Is it always valid to say that the Taylor series of some function f(x) about a point x=a equivalent to the the Maclaurin series of another function h(x)=f(x+a)?

Mark Rosales

Mark Rosales

Answered question

2022-11-02

Is it always valid to say that the Taylor series of some function f ( x ) about a point x = a equivalent to the the Maclaurin series of another function h ( x ) = f ( x + a )?

Answer & Explanation

reinmelk3iu

reinmelk3iu

Beginner2022-11-03Added 21 answers

No. If we consider an infinitely differentiable function f ( x ), then we can write down it's Taylor series about the point x = a as
n = 0 f ( n ) ( a ) n ! ( x a ) n .
Now, if we consider the translation h ( x ) = f ( x + a ) and compute its Maclaurin series, we get
n = 0 h ( n ) ( 0 ) n ! x n = n = 0 f ( n ) ( a ) n ! x n .
Note that
n = 0 f ( n ) ( a ) n ! ( x a ) n n = 0 f ( n ) ( a ) n ! x n
because they clearly disagree at x = a. What we can say, however, is that the Taylor series for f ( x ) is the Maclaurin series for g ( x ) = f ( x + a ) shifted to the right by a.

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