Taylor Series and Maclaurin Series Doubt

sarvesh0303
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Homework Statement


If I take a function f(x) and its taylor series, then will the infinite series give me the value of the function at any x value or will it only give proper values for x≈a?

For example, If I take a maclaurin series for a function will it give me proper values for all x values or only if x≈a=0?


Homework Equations





The Attempt at a Solution

 
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sarvesh0303 said:

Homework Statement


If I take a function f(x) and its taylor series, then will the infinite series give me the value of the function at any x value or will it only give proper values for x≈a?

For example, If I take a maclaurin series for a function will it give me proper values for all x values or only if x≈a=0?

It always works for x = a. It is possible that is the only value for which it works. But more usually there is a radius of convergence where it works for |x-a|<r and the series diverges for |x-a|>r. Or it may converge and equal the function for all x. Generally, the farther you are from x = a, the more terms you need for a given accuracy. There are examples where the series converges for all x but doesn't equal the function except at x = a.
 
There exist functions for which the Taylor series exist and converges for all x but converges to the value of the function only at the given point. The function defined as f(x)= e^{-1/x^2} for x not equal to 0, 0 for x= 0, has derivatives of all order and they are all 0 at x= 0 so the Maclaurin series (Taylor series with a= 0) is just 0 for all x. But f is 0 only at x= 0.

(Functions that have the property that they are equal to their Taylor series at every point on a set are called 'anlytic' on that set. Those are especially important in functions of complex numbers.)
 
So how could we find out whether the sum will give me the approximate value of the function or not?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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