# 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:

$\sum_{k=k_0}^\infty a_k\lambda^kx^k$

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

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

$\sum_{k=k_0}^\infty a_ku^k$

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:

$\sum_{k=k_0}^\infty a_kx^{mk}$

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

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

$\sum_{k=k_0}^\infty a_ku^k$

### 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:

$\sum_{k=k_0}^\infty a_k\lambda^kx^{mk}$

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

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

$\sum_{k=k_0}^\infty a_ku^k$

## 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)
$\sum_{k=0}^\infty \frac{x^{3k}}{k!}$ $u = x^3$ power $\sum_{k=0}^\infty \frac{u^k}{k!}$ $e^u$ $e^{x^3}$
$\sum_{k=1}^\infty \frac{x^{5k}}{k}$ $u = x^5$ power $\sum_{k=1}^\infty \frac{u^k}{k}$ $-\ln(1 - u)$ $-\ln(1 - x^5)$ (note that the interval of convergence is not all of $\R$, but rather is $[-1,1)$ -- however, that is not the focus here)
$\sum_{k=0}^\infty \frac{2^kx^k}{k!}$ $u = 2x$ scalar multiple $\sum_{k=0}^\infty \frac{u^k}{k!}$ $e^u$ $e^{2x}$
$\sum_{k=0}^\infty \frac{3^{2k}x^{3k}}{k!}$ $u = 9x^3$ combined $\sum_{k=0}^\infty \frac{u^k}{k!}$ $e^u$ $e^{9x^3}$
$\sum_{k=0}^\infty \frac{x^{14k}}{(2k)!}$ $u = x^7$ (we want the exponent on it to match the expression under the factorial) power $\sum_{k=0}^\infty \frac{u^{2k}}{(2k)!}$ $\cosh u$ $\cosh(x^7)$

### 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)
$\sum_{k=0}^\infty \frac{x^{3k + 1}}{k!}$ first pull out a factor of $x$, then do $u = x^3$ power $x \sum_{k=0}^\infty \frac{u^k}{k!}$ $xe^u$ $xe^{x^3}$
$\sum_{k=0}^\infty \frac{x^{14k + 2}}{(2k + 1)!}$ first multiply/divide by $x^5$ (to make the exponent $7(2k + 1)$), then do $u = x^7$ power $\frac{1}{x^5} \sum_{k=0}^\infty \frac{u^{2k + 1}}{(2k + 1)!}$ $\frac{\sinh u}{x^5}$ $\frac{\sinh(x^7)}{x^5}$ (Note: The expression is not valid at $x = 0$ though its limit equals the power series sum at 0; both the limit and the power series sum at 0 are equal to 0)

### Example involving a fractional power substitution

Consider the power series summation problem:

$\sum_{k=0}^\infty \frac{x^k}{(2k)!}$

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 $x = u^2$ $\sum_{k=0}^\infty \frac{u^{2k}}{(2k)!}$ $\cosh u$ $\cosh \sqrt{x}$
negative $u = \sqrt{-x}$, so $x = -u^2$ $\sum_{k=0}^\infty \frac{(-1)^ku^{2k}}{(2k)!}$ $\cos u$ $\cos \sqrt{-x}$

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

$\sum_{k=0}^\infty \frac{x^k}{(2k)!} = \left\lbrace \begin{array}{rl} \cosh \sqrt{x}, & x \ge 0 \\ \cos \sqrt{-x}, & x < 0 \\\end{array}\right.$

## Interval of validity of power series summation

### Transforming the interval of validity along with the substitution

Consider the $u$-substitution $u = \lambda x^m$, giving:

$\sum a_k\lambda^kx^{mk} \to \sum a_ku^k$

We use some formula to sum up the power series rewritten in terms of $u$. In the typical scenario, the formula we use is valid over the entire interval of convergence of $u$, though we may sometimes have a situation where the formula we've obtained is valid only on part of the interval. We then plug back $u = x^{mk}$ and get a formula for the summation in terms of $x$.

The final formula is valid precisely for those $x$ for which $x^{mk}$ is in the set of values of $u$ for which the formula in terms of $u$ is valid.

In the good case that the formula is valid over the entire interval of convergence for the power series in $u$, we have:

• If the power series in $u$ converges everywhere to the function we have obtained, then the power series in $x$ also converges everywhere to the function we have obtained.
• If the radius of convergence for the power series in $u$ is $R$, the radius of convergence of the power series in $x$ is $(R/|\lambda|)^{1/m}$.
• To determine whether the endpoints are included, apply the map $x \mapsto \lambda x^m$ to the values $x = \pm (R/|\lambda|)^{1/m}$ and check whether the image of the map is in the interval of convergence for $u$.

Let's now consider some of the finite cases more explicitly:

Sign of $\lambda$ Parity of $m$ What can we say?
positive odd The negative endpoint $-(R/|\lambda|)^{1/m}$ is in the interval of convergence for $x$ if and only if $-R$ is in the interval of convergence for $u$, and the positive endpoint $(R/|\lambda|)^{1/m}$ is in the interval of convergence for $x$ if and only if $R$ is in the interval of convergence for $u$.
positive even Both endpoints $\pm (R/|\lambda|)^{1/m}$ are in the interval of convergence for $x$ if and only if $R$ is in the interval of convergence for $u$. Otherwise, neither endpoint is included.
negative odd The negative endpoint $-(R/|\lambda|)^{1/m}$ is in the interval of convergence for $x$ if and only if $R$ is in the interval of convergence for $u$, and the positive endpoint $(R/|\lambda|)^{1/m}$ is in the interval of convergence for $x$ if and only if $-R$ is in the interval of convergence for $u$.
negative even Both endpoints $\pm (R/|\lambda|)^{1/m}$ are in the interval of convergence for $x$ if and only if $-R$ is in the interval of convergence for $u$. Otherwise, neither endpoint is included.

### Multiplication and division

If the substitution method is combined with multiplying/dividing by a power of $x$, then we also need to worry about the case $x = 0$. In this case, the power series may still converge, but the formal expression we obtained may not be valid, though its limit would still be the right value.