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Math Help - Find derivative using Taylor Series

  1. #1
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    Find derivative using Taylor Series

    I am trying to calculate the 11th derivative, f^{(11)}(1), of a function f(x) = \frac{x^{2}}{2+3x} using the Taylor Series centred at one.

    I arrived to this stage where I form it into a geometry series format: \frac{x^{2}}{5(1-(-\frac{3(x-1)}{5}))}

    From here, I believe that the -\frac{3(x-1)}{5}) is essentially the r in a geometry series formula.

    So I put that into a summation: \frac{x^{2}}{5}\sum_{n=0}^{\infty}(-\frac{3(x-1)}{5})

    I cleaned it up a little bit and manipulated the algebra and arrived at: <br />
\sum_{n=0}^{\infty}\frac{1}{5}(-\frac{3}{5})^{n}x^{2}(x-1)^{n}

    Now, this looks like a Taylor Series format to me with (x-1)^{n} at the back and \frac{1}{5}(-\frac{3}{5})^{n}x^{2} as my co-efficient.

    Since I got the co-efficient of the series, I thought I could just do this: \frac{f^{(11)}(1)}{11!} = \frac{1}{5}(-\frac{3}{5})^{n}x^{2}
    and then get: f^{(11)}(1) = 11!\frac{1}{5}(-\frac{3}{5})^{n}x^{2}

    However, my answer is wrong and when I do it on a calculator, \frac{\partial^{11} }{\partial x^{11}}(\frac{x^{2}}{2+3x}) \neq 11!\frac{1}{5}(-\frac{3}{5})^{n}x^{2}

    What should I do after this? I thought I can get the answer once I turn it into a format that looks like the original Taylor series format and find the answer from the co-efficient?

    thanks!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by xEnOn View Post
    I am trying to calculate the 11th derivative, f^{(11)}(1), of a function f(x) = \frac{x^{2}}{2+3x} using the Taylor Series centred at one.

    I arrived to this stage where I form it into a geometry series format: \frac{x^{2}}{5(1-(-\frac{3(x-1)}{5}))}

    From here, I believe that the -\frac{3(x-1)}{5}) is essentially the r in a geometry series formula.

    So I put that into a summation: \frac{x^{2}}{5}\sum_{n=0}^{\infty}(-\frac{3(x-1)}{5})

    I cleaned it up a little bit and manipulated the algebra and arrived at: <br />
\sum_{n=0}^{\infty}\frac{1}{5}(-\frac{3}{5})^{n}x^{2}(x-1)^{n}

    Now, this looks like a Taylor Series format to me with (x-1)^{n} at the back and \frac{1}{5}(-\frac{3}{5})^{n}x^{2} as my co-efficient.

    Since I got the co-efficient of the series, I thought I could just do this: \frac{f^{(11)}(1)}{11!} = \frac{1}{5}(-\frac{3}{5})^{n}x^{2}
    and then get: f^{(11)}(1) = 11!\frac{1}{5}(-\frac{3}{5})^{n}x^{2}

    However, my answer is wrong and when I do it on a calculator, \frac{\partial^{11} }{\partial x^{11}}(\frac{x^{2}}{2+3x}) \neq 11!\frac{1}{5}(-\frac{3}{5})^{n}x^{2}

    What should I do after this? I thought I can get the answer once I turn it into a format that looks like the original Taylor series format and find the answer from the co-efficient?

    thanks!
    What you have done is fairly clever, and I commend you for it. I don't know if you could save this or not. (You may have to work on the dreary task of working out the needed series term by term until you can find a formula for it.) But for now you need to know that \displaystyle \frac{1}{5}(-\frac{3}{5})^{n}x^{2} cannot possibly be your coefficient because it depends on x.

    -Dan
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  3. #3
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    hmm...So the coefficient cannot have x in it.
    I eventually reach my current stage from what I had in my previous post: \sum_{n=0}^{\infty}\frac{1}{5}(-\frac{3}{5})^{n}[(x-1)^{n+2} + 2(x-1)^{n+1} +(x-1)^{n})]
    But I cannot figure out how to carry on from this to the given final answer, which is: f^{(11)}(1) = 11! [\frac{1}{5}(-\frac{3}{5})^{11} + 2\frac{1}{5}(-\frac{3}{5})^{10} + \frac{1}{5}(-\frac{3}{5})^{9}]

    Any idea how I could arrive the final answer from my current stage?
    thanks!
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  4. #4
    Forum Admin topsquark's Avatar
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    What I would do is this.
    \displaystyle f(x) = \frac{x^2}{3x + 2}

    \displaystyle f^{\prime}(x) = \frac{3x^2 + 4x}{(3x + 2)^2}

    \displaystyle f^{\prime \prime}(x) = \frac{8}{(3x + 2)^3}

    Now consider this last derivative. Can you come up with a series solution for the Taylor expansion of
    \displaystyle \frac{1}{(x + a)^3}

    The 11th derivative term should be easy to come by now.

    -Dan
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  5. #5
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    I could see that the denominator is something like (3x+2)^{n-1} but I am not sure about the numerator because the first derivative has x in it and the second derivative has no more x in it?
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    A useful formula is that, if f(x) = (ax+b)^m, then:


    \displaystyle f^{(n)}(x) = a^n\left(ax+b\right)^{m-n}\prod_{1 \le k \le n}\left(m-k+1\right).

    You can write \displaystyle \frac{x^2}{3x + 2} = \frac{x}{3}+\frac{4}{9 (3 x+2)}-\frac{2}{9},
    and apply it (the two fractions on the sides vanish!).
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    Quote Originally Posted by xEnOn View Post
    I am trying to calculate the 11th derivative, f^{(11)}(1), of a function f(x) = \frac{x^{2}}{2+3x} using the Taylor Series centred at one.

    I arrived to this stage where I form it into a geometry series format: \frac{x^{2}}{5(1-(-\frac{3(x-1)}{5}))}

    From here, I believe that the -\frac{3(x-1)}{5}) is essentially the r in a geometry series formula.

    So I put that into a summation: \frac{x^{2}}{5}\sum_{n=0}^{\infty}(-\frac{3(x-1)}{5})

    I cleaned it up a little bit and manipulated the algebra and arrived at: <br />
\sum_{n=0}^{\infty}\frac{1}{5}(-\frac{3}{5})^{n}x^{2}(x-1)^{n}

    Now, this looks like a Taylor Series format to me with (x-1)^{n} at the back and \frac{1}{5}(-\frac{3}{5})^{n}x^{2} as my co-efficient.

    Since I got the co-efficient of the series, I thought I could just do this: \frac{f^{(11)}(1)}{11!} = \frac{1}{5}(-\frac{3}{5})^{n}x^{2}
    and then get: f^{(11)}(1) = 11!\frac{1}{5}(-\frac{3}{5})^{n}x^{2}

    However, my answer is wrong and when I do it on a calculator, \frac{\partial^{11} }{\partial x^{11}}(\frac{x^{2}}{2+3x}) \neq 11!\frac{1}{5}(-\frac{3}{5})^{n}x^{2}

    What should I do after this? I thought I can get the answer once I turn it into a format that looks like the original Taylor series format and find the answer from the co-efficient?

    thanks!
    Setting x=1+t the function becomes...

    \displaystyle f(t)= \frac{(1+t)^{2}}{5 + 3\ t} = \frac{1}{5}\ (1+2\ t + t^{2})\ \sum_{n=0}^{\infty} (-\frac{3\ t}{5})^{n} (1)

    Now the coefficient of the term of degree 11 in (1) is...

    \displaystyle a_{11}= -\frac{3^{9}}{5^{10}}\ (\frac{9}{25} - \frac{6}{5} + 1) = - \frac{2^{2}\ 3^{9}}{5^{12}} (2)

    ... so that the requested value should be...

    \displaystyle f^{(11)} (1)=  - \frac{11!\ 2^{2}\ 3^{9}}{5^{12}} (3)

    Because in pure calculus I'm a little poor [...] it would be better if some young man or woman controls my computation...

    Kind regards

    \chi \sigma
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  8. #8
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    Quote Originally Posted by chisigma View Post
    Because in pure calculus I'm a little poor [...] it would be better if some young man or woman controls my computation...
    Dear Chisigma, your computation is perfect, as is your knowledge of pure calculus! I get the same.
    (Albeit I did not use power series -- it's first time I've seen it done this way; it's quite impressive).
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  9. #9
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    Quote Originally Posted by chisigma View Post
    Setting x=1+t the function becomes...

    \displaystyle f(t)= \frac{(1+t)^{2}}{5 + 3\ t} = \frac{1}{5}\ (1+2\ t + t^{2})\ \sum_{n=0}^{\infty} (-\frac{3\ t}{5})^{n} (1)

    Now the coefficient of the term of degree 11 in (1) is...

    \displaystyle a_{11}= -\frac{3^{9}}{5^{10}}\ (\frac{9}{25} - \frac{6}{5} + 1) = - \frac{2^{2}\ 3^{9}}{5^{12}} (2)

    ... so that the requested value should be...

    \displaystyle f^{(11)} (1)=  - \frac{11!\ 2^{2}\ 3^{9}}{5^{12}} (3)

    I could understand the line (1). But I still couldn't figure out how to derive from (1) to line (2) to get the 11th term's coefficient. What was being substituted into the t? If I use back x=t+1, t=x-1, and with centred at 1, t would become 1-1=0?

    thanks. it looks like a neat method. really wish to try it.
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    Quote Originally Posted by TheCoffeeMachine View Post
    A useful formula is that, if f(x) = (ax+b)^m, then:


    \displaystyle f^{(n)}(x) = a^n\left(ax+b\right)^{m-n}\prod_{1 \le k \le n}\left(m-k+1\right).

    You can write \displaystyle \frac{x^2}{3x + 2} = \frac{x}{3}+\frac{4}{9 (3 x+2)}-\frac{2}{9},
    and apply it (the two fractions on the sides vanish!).
    hmm...please pardon me, what does the big "pie" stands for with the conditions at the bottom? It is a summation sign too? because I couldn't seem to find a way to fit the equation into the formula.
    sorry. i am pretty new to the topic on series.
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  11. #11
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by xEnOn View Post
    I could understand the line (1). But I still couldn't figure out how to derive from (1) to line (2) to get the 11th term's coefficient. What was being substituted into the t? If I use back x=t+1, t=x-1, and with centred at 1, t would become 1-1=0?

    thanks. it looks like a neat method. really wish to try it.
    Remember that is...

    \displaystyle f(t)= \sum_{n=0}^{\infty} \frac{f^{(n)} (0)}{n!}\ t^{n}= \sum_{n=0}^{\infty} a_{n}\ t^{n} (1)

    ... so that...

    \displaystyle f^{(n)}(0)= a_{n}\ n! (2)

    Setting x-1=t in (1) You obtain with (2) f^{(n)} (1)...

    Kind regards

    \chi \sigma
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by xEnOn View Post
    hmm...please pardon me, what does the big "pie" stands for with the conditions at the bottom? It is a summation sign too? because I couldn't seem to find a way to fit the equation into the formula.
    sorry. i am pretty new to the topic on series.
    "pie"??!? [rant]The Greek letter is "pi."[/rant]

    The product \displaystyle \prod_{i = 0}^N a_i works the same way as a summation \displaystyle \sum_{i = 0}^N a_i except whereas the summation adds values:
    \displaystyle \sum_{i = 0}^N a_i = a_0 + a_1 + a_2 +~...~+a_N

    the product multiplies them:
    \displaystyle \prod_{i = 0}^N a_i = a_0 \cdot a_1 \cdot a_2 \cdot ~...~ \cdot a_N

    -Dan
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  13. #13
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    Quote Originally Posted by chisigma View Post
    Remember that is...

    \displaystyle f(t)= \sum_{n=0}^{\infty} \frac{f^{(n)} (0)}{n!}\ t^{n}= \sum_{n=0}^{\infty} a_{n}\ t^{n} (1)

    ... so that...

    \displaystyle f^{(n)}(0)= a_{n}\ n! (2)

    Setting x-1=t in (1) You obtain with (2) f^{(n)} (1)...

    Kind regards

    \chi \sigma
    So from: \frac{1}{5}(1+2t+t^{2})\sum_{n=0}^{\infty}(-\frac{3}{5}t)^{n}

    I substitute t=x-1 into t and I get: \frac{1}{5}[1+2(x-1)+(x-1)^{2}]\sum_{n=0}^{\infty}(-\frac{3}{5}(x-1)^{n}

    Then I bring it into the summation and expand it and get: \sum_{n=0}^{\infty}(-\frac{3}{5})^{n}\frac{1}{5}[(x-1)^{n}+2(x-1)^{2}+(x-1)^{3}]

    Since my coefficient cannot have X in it, and I have to make the whole thing look like only (x-1)^{n} to match the Taylor Series, I need to somehow manipulate the algebra to look like that. But I couldn't seem to figure out how to do that because there is like so many (x-1) in [(x-1)^{n}+2(x-1)^{2}+(x-1)^{3}]. How do I continue from here?


    Edit:
    Oh! I figure it out! Thanks! What a neat method! Thank you so much!
    Last edited by xEnOn; February 17th 2011 at 03:18 PM.
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  14. #14
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    Quote Originally Posted by xEnOn View Post
    hmm...please pardon me, what does the big "pie" stands for with the conditions at the bottom? It is a summation sign too? because I couldn't seem to find a way to fit the equation into the formula.
    sorry. i am pretty new to the topic on series.
    Like Topsquark explained, \displaystyle  \prod_{1\le k \le n}\left(m-k+1\right) means the product (m-k+1) as k ranges from 1 to n.

    You can write it as  m(m-1)(m-2)\cdots(m-n+1)  if you are not comfortable with that notation.

    For your problem, we have \displaystyle \frac{x^2}{3x + 2} = \frac{x}{3}+\frac{4}{9 (3 x+2)}-\frac{2}{9}.

    Both \frac{2}{9} and \frac{x}{3} vanish after the 1st and 2nd derivative respectively.
    So we only need to calculate the 11th derivative of \frac{4}{9(3x+2)}.

    If f(x) = \frac{4}{9(3x+2)}, then:

    \displaystyle \begin{aligned} & f^{(n)}(x) = \frac{4}{9}\times 3^n\left(3x+2\right)^{-1-n}\prod_{1 \le k \le n}\left(1-k+1\right) \\& \Rightarrow f^{(11)}(x) = \frac{2^2}{3^2}\times 3^{11}\left(3x+2\right)^{-12}\prod_{1 \le k \le 11}\left(-k\right) \\& \Rightarrow f^{(11)}(x) = -\frac{2^2\times 3^{9}\times 11!}{\left(3x+2\right)^{12}} \\& \Rightarrow f^{(11)}(1)  =  -\frac{2^2\times 3^{9}\times 11!}{5^{12}}.\end{aligned}
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  15. #15
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    Quote Originally Posted by TheCoffeeMachine View Post
    Like Topsquark explained, \displaystyle  \prod_{1\le k \le n}\left(m-k+1\right) means the product (m-k+1) as k ranges from 1 to n.

    You can write it as  m(m-1)(m-2)\cdots(m-n+1)  if you are not comfortable with that notation.

    For your problem, we have \displaystyle \frac{x^2}{3x + 2} = \frac{x}{3}+\frac{4}{9 (3 x+2)}-\frac{2}{9}.

    Both \frac{2}{9} and \frac{x}{3} vanish after the 1st and 2nd derivative respectively.
    So we only need to calculate the 11th derivative of \frac{4}{9(3x+2)}.

    If f(x) = \frac{4}{9(3x+2)}, then:

    \displaystyle \begin{aligned} & f^{(n)}(x) = \frac{4}{9}\times 3^n\left(3x+2\right)^{-1-n}\prod_{1 \le k \le n}\left(1-k+1\right) \\& \Rightarrow f^{(11)}(x) = \frac{2^2}{3^2}\times 3^{11}\left(3x+2\right)^{-12}\prod_{1 \le k \le 11}\left(-k\right) \\& \Rightarrow f^{(11)}(x) = -\frac{2^2\times 3^{9}\times 11!}{\left(3x+2\right)^{12}} \\& \Rightarrow f^{(11)}(1)  =  -\frac{2^2\times 3^{9}\times 11!}{5^{12}}.\end{aligned}
    Thanks!
    Is there a particular feature to look out for when you try to derive from \frac{x^2}{3x + 2} to \frac{x}{3}+\frac{4}{9 (3 x+2)}-\frac{2}{9}? I wouldn't have thought of splitting the equation into this format.
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