The Taylor series for f(x)=x^3 at -4 is sum_(n=0)^(oo) c_n(x+4)^n. Find the first few coefficients. c_0=? c_1=? c_2=? c_3=? c_4=?

Aryan Lowery

Aryan Lowery

Answered question

2022-09-29

The Taylor series for f(x)=x^3 at -4 is \sum_{n=0}^{\infty} c_n(x+4)^n. Find the first few coefficients.
c_0=?
c_1=?
c_2=?
c_3=?
c_4=?

Answer & Explanation

Roger Clements

Roger Clements

Beginner2022-09-30Added 4 answers

f ( x ) = x 3           a = 4           n = 0 c n ( x + 4 ) n
Series will be n = 0 f ( n ) ( a ) b ! ( x a ) n
So, c n = f ( n ) ( 4 ) n !
f ( 0 ) = x 3                       f ( 0 ) ( 4 ) = 64 f ( 1 ) = 3 x 2                       f ( 1 ) ( 4 ) = 48 f ( 2 ) = 6 x                       f ( 2 ) ( 4 ) = 24 f ( 3 ) = 6                       f ( 3 ) ( 4 ) = 6 f ( 4 ) = 0                       f ( 4 ) ( 4 ) = 0
c 0 = 64 c 1 = 48 1 = 48 c 2 = 24 2 ! = 12 c 3 = 6 2 × 3 × 1 = 1 c 4 = 0

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