Taylor-polynomial of function $f(x) = e^{x}*\sin(2x)$

Question

This is not homework, I'm asking to learn for an exam which I'll write in 2.5 months.

Count the Taylor-polynomial 3th grade of the function $f: \mathbb{R} \rightarrow \mathbb{R}, f(x) = e^{x}\sin(2x)$ in $x_{0} = 0$

3th grade means we will need the derivations $0,1,2,3$ of $f$: $$\begin{align} &f^{(0)}(x) = e^{x}\sin(2x)\\ &f^{(1)}(x) = e^{x}(\sin(2x)+2\cos(2x))\\ &f^{(2)}(x) = e^{x}(-3\sin(2x)+4\cos(2x))\\ &f^{(3)}(x) = e^{x}(-11\sin(2x)-2\cos(2x)) \end{align}$$ which implies $$\begin{align} e^{x}\sin(2x) &= \frac{f^{0}(0)}{0!}(x-0)^{0}+\frac{f^{1}(0)}{1!}(x-0)^{1}+\frac{f^{2}(0)}{2!}(x-0)^{2}+\frac{f^{3}(0)}{3!}(x-0)^{3}+\cdots\\ &= \frac{0}{1}\cdot 1+ \frac{2}{1}x + \frac{4}{2}x^2 - \frac{2}{6}x^{3}+\cdots\\ &= 2x +2x^{2}- \frac{x^{3}}{3}+\cdots \end{align}$$

Now my questions:

1) Did I do it correctly?

2) If it says 3rd grade in the task, am I only allowed to derivate 3 times then?

3) How can I create the series in the end with $2x +2x^{2}- \frac{x^{3}}{3}+\cdots$ if there is much info missing (without more derivations, I cannot see what will come next and thus I cannot create the complete series. BUT the task only said 3th grade, so it should have been enough, no?

Answer 1

For starters, you have a mistake, this is how things should be: $$f'(0)=2 \quad \text{and} \quad f''(0)=4$$

$\textbf{Another way to proceed}$: if you know the series expansion of both $e^x$ and $\sin(2x)$: $$e^x=1+x+\frac {x^2}{2}+\frac {x^3}6+\cdots$$ $$\sin(2x)=2x-\frac {4x^3}3+\cdots$$

Then $$e^x\sin(2x)=\left(1+x+\frac {x^2}{2}+\frac {x^3}6+\cdots\right)\left(2x-\frac {4x^3}3+\cdots\right)$$ $$=2x-\frac {4x^3}3+2x^2-\frac{4x^4}3+{x^3}-\frac{4x^5}6+\cdots$$ $$=2x+2x^2-\frac13x^3+\cdots$$

$\textbf{If you want to find the general expression}:$

We're only interested in the coefficient of the $\cos(2x)$ in the $n^{th}$ derivative so let $a(n)$ be that coefficient and $b(n)$ the coefficient of $\sin(2x)$. We notice that if $g^{(n)}(x)=e^xf(x)$ then $g^{(n+1)}(x)=e^x(f(x)+f'(x))$ this implies that the coefficients in the $(n+1)^{st}$ derivative are related to those of the previous as follows: $$a(n+1)=a(n)+2b(n)$$$$b(n+1)=b(n)-2a(n)$$

If we multiply the second equation by two, then subtract it from the first, we get: $$a(n+1)-2b(n+1)=5a(n)$$ or $$a(n)-2b(n)=5a(n-1)$$

Replacing $2b(n)$ in the first equation yields $$a(n+1)=2a(n)-5a(n-1)$$ with $a(0)=0$ and $a(1)=2$. This is a way of generating each coefficient in the Taylor series. Solving the recurrence relation using the characteristic equation results in: $$a(n)=\frac 12i((1-2i)^n-(1+2i)^n)$$ So that generally $$\color{blue}{\boxed{\quad e^x\sin(2x)=\sum_{k=0}^\infty \frac 12i((1-2i)^k-(1+2i)^k)\frac{x^k}{k!}\quad}}$$

But honestly, I don't think they were asking for this general solution.

Answer 2

You miscalculated the first and second order terms, remember that $sin(0) = 0$ and $ cos(0) = 1$.

Answer 3

Based on your work above $f''(0) = 4$ check this.

Where did the $x^1$ term go?

$f(x) = \frac {0}{0!} + \frac {2}{1!}x + \frac {4}{2!}x^2 + \frac {-2}{3!} x^3 \cdots$

I would say $3^{rd}$ degree polynomial, and not "grade." But that would be correct, the highest exponent of $x = 3.$ And, the highest derivative you would need is 3.

What about the remaining terms of the polynomial? If $x$ is small, then $x^2$ is smaller and the higher degree terms are smaller than that. Depending on the size of your interval, the higher degrees of the polynomial are irrelevant.

Or, more precisely, we can find a bound for the error, and decide if our estimate is sufficiently precise for our needs.