Power Series
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BigGlenntheHeavy
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Instead of using a known Maclaurin series, thanks to my trusty TI-89, I took the derivative of f(x) ten times and evaluated each derivative at zero. I then tried to recognize a pattern for these numbers and came up with:
\(\displaystyle f(x) \ = \ \frac{x^{3}}{(x-2)^{2}} \ = \ \sum_{n=1}^{\infty} \frac{(2n-1)x^{2n+1}+nx^{2n+2}}{4^{n}}, \ converges \ at \ (-2,2).\)
Addendum: I could have just as easily, in fact easier, used my Taylor series on my trusty TI-89 (F3-9) to arrive at the same conclusion.
\(\displaystyle f(x) \ = \ \frac{x^{3}}{(x-2)^{2}} \ = \ \sum_{n=1}^{\infty} \frac{(2n-1)x^{2n+1}+nx^{2n+2}}{4^{n}}, \ converges \ at \ (-2,2).\)
Addendum: I could have just as easily, in fact easier, used my Taylor series on my trusty TI-89 (F3-9) to arrive at the same conclusion.
D
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sraz said:
\(\displaystyle \sum \frac{x^3}{(x-2)^2} \, = \, \sum x \cdot \frac{1}{(1-\frac{2}{x})^2} \, = \,\)
strong edit: the below is NOT a power series, hah. i need to learn to read. i have no idea why i did this.
\(\displaystyle \frac{x^3}{(x-2)^2} = \frac{x^3}{x-2}\frac{1}{x-2} = -\frac{x^3}{2(x-2)} \cdot \frac{1}{1 - \frac{x}{2}}\)
We make sure to note that we need \(\displaystyle |\frac{x}{2}| < 1\) then we get:
\(\displaystyle \therefore \frac{x^3}{(x-2)^2} = \frac{-x^3}{2(x-2)}\sum_{k=0}^{\infty} (\frac{x}{2})^k = \sum_{k=0}^{\infty} \frac{x^{k+3}}{2^{k+1}(2-x)}\)
Then the requirement is of course:
\(\displaystyle |\frac{x}{2}| < 1 \implies |x| < 2 \implies x \in (-2,2)\)
\(\displaystyle \frac{x^3}{(x-2)^2} = \frac{x^3}{x-2}\frac{1}{x-2} = -\frac{x^3}{2(x-2)} \cdot \frac{1}{1 - \frac{x}{2}}\)
We make sure to note that we need \(\displaystyle |\frac{x}{2}| < 1\) then we get:
\(\displaystyle \therefore \frac{x^3}{(x-2)^2} = \frac{-x^3}{2(x-2)}\sum_{k=0}^{\infty} (\frac{x}{2})^k = \sum_{k=0}^{\infty} \frac{x^{k+3}}{2^{k+1}(2-x)}\)
Then the requirement is of course:
\(\displaystyle |\frac{x}{2}| < 1 \implies |x| < 2 \implies x \in (-2,2)\)
BigGlenntheHeavy
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daon, why isn't your analysis a "power series"?
I plugged it into Maple 8 and everything worked, the convergence (-2,2) and divergence (?) rang true.
I plugged it into Maple 8 and everything worked, the convergence (-2,2) and divergence (?) rang true.
I know its correct, but it isn't in the form:
\(\displaystyle \sum_n a_n (x-k)^n\)
I thought of a way (at least i think i did) to make it into one from where i left off, but the process would have been much longer and difficult than called for
\(\displaystyle \sum_n a_n (x-k)^n\)
I thought of a way (at least i think i did) to make it into one from where i left off, but the process would have been much longer and difficult than called for
BigGlenntheHeavy
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Yes, daon, I played with it too. Its that (2-x) in the denominator that's the stumbling block.
Have you all tried looking at the pattern in the nth derivatives of the Taylor series, then letting x=0.
\(\displaystyle f(x)=\frac{x^{3}}{(x-2)^{2}}\)
\(\displaystyle f'(x)=\frac{x^{3}-6x^{2}}{(x-2)^{3}}\)
\(\displaystyle f^{(2)}(x)=\frac{8(3x+0)}{(x-2)^{4}}\)
\(\displaystyle f^{(3)}(x)=\frac{-24(3x+2)}{(x-2)^{5}}\)
\(\displaystyle f^{(4)}(x)=\frac{96(3x+4)}{(x-2)^{6}}\)
\(\displaystyle f^{(5)}(x)=\frac{-480(3x+6)}{(x-2)^{7}}\)
\(\displaystyle f^{(6)}(x)=\frac{2880(3x+8)}{(x-2)^{8}}\)
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There is a definite pattern. Now, to find the general pattern. Just a thought.
\(\displaystyle f(x)=\frac{x^{3}}{(x-2)^{2}}\)
\(\displaystyle f'(x)=\frac{x^{3}-6x^{2}}{(x-2)^{3}}\)
\(\displaystyle f^{(2)}(x)=\frac{8(3x+0)}{(x-2)^{4}}\)
\(\displaystyle f^{(3)}(x)=\frac{-24(3x+2)}{(x-2)^{5}}\)
\(\displaystyle f^{(4)}(x)=\frac{96(3x+4)}{(x-2)^{6}}\)
\(\displaystyle f^{(5)}(x)=\frac{-480(3x+6)}{(x-2)^{7}}\)
\(\displaystyle f^{(6)}(x)=\frac{2880(3x+8)}{(x-2)^{8}}\)
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There is a definite pattern. Now, to find the general pattern. Just a thought.
BigGlenntheHeavy
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Taylor series centered at c = 0 (Maclaurin Series).
\(\displaystyle f(x) \ = \ \frac{x^{3}}{(x-2)^{2}} \ = \ \frac{x^{3}}{4}+\frac{x^{4}}{4}+\frac{3x^{5}}{16}+\frac{x^{6}}{8}+\frac{5x^{7}}{64}+\frac{3x^{8}}{64}+\frac{7x^{9}}{256}+\frac{x^{10}}{64}+\frac{9x^{11}}{1024}+\frac{5x^{12}}{1024}+...\)
\(\displaystyle Do \ you \ see \ a \ pattern?\)
\(\displaystyle f(x) \ = \ \frac{x^{3}}{(x-2)^{2}} \ = \ \frac{x^{3}}{4}+\frac{x^{4}}{4}+\frac{3x^{5}}{16}+\frac{x^{6}}{8}+\frac{5x^{7}}{64}+\frac{3x^{8}}{64}+\frac{7x^{9}}{256}+\frac{x^{10}}{64}+\frac{9x^{11}}{1024}+\frac{5x^{12}}{1024}+...\)
\(\displaystyle Do \ you \ see \ a \ pattern?\)
Using the info from my last post, with q fixed, one can see it is the sum of the following:
edit: the ith row represents replacing q with q+i, and the jth column represents p=j. 1st row/col is the "0th"
....\(\displaystyle \text{ } \frac{x^{q+3}}{2^{q+2}} + \frac{x^{q+4}}{2^{q+3}} + \frac{x^{q+5}}{2^{q+4}} + ...\)
\(\displaystyle \text{ +} \frac{x^{q+4}}{2^{q+3}} + \frac{x^{q+5}}{2^{q+4}} + \frac{x^{q+6}}{2^{q+5}} + ...\)
\(\displaystyle \text{ +} \frac{x^{q+5}}{2^{q+4}} + \frac{x^{q+6}}{2^{q+5}} + \frac{x^{q+7}}{2^{q+6}} + ...\)
\(\displaystyle \vdots\)
Pattern here is that the diagonals are all the same, so we get the sum:
\(\displaystyle \sum_{q=0}^{\infty}\sum_{p=0}^{\infty}\frac{x^{p+q+3}}{2^{p+q+2}} = \sum_{n=0}^{\infty} \frac{(n+1)x^{n+3}}{2^{n+2}} = \sum_{n=0}^{\infty} \Big{[} \frac{n+1}{2^{n+2}} \Big{]} x^{n+3}\)
edit: the ith row represents replacing q with q+i, and the jth column represents p=j. 1st row/col is the "0th"
....\(\displaystyle \text{ } \frac{x^{q+3}}{2^{q+2}} + \frac{x^{q+4}}{2^{q+3}} + \frac{x^{q+5}}{2^{q+4}} + ...\)
\(\displaystyle \text{ +} \frac{x^{q+4}}{2^{q+3}} + \frac{x^{q+5}}{2^{q+4}} + \frac{x^{q+6}}{2^{q+5}} + ...\)
\(\displaystyle \text{ +} \frac{x^{q+5}}{2^{q+4}} + \frac{x^{q+6}}{2^{q+5}} + \frac{x^{q+7}}{2^{q+6}} + ...\)
\(\displaystyle \vdots\)
Pattern here is that the diagonals are all the same, so we get the sum:
\(\displaystyle \sum_{q=0}^{\infty}\sum_{p=0}^{\infty}\frac{x^{p+q+3}}{2^{p+q+2}} = \sum_{n=0}^{\infty} \frac{(n+1)x^{n+3}}{2^{n+2}} = \sum_{n=0}^{\infty} \Big{[} \frac{n+1}{2^{n+2}} \Big{]} x^{n+3}\)
BigGlenntheHeavy
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Good show, daon, I am not familiar with that approach, but it is better than my final way, see above.
I think I do.
Just for kicks, I have a pattern I came up with using the Taylor series.
Starting at n=2 (second derivative) for the derivatives of \(\displaystyle \frac{x^{3}}{(x-2)^{2}}\):
\(\displaystyle f^{(n)}(x)=\frac{(-1)^{n}4n!\cdot (3x+2(n-2))}{(x-2)^{n+2}}\)
Dividing by the corresponding n!, as we do when using Taylor series, we get:
\(\displaystyle \frac{(-1)^{n}\cdot 4(3x+2(n-2))}{(x-2)^{n+2}}=(-1)^{n}\left(\frac{12}{(x-2)^{n+1}}+\frac{8(n+1)}{(x-2)^{n+2}}\right)\)
This gives the correct coefficients from n=2 on up. The series begins at n=3 anyway. I get -1 for the first derivative, though, using the formula
.But the first derivative is obviously 0 if we just sub in x=0 directly into f'(x). Same as f(x) itself.
For instance, Let x=0 throughout and n=3 we get 1/4; n=4, we get 1/4; n=5, we get 3/16; n=6, we get 1/8; n=7, we get 5/64, and so on and so on. They all check with Glenn's series above.
Just a thought as I was playing with it for a little while after I finished mowing grass
. For what it's worth.