help cal 2 homework

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sheilaw

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find a power series representation for f(x)=x^3/(x-2)^2

i dont no how to start it, please give me a hint for this question.Thank you
 
sheilaw said:
find a power series representation for f(x)=x^3/(x-2)^2

i dont no how to start it, please give me a hint for this question.Thank you
Click to expand...

Start with power series representation of (1+a)[sup:11nk3g3l]-n[/sup:11nk3g3l]
 
f(x) = x3(x−2)2, f(0) = 0.\displaystyle f(x) \ = \ \frac{x^{3}}{(x-2)^{2}}, \ f(0) \ = \ 0.

f1(x) = x2(x−6)(x−2)3, f1(0) = 0.\displaystyle f^{1}(x) \ = \ \frac{x^{2}(x-6)}{(x-2)^{3}}, \ f^{1}(0) \ = \ 0.

f2(x) = 24x(x−2)4, f2(0) = 0.\displaystyle f^{2}(x) \ = \ \frac{24x}{(x-2)^{4}}, \ f^{2}(0) \ = \ 0.

f3(x) = −24(3x+2)(x−2)5, f3(0) = 32.\displaystyle f^{3}(x) \ = \ \frac{-24(3x+2)}{(x-2)^{5}}, \ f^{3}(0) \ = \ \frac{3}{2}.

f4(x) = 96(3x+4)(x−2)6, f4(0) = 6.\displaystyle f^{4}(x) \ = \ \frac{96(3x+4)}{(x-2)^{6}}, \ f^{4}(0) \ = \ 6.

f5(x) = −1440(x+2)(x−2)7, f5(0) = 452.\displaystyle f^{5}(x) \ = \ \frac{-1440(x+2)}{(x-2)^{7}}, \ f^{5}(0) \ = \ \frac{45}{2}.

f6(x) = 2880(3x+8)(x−2)8, f6(0) = 90.\displaystyle f^{6}(x) \ = \ \frac{2880(3x+8)}{(x-2)^{8}}, \ f^{6}(0) \ = \ 90.

f7(x) = −20160(3x+10)(x−2)9. f7(0) = 15754.\displaystyle f^{7}(x) \ = \ \frac{-20160(3x+10)}{(x-2)^{9}}. \ f^{7}(0) \ = \ \frac{1575}{4}.

f8(x) = 483840(x+4)(x−2)10, f8(0) = 1890.\displaystyle f^{8}(x) \ = \ \frac{483840(x+4)}{(x-2)^{10}}, \ f^{8}(0) \ = \ 1890.

f9(x) = −1451520(3x+14)(x−2)11, f9(0) = 198452.\displaystyle f^{9}(x) \ = \ \frac{-1451520(3x+14)}{(x-2)^{11}}, \ f^{9}(0) \ = \ \frac{19845}{2}.

f10(x) = 14515200(3x+16)(x−2)12, f10(0) = 56700.\displaystyle f^{10}(x) \ = \ \frac{14515200(3x+16)}{(x-2)^{12}}, \ f^{10}(0) \ = \ 56700.

Definition of a Maclaurin polynomial:

Pn(x) = f(0)+f1(0)x+f2(0)x22!+f3(0)x33!+...+fn(0)xnn!.\displaystyle P_n(x) \ = \ f(0)+f^{1}(0)x+\frac{f^{2}(0)x^{2}}{2!}+\frac{f^{3}(0)x^{3}}{3!}+...+\frac{f^{n}(0)x^{n}}{n!}.

Hence, P10(x) = 0+0+0+x34+x44+3x516+x68+5x764+3x864+7x9256+x1064.\displaystyle Hence, \ P_{10}(x) \ = \ 0+0+0+\frac{x^{3}}{4}+\frac{x^{4}}{4}+\frac{3x^{5}}{16}+\frac{x^{6}}{8}+\frac{5x^{7}}{64}+\frac{3x^{8}}{64}+\frac{7x^{9}}{256}+\frac{x^{10}}{64}.

Revising, we get: P10(x) = x3+x422+3x5+2x624+5x7+3x826+7x9+4x1028\displaystyle Revising, \ we \ get: \ P_{10}(x) \ = \ \frac{x^{3}+x^{4}}{2^{2}}+\frac{3x^{5}+2x^{6}}{2^{4}}+\frac{5x^{7}+3x^{8}}{2^{6}}+\frac{7x^{9}+4x^{10}}{2^{8}}

Ergo, we have a pattern, therefore;

x3(x−2)2 = ∑n=0∞(2n+1)x2n+3+(n+1)x2n+422n+2\displaystyle \frac{x^{3}}{(x-2)^{2}} \ = \ \sum_{n=0}^{\infty}\frac{(2n+1)x^{2n+3}+(n+1)x^{2n+4}}{2^{2n+2}}

Note: converges on (-2,2).

Also note: all this protractedness could have been avoided by using my trusty TI-89 for Taylor series.(F3-9)
 
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