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Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that Rn(x) → 0.] f(x) = x4 − 2x2 + 1, a = 2 ∞ f n(2) n! (x − 2)n n = 0 = 24 + 9(x − 2) + 22(x − 2)2 + 8(x − 2)3 + (x − 2)4 ∞ f n(2) n! (x − 2)n n = 0 = −9 + 24(x − 2) + 22(x − 2)2 + 8(x − 2)3 + (x − 2)4 ∞ f n(2) n! (x − 2)n n = 0 = 24 + 9(x − 2) + 8(x − 2)2 + 22(x − 2)3 + (x − 2)4 ∞ f n(2) n! (x − 2)n n = 0 = 9 + 24(x − 2) + 22(x − 2)2 + 8(x − 2)3 + (x − 2)4 ∞ f n(2) n! (x − 2)n n = 0 = 9 + 24(x − 2) + 8(x − 2)2 + 22(x − 2)3 + (x − 2)4

Respuesta :

LammettHash

[tex]f(x)=x^4-2x^2+1\implies f(2)=9[/tex]

[tex]f'(x)=4x^3-4x\implies f'(2)=24[/tex]

[tex]f''(x)=12x^2-4\implies f''(2)=44[/tex]

[tex]f'''(x)=24x\implies f'''(2)=48[/tex]

[tex]f^{(4)}(x)=24\implies f^{(4)}(2)=24[/tex]

[tex]f^{(n)}(x)=0\text{ for }n\ge5\implies f^{(n)}(2)=0[/tex]

The Taylor expansion around [tex]x=2[/tex] is then

[tex]T(x)=\dfrac{f(0)(x-2)^0}{0!}+\dfrac{f'(0)(x-2)^1}{1!}+\dfrac{f''(0)(x-2)^2}{2!}+\dfrac{f'''(0)(x-2)^3}{3!}+\dfrac{f^{(4)}(0)(x-2)^4}{4!}[/tex]

[tex]T(x)=9+24(x-2)+22(x-2)^2+8(x-2)^3+(x-2)^4[/tex]

ashishdwivedilVT

Taylor's series expansion of f(x) about x=2 was written using the expansion formula.

[tex]f(x)=x^4-2x^{2} +1[/tex]

[tex]f(2)=9[/tex]

[tex]f'(x)=4x^3-4x[/tex]

[tex]f'(2)=24[/tex]

[tex]f''(x)=12x^{2} -4[/tex]

[tex]f''(2)=44[/tex]

[tex]f'''(x)=24x[/tex]

[tex]f'''(2)=48[/tex]

What is Taylor's series expansion of f(x) about x=c?

Taylor's series expansion of f(x) about x=c is given by:

[tex]f(x)=f(c)+f'(c)(x-c)+\frac{f''(c)}{2!} (x-c)^2+\frac{f'''(c)}{3!} (x-c)^3+....[/tex]

So, Taylor's series expansion about x=2 will be given by

[tex]f(x)=f(2)+f'(2)(x-2)+\frac{f''(2)}{2!} (x-2)^2+\frac{f'''(2)}{3!} (x-2)^3+....[/tex]

[tex]f(x)=9+24(x-2)+21(x-2)^2+8(x-2)^3+....[/tex]

Hence, Taylor's series expansion of f(x) about x=2 was written using the expansion formula.

To get more about Taylor's series visit:

https://brainly.com/question/24188700

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