Geometric series are special instances of power series. As we will see later, the theory of power series is of importance for approximating differentiable functions in the neighbourhood of a point by means of a polynomial.
A power series is a series of the form \[ \displaystyle\sum_{k=0}^{\infty}a_k(x-p)^k\]where #x# and #p# are real numbers and #a_0, a_1, a_2, \ldots# is a sequence of real numbers.
The number #p# is called the center of the power series and #a_k# the #k#-th coefficient.
The interval of convergence of a power series is the set of all values of #x# for which the power series converges.
The function determined by a power series is the real function #f# on the interval of convergence of the power series that assigns to each #x# the value #f(x)# of the power series at #x#.
The radius of convergence of the power series is the number #r# such that the series converges for all values of #x# in the open interval #\ivoo{p-r}{p+r}# and diverges for all values of #x# outside the closed interval #\ivcc{p-r}{p+r}#. In the case where the power series converges for all #x#, we set #r=\infty#.
Consider the geometric power series
\[\sum_{n=0}^{\infty} {{\left(x-5\right)^{n}}\over{10^{n}}} \]
The series' interval of convergence is \(\ivoo{-5}{15}\). In order to show this, we will use the fact that the radius of convergence of the geometric series # \sum_{n=0}^\infty c^n# is equal to # 1#. To this end we write \[\sum_{n=0}^{\infty}{{\left(x-5\right)^{n}}\over{10^{n}}}= \sum_{n=0}^{\infty} \left({{x-5}\over{10}}\right)^n \] In view of the theorem Infinite geometric series formula, this shows that the series is geometric and that the common ratio of the series is #{{{x-5}\over{10}}}#. As a consequence, the interval of convergence of the given series is determined by \[\left|{{{x-5}\over{10}}}\right|\lt 1\] By definition of absolute value, this inequality is equivalent to \[{{x-5}\over{10}}\gt -1\land {{x-5}\over{10}}\lt 1 \] which can be rewritten as \[x\gt -5\land x\lt 15\]We conclude that the interval of convergence is \(\ivoo{-5}{15}\). By the infinite geometric series formula, we find that on \(\ivoo{-5}{15}\), the interval of convergence, the function rule of #f# is \[f(x) = \frac{1}{1-{{x-5}\over{10}}} ={{10}\over{15-x}}\]
The radius of convergence #r# does not tell us what happens to points whose distance to the center is exactly #r#. For example, consider the series \[s(x) =x+\frac12 x^2+\frac13 x^3 + \cdots =\sum_{k=1}^\infty\frac{1}{k}x^{k}\]
Here, the center is #0#, and #a_k = \frac{1}{k}# for #k\ge1#. The value of this series at the center is #a_0=0#.
At #x=1#, the series becomes
\[\begin{array}{rcl}s(1) &=& \displaystyle 1+\frac12 +\frac13+\frac14 + \cdots \\&=&\displaystyle1+\frac12 +\left(\frac13 +\frac14\right)+\left( \frac15+\frac16 +\frac17+\frac18\right)+\cdots\\ &\ge&\displaystyle 1+\frac12 +2\cdot \frac14+4\cdot \frac18+8\cdot\frac{1}{16}+\cdots\\ &=&\displaystyle 1+\frac12+\frac12+\frac12+\cdots\end{array}\]
so it diverges.
At #x=-1#, the series evaluates to a sum of negative terms #\frac{-1}{2n\cdot(2n-1)}# which tend to #0# faster than #\frac{1}{n^2}#:
\[\begin{array}{rcl}s(-1) &=& \displaystyle\sum_{k=1}^\infty\frac{1}{k}(-1)^{k} \\&=&\displaystyle\sum_{n=1}^\infty\left(\frac{1}{2n-1}(-1)^{2n-1}+\frac{1}{2n}(-1)^{2n}\right)\\ &=&\displaystyle\sum_{n=1}^\infty\frac{-1}{2n\cdot(2n-1)}\end{array}\]
This implies that #s(-1)# converges. Later, we will see that the function determined by this series is \(-\ln(1-x)\) and so #s(-1) = -\ln(2)#.
The radius of convergence of #s(x)# is #1#. The example shows that at distance #1# to the center, there is a point (#x=1#) at which the series diverges and a point (#x=-1#) at which the series converges.
Suppose that there are nonzero constants #c# and #d# such that #a_k=d\cdot c^k# for #k=0,1,2,\ldots#. Then the power series is a geometric series for each value of #x#. Indeed, then
\[\displaystyle\sum_{k=0}^{\infty}a_k(x-p)^k=\displaystyle\sum_{k=0}^{\infty}d\cdot c^k (x-p)^k=\displaystyle\sum_{k=0}^{\infty}d\cdot \left(c\cdot (x-p)\right)^k\]
so the common ratio equals #c\cdot (x-p)#. In particular, the series converges if and only if #\left|c\cdot (x-p)\right|\lt 1#, so the interval of convergence is the set of all #x# such that #x\gt p-\frac{1}{|c|}# and #x\lt p+\frac{1}{|c|}#. In other words,
- the interval of convergence is #\ivoo{p-\frac{1}{|c|}}{p+\frac{1}{|c|}}#, and
- the radius of convergence is #\frac{1}{|c|}#.
By the infinite geometric series formula, the function #f# determined by the geometric power series satisfies
\[ f(x) = \frac{d}{1-c\cdot (x-p)} =\frac{d}{p\cdot c +1-c\cdot x} \phantom{xxx}\text{ for } \phantom{xxx} x\text{ in }\ivoo{p-\frac{1}{|c|}}{p+\frac{1}{|c|}}\]
If #c=d=1# and #p=0#, we recover the geometric series \(\sum_{k=0}^{\infty}x^k\). Substituting these values for #p# and #c# in the interval of convergence, we find again that this geometric series converges if #x# lies in the open interval #\ivoo{-1}{1}#, that is, if and only if #|x|\lt1#.
The natural exponential function \[\exp(x) = \sum_{k=0}^{\infty}\frac{1}{k!}x^k\] is defined for all real #x# and satisfies #\exp(x) = {\e}^x#.
The series appearing in the function rule for #\exp# can be shown to converge for all #x#, so #\exp# is the function defined by the power series. Later, when dealing with Taylor series, we will see how, starting from the well-known property that the exponential function is equal to its derivative, we can find the series.
At the center #p#, the value of the series is #a_0#. Suppose that all coefficients #a_k# lie in an interval #\ivcc{-M}{M}# for some positive real number #M#. We then say that the sequence #a_k# #(k=0,1,2,\ldots)# of coefficients is bounded. Take an arbitrary small number #\varepsilon\in\ivoo{0}{\frac12}#. For all #x# in the interval #\ivco{p-\varepsilon}{p+\varepsilon}#, we have
\[\begin{array}{rcl}\displaystyle\sum_{k=0}^{\infty}a_k(x-p)^k&=& a_0+a_1(x-p)+\sum_{k=2}^{\infty}a_k(x-p)^k \\ &\le&\displaystyle a_0+a_1\cdot(x-p)+\sum_{k=2}^{\infty}M\cdot \varepsilon^k \\ &=&\displaystyle a_0+a_1(x-p)+M\cdot \dfrac{\varepsilon^2}{1-\varepsilon}\\ &\le&\displaystyle a_0+a_1(x-p)+2M\cdot {\varepsilon^2}\end{array}\]and, similarly, \(\displaystyle\sum_{k=0}^{\infty}a_k(x-p)^k\ge a_0+a_1(x-p)-2M\cdot {\varepsilon^2}\).
This shows that, if we take #x# to be at distance at most #\varepsilon# to the center and we are prepared to neglect higher products of powers of #\varepsilon# by #2M#, then we can approximate the function determined by the series by its linearization #a_0+a_1(x-p)#. A similar argument holds for higher powers of #x#.
The power series #\sum_{k=0}^{\infty}a_k(x-p)^k# converges at #c# if and only if the power series #\sum_{k=0}^{\infty}a_kx^k# converges at #c-p#. Therefore, we can reduce the study of convergence to the case where #p=0#.
Clearly, the power series #s(x) = \sum_{k=0}^{\infty}a_kx^k# converges for #x=0#. Suppose that the power series #\sum_{k=0}^{\infty}a_kx^k# converges for #x=c#, where #c\ne0#. It can be shown that then #s(x)# converges for all #x# with #\abs{x}\lt \abs{c}#. Moreover, if #s(x)# diverges for #x=c#, then #s(x)# diverges for all #x# with #\abs{x}\gt\abs{c}#. This explains the existence of a radius of convergence, and the fact that, if the radius of convergence is equal to #r#, then the interval of convergence of #s(x)# is one of the intervals #\ivoo{-r}{r}#, #\ivoc{-r}{r}#, #\ivco{-r}{r}#, #\ivcc{-r}{r}#.
In the case of the general power series #\sum_{k=0}^{\infty} a_k(x-p)^k# and radius of convergence #r#, the interval of convergence will be one of #\ivoo{p-r}{p+r}#, #\ivoc{p-r}{p+r}#, #\ivco{p-r}{p+r}#, #\ivcc{p-r}{p+r}#.
Once we know the power series determining a function #f#, the term-by-term derivative of the power series determines #f'# and the term-by-term anti-derivative of the series determines an anti-derivative of #f#. Also sums and products of power series are well behaved.
Let #f# be the function determined by the power series \[\displaystyle\sum_{k=0}^{\infty}a_k(x-p)^k\]where #x# and #p# are real numbers and #a_0, a_1, a_2, \ldots# is a sequence of real numbers. Suppose that #r# is a positive number (possibly infinite) less than or equal to the radius of convergence.
- The power series \( \sum_{k=1}^{\infty}k\cdot a_k(x-p)^{k-1}\) converges on #\ivoo{p-r}{p+r}# and determines the function #f'# on this interval.
- The power series \( \sum_{k=0}^{\infty}\frac{ a_k}{k+1}(x-p)^{k+1}\) converges on #\ivoo{p-r}{p+r}# and determines the anti-derivative of the function #f# on this interval whose value at #p# is #0#.
Suppose that, in addition, #g# is the function determined by the power series \[\displaystyle\sum_{k=0}^{\infty}b_k(x-p)^k\]where #p# is as above, #x# is a real number, and #b_0, b_1, b_2, \ldots# is a sequence of real numbers. Assume further that its radius of convergence is at least #r#.
- The power series obtained by expanding the sum of the power series corresponding to #f# and #g# converges on #\ivoo{p-r}{p+r}# and determines the function #f+ g#. Its #k#-th coefficient is \[ a_k + b_{k}\]
- The power series obtained by expanding the product of the power series corresponding to #f# and #g# converges on #\ivoo{p-r}{p+r}# and determines the function #f\cdot g#. Its #k#-th coefficient is \[\sum_{j=0}^k a_j\cdot b_{k-j}\]
The first statement means that the terms of the power series determining the derivative of a function #f# are the derivatives of the terms of the power series determining #f#.
The second statement means that the terms of the power series determining the anti-derivative of a function #f# are anti-derivatives of the terms of the power series determining #f#.
The third statement means that the sum of the power series with the same center determining the functions #f# and #g# is a power series determining #f+g#.
The fourth statement means that the product of the power series with the same center determining the functions #f# and #g# is a power series determining #f\cdot g#.
Consider the function #\frac{1}{x}#. It can be written as #\frac{1}{1-(1-x)}#, which, by the infinite geometric series formula, is the value of geometric series \[\sum_{k=0}^\infty {(1-x)^k} \] for #\abs{x-1}\lt 1#, that is #x\in\ivoo{0}{2}#. Since #\ln(x)# is the anti-derivative of #\frac{1}{x}# whose value at #1# is equal to #0#, the theorem gives
\[\ln (x)= \sum_{k=0}^{\infty}\frac{ -1}{k+1}(1-x)^{k+1}\phantom{xxx}\text{ for } \phantom{xxx}x\in\ivoo{0}{2}\]
Replacing #x# by #1-x# and #k# by #k-1#, we can rewrite this formula as
\[\ln (1-x)=- \sum_{k=1}^{\infty}\frac{ 1}{k}x^{k}\phantom{xxx}\text{ for } \phantom{xxx}x\in\ivoo{-1}{1}\]Earlier, we saw that this series converges for #x=-1# and diverges for #x=1#. The divergence is in accordance with our knowledge of the domain of #\ln#, which is #\ivoo{0}{\infty}#.
As we will see later, the trigonometric function \(\sin\) is determined by the following power series, which converges everywhere.
\[\sin(x) = \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}x^{2k+1}\]
By differentiating both sides of the equality, we find
\[\cos(x) = \sum_{k=0}^{\infty}\frac{(-1)^k(2k+1)}{(2k+1)!}x^{2k} = \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k)!}x^{2k}\]
By composing with various functions, we can use the theory of power series and geometric series for studying a wider variety of series. In the examples below, we will work with functions #x^2# and #\sqrt{x}# and substitute these for #x# in a power series.
Consider the series
\[\sum_{n=0}^{\infty} {{\left(x-2\right)^{n}}\over{3^{n}}} \]
Find the series' interval of convergence and, within this interval, the sum of the series as a function of #x#.
- The interval of convergence is \(\ivoo{-1}{5}\)
- The function #f# determined by the series on this interval has function rule \(f(x) ={{3}\over{5-x}}\)
We will use the fact that the radius of convergence of the geometric series # \sum_{n=0}^\infty x^n# is equal to # 1#. To this end we write
\[\sum_{n=0}^{\infty} {{\left(x-2\right)^{n}}\over{3^{n}}}= \sum_{n=0}^{\infty} \left( {{x-2}\over{3}}\right)^n \]
This shows that the series is
geometric and that the common ratio of the series is #{{{x-2}\over{3}}}#. As a consequence, the interval of convergence of the given series is determined by
\[\left|{{{x-2}\over{3}}}\right|\lt 1\]
By definition of absolute value, this inequality is equivalent to \[ {{x-2}\over{3}}\gt -1\land {{x-2}\over{3}}\lt 1\] which can be rewritten as \[x\gt -1\land x\lt 5 \]We conclude that the interval of convergence is \(\ivoo{-1}{5}\).
By the
infinite geometric sum formula, we find that on \(\ivoo{-1}{5}\), the interval of convergence, the function rule of #f# is \[f(x) = \frac{1}{1-{{x-2}\over{3}}} = {{3}\over{5-x}}\]
In the figure below, the graphs of the function #f(x) = {{3}\over{5-x}} # (in red) and the sum #\sum_{k=0}^3\left( {{x-2}\over{3}}\right)^k# of the first four terms of the series (in blue) are drawn. It illustrates that the head of the series approximates the function #f# well near the point #x = 2#.
Power series
