Power series

Newton is the great virtuoso of infinite power series – what he did with them is astounding, e.g. by age 23 in 1665 found the sine series, cosine series, arc sine series, logarithmic series, binomial series AND the exponential series “which may…be the most important series in mathematics.” Although Indian mathematicians mostly got there hundreds of years before! But no-one in Europe knew that until much later.

The simplest power series is:

(1)   \begin{equation*} \boxed{1+x+x^2+x^3+x^4+\dots = \frac{1}{1-x}} \end{equation*}

1o1mx and ps
Such infinite series are easy to construct, but still I find it kind of mysterious that they add to the fraction on the other side. Well, it’s due to the telescoping.

To build one from scratch, start by deciding what you want the fraction to be, in this case \dfrac{1}{1-x}.
Then take the denominator over to the other side and start writing in terms to make the RHS add to 1. i.e.
1 = (1-x)(1+\dots and multiply this out on another line:
= 1 - x
Now there’s a -x to get rid of, so write in a +x
1 = (1-x)(1+x\dots
= 1-\not{x}+\not{x}-x^2+\dots and the xs cancel. Next an x^2:

1 = (1-x)(1+x+x^2+\dots
= 1-\not{x^2}+\not{x^2}-x^3+\dots and the x^2s cancel.
Not hard to see that this will go on forever in exactly the same way.
And we are left with 1+x+x^2+x^3+x^4+\dots = \dfrac{1}{1-x}.
IF x^n\to0 as n\to \infty! i.e. the series converges. For x\geq 1, that evidently won’t happen.

This is shown visually with the 2 similar triangles in this picture: ST/PS = PQ/QR
GS 2
I came across this picture (from Roger Nelsen’s book Proofs Without Words) after exploring power series, generating functions, continued fractions etc for a while – it came as a revelation to me. I hadn’t imagined there could be a simple drawing showing how it works! Why hadn’t I seen it before?! There are too many maths books without pictures! Although not so many these days; people realise the huge value of a good diagram.

geom sum

This one shows the more general case of \dfrac{a}{1-r} :

GS 1

All kinds of interesting results pop out when you plug values in, e.g if x=\frac{2}{3},

\dfrac{1}{1-\frac{2}{3}} = 3 = 1 + \dfrac{2}{3} + \left(\dfrac{2}{3}\right)^2 + \left(\dfrac{2}{3}\right)^3\dots

This pic from Proofs Without Words II :

Screen shot 2014-09-27 at 2.39.55 PM

I think x must be -1<x<1, otherwise the series diverges to infinity, producing some crazy equation. (Although Euler didn’t mind.)

Using \quad x=-\dfrac{1}{2}\quad\, produces \quad\dfrac{2}{3}=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}\dots

Plugging x^2 for x into (1) produces:

(2)   \begin{equation*} 1+x^2+x^4+x^6+\dots+x^{2n}+\dots = \frac{1}{1-x^2}\qquad \text{ if } |x|<1. \end{equation*}

Multiplying both sides of (2) by x gets:

    \[x+x^3+x^5+x^7+\dots+x^{2n+1}+\dots = \frac{x}{1-x^2}\qquad \text{ if } |x|<1.\]

Replacing x by -x in (1), we get:

(3)   \begin{equation*} 1-x+x^2-x^3+x^4-\dots+(-1)^nx^n+\dots = \frac{1}{1+x}\qquad \text{ if } |x|<1. \end{equation*}

Replacing x by x^2 in (3) gets:

(4)   \begin{equation*} 1-x^2+x^4-x^6+x^8-\dots+(-1)^nx^{2n}+\dots = \frac{1}{1+x^2}\qquad \text{ if } |x|<1. \end{equation*}

Multiplying both sides of (4) by x, we get:

(5)   \begin{equation*} x-x^3+x^5-x^7+x^9-\dots+(-1)^nx^{2n+1}+\dots = \frac{x}{1+x^2}\qquad \text{ if } |x|<1. \end{equation*}

Replacing x with 2x in (2) produces:

(6)   \begin{equation*} 1+4x^2+16x^4+64x^6+\dots+4^n x^{2n}+\dots = \frac{1}{1-4x^2}\qquad \text{ if } |2x|<1. \end{equation*}

Differentiating (1) gives:

    \[1+2x+3x^2+4x^3+\dots+nx^{n-1}+\dots = \frac{1}{(1-x)^2}\qquad \text{ if } |x|<1.\]

Integrating (3) gives:

    \[x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots+\frac{(-1)^{n-1}x^n}{n}+\dots = \log (1+x)\qquad\text{ if }  -1<x\les 1.\]

(Mercator, Brouncker 1668)
Integrating (4) gives:

    \[x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\dots+\frac{(-1)^{n-1}x^{2n-1}}{2n-1}+\dots = \arctan x\quad\text{ (Gregory 1671)}\]

Plugging in x=1 to this gives:


Reciprocal of a power series

If f(x)=\sum\limits_{n=0}^\infty a_n x^n \quad \text{with }|x|<R \text{ and } a_0\neq 0, there is a power series such that

    \[\fr{f(x)}=\sum\limits_{n=0}^\infty b_n x^n\]

with coefficients b_n determined recursively

    \[b_0=1/a_0, \quad b_n=-\fr{a_0} \sum\limits_{k=0}^{n-1} a_{n-k}b_k, \quad \text{ for } n\ges 1.\]

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