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Find the radius of convergence for the Taylor series of π of π₯ is equal to one divided by two π₯ plus one about π₯ is equal to one.
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Weβre given a rational function π of π₯, and weβre asked to find the radius of convergence of the Taylor series for this rational function π of π₯ about π₯ is equal to one.
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We have several different methods we could use to answer this question.
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For example, we could directly find the Taylor series for π of π₯ about π₯ is equal to one by using either the chain rule or the general power rule.
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However, to use this method, we would need to find an expression for the πth derivative of π evaluated at one.
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And then we would need to use something like the ratio test to evaluate the radius of convergence.
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So this method will require a lot of work.
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However, there is an easier method involving what we know about infinite geometric series.
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We need to recall the following facts about the sum of an infinite geometric series.
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If our initial value is π and ratio of successive terms is π, then the sum from π equals zero to β of π times π to the πth power will be equal to π divided by one minus π, provided the absolute value of π is less than one.
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And if the absolute value of π is greater than or equal to one, then this series will be divergent.
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And itβs also worth pointing out, technically, this assumes the value of π is nonzero.
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We want to use this to find the Taylor series of our function π of π₯ centered around π₯ is equal to one.
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The first thing we need to remember is our Taylor series being centered around π₯ is equal to one means that our power series needs to be in terms of π₯ minus one.
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So what we want to do is rewrite our function π of π₯ in the form π divided by one minus π.
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However, π needs to include π₯ minus one because then our power series will have powers of π₯ minus one.
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So letβs discuss how weβre going to do this.
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Well, we want to write π₯ minus one in our dominator.
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If we were to just write this directly, we would get two times π₯ minus one.
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Of course if we distribute this, we see we get two π₯ minus two.
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However, our denominator is supposed to be two π₯ plus one.
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So of course, we need to add an extra value of three.
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So we have one over two π₯ plus one is equal to one over two times π₯ minus one plus three.
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Remember though, weβre trying to write this in the form π divided by one minus π.
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To make this easier to see, letβs start by rearranging the two terms in our denominator.
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This gives us the following expression.
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Now, we see in our denominator we need a constant one.
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So weβre going to need to take a factor of three outside of our denominator.
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Taking out a factor of three from our denominator, our new denominator will be three times one plus two-thirds multiplied by π₯ minus one.
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And now weβre almost in the required form.
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However, instead of dividing by three in our denominator, weβll instead move one-third into our numerator.
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So by doing this, weβve rewritten π of π₯ to be equal to one-third divided by one plus two-thirds times π₯ minus one.
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And now this is exactly in the required form.
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Our value of π will be equal to one-third, and our value of π will be equal to two-thirds times π₯ minus one.
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All we need to do then is substitute our value of π and π into our formula for the sum of an infinite geometric series.
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This gives us that π of π₯ is equal to the sum from π equals zero to β of one-third times two-thirds multiplied by π₯ minus one all raised to the πth power, which of course will be convergent provided the absolute value of our ratio two-thirds times π₯ minus one is less than one and will be divergent if the absolute value of two-thirds times π₯ minus one is greater than or equal to one.
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So letβs discuss exactly what weβve just done.
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We found a power series representation for our function π of π₯.
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And we can see itβs centered at π₯ is equal to one.
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But we know all power series representations centered about a value for a function must be equal to their Taylor series.
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So this must be the Taylor series of our function π of π₯ centered at π₯ is equal to one.
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So we just need to find the radius of convergence of this series.
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And in fact, we already have information about the radius of convergence of this series.
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We can see that our power series converges if the absolute value of two-thirds times π₯ minus one is less than one and will diverge if the absolute value of two-thirds times π₯ minus one is greater than or equal to one.
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So all we need to do is rearrange these inequalities to help us find the radius of convergence.
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Weβll start with our first inequality.
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We know two-thirds is a positive constant.
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So we can take it just outside of our absolute value sign.
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But two-thirds is just a positive constant.
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So we can just divide through by two over three or alternatively multiply through by three over two.
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So we were able to rewrite this inequality as the absolute value of π₯ minus one must be less than three over two.
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And we can do exactly the same thing to find the values of π₯ where this is divergent.
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We see our power series is divergent when the absolute value of π₯ minus one is greater than or equal to three over two.
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And this is exactly what it means to say the radius of convergence of our power series is three over two.
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Therefore, by using what we know about geometric series, we were able to show the Taylor series of π of π₯ is equal to one over two π₯ plus one about π₯ is equal to one has a radius of convergence three over two.