# Derivation of Power Series

1. Apr 1, 2005

### bomba923

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! :grumpy: ; how do you write the shortest and fastest derivation for it--briefly and fluently??)

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2. Apr 1, 2005

### whozum

Did you not attach the answer to your question?

3. Apr 1, 2005

### Moo Of Doom

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...

4. Apr 1, 2005

### whozum

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 :)

5. Apr 1, 2005

### bomba923

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) = $$c_0 + c_1 x + c_2 x^2 + c_3 x^3 +...$$
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 $$(-1)^n$$ and $$x^{2n}$$
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 $$cos(x) = \sum\limits_{n = 0}^\infty {\frac{{\left( {-1} \right)^n x^{2n} }}{{\left( {2n} \right)!}}}$$

My question is: HOW to constrict/shorten this procedure??
Are there parts here I can connect or skip??

Last edited: Apr 2, 2005
6. Apr 2, 2005

### tongos

maybe the squareroot algorithm

7. Apr 2, 2005

### Data

Do you know Euler's identity,

$$e^{i\theta} = \cos \theta + i \sin \theta$$

?

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

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

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