What is this equal to?

  • #1
[tex]\lim_{N\rightarrow\infty} \lim_{h\rightarrow 0} \frac{1}{N}\sum_{k=0}^N((kh)^h)^k=[/tex]

and
Nh=L
h = L/N
 

Answers and Replies

  • #2
22,129
3,297
Hi RandomMystery! :smile:

[tex]\lim_ {N\rightarrow +\infty}{\lim_{h\rightarrow 0}{\frac{1}{N}\sum_{k=0}^N{(kh)^{hk}}}}=\lim_ {N\rightarrow +\infty}{\frac{1}{N}\sum_{k=0}^N{\left(\lim_{h \rightarrow 0}{(kh)^h}\right)^k}}[/tex]

So, the first thing you need to do is evaluate

[tex]\lim_{h\rightarrow 0}{(kh)^h}[/tex]
 
  • #3
It seems to go toward 1
 
  • #4
22,129
3,297
Except if k=0, of course. So that leaves us with

[tex]\lim_{N\rightarrow +\infty}{\frac{1}{N}\sum_{k=1}^N{1}}[/tex]

So, can you evaluate this?
 
  • #5
[tex]\lim_{N\rightarrow +\infty}{\frac{1}{N}\sum_{k=1}^L{1^k}}=1[/tex]

L=Nh (it doesn't let me put two characters on top of the sigma)

It's 1^k right? nvm it always equals 1.
Oh, I think that what I meant was from k=1 or 0 to k=Nh not N on the sigma sign.

So, is what I wrote above right? It can't be right though.

[tex]\lim_ {N\rightarrow +\infty}{\lim_{h\rightarrow 0}{\frac{1}{N}\sum_{k=0}^N{(kh)^{hk}}}}=\lim_ {N\rightarrow +\infty}{\frac{1}{N}\sum_{k=0}^N{\left(\lim_{h \rightarrow 0}{(kh)^h}\right)^k}}= \frac{\int_0^{L}{x^x}dx}{L}[/tex]

Nh=L

I know that the average value of that function can not be 1 all the time.

I don't know how to prove this though (I just learned series yesterday remember).

Maybe this will fix it...

[tex]\lim_ {N\rightarrow +\infty}{\lim_{h\rightarrow 0}{\frac{1}{N}\sum_{k=0}^N{(kh)^{hk}}}}=\lim_ {N\rightarrow +\infty}{\frac{1}{N}\sum_{k=0}^N{(\frac{L}{N}k)^{(\frac{Lk}{N})}}}= \frac{\int_0^{L}{x^x}dx}{L}[/tex]
 
Last edited:
  • #6
Mute
Homework Helper
1,388
10
Hi RandomMystery! :smile:

[tex]\lim_ {N\rightarrow +\infty}{\lim_{h\rightarrow 0}{\frac{1}{N}\sum_{k=0}^N{(kh)^{hk}}}}=\lim_ {N\rightarrow +\infty}{\frac{1}{N}\sum_{k=0}^N{\left(\lim_{h \rightarrow 0}{(kh)^h}\right)^k}}[/tex]

So, the first thing you need to do is evaluate

[tex]\lim_{h\rightarrow 0}{(kh)^h}[/tex]

The N and h limits are not independent. As stated in the first post, Nh = L = const. The limits cannot be performed separately.

[tex]\lim_{N\rightarrow +\infty}{\frac{1}{N}\sum_{k=1}^L{1^k}}=1[/tex]

L=Nh (it doesn't let me put two characters on top of the sigma)

It's 1^k right? nvm it always equals 1.
Oh, I think that what I meant was from k=1 or 0 to k=Nh not N on the sigma sign.

So, is what I wrote above right? It can't be right though.

[tex]\lim_ {N\rightarrow +\infty}{\lim_{h\rightarrow 0}{\frac{1}{N}\sum_{k=0}^N{(kh)^{hk}}}}=\lim_ {N\rightarrow +\infty}{\frac{1}{N}\sum_{k=0}^N{\left(\lim_{h \rightarrow 0}{(kh)^h}\right)^k}}= \frac{\int_0^{L}{x^x}dx}{L}[/tex]

Nh=L

I know that the average value of that function can not be 1 all the time.

I don't know how to prove this though (I just learned series yesterday remember).

Maybe this will fix it...

[tex]\lim_ {N\rightarrow +\infty}{\lim_{h\rightarrow 0}{\frac{1}{N}\sum_{k=0}^N{(kh)^{hk}}}}=\lim_ {N\rightarrow +\infty}{\frac{1}{N}\sum_{k=0}^N{(\frac{L}{N}k)^{(\frac{Lk}{N})}}}= \frac{\int_0^{L}{x^x}dx}{L}[/tex]

The sum should indeed give you that integral. The integral cannot be performed in terms of elementary functions, so the sum similarly can't be done in closed form.
 
  • #7
DArN iT!
I thought I had found a way around it though.
Have you at least tried it this way?

I have another hunch:

15f5460b0d41750d9f3f23f47e0ba5fd.png


and look!
[tex]\lim_ {N\rightarrow +\infty}{\frac{1}{N}\sum_{k=0}^N{(\frac{L}{N}k)^{(\frac{Lk}{N})}}}[/tex]

Which is so close to:

[tex]\lim_ {N\rightarrow +\infty}{\frac{1}{N}\sum_{k=0}^N{(1+\frac{L}{N}k)^{(\frac{Lk}{N})}}}[/tex]

Maybe we need to define some new function/constant?

[tex]\lim_ {N\rightarrow \infty}(\frac{L}{N}k)^\frac{Lk}{N}=[/tex] "f" or "e II"
 
Last edited:
  • #8
Mute
Homework Helper
1,388
10
Maybe we need to define some new function/constant?

[tex]\lim_ {N\rightarrow \infty}(\frac{L}{N}k)^\frac{Lk}{N}=[/tex] "f" or "e II"

That limit is equivalent to [itex]\lim_{x \rightarrow 0} x^x[/itex], which is 1.

I don't mean to try and stomp on your enthusiasm or creativity, but you should know that there are theorems that show that certain functions, such as x^x, cannot be integrated in terms of elementary functions. See, for example, Liouville's Theorem (differential algebra) and the Risch algorithm. How exactly one shows this I don't know; I haven't really studied these theorems, I am merely aware of their existence.

So, keep playing around with math like this - you likely won't be discovering a closed form integral for x^x in terms of finitely many elementary functions, but maybe you'll discover something else that's cool.
 

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