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Math Help - function for Taylor series and find the 100th derivative of this function

  1. #1
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    function for Taylor series and find the 100th derivative of this function

    Hey, again ... My neighbour send me again one exercise for his exam for Calculus 1. He needs to find the Taylor series and the 100th derivative of this function at x = -1:
    f(x) = (x^2 + 3x + 1)*e^{2x-1} . Actually, I really forgot a little bit what you must do here, so please help me to solve...
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    Re: function for Taylor series and find the 100th derivative of this function

    The Taylor's series for e^x, about x= 0, is \sum_{n=0}^\infty \frac{x^n}{n!}. Also e^{2x-1}= e^{-1}e^{2x} so the Taylor's series for e^{2x-1} is \sum_{n=0}^\infty e^{-1}2^n \frac{x^n}{n!}. To get the Taylor's series for (x^2+ 3x+ 1)e^{2x-1} multiply that series by x^2+ 3x+ 1}:
    x^2(\sum_{n=0}^\infty e^{-1}2^n\frac{x^n}{n!})+ 3x(\sum_{n=0}^\infty e^{-1}2^n\frac{x^n}{n!})+ (\sum_{n=0}^\infty e^{-1}2^n\frac{x^n}{n!}) = (\sum_{n=0}^\infty e^{-1}2^n\frac{x^{n+2}}{n!}) + (\sum_{n=0}^\infty 3e^{-1}2^n\frac{x^{n+1}}{n!}) + (\sum_{n=0}^\infty e^{-1}2^n\frac{x^n}{n!}).

    In order to write that as a single sum, you will want to "adjust" the indices. In the first sum let j= n+ 2 so that n= j- 2 and the sum becomes \sum_{j= 2}^\infty e^{-1}2^{j- 2}(j)(j-1)\frac{x^j}{j!}. For the second, let j= n+ 1 so that n= j- 1 and the sum becomes \sum_{j=1}^\infty 3e^{-1}2^{j- 1}j\frac{x^j}{j!}. In the final sum let j= n so it is the same sum, just indexed by j rather than n.
    Last edited by HallsofIvy; July 30th 2014 at 05:47 AM.
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    Re: function for Taylor series and find the 100th derivative of this function

    Ok, thanks for that. But actually, you gave me that for x = 0. I need for x = -1. Would it be the same way?
    How do I find the 100th derivative f^{100}(-1) from this Taylor series? Or I need to find first the n-th derivative and then insert the value n = 100 or can I get the 100-th derivative directly from Taylor series?
    I will discuss with my neighbour tomorrow.
    For x = -1 I get: e^{2x-1} = e^{2(x + 1) - 3}. Can this even help here?
    Last edited by lebdim; July 30th 2014 at 07:06 AM.
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    Re: function for Taylor series and find the 100th derivative of this function

    And my neighbour just send me a couple of exercises for his exam (examples for previous exams):

    1. We have the rational function f(x) = \frac{x^2 + x + 1}{-x^3 + 2x^2 - 3x + 6}. Also we have to find the Taylor series for this rational function at point x = 0 and find the 88-th derivative f^{(88)}(0).

    2. We have to find the indefinite integral for: \int{\sqrt{\frac{ch^2(x) - 1}{ch(x)}}}dx, where is the ch(x) = \frac{e^x + e^{-x}}{2}.

    3. We have to find the indefinite integral for: \int{\frac{6x^3 + 9x^2 + 15x + 5}{\sqrt{x^2 + x + 2}}}dx.

    My neighbour just wrote me: "If you know, how to find the solution, please let me know..." If anybody here on the forum knows, please show me way ...
    Last edited by lebdim; July 30th 2014 at 07:39 AM.
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    Re: function for Taylor series and find the 100th derivative of this function

    e^{2x- 1}= e^{2(x- 1)+ 1}= e(e^{2(x- 1)}). So just change my "x" to "x- 1" and multiply by e, which will just cancel the " e^{-1}" in my formula.
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    Re: function for Taylor series and find the 100th derivative of this function

    Let g(x) = (ax^2+bx+c)e^{2x-1}. Then g'(x) = (2ax+b)e^{2x-1}+2(ax^2+bx+c)e^{2x-1} = \left( 2ax^2+(2a+b)x+b+2c \right)e^{2x-1}. In other words, g'(x) has the same terms with different coefficients as g(x). So, let:

    \dfrac{d^n}{dx^n}\left(f(x)\right) = \left(a_nx^2+b_nx+c_n\right)e^{2x-1}

    The problem gives us that a_0=1, b_0=3, c_0=1. Then:

    \begin{align*} \left( a_{n+1}x^2+b_{n+1}x+c_{n+1} \right)e^{2x-1} & = \dfrac{d^{n+1}}{dx^{n+1}}\left( f(x) \right) \\ & = \dfrac{d}{dx}\left[ \dfrac{d^n}{dx^n}\left( f(x) \right) \right] \\ & = \dfrac{d}{dx}\left[\left( a_nx^2 + b_nx + c_n \right)e^{2x-1} \right] \\ & = a_n\dfrac{d}{dx}\left(x^2e^{2x-1}\right) + b_n\dfrac{d}{dx}\left(xe^{2x-1}\right) + c_n\dfrac{d}{dx}\left(e^{2x-1}\right) \\ & = a_n\left( 2x^2e^{2x-1}+2xe^{2x-1} \right) + b_n\left( 2xe^{2x-1} + e^{2x-1} \right) + c_n\left( 2e^{2x-1} \right) \\ & = \left(2a_nx^2 + (2a_n+2b_n)x + (b_n+2c_n)\right)e^{2x-1}\end{align*}

    Equating coefficients gives the following relations:

    a_{n+1} = 2a_n
    b_{n+1} = 2a_n+2b_n
    c_{n+1} = b_n+2c_n

    Since a_0=1, we have a_n = 2^n.
    Plugging in, we get:

    b_{n+1} = 2\cdot 2^n+2b_n

    Since b_0 = 3, we have b_n = 2^n(n+3)

    Plugging that in, we get:

    c_{n+1} = 2^n(n+3) + 2c_n

    Since c_0=1, we have c_n = 2^{n-2}(n+1)(n+4)

    This gives:

    \begin{align*}\dfrac{d^{100}}{dx^{100}}\left( f(x) \right) & = \left(a_{100}x^2 + b_{100}x + c_{100}\right)e^{2x-1} \\ & = \left( 2^{100}x^2 + 103\cdot 2^{100}x + 101\cdot 104\cdot 2^{98} \right)e^{2x-1} \\ & = 2^{100}\left( x^2 + 103x + 2626 \right)e^{2x-1}\end{align*}

    So, f^{(100)}(-1) = 2^{100}\left( (-1)^2 + 103(-1) + 2626 \right)e^{2(-1)-1} = 2524\cdot 2^{100}e^{-3}
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    Re: function for Taylor series and find the 100th derivative of this function

    Ok, thanks. I understand all the ways to the solution. What about the other 3 exercises? Does anybody know?
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    Re: function for Taylor series and find the 100th derivative of this function

    Quote Originally Posted by lebdim View Post
    And my neighbour just send me a couple of exercises for his exam (examples for previous exams):

    1. We have the rational function f(x) = \frac{x^2 + x + 1}{-x^3 + 2x^2 - 3x + 6}. Also we have to find the Taylor series for this rational function at point x = 0 and find the 88-th derivative f^{(88)}(0).

    2. We have to find the indefinite integral for: \int{\sqrt{\frac{ch^2(x) - 1}{ch(x)}}}dx, where is the ch(x) = \frac{e^x + e^{-x}}{2}.

    3. We have to find the indefinite integral for: \int{\frac{6x^3 + 9x^2 + 15x + 5}{\sqrt{x^2 + x + 2}}}dx.

    My neighbour just wrote me: "If you know, how to find the solution, please let me know..." If anybody here on the forum knows, please show me way ...
    1. f(x) = \dfrac{x^2+x+1}{-x^3+2x^2-3x+6} = \dfrac{x^2+x+1}{(2-x)(x+i\sqrt{3})(x-i\sqrt{3})} = \dfrac{A}{2-x} + \dfrac{B}{x+i\sqrt{3}} + \dfrac{C}{x-i\sqrt{3}}

    Solving this partial fractions, you find:

    f(x) = \dfrac{1}{2-x} - \dfrac{i\sqrt{3}}{6(x+i\sqrt{3})} + \dfrac{i\sqrt{3}}{6(x-i\sqrt{3})}

    \dfrac{d^n}{dx^n}\left( \dfrac{1}{2-x} \right) = \dfrac{n!}{(2-x)^n}

    \dfrac{d^n}{dx^n}\left( \dfrac{ i\sqrt{3} }{6(x+i\sqrt{3})} \right) = \dfrac{i\sqrt{3}}{6}\cdot \dfrac{(-1)^n n!}{(x+i\sqrt{3})^n}

    \dfrac{d^n}{dx^n}\left( \dfrac{ i\sqrt{3} }{6(x-i\sqrt{3})} \right) = \dfrac{i\sqrt{3}}{6}\cdot \dfrac{(-1)^n n!}{(x-i\sqrt{3})^n}

    So, plugging in, this gives:

    f^{(n)}(x) = \dfrac{n!}{(2-x)^n} + \dfrac{i\sqrt{3}}{6}\left( \dfrac{(-1)^{n+1}n!}{(x+i\sqrt{3})^n} + \dfrac{(-1)^n n!}{(x-i\sqrt{3})^n} \right)

    \begin{align*}f^{(n)}(0) & = \dfrac{n!}{2^n} + \dfrac{ i\sqrt{3} }{6}\left( \dfrac{ (-1)^{n+1} n! }{ (i\sqrt{3})^n } + \dfrac{ (-1)^n n! }{ (-i\sqrt{3})^n } \right) \\ & = \dfrac{n!}{2^n} + \dfrac{ i\sqrt{3} }{6}\left( (-1)^n n! \dfrac{ (i\sqrt{3})^n-(-i\sqrt{3})^n }{ (i\sqrt{3})^n(-i\sqrt{3})^n } \right) \\ & = \dfrac{n!}{2^n} + \dfrac{ i\sqrt{3} }{6}\left( (-1)^n n! \dfrac{ (i\sqrt{3})^n-(-i\sqrt{3})^n }{3^n} \right) \\ & = \begin{cases}\dfrac{n!}{2^n} & n\equiv 0\pmod{2} \\ \dfrac{n!}{2^n} + \dfrac{n!}{ (\sqrt{3})^{n+1} } & n\equiv 1\pmod{4} \\ \dfrac{n!}{2^n} - \dfrac{n!}{ (\sqrt{3})^{n+1} } & n\equiv 3\pmod{4} \end{cases}\end{align*}

    Hence,
    \begin{align*}f(x) & = \sum_{n\ge 0}\dfrac{f^{(n)}(0)}{n!}x^n \\ & = \sum_{n\ge 0} \dfrac{(2n)!}{2^{2n} (2n)!}x^{2n} + \sum_{n\ge 0}\dfrac{\dfrac{(2n+1)!}{2^{2n+1}} + (-1)^n\dfrac{(2n+1)!}{ 3^{n+1} } }{ (2n+1)! } x^{2n+1} \\ & = \sum_{n\ge 0} \dfrac{x^{2n}}{4^n} + \sum_{n\ge 0}\dfrac{2\cdot 3^{n+1} + (-1)^n4^{n+1}}{12^{n+1}}x^{2n+1}\end{align*}

    Using complex numbers, we have:

    f(x) = \sum_{n\ge 0}\dfrac{2\cdot 3^{(n+1)/2}-2^n(i^{n+1}+(-i)^{n+1})}{12^{(n+1)/2}} x^n

    Oh, and f^{(88)}(0) = \dfrac{88!}{2^{88}} (see above)

    2. You need to simplify the expression. \text{ch}^2(x) - 1 = \text{sh}^2(x) where \text{sh}(x) = \dfrac{e^x-e^{-x}}{2}. So, the expression becomes:

    \int \sqrt{ \dfrac{ \text{ch}^2(x) - 1 }{ \text{ch}(x) } }dx = \int \dfrac{ \text{sh}(x) }{\sqrt{ \text{ch}(x) } }dx

    Let u = \text{ch}(x). Then du = \text{sh}(x)dx. This gives:

    \int \dfrac{ \text{sh}(x) }{ \sqrt{ \text{ch}(x) } }dx = \int u^{-1/2}du = 2u^{1/2}+C = 2\sqrt{\text{ch}(x)}+C

    3. First, we want to use the division algorithm to find q(x) such that 6x^3+9x^2+15x+5 = q(x)(x^2+x+2)+r(x), \text{deg}(r(x))<2. 6x^3+9x^2+15x+5 - 6x(x^2+x+2) = 3x^2+3x+5, so let's try q(x) = 6x+3. Then 6x^3+9x^2+15x+5 = (6x+3)(x^2+x+2) - 1, so r(x)=-1 is definitely a polynomial with degree less that 2. Let's rewrite the integral:

    \begin{align*}\int \dfrac{6x^3+9x^2+15x+5}{\sqrt{x^2+x+2} }dx & = \int \dfrac{(6x+3)(x^2+x+2)-1}{\sqrt{x^2+x+2}}dx \\ & = \int 3(2x+1)\sqrt{x^2+x+2}dx - \int \dfrac{dx}{\sqrt{\left(x+\dfrac{1}{2}\right)^2 + \dfrac{7}{4} } }\end{align*}

    For the first integral, use the substitution u = x^2+x+2, du = (2x+1)dx. For the second, use the substitution x+\dfrac{1}{2} = \dfrac{\sqrt{7}}{2}\tan \theta, dx = \dfrac{\sqrt{7}}{2}\sec^2\theta d\theta.

    \begin{align*}\int 3(2x+1)\sqrt{x^2+x+2}dx - \int \dfrac{dx}{\sqrt{\left(x+\dfrac{1}{2}\right)^2 + \dfrac{7}{4} } } & = 3\int u^{1/2}du - \dfrac{\sqrt{7}}{2}\int \dfrac{\sec^2\theta}{\sqrt{\dfrac{7}{4}\left( \tan^2 \theta +1\right) } }d\theta \\ & = 2u^{3/2} - \int \sec \theta d\theta \\ & = 2(x^2+x+2)^{3/2} - \ln|\sec \theta + \tan \theta| + C\end{align*}

    Now, \tan \theta = \dfrac{\text{opp} }{\text{adj} } = \dfrac{2x+1}{\sqrt{7}}, so \sec \theta = \dfrac{\text{hyp}}{\text{adj} } = \dfrac{2\sqrt{x^2+x+2}}{\sqrt{7}}. Plugging in, we get:

    \begin{align*}& 2(x^2+x+2)^{3/2} - \ln\left| \sec \theta + \tan \theta \right| + C \\ = & 2(x^2+x+2)^{3/2} - \ln\left| \dfrac{2x+1 + 2\sqrt{x^2+x+2} }{\sqrt{7} } \right| + C\end{align*}

    Simplifying (using logarithm rules), we have:

    \begin{align*}& 2(x^2+x+2)^{3/2} - \ln\left| \dfrac{2x+1 + 2\sqrt{x^2+x+2} }{\sqrt{7} } \right| + C \\ = & 2(x^2+x+2)^{3/2} - \ln\left| 2x+1 + 2\sqrt{x^2+x+2} \right| + C'\end{align*}

    where C' = C+\ln(\sqrt{7}).
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