# Thread: Find derivative using Taylor Series

1. ## Find derivative using Taylor Series

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

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

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

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

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

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

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

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

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!

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

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

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

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

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

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

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

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

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 $\displaystyle \displaystyle \frac{1}{5}(-\frac{3}{5})^{n}x^{2}$ cannot possibly be your coefficient because it depends on x.

-Dan

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

Any idea how I could arrive the final answer from my current stage?
thanks!

4. What I would do is this.
$\displaystyle \displaystyle f(x) = \frac{x^2}{3x + 2}$

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

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

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

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

-Dan

5. I could see that the denominator is something like $\displaystyle (3x+2)^{n-1}$ 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?

6. A useful formula is that, if $\displaystyle f(x) = (ax+b)^m$, then:

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

You can write $\displaystyle \displaystyle \frac{x^2}{3x + 2} = \frac{x}{3}+\frac{4}{9 (3 x+2)}-\frac{2}{9},$
and apply it (the two fractions on the sides vanish!).

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

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

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

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

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

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

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

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

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 $\displaystyle x=1+t$ the function becomes...

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

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

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

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

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

$\displaystyle \chi$ $\displaystyle \sigma$

8. Originally Posted by chisigma
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).

9. Originally Posted by chisigma
Setting $\displaystyle x=1+t$ the function becomes...

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

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

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

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

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

10. Originally Posted by TheCoffeeMachine
A useful formula is that, if $\displaystyle f(x) = (ax+b)^m$, then:

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

You can write $\displaystyle \displaystyle \frac{x^2}{3x + 2} = \frac{x}{3}+\frac{4}{9 (3 x+2)}-\frac{2}{9},$
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.

11. Originally Posted by xEnOn
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...

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

... so that...

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

12. Originally Posted by xEnOn
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 $\displaystyle \displaystyle \prod_{i = 0}^N a_i$ works the same way as a summation $\displaystyle \displaystyle \sum_{i = 0}^N a_i$ except whereas the summation adds values:
$\displaystyle \displaystyle \sum_{i = 0}^N a_i = a_0 + a_1 + a_2 +~...~+a_N$

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

-Dan

13. Originally Posted by chisigma
Remember that is...

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

... so that...

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

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

Kind regards

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

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

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

Since my coefficient cannot have X in it, and I have to make the whole thing look like only $\displaystyle (x-1)^{n}$ 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 $\displaystyle [(x-1)^{n}+2(x-1)^{2}+(x-1)^{3}]$. How do I continue from here?

Edit:
Oh! I figure it out! Thanks! What a neat method! Thank you so much!

14. Originally Posted by xEnOn
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, $\displaystyle \displaystyle \prod_{1\le k \le n}\left(m-k+1\right)$ means the product $\displaystyle (m-k+1)$ as $\displaystyle k$ ranges from $\displaystyle 1$ to $\displaystyle n$.

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

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

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

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

\displaystyle \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}

15. Originally Posted by TheCoffeeMachine
Like Topsquark explained, $\displaystyle \displaystyle \prod_{1\le k \le n}\left(m-k+1\right)$ means the product $\displaystyle (m-k+1)$ as $\displaystyle k$ ranges from $\displaystyle 1$ to $\displaystyle n$.

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

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

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

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

\displaystyle \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 $\displaystyle \frac{x^2}{3x + 2}$ to $\displaystyle \frac{x}{3}+\frac{4}{9 (3 x+2)}-\frac{2}{9}$? I wouldn't have thought of splitting the equation into this format.

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