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 Trouble with [5.5] 
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Post Trouble with [5.5]
The 'addition-to-multiplication' property of e^{(a+b)}=e^ae^b is meant to be checked by using the given facts of the binomial theorem and the series for e^z, right?

So, the outline of the solution should look like this:

e^ae^b=\sum _{i=0}^{\infty } \frac{a^i}{i!}\times \sum _{i=0}^{\infty } \frac{b^i}{i!}

{something here}

\therefore  e^ae^b=e^{a+b}

I can't figure out how to use the given facts of the binomial theorem to solve this.

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05 Apr 2008, 15:13

Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Re: Trouble with [5.5]
I would prefer to give an outline instead of a formal proof.
(e^a)(e^b)=(1+a/1!+a^2/2!+...)(1+b/1!+b^2/2!+...)
Now there will be terms in the product equal to 1,a and b.Also there will be a term called ab/1! which can be written as 2ab/2!.This term can be added to a^2/2! +b^2/2! which will also be present in the product to yield
1+(a+b)/1!+(a+b)^2/2!
More terms containing higher powers can be made in a similar way.Thus we can prove that any term of e^(a+b) will be present in the product of the separate expansions.
Now we prove there exists no term in the expansion of (e^a)(e^b) which can't be in the expansion of e^(a+b).
Assume such a term exists.Any term in the product of the expansions (including this) would be of the form

a^(m)b^(k)/m!k!
where m and k are whole numbers.
Put m+k=n,then this term can be represented as
a^(n-k)b^(k)/(n-k)!k!
Multiplying both numerator and denominator by n! we can write this term as,
1/n!{n!a^(n-k)b^(k)/(n-k)!k!}
But this term is present in the expansion of e^(a+b)as it is present in the term (a+b)^n/n! contradicting our assumption.
Reductio ad absurdum!


06 Apr 2008, 10:38
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Joined: 19 Mar 2008, 14:09
Posts: 36
Post Re: Trouble with [5.5]
We define:

e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}

Therefore:

e^ae^b = \left(\sum_{n=0}^{\infty}a_n\right) \left(\sum_{n=0}^{\infty}b_n\right) = \left(\sum_{n=0}^{\infty}c_n\right)

where, c_n = a_0b_n + a_1b_{n-1} + ... + a_{n-1}b_1 + a_nb_0 which is a property of the combination of power series. Also a_n = \frac{a^n}{n!} and b_n = \frac{b^n}{n!}.

Thus:

c_0 = a_0b_0 = 1
c_1 = a_0b_1 + a_1b_0 = a + b
c_2 = a_0b_2 + a_1b_1 + a_2b_0 = \frac{b^2}{2!} + ab + \frac{a^2}{2!}=\frac{1}{2!}(a+b)^2
.
.
.
c_n = \frac{b^n}{n!} + a \frac{b^{n-1}}{(n-1)!} + .... + \frac{a^{n-1}}{(n-1)!} b + \frac{a^n}{n!} = \frac{(a+b)^n}{n!}

and so we obtain:

e^ae^b = \sum_{n=0}^{\infty} c_n =  \sum_{n=0}^{\infty} \frac{(a+b)^n}{n!} = e^{(a+b)}


09 Apr 2008, 15:13
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Post Re: Trouble with [5.5]
Thank you Kurdt!

Btw. Is Sameed's proposed solution correct? I have tried to follow his argument, but I just don't get it.

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09 Apr 2008, 16:06
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Joined: 19 Mar 2008, 14:09
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Post Re: Trouble with [5.5]
Shaun Culver wrote:
Thank you Kurdt!

Btw. Is Sameed's proposed solution correct? I have tried to follow his argument, but I just don't get it.

To be honest I haven't really looked yet. I'm quite lazy when it comes to reading and I don't like deciphering maths written in that form. When I get a minute I'll give it a go.

I didn't really say anything about the binomial theorem in my solution but it is in there if people look for it.


09 Apr 2008, 16:16
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Post Re: Trouble with [5.5]
Kurdt: I have copied your solution to the "Solutions" forum. I think that the outline is good, but it may need some modifications (e.g. showing the use of the facts of the binomial theorem given by Prf. Penrose's).

Sameed: Could you please (if you have time, of course) restate your argument so that it is easier to understand (and preferably in an easier-to-read format.)

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09 Apr 2008, 16:41
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Joined: 19 Mar 2008, 14:09
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Post Re: Trouble with [5.5]
Shaun Culver wrote:
Kurdt: I have copied your solution to the "Solutions" forum. I think that the outline is good, but it may need some modifications (e.g. showing the use of the facts of the binomial theorem given by Prf. Penrose's).


Thanks Shaun. I totally forgot about the binomial theorem till I posted later on and thought I'd mention it. When I get a bit more time I'll try and work something in.


09 Apr 2008, 16:47

Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Re: Trouble with [5.5]
Shaun,
I am really sorry that I have been too busy lately.The 'gist' of my argument is as follows:
e^a=1+\frac{a}{1!}+\frac{a^2}{2!}+\frac{a^3}{3!}+...
and
e^b=1+\frac{b}{1!}+\frac{b^2}{2!}+\frac{b^3}{3!}+...
The expansion of
e^a\ e^b=(1+\frac{a}{1!}+\frac{a^2}{2!}+\frac{a^3}{3!}+...)(1+\frac{b}{1!}+\frac{b^2}{2!}+\frac{b^3}{3!}+...)
=1(1+\frac{b}{1!}+\frac{b^2}{2!}+\frac{b^3}{3!}+...)+\frac{a}{1!}(1+\frac{b}{1!}+\frac{b^2}{2!}+\frac{b^3}{3!}+...)+\frac{a^2}{2!}(1+\frac{b}{1!}+\frac{b^2}{2!}+\frac{b^3}{3!}+...)+...
Remember this expression.Now the RHS definitely contains 1 (as the first term) and \frac{b}{1!}(as the second term).
It also contains \frac{a}{1!}.
Couple \frac{a}{1!}and \frac{b}{1!}into\frac{(a+b)}{1!}
The expression I told you to remember also has \frac{a^2}{2!}and\frac{b^2}{2!} on the RHS.
Little thought reveals that there is also a term \frac{ab}{1!}on the RHS.(which is the product of \frac{a}{1!} and \frac{b}{1!}).This term can also be written as\frac{2ab}{2!}.This term along with the two mentioned above can be collectively written as\frac{(a+b)^2}{2!}.In a similar manner (using mathematical induction) we can prove that every term of the expansion ofe^{a+b} is present in the expansion ofe^a\ e^b.
Understand carefully what this means. We have not proved that the two expressions are identical.We have merely proved that
e^a\ e^b=e^{a+b}\ +Q(a,b) where Q is a function of a & b which contains terms not in the expansion ofe^{a+b}.We now have to prove that Q=0,that is there is no term in the expansion ofe^a\ e^b which is not present in the expansion of e^{a+b}, which is where the binomial theorem comes into play.
We seek to prove this by contradiction.Assume Q is not zero.Then Q must be the sum of terms of the form
\frac{a^p}{p!}\frac{b^q}{q!}(as it is a collection of terms in the expansion ofe^a\ e^b)where p & q are whole numbers.
Consider each such term of Q separately.Put p+q=n
\frac{a^p}{p!}\frac{b^q}{q!}=\frac{a^p}{p!}\frac{b^{n-p}}{(n-p)!}=\frac{1}{n!}\frac{n!a^{p}b^{n-p}}{p!(n-p!)}
But this term is present in the binomial expansion of\frac{(a+b)^n}{n!}and is hence present in the expansion of e^{a+b}contradicting our defintion of Q according to which Q should not contain any term which is in the expansion fore^{a+b}.


16 Apr 2008, 09:29
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Post Re: Trouble with [5.5]
Sameed,

Thank you.

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17 Apr 2008, 17:34

Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Re: Trouble with [5.5]
Shaun,
I have discovered a remarkable proof for this problem,in fact a "proof without words".However,to post it I need to generate an array(a 5x5 array would do).I am having trouble generating it using latex.


27 May 2008, 09:49
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Post Re: Trouble with [5.5]
Sameed,
You will find links to tutorials towards the bottom of this page:
http://roadtoreality.info/viewtopic.php?f=4&t=28

I have Mathematica on my pc - this makes things easier. If you could type out your proof here, then I'll post it under your name in the "Solutions" section using LaTeX.

Use this template to type in the matrix entries (Insert the LaTeX code between the brackets "[" & "]" ):

[A_1 1];[A_1 2];[A_1 3];[A_1 4];[A_1 5]
[A_2 1];[A_2 2];[A_2 3];[A_2 4];[A_2 5]
[A_3 1];[A_3 2];[A_3 3];[A_3 4];[A_3 5]
[A_4 1];[A_4 2];[A_4 3];[A_4 4];[A_4 5]
[A_5 1];[A_5 2];[A_5 3];[A_5 4];[A_5 5]

This will make it easier for me.

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Our imagination is stretched to the utmost, not, as in fiction, to imagine things which are not really there, but just to comprehend those things which are there.


27 May 2008, 13:44

Joined: 22 May 2008, 19:08
Posts: 18
Post Re: Trouble with [5.5]
May be it's easier to show that both expressions, \left(\sum_{n=0}^{\infty}\frac{a^n}{n!}\right)\left(\sum_{n=0}^{\infty}\frac{b^n}{n!}\right) and \sum_{n=0}^{\infty}\frac{(a+b)^n}{n!}\right, are in fact equal to \sum_{p,q=0}^{\infty}\frac{a^pb^q}{p!q!}.

And therefore equal between each other, although I'm not completely sure if the kind of notation used on the last sum is formally correct.


27 May 2008, 23:04

Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Re: Trouble with [5.5]
"The proof without words"
[];[1];[\frac{a}{1!}];[\frac{a^2}{2!}];[\frac{a^3}{3!}];
[1];[1];[\frac{a}{1!}];[\frac{a^2}{2!}];[\frac{a^3}{3!}];
[\frac{b}{1!}];[\frac{b}{1!}];[\frac{2ab}{2!}];[\frac{3a^{2}b}{3!}];[];
[\frac{b^2}{2!}];[\frac{b^2}{2!}];[\frac{ab^2}{3!}];[];[];
[\frac{b^3}{3!}];[\frac{b^3}{3!}];[];[];[];

Shaun,
There are some entries which are meant to be blank.I have kept them as such.


31 May 2008, 10:12
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Post Re: Trouble with [5.5]
The 5 by 5 matrix:
\left(<br />\begin{array}{ccccc}<br />   & 1 & \frac{a}{1!} & \frac{a^2}{2!} & \frac{a^3}{3!} \\<br /> 1 & 1 & \frac{a}{1!} & \frac{a^2}{2!} & \frac{a^3}{3!} \\<br /> \frac{b}{1!} & \frac{b}{1!} & \frac{2a b}{2!} & \frac{3a^2b}{3!} &   \\<br /> \frac{b^2}{2!} & \frac{b^2}{2!} & \frac{a b^2}{3!} &   &   \\<br /> \frac{b^3}{3!} & \frac{b^3}{3!} &   &   &  <br />\end{array}<br />\right)

The LaTeX source code:
Code:
\left(
\begin{array}{ccccc}
   & 1 & \frac{a}{1!} & \frac{a^2}{2!} & \frac{a^3}{3!} \\
1 & 1 & \frac{a}{1!} & \frac{a^2}{2!} & \frac{a^3}{3!} \\
\frac{b}{1!} & \frac{b}{1!} & \frac{2a b}{2!} & \frac{3a^2b}{3!} &   \\
\frac{b^2}{2!} & \frac{b^2}{2!} & \frac{a b^2}{3!} &   &   \\
\frac{b^3}{3!} & \frac{b^3}{3!} &   &   & 
\end{array}
\right)

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Our imagination is stretched to the utmost, not, as in fiction, to imagine things which are not really there, but just to comprehend those things which are there.


31 May 2008, 13:18
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