Question

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

Answer 1

We need to find the Maclaurin series for ${e^x}\sin x$. We know that Maclaurin series is given by $f(x) = f(0) + \dfrac{{{f{'}}(0)}}{{1!}}x + \dfrac{{{f{''}}(0)}}{{2!}}{x^2} + \dfrac{{{f^{'''}}(0)}}{{3!}}{x^3} + ... + \dfrac{{{f^n}(0)}}{{n!}}{x^n} + ...$
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 $f(x) = f(0) + \dfrac{{{f{'}}(0)}}{{1!}}x + \dfrac{{{f{''}}(0)}}{{2!}}{x^2} + \dfrac{{{f^{'''}}(0)}}{{3!}}{x^3} + ... + \dfrac{{{f^n}(0)}}{{n!}}{x^n} + ... - - - (2)$
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,
$\Rightarrow \dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$
Differentiating $(1)$ with respect to $x$ using the product rule of differentiation, we get
$\Rightarrow f'(x) = \dfrac{d}{{dx}}\left( {{e^x}\sin x} \right)$
$\Rightarrow f'\left( x \right) = \left( {{e^x}} \right)\left( {\dfrac{d}{{dx}}\sin x} \right) + \left( {\dfrac{d}{{dx}}{e^x}} \right)\left( {\sin x} \right)$
On simplification, we get
$\Rightarrow f'\left( x \right) = {e^x}\cos x + {e^x}\sin x$
Putting $x = 0$, we get
$\Rightarrow f'\left( 0 \right) = {e^0}\cos \left( 0 \right) + {e^0}\sin \left( 0 \right)$
Putting the value of ${e^0} = 1$, $\cos \left( 0 \right) = 1$ and $\sin \left( 0 \right) = 0$, we get
$\Rightarrow f'\left( 0 \right) = 1$
Now, differentiating $f'\left( x \right)$ with respect to $x$, we get
$\Rightarrow f''\left( x \right) = \dfrac{d}{{dx}}\left( {{e^x}\cos x} \right) + \dfrac{d}{{dx}}\left( {{e^x}\sin x} \right)$
$\Rightarrow f''\left( x \right) = \left( {{e^x}} \right)\dfrac{d}{{dx}}\left( {\cos x} \right) + \left( {\cos x} \right)\dfrac{d}{{dx}}\left( {{e^x}} \right) + \left( {{e^x}} \right)\dfrac{d}{{dx}}\left( {\sin x} \right) + \left( {\sin x} \right)\dfrac{d}{{dx}}\left( {{e^x}} \right)$
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
$\Rightarrow f''\left( x \right) = 2{e^x}\cos x$
Putting $x = 0$, we get
$\Rightarrow f''\left( 0 \right) = 2{e^0}\cos \left( 0 \right)$
$\Rightarrow f''\left( 0 \right) = 2$
Now, differentiating $f''\left( x \right)$ with respect to $x$, we get
$\Rightarrow f'''\left( x \right) = 2\dfrac{d}{{dx}}\left( {{e^x}\cos x} \right)$
$\Rightarrow f'''(x) = 2\left( {\left( {{e^x}} \right)\dfrac{d}{{dx}}\left( {\cos x} \right) + \left( {\dfrac{d}{{dx}}{e^x}} \right)\left( {\cos x} \right)} \right)$
On simplification, we get
$\Rightarrow f'''(x) = - 2{e^x}\sin x + 2{e^x}\cos x$
Putting $x = 0$, we get
$\Rightarrow f'''(0) = - 2{e^0}\sin \left( 0 \right) + 2{e^0}\cos \left( 0 \right)$
$\Rightarrow f'''(0) = 2$
From $(1)$ and $(2)$, we get
${e^x}\sin x = 0 + \dfrac{1}{{1!}}x + \dfrac{2}{{2!}}{x^2} + \dfrac{2}{{3!}}{x^3} + ...$
On simplification, we get
${e^x}\sin x = x + {x^2} + \dfrac{1}{3}{x^3} + ...$
Therefore, the Maclaurin series for ${e^x}\sin x$ is $x + {x^2} + \dfrac{1}{3}{x^3} + ...$.

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: $f(x) = f(a) + \dfrac{{{f{'}}(a)}}{{1!}}\left( {x - a} \right) + \dfrac{{{f{''}}(a)}}{{2!}}{\left( {x - a} \right)^2} + \dfrac{{{f^{'''}}(a)}}{{3!}}{\left( {x - a} \right)^3} + ... + \dfrac{{{f^n}(a)}}{{n!}}{\left( {x - a} \right)^n} + ...$
If we put $a = 0$, then we will obtain the Maclaurin series.
$\Rightarrow f(x) = f(0) + \dfrac{{{f{'}}(0)}}{{1!}}\left( {x - 0} \right) + \dfrac{{{f{''}}(0)}}{{2!}}{\left( {x - 0} \right)^2} + \dfrac{{{f^{'''}}(0)}}{{3!}}{\left( {x - 0} \right)^3} + ... + \dfrac{{{f^n}(0)}}{{n!}}{\left( {x - 0} \right)^n} + ...$
On simplifying, we get
$\Rightarrow f(x) = f(0) + \dfrac{{{f{'}}(0)}}{{1!}}x + \dfrac{{{f{''}}(0)}}{{2!}}{x^2} + \dfrac{{{f^{'''}}(0)}}{{3!}}{x^3} + ... + \dfrac{{{f^n}(0)}}{{n!}}{x^n} + ...$ which is the Maclaurin series.