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Problem : Compute the Taylor series for f (x) = 1/(1 + x).

The first few derivatives of the function are


f'(x)=  
f''(x)=  
f(3)(x)=  

so f (0) = 1, f'(0) = - 1, f''(0) = 2, f(3)(0) = - 6. The general case is clearly that f(n)(0) = (- 1)nn!. Hence the Taylor series for f (x) is


p∞(x)=xn  
 =(- 1)nxn  
 =1 - x + x2 - x3 + ...  

Problem : What is the Taylor series of a polynomial p(x) = anxn + an-1xn-1 + ... + a0?

It is easy to check that the Taylor series of a polynomial is the polynomial itself! (All the coefficients of higher order terms are equal to 0.)

Problem : Find the Taylor series for the function g(x) = 1/ about x = 1.

The first couple derivatives of the function are


g'(x)=x-3/2  
g''(x)=x-5/2  
g(3)(x)=x-7/2  

so g(1) = 1, g'(1) = - 1/2, g''(1) = (- 1/2)(- 3/2). We deduce that

g(n)(1) = (1)(3) ... (2n - 1) =    

Hence the Taylor series for g(x) is


xn    

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