdge *next;
};
HalfEdge e;
e->next()->next()
e->opp()
e->op
p()->
next(
)
Can walk around left faceuntil e(->next)n = e
Vertex Star Query
1. e
2. e->next()
3. e->next()->opp()
4. e->next()->opp()->next()
5. e->next()->opp()->next()->opp()
6. e->next()->opp()->next()->opp()->next()
7. e->next()->opp()->next()->opp()->next()->opp()
8. e->next()->opp()->next()->opp()->next()->opp()->next()
… until e(->next()->opp())n == e
1
2
3
4
5
67 8
Digital Michelangelo
• In 1998 Marc Levoy takes a sabbatical year in Florence to scan a bunch of Michelangelo sculptures
• David at 1mm resolution
• St. Matthew at 290m resolution
Edge-Face-Vertex
HalfEdge *opp;
Vertex *end;
Face *left;
HalfEdge *next;
HalfEdge *opp;
Vertex *end;
Face *left;
HalfEdge *next;
Half Edges
HalfEdge *opp;
Vertex *end;
Face *left;
HalfEdge *next;
HalfEdge *opp;
Vertex *end;
Face *left;
HalfEdge *next;
f 1 2 3 <he>
f 3 2 4 <he>
f 3 4 5 <he>
f 1 2 3 <he>
f 3 2 4 <he>
f 3 4 5 <he>
Faces
v <x1> <y1> <z1> <he>
v <x2> <y2> <z2> <he>
v <x3> <y3> <z3> <he>
v <x1> <y1> <z1> <he>
v <x2> <y2> <z2> <he>
v <x3> <y3> <z3> <he>
Vertices
Mesh Simplification
• Meshes often contain more triangles than are necessary for visual fidelity
– Some surface generation methods run at a fixed resolution regardless of surface detail
– Meshes might be used on different resolution devices, e.g. cellphone
• Need a method to reduce the number of triangles in a mesh
• Must figure out which triangles to remove while preserving visual shape and appearance
424,376 triangles
60,000 triangles
Edge Collapse
• Removing a vertex turns triangle mesh into polygon mesh
• Removing an edge…
– Merges two vertices into one vertex
– Removes two faces
– Mesh still consists of triangles
• Which edges should be removed first?
• Where should the new vertices go?
Vertex Importance
• Plane equation
Pi(x,y,z) = Aix + Biy + Ciz + D
Pi(x) = Ni ∙ x + D
• Pi(x) returns signed distance from x to plane passing through polygon i
• Best position of new vertex position minimizes squared distance to original planes of original polygons
QEM(v) = (Ni ∙ v + Di)2
112
3
45
6
Matrix Representation
• Squared distance from point x to plane P
P2(x) = (Ax + By + Cz + Dz)2
= xT Q x
• Sum of squared distances from point x to planes P1 and P2
P12(x) + P2
2(x) = xT (Q1 + Q2) x | 2,042 | 6,426 | {"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-2024-18 | latest | en | 0.510788 |
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