Taylor Series Expansion Explanation

In summary, mbeaumont99 is asking for help with finding the summation of a function involving infinite summation. After some discussion, it is agreed upon that the summation is equal to \(a^x\) and that the restriction \(a > 0\) should be imposed. The conversation also touches on using a Taylor series and the importance of choosing a branch of the logarithm for convergence.
  • #1
Sudharaka
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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,

\[f(x)=\sum_{n=0}^{\infty}\frac{(\ln(a))^n}{n!}\, x^{n}=\sum_{n=0}^{\infty}t_{n}\]
 
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  • #2
Sudharaka said:
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
 
  • #3
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

$\chi$ $\sigma$
 
  • #4
chisigma said:
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
 
Last edited:
  • #5


This is the Taylor series expansion of \(f(x)=a^{x}\) around the neighborhood \(b=0\). Therefore, your general statement is correct and can be proved using the Taylor series expansion. I hope this helps. Let me know if you have any further questions.

Best,
 

FAQ: Taylor Series Expansion Explanation

1. What is a Taylor Series expansion?

A Taylor Series expansion is a mathematical representation of a function as an infinite sum of polynomials. It is a way to approximate a complicated function with a simpler one, making it easier to work with and analyze.

2. How is a Taylor Series expansion calculated?

A Taylor Series expansion is calculated by taking the derivatives of a function at a specific point and plugging them into a formula that involves powers of the variable and a constant. This process can be repeated to get more precise approximations.

3. Why is a Taylor Series expansion important?

A Taylor Series expansion is important because it allows us to approximate a complex function with a simpler one, making it easier to understand and manipulate. It is also used in many areas of science and engineering, such as calculus, physics, and computer science.

4. What is the difference between a Taylor Series expansion and a Maclaurin Series expansion?

A Maclaurin Series expansion is a special case of a Taylor Series expansion, where the function is expanded around the point x = 0. This means that the constant term in the formula for the series is equal to the value of the function at x = 0. In a general Taylor Series expansion, the series is expanded around any point x = a.

5. How can a Taylor Series expansion be used to approximate a function?

A Taylor Series expansion can be used to approximate a function by truncating the infinite series to a finite number of terms. The more terms that are included, the more accurate the approximation will be. This is useful in situations where the exact value of a function is difficult to calculate, but an approximation is sufficient.

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