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July 22, 2014

July 22, 2014

Posted by **Heather** on Wednesday, November 28, 2012 at 9:19pm.

- Calculus 2 -
**Count Iblis**, Wednesday, November 28, 2012 at 11:19pmUsually, the efficient way to compute Taylor series is by using the Taylor series of the known standard fucntions like sin(x), cos(x) etc., instead of using the definition of the Taylor series in terms of the derivatives of the function.

In this case, you can write:

tan(x) = sin(x)/cos(x)

If we put:

tan(x) = c1 x + c3 x^3 + c5 x^5 + ...

(note that because tan(x) is n odd fucntion, it can only have odd powers of x in the Taylor expansion), then we can find the ck from:

sin(x) = [c1 x + c3 x^3 + c5 x^5...] * cos(x)

if we substitute for sin(x) and cos(x) their respective series expansions and equate equal coefficients of x:

x - x^3/3! + x^5/5! - ... =

[c1 x + c3 x^3 + c5 x^5...] *

[1 - x^2/2! + x^4/4! - ...]

The coefficient of x on the r.h.s. is c1 and this has to be 1, so c1 = 1.

Equating the coefficient of x^3 on both sides gives:

-1/6 = -1/2 + c3 --->

c3 = 1/3

Equating the coefficient of x^5 on both sides gives:

1/120 = 1/24 -1/6 + c5 ---->

c5 = 2/15

So, we have:

tan(x) = x + x/3 + 2/15 x^5 + ...

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