Representing Functions as Power Series - AP Calculus BC Study Guide

The Core Idea: Representing Functions as Power Series

The fundamental concept of this topic is that many common functions can be expressed as an infinite polynomial, known as a power series. This representation is incredibly powerful, allowing us to approximate function values, evaluate difficult integrals, and solve differential equations. The starting point for this entire process is a single, foundational tool: the formula for the sum of an infinite geometric series.

By algebraically manipulating a given function to match the structure of the geometric series sum formula, $$f(x)=\frac{a}{1-r}$$, we can directly write down its power series representation, $$\sum _{n=0}^{\infty }a{r}^n$$. Furthermore, we can generate new power series from known ones through techniques like substitution, multiplication, division, term-by-term differentiation, and term-by-term integration. This allows us to build a library of power series representations for a wide variety of functions, all stemming from one basic geometric series.

Key Formulas

The entire process of representing functions as power series is built upon the formula for the sum of a convergent infinite geometric series.

The Geometric Series Sum Formula

A geometric series with first term $a$ and common ratio $r$ is given by:

$$\sum _{n=0}^{\infty }a{r}^n=a+ar+a{r}^2+a{r}^3+\dots$$

This series converges to a sum if and only if the absolute value of the common ratio is less than 1. The sum is given by the formula:

$$\frac{a}{1-r}=\sum _{n=0}^{\infty }a{r}^n,\text{for }|r|<1$$

In the context of functions, we treat an expression involving $x$ as the common ratio $r$. The goal is to make a given function $f(x)$ look like the left side of the equation, $$\frac{a}{1-r}$$, to find its power series representation on the right side.

Understanding the Manipulations

A power series for a function can be found by manipulating a known power series. The primary known series we use is the geometric series. The key manipulations are substitution, multiplication/division, differentiation, and integration.

1. Substitution

If we have a known series for $$f(u)=\sum _{n=0}^{\infty }{c}_n{u}^n$$, we can find the series for $$g(x)=f(h(x))$$ by substituting $$u=h(x)$$ into the series expression. For our base case, we start with $$\frac{1}{1-u}=\sum _{n=0}^{\infty }{u}^n$$ and substitute an expression in $x$ for $u$.

2. Multiplication and Division

A known power series can be multiplied or divided by a constant and/or a variable. This operation is applied to every term in the series. For example, if $$f(x)=\sum _{n=0}^{\infty }{c}_n{x}^n$$, then $$x^{k}f(x)=x^k\sum _{n=0}^{\infty }{c}_n{x}^n=\sum _{n=0}^{\infty }{c}_n{x}^{n+k}$$.

3. Differentiation and Integration

A power series can be differentiated or integrated term-by-term within its interval of convergence.

• Differentiation: If $$f(x)=\sum _{n=0}^{\infty }{c}_n{x}^n$$, then the series for its derivative is $$f^{\prime }(x)=\frac{d}{dx}\left(\sum _{n=0}^{\infty }{c}_n{x}^n\right)=\sum _{n=0}^{\infty }\frac{d}{dx}(c_n{x}^n)=\sum _{n=1}^{\infty }{c}_{n}n{x}^{n-1}$$.

• Integration: If $$f(x)=\sum _{n=0}^{\infty }{c}_n{x}^n$$, then the series for its integral is $$\int f(x)dx=\int \left(\sum _{n=0}^{\infty }{c}_n{x}^n\right)dx=C+\sum _{n=0}^{\infty }\int (c_n{x}^n)dx=C+\sum _{n=0}^{\infty }{c}_n\frac{x^{n+1}}{n+1}$$. The constant of integration, $C$, must be determined, typically by evaluating the function and the series at $x=0$.

Core Concepts & Rules

• Foundation: The geometric series formula, $$\frac{a}{1-r}=\sum _{n=0}^{\infty }a{r}^n$$ for $$|r|<1$$, is the fundamental tool used to generate power series for functions.

• Manipulation Goal: To find a power series for a function $f(x)$, you must first algebraically manipulate $f(x)$ into the form $$\frac{a}{1-r}$$.

• Substitution: A power series for a new function can be found by substituting an expression (e.g., $-x$, $x^2$, $3x$) for the variable in a known power series.

• Scaling: A power series can be multiplied or divided by a constant or a variable term (e.g., $x^k$) to create a new series.

• Term-by-Term Calculus: A power series can be differentiated or integrated term-by-term to find the power series for the derivative or integral of the function it represents.

• Constant of Integration: When finding a power series via integration, you must include the constant of integration $C$ and solve for it using an initial condition (e.g., evaluating at $x=0$).

Step-by-Step Example 1: Representation using Substitution and Multiplication

Problem: Find a power series representation for the function $$f(x)=\frac{x^2}{1+4x}$$.

Step 1: Isolate the geometric series structure.

Rewrite the function to separate the part that does not fit the $$\frac{a}{1-r}$$ form.

$$f(x)=x^2\cdot \frac{1}{1+4x}$$

Our goal is now to find the series for $$\frac{1}{1+4x}$$.

Step 2: Manipulate the expression into the form $$\frac{a}{1-r}$$.

The denominator must be in the form $$1-(\text{something})$$.

$$\frac{1}{1+4x}=\frac{1}{1-(-4x)}$$

Step 3: Identify $a$ and $r$ and apply the geometric series formula.

From the form $$\frac{1}{1-(-4x)}$$, we can identify:

• First term, $a=1$

• Common ratio, $r=-4x$

Now, substitute these into the series formula $$\sum _{n=0}^{\infty }a{r}^n$$:

$$\frac{1}{1-(-4x)}=\sum _{n=0}^{\infty }(1)(-4x{)}^n=\sum _{n=0}^{\infty }(-4{)}^n{x}^n$$

This representation is valid for $$|r|<1$$, which means $$|-4x|<1$$, or $$|x|<\frac{1}{4}$$.

Step 4: Perform the final multiplication.

Now, multiply the series from Step 3 by the $$x^2$$ term we isolated in Step 1.

$$f(x)=x^2\sum _{n=0}^{\infty }(-4{)}^n{x}^n=\sum _{n=0}^{\infty }{x}^2\cdot (-4{)}^n{x}^n$$

$$f(x)=\sum _{n=0}^{\infty }(-4{)}^n{x}^{n+2}$$

Final Answer: The power series representation for $$f(x)=\frac{x^2}{1+4x}$$ is $$\sum _{n=0}^{\infty }(-4{)}^n{x}^{n+2}$$.

Step-by-Step Example 2: Representation using Integration

Problem: Find a power series representation for $$g(x)=\ln (1-x)$$.

Step 1: Relate the function to a known series via calculus.

We do not have a direct way to represent $$\ln (1-x)$$ using the geometric series formula. However, we can recognize that its derivative is a function we can represent.

$$g^{\prime }(x)=\frac{d}{dx}\ln (1-x)=\frac{1}{1-x}\cdot (-1)=-\frac{1}{1-x}$$

This means that $$g(x)=\int \left(-\frac{1}{1-x}\right)dx$$.

Step 2: Find the power series for the derivative.

The function $$-\frac{1}{1-x}$$ is in the form $$\frac{a}{1-r}$$ with $a=-1$ and $r=x$.

Using the geometric series formula:

$$g^{\prime }(x)=-\frac{1}{1-x}=\sum _{n=0}^{\infty }(-1)x^n=-\sum _{n=0}^{\infty }{x}^n$$

This is valid for $$|x|<1$$.

Step 3: Integrate the power series term-by-term.

Now we integrate the series for $g^{\prime }(x)$ to find the series for $g(x)$.

$$g(x)=\int \left(-\sum _{n=0}^{\infty }{x}^n\right)dx=-\int \left(\sum _{n=0}^{\infty }{x}^n\right)dx$$

$$g(x)=C-\sum _{n=0}^{\infty }\left(\int x^{n}dx\right)=C-\sum _{n=0}^{\infty }\frac{x^{n+1}}{n+1}$$

Step 4: Solve for the constant of integration, $C$.

To find $C$, we evaluate both the function and the series at a convenient point, usually $x=0$.

• Function value: $$g(0)=\ln (1-0)=\ln (1)=0$$.

• Series value: At $x=0$, the series is $$C-\sum _{n=0}^{\infty }\frac{0^{n+1}}{n+1}=C-0=C$$.

Equating the two, we find $$0=C$$.

Step 5: Write the final power series.

Substitute C=0 back into the expression from Step 3.

$$g(x)=\ln (1-x)=-\sum _{n=0}^{\infty }\frac{x^{n+1}}{n+1}$$

Final Answer: The power series representation for $$g(x)=\ln (1-x)$$ is $$-\sum _{n=0}^{\infty }\frac{x^{n+1}}{n+1}$$.

Using Your Calculator

This topic involves purely analytical methods for generating power series. A calculator is not used to find the series representation itself.

However, a graphing calculator is an excellent tool for verifying your answer. You can check if your derived power series seems correct by comparing the graph of the original function to the graph of a partial sum of your series.

Example Check: To verify the result from Example 2, \ln(1-x) = -\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}

The Core Idea: Representing Functions as Power Series

The fundamental concept of this topic is that many common functions can be expressed as an infinite polynomial, known as a power series. This representation is incredibly powerful, allowing us to approximate function values, evaluate difficult integrals, and solve differential equations. The starting point for this entire process is a single, foundational tool: the formula for the sum of an infinite geometric series.

By algebraically manipulating a given function to match the structure of the geometric series sum formula, $$f(x)=\frac{a}{1-r}$$, we can directly write down its power series representation, $$\sum _{n=0}^{\infty }a{r}^n$$. Furthermore, we can generate new power series from known ones through techniques like substitution, multiplication, division, term-by-term differentiation, and term-by-term integration. This allows us to build a library of power series representations for a wide variety of functions, all stemming from one basic geometric series.

Key Formulas

The entire process of representing functions as power series is built upon the formula for the sum of a convergent infinite geometric series.

The Geometric Series Sum Formula

A geometric series with first term $a$ and common ratio $r$ is given by:

$$\sum _{n=0}^{\infty }a{r}^n=a+ar+a{r}^2+a{r}^3+\dots$$

This series converges to a sum if and only if the absolute value of the common ratio is less than 1. The sum is given by the formula:

$$\frac{a}{1-r}=\sum _{n=0}^{\infty }a{r}^n,\text{for }|r|<1$$

In the context of functions, we treat an expression involving $x$ as the common ratio $r$. The goal is to make a given function $f(x)$ look like the left side of the equation, $$\frac{a}{1-r}$$, to find its power series representation on the right side.

Understanding the Manipulations

A power series for a function can be found by manipulating a known power series. The primary known series we use is the geometric series. The key manipulations are substitution, multiplication/division, differentiation, and integration.

1. Substitution

If we have a known series for $$f(u)=\sum _{n=0}^{\infty }{c}_n{u}^n$$, we can find the series for $$g(x)=f(h(x))$$ by substituting $$u=h(x)$$ into the series expression. For our base case, we start with $$\frac{1}{1-u}=\sum _{n=0}^{\infty }{u}^n$$ and substitute an expression in $x$ for $u$.

2. Multiplication and Division

A known power series can be multiplied or divided by a constant and/or a variable. This operation is applied to every term in the series. For example, if $$f(x)=\sum _{n=0}^{\infty }{c}_n{x}^n$$, then $$x^{k}f(x)=x^k\sum _{n=0}^{\infty }{c}_n{x}^n=\sum _{n=0}^{\infty }{c}_n{x}^{n+k}$$.

3. Differentiation and Integration

A power series can be differentiated or integrated term-by-term within its interval of convergence.

• Differentiation: If $$f(x)=\sum _{n=0}^{\infty }{c}_n{x}^n$$, then the series for its derivative is $$f^{\prime }(x)=\frac{d}{dx}\left(\sum _{n=0}^{\infty }{c}_n{x}^n\right)=\sum _{n=0}^{\infty }\frac{d}{dx}(c_n{x}^n)=\sum _{n=1}^{\infty }{c}_{n}n{x}^{n-1}$$.

• Integration: If $$f(x)=\sum _{n=0}^{\infty }{c}_n{x}^n$$, then the series for its integral is $$\int f(x)dx=\int \left(\sum _{n=0}^{\infty }{c}_n{x}^n\right)dx=C+\sum _{n=0}^{\infty }\int (c_n{x}^n)dx=C+\sum _{n=0}^{\infty }{c}_n\frac{x^{n+1}}{n+1}$$. The constant of integration, $C$, must be determined, typically by evaluating the function and the series at $x=0$.

Core Concepts & Rules

• Foundation: The geometric series formula, $$\frac{a}{1-r}=\sum _{n=0}^{\infty }a{r}^n$$ for $$|r|<1$$, is the fundamental tool used to generate power series for functions.

• Manipulation Goal: To find a power series for a function $f(x)$, you must first algebraically manipulate $f(x)$ into the form $$\frac{a}{1-r}$$.

• Substitution: A power series for a new function can be found by substituting an expression (e.g., $-x$, $x^2$, $3x$) for the variable in a known power series.

• Scaling: A power series can be multiplied or divided by a constant or a variable term (e.g., $x^k$) to create a new series.

• Term-by-Term Calculus: A power series can be differentiated or integrated term-by-term to find the power series for the derivative or integral of the function it represents.

• Constant of Integration: When finding a power series via integration, you must include the constant of integration $C$ and solve for it using an initial condition (e.g., evaluating at $x=0$).

Step-by-Step Example 1: Representation using Substitution and Multiplication

Problem: Find a power series representation for the function $$f(x)=\frac{x^2}{1+4x}$$.

Step 1: Isolate the geometric series structure.

Rewrite the function to separate the part that does not fit the $$\frac{a}{1-r}$$ form.

$$f(x)=x^2\cdot \frac{1}{1+4x}$$

Our goal is now to find the series for $$\frac{1}{1+4x}$$.

Step 2: Manipulate the expression into the form $$\frac{a}{1-r}$$.

The denominator must be in the form $$1-(\text{something})$$.

$$\frac{1}{1+4x}=\frac{1}{1-(-4x)}$$

Step 3: Identify $a$ and $r$ and apply the geometric series formula.

From the form $$\frac{1}{1-(-4x)}$$, we can identify:

• First term, $a=1$

• Common ratio, $r=-4x$

Now, substitute these into the series formula $$\sum _{n=0}^{\infty }a{r}^n$$:

$$\frac{1}{1-(-4x)}=\sum _{n=0}^{\infty }(1)(-4x{)}^n=\sum _{n=0}^{\infty }(-4{)}^n{x}^n$$

This representation is valid for $$|r|<1$$, which means $$|-4x|<1$$, or $$|x|<\frac{1}{4}$$.

Step 4: Perform the final multiplication.

Now, multiply the series from Step 3 by the $$x^2$$ term we isolated in Step 1.

$$f(x)=x^2\sum _{n=0}^{\infty }(-4{)}^n{x}^n=\sum _{n=0}^{\infty }{x}^2\cdot (-4{)}^n{x}^n$$

$$f(x)=\sum _{n=0}^{\infty }(-4{)}^n{x}^{n+2}$$

Final Answer: The power series representation for $$f(x)=\frac{x^2}{1+4x}$$ is $$\sum _{n=0}^{\infty }(-4{)}^n{x}^{n+2}$$.

Step-by-Step Example 2: Representation using Integration

Problem: Find a power series representation for $$g(x)=\ln (1-x)$$.

Step 1: Relate the function to a known series via calculus.

We do not have a direct way to represent $$\ln (1-x)$$ using the geometric series formula. However, we can recognize that its derivative is a function we can represent.

$$g^{\prime }(x)=\frac{d}{dx}\ln (1-x)=\frac{1}{1-x}\cdot (-1)=-\frac{1}{1-x}$$

This means that $$g(x)=\int \left(-\frac{1}{1-x}\right)dx$$.

Step 2: Find the power series for the derivative.

The function $$-\frac{1}{1-x}$$ is in the form $$\frac{a}{1-r}$$ with $a=-1$ and $r=x$.

Using the geometric series formula:

$$g^{\prime }(x)=-\frac{1}{1-x}=\sum _{n=0}^{\infty }(-1)x^n=-\sum _{n=0}^{\infty }{x}^n$$

This is valid for $$|x|<1$$.

Step 3: Integrate the power series term-by-term.

Now we integrate the series for $g^{\prime }(x)$ to find the series for $g(x)$.

$$g(x)=\int \left(-\sum _{n=0}^{\infty }{x}^n\right)dx=-\int \left(\sum _{n=0}^{\infty }{x}^n\right)dx$$

$$g(x)=C-\sum _{n=0}^{\infty }\left(\int x^{n}dx\right)=C-\sum _{n=0}^{\infty }\frac{x^{n+1}}{n+1}$$

Step 4: Solve for the constant of integration, $C$.

To find $C$, we evaluate both the function and the series at a convenient point, usually $x=0$.

• Function value: $$g(0)=\ln (1-0)=\ln (1)=0$$.

• Series value: At $x=0$, the series is $$C-\sum _{n=0}^{\infty }\frac{0^{n+1}}{n+1}=C-0=C$$.

Equating the two, we find $$0=C$$.

Step 5: Write the final power series.

Substitute C=0 back into the expression from Step 3.

$$g(x)=\ln (1-x)=-\sum _{n=0}^{\infty }\frac{x^{n+1}}{n+1}$$

Final Answer: The power series representation for $$g(x)=\ln (1-x)$$ is $$-\sum _{n=0}^{\infty }\frac{x^{n+1}}{n+1}$$.

Using Your Calculator

This topic involves purely analytical methods for generating power series. A calculator is not used to find the series representation itself.

However, a graphing calculator is an excellent tool for verifying your answer. You can check if your derived power series seems correct by comparing the graph of the original function to the graph of a partial sum of your series.

Example Check: To verify the result from Example 2, , follow these steps: 1. In `Y1`, enter the original function: `Y1 = ln(1-x)`. 2. In `Y2`, enter a partial sum of the series (the first few terms). Let's use the terms up to `n=3`: `Y2 = -(x^(0+1)/(0+1) + x^(1+1)/(1+1) + x^(2+1)/(2+1) + x^(3+1)/(3+1))` `Y2 = -(x + x^2/2 + x^3/3 + x^4/4)` 3. Graph both functions. Near $x=0$ (the center of the series), the graphs of `Y1` and `Y2` should be nearly identical. The approximation gets better as you add more terms to the partial sum in `Y2`. ## AP Exam Quick Hit ### Common Question Types - **Direct Manipulation and Substitution:** You will be given a function that is a variation of `$\frac{1}{1-x} and asked for its power series.

-   *Example:* Find the first four nonzero terms and the general term for the power series for $$f(x)=\frac{3}{1+x^2}$$.

• Representation via Calculus: You will be given a function whose derivative or integral is a variation of the geometric series sum, such as $$\arctan (x)$$ or $$\ln (1+x)$$.

  • Example: The function $g$ is given by $$g(x)=\frac{d}{dx}\left(\frac{1}{1-x^3}\right)$$. Find the power series for $g(x)$.

• Building a New Series from a Given One: You may be given a power series for a function $f(x)$ and asked to find the series for a related function like $$x^{2}f(x)$$ or $$f^{\prime }(x)$$.

  • Example: Let $$f(x)=\sum _{n=0}^{\infty }\frac{x^n}{n!}$$. Find the power series for $${\int }_0^{x}f(t)dt$$.

Common Mistakes

• Incorrect Identification of r: For a function like `$f(x)=\frac{1}{4-x}Formula[61]\frac{1}{4(1-x/4)}Formula[62]a=1/4Formula[63]r=x/4Formula[64]\frac{1}{1+x}=\frac{1}{1-(-x)}Formula[65](-1{)}^{n}Formula[66]r^n=(-x{)}^n=(-1{)}^n{x}^{n}Formula[67](2x^3{)}^{n}Formula[68]2^n{x}^{3n}Formula[69]2x^{3n}Formula[70]|r|<1$$. All manipulations are performed under this assumption.