# Calculating Power Series Coefficients with Differentiability

In summary, the conversation discusses the relationship between power series, infinitely differentiable functions, and Taylor's formula. It explains that given a power series f(x), if it is infinitely differentiable, then the coefficients a(n) can be found using the formula n!a(n)=D^{n}f(0). However, if we start with a C^\infty function and try to find its power series using the same formula, the resulting series may not necessarily be equal to the original function. The conversation also mentions the importance of the Lagrange remainder in determining when the power series and the original function are equal.
Let be the powe series:

$$f(x)=\sum_{n=0}^{\infty}a(n)x^{n}$$

then if f(x) is infinitely many times differentiable then for every n we have:

$$n!a(n)=D^{n}f(0)$$ (1) of course we don't know if the series above is

of the Taylor type, but (1) works nice to get a(n) at least for finite n.

Do you have a question?

Btw - if the radius of convergence of

$$\sum_{n=0}^{\infty}a(n)x^{n}$$

is R, then for all -R<x<+R, define

$$f(x)=\sum_{n=0}^{\infty}a(n)x^{n}$$

Then automatically, f defined in this way is $C^{\infty}$ because that is a property of power series. and additionally, the coefficients are related to f by $n!a(n)=D^{n}f(0)$.

On the other hand, if you start with a function f(x), that you know is $C^{\infty}$ on some interval (-a,a), and you define the series

$$\sum_{n=0}^{\infty}\frac{D^{n}f(0)}{n!}x^{n}$$

then the function g(x), defined by the value of the series where it converges, i.e.

$$g(x)=\sum_{n=0}^{\infty}\frac{D^{n}f(0)}{n!}x^{n} \ \ \ x\in(-R,R)$$

is not necessarily f(x). Although by the above, we do have that $n!a(n)=D^{n}g(0)$*, as you pointed out in your thread. It's a little subtlety.

*(which in this case expresses the equality $D^{n}g(0)=D^{n}f(0)$)

There are theorems however that give conditions for the equality of g(x) to f(x). One of them requires that the Lagrange remainder from Taylor's formula vanishes as $n\rightarrow \infty$.

Last edited:
Let be the powe series:

$$f(x)=\sum_{n=0}^{\infty}a(n)x^{n}$$

then if f(x) is infinitely many times differentiable then for every n we have:

$$n!a(n)=D^{n}f(0)$$ (1) of course we don't know if the series above is

of the Taylor type, but (1) works nice to get a(n) at least for finite n.
The power series expansion, about $x= x_0$ of a $C^\infty$ function is unique. It doesn't matter how you get it, it will be identical to the Taylor's series for the function about that point.

## 1. What is a power series and how is it used in calculations?

A power series is an infinite series of the form ∑n=0∞ cn(x-a)n, where cn are coefficients, x is the variable, and a is a constant. It is used in calculations to approximate functions by expressing them as a sum of powers of the variable.

## 2. How can I calculate the coefficients of a power series if I know the function and its differentiability?

The coefficients of a power series can be calculated using the formula cn = f(n)(a)/n!, where f(n)(a) is the nth derivative of the function evaluated at the constant a. This formula is derived from the Taylor series expansion of a function.

## 3. Can I use power series to approximate any function?

No, power series can only be used to approximate functions that are infinitely differentiable at a specific point. If a function has a discontinuity or singularity at that point, the power series will not converge to the function.

## 4. Is it possible to calculate the coefficients of a power series using a computer program?

Yes, there are many computer programs and mathematical software that have built-in functions to calculate power series coefficients. These programs use numerical methods to approximate the coefficients and can handle complex functions with ease.

## 5. How accurate are power series approximations compared to the original function?

The accuracy of a power series approximation depends on the number of terms used in the series. The more terms included, the closer the approximation will be to the original function. However, it is important to note that power series are only valid within a certain radius of convergence, so the accuracy may decrease as the distance from the center point increases.

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