Wikipedia:Reference desk/Archives/Mathematics/2007 October 8

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October 8

Mathematical Induction

How would one show that 7^k-4^k-3^k is divisible by 12 by math induction?

I know how to do all the pre-requisite stuff... but I am stuck at the point where I have to show that Pk+1 is true (k+1 should be on a lower level).

So now, I sub all the k's with (k+1)'s. What is the next step I should do?

Thanks. Acceptable 02:52, 8 October 2007 (UTC)

If you're trying to prove f(x) = g(x), at that stage you'd be trying to prove that the f(k+1) case is true assuming that case f(k) is true, so, by your assumption:

f(k) = g(k)

So now you're trying to prove:

f(k+1) = g(k+1)

So what you can do now, is try to group terms of f(k+1) into something that is equal to f(k), at which point you can substitute g(k) for f(k), since you've already assumed that. After doing this, it should be easier to prove that whatever's left over is true.

Take a look at the example at Mathematical induction. Ask again if you still have trouble, maybe show your work. (Help:Formula might be useful.) - Rainwarrior 03:19, 8 October 2007 (UTC)

Wouldn't it be much easier to prove that 7^k-4^k-3^k is divisible by 12 for all integers k through modular arithmetic?? —Preceding unsigned comment added by 69.54.140.201 (talk) 06:08, 8 October 2007 (UTC)

That would be easier, but presumably this is an exercise in which the student is required to use induction. In this case it helps to prove something formally stronger (by induction):

for k ≥ 1, 7^k − 4^k − 3^k is divisible by 12 AND 3·4^k + 4·3^k is divisible by 12.

Then, in the induction step, in proving that 7^k+1 − 4^k+1 − 3^k+1 is divisible by 12, you can also use the assumption (part of the induction hypothesis) that 3·4^k + 4·3^k is divisible by 12.  --Lambiam 06:16, 8 October 2007 (UTC)

On second thoughts, taking "3·4^k + 4·3^k is divisible by 12" along in the induction proof is quite unnecessary, since this can be easily proved (for k ≥ 1) in an entirely straightforward algebraic way.  --Lambiam 19:34, 8 October 2007 (UTC)

ASSISTANCE

I THINK THAT I HAVE COME UP WITH ANEW THEORY ABOUT REAL NUMBERS.I NEED SOME HELP TO ASSESS AND FIX THE MATHEMATICAL CONSTRUCTOR OF THIS THEORY.I WENT TO SOME SCHOOLS OF MATH.WHERE I LIVE BUT I DIDNOT GET ANY UNDERSTANDING.MAY I GET ANY HELP HERE?80.123.226.186 12:06, 8 October 2007 (UTC)h.dyeni

I don't know about assistance, but all caps is the equivalent of shouting in these forums, normal English syntax is preferred.—Cronholm¹⁴⁴ 12:26, 8 October 2007 (UTC)

That said, if you describe your theory we can express our opinion of it. -- Meni Rosenfeld (talk) 12:54, 8 October 2007 (UTC)

Your name is not JSH by any chance is it? 211.28.129.8 14:30, 8 October 2007 (UTC)

I don't believe JSH would have said "I think that I have ...". PrimeHunter 22:43, 8 October 2007 (UTC)

I feel like I'm missing something important here... care to spill the beans? Algebraist 23:11, 8 October 2007 (UTC)

From what I can gather fromm google searches, he looks to be a notorious character in Google Groups, pushing "out there" theories. I could be wrong though. --Ybbor^Talk 23:25, 8 October 2007 (UTC)

Right. James S. Harris aka JSH is probably the most notorious crank in the usenet group sci.math (which Google Groups has an interface to). He often claims to have made great discoveries and usually expresses no doubt about his own claims, despite his less than impressive track record. PrimeHunter 00:18, 9 October 2007 (UTC)

And JSH is now a disambig, in case this comes up again. Algebraist 13:30, 9 October 2007 (UTC)

I would be interested in helping, please do show us what you have so far. —Preceding unsigned comment added by 69.54.140.201 (talk) 23:37, 10 October 2007 (UTC)

Variance problem!

i'd like to ask you about the variance..

why cant we sum variance's normally?? without using the covariance??

thax.. —Preceding unsigned comment added by 86.108.109.227 (talk) 16:45, 8 October 2007 (UTC)

Perhaps our article Variance will help. A better question would be "why should we be able to sum variances normally?". There is just no reason why this should be possible.

Also, consider the following 3 scenarios. In all of them, X is a variable with variance 1. Y can be either equal to $X$, equal to $-X$ or an independent variable which also has variance 1. In the first case, $X+Y$ is equal to $2X$, so its variance is 4 (since in calculating variance you take squares). In the second case, $X+Y$ is the constant 0 so its variance is 0. And in the third case you do get the variance to be equal to 2, the sum of variances. -- Meni Rosenfeld (talk) 17:30, 8 October 2007 (UTC)

I fear you have your cases in the wrong order there. Algebraist 19:00, 8 October 2007 (UTC)

Sorry, thanks. Fixed. -- Meni Rosenfeld (talk) 21:33, 8 October 2007 (UTC)

This can be derived as a simple consequence of the definitions of variance and covariance, together with the property of the expected value operator that it distributes over addition. To simplify things, assume we have two random variables X and Y, both with expected value 0 – possibly by taking some other random variables minus their respective expected values, which does not change their variances or covariance. Then

Var(X + Y) = E((X + Y)²) = E(X² + Y² + 2XY) = E(X²) + E(Y²) + 2E(XY) = Var(X) + Var(Y) + 2Cov(X, Y).

The covariance term vanishes precisely when X and Y are statistically independent.  --Lambiam 19:59, 8 October 2007 (UTC)

The covariance term vanishes with independence, yes, but surely 'precisely' is wrong there: events can be uncorrelated without being independent. Algebraist 23:11, 8 October 2007 (UTC)

I stand corrected.  --Lambiam 05:23, 9 October 2007 (UTC)