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Math Help - Maclaurin Expansion

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    Maclaurin Expansion

    Can someone show me how the \tan x expansion is derived?

    Thanks in advance.
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Air View Post
    Can someone show me how the \tan x expansion is derived?

    Thanks in advance.
    \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!} = x-\frac{x^3}{6}+\frac{x^5}{120}-...

    \cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}=1-\frac{x^2}{2}+\frac{x^4}{24}-...

    Now we use long division

    Maclaurin Expansion-capture.jpg


    The more terms we write out and use in the long division the more terms we get in the series of tangent.
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  3. #3
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    Quote Originally Posted by Air View Post
    Can someone show me how the \tan x expansion is derived?

    Thanks in advance.
    tan(x) = \sum_{n = 0}^{\infty}\frac{1}{n!}(tan(0))^{(n)}x^n
    where (tan(0))^{(n)} is the nth derivative of tan(x) evaluated at x = 0.

    tan(0) = 0

    (tan(0))^{(1)} = 1

    (tan(0))^{(2)} = 0

    etc. The derivatives are something of a pain, but with a little effort you should be able to find that
    tan(x) = x + \frac{x^3}{3} + \frac{2x^5}{15} + ~...

    I'll leave it to you to find a general expression for the coefficient of the nth power of x.

    -Dan
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