How to Derive the Cosine Power Series Using Euler's Identity?

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Discussion Overview

The discussion revolves around the derivation of the cosine power series, specifically exploring methods to achieve a concise and fluent derivation. Participants consider various approaches, including the use of Euler's identity and the Taylor series for sine.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant seeks a shorter and more fluent derivation of the cosine power series, expressing dissatisfaction with their current lengthy method.
  • Another participant suggests that differentiating the power series for sine could provide a quicker derivation for cosine.
  • A different viewpoint proposes that if differentiation of sine is not allowed, an alternative fast derivation method should be considered.
  • One participant outlines their current method involving the Taylor series for cosine, detailing the steps taken to derive the series and expressing a desire to streamline the process.
  • Another participant mentions the possibility of using the square root algorithm as a potential method, though details are not provided.
  • One participant introduces Euler's identity and suggests that substituting into the Taylor series for \( e^x \) could lead to a straightforward derivation of the cosine series.

Areas of Agreement / Disagreement

Participants express differing opinions on the best method for deriving the cosine power series, with no consensus reached on a single optimal approach. Some participants agree on the utility of Euler's identity, while others focus on alternative methods.

Contextual Notes

The discussion includes various assumptions about the familiarity with Taylor series and Euler's identity, which may affect the accessibility of proposed methods. Additionally, the effectiveness of suggested shortcuts remains unverified.

Who May Find This Useful

This discussion may be useful for students and educators in mathematics or physics who are interested in power series, Taylor series, and alternative derivation methods for trigonometric functions.

bomba923
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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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Did you not attach the answer to your question?
 
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...
 
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 :)
 
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??
 
Last edited:
maybe the squareroot algorithm
 
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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