Asked by W
Find a power series, centered @ x=0, for function f(x)=x/(1+2x).
I know this is a maclaurin series, but my work doesn't get the right answer. Can you please show steps? Also,do all power series start with a 1, as in (1+2x+4x^2+...)?
Thanks in advance!
I know this is a maclaurin series, but my work doesn't get the right answer. Can you please show steps? Also,do all power series start with a 1, as in (1+2x+4x^2+...)?
Thanks in advance!
Answers
Answered by
Steve
No, take a look at the definition of the Maclaurin series. It starts with f(0). very power series starts with the first term of its defined sequence.
f = x/(1+2x)
f' = 1/(1+2x)^2
f<sup>(2)</sup> = -4/(1+2x)^3
f<sup>(3)</sup> = 24/(1+2x)^4
f<sup>(4)</sup> = -192/(1+2x)^5
from then on it's just the power rule:
f(n) = (-2)^(n-1) n! (1+2x)^-(n+1)
so, we have
f(0) + 1/1! f<sup>(1)</sup>(0) + 1/2! f<sup>(2)</sup>(0) + ...
= 0 + 1x - 4/2!x^2 + 24/3!x^3 - 192/4!x^4 + ...
= 0 + x - 2x^2 + 4x^3 - 8x^4 + ...
f = x/(1+2x)
f' = 1/(1+2x)^2
f<sup>(2)</sup> = -4/(1+2x)^3
f<sup>(3)</sup> = 24/(1+2x)^4
f<sup>(4)</sup> = -192/(1+2x)^5
from then on it's just the power rule:
f(n) = (-2)^(n-1) n! (1+2x)^-(n+1)
so, we have
f(0) + 1/1! f<sup>(1)</sup>(0) + 1/2! f<sup>(2)</sup>(0) + ...
= 0 + 1x - 4/2!x^2 + 24/3!x^3 - 192/4!x^4 + ...
= 0 + x - 2x^2 + 4x^3 - 8x^4 + ...
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