How to factor out the b from e^(ab)

  • #1

Main Question or Discussion Point

I want to know if there is a way to factor out the b from e^(ab) so that:

e^(ab) = e^(a)[f(b)] or e^a + f(b)

or

e^(ab) = some other function where a and b are separated.
 
Last edited:

Answers and Replies

  • #2
22,097
3,283
Hi RandomMystery! :smile:

Could you perhaps tell us why you want this? If we know what you're after, then we are in a better posiiton to give suggestions...
 
  • #3
Well...

I have two limits:
h-> 0
N-> infinity

When I multiply these limits together, I get a constant:
Nh = L

I have the following "thing" (since it's not a function or is it?):
e^(h) + e^(2h) +...e^(Nh)
N​

If we can factor out the h from ^ that ^, then we can put it to the side and use L'Hopital's rule for the remaining fraction, whose denominator and numerators should both be approaching infinity.

Ohh, and Thanks for replying!
 
Last edited:
  • #4
Mute
Homework Helper
1,388
10
Well...

I have two limits:
h-> 0
N-> infinity

When I multiply these limits together, I get a constant:
Nh = L

I have the following "thing" (since it's not a function or is it?):
...e^(Nh)
N​

If we can factor out the h from ^ that ^, then we can put it to the side and use L'Hopital's rule for the remaining fraction, whose denominator and numerators should both be approaching infinity.

Ohh, and Thanks for replying!
Is there anything else in that expression that depends on N or h? As given, since Nh = L, a constant, you just have

[tex]\lim_{N \rightarrow \infty} \frac{e^L}{N} \rightarrow 0[/tex]

In general, if you have two limits that are related by a relation like Nh = constant = L, then just replace all instances of Nh with L, and replace all h with h = L/N (or use N = L/h). Then you have a single limit to take.
 
  • #5
Is there anything else in that expression that depends on N or h? As given, since Nh = L, a constant, you just have

[tex]\lim_{N \rightarrow \infty} \frac{e^L}{N} \rightarrow 0[/tex]

In general, if you have two limits that are related by a relation like Nh = constant = L, then just replace all instances of Nh with L, and replace all h with h = L/N (or use N = L/h). Then you have a single limit to take.
What was that? I think I had the mute button on. Here's what I meant
e^(h) + e^(2h).... e^(Nh)
N​
I want to factor out the h's, so that I have something like:

[e^(1) + e^(2).... e^(N)][something]
N​
 
Last edited:
  • #6
22,097
3,283
What was that? I think I had the mute button on. Here's what I meant
e^(h) + e^(2h).... e^(Nh)
N​
I want to factor out the h's.
[tex]\frac{1}{N}\sum_{k=1}^N{(e^h)^k}=\frac{1}{N} e^h \frac{1-e^{hN}}{1-e^h}[/tex]

This will be a bit easier to handle probably...
 
  • #7
[tex]\frac{1}{N}\sum_{k=1}^N{(e^h)^k}=\frac{1}{N} e^h \frac{1-e^{hN}}{1-e^h}[/tex]

This will be a bit easier to handle probably...
I'm not sure how you got the right part of that equation (induction?). Could you show me the logic behind it or at least tell me what math course that you take to learn that sort of stuff?
Using whatever you have there, Can you prove that it equals:
e^L
L​
?
 
Last edited:
  • #8
22,097
3,283
I'm not sure how you got the right part of that equation (induction?).
Using whatever you have there. Can you prove that it equals:
e^L
L​
It's just a geometric sum:

[tex]1+r+r^2+r^3+...+r^n=\frac{1-r^{n+1}}{1-r}[/tex]

and we can prove this by induction, indeed...
 
  • #9
[tex]\frac{1}{N}\sum_{k=1}^N{(e^h)^k}=\frac{1}{N} e^h \frac{1-e^{hN}}{1-e^h}=\frac{e^L}{L}[/tex]
and
Nh=L

I know that it should equal (e^L)/L but I don't know why.

Here's another situation:

2^(h)+2^(2h)...2^(Nh)
N​

Nh=L

(based off of what you had-I don't know if this is right though)

[tex]\frac{1}{N}\sum_{k=1}^N{(2^h)^k}=\frac{1}{N} 2^h \frac{1-2^{hN}}{1-2^h}=\frac{2^L}{log(2)L}[/tex]

I think we have to use log's, but I don't know how to use logs to get ^that^ answer.
 
  • #10
22,097
3,283
[tex]\frac{1}{N}\sum_{k=1}^N{(e^h)^k}=\frac{1}{N} e^h \frac{1-e^{hN}}{1-e^h}=\frac{e^L}{L}[/tex]
and
Nh=L

I know that it should equal (e^L)/L but I don't know why.
Firstly, I think you mean to say that this is true "in the limit"? Writing that this equal (e^L)/L probably isn't what you mean.

Secondly, it isn't even true in the limit. A lot depends on how h and N behave with respect to each other. If you pick h=1/N and L=1, then the limit of the above sequence is -1. So it's certainly not equal to (e^L)/L=e...

Why would you think it should equal (e^L)/L anyway?
 
  • #11
Mute
Homework Helper
1,388
10
[tex]\frac{1}{N}\sum_{k=1}^N{(e^h)^k}=\frac{1}{N} e^h \frac{1-e^{hN}}{1-e^h}=\frac{e^L}{L}[/tex]
and
Nh=L

I know that it should equal (e^L)/L but I don't know why.
As I said before, put everything in terms of h and L:

[tex]\frac{h}{L}\frac{e^h}{1-e^h}(1-e^{L}).[/tex]

Now take the limit as h tends to zero. You'll get an indeterminate form, which should work out to -1. So, this gives you [itex](e^L-1)/L[/itex]. I suppose if L is large you get e^L/L to leading order.

Here's another situation:

2^(h)+2^(2h)...2^(Nh)
N​

Nh=L

(based off of what you had-I don't know if this is right though)

[tex]\frac{1}{N}\sum_{k=1}^N{(2^h)^k}=\frac{1}{N} 2^h \frac{1-2^{hN}}{1-2^h}=\frac{2^L}{log(2)L}[/tex]

I think we have to use log's, but I don't know how to use logs to get ^that^ answer.
Follows pretty much the same way, except evaluating the indeterminate form will give you the logarithm, and you really get [itex](2^L-1)/(\log (2) L)[/itex].
 
Last edited:
  • #12
Firstly, I think you mean to say that this is true "in the limit"? Writing that this equal (e^L)/L probably isn't what you mean.

Secondly, it isn't even true in the limit. A lot depends on how h and N behave with respect to each other. If you pick h=1/N and L=1, then the limit of the above sequence is -1. So it's certainly not equal to (e^L)/L=e...

Why would you think it should equal (e^L)/L anyway?
I'm trying to find the average of an infinite series.
We have an infinite number of subdivisions: N subdivisions
All of the subdivision's lengths are equal and virtually zero: h length

If we multiply the subdivisions (h) by the number of them (N) we should get the length of all of them together (L)

One more note: How do I add the lim (h -> 0) to the equation above and replace N with infinity (sideways 8)?


As I said before, put everything in terms of h and L:

[tex]\frac{h}{L}\frac{e^h}{1-e^h}(1-e^{L}).[/tex]

Now take the limit as h tends to zero. You'll get an indeterminate form, which should work out to -1. So, this gives you [itex](e^L-1)/L[/itex]. I suppose if L is large you get e^L/L to leading order.



Follows pretty much the same way, except evaluating the indeterminate form will give you the logarithm, and you really get [itex](2^L-1)/(\log (2) L)[/itex].
Okay, let's say that the k=0 and so e^0 is the first term. Will that eliminate the -1?
 
Last edited:
  • #13
Has anyone here heard of discrete calculus? I think that's what I'm trying to do, but the most advanced math course I've taken is Calculus I so I don't understand how you reduced the original Sigma equation to this:

[tex]\frac{1}{N} e^h \frac{1-e^{hN}}{1-e^h}[/tex]

?

The guy at this website talks about it, however, I don't understand what he's writing at point 2 and beyond.
http://www.math.uic.edu/~kauffman/DCalc.pdf
 
  • #14
Mute
Homework Helper
1,388
10
Okay, let's say that the k=0 and so e^0 is the first term. Will that eliminate the -1?
[strike]Yes.[/strike] No. At first I accidentally thought starting the sum at k = 0 would amount to adding 1/L to the result, but you're actually adding 1/N = h/L, which tends to zero in the desired limit.

Has anyone here heard of discrete calculus? I think that's what I'm trying to do, but the most advanced math course I've taken is Calculus I so I don't understand how you reduced the original Sigma equation to this:

[tex]\frac{1}{N} e^h \frac{1-e^{hN}}{1-e^h}[/tex]

?

The guy at this website talks about it, however, I don't understand what he's writing at point 2 and beyond.
http://www.math.uic.edu/~kauffman/DCalc.pdf
You don't need "discrete calculus" to compute the sum. It's a finite geometric series.

Consider:

[tex]S \equiv \sum_{k=a}^{a+N} x^k = x^a + x^{a+1} + \dots + x^{a+N} = x^a + x(x^a + x^{a+1} + \dots + x^{a + N - 1}) = x^a + x(S-x^{a+N})[/tex]

Thus, [itex]S = x^a + x(S-x^{a+N})[/itex]. You can solve this equation for S, which gives you the value of the sum:

[tex]S = \frac{x^a - x^{a+N+1}}{1-x}.[/tex]

So, this is the result that micromass used to convert your original expression into the closed form expression that we did the limit for.
 
Last edited:
  • #15
That's Awsome! Thanks for showing that to me Mute.
The following is true, right?:

[tex]^{lim}_{N\rightarrow\infty}\frac{1}{N}\sum_{k=0}^N{^{lim}_{h\rightarrow0}(2^h)^k}=\frac{1}{N} 2^h \frac{1-2^{hN}}{1-2^h}=\frac{h}{L}\frac{e^h}{1-e^h}(1-e^{L})=\frac{2^L}{log(2)L}[/tex]

(Nh=L)

I have a feeling that this part of the equation above is not right since I didn't fix it for k being zero instead of 1:

...Correct.. = ...Not Right...= ...Not Right...= ...Correct..
 
Last edited:
  • #16
Mute
Homework Helper
1,388
10
I made a mistake - changing the sum to start from k = 0 won't change anything. I mistakenly thought it amounted to changing the final result by 1/L, but really you're adding in 1/N = h/L which tends to zero, so this makes no difference and you get the same result.

The following is the derivation for k = 0 as the starting point. You'll see it makes no difference.

[tex]\lim_{N\rightarrow\infty} \lim_{h\rightarrow 0} \frac{1}{N}\sum_{k=0}^N(2^h)^k=\lim_{N\rightarrow \infty} \lim_{h\rightarrow 0}\frac{1}{N}\frac{1-2^{hN+h}}{1-2^h}= \lim_{h\rightarrow 0} \frac{h}{L}\frac{1}{1-2^h}(1-2^{L+h})=\frac{2^L-1}{\log(2)L}[/tex]
having set N = L/h.
 
Last edited:
  • #17
I must have done something wrong. There shouldn't be a -1 next to the 2^L.
You wouldn't happen to know a a formula that would get (2^L)/(log(2)L) or how we can edit the original conditions to remove that -1 in the end.

I have hunch that dividing the original sigma by N+1 instead of N might fix this.
 
Last edited:
  • #18
Mute
Homework Helper
1,388
10
I must have done something wrong. There shouldn't be a -1 next to the 2^L.
You wouldn't happen to know a a formula that would get (2^L)/(log(2)L) or how we can edit the original conditions to remove that -1 in the end.

I have hunch that dividing the original sigma by N+1 instead of N might fix this.
Changing N to N+1 won't "fix" it. For large N the difference between N and N+1 is imperceptible. Similarly, however, if L is large, 2^L is much larger than 1 and you may neglect the 1.

Perhaps you should explain exactly why you think the -1 shouldn't be there, because it really is there. Another way to get the result is to realize that the sum

[tex]\lim_{h \rightarrow 0} \frac{h}{L} \sum_{k=0}^{L/h} 2^{hk}[/tex]
is really a Reimann sum - that is, it is equivalent to the integral

[tex]\frac{1}{L}\int_0^L dx~2^x[/tex]
which is the average of the function [itex]2^x[/itex] over the interval [0,L]. You can check that it gives the same result, [itex](2^L-1)/(L\log 2)[/itex].
 
  • #19
Yeah, you're right. So, how many other ways are there of doing integrals and derivative?

You can ignore the following rant:

"I didn't like the whole "work backwards" process for finding the integrals of function, so I tried to find the average value then multiply that value by the length of the interval.

The summation method is a little better, but I still don't completely comprehend the mechanics of every integral and derivative, except that of linear functions. I've seen the algebraic and calculus proofs, but I kind of wish there was a geometric proof of why the area under the curve of x^2 is one third of x^3.

It's just not as intuitive as the areas of triangles, rectangles, and squares (which would be linear functions)."


My last question now is this:

If we have the following series:

e^(h) + e^(2h) +...e^(Nh)​
N​

in which Nh=L ... limits....you know the rest...

Is it possible to take the derivative of both the top and the bottom to where last term on top is the derivative of the top:

d(^(h) + e^(2h) +...e^(Nh)) = e^(Nh)

If this is the case, I don't know what (?) the derivative is respect to. I also want to know how the derivate of N with respect to that same derivative (?) would be.

It should give us the same answer that was shown above.

P.S. - I've done this with polynomials and it worked. I'll show it next post.
 
  • #20
I'm trying to find the average value of x^4 from 0 to L
So, I simply made an infinite number of function values be divided by the number of those function values:

(h)^(4) + (2h)^(4) +...(Nh)^(4)
N​
P.S: Nh=L​

I factored out the h^4...:

[(1)^(4) + (2)^(4) +...(N)^(4)](h)^(4)
N​

Replaced h with L/N:

[(1)^(4) + (2)^(4) +...(N)^(4)](L)^(4)
N^5​

Now I took the derivative of the top and bottom (with respect to N I think).
Note:Does it matter what I take the derivative with respect to?​

(N)^(4)(L)^(4)
5N^(4)​

and now we just cancel out!

L^4
5​

See how easy that was!
I tried to do something similar with e^x to see if it works. If you can some how take the derivate of the top and bottom with respect to something that will make the last term the derivative, maybe it and the bottom derivative will work out to equal:

(e^L - 1)
L​

* I would just like to thank you for taking your time to help me with this.
 
  • #21
Mute
Homework Helper
1,388
10
My last question now is this:

If we have the following series:

e^(h) + e^(2h) +...e^(Nh)​
N​

in which Nh=L ... limits....you know the rest...

Is it possible to take the derivative of both the top and the bottom to where last term on top is the derivative of the top:

d(^(h) + e^(2h) +...e^(Nh)) = e^(Nh)

If this is the case, I don't know what (?) the derivative is respect to. I also want to know how the derivate of N with respect to that same derivative (?) would be.

It should give us the same answer that was shown above.

P.S. - I've done this with polynomials and it worked. I'll show it next post.
No, that doesn't work. You're not taking the derivative of anything. You're just throwing away all but the last term. You could define an operator that throws away all terms but the last one, but that's not really taking a derivative.

I'm trying to find the average value of x^4 from 0 to L
So, I simply made an infinite number of function values be divided by the number of those function values:

(h)^(4) + (2h)^(4) +...(Nh)^(4)
N​
P.S: Nh=L​

I factored out the h^4...:

[(1)^(4) + (2)^(4) +...(N)^(4)](h)^(4)
N​

Replaced h with L/N:

[(1)^(4) + (2)^(4) +...(N)^(4)](L)^(4)
N^5​

Now I took the derivative of the top and bottom (with respect to N I think).
Note:Does it matter what I take the derivative with respect to?​

(N)^(4)(L)^(4)
5N^(4)​

and now we just cancel out!

L^4
5​

See how easy that was!
Unfortunately, all you've really done is performed some symbolic manipulations which give you the correct answer, but those manipulations don't really have a well defined meaning. The trick happened to work for a power of x, but I think it's a bit of a coincidence that it worked out at all. Essentially, the sums you started with are Reimann sums, so they have to give you the same value as the integral, but your manipulations to try and shortcut doing the integral (or to do the integral when it otherwise can't be done) won't work in general. You can try to formalize your rules a bit more, but other than being a good exercise for yourself I don't think they'll give you what you want them to.
 
  • #22
But Hey! According to wikipedia:
http://en.wikipedia.org/wiki/The_fundamental_theorem_of_calculus


,this image:
500px-FTC_geometric.svg.png




,and this formula:
3198baf3dc56efafad5cd277b2457dad.png


When you take the derivative of a remain sums, the last sum times dx should be your answer.

In this case, I did not have a dx because I'm adding the infinite number of function values between 0 and L together and I am dividing it by the infinite number of function values to find the average value between 0 and L for that function:



frac{\int_{0}^{Ndx}f(x)}{N}=\frac{\int_{0}^{Ndx}f(x)dx}{Ndx}=\frac{\int_{0}^{L}f(x)d&space;x}{L}.gif



If we take the derivative:



{L-0}=\frac{d}{dx}\frac{f(dx&space;..or..&space;0)+f(2dx)...f(Ndx&space;..or..&space;L)}{N}.gif



So I was wrong, the rate of change of a summation is the last summation minus the first summation (which in this case was zero). Which makes sense to me. This should work for the function e^x I'm just having trouble doing it though.

P.S. My computer said the security certificate is missing? and that I should contact the administrator when I tried to post this the first time.
 
Last edited:

Related Threads on How to factor out the b from e^(ab)

  • Last Post
Replies
2
Views
5K
Replies
3
Views
13K
  • Last Post
Replies
8
Views
2K
Replies
4
Views
1K
Replies
6
Views
4K
Replies
17
Views
52K
Replies
4
Views
5K
Replies
9
Views
2K
Replies
2
Views
3K
  • Last Post
Replies
2
Views
2K
Top