Question

How do you find the Maclaurin series for $$e^x\sin x$$?

Answer 1

Hint: We need to find the Maclaurin series for $$e^x\sin x$$. We know that Maclaurin series is given by $$f(x)=f(0)+{\displaystyle \frac{f{}^{\prime }(0)}{1!}}x+{\displaystyle \frac{f{}^{″}(0)}{2!}}{x}^2+{\displaystyle \frac{f^{{}^{‴}}(0)}{3!}}{x}^3+...+{\displaystyle \frac{f^n(0)}{n!}}{x}^n+...$$
To calculate the Maclaurin series, we need to calculate the first, second, third up to $$n^{th}$$ derivative of the given equation. After finding the derivative we need to calculate the value of the derivatives at zero. Putting all the values in the Maclaurin series of $$f(x)$$, we will get the Maclaurin series for $$e^x\sin x$$.

Complete step by step answer:
We have to find the Maclaurin series of $$e^x\sin x$$.
Let $$f(x)=e^x\sin x---(1)$$.
As we know that the Maclaurin series is given by $$f(x)=f(0)+{\displaystyle \frac{f{}^{\prime }(0)}{1!}}x+{\displaystyle \frac{f{}^{″}(0)}{2!}}{x}^2+{\displaystyle \frac{f^{{}^{‴}}(0)}{3!}}{x}^3+...+{\displaystyle \frac{f^n(0)}{n!}}{x}^n+...---(2)$$
Putting $$x=0$$ in $$(1)$$, we get
$$\Rightarrow f(0)=e^0\sin \left(0\right)$$
On simplifying, we get
$$\Rightarrow f(0)=0$$
As we know, from the product rule of differentiation,
$$\Rightarrow {\displaystyle \frac{d}{dx}}\left(uv\right)=u{\displaystyle \frac{dv}{dx}}+v{\displaystyle \frac{du}{dx}}$$
Differentiating $$(1)$$ with respect to $$x$$ using the product rule of differentiation, we get
$$\Rightarrow f^{\prime }(x)={\displaystyle \frac{d}{dx}}\left(e^x\sin x\right)$$
$$\Rightarrow f^{\prime }\left(x\right)=\left(e^x\right)\left({\displaystyle \frac{d}{dx}}\sin x\right)+\left({\displaystyle \frac{d}{dx}}{e}^x\right)\left(\sin x\right)$$
On simplification, we get
$$\Rightarrow f^{\prime }\left(x\right)=e^x\cos x+e^x\sin x$$
Putting $$x=0$$, we get
$$\Rightarrow f^{\prime }\left(0\right)=e^0\cos \left(0\right)+e^0\sin \left(0\right)$$
Putting the value of $$e^0=1$$, $$\cos \left(0\right)=1$$ and $$\sin \left(0\right)=0$$, we get
$$\Rightarrow f^{\prime }\left(0\right)=1$$
Now, differentiating $$f^{\prime }\left(x\right)$$ with respect to $$x$$, we get
$$\Rightarrow f^{″}\left(x\right)={\displaystyle \frac{d}{dx}}\left(e^x\cos x\right)+{\displaystyle \frac{d}{dx}}\left(e^x\sin x\right)$$
$$\Rightarrow f^{″}\left(x\right)=\left(e^x\right){\displaystyle \frac{d}{dx}}\left(\cos x\right)+\left(\cos x\right){\displaystyle \frac{d}{dx}}\left(e^x\right)+\left(e^x\right){\displaystyle \frac{d}{dx}}\left(\sin x\right)+\left(\sin x\right){\displaystyle \frac{d}{dx}}\left(e^x\right)$$
On simplification, we get
$$\Rightarrow f^{″}\left(x\right)=-e^x\sin x+e^x\cos x+e^x\cos x+e^x\sin x$$
On simplification, we get
$$\Rightarrow f^{″}\left(x\right)=2e^x\cos x$$
Putting $$x=0$$, we get
$$\Rightarrow f^{″}\left(0\right)=2e^0\cos \left(0\right)$$
$$\Rightarrow f^{″}\left(0\right)=2$$
Now, differentiating $$f^{″}\left(x\right)$$ with respect to $$x$$, we get
$$\Rightarrow f^{‴}\left(x\right)=2{\displaystyle \frac{d}{dx}}\left(e^x\cos x\right)$$
$$\Rightarrow f^{‴}(x)=2\left(\left(e^x\right){\displaystyle \frac{d}{dx}}\left(\cos x\right)+\left({\displaystyle \frac{d}{dx}}{e}^x\right)\left(\cos x\right)\right)$$
On simplification, we get
$$\Rightarrow f^{‴}(x)=-2e^x\sin x+2e^x\cos x$$
Putting $$x=0$$, we get
$$\Rightarrow f^{‴}(0)=-2e^0\sin \left(0\right)+2e^0\cos \left(0\right)$$
$$\Rightarrow f^{‴}(0)=2$$
From $$(1)$$ and $$(2)$$, we get
$$e^x\sin x=0+{\displaystyle \frac{1}{1!}}x+{\displaystyle \frac{2}{2!}}{x}^2+{\displaystyle \frac{2}{3!}}{x}^3+...$$
On simplification, we get
$$e^x\sin x=x+x^2+{\displaystyle \frac{1}{3}}{x}^3+...$$
Therefore, the Maclaurin series for $$e^x\sin x$$ is $$x+x^2+{\displaystyle \frac{1}{3}}{x}^3+...$$.

Note:
To find the Taylor series and Maclaurin series, the function must be infinitely times differentiable and continuous. Also, note that the Maclaurin series is just the Taylor series about the point $$0$$.
The Taylor series of any function $$f\left(x\right)$$ about the point $$x=a$$ is given by the following expression: $$f(x)=f(a)+{\displaystyle \frac{f{}^{\prime }(a)}{1!}}\left(x-a\right)+{\displaystyle \frac{f{}^{″}(a)}{2!}}{\left(x-a\right)}^2+{\displaystyle \frac{f^{{}^{‴}}(a)}{3!}}{\left(x-a\right)}^3+...+{\displaystyle \frac{f^n(a)}{n!}}{\left(x-a\right)}^n+...$$
If we put $$a=0$$, then we will obtain the Maclaurin series.
$$\Rightarrow f(x)=f(0)+{\displaystyle \frac{f{}^{\prime }(0)}{1!}}\left(x-0\right)+{\displaystyle \frac{f{}^{″}(0)}{2!}}{\left(x-0\right)}^2+{\displaystyle \frac{f^{{}^{‴}}(0)}{3!}}{\left(x-0\right)}^3+...+{\displaystyle \frac{f^n(0)}{n!}}{\left(x-0\right)}^n+...$$
On simplifying, we get
$$\Rightarrow f(x)=f(0)+{\displaystyle \frac{f{}^{\prime }(0)}{1!}}x+{\displaystyle \frac{f{}^{″}(0)}{2!}}{x}^2+{\displaystyle \frac{f^{{}^{‴}}(0)}{3!}}{x}^3+...+{\displaystyle \frac{f^n(0)}{n!}}{x}^n+...$$ which is the Maclaurin series.