Substitution method for power series summation

Description of the method

The substitution method for power series summation is a method that can be used to convert one power series summation problem into another one. It is typically done with the goal of making the summation easier to obtain an explicit closed-form expression for.

Scalar multiple substitution

This is a substitution of the form $u=\lambda x$ for $\lambda$ a constant. Explicitly, consider a power series of the form:

$\sum _{k=k_0}^{\mathrm{\infty }}{a}_k{\lambda }^k{x}^k$

(Note that the starting point $k_0$ could be 0, 1, or anything).

Then, with the substitution $u=\lambda x$, this becomes:

$\sum _{k=k_0}^{\mathrm{\infty }}{a}_k{u}^k$

We can use scalar multiple substitutions in order to get rid of purely exponential parts of the coefficients.

Power substitution

This is a substitution of the form $u=x^m$ for $m$ a constant. Explicitly, consider a power series of the form:

$\sum _{k=k_0}^{\mathrm{\infty }}{a}_k{x}^{mk}$

(Note that the starting point $k_0$ could be 0, 1, or anything).

Then, with the subtsitution $u=x^m$, this becomes:

$\sum _{k=k_0}^{\mathrm{\infty }}{a}_k{u}^k$

Scalar multiple of power substitution

This combines the previous two substitution ideas, with a substitution of the form $u=\lambda x^m$ for $\lambda ,m$ constants. Explicitly, consider a power series of the form:

$\sum _{k=k_0}^{\mathrm{\infty }}{a}_k{\lambda }^k{x}^{mk}$

(Note that the starting point $k_0$ could be 0, 1, or anything).

Then, with the subtsitution $u=\lambda x^m$, this becomes:

$\sum _{k=k_0}^{\mathrm{\infty }}{a}_k{u}^k$

Application

Goal of the substitution

The substitution method is typically used for two purposes:

• Get rid of unnecessary multiplicative exponential terms in the coefficients (the scalar multiple part takes care of this)
• Try to scale the exponent so that it better matches the coefficients (the power part takes care of this): The general rule is that, at the end of the substitution, the exponent should match, as closely as possible, any term that is in a denominator or whose factorial is in the denominator.

Combination with multiplication

Substitution can be combined with another common technique for power series manipulation: multiply and divide by $x$ in order to make the exponent better match the coefficient.

Examples

Simple examples

Power series in $x$	$u$-substitution	Scalar multiple or power or both?	New power series in $u$	Sum in term of $u$	Sum in terms of $x$ (need to substitute back)
$\sum _{k=0}^{\mathrm{\infty }}\frac{x^{3k}}{k!}$	$u=x^3$	power	$\sum _{k=0}^{\mathrm{\infty }}\frac{u^k}{k!}$	$e^u$	$e^{x^3}$
$\sum _{k=1}^{\mathrm{\infty }}\frac{x^{5k}}{k}$	$u=x^5$	power	$\sum _{k=1}^{\mathrm{\infty }}\frac{u^k}{k}$	$-\ln (1-u)$	$-\ln (1-x^5)$
$\sum _{k=0}^{\mathrm{\infty }}\frac{2^k{x}^k}{k!}$	$u=2x$	scalar multiple	$\sum _{k=0}^{\mathrm{\infty }}\frac{u^k}{k!}$	$e^u$	$e^{2x}$
$\sum _{k=0}^{\mathrm{\infty }}\frac{3^{2k}{x}^{3k}}{k!}$	$u=9x^3$	combined	$\sum _{k=0}^{\mathrm{\infty }}\frac{u^k}{k!}$	$e^u$	$e^{9x^3}$
$\sum _{k=0}^{\mathrm{\infty }}\frac{x^{14k}}{(2k)!}$	$u=x^7$ (we want the exponent on it to match the expression under the factorial)	power	$\sum _{k=0}^{\mathrm{\infty }}\frac{u^{2k}}{(2k)!}$	$\cosh u$	$\cosh (x^7)$

Examples that involve some combination of substitution and multiplying/dividing

Power series in $x$	manipulation + $u$-substitution	Scalar multiple or power or both?	New power series in $u$	Sum in term of $u$	Sum in terms of $x$ (need to substitute back)
$\sum _{k=0}^{\mathrm{\infty }}\frac{x^{3k+1}}{k!}$	first pull out a factor of $x$, then do $u=x^3$	power	$x\sum _{k=0}^{\mathrm{\infty }}\frac{u^k}{k!}$	$x{e}^u$	$x{e}^{x^3}$
$\sum _{k=0}^{\mathrm{\infty }}\frac{x^{14k+2}}{(2k+1)!}$	first multiply/divide by $x^5$ (to make the exponent $7(2k+1)$, then do $u=x^7$	power	$\frac{1}{x^5}\sum _{k=0}^{\mathrm{\infty }}\frac{u^{2k+1}}{(2k+1)!}$	$\frac{\sinh u}{x^5}$	$\frac{\sinh (x^7)}{x^5}$

Example involving a fractional power substitution

Consider the power series summation problem:

$\sum _{k=0}^{\mathrm{\infty }}\frac{x^k}{(2k)!}$

We want to do a $u$-substitution that makes the exponent $2k$ so as to match the denominator. In order to do this, we would need to put $u=x^{1/2}$. This, however, is problematic since we don't know the sign of $x$. Thus, we make cases:

Case on sign of $x$	$u$-substitution	Power series in terms of $u$	Sum in terms of $u$	Sum in terms of $x$
positive	$u=\sqrt{x}$, so $x=u^2$	$\sum _{k=0}^{\mathrm{\infty }}\frac{u^{2k}}{(2k)!}$	$\cosh u$	$\cosh \sqrt{x}$
negative	$u=\sqrt{-x}$, so $x=-u^2$	$\sum _{k=0}^{\mathrm{\infty }}\frac{(-1{)}^k{u}^{2k}}{(2k)!}$	$\cos u$	$\cos \sqrt{-x}$

At $x=0$, either description fits. Overall, we have a piecewise definition of function for the sum of the series:

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