What Are the Taylor Polynomials for Specific Functions?

Jimmy84
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Homework Statement


Obtain the Taylor polynomials Tnf(x) as indicated. In each case, it
is understood that f(x) is defined for a11 x for which f(x) is meaningful.


Problem one
Tn = (a^x) = sigma from k = 0 to n of ((log a)^k)/k! x^k

Problem two
Tn = (1/(1+x)) = sigma from k = o to n of (-1)^k x^k


Homework Equations





The Attempt at a Solution



Im totally lost and I don't know where to start can anyone help me please?
 
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Did you really have to use "a11" when you have a perfectly good "l" key?

They Taylor series for function f(x) is \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n where f^{(n)}(0) indicates the nth derivative of f evaluated at 0.

(Actually, that is the "Taylor series at 0" or "MacLaurin" series.)

So for a^x you only need to find the derivatives of a^x and evaluate at x= 0. That can be done most efficiently by writing a^x= e^{ln a^x}= e^{x ln a}.

It is also true that any power series equal to a given function is a Taylor series. If you remember that the sum of the geometric series \sum_{n=0}^\infty r^n is equal to 1/(1- r), then the Taylor series for 1/(1+x) should be obvious.
 
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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