# Thread: function for Taylor series and find the 100th derivative of this function

1. ## 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...

2. ## 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.

3. ## 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?

4. ## 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 ...

5. ## 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.

6. ## 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, $f^{(100)}(-1) = 2^{100}\left( (-1)^2 + 103(-1) + 2626 \right)e^{2(-1)-1} = 2524\cdot 2^{100}e^{-3}$

7. ## 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?

8. ## Re: function for Taylor series and find the 100th derivative of this function

Originally Posted by lebdim
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 $C' = C+\ln(\sqrt{7})$.