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Math Help - Find the taylor series

  1. #1
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    Find the taylor series

    Hello,
    I have been trying to find the taylor series for the follwing problems for a couple of hours but I have made no progress....please show how I can solve them......

    1. f(x)=cos((x+1)^1/2) at x=0
    2. f(x)=(x+1)^1/3 at x=0
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by gichfred View Post
    Hello,
    I have been trying to find the taylor series for the follwing problems for a couple of hours but I have made no progress....please show how I can solve them......

    1. f(x)=cos((x+1)^1/2) at x=0
    2. f(x)=(x+1)^1/3 at x=0
    taylor

    \cos (\sqrt{1+x}) at x=0

    let u= \sqrt{x+1}

    find taylor at u=1
    \cos u

    f(u) = \cos u
    f(1) = \cos (1)
    f'(u) = -\sin u \Rightarrow f'(1) =-\sin (1)
    f''(u) = -\cos u \Rightarrow f''(1) = -\cos (1)
    f'''(u) = \sin u \Rightarrow f'''(1) = \sin (1)

    f(u)=\cos u = f(1) + f'(1)(u-1) + \frac{f''(1)(u-1)^2}{2} + \frac{f'''(1)(u-1)^3}{3!} + ...

    f(u)=\cos u= \cos (1) - \sin (1)\cdot  (u-1)-\frac{\cos (1) \cdot (u-1)^2 }{2} + \frac{\sin (1) \cdot (u-1)^3 }{3!}  + \frac{\cos (1) \cdot (u-1)^4 }{4!}- \frac{\sin (1) \cdot (u-1)^5}{5!} + ...

    f(u) = \cos u = \sum_{k=0}^{\infty} \frac{(-1)^k \cos (1) (u-1)^{2k}}{(2k)!} + \sum_{k=0}^{\infty} \frac{(-1)^{k+1} \sin (1) (u-1)^{2k+1} }{(2k+1)!}

    sub u value

    \cos \sqrt{x+1} =\sum_{k=0}^{\infty} \frac{(-1)^k \cos (1) (\sqrt{x+1}-1)^{2k}}{(2k)!} + \sum_{k=0}^{\infty} \frac{(-1)^{k+1} \sin (1) (\sqrt{x+1}-1)^{2k+1} }{(2k+1)!}


    f(x) = (x+1)^{\frac{1}{3}} at x=0

    f(0) =1
    f'(x) = \frac{(x+1)^{\frac{-2}{3}}}{3}\Rightarrow f'(0)= \frac{1}{3}
    f''(x) = \frac{-2\cdot (x+1)^{\frac{-5}{3}}}{3^2 } \Rightarrow f''(1) = \frac{-2}{3^2}
    f'''(x) = \frac{-2\cdot -5 \cdot (x+1)^{\frac{-8}{3}}}{3^3}\Rightarrow \frac{2\cdot 5 }{3^3}
    f''''(1) = -\frac{2\cdot 5 \cdot 8 }{3^4}

    f(x) = (x+1)^{\frac{1}{3}} = 1 + \frac{x}{3\cdot 1!} - \frac{2x^2}{3^2 \cdot 2!} + \frac{2\cdot 5\cdot x^3 }{3^3 \cdot 3!} - \frac{2\cdot 5 \cdot 8 \cdot x^4}{3^4 \cdot 4!} + ...

    you can use \prod to collect 2.5.8...
    can you find it now ?
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  3. #3
    MHF Contributor Amer's Avatar
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    I make a mistake in cos sqrt{x+1} the Taylor expansion should not contains roots so it will come like this


    \cos \sqrt{x+1} = \sum_{k=0}^{\infty} \frac{(-1)^k (\sqrt{x+1})^{2k}}{(2k)!}

    \cos \sqrt{x+1} = \sum_{k=0}^{\infty} \frac{(-1)^k (x+1)^k}{(2k)!}

    but

    (x+1)^k = \sum_{i=0}^{k} com(k,i) x^i

    com(k,i)= \frac{k!}{i! (k-i)!}

    \cos \sqrt{x+1} = \sum_{k=0}^{\infty} \frac{(-1)^k \cdot \sum_{i=0}^{k} com(k,i) x^i }{(2k)!}

    simplify it, sub k values and take x,x^2,x^3,x^4,... as factors

    This solution for my prof not me
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