What is the first step to find the slope of a curve represented by parametric equations $x(t)$ and $y(t)$ at a specific point $t$?

Find $\frac{dy}{dt}$ and $\frac{dx}{dt}$ and then take their quotient.

Integrate both $x(t)$ and $y(t)$ with respect to $t$.

Differentiate $x(t) \cdot y(t)$ with respect to $t$.

Solve for $t$ in terms of $x$ and then differentiate $y$ with respect to $x$.

How can you accurately describe changes to both radius and interval convergence caused by replacing every term within an existing power series containing variable "X" into something like "(k*x)"?

Multiplication comon ratio, thus altering the condensed area where a specific range becomes valid

Which advanced method would most effectively approximate the boundary point's behavior for interval determination on a complex plane for $\sum_{k=0}^{\infty}(z+i)^k$, where z is complex?

Abelian theorem predicts continuity up until but not including boundary points marked by unit circle around -i.

Considering a pharmaceutical company testing drug efficacy over time using a power series model $\sum_{n=0}^\infty c_n(d-100\text{ days})^n$, if their tests reveal that side effects vary significantly outside an interval with a radius equal to half-a-year from day one hundred after administration, how long post-administration do side effect patterns remain predictable based on this model?

Patterns remain predictable up to day one hundred plus or minus half-a-year (182 days).

Patterns remain predictable indefinitely as long as administered after day one hundred.

Patterns remain unpredictable before day two hundred fifty-two post-administration.

Side effect patterns become instantly unpredictable after administration regardless of time frame.

For the alternating series $\sum_{n=1}^{\infty} (-1)^{n+1}\frac{(x+5)^n}{(3n)!}$, which application provides a valid alternative for determining its interval of convergence without direct computation?

The alternating series test implies it converges for all real numbers due to factorial growth in the denominator.

Applying L'Hôpital's rule to assess limit behavior at endpoints.

Utilizing integration to identify divergence or conditional convergence.

Employing differentiation to examine changes in terms across intervals.

Considering a function represented by a power series $\sum_{n=0}^{\infty}a_n(x-1)^n$ converges absolutely at x=3, how would you describe its behavior at x=-1?

It may either converge or diverge without further information.

It must converge absolutely because it's within the radius of convergence.

It must diverge since it lies outside the known interval of absolute convergence.

It must converge conditionally based on symmetry around x=1.

A climate model uses the power series $g(t) = \sum_{k=0}^{\infty}a_k (t-t_0)^k$, where $t_0$ is present time; to predict future carbon dioxide levels in parts per million (ppm), if it’s known that the radius of convergence is exactly 50 years, what is the furthest year into the future where predictions made by this model would be considered reliable?

Predictions are reliable up to year $t_0 +50$.

Predictions are reliable up to year $t_0 +25$.

Predictions are reliable indefinitely beyond year $t_0$.

No predictions can be reliably made using this model.

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10.13 Radius and Interval of Convergence of Power Series

1 min read•june 7, 2020

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Oooookay, that title definitely had a lot of buzzwords… namely, radius of convergence, interval of convergence, and power series. You haven’t seen them in any of the previous study guides, either, so they’re definitely new to you. Let’s define them one by one! 😉

👊 What’s a Power Series?

A power series is a series of the form $\sum^{\infty}_{n=0}a_n(x-r)^n$ , where n is a non-negative integer, $a_n$ is a sequence of real numbers, and r is a real number.

In this case, $a_n$ can be any sequence, most of which you’ve already seen in previous study guides! r refers to where we center our power series function (e.g., a “power series centered at x = 3” will give us (x - 3) at that part of the series).

🔵 Radius and Interval of Convergence

One of the questions we have about power series approximations of functions is where the approximation is valid, or in other words, where the power series converges. For a given x, we can find the radius, and then the interval of convergence for a power series.

For a Taylor series centered at x = r, the only place where we are entirely sure that it converges to is at x = r, but we can expand this to a greater range using our knowledge of the ratio test. Let’s make an example to demonstrate this!

Recap: The Ratio Test

For a series $\sum a_n$ , let $L=|\lim\limits_{n→ \infty}\frac{a_{n+1}}{a_n}|$ .

If L < 1, then the series converges.
If L > 1, then the series diverges.
If L = 1, then the test is indeterminate.

To review the ratio test in depth, check out 10.8 Ratio Test for Convergence.

❓ Power Series Practice Question

Find the interval, radius of convergence, and the center of the interval of convergence for the series:

\sum^{\infty}_{n=0}\frac{2^n}{n}(4x-8)^n

First, we use the Ratio Test and identify our $a_n$ and $a_{n+1}$ :

a_{n}=\frac{2^{n}}{n}(4x-8)^{n}

a_{n+1}=\frac{2^{n+1}}{n+1}(4x-8)^{n+1}=\frac{2^n*2}{n+1}(4x-8)^n(4x-8)

Then, we find L:

L=\lim\limits_{n→ \infty}|\frac{a_{n+1}}{a_n}|=\lim\limits_{n→ \infty}|\frac{\frac{2^n*2}{n+1}(4x-8)^n(4x-8)}{\frac{2^{n}}{n}(4x-8)^{n}}|

Seems complicated at first but we actually cancel out a couple terms: $2^n,(4x-8)^n$ . This leaves us with:

L=\lim\limits_{n→ \infty}|\frac{2n}{n+1}(4x-8)|

Since (4x - 8) is not in terms of n, we can factor it out of the limit term. We, then, evaluate the limit using L’hopital’s Rule, which gives us 2:

L=(4x-8)*\lim\limits_{n→ \infty}|\frac{2n}{n+1}|=(4x-8)*2

Remember, the series converges when L < 1, which we can simplify:

|L|<1

|2(4x-8)|<1

|2*4(x-2)|<1

|x-2|<\frac{1}{8}

This gives us a radius of converge of 1/8, a center of the interval of convergence at x = 2

To find the interval of convergence, we use our knowledge of absolute values and the expression $|x-2|<\frac{1}{8}$ to get:

x-2<\frac{1}{8}=x<\frac{1}{8}+\frac{16}{8}=x<\frac{17}{8}

x-2>-\frac{1}{8}=x>-\frac{1}{8}+\frac{16}{8}=x>\frac{15}{8}

One last thing: we need to test these endpoints—namely, the two extreme points of a line segment or interval—by plugging the values into the original series to see if they are included in our solution or not:

At x = 15/8:

\sum^{\infty}_{n=0}\frac{2^n}{n}(4(\frac{15}{8})-8)^n

\sum^{\infty}_{n=0}\frac{2^n}{n}(\frac{15}{2}-8)^n

\sum^{\infty}_{n=0}\frac{2^n}{n}(\frac{-1}{2})^n

\sum^{\infty}_{n=0}\frac{2^n}{n}(\frac{(-1)^n}{2^n})

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

From our previous encounter with the alternate harmonic series above, we can say that the series converges at x = 15/8. In other words, 15/8 is included in our interval of convergence. What about x = 17/8?

At x = 17/8:

\sum^{\infty}_{n=0}\frac{2^n}{n}(4(\frac{17}{8})-8)^n

\sum^{\infty}_{n=0}\frac{2^n}{n}(\frac{17}{2}-8)^n

\sum^{\infty}_{n=0}\frac{2^n}{n}(\frac{1}{2})^n

\sum^{\infty}_{n=0}\frac{2^n}{n}(\frac{1}{2^n})

\sum^{\infty}_{n=0}\frac{1}{n}

Another familiar face: the harmonic series! We can, thus, say that the series diverges at x = 17/8. In other words, 17/8 is not included in our interval of convergence.

Altogether, our interval of convergence is $[\frac{15}{8},\frac{17}{8})$ or $\frac{15}{8}\leq x <\frac{17}{8}.$

⭐ Summing Up Power Series

To summarize what we introduced above:

To make things easier for you, here’s a quick guide on what you should do when you encounter a power series problem that asks you to find the radius & interval of convergence:

Apply the ratio test!
Use the ratio test to find your radius of convergence and endpoints.
Plug endpoints back into your original series to see if they are included in the solution or not… this’ll help you finalize your interval of convergence!
- If the endpoint series converges, that endpoint is included.
- If the endpoint series diverges, that endpoint is not included.
- Use your “series and tests” (study guides 10.3 to 10.9) toolkit!

That’s it! While the journey to the answer seems long and arduous, you’ll notice that the building blocks from earlier study guides and math courses like the p-series test, the ratio test, and even absolute values all come together in the concept of power series. As always, mastery comes with practice, and becoming an expert at this topic will help you brush up on the prior concepts in this unit as well.

Good luck! 🎈

Key Terms to Review (5)

Endpoints

: The endpoints are the two extreme points of a line segment or interval.

Interval of Convergence

: The interval of convergence is the range of x-values for which a power series converges, meaning it approaches a finite value.

Power Series

: A power series is an infinite series that represents a function as an infinite polynomial expression.

Ratio Test

: The ratio test is used to determine whether an infinite series converges absolutely, converges conditionally, or diverges by examining the limit of the ratio of consecutive terms.

Taylor Series

: A Taylor series is an expansion of a function into an infinite sum of terms, where each term represents the contribution from different derivatives of the function at a specific point.

10.13 Radius and Interval of Convergence of Power Series

1 min read•june 7, 2020

Attend a live cram event

Review all units live with expert teachers & students

👊 What’s a Power Series?

A power series is a series of the form $\sum^{\infty}_{n=0}a_n(x-r)^n$ , where n is a non-negative integer, $a_n$ is a sequence of real numbers, and r is a real number.

🔵 Radius and Interval of Convergence

Recap: The Ratio Test

For a series $\sum a_n$ , let $L=|\lim\limits_{n→ \infty}\frac{a_{n+1}}{a_n}|$ .

If L < 1, then the series converges.
If L > 1, then the series diverges.
If L = 1, then the test is indeterminate.

To review the ratio test in depth, check out 10.8 Ratio Test for Convergence.

❓ Power Series Practice Question

Find the interval, radius of convergence, and the center of the interval of convergence for the series:

\sum^{\infty}_{n=0}\frac{2^n}{n}(4x-8)^n

First, we use the Ratio Test and identify our $a_n$ and $a_{n+1}$ :

a_{n}=\frac{2^{n}}{n}(4x-8)^{n}

a_{n+1}=\frac{2^{n+1}}{n+1}(4x-8)^{n+1}=\frac{2^n*2}{n+1}(4x-8)^n(4x-8)

Then, we find L:

L=\lim\limits_{n→ \infty}|\frac{a_{n+1}}{a_n}|=\lim\limits_{n→ \infty}|\frac{\frac{2^n*2}{n+1}(4x-8)^n(4x-8)}{\frac{2^{n}}{n}(4x-8)^{n}}|

Seems complicated at first but we actually cancel out a couple terms: $2^n,(4x-8)^n$ . This leaves us with:

L=\lim\limits_{n→ \infty}|\frac{2n}{n+1}(4x-8)|

Since (4x - 8) is not in terms of n, we can factor it out of the limit term. We, then, evaluate the limit using L’hopital’s Rule, which gives us 2:

L=(4x-8)*\lim\limits_{n→ \infty}|\frac{2n}{n+1}|=(4x-8)*2

Remember, the series converges when L < 1, which we can simplify:

|L|<1

|2(4x-8)|<1

|2*4(x-2)|<1

|x-2|<\frac{1}{8}

This gives us a radius of converge of 1/8, a center of the interval of convergence at x = 2

To find the interval of convergence, we use our knowledge of absolute values and the expression $|x-2|<\frac{1}{8}$ to get:

x-2<\frac{1}{8}=x<\frac{1}{8}+\frac{16}{8}=x<\frac{17}{8}

x-2>-\frac{1}{8}=x>-\frac{1}{8}+\frac{16}{8}=x>\frac{15}{8}

At x = 15/8:

\sum^{\infty}_{n=0}\frac{2^n}{n}(4(\frac{15}{8})-8)^n

\sum^{\infty}_{n=0}\frac{2^n}{n}(\frac{15}{2}-8)^n

\sum^{\infty}_{n=0}\frac{2^n}{n}(\frac{-1}{2})^n

\sum^{\infty}_{n=0}\frac{2^n}{n}(\frac{(-1)^n}{2^n})

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

At x = 17/8:

\sum^{\infty}_{n=0}\frac{2^n}{n}(4(\frac{17}{8})-8)^n

\sum^{\infty}_{n=0}\frac{2^n}{n}(\frac{17}{2}-8)^n

\sum^{\infty}_{n=0}\frac{2^n}{n}(\frac{1}{2})^n

\sum^{\infty}_{n=0}\frac{2^n}{n}(\frac{1}{2^n})

\sum^{\infty}_{n=0}\frac{1}{n}

Another familiar face: the harmonic series! We can, thus, say that the series diverges at x = 17/8. In other words, 17/8 is not included in our interval of convergence.

Altogether, our interval of convergence is $[\frac{15}{8},\frac{17}{8})$ or $\frac{15}{8}\leq x <\frac{17}{8}.$

⭐ Summing Up Power Series

To summarize what we introduced above:

To make things easier for you, here’s a quick guide on what you should do when you encounter a power series problem that asks you to find the radius & interval of convergence:

Apply the ratio test!
Use the ratio test to find your radius of convergence and endpoints.
Plug endpoints back into your original series to see if they are included in the solution or not… this’ll help you finalize your interval of convergence!
- If the endpoint series converges, that endpoint is included.
- If the endpoint series diverges, that endpoint is not included.
- Use your “series and tests” (study guides 10.3 to 10.9) toolkit!

Good luck! 🎈

Key Terms to Review (5)

Endpoints

: The endpoints are the two extreme points of a line segment or interval.

Interval of Convergence

: The interval of convergence is the range of x-values for which a power series converges, meaning it approaches a finite value.

Power Series

: A power series is an infinite series that represents a function as an infinite polynomial expression.

Ratio Test

: The ratio test is used to determine whether an infinite series converges absolutely, converges conditionally, or diverges by examining the limit of the ratio of consecutive terms.

Taylor Series

: A Taylor series is an expansion of a function into an infinite sum of terms, where each term represents the contribution from different derivatives of the function at a specific point.

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AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.

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