Introduction to Power Series: Videos & Practice Problems

To find the first four terms of the sequence defined by $a_n=n²n_{-}!$, substitute $n$ values from 1 to 4. The calculations yield: $a_1=1$, $a_2=4$, $a_3=18$, and $a_4=96$. Thus, the sequence begins with 1, 4, 18, and 96.

Intro to Power Series

To find the first four terms of the sequence defined by \( a_n = n^2 \times (n-1)! \), we will evaluate the expression for \( n = 1 \) through \( n = 4 \).

Starting with \( n = 1 \):

We have:

\( a_1 = 1^2 \times (1-1)! \)

Calculating this gives:

\( a_1 = 1 \times 0! \)

Since \( 0! = 1 \), we find:

\( a_1 = 1 \times 1 = 1 \)

Next, for \( n = 2 \):

We calculate:

\( a_2 = 2^2 \times (2-1)! \)

This simplifies to:

\( a_2 = 4 \times 1! \)

And since \( 1! = 1 \), we have:

\( a_2 = 4 \times 1 = 4 \)

Now, for \( n = 3 \):

We find:

\( a_3 = 3^2 \times (3-1)! \)

This becomes:

\( a_3 = 9 \times 2! \)

Calculating \( 2! = 2 \times 1 = 2 \), we get:

\( a_3 = 9 \times 2 = 18 \)

Finally, for \( n = 4 \):

We compute:

\( a_4 = 4^2 \times (4-1)! \)

This simplifies to:

\( a_4 = 16 \times 3! \)

Since \( 3! = 3 \times 2 \times 1 = 6 \), we find:

\( a_4 = 16 \times 6 = 96 \)

Thus, the first four terms of the sequence are \( 1, 4, 18, \) and \( 96 \). This sequence can be extended further by continuing to apply the same formula for higher values of \( n \).

Intro to Power Series Example 1

To find the first four terms of the sequence defined by \( a_n = n^2 \times (n-1)! \), we will evaluate the expression for \( n = 1 \) through \( n = 4 \).

Starting with \( n = 1 \):

We have:

\( a_1 = 1^2 \times (1-1)! \)

Calculating this gives:

\( a_1 = 1 \times 0! \)

Since \( 0! = 1 \), we find:

\( a_1 = 1 \times 1 = 1 \)

Next, for \( n = 2 \):

We calculate:

\( a_2 = 2^2 \times (2-1)! \)

This simplifies to:

\( a_2 = 4 \times 1! \)

And since \( 1! = 1 \), we have:

\( a_2 = 4 \times 1 = 4 \)

Now, for \( n = 3 \):

We find:

\( a_3 = 3^2 \times (3-1)! \)

This becomes:

\( a_3 = 9 \times 2! \)

Calculating \( 2! = 2 \times 1 = 2 \), we get:

\( a_3 = 9 \times 2 = 18 \)

Finally, for \( n = 4 \):

We compute:

\( a_4 = 4^2 \times (4-1)! \)

This simplifies to:

\( a_4 = 16 \times 3! \)

Since \( 3! = 3 \times 2 \times 1 = 6 \), we find:

\( a_4 = 16 \times 6 = 96 \)

Thus, the first four terms of the sequence are \( 1, 4, 18, \) and \( 96 \). This sequence can be extended further by continuing to apply the same formula for higher values of \( n \).

Intro to Power Series Example 2

To find the first four terms of the sequence defined by \( a_n = n^2 \times (n-1)! \), we will evaluate the expression for \( n = 1 \) through \( n = 4 \).

Starting with \( n = 1 \):

We have:

\( a_1 = 1^2 \times (1-1)! \)

Calculating this gives:

\( a_1 = 1 \times 0! \)

Since \( 0! = 1 \), we find:

\( a_1 = 1 \times 1 = 1 \)

Next, for \( n = 2 \):

We calculate:

\( a_2 = 2^2 \times (2-1)! \)

This simplifies to:

\( a_2 = 4 \times 1! \)

And since \( 1! = 1 \), we have:

\( a_2 = 4 \times 1 = 4 \)

Now, for \( n = 3 \):

We find:

\( a_3 = 3^2 \times (3-1)! \)

This becomes:

\( a_3 = 9 \times 2! \)

Calculating \( 2! = 2 \times 1 = 2 \), we get:

\( a_3 = 9 \times 2 = 18 \)

Finally, for \( n = 4 \):

We compute:

\( a_4 = 4^2 \times (4-1)! \)

This simplifies to:

\( a_4 = 16 \times 3! \)

Since \( 3! = 3 \times 2 \times 1 = 6 \), we find:

\( a_4 = 16 \times 6 = 96 \)

Thus, the first four terms of the sequence are \( 1, 4, 18, \) and \( 96 \). This sequence can be extended further by continuing to apply the same formula for higher values of \( n \).

Radius of Convergence

To find the first four terms of the sequence defined by \( a_n = n^2 \times (n-1)! \), we will evaluate the expression for \( n = 1 \) through \( n = 4 \).

Starting with \( n = 1 \):

We have:

\( a_1 = 1^2 \times (1-1)! \)

Calculating this gives:

\( a_1 = 1 \times 0! \)

Since \( 0! = 1 \), we find:

\( a_1 = 1 \times 1 = 1 \)

Next, for \( n = 2 \):

We calculate:

\( a_2 = 2^2 \times (2-1)! \)

This simplifies to:

\( a_2 = 4 \times 1! \)

And since \( 1! = 1 \), we have:

\( a_2 = 4 \times 1 = 4 \)

Now, for \( n = 3 \):

We find:

\( a_3 = 3^2 \times (3-1)! \)

This becomes:

\( a_3 = 9 \times 2! \)

Calculating \( 2! = 2 \times 1 = 2 \), we get:

\( a_3 = 9 \times 2 = 18 \)

Finally, for \( n = 4 \):

We compute:

\( a_4 = 4^2 \times (4-1)! \)

This simplifies to:

\( a_4 = 16 \times 3! \)

Since \( 3! = 3 \times 2 \times 1 = 6 \), we find:

\( a_4 = 16 \times 6 = 96 \)

Thus, the first four terms of the sequence are \( 1, 4, 18, \) and \( 96 \). This sequence can be extended further by continuing to apply the same formula for higher values of \( n \).

Radius of Convergence Example 3

To find the first four terms of the sequence defined by \( a_n = n^2 \times (n-1)! \), we will evaluate the expression for \( n = 1 \) through \( n = 4 \).

Starting with \( n = 1 \):

We have:

\( a_1 = 1^2 \times (1-1)! \)

Calculating this gives:

\( a_1 = 1 \times 0! \)

Since \( 0! = 1 \), we find:

\( a_1 = 1 \times 1 = 1 \)

Next, for \( n = 2 \):

We calculate:

\( a_2 = 2^2 \times (2-1)! \)

This simplifies to:

\( a_2 = 4 \times 1! \)

And since \( 1! = 1 \), we have:

\( a_2 = 4 \times 1 = 4 \)

Now, for \( n = 3 \):

We find:

\( a_3 = 3^2 \times (3-1)! \)

This becomes:

\( a_3 = 9 \times 2! \)

Calculating \( 2! = 2 \times 1 = 2 \), we get:

\( a_3 = 9 \times 2 = 18 \)

Finally, for \( n = 4 \):

We compute:

\( a_4 = 4^2 \times (4-1)! \)

This simplifies to:

\( a_4 = 16 \times 3! \)

Since \( 3! = 3 \times 2 \times 1 = 6 \), we find:

\( a_4 = 16 \times 6 = 96 \)

Thus, the first four terms of the sequence are \( 1, 4, 18, \) and \( 96 \). This sequence can be extended further by continuing to apply the same formula for higher values of \( n \).

Radius of Convergence Example 4

To find the first four terms of the sequence defined by \( a_n = n^2 \times (n-1)! \), we will evaluate the expression for \( n = 1 \) through \( n = 4 \).

Starting with \( n = 1 \):

We have:

\( a_1 = 1^2 \times (1-1)! \)

Calculating this gives:

\( a_1 = 1 \times 0! \)

Since \( 0! = 1 \), we find:

\( a_1 = 1 \times 1 = 1 \)

Next, for \( n = 2 \):

We calculate:

\( a_2 = 2^2 \times (2-1)! \)

This simplifies to:

\( a_2 = 4 \times 1! \)

Knowing that \( 1! = 1 \), we have:

\( a_2 = 4 \times 1 = 4 \)

Now, for \( n = 3 \):

We find:

\( a_3 = 3^2 \times (3-1)! \)

This becomes:

\( a_3 = 9 \times 2! \)

Since \( 2! = 2 \), we calculate:

\( a_3 = 9 \times 2 = 18 \)

Finally, for \( n = 4 \):

We compute:

\( a_4 = 4^2 \times (4-1)! \)

This simplifies to:

\( a_4 = 16 \times 3! \)

Knowing that \( 3! = 6 \), we find:

\( a_4 = 16 \times 6 = 96 \)

Thus, the first four terms of the sequence are \( 1, 4, 18, \) and \( 96 \). This sequence can be extended further by continuing to apply the same formula for higher values of \( n \).

Radius of Convergence Example 5

To find the first four terms of the sequence defined by \( a_n = n^2 \times (n-1)! \), we will evaluate the expression for \( n = 1 \) through \( n = 4 \).

Starting with \( n = 1 \):

We have:

\( a_1 = 1^2 \times (1-1)! \)

Calculating this gives:

\( a_1 = 1 \times 0! \)

Since \( 0! = 1 \), we find:

\( a_1 = 1 \times 1 = 1 \)

Next, for \( n = 2 \):

We calculate:

\( a_2 = 2^2 \times (2-1)! \)

This simplifies to:

\( a_2 = 4 \times 1! \)

And since \( 1! = 1 \), we have:

\( a_2 = 4 \times 1 = 4 \)

Now, for \( n = 3 \):

We find:

\( a_3 = 3^2 \times (3-1)! \)

This becomes:

\( a_3 = 9 \times 2! \)

Calculating \( 2! = 2 \times 1 = 2 \), we get:

\( a_3 = 9 \times 2 = 18 \)

Finally, for \( n = 4 \):

We compute:

\( a_4 = 4^2 \times (4-1)! \)

This simplifies to:

\( a_4 = 16 \times 3! \)

Since \( 3! = 3 \times 2 \times 1 = 6 \), we find:

\( a_4 = 16 \times 6 = 96 \)

Thus, the first four terms of the sequence are \( 1, 4, 18, \) and \( 96 \). This sequence can be extended further by continuing to apply the same formula for higher values of \( n \).

Interval of Convergence

To find the first four terms of the sequence defined by \( a_n = n^2 \times (n-1)! \), we will evaluate the expression for \( n = 1 \) through \( n = 4 \).

Starting with \( n = 1 \):

We have:

\( a_1 = 1^2 \times (1-1)! \)

Calculating this gives:

\( a_1 = 1 \times 0! \)

Since \( 0! = 1 \), we find:

\( a_1 = 1 \times 1 = 1 \)

Next, for \( n = 2 \):

We calculate:

\( a_2 = 2^2 \times (2-1)! \)

This simplifies to:

\( a_2 = 4 \times 1! \)

Knowing that \( 1! = 1 \), we have:

\( a_2 = 4 \times 1 = 4 \)

Now, for \( n = 3 \):

We find:

\( a_3 = 3^2 \times (3-1)! \)

This becomes:

\( a_3 = 9 \times 2! \)

Since \( 2! = 2 \), we calculate:

\( a_3 = 9 \times 2 = 18 \)

Finally, for \( n = 4 \):

We compute:

\( a_4 = 4^2 \times (4-1)! \)

This simplifies to:

\( a_4 = 16 \times 3! \)

Knowing that \( 3! = 6 \), we find:

\( a_4 = 16 \times 6 = 96 \)

Thus, the first four terms of the sequence are \( 1, 4, 18, \) and \( 96 \). This sequence can be extended further by continuing to apply the same formula for higher values of \( n \).

Interval of Convergence Example 6

To find the first four terms of the sequence defined by \( a_n = n^2 \times (n-1)! \), we will evaluate the expression for \( n = 1 \) through \( n = 4 \).

Starting with \( n = 1 \):

We have:

\( a_1 = 1^2 \times (1-1)! \)

Calculating this gives:

\( a_1 = 1 \times 0! \)

Since \( 0! = 1 \), we find:

\( a_1 = 1 \times 1 = 1 \)

Next, for \( n = 2 \):

We calculate:

\( a_2 = 2^2 \times (2-1)! \)

This simplifies to:

\( a_2 = 4 \times 1! \)

Knowing that \( 1! = 1 \), we have:

\( a_2 = 4 \times 1 = 4 \)

Now, for \( n = 3 \):

We find:

\( a_3 = 3^2 \times (3-1)! \)

This becomes:

\( a_3 = 9 \times 2! \)

Since \( 2! = 2 \), we calculate:

\( a_3 = 9 \times 2 = 18 \)

Finally, for \( n = 4 \):

We compute:

\( a_4 = 4^2 \times (4-1)! \)

This simplifies to:

\( a_4 = 16 \times 3! \)

Knowing that \( 3! = 6 \), we find:

\( a_4 = 16 \times 6 = 96 \)

Thus, the first four terms of the sequence are \( 1, 4, 18, \) and \( 96 \). This sequence can be extended further by continuing to apply the same formula for higher values of \( n \).

Interval of Convergence Example 7

To find the first four terms of the sequence defined by \( a_n = n^2 \times (n-1)! \), we will evaluate the expression for \( n = 1 \) through \( n = 4 \).

Starting with \( n = 1 \):

We have:

\( a_1 = 1^2 \times (1-1)! \)

Calculating this gives:

\( a_1 = 1 \times 0! \)

Since \( 0! = 1 \), we find:

\( a_1 = 1 \times 1 = 1 \)

Next, for \( n = 2 \):

We calculate:

\( a_2 = 2^2 \times (2-1)! \)

This simplifies to:

\( a_2 = 4 \times 1! \)

Knowing that \( 1! = 1 \), we have:

\( a_2 = 4 \times 1 = 4 \)

Now, for \( n = 3 \):

We find:

\( a_3 = 3^2 \times (3-1)! \)

This becomes:

\( a_3 = 9 \times 2! \)

Since \( 2! = 2 \), we calculate:

\( a_3 = 9 \times 2 = 18 \)

Finally, for \( n = 4 \):

We compute:

\( a_4 = 4^2 \times (4-1)! \)

This simplifies to:

\( a_4 = 16 \times 3! \)

Knowing that \( 3! = 6 \), we find:

\( a_4 = 16 \times 6 = 96 \)

Thus, the first four terms of the sequence are \( 1, 4, 18, \) and \( 96 \). This sequence can be extended further by continuing to apply the same formula for higher values of \( n \).

Representing Functions as Power Series

To find the first four terms of the sequence defined by \( a_n = n^2 \times (n-1)! \), we will evaluate the expression for \( n = 1 \) through \( n = 4 \).

Starting with \( n = 1 \):

We have:

\( a_1 = 1^2 \times (1-1)! \)

Calculating this gives:

\( a_1 = 1 \times 0! \)

Since \( 0! = 1 \), we find:

\( a_1 = 1 \times 1 = 1 \)

Next, for \( n = 2 \):

We calculate:

\( a_2 = 2^2 \times (2-1)! \)

This simplifies to:

\( a_2 = 4 \times 1! \)

And since \( 1! = 1 \), we have:

\( a_2 = 4 \times 1 = 4 \)

Now, for \( n = 3 \):

We find:

\( a_3 = 3^2 \times (3-1)! \)

This becomes:

\( a_3 = 9 \times 2! \)

Calculating \( 2! = 2 \times 1 = 2 \), we get:

\( a_3 = 9 \times 2 = 18 \)

Finally, for \( n = 4 \):

We compute:

\( a_4 = 4^2 \times (4-1)! \)

This simplifies to:

\( a_4 = 16 \times 3! \)

Since \( 3! = 3 \times 2 \times 1 = 6 \), we find:

\( a_4 = 16 \times 6 = 96 \)

Thus, the first four terms of the sequence are \( 1, 4, 18, \) and \( 96 \). This sequence can be extended further by continuing to apply the same formula for higher values of \( n \).

Representing Functions as Power Series Example 8

To find the first four terms of the sequence defined by \( a_n = n^2 \times (n-1)! \), we will evaluate the expression for \( n = 1 \) through \( n = 4 \).

Starting with \( n = 1 \):

We have:

\( a_1 = 1^2 \times (1-1)! \)

Calculating this gives:

\( a_1 = 1 \times 0! \)

Since \( 0! = 1 \), we find:

\( a_1 = 1 \times 1 = 1 \)

Next, for \( n = 2 \):

We calculate:

\( a_2 = 2^2 \times (2-1)! \)

This simplifies to:

\( a_2 = 4 \times 1! \)

Knowing that \( 1! = 1 \), we have:

\( a_2 = 4 \times 1 = 4 \)

Now, for \( n = 3 \):

We find:

\( a_3 = 3^2 \times (3-1)! \)

This becomes:

\( a_3 = 9 \times 2! \)

Since \( 2! = 2 \), we calculate:

\( a_3 = 9 \times 2 = 18 \)

Finally, for \( n = 4 \):

We compute:

\( a_4 = 4^2 \times (4-1)! \)

This simplifies to:

\( a_4 = 16 \times 3! \)

Knowing that \( 3! = 6 \), we find:

\( a_4 = 16 \times 6 = 96 \)

Thus, the first four terms of the sequence are \( 1, 4, 18, \) and \( 96 \). This sequence can be extended further by continuing to apply the same formula for higher values of \( n \).

Find the power series representation centered at $x=0$ of the following function. Give the interval of convergence for the resulting series. 
$f\left(x\right)=\frac{1}{1-x^3}$

• A. $\sum_{n=0}^{\infty}x^{3n}$; converges on (-∞,∞)
• B. ${\sum }_{n=0}^{\infty }{x}^{3n}$; converges on (-1,1)
• C. ${\sum }_{n=0}^{\infty }\frac{1}{x^{3n}}$; converges on (-1,1)
• D. ${\sum }_{n=0}^{\infty }\frac{1}{x^{3n}}$; diverges everywhere

Answer: (B)

Here’s what students ask on this topic:

What is a power series, and how is it defined?

A power series is an infinite series of the form:

$n=0\infty a_n{x}^n$

Here, $a_n$ represents the coefficients of the series, and $x$ is the variable. The series is centered at $x=0$, but it can also be centered at another value, $c$, in which case it takes the form:

$n=0\infty a_n{(x-c)}^n$

Power series are used in calculus to represent functions, approximate values, and solve differential equations. The interval of convergence determines the values of $x$ for which the series converges.

How do you determine the radius and interval of convergence for a power series?

To determine the radius and interval of convergence for a power series, follow these steps:

1. Use the Ratio Test or Root Test to find the radius of convergence. For a power series:

$n=0\infty a_n{(x-c)}^n$

apply the Ratio Test:

$\frac{\left|a_{n+1}{(x-c)}^{n+1}\right|}{\left|a_n{(x-c)}^n\right|}<1$

2. Solve for $x$ to find the radius of convergence, $R$.

3. Test the endpoints of the interval $[c-R,c+R]$ to determine whether the series converges at those points. This gives the interval of convergence.

What is the difference between a Taylor series and a Maclaurin series?

A Taylor series is a power series representation of a function centered at a point $c$. It is given by:

$n=0\infty \frac{f^{\left(}\left(c\right)}{n!}{(x-c)}^n$

where $f^{\left(}$ is the $n$th derivative of $f$ evaluated at $c$.

A Maclaurin series is a special case of the Taylor series where the center is $c=0$. It is given by:

$n=0\infty \frac{f^{\left(}\left(0\right)}{n!}{x}^n$

In summary, the difference lies in the center of the series: Taylor series are centered at any point $c$, while Maclaurin series are specifically centered at $c=0$.

How do you find the first four terms of a sequence involving factorials?

To find the first four terms of a sequence involving factorials, follow these steps:

1. Write the general formula for the sequence. For example, consider:

$a_n=n^2\cdot (n-1)!$

2. Substitute $n$ values from 1 to 4 into the formula:

- For $n=1$: $a_1=1^2\cdot 0!=1$

- For $n=2$: $a_2=2^2\cdot 1!=4$

- For $n=3$: $a_3=3^2\cdot 2!=18$

- For $n=4$: $a_4=4^2\cdot 3!=96$

3. The first four terms are: 1, 4, 18, and 96.

What is the significance of power series in calculus?

Power series are significant in calculus because they provide a way to represent functions as infinite sums of terms. This representation is useful for:

1. **Approximating Functions**: Power series can approximate functions to a desired degree of accuracy, especially near the center of the series.

2. **Solving Differential Equations**: Many differential equations can be solved using power series, especially when standard methods fail.

3. **Analyzing Convergence**: Power series help in understanding the behavior of functions within their interval of convergence.

4. **Expanding Functions**: Functions like $e{x}^{}$, $\sin \left(x\right)$, and $\ln \left(x\right)$ can be expressed as power series, making them easier to work with in computations.

Overall, power series are a foundational tool in both theoretical and applied mathematics.