You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
help cal 2 homework
- Thread starter sheilaw
- Start date
D
Deleted member 4993
Guest
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
Start with power series representation of (1+a)[sup:11nk3g3l]-n[/sup:11nk3g3l]
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle f(x) \ = \ \frac{x^{3}}{(x-2)^{2}}, \ f(0) \ = \ 0.\)
\(\displaystyle f^{1}(x) \ = \ \frac{x^{2}(x-6)}{(x-2)^{3}}, \ f^{1}(0) \ = \ 0.\)
\(\displaystyle f^{2}(x) \ = \ \frac{24x}{(x-2)^{4}}, \ f^{2}(0) \ = \ 0.\)
\(\displaystyle f^{3}(x) \ = \ \frac{-24(3x+2)}{(x-2)^{5}}, \ f^{3}(0) \ = \ \frac{3}{2}.\)
\(\displaystyle f^{4}(x) \ = \ \frac{96(3x+4)}{(x-2)^{6}}, \ f^{4}(0) \ = \ 6.\)
\(\displaystyle f^{5}(x) \ = \ \frac{-1440(x+2)}{(x-2)^{7}}, \ f^{5}(0) \ = \ \frac{45}{2}.\)
\(\displaystyle f^{6}(x) \ = \ \frac{2880(3x+8)}{(x-2)^{8}}, \ f^{6}(0) \ = \ 90.\)
\(\displaystyle f^{7}(x) \ = \ \frac{-20160(3x+10)}{(x-2)^{9}}. \ f^{7}(0) \ = \ \frac{1575}{4}.\)
\(\displaystyle f^{8}(x) \ = \ \frac{483840(x+4)}{(x-2)^{10}}, \ f^{8}(0) \ = \ 1890.\)
\(\displaystyle f^{9}(x) \ = \ \frac{-1451520(3x+14)}{(x-2)^{11}}, \ f^{9}(0) \ = \ \frac{19845}{2}.\)
\(\displaystyle f^{10}(x) \ = \ \frac{14515200(3x+16)}{(x-2)^{12}}, \ f^{10}(0) \ = \ 56700.\)
Definition of a Maclaurin polynomial:
\(\displaystyle P_n(x) \ = \ f(0)+f^{1}(0)x+\frac{f^{2}(0)x^{2}}{2!}+\frac{f^{3}(0)x^{3}}{3!}+...+\frac{f^{n}(0)x^{n}}{n!}.\)
\(\displaystyle Hence, \ P_{10}(x) \ = \ 0+0+0+\frac{x^{3}}{4}+\frac{x^{4}}{4}+\frac{3x^{5}}{16}+\frac{x^{6}}{8}+\frac{5x^{7}}{64}+\frac{3x^{8}}{64}+\frac{7x^{9}}{256}+\frac{x^{10}}{64}.\)
\(\displaystyle Revising, \ we \ get: \ P_{10}(x) \ = \ \frac{x^{3}+x^{4}}{2^{2}}+\frac{3x^{5}+2x^{6}}{2^{4}}+\frac{5x^{7}+3x^{8}}{2^{6}}+\frac{7x^{9}+4x^{10}}{2^{8}}\)
Ergo, we have a pattern, therefore;
\(\displaystyle \frac{x^{3}}{(x-2)^{2}} \ = \ \sum_{n=0}^{\infty}\frac{(2n+1)x^{2n+3}+(n+1)x^{2n+4}}{2^{2n+2}}\)
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)
\(\displaystyle f^{1}(x) \ = \ \frac{x^{2}(x-6)}{(x-2)^{3}}, \ f^{1}(0) \ = \ 0.\)
\(\displaystyle f^{2}(x) \ = \ \frac{24x}{(x-2)^{4}}, \ f^{2}(0) \ = \ 0.\)
\(\displaystyle f^{3}(x) \ = \ \frac{-24(3x+2)}{(x-2)^{5}}, \ f^{3}(0) \ = \ \frac{3}{2}.\)
\(\displaystyle f^{4}(x) \ = \ \frac{96(3x+4)}{(x-2)^{6}}, \ f^{4}(0) \ = \ 6.\)
\(\displaystyle f^{5}(x) \ = \ \frac{-1440(x+2)}{(x-2)^{7}}, \ f^{5}(0) \ = \ \frac{45}{2}.\)
\(\displaystyle f^{6}(x) \ = \ \frac{2880(3x+8)}{(x-2)^{8}}, \ f^{6}(0) \ = \ 90.\)
\(\displaystyle f^{7}(x) \ = \ \frac{-20160(3x+10)}{(x-2)^{9}}. \ f^{7}(0) \ = \ \frac{1575}{4}.\)
\(\displaystyle f^{8}(x) \ = \ \frac{483840(x+4)}{(x-2)^{10}}, \ f^{8}(0) \ = \ 1890.\)
\(\displaystyle f^{9}(x) \ = \ \frac{-1451520(3x+14)}{(x-2)^{11}}, \ f^{9}(0) \ = \ \frac{19845}{2}.\)
\(\displaystyle f^{10}(x) \ = \ \frac{14515200(3x+16)}{(x-2)^{12}}, \ f^{10}(0) \ = \ 56700.\)
Definition of a Maclaurin polynomial:
\(\displaystyle P_n(x) \ = \ f(0)+f^{1}(0)x+\frac{f^{2}(0)x^{2}}{2!}+\frac{f^{3}(0)x^{3}}{3!}+...+\frac{f^{n}(0)x^{n}}{n!}.\)
\(\displaystyle Hence, \ P_{10}(x) \ = \ 0+0+0+\frac{x^{3}}{4}+\frac{x^{4}}{4}+\frac{3x^{5}}{16}+\frac{x^{6}}{8}+\frac{5x^{7}}{64}+\frac{3x^{8}}{64}+\frac{7x^{9}}{256}+\frac{x^{10}}{64}.\)
\(\displaystyle Revising, \ we \ get: \ P_{10}(x) \ = \ \frac{x^{3}+x^{4}}{2^{2}}+\frac{3x^{5}+2x^{6}}{2^{4}}+\frac{5x^{7}+3x^{8}}{2^{6}}+\frac{7x^{9}+4x^{10}}{2^{8}}\)
Ergo, we have a pattern, therefore;
\(\displaystyle \frac{x^{3}}{(x-2)^{2}} \ = \ \sum_{n=0}^{\infty}\frac{(2n+1)x^{2n+3}+(n+1)x^{2n+4}}{2^{2n+2}}\)
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)