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Conditional Probability problem
September 19th 2011, 10:52 AM #1
Sep 2011
Conditional Probability problem
Let S be a sample space, with A is a subset S and B is a subset S. If P(A) = .6 what can be said about, P(A ∩ B),
(a) A and B are mutully exclusive?
(b) A is a subset B
(c) B is a subset A
(d) A′ is a subset B′
(e) A is a subset B′
Not entirely sure how to approach this problem...let me know if the first is on the right track.
a) If A and B are mutually exclusive, P(A ∩ B)= P(A)P(B)
Last edited by mike2208; September 19th 2011 at 11:03 AM.
Re: Conditional Probability problem
Let S be a sample space, with A is a subset S and B is a subset S. If P(A) = .6 what can be said about, P(A ∩ B),
(a) A and B are mutully exclusive?
(b) A is a subset B
(c) B is a subset A
(d) A′ is a subset B′
(e) A is a subset B′
Not entirely sure how to approach this problem...let me know if the first is on the right track.
a) If A and B are mutually exclusive, P(A ∩ B)= P(A)P(B)
No, $\mathcal{P}(A\cap B)=0$.
Mutually exclusive means $A\cap B=\emptyset.$
Just as $A\subseteq B$ means $A\cap B= A$
Re: Conditional Probability problem
That's right, if they are mutually exclusive then the probability they intersect is zero.
So would B subset of A mean B ∩ A = B?
Re: Conditional Probability problem
Re: Conditional Probability problem
Ok, I have all but the last.
If A' is subset of B', then B is subset of A, so P(A ∩ B) = P(B).
However, I still have not found any rules on the final question, A is subset of B'.
Last edited by mike2208; September 19th 2011 at 01:07 PM.
Re: Conditional Probability problem
Re: Conditional Probability problem
Oh thanks. I haven't seen that one before. I'm looking for the proof online. It's not listed in my textbook.
I can see that easily from a diagram, but I'm trying to find a list of important set theory laws and indentities.
Re: Conditional Probability problem
September 19th 2011, 11:25 AM #2
September 19th 2011, 11:29 AM #3
Sep 2011
September 19th 2011, 11:46 AM #4
September 19th 2011, 12:08 PM #5
Sep 2011
September 19th 2011, 12:59 PM #6
September 19th 2011, 01:09 PM #7
Sep 2011
September 19th 2011, 01:13 PM #8
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Having trouble with the sqrt() function.
Having trouble with the sqrt() function.
I'm getting strage output with the sqrt() function in a quadratic formula program. I'll also add that I had the same problem using pow(x,.5). It's definitely the function that's causing the
problem, I've made sure of it.
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
void quad(double, double, double, double&, double&);
int main() {
double a, b, c, x1, x2;
cout.setf(ios::fixed, ios::floatfield);
cout << "Enter the values of A, B, and C for the quadratic equation."
<< "\n>";
cin >> a >> b >> c;
cout << setprecision(2) << "X = " << x1 << " or " << x2 << ".";
return 0;
void quad(double a, double b, double c, double& xPos, double& xNeg) {
if (2*a != 0 && (b*b - 4*a*c) < 0) {
xPos = (-b + sqrt(b*b - 4*a*c))/(2*a);
xNeg = (-b - sqrt(b*b - 4*a*c))/(2*a);
else {
xPos = 0;
xNeg = 0;
Enter the values of A, B, and C for the quadratic equation.
>3 4 5
X = -1.#J or -1.#J.
b is 4. b*b is 16. a is 3, c is 5. 4*3*5 is 60. 16-60 is -44. You are taking the square root of a negative number.
The < in your if should probably be >.
A better test case might be 1 -5 6.
Hey, good call. A can't believe a genuine typo wasted 15 minutes of my life. What's the garbage that sqrt() returns if you attemp to take the sqrt of a negetive number. I know it's impossible in
real life, but how does a computer come to the answer -.#J?
I know it's impossible in real life, but how does a computer come to the answer -.#J?
It is not impossible.
It is called ‘i’ which is short for imaginary.
‘i’ is just the square root of negative one.
Square root of -1 X square root of 4 is the same thing as the square root of -4.
Square root of 4 is two, square root of -1 is i, so the square root of -4 = 2i
You could just check what is under the root, if it is less than 0 then take the absolute value of it (meaning positive) evaluate the square root and tack on a ‘i’ to it.
I think the result is just a general error/problem message.
Yes, I know the value 'i', but I was refering of the spectrum of real numbers. The computer can't find a real number answer so it spits out garbage. ...but where does it find the garbage?
Since sqrt is only defined to work with non-negative numbers, you cannot expect valid output when you use it incorrectly. However, that isn't really garbage, that is probably just your
implementation's way of expressing NaN (not a number).
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Fun Code of the Week #6: Primality Checking
I decided to give KhanAcademy a spin, and watched this set of videos:
And that let to me writing this code:
function IsPrime(const x: integer): Boolean;
i: integer;
i := 2;
if X mod i = 0 then
Result := False;
until i > Sqrt(x);
Result := True;
I know that this can be optimized (which I'll do if I end up watching the next video :-) ), but I don't often write really geeky code like this, so I thought I'd post it so you all can write lots of
comments like you always do when I post code. ;-)
UPDATE #1: There was a bug if you passed in a negative number.
function IsPrime(const x: integer): Boolean;
i: integer;
if (x <= 2) then
Result := False;
i := 3;
if X mod i = 0 then
Result := False;
until i > Sqrt(x);
Result := True;
UPDATE #2: Took the Sqrt() call out of the loop.
function IsPrime(const x: integer): Boolean;
i: integer;
Wall: double;
if (x <= 2) then
Result := False;
i := 3;
Wall := Sqrt(x);
if X mod i = 0 then
Result := False;
until i > Wall;
Result := True;
UPDATE #3: The learning and optimizing continues! I updated the code based on more suggestions, but the special case of 2 is irritating me. I confess I thought a while (I’m not cheating by looking up
implementations on Google…) and so I came up with something of a hack. Also, just for Julian, all the x’s are lowercase now. 1 is not a prime either, by definition. Interesting. Any more
optimizations out there? Is there a “correct” way to handle 2? The end goal here is to have a really nice, really efficient IsPrime function.
function IsPrime(const x: integer): Boolean;
i: integer;
Wall: integer;
if x = 2 then
Result := True;
i := 3;
Result := not ((x < i) or (x mod 2 = 0));
if not Result then Exit;
Wall := Trunc(Sqrt(x));
Result := not (X mod i = 0);
if not Result then Exit;
Inc(i, 2);
until i > Wall;
Result := True;
10/27/2012 7:15:23 PM #
I would store sqrt(x) instead of repeatedly calculating it in real world code.
For example, check once for X mod 2 outside the loop using ((x and 1)=0), then loop starting with I=3, adding 2 to I each time, only use odd numbers in the loop, you would roughly double the speed of
your primality testing.
10/27/2012 9:14:50 PM #
Warren --
Good point.
10/27/2012 10:01:31 PM #
I notice that you only took one of Warren's recommendations to heart (but in doing so, introduced another bug: 2 is prime). I would also calculate the root of x as an integer to slightly improve the
speed of the comparison at the end of the repeat loop (you're comparing an integer value to a double, so one value has to be converted to the other type).
Also, one reason I now don't use Pascal (er, Delphi) much: you have x, x, X, x. Meh. ;)
Cheers, Julian
10/28/2012 5:18:47 AM #
Check out the Euler Project
10/28/2012 6:57:01 AM #
I was going to describe in words, but copy and paste works fine ;)
Thanks for previous contributors
function IsPrime(const x: integer): Boolean;
i: integer;
Wall: integer;
i := 3;
Result := not ((x < i) OR (X mod 2 = 0));
if not Result then Exit;
Wall := Trunc(Sqrt(x));
Result := not (X mod i = 0);
if not Result then Exit;
until i > Wall;
Result := True;
10/28/2012 3:00:13 PM #
Just to be complete ... Sieve of Eratosthenes and/or store the prime numbers found into a file.
10/28/2012 3:55:19 PM #
Also you can change
if x = 2 then
Result := True;
if x = 2 then
10/29/2012 4:18:18 AM #
The new exit syntax can be quite a line saver here. It also spares setting and checking for the result variable.
Another optimization is to replace the x mod 2 = 0 with a not odd(x). The compiled code looks much shorter (XE2 Win32):
Project191.dpr.24: if x mod 2 = 0 then Exit(false);
0040512C 8B45FC mov eax,[ebp-$04]
0040512F 2501000080 and eax,$80000001
00405134 7905 jns $0040513b
00405136 48 dec eax
00405137 83C8FE or eax,-$02
0040513A 40 inc eax
0040513B 85C0 test eax,eax
0040513D 7506 jnz $00405145
0040513F C645FB00 mov byte ptr [ebp-$05],$00
00405143 EB0A jmp $0040514f
Project191.dpr.25: if not Odd(x) then Exit(false);
00405145 F645FC01 test byte ptr [ebp-$04],$01
00405149 7504 jnz $0040514f
0040514B C645FB00 mov byte ptr [ebp-$05],$00
Putting all together and shuffling a bit, this is my small contribution (keeping the algorithm):
function IsPrime(const x: integer): Boolean;
i: integer;
Wall: integer;
if Odd(x) then begin
if x < 3 then Exit(false);
i := 3;
Wall := Trunc(Sqrt(x));
if (x mod i = 0) then Exit(false);
Inc(i, 2);
until i > Wall;
Result := True;
else begin
Result := (x = 2);
10/29/2012 10:46:40 AM #
Uwe -- nice!
11/3/2012 7:52:40 PM #
In fact the
until i > Wall;
should be replaced by a
while i <= Wall do
which also prevents that 3 is mistakenly rejected as a prime.
To furthermore minimize the calculations, it is actually only necessary to check division by prime numbers instead of all odd numbers. However I guess that the housekeeping needed to only divide by
primes may actually take more computing power than dividing by all odd numbers.
10/29/2012 12:13:42 PM #
> The end goal here is to have a really nice, really efficient IsPrime function.
Check the one in DWScript's dwsMathFunctions, it uses a slightly improved algorithm, and when benchmarking on
for i:=1 to 1000000 do
using your Update #3 (which btw fails for 3) I see 0.49 sec in Delphi using Integer
The same loop runs in 0.39 sec in DWScript (using Int64 in a 32bit build).
10/30/2012 5:56:23 PM #
Eric --
Will do. I should have thought to look there. You don't let inefficient code into your libraries.
11/10/2012 3:57:26 PM #
The DWScript code skips every third odd number, avoiding all odd numbers divisible by 3. This males the routine about a third faster. Applying this idea to Uwe's code gets:
function TForm1.IsPrime(const x: int64): Boolean;
i: Int64;
Wall: Int64;
if Odd(x) then begin
if x < 3 then Exit(false);
if ((x mod 3) = 0) then Exit(false);
i := 5;
Wall := Trunc(Sqrt(x));
if (x mod i = 0) then Exit(false);
Inc(i, 2);
if (x mod i = 0) then Exit(false);
Inc(i, 4);
until i > Wall;
Result := True;
else begin
Result := (x = 2);
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Re: st: Problem with margins after logit on a person period data
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Re: st: Problem with margins after logit on a person period data
From Urmi Bhattacharya <ub3@indiana.edu>
To statalist@hsphsun2.harvard.edu
Subject Re: st: Problem with margins after logit on a person period data
Date Thu, 9 Jun 2011 13:36:18 -0400
Hi Austin,
I wonder if I could circumvent this problem you mentioned by
considering the clogclog regression instaed of logit because then what
I am modeling is the discrete hazard of dropping out in an interval
conditional upon having survived till before that.
So I could do the following
clogclog school_left childage i.childfemale i.urban i.scstobc
i.casteother i.dadp i.dadm i.momp i.momm wagep wage5 wage8 wage9 distp
distm disth percapcons durat1 durat2 durat3 durat4 durat5 durat6
durat7 durat8 durat9 durat10 durat11, nocons nolog
My goal is to find investigate the if the effect of mothers
education(momp) on the probability of dropping out varies with
So if I find the predicted probability when momp=1 and durat1=1 and
the predicted probability when momp=1 and durat2=1, then look at the
estimated change in probability and see that the estimated change is
significantly positive, then could I use this as evidence to say that
in risk period 2, momp=1 matters more than in risk period1 in terms of
interval hazard of dropping out?
On Thu, Jun 9, 2011 at 12:17 PM, Austin Nichols <austinnichols@gmail.com> wrote:
> Urmi Bhattacharya <ub3@indiana.edu> :
> This whole exercise is highly suspect--you are computing marginal
> effects over a sample of periods at risk, not people. Note that
> people are in your model for very different numbers of periods, but
> you are averaging over all periods; what is the goal here? You said
> you are "interested in the marginal effects of the variables on the
> probability of hazard" which I think means you want to measure the
> marginal effects of the variables on the conditional probability of
> leaving school (conditional on not having left yet) at different
> durations, which means you should calculate marginal effects for each
> sample of people still at risk, at different durations. These are
> unlikely to be very informative in the probability metric, however;
> odds ratios are used for a reason for such applications.
> On Wed, Jun 8, 2011 at 10:31 PM, Urmi Bhattacharya <ub3@indiana.edu> wrote:
>> Hi,
>> I dropped one of the duration dummies durat1 and ran the following
>> logit regression
> ....
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
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MarketingExperiments Blog: Research-driven optimization, testing, and marketing ideas
Online Marketing Tests: A data analyst’s view of balancing risk and reward
October 3rd, 2012
The answer “yes” to the question, “Is this test statistically significant?” — in Daniel Burstein’s MarketingExperiments blog post, “Online Marketing Tests: How could you be sure?” — is not only
“misleading” as he mentions, but it is also incomplete.
In fact, if someone asked me that question, as a data analyst, instead of responding with a dichotomous “Yes/No” answer, I would ask, “At what level of significance?” The reason statistical
significance is so important in experimental testing is it quantifies the amount of risk that one is willing to take in drawing inferences from a test.
From some recent conversations I have had with marketers running A/B tests, it seems like many of them are really interested in knowing the ‘roots’ of how the testing works, and they find it helpful
to understand it so they can draw better conclusions.
So, in today’s MarketingExperiments blog post, I’ll give you a look at what some of the numbers and values you encounter in testing really mean.
Testing nomenclature
In A/B testing, you’re essentially trying to determine if A and B are different. Let me show you how the math works.
In a classical inferential experiment (that is, an experiment in which we’d like to infer a conclusion from), the researcher prepares two hypotheses: a null hypothesis (H[0]) and an alternative
hypothesis (H[a]).
It is necessary that these two hypotheses are mutually exclusive and exhaustive – meaning every possible outcome must satisfy one or the other, but not both.
For instance, a null hypothesis may be that the conversion rates for two landing pages (Page A and Page B) are the same. The alternative hypothesis would be the complement of the null – namely, the
conversion rates for Page A and Page B are NOT the same.
By convention, the alternative hypothesis is the one the researcher would ‘like’ to be able to support at a certain level of significance, and the null hypothesis is something we are ‘stuck with’ if
the experimental data fails to reach the level of significance.
What are the chances I’ll get the wrong answer?
Since the tests rely only on a small sample and not the whole population, there is always a risk that the group of members we choose is a poor representation of the population.
Returning to the example cited in Burstein’s prior post – suppose we would like to determine if the child automobile safety restraints really do reduce child fatalities compared to seat belts alone.
The null hypothesis (H[0]) would be, “The death rate among children in accidents who are secured by child restraints IS the same as that when only seat belts are used.” So then, the alternative
hypothesis (H[a]) would be, “The death rate among children in accidents who are secured by child restraints is NOT the same as when only seat belts are used.”
Further, we decide that if there is any more than 5% chance that, from our sample data, any observed difference in the rates is just due to having picked wrong accident cases for our sample or random
chance, then we will have to presume that they are not actually different.
Once sample data is collected and analyzed, there are only two possible valid inferences:
1. There is less than 5% chance that the observed difference in rates is merely due to ‘sampling error,’ in which case the null hypothesis (H[0]) will be rejected (i.e., we do NOT believe that both
rates are the same, and it seems likely that the child restraints do make a difference).
2. There is 5% or greater chance that the observed difference is due to ‘sampling error,’ in which case we must concede that not enough evidence exists to compel us to believe that the child
restraints actually do reduce fatalities.
If these policies result in inferring a difference that really doesn’t exist, we have fallen victim to Type I Error. Type I Error is a term used in statistics that simply means a true null hypothesis
was incorrectly rejected.
The probability of a Type I Error is what the significance level measures.
There is also a chance that, due to the standards we have chosen, we may fail to detect the difference that really does exist (i.e., mistakenly failing to reject a false null hypothesis). This
probability is given the name Type II Error.
It is a longstanding and widely held convention that researchers take much greater precautions to avoid committing a Type I Error than a Type II Error.
Balancing risk and reward (in theory)
Type I Error is considered a bigger ‘no-no’ because that it is the alternative hypothesis — if accepted over the null hypothesis — that challenges the status quo and signals the need for change. So,
mistakenly provoking a change that was unnecessary (or possibly even damaging) is considered a more grievous sin than failing to recommend a change that might have made things better.
Think about it as erring on the side of caution. It’s probably safer to falsely assume that your landing page treatment will not generate a lift, and therefore still use a reliable control, than to
falsely assume that the landing page treatment does generate a lift, when in fact replacing the control will result in a reduction in conversion.
So how does a rational person work out a risk-reward decision based on the information provided by inferential testing?
Consider a fair coin is tossed five times, and I offer to give you $500 if we observe five straight heads. If not, you give me only $50.
Here, even if the reward following the five straight heads is much higher for you, the likelihood of such an event is comparatively low (about 3%). While you have only about a 3% chance of winning
$500, I have about a 97% chance of winning $50.
So, I am willing to take the 3% risk of losing $500 to win $50.
Now, if we toss the coin only four times, and I had to give $10,000 for four straight heads, my likelihood of winning $50 declines to about 94% and your chances of walking away with $10,000 roughly
doubles from 3% to 6%.
In this case, I have some soul searching (not to mention wallet searching) to do. Presuming that I could actually cover the $10,000 bet, am I willing to risk that amount, where there is a 6% chance
of losing $10,000 vs. a 94% chance of winning $50?
It turns out, there is a comparatively simple way – used by ‘gamblers’ in hundreds of professions from card sharks to hedge fund managers to insurance actuaries – called ‘Expected Value’ that
provides the ideal “rational person’s answer.”
Balancing risk and reward (in practice)
Suppose now, in the child restraint study I mentioned above, we collected a random sample that supported the alternative hypothesis (at a 0.05 level of significance) that the fatality rates are NOT
the same – in fact, the data suggests that the rate is lower with child restraints.
How should legislators decide whether to enact the law that mandates the use of such restraints, including fines for non-compliance?
Sadly, this is one of those irksome real-world situations in which it is necessary to take a morally reprehensive and emotionally distasteful step – namely assigning a dollar value to human life (and
children’s lives at that).
The coward’s way out would be to simply abdicate and set the number to ‘infinity.’ That would inevitably lead to the conclusion that “… there ought to be a law!” regardless of the cost, “even if it
saves just one life.”
In real life, though, such cowardice (or laziness) comes with a high risk of unintended consequences, stemming mainly from the absence of infinite resources. That is, the enactment of one
exceptionally expensive mandate on these grounds could derail many other subsequent policy initiatives that would have had a much greater collective impact on lives saved.
So, while they seldom speak of them publicly, insurance actuaries and automobile companies, members of Congress and courts of law, the medical profession and the military all maintain finite
‘provisional’ figures for this invaluable practical purpose.
Happily, as professional marketers, we are seldom faced with such distasteful decisions, though we can use the same statistical methods and tools in making our own tough decisions about which road to
take by quantifying the risks and rewards of each through inferential testing.
Traditionally, the significance level is determined by the researcher before all the data-crunching.
However, for our practical purposes, we can determine whether or not to conclude the test valid by analyzing how close our result came to the designated significance level and amount of risk involved
in implementing the changes supported by the test.
Suppose if a test would have validated at a 0.06 significance level, but came inconclusive at a preset 0.05 significance level. Depending on how willing one is to take the extra 1% “risk” of error,
it might not be a bad idea after all to reset the significance level to 0.06 from 0.05 and validate the test rather than rerunning the test with other treatments to gain a tiny fraction of additional
difference to reach the level of confidence previously set.
Related Resources:
Marketing Optimization: You can’t find the true answer without the right question
Optimization 101: How to get real results from A/B testing
Online Marketing Tests: How do you know you’re really learning anything?
Categories: Analytics & Testing a/b testing, level of significance, online testing, validity
1. October 4th, 2012 at 14:29 | #1
Anuj! Excellent post dude. I needed to brush up on graduate statistics. That stuff is definitely perishable, and vitally important if you want to do it correctly.
2. January 2nd, 2013 at 20:14 | #2
I see several times you note a significance level of 0.05. I understand in statistics they often teach the 95% confidence level, but what about in real world marketing applications where there
isn’t enough data to reach that level in a comfortable period of time? Have you gone down to 90%, even 80% in experiments?
□ January 3rd, 2013 at 08:51 | #3
You bring up a very good and important point Dave. For practical purposes yes it is acceptable to drop the level of confidence to 90% occasionally depending on how much risk and/or cost is
involved in implementing the intervention following the test. That is why every time we talk about winning or losing treatment in marketing it is ALWAYS followed by “with x% confidence”.
Academically however, it is a common practice to aim for the 95% confidence. In some fields such as healthcare and public safety, where intervention may be too costly or even risky, it is not
uncommon to aim for even a 99% confidence instead of 95%.
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MARK 203 1 & 2
NAME: ________________________
5 Written Questions
5 Matching Questions
1. survey design? I M Q S W SL
2. types of surveys? 5
3. problems are depicted on a?
4. factors affecting survey method?
5. relevance?
1. a 1) info needed- exploratory research to gain insight- who? where? how?
-consider target market
2) method
3) questions
-that overcome unwillingnes to respond
-avoid double barrelled questions
-link to research
4) structure-open vs close ended, structured
5) wording-simple, appropriate length, non leading
6) sequence and layout- gernal to specific, senstive questions later
2. b reliability x validity
3. c personal
4. d sampling
population type
question form & content
response rate
5. e decison tree
5 Multiple Choice Questions
1. problem and opportunities
2. numbers serve as labels for classifying
-mutually exclusive e.g male=1 female=2
-based on frequency counts
3. sequential variables that follow numerical order
4. two or more questions combined into one
-always break into two questions
5. list or rating scale alternative responses provided
-ad: les potential error, comparable responses, easy answers for interviewer
-disad: meaningless results, limited self expression, reponse creation (bias/limitation)
5 True/False Questions
1. research design depends on? → nature of problem
target market
cost and time
2. reliability? → consistent measures across time and items
-internal consistency- highly correlated
-stability- same results over time
3. measurement and scaling impacts on? → nature of problem
target market
cost and time
4. step 3 exploratory? → intensity and direction relating to a stimuli aspect
5. influences on survey types? 4 → target
response rate
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Woodland Park, NJ Algebra 2 Tutor
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Contents of this Issue
Other Issues
ELibM Journals
ELibM Home
EMIS Home
Remarks on numerically positive line bundles on normal surfaces
Erika Emura and Hideo Kojima
Niigata High School, Niigata, Japan; Department of Information Engineering, Faculty of Engineering, Niigata University, Niigata 950-2181, Japan, e-mail: kojima@ie.niigata-u.ac.jp
Abstract: Let $L$ be a numerically positive Cartier divisor on a normal complete algebraic surface $X$. We prove that $L$ is ample if $g(L) \leq 1$.
Classification (MSC2000): 14C20; 14J26
Full text of the article:
Electronic version published on: 27 Jan 2010. This page was last modified: 28 Jan 2013.
© 2010 Heldermann Verlag
© 2010–2013 FIZ Karlsruhe / Zentralblatt MATH for the EMIS Electronic Edition
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Is the probability of an element in Set B also being an elem
Question Stats:
60%40% (01:09)based on 70 sessions
Is the probability of an element in Set B also being an element in Set A equal to the probability of an element in Set A also being an element in Set B?
(1) Set A consists of all integers from 1 through 5 and Set B consists of all prime numbers less than 10.
(2) Set A consists of all odd numbers between 1 and 10, both inclusive, and Set B consists of all even numbers between 1 and 10, both inclusive.
If this post helped you in your GMAT prep in anyway, please take a moment and hit the "Kudos" button.
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Multiplication Of Large Numbers..
I have got a unique problem while handling large number multiplications.I am using Currency datatype for all my multiplications. for example if the first number is 12345678.1234 and the second number
is 123345678.1234 and the third number is 1234567.123 .Multiping these 3 nos gives me a error.since Iam involving in the development for an Indonasian Package where the currency handling is
more.Please give me some soln at the earliest.Praveen B
View Complete Forum Thread with Replies
See Related Forum Messages: Follow the Links Below to View Complete Thread
Multiplication Of Large Numbers..
I have got a unique problem while handling large number multiplications.
I am using Currency datatype for all my multiplications. for example if the first number is 12345678.1234 and the second number is 123345678.1234 and the third number is 1234567.123 .Multiping these
3 nos gives me a error.
since Iam involving in the development for an Indonasian Package where the currency handling is more.
Please give me some soln at the earliest.
Praveen B
Large Matrix Multiplication
I am writing a VB6 program for a scientific application that needs to multiply two very large matrices (1000x1000). VB6 is hopelessly slow at this. I have used a Fortran DLL which speeds things up
considerably, but is still slow, probably because I am using a simple code without any tricks to do the multiplication. Does anyone know a really fast matrix multiplication routine in VB6, Fortran or
C++ that I can use to solve this problem?
Multiplication With 2's Complement Numbers?
Hi, all,
I need to do a multiplication with two 54-bit 2's complement numbers and get a final product in a 108-bit 2's complement number.
Is it possible to convert a number (may up to 1.0*E40) in double data-type into a 108-bit 2's complement number? I always encounter over-flow errors.
I need the function to verify the functionality of the IC circuit in my final year project.
Thank you!
Multiplication With Two 54-bit 2's Complement Numbers?
Hi folks,
I need to do a multiplication with two 54-bit 2's complement numbers and get the final product in a 108-bit 2's complement number.
Anyone has good ideas on this ?
Thanks a lot in advance.
Splitting Large Numbers Into Smaller Numbers
i want to divide a decimal into smaller decimals in an array form.
starting with 0.22335566
array(0) = 0.22
array(1) = 0.0033 or 3.3 *10-2
array(2) = 0.000055 or 5.5*10-4
can anyone help?
thanx a million
Very Large Numbers
I'm currently working on a program that uses VERY large numbers, and I need a way to get around overflow errors. The current problem is this line:
Param1 = X And 4294967295#
where X is anything at all. I need to be able to use "And &HFFFFFFF" on variables a lot in this program, but &HFFFFFFFF returns -1 because longs are signed. Anyone know how to fix this?
Large Numbers
why following gives me overflow error?
Dim I as Long
I = 350 * 350
i also need to multiply this:
I = 150000 * 150000 * 1000000 (same error)
or at least I = 150 * 150 * 1000
PLZ help!
Add Two Large Numbers
i want to multiply two numbers in two textboxes
and give the equation
when i multiply two large numbers
the result will like 34.343434343 + E
is there any other way to multiply large numbers so that i get the exact value
Large Numbers
does anyone know how I can do math algorithms that can handle numbers way bigger than long, double, or currency? I am talking trillions, maybe bigger. If the calculator built into windows can handle
them, why can't VB? Help?
Large Numbers
Hi, I'm writing an application in which I need to log the start and end times of an event and the subsequent difference in time time measured in seconds. I have used text boxes for the times (1 for
hour, 1 for min etc). Then when all is completed, I check all the boxes and work out the times.
Ideally I just take the start units from the end, and if the result is less than zero, ie starts at 23:58:01 and ends at 00:12:34 then I effectively add 24hours to get the number. However, this
doesn't even come into question because unless there is only a few mins difference and the start hour is less than the end, then I get overflow errors.
My assumption was that "integer" was < 32k (two bytes) and that "long" was <2b (four bytes).
If this is the case, why do I get overflow errors when a "long" variable is assigned to 86k ??
If long is not 4 bytes, how do deal with decently large numbers ? Does VB have an unsigned long or some other way of manipulating large numbers.
Code Extract:
lngSum = ((CLng(txtFH) - CLng(txtSH)) * 3600) + ((CLng(txtFM) - CLng(txtSM)) * 60) + (CLng(txtFS) - CLng(txtSS))
lngDay = 24 * 3600
If lngSum < 0 Then
lngTotal = lngDay + lngSum
lngTotal = lngSum
End If
MsgBox lngTotal 'Only here as a debug
Very Large Numbers
I am looking for some functions to do addition, subtraction, multiplication and division for numbers of any length. Can anybody help?
Large Random Numbers
Public Function Random(Lowerbound As Double, Upperbound As Double) As Double
Random = Int(Rnd * Upperbound) + Lowerbound
End Function
Now if I use 'MsgBox Random(50000, 100000)' it will give me numbers bigger than 100000! Any solution to this?
Handling Large Numbers In VB6
Hi to all, i am a new entry and i have already a question for u.
I am implementing a cryptographic software based on some well known algorithms and other code written by myself.
The real problem in vb is handling very large numbers.
I use the following function to determine whenever a number is prime or not (for use in RSA enc) but they works only for smaller numbers.
Public Function IsPrime(ByVal N As Double) As Boolean
If Modulus(SQRN(2, (N - 1)), N) = 1 Then
IsPrime = True
IsPrime = False
End If
End Function
Private Function Modulus(ByVal X As Double, Y As Double) As Double
Dim Z As Double
Z = Fix(X / Y)
Modulus = X - (Y * Z)
End Function
Public Function SQRN(X As Double, N As Double) As Double
Dim r As Double
Dim Y As Double
r = 1
Y = X
Do While N > 1
If Modulus(N, 2) <> 0 Then r = r * Y
N = Fix(N / 2)
Y = Y * Y
r = r * Y
SQRN = r
End Function
There is someone who can help me?
Thank You
Work With Large Numbers
I need to be able to devide numbers of at least 13 digits (1 trillion).
However, I constantly get a "Overflow" error. I am using a Long variable type, though I have tried a double as well.
Any suggestions would be greatly appreciated
Working With Very Large Numbers In VB?
I am trying to find the sum of all the digits in the number:
How do I access the digits that aren't being shown because of the scientific notation?
Problem With Large Numbers
I am working on a project which involves me doing math and other types of manipulation on numbers that are extremely large. For example when I try to define a variable with the value of 7544824877 as
soon as I go to the next line it goes to value of 7544824877# or when I generate a value of 900000000000000 in a calculation it goes to a value of 9E+17, I know that they are the same but when I go
an try to do furthur math on them I get errors on the +'s. What do I need to do so that the numbers stay their original value and not change to ...# or to ..E+17 etc.
thanks to all that reply.
Subtracting Large Numbers
Hi there,
The following code currently adds two very large integer numbers (which needs to be up to 60 digits long). But my problem is that I need to be able to subtract two numbers, which can be up to the
same length. Can somebody please help me by giving me an idea on how to do this.
Thanks for your help!
Private Sub Form_Load()
Combo1.AddItem "+"
Combo1.AddItem "-"
Combo2.AddItem "+"
Combo2.AddItem "-"
End Sub
If Combo1.Text = "+" And Combo2.Text = "+" Then
If Len(a) >= Len(b) Then
largestring = a
smallstring = b
largestring = b
smallstring = a
End If
For x = 1 To Len(largestring)
topnumber = Mid(largestring, Len(largestring) - (x - 1), 1)
If Len(smallstring) >= x Then
bottomnumber = Mid(smallstring, Len(smallstring) - (x - 1), 1)
bottomnumber = ""
End If
lresult = Val(topnumber) + Val(bottomnumber) + Val(scarryover)
sTotalString = Right$(CStr(lresult), 1) & sTotalString
If Len(CStr(lresult)) = 2 Then
scarryover = Left$(CStr(lresult), 1)
scarryover = ""
End If
Next x
Text3.Text = scarryover & sTotalString
End If
End Sub
Very Large Numbers And The Mod Operator
Having a bit of a prob with finding the remainder with REALLY big numbers, I have had a look around the threads and tried the functions. However the problem I have is that the numbers I am getting
are currently 25 digits long and will be much larger eventually. DP doesnt seem to put these to enough places to allow for the remainder to be calculated. The mantissa doesnt seem to be large enough.
Therefore I get a remainder 0 when it should be one. When I try to store the numbers as strings it conversts them to DP for some reason. Any thoughts?
Calculating Large Numbers
I have to multiply two numbers whose length is Number(25,4)
when i use Double after 15 digits it is being converted into Exponential Form. I am not able to get the precision. Is there any other method to achieve this.
dim x as double
dim y as double
I want to store the Result of x*y without losing Data.
Calculating Large Numbers
I have to multiply two numbers whose length is Number(25,4)
when i use Double after 15 digits it is being converted into Exponential Form. I am not able to get the precision. Is there any other method to achieve this.
dim x as double
dim y as double
I want to store the Result of x*y without losing Data.
Large Numbers In Access
I need to add some data to an access database, but unfortunately one of the bits of data is a 12 digit number.
Does anyone how I can get this 12 digit number into my table and out using VB?
I have written most of my code dynamically, so using Val(textfield) on the order by is not an option!
Large (135 Digit) Numbers
I want to be able to accurately add, subtract, multiply, and divide large (135 digit) numbers. None of the built in data types will do this. Does anyone have any suggestions?
Computation With Large Numbers In VB
Hi All,
I look forward your help for computing large numbers in VB.
I have to multiply two numbers whose length is Number(25,4)
when i use Double after 15 digits it is being converted into ExponentialForm. I am not able to get the precision. Is there any other method toachieve this.
dim x as double
dim y as double
I want to store the Result of x*y without losing Data.
Thanks in advance
- Bala
Computation Of Large Numbers
Dear All,
We have table that containing column qty n(13,4) & rate (25,4) as theapplication requires this much width. During computation of these fieldssay qty*rate - results in exponential form or overflow
error. How tohandle the situation? Any suggestion/solution is much appreciated.
Thanks & Regards
Loosing Precision With Very Large Numbers
OK, I wrote this function to create a 24 character Hex String (FFFF FFFF FFFF FFFF FFFF FFFF Max). Well, supposed to anyway. If it worked...
The Problem I am having is when you get close to it's limits, I start loosing precision. It happens when I take the Input number (maximum of a Decimal) and divide it by my Constant to split the
number apart.
The problem is like this.
When I take the MAX Input which is 79,228,162,514,264,337,593,543,950,335 (or anything near that) and divide it with my Const (268435456) I am supposed to get:
But I get 295147905179352825856 with the code. (It rounds it off)
If I use 79,228,162,514,264,337,593,543,950,334 (1 less) I get
Instead of 295147905179352825855.99999999225 (Like It's supposed to calculate out.
Mind you I have to use the CDec() function...
I can't figure out how to get this working. I have WRACKED by brain... Can anyone help me figure out how to use these large numbers without loosing any precision?
Here is the Function (Don't laugh at my messy code)
Const MAX = 268435456 ' Hex 10000000
Const MAXDEC = "79228162514264337593543950335" ' Upper Limits of Decimal
Private Function LongHex(strNum As String) As String
' The largest value this function can calculate is 79,228,162,514,264,337,593,543,950,335
' Which calculates to FFFF FFFF FFFF FFFF FFFF FFFF (Or supposed to)
Dim i As Integer ' Loop variable
Dim Hex1 As Variant ' Left of the decimal point
Dim Hex2 As Variant ' Right of the decimal point
Dim sHex As String ' String to output
On Error Resume Next
' Exit if it's not a number
If Not IsNumeric(strNum) Then
Exit Function
End If
' Get rid of any comma's that may have been inputted
strNum = Replace(strNum, ",", "")
If strNum > MAXDEC Then
Exit Function
End If
' Max will calculate out to 1.0 Split it into two parts, Hex1 and Hex2
' Get the value of anything over MAX
' RIGHT HERE is where I am loosing precision with Large numbers...
Hex1 = CDec(strNum) / MAX
' Give me 28 places to the right of the decimal
Hex1 = Format(Hex1, "#.0000000000000000000000000000")
' Multiply the decimal with max to calculate anything less than Max
Hex2 = CDec(("." & Split(CStr(Hex1), ".")(1)) * MAX)
' Get rid of decimals and Hex the values
For i = 1 To Len(Hex1)
If Mid(Hex1, i, 1) = "." Then
Hex1 = Mid(Hex1, 1, i - 1)
Exit For
End If
' Hex the values
' If Hex1 value is larger than MAX (and not ""), Call the function again to break it apart.
If Hex1 > MAX And Hex1 <> "" Then
Hex1 = LongHex(CStr(Hex1))
Hex1 = Hex$(Hex1)
End If
Hex2 = Hex$(Hex2)
' Pad the trailing Hex
Hex2 = Left("0000000", 7 - Len(Hex2)) & Hex2
' Combine the hex values
sHex = Hex1 & Hex2
' Trim any leading 0's
For i = 1 To Len(sHex)
If Mid(sHex, i, 1) <> "0" Then
sHex = Mid(sHex, i, Len(sHex) - i + 1)
Exit For
End If
LongHex = sHex
End Function
Double And Working With Large Numbers!
I've made a proggy using very large numbers. However, I get an overflow error when calculating with large doubles, even though they are in the "valid" range for doubles. Can't Visual Basic calculate
stuff like this:
Dim val1 as double
Dim val2 as double
Dim sum as double
val1 = 6.7 * 10^28
val2 = 2.1 * 10^15
sum = val1 + val2
I get an error when doing something like this... I also use Xor and Mod, but MSDN says they can calculated doubles as well....
please help...
No Scientific Notation For Large Numbers
Is there a way to stop VB displaying large numbers in scientific notation (eg 1.2322343242E+15)?
I've got the caption of a label set to a value in an excel sheet which happens to be a 16 digit number.
How do i force it to show the full number normally?
Trying To Find A Way To Compute Large Numbers
I'm still busy working on my prime number program and i have ran into another problem. I want my program to be able to process large numbers like '4364365475467658658768769879797' for example. Now
the maximum capacity of the largest numeric type, named currency, is 922337203685477,5808. Thus trying to place a larger number than the maximum will result into an overflow error.
And im using dynamic currency arrays for the Sieve of Eratosthenes which is also included in my program. A currency array is restricted to using a maximum of 10000000 elements per array, Because i
want the Sieve of Eratosthenes to be able to process large numbers aswell, i thought of processing the large number per 10000000 elements.
This is how i do it:
dblDeling = Val(txtInvoer.Text) / 10000000
dblRest = dblDeling - Int(dblDeling)
Opslag = Round(CDbl(dblRest * 10000000), 0)
VermenigVuldiger = Round(dblDeling, 0)
now i know how many times i have to repeat processing the number per 10000000.
then i keep declaring dynamic arrays kind of like this:
for i = 1 to vermenigvuldiger
redim number(10000000 * i to 10000000 * (i+1))
bereken number
next i
But there is a downside to this method, i have discovered that whenever you declare a currency array with borders (2130000000 to 2140000000) it will generate an overflow aswell because the borders
are to high...
So this means my program will eventually overflow and i dont like that thought.
So is there perhaps a way through the use of API commands (which im still unfamiliar with ) to designate enough memory to hold a large number or an array with large bounds or allot of elements?
I want my program to be able to process numbers that are 2.2 million characters long (the capacity of a variable string type).
Any idea's?
Thanks in advance
Mathematical Problem (Large Numbers)
Can anybody explain me (or give me the right address) how to do same main mathematical calculations with numbers greater then Integer and Long can hold (e.g. N= 10 ^ 41324)? Also, in case that you
know, how can I calculates: 2 ^ N ?
Thanks in advance.
P.S. I heard that this can be solved by using arrays, but I don’t know how.
Data Type For Large Numbers
If I want to read a number eg 4,123,456,789 using VB6 what type would I use.
I have tried a long but I think that only goes up to 2,xxx,xxx,xxx and I get overflow error. I have used double but get type mismatch !
How To Perform XOR On Two Large Numbers (V Urgent)
I am trying to execute the following code.
Res = CDbl(20041117101515) Xor CDbl(1250000)
I am getting the Overflow error.
Can any body have the solution.
Putting Commas In Large Numbers
What's the best way to break up a large number with commas?
For example, how would I turn 30892457245 into 30,892,457,245?
Converting LARGE Hex Numbers Into Decimal
Does anyone know of any function to convert a large hex number (held in a string) into a decimal number (held in a string). The hex numbers I am wishing to convert may be from 10 - 20 digits, this
will make a big decimal number that must be held in a string.
Help With Storing Extremely Large Numbers
Dear Members,
I want to store a value which has 80 to 150 Number of digits. The Problem is to store them, could u help me to solve this issue. for example I need to generate 100 Digits long Prime Nos. etc.
Just A Query... {Large Numbers In The Calculator Screen}
hi everyone
i was wondering, u know if you make a calculator for instance, and if the number get really big, the display says 1.233438 E+10, or summit like that. Is there any way of making the display actually
show the entire number, rather then one using E?
Extracting Single Numbers From A Large Number
My program outputs an angle eg. 267.9135843
I then need to specify the varialbes a, b, c and d as the first 4 digits of this angle,
So that I can then input them into a command string to an external device.
Is there a function in VB that will allow me to do this?
Many Thanks
Math Calculations...Multiplying Large Numbers
Hi all I have a program that does some math, I never knew VB had such a limit (I always thought you were just limited by the size of your variable)
MsgBox 62 * 532
Dim Bob as currency
bob = 62*532
How do I get this to compute? probably some simple solution.
(urgent) Mathematical Operation On Very Large Numbers
hi all,
I want to do mathematical operations on a number which could be as large as of 80,000 digits.
Can anyone tell me the required data type to support that.
Any Idea other then that......
Why do I get an overflow error on this operation?
Dim lngyPos As Long
lngxPos = 240 * 8192 <---------???
I have Hscroll1 with min=0 and max=5.
I'll make operation text1=hscroll1.value*10000
But why if value=4 or 5 then i get error message.
Can anybody help me?
Thank before...
VB6 Multiplication Bug ?
i was trying the following when i encountered an overflow error !
Private Sub Command1_Click()
Text1.Text = 90 * 600
End Sub
Does anyone know why ??
Multiplication In Vb6
hi i am trying to make a calculator in vb but then realized that i have never stuided multiplication or any type of sums lol....so can any one help me...eg below....
thanks pip..
2 To 12 Multiplication
how do u go about writing a program which displays the multiplication tables 2 to 12 in a text box using a for...next loop in the form load event. could any one tell me how to go about this?
How do I multiply values in a list box by 100 and put them into another list box.Sounds easy enoughf but can't seem to do it. Thanks in advance
Multiplication Causes Overflow.
Dont know if anyone can help me with this, basically I have 2 Long variables, both are 8 digits, if I try and multiple them together I get an overflow even if I use a double?
Private lngKeyMod1 As Long
Private lngKeyMod2 As Long
Public Sub process(tmpKeyMod1 As Long, tmpKeyMod2 As Long)
Dim tmpKeyMod3 As Double
lngKeyMod1 = ((tmpKeyMod1 + tmpKeyMod2) Xor 17185) Xor tmpKeyMod2
tmpKeyMod3 = CDbl(lngKeyMod1 ) * CDbl(lngKeyMod1 ) 'OVERFLOW HERE
lngKeyMod2 = Clng(lngKeyMod3 And 4294967295#) ' I Expect an overflow here to
End Sub
lngKeyMod1 ends up at a value 8 digits long (11,000,000ish)
Anyone know what I can do to use longer numbers?
Date Multiplication
I work with SQL Server and I have a datetime field to store time for a process.
so I have the date 01/01/1900 00:41:38 and I want to multiply the 41:38 by 50 and the and then add it to a date
How can I multiply time ??
Multiplication Table
Dim Row As Integer
Dim Column As Integer
Dim MakeRow As Integer
Dim MakeColumn As Integer
Private Sub cmdMult_Click()
Row = Val(lblRow)
Column = Val(lblColumn)
For MakeRow = 1 To Row
For MakeColumn = 1 To Column
picDisplay.Print MakeColumn;
End Sub
I made this code to display a multiplication table but I don't seem to know how to muliply the row by the column... The output I get is... if lblRow = 5 and lblColumn = 4
How can I make it go like...
Multiplication Table
i have a question for you though, okay, i am trying to make a button that will display, in a listbox, the multiplication tables of 1x1, 1x2 ....2x1, 2x2 ... all the way to 25x25 ... do you think you
could help get me started on that, i know it's going to take 2 variables .. and it's going to be a for, next statement, correct? maybe someone can help me out wtih this ... thanks..
i wrote a multiplication and division function that can handle very large numbers (hundreds of digits) but i need help making it work with decimal places...and division routine also can't divided a
larger number into a smaller number...any help would be useful
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|
Number of results: 214,135
Logs are stacked up in apile as shown in the figure. The top row has 15 logs and the bottom row has 21 logs. How many logs are in the stack? a1= 1+14=15 an= 7+14=21 s1=(n/2)(a1+an) (n/2)(15+21) = 18
How do I figure out the n?
Tuesday, May 4, 2010 at 4:37pm by Abbey(Please help)
math - natural logs
Find the EXACT solution to e^(1-x) = 4^x You must use logs, not just a calculator.
Monday, October 31, 2011 at 10:10pm by BJ
you will have to use logs 4( 7^(x-2) ) = 8 divide both sides by 4 7^(x-2) = 2 take logs of both sides log (7^(x-2) ) = log 2 by rules of logs (x-2)(log 7) = log 2 x-2 = log 2/log 7 x = 2 + log2/log7
= appr 2.3562
Thursday, January 23, 2014 at 10:06am by Reiny
Well I know that a product law of logs is log(a) + log(b) = log(ab) and the quotient law of logs is log(a)-log(b)=log(a/b) But how do I change those laws, or relate it to prove the law of exponents.
I know what the law of exponents are, but I just need help proving it IN ...
Saturday, May 23, 2009 at 5:07pm by Jus
math logs
solve 3^(2x+1) = 9^(2x-1) using base 3 logs
Saturday, November 12, 2011 at 9:03pm by Sejul
By the way, in the old days like when I went sailboat racing, we had to use logs of trig functions to do these products by adding and subtracting logs because we did not have calculators. This was a
total mess and resulted in complicated things like using half the angles so ...
Friday, January 4, 2008 at 3:49am by Damon
.5=e^(-20t) I am assuming this involves logs but we did't really learn logs. Anyone care to walk me thru with a calculator. Have the Ti-83
Thursday, March 13, 2008 at 7:08pm by Trevor
start by collecting logs: log5 (x+3) + log5(x-1) = 1 now recall that sum of logs is log of product log5 [(x+3)(x-1)] = 1 raise 5 to the power of both sides: (x+3)(x-1) = 5 x^2 + 2x - 3 = 5 x^2 + 2x -
8 = 0 (x+4)(x-2) = 0 x = 2 or -4 Since logs of negative numbers do not exist...
Monday, January 9, 2012 at 11:07am by Steve
another pre cal (logs)
according to your basic laws of logs your expression is log3 ((3x-6)(x^2 - 4)/81)
Monday, March 31, 2008 at 1:14am by Reiny
the base doesn't really matter. All logs follow the same rules. (a) log(x^2-9) - 9log(x) but your parentheses are not balanced, so that's a guess. Could also be 3(log(x^2-9) - 3log(x)) (b) log(5) +
4log(a) Now, it helps to know that we're using logs base a, since then we have ...
Monday, April 8, 2013 at 12:25am by Steve
"Yule logs are simulated wood logs make of compressed paper that burn with a red and green flame.How would you produce a similar yule log?"
Wednesday, July 9, 2008 at 1:11am by jennifer
Pre Cal (logs)
I need help with these logs 1: Log underscore4 16X 2: 2 Ln ( X(sqr. root)e 3:5^[2logUnderscore5(3x)] 4:log underscore2 ^[1280-logunderscore2 5) I don't understand how to do them thanks
Sunday, March 30, 2008 at 11:25pm by Deb
You will have to know the basic rules of logs log(A/B) = logA - logB comes into play here log6 7 - log6 x = 4 log6 (7/x) = 4 by definition of logs 6^4 = 7/x 1296 = 7/x x = 7/1296
Sunday, December 6, 2009 at 11:11am by Reiny
Mr Logan can saw 8 logs in 3 hours. How many logs can Mr Logan saw in one hour?
Wednesday, August 15, 2012 at 1:15pm by Andrea
5^(logx) + x^(log5) = 50 Note that log(5^logx)) = logx * log5 log(x^log5)) = log5 * logx So, the two are equal. That means 2*5^logx = 50 5^logx = 25 = 5^2 logx = 2 x = b^2 where logs are base b. So,
if natural logs, x = e^2 if common logs, x = 10^2 = 100 check (common logs): 5...
Wednesday, March 19, 2014 at 3:21pm by Steve
Maths (Logs)
1.Write the following equation without logs log y = mx+ c This would be y = 10^(mx+c) right? 2. Explain why this fits the exponential model (in other words, it can be written in the form y = a x b^x
where a and b are real numbers).
Monday, April 7, 2008 at 10:23pm by BP
misplaced modifier
Were you hunting behind a pile of logs? -- or Was the fire extinguisher behind a pile of logs? We can't tell from this sentence.
Saturday, September 24, 2011 at 4:32pm by Ms. Sue
Take logs of both sides. I'll use logs to base 10, but it doesn't matter what base is used as long as it is consistent. log 14 = n log 1.4 n = log 14/log 1.4 = 1.14613/.14613 = 7.843
Tuesday, February 26, 2008 at 2:01am by drwls
n/2(26) = 18 26n = 36 n = 36/26 something very wrong with your equation, since n has to a whole number where does the 18 come from? you seem to be using the formula S(n) = n/2(first + last) so 18
would be the sum of the logs ???? that makes no sense, since the bottom row alone...
Tuesday, May 4, 2010 at 4:37pm by Reiny
math logs
3^(2x+1) = 3^2(2x-1) easy way 2x+1 = 4x-2 2z=3 x= 3/2 using logs base 3 2x+1 = (2x-1)log3(9) 2x+1 = (2x-1)log3(3^2) 2x+1 = (2x-1)(2) log3(3) 2x+1 = 2(2x-1) same as above
Saturday, November 12, 2011 at 9:03pm by Damon
Hmmm... it may require some knowledge of how some functions work. Take 1/x. You can plug any value into x and it will be defined unless x is 0. You cannot divide something by 0. 0 no matter how many
times will not be equal to 1. Another common one is sqrt(x). It's no longer a ...
Wednesday, April 2, 2008 at 8:19pm by Anonymous
what's ln vs lg? Assuming all logs are the same, and taking powers of e, e^(ln 4x) = e^(ln3 * ln5) (e^ln3)^(ln5) 4x = 3^(ln5) x = 1/4 * 3^(ln5) = 1.46 If the logs are different bases, then adjust the
powers accordingly
Saturday, July 6, 2013 at 11:59pm by Steve
Alegbra (logs)
if logs are the same, the arguments are the same, so x-16 = 9x x = -2 But, log(x) is not defined for x<0, regardless of the base if you mean the entire exponents is 1+ln(7), then e^(1+ln(7)) = e^1 *
e^ln(7) = 7e by definition, e^ln(n) = n
Monday, April 22, 2013 at 3:39pm by Steve
for the first you don't even need logs 3^(x^3)=9^x 3^(x^3)=(3^2)^x 3^(x^3) = 3^(2x) so x^3 = 2x x^3 - 2x = 0 x(x^2 - 2) = 0 x = 0 or x = ±√2 for second, take logs of both sides log(2^x) = log 10
xlog2 = 1 x = 1/log2 for the last 4^x = 2^x 2^(2x) = 2^x 2x = x x = 0
Monday, December 1, 2008 at 3:50pm by Reiny
I don't see why you have to convert it; they are indeed the same thing. I want to draw a graph for this, but I can't :-) In general, you can use the rule that ln(1/x) = -ln(x), so you can simply
declare that and be done. I'll try to show an equivalence so it might make a bit ...
Sunday, September 20, 2009 at 4:59am by jim
The fraction (1/6) that remains after 70 years satisfies the relation: 1/6 = 2^(-t/T), where T is the half life and t = 70 years. Take logs of both sides to solve for T. I'll use logs to base e but
it doesn't matter -1.7918 = -70/T ln 2 = -48.52/T T = 27.1 years
Wednesday, May 21, 2008 at 5:51am by drwls
How do you graph the following logs? f(x)=log5 (x-2) f(x)=log5 x-2 f(x)= log5 x f(x)=log5 (x+2)
Tuesday, October 23, 2007 at 11:11pm by Jayme
College Algebra
The purpose of logs is that when multiplying, logs add. Thence the slide rule, and a mydrid of mechanical computing devices. The other use, is that many natural processes in nature are logarithimic
in nature (population growth, interest, radioactiveity, in fact, anything whose...
Thursday, October 21, 2010 at 10:54am by bobpursley
Quantative Method
Generate your Unit transportation costs (in excel) 1. Ship construction logs from 8 provinces in Canada to 50 states in the U.S. Minimize total shipping cost. 2. Set up the demand in each state of
the U.S. 3. Set up availability of logs in the 8 provinces of Canada
Friday, June 14, 2013 at 10:36pm by Anita
more pre cal logs
sorry I am so lost I missed the class that talked about logs since my sister was in the hospital giving birth to a still born baby.... now I am all lost... Expand the log Ln 3sqrootx^2y/x+3 (the 3 is
part of the sqroot not infront like the 3rd power) I hope that makes since
Monday, March 31, 2008 at 1:52am by Deb
By the way, at the vertical asymptote where x = -1/3, the value for y gets infinitely negative as you get closer and closer to that line (since the values in the parenthesis are getting smaller and
smaller as you approach the minimum value for x, -1/3). Remember that it's ...
Wednesday, March 11, 2009 at 9:57pm by Toothpicks
Math: Calculus
changing the base of a log: Alg II http://www.sosmath.com/algebra/logs/log4/log43/log43.html
Monday, October 17, 2011 at 7:37pm by bobpursley
Math - logs
Saturday, January 24, 2009 at 5:49pm by Damon
Monday, December 21, 2009 at 4:32pm by Anonymous
Math (logs)
Sunday, December 8, 2013 at 8:40pm by Steve
solve with logs 3^(1-2x)=4^(x)
Monday, November 24, 2008 at 8:38pm by kayla
Alg 2 (HS)
take logs of both sides (x+2) log 4 = log 160 x+2 = log160/log4 x = log160/log4 - 2 that is also log4160 - 2 but most calculators don't do logs to various bases, but do have log
Saturday, September 22, 2012 at 1:15am by Steve
how do i solve these logs? log1/2 (3x+1)^1/3= -2 3^(x^3)= 9^x
Thursday, January 8, 2009 at 6:15pm by kayla
hint - take logs
Monday, November 29, 2010 at 8:36am by Dr Russ
take the log base 5 of each side x-1=log5 9=log10 9 /( log10 5) http://www.sosmath.com/algebra/logs/log4/log43/log43.html
Sunday, October 5, 2008 at 5:12pm by bobpursley
more pre cal logs
In your text or in your notes you should find 3 main rules for logs. 1. log(AB) = logA + logB 2. log (A/B) = logA - logB 3. log (A^n) = nlogA use them in this question. You did not use brackets to
establish the correct order of operations here, so I will let you finish it. You...
Monday, March 31, 2008 at 1:52am by Reiny
Now on this one I am not sure what you mean. log base 2 (x) or log (2x) I will assume base 2 then log x + log (4-x) = 0 where logs are base 2 adding logs is multiplying log (x(4-x)) = 0 x(4-x) = 2^0
which is one 4x - x^2 = 1 x^2 -4x +1 = 0 solve for x
Wednesday, July 22, 2009 at 4:59pm by Damon
college algebra
log(a^b) = b log(a) log(ab) = log(a) + log(b) since 1/b = b^-1, log(a/b) = log(a) - log(b) since √a = a^(1/2), log √a = 1/2 log a log x^4y^7/4 = 4logx + 7logy - log4 so you should be able to do the
other if these problems involve logs base 4, that doesn't matter. ...
Wednesday, April 17, 2013 at 4:36pm by Steve
Do you need to solve for t? Hint - take logs.
Tuesday, March 11, 2008 at 2:29pm by DrRuss
math- logarithms
how do i solve these logs? log1/2 (3x+1)^1/3= -2 3^(x^3)= 9^x
Thursday, January 8, 2009 at 7:16pm by kayla
Math - logs
3^log_3(2x - 1) = 3^2 but 3^log3 a = a for any a so 2 x - 1 = 3^2 = 9 2 x = 10 x = 5
Saturday, January 24, 2009 at 5:49pm by Damon
how do you solve for x using logs: [(a + b)^x]/b^x I am not sure of the steps to get there.
Tuesday, January 19, 2010 at 8:59pm by David
Math 12 (Logs)
Solve: 3^(2a+3)=7^(a-4) Please Help and Thank you
Sunday, April 15, 2012 at 7:34pm by Anonymous
math- exponents and logs
Yes, I am getting 2.063664032
Monday, March 4, 2013 at 5:07pm by Dr. Jane
help required in calculating the pH value of a .84 mols per litre solution of sulfuric acid. I understand that pH = -log 10 (2x 0.84), I am not very experienced with logs and need assistance as to
which keys to press on the scientific calculator in relation to the logs. thank ...
Tuesday, March 6, 2007 at 5:45am by ruby
math - logs
if logx3 = a and logx25 = b, determine an expression for logx(9/5).
Thursday, December 10, 2009 at 10:36pm by Cyn
Math Repost (Logs)
No. 10^.704=100/x solve for x
Sunday, February 21, 2010 at 4:54pm by bobpursley
Require that X*(1.025)^16 = 15,000. X is the initial amount that you will need to invest. The 1.025 is the factor by which balance increases by every 3 months. It does this 16 times in 4 years. You
can use logs to get the answer. I will use base 10 logs. (The base does not ...
Sunday, November 15, 2009 at 12:56pm by drwls
Which of the following materials will burn the fastest in open air? A. a log, two feet in diameter B. two logs, each one foot in diameter C. a pile of small splinters made from a two-foot diameter
log D. Both logs and the splinters will burn at the same rate.
Thursday, March 8, 2012 at 11:41am by Samantha
Which of the following materials will burn the fastest in open air? A. a log, two feet in diameter B. two logs, each one foot in diameter C. a pile of small splinters made from a two-foot diameter
log D. Both logs and the splinters will burn at the same rate.
Monday, March 12, 2012 at 4:47pm by Samantha
I see y = (2x+1)(x-2)^.5 / [(x-3)^(2/3)] There are many ways to do this to find dy/dx one is to use the quotient rule and just slug it out. If you know logs, there is a nice way: using the rules of
logs we get ln y = (1/2)ln(x-2) + ln(2x+1) - (2/3)ln(x-3) then (dy/dx) / y = 1...
Thursday, November 8, 2007 at 6:33pm by Reiny
The log button provides the logarithm (base 10) of any number punched into the calculator. (There is also an ln button on most calculators that provides the logarithm (base e) of any number punched
into the calculator. There is probably a way of solving that equation without ...
Saturday, November 29, 2008 at 3:51pm by DrBob222
Math - Working with Logs!!
So, you don't use the equation: M = log(I_i / I_o) ?
Tuesday, December 11, 2007 at 10:30pm by Anonymous
How would I simplify this problem using logs? (25/16) -3/2
Monday, June 2, 2008 at 7:26pm by Jenna
solve for n ... 2000 = 1000(1.035)^n YOu will have to use logs
Tuesday, November 9, 2010 at 10:49pm by Reiny
Algebra 2 ..
256^2 = (4^4)^2 = 4^8 so, you are correct. But, you can't be smart IF YOU SHOUT IN CAPS!! log 45^x = x log45, so you are correct again. log 12^(x+1) = (x+1) log 12/log 6, so you are right again.
Don't know whether you're smart, but you seem to understand logs. Better than I ...
Friday, February 22, 2013 at 10:19pm by Steve
Could you check these pleas? Firewood usually sold by a measure known as a cord. A full cord may be a stack 8x4x4ft or a stack 8x8x2ft 10. what is the volume of a full cord? 8*4*4=(128ft) 11. A short
cord or a face cord of wood is 8x4xThe lenght of the logs. What is the volume...
Tuesday, September 7, 2010 at 8:19pm by TiffanyJ
just as powers of products add (6^2 * 6^3 = 6^5) and subtract (6^8 / 6^3 = 6^5) log of a product is the sum of the logs, and log of a quotient is difference of logs. That's because of the definition
of log: log6N is the power of 6 needed to get N. Even though it looks weird, a...
Thursday, March 1, 2012 at 4:00am by Steve
Pre calculus
Usually you take logs to the base of the number having exponents. In your example, you have e^(3x-7), so take logs base e. If you had written 2^(x+3) * 2^(x-7) = 32 then you'd take logs base 2. As
for graphing, just recall the all exponent/log graphs look the same, except for ...
Tuesday, November 20, 2012 at 10:48am by Steve
This is an exponential function of the form y=ae^x Substituting x,y values (20,45)into the equation, 20=160*e^(45k) take logs on both sides and solve for k k=-(1/45)ln(8) =-0.04621 So the
mathematical model becomes: y(t)=y0*e^(-0.04621t) t=time in hours y0=initial mass y(t)=...
Sunday, June 23, 2013 at 12:30pm by MathMate
Math - logs
thank you! so if I get a problem like that, all I do is raise the base to the power of the log?
Saturday, January 24, 2009 at 5:49pm by Anonymous
also a problem that i cant figure out is how to do inverses of logs f(x)= 2log(base2)X
Friday, March 2, 2012 at 10:16am by Diana
3hours=8 logs 1hour =? if less, more will divide. 1/3 multiplys 8logs =8/3 =2 2/3
Wednesday, August 15, 2012 at 1:15pm by Vanessa
I need to use logs, sorry. 1.0275^n = 50,000/6250
Monday, January 14, 2013 at 6:56pm by Damon
Math (logs)
In x = -3 find x. I worked it out like this e^inx= e^-3 then I got x = 1/e^3. Is that correct?
Sunday, December 8, 2013 at 8:40pm by Wilson
HI THE QUESTION STATES THAT ONE SHOULD CALCULATE THE LOGS WITHOUT THE USE OF A CALCULATOR...OKAY MY PROBLEM IS THIS... 2 3 2log 8 + 2log 8 what should i do? i had come to the piont of... 4 5 log 8 +
log 8 But what's to do now i dont know..Think u can help?... cheers michelle ...
Wednesday, October 25, 2006 at 8:15am by TIGGER
Mr.Reiny, We don't use logs yet, so how do I find it without using log?
Thursday, April 18, 2013 at 7:48pm by Nancy
college algebra
You are going to use 3 man rules of logs 1. log(A/B) = logA - logB 2. Log(AB) = logA + logB 3. log A^n = nlogA log 3√x + log x^4 - log x^3 = log 3 + log x^(1/2) + 4logx - 3logx = log3 + (1/2)logs +
logx = log3 + (3/2)logx 2. 4 log (x+3) - 5 log (x^2+4) + 1/3 log y = log...
Wednesday, April 17, 2013 at 8:57am by Reiny
Algebra 2
Addition/subtraction of logs means product/division of numbers, while product of logs means power of numbers.Log m to the base b x Log n to the base b=log m^n. On keyboard,select the character to be
subscripted, press Cntrl+=. Or else select the character, go to Fonts Menu-...
Sunday, July 21, 2013 at 5:58pm by Anonymous
5 divided by (3-e to the -x power)=4 5/(3- e^-x) = 4 5 = 12 - 4 e^-x 4 e^-x = 7 e^-x = 7/4 e^x = 4/7 Take natural logs of both sides, x = ln (4/7) = -0.5596
Friday, December 8, 2006 at 1:00am by Anonymous
OMG!! THANKS!!! I totally kept adding in logs where I shouldn't have!! AWESOME!! Thanks!
Wednesday, January 9, 2008 at 10:55pm by Math genius! Urgent!
4^2.3 = 24.251... Multiply that by 3 for the answer. I used a hand calculator to get 4^2.3 You could also use logs.
Wednesday, May 28, 2008 at 11:28pm by drwls
solve for x without using a calculator: log5(3^x) = log25(9^(1-2x)) I don't know how to get rid of the logs for this one. Help? Thank you!
Sunday, November 15, 2009 at 6:18pm by Momo
Math (check please)
Could you check these please? Firewood is usually sold by a measure known as a cord. A full cord may be a stack 8x4x4ft or a stack 8x8x2ft 10. what is the volume of a full cord? 8*4*4=(128ft) 11. A
short cord or a face cord of wood is 8x4xThe lenght of the logs. What is the ...
Tuesday, September 7, 2010 at 8:51pm by TiffanyJ
How would you condense a logarithm expression into a single quantity if the logs have different bases?
Tuesday, November 20, 2012 at 1:20am by Jamie
How do I solve this using logs? I am having a total math meltdown tonight! 10=2^10/n Thanks again!
Wednesday, January 9, 2008 at 10:55pm by Math genius! Urgent!
Algebra II
If 4^x = 4+√19, then take logs, recalling that log(x^n) = n*log(x) x*log(4) = log(4+√19) and so on log(x^2) = (log x)^2 2 logx = (logx)^2 logx(2-logx) = 0 logx = 2 or 0 x = b^2 or 1 to whatever base
b you are using for the logs.
Saturday, February 15, 2014 at 9:17pm by Steve
College Algebra
#1. assuming base 8, just for ease of readability, 2logx = log(4x+12) log(x^2) = log(4x+12) so, if the logs are equal, so are the numbers: x^2 = 4x+12 x^2-4x-12=0 (x-6)(x+2)=0 x=6 only, since log(-2)
is not real. #2. so far, so good. Recall that adding/subtracting logs means ...
Wednesday, December 4, 2013 at 1:17pm by Steve
domain is your choice of x's that you can use in your equation For logs, one of the main properties is that I can only take the log of a positive number so for y = 4 + lnx , the domain is x > 0
however, the result of taking such logs results in any real number, so the range...
Sunday, April 7, 2013 at 10:17pm by Reiny
1. Given the product law of logarithms, prove the product law of exponents. 2. Given the quotient law of logarithms, prove the quotient law of exponents. 3. Apply algebraic reasoning to show that a=b
^(loga/logb) for any a,b>0 Please explain these to me. All I know is that ...
Monday, May 25, 2009 at 5:38pm by Jus
MATH please help!
adding logs is same as multplying log((2^3)/x)=log32 take the antilog of each side. 8/x=32 solve for x
Thursday, December 16, 2010 at 4:41pm by bobpursley
Thank you, Ms. Sue and Vanessa! I thought this might be a trick question, and that perhaps the correct answer was 2 (keeping logs whole).
Wednesday, August 15, 2012 at 1:15pm by Andrea
misplaced nodifier
Modifiers (adjectives and adjective phrases mostly) need to be placed as close as possible to whatever they're describing/modifying. Which one, do you think, is the corrected version of what you
posted? ~~> We had been hunting for the fire extinguisher behind a pile of logs...
Saturday, September 24, 2011 at 4:32pm by Writeacher
Subtracting logs is the equivalent of dividing. log 9 - log 300 = ??
Tuesday, July 28, 2009 at 10:17pm by DrBob222
change logs to numbers. 1E6(1+-.02) now, go to logs. remember that log(ab)=log a+logb Intensity=6 +log.98 to 6 +log(1.02) The error is the difference, so error=log1.02-log.98= .009+.009=.018 You can
work it more accurately. Now differentials: Error=highentensity-lowintensity ...
Wednesday, November 7, 2012 at 5:24pm by bobpursley
Math - logs
the log(10^ -1)of 3.66 = of 9.86 = of -2.8 = thank you
Saturday, January 19, 2013 at 12:57pm by Needs log help
thanks so much! i was sure it had logs involved but wasn't sure. thanks drwls and quidditch!
Wednesday, July 23, 2008 at 12:09am by Miley
solve for n 8000 = 4000(1.0785)^n 2 = (1.0785)^n you will have to use logs
Wednesday, April 21, 2010 at 4:53pm by Reiny
math. logs
anyone know how to solve this step by step? e^(3ln2)
Sunday, September 28, 2008 at 9:50pm by Kay
Thank you so much for your detailed reply. My problem is that we haven't done logs or "e" yet. Is there any other way arrive at an answer without knowing these? I really appreciate your help.
Tuesday, December 1, 2009 at 12:47am by Jim
Math logs
Solve the system of equations: y= (lnx)^2 + 2 (lnx^2) y = 3ln(1/x^2)+24 Im not sure what to do..
Wednesday, October 14, 2009 at 12:53am by Mike
math - logs
2 ln (x) = ln (20-x)
Monday, October 31, 2011 at 10:34pm by Sebastian
Please help me!!!! maths
I am not sure if the 5 , 8 , and 6 are bases of the logs or just multipliers. Let's hope they are just multipliers, or else it would be a terrible terrible mess by the laws of logs log5(x-2)+log8
(x-4) = log6(x-1) log [40(x-2)(x-4)] = log 6(x-1) anti-log it 40(x^2 - 6x + 8) = ...
Wednesday, April 3, 2013 at 5:33am by Reiny
Pre Calculus
Major Error ! How did you mathemagically change 5^x to 20^4x ? Did you do something like (5^x)(4^4) = 20^4x ?? If so then something like this should work : (3^5)(4^2) = 12^10 which is certainly false
You can only do this type of question if you know logs. take the log of both ...
Sunday, July 26, 2009 at 11:22am by Reiny
ln (24/6) using the law of logs or lns where subtraction becomes division. so you have ln 4
Tuesday, March 5, 2013 at 4:56pm by Dr. Jane
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September 17th 2007, 05:45 AM #1
I have two questions:
The first is that I can't seem to get the answer to the attached equation. The book says x = 1, or x = 1/2. Though this is probably because of my second question:
The second question is, is there a method to work out how to factorise? Sometimes it seems impossible by looking at the numbers. Is there more than one method to do this?
I have two questions:
The first is that I can't seem to get the answer to the attached equation. The book says x = 1, or x = 1/2. Though this is probably because of my second question:
The second question is, is there a method to work out how to factorise? Sometimes it seems impossible by looking at the numbers. Is there more than one method to do this?
Let me just go through that working again:
$x=1$ or $x=\frac{1}{2}$
Some extra info:
If you look at the discriminant, $\Delta=\sqrt{b^2-4ac}$, of a quadratic expression $ax^2+bx+c$, you can work out:
1. How many solutions it has
2. Whether or not it can be factored
For 1.
If $\Delta>0$, there are 2 real solutions.
If $\Delta = 0$, there is 1 real solution.
If $\Delta < 0$, there are no real solutions.
For 2.
If $\Delta$ is a perfect square, that is, if $\Delta=n^2,n \in Z$, then the expression can be factored, else it can't be.
For your original incorrect expression $2x^2-6x+1=0$, $\Delta=\sqrt{(-6)^2-4 \times 2 \times 1}=\sqrt{28}$, and since 28 isn't a perfect square it can't be factorised.
Whereas $2x^2-3x+1$ has discriminant $\Delta = \sqrt{(-3)^2-4 \times 2 \times 1}=\sqrt{1}$, which is a perfect square so it can.
There are other methods of finding roots, the first is completing the square, and the second is the quadratic formula:
The quadratic formula is very useful and you probably should remember it. For any expression $ax^2+bx+c=0$, the solutions are
$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-b \pm \Delta}{2a}$
Completing the square
In attached I don't understand the the last line. Could you explain it please?
How can we replace the first 2 terms in that way? I can't quite see it.
Thank you.
To form a square with the first two terms of a quadratic expression, add and subtract the square of half the coefficient of the x term (line 2). Then factorise (line 3).
In general:
$=ax^2+bx+\left(\frac{b}{2} \right)^2-\left(\frac{b}{2} \right)^2$
$=\left(a+\frac{b}{2} \right)^2-\left(\frac{b}{2} \right)^2$
September 17th 2007, 06:14 AM #2
September 17th 2007, 09:18 AM #3
September 17th 2007, 08:39 PM #4
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I think this has no answer.. There are 50 horses and 9 rooms.Accomodate the horses in the rooms in such a way that no room contains even number of horses...Any idea?
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If it were 10 rooms,the question would become easier.
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@ashwinjohn3 if it was 10 i wouldnt have posted it here !lol
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@ashwinjohn3 you are from mavelikkara?I am from Thrissur
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@Krishnadas NICE!!!!!Finally a keralite in openstudy!!! in which class r u studying?
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and you?
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@Krishnadas going 2 pc thomas classes?
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@ashwinjohn3 no ...I am counting on self study
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@Krishnadas Well if u want IIT u must go to coaching classes
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I don't think it is solvable. I'm not sure how to prove it yet.
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Start by putting 1 horse horse in each room, and then the remaining 41 horses, in any room. You have 8 rooms with 1 horse and 1 room with 42 horses.
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Obviously in this setup, we fail because room 9 has an even number of horses.
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Now, from this set up, we could reach any other possible set up by moving a horse from any room to any other room.
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@wio move??I didnt get that
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I mean, we can reach any other possible set up by a series of 'steps' in which each step is moving a single horse from 1 room to another.
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its about accomodating...not moving them around
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I understand that. What I mean is this... suppose you start off by putting all of the horses in one room, got it? We can eventually get to any other accommodation by moving one horse at a time to
some room.
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The motivation for this is to show there is an underlying property which CAN'T be violated regardless of your accommodation. Essentially it is proof by induction.
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but @wio each room should be accomodated
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Yeah, hold your horses and let me explain.
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Suppose you start out with 50 horses in one room and 0 in the rest. ok?
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You move a horse from one room to another. So you have a room with 49, a room with 1, and the rest have 0 horses.
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Whenever you move a horse between a room, both rooms change parity (change from even to odd or odd to even).
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This means that when you move a horse, you must change the parity of two rooms at a time.
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Does that make sense so far?
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There are four ways we can move a horse between rooms: 1) Move the horse from an odd room to an even one. In this case the number of even rooms stays the same, because the odd room becomes even
and vise versa 2) Move the horse from an odd room to an odd one. In this case the number of even rooms increases by 2, because both rooms now have even number of horses. 3) Move the horse from an
even room to an even one. In this case both rooms become odd, so the number of even rooms decreases by 2 4) Move the horse from an even room to an odd one. In this case, just like case 1, the
number of even rooms stays the same.
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So, by moving a horse between rooms, the number of even rooms either: 1) doesn't change 2) increases by 2 3) decreases by 2
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So we start with 50 horses in one room. We have 9 even rooms. Your goal is to have 0 even rooms, but the number of even rooms must go down by 2, up by 2, or not change... so it's impossible to
get 0 even rooms.
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Suppose someone came up with an accommodation that had 0 even rooms. We would be able to move every horse, one at a time, from the 8 other rooms into the 9th room. The number of even rooms could
only change by 2 or 0 though, so it's impossible for us to end up with 9 even rooms... yet they would have to be able to do it... So basically, this is why it is impossible.
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@wio thanks
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Do you really understand it?
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Sometimes I explain things poorly.
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x1+x2+x3+x4+x5+x6+x7+x8+x9=50 Let x1,x2,x3,x4,x5,x6,x7,x8 be odd numbers then x1+x2 =even number x3+x4 =even number x5+x6 =even number x7+x8= even number And even number + even number = even
number SO, x1+x2+x3+x4+x5+x6+x7+x8 = even number Now, Since 50 is even number x9 must also be even as x1+x2+x3+x4+x5+x6+x7+x8 is even. Thus, NO SOLUTION
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What Exactly is a Fraction?
Date: 10/15/2001 at 07:17:48
From: Sridhar Rajagopalan
Subject: What (exactly) are fractions? Is 3 a fraction?
What is a fraction? Is 3/1 a fraction? Is 5/sqrt(2) a fraction?
I tried to check some internet glossaries, and there does not seem to
be unanimity. The TUSD Math Glossary says:
"Fraction: a numeral representing some part of a whole; a numeral of
the form a/b (meaning a divided by b) where b is not zero."
The _Math to Build On_ glossary says:
"Fraction - A number that expressed a portion of a whole. The
denominator of a fraction represents the number of the portions the
whole has been divided into, and the numerator expresses the number
of the portions measured. The fraction 1/4 could be stated as 1 our
of 4 parts of the whole."
The MathPro Press On-line Mathematics Dictionary says:
"Fraction: An expression of the form a/b."
(This reminds me of the definition of a trapezium. Some say that a
trapezium is a quadrilateral with one pair of sides parallel. Some
say that plus, the other pair of opposite sides should not be
Personally, I am comfortable with having identical definitions of
fractions and rational numbers. But I have seen many sources defining
fractions in such a way that -3/5 is not a fraction, only a rational
Date: 10/15/2001 at 08:37:20
From: Doctor Rick
Subject: Re: What (exactly) are fractions? Is 3 a fraction?
Hello, Sridhar.
You've asked a good question, but since there is no official
international governing body for mathematical definitions (as far as I
know), you'll only get another viewpoint from me, not a definitive
I am not particularly comfortable with having identical definitions of
fraction and rational number, for the simple reason that then we would
have no reason to keep both words in our lexicon.
I am not entirely pleased with any of the definitions you list,
because they all omit an essential requirement. I would go with this
A fraction is a representation for a rational number, in the form
a/b, where a is an integer, and b is a *positive* integer. The
number so represented is the result of dividing a by b.
Do you see what was missing in your definitions? This is NOT a
Your first definition comes closest to mine. I like the word "numeral"
in that definition, as distinct from "number." The same *number* can
be represented by the *numerals* 3/2 or 1.5; only the first is a
fraction. Likewise, 3 can be represented by the numeral (fraction)
3/1, but the number itself is not therefore a fraction.
I have no problem with calling -2/5 or 3/2 a fraction; both are
improper fractions, representing numbers outside the range 0 to 1.
This makes me somewhat uncomfortable with the phrase "representing
some part of a whole," which strictly limits the definition to proper
fractions. This description is helpful in introducing the concept of
fractions, but once improper fractions are introduced, it no longer
belongs in the definition.
I see no need to have the definition so broad that 2/-5 is a fraction.
As it stands, every rational number can be represented as a fraction.
(In fact, *only* rational numbers can be represented as a fraction, so
this makes for a simple definition of rational numbers.)
I see no need, on the other hand, to restrict the definition so that
every rational number has a *unique* representation as a fraction.
Thus, 4/6 and 2/3 represent the same rational number. There is,
however, a unique representation as a fraction *in lowest terms*.
These are my opinions. I will leave the question for other Math
Doctors to respond to if they have different or additional viewpoints.
- Doctor Rick, The Math Forum
Date: 10/15/2001 at 09:29:37
From: Doctor Peterson
Subject: Re: What (exactly) are fractions? Is 3 a fraction?
Hi, Sridhar.
I would like to take a slightly different perspective from Dr. Rick's
on this. I fully agree with him on the most precise definition of
"fraction": it is a particular representation of a rational number.
But it is worth noting that, like many other words, it actually has
several different meanings depending on the context.
The root meaning of "fraction" is "broken piece," which is the source
of the idea that it must be a part of a whole. This is common in
informal use; if I say "only a fraction of the people here understand
what the word means," I mean "less than the whole," and probably
much less. I am saying nothing about whether those people are
"rational." ;-) In mathematical terms, this concept arises in the
phrase "proper fraction" - that is, a mathematical fraction that fits
the informal sense that it should be less than one.
Once we get into the mathematical realm, a fraction always refers to a
way of writing a number, using numerator and denominator. Most
narrowly, these must be whole numbers (or integers, once children are
introduced to negative numbers). This kind of fraction is more fully
called a "common" or "vulgar" fraction, and this is what Dr. Rick
defined for you. When we use the word "fraction" alone, we usually
mean this.
From here we find a broader definition, given in my American Heritage
dictionary as "an indicated quotient of two quantities." This retains
the concept that a fraction is a way of writing something, as distinct
from its actual value, but ignores questions as to what sort of
quantities are being divided. This allows for "fractions" like 1.2/3.4
or (x+1)/(y-1). The term is in fact used in these senses; the latter
might be called an algebraic fraction. In this realm we can
distinguish between "simple fractions" and "complex or compound
fractions" like (1/2)/(3/4); common fractions are always simple.
Finally, we see the phrase "decimal fraction" used for non-integral
decimal numbers; this drops the "indicated quotient" concept and
retains only the "broken" (non-integral) aspect of the most basic
definition. I don't think this meaning is ever intended when we use
the word "fraction" without qualification; in fact, we more often drop
the word "fraction" and just call it a "decimal," which can be a
dangerous practice!
Clearly the term "fraction" has a somewhat different meaning in each
To answer your specific questions, 3/1 is a fraction (specifically a
simple, common, but improper fraction), while 5/sqrt(2) would only be
called a fraction in an algebraic context. On the other hand, 3 is not
a fraction, even though it is a rational number.
- Doctor Peterson, The Math Forum
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New to CompSci and Java
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Hi i just started my CompSci class and for hw my teacher gave us the question:
"Write an expression that, given a positive integer n, computes a new integer
in which the units and tens digits have swapped places. For example, if
n = 123, the result should be 132; if n = 3, the tens digit is zero and the result should be 30."
Im not really sure how i would do this but my guess is that it is more of a math question than a java question. If you have any ideas that might help please let me know.
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There are two ways to approach this question.
A) Convert the int into a String and manipulate the String chars using the String method charAt(int i).
B) Mathematically isolate your tens and ones digits and all the digits from the hundreds on up. This is probably what your teacher wants you to do. To do this, look up "integer division" and
also the modulus operator. HTH.
Here is a the solution in the first way that Fubarable explain,
Java Code:
Scanner scn = new Scanner(System.in);
String result = null;
System.out.println("Enter the value: ");
String str = scn.nextLine();
if(str.length() == 1)
result = str + "0";
result = str.substring(0, (str.length() - 2)) +
str.charAt(str.length() - 1) + str.charAt(str.length() - 2);
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And here is the other way:
Java Code:
public int swapTensAndUnits(int n) {
int tens = (n/10)%10;
int units = n%10;
int rest = n-n%100;
return rest+units*10+tens;
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I would always have a go first, cause its going to get harder you need the basics. Good luck in your studies.
Ian J.
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Heh, that question totally blew my mind, you guys thought of the answer pretty quickly, do questions like that get easier over time?
I'm not clear what you are trying to say here.
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We'll I was pretty lost as that guy about the question because that isn't a normal math question and you have to think differently then normally does it get easier. I don't know if I can
explain it any easier, maybe it's a dumb question?
If you are looking in programing way, post #3 have the solution. Or else if you are looking on mathematical way, post #4 is the solution.
I don't know what you mean by normal math question. There are no single pattern in maths. In different applications it takes in different ways. Quite similar thing is done in image
processing, when converting an image into black and white, and many more usages are there.
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1. By annoyingzhang in forum New To Java
Last Post: 10-12-2008, 09:02 PM
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Great Prelim Question
Roy is coming to plant flowers in Zoe’s garden. Zoe loves flowers, her utility for a garden with $x$ flowers is
$z(x) = x.$
Roy plants a unit mass of seeds and the fraction of these that will bloom into flowers depends on how attentive Roy is as a gardener. Roy’s attentiveness is his type $\theta$. In particular when
Roy’s type is $\theta$, absent any sabotage by Zoe, there will be $\theta$ flowers in Zoe’s garden in Spring. Roy’s attentiveness is unknown to everyone and it is believed by all to be uniformly
distributed on the unit interval.
Jane, Zoe’s neighbor, is looking for a gardener for the following Spring. Jane has high standards, she will hire Roy if and only if he is sufficiently attentive. In particular, Jane’s utility for
hiring Roy when his true type is $\theta$ is given by
$j(\theta) = \theta - 2/3.$
(Her utility is zero if she does not hire Roy.)
Roy tends to one and only one garden per year. Therefore Roy will continue to plant flowers in Zoe’s garden for a second year if and only if Jane does not hire him away from her.
Consequently, Zoe is contemplating sabotaging Roy’s flowers this year. If Zoe destroys a fraction $1 - \alpha$ of Roy’s seeds then the total number of flowers in Zoe’s garden when Spring arrives will
be $x = \alpha\theta$. Of course sabotage is costly for Zoe because she loves flowers.
There will be no sabotage in the second year because after two years of gardening Roy goes into retirement. Therefore, if Zoe destroys $1-\alpha$ in the first year and Roy continues to work for Zoe
in the second year, Zoe’s total payoff will be
$z(\alpha\theta) + z(\theta)$
whereas if Roy is hired away by Jane, then Zoe’s total payoff is just $z(\alpha\theta)$.
This is a two-player (Zoe and Jane) extensive-form game with incomplete information. The timing is as follows. First, Roy’s type is realized. Nobody observes Roy’s type. Zoe moves first and chooses $
\alpha \in [0,1]$. Then Spring arrives and the flowers bloom. Jane does not observe $\alpha$ but does observe the number of flowers in Zoe’s garden. Then Jane chooses whether or not to hire Roy away
from Zoe. Then the game ends.
Describe the set of all Perfect Bayesian Equilibria.
6 comments
Is it the empty set?
After some clumsy crowdsourcing, 6493 provides a promising answer
To the comment above:
- Jane’s Utility decreasing by 2/3 implies that she only hires Roy if she thinks that thetha>2/3. Remember that not hiring gives Jane zero utility and hiring Roy with thetha2/3, then Roy will be
hired in the second period, Zoe gets 5/6
if thetha1/2
So destroying the flowers is a dominated strategy… so much for a deterministic solution.
Zoe will randomize using f(y)=P(alpha=y) over (0,1] so that it will make Jane more uninformed, that is, x is still a signal of thetha, but is less precise, since now it is made of the product of 2
random variables, thetha and p.
The cheapest way to do so is to find a function f(y) such that the signal x is uninformative, and the posterior is such that Jane is indifferent between hiring Roy or not, that is E(thetha-2/3/x,f)=E
What Are Your Thoughts?
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Algorithmic graph theory and perfect graphs, volume 57 of Annals of Discrete Mathematics
Results 1 - 10 of 906
- J. Algorithms , 1998
"... The notion of treewidth has seen to be a powerful vehicle for many graph algorithmic studies. This survey paper wants to give an overview of many classes of graphs that can be seen to have a
uniform upper bound on the treewidth of graphs in the class. Also, some mutual relations between such classes ..."
Cited by 255 (38 self)
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The notion of treewidth has seen to be a powerful vehicle for many graph algorithmic studies. This survey paper wants to give an overview of many classes of graphs that can be seen to have a uniform
upper bound on the treewidth of graphs in the class. Also, some mutual relations between such classes are discussed.
- In Proceedings of ISMB 2002 , 2002
"... In gene expression data, a bicluster is a subset of the genes exhibiting consistent patterns over a subset of the conditions. We propose a new method to detect significant biclusters in large
expression datasets. Our approach is graph theoretic coupled with statistical modelling of the data. Under p ..."
Cited by 190 (4 self)
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In gene expression data, a bicluster is a subset of the genes exhibiting consistent patterns over a subset of the conditions. We propose a new method to detect significant biclusters in large
expression datasets. Our approach is graph theoretic coupled with statistical modelling of the data. Under plausible assumptions, our algorithm is polynomial and is guaranteed to find the most
significant biclusters. We tested our method on a collection of yeast expression profiles and on a human cancer dataset. Cross validation results show high specificity in assigning function to genes
based on their biclusters, and we are able to annotate in this way 196 uncharacterized yeast genes. We also demonstrate how the biclusters lead to detecting new concrete biological associations. In
cancer data we are able to detect and relate finer tissue types than was previously possible. We also show that the method outperforms the biclustering algorithm of Cheng and Church (2000).
- Journal of Algorithms , 1985
"... This is the nineteenth edition of a (usually) quarterly column that covers new developments in the theory of NP-completeness. The presentation is modeled on that used by M. R. Garey and myself
in our book ‘‘Computers and Intractability: A Guide to the Theory of NP-Completeness,’ ’ W. H. Freeman & Co ..."
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This is the nineteenth edition of a (usually) quarterly column that covers new developments in the theory of NP-completeness. The presentation is modeled on that used by M. R. Garey and myself in our
book ‘‘Computers and Intractability: A Guide to the Theory of NP-Completeness,’ ’ W. H. Freeman & Co., New York, 1979 (hereinafter referred to as ‘‘[G&J]’’; previous columns will be referred to by
their dates). A background equivalent to that provided by [G&J] is assumed, and, when appropriate, cross-references will be given to that book and the list of problems (NP-complete and harder)
presented there. Readers who have results they would like mentioned (NP-hardness, PSPACE-hardness, polynomial-time-solvability, etc.) or open problems they would like publicized, should
, 1999
"... Recently, there has been much interest in reverse engineering genetic networks from time series data. In this paper, we show that most of the proposed discrete time models — including the
boolean network model [Kau93, SS96], the linear model of D’haeseleer et al. [DWFS99], and the nonlinear model of ..."
Cited by 157 (1 self)
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Recently, there has been much interest in reverse engineering genetic networks from time series data. In this paper, we show that most of the proposed discrete time models — including the boolean
network model [Kau93, SS96], the linear model of D’haeseleer et al. [DWFS99], and the nonlinear model of Weaver et al. [WWS99] — are all special cases of a general class of models called Dynamic
Bayesian Networks (DBNs). The advantages of DBNs include the ability to model stochasticity, to incorporate prior knowledge, and to handle hidden variables and missing data in a principled way. This
paper provides a review of techniques for learning DBNs. Keywords: Genetic networks, boolean networks, Bayesian networks, neural networks, reverse engineering, machine learning. 1
- Uncertainty in Artificial Intelligence , 1990
"... In this paper, we describe an abstract framework and axioms under which exact local computation of marginals is possible. The primitive objects of the framework are variables and valuations. The
primitive operators of the framework are combination and marginalization. These operate on valuations. We ..."
Cited by 137 (17 self)
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In this paper, we describe an abstract framework and axioms under which exact local computation of marginals is possible. The primitive objects of the framework are variables and valuations. The
primitive operators of the framework are combination and marginalization. These operate on valuations. We state three axioms for these operators and we derive the possibility of local computation
from the axioms. Next, we describe a propagation scheme for computing marginals of a valuation when we have a factorization of the valuation on a hypertree. Finally we show how the problem of
computing marginals of joint probability distributions and joint belief functions fits the general framework. 1.
- Artificial Intelligence , 1992
"... Representing and reasoning about incomplete and indefinite qualitative temporal information is an essential part of many artificial intelligence tasks. An interval-based framework and a
point-based framework have been proposed for representing such temporal information. In this paper, we address ..."
Cited by 137 (6 self)
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Representing and reasoning about incomplete and indefinite qualitative temporal information is an essential part of many artificial intelligence tasks. An interval-based framework and a point-based
framework have been proposed for representing such temporal information. In this paper, we address two fundamental reasoning tasks that arise in applications of these frameworks: Given possibly
indefinite and incomplete knowledge of the relationships between some intervals or points, (i) find a scenario that is consistent with the information provided, and (ii) find the feasible relations
between all pairs of intervals or points. For the point-based framework and a restricted version of the intervalbased framework, we give computationally efficient procedures for finding a consistent
scenario and for finding the feasible relations. Our algorithms are marked improvements over the previously known algorithms. In particular, we develop an O(n 2 ) time algorithm for finding one co...
, 1992
"... Chaitin and his colleagues at IBM in Yorktown Heights built the first global register allocator based on graph coloring. This thesis describes a series of improvements and extensions to the
Yorktown allocator. There are four primary results: Optimistic coloring Chaitin's coloring heuristic pessimis ..."
Cited by 135 (4 self)
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Chaitin and his colleagues at IBM in Yorktown Heights built the first global register allocator based on graph coloring. This thesis describes a series of improvements and extensions to the Yorktown
allocator. There are four primary results: Optimistic coloring Chaitin's coloring heuristic pessimistically assumes any node of high degree will not be colored and must therefore be spilled. By
optimistically assuming that nodes of high degree will receive colors, I often achieve lower spill costs and faster code; my results are never worse. Coloring pairs The pessimism of Chaitin's
coloring heuristic is emphasized when trying to color register pairs. My heuristic handles pairs as a natural consequence of its optimism. Rematerialization Chaitin et al. introduced the idea of
rematerialization to avoid the expense of spilling and reloading certain simple values. By propagating rematerialization information around the SSA graph using a simple variation of Wegman and
Zadeck's constant propag...
- NETWORKS , 1995
"... Unit disk graphs are intersection graphs of circles of unit radius in the plane. We present simple and provably good heuristics for a number of classical NP-hard optimization problems on unit
disk graphs. The problems considered include maximum independent set, minimum vertex cover, minimum coloring ..."
Cited by 126 (6 self)
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Unit disk graphs are intersection graphs of circles of unit radius in the plane. We present simple and provably good heuristics for a number of classical NP-hard optimization problems on unit disk
graphs. The problems considered include maximum independent set, minimum vertex cover, minimum coloring and minimum dominating set. We also present an on-line coloring heuristic which achieves a
competitive ratio of 6 for unit disk graphs. Our heuristics do not need a geometric representation of unit disk graphs. Geometric representations are used only in establishing the performance
guarantees of the heuristics. Several of our approximation algorithms can be extended to intersection graphs of circles of arbitrary radii in the plane, intersection graphs of regular polygons, and
to intersection graphs of higher dimensional regular objects.
- In International Symposium on Low Power Electronics and Design , 1996
"... Abstract|We present a dynamic programming technique for solving the multiple supply voltage scheduling problem in both non-pipelined and functionally pipelined data-paths. The scheduling problem
refers to the assignment of a supply voltage level (selected from a xed and known number of voltage level ..."
Cited by 120 (5 self)
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Abstract|We present a dynamic programming technique for solving the multiple supply voltage scheduling problem in both non-pipelined and functionally pipelined data-paths. The scheduling problem
refers to the assignment of a supply voltage level (selected from a xed and known number of voltage levels) to each operation in a data ow graph so as to minimize the average energy consumption for
given computation time or throughput constraints or both. The energy model is accurate and accounts for the input pattern dependencies, re-convergent fanout induced dependencies, and the energy cost
of level shifters. Experimental results show that using three supply voltage levels on a number of standard benchmarks, an average energy saving of 40.19% (with a computation time constraint of 1.5
times the critical path delay) can be obtained compared to using a single supply voltage level.
- Journal of the ACM , 2000
"... We present a general framework for solving resource allocation and scheduling problems. Given a resource of fixed size, we present algorithms that approximate the maximum throughput or the
minimum loss by a constant factor. Our approximation factors apply to many problems, among which are: (i) real ..."
Cited by 119 (18 self)
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We present a general framework for solving resource allocation and scheduling problems. Given a resource of fixed size, we present algorithms that approximate the maximum throughput or the minimum
loss by a constant factor. Our approximation factors apply to many problems, among which are: (i) real-time scheduling of jobs on parallel machines, (ii) bandwidth allocation for sessions between two
endpoints, (iii) general caching, (iv) dynamic storage allocation, and (v) bandwidth allocation on optical line and ring topologies. For some of these problems we provide the first constant factor
approximation algorithm. Our algorithms are simple and efficient and are based on the localratio technique. We note that they can equivalently be interpreted within the primal-dual schema.
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Next: About this document Up: SDPpack User's Guide Version Previous: Examples
This appendix provides benchmarks of SDPpack Version 0.9 BETA on some randomly generated test problems, a set of 16 LMI problems from control applications (some of which apparently are infeasible),
and a set of 8 problems from truss topology design. The control and truss design problems are available as mat files from the SDPpack home page.
All problems were solved using sql.m. The random problems and the truss design problems used the default option values seet by setopt.m, i.e.:
The control LMI problems were solved with the same parameter settings above, but with
Table 3 shows a set of random problems with semidefinite blocks only. Each problem has three semidefinite blocks of equal size. The number n shown in the table is the sum of the block sizes. Thus,
for example, the largest problem shown has three blocks each of size 100.
Table 4 shows a set of random problems with quadratic blocks only. Each problem has three quadratic blocks of equal size. The number n shown in the table is the sum of the block sizes. Thus, for
example, the largest problem shown has three blocks each of size 250.
Tables 5 and 6 show the truss and control LMI results respectively. When the symbol * appears next to the value of termflag, this indicates that a restart was necessary.
Table 3: Randomly generated with semidefinite blocks only
Table 4: Randomly generated with quadratic blocks only
Table 5: Benchmarks on problems from truss topology design
Table 6: Benchmarks on LMI problems from control applications
Next: About this document Up: SDPpack User's Guide Version Previous: Examples Madhu Nayakkankuppam
Wed Jun 25 18:01:54 EDT 1997
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Beacon Lesson Plan Library
Drawing Bugs Game
Michal Dunnivant
Students explore probability by predicting the likelihood of rolling any one number on a fair die, graphing data, and analyzing the results of playing a drawing game.
Florida Sunshine State Standards
The student writes notes, comments, and observations that reflect comprehension of content and experiences from a variety of media.
The student solves problems by generating, collecting, organizing, displaying, and analyzing data using histograms, bar graphs, circle graphs, line graphs, pictographs, and charts.
The student predicts the likelihood of simple events occurring.
Florida Process Standards
Effective Communicators
02 Florida students communicate in English and other languages using information, concepts, prose, symbols, reports, audio and video recordings, speeches, graphic displays, and computer-based
Numeric Problem Solvers
03 Florida students use numeric operations and concepts to describe, analyze, communicate, synthesize numeric data, and to identify and solve problems.
- Graph Paper (see Attached File)
- Overhead transparency of Graph Paper
- Standard die (one per student or pair of students)
- Overhead pen
- Data Diary (one per student)
- Overhead projector
- Crayons or other marking pencils
- Overhead transparency grid for class graph or butcher paper
- Graph Criteria poster (see Attached File)
- Product Rubric (see Attached File)
- Drawing Bug Direction Sheet (see Attached File)
- Online student lessons (see Web Links)
The teacher needs to:
1. Duplicate Graph Paper one per student or per pair of students from the master in the Attached File.
2. Make one transparency from the Graph Paper master.
3. Make one Data Diary (or journal) per student by stapling five sheets of folded 11 x 17 copy paper book-style together with a construction paper cover.
4. Preview the lessons listed in the Weblinks that reinforce the concepts taught in this lesson.
5. Prepare an overhead transparency or butcher paper to organize the data into a class graph.
1. Review other simple events and experiments conducted with spinners and coins thus far in the unit. (see Extensions)
2. Ask students what they know about a die or dice and why they think people use dice (or number cubes).
3. Tell students that today we will explore probability by playing a drawing game with a die. Discuss the importance of understanding how likelihood works in probability for decision-making in life.
4. Ask: When you roll a die, what are the possible outcomes? Draw a tree diagram to illustrate each of the possible outcomes.
5. Ask: What is the likelihood of rolling any one number on the die? During this discussion, ask a number of questions and refer back to previous experiments.
-Is a standard die fair?
-Why do you say so?
-Will any one number be rolled more than another?
-What are the chances of rolling one number rather than another?
6. Ask students to predict which number will win? Chart their responses. Ask for clarification or justification based on their past experiences.
7. Tell students that today we will play the Drawing Bugs game to help us answer the question, What is the likelihood of rolling any one number on a die? Establish this as the problem.
8. Model one round of the game for the class and record the results on the Drawing Bug Directions Sheet overhead transparency.
9. Distribute the Drawing Bugs Direction Sheet. Ask the class: How many dice rolls will it take to make a complete bug? Will the bug have more or less of the body parts it needs to be complete?
Discuss their predictions as they record in their Data Diary.
10. Check to make sure students understand how to play the game and distribute the materials: One die per student or pair of students, Drawing Bugs Direction Sheet, and Data Diaries or journal paper.
Remind students to keep a tally of the each and every roll of the die as they complete their drawing. (This is very important to minimize confusion. Students record each roll of the die as a tally
mark, but they are only allowed to draw a certain number of each body part.)
11. Allow time for students to conduct experiments and record data by tallying in the box at the bottom of the sheet. Then, they can draw the number of bug body parts that coincides with the number
12. During the round, pause and analyze the results.
-What is the data telling us?
-Would you like to change your prediction? Why?
-What true statements can you say about the data? (A true statement is a fact expressed by the data, i.e., six is rolled 10 times, two was rolled more times than three of the other numbers, etc.)
Emphasize the importance of recording the data accurately.
13. Ask students to make predictions for class results of the game, i.e., Which insect body part will have the most wins? Why?
14. After students complete their games, model the process of deciding how to organize the data so we can analyze the results. Ask,
-Do we know what the likelihood is for rolling any one number on a die from the results of the game?
-Why or why not?
-How can we easily determine this with the data we have?
Lead students to realize that the data needs to be organized first.
15. Tell students that they are going to create a graph to organize the results. Model how to create a graph. Graphs should include appropriate labels, scale, and title. As you create a graph on the
overhead, review the components of a graph as shown on the Product Rubric written in student wording (see Associate File).
16. Allow students time for creating their graphs to display the results.
17. Encourage students to check their predictions with the data. Ask
-What do we know about the likelihood of rolling any one number on the die after conducting our experiments?
-How close were our predictions to the actual results of the experiment?
-Was any one number rolled more times than another? Why?
-Did everyone get the same results? Why or why not?
18. After the discussion, students then record what they know about probability based upon previous experiments in their Data Diary. Their entry should also explain their initial prediction and how
close it was to the actual results. Here are some other questions to which students might respond.
-Did you change your prediction during the experiments? Why?
-What is the likelihood of rolling any one number on the die? How do you know?
-Is the die fair? How do you know?
Then show the students a non-standard die, i.e., maybe an eight-sided die or one with fewer than six sides, or maybe one with different markings. Ask: What is the likelihood of rolling any one number
on this die?-
NOTE: The real-life bug with the same number of body parts is the honeybee.
Assess the student-created graphs in this lesson as a product and the Data Diary (or journal) entry for understanding content.
Introduce the Product Rubric (for students) that is in the Attached File as you assist students with self-assessment. The purpose of this assessment is not for a grade, but to guide students to reach
for higher quality products and understanding. If you need to take a grade at this point, equate a value to the levels of performance on the rubric, but allow students to improve their work through
self-assessment before you actually score it. It is also helpful for students to score their own work. This will help them evaluate the quality of their work based upon the criteria and begin to see
where they need to adjust to improve next time.
In this lesson, students examine likelihood. This concept can be extended to express likelihood as ratios in fraction form as found in the lesson entitled Notes to a Mathematician.
Web Links
Interactive Student Web Lesson
Pin the Tail on the Tiger
Interactive Student Web Lesson
Lions and Tigers and Probability, Oh My!
Return to the Beacon Lesson Plan Library.
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New to CompSci and Java
Join Date
Oct 2008
Rep Power
Hi i just started my CompSci class and for hw my teacher gave us the question:
"Write an expression that, given a positive integer n, computes a new integer
in which the units and tens digits have swapped places. For example, if
n = 123, the result should be 132; if n = 3, the tens digit is zero and the result should be 30."
Im not really sure how i would do this but my guess is that it is more of a math question than a java question. If you have any ideas that might help please let me know.
Join Date
Jun 2008
Blog Entries
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There are two ways to approach this question.
A) Convert the int into a String and manipulate the String chars using the String method charAt(int i).
B) Mathematically isolate your tens and ones digits and all the digits from the hundreds on up. This is probably what your teacher wants you to do. To do this, look up "integer division" and
also the modulus operator. HTH.
Here is a the solution in the first way that Fubarable explain,
Java Code:
Scanner scn = new Scanner(System.in);
String result = null;
System.out.println("Enter the value: ");
String str = scn.nextLine();
if(str.length() == 1)
result = str + "0";
result = str.substring(0, (str.length() - 2)) +
str.charAt(str.length() - 1) + str.charAt(str.length() - 2);
Join Date
Oct 2008
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And here is the other way:
Java Code:
public int swapTensAndUnits(int n) {
int tens = (n/10)%10;
int units = n%10;
int rest = n-n%100;
return rest+units*10+tens;
Join Date
Oct 2008
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I would always have a go first, cause its going to get harder you need the basics. Good luck in your studies.
Ian J.
Join Date
Jun 2008
Blog Entries
Rep Power
Join Date
Oct 2008
Rep Power
Heh, that question totally blew my mind, you guys thought of the answer pretty quickly, do questions like that get easier over time?
I'm not clear what you are trying to say here.
Join Date
Oct 2008
Rep Power
We'll I was pretty lost as that guy about the question because that isn't a normal math question and you have to think differently then normally does it get easier. I don't know if I can
explain it any easier, maybe it's a dumb question?
If you are looking in programing way, post #3 have the solution. Or else if you are looking on mathematical way, post #4 is the solution.
I don't know what you mean by normal math question. There are no single pattern in maths. In different applications it takes in different ways. Quite similar thing is done in image
processing, when converting an image into black and white, and many more usages are there.
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Key Enumeration
The topic you requested is included in another documentation set. For convenience, it's displayed below. Choose Switch to see the topic in its original location.
Key Enumeration
Specifies the possible key values on a keyboard.
Namespace: System.Windows.Input Assembly: System.Windows
(in System.Windows.dll)
<object property="enumMemberName"/>
Member name Description
None A special value indicating no key.
Back The BACKSPACE key.
Tab The TAB key.
Enter The ENTER key.
Shift The SHIFT key.
Ctrl The CTRL (control) key.
Alt The ALT key.
CapsLock The CAPSLOCK key.
Escape The ESC (also known as ESCAPE) key.
Space The SPACE key.
PageUp The PAGEUP key.
PageDown The PAGEDOWN key.
End The END key.
Home The HOME key.
Left The left arrow key.
Up The up arrow key.
Right The right arrow key.
Down The down arrow key.
Insert The INSERT key.
Delete The DEL (also known as DELETE) key.
D0 The 0 (zero) key.
D1 The 1 key.
D2 The 2 key.
D3 The 3 key.
D4 The 4 key.
D5 The 5 key.
D6 The 6 key.
D7 The 7 key.
D8 The 8 key.
D9 The 9 key.
A The A key.
B The B key.
C The C key.
D The D key.
E The E key.
F The F key.
G The G key.
H The H key.
I The I key.
J The J key.
K The K key.
L The L key.
M The M key.
N The N key.
O The O key.
P The P key.
Q The Q key.
R The R key.
S The S key.
T The T key.
U The U key.
V The V key.
W The W key.
X The X key.
Y The Y key.
Z The Z key.
F1 The F1 key.
F2 The F2 key.
F3 The F3 key.
F4 The F4 key.
F5 The F5 key.
F6 The F6 key.
F7 The F7 key.
F8 The F8 key.
F9 The F9 key.
F10 The F10 key.
F11 The F11 key.
F12 The F12 key.
NumPad0 The 0 key on the number pad.
NumPad1 The 1 key on the number pad.
NumPad2 The 2 key on the number pad.
NumPad3 The 3 key on the number pad.
NumPad4 The 4 key on the number pad.
NumPad5 The 5 key on the number pad.
NumPad6 The 6 key on the number pad.
NumPad7 The 7 key on the number pad.
NumPad8 The 8 key on the number pad.
NumPad9 The 9 key on the number pad.
Multiply The * (MULTIPLY) key.
Add The + (ADD) key.
Subtract The - (SUBTRACT) key.
Decimal The . (DECIMAL) key.
Divide The / (DIVIDE) key.
Unknown A special value indicating the key is out of range of this enumeration.
To detect key combinations or modifier keys, use techniques as documented in the topic Keyboard Support.
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Spline Interpolation
Spline Interpolation
December 13, 2007 - 17:26
Is the Interpolator that is returned from Interpolators.getSplineInstance() supposed to be compatible with the SMIL standard? This would be important to know for any conversion to or from SVG. From
my tests, it seems to be more or less compatible with the Timing Framework's behavior, but not the SMIL standard. Maybe I am missing something.
Re: Spline Interpolation
January 15, 2008 - 12:26
Is there a different place that I should have sent the signed SCA to other than the official one? I emailed a scanned copy almost 4 weeks ago, and faxed it in 1 week ago, and I still am not showing
up on the list.
Re: Spline Interpolation
January 15, 2008 - 12:53
Hi David,
Sorry for the delays; I don't know what the cause is, but I'm looking
into it right now.
scenario@javadesktop.org wrote:
> Is there a different place that I should have sent the signed SCA to other than the official one? I emailed a scanned copy almost 4 weeks ago, and faxed it in 1 week ago, and I still am not showing
up on the list.
> [Message sent by forum member 'davidbrowne' (davidbrowne)]
> http://forums.java.net/jive/thread.jspa?messageID=254048
> ---------------------------------------------------------------------
> To unsubscribe, e-mail: dev-unsubscribe@scenegraph.dev.java.net
> For additional commands, e-mail: dev-help@scenegraph.dev.java.net
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Re: Spline Interpolation
January 15, 2008 - 13:50
Hi David,
It seems there was a backlog from winter break, which has now been processed. Your name is now on the list, so we can finally take a look at your code.
Re: Spline Interpolation
January 15, 2008 - 14:47
Thanks Chris. I hope it is useful to you.
Re: Spline Interpolation
February 12, 2008 - 19:49
Hi David,
I've finally (finally!) committed the code that you contributed, as well as a test case based on the one you posted in this thread. I did some minor code formatting/doc cleanup, as well as renamed
the class. Hopefully you and others can pound on it in the coming weeks and will let us know if there are any concerns. Thanks again for the contribution, and sorry for taking so long to get around
to this.
Re: Spline Interpolation
February 12, 2008 - 20:31
No worries, Chris. Glad to help.
Re: Spline Interpolation
December 13, 2007 - 17:26
Hi David,
On Dec 13, 2007, at 4:26 PM, scenario@javadesktop.org wrote:
> Is the Interpolator that is returned from
> Interpolators.getSplineInstance() supposed to be compatible with
> the SMIL standard?
It should be, yes.
> This would be important to know for any conversion to or from
> SVG. From my tests, it seems to be more or less compatible with
> the Timing Framework's behavior, but not the SMIL standard. Maybe
> I am missing something.
Can you provide a testcase that demonstrates the differences that
you're seeing? It's hard to guess what the problem could be otherwise.
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Re: Spline Interpolation
December 13, 2007 - 22:08
Hi Chris,
I should first point you to this discussion on the Timing Framework forum, http://forums.java.net/jive/thread.jspa?threadID=33844&tstart=0
The Scene Graph numbers are similar to the Timing Framework numbers, so I am assuming that you are using a similar approach. To quote pepe, "It looks like SplineInterpolator acts as a constant
velocity path following algorithm, which is different from what i understood of the smil spec. It actually returns Y based on calculations of the length of the curve instead of 'just' returning the Y
of projection of the factor on X axis onto the curve."
Both Mozilla and Batik come pretty close to the SMIL numbers.
Mozilla uses Newton-Raphson on the power series version of the Bezier equation, starting with a reasonable first guess. See section 6.4 of http://brian.sol1.net/svg/report/report.pdf for a more
detailed algorithm description.
Batik does not use Newton-Raphson nor does it start with a good first guess. It just iterates until it converges to an answer within a tolerance.
In the code below, I get the SMIL interpolated values (which are admittedly approximate) from the table after Figure 7 at http://www.w3.org/TR/SMIL/animation.html#animationNS-OverviewSpline
import java.text.NumberFormat;
import com.sun.scenario.animation.Interpolator;
import com.sun.scenario.animation.Interpolators;
public class SplineInterpolationTest
private static final NumberFormat nf = NumberFormat.getInstance();
static {
private final float x1, y1, x2, y2;
private final float[] smilData;
public static String scaleFrom10To20(float normalizedVal) {
return nf.format(10 + normalizedVal*10);
public SplineInterpolationTest(float x1, float y1, float x2, float y2,
float[] smilData) {
this.x1 = x1;
this.y1 = y1;
this.x2 = x2;
this.y2 = y2;
this.smilData = smilData;
public void runComparison() {
Interpolator splineInterpolator =
Interpolators.getSplineInstance(this.x1, this.y1, this.x2, this.y2);
System.out.println("\nControl points: (" + x1 + ", " + y1 + "), (" + x2 +
", " + y2 + ")");
float increment = 1f / (smilData.length - 1);
for (int i = 0; i < smilData.length; ++i) {
float x = i * increment;
System.out.println(x + "\t" + i + "\t" + nf.format(smilData[i]) + "\t" +
public static void main(String[] args) {
new SplineInterpolationTest(0f, 0f, 1f, 1f,
new float[]{10f, 12.5f, 15f, 17.5f, 20f}).runComparison();
new SplineInterpolationTest(0.5f, 0f, 0.5f, 1f,
new float[]{10f, 11f, 15f, 19f, 20f}).runComparison();
new SplineInterpolationTest(0f, 0.75f, 0.25f, 1f,
new float[]{10f, 18f, 19.3f, 19.8f, 20f}).runComparison();
new SplineInterpolationTest(1f, 0f, 0.25f, 0.25f,
new float[]{10f, 10.1f, 10.6f, 16.9f, 20f}).runComparison();
Re: Spline Interpolation
December 14, 2007 - 12:51
Hi David,
The current Scenario implementation of Interpolators.Spline uses
something called Path behind the scenes, which was written to provide
interpolation for arbitrary n-dimensional Bezier curves. (We also
use Path for spatial interpolation via MotionPath, but the
Interpolators.Spline case is just a trivial 2-dimensional use case.)
Anyway, the basic approach is similar to what was in TimingFramework,
in that we subdivide/flatten the curve and then iterate through the
segments to find the answer. This may not be the right approach for
compatibility with SMIL (nor is it the most optimal). Would you or
someone else like to look into this discrepancy and perhaps suggest a
different implementation? The approach used in the Mozilla paper
might be a good place to start, although I haven't looked at it in
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Re: Spline Interpolation
December 20, 2007 - 12:14
Hi Chris,
Ok, here is an implementation that I am pretty happy with.
Description: This is an implementation of a spline interpolator for
* spline animation that tries to follow the specification referenced by
* http://www.w3.org/TR/SMIL/animation.html#animationNS-OverviewSpline
* Basically, a cubic Bezier curve is created with start point (0,0) and
* endpoint (1,1). The other two control points (px1, py1) and (px2, py2) are
* given by the user, where px1, py1, px1, and px2 are all in the range [0,1].
* A property of this specially constrained Bezier curve is that it is strictly
* monotonically increasing in both X and Y with t in range [0,1].
* The interpolator works by giving it a value for X. It then finds what
* parameter t would generate this X value for the curve. Then this t parameter
* is applied to the curve to solve for Y. As X increases from 0 to 1, t also
* increases from 0 to 1, and correspondingly Y increases from 0 to 1. The
* X-to-Y mapping is not a function of path/curve length.
* @author David C. Browne
public class BezierInterpolator {
* the coordinates of the 2 2D control points for a cubic Bezier curve,
* with implicit start point (0,0) and end point (1,1) -- each individual
* coordinate value must be in range [0,1]
private final float x1, y1, x2, y2;
* do the input control points form a line with (0,0) and (1,1), i.e.,
* x1 == y1 and x2 == y2 -- if so, then all x(t) == y(t) for the curve
private final boolean isCurveLinear;
* power of 2 sample size for lookup table of x values
private static final int SAMPLE_SIZE = 16;
* difference in t used to calculate each of the xSamples values -- power of
* 2 sample size should provide exact representation of this value and its
* integer multiples (integer in range of [0..SAMPLE_SIZE]
private static final float SAMPLE_INCREMENT = 1f / SAMPLE_SIZE;
* x values for the bezier curve, sampled at increments of 1/SAMPLE_SIZE --
* this is used to find the good initial guess for parameter t, given an x
private final float[] xSamples = new float[SAMPLE_SIZE + 1];
* constructor -- cubic bezier curve will be represented by control points
* (0,0) (px1,py1) (px2,py2) (1,1) -- px1, py1, px2, py2 all in range [0,1]
* @param px1 is x-coordinate of first control point, in range [0,1]
* @param py1 is y-coordinate of first control point, in range [0,1]
* @param px2 is x-coordinate of second control point, in range [0,1]
* @param py2 is y-coordinate of second control point, in range [0,1]
public BezierInterpolator(float px1, float py1, float px2, float py2) {
// check user input for precondition
if (px1 < 0 || px1 > 1 || py1 < 0 || py1 > 1 ||
px2 < 0 || px2 > 1 || py2 < 0 || py2 > 1) {
throw new IllegalArgumentException("control point coordinates must " +
"all be in range [0,1]");
// save control point data
x1 = px1;
y1 = py1;
x2 = px2;
y2 = py2;
// calc linearity/identity curve
isCurveLinear = ((x1 == y1) && (x2 == y2));
// make the array of x value samples
if (!isCurveLinear) {
for (int i = 0; i < SAMPLE_SIZE + 1; ++i) {
xSamples[i] = eval(i * SAMPLE_INCREMENT, x1, x2);
} // BezierInterpolator()
* get the y-value of the cubic bezier curve that corresponds to the x input
* @param x is x-value of cubic bezier curve, in range [0,1]
* @return corresponding y-value of cubic bezier curve -- in range [0,1]
public float interpolate(float x) {
// check user input for precondition
if (x < 0 || x > 1) {
throw new IllegalArgumentException("x must be in range [0,1]");
// check quick exit identity cases (linear curve or curve endpoints)
if (isCurveLinear || x == 0 || x == 1) {
return x;
// find the t parameter for a given x value, and use this t to calculate
// the corresponding y value
return eval(findTForX(x), y1, y2);
} // interpolate()
* use Bernstein basis to evaluate 1D cubic Bezier curve (quicker and more
* numerically stable than power basis) -- 1D control coordinates are
* (0, p1, p2, 1), where p1 and p2 are in range [0,1], and there is no
* ordering constraint on p1 and p2, i.e., p1 <= p2 does not have to be true
* @param t is the paramaterized value in range [0,1]
* @param p1 is 1st control point coordinate in range [0,1]
* @param p2 is 2nd control point coordinate in range [0,1]
* @return the value of the Bezier curve at parameter t
private float eval(float t, float p1, float p2) {
// Use optimzied version of the normal Bernstein basis form of Bezier:
// (3*(1-t)*(1-t)*t*p1)+(3*(1-t)*t*t*p2)+(t*t*t), since p0=0, p3=1
// The above unoptimized version is best using -server, but since we are
// probably doing client-side animation, this is faster.
float compT = 1 - t;
return t * (3 * compT * (compT * p1 + t * p2) + (t * t));
} // eval()
* evaluate Bernstein basis derivative of 1D cubic Bezier curve, where 1D
* control points are (0, p1, p2, 1), where p1 and p2 are in range [0,1], and
* there is no ordering constraint on p1 and p2, i.e., p1 <= p2 does not have
* to be true
* @param t is the paramaterized value in range [0,1]
* @param p1 is 1st control point coordinate in range [0,1]
* @param p2 is 2nd control point coordinate in range [0,1]
* @return the value of the Bezier curve at parameter t
private float evalDerivative(float t, float p1, float p2) {
// use optimzed version of Berstein basis Bezier derivative:
// (3*(1-t)*(1-t)*p1)+(6*(1-t)*t*(p2-p1))+(3*t*t*(1-p2)), since p0=0, p3=1
// The above unoptimized version is best using -server, but since we are
// probably doing client-side animation, this is faster.
float compT = 1 - t;
return 3 * (compT * (compT * p1 + 2 * t * (p2 - p1)) + t * t * (1 - p2));
} // evalDerivative()
* find an initial good guess for what parameter t might produce the x-value
* on the Bezier curve -- uses linear interpolation on the x-value sample
* array that was created on construction
* @param x is x-value of cubic bezier curve, in range [0,1]
* @return a good initial guess for parameter t (in range [0,1]) that gives x
private float getInitialGuessForT(float x) {
// find which places in the array that x would be sandwiched between,
// and then linearly interpolate a reasonable value of t -- array values
// are ascending (or at least never descending) -- binary search is
// probably more trouble than it is worth here
for (int i = 1; i < SAMPLE_SIZE + 1; ++i) {
if (xSamples[i] >= x) {
float xRange = xSamples[i] - xSamples[i-1];
if (xRange == 0) {
// no change in value between samples, so use earlier time
return (i - 1) * SAMPLE_INCREMENT;
} else {
// linearly interpolate the time value
return ((i - 1) + ((x - xSamples[i-1]) / xRange)) *
// shouldn't get here since 0 <= x <= 1, and xSamples[0] == 0 and
// xSamples[SAMPLE_SIZE] == 1 (using power of 2 SAMPLE_SIZE for more
// exact increment arithmetic)
return 1;
} // getInitialGuessForT()
* find the parameter t that produces the given x-value for the curve --
* uses Newton-Raphson to refine the value as opposed to subdividing until
* we are within some tolerance
* @param x is x-value of cubic bezier curve, in range [0,1]
* @return the parameter t (in range [0,1]) that produces x
private float findTForX(float x) {
// get an initial good guess for t
float t = getInitialGuessForT(x);
// use Newton-Raphson to refine the value for t -- for this constrained
// Bezier with float accuracy (7 digits), any value not converged by 4
// iterations is cycling between values, which can minutely affect the
// accuracy of the last digit
final int numIterations = 4;
for (int i = 0; i < numIterations; ++i) {
// stop if this value of t gives us exactly x
float xT = (eval(t, x1, x2) - x);
if (xT == 0) {
// stop if derivative is 0
float dXdT = evalDerivative(t, x1, x2);
if (dXdT == 0) {
// refine t
t -= xT/ dXdT;
return t;
} // findTForX()
} // class BezierInterpolator
Re: Spline Interpolation
December 20, 2007 - 12:35
Hi David,
Thanks for digging into this. Unfortunately I didn't see your name on the SCA signatories list:
and we can't look at your code until you sign the SCA:
Once you sign, we'd be happy to discuss it more.
Sorry for the hassle,
Re: Spline Interpolation
December 20, 2007 - 13:16
I just emailed my signed scanned copy of the agreement to the sca email address.
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|
Boltzmanns entropy equation
In my textbook they claim that:
They lost me when they said "thermodynamic state". Do they mean thermodynamic system? I know that a system such as a container of gas will have a very large number of microstates considering the
amount of gas molecules present and that statistical methods would be required to calculate the number of microstates. I know that a system like this also has macrostates such as pressure, volume,
temperature etc. but I don't know what they mean when they say a thermodynamic "state" has a number of microstates associated with it.
The thermodynamic state refers to a set of macroscopically measured parameters of a substance in thermal equilibrium. Any other system in thermal equilibrium having those same parameters will be
thermodynamically "the same".
For an ideal gas, the thermodynamic state is defined by Pressure, Volume and Temperature. Any other ideal gas at the same Pressure, Volume and Temperature will be thermodynamically equivalent.
For a paramagnetic system, magnetic intensity and magnetization as well as pressure, volume and temperature define the thermodynamic state.
A thermodynamic system may consist of different components that are in thermal equilibrium with themselves but are not in thermal equilibrium with each other. So a thermodynamic system may not have a
single thermodynamic state.
What macroscopic states are they talking about here? Are they saying macrostates like pressure or volume have microstates associated with them?
Even a small quantity of an ideal gas (in thermal equilbrium) consists of a huge number of molecules all moving in different directions and speeds. The microstate describes the motions of all the
molecules. Two samples of an ideal gas having the same parameters of P, V and T will not have the same microstates (ie their molecules are not all moving identically at any given time). Indeed, the
microstates of each sample are continually changing. However, since their thermodynamic parameters remain unchanged, for thermodynamic purposes, they are the same.
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[SciPy-User] Confidence interval for bounded minimization
josef.pktd@gmai... josef.pktd@gmai...
Thu Feb 23 00:23:44 CST 2012
On Thu, Feb 23, 2012 at 1:10 AM, <josef.pktd@gmail.com> wrote:
> On Thu, Feb 23, 2012 at 12:09 AM, Christopher Jordan-Squire
> <cjordan1@uw.edu> wrote:
>> On Wed, Feb 22, 2012 at 2:02 PM, Nathaniel Smith <njs@pobox.com> wrote:
>>> On Wed, Feb 22, 2012 at 8:48 PM, <josef.pktd@gmail.com> wrote:
>>>> On Wed, Feb 22, 2012 at 3:26 PM, Greg Friedland
>>>> <greg.friedland@gmail.com> wrote:
>>>>> Hi,
>>>>> Is it possible to calculate asymptotic confidence intervals for any of
>>>>> the bounded minimization algorithms? As far as I can tell they don't
>>>>> return the Hessian; that's including the new 'minimize' function which
>>>>> seemed like it might.
>>>> If the parameter ends up at the bounds, then the standard statistics
>>>> doesn't apply. The Hessian is based on a local quadratic
>>>> approximation, which doesn't work if part of the local neigborhood is
>>>> out of bounds.
>>>> There is some special statistics for this, but so far I have seen only
>>>> the description how GAUSS handles it.
>>>> In statsmodels we use in some cases the bounds, or a transformation,
>>>> just to keep the optimizer in the required range, and we assume we get
>>>> an interior solution. In this case, it is possible to use the standard
>>>> calculations, the easiest is to use the local minimum that the
>>>> constraint or transformed optimizer found and use it as starting value
>>>> for an unconstrained optimization where we can get the Hessian (or
>>>> just calculate the Hessian based on the original objective function).
>>> Some optimizers compute the Hessian internally. In those cases, it
>>> would be nice to have a way to ask them to somehow return that value
>>> instead of throwing it away. I haven't used Matlab in a while, but I
>>> remember running into this as a standard feature at some point, and it
>>> was quite nice. Especially when working with a problem where each
>>> computation of the Hessian requires an hour or so of computing time.
>> Are you talking about analytic or finite-difference gradients and
>> hessians? I'd assumed that anything derived from finite difference
>> estimations wouldn't give particularly good confidence intervals, but
>> I've never needed them so I've never looked into it in detail.
> statsmodels has both, all discrete models for example have analytical
> gradients and hessians.
> But for models with a complicated log-likelihood function, there isn't
> much choice, second derivatives with centered finite differences are
> ok, scipy.optimize.leastsq is not very good. statsmodels also has
> complex derivatives which are numerically pretty good but they cannot
> always be used.
> I think in most cases numerical derivatives will have a precision of a
> few decimals, which is more precise than all the other statistical
> assumptions, normality, law of large numbers, local definition of
> covariance matrix to calculate "large" confidence intervals, and so
> on.
> One problem is that choosing the step size depends on the data and
> model. numdifftools has adaptive calculations for the derivatives, but
> we are not using it anymore.
> Also, if the model is not well specified, then the lower precision of
> finite difference derivatives can hurt. For example, in ARMA models I
> had problems when there are too many lags specified, so that some
> roots should almost cancel. Skipper's implementation works better
> because he used a reparameterization that forces some nicer behavior.
> The only case in the econometrics literature that I know is that early
> GARCH models were criticized for using numerical derivatives even
> though analytical derivatives were available, some parameters were not
> well estimated, although different estimates produced essentially the
> same predictions (parameters are barely identified)
> Last defense: everyone else does it, maybe a few models more or less,
> and if the same statistical method is used, then the results usually
> agree pretty well.
> (But if different methods are used, for example initial conditions are
> treated differently in time series analysis, then the differences are
> usually much larger. Something like: I don't worry about numerical
> problems at the 5th or 6th decimal if I cannot figure out what these
> guys are doing with their first and second decimal.)
> (maybe more than anyone wants to know.)
In case it wasn't clear: analytical derivatives are of course much
better, and I would be glad if the scipy.stats.distributions or sympy
had the formulas for the derivatives of the log-likelihood functions
for the main distributions. (but it's work)
> Josef
> .
>> -Chris
>>> -- Nathaniel
>>> _______________________________________________
>>> SciPy-User mailing list
>>> SciPy-User@scipy.org
>>> http://mail.scipy.org/mailman/listinfo/scipy-user
>> _______________________________________________
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A Body Oscillates A Body Oscillates With Simple ... | Chegg.com
A Body Oscillates A body oscillates with simple harmonic motion according to the equation
x = (3.0 m) cos[(5p rad/s)t + p/6 rad].
(a) At t = 10.0 s, what is the displacement?
1 m
(b) At t = 10.0 s, what is the velocity?
2 m/s
(c) At t = 10.0 s, what is the acceleration?
3 m/s2
(d) At t = 10.0 s, what is the phase of the motion?
4 rad
(e) At t = 10.0 s, what is the frequency of the motion?
5 Hz
(f) At t = 10.0 s, what is the period of the motion?
6 s
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Math Forum Discussions - RE: Probability of a Triangle
Date: Sep 15, 2004 11:44 AM
Author: Pat ballew
Subject: RE: Probability of a Triangle
And just as a note, Is the third solution shown here (the one for question B
below) correct?
Pat Ballew
Lakenheath, UK
MathWords http://www.pballew.net/etyindex.html
-----Original Message-----
From: owner-geometry-puzzles@mathforum.org
[mailto://owner-geometry-puzzles@mathforum.org] On Behalf Of Alexander
Sent: Tuesday, September 14, 2004 2:59 AM
To: geometry-puzzles@mathforum.org
Subject: Re: Probability of a Triangle
On May 29 15:45:21 1996, Pat Ballew wrote:
> a) If a unit length segment is randomly broken at two points
> its length, what is the probability that the three pieces
created in
> this fashion will form a triangle?
> b) If the length is broken at a random point, and then one of
the two
> pieces is randomly selected and broken at a random point on its
> what is the probability that the three pices will form a
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Mathematics Magazine - June 2008
The Mathematics of Helaman Ferguson's Four Canoes
Melissa Shepard Loe and Jenny Merrick Borovsky
In 1997 the University of St. Thomas installed a massive granite sculpture by Helaman Ferguson outside of the new Science and Engineering Center. The sculpture consists of two linked six-foot
“donuts,” each weighing more than three tons. Two pedestals support the donuts, and thirty jagged hexagons tile the ground beneath the sculpture. Why did Ferguson entitle this sculpture Four Canoes
? How are the tiles and sculpture related? What rule govern placement of the tiles? And what keeps the donuts from rocking off their pedestals? In this paper, we provide an explanation of the unusual
name and explore the mathematical significance of the sculpture. We discuss the link between the sculpture and the tiles, the rules governing placement of the tiles, and the question of periodicity
of the tessellation. Lastly, we examine the mathematics in the construction and “fit” of the sculpture, pedestals, and tiling.
A Primer on Bernoulli Numbers and Polynomials
Tom M. Apostol
Although a large literature exists on Bernoulli numbers and Bernoulli polynomials, much of it is in widely scattered books and journals. This article serves as a brief primer on the subject, bringing
together basic results (most of which are well known), together with proofs, in a manner readily accessible to those with a knowledge of elementary calculus. Some new formulas are also derived
Somewhat More Than Governors Need to Know About Trigonometry
Skip Garibaldi
In school, some of us had to memorize the values of sine and cosine at the angles 0, 30, 45, 60, and 90 degrees. Why those angles and not others? If you had to make your own table of trig values from
scratch, what angles might you include? This paper shows how these questions are answered by Galois theory from standard textbooks. And it explains what this all has to do with former Florida
Governor Jeb Bush.
Pi to thousands of digits from Vieta's formula
Rick Kreminski
Vieta's 16th century infinite product formula for pi, expressed in terms of nested radicals of 2, converges surprisingly quickly: the partial product of n terms gives pi to almost exactly 0.6n digits
of accuracy. From our formulation of the error, successive cancellation of smaller and smaller error terms using linear combinations of partial products computes pi to well over 0.6n(k+1) digits of
accuracy, where k is the number of iterations of the algorithm. For n and k about 600, we computed pi to over 300,000 digits.
Sum Kind of Asymptotic Trouble
George W. Benthien, Keith J. Coates
When analyzing the asymptotic behavior of the sum of two or more functions, it can be easy to hoodwink oneself into making an incorrect approximation of the sum. The authors of this note found, in a
well-regarded analysis book, a seemingly straightforward approximation of the sum of two terms that contains an elementary but rather subtle error. The note explains the error, how it was likely to
have come about, and how to correct it.
Why are the Gergonne and Soddy Lines Perpendicular? A Synthetic Approach
Zuming Feng
In any scalene triangle the three points of tangency of the incircle together with the three vertices can be used to define three new points which are, remarkably, always collinear. This line is
called the Gergonne Line. Moreover cevians through these tangent points are always concurrent at a common point that, together with the incenter, defines a second line, the Soddy Line. Why should
these lines be perpendicular? Such a geometric gem deserves a synthetic geometric proof. We use the classical theorems of Ceva and Menelaus to define these lines and then establish their
perpendicularity by using a certain inversion.
Golden Matrix Ring Mod p
Kung–Wei Yang
We show how to use the golden matrix ring Z(A) generated by
Geometric Proofs of the Weitzenböck and Hadwiger-Finsler Inequalities
Claudi Alsini and Roger B. Nelsen
Let a, b, and c denote the sides and T the area of a triangle. Weitzenböck’s inequality states that F that minimizes the sum AF + BF + CF ) to present simple geometric proofs of both inequalities.
Proof Without Words
Area of a Parabolic Segment
Carl R. Seaquist
The area under the segment of a parabola is found via the volume of a pyramid.
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Text Processing (Part 1): Entity Recognition
Entity recognition is commonly used to parse unstructured text document and extract useful entity information (like location, person, brand) to construct a more useful structured representation. It
is one of the most common text processing to understand a text document.
I am planning to write a blog series on text processing. In this first blog of a series of basic text processing algorithm, I will introduce some basic algorithm for entity recognition.
Given the sentence: Ricky is living in CA USA.
Output: Ricky/SP is/NA living/NA in/NA CA/SL USA/CL
Basically we want to tag each word with the entity, whose definition is domain specific. In this example, we define the following tags
• NA - Not Applicable
• SP - Start of a person name
• CP - Continue of a person name
• SL - Start of a location name
• CL - Continue of a location name
Hidden Markov Model
Lets say there is a sequence of state, lets look at multiple probabilistic graph.
However, in our tagging example, we don't directly observe the tag. Instead, we only observe the words. In this case, we can use a hidden markov model (ie: HMM).
Now the tagging problem can be structured as follows.
Given a sequence of words, we want to predict the most likely tag sequence.
Find a tag sequence t1, t2, ... tn that maximize the probability of P(t1, t2, .... | w1, w2 ...)
Using Bayes rules,
P(t1, t2, .... | w1, w2 ...) = P(t1, t2, ... tn, w1, w2, ... wn) / P(w1, w2, ... wn)
Since the sequence w1, w2 ... wn is observed and constant among all tag sequence. This is equivalent to maximize P(t1, t2, ... tn, w1, w2, ... wn) which is equal to P(t1|S)*P(t2|t1)…P(E|tn)*P(w1|t1)
Now, P(t1|S), P(t2|t1) ... can be estimated by counting the occurrence within the training data.
P(t2|t1) = count(t1, t2) / count(*, t2)
Similarly, P(w1|t1) = count(w1, t1) / count(*, t1)
Viterbi Algorithm
Now the problem is find a tag sequence t1, ... tn to maximize
A naive method is to find all possible combination of tag sequence and then evaluate the above probability. The order of complexity will be O(|T|^n) where T is the number of possible tag values.
Notice that this is exponential complexity with respect to the length of the sentence.
However, there is a more effective Viterbi algorithm that leverage the Markov chain properties as well as dynamic programming technique.
The key element is M(k, L) which indicates the max probability of any length k sequence that ends at tk = L. On the other hand, M(k, L) is computed by looking back different choices of S of the
length k-1 sequence, and pick the one that gives the maximum M(k-1, S)*P(tk=L | tk-1=S)*P(wk|tk=L). The complexity of this algorithm is O(n*|T|^2).
To find the actual tag sequence, we also maintain a back pointer from every cell to S which leads to the cell. Then we can trace back the path from the max cell M(n, STOP) where STOP is the end of
Notice that for some rare words that is not observed from the training data, P(wk|tk=L) will be zero and cause the whole term to be zero. Such words can be numbers, dates. One way to solve this
problem is to group these rare words into patterns (e.g. 3 digits, year2012 ... etc) and then compute P(group(wk) | tk=L). However, such grouping is domain specific and has to be hand-tuned.
NLP course from Michael Collins of Columbia Unversity
Published at DZone with permission of Ricky Ho, author and DZone MVB. (source)
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Iterative Flattening Search for the Flexible Job Shop Scheduling Problem
Angelo Oddi, Riccardo Rasconi, Amedeo Cesta, Stephen F. Smith
This paper presents a meta-heuristic algorithm for solving the Flexible Job Shop Scheduling Problem (FJSSP). This strategy, known as Iterative Flattening Search (IFS), iteratively applies a
relaxation-step, in which a subset of scheduling decisions are randomly retracted from the current solution; and a solving-step, in which a new solution is incrementally recomputed from this partial
schedule. This work contributes two separate results: (1) it proposes a constraint-based procedure extending an existing approach previously used for classical Job Shop Scheduling Problem; (2) it
proposes an original relaxation strategy on feasible FJSSP solutions based on the idea of randomly breaking the execution orders of the activities on the machines and opening the resource options for
some activities selected at random. The efficacy of the overall heuristic optimization algorithm is demonstrated on a set of well-known benchmarks.
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Without use of any calculator
Never mind, forget what I wrote above. I was being naughty and used a calculator.
But I got it now. Here’s how.
By considering the sum of the complex roots of
, you have
But .
Note that . Hence
And . And there you have it.
Last edited by JaneFairfax (2008-07-25 23:27:41)
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fits significant sine waves to data (continuous data) using overlapping windows.
function [datac,datafit,Amps,freqs]=rmlinesmovingwinc(data,movingwin,tau,params,p,plt,f0)
fits significant sine waves to data (continuous data) using overlapping windows.
Usage: [datac,datafit]=rmlinesmovingwinc(data,movingwin,tau,params,p,plt)
Note that units of Fs, fpass have to be consistent.
data (data in [N,C] i.e. time x channels/trials or as a single vector) - required.
movingwin (in the form [window winstep] i.e length of moving
window and step size)
Note that units here have
to be consistent with
units of Fs - required
tau parameter controlling degree of smoothing for the amplitudes - we use the
function 1-1/(1+exp(-tau*(x-Noverlap/2)/Noverlap) in the region of overlap to smooth
the sinewaves across the overlap region. Noverlap is the number of points
in the overlap region. Increasing tau leads to greater overlap smoothing,
typically specifying tau~10 or higher is reasonable. tau=1 gives an almost
linear smoothing function. tau=100 gives a very steep sigmoidal. The default is tau=10.
params structure containing parameters - params has the
following fields: tapers, Fs, fpass, pad
tapers : precalculated tapers from dpss or in the one of the following
(1) A numeric vector [TW K] where TW is the
time-bandwidth product and K is the number of
tapers to be used (less than or equal to
(2) A numeric vector [W T p] where W is the
bandwidth, T is the duration of the data and p
is an integer such that 2TW-p tapers are used. In
this form there is no default i.e. to specify
the bandwidth, you have to specify T and p as
well. Note that the units of W and T have to be
consistent: if W is in Hz, T must be in seconds
and vice versa. Note that these units must also
be consistent with the units of params.Fs: W can
be in Hz if and only if params.Fs is in Hz.
The default is to use form 1 with TW=3 and K=5
Note that T has to be equal to movingwin(1).
Fs (sampling frequency) -- optional. Defaults to 1.
fpass (frequency band to be used in the calculation in the form
[fmin fmax])- optional.
Default all frequencies between 0 and Fs/2
pad (padding factor for the FFT) - optional (can take values -1,0,1,2...).
-1 corresponds to no padding, 0 corresponds to padding
to the next highest power of 2 etc.
e.g. For N = 500, if PAD = -1, we do not pad; if PAD = 0, we pad the FFT
to 512 points, if pad=1, we pad to 1024 points etc.
Defaults to 0.
p (P-value to calculate error bars for) - optional.
Defaults to 0.05/Nwin where Nwin is length of window which
corresponds to a false detect probability of approximately 0.05.
plt (y/n for plot and no plot respectively) - default no
f0 frequencies at which you want to remove the
lines - if unspecified the program uses the f statistic
to determine appropriate lines.
datafit (fitted sine waves)
datac (cleaned up data)
This function calls:
• fitlinesc fits significant sine waves to data (continuous data).
• mtspectrumsegc Multi-taper segmented spectrum for a univariate continuous process
This function is called by:
SOURCE CODE
0001 function [datac,datafit,Amps,freqs]=rmlinesmovingwinc(data,movingwin,tau,params,p,plt,f0)
0002 % fits significant sine waves to data (continuous data) using overlapping windows.
0003 %
0004 % Usage: [datac,datafit]=rmlinesmovingwinc(data,movingwin,tau,params,p,plt)
0005 %
0006 % Inputs:
0007 % Note that units of Fs, fpass have to be consistent.
0008 % data (data in [N,C] i.e. time x channels/trials or as a single vector) - required.
0009 % movingwin (in the form [window winstep] i.e length of moving
0010 % window and step size)
0011 % Note that units here have
0012 % to be consistent with
0013 % units of Fs - required
0014 % tau parameter controlling degree of smoothing for the amplitudes - we use the
0015 % function 1-1/(1+exp(-tau*(x-Noverlap/2)/Noverlap) in the region of overlap to smooth
0016 % the sinewaves across the overlap region. Noverlap is the number of points
0017 % in the overlap region. Increasing tau leads to greater overlap smoothing,
0018 % typically specifying tau~10 or higher is reasonable. tau=1 gives an almost
0019 % linear smoothing function. tau=100 gives a very steep sigmoidal. The default is tau=10.
0020 % params structure containing parameters - params has the
0021 % following fields: tapers, Fs, fpass, pad
0022 % tapers : precalculated tapers from dpss or in the one of the following
0023 % forms:
0024 % (1) A numeric vector [TW K] where TW is the
0025 % time-bandwidth product and K is the number of
0026 % tapers to be used (less than or equal to
0027 % 2TW-1).
0028 % (2) A numeric vector [W T p] where W is the
0029 % bandwidth, T is the duration of the data and p
0030 % is an integer such that 2TW-p tapers are used. In
0031 % this form there is no default i.e. to specify
0032 % the bandwidth, you have to specify T and p as
0033 % well. Note that the units of W and T have to be
0034 % consistent: if W is in Hz, T must be in seconds
0035 % and vice versa. Note that these units must also
0036 % be consistent with the units of params.Fs: W can
0037 % be in Hz if and only if params.Fs is in Hz.
0038 % The default is to use form 1 with TW=3 and K=5
0039 % Note that T has to be equal to movingwin(1).
0040 %
0041 % Fs (sampling frequency) -- optional. Defaults to 1.
0042 % fpass (frequency band to be used in the calculation in the form
0043 % [fmin fmax])- optional.
0044 % Default all frequencies between 0 and Fs/2
0045 % pad (padding factor for the FFT) - optional (can take values -1,0,1,2...).
0046 % -1 corresponds to no padding, 0 corresponds to padding
0047 % to the next highest power of 2 etc.
0048 % e.g. For N = 500, if PAD = -1, we do not pad; if PAD = 0, we pad the FFT
0049 % to 512 points, if pad=1, we pad to 1024 points etc.
0050 % Defaults to 0.
0051 % p (P-value to calculate error bars for) - optional.
0052 % Defaults to 0.05/Nwin where Nwin is length of window which
0053 % corresponds to a false detect probability of approximately 0.05.
0054 % plt (y/n for plot and no plot respectively) - default no
0055 % plot.
0056 % f0 frequencies at which you want to remove the
0057 % lines - if unspecified the program uses the f statistic
0058 % to determine appropriate lines.
0059 %
0060 % Outputs:
0061 % datafit (fitted sine waves)
0062 % datac (cleaned up data)
0063 if nargin < 2; error('Need data and window parameters'); end;
0064 if nargin < 4 || isempty(params); params=[]; end;
0066 if length(params.tapers)==3 & movingwin(1)~=params.tapers(2);
0067 error('Duration of data in params.tapers is inconsistent with movingwin(1), modify params.tapers(2) to proceed')
0068 end
0070 [tapers,pad,Fs,fpass,err,trialave,params]=getparams(params); % set defaults for params
0071 clear err trialave
0072 if nargin < 6; plt='n'; end;
0073 %
0074 % Window,overlap and frequency information
0075 %
0076 data=change_row_to_column(data);
0077 [N,C]=size(data);
0078 Nwin=round(Fs*movingwin(1)); % number of samples in window
0079 Nstep=round(movingwin(2)*Fs); % number of samples to step through
0080 Noverlap=Nwin-Nstep; % number of points in overlap
0081 %
0082 % Sigmoidal smoothing function
0083 %
0084 if nargin < 3 || isempty(tau); tau=10; end; % smoothing parameter for sigmoidal overlap function
0085 x=(1:Noverlap)';
0086 smooth=1./(1+exp(-tau.*(x-Noverlap/2)/Noverlap)); % sigmoidal function
0087 smooth=repmat(smooth,[1 C]);
0088 %
0089 % Start the loop
0090 %
0091 if nargin < 5 || isempty(p); p=0.05/Nwin; end % default for p value
0092 if nargin < 7 || isempty(f0); f0=[]; end; % empty set default for f0 - uses F statistics to determine the frequencies
0093 params.tapers=dpsschk(tapers,Nwin,Fs); % check tapers
0094 winstart=1:Nstep:N-Nwin+1;
0095 nw=length(winstart);
0096 datafit=zeros(winstart(nw)+Nwin-1,C);
0097 Amps=cell(1,nw);
0098 freqs=cell(1,nw);
0099 for n=1:nw;
0100 indx=winstart(n):winstart(n)+Nwin-1;
0101 datawin=data(indx,:);
0102 [datafitwin,as,fs]=fitlinesc(datawin,params,p,'n',f0);
0103 Amps{n}=as;
0104 freqs{n}=fs;
0105 datafitwin0=datafitwin;
0106 if n>1; datafitwin(1:Noverlap,:)=smooth.*datafitwin(1:Noverlap,:)+(1-smooth).*datafitwin0(Nwin-Noverlap+1:Nwin,:);end;
0107 datafit(indx,:)=datafitwin;
0108 end;
0109 datac=data(1:size(datafit,1),:)-datafit;
0110 if strcmp(plt,'y');
0111 [S,f]=mtspectrumsegc(data,movingwin(1),params);
0112 [Sc,fc]=mtspectrumsegc(datac,movingwin(1),params);
0113 plot(f,10*log10(S),fc,10*log10(Sc));
0114 end;
Generated on Fri 28-Sep-2012 12:34:30 by m2html © 2005
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Euler's totient function
December 8th 2008, 05:54 PM #1
Junior Member
Oct 2008
Euler's totient function
Hi all again,
Now the problem is with Euler's Function. See:
1. Find all positive and integer $n$ such that $\varphi(n) = pq$, where p and q are prime.
2. Find all positive numbers $n$ such that $\varphi(n) = 8$
How do I approach these? Any tips are welcome.
Thanks in advance,
Write $n=p_1^{a_1}...p_k^{a_k}$ and now apply formula for $\phi (n)$.
After that compare sides and see what $n$ got to be.
Also for (1) it seems that $p,q$ are odd primes because $\phi(n) = 14$ has no solutions.
December 8th 2008, 08:53 PM #2
Global Moderator
Nov 2005
New York City
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Solution of equation
June 19th 2006, 07:31 AM #1
I was just wondering how could one find the solution of equations involving trigonometric and polynomial expressions.
For example, how could we solve $x - cosx=0$?
Keep Smiling
I was just wondering how could one find the solution of equations involving trigonometric and polynomial expressions.
For example, how could we solve $x - cosx=0$?
Keep Smiling
I can proof that a unique solution exists.
You can approximate solutios with "Newton-Raphson Alogorthm".
I would not be supprised that the solution is transcendental
I was just wondering how could one find the solution of equations involving trigonometric and polynomial expressions.
For example, how could we solve $x - cosx=0$?
Keep Smiling
Most solutions to mixed algebraic/tanscendetal equations are do
not have simple closed forms in terms of the usual elementary functions
(no don't ask me to prove it). So do however have such solutions, see
this thread
June 19th 2006, 07:44 AM #2
Global Moderator
Nov 2005
New York City
June 19th 2006, 07:52 AM #3
Grand Panjandrum
Nov 2005
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Derivative qus.
April 29th 2006, 09:22 AM #1
Derivative qus.
Hey guys,
Finding hard to solve these derivative:
1) $q = 2e^{-t/2}cos2t$ (don't know how to deal with the index -t/2)
2) $x = e^{2t}t^3(2-t)^4$ (this seem confusing as don't know what rule to apply)
3) $e^{x+y} + sin3x = 0$ (not to sure about index agian)
I hope someone can help. Thanks
1) $q = 2e^{-t/2}\cos(2t)$ (don't know how to deal with the index -t/2)
Start with the product rule:
$<br /> \frac{dq}{dt}=2 \left[\left(\frac{d}{dt}e^{-t/2}\right)\cos(2t)+e^{-t/2}\left(\frac{d}{dt} \cos(2t)\right)\right]<br />$
so using the chain rule to do the derivatives:
$<br /> \frac{dq}{dt}=2 \left[-\frac{1}{2}e^{-t/2}\cos(2t)-2e^{-t/2}\sin(2t)\right]<br />$
$<br /> \frac{dq}{dt}=-e^{-t/2}[\cos(2t)+4\sin(2t)]<br />$
Hey guys,
Finding hard to solve these derivative:
1) $q = 2e^{-t/2}cos2t$ (don't know how to deal with the index -t/2)
2) $x = e^{2t}t^3(2-t)^4$ (this seem confusing as don't know what rule to apply)
3) $e^{x+y} + sin3x = 0$ (not to sure about index agian)
I hope someone can help. Thanks
What you seem to be having a problem with in 1) and 3) is:
$<br /> \frac{d}{dx}e^{f(x)}<br />$
By the chain rule this is:
$<br /> \frac{d}{dx}e^{f(x)}=f'(x)\ \frac{d}{dx}e^{y}\left\vert_{f(x)}\frac{}{}=f'(x)\ e^{f(x)}<br />$
Thanks for the quick response and the general rule!
So for 2) is it product and chain together?
Thanks for the quick response and the general rule!
So for 2) is it product and chain together?
Yes, though you will have the product:
$<br /> x = \{e^{2t}\}\ \{t^3(2-t)^4\}<br />$
so you may need to use the product rule again for the derivative of the
second term on the RHS.
3) $e^{x+y} + sin3x = 0$ (not to sure about index agian)
This is an example of implicit differentiation (and the chain rule):
$<br /> \frac{d}{dx} e^{x+y}+\frac{d}{dx}\sin(3x)=0<br />$,
$<br /> e^{x+y}\ \frac{dy}{dx} + 3\cos(3x)=0\ \ \ \dots(1)<br />$
but from the original equation we know that:
$<br /> e^{x+y}=-\sin(3x)<br />$,
so substituting this into equation $(1)$ gives:
$<br /> -\sin(3x)\frac{dy}{dx}+3\cos(3x)=0<br />$
or rearranging:
$<br /> \frac{dy}{dx}=3\frac{\cos(3x)}{\sin(3x)}=3\cot(3x)<br />$
Hi Thanks again!
Both methods should give the same answer?
The implicit way is a much neater, when applying the product and chain rule it gets abit messy that's what I have been trying!
April 29th 2006, 09:37 AM #2
Grand Panjandrum
Nov 2005
April 29th 2006, 09:48 AM #3
Grand Panjandrum
Nov 2005
April 30th 2006, 12:57 AM #4
April 30th 2006, 01:28 AM #5
Grand Panjandrum
Nov 2005
April 30th 2006, 01:38 AM #6
Grand Panjandrum
Nov 2005
April 30th 2006, 08:08 AM #7
April 30th 2006, 08:23 AM #8
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Number theory textbook with an algebraic perspective
up vote 8 down vote favorite
Most of the number theory textbooks I've dealt with take a very classical approach to the subject. I'm looking for a textbook that's something like a first course in number theory for people who have
a decent command of modern algebra (at the level of something like Lang's Algebra). Does such a book exist, and if it does, what is it called?
Edit: As I posted in a comment below:
In the introduction to Ireland and Rosen, they note something that was bugging me for a while, "Nevertheless it is remarkable how a modicum of group and ring theory introduces unexpected order into
the subject."
This is precisely the perspective I was looking for, so if anyone passes by this topic looking for a book that approaches number theory in this way, I feel like this quote should point him (her?) in
the right direction.
soft-question nt.number-theory books
fpqc--If you'd like to come by my office and take a look at some of these books, I have many of them on my shelf, feel free to send me an e-mail and stop by. – Ben Weiss Dec 7 '09 at 20:42
add comment
4 Answers
active oldest votes
There are probably many such books, for instance "Fundamentals of Number Theory" by LeVeque, "Elementary Number Theory" by Bolker and "A Classical Introduction to Modern
up vote 8 down vote Number Theory" by Ireland and Rosen.
Do you have a recommendation of one in particular? – Harry Gindi Dec 7 '09 at 10:39
I like all three; LeVeque is a good and cheap, though I'm not sure what level you aim at. – lhf Dec 7 '09 at 10:43
I'm looking through Ireland and Rosen right now, and it seems to be what I was looking for. – Harry Gindi Dec 7 '09 at 10:51
4 I like Ireland-Rosen book very much too. – mathreader Dec 7 '09 at 11:21
Seconded. I always wish to have read Ireland-Rosen much earlier. – Ho Chung Siu Dec 7 '09 at 13:40
show 1 more comment
Well, it depends on the actual subject you want to approach and the "decent command of modern algebra" already assumed; without knowing more, I would recommend:
• Number fields (1995) by Marcus. Universitext. Just as the title says, a (great!) introduction to number fields.
up vote 3 • Algebraic Number Theory (1986) by Cassels and Frölich. Academic Press. It explains the basics (class field theory, zeta functions) to understand the Langlands Program.
down vote
• A course on arithmetic (1996) by Serre. Graduate Texts in Mathematics. P-adic fields, quadratic forms, zeta functions and modular forms.
I was looking not for a book on algebraic number theory, but an algebraic book on number theory, if you understand the distinction I'm trying to make. – Harry Gindi Dec 7 '09 at
Not really. Can you elaborate, please? Maybe I can suggest some other kind of books! – Jose Brox Dec 7 '09 at 11:33
1 In the introduction to Ireland and Rosen, they note something that was bugging me for a while, "Nevertheless it is remarkable how a modicum of group and ring theory introduces
unexpected order into the subject." This is precisely the perspective I was looking for. – Harry Gindi Dec 7 '09 at 11:50
Serre is a very specific book; I wouldn't call it a first course at all. – Qiaochu Yuan Dec 7 '09 at 14:30
2 I wish Marcus's "Number fields" had a new edition, typeset, not typewritten. Otherwise, it's a great book with lots of hands-on material. – lhf Dec 7 '09 at 15:19
show 1 more comment
If I understand it well now, what you want are books about basic number theory with a good algebraic foundation. I can recommend the following:
• Elementary methods in number theory (2000), by Nathanson. Graduate Texts in Mathematics. It starts low, but it reaches quite high.
• Elementary Number Theory with Applications (2007) by Koshy. Elsevier. Truly basic. Not very very algebraic, but a really nice textbook.
up vote 3 down
vote • Algebra and number theory by Andrew Baker. Online notes. Fairly basic.
• A computational introduction to number theory and algebra (2005) by Shoup. Cambridge University Press / Online Free Version. Despite the title, I think it satisfies the
conditions you are looking for.
Numbers one and three look interesting, while numbers two and four look awful. – Harry Gindi Dec 7 '09 at 12:36
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up vote 3 down vote
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FOM: Effective Bounds in Core Mathematics
Fred Richman richman at fau.edu
Wed Jun 28 11:00:30 EDT 2000
Harvey Friedman wrote:
> Under my suggestion, a constructivist would be defined as "someone who
> studies constructivity in mathematics, or constructive proofs". This is
> analogous to a proof theorist as "someone who studies formal proofs" or a
> topologist as "someone who studies topological spaces".
Is this a simple terminological dispute? I wasn't thinking of that at
all. I would be inclined to say that a constructivist is a
mathematician who systematically proves theorems without appeal to the
law of excluded middle. I might note that although I have never heard
of "proof theorism" or "topologism", people do seem to talk about
"constructivism", referring to a view of mathematics.
> I still don't see why it is attractive, or even reasonable, to take the
> point of view that I put in between *'s above. As I said before, it is
> counterproductive and needlessly provocative. Furthermore, it almost
> guarantees that most logicians and mathematicians will not study
> constructivity.
I don't see why the term "proof" must of necessity be reserved for
classical proof. In a thoroughgoing constructive development of
mathematics, the proofs would all be constructive. Theorems whose
proofs require the law of excluded middle would be so flagged in much
the same way as we sometimes flag theorems that use the axiom of
choice, the continuum hypothesis, or the Riemann hypothesis. The
constructivist program is to develop mathematics in exactly that way.
In such a context, the starred point of view is practically mandatory.
As for the last remark, constructivists are not particularly
interested in studying constructivity. They are interested in doing
mathematics in a constructive manner.
> Taking a counterproductive and needlessly provocative stance like * needs a
> justification such as this. Otherwise, it is just counterproductive and
> needlessly provocative.
May I point out that using the phrase "counterproductive and
needlessly provocative" so often is counterproductive and needlessly
More information about the FOM mailing list
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Without use of any calculator
Never mind, forget what I wrote above. I was being naughty and used a calculator.
But I got it now. Here’s how.
By considering the sum of the complex roots of
, you have
But .
Note that . Hence
And . And there you have it.
Last edited by JaneFairfax (2008-07-25 23:27:41)
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Differentiate from first principles?
August 5th 2008, 12:51 PM #1
Differentiate from first principles?
Q1) Given that f(x) = 2x^3 + x , find ;
(a) the derivative of f(x) with respect to x, from first principles
can somebody explain the answers that are given here (pdf file) and it is question 1 on the exam paper, and also what exactly is the difference between first principles and regular
differentiation? Thanks A LOT!
Q1) Given that f(x) = 2x^3 + x , find ;
(a) the derivative of f(x) with respect to x, from first principles
can somebody explain the answers that are given here
What is not clear in this answer?
What exactly is the difference between first principles and regular differentiation?
Finding $f'$ from first principles means using the definition of the derivative : $f'(x)=\lim_{h\to0}\tfrac{f(x+h)-f(x)}{h}$. If you had been allowed to use "regular differentiation" you could
have used results which are derived from the definition of the derivative. (for example $(x^n)'=nx^{n-1}$)
August 5th 2008, 01:11 PM #2
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Pi Day Resources through LiveBinders
The University of Kansas has a wonderful math website designed for middle school students called Matrix Learning. In their words, Matrix Learning "provides resources to improve middle school
reading and mathematics achievement through the development of interactive educational games." There are several interactive games/activities that are not only educational and engaging, but fun!
There also are videos and mulitplayer games. Many of the activities would present well on the SB and some such as Definition Training- similar to hang man- would work well on the SB. I also would
encourage you to check out the videos from the University of New Mexico that are on bottom left of the page. So if you are a middle school math teacher I would highly recommend that you take time
to scope out this very worthwhile site. ENJOY!
Cassie Banka on 16 Nov 09
Game Classroom is an educational games website catering to the K-6 market. Game Classroom offers mathematics games and language arts games. Games can be found by selecting a grade level and then
a subject area. Both the mathematics and language arts categ
Loads of great teaching resources from the UK. Many have downloadable media such as interactive flash files, spreadsheets, word docs, powerpoint, iwb files (in various formats), and more.
Interactive, animated maths dictionary for kids with over 600 common math terms explained in simple language. Math glossary with math definitions, examples, math practice interactives,
mathematics activities and math calculators. © Jenny Eather 2007.
Lots of real life interactive examples in between....
1 - 7 of 7
items per page
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Spectral asymptotics of self-adjoint fourth order boundary value problems with eigenvalue parameter dependent boundary conditions
A regular fourth order differential equation with λ-dependent boundary conditions is considered. For four distinct cases with exactly one λ-independent boundary condition, the asymptotic eigenvalue
distribution is presented.
MSC: 34L20, 34B07, 34B08, 34B09.
fourth order boundary value problems; self-adjoint; boundary conditions; eigenvalue distribution; pure imaginary eigenvalues; spectral asymptotics
1 Introduction
Sturm-Liouville problems have attracted extensive attention due to their intrinsic mathematical challenges and their applications in physics and engineering. However, apart from classical
Sturm-Liouville problems, also higher order ordinary linear differential equations occur in applications, with or without the eigenvalue parameter in the boundary conditions. Such problems are
realized as operator polynomials, also called operator pencils. Some recent developments of higher order differential operators whose boundary conditions depend on the eigenvalue parameter, including
spectral asymptotics and basis properties, have been investigated in [1-4]. General characterizations of self-adjoint boundary conditions have been presented in [5,6] for singular and (quasi-)regular
problems. In all these cases, the minimal operator associated with an nth order differential equation must be symmetric, see [7,8] for necessary and sufficient conditions. A more general discussion
on the spectra of fourth order differential operators can be found in [9,10].
The generalized Regge problem is realized by a second order differential operator which depends quadratically on the eigenvalue parameter and which has eigenvalue parameter dependent boundary
conditions, see [11]. The particular feature of this problem is that the coefficient operators of this pencil are self-adjoint, and it is shown in [11] that this gives some a priori knowledge about
the location of the spectrum. In [12] this approach has been extended to a fourth order differential equation describing small transversal vibrations of a homogeneous beam compressed or stretched by
a force g. Separation of variables leads to a fourth order boundary problem with eigenvalue parameter dependent boundary conditions, where the differential equation
depends quadratically on the eigenvalue parameter. This problem is represented by a quadratic operator pencil, in a suitably chosen Hilbert space, whose coefficient operators are self-adjoint. In [13
] we have investigated a class of boundary conditions for which necessary and sufficient conditions were obtained such that the associated operator pencil consists of self-adjoint operators, while in
[14] we have continued the work of [13] in the direction of [12] to derive eigenvalue asymptotics associated with boundary conditions which lead to self-adjoint operator representations. We have
considered the particular case of boundary conditions which do not depend on the eigenvalue parameter at the left endpoint and depend on the eigenvalue parameter at the right endpoint.
In this paper, we extend the work of [14] to a class of boundary conditions where exactly one of the left endpoint boundary conditions does not depend on the eigenvalue parameter, while the remaining
boundary conditions depend on the eigenvalue parameter.
We define the operator pencil in Section 2 and we discuss which boundary conditions are considered. In Section 3, the eigenvalue asymptotics for the case are derived. In Section 4, it is shown that
the boundary value problems under consideration are Birkhoff regular, which implies that the eigenvalues for general g are small perturbations of the eigenvalues for . Hence, in Section 5, the first
four terms of the eigenvalue asymptotics are found and are compared to those obtained in [14].
2 The quadratic operator pencil L
On the interval , we consider the boundary value problem
where , is a real valued function and (2.2) are separated boundary conditions where the are constant or depend on λ linearly. The boundary conditions (2.2) are taken at the endpoint 0 for and at
the endpoint a for . Further, we assume for simplicity that either or , where for , for , and . We recall that the quasi-derivatives associated with (2.1) are given by
see [[8], p.26].
Assumption 2.1The numbers , , for are distinct as well as the numbers , , for .
We denote by U the collection of the boundary conditions (2.2) and define the following operators related to U:
We put and consider the linear operators , K and M in the space with domains
given by
It is easy to check that , , and . We associate a quadratic operator pencil
in the space with the problem (2.1), (2.2).
The conditions under which the differential operator is self-adjoint are given in
Theorem 2.2 ([13], Theorem 1.2)
Denote by the set ofpin for theλ-independent boundary conditions and by the corresponding set for . Then the differential operator associated with this boundary value problem is self-adjoint if and
only if for all boundary conditions of the form and ifqis even in case or odd in case , otherwise, , , and .
Proposition 2.3The operator pencil is a Fredholm valued operator function with index 0. The spectrum of the Fredholm operator consists of discrete eigenvalues of finite multiplicities, and all
eigenvalues of , , lie in the closed upper half-plane and on the imaginary axis and are symmetric with respect to the imaginary axis.
Proof As in [[12], Section 3], we can argue that for all , is a relatively compact perturbation of , where is well known to be a Fredholm operator. The statement on the location of the spectrum
now follows as in [[12], Lemma 3.1].□
We now consider the particular cases that exactly one of the boundary conditions at 0 depends on λ, whereas both boundary conditions at a depend on λ. Therefore, taking Assumption 2.1 and Theorem 2.2
into account, we have the four boundary conditions
where , , , , and , while and . Thus, we have 8 and 4 possible sets of boundary conditions at the endpoint 0 and a, respectively. Whence there are 32 different sets of boundary conditions.
Recall that the parameter λ emanates from derivatives with respect to the time variable in the original partial differential equation, and it is reasonable that the highest space derivative occurs in
the term without time derivative. Thus, the most relevant boundary conditions would have , and . This leaves us with four different cases for the boundary conditions .
These four cases are uniquely determined by the value of , so that we will consider
Case 1: ; Case 2: ; Case 3: ; Case 4: .
The corresponding boundary operators are then
3 Asymptotics of eigenvalues for
In this section, we consider the boundary value problem (2.1), (2.2) with . We count all eigenvalues with their proper multiplicities and develop a formula for the asymptotic distribution of the
eigenvalues, which is used to obtain the corresponding formula for general g. Observe that for , the quasi-derivatives coincide with the standard derivatives . We take the canonical fundamental
system , , of (2.1) with if for . It is well known that the functions are analytic on ℂ with respect to λ. Putting
the eigenvalues of the boundary value problem (2.1), (2.2) are the eigenvalues of the analytic matrix function M, where the corresponding geometric and algebraic multiplicities coincide, see [[15],
Theorem 6.3.2].
it is easy to see that
The second row of has exactly two non-zero entries (for ), and these non-zero entries are:
Since the first row of has exactly one entry 1 and all other entries zero, an expansion of with respect to the second row shows that , where
In view of (2.7), (2.8) this gives
Each of the summands in ϕ is a product of a power in μ and a product of two sums of a trigonometric and a hyperbolic functions. The terms with the highest μ-powers in are non-zero constant multiples
For the above four cases, we obtain:
We next give the asymptotic distributions of the zeros of with proper counting.
Lemma 3.1 Case 1: has a zero of multiplicity 8 at 0, simple zeros at
simple zeros at , and for , and no other zeros.
Case 2: has a zero of multiplicity 4 at 0, exactly one simple zero in each interval for positive integerskwith asymptotics
simple zeros at , and for , and no other zeros.
Case 3: has a zero of multiplicity 6 at 0, simple zeros at
simple zeros at , and for , and no other zeros.
Case 4: has a zero of multiplicity 6 at 0, exactly one simple zero in each interval for positive integerskwith asymptotics
simple zeros at , and for , and no other zeros.
Proof The result is obvious in Cases 1 and 3. Cases 2 and 4 only differ in the factor with the power of μ, and the multiplicity of the corresponding zero of at 0 is easy to verify. The choice of the
indexing for the non-zero zeros of in each case will become apparent later.
It, therefore, remains to describe the behavior of the non-zero zeros of in Case 2. First, we are going to find the zeros of on the positive real axis. One can observe that for , implies and ,
whence the positive zeros of are those for which . Since and for all where the functions are defined, the function is increasing with a positive derivative on each interval , . On each of
these intervals, the function moves from −∞ to ∞, thus we have exactly one simple zero of in each interval , where k is a positive integer, and no zero in . Since as , we have
The location of the zeros on the other three half-axes follows by repeated application of .
The proof will be complete if we show that all zeros of lie on the real or the imaginary axis. To this end, we observe that the product-to-sum formula for trigonometric functions gives
Putting , , it follows for that
Since has a positive derivative on , this function is strictly increasing, and therefore, implies by (3.2) that and thus . Then
is either real or pure imaginary.□
Proposition 3.2For , there exists a positive integer such that the eigenvalues , counted with multiplicity, of the problem (2.1), (2.5)-(2.8), where in Cases 1 and 2 and in Cases 3 and 4, can be
enumerated in such a way that the eigenvalues are pure imaginary for , and for . For , we can write , where the have the following asymptotic representation as :
In particular, the number of pure imaginary eigenvalues is even in Cases 1 and 2 and odd in Cases 3 and 4.
Proof Case 4: A straightforward calculation gives
Up to the constant factor , the second term equals . It follows that for μ outside the zeros of , and , we have
Fix and for let be the squares determined by the vertices , . These squares do not intersect due to . Since if and only if and , it follows from the periodicity of tan that the number
is positive and independent of ε. Since uniformly in the strip as , there is an integer such that
By periodicity, there are numbers and such that and for all and all k. Observing , it follows that there is such that for all μ on the squares with the estimate holds. Further, we may
assume by Lemma 3.1 that is inside for and that no other zero of has this property. Hence, it follows by Rouché’s theorem that there is exactly one (simple) zero of ϕ in each for . Replacing μ
with iμ only changes the sign of the second term in (3.3) and thus the sign of . Hence, the same estimates apply to corresponding squares along the other three half-axes, and we therefore have that
ϕ has zeros , for with the same asymptotic behavior as the zeros , of as discussed in Lemma 3.1.
Next, we are going to estimate on the squares , , whose vertices are . For and ,
Therefore, we have for , where and , that
uniformly in y as . Hence, there exists such that for all , and ,
It follows from (3.6) and (3.7) for , , and that
Furthermore, we are going to make use of the estimates
which hold for all with and all . Therefore, it follows from (3.6), (3.8)-(3.10) and the corresponding estimates with μ replaced by iμ that there is such that for all with . Again from the
definition of in (3.4) and Rouché’s theorem, we conclude that the functions and ϕ have the same number of zeros in the square , for with .
Since has zeros inside and thus zeros inside , it follows that ϕ has no large zeros other than the zeros found above for sufficiently large, and that account for all eigenvalues of the
problem (2.1)-(2.2) since each of these eigenvalues gives rise to two zeros of ϕ, counted with multiplicity. By Proposition 2.3, all eigenvalues with non-zero real part occur in pairs , , which
shows that we can index all such eigenvalues as . Since there is an odd number of remaining indices, the number of pure imaginary eigenvalues must be odd.
Case 2: The function ϕ in this case is
Then all the estimates are as in Case 4, and the result in Case 2 immediately follows from that in Case 4 if we observe that each for k large enough contains two fewer zeros of ϕ than in Case 4.
Case 1: A straightforward calculation gives
The result follows with reasonings and estimates as in the proof of Case 4, replacing μ by and , respectively.
Case 3: The function ϕ in this case is
and a reasoning as in Case 1 completes the proof.□
4 Birkhoff regularity
We refer to [[15], Definition 7.3.1] for the definition of the Birkhoff regularity.
Proposition 4.1The boundary value problem (2.1), (2.5)-(2.8) is Birkhoff regular for with respect to the eigenvalue parameterμgiven by .
Proof The characteristic function of (2.1) as defined in [[15], (7.1.4)] is , and its zeros are , . We can choose
according to [[15], Theorem 7.2.4.A]. The boundary condition (2.5)-(2.8) can be written in the form
and where denotes the νth unit vector in . Thus the boundary matrices defined in [[15], (7.3.1)] are given by
Choosing , it follows that , where
for Case r and for Cases 1 and 2, while for Cases 3 and 4. The Birkhoff matrices are
where , are the diagonal matrices with 2 consecutive ones and 2 consecutive zeros in the diagonal in a cyclic arrangement, see [[15], Definition 7.3.1 and Proposition 4.1.7]. It is easy to see
that after a permutation of columns, the matrices (4.1) are block diagonal matrices consisting of blocks taken from two consecutive columns (in the sense of cyclic arrangement) of the first two rows
of and the last two rows of , respectively. Hence the determinants of the Birkhoff matrices (4.1) are
in Cases 1 and 2, i.e., , whereas
in Cases 3 and 4. Thus, the problem (2.1), (2.5)-(2.8) is Birkhoff regular.□
5 Asymptotic expansions of eigenvalues
Let D, as a function of μ with , be the characteristic function of the problem (2.1), (2.5)-(2.8) with respect to the fundamental system , , with for , where δ is the Kronecker delta. Denote by
the corresponding characteristic function for . Note that the characteristic functions and considered in Section 3 have the same zeros counted with multiplicity. Due to the Birkhoff regularity, g
only influences lower order terms in D. Therefore, it can be inferred that outside the interior of the small squares , , , around the zeros of , if is sufficiently large. Since the fundamental
system , , depends analytically on μ, also D and are analytic functions. Hence, applying Rouché’s theorem both to the large squares and to the small squares which are sufficiently far away from
the origin, it follows that the eigenvalues of the boundary value problem for general g have the same asymptotic distribution as for . Whence Proposition 3.2 leads to
Proposition 5.1For , there exists a positive integer such that the eigenvalues , counted with multiplicity, of the problem (2.1), (2.5)-(2.8), where in Cases 1 and 2 and in Cases 3 and 4, can be
enumerated in such a way that the eigenvalues are pure imaginary for , and for . For , we can write , where the have the following asymptotic representation as :
In particular, the number of pure imaginary eigenvalues is even in Cases 1 and 2 and odd in Cases 3 and 4.
In the remainder of the section, we are going to establish more precise asymptotic expansions of the eigenvalues. According to [[15], Theorem 8.2.1], (2.1) has an asymptotic fundamental system of
the form
and means that we omit those terms of the Leibniz expansion which contain a function with . Since the coefficient of in (2.1) is zero, we have , see [[15], (8.2.3)].
We will now determine the functions and . In this regard, observe that and in the notation of [[15], (8.1.2) and (8.1.3)], see [[15], Theorem 8.1.2]. From [[15], (8.2.45)], we know that
where is the νth unit vector in , , and are matrices given by [[15], (8.2.28), (8.2.33) and (8.2.34)], that is, ,
where , , and . Let . A lengthy but straightforward calculation gives
and thus
for , where means that the estimate is uniform in x.
The characteristic function of (2.1), (2.5)-(2.8) is
Note that
where , , , , , and each of the functions has asymptotic representations of the form .
It follows from (5.9) that
where , , , . If , we have for and the terms for can be absorbed by as they are of the form for any integer s. Hence, for ,
A straightforward calculation gives
For the other two factors in (5.12) and (5.13), we have to consider the four different cases.
Thus, we have
Hence, we get
Thus, we have
We already know by Proposition 5.1 that the zeros of D satisfy the asymptotic representations as . In order to improve on these asymptotic representations, write
Because of the symmetry of the eigenvalues, we will only need to find the asymptotic expansions as . We know from Proposition 5.1, and it is our aim to find and . To this end, we will substitute
(5.24) into and we will then compare the coefficients of , and .
Observe that
Using (5.11), can be written as
where γ is the highest μ-power in and . Substituting (5.25) and (5.26) into (5.27) and comparing the coefficients of , and , we get
Theorem 5.2For , there exists a positive integer such that the eigenvalues , , counted with multiplicity, of the problem (2.1), (2.5)-(2.8), where in Cases 1 and 2 and in Cases 3 and 4, can be
enumerated in such a way that the eigenvalues are pure imaginary for , and for , where and the have the asymptotic representations
and the numbers , , are as follows:
In particular, the number of pure imaginary eigenvalues is even in Cases 1 and 2 and odd in Cases 3 and 4.
Remark 5.3 In [14] we have considered the differential equation (2.1) with the same boundary conditions , at a as in this paper but with λ-independent boundary conditions at 0, that is, the
boundary conditions also occur in [14]. Whereas in [14] the number of pure imaginary eigenvalues is odd in each case, this number is even in Cases 1 and 2 of this paper. We observe that in Cases 1
and 2, the λ-dependent part is the ‘dominating’ part of the boundary condition , in the sense that it has the highest μ-power arising as from , whereas in Cases 3 and 4 the λ-independent part is
dominating. It may be interesting to investigate if, in general, the parity of the number of pure imaginary eigenvalues can be determined by the number of dominating λ-dependent parts in the boundary
We can observe that the functions in the Cases 3 and 4 are respectively the same as in [14] since the corresponding dominating terms in the boundary conditions coincide. However, the numbers and
differ from those of [14] in each case, which is due to the λ-term in the boundary condition .
Authors’ contributions
The subject of this paper is part of the PhD thesis of BZ. The subject has been suggested and supervised by MM, and the initial version of the paper has been written by BZ. The submitted version has
been verified and discussed by MM and BZ.
This research was partially supported by a grant from the NRF of South Africa, Grant number 69659. Various of the above calculations have been verified with Sage.
1. Kerimov, NB, Aliev, ZS: Basis properties of a spectral problem with a spectral parameter in the boundary condition (Russian). Mat. Sb.. 197(10), 65–86 translation in Sb. Math. 197, 1467-1487
(2006) (2006)
translation in Sb. Math. 197, 1467-1487 (2006)
Publisher Full Text
2. Kerimov, NB, Aliev, ZS: On the basis property of the system of eigenfunctions of a spectral problem with a spectral parameter in the boundary condition (Russian). Differ. Uravn.. 43, 886–895
translation in Differ. Equ. 43, 905-915 (2007) (2007)
3. Marletta, M, Shkalikov, A, Tretter, C: Pencils of differential operators containing the eigenvalue parameter in the boundary conditions. Proc. R. Soc. Edinb., Sect. A, Math.. 133, 893–917 (2003).
Publisher Full Text
4. Shkalikov, AA: Boundary problems for ordinary differential equations with parameter in the boundary conditions (Russian). Tr. Semin. Im. I.G. Petrovskogo. 9, 190–229 translation in J. Sov. Math.
33, 1311-1342 (1986) (1983)
5. Wang, A, Sun, J, Zettl, A: The classification of self-adjoint boundary conditions: separated, coupled, and mixed. J. Funct. Anal.. 255, 1554–1573 (2008). PubMed Abstract | Publisher Full Text
6. Wang, A, Sun, J, Zettl, A: Characterization of domains of self-adjoint ordinary differential operators. J. Differ. Equ.. 246, 1600–1622 (2009). Publisher Full Text
7. Möller, M, Zettl, A: Symmetric differential operators and their Friedrichs extension. J. Differ. Equ.. 115, 50–69 (1995). Publisher Full Text
8. Behncke, H: Spectral analysis of fourth order differential operators. I. Math. Nachr.. 279, 58–72 (2006). Publisher Full Text
9. Behncke, H: Spectral analysis of fourth order differential operators. II. Math. Nachr.. 279, 73–85 (2006). Publisher Full Text
10. Pivovarchik, V, van der Mee, C: The inverse generalized Regge problem. Inverse Probl.. 17, 1831–1845 (2001). Publisher Full Text
11. Möller, M, Pivovarchik, V: Spectral properties of a fourth order differential equation. J. Anal. Appl.. 25, 341–366 (2006)
12. Möller, M, Zinsou, B: Self-adjoint fourth order differential operators with eigenvalue parameter dependent boundary conditions. Quaest. Math.. 34, 393–406 doi:10.2989/16073606.2011.622913 (2011)
Publisher Full Text
13. Möller, M, Zinsou, B: Spectral asymptotics of self-adjoint fourth order differential operators with eigenvalue parameter dependent boundary conditions. Complex Anal. Oper. Theory. 6, 799–818
doi:10.1007/s11785-011-0162-1 (2012)
Publisher Full Text
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MathGroup Archive: August 2008 [00532]
[Date Index] [Thread Index] [Author Index]
Partial differential equation with evolving boundary conditions
• To: mathgroup at smc.vnet.net
• Subject: [mg91477] Partial differential equation with evolving boundary conditions
• From: "Ingolf Dahl" <ingolf.dahl at telia.com>
• Date: Sat, 23 Aug 2008 01:41:41 -0400 (EDT)
• Organization: Goteborg University
• Reply-to: <ingolf.dahl at telia.com>
Best friends!
I am trying to solve a partial differential equation (in principle the heat
equation) with boundary conditions that also evolve with time. For instance
this works for me:
\*SubscriptBox[\(\[PartialD]\), \(y, y\)]\ \(\[Theta][y,
\*SubscriptBox[\(\[PartialD]\), \(t\)]\(\[Theta][y, t]\)\),
\*SubscriptBox[\(\[PartialD]\), \(y1\)]\(\[Theta][y1,
t]\)\)/.{y1->0})*0.1+\[Theta][0,t]==0,+( \!\(
\*SubscriptBox[\(\[PartialD]\), \(y1\)]\(\[Theta][y1,
but this does not work
\*SubscriptBox[\(\[PartialD]\), \(y, y\)]\ \(\[Theta][y,
\*SubscriptBox[\(\[PartialD]\), \(t\)]\(\[Theta][y, t]\)\),
\*SubscriptBox[\(\[PartialD]\), \(t\)]\(\[Theta][0,
\*SubscriptBox[\(\[PartialD]\), \(t\)]\(\[Theta][1, t]\)\),
I obtain the error message
NDSolve::bdord: Boundary condition \[Theta][0,t]+1. (\[Theta]^(0,1))[0,t]
should have derivatives of order lower than the differential order of the
partial differential equation. >>
plus a lot more. As I understand it, the second case should be solvable in
principle. Is this error a deficiency of NDSolve, or have I made some
mistake? This problem can be seen as a partial differential equation (in y
and t) coupled to two ordinary differential equations (for the boundary
conditions, only in t). What is the best way to solve such problems?
This is the simplification of a more involved case, where the boundary
conditions also are coupled to the partial differential equation, but I have
tried to boil down the problem here.
Best regards
Ingolf Dahl
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Name the lengths of the sides of three rectangles that have perimeters of 14 units
Name the lengths of the sides of three rectangles that have perimeters of 14 units.
This question still have no answer summary yet.
1 answers.
#1ALAINIAnswered at 2013-02-19 15:06:48
P = 2L + 2W
Some possible rectangles:
2 by 5
1 by 6
3 by 4
Did this answer your question? If not, ask a new question.
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Rythmos::IntegrationObserverBase< Scalar >
Base class for strategy objects that observe and time integration by observing the stepper object. More...
#include <Rythmos_IntegrationObserverBase.hpp>
Public Member Functions
virtual RCP
< IntegrationObserverBase
< Scalar > > cloneIntegrationObserver () const =0
Clone this integration observer if supported .
virtual void resetIntegrationObserver (const TimeRange< Scalar > &integrationTimeDomain)=0
Reset the observer to prepair it to observe another integration.
virtual void observeStartTimeIntegration (const StepperBase< Scalar > &stepper)
Observe the beginning of a time integration loop.
virtual void observeEndTimeIntegration (const StepperBase< Scalar > &stepper)
Observe the end of a time integration loop.
virtual void observeStartTimeStep (const StepperBase< Scalar > &stepper, const StepControlInfo< Scalar > &stepCtrlInfo, const int timeStepIter)
Observer the beginning of an integration step.
virtual void observeCompletedTimeStep (const StepperBase< Scalar > &stepper, const StepControlInfo< Scalar > &stepCtrlInfo, const int timeStepIter)=0
Observe a successfully completed integration step.
virtual void observeFailedTimeStep (const StepperBase< Scalar > &stepper, const StepControlInfo< Scalar > &stepCtrlInfo, const int timeStepIter)
Observer a failed integration step.
Detailed Description
template<class Scalar>
class Rythmos::IntegrationObserverBase< Scalar >
Base class for strategy objects that observe and time integration by observing the stepper object.
ToDo: Finish Implementation!
Definition at line 24 of file Rythmos_IntegrationObserverBase.hpp.
Member Function Documentation
template<class Scalar >
virtual RCP<IntegrationObserverBase<Scalar> > Rythmos::IntegrationObserverBase< Scalar >::cloneIntegrationObserver ( ) const [pure virtual]
template<class Scalar >
virtual void Rythmos::IntegrationObserverBase< Scalar >::resetIntegrationObserver ( const TimeRange< Scalar > & integrationTimeDomain ) [pure virtual]
Reset the observer to prepair it to observe another integration.
integrationTimeDomain [in] The time domain over which the integration will be defined.
• integrationTimeDomain.length() > 0.0
Add initial guess as an argument
Implemented in Rythmos::CompositeIntegrationObserver< Scalar >, Rythmos::ForwardResponseSensitivityComputerObserver< Scalar >, and Rythmos::LoggingIntegrationObserver< Scalar >.
template<class Scalar >
void Rythmos::IntegrationObserverBase< Scalar >::observeStartTimeIntegration ( const StepperBase< Scalar > & stepper ) [virtual]
Observe the beginning of a time integration loop.
stepper [in] The stepper object.
Warning! This function is *NOT* stateless. It should be called once and only once at the beginning of getFwdPoints().
NOTE: The function resetIntegrationControlStrategy() must be called prior to even the first call to function.
NOTE: This method should be pure virtual but has been given a default implementation for backwards compatibility. We will make this pure virtual in the future.
Reimplemented in Rythmos::CompositeIntegrationObserver< Scalar >, and Rythmos::LoggingIntegrationObserver< Scalar >.
Definition at line 244 of file Rythmos_IntegrationObserverBase.hpp.
template<class Scalar >
void Rythmos::IntegrationObserverBase< Scalar >::observeEndTimeIntegration ( const StepperBase< Scalar > & stepper ) [virtual]
Observe the end of a time integration loop.
stepper [in] The stepper object.
Warning! This function is *NOT* stateless. It should be called once and only once at the end of getFwdPoints().
NOTE: The function resetIntegrationControlStrategy() must be called prior to even the first call to function.
NOTE: This method should be pure virtual but has been given a default implementation for backwards compatibility. We will make this pure virtual in the future.
Reimplemented in Rythmos::CompositeIntegrationObserver< Scalar >, and Rythmos::LoggingIntegrationObserver< Scalar >.
Definition at line 251 of file Rythmos_IntegrationObserverBase.hpp.
template<class Scalar >
void Rythmos::IntegrationObserverBase< Scalar >::observeStartTimeStep ( const StepperBase< Scalar > & stepper,
const StepControlInfo< Scalar > & stepCtrlInfo,
const int timeStepIter
) [virtual]
Observer the beginning of an integration step.
stepper [in] The stepper object.
stepCtrlInfo [in] The info for the time step about to be taken.
timeStepIter [in] The time step iteration counter. In the first call to this function, this should be timeStepIter==0 and it should be incremented on each call only once. While the concrete
implementation of *this could keep track of the this counter, putting it in the argument list helps to simplify logic and helps to validate correct usage.
Warning! This function is *NOT* stateless. It should be called once and only once at the beginning of each time step.
NOTE: The function resetIntegrationControlStrategy() must be called prior to even the first call to function.
NOTE: This method should be pure virtual but has been given a default implementation for backwards compatibility. We will make this pure virtual in the future.
Reimplemented in Rythmos::CompositeIntegrationObserver< Scalar >, and Rythmos::LoggingIntegrationObserver< Scalar >.
Definition at line 258 of file Rythmos_IntegrationObserverBase.hpp.
template<class Scalar >
virtual void Rythmos::IntegrationObserverBase< Scalar >::observeCompletedTimeStep ( const StepperBase< Scalar > & stepper,
const StepControlInfo< Scalar > & stepCtrlInfo,
const int timeStepIter
) [pure virtual]
Observe a successfully completed integration step.
stepper [in] The stepper object that was just stepped forward once to integrate the transient ODE/DAE equations. On the very first call and every other call, this stepper should have a
finite time range for a successfull step.
stepCtrlInfo [in] The info for the actual time step that was just completed.
timeStepIter [in] The time step iteration counter. In the first call to this function, this should be timeStepIter==0 and it should be incremented on each call only once. While the concrete
implementation of *this could keep track of the this counter, putting it in the argument list helps to simplify logic and helps to validate correct usage.
Warning! This function is *NOT* stateless. It should be called once and only once per time step iteration.
NOTE: The function resetIntegrationControlStrategy() must be called prior to even the first call to function.
NOTE: If isInitialTimeStep(stepper->getTimeRange(), fullTimeRange) == true then this is the first time step (where fullTimeRange was passed into resetIntegrationObserver().
NOTE: If isFinalTimeStep(stepper->getTimeRange(), fullTimeRange) == true then this is the last time step (where fullTimeRange was passed into resetIntegrationObserver().
Implemented in Rythmos::CompositeIntegrationObserver< Scalar >, Rythmos::ForwardResponseSensitivityComputerObserver< Scalar >, and Rythmos::LoggingIntegrationObserver< Scalar >.
template<class Scalar >
void Rythmos::IntegrationObserverBase< Scalar >::observeFailedTimeStep ( const StepperBase< Scalar > & stepper,
const StepControlInfo< Scalar > & stepCtrlInfo,
const int timeStepIter
) [virtual]
Observer a failed integration step.
stepper [in] The stepper object that was just stepped forward once to integrate the transient ODE/DAE equations. On the very first call and every other call, this stepper should have a
finite time range for a successfull step.
stepCtrlInfo [in] The info for the actual time step that was just attempted.
timeStepIter [in] The time step iteration counter. In the first call to this function, this should be timeStepIter==0 and it should be incremented on each call only once. While the concrete
implementation of *this could keep track of the this counter, putting it in the argument list helps to simplify logic and helps to validate correct usage.
Warning! This function is *NOT* stateless. It should be called once and only once per failed time step iteration.
NOTE: The function resetIntegrationControlStrategy() must be called prior to even the first call to function.
NOTE: If isInitialTimeStep(stepper->getTimeRange(), fullTimeRange) == true then this is the first time step (where fullTimeRange was passed into resetIntegrationObserver().
NOTE: If isFinalTimeStep(stepper->getTimeRange(), fullTimeRange) == true then this is the last time step (where fullTimeRange was passed into resetIntegrationObserver().
NOTE: This method should be pure virtual but has been given a default implementation for backwards compatibility. We will make this pure virtual in the future.
Reimplemented in Rythmos::CompositeIntegrationObserver< Scalar >, and Rythmos::LoggingIntegrationObserver< Scalar >.
Definition at line 269 of file Rythmos_IntegrationObserverBase.hpp.
The documentation for this class was generated from the following file:
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A One-in-a-Million Baseball Play
23 September 2011, 9:00 am
As the 2011 MLB season winds down, there is a slim chance of something very unusual happening: a three-way tie for the wild card playoff birth!
It seems highly unlikely that the Red Sox, Rays, and Angels will actually all finish in a dead-heat, but if they do, it will pose a lot of problems for playoff scheduling.
This is a fun, if complicated, math question to think about: what are the chances that after a 162-game season, three of the eleven teams ultimately vying for the wild card end up with identical
To investigate, the first thing I’d do is simplify the situation. I’d reduce the number of teams and the number of games, give every team a 50/50 chance to win every game, and then see what happens.
After I’d explored a bit, I’d then consider complicating matters by using more teams, more games, and more realistic winning percentages.
A math challenge that any Strat-o-matic player could love!
Click here to see more in Sports.
• MrHonner on Balancing Act
Filed under Probability, Sports, Statistics
Comment (RSS) | Permalink
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Tamburini: Neutrinos Are Majorana Particles, Relativity Is OK
TEDx Bologna
on October 15th on his theory of photon vortices. Anyway, back to neutrinos.
The recent result by Opera which hints at a possible superluminal motion of neutrinos produced by CERN and shot underground to the Gran Sasso mine in central Italy raised Tamburini's interest.
Together with Marco Laveder, a colleague of the Physics Department of Padova University, Tamburini recently wrote an interesting paper where he explains that Einstein's Relativity Theory needs not be
put in discussion by the measurement: rather, the fact that neutrinos could exhibit an apparent superluminal motion is inherent in the theory written as far back as 1932 by Ettore Majorana, the
talented and mysterious Italian physicist who disappeared shortly thereafter -probably to live in South America under false identity.
Tamburini and Laveder's paper (which you can
download from the arxiv
) explains how a fictitious "imaginary mass" term in the solution of Majorana equations for neutrino propagation may be responsible for the observed faster-than-light travel of neutrinos in a dense
Specifically, the authors start from the hypothesis that the observed superluminal speed of neutrinos may be caused by matter effects, and they work out the effective imaginary mass which corresponds
to each of the experimental measurements.
What I find interesting is that Tamburini and Laveder do not stop at discussing the theoretical interpretation of the alleged superluminal motion, but put their hypothesis to the test by comparing
known measurements of neutrino velocity on a graph, where the imaginary mass is computed from the momentum of neutrinos and the distance traveled in a dense medium. The data show a very linear
behaviour, which may constitute an explanation of the Opera effect:
This can't be right. Lorentz invariance is a purely geometric effect - there's no interaction term in a Lagrangian that can make a particle propagate superluminally, and the "fictitious" imaginary
mass is an effective field theory approximation to some underlying interaction.
You have to get rid of Lorentz invariance to explain this result. Otherwise you get CTCs and grandfather paradoxes.
Foster Boondoggle (not verified) | 10/10/11 | 08:58 AM
That would imply a remarkably strong coupling between neutrinos and matter. Neutrinos have a 10^-10 chance to interact with matter directly over the distance of 700 km, but their propagation speed is
somehow changed by a factor of 10^-5?
Nameless (not verified) | 10/10/11 | 14:07 PM
Well, Nameless, while I share your intuition, let's admit that there are holes in your argument and it's far from waterproof. Light isn't basically absorbed in glass either - but it's still slower by
1/3 than it is in the vacuum.
Luboš Motl (not verified) | 10/12/11 | 11:08 AM
Vladimir Kalitvia... | 10/12/11 | 12:43 PM
Vladimir, the quantities expressing numerical values of couplings may have various units but "percent" is not among them. Moreover, light works in the same way when it's diluted down to individual
photons in which case it's completely analogous to individual neutrinos so your "collective" comments have nothing to do with the actual physics, either. Return to the asylum, please.
Luboš Motl (not verified) | 10/12/11 | 12:25 PM
Yes, I can express coupling in "percent". For example, if the glass thickness is smaller than necessary for 100% coupling, there will be incomplete refraction, you did not know that?
Also, these collective phenomena with light in the glass are well "linear" or photon-number independent, to you information. What is valid for one photon, is valid for a beam. Only very high
intensity light may cause non-linear effects which are also collective. In other words, photon in the glass is a quasi-particle, like phonon, plasmon, magnon, etc.
I did not get you phrase about asylum. What do you mean with it?
Vladimir Kalitvia... | 10/13/11 | 09:06 AM
> light works in the same way when it's diluted down to individual photons
I must be in a bad mood today, but holy cow. "diluted down" ... how can people do physics when they don't even have a way to express the situation whereby the photon field only has enough energy to
yield a single photon?
Anonymous (not verified) | 10/16/11 | 12:41 PM
I don't think that you're in a bad mood today. I think that your brain is in a very bad shape in this particular life of yours. I have described the situation very crisply: the electromagnetic waves
may be diluted down to very low energy densities so that the photons are propagating one-by-one. When it's so, their mutual interactions (and therefore also their statistics) are negligible and their
behavior is dictated by the wave equation describing individual particles which is fully analogous to the equation describing the neutrinos. The photon-photon interactions are actually negligible
even if the energy density is high.
If you're not capable of understanding these simple ideas, please realize that the problem is on your side and this problem gives you no right to contaminate the public spaces with your utter
Luboš Motl | 10/16/11 | 12:53 PM
Vladimir Kalitvia... | 10/16/11 | 13:19 PM
Looks like my second comment has disappeared AGAIN, so here's the condensed version of the idea.
Consider a glass prism. When you shoot a photon at the prism, it comes out at an angle to the incident direction. Therefore, even if it's not absorbed, it still interacts with glass. Because glass
really couples quite strongly with the EM field. And that is a prerequisite to being able to refract light.
I know that my argument is not waterproof, but it still compares quite positively to Tamburini's argument, which is basically one big hole. :)
Nameless (not verified) | 10/14/11 | 07:30 AM
Tommaso Dorigo | 10/14/11 | 09:12 AM
Tommaso Dorigo | 10/10/11 | 15:49 PM
Vladimir Kalitvia... | 10/14/11 | 09:42 AM
How is the Tamburini model reconciled with the Cohen-Glashow paper at arXiv 1109.6562 based on “… pair bremsstrahllung …[which]… proceeds through the neutral current interaction …" ?
Is there some way in which a Tamburini superluminal neutrino avoids emitting neutral current pair bremsstrhllung and the consequent loss of energy going from CERN to Gran Sasso?
Tony Smith (not verified) | 10/10/11 | 17:44 PM
Tommaso Dorigo | 10/11/11 | 01:59 AM
Tommaso, you say that you "see no reason why bremsstrahlung should affect more the superluminal component".
Cohen and Glashow in arXiv 1109.6562 say "... in all cases of superluminal propagation, certain otherwise forbidden processes are kinematically permitted, even in vacuum ...".
I do not see a detailed discussion or reference about that quote from their paper
it may be relevant that they do say:
"... The threshold energy for ... pair bremsstrahllung ... is ...
2 m_e / sqrt( v_nu^2 - v_e^2 )
where v_e is the maximal attainable velocity of an electron and m_e its mass ...
we know that v_e = 1 [the speed of light] to a precision of at least 10^(-15) ...".
Maybe Cohen and Glashow are saying that since the electron cannot go faster than light,
neither can the neutrino unless it begins to emit bremsstrahllung electron-positron pairs.
Tony Smith (not verified) | 10/11/11 | 09:26 AM
Tommaso Dorigo | 10/11/11 | 13:30 PM
Lorenz Invariance is already broken when the neutrino is traveling through a medium, the medium is a preferred frame.
The paper however doesn't give any reason why mass term is created in a neutrino traveling through matter.
BDOA (not verified) | 10/10/11 | 18:23 PM
>the medium is a preferred frame
What the hell?
Anonymous (not verified) | 10/16/11 | 12:26 PM
All these Italians with their faster-than-light neutrinos... nothing more than a big ploy to direct attention away from the economic state of affairs in Italy...
Anton (not verified) | 10/10/11 | 21:44 PM
My question is exactly the same as Tony's. I don't see anything that would avoid the energy loss of known electroweak processes (e+ e- emission), something which was not seen in the OPERA final state
mean energy distributions?!
Can this "imaginary mass" term somehow effect the *real* kinematics of the neutrinos?
Roy Johnstone (not verified) | 10/10/11 | 21:47 PM
In my own opinion you may not agree with me , so their we go.
Breaking the speed of light is really a not pleasent news and is more like a big mistake that an actual discovery, Why.
1. Speed of ligth on the case of navigation pourpuse is totaly worthless effort, because it used on a real navigation may not only alters the magnetic fields on space causing a distress on it and may
change it structure not for future discovery if not for future distruction of ballance. because it will affect only one side of it.
2. to control for example a traveling device at that speed will be very dificult because the universe change every milliseond and what do you see if not what is there.
3. and mos important, Knowing the way we do test and challenge nature today i can,t imagine what may be the result of something that we cant control outside of a laboratory.
So my point is instead o forsing the speed of ligth why not follow it in using natural magnetic fields.
Ill bet this sports for wining novel prices will end on a problem that we may not be able to fix.
like we always do.
abdel rosado (not verified) | 10/11/11 | 02:55 AM
If these neutrinos are travelling at superluminal speed, should they not be travelling backwards through time?
If they are travelling backwards through time, how can someone travelling forwards through time observe them for more than an infinitesimally short period of time?
Bill S. (not verified) | 10/11/11 | 13:02 PM
>If these neutrinos are travelling at superluminal speed, should they not be travelling backwards through time?
Only in some reference frames moving relative to the one in which the neutrino's superspeed™ is being measured. Yes that yields trouble, as they will arrive before being sent, so it will look as if
the receiver end influences the sender end. Nobody wants that.
Anonymous (not verified) | 10/16/11 | 12:30 PM
Refractive index if "formed" as a result of collective effects. Remember slow neutrons that can be reflected from some media. It is also a collective interaction effect. Fast neutrons do not manifest
that because of too short de Broglie wave length.
The wave length of those energetic neutrinos is so short that it cannot result in collective effects.
Vladimir Kalitvia... | 10/11/11 | 14:06 PM
May be I produced some pure nonsense, but I tried to write the Dirac equation and the Majorana equation as pure quaternionic equations, rather than as complex equations that use matrices and spinors.
The matrices and spinors immitate the behavior of quaternions and the sign selections that occur in quaternionic (skew) fields due to their conjugation and handedness.
The quaternionic Dirac equation runs: (ψ ⇒ψ*)
∇ψ = m ψ*
(No spinors, no α, β, or γ matrices, just quaternions!)
ψ consists of a one dimensional real part and a 3D imaginary part. Here the star means conjugation. It changes the sign of all imaginary base vectors of the quaternion. This action switches the
handedness of the field. ψ* is the antiparticle of ψ.
The covariant derivative D delivers
Dψ = m ψ* + A ψ
A derived equation is
∇(ψ ψ) = m (ψ ψ*) = 2 m|ψ|²
Another derived equation is
∇(ψ ψ*) = 2 m Re(ψ ψ)
With the interpretation of (ψ ψ*) as a presence density follows:
∫˯ (ψ ψ*) dV=1
and as a consequence
∫˯∇(ψ ψ) dV =2m
The quaternionic Majorana equation runs: (ψ ⇒ ψˤ)
∇ψ = m ψˤ
In ψˤ two of the imaginary base vectors switch sign. This action DOES NOT switch the handedness. (ψ ψˤ) can no longer be interpreted as presence density. It is not even real!
∫˯ (ψ ψˤ) dV is not equal to one. It is not even real. So the factor m in the Majorana equation may be quaternionic rather than real.
For more details see: http://www.crypts-of-physics.eu/Quaternionic_continuity_equation_for_charges.pdf .
If you think, think twice
Hans van Leunen | 10/11/11 | 15:01 PM
Hans, if you've made an equation where the probablities don't sum to one, you've made an equation that can't describe any quantum particle. And its not suprising that the Majorana equation would not
work with quaternions, it normally made with a two component weyl spinor, half as many components as a Dirac particle.
BDOA (not verified) | 10/16/11 | 23:36 PM
Hans van Leunen | 10/17/11 | 03:33 AM
I adapted the article such that it contains an equation of (free) motion for each elementary particle of the standard model. With each equation of motion comes an extra equation that enables the
computation of the coupling factor (=rest mass) of the particle. If this is true, then there is no need for a Higgs field.
If you think, think twice
Hans van Leunen | 10/22/11 | 14:37 PM
There is gap between, "Proton Collision ->Decay to Muons and Muon Neutrinos ->Tau Neutrino ->[gap] tau lepton may travel some tens of microns before decaying back into neutrino and charged tracks."
Use the case of Relativistic Muons?
Plato (not verified) | 10/12/11 | 09:41 AM
Superluminal neutrinos is correct.
Is an effect of the existence of 3 basic lengths at the quantum level.
The neutrino breaking the CPT violation, with a nonzero angle theta13, is bounded breaks Lorentz invariance at the quantum level as they move along roads in extra dimensions whose length is less than
that corresponding to particles that do not violate CPT
number26 (not verified) | 10/13/11 | 01:20 AM
Wouldn't it be cool if people going into extra dimensions and New Physics at the tip of a hat learned to correctly use punctuation marks, spelling, and the basic grammatical structure of English?
Jesus Christ.
Anonymous (not verified) | 10/16/11 | 12:35 PM
Anon, if you refer to the comments above, I think your troll contribution is unwelcome here. Please remind yourself to look at a world map and figure out that English is not the only one, nor the
most common language in the world. People may need to talk in English here and elsewhere to communicate, but that makes them no inferior to you. How many languages do you master (with correct
punctuation) ? Should we continue it in English, o ci mettiamo a parlare in italiano ? H mhpws sta Ellhnika ?
Get down that pedestal, you don't look any better there.
Tommaso Dorigo | 10/16/11 | 14:05 PM
Appreciated, Tommaso. You should force the visitors to communicate in Czech. By the way, I liked your visit to Malta:
Luboš Motl | 10/17/11 | 00:49 AM
Tommaso Dorigo | 10/17/11 | 03:32 AM
Luboš Motl | 10/17/11 | 05:47 AM
"Only in some reference frames moving relative to the one in which the neutrino's superspeed™ is being measured"
Because movement is relative, does this not mean that every frame of reference that is not the F of R of the neurino could be considered as moving relative to that of the neutrino.
BTW, Tommaso, I applaud your comments about the use of English. I post on a forum or two and struggle with the science, I dread to think what it must be like trying to do that in another language. I
have great admiration for those brave people who do.
Bill S. (not verified) | 10/16/11 | 15:27 PM
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Roman numerals
1. Roman numerals are defined as combinations of the letters I, V, X, L, C, D and M which are used in various orders to stand for a specific number.
An example of a Roman numeral is IX which stands for the number 9.
Roman numerals for the number eight.
Roman numerals
the Roman letters used as numerals until the 10th cent. : in Roman numerals I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, and M = 1,000 Other numbers are formed from these by adding or subtracting:
the value of a symbol following another of the same or greater value is added (e.g., III = 3, XV = 15); the value of a symbol preceding one of greater value is subtracted (e.g., IX = 9); and the
value of a symbol standing between two of greater value is subtracted from that of the second, the remainder being added to that of the first (e.g., XIX = 19): Roman numerals are commonly written in
capitals, though they may be written in lowercase letters, as in numbering subdivisions (e.g., Act IV, scene iii) A bar over a letter indicates multiplication by 1,000 (e.g., = 5,000)
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Statistics Division, Linköping University
Next research seminar is on Tuesday, May 22, 3.15 pm
Topic: Bayesian inference in Structural Second-Price Auctions With Both Private-value and Common-value Bidders
Presenter: Bertil Wegmann, Statistics, Linköping University.
Abstract: Auctions with asymmetric bidders have been actively studied in recent years. Tan and Xing (2011) show the existence of monotone pure-strategy equilibrium in auctions with both private-value
and common-value bidders. The equilibrium bid function is given as the solution to an ordinary differential equation (ODE). We approximate the ODE and obtain a very accurate, approximate inverse bid
as an explicit function of a given bid. We propose a model where the valuations of both common-value and private-value bidders are functions of covariates. The probability of being a common-value
bidder is modeled by a logistic regression model with Bayesian variable selection. The model is estimated on a dataset of eBay coin auctions. We analyze the model using Bayesian methods implemented
via a Metropolis-within-Gibbs algorithm.
Location: Alan Turing.
for info about past and future seminars.
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Frank Thorne
University of South Carolina
Department of Mathematics
400-G LeConte College
1523 Greene Street
Columbia, SC 29201
thorne [at] math [dot] sc [dot] edu
Math 788, The Geometry of Numbers, Spring 2014.
Math 580, Elementary Number Theory, Fall 2013.
Math 142, Calculus II, Fall 2013.
Math 788p, Topics in Algebraic Number Theory, Spring 2013.
Reading seminars: Complex analysis, Mondays at 10:00 am; Algebraic geometry, Wednesdays at 9:30 am.
Math 531, Foundations of Geometry, Fall 2012.
Math 141, Calculus I, Fall 2012.
Math 574, Discrete Math, Spring 2012.
Math 141, Calculus I, Fall 2011.
Math 782, Analytic Number Theory, Fall 2011.
I work in number theory. Some of my particular interests:
• Classical analytic number theory. My first task in graduate school was to read all of Davenport. I am interested in sieve methods, L-functions, and related topics, and their classical
applications -- as well as applications beyond their traditional settings.
• The distribution of number fields, class group torsion, and related topics. I am especially interested in these from an analytic point of view involving Shintani zeta functions associated to
prehomogeneous vector spaces. I am also interested in other perspectives on these questions: class field theory and Kummer theory; the Scholz reflection principle; algebro-geometric perspectives
on these topics; and other related perspectives.
Graduate Students: Richard Oh and Daniel Kamenetsky.
My talk at the Joint Meetings: 1 + 2 + 3 + 4 + ...
Dartmouth: 1 + 2 + 3 + 4 + ...
Seminars at the University of South Carolina:
Number Theory Seminar, Spring 2013
South Carolina Algebraic Geometry, Commutative Algebra, Number Theory, Fall 2013
Publications and Preprints:
1. Bounded gaps between products of primes with applications to elliptic curves and ideal class groups.
International Mathematics Research Notices (2008), 41 pp.
2. Irregularities in the distributions of primes in function fields.
Journal of Number Theory 128 (2008), 1784-1794.
3. Bubbles of congruent primes.
Mathematical Proceedings of the Cambridge Philosophical Society, to appear.
4. An uncertainty principle for function fields.
Journal of Number Theory 131 (2011), 1363-1389.
5. Maier matrices beyond Z.
Proceedings of the Integers Conference 2007.
6. Analytic properties of Shintani zeta functions.
Proceedings of the RIMS Symposium on automorphic forms, automorphic representations, and related topics (Kyoto, 2010).
7. The secondary term in the counting function for cubic fields, with Takashi Taniguchi.
Duke Mathematical Journal, to appear.
8. Shintani's zeta function is not a finite sum of Euler products.
Proceedings of the American Mathematical Society, to appear.
9. Orbital L-functions for the space of binary cubic forms, with Takashi Taniguchi.
Canadian Journal of Mathematics, to appear.
10. Four perspectives on secondary terms in the Davenport-Heilbronn theorems.
Integers Volume 12B, Proceedings of the Integers Conference 2011.
11. An error estimate for counting S_3-sextic number fields, with Takashi Taniguchi.
International Journal of Number Theory, to appear.
12. Book review of Opera de Cribro and An introduction to sieve methods and their applications.
Bulletin of the American Mathematical Society, to appear.
13. On the existence of large degree Galois representations for fields of small discriminant, with Jeremy Rouse.
14. Dirichlet series associated to cubic fields with given quadratic resolvent,with Henri Cohen.
Michigan Math Journal, to appear.
15. Dirichlet series associated to quartic fields with given resolvent, with Henri Cohen.
16. Zeros of L-functions outside the critical strip, with Andrew Booker.
Several more in preparation.
Upcoming/Recent Professional Activities:
Palmetto Number Theory Series, Clemson, December 7-8, 2013.
Pro p-groups and arithmetic, Besançon, December 16-20, 2013.
Automorphic Forms and Arithmetic, Gottingen, February 10-14, 2014.
University of Bonn, March 4-14, 2014.
Zürich Number Theory Days, March 7-8, 2014.
Arizona Winter School, Tucson, March 15-19, 2014.
Southeast Regional Meeting on Numbers, Wofford College, April 19-20, 2014.
Ursinus College, early May.
(Kobe and Kyoto? TBD)
Canadian Number Theory Association XIII (maybe), June 16-20, 2014.
Counting Arithmetic Objects, Montreal, June 23-July 4, 2014.
Richard Oh's thesis defense, July 7-9, 2014. (Exact date TBD)
Emerging Leaders and Evolving Frontiers in Analytic Number Theory, Bonn, July 14-18, 2014.
Gangnam Style, August 13-21, 2014.
Emory University, August 27, 2014.
Workshop on Counting Arithmetic Objects, Montreal, November 10-14, 2014.
Analysis, Spectra, and Number Theory, Princeton, December 15-19, 2014.
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Reviews for Maths | FictionPress
Reviews for Maths
chapter 1 . 1/29/2004
Hehe. I personally hate maths, despise it with every fibre of my being, but my friend hate this poem. Math geek.
Ame Kenshin chapter 1 . 1/29/2004
Is this what happens in your math classes or something? It was a good poem.
deadly-angel-53 chapter 1 . 1/29/2004
Yep Maths is the Worst subject ever. Although in my class we all muck about our teacher can't control us and is just a S.O.B. Good poem.
chapter 1 . 1/29/2004
ok, really dont mean to write flames, but this is crap. really, just crap.
Infinity Plus One chapter 1 . 1/29/2004
I love maths, it's one of my favourite subjects and I worked out 26 factorial without a calculator (because I use my head instead lol) and I got around 43 trillion trillion.
Have a nice day.
Hope maths gets better and that your teacher gets fired.
Bye bye.
Florence says "smeg off".
Florence says "I didn't say that Olympia."
Olympia decides that she is still a bit bonkers.
CrimsonSoulTears chapter 1 . 1/29/2004
You're RIGHT! Maths is the worst lesson in the world. What sad loner invented all those equations?
Except last maths lesson I had I understood what we had to do which was amazing.
And to think that they put me in the top group orginaly (I failed a test on purpose - very badly. The teachers had a go at me, but I go moved down which was the aim of the entire test failing)
Seriously, some of the stuff we have to learn... why? How is it going to benefit our lives? Basic maths fairplay, but Trigonometry? Why do I need to know how to find an accurate length or angle or
whatever it is (I don't understand it) of a right angled triangle? Can't I just measure it?
Yes, well, I've ranted for long enough now.
I concur!
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Teaching with R: the switch
October 21, 2011
By Luis
There are several blog posts, websites (and even books) explaining the transition from using another statistical system (e.g. SAS, SPSS, Stata, etc) to relying on R. Most of that material treats the
topic from the point of view of i- an individual user and ii- a researcher. This post explains some of the issues involved in, first, moving several users and, second, with an emphasis in teaching.
I have made part of this information available before, but I wanted to update it and keep it together with all the other posts in Quantum Forest. The process started in March 2009.
March 2009
I started explaining to colleagues my position on using R (and R commander) for teaching purposes. Some background first: forestry deals with variability and variability is the province of
statistics. The use of statistics permeates forestry: we use sampling for inventory purposes, we use all sort of complex linear and non-linear regression models to predict growth, linear mixed models
are the bread and butter of the analysis of experiments, etc.
I think it is fair to expect foresters to be at least acquainted with basic statistical tools, and we have two courses covering ANOVA and regression. In addition, we are supposed to introduce/
reinforce statistical concepts in several other courses. So far so good, until we reached the issue of software.
During the first year of study, it is common to use MS Excel. I am not a big fan of Excel, but I can tolerate its use: people do not require much training to (ab)use it and it has a role to introduce
students to some of the ’serious/useful’ functions of a computer; that is, beyond gaming. However, one can hit Excel limits fairly quickly which–together with the lack of audit trail for the analyses
and the need to repeat all the pointing and clicking every time we need an analysis–makes looking for more robust tools very important.
Until the end of 2009 SAS (mostly BASE and STAT, with some sprinkles of GRAPH) was our robust tool. SAS was introduced in second year during the ANOVA and regression courses. SAS is a fine product,
• We spent a very long time explaining how to write simple SAS scripts. Students forgot the syntax very quickly.
• SAS’s graphical capabilities are fairly ordinary and not at all conducive to exploratory data analysis.
• SAS is extremely expensive.
• SAS tends to define the subject; I mean, it adopts new techniques very slowly, so there is the tendency to do only what SAS can do. This is unimportant for undergrads, but it is relevant for
• Users sometimes store data in SAS’s own format, which introduces another source of lock-in.
At the time, in my research work I used mostly ASReml (for specialized genetic analyses) and R (for general work); since thenI have moved towards using asreml-R (an R library that interfaces ASReml)
to have a consistent work environment. For teaching I was using SAS to be consistent with second-year material.
Considering the previously mentioned barriers for students I started playing with R-commander (Rcmdr), a cross-platform GUI for R created by John Fox (the writer of some very nice statistics books,
by the way. As I see it:
• R in command mode is not more difficult (but not simpler either) for students than SAS. I think that SAS is more consistent and they have worked hard at keeping a very similar structure between
• We can get R-commander to start working right away with simple(r) methods, while maintaining the possibility of moving to more complex methods later by typing commands or programming.
• It is free, so our students can load it into their laptops and keep on using it when they are gone. This is particularly true with international students: many of them will never see SAS again in
their home countries.
• It allows an easy path to data exploration (pre-requisite for building decent models) and high quality graphs.
• R is open source (nice, but not a deal breaker for me) and easily extensible (this one is really important for me).
At the time I thought that R would be an excellent fit for teaching; nevertheless, there could be a few drawbacks, mostly when dealing with postgrads:
• There are restrictions to the size of datasets (they have to fit in memory), although there are ways to deal with some of the restrictions. On the other hand, I have hit the limits of PROC GLM
and PROC MIXED before and that is where ASReml shines. In two years this has never been a problem.
• Some people have an investment in SAS and may not like the idea of using a different software. This was a problem the first few months.
As someone put it many years ago–there is always resistance to change:
It must be remembered that there is nothing more difficult to plan, more doubtful of success, nor more dangerous to manage, than the creation of a new system. For the initiator has the enmity of
all who would profit by the preservation of the old institutions and merely lukewarm defenders in those who would gain by the new ones.—Niccolò Machiavelli, The Prince, Chapter 6
Five months later: August 2009
At the department level, I had to spend substantial time compiling information to prove that R could satisfy my colleagues’ statistical needs. Good selling points were nlme/lme4, lattice/ggplot2 and
pointing my most statistically inclined colleagues to CRAN. Another important issue was the ability to have a GUI (Rcmdr) that could be adapted to our specific needs. At that time the School of
Forestry adopted R as the default software for teaching any statistical content during the four years of the curriculum.
At the university level, my questions to the department of Mathematics and Statistics sparkled a lot of internal discussion, which resulted in R being adopted as the standard software for the ANOVA
and regression second year courses (it was already the standard for many courses in 3rd and 4th year). The decision was not unanimous, particularly because for statisticians SAS is one of those ‘must
be in the CV’ skills, but they went for change. The second year courses are offered across colleges, which makes the change very far reaching. These changes implied that many computers in the
university labs now come with R pre-installed.
A year later: April 2010
R and R-commander were installed in our computer labs and we started using them in our Research Methods course. It is still too early to see what will be the effect of R versus SAS, but we expect to
see an increase on the application of statistics within our curriculum.
One thing that I did not properly consider in the process were the annoying side-effects of the university’s computer policies. Students are not allowed to install software in the university
computers and R packages fall within that category. We can either stay with the defaults + R commander (our current position) or introduce an additional complication for students, pushing them to
define their own library location. I’d rather teach ggplot2 than lattice, but ggplot2 is an extra installation. Choices, choices… On the positive side, the default installation for some of the
computer labs install all the packages by default.
Two years later: March 2011
Comments after teaching a regression modeling course using R-commander:
• Some students really appreciate the possibility of using R-commander as their ‘total analysis system’. Most students that have never used a command line environment prefer it.
• Students that have some experience with command-line work do not like much R-commander as they find it confusing, particularly when it is possible to access the R console through two points:
Rcmdr and the default console. Some of them could not see the point of using an environment with a limited subset of functionality.
• Data transformation facilities in R-commander are somewhat limited to the simplest cases.
• Why is that the linear regression item does not accept categorical predictors? That works under ‘linear models’, but it is such an arbitrary separation.
• The OS X version of R-commander (under X Windows) is butt ugly. This is not John Fox’s fault, but just a fact of life.
In general, R would benefit of having a first-class Excel import system that worked across platforms. Yes, I know that some people say that researchers should not use Excel; however, there is a
distinction between normative and positive approaches to research. People do use Excel and insisting that they should not is not helpful.
I would love to hear anyone else’s experiences teaching basic statistics with R. Any comments?
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Even Perfect numbers $n$ with $n+1$ prime
up vote 3 down vote favorite
The set $S$ of even perfect numbers $n$ such that $n+1$ is a prime number contains $$ 6,28,33550336,137438691328 $$
Latter number found by Joerg Arndt, corresponds to $M_{19}$ (mersenne)
Question: Is $S$ reduced to these $4$ numbers.
New: Joerg Arndt checked up to exponent $110503$ that the corresponding number $n+1$ is composite. (Improved $19$ to $110503$).
Which function of $x$ migh describe well the size of the set of elements in $S$ less than $x$
by the size of the set of all even perfect numbers less than $x$; mainly with big $x.$
So, I am asking for relative size not absolute size. E.g., if I were asking for relative density of the prime numbers congruent to $3$ modulo $4$: I do not want to use the big machinery of the prime
number theorem, or Dirichlet's Theorem to deduce how many should be there. I just want (in these case) to know how to describe in terms of $x$
number of primes congruent to $3$ modulo $4$ and less than $x$
divided by
number of primes less than $x$
How many such numbers $n$ we may expect inside the known 47 perfect numbers ?
1 One reason not to consider the- 1 version is that almost all such numbers are multiples of 3. It might be prudent to check a few more congruences and see what that says about the Mersenne exponent
n before going much farther. Gerhard "Ask Me About System Design" Paseman, 2011.04.23 – Gerhard Paseman Apr 24 '11 at 3:43
If the Mersenne exponent is 7 mod 10; your number is composite. This takes out about a third of the known perfect numbers. 2 mod 3 takes out about another third. There are other congruences on the
exponent that show some of your numbers aren't prime; you can use them to get an upper bound or even a density estimate. Gerhard "Ask Me About System Design" Paseman, 2011.04.25 – Gerhard Paseman
Apr 25 '11 at 7:20
1 Simulation suggests the next candidate corresponds to Mersenne exponent 61, with at most five more candidates among the known perfect numbers after that. Gerhard "Ask Me About System Design"
Paseman, 2011.04.25 – Gerhard Paseman Apr 25 '11 at 8:04
Based on Tapio Rajala's statements to a parallel (not quite duplicate) question, I now guess that there are four prime candidates among the 47. Gerhard "12 Guesses For 10 Cents" Paseman,
2011.05.07 – Gerhard Paseman May 7 '11 at 17:32
add comment
4 Answers
active oldest votes
There's a conjecture (for which I can't find a source now) that the number of Mersenne primes $2^n-1$ with $n < x$ is $c \log x$ for some constant $c$. Differentiating this, the
"probability" that $2^n-1$ is prime is about $c/n$. (This is unconditional; that is, I'm not assuming $n$ is prime.)
The even perfect numbers are exactly of the form $2^{n-1}(2^n-1)$ with $2^n-1$ prime.
up vote 10 So the "probability" that $2^n-1$ and $2^{n-1} (2^n-1) + 1$ is prime, assuming independence, is $c/n$ times the probability that $2^{n-1} (2^n-1) + 1$ is prime. $2^{n-1} (2^n-1) + 1$ is
down vote roughly $2^{2n}$, so by the prime number theorem its "probability" of being prime is about $1/log(2^{2n})$, or again a constant divided by $n$. That is, the "probability" that $2^n-1$
accepted is prime and the corresponding number is one less than a prime is $c/n^2$; since $\sum_{n \ge 1} cn^{-2}$ is finite this leads us to suspect that there are finitely many solutions.
Of course none of this is anywhere near being a proof...
1 Based on these heuristics. How many such numbers $n$ we may expect inside the known 47 perfect numbers ? – Luis H Gallardo Apr 25 '11 at 2:23
add comment
For dealing with large potential primes a good choice is openpfgw
Using openpfgw I finished the list to $1 3466 917$ in about 20 minutes without finding new primes.
[S:[added] The only prime perfect + 1 candidate from the known Mersenne primes is for $M_{20996011}$ - I am running ECM factoring on it.:S]
[later] François Brunault found that $M_{20996011}$ is divisible by $1552147$ which settles the question for the known perfect numbers.
Here is the log:
./pfgw64 -f10 -lmer1log.txt /tmp/mer.txt
2^0*(2^1-1)+1 is trivially prime!: 2
2^1*(2^2-1)+1 is trivially prime!: 7
2^2*(2^3-1)+1 is trivially prime!: 29
2^4*(2^5-1)+1 trivially factors as: 7*71
2^6*(2^7-1)+1 trivially factors as: 11*739
2^12*(2^13-1)+1 is trivially prime!: 33550337
2^16*(2^17-1)+1 has factors: 7
2^18*(2^19-1)+1 is 3-PRP! (0.0000s+0.0009s)
2^30*(2^31-1)+1 has factors: 29
2^60*(2^61-1)+1 is composite: RES64: [36E090A8C361AD6C] (0.0000s+0.0003s)
2^88*(2^89-1)+1 has factors: 7
2^106*(2^107-1)+1 has factors: 7
2^126*(2^127-1)+1 has factors: 11
2^520*(2^521-1)+1 has factors: 7
2^606*(2^607-1)+1 has factors: 11
2^1278*(2^1279-1)+1 is composite: RES64: [570A6B3FD91E6339] (0.8700s+0.0011s)
2^2202*(2^2203-1)+1 is composite: RES64: [ECB4FE924C674723] (4.6906s+0.0010s)
up vote 4 down vote 2^2280*(2^2281-1)+1 has factors: 197
2^3216*(2^3217-1)+1 has factors: 11
2^4252*(2^4253-1)+1 has factors: 7
2^4422*(2^4423-1)+1 is composite: RES64: [F3603EEF4BD4F197] (17.0237s+0.0031s)
2^9688*(2^9689-1)+1 has factors: 7
2^9940*(2^9941-1)+1 has factors: 7
2^11212*(2^11213-1)+1 has factors: 7
2^19936*(2^19937-1)+1 has factors: 7
2^21700*(2^21701-1)+1 has factors: 7
2^23208*(2^23209-1)+1 has factors: 35603
2^44496*(2^44497-1)+1 has factors: 11
2^86242*(2^86243-1)+1 has factors: 7
2^110502*(2^110503-1)+1 has factors: 491
2^132048*(2^132049-1)+1 is composite: RES64: [1B3B60AEC3578817] (744.2790s+111.7145s)
2^216090*(2^216091-1)+1 has factors: 4673
2^756838*(2^756839-1)+1 has factors: 7
2^859432*(2^859433-1)+1 has factors: 7
2^1257786*(2^1257787-1)+1 has factors: 11
2^1398268*(2^1398269-1)+1 has factors: 7
2^2976220*(2^2976221-1)+1 has factors: 7
2^3021376*(2^3021377-1)+1 has factors: 7
2^6972592*(2^6972593-1)+1 has factors: 7
2^13466916*(2^13466917-1)+1 has factors: 11
The rest Mersenne primes lead to small factors:
Format ($p$,factor) (24036583,149),(25964951,7),( 30402457,11),( 32582657,7),( 37156667,7),( 42643801,3593),( 43112609,7)
The only remaining candidate seems to be divisible by p=1552147 (computed used fast exponentiation mod p). – François Brunault Nov 9 '11 at 16:08
@François Indeed, will update the answer. – joro Nov 9 '11 at 16:17
It's good to confirm Tapio Rajala's findings. Gerhard "Ask Me About System Design" Paseman, 2011.11.09 – Gerhard Paseman Nov 9 '11 at 16:37
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The numbers involved are pretty huge - have you tried all the Mersenne primes' perfect numbers yet?
The other answer might be referring to Wagstaff's conjecture about the number of these primes being less than $e^{\gamma}/\log(2) *\log(\log(x))$; see e.g. here, here, or here for
up vote 3 down some references (some better than others).
I would imagine that this would be helpful in solving this, but gives a sense of just how hard it would be to prove anything.
Based on these heuristics. How many such numbers $n$ we may expect inside the known 47 perfect numbers ? – Luis H Gallardo Apr 25 '11 at 2:24
add comment
I used an awk program to generate congruences on n such that, if the Mersenne exponent n satisfied such a congruence, then the corresponding candidate had a small prime factor, which
usually was smaller than the candidate prime. Using moduli up to 4800, I found that the candidate corresponding to 216091 exponent was a multiple of 4673, most of the remaining candidates
were divisible by smaller primes. At this writing, 61, 1279, 23209, and 20996011, are the exponents whose corresponding candidates may be prime, if I didn't foul up the coding. So my guess
up vote 1 is: at most eight.
down vote
Gerhard "Ask Me About System Design" Paseman, 2011.04.25
Can I change my guess to at most 5? Gerhard "Ask Me About System Design" Paseman, 2011.04.25 – Gerhard Paseman Apr 25 '11 at 8:49
Well, the coding might have been good, but the filtering wasn't: some composite moduli slipped in. The latest version (morally just trial factorization) gives pairs $(m,f)$ with $f$ a
factor of candidate using exponent $m$: (20996011,1552147), (110503,491), (216091,4673), (23209,35603), (42643801,3593),(4423,2163571) (2203,60449). Most of the rest of the candidates
have small factors or are already seen to be prime. Still to crack are 61, 1279, and 132049. Gerhard "Ask Me About System Design" Paseman, 2011.05.03 – Gerhard Paseman May 3 '11 at 18:42
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Mathematics and Statistics Models
This resource received an Accept or Accept with minor revisions rating from a Panel Peer Review process
These materials were reviewed using face-to-face NSF-style review panel of geoscience and geoscience education experts to review groups of resources addressing a single theme. Panelists wrote reviews
that addressed the criteria:
1. scientific accuracy and currency
2. usability and
3. pedagogical effectiveness
Reviewers rated the resources:
1. Accept
2. Accept with minor revisions
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4. Reject.
They also singled out those resources they considered particularly exemplary, which are given a gold star rating.
Following the panel meetings, the conveners wrote summaries of the panel discussion for each resource; these were transmitted to the creator, along with anonymous versions of the reviews. Relatively
few resources were accepted as is. In most cases, the majority of the resources were either designated as 1) Reject or 2) Accept with major revisions. Resources were most often rejected for their
lack of completeness to be used in a classroom or they contained scientific inaccuracies.
This page first made public: Dec 21, 2006
This material was originally created for
Starting Point:Introductory Geology
and is replicated here as part of the
SERC Pedagogic Service.
What are Mathematical and Statistical Models
These types of models are obviously related, but there are also real differences between them.
Mathematical Models:
grow out of equations that determine how a system changes from one state to the next (differential equations) and/or how one variable depends on the value or state of other variables (state
equations) These can also be divided into either numerical models or analytical models.
Statistical Models:
include issues such as statistical characterization of numerical data, estimating the probabilistic future behavior of a system based on past behavior, extrapolation or interpolation of data
based on some best-fit, error estimates of observations, or spectral analysis of data or model generated output.
As a way to clarify the above ideas, here is an example of the development of a simple mathematical model.
Why use mathematical and statistical models to teach introductory courses?
Mathematical and Statistical models can be used to help students obtain a better grasp on a variety of topics.
Learn more about the benefits of using mathematical and statistical models
How can these models be used effectively in class?
In addition to the general discussion about how to use models effectively, there are a number of considerations, both pedagogical and technical, that have to do with using mathematical and
statistical models specifically.
More about How to Use Mathematical and Statistical Models
Issues Related to Teaching Quantitative Skills: This page is part of the Teaching Quantitative Skills in the Geosciences website and includes a discussion of the relevance of mathematics and
statistics in geoscience education.
Zachary, D. (1989). Teaching, learning, and using mathematical models in physics. Physics Education. 24(6), 339-343.
Edelstein-Keshet, K. (1988). Mathematical Models in Biology. McGraw-Hill. ISBN: 0075549506
Building the Quantitative Skills of Students in Geoscience Courses: This special issue (v48 n4, edited by Macdonald, Srogi, and Stracher) of the Journal of Geoscience Education has several examples
of how mathematics and statistics can be used in geoscience courses.
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How do I calculate g of solute in a given volume of solution using the percent concentration?
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How do I calculate g of solute in a given volume of solution using the percent concentration?
If I have a solution that is 5%(w/v) of an alcohol in water, what mass of alcohol is present in 1.0 L of this solution?
Kathleen 10/08/99
The percentage tells you that there are 5 grams of alcohol in every 100 mL of solution. You have 1.0 L, which is 10 100 mL portions, so you have 50 grams of alcohol.
You can also treat this as a simple unit conversion problem and solve it with the five step plan for converting units:
1. Identify the unknown, including units. You want to find grams of alcohol.
2. Choose a starting point. You have two pieces of information: the volume of the solution, and its concentration. "g alcohol" is not a fraction, so your starting point shouldn't be a fraction
either. 5% (w/v) alcohol means 5 g alcohol per 100 mL of solution- it should be used as a conversion factor but not as a starting point. Your strategy is
3. List the connecting conversion factors. Conversion factor are stepping stones that connect the starting point and the unknown. The concentration connects a volume of solution with a mass of
1.0 L solution mL solution 100 mL solution = 5 g alcohol g alcohol
4. Multiply the starting measurement by conversion factors.
1.0 L solution ( 1000 mL solution1 L solution ) ( 5 g alcohol100 mL solution ) = 50 g alcohol
5. Check the result: does the answer make sense? Don't forget this step!
Author: Fred Senese senese@antoine.frostburg.edu
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[Tutor] a question about passing values between functions
kristinn didriksson kdidriksson at yahoo.com
Fri Nov 17 19:05:02 CET 2006
I am still wrestling with the concept of values going
between functions. (just starting out)The program
below works seems to work, but here is my question.
In my understanding, return area in the first routine
makes the value of area an instance of areaCirc and I
use areaCirc in the other program to call the value
for the area that was calculated in the first routine.
Is that the right way to think of this program? Did I
use return area correctly?
This is a new vocabulary, so I hope I am making sense.
Basically, I am wondering if I needed to use the
return area statement.
# this program is a redo of ex2 ch3 with a
2 # use two functions--one to compute the area
of a pizza, and one to
3 # to compute cost per square inch.
4 # Given are the diameter and the price. A =
6 # define the function that computes the area
7 import math
8 def areaCirc():
9 diameter = input("Please input the
diameter of the pizza: ")
10 area = 4*(math.pi)*(diameter/2)**2
11 print "The area of the pizza is %0.2f" %
12 return area
15 def unitCost():
16 price = input("Please input the cost of
the pizza per square inch: ")
17 area = areaCirc()
18 cost = area * price
19 print "The cost of the pizza is %0.2f"
21 unitCost()
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An Adventure in the
An Adventure in the Nth Dimension
On the mystery of a ball that fills a box, but vanishes in the vastness of higher dimensions
The Master Formula
An n-ball of radius 1 (a “unit ball”) will just fit inside an n-cube with sides of length 2. The surface of the ball kisses the center of each face of the cube. In this configuration, what fraction
of the cubic volume is filled by the ball?
The question is answered easily in the familiar low-dimensional spaces we are all accustomed to living in. At the bottom of the hierarchy is one-dimensional geometry, which is rather dull: Everything
looks like a line segment. A 1-ball with r=1 and a 1-cube with s=2 are actually the same object—a line segment of length 2. Thus in one dimension the ball completely fills the cube; the volume ratio
is 1.0.
In two dimensions, a 2-ball inside a 2-cube is a disk inscribed in a square, and so this problem can be solved with one of my childhood formulas. With r=1, the area πr^2 is simply π, whereas the area
of the square, s^2, is 4; the ratio of these quantities is about 0.79.
In three dimensions, the ball’s volume is 4/3π, whereas the cube has a volume of 8; this works out to a ratio of approximately 0.52.
On the basis of these three data points, it appears that the ball fills a smaller and smaller fraction of the cube as n increases. There’s a simple, intuitive argument suggesting that the trend will
continue: The regions of the cube that are left vacant by the ball are the corners. Each time n increases by 1, the number of corners doubles, so we can expect ever more volume to migrate into the
nooks and crannies near the cube’s vertices.
To go beyond this appealing but nonquantitative principle, I would have to calculate the volume of n-balls and n-cubes for values of n greater than 3. The calculation is easy for the cube. An n-cube
with sides of length s has volume s^n. The cube that encloses a unit ball has s=2, so the volume is 2^n.
But what about the n-ball? As I have already noted, my early education failed to equip me with the necessary formula, and so I turned to the Web. What a marvel it is! (And it gets better all the
time.) In two or three clicks I had before me a Wikipedia page titled “Deriving the volume of an n-ball.” Near the top of that page was the formula I sought:
Later in this column I’ll say a few words about where this formula came from, both mathematically and historically, but for now I merely note that the only part of the formula that ventures beyond
routine arithmetic is the gamma function, Γ, which is an elaboration on the idea of a factorial. For positive integers, Γ(n+1)=n!=1×2×3×...×n. But the gamma function, unlike the factorial, is also
defined for numbers other than integers. For example, Γ(½) is equal to the square root of π.
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Mutually Exclusive Events
March 10th 2012, 08:34 AM #1
Oct 2009
Mutually Exclusive Events
Hello people, i need some help
This is the Question:
Check the truth of the following statement: if P(A)=2/3 P(B) , P(B)=3/8 P(C) and P(C) = 2/3, then the events are mutually exclusive per two. To be honest, i cant understand that "per two" phrase.
So , two events in order to be mutual exclusive, A /\ B = 0 , P(A\/B) = P(A) + P(B)
I am not sure how to begin solving this. i noticed also that P(A) + P(B) + P(C) > 1 (13/12).
Thanks in advance ;-)
Re: Mutually Exclusive Events
This is the Question:
Check the truth of the following statement: if P(A)=2/3 P(B) , P(B)=3/8 P(C) and P(C) = 2/3, then the events are mutually exclusive per two. To be honest, i cant understand that "per two" phrase.
So , two events in order to be mutual exclusive, A /\ B = 0 , P(A\/B) = P(A) + P(B)
I am not sure how to begin solving this. i noticed also that P(A) + P(B) + P(C) > 1 (13/12).
You ought to know that:
$1\ge\mathcal{P}(A\cup B\cup C)$.
What if $A\cap B=\emptyset,~A\cap C=\emptyset,~\&~B\cap C=\emptyset~?$
Re: Mutually Exclusive Events
Hey Plato, thanks for your response ;-)
I know that, apparently every probabillity must be <=1
What if $A\cap B=\emptyset,~A\cap C=\emptyset,~\&~B\cap C=\emptyset~?$
Err, that means that the above are mutually exclusive.
But, how am i supposed to prove this nice? What steps should i take ?
Last edited by primeimplicant; March 10th 2012 at 01:50 PM.
Re: Mutually Exclusive Events
This is a basic axiom of probability.
If $\left\{ {{A_n}} \right\}$is any collection of pair-wise disjoint evednts then
$\mathcal{P}\left( {\bigcup\limits_n {{A_n}} } \right) = \sum\limits_n {\mathcal{P}\left( {{A_n}} \right)} \le 1$
Re: Mutually Exclusive Events
Ok cool,
After some thought....
${P}(A\cup B\cup C)$ = P(A)+P(B)+P(C) - P(A/\B) -P(A/\C)-P(B/\C) +P(A/\B/\C)
So, if the events are mutually exclusive, then P(A/\B)=P(A/\C)=P(B/\C)=P(A/\B/\C) $=\emptyset.~$ Is it alright up to this point ?
Then, i conclude that : ${P}(A\cup B\cup C)$ = P(A)+P(B)+P(C) .
BUT P(A)+P(B)+P(C) = 13/12 . So if my calculation is right, P(A)+P(B)+P(C) > 1 implies that ${P}(A\cup B\cup C)$ > 1 . Thats against the basic axiom of probability.
Thus, there is no truth on the given statement.
Its this right, or i am missing something?
Re: Mutually Exclusive Events
March 10th 2012, 08:52 AM #2
March 10th 2012, 01:35 PM #3
Oct 2009
March 10th 2012, 01:44 PM #4
March 10th 2012, 02:10 PM #5
Oct 2009
March 10th 2012, 02:15 PM #6
March 10th 2012, 02:20 PM #7
Oct 2009
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Use the triangle inequality to show...
October 22nd 2009, 09:02 PM #1
Sep 2009
Use the triangle inequality to show...
Use the triangle inequality to show that if a and be are in R, with a not equal to b, then there exists an open interval U centered at a and V centered at b, both of radius epsilon= 1/2 abs[a-b]
, with U/\V= 0 (intersection)
I really don't know what I wanna do here. I'm imagining a line centered at A and another at b with radius epsilon something like this =>
----- (-e)-------a-------(e)------
Use the triangle inequality to show that if a and be are in R, with a not equal to b, then there exists an open interval U centered at a and V centered at b, both of radius epsilon= 1/2 abs[a-b]
, with U/\V= 0 (intersection)
I really don't know what I wanna do here. I'm imagining a line centered at A and another at b with radius epsilon something like this =>
----- (-e)-------a-------(e)------
We may assume wlog that $a<b$. Thus the open interval around $a$ is $U=\left(a-\frac{1}{2}(b-a),a+\frac{1}{2}(b-a)\right)=\left(\frac{1}{2}(3a-b),\frac{1}{2}(a+b)\right)$
The open interval around $b$ is $V=\left(b-\frac{1}{2}(b-a),b+\frac{1}{2}(b-a)\right)=\left(\frac{1}{2}(a+b),\frac{1}{2}(-a+3b)\right)$
From this it's clear that $U$ and $V$ are disjoint, so $U\cap V=\emptyset$.
what about showing the interval for both U and V to be of radius epsilon= 1/2*absolute value[a-b]?
I don't really follow...how come we can assume that a < b ?
and we make this assumption so we can relate U and V?
I made the assumption so I didn't have to deal with the absolute value sign.
ok I follow this, but if I need to use the triangle property how do i go about using it?
A friend told me:
let u= (a-e, a+e) = absolute value(c-a) < e for c in R
let v= (b-e, b+e) = absolute value(d-a) < e for d in R
then absloutevalue(x-a) < e = 1/2absloute(a-b)
absloute value(x-b) < e = 1/2absloute(a-b)
is this correct? and if so....how does he go from abs(c-a) < e to
abs(x-a) < e
and then to = 1/2 abs(a-b)
ok i should have written
{ c in R : abs(c-x) < e }
ok I follow this, but if I need to use the triangle property how do i go about using it?
A friend told me:
let u= (a-e, a+e) = absolute value(c-a) < e for c in R
let v= (b-e, b+e) = absolute value(d-a) < e for d in R
then absloutevalue(x-a) < e = 1/2absloute(a-b)
absloute value(x-b) < e = 1/2absloute(a-b)
is this correct? and if so....how does he go from abs(c-a) < e to
abs(x-a) < e
and then to = 1/2 abs(a-b)
Your friend introduced more variables than are necessary (namely $c$ and $d$). If $U=\{x\in\mathbb{R}:|a-x|<\epsilon\}$ and $V=\{x\in\mathbb{R}:|x-b|<\epsilon\}$, then assume for a contradiction
that $U\cap Veq\emptyset$; that is, $\exists x$ such that $x\in U$ and $x\in V$. Then $|a-x|<\epsilon$ and $|x-b|<\epsilon$.
Then by the triangle inequality $|a-b|\leq|a-x|+|x-b|<\epsilon+\epsilon=2\epsilon=|a-b|$.
So we have $|a-b|<|a-b|$, which is a contradiction. So $U\cap V=\emptyset$.
October 22nd 2009, 09:47 PM #2
October 22nd 2009, 09:51 PM #3
Sep 2009
October 22nd 2009, 09:55 PM #4
October 22nd 2009, 10:01 PM #5
Sep 2009
October 22nd 2009, 10:04 PM #6
October 22nd 2009, 10:10 PM #7
Sep 2009
October 22nd 2009, 10:11 PM #8
October 22nd 2009, 10:32 PM #9
Sep 2009
October 22nd 2009, 10:37 PM #10
October 22nd 2009, 10:42 PM #11
Sep 2009
October 22nd 2009, 11:06 PM #12
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Circuit optimization
10-18-2009, 02:00 AM
Circuit optimization
In my computer science class, I learned about boolean circuit. I know how two one-digit inputs add or subtract and get one output or borrow or carry.But if you add two numbers that contain N
binary digits rather than just one, what do you do? output is the sum. In order to use any boolean algebra identities, like distributive law, I have to first know their general formulas, For
example, a=0,b=1,c=0, sum is 1, the sum of product is then not'a'*b*not'0'+.... etc. But I don't know the case here for N. Thank you.
10-18-2009, 02:14 AM
My memory is dated, and I certainly don't want to give a full tutorial. Briefly, what you have been taught is probably a half adder. This circuit will sum two operands, yielding a sum and a
You can make a full adder with two half adders. This circuit will have a carry input and a carry output that can be cascaded into another full adder.
This creates what is called a ripple adder, that is, the output isn't valid until the carrys all propagate to the last cell. But a look ahead adder can speed up this propagation. This is beyond
the scope of your question
Do a search for half adders and full adders. Wikipedia has a reasonably comprehensive article.
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determined that they will be able to borrow the majority of the money nee...
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Wrong. B is not the Answer. I got it wrong on my Penn Foster Exam. 9/23/2013 * Wrong Answer. @Ms.Sue
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A park ranger wants to shoot a monkey hanging from a branch of a tree with a tranquilizing dart. The ranger aims directly at the monkey, not realizing that the dart will follow a parabolic path and
thus will fall below the monkey. The monkey, however, sees the dart leave the g...
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To combat wasteful habits, we often speak of conserving energy, by which we mean turning off lights, heating or cooling systems, and hot water when not being used. In this chapter, we also speak of
energy conservation. Distinguish between these two usag...
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+ 3
math 7th
Thanks Ms. Sue I see what I forgot to do.
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On the nature of the "infinite" fall toward the EH
Taking it as given that we are not talking about visually seeing but rather calculating through the metric, how do you calculate that the distant static clock falls behind the inertial clock
approaching the horizon???
Within limits, we are talking about seeing. In the case of supermassive BH, conditions on event horizon crossing are not extreme in any way.
Could you explain your statement above regarding time on the static clock at infinity??
DO you think that the geometry that the falling clock is passing through has no effect on the periodicity of this clock??
For a supermassive BH, there is minimal curvature at the horizon.
That it would not be red shifted relative to the distant clo9ck equivalent to a proximate static clock?
That the integrated proper times of the relative clocks would not be related by the metric?
That dt=d[itex]\tau[/itex]/(1-2M/r)^1/2(1-v^2/c^2 would not apply???
In the above, you have two limits competing. Remember, v is relative to an adjacent static observer. For any infaller, v->c as horizon is approached. The limit of the product is always finite, and
for free fall from infinity represents a redshift at the horizon. Inside the horizon, this formula loses
validity because there are no static observers. However, there is a uniform approach to redshift and clock comparison that I have explained several times on this thread. Using the general method
(aside: it is
never necessay
to use gravitational redshift - that is computational convenience for the very special case of static spacetime - which doesn't exist inside the horizon; it also doesn't exist for to co-orbiting
neutron stars), redhshift perceived by an inside horizon observer remains finite, and (for free fall from far away from BH) reshifted up to singularity.
I was under the impression that it was an implicit assumption of valid coordinate systems that relative velocity was symmetric and reciprocal.
Relative velocity at a distance is
in GR. Only relative velocity for nearby observers is defined. Coordinate velocity is not relative velocity. It is a purely arbitrary convention.
That the velocity of the faller relative to the distant observer is the same as the velocity of the distant observer relative to the faller.
Does this not hold in Sc coordinates???
There is no such thing as relative velocity for distant observers in GR, at all. The basic issue is that if you bring one 4-velocity over a distance to another, you get a different result depending
on what path you choose. That is at the core of the definition of curvature. There is no physical basis to choose one path over another. Thus curvature
giving meaning to relative velocity at a distance.
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Help me understand something please
Ad Astra Per Aspera
AndrewSeven wrote:
I'm using "old" algs to orient the last layer (OLL) while I learn the PLL, which is much smaller than 70.
However, if you believe that the problem is more mechanical, then it may be of some advantage to do a bit of sanding on the inner parts of the cube.
Each cubie has the main cube and the extra parts. Just go through and make sure that the extra parts don't have any edges to hang on, and then round out the parts that other cubies will slide along.
Sometimes the tolerances will be off and the piece, no matter how perfect, will be squeezed enough to inhibit motion. And if your cube, after sanding everything else, is still sticking, I would
recommend sanding the sides of the center pieces to allow more tolerance.
After all that sanding, make sure to reapply the silicone.
Those finger tricks were cool. I'll have to start using those when I get my cube to turn smoother. Right now it is hard to turn it. I don't see how using that silicon spray or the KY liquid stuff
will make it turn easier. It seems like it's more of an internal problem. I took the top layer cubes out except for one and turned it around and found that once it hit the orange side it was hard to
turn it. Is 1:30 - 2:00 good for a non-smooth cube and for using that simple method?
http://www.math.leidenuniv.nl/~jnoort/i ... =tutorial6
The first layer sucks, finding edges for the cross, the R'D'RD, and the URUiRiUiFiUF or UiFiUFURUiRi to get the middle layer edges seems like it takes me 2 or 3 minutes but when I'm focused I
probably do that part in 50 secs to 1 minute.
Yeah, I'm planning to stick with the method I'm using to do F2L and learn the last layer stuff and after I'm good with that I guess I'll learn the corner edges thing.
About learning Jasmine's method....am I going to have a hard time switching from mine to hers and then switching to learning those 78 algorithms for the last layer?
http://www.geocities.com/jasmine_ellen/ ... ution.html
http://www.speedcubing.com/final_layer_ ... flips.html
Is it correct, then, that one would need to memorize all 70 moves in order to get quicker on the LL so that they can get away from using the R'D'RD method?
Absolutely not!
The group of 70 algorithms is for a 2 look last layer. For instance, you look at the last layer once, and then orient the corners AND edges with an alg. Then you look at it again, and then permute
the corners AND edges with a second alg.
Many people use a 3 or a 4 look last layer.
The 3 and 4 look last layers use specific algs taken from the list of 70.
For a true 4 look(where you Don't repeat any algs), you would need:
3 that flip edges:
straight line across - F (RU R'U') F'
little "L" - F (RU R'U')(RU R'U') F'
no edges correct - LFR (U2 R')(U2 R)(U2 R'F'L')
7 that twist corners (sune, antisune, double sune, the cross "on heel", 2 to one side, 2 to opposite sides, figure 8)
Sune - R U R' U R U2 R'
Antisune - R U2 R' U' R U' R'
Double Sune - R U R' U R U' R' U R U2 R'
Cross on heel - R U2 R2 U' R2 U' R2 U2 R
2 to one side - F2D'FU2F'DFU2F
2 to opposite - R'F'LFRF'L'F
figure 8 - R'F'L'FRF'LF
4 for permuting edges (M' means the vertical slice in the front goes UP):
switch across -M2 U' M2 U2 M2 U' M2
switch adjacent - R B' R' B F R' B' F R' B R F2 U
3 cycle CW - F2 U M' U2 M U F2
3 cycle CW - F2 U' M' U2 M U' F2
3 for permuting corners
3 cycle CW - R' F R' B2 R F' R' B2 R2
3 cycle CCW - R2 B2 R F R' B2 R F' R
Adjacent swap (I honestly just do 3cycle CCW, turn cube Counter clockwise, 3 cycle CCW)
For a grand total of 17 algs. (well, really 16 the way I did it)
To figure out what you're actually looking for, just do the alg over and over till it solves itself. Whatever it looked like before it was solved is what you're looking for.
TBTTyler Fox
TBTTyler wrote:
4 for permuting edges (M' means the vertical slice in the front goes UP):
switch across -M2 U' M2 U2 M2 U' M2
Um, isn't U2 sufficient?
LOL, Nice catch
I made the mistake of putting the edges before the corners, which I usually do the other way around.
then I guess I should add to the corner perms
Switch Diagonal Corners
M2 U' M2 U2 M2 U' M2 U2
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Spin dependent tunneling spectroscopy in 1.2 nm dielectrics
FIG. 1.
curves before and after high electric field stressing.
FIG. 2.
vs , where is the gate current density prestress and is the gate current density poststress minus . The peak in the curve is caused by a trap assisted tunneling current in the stressed measurement of
Fig. 1.
FIG. 3.
Representative SDT measurement taken with biased to correspond to the peak in the curve of Fig. 2. The measurement was taken with the magnetic field parallel to the Si/dielectric interface normal.
The zero crossing .
FIG. 4.
In this trace the sample is rotated in the magnetic field so that the Si/dielectric interface normal is perpendicular to the magnetic field. Note that the spectrum zero crossing g does not change,
within experimental error, from the g with the interface normal parallel to the magnetic field as shown in Fig. 3.
FIG. 5.
Comparison between the normalized SDT intensities as a function of (a) and the vs (b) plot of Fig. 2. The normalization of (a) is achieved by dividing the spin dependent modification to the tunneling
current by the total dc current (I). The SDT response very closely follows the characteristic trap assisted tunneling peak of (b).
FIG. 6.
SDT spin dependent modification to the tunneling current as a function of . Note that it peaks at about indicating the peak at in the SDT is shifted downward because direct tunneling overwhelms the
trap assisted tunneling process at higher voltages.
FIG. 7.
Energy band diagrams for the sample at three different values of . Note that the only plausible explanation for the tunneling current must involve electron tunneling through defects with levels
corresponding to the range of the silicon band gap. The simplified sketch illustrates two dielectric defect levels, consistent with the experimental result.
FIG. 8.
(a) The SDT response as a function of interface , (b) a crude schematic representation of K center DOS, and (c) a cartoon representation of the charge states of the K centers.
FIG. 9.
Schematic illustration of the DOS for an array of precisely identical defects with precisely identical energy levels.
FIG. 10.
(a) A more physically reasonable DOS in which each of the levels of Fig. 9 is broadened to take into account disorder. (b) The SDT response from the levels of (a). (c) Schematic illustration of the
derivative of the SDT amplitude vs energy response of (b). (d) The absolute value of the derivative (c). The plot illustrated in (d) is, as discussed in the text, an approximation of the defect DOS.
FIG. 11.
SDT signal intensity vs square root of microwave power. Note that the signal intensity does not saturate at the highest power level available in our measurements. This indicates that far higher
sensitivities are possible.
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27.2 Some Special Tricks
Home | 18.013A | Chapter 27 Tools Glossary Index Up Previous Next
27.2 Some Special Tricks
There are functions that are not integrable in general but for which integrals between certain specific endpoints can be evaluated.
These are often integrals that can be rewritten, either by adding a known integral, or by using symmetry, or by some other trick, as an integral over a closed path in the complex plane.
Such integrals can be evaluated by use of the Residue Theorem, which states that the integral of a function f(z) counterclockwise around a simple closed path C is 2
The residue of a function with an isolated singularity is the coefficient of its minus first power at the singular point.
Thus for example,
We know that the function
Here is another way to deduce this fact. We can use partial fractions to write
Thus this function has a singularity at i in the upper half plane and a singularity at -i in the lower half plane, with residues
The value of the integral will therefore be
The integrand will behave like
This tells us that the value of the integral from -R to R on the real line will go to
This gives us an integral we know.
However the same technique applies to much more complicated integrands and allows us to do lots of integrals again from -R to R as R approaches infinity.
We give two examples.
One is the so called Fourier transform of
where C is the semicircle of radius R in the upper half plane, and again the integral on C goes to zero as R increases.
Now we use this method to sum a series. The function cot x is singular at x = 0 and is periodic with period
Moreover, if you wander far off the real line, it quickly approaches either i or -i, since it is
and the second terms in the numerator and denominator will dominate in the upper half plane making the integrand approach -i and the first terms will dominate in the lower half plane so that it
approaches i as |y| increases there.
This implies that an integral of
This further implies that the sum of the residues of this function must go to zero inside this circle. But for each positive or negative integer j the residue of this function at
The residue of this integrand at z = 0 can be computed as half the second derivative of z cot z at z = 0. (We factor out the singular term which is here z^-3 and expand the rest of the integrand in a
Taylor series to get z^-1 coefficient.)
Since sin z goes as
The residue of
You can actually sum the first 128 (or 1024) terms of this sum on a spreadsheet and extrapolate by comparing the sum up to different powers of 2. If you extrapolate first forming S[2](k) = S(2^k)-S(2
^k-1), then S[3](k) = (4S[2](k)-S[2](k-1))/3 then S[4](k) = (8S[3](k)-S[3](k-1))/7, etc. You can get this answer to enormous accuracy numerically and verify this conclusion.
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Investigating Isotopes: Using M&M's as a Model for Calculating Average Atomic Mass
In this chemistry laboratory activity, students will be given a random sample of the fictitious element "M&Mium." This sample contains at least three different "isotopes" of M&Mium (examples include
plain, peanut, almond, peanut butter, etc). The students will design and carry out a procedure to determine the average "atomic" mass of the element M&Mium.
Learning Goals
This activity is designed for students to gain an understanding of the relationship between the mass and relative abundance of an isotope and its effect on the average atomic mass of the element. The
activity will require students to design a procedure and utilize critical thinking skills. Students will also have to work cooperatively with other students, collect data, and analyze data. In order
for students to complete this activity, they need to have a clear understanding of the concept of average atomic mass; average atomic mass is not simply the mathematical average of the masses of each
isotope, but instead depends upon the amount of isotope present.
Vocabulary words used throughout this activity include isotope, percent abundance, relative abundance, and average atomic mass.
Context for Use
This lab activity is appropriate for mainstream or honors high school chemistry students (grades 11-12). The activity can be completed in one class period. Necessary equipment includes an electronic
balance (2 or 3 are ideal with larger classes) and small paper cups. Prior to this activity, students should have an understanding of the atom, its constituent particles, and the masses of each.
Students should also understand the term isotope. Students should have previous experience calculating percents. They should also understand the relationship between percent abundance and relative
abundance. This activity should be used after the concept of calculating average atomic mass has been introduced. This activity would be used early in a chemistry course—typically in the first couple
of units. This activity could be adapted for lower ability students by providing additional procedural steps (rather than having the students develop them on their own).
Subject: Chemistry:General Chemistry:Atomic Structure
Resource Type: Activities:Lab Activity, Classroom Activity
Grade Level: High School (9-12)
Description and Teaching Materials
To set up the activity, place electronic balances around the room. Purchase Dixie cups or other small paper or plastic cups—four cups are required for each pair of students. In a large (gallon sized)
plastic bag, combine 2 large bags of plain M&Ms, 2 large bags of peanut M&Ms, and 1 large bag of almond M&Ms. Mix well. This bag of M&Ms should be enough for approximately 70 students working in
To introduce the activity, review with the students a calculation where average atomic mass is determined from an actual element's isotope data. Discuss with the students how a mass spectrometer is
used to determine isotope data (the number of isotopes, the mass of the isotopes, and the percent abundance of each). Because of the cost and lack of availability of a mass spectrometer, explain to
students that a model will have to be used to study isotopes. Then show the students the bag of M&Ms and explain that each M&M piece will serve as an "atom" and that our sample "element" contains
three common isotopes: plain, peanut, and almond.
As listed in a handout for students, review the goals of the activity: the overall goal is to calculate the average atomic mass of an M&Mium "atom" (like the example calculation students reviewed
earlier in the lesson). In order to complete this calculation, students will have to design a procedure to determine the average mass of each isotope and the percent abundance of each isotope. As the
students determine the method they will use, they should record the steps or procedure, and create a data table to organize their data. Remind students that they are allowed to eat the sample of M&
Mium when they are finished collecting data, so the M&Ms must remain on a paper towel or in a cup, but never on the surface of the lab tables.
After this introduction, students may come up and obtain a small cup of M&Ms and three extra cups and begin designing their procedure.
For teacher reference, a suggested procedure is a follows:
1. Separate the almond, peanut, and plain isotopes.
2. Determine the total mass of each isotope (the mass of all the almond isotopes together, the mass of all the peanut isotopes, etc.) and record.
3. Count the number of "atoms" of each isotope and record.
4. Divide the total mass of each type of isotope by the number of "atoms" of that isotope to determine the average mass of each isotope. Record.
5. Determine the percent of each isotope present in the sample by dividing the number of atoms of an isotope by the total number of M&Ms and multiplying by 100.
6. Express the percent abundance as relative abundance (decimal percent).
7. Calculate the average atomic mass of M&Mium by using the average mass and the relative abundance.
As students are working, it may be necessary to provide hints to students who have trouble determining the procedure. Once the students have completed the calculation, they should record their
average atomic mass on the board for comparison with other groups. In their conclusion, students should address the two following questions: Explain any differences between the atomic mass of your M&
Mium sample and that of your neighbor. Explain why the difference would be smaller if larger samples were used.
Teaching Notes and Tips
A common area of confusion is the final calculation of average atomic mass. Two mistakes are often encountered: 1) students will try to take a mathematical average of the three isotope masses rather
than taking into account the abundance of each isotope; and 2) students have a tendency to use the total mass of the isotope rather than the average mass of each isotope. This will lead to an average
atomic mass that is much too large.
One point that needs to be reinforced and clarified: students have a hard time changing the percent abundance into relative abundance, even though it only requires moving the decimal two places to
the left.
In terms of safety, eating in the lab is usually not permitted; therefore if students are going to eat the M&Ms, they need to be careful not to allow the M&Ms to come into contact with the surfaces
of the lab benches or directly with the pans on the electric balances. If peanut butter or crispy M&Ms are used in place of plain, peanut, or almond, it may be more difficult for students to
distinguish the isotopes by size.
In the past, I have used a similar activity but provided students the actual procedure to be used. The current activity is more inquiry based because it requires the students to design the procedure;
this change results in a higher level of thinking and a deeper understanding of the concepts involved.
Throughout the activity, I would walk around the room to assess students understanding of the activity and provide guidance where necessary. Students may be provided with a handout to complete the
activity and record their data, or they may complete the activity in a laboratory notebook. Students will be expected to record their materials list, procedure, and organize their data into a table.
They should also clearly show how they calculated average atomic mass. As students post their results on the board, it will be apparent which groups may need assistance with their calculations. A
conclusion paragraph should be written that addresses the questions mentioned earlier.
9-12 II.A.3—Properties of an element and its isotopes
References and Resources
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algorithm or suggestion on a problem
Have the following problem (simplified)
In a X by Y 'town', I have to place some 'houses'. The 'houses' can be 1x1, 2x2 and 3x3. There is also a minimum number of 'houses' to be placed from each size (let's say 2 3x3, 12 2x2 and 11 1x1).
I have to find out the maximum number of 'houses' that can be placed in the 'town'.
Problem is, that each house, has to have an 1x1 'power generator' attached to it. One generator can serve as many houses as needed. Only left/right/up/down is considered as attached.
Without the generator part, it looked like a knapsack problem (although not 100% sure)
The most problems I had was with the different size of the houses...
I had a brute force algorithm, but using recursion. Looked good in the beginning, but when trying it out with a decent town size, it gave a nice little stack overflow
Before re-factoring using loops, I wanted to check again if there aren't any alternate solutions.
Any ideas or suggestions are welcomed...
What's the town size?
Genetic algorithms are usually fun to use *and* efficient.
edit: http://watchmaker.uncommons.org/ is good if you do Java
koala wrote:
What's the town size?
Town size is not fixed (from 10x10, but i think that it won't get above 20x20)
The original problem is not a knapsack problem because it has geometry. Instead, it's a packing problem and because your shapes are irregular and variable and you have a constraint on minimum
quantity, I suspect it's a very hard problem to solve analytically. Fortunately, your region may be small enough that you can brute force it or use some other search algorithm instead.
I'm wondering if you can solve this on a small scale (or for a few small sizes) and then just tesselate those solutions into any larger grid.
Do you count houses as discrete units (e.g. a 1x1 and a 2x2 both count as '1') or are you trying to maximize house-area (so a 2x2 would count as 4)? In the former case, you want to place your
minimums and then tile in 1x1 houses to increase your counts. In the latter case (which seems more likely), you want to maximize the number of 3x3 houses because they're the most efficient use of
space (only require one generator for 9 units), and use generators as efficiently as possible.
The minimums for different sizes throws a wrench into that approach, but it might be moot if you have the components to reach those goals easily. Once everything is block-shaped the problem becomes
Using the constraints (X by Y town, a minimum of m3 3x3 houses, a minimum of m2 2x2 houses, a minimum of m1 1x1 houses, each house must have at least one side shared [in part for larger houses] with
a side of a 1x1 power generator), one can establish the minimum and maximum number of power generators.
The maximum number (MAXG) of power generators is X*Y - (m3*9) - (m2*4) - m1.
The required power generators (RG) for the minimum number of houses is ceiling((m3+m2+m1)/4). This leaves (m3+m2_m1) mod 4 potential attachments for 1x1 houses (X1).
Since the maximum number of houses per generator is 4 (so 5 spaces per generator is the optimal), the minimum number (mG) of power generators under the condition of maximizing the number of houses is
RG + ceiling((MAXG - RG - X1)/5)
Since using more larger houses cannot increase the number of houses (e.g., a 2x2 house could be replaced by a 1x1 house and 3 'free' generators [connections for 6 additional houses]), the minimum
number of larger houses should be used.
Since power generators are just wasted space (not counting toward any minimum nor increasing the count of houses), the goal is to minimize the number of power generators (where any space not occupied
by a house is occupied by a generator).
I am not certain if the fact that a corner-positioned generator wastes 2 connections means that the corners should be filled with the 3x3 houses first (then 2x2 houses if m3 3x3 houses have been
placed). Likewise the edges should have the minimum number of 1x1 houses or generators (since an edge-positioned generator can only connect to three houses), but one cannot have both the edge houses
next to the corner house be the same size as (or larger than) the corner house (or the corner house could not connect to a generator). (Any town which is too small to have a 3x3 house in each corner
will necessarily have m3 less than the number of 3x3 houses needed to fill the corners.)
These constraints would seem to significantly reduce the effort for a brute force solution.
Alamout wrote:
I'm wondering if you can solve this on a small scale (or for a few small sizes) and then just tesselate those solutions into any larger grid.
That sounds like the approach I would take. Pre-generate some configurations and use those configurations as the smallest blocks to fill in the grid. Paradoxically, you'd want to place the largest
houses first and fill in the gaps using smaller patterns.
This is just a strange version of Tetris using rotated versions the following shapes:
The rest is just a packing algorithm (or possible a genetic algorithm).
The most efficient pattern for using the least amount of generators, a diamond, is also the worst for using all the spaces. Altering the diamond, the 2nd and 3rd patterns below are my guess for the
most balanced pattern.
Using this, a 3x3 square can have 6 houses at most (so a 3x6 can have 12). This means a 3x7 can have at most 14. The best case is a 2 to 3 ratio for any shape. This specific pattern, however,
requires the height dimension (in this orientation) to be odd to reach the 2x3 ratio.
Any 1x or 2x shape then becomes trivial to make since you've almost no choices for doing it.
So, packing the smallest houses around a 2x2 or a 3x3 means you just your pre-computed patterns to fill in the gaps.
I had a response written up and then Chrome got all weird on me about malware. I'll see if I can remember it.
Since using more larger houses cannot increase the number of houses (e.g., a 2x2 house could be replaced by a 1x1 house and 3 'free' generators [connections for 6 additional houses]), the minimum
number of larger houses should be used.
This assumes you care about number of discrete houses rather than total area of housing (the OP was unclear and is now MIA). If you care more about total housing area, then the larger houses are
better because they minimize the house-to-generator ratio.
The most efficient pattern for using the least amount of generators, a diamond, is also the worst for using all the spaces. Altering the diamond, the 2nd and 3rd patterns below are my guess for the
most balanced pattern.
The diamond is fine if you tesselate it correctly. This pattern is pretty good and tiles forever:
It has a ratio of roughly 1 generator to 3 houses (if you expand it infinitely).
Hello guys,
I was indeed out for a couple of days, because this problem is keep getting nastier
but first, about the posts: (in random order)
@Solomonoff's Secret: my bad, I'm kinda new to finding algorithm patterns in problems, and wasn't really paying much attention
@Jackass JoeJoe:
This is just a strange version of Tetris using rotated versions the following shapes
It's not really that, because the following shapes can fit into a 5x2 area (In Tetris, i guess it would be 7x2, if you can rotate them)
YY YY X X
YYG GYY G G
Do you count houses as discrete units (e.g. a 1x1 and a 2x2 both count as '1') or are you trying to maximize house-area (so a 2x2 would count as 4)?
Will get back to this at the end of the post...
The diamond is fine if you tesselate it correctly
It has a ratio of roughly 1 generator to 3 houses (if you expand it infinitely).
This is my version of the diamond pattern
If the town is large enough, and if you have only 1x1 houses, the ratio could be closer to 1 to 4 houses than the other pattern. (but I might be wrong)
You can also do the following pattern for 2x2 houses, but I can't figure out how to move from one pattern to the other if needed.
Hope that I covered everything...
And regarding the changes...
It seems that the houses will be weighted
I will not include this in the problem, but rather will try to calculate the needed houses, and change the minimum required houses so they will fill the town as much as possible
But because of this, the most efficient distribution of the given houses becomes the most pressing point of the problem. The rest of the space can be just left blank (if i see a large number of
blanks, i will just re-run the algorithm with an increased number of minimum houses)
So I believe that the pattern approach will not give the required result (most efficient distribution trumps time needed to solve the problem)
(most efficient distribution trumps time needed to solve the problem)
Brute force flood fill from (WoLoG) the top left corner of the grid, co-ords [0,0].
base case: assume an infinitely small piece with an 'edge' of { [0,0] }
nth case:
iterate through every possible placement on the edge of the existing layout class[1] this a 'slot'
for each slot iterate through every posssible placement of a piece that would be legal (i.e. not leave a house without a generator, not overlap, not go off the edge)
if grid filled check you have satisfied the constraints and return the 'score' for the solution if so
recurse, you know for the problem space as described you won't blow the stack on anything reasonable
You can do better by early exiting at the 'slot legality' check if you would violate a constraint on the necessary number of homes in certain starting conditions, but in others it is likely not worth
the additional check cost at each slot test compared to end case validation. This is an optimisation, therefore, profile accordingly!
The bits that are glossed over but 'tricky' are:
Iterating over the edge of the existing layout (this is made vastly easier if you have a Persistent Catenable Deque, search based on your programming language of choice for one of these) since all
you have to do is 'cut' the existing edges into three pieces (the middle bit being where your piece shares any edges) and dealing with hole filling in/creation.
Testing for legality (this actually isn't that hard at all, given the sizes involved you could do this with bit masks pretty trivially but quad trees (again helpfully a nice persistent data
structure) are fully general and just as simple to use, checking that a house has a generator is pretty easy with an immutable structure as you only need to test each piece you touch when you add
another, plus the added piece (which is a max of 13 in worst case). Dealing width the boundaries as special nodes in the quad tree may be the easiest way to do them.
Testing for the constraints at the end should be pretty trivial
Scoring at the end should be trivial
It is possible that this will result in no successful solution (but that's okay, obviously this is the case, your result is 'no solution'). You may wish to do a very fast sanity check as to whether
the constraints fall outside the boundaries PAC-humanoid went through since you can do so in constant time.
Very roughly speaking you're talking factorial complexity in n but the requirement to keep a generator connected to any house as you go along is going to prune an awful lot of the tree as you go
along. If someone wants it to go much beyond 20x20 this isn't going to scale of course. but it should be a doddle to code up and is likely to be instructive in going further.
1. this is a list of lists since this algorithm will result in 'holes' which themselves need filling in.
Alamout wrote:
Since using more larger houses cannot increase the number of houses (e.g., a 2x2 house could be replaced by a 1x1 house and 3 'free' generators [connections for 6 additional houses]), the minimum
number of larger houses should be used.
This assumes you care about number of discrete houses rather than total area of housing (the OP was unclear and is now MIA). If you care more about total housing area, then the larger houses are
better because they minimize the house-to-generator ratio.
You are correct. In a later posting the OP indicates houses will be weighted. I assumed "find out the maximum number of 'houses' that can be placed in the 'town'" meant maximum number of houses not
maximum area covered by houses (or some more complex weighting). I think this mostly just made an 'ass' out of 'me'.
loozer wrote:
This is my version of the diamond pattern
If the town is large enough, and if you have only 1x1 houses, the ratio could be closer to 1 to 4 houses than the other pattern. (but I might be wrong)
If you try to expand that pattern (on both axes) I think it won't be as good as it seems--you're using generators less efficiently.
I think this problem is solvable (or at least you can get within some factor of optimal), but if you really just need to hit some minimums then you might as well do something that's faster.
Using a CSP solver might be one way to get a reasonably close answer. It might even solve the whole thing, if you can come up with a clever way to code the constraints. edit: Except I suppose it
won't maximize, it'll just find a way to fill the grid.
The pattern
leads to four 1x1 houses for every generator. I don't think it gets better than that. This is exactly the pattern to use for sugar cane fields in Minecraft.
Hm, right you are--the next row fills in the gaps. I hadn't tried to expand upwards and downwards enough.
Figures the Minecraft people would have figured something like that out.
If only minecraft people could figure out how to join
That's the same pattern.
· X··
··X· ····X
·X··· ··X··
···X· X····
·X·· ··X
bames53 wrote:
That's the same pattern.
I don't see what you're referring to ? If it's my post, then the first example is for the 1x1 houses, and the 2nd is for the 2x2
Also, In your example 'town', either make a definition on which is 1x1 and which is 2x2, or note that there are dots without an attached x (on the edges, repeating 5 by 5)
I don't see what you're referring to ? If it's my post, then the first example is for the 1x1 houses, and the 2nd is for the 2x2
Your second example is just your first example rotated 90deg. And it works for 1x1 or 2x2, or a mix.
loozer wrote:
bames53 wrote:
That's the same pattern.
I don't see what you're referring to ? If it's my post, then the first example is for the 1x1 houses, and the 2nd is for the 2x2
Ah, I see. I took it to be a challenge saying that expanding the pattern of 1x1 houses didn't actually work.
In any case it's similar for 2x2
and 3x3, and NxM actually.
┌─┐│ │
│ │└─┘
┌─┐│ │
│ │└─┘
Also, In your example 'town', either make a definition on which is 1x1 and which is 2x2, or note that there are dots without an attached x (on the edges, repeating 5 by 5)
Yes, that block shows part of an infinitely repeating grid. The pattern doesn't fit neatly into a square (which is why my sugarcane fields are always diamonds). This is the only reason I said earlier
that I didn't think there was a way to get better than this pattern. Rather than make that claim definitively I considered the possibility that some interaction with the edges of a square grid might
provide opportunity for an improvement, though I don't know of any such.
Yes, that block shows part of an infinitely repeating grid.
This is what I've missed, I thought it was only for the specific square...
The actual challenge is mixing the 1x1 houses with the 2x2 houses
I feel that there was more accent placed on defining a good pattern, instead of finding a way to get a solution... so if there are any ideas on how to implement this into a working algorithm
The actual challenge is mixing the 1x1 houses with the 2x2 houses
I'd say they mesh pretty well.
You can fit them in but it results in an inefficient placement of generators; each 2x2 area abuts 4 generators.
I would think that if the purpose is to get as many houses as possible, that means the 1x1 houses are the most "valuable" in terms of "goal". Thus, you'd place the minimum of the rest of the sizes
until you had only 1x1 houses.
With that in mind, a 1x1 generator will always have 4 houses attached - at least until the second half of this logic - in a "pinwheel" configuration. (Showed pretty elegantly above, but will write
again just to be absolutely clear)
(Pinwheel of 3x3:)
You'd place as many of them as possible - probably "flipping" horizontally the pattern - which leaves 2x6 "gaps".
Second "half"
Now, you'd fill those gaps with 2 2x2 and a generator, twice. The two open spaces will take a 1x1 nicely.
Finally, you'd fill the remaining space on the grid with 1x1 "plus" patterned houses staggered so it's two houses side by side each "axis".
I can't think of a more efficient way to pack that. Assuming you had more 2x2 houses than 3x3 leaves gaps, you'd just pack them in a pinwheel with flipped axis too. The resultant gaps would be 2x4
and you could pack 5 houses, 2 generators and one "space" in to that area.
That does leave open spaces. I wonder how it'd fit if they weren't flipped as they were placed. I'll have to look in to it more when I'm not supposed to be working. =)
I would think that if the purpose is to get as many houses as possible, that means the 1x1 houses are the most "valuable" in terms of "goal". Thus, you'd place the minimum of the rest of the sizes
until you had only 1x1 houses.
That's not the goal.
The 3x3 pinwheel actually tiles fairly nicely--if you don't have large minimums on the other house sizes, that might a good way to start:
Perhaps you could just fill as much space as possible with that pattern, and then fill in the edges with 2x2s and 1x1s.
Alamout wrote:
That's not the goal.
loozer wrote:
I have to find out the maximum number of 'houses' that can be placed in the 'town'.
Then what is the goal?
The houses are weighted by size (at least, it appears that he says this in a later post), so 4 1x1 houses is not better than 1 2x2 house.
He never clarified that point, so I can't act upon it. He even goes on to (sort of) contradict it by saying he would "fill it up" with 1x1 houses after the first "pass" with the minumum number of
Alamout wrote:
The 3x3 pinwheel actually tiles fairly nicely--if you don't have large minimums on the other house sizes, that might a good way to start:
Perhaps you could just fill as much space as possible with that pattern, and then fill in the edges with 2x2s and 1x1s.
In general, I would think the algorith would be a loop and the loop would be:
o Find smallest area with at least 2 squares
o Fit biggest pattern in to it
o If biggest pattern can fit in multiple places, favor those that have maximum number of shared edges
Dracorat wrote:
He never clarified that point, so I can't act upon it. He even goes on to (sort of) contradict it by saying he would "fill it up" with 1x1 houses after the first "pass" with the minumum number of
in one of the eariler posts, there was a comment:
It seems that the houses will be weighted
I will not include this in the problem, but rather will try to calculate the needed houses, and change the minimum required houses so they will fill the town as much as possible
But because of this, the most efficient distribution of the given houses becomes the most pressing point of the problem. The rest of the space can be just left blank (if i see a large number of
blanks, i will just re-run the algorithm with an increased number of minimum houses)
so if i have a 10x10 town, there will be no chance that the minimum requirement would be one of each type...
if i say (random example with random numbers) the minimum requirement is 1 3x3 house, 10 2x2 and 11 1x1, and i see that the algorithm placed those in the town, and there are still 13 spaces
remaining, i will change the minimum requirements to let's say 1, 12, 12 and run it again...
ronelson wrote:
I'd say they mesh pretty well.
they mesh pretty well, but the example you gave me is for simply putting the 2x2 houses in the optimal grid for 1x1 houses
if you consider every dot part of a 2x2 house, then your arrangement can be optimized fairly easily...
in your example, the minimum path between generators, is 3 squares (1 up, 2 right, or 2 up, one left) but in an optimal arrangement of 2x2 houses, the minimum path is 5
The part that's still missing is a definition of an "efficient" use of the space if it's not filled up and why you wouldn't want the program to just pack as many as you could otherwise, rather than
having to hand-tune it.
If a definition of what constitutes "efficient space" would definitely help understand the alternate direction.
Dracorat wrote:
The part that's still missing is a definition of an "efficient" use of the space if it's not filled up and why you wouldn't want the program to just pack as many as you could otherwise, rather than
having to hand-tune it.
If a definition of what constitutes "efficient space" would definitely help understand the alternate direction.
There's indeed a problem with the definition of 'efficient space' because of the problem with weighting the houses (which would then make the problem even more complex)
Let's say that the following is a clarification that would make the definition of 'efficient' more obvious:
We can assume that the minimum number of houses is actually the desired number of houses. Meaning if the required houses are 3 3x3, 12 2x2 and 10 1x1, the algorithm should only try to place those
houses in the town.
This way the user will be able to juggle with the type of houses, and find the optimally weighted solution by it's own (more trouble for him, less for the algorithm)
So the main focus is not how to put 2 3x3 houses in a 12x12 town and then fill the spaces, but to try to figure out how to put 10 3x3 and 10 2x2 houses in a 12x12 town (if possible)
I think the problem could be solved by learning how to use wires to allow generators to power houses that are not directly adjacent
loozer wrote:
Dracorat wrote:
The part that's still missing is a definition of an "efficient" use of the space if it's not filled up and why you wouldn't want the program to just pack as many as you could otherwise, rather than
having to hand-tune it.
If a definition of what constitutes "efficient space" would definitely help understand the alternate direction.
There's indeed a problem with the definition of 'efficient space' because of the problem with weighting the houses (which would then make the problem even more complex)
Let's say that the following is a clarification that would make the definition of 'efficient' more obvious:
We can assume that the minimum number of houses is actually the desired number of houses. Meaning if the required houses are 3 3x3, 12 2x2 and 10 1x1, the algorithm should only try to place those
houses in the town.
This way the user will be able to juggle with the type of houses, and find the optimally weighted solution by it's own (more trouble for him, less for the algorithm)
So the main focus is not how to put 2 3x3 houses in a 12x12 town and then fill the spaces, but to try to figure out how to put 10 3x3 and 10 2x2 houses in a 12x12 town (if possible)
Additional clarifying question: If said houses are placed and are all "grouped" together, then the rest of the board is open space is that an acceptable situation for your algorithm?
(I worked quite a bit on this last night, but not really on the problem. I put together a gridding / mapping solution that probably a lot more complex than I should do, but hey that's how I work!)
Dracorat wrote:
Additional clarifying question: If said houses are placed and are all "grouped" together, then the rest of the board is open space is that an acceptable situation for your algorithm?
(I worked quite a bit on this last night, but not really on the problem. I put together a gridding / mapping solution that probably a lot more complex than I should do, but hey that's how I work!)
Yes, this is an acceptable situation, since then the user gets a positive answer to the question "do any other houses fit on the board / in the town"
This is if the grouping follows the 'each house needs to have a power generator attached' rule
So the main focus is not how to put 2 3x3 houses in a 12x12 town and then fill the spaces, but to try to figure out how to put 10 3x3 and 10 2x2 houses in a 12x12 town (if possible)
You can, but holy hell was it annoying to figure out (manually, because I'm bored).
Spoiler: show
Alamout wrote:
So the main focus is not how to put 2 3x3 houses in a 12x12 town and then fill the spaces, but to try to figure out how to put 10 3x3 and 10 2x2 houses in a 12x12 town (if possible)
You can, but holy hell was it annoying to figure out (manually, because I'm bored).
I might get some angry words from you, but that example (with 10 3x3 and 10 2x2 in a 12x12 town) was just that, an example... I appreciate that you gave me a possible solution for it, but what I'm
looking for is an algorithm for x 3*3, y 2*2 and z 1*1 houses in a t*t town
It seems that next week I will start with the code, most probably using brute force after checking what I can borrow form the packing algorithm
I'm not sure how useful the packing pattern will be. I think all the packing pattern does is give you larger 'blocks' to place but without actually reducing the complexity. If you have a large number
of houses of a particular size relative to the town area and the total number of houses you have to place then the solution will probably involve that pattern, but it's really just one special case.
Really you'll have to figure out how to take a problem and find sub-problems where the solutions can be composed.
You might be able to get some traction with a dynamic programming approach, that breaks the available space into subcomponents and tries to optimize each. Naively, it will be off optimal because of
things like a disregard for generators along edges, but that might be fixable with some cleverness.
I completed the problem along with a nice display for all of it. The code's (c) me and not open source, but you can use it for your personal use all you want.
I will probably release an open source version of the significant portions of the code once I make it a v2.
At any rate, the code is here:
And here's a screenshot of it in action:
I brute force calculate the best position for each shape by basically:
Move a hypothetical generator to each open spot on the board.
Place up to 4 hypothetical shapes next to each in every possible configuration.
Calculate the number of shared edges with: edges shared between shapes -- more valuable than edges at the edge of the map.
Choose the highest spot by edge count score.
Place generator there.
Place shapes next to it.
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How to Graph Linear Equations With Two Variables | The Classroom | Synonym
A linear equation can be written in the form ax + by = c, where x and y are variables and a, b, and c are real numbers. Although many modern graphing calculators can graph linear equations (in any
form) with minimal work on your part, graphing linear equations is a basic math skill that you should learn.
Use the Solutions to Graph the Equation
Step 1
Substitute x=0 in the equation and find a value for y that makes the resulting equation true. For example, in the equation x-2y=2, when x=0, the equation becomes -2y=2. Since y=-1 makes this equation
true, one point on the line is (0,-1). The x-value is the first coordinate. The y-value is the second coordinate.
Step 2
Substitute y=0 in the equation and find a value for x that makes the resulting equation true. Following the example in Step 1, when y=0, the equation x-2y=2 becomes x=2. Thus, another point on the
line is (2,0).
Step 3
Find a third point on the line. Substitute any value for x or y that is not already listed and solve for the other variable. For instance, when y=1, the equation x-2y=2 reduces to x-2=2 or x=4. A
third point on the line is (4,1).
Step 4
Plot the three points on a coordinate plane. Draw a line through the three points.
Style Your World With Color
• You only need two points to draw a line. However, the third point serves as a check—-if you found the points correctly, the three points form a line.
• If Step 2 gives you the same point as Step 1, try another value for y.
• "Integrated Algebra 1"; Ann Xavier Gantert; 2007
Photo Credits
• Jupiterimages, Brand X Pictures/Brand X Pictures/Getty Images
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FOM: Long sequence and the Friedman-Shoenfield debate on Incompleteness
Joe Shipman shipman at savera.com
Mon Sep 21 15:18:09 EDT 1998
Harvey points out that n(3), defined as
"the longest length of a finite sequence from 3 letters in which no
block x_i,...,x_2i
is a subsequence of any other block x_j,...,x_2j.",
is a very short description of a very large integer.
But he's calculated an upper bound for n(3) which implies that an even
shorter description of an even larger integer is
"A_9(9), where A_1(m) = 2m, A_n+1(m) =A_nA_n...A_n(1), where there are m
applications of A_n."
However, if you change "3 letters" to "4 letters" in the first
description you get
(according to Harvey) a MUCH larger integer, which is not even
describable by a term of feasible size involving the A_*(*) operator.
So an "objective" criterion for the significance of his result is that
no one has ever named a larger integer with a shorter description (here
I shall define "description" informally to mean "a sentence that is
sufficiently clear that an ordinary student at M.I.T., given the
description, could easily write a program to output the integer", though
you could make this more precise in many ways without affecting the
truth of "no one has ever named a larger integer with a shorter
Can anyone think of a counterexample to this claim? (No fair using
noncomputable functions like the Busy Beaver function and saying "the
largest integer put out by any 99-quintuple Turing machine that halts on
blank input"; by the way, does anyone know how the current Busy Beaver
record-holders compare to Harvey's n(3) and n(4)?) The shortest
description I can think of for an integer that transcends the n(k)
methods of feasible expression would involve Goodstein sequences, which
are only mildly complicated to describe; can anyone do better?
The debate on Incompleteness seems to boil down to "who should care
about the unprovability of a statement"? Shoenfield says that
specialists in the field are the most important judges (e.g. for set
theory, Solovay's opinion matters more than 20 Fields medalists
[presumably not including P.J. Cohen!]) -- Friedman says that if you can
get specialists in other areas of mathematics interested ("general
mathematical interest") there is much more (and longer-lasting)
significance to the result, and he has clearly accomplished this much.
He further says that if you can get high school students or the educated
man on the street interested ("general intellectual interest") there is
a great deal more significance still. I agree that his n(k) result
meets the g.i.i. test but it is not independent; and his independent
finite combinatorial results meet the g.mi. test but do not *directly*
meet the g.i.i. test (they do indirectly because it is of g.i.i. that
ordinary professional mathematicians are beginning to need to upgrade
their foundations). When his independent statements reach the "high
school" level of simplicity and immediacy that "n(k) exists" has, a
revolution will have taken place.
-- Joe Shipman
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Fractals are geometrically self-similar shapes that are often fascinating to behold, such as the one below. They are created by iterating a mathematical function or geometric transformation, but
because of certain qualities they can sometimes be more useful for describing natural phenomena such as leaves and coastlines than the combinations of regular geometric shapes such as squares and
To understand fractals, you must begin with understanding the principles of iteration.
For instance, take a certain function, like the square function, f(x) = x^2.
What happens as you iterate any function is that the result most often does one of two things: It will either
1. spiral to infinity, or
2. approach a fixed point.
For example, iterate the squared function, f(x) = x2. If you choose x = 4,
4^2 = 16
16^2 = 256
256^2 = 65536… it will spiral to infinity.
But, begin with the number 1, 1^2 = 1, and the function remains at 1. This makes 1 a fixed point. Similarly, begin with a number between one and zero, such as x = .5,
.5^2 = .25
.25^2 = .0625
.625^2 = .00390625...
nd the result will eventually approach zero. Therefore, for the squared function, the points 1 and 0 are fixed points. The definition of a fixed point is a point where F(x) = x.
For some functions, points will oscillate between different fixed points rather than going towards a single fixed point. This is known as an orbit.
For example, in the function f(x) = -x, zero is a single fixed point but all other points lie on a orbit of period 2. To demonstrate,
f(2) = -2
f(-2) = 2
so the function bounces from one point to another. Orbits can have any number of fixed points.
Chaotic Quadratics:
In the quadratic equation, f(x) = x2 + c , certain ranges of c values display a behavior where the number of fixed points increases exponentially, for instance one range of c values will have 2
fixed points, the next range 4, and then 8, and so on. This behavior swiftly diverges towards infinity, so certain c values produce an infinite number of fixed points. This is bifurcating,
chaotic behavior.
Julia Sets:
Points that are attracted to a cycle of fixed points, regardless of how many fixed points are in the cycle, are in the basin of attraction. A Julia set is defined as the set of points right on
the boundary between the points in the basin of attraction and the points that escape towards infinity. It is plotted by coloring a point black if it is attracted to a fixed point, and white if
it is not.
The Mandlebrot set
The Mandlebrot set is the grandaddy of all fractals in the sense that it connects chaotic behavior with the behavior of Julia sets. To understand a Julia set, one can analyze the behavior of the
critical point, zero . Zero is the ciritcal point because it is where the slope is zero for the quadratic equation. If, within the iterated function, an input of zero diverges to infinity, the
Julia set is called fractal dust . If the orbit of zero is attracted, the Julia set is connected . The Mandlebrot set is defined as the set of c-values for the function f(x) = x^2 + c for which
the critical orbit does not diverge to infinity. Each point that is painted black corresponds to a Julia set that is connected.
A note about coloring: rainbow fractals are colored so that while each black point corresponds to a set that is connected or attracted, the color of the rainbow area depends on how swiftly the
input diverges to infinity. Whereas in black and white fractals this area is left white.
To see how each point on the Mandelbrot Set correlates to a Julia Set, check out The Mandelbrot and Julia Set Explorer. This awesome site depicts the entire range of beautiful fractals possible
within the Mandelbrot set.
Fractals are geometrical shapes that follow this same iterative process, on a geometrical level rather than algebraic. They are defined as having two special properties: 1. They are self-
similar. 2. They have fractal dimension.
The Koch snowflake is an example of the same, but with successive additions of an equilateral triangle on every side.
One amazing property of fractals like this is that they have a finite area but their perimeter is infinite . This had lead many observers to surmise, as Mandelbrot did, that fractals could form
an excellent description of such natural phenomena as coastlines, ferns, leaves, and the bifurcation of blood veins, that all have self-similar qualities as one descends through scale. Especially
in the development of organisms, it makes sense for purposes of diffusion to have a nearly infinite surface perimeter in a finite area, such as the development of circulatory systems.
For examples of fractal images in nature, look at The Center for Polymer Studies' Image Galleries.
The classic example of fractals in plant-life is that of the fern, where each smaller leaf resembles the larger one. Examining Microcladia coulteri is reminiscent of examining ferns, as each
smaller bifurcation appears to retain the same angle of branching and alternate pattern. If one were to describe its surface with a function, it is easy to imagine an iterative function forming
the basis for all of the development from the tiny tips to the main thalli.
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[SciPy-User] memory error - numpy mean - netcdf4
srean srean.list@gmail....
Fri Aug 26 19:54:41 CDT 2011
On Fri, Aug 26, 2011 at 2:33 PM, Phil Morefield <philmorefield@yahoo.com>wrote:
> The formula you have written looks like you're collapsing everything into a
> single value. I think he's trying to average a bunch of 2D arrays into a
> single 2D array.
You are correct, the form that I posted can be read as if it is for
updating single mean vector \mu, but you can use the same for an nd-array
trivially. Just have \mu and t as nd-arrays. m can be one too. Numpy
broadcasting will take care of the rest.
One advantage is that it requires only a constant amount of memory for the
computation, you can even read the data in from an infinite pipe or
generator that yields a single vector or a matrix at a time (or bundles them
up m at a time). It will always be uptodate with the current estimate of the
means. In fact will work for any moment too.
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Elasticity is the property of materials to return to their original size and shape after being deformed (that is, after the deforming force has been released). Since it is really a property of
materials, a more complete discussion will have to wait until later. For now we'll stick to a very simple elastic system (the coil spring) and a very simple law (Hooke's law).
Hooke's law isn't about hooks. It's about springs — coil springs — the kind of spring found in a car's suspension or a retractable pen, the kind that look like a pig's tail or a lock of curly hair.
Coil springs are also known as helical springs since the mathematical name for this kind of shape is a helix. The law is named in honor of its discover, the English scientist, mathematician, and
architect Robert Hooke (1635–1703). Although Hooke's name is now usually associated with elasticity and springs, he was interested in many aspects of science and technology. His most famous written
work is probably the Micrographia, a compendium of drawings he made of objects viewed under a magnifying glass. In this book, he was the first to use the word "cell" to described the walled-in
regions he saw when looking at a magnified slice of plant tissue (in Hooke's case, a slice of cork). The standard story is that he compared these walled-in regions to the cells in a prison or
monastery, but I could find no mention of this in the Micrographia. He also compared biological cells to pores, pumice, and honeycombs, but cell was the word that stuck.
As I said, Hooke was interested in many areas of science. He not only looked through magnifying glasses and microscopes, but also through telescopes. Like Galileo he pointed his telescope at the sun,
and like Galileo he did not look at the sun with his eye. That would have been stupid. Instead, like every sensible person since Galileo, he placed a sheet of white paper several inches in front of
the eyepiece and looked at the projected image of the sun. Such a device is called a helioscope. (In Greek, ήλιος "elios" is the sun and σκοπεῖν "skopein" is to observe.) In 1675, Hooke wrote a book
on the helioscope and added this little bit of text to fill up the white space leftover at the bottom of the last page …
To fill the vacancy of the enſuing page, I have added a decimate centeſme [a thousandth] of the Inventions I intend to publiſh, though poſſibly not in the ſame order, but as I can get opportunity
and leaſure; moſt of which, I hope, will be as uſeful to Mankind, as they are yet unknown and new.
He then went on to list ten inventions and discoveries he had made. (This was not followed by any later list with the remaining 990 inventions he promised, by the way.) These included a way to
regulate pendulum clocks, a method for constructing arches, and other inventions in optics, hydraulics, and mechanical engineering. The third item on his list is of importance to us right now.
3. The true Theory of Elaſticity or Springineſs, and a particular Explication thereof in ſeveral Subjects in which it is to be found: And the way of computing the velocity of Bodies moved by
them. ceiiinosssttuu
That weird bit that looks like someone fell asleep on their computer keyboard is not a mistake. It's an anagram. In the time before patents and other intellectual property rights, publishing an
anagram was a way to announce a discovery, establish priority, and still keep the details secret long enough to develop it fully. Hooke was hoping to apply his new theory to the design of timekeeping
devices and didn't want the competition profiting off his discovery. He was successful in this regard and in 1678 Hooke made the solution to the anagram, and the true theory of springiness that now
bears his name, public knowledge.
About two years ſince I printed this Theory in an Anagram at the end of my Book of the Deſcriptions of Helioſcopes, viz. ceiinosssttuu, that is Ut tenſio sic vis; That is, The Power of any Spring
is in the ſame proportion with the Tenſion thereof: That is, if one power ſtretch of bend it one ſpace, two will bend it two, and three will bend it three, and ſo forward. Now as the Theory is
very ſhort, ſo the way of trying it is very eaſie.
The Latin …
Ut tensio sic vis
literally translated into English would read something like …
As extension, so is force
but in contemporary English, we would probably say something more like …
Extension is directly proportional to force.
The remainder of the quoted passage that follows his Latin phrase is a description of what is means for two things to be directly proportional. Try not to get confused with his apparent misuse of
words, however. Scientific terminology in the English language is much more precise now than it was in the Seventeenth Century. By "tension" he means extension and by "power" he means force. The
directly proportional relationship is between extension and force, not tension and power.
We can write Hooke's law as a proportionality statement in mathematical shorthand like this …
Δx ∝ F
where …
F = force, spring force, elastic force, applied force, deforming force, …. You get the idea. Versions with subscripts are also common (F[s], F[e], etc.).
Δx = extension or compression of the spring; that is, the change in length from the spring's relaxed, natural, or original length (x[0]). Use of ∆ [delta] is optional as the idea of "change" is
Since equations are so popular nowadays (meaning the last 150 years or so) we should probably finish by writing Hooke's law as an equation …
F = − kΔx
The constant of proportionality (k), which is needed to make the units work out right, is called the spring constant — an apt name since it is a constant that goes with a particular spring. It is not
a constant that goes with a particular material. Materials don't have constants in elasticity, they have moduli (plural of modulus). Hooke's law is now recognized as being approximately true for a
variety of elastic applications, not just springs, but as I said earlier, a more complete discussion of this will have to wait until later in this book.
The SI unit of the spring constant is the newton per meter, which has no special name.
k = F ⇒ ⎡ N = N ⎤
x ⎣ m m ⎦
Since most springs would never stretch anything close to a meter, other units like the newton per centimeter [N/cm] or newton per millimeter [N/mm] are also common.
You may have noticed a negative sign in the equation above. This gives the spring force its direction. If the spring is stretched in the positive direction (+x) the spring force pulls back in the
negative direction (−F). If the spring is compressed in the negative direction (−x), the spring force pushes back in the positive direction (+F).
Newton and Hooke. Hooke and Newton. Reputed to be the ugliest scientist of all times, no portrait of Hooke is known to exist (no undisputed, original portrait). Hooke was also short and Newton mocked
him with his famous "shoulders of giants" line.
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Tuesday, June 25, 2013 at 1:14pm
PLS HELP, BEEN STUCK FOREVER ON TWO QUESTIONS, THIS IS ONE : There are usually no costs for the first 3 years, but thereafter maintenance is re- quired for restriping, weed control, light
replacement, shoulder repairs, etc. For one section of a particular highway, these costs ...
Monday, June 24, 2013 at 2:22am
Engineering science
a=(vf-Vi)/time change km/hr to m/s force=mass*a
Friday, June 21, 2013 at 10:39am
Engineering science
A lorry has a mass of 5640kg and accelerates uniformly from rest to reach a velocity of 100km/h in 10s. Determine the acceleration of the lorry. Determine the force needed to accelerate the lorry
Friday, June 21, 2013 at 10:26am
Mechanical Engineering
This is engineering? Hmmmm. Vf^2=Vi^2 + 2 ad Vf^2=12f/s+2*-32f/s^2*-4.5 Vf^2=288 Vf=12 sqrt2 fps check that
Friday, June 21, 2013 at 4:50am
Mechanical Engineering
a tennis ball was held 4.5 ft above the surface of a tennis court and was thrown vertically downward with an initial velocity whose magnitude was 12 fps. determine the magnitude of the velocity of
the ball just before it struck the surface of the court
Friday, June 21, 2013 at 12:38am
Chemical Engineering
Friday, June 14, 2013 at 4:56pm
petroleum engineering physics
Vector A has a magnitude of 5.00units,and B has a magnitude of 9.00units.The two vectors make an angle of 50.0degrees with each other .Find A.B.
Wednesday, June 12, 2013 at 6:48pm
Wednesday, June 12, 2013 at 4:01pm
HW6_1B: (2*Q*L^2)/(G*pi*t*d_A^2*(L+x)^2) Why? You have to start by describing the linear function of diameter/radius variations (whatever) r(x)=(d_A/2L)*x+d_A/2 d_A(x)=(d_A/L)*x+d_A We can obtain I
(x)=pi*t*d_A^3*(x+L)^3/(4*L^3) We know that gamma=r(x)*(d_Phi/d_x) and d_Phi/d_x...
Wednesday, June 12, 2013 at 3:40pm
anyone has hw6_1B answer?
Tuesday, June 11, 2013 at 9:59am
6_3 = 8*phi_m*G*I_p/(L^2)
Tuesday, June 11, 2013 at 7:25am
6_2A= -t_0*x 6_2B= -t_0*(x^2-L^2)/(8*pi*G_0*R^4) 6_2C= -L^2*t_0/(8*pi*G_0*R^4) 6_2D= 5*t_0*L/(4*pi*R^3) 6_3= 8*psi*G*I_o/(L^2) i need 6_1B
Tuesday, June 11, 2013 at 12:38am
6.1B, 6.1C, 6_2a to 6_2d, 6_3
Tuesday, June 11, 2013 at 12:19am
HW6_1A : 2*d_A^3*pi*t*G/(3*L )
Monday, June 10, 2013 at 3:54pm
tapered thin-wall circular shaft has constant wall thickness, t, length L, and diameters linearly varying between dA at the support A(x=0) and dB at its free end B(x=L). The shaft is homogeneous with
shear modulus G HW6_1A : 20.0 POINTS Obtain a symbolic expression for the ...
Saturday, June 8, 2013 at 3:24pm
gamma= r*Q/(L*Kt)
Saturday, June 8, 2013 at 5:13am
There is no figure.
Friday, June 7, 2013 at 2:37pm
Help with the rest?
Friday, June 7, 2013 at 2:07pm
A 10.0 Kg block is released from point A in the figure below. The block travels along a smooth frictionless track. It makes a collision with stationary 10 kg ball. The ball moves forward with speed
of 2 m/s. a) What is the speed of the block at point C? b) What is the velocity...
Friday, June 7, 2013 at 1:39pm
Let s suppose the message x(t) is an analog signal shown in the figure below. a) Draw the block diagrams of the system which converts the analog message signal x(t) to a pulse modulated signal. b)
Draw the sampled signal with the sampling frequency fs=2khz c) Carry out ...
Thursday, June 6, 2013 at 1:01pm
Thursday, June 6, 2013 at 2:14am
A tapered thin-wall circular shaft has constant wall thickness, t, length L, and diameters linearly varying between dA at the support A(x=0) and dB at its free end B(x=L). The shaft is homogeneous
with shear modulus G HW6_1A : 20.0 POINTS Obtain a symbolic expression for the ...
Wednesday, June 5, 2013 at 5:04pm
can anyone post the solutions pls!!
Tuesday, May 28, 2013 at 5:29am
All you guys are awesome !! Thanks a billion !! :D
Tuesday, May 28, 2013 at 3:42am
Q1_2_3 (25/12*W*H)/A_0*E_0
Monday, May 27, 2013 at 4:17pm
Q1_2 _1 400 500 699.99 665.68
Monday, May 27, 2013 at 3:56pm
Engineering science
Hints: Kinetic energy, KE=(1/2)mv² where m=mass of object (kg) v=velocity (m/s) Potential energy =mgh m=mass in kg g=acceleration due to gravity, m/s-2 h=height above datum in m. To calculate h, you
need the distance travelled. For uniform acceleration, distance d, is the...
Sunday, May 26, 2013 at 10:48am
Engineering science
A vehicle with a mass of 1200 kg accelerates uniformly from rest upwards with an incline of 1:25 and reaches a speed of 72 km/h after 2 minutes,calculate: 1.The kineyic energy of the vehicle after 2
minutes 2.The potential energy of the vehicle after 2 minutes
Sunday, May 26, 2013 at 10:16am
Please post 1_4 all parts
Friday, May 24, 2013 at 11:14pm
Obtain symbolic expressions for the maximum tensile and compressive stresses ( σmax,Tn and σmax,Cn )and their locations (coordinates xmax,T and xmax,C ) in terms of ρ, g, L (with ρ written as rho).
(Note: enter the expressions for the stresses with their ...
Friday, May 24, 2013 at 7:45pm
I need all four parts for 1-1-4
Friday, May 24, 2013 at 6:53pm
I need Tn and Cn
Friday, May 24, 2013 at 6:52pm
I need Tn and Cn
Friday, May 24, 2013 at 6:09pm
Q1_1_4 I only have the locations T=0 C= 3*L
Friday, May 24, 2013 at 5:57pm
Q1_1_4 I only have the locations T=0 C= 3*L
Friday, May 24, 2013 at 5:57pm
Q1_1_4 : 80.0 POINTS Obtain symbolic expressions for the maximum tensile and compressive stresses ( σmax,Tn and σmax,Cn )and their locations (coordinates xmax,T and xmax,C ) in terms of ρ, g, L (with
ρ written as rho). (Note: enter the expressions for the ...
Friday, May 24, 2013 at 5:45pm
Friday, May 24, 2013 at 5:38pm
1000 bone cookies dog
Friday, May 24, 2013 at 5:33pm
1_1_2 -A*g*L*rho
Friday, May 24, 2013 at 5:30pm
1_1_2 -A*g*L*rho
Friday, May 24, 2013 at 5:30pm
Q1_1_2 : 100.0 POINTS Use the force method for statically indeterminate structures to obtain a symbolic expression for the reaction RCx at the support C in terms of ρ, g, A, L (with ρ written as
rho): Obtain symbolic expressions for the maximum tensile and ...
Friday, May 24, 2013 at 5:15pm
1-1-2, 1-1-4
Friday, May 24, 2013 at 5:06pm
Great dog
Friday, May 24, 2013 at 4:47pm
Thanks dog
Friday, May 24, 2013 at 4:43pm
UX(=2l) (3*g*L^2*rho)/(8*E)
Friday, May 24, 2013 at 4:35pm
1_2 also
Friday, May 24, 2013 at 4:11pm
Please post the answers for 1_4 and 1_5
Friday, May 24, 2013 at 3:26pm
Help with Question 1_1_2, 1-1-4, 1-1-5? SERIOUS ANSWERS NEEDED Use the force method for statically indeterminate structures to obtain a symbolic expression for the reaction RCx at the support C in
terms of ρ, g, A, L (with ρ written as rho): Obtain symbolic ...
Friday, May 24, 2013 at 2:21pm
Q1_1_3 0<x<L rho*g*A*(L-x) L<x<3L (rho*g*A)/2*(L-x)
Friday, May 24, 2013 at 2:13pm
Q11 2 always false
Friday, May 24, 2013 at 1:51pm
Q.1_1_1 rho*g*A 1/2*rho*g*A Q.1_1_2 -2*rho*g*A*L
Friday, May 24, 2013 at 1:37pm
No it is not good Jason
Friday, May 24, 2013 at 12:57pm
Q1_1_2. My answer is -4*rho*A*g*L
Friday, May 24, 2013 at 12:50pm
Use the force method for statically indeterminate structures to obtain a symbolic expression for the reaction RCx at the support C in terms of ρ, g, A, L (with ρ written as rho): RCx= incorrect
34⋅ρ⋅g⋅A⋅L You have used 1 of 3 submissions ...
Friday, May 24, 2013 at 12:16pm
Sorry, just noticed Q1_1_1: 3/2*rho*g*A is WRONG! It is 0.5*rho*g*A as trustee says.
Friday, May 24, 2013 at 11:43am
Anyone for the rest please?
Friday, May 24, 2013 at 11:33am
Q1_2_4= 300 mm^2
Friday, May 24, 2013 at 11:33am
These are my results (done by me): Q1_1_1 rho*g*A 3/2*rho*g*A Q1_2_1 400 500 700 565.685424 Q1_2_2 -800 -300 0 700 Q1_2_3 ((W*1.25)*(5/3*H))/(A_0*E_0) need the rest...
Friday, May 24, 2013 at 11:11am
In my case it was 500 guys, just try it.
Friday, May 24, 2013 at 10:34am
q1-2 wrong
Friday, May 24, 2013 at 10:34am
Thanks simonsay, in q1_2_1: 300 is wrong for me, is there maybe a -300?
Friday, May 24, 2013 at 10:30am
q1_2_1 400, 300, 700, 565.685 q1_2_2 -800, -300, 0, 700
Friday, May 24, 2013 at 10:19am
Trustee thanks the first two answers right but third one is wrong, can you check please?
Friday, May 24, 2013 at 9:36am
3/4*rho*g*A*L is not good answer
Friday, May 24, 2013 at 9:29am
1. rho*g*A 2, 0.5*rho*g*A 3. 3/4*rho*g*A*L
Friday, May 24, 2013 at 8:44am
Friday, May 24, 2013 at 8:23am
Any help please?
Friday, May 24, 2013 at 4:19am
Anyone please?
Thursday, May 23, 2013 at 4:32pm
Please help!
Thursday, May 23, 2013 at 3:34pm
The bar in the figure has constant cross sectional area A. The top third of the bar, of length L, is made of a material with mass density ρ and Young's modulus E. The bottom two thirds of the bar
(length 2L) is made of a different material, with density ρ/2 and ...
Thursday, May 23, 2013 at 2:00pm
Engineering Physics
An ideal gas at 12.1 °C and a pressure of 2.38 x 105 Pa occupies a volume of 2.88 m3. (a) How many moles of gas are present? (b) If the volume is raised to 4.70 m3 and the temperature raised to 28.3
°C, what will be the pressure of the gas?
Wednesday, May 22, 2013 at 9:01pm
Mechanical Engineering
A load W=2 kN is applied vertically to joint C of truss ABCDE as indicated. You will use the method of joints to obtain the axial forces in the bars and reactions at the supports A, E . We start with
the classification of the degrees of freedom as "free" or "...
Sunday, May 19, 2013 at 9:13pm
Name of physics book for 11 and 12.to catch engineering exam and other competetive exam
Monday, May 13, 2013 at 2:55pm
use fomular P*v=n*R*T cause temperature and volumn not change: n(initial)=(P(initial)*V)/(R*T) n(final)=(P(final)*V)/(R*T) we want the final n, therefore: n(final)=P(final)*n(initial)/P(initial) =
Wednesday, May 8, 2013 at 8:34am
Chemical engineeing
Can someone help me with online exam ( fundamental of chemical engineering class ) ? Please eamil me ( davd_jo(at)(ya h o o )dot (com ) I'll offer money
Monday, May 6, 2013 at 5:45pm
design a hydraulic jack with 1000kn and which can left 0
Sunday, April 28, 2013 at 4:55pm
m*v^2/R=mg centripetal force=the weight under the condition that traveled without losing contact with the sphere --> g= v^2/R--> v=(2gR)^(1/2)=(2*9.8*13.4)^(1/2)=16.206
Sunday, April 28, 2013 at 12:39pm
A physics major is working to pay his college tuition by performing in a traveling carnival. He rides a motorcycle inside a hollow transparent plastic sphere. After gaining sufficient speed, he
travels in a vertical circle with a radius of 13.4 . The physics major has a mass ...
Sunday, April 28, 2013 at 11:16am
An elevator cable can withstand a maximum tension of 1.89×104N before breaking. The elevator has a mass of 480kg and a maximum acceleration of 2.25m/s^2 . Engineering safety standards require that
the cable tension never exceed two-thirds of the breaking tension.
Friday, April 19, 2013 at 9:28pm
Need help with some engineering math problems? 1. You wish to determine the total system efficiency of your hydraulic press that is powered by an electric motor. The characteristics of the hydraulic
press are as follows: Electric motor utilizes 5 amps under load & ...
Thursday, April 18, 2013 at 1:07am
F = 9C/5 + 32 = 80 9c/5 = 80-32 = 48 9c = 240 C = 26.67o Celsius.
Tuesday, April 9, 2013 at 9:17pm
The power plant is to be operated in a region where the same company operates several other coal fired power plants that were all built in the 1960s. The new power plant is planned to have 500 MW
energy generating capacity. The next largest plant is 50 MW and each is emitting ...
Tuesday, April 9, 2013 at 5:50pm
The power plant is to be operated in a region where the same company operates several other coal fired power plants that were all built in the 1960s. The new power plant is planned to have 500 MW
energy generating capacity. The next largest plant is 50 MW and each is emitting ...
Tuesday, April 9, 2013 at 5:47pm
Air pollution Engineering
The power plant is to be operated in a region where the same company operates several other coal fired power plants that were all built in the 1960s. The new power plant is planned to have 500 MW
energy generating capacity. The next largest plant is 50 MW and each is emitting ...
Tuesday, April 9, 2013 at 5:03pm
stat probability
At a certain college, there were 780 science majors, 320 engineering majors, and 390 business majors. If one student was selected at random, the probability that they are a business major is?
Monday, April 8, 2013 at 7:38pm
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Chemistry: Mixture and Mass
I'm not sure how you can do it using percent compositions, at least I can't see it.
How many moles of Na do you have? Since the elements in each compound are 1 to 1, the moles of Na and Br added together equal the total number of moles of Na in the mixture.
What's the mass of the Br and Cl together? The moles of each times their atomic weight, added together will equal that mass.
I would try solving this as a system of two equations with the two variables being the moles of Cl and Br in the mixture.
I did: [tex]\frac{.76 g Na}{1}[/tex] x [tex]\frac{mol Na}{22.99 g}[/tex] = .033 mol Na
2.08 g
- .76 g
= 1.32 g
[Cl + Br]
I then tried: [tex]\frac{.033 mol Na}{1}[/tex] x [tex]\frac{mol Br}{2 mol Na}[/tex] x [tex]\frac{79.909 g}{mol Br}[/tex] = 1.32 g Br
That mass of Br cannot be true, so apparently I am doing something wrong.
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Ars Mathematica
More Lévy
According to Wikipedia, the analogue of a random walk for a Lévy distribution is called a Lévy flight. I presume it’s called a “flight” because paths are no longer guaranteed to be continuous, but
can have sudden jumps.
I also spotted a survey paper, More “Normal” than Normal: Scaling Distributions and Complex Systems, which argues that in physical applications, heavy-tailed distributions such as the Lévy
distribution are more natural than the normal distribution. This seems to be emerging conventional wisdom in some circles, but I don’t know how true it is.
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Formal Analysis, September 11 2001
Note added in 2009: These analyses and descriptions were written in 2001, and we have in the meantime learned more. On the one hand, it is clear that the formal prediction for the deviation of means
gives only part of the picture -- the deviation that begins during this period apparently continues for two days. But our analysis of the poorly defined variance question can be much refined now, and
a conservative reading suggests that measure is just marginally significant. 9/11 was a powerful event, but even so, should not be taken by itself as proof of a global consciousness effect -- that
requires the patient accumulation of replications.
The following material shows the behavior of the Global Consciousness Project's network of 37 REG devices called "eggs" placed around the world as they responded during the periods of time specified
in formal predictions for the events of September 11 2001.
Deviation of Means
The Global Consciousness Project has a standard protocol for testing the hypothesis that great events in the world may affect the eggs in a way that can be detected by statistical analysis. The data
from all the eggs are first combined in a single (Stouffer) Z-score representing the signed deviation of the mean across all eggs for each second. These Z-scores are squared and summed to yield a
Chisquare for the whole period. The departure from expectation of this Chisquare reflects a correlated response across the eggs. Our prediction is for a positive deviation of the Chisquare, which
would indicate a tendency toward increased deviations of the individual egg scores from theoretical expectation. More information about the analysis procedures is given in the Methodology section of
the GCP Web site.
The first formal prediction for September 11 was essentially the same as that made for the terrorist bombing of American Embassies in Africa in August 1998. That specified a period beginning a few
minutes before the bombing and included an aftermath period of three hours. Following that model, I specified a period beginning 10 minutes before the first plane crashed into the WTC tower, and
ending four hours after, thus defining a similar aftermath period.
To visualize the data, we plot the cumulative deviation of the Chisquare sequence from chance expectation. If there is no effect, such a plot will show a random walk (sometimes called a "drunkard's
walk") around the horizontal line of expectation. That is, the trace will wander up and down but will have no clear trend.
The graph of data from the formal prediction for September 11 shows a fluctuating deviation throughout the moments of the five major events, during which ever-increasing numbers of people around the
world are hearing the news and watching in stunned disbelief. Times of the major events are marked by boxes on the line of zero deviation. The uncertain fluctuation of the EGG data continues for
almost half an hour after the fall of the second WTC tower. Then, at about 11:00, the cumulative deviation takes on a strong trend that continues through the aftermath period and ultimately exceeds
the significance criterion. There were 37 eggs reporting on September 11, and over the 4 hours and 10 minutes of the prediction period, their accumlated Chisquare was 15332 on 15000 degrees of
freedom. The final probability for the formal hypothesis test was 0.028, which is equivalent to an odds ratio of 35 to one against chance.
Deviation of Variance
The next figure shows the cumulative deviation of a measure of the variability of scores (variance) among the 37 eggs over the course of the day of September 11. It was generated as a test of Dean
Radin's prediction that the variance would show strong fluctuations: "I'd predict something like ripples of high and low variance, as the emotional shocks continue to reverberate for days and weeks."
Although this was only a partial specification it is effectively a prediction that the variance around the time of the disaster should deviate from expectation. I added the necessary specifications
for a formal prediction.
The variance measure shows a normal fluctuation around the horizontal line of expectation until about three or four hours before the attack, and then a steep and persistent rise indicating a great
excess of variance, continuing until about 11:00. Shortly thereafter, a long period begins during which the data show an equally precipitous decrease of variance. It is difficult to make a direct
calculation of probability for this figure, but the extreme excursion in Dean Radin's similar analysis reaches a level of more than three sigma, which corresponds to odds of less than 1 in 1000. An
additional analysis using permutation of the data to determine how often such an extreme excursion occurs in randomly ordered sequences, to compare with the original temporal sequence leads to an
estimate of p = 0.0048, based on 10,000 iterations. (A more recent computation using the mean of variance as the standard rather than theory yields p = 0.0009.)
A more conservative estimate is included in the formal database. It is based on assessing the fast rise and the fast fall of the variance measure surrounding the period of the attacks. The
probability for each was calculated by extrapolation of the probability envelope as far (as many seconds) as would be needed to achieve the extreme rise or fall by chance, compared to the much
shorter envelope that covers the time of the actual rise or fall. The ratios of these times were divided by the square root of 2 to compensate for the fact that a cumulative deviation trace reaching
the terminal significance level at some prior time during the cumulation is twice as likely as the terminal probability itself. The resulting estimate is p = 0.096.
For a visual indication of the likelihood that this is merely a random fluctuation, a comparison can be made with pseudo-data generated for September 11, 2001, and plotted in the same format. In
contrast to the real data, there are no long-sustained periods of strong deviation in the algorithmically generated data. This comparison with the pseudo-data indicates that the variance measure is
unusual around the time of the attacks.
In this figure, the times on the X-axis are Eastern Daylight Time, allowing a direct assessment of the timing of the strong deviations. As in the first figure showing the cumulative deviation of the
Chisquare, there is an indication that the effects registered for this horrendous event might have begun several hours prior to the first attack. Again, the pseudo data are used for a direct
comparison. More on this topic, in the context of exploratory analyses, can be found on the 9/11 Variance page and on the extended analysis page.
Silent Prayer
A third formal prediction was made with direct reference to the September 11 events. Since the horrible event, innumerable calls for prayer have been made. On the 14th of September there was a
special emphasis on such collective spiritual moments, including major organized periods of silence in Europe and America. Doug Mast made a specific formal prediction for a deviation of the Chisquare
"over the time periods 1000 to 1003 GMT, corresponding to a European organized mourning and the time period 1200 to 1203 EDT (1600 to 1603 GMT) corresponding to the beginning of the Washington
service and many organized mourning events in the Eastern US." Here is the resulting graph.
The picture is compelling, although it does not confirm the formal prediction. Instead, the trend shows a marginally significant decrease in the deviations of the egg data. The Chisquare is 150.68 on
180 degrees of freedom, with probability 0.9455. The trend is steadily opposite to the specified direction.
Because this was an extraordinary event, we have done a great deal of contextual analysis. The questions asked are necessarily post hoc but our intention has been to determine what the data have to
say, primarily by looking for structure in what should be truly random data. Our purpose has not been to "prove" a point or to support a thesis, but instead to try to learn from the empirical
indicators. The exploratory analysis page shows the main extensions beyond the formal analyses. An extended analysis page duplicates these and adds others that were part of the developing analysis
program over the first few days following the tragedy. For more details about the project itself, you can go to the GCP home page where you will find links to all aspects.
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Algebra Archive | November 23, 2010 | Chegg.com
Algebra Archive: Questions from November 23, 2010
• Anonymous asked
1 answer
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0 answers
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1 answer
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2 answers
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1 answer
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1 answer
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I need to show steps:
A trucking company has three size trucks: Large (L), medium (M), and small (S).... Show more
I need to show steps:
A trucking company has three size trucks: Large (L), medium (M), and small (S). Experience has shown that the large truck can carry 7 of container A, 6 of container B, and 4 of container C. The
medium truck can carry 6 of A, 3 of B, and 2 of C. The small truck can carry 8 of A, 1 of B, and 2 of C. How many trucks of the three sizes are needed to deliver 64 of container A, 33 of
container B, and 26 of container C? • Show less
1 answer
• Anonymous asked
Below is a problem from a Linear Algebra assignment which needs to be partly done through the MA... Show more
Below is a problem from a Linear Algebra assignment which needs to be partly done through the MATLAB program. I am having trouble understanding what the questions are stating, i.e I don't
understand the wording. I would seriously appreciate it if anyone was willing to decode these questions as I know it's hard to post things from MATLAB into the browser. I'm not looking so much
for the actual answer as I want to understand what the question is asking and I know I can move along from there, although the answer to part a would be immensely appreciated.
x and y are column vectors
f transforms x into y f(x)=y
"The following data is a selection of apartments sold around SoHo/Greenwich Village
area in New York City from August 5, 2010 to September 21, 2010 taken from
zillow.com. It lists the apartments’ areas in square feet and prices in thousands of US
dollars. In this project we will try to fit the data with polynomials of various degrees
and see whether they work as good predictions for the relation between the areas and
the prices of the apartments from that area.
sqft 467 1443 830 402 553 2200 440 2670 1837 772 1435
price 540 2147 1400 495 800 2630 470 3750 1620 1120 1780
Since we plan to work with polynomials, we will have to raise some of the above
numbers to powers, and the numbers will blow up once the powers are larger than 2
or 3. So, it makes sense to normalize everything first to make all the numbers not
too far from 1, i.e. divide all of them by 1000. This produces the vector x of areas in
thousands of square feet and the vector y of prices in millions of dollars.
y =(
Suppose some function f maps each xj to yj , i.e. f(xj) = yj , 1 < j < 11. Then
we have eleven equations. In order to hope for existence of a solution, we need 11
unknowns. Let f(x) be a polynomial of degree 10, i.e.
f(x) = a1 + a2x + a3x^2 + ... + a11x^10
a)Write down the system of linear equations with a1; ... ; a11
as unknowns such that f(xj) = yj , 1 < j < 11.
b)Write the coefficient matrix
T for the system from from part a, where T stands for power ten.
c)Solve the system Ta = y in MATLAB using row reduction. Name your
answer a.
Thanks ahead, I will rate LIFE SAVER for sure • Show less
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• Anonymous asked
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1 answer
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Proof dealing with a closed set
February 11th 2011, 05:33 PM #1
Nov 2009
Show that the set E' of points of accumulation of any set E must be closed.
Not too sure what to do with this; I understand why it is but really don't have a clue how to show it. Help please?
What is the definition? What happens if you let x be an element of the boundry? for all epsilon>0. we have that the intersection between that neighbourhood and the set E is nonempty. So...
What sort of space are we working with?
I deleted a reply after realizing I had made an unjustified assumption.
There are topological spaces in which this problem is a false statement.
Plato, I'm not quite sure how to answer your question. I am not in a typology class, this is for my real analysis class. This problem is just under a section dealing with open and closed sets.
I think that he's just talking about subsets of the real numbers.
A real analysis class would at the very least require a metric space.
In order for this to be true we should have that if $pot= q$ then there are disjoint open balls one containing p and the other containing q.
If $p$ is a limit point of $E'$ and there is an open ball $p\in \mathcal{B}$ then by definition there is $q\in \mathcal{B}\cap E'$ such that $qot= p$.
Now there is an open ball $\mathcal{C}$ such that $q\in \mathcal{C}\subset \mathcal{B}\setminus\{p\}$.
Now you should be able to complete the proof.
Recall the definition of limit point of the set $E$.
I'm not really an analysis guy, so this could be wrong...
But can't you just write $E^{\prime}=\overline{E}\cap\overline{X\setminus E}$?
Simplified this and redid it below with original post. Didn't know how to delete this.
Last edited by Hartlw; February 12th 2011 at 08:40 PM.
$E^'$ = $E_0^c \cap \overline{E}$
$E_0^c$ is the complement of the interior (open) and so is closed
$\overline{E}$ is the closure of E and so is closed.
Intersection of closed sets is closed.
If $E=[0,1)$ then $E_0^c \cap \overline{E}=\{0,1\}$ the boundary.
BUT $E'=[0,1].$
The complement of the interior is all points except the interior.
The closure is the interior E0 plus the boundary E'. The intersection is the boundary.
if E=[0,1), then E' = 0 and 1
E' is the set of all accumulation points of E, which is the same for (0,1) or [0,1) or [0,1]: 0 and 1
Last edited by Hartlw; February 14th 2011 at 08:28 AM. Reason: A is E, and asdd last sentence
All of that is very true. But it has nothing to do with this OP.
The question is about the derived set, $A'$, the set of all accumulation points of $A$.
It is not about boundary points.
The problem is to show that $A'$ is closed in a metric space.
Your proof does not do that. It leaves out any accumulation point in $A^o$.
P.S BTW. Not every boundary point is an accumulation point.
E' is the set of all accumulation points of E, which by definition is the boundary.
If $E=[0,1)\cup \{2,3\}$ then $E'=[0,1]$.
I think that you are confused about the definition of accumulation point.
You are correct of course. Interior points are also accumulation points. All boundary points are accumulation points but not the other way around.
The complement of the interior includes the boundary and exterior points, and my original proof was correct.
Unfortunateley I can't copy it and I just don't feel like plowing through the code again, so it will have to remain buried (#9).
My original statement:
"E' is the set of all accumulation points of E, which by definition is the boundary" shoud read:
E' is the set of all accumulation points of E, which by definition includes the boundary.
Last edited by Hartlw; February 14th 2011 at 09:08 AM. Reason: add original post number
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ENGR 2001 - Engineering Computing Applctns, 1 Credit
Prerequisite(s): MATH 1084 with D or better
Level: Lower
An introductory, software-oriented, engineering computing course using an interactive, high-performance, scientific and engineering software package which integrates computation and visualization in
a programming environment to solve engineering application problems. Topics will include embedded mathematical functions, complex numbers, matrix manipulation, plotting, user defined script and
function files, matrix algebra, numerical techniques and graphical user interfaces.
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Solving ax + b is less than cx + d
Date: 12/11/2001 at 23:54:35
From: Sara
Subject: Solving ax + b < cx + d
Here is a problem:
6-9x is less than or equal to 5x-1
Please teach me the process in English, as if you were explaining to a
Date: 12/12/2001 at 08:50:36
From: Doctor Rick
Subject: Re: Solving ax + b < cx + d
Hi, Sara,
I assume you are learning algebra, but I'll write as if you had never
heard of algebra.
Imagine that you have a balance scale - one of those things with two
pans hanging by chains from opposite sides of a rod. "Justice" is
often pictured holding one of these. You put things on each pan, and
if the right side goes down, then what's in the right pan is heavier
than what's in the left pan.
Now, we put 6-9x in the left pan and 5x-1 in the right pan, and the
right pan goes down - or maybe the pans balance, I can't see it that
clearly. In any case, the left pan DOESN'T go down! This is what "less
than or equal" means: the left side is NOT heavier than the right
6-9x <= 5x-1
Now, if I add the same amount to each pan, it won't change which pan
is heavier. Let's add 1 to both sides. We get
6-9x+1 <= 5x-1+1
The left side is the same as 6+1-9x (we can move the 3 things in the
pan around without changing the total weight). Combine the 6 and
the 1: it's 7-9x.
On the right side, -1+1=0, so all we have is 5x. Thus our balance now
7-9x <= 5x
Let's do the same thing again, only this time I will add 9x to each
7-9x+9x <= 5x+9x
On the left, even though we don't know how much 9x is (since we don't
know what number x stands for), we do know that -9x+9x = 0, just as
-1+1 = 0. On the right side, we have 5 somethings and 9 somethings;
regardless of what "something" is, that's 14 somethings. (This is the
distributive property, for a 13-year-old!) Now we have
7 <= 14x
The last step is harder to visualize with a balance scale. We have to
take what is in each pan, and break it into 14 equal pieces. (Each
of the pieces in the right pan will be x.) Then if all 14 pieces in
the left pan weigh no more than all 14 pieces in the right pan, we
know that one of the pieces from the left pan will also weigh no more
than one of the pieces from the right pan. In other words, we can
divide each side by 14:
7/14 <= 14x/14
1/2 <= x
Now I'm talking to a 13-year-old again. The one tricky part in working
with inequalities is that, unlike when solving equations, you aren't
allowed to divide or multiply by a negative number. Actually, you can
do this, but you have to switch the direction of the inequality. If
this confuses you, you can always find a way to solve the inequality
without multiplying or dividing by a negative number. I did this by
moving the -9x over to the other side, so that I didn't have a
negative coefficient of x.
Does this help? Feel free to ask me about anything that still doesn't
make sense to you.
- Doctor Rick, The Math Forum
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[Rd] Design choice of plot.design for formulas
Thaler,Thorn,LAUSANNE,Applied Mathematics Thorn.Thaler at rdls.nestle.com
Mon Mar 19 15:44:37 CET 2012
Dear all,
Today I figured out that the formula interface of plot.design is kind of
counter intuitive. Suppose the following setting
ddf <- expand.grid(a=factor(1:3), b=factor(1:3))
ddf$y <- rnorm(9)
plot.design(y ~ a + b, data=ddf)
which does what it should do, basically printing the means for the
respective levels of the factors. I had to learn that the function does
not care at all whether I specify a variable at the LHS or the RHS of
the formula. Thus, the following commands are all equivalent
plot.design(~ y + a + b, data=ddf)
plot.design(a ~ y + b, data=ddf)
plot.design(b ~ y + a, data=ddf)
A closer look into the code revealed that the function basically looks
whether a variable is numeric or a factor. All factors are supposed to
be stratification factors, while all numerical variables are supposed to
be responses. While the former assumption makes sense, the latter is
misleading in conjunction with the formula interface:
ddf$z <- sample(3, 9, TRUE)
plot.design(y ~ a + z, data=ddf)
In my reading that should produce a plot where a and z are regarded as
stratification factors, while y is the response. Instead the function
regards y and z as responses.
So my question: is there a particular reason why the formatting of a
variable in a data frame (factor vs. numerical) takes precedence over
the specification in the formula interface of plot.design? Is it the
case that one cannot specify multiple responses otherwise? In this case,
I was wondering whether an approach like in lattice where one can
specify multiple responses would be useful:
ddf$y.new <- rnorm(9)
lattice:::xyplot(y + y.new ~ a, data = ddf, pch = 15)
Thanks for your feedback.
Kind regards,
More information about the R-devel mailing list
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word problem division in to zero [Archive] - Free Math Help Forum
View Full Version : word problem division in to zero
why is division into zero or 0/x is equal to zero? Because any number you divide into zero will always give you zero.
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Fuzzy pizza
Fuzziness is usually the last thing you would want to associate with a pizza, as it conjures up images of quietly decaying remnants of last week's dinner. However, researchers from University
College, Dublin are relying on fuzziness to help deliver you the perfect pizza.
As pizza has become more popular, producers have had to develop their manufacturing methods to ensure efficiency and quality - the last thing hungry customers want is a wonky non-circular base and
sloppily spread sauce. To this end, Da-Wen Sun and Tadhg Brosnan have used a combination of computer vision and fuzzy logic to monitor the quality of pizza bases and how sauce is spread in pizza
By analysing images of pizza bases, they used computer methods to calculate measures of the quality of the base, such as overall diameter and circularity. Similarly, images of bases spread with sauce
provided the information needed to calculate the sauce area percentage and other relevant sauce quality measures.
From this data they then established fuzzy logic rules for the computer to decide whether a photographed pizza was of an acceptable standard. "Fuzzy logic embodies the nature of [the human] mind in
some sense," the researchers said in their paper in Journal of Food Engineering. This is because fuzzy logic can use approximate information and uncertainty to generate decisions, just as we can.
Fuzzy logic is based on a mathematical process of deciding to which fuzzy set (e.g. the set of acceptable pizzas, the set of unacceptable pizzas) a fuzzy variable (in this case, the percentage of the
pizza base covered in sauce) belongs. The flexibility of fuzzy logic comes from the blurred, or fuzzy, boundaries of these sets - an object can partially belong to a set and the sets can overlap. The
sets are described by a mathematical equation called a Membership Function (MF), which gives every object a membership value between 1 (completely in the set) and 0 (completely out of the set). This
makes fuzzy logic much more similar to a human assessment of such questions than strict Boolean logic. In Boolean logic an object is either in (an MF of 1) or out (an MF of 0) of a set, whereas in
the fuzzy universe, not only can objects be completely in the set, or completely out, but they can "sort of be", or "be just a bit" in the set (an MF somewhere between 0 or 1).
In fact, according to their trials, the researchers' methods of fuzzy logic gave a 92% accurate assessment of sauce spread on pizza bases, compared to human monitoring. An encouraging result for
discerning pizza connoisseurs, who may now be asking for an extra topping of fuzzy logic with their next pizza.
Submitted by Anonymous on November 24, 2012.
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Math Forum: Teacher2Teacher - Q&A #101
View entire discussion
From: Sarah (for Teacher2Teacher Service)
Date: Jul 19, 1998 at 12:01:25
Subject: Re: Graphing Calculator Courses
Hi Veronica,
Here are some more sites on the Web to look at, found by searching the Math
Forum archives using the form at
using the keywords graphing calculator (select "that exact phrase"):
Math Teacher Link:
Algebra Through Modeling with the TI-82 (or TI-83) Calculator
The purpose of this module is to familiarize mathematics teachers at the
high school and lower division college level with the content and teaching
strategies for a promising new approach to teaching advanced algebra. This
approach stresses modeling and solving real world problems and develops
skills and concepts of algebra as needed for this modeling process. In
this course, the students work in a collaborative learning environment and
make extensive use of a graphing calculator with data analysis and
sequence capabilities such as those found on the TI-82 or TI-83.
Advanced Algebra Through Data Exploration - A Graphing Calculator Approach (a
textbook from Key Curriculum Press)
"an all-new text that combines data exploration, hands-on activities, and
the power of graphing calculators to enhance the content of advanced
algebra. Embracing the vision and spirit of the National Council of
Teachers of Mathematics (NCTM) Curriculum and Evaluation Standards for
School Mathematics, this innovative text emphasizes conceptual
understanding rather than algebraic technique."
The Graphing Calculator: Teachers and Students Learning Together
A project from the Oklahoma State University Department of Mathematics,
Stillwater, OK. Teacher Learning Systems for the TI-82 and TI-85 graphic
Mathematics Education 3103 - David Royster
A series of lessons for secondary school mathematics teachers that focus
on a variety of tools and issues for math education. Included are:
introduction to the TI-82 graphing calculator, linear equations on a
calculator, using a graphing calculator in teaching Algebra I and II,
using a graphing calculator to explore polynomial functions, introduction
to data analysis using the TI-82 calculator, box plots, two variable data
and linear relationships, introduction to TI-82 programming, programming
concepts, introduction to Maple, comments on using calculators, data
resources for the course, and a bibliography. Lesson plans cover basic
concepts, pose questions, and suggest activities.
NuCalc Graphing Calculator for Mac and Windows - Pacific Tech
NuCalc 2.0: A Graphing Calculator for Macintosh and Windows that allows
the user to calculate and view two- and three-dimensional mathematical
objects easily. The web site includes a list of features, a guided tour, a
picture gallery, a free demo, and ordering capabilities, as well as
similar information on the company's other products.
For more, see the UTK Math Archives graphing calculator listings at
- Sarah, for the Teacher2Teacher Service
Post a public discussion message
Ask Teacher2Teacher a new question
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Re: Conversion load to mass
Hi dear
why there is a difference between the modal mass and dynamic mass?
And one more problem
the Base shear coefficient that the robot 2011 is calculating is too big , about 2.5 in dynamic mass or sometimes 0.55 in modal mass but the real value is about 0.35
I have used ubc 97 for applying seismic load.
but v didn't calculate from multiply c.w from the value for c and w that the robot 2011 has calculated.
Please help me
thank you .
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Balloon Payment Calculator
The Balloon Payment Calculator is a very fast, easy-to-use and flexible loan calculator which also handles balloon payments. The Balloon Payment Calculator will calculate what the regular payment
amount needs to be to result in a given balloon for a particular payment number. It will also calculate the balloon payment amount when the user specifies the regular payment.
For example, assume that you have taken out a loan for $500,000, with an annual interest rate of 8 1/2%. If, after the 48th period, you want a $400,000 Balloon Payment due, you must make regular
payments of $5,298.16 assuming a monthly payment period with monthly compounding and payments made in arrears.
Our new, enhanced Balloon Payment Calculator will solve for any one of five unknowns:
• Regular periodic payment amount
• Amount of loan
• Final balloon payment amount
• Period balloon payment is due
• Interest rate
The Balloon Payment Calculator is one of our most popular calculators used by borrowers when they want to design their own custom loan. It makes what-if financial planning for loans fast and easy.
Our Flexible Amortization Schedule will produce an amortization table for a balloon loan.
Here's our online balloon payment calculator.
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Comparing Singapore Math to Saxon Math
The math wars of the 1990′s have quieted down and are almost a thing of the past.
With the release of their 2006 guidelines, the National Council for Teachers of Mathematics (NCTM) effectively ended 17 years of promoting ‘reform math’ programs and acknowledged the need for schools
to return to more traditional math programs.
Teachers and Homeschool parents can now move on to the next question. Which traditional math curriculum is best? The correct answer depends largely on the needs and preferences of each teacher and
There are many traditional math programs to choose from. This provides a brief review of two of the most popular programs, Singapore Math and Saxon Math. These two curriculums have some things in
• Both are used in public schools, private schools and homeschools
• Both have clear track records of improving standardized test scores
• Both share the traditional math emphasis on math facts as the building blocks of all math concepts
• Both have proven to be effective with a wide range of students
They also have some important differences:
Cost Comparison -Saxon Math books are more expensive than Singapore Math books because Saxon has a lot more pages. Saxon student books are hard cover from 8th grade and up.
Saxon Emphasizes Practice – Saxon Math puts more emphasis on doing practice exercises while Singapore Math puts more emphasis on critically thinking through concepts. After concepts are introduced,
Saxon moves immediately into practice exercises to help cement the concept in the student’s mind. Saxon requires students to memorize formulas, achieve fluent recall of math facts and apply
algorithms to solve problems.
Singapore Emphasizes Thinking – Singapore teachers spend more time helping students to think through and verbally discuss each component of the concept. Singapore Math avoids reliance on memorized
formulas and algorithms so there is not as much emphasis on repetitive practice exercises. Instead, Singapore strives to give students an understanding of math concepts by walking students through
each component of a problem, and then presenting them with the whole problem to solve. This way, students are trained to think actively as they work through each step of a problem instead of plugging
the problem into a formula.
Saxon is More Structured – Saxon Math is more structured, making it easier for teachers and students to follow the road map. Each new concept is followed by practice exercises. Review questions are
provided after every 10 lessons.
Singapore Requires More Teaching – Singapore Math is less structured, using an approach which is less familiar to anyone who learned math in the U.S. As a result, Singapore Math can be more
challenging for U.S. teachers and students, especially older students who are already familiar with U.S. math programs. Singapore’s approach puts more burden on teachers to:
• spend more time teaching new concepts, breaking the concepts into components to ensure students are understanding
• stimulate verbal discussion of the concepts
• supplement the material as needed with flash cards, manipulative items, and extra drills
• continually assess how well students are grasping concepts then provide additional assistance as needed
Singapore is More Focused – Saxon Math practice exercises blend previously covered concepts together with new concepts, forcing students to continually review previous concepts. The rationale is
repetitive practice over time is necessary to grasp the concepts and to achieve quick and effortless recall of math facts. Singapore focuses on one concept at a time, seeking mastery of each concept
before moving on to the next one. One of the reasons why the NCTM liked the Singapore curriculum is because it focuses only on a few key concepts for each school year. The NCTM recognized a key
weakness with some U.S. programs is having too many objectives, making them incoherent and difficult for students to master anything.
Recommendations – For school teachers who are willing to try something new and put more effort into teaching, my recommendation is Singapore Math. Singapore students lead the world in math test
scores and your students can do the same. For Homeschool parents who are pressed for time and need a program that allows students to work more independently, my recommendation is Saxon Math. Saxon
also offers CD ROM teaching videos to enable students to work even more independently. Saxon is also a great choice for U.S. school teachers looking for a program with a more familiar approach. Saxon
Math has proven success with a wide range of students, even turning struggling math students into math lovers!
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