Differentiating and integrating power series Find the power se...

Question

Differentiating and integrating power series Find the power series representation for g centered at 0 by differentiating or integrating the power series for f (perhaps more than once). Give the interval of convergence for the resulting series.

g(x) = − 1/(1 + x)² using f(x) = 1/(1 + x)

g(x) = − 1/(1 + x)² using f(x) = 1/(1 + x)

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says let PXX be equal to 1 divided by the quantity of 1 + 2 X in quantity squared. Find the power series representation for PFX centered at 0 by differentiating the power series for Q of X, which is equal to 1 divided by the quantity of 1 + 2 X in quantity. What is the interval of convergence? Now, if you look at our function Q of X, we see that it's very similar to the result that we get from a geometric series. So, we call for a geometric series that the sum. Of K equal to 0 to infinity. Of R rated to the power K is going to be equal to 1 divided by the quantity of 1 minus R in quantity if the absolute value of R is less than 1. Now, if we look at our function Q of X, Which is equal to 1 divided by the quantity, and here we're going to rewrite our denominator as 1 minus the quantity of -2 X in quantity. And by rewriting this way, we notice that this looks like a Geometric series with R is equal to minus 2 X. So that means that this function is going to be equal to the sum. Of K equal to 0 to infinity. Of the quantity of minus 2 X in quantity raised to the power K. And this is valid if the absolute value of minus 2 X. Is less than one. Or if we solve this for the absolute value of X, that means that the absolute value of X has to be less than 1/2. So next we need to differentiate our function Q ofX, so we need to actually differentiate the power series. So that means we're gonna be looking for Q, which is going to be the derivative of Q of X with respect to X, which is going to be equal to the derivative with respect to X of the sum of K equal to 0 to infinity. Of the quantity of minus 2 in quantity rates to the power K multiplied by X rates to the K. So here we've just broken up. The quantity that was raised to the power K into two separate quantities, one that is a constant and one that is a variable X. So when we take the derivative of this quantity, we see that Q X is going to be equal to the sum of K equal to 1 to infinity. Of quantity of minus 2 in quantity rates of the power k. Multiplied by K, multiplied by X raises to the quantity of K minus 1 in quantity. So since the first term in our series expansion for Q of X is a constant, when K is equal to 0, it's just the first term is just going to be 1. When you take that derivative, it is 0. So that's why the summation for our derivative here starts at K equal to 1. Now we need to relate our Q of x to our function P. Now if we look at q of X, we see that that is going to be equal to, and if we just differentiate our function, that's going to be -2 divided by the quantity of 1 + 2 X in quantity squared, which is just going to be equal to minus 2 multiplied by p of X. So this means that P X It's just going to be equal to -12 multiplied by Q X. So substituting in our series expansion for Q X, we see that P of X is going to be equal to minus 12 multiplied by the sum. Of K equal to 1 to infinity of the quantity of minus 2 in quantity rates of the power K multiplied by K multiplied by X rates of the quantity of K minus 1. Now we can distribute the minus 1 half factor. In doing so, we see that this is going to be equal to the sum of K equal to 1 to infinity. Of the quantity of 1 in quantity raised to the quantity of K + 1. Multiplied by 2, raise to the quantity of K minus 1. Multiplied by in quantity, multiplied by K, multiplied by X, raises the quantity of K minus 1. Now, what we can do is reindex this by setting N equal to K minus 1. So with N equal to K minus 1, that means that K is going to be equal to N + 1. So, reindexing our summation here, we see that P of X is going to be equal to the sum. Of N equal to 0. To infinity Of the quantity of 1 in quantity raised to the quantity of N + 2. In quanti Multiplied by 2 Raised to the power n multiplied by the quantity of N + 1 in quantity multiplied by X raises to the power n. Now, since -1 squad is just equal to 1, we can simplify this further, and so this is just going to be equal to the sum. Of N equal to 0. To infinity of the quantity of minus 2 in quantity raised to the power n multiplied by the quantity of n + 1 in quantity. Multiplied by X, raised to the power N, and this is still only valid if the absolute value of X is less than 12. So this here is the series expansion that we're looking for for this problem. Now, we do need to find the interval of convergence. And since the absolute value of access to be less than 12, that means we actually need to check the endpoints, so we need to check two different cases. So we need to check the case. When X is equal to 1/2 and see if our series converges. So if X is equal to -12, we will have the sum of N equal to 0 to infinity of the quantity of minus 2 in quantity raised to the power n multiplied by the quantity of N + 1 in quantity multiplied by the quantity of minus 1/2. In Kwati raise to the power in. Now we can combine the first and last terms in our product here, and this is just going to be equal to. The sum Of N equal to 0 to infinity of 1 raised to the power n. Multiplied by the quantity of N + 1, which is just going to be equal to the sum. Of N equal to 0 to infinity of quantity of n plus 1 in quantity. Now, as it approaches infinity, that means the argument there of our summation is going to also approach infinity. So this here is actually going to diverge. And since this summation diverges, that means that we will not include this end point in our interval of convergence. Next, we need to look at the case. When X is equal to 1/2, this is our other end point. And here, when we substitute an X equal to 12, we have the sum. Of n equal to 0 to infinity. Of the quantity of minus 2 and the quantity raised to the power N multiplied by the quantity of N + 1 in quantity, multiplied by the quantity of 1/2, in quantity raised to the power N. And so, again, we can supply this expression, and this is just going to be equal to the sum. Of N equal to 0 to infinity of the quantity of minus 1 in quantity raised to the power N, multiplied by the quantity of N + 1 in quantity. Now here we have an alternating series, but the quantity that is multiplying the quantity of -1 and quantity rates of the power in, which is N + 1. Diverges to purchase infinity as in approaches infinity. So that means that this series is also going to diverge, and since it diverges, That means that we're not going to include this endpoint in our interval of convergence. So that means our interval of convergence for our series is going to be the open interval from -12 to 1/2. So that's going to be the answer to the second part of this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Differentiating and integrating power series Find the power series representation for g centered at 0 by differentiating or integrating the power series for f (perhaps more than once). Give the interval of convergence for the resulting series.

g(x) = − 1/(1 + x)² using f(x) = 1/(1 + x)

Key Concepts

Power Series Representation

Term-by-Term Differentiation and Integration of Power Series

Interval of Convergence

Watch next

Differentiating and integrating power series Find the power series representation for g centered at 0 by differentiating or integrating the power series for f (perhaps more than once). Give the interval of convergence for the resulting series.g(x) = − 1/(1 + x)² using f(x) = 1/(1 + x)

Key Concepts

Power Series Representation

Term-by-Term Differentiation and Integration of Power Series

Interval of Convergence

Watch next

Differentiating and integrating power series Find the power series representation for g centered at 0 by differentiating or integrating the power series for f (perhaps more than once). Give the interval of convergence for the resulting series.

g(x) = − 1/(1 + x)² using f(x) = 1/(1 + x)