Taylor series and interval of convergence

c. ...

Question

Taylor series and interval of convergence

c. Determine the interval of convergence of the series.

f(x)=3ˣ, a=0

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says determine the interval of convergence of the the co series for the function F of X, which is equal to 5 rates to the power X, and we're given 4 possible choices as our answers. For choice A, we have the open interval from minus infinity to infinity. For choice B, we have the open interval from -1 to 1. For choice C, we have the closed interval from -1 to 1. And for choice D, we have the open interval from -3 to 1. Now the first thing we're going to do is actually rewrite our function. So we're gonna have F of X. And this is going to be equal to, I'm going to rewrite this in terms of the exponential. So we'll have E raised to the quantity of X multiplied by the natural log of 5. So here we're just rewriting our function here with a base of E instead of a base of 5. And the reason why we want to do this is now we can use the McLaurin series expansion for our exponential. And so recall that that is going to be equal to the sum. Of N equal to 0 to infinity of the quantity of X multiplied by the natural log of 5, in quantity raised to the power N divided by N factorial. And we can just simplify this as being equal to the sum of N equal to 0 to infinity of the quantity of the natural log of 5 in quantity rates to the power N divided by N factoral multiplied by X rated to the power n. And so, in order to determine our interval of convergence, we're going to use the ratio test. As a recall for the ratio test that we want to calculate L, which is going to be equal to the limit, as N approaches infinity of the absolute value of the quantity of a n + 1 divided by a N and quantity, where in our case a subn is going to be equal to the quantity of the natural log of 5 in quantity rates to the power n divided by In factoral multiplied by X-rays to the power N. So using the expression for ace of N, we see that L is going to be equal to the limit as N approaches infinity of the absolute value. Of the quantity of the quantity of the natural log of 5 in quantity raises to the quantity of N + 1 in quantity. Divided by the quantity of n + 1 in quantity factorial multiplied by x rates to the quantity of n + 1 in quantity, and that is all divided by the quantity of the quantity of the natural log of 5 in quantity raised to the power n divided by in factoral multiplied by x raised to the power n in quantity. So now we need to do is to simplify this expression. And so simplifying this expression, we see that this is going to be equal to the limit. As in approaches infinity of the absolute value. Of the natural log of 5, divided by the quantity of N + 1 in quantity, multiplied by X in quantity. Now, as n approaches infinity, the natural log of 5 divided by the quantity of n + 1 n quantity is going to approach 0. So that means that this limit is just going to be equal to 0. And recall that if our limit here is less than 1, then our series converges. And since our limit here doesn't have an X dependence, that means it's going to converge to 0 for all values of X. That means that our interval of convergence is going to be the open interval from minus infinity to infinity. And so that here is the answer that we are looking for for this problem. And if we compare our answer here to our answer choices, we see that the correct answer choice corresponds to answer choice A. So I hope that this explanation has been useful, and I'll see you in the next video.

Taylor series and interval of convergence

c. Determine the interval of convergence of the series.

f(x)=3ˣ, a=0

Key Concepts

Taylor Series Expansion

Radius and Interval of Convergence

Ratio Test for Convergence

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