Differentiation and Integration of Power Series
Solved Problems
Example 3.
Find a power series for \[\frac{{6x}}{{5{x^2} - 4x - 1}}.\]
Solution.
First we find the partial fraction decomposition for this function. The quadratic function in the denominator can be written as
\[5{x^2} - 4x - 1 = \left( {5x + 1} \right)\left( {x - 1} \right),\]
so we can set:
\[\frac{{6x}}{{5{x^2} - 4x - 1}} = \frac{A}{{5x + 1}} + \frac{B}{{x - 1}}.\]
Multiply both sides of the expression by \(5{x^2} - 4x - 1 \) \(= \left( {5x + 1} \right)\left( {x - 1} \right)\) to obtain
\[
6x = A\left( {x - 1} \right) + B\left( {5x + 1} \right),\;\; \Rightarrow
6x = Ax - A + 5Bx + B,\;\; \Rightarrow
6x = \left( {A + 5B} \right)x + \left( { - A + B} \right),\;\; \Rightarrow
\left\{ {\begin{array}{*{20}{l}}
{A + 5B = 6}\\
{ - A + B = 0}
\end{array}} \right..\]
The solution of this system of equations is \(A = 1,\) \(B = 1.\) Hence, the partial decomposition of the given function is
\[\frac{{6x}}{{5{x^2} - 4x - 1}} = \frac{1}{{5x + 1}} + \frac{1}{{x - 1}} = \frac{1}{{1 + 5x}} - \frac{1}{{1 - x}}.\]
Both fractions are the sums of the infinite geometric series:
\[\frac{1}{{1 + 5x}} = \frac{1}{{1 - \left( { - 5x} \right)}} = 1 - 5x + {\left( { - 5x} \right)^2} + {\left( { - 5x} \right)^3} + \ldots = \sum\limits_{n = 0}^\infty {{{\left( { - 5x} \right)}^n}} ,\]
\[\frac{1}{{1 - x}} = 1 + x + {x^2} + {x^3} + \ldots = \sum\limits_{n = 0}^\infty {{x^n}} .\]
Hence, the power series expansion of the initial function is
\[\frac{{6x}}{{5{x^2} - 4x - 1}} = \sum\limits_{n = 0}^\infty {{{\left( { - 5x} \right)}^n}} - \sum\limits_{n = 0}^\infty {{x^n}} = \sum\limits_{n = 0}^\infty {\left[ {{{\left( { - 5x} \right)}^n} - {x^n}} \right]} = \sum\limits_{n = 0}^\infty {\left[ {{{\left( { - 5} \right)}^n} - 1} \right]{x^n}} .\]
Example 4.
Find a power series representation for the function \[\ln \left( {1 + x} \right),\;\left| x \right| \lt 1.\]
Solution.
In Example \(1\) we found the power series expansion
\[\frac{1}{{1 + x}} = 1 - x + {x^2} - {x^3} + \ldots = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{x^n}}, \;\; \left| x \right| \lt 1.\]
Integrating this series term-by-term on the interval \(\left[ {0,x} \right],\) we find that
\[\ln \left( {1 + x} \right) = \int\limits_0^x {\frac{{dt}}{{1 + t}}} = \int\limits_0^x {\left( {1 - t + {t^2} - {t^3} + \ldots } \right)dt} = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} + \ldots = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^{n + 1}}}}{{n + 1}}} = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}{x^n}}}{n}} .\]
Example 5.
Represent the integral \[\int\limits_0^x {{\frac{{\ln \left( {1 + t} \right)}}{t}} dt}\] as a power series expansion.
Solution.
In the previous problem (Example \(4\)) we have found the power series expansion for logarithmic function:
\[\ln \left( {1 + t} \right) = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}{t^n}}}{n}} = t - \frac{{{t^2}}}{2} + \frac{{{t^3}}}{3} - \frac{{{t^4}}}{4} + \ldots ,\;\; \left| t \right| \lt 1.\]
Then we can write:
\[\frac{{\ln \left( {1 + t} \right)}}{t} = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}{t^{n - 1}}}}{n}} = 1 - \frac{t}{2} + \frac{{{t^2}}}{3} - \frac{{{t^3}}}{4} + \ldots\]
Integrating this series term-by-term on the interval \(\left[ {0,x} \right],\) we obtain
\[\int\limits_0^x {\frac{{\ln \left( {1 + t} \right)}}{t}dt} = \int\limits_0^x {\left[ {1 - \frac{t}{2} + \frac{{{t^2}}}{3} - \frac{{{t^3}}}{4} + \ldots } \right]dt} = x - \frac{{{x^2}}}{{2 \cdot 2}} + \frac{{{x^3}}}{{3 \cdot 3}} - \frac{{{x^4}}}{{4 \cdot 4}} + \ldots = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}{x^n}}}{{{n^2}}}} .\]
Example 6.
Obtain a power series representation for the exponential function \({e^x}.\)
Solution.
Obtain a power series representation for the exponential function \({e^x}.\)
\[f\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \ldots ,\]
that converges for all \(x.\)
Differentiating it term-by-term, we have
\[f'\left( x \right) = \frac{d}{{dx}}1 + \frac{d}{{dx}}x + \frac{d}{{dx}}\frac{{{x^2}}}{{2!}} + \frac{d}{{dx}}\frac{{{x^3}}}{{3!}} + \ldots = 0 + 1 + x + \frac{{{x^2}}}{{2!}} + \ldots = f\left( x \right).\]
Hence, the function \(f\left( x \right)\) satisfies the differential equation \(f' = f.\) The general solution of this equation has the form \(f\left( x \right) = c{e^x},\) where \(c\) is a constant. Substituting the initial value \(f\left( 0 \right) = 1,\) we find that \(c = 1.\) Thus, we obtain the following power series expansion for \({e^x}:\)
\[f\left( x \right) = {e^x} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \ldots \]
Example 7.
Obtain a power series representation for the exponential function \({e^x}.\)
Solution.
Since \(\sinh x = {\frac{{{e^x} - {e^{ - x}}}}{2}},\) we can use power series representations for \({e^x}\) and \({e^{-x}}.\)
In the previous example we obtained the formula
\[{e^x} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \ldots \]
Substituting \(-x\) for \(x,\) we get
\[{e^{ - x}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - x} \right)}^n}}}{{n!}}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^n}}}{{n!}}} = 1 - x + \frac{{{x^2}}}{{2!}} - \frac{{{x^3}}}{{3!}} + \ldots \]
Then the expansion for the hyperbolic sine function has the form:
\[\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2} = \frac{1}{2}\left[ {\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} - \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - x} \right)}^n}}}{{n!}}} } \right] = \frac{1}{2}\left[ {\left( {1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \ldots } \right) - \left( {1 - x + \frac{{{x^2}}}{{2!}} - \frac{{{x^3}}}{{3!}} + \ldots } \right)} \right] = \frac{1}{2}\left[ {2\left( {x + \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} + \ldots } \right)} \right] = x + \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} + \ldots = \sum\limits_{n = 0}^\infty {\frac{{{x^{2n + 1}}}}{{\left( {2n + 1} \right)!}}} .\]
