Prove that if f is defined for |x|<r and if there exists a constant B such that |f^n(x)|≤B for all |x|<r and n in N, then the Taylor series expansion : sum_(n=0)^oo (f^(n)(0)x^n)/(n!) converges to f(x) for |x|<r.

Davirnoilc

Davirnoilc

Answered question

2022-11-16

Prove that if f is defined for | x | < r and if there exists a constant B such that
| f n ( x ) | B
for all | x | < r and n N , then the Taylor series expansion :
n = 0 f ( n ) ( 0 ) x n n !
converges to f ( x ) for | x | < r.

Answer & Explanation

Cseszteq5j

Cseszteq5j

Beginner2022-11-17Added 17 answers

A Taylor series converges if and only if its remainder term, R n ( x ) tends to zero. This is just a consequence from taylors theorem
Theorem: If f ( x ) = T n ( x ) + R n ( x ) , where T n is the nth degree Taylor polynomial of f at a and lim n R n ( x ) = 0for | x a | < R , then f is equal to the sum of its taylor series on the interval | x a | < R That is because, as you suggested, we have
| R n ( x ) | B x n + 1 ( n + 1 ) !
and
Taylors Inequality: If | f n + 1 ( x ) | M for | x a | d , then the remainder, R n ( x ) of the Taylor series satisfies the inequality | R n ( x ) | M ( n + 1 ) ! | x a | n + 1 for | x a | d
so now you must consider this,
lim n B x n + 1 ( n + 1 ) !
and we that
B lim n x n + 1 ( n + 1 ) ! = B 0 = 0
thus we have that | R n ( x ) | 0 i.e., the remainder will tend to zero as n approaches infinity , and the proof is finished.

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