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Derivation of Power Series

 
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Apr1-05, 09:17 PM   #1
 

Derivation of Power Series


How do I write a derivation of the cosine power series?

(I understand and can derive it, but it takes much space and is disjointed! ; how do you write the shortest and fastest derivation for it--briefly and fluently??)
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Apr1-05, 09:21 PM   #2
 
Did you not attach the answer to your question?
Apr1-05, 09:24 PM   #3
 
I believe he means whether there was a faster way to come up with a power series than equating derivatives.

I do not believe there is...
Apr1-05, 09:51 PM   #4
 

Derivation of Power Series


You can define it as the derivative of the power series for sin(x). Thats the shortest derivation of cosin, but then you'd have to define sin :)
Apr1-05, 11:46 PM   #5
 
Of course, that would be much faster--->just differentiate the sum formula for sin(x).
But suppose we can't do that!-->what would be another fast way to derive cos(x)?

Currently I start with cos(x), and then explain that as a Taylor series, cos(x-a) in this case is really the McLaurin series for cos (x), where a=0.
Then, I write cos(x) = [tex] c_0 + c_1 x + c_2 x^2 + c_3 x^3 +... [/tex]
Next, I express each derivative of cos (x) as the same series, except I reduce the powers of x and subunits of the constants appropriately.
Then, I substitute x=0 (b/c it is McLaurin) and express each derivative of cos(0) as the constant "c" with the appropriate subunits.
Next, I show that the odd derivatives of cos(0) are zero, and explain why the sum formula includes [tex] (-1)^n [/tex] and [tex] x^{2n} [/tex]
Then, I show why, as a power series, the factorial (2n)! is needed in the denominator so that the terms will match the derivatives when multiplied out.

Finally, I write the formula (I wish I knew LaTex...better!) as [tex] cos(x) = \sum\limits_{n = 0}^\infty {\frac{{\left( {-1} \right)^n x^{2n} }}{{\left( {2n} \right)!}}} [/tex]

My question is: HOW to constrict/shorten this procedure??
Are there parts here I can connect or skip??
Apr2-05, 11:55 AM   #6
 
maybe the squareroot algorithm
Apr2-05, 12:06 PM   #7
 
Do you know Euler's identity,

[tex] e^{i\theta} = \cos \theta + i \sin \theta[/tex]

?

If so, the Taylor series for [itex]e^x[/itex] is obviously really easy to derive. Just substitute in [itex]x = i \theta[/itex], and note that

[tex] \cos \theta = \mbox{Re} \left[ e^{i \theta} \right] = \frac{ e^{i \theta} + e^{-i \theta}}{2}[/tex]

from which you can just take the series for each of those and add them.
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