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Exponential limit

Jul 22, 2012
35
Show that $$ \lim_{n \to \infty} \prod_{1 \le k \le n}\left(1+\frac{kx}{n^2}\right) = e^{x/2}$$​
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
We are given:

$\displaystyle \lim_{n\to\infty}\prod_{k=1}^n\left(1+\frac{kx}{n^2} \right)=L$

$\displaystyle \lim_{n\to\infty}\ln\left(\prod_{k=1}^n\left(1+ \frac{kx}{n^2} \right) \right)=\ln(L)$

$\displaystyle \sum_{k=1}^n\lim_{n\to\infty}\frac{\ln\left(\left(1+\frac{kx}{n^2} \right)^{n^2} \right)}{n^2}=\ln(L)$

$\displaystyle \sum_{k=1}^n\lim_{n\to\infty}\frac{kx}{n^2}=\ln(L)$

$\displaystyle x\lim_{n\to\infty}\frac{n^2+n}{2n^2}=\ln(L)$

$\displaystyle \frac{x}{2}=\ln(L)$

$\displaystyle L=e^{\frac{x}{2}}$
 
Jul 22, 2012
35
Brilliant! (Clapping)

It's much easier than the solution I know.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
If you are so inclined, I would be interested in seeing another method. My method tacitly assumes the limit exists. I suspect your method is more rigorous.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
We are given:

$\displaystyle \lim_{n\to\infty}\prod_{k=1}^n\left(1+\frac{kx}{n^2} \right)=L$

$\displaystyle \lim_{n\to\infty}\ln\left(\prod_{k=1}^n\left(1+ \frac{kx}{n^2} \right) \right)=\ln(L)$

$\displaystyle \sum_{k=1}^n\lim_{n\to\infty}\frac{\ln\left(\left(1+\frac{kx}{n^2} \right)^{n^2} \right)}{n^2}=\ln(L)$

$\displaystyle \sum_{k=1}^n\lim_{n\to\infty}\frac{kx}{n^2}=\ln(L)$

$\displaystyle x\lim_{n\to\infty}\frac{n^2+n}{2n^2}=\ln(L)$

$\displaystyle \frac{x}{2}=\ln(L)$

$\displaystyle L=e^{\frac{x}{2}}$
Hi MarkFL, :)

I don't know if I am missing something here, but I don't quite understand how you got,

\[\sum_{k=1}^n\lim_{n\to\infty}\frac{\ln\left(\left( 1+\frac{kx}{n^2} \right)^{n^2} \right)}{n^2}=\ln(L)\]

from,

\[\lim_{n\to\infty}\sum_{k=1}^n\frac{\ln\left(\left( 1+\frac{kx}{n^2} \right)^{n^2} \right)}{n^2}=\ln(L)\]

Can you please elaborate? :)

Kind Regards,
Sudharaka.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,705
My method tacitly assumes the limit exists. I suspect your method is more rigorous.
I suspect that your argument can be modified to avoid the assumption that the limit exists (and also avoiding the fancy footwork with interchanging limits and taking partial limits (Wondering)).

Start by using the Taylor expansion of $\ln(1+t)$ to deduce that $\ln(1+t) = t + O(t^2)$ as $t\to0$. Then $$\begin{aligned}\sum_{k=1}^n \ln\Bigl(1 + \frac{kx}{n^2}\Bigr) &= \sum_{k=1}^n \Bigl(\frac{kx}{n^2} + O\bigl(\tfrac{k^2}{n^4}\bigr)\Bigr) \\ &= \frac{\frac12n(n+1)x}{n^2} + n\cdot O\bigl(\tfrac1{n^2}\bigr) \\ &= \tfrac12x + O\bigl(\tfrac1{n}\bigr). \end{aligned} $$ Now let $n\to\infty$ to get $\displaystyle \lim_{n\to\infty}\sum_{k=1}^n \ln\Bigl(1 + \frac{kx}{n^2}\Bigr) = \frac x2$ and hence (because the exponential function is continuous) $\displaystyle \lim_{n \to \infty} \prod_{k=1}^n \Bigl(1+\frac{kx}{n^2}\Bigr) = e^{x/2}.$

I'm not entirely confident about that argument because it's all too easy to use the "big O" notation to disguise lapses in rigour. But it looks to me as though it's basically correct.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hi MarkFL, :)

I don't know if I am missing something here, but I don't quite understand how you got,

\[\sum_{k=1}^n\lim_{n\to\infty}\frac{\ln\left(\left( 1+\frac{kx}{n^2} \right)^{n^2} \right)}{n^2}=\ln(L)\]

from,

\[\lim_{n\to\infty}\sum_{k=1}^n\frac{\ln\left(\left( 1+\frac{kx}{n^2} \right)^{n^2} \right)}{n^2}=\ln(L)\]

Can you please elaborate? :)

Kind Regards,
Sudharaka.
I suppose it was unnecessary to take the summation outside the limit, but I was using the property that the limit of a sum is the sum of the limits.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
I suppose it was unnecessary to take the summation outside the limit, but I was using the property that the limit of a sum is the sum of the limits.
That is true but note the the summation includes \(n\) as the upper limit. In that case I don't think that this property generally holds. For example,

\[2=\lim_{n\rightarrow\infty}\sum_{k=1}^{n}2^{k-n}\neq\sum_{k=1}^{n}\lim_{n\rightarrow\infty}2^{k-n}=0\]

Kind Regards,
Sudharaka.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Very good point...thank you!

Rather than edit my initial post, I will rewrite it so that others at my level might see the mistake and learn from it as well.

We are given:

$\displaystyle \lim_{n\to\infty}\prod_{k=1}^n\left(1+\frac{kx}{n^2} \right)=L$

$\displaystyle \lim_{n\to\infty}\ln\left(\prod_{k=1}^n\left(1+ \frac{kx}{n^2} \right) \right)=\ln(L)$

$\displaystyle \lim_{n\to\infty}\sum_{k=1}^n\frac{\ln\left(\left(1+\frac{kx}{n^2} \right)^{n^2} \right)}{n^2}=\ln(L)$

$\displaystyle \lim_{n\to\infty}\sum_{k=1}^n\frac{kx}{n^2}=\ln(L)$

$\displaystyle x\lim_{n\to\infty}\frac{n^2+n}{2n^2}=\ln(L)$

$\displaystyle \frac{x}{2}=\ln(L)$

$\displaystyle L=e^{\frac{x}{2}}$
 
Jul 22, 2012
35
Not much of a mistake, tbf. In this case interchanging the limit and the sum works; it's the justification that's missing.

In your rewrite, I'm not sure how you reduced the summand to $\frac{kx}{n^2}$ without interchanging the sum and the limit, though?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Good point. :)
 
Jul 22, 2012
35
Using Bernoulli's Inequality we have:

$ \begin{aligned} \displaystyle & \hspace{0.5in}\bigg(1+\frac{x}{n}\bigg)^{\frac{k}{n}} \le \bigg(1+\frac{kx}{n^2}\bigg) \le \bigg(1+\frac{x}{n^2}\bigg)^k \\& \iff \prod_{ 0\le k \le n}\bigg(1+\frac{x}{n}\bigg)^{\frac{k}{n}} \le \prod_{ 0\le k \le n}\bigg(1+\frac{kx}{n^2}\bigg) \le \prod_{ 0\le k \le n}\bigg(1+\frac{x}{n^2}\bigg)^k \\& \iff \hspace{0.1in} \bigg(1+\frac{x}{n}\bigg)^{ \frac{1}{2}(n+1)} \le \prod_{0 \le k \le n}\bigg(1+\frac{kx}{n^2}\bigg) \le \bigg(1+\frac{x}{n^2}\bigg)^{ \frac{1}{2} n(n+1)} \\& \iff \lim_{n \to \infty}\bigg(1+\frac{x}{n}\bigg)^{ \frac{1}{2}(n+1)} \le \lim_{n \to \infty}\prod_{0 \le k \le n}\bigg(1+\frac{kx}{n^2}\bigg) \le \lim_{n \to \infty}\bigg(1+\frac{x}{n^2}\bigg)^{ \frac{1}{2} n(n+1)} \\& \iff e^{x/2} \le \lim_{n \to \infty}\prod_{0 \le k \le n}\bigg(1+\frac{kx}{n^2}\bigg) \le e^{x/2}.\end{aligned}$

Therefore $\displaystyle \lim_{n \to \infty}\prod_{0 \le k \le n}\bigg(1+\frac{kx}{n^2}\bigg) = e^{x/2}$.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Using Bernoulli's Inequality we have:

$ \begin{aligned} \displaystyle & \hspace{0.5in}\bigg(1+\frac{x}{n}\bigg)^{\frac{k}{n}} \le \bigg(1+\frac{kx}{n^2}\bigg) \le \bigg(1+\frac{x}{n^2}\bigg)^k \\& \iff \prod_{ 0\le k \le n}\bigg(1+\frac{x}{n}\bigg)^{\frac{k}{n}} \le \prod_{ 0\le k \le n}\bigg(1+\frac{kx}{n^2}\bigg) \le \prod_{ 0\le k \le n}\bigg(1+\frac{x}{n^2}\bigg)^k \\& \iff \hspace{0.1in} \bigg(1+\frac{x}{n}\bigg)^{ \frac{1}{2}(n+1)} \le \prod_{0 \le k \le n}\bigg(1+\frac{kx}{n^2}\bigg) \le \bigg(1+\frac{x}{n^2}\bigg)^{ \frac{1}{2} n(n+1)} \\& \iff \lim_{n \to \infty}\bigg(1+\frac{x}{n}\bigg)^{ \frac{1}{2}(n+1)} \le \lim_{n \to \infty}\prod_{0 \le k \le n}\bigg(1+\frac{kx}{n^2}\bigg) \le \lim_{n \to \infty}\bigg(1+\frac{x}{n^2}\bigg)^{ \frac{1}{2} n(n+1)} \\& \iff e^{x/2} \le \lim_{n \to \infty}\prod_{0 \le k \le n}\bigg(1+\frac{kx}{n^2}\bigg) \le e^{x/2}.\end{aligned}$

Therefore $\displaystyle \lim_{n \to \infty}\prod_{0 \le k \le n}\bigg(1+\frac{kx}{n^2}\bigg) = e^{x/2}$.
Hi QuestForInsight, :)

Thank you for the solution. I think this proof is valid only for \(x>-1\) since the Bernoulli's inequality is defined as such. Isn't? :)
 

Poirot

Banned
Feb 15, 2012
250
Why do you say that assuming the limit exists is sloppy? Since the implications all reverse, the result is fully justified.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
What if the limit didn't exist at all, and you arrived at such 'solution'? It may seem a trivial matter, but beforehand you couldn't affirm that the limit exists. What if the question was posed as "Find out whether the limit exists or not"?

You cannot assume the existence of certain objects a priori, that's why the argument was labeled as sloppy. (Wondering)
 

Poirot

Banned
Feb 15, 2012
250
What if the limit didn't exist at all, and you arrived at such 'solution'? It may seem a trivial matter, but beforehand you couldn't affirm that the limit exists. What if the question was posed as "Find out whether the limit exists or not"?

You cannot assume the existence of certain objects a priori, that's why the argument was labeled as sloppy. (Wondering)
As I say, if the implications reverse then it is both neccesary and sufficent that the answer is the limit. In plain language, it is the limit.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
please excuse my denseness, but i am troubled by a few points.

first of all, for which x is this supposed to hold? the reason i ask is because of the questions raised by Sudharaka regarding the Bernoulli's inequality proof, and because of MarkFL's liberal use of the logarithm function (which is only defined for positive real numbers).

i am also troubled by the point raised in post #8. i can't see that this objection has been fully answered. to justify a method by "it gives the right answer" seems poor reasoning:

after all, 2*2 = 2+2, but we cannot conclude from this that if we need to square a number, we can get by with doubling it.

i am also troubled by the inequalities in line 2 of post #12, they seem to tacitly assume each term in the n-fold products is non-negative (am i missing something?). i fear things may go very badly if x = -100, for example.

i understand the general strategy here: we want to leverage a known limit for e into something we can use. since both sides of the originally posted limit make sense for all real x, i would feel better about a proof which does not restrict what x may be (in all fairness, some of these concerns might be allayed by requiring n be at least large enough to ensure my concerns are invalid, but this point seems glossed over in the preceding posts).

please enlighten me, for my own peace of mind :)
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Maybe you could try using Weierstrass Product/Factorization Theorem.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,705
please excuse my denseness, but i am troubled by a few points.

first of all, for which x is this supposed to hold? the reason i ask is because of the questions raised by Sudharaka regarding the Bernoulli's inequality proof, and because of MarkFL's liberal use of the logarithm function (which is only defined for positive real numbers).

i am also troubled by the point raised in post #8. i can't see that this objection has been fully answered. to justify a method by "it gives the right answer" seems poor reasoning:

after all, 2*2 = 2+2, but we cannot conclude from this that if we need to square a number, we can get by with doubling it.

i am also troubled by the inequalities in line 2 of post #12, they seem to tacitly assume each term in the n-fold products is non-negative (am i missing something?). i fear things may go very badly if x = -100, for example.

i understand the general strategy here: we want to leverage a known limit for e into something we can use. since both sides of the originally posted limit make sense for all real x, i would feel better about a proof which does not restrict what x may be (in all fairness, some of these concerns might be allayed by requiring n be at least large enough to ensure my concerns are invalid, but this point seems glossed over in the preceding posts).

please enlighten me, for my own peace of mind :)
When looking at the limit (if it exists) $\displaystyle \lim_{n \to \infty} \prod_{k=1}^n \Bigl(1+\frac{kx}{n^2}\Bigr)$, we are only interested in what happens when $n$ gets large. So we might as well assume from the start that we are only going to look at values of $n$ large enough to ensure that $|x/n|$ is small. In particular, if $|x/n|<1$ then $1-\dfrac{kx}{n^2}>0$ (for $1\leqslant k\leqslant n$), and so it will be legitimate to take its logarithm. If you then use the argument in my post #6, you can check that the limit exists and equals $e^{\,x/2}$, for all real $x$.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
yes, i thought that might be the case (which answers both the concerns i had about MarkFL's logs and QuestForInsight's products).

and it is important for post #6, because of the comparatively small region of convergence for the taylor series of log(1+t) (presumably centered at 0).

the convergence is important (in my mind, anyway) so that we know the big O terms can indeed be disregarded.

and yes, that does avoid the limit/sum interchange the proof of which seems to me problemmatic.

so far, your (Opalg's) proof is the only one that "makes sense" to me.

i am also puzzled by one of the inequalities in QuestForInsight's post (#12). try as i might, i cannot obtain the lower bound, and it seems to me that bernoulli's inequality actually tells us:

\(\displaystyle \left(1 + \frac{kx}{n^2}\right) \leq \left(1 + \frac{x}{n}\right)^{\frac{k}{n}}\)

again, i may just be old and senile, but i find this particular problem rather interesting, and would like to understand it better.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
i am also puzzled by one of the inequalities in QuestForInsight's post (#12). try as i might, i cannot obtain the lower bound, and it seems to me that bernoulli's inequality actually tells us:

\(\displaystyle \left(1 + \frac{kx}{n^2}\right) \leq \left(1 + \frac{x}{n}\right)^{\frac{k}{n}}\)

again, i may just be old and senile, but i find this particular problem rather interesting, and would like to understand it better.
Hi Deveno, :)

I think that the inequality in post #12 is correct. We have to use the Generalized Bernoulli's inequality to show that.

Kind Regards,
Sudharaka.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967