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 \[f(x) = f(0) + \dfrac{{{f{'}}(0)}}{{1!}}x + \dfrac{{{f{''}}(0)}}{{2!}}{x^2} + \dfrac{{{f^{'''}}(0)}}{{3!}}{x^3} + ... + \dfrac{{{f^n}(0)}}{{n!}}{x^n} + ...\]
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.