Problem

Is it possible to find a power series whose inter…

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Problem 32 Hard Difficulty

Let $ p $ and $ q $ be real numbers with $ p < q. $ Find a power series whose interval of convergence is
(a) $ (p, q) $
(b) $ (p, q] $
(c) $ [p, q) $
(d) $ [p, q] $

Answer

a) $\sum \frac{(x-a)^{n}}{b^{n}}$
b) $\sum \frac{(a-x)^{n}}{n b^{n}}$
c) $\sum \frac{(a-x)^{n}}{n b^{n}}$
d) $\sum \frac{(x-a)^{2 n}}{n b^{2 n}}(-1)^{n}$

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 8

Power Series

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Video Transcript

Okay, So for part A, let's assume that the sum we're working with take this type of form if our interval of convergence is from Pete Acute and there has to be some type of nice symmetry relationship happening there. So if X is equal to P here, we would have p minus a certainly P minus. A cannot be equal that Q minus A. Because then people be equal to Q. But it is possible for P minus A to be equal Tio lips, eh? Minus Q pain. If we saw for the a value that make this happen, then we get that is equal to que plus p over two. Okay, and now we're going to find another term here. Let be be equal to Q minus a. And if B is equal to Cuban us eh, then we get that p minus is minus b. Okay, so now if we consider the sum from I equals one to infinity of X minus a to the end over be the end. We see that for any value, acts between P and Q would have that X minus a divided by B ears less than one in absolute value. So therefore we'd get convergence there. And if we have that X is equal to P, then we would have P minus a which is minus B. So this would be minus B to the n divided by B to the end. So then we'd have minus one to the end, which diverges which is good because we don't want to include p here. And if you look at X equals, Q then we'LL get just some from n equals one to infinity of one, which is also certainly going to diverge. Okay, so this some works for partner and then we just need to make some modifications for part B. So for part B, we can consider the sum from n equals one to infinity of a minus X to the end, divided by in times be the end. Okay, So if X is equal to P, then we have a minus p. Okay, so a minus p is minus P minus a P minus A was minus B. So this is B Okay, so if we plug in X equals P, then we get to the end over B to the end. So then we'd just be summing up from in equals. One to infinity of one over ends would get divergence there. Which is good. We don't want include p here. Okay, so that's good. And what about when X is equal? That Q So then we would have a minus. Q. Okay, Q minus is equal to be so a minus. Q is minus B. So if you plug in X equals Q then we'd have minus B to the end over be the end. So then we would have minus one to the end over in, and this would converge by the alternating signed test. Okay, so indeed, this is going to work here so that'LL work for party and then we do something similar for apart. See, for part C, we just need to do it the other way around. Now we have a minus X to the end, divided by in B to the end. And this is going to work for the same reason, right? We just switched the roles of a an X. So now when we plug in P, we're going to get that alternating son and get convergence. And when we plug in queue, then we'LL get divergence because we'LL get the harmonic Siri's Okay, so that's going to work. And now for D we'LL make some other modification now for D. Well, consider X minus A to the two n So now it doesn't matter if we plug in P. R. Q. Since we're squaring it, it's going to get rid of the minus sign. So plugging in p ork you here, we'LL get the same thing. So that's good, because we want to get convergence for X equals P and for X equals Q. And now we're going to divide this by in a times B to the two n. Okay, so now if we plugged in X equals P R. Q, we would get one over in, which would be divergent. But if we had this minus one to the end, then plugging and pee or Q. We get the alternating sign of one over end, so alternating signed test gives us convergence there, so plug in and peor que works here, plugging in anything larger than Pierre queue. Then the ratio X minus a over B is going to be bigger than one, and then you can see that you'LL get divergence there because the terms will not go to zero

Gabriel R.

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 8

Power Series

Problem

Is it possible to find a power series whose inter…

Problem 32 Hard Difficulty

Let $ p $ and $ q $ be real numbers with $ p < q. $ Find a power series whose interval of convergence is
(a) $ (p, q) $
(b) $ (p, q] $
(c) $ [p, q) $
(d) $ [p, q] $

Answer

a) $\sum \frac{(x-a)^{n}}{b^{n}}$
b) $\sum \frac{(a-x)^{n}}{n b^{n}}$
c) $\sum \frac{(a-x)^{n}}{n b^{n}}$
d) $\sum \frac{(x-a)^{2 n}}{n b^{2 n}}(-1)^{n}$

Related Courses

Calculus: Early Transcendentals

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Discussion

Top Calculus 2 / BC Educators

Anna Marie V.

Heather Z.

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Watch More Solved Questions in Chapter 11

Video Transcript

Calculus: Early Transcendentals

Anna Marie V.

Heather Z.

Caleb E.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

a) $\sum \frac{(x-a)^{n}}{b^{n}}$b) $\sum \frac{(a-x)^{n}}{n b^{n}}$c) $\sum \frac{(a-x)^{n}}{n b^{n}}$d) $\sum \frac{(x-a)^{2 n}}{n b^{2 n}}(-1)^{n}$

Calculus: Early Transcendentals

Anna Marie V.

Heather Z.

Caleb E.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

Anna Marie V.

Heather Z.

Caleb E.

Joseph L.

a) $\sum \frac{(x-a)^{n}}{b^{n}}$
b) $\sum \frac{(a-x)^{n}}{n b^{n}}$
c) $\sum \frac{(a-x)^{n}}{n b^{n}}$
d) $\sum \frac{(x-a)^{2 n}}{n b^{2 n}}(-1)^{n}$