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[SOLVED] Taylor Series Expansion Explanation

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
mbeaumont99's question from Math Help Forum,

This is to do with some infinite summation work that we are going through at college at the moment. We have the function \(\displaystyle t_{n} =\frac{(x\ln a)^{n}}{n!}\) and have been substituting in different x and a values to determine a general statement for the infinite summation of the function. I have found that \(\displaystyle S_{n}=a^{x}\)

I need to do a formal proof for this general statement and heard that a Taylor series would be able to do that. If anyone would be able to start me off on this, send me in a different direction, or simply contribute to this thread then I would be extremely appreciative.

Thanks,
mbeaumont99
Hi mbeaumont99,

One thing you can do is to find the Taylor series expansion of \(f(x)=a^{x}\) and see whether it is \(\displaystyle \sum t_{n}\). The Taylor series for the function \(f \) around a neighborhood \(b\) is,

\[f(x)=\sum_{n=0}^{\infty}\frac {f^{(n)}(b)}{n!} \, (x-b)^{n}\]

Of course I am assuming here that the function \(f\) can be expressed as a Taylor series expansion around a neighborhood of \(b\) (that is \(f\) is analytic). To get a more detailed idea about what functions are analytic read this and this. We shall use \(b=0\) so that we get the Maclaurin's series.

\[f(x)=\sum_{n=0}^{\infty}\frac {f^{(n)}(0)}{n!} \, x^{n}\]

Now we have to find out \(f^{(n)}(0)\) with regard to the function \(f(x)=a^{x}\). Differentiating \(f\) a couple of times we can "feel" that \(f^{n}(x)=a^x(\ln(a))^n\,\forall\,n\in\mathbb{N}=\mathbb{Z}\cup\{0\}\). To prove this in a formal manner we shall use mathematical induction.

When \(n=0\), the result is obvious. We shall assume that the result is true for \(n=p\in\mathbb{N}\). That is,

\[f^{(p)}(x)=a^{x}(\ln(a))^p\]

Now consider, \(f^{(p+1)}(x)\).

\[f^{(p+1)}(x)=\frac{d}{dx}f^{(p)}(x)=(\ln(a))^p \frac{d}{dx}a^x=a^x(\ln(a))^{p+1}\]

Therefore by Mathematical induction, \(f^{n}(x)=a^x(\ln(a))^n\,\forall\,\in\mathbb{N}\)

\[\therefore f^{n}(0)=(\ln(a))^n\,\forall\,\in\mathbb{N}\]

Hence,

\[f(x)=\sum_{n=0}^{\infty}\frac{(\ln(a))^n}{n!}\, x^{n}=\sum_{n=0}^{\infty}t_{n}\]
 

CaptainBlack

Well-known member
Jan 26, 2012
890
mbeaumont99's question from Math Help Forum,

mbeaumont99's question from Math Help Forum,

This is to do with some infinite summation work that we are going through at college at the moment. We have the function \(\displaystyle t_{n} =\frac{(x\ln a)^{n}}{n!}\) and have been substituting in different x and a values to determine a general statement for the infinite summation of the function. I have found that \(\displaystyle S_{n}=a^{x}\)

I need to do a formal proof for this general statement and heard that a Taylor series would be able to do that. If anyone would be able to start me off on this, send me in a different direction, or simply contribute to this thread then I would be extremely appreciative.

Thanks,
mbeaumont99
This is asking for the summation:

\( \displaystyle S=\sum_{n=0}^{\infty} \frac{(x\ln(a))^n}{n!} \)

We note the series expansion for the exponential function:

\( \displaystyle e^u=\sum_{n=0}^{\infty} \frac{u^n}{n!} \)

which is convergent for all real or complex \(u\). Put \(u=x\ln(a) \) to get:

\( \displaystyle e^{x\ln(a)}=\sum_{n=0}^{\infty} \frac{(x\ln(a))^n}{n!}=S \)

Now \( x\ln(a) = \ln(a^x) \) so:

\( \displaystyle a^x=e^{\ln(a^x)}=\sum_{n=0}^{\infty} \frac{(x\ln(a))^n}{n!}=S \)

CB
 

chisigma

Well-known member
Feb 13, 2012
1,704
All right!... it seems that we all agree on the identity...

$\displaystyle a^{x}= e^{x\ \ln a}= \sum_{n=0}^{\infty} \frac{(x\ \ln a)^{n}}{n!}$ (1)

Nobody however has imposed constraints on a, so that can a be anything we like?... but in this case what does it happen when is $a=0?$... or when is $a<0$?... better is to avoid problems and impose $a>0$ or critically examine the general case of any real value for a?... a nice question!...

Kind regards

$\chi$ $\sigma$
 

CaptainBlack

Well-known member
Jan 26, 2012
890
All right!... it seems that we all agree on the identity...

$\displaystyle a^{x}= e^{x\ \ln a}= \sum_{n=0}^{\infty} \frac{(x\ \ln a)^{n}}{n!}$ (1)

Nobody however has imposed constraints on a, so that can a be anything we like?... but in this case what does it happen when is $a=0?$... or when is $a<0$?... better is to avoid problems and impose $a>0$ or critically examine the general case of any real value for a?... a nice question!...

Kind regards

$\chi$ $\sigma$
I expect the implied restriction is that \(a>0\), but I am reasonably sure that once one picks a branch of the logarithm the series converges to the given sum (assuming \( a\ne 0\) ).

CB
 
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