Wikipedia:Reference desk/Archives/Mathematics/2009 February 11

= February 11 =

Help me grok e
This is a result that mathematicians everywhere seem to take for granted:


 * $$\lim_{n\to\infty} (1 + \frac{r}{n})^n = (\lim_{n\to\infty} (1 + \frac{1}{n})^n)^r = e^r$$

Can someone provide me with a proof of it? --Tigerthink (talk) 01:02, 11 February 2009 (UTC)


 * Which equals sign are you looking for a proof of? The second one can be taken as a definition of e. There are various alternative definitions, which you can prove are equivalent - is there a particular definition you favour? The first equality is intuitively obvious, I think (just expand out brackets for the first few terms), it may require a little more work to make it rigorous (I can't do rigorous analysis at 1:20am...). --Tango (talk) 01:21, 11 February 2009 (UTC)


 * Take a logarithm. Find the limit of the result. Re-exponentiate. Ray (talk) 02:18, 11 February 2009 (UTC)

Assuming that you already know that $$\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n=e$$ then

$$\lim_{n\rightarrow\infty}\left(1+\frac{r}{n}\right)^n=\lim_{j\rightarrow\infty}\left(1+\frac{1}{j}\right)^{rj}$$ by making the substitution jr=n

$$=\lim_{j\rightarrow\infty}\left(\left(1+\frac{1}{j}\right)^j\right)^r=\left(\lim_{j\rightarrow\infty}\left(1+\frac{1}{j}\right)^j\right)^r=e^r$$. This should take care of both equalities.-Looking for Wisdom and Insight! (talk) 07:25, 11 February 2009 (UTC)


 * If you want to do it well and in elementary way, here is the program.
 * Program: prove that for any real number r the sequence $$\lim_{n\to\infty} (1 + \frac{r}{n})^n$$ is increasing as soon as n>|r|.  Prove that it is bounded. So it is convergent: define exp(r) its limit. Prove that exp(r+s)=exp(r)exp(s) for all real numbers. Prove the equality  with the exponential series. etc.--194.95.184.74 (talk) 09:39, 11 February 2009 (UTC)

I don't understand this proof
This was a proof for a statement on the binomial coefficient page.

Furthermore,
 * $$ \binom n k \equiv 0 \pmod{n} $$

for all 0 < k < n if and only if n is prime.

We can prove this as follows: When p is prime, p divides
 * $$ \binom p k = \frac{p \cdot (p-1) \cdots (p-k+1)}{k \cdot (k-1) \cdots 1} $$ for all 0 < k < p

because it is a natural number and the numerator has a prime factor p but the denominator does not have a prime factor p. So $$ \tbinom p k $$&equiv;0 (mod p)

Unfortunately, I don't understand how the conclusion that $$ \tbinom p k $$&equiv;0 (mod p) is reached. It's probably very simple, and I'm just missing it, but could someone explain it please. Btw I'm somewhat familiar with modular arithmatic, but I would prefer it was explained in another context (i.e. by divisbility). Thanks,. —Preceding unsigned comment added by 65.92.237.46 (talk) 06:12, 11 February 2009 (UTC)


 * The numerator is $$p \cdot (p-1) \cdots (p-k+1)$$ with k > 0 so it is clear that this is a multiple of p. The denominator is $$k \cdot (k-1) \cdots 1$$ and this is not a multiple of p because k is less than p and p is prime - if the denomiantor were a multiple of p then we could find a non-trivial factorisation of p, which is impossible. So we are dividing a multiple of p by something that is not a multiple of p - we know the result is an integer, but it must also be a multiple of p because there is no factor of p in the denominator to cancel the factor of p in the numerator. Gandalf61 (talk) 09:24, 11 February 2009 (UTC)
 * In case it's the last step that's troubling you, be aware that 'a&equiv;0 (mod n)' is just a fancy way of saying that n divides a. Algebraist 11:51, 11 February 2009 (UTC)

Many thanks. —Preceding unsigned comment added by 65.92.237.46 (talk) 11:53, 11 February 2009 (UTC)

Degree of Relatedness
Not sure if this is more math or biology, but here goes. What is the degree of relatedness between two children of an incestuous relationship between a half-sister and half-brother, I feel like it's 0.75, but I'm not, any ideas? 169.229.75.128 (talk) 07:12, 11 February 2009 (UTC)


 * The expected value of the degree of relatedness of the offspring of two half-siblings (that is, sharing one parent) is 9/16. To get an expected value of degree of relatedness of 3/4 you need to consider the offspring of two clones.


 * A few things: the degree of relatedness of two offspring is a distribution (in theory, the offpsring could be identical, or could have no genes in common, although these two extremes are extremely unlikely), so properly you should be talking about the expected value of their degree of relatedness.


 * Secondly, I find the term "degree of relatedness" a bit confusing and loaded. For some species, such as ants, where different individuals may have different numbers of genes, it is possible for the degree of relatedness of A and B to be different from the degree of relatedness of B and A.  It is somewhat clumsier but more descriptive to say:  "given a gene of A, what is the probability that that gene is found in B?".  Thinking about the problem in that manner may also make it easier to go about solving it.  Eric.  131.215.158.184 (talk) 10:13, 11 February 2009 (UTC)


 * How did you get 9/16, though. 169.229.75.128 (talk) 16:43, 11 February 2009 (UTC)


 * I believe that was explained on another Desk. I agree that it's correct.  Posting on multiple Desks isn't generally allowed, as it leads to us having to repeat ourselves.  StuRat (talk) 00:01, 12 February 2009 (UTC)


 * By the way, just in case someone finds their way back to this thread in the future, I was in fact using the wrong definition of coefficient of relationship and the answer is actually 5/8. Eric.  131.215.158.184 (talk) 07:11, 14 February 2009 (UTC)

Absolute Integrability
This issue came up defining Fourier transform in my PDE class. Let us say that a function f is absolutely integrable if $$\int_{-\infty}^{\infty}|f|$$ is a finite number. My question is, if a real function f is absolutely integrable, then can't we already say (without assuming anything extra) that $$\lim_{x\rightarrow\infty}f(x)=0$$. I mean how can a function have a nonzero limit as x grows without bound and still be absolutely integrable. If the limit is any nonzero number, then wouldn't the absolute integral be infinite? The same will be true as x goes to negative infinity. That limit must be zero as well. Is my reasoning correct or wrong?-Looking for Wisdom and Insight! (talk) 07:33, 11 February 2009 (UTC)
 * The limit could simply fail to exist. Imagine (a smoothed version of) Σ n χ[n,n+1/n3).  It is a non-negative function with integral Σ 1/n2 ≤ 2, where all sums are over the positive integers.  However, its limsup on every interval [a,∞) is +∞, its liminf on every interval [a,∞) is 0, and the limit as x → ∞ does not exist. JackSchmidt (talk) 09:09, 11 February 2009 (UTC)
 * By the way, such functions are normally called simply integrable (or L1). Algebraist 11:49, 11 February 2009 (UTC)

Or they're called "Lebesgue-integrable".

So say you have a non-negative function with a pulse of height 1 at 1, another at 2, another at 3, and so on. But the pulses keep getting narrower, so the sum of all the areas under them is a convergent series. Then the integral from 0 to &infin; is finite but the function does not approach 0 at &infin;. Michael Hardy (talk) 21:12, 11 February 2009 (UTC)


 * But you can always say without any extra assumption that $$\liminf_{x\to\pm\infty} |f| =0$$. Unfortunately, it is possible that $$\limsup_{x\to\pm\infty} |f| >0$$ (Igny (talk) 05:09, 15 February 2009 (UTC))

GENERAL QUESTION>
I'm not asking for an answer for this question so it ain't homework i just couldn't figure out a logic to go about this sum.It's as follows ->Calculate the total natural numbers that exist from 0 to 2000 which has a square such that its sum of digits is 21. .Now no need to tell me any answer but just tell me a basic logic i can use to go about doing this sum. —Preceding unsigned comment added by Vineeth h (talk • contribs) 14:32, 11 February 2009 (UTC)
 * It's a trick question. Algebraist 14:52, 11 February 2009 (UTC)


 * The only sum of digits you get are: 0, 1, 4, 7, 9, 10, 13, 16, 18, 19, 25, 27, 28, 31, 34, 36, 37, 40, 43, 45, 46, 49. -- SGBailey (talk) 15:05, 11 February 2009 (UTC)
 * You missed 22. Algebraist 15:15, 11 February 2009 (UTC)
 * Consider divisibility rules. — Emil J. 15:11, 11 February 2009 (UTC)

SGBailey-> you're right,turns out 21 isn't divisible by any number so the final answer turns out to be 0 natural numbers.So could you please tell me how you generalised that the only sum of digits you get are the ones you mentioned?Whats the logic behind that?Cause one can't possibly remember all the numbers you mentioned there just to solve a sum like this!Vineeth h (talk) 17:12, 11 February 2009 (UTC)


 * I must have deleted 22 by mistake when doing the previous edit. The "algorithm" I used was to do the calculation 2001 times x -> x^2 -> sum(digits(x^2)). -- SGBailey (talk) 20:11, 11 February 2009 (UTC)
 * No need for scare quotes - brute force is a perfectly legitimate algorithm! --Tango (talk) 20:13, 11 February 2009 (UTC)


 * You don't need to know the list. Just remember the divisibility criteria for 3 and 9: if a number has sum of digits 21, then it is divisible by 3, but not 9, and that's impossible for a square. — Emil J. 17:23, 11 February 2009 (UTC)


 * This BASIC code prints out the sums of digits that occur and how often they occur. Sorry, the sum 21 never occurs. Cuddlyable3 (talk) 15:09, 12 February 2009 (UTC)

deflng a-z dim f(49) for x=0 to 2000 x2=x^2 sd=0 for p=6 to 0 step -1 'powers of 10 dp=int(x2/10^p) sd=sd+dp x2=x2-dp*10^p next p incr f(sd) next x for n=0 to 49 if f(n)>0 then print n;f(n) next n