WEBVTT
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In this video, weβll learn how to find a Taylor representation of an arbitrary function by using the Taylor series expansion.
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Weβll consider how we can find the radius of convergence of these series as well as intervals of convergence, primarily using the ratio test for convergence.
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We begin by supposing that π is a function that can be represented by a power series π of π₯ equals π zero plus π one π₯ minus π plus π two π₯ minus π all squared and so on, where the absolute value of π₯ minus π is less than some value capital π
.
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We can begin to work out what ππ, the coefficients, must be in terms of π.
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For example, if we let π₯ be equal to π, we see that π₯ minus π becomes π minus π, which is zero.
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And then we find that π of π equals π zero.
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Okay, well, thatβs great.
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But thereβs nothing else we can do with this function as it stands to help us represent the other coefficients in terms of π.
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So instead, we differentiate our function π with respect to π₯, remembering that we can do so term by term.
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And π zero is a constant so its derivative is zero.
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Then the derivative of π one times π₯ minus π is just π one.
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And we could achieve this result by using the product rule or distributing our parentheses and differentiating as normal.
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Similarly, we find the derivative of π two times π₯ minus π all squared to be two times π two times π₯ minus π.
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The derivative of π three times π₯ minus π all cubed is three π three times π₯ minus π all squared and so on.
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Once again, we let π₯ be equal to π.
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And this time we obtain π prime of π to be equal to π one.
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Now, it follows that we can repeat this process.
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We differentiate our function once again with respect to π₯.
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π double prime of π₯ is two π two plus two times π three times π₯ minus π plus another term, three times four π four times π₯ minus π all squared and so on.
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And this time when we let π₯ be equal to π we find that π double prime of π is equal to two π two.
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Weβll repeat this one more time to help us establish a pattern.
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The third derivative is two times three π three plus two times three times four π four times π₯ minus π plus three times four times five π five times π₯ minus π all squared and so on.
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And letting π₯ be equal to π now, we see that π triple prime of π, the third derivative of our function with respect to π₯, evaluated at π₯ is equal to π is two times three times π three, which we could write alternatively as three factorial π three.
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Now, if we were to continue to different and let π₯ be equal to π, we would find that the πth derivative of our function evaluated at π is two times three times four times five, all the way up to π lots of ππ.
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In other words, itβs π factorial times ππ.
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Weβre going to solve the equation the πth derivative of π evaluated at π equals π factorial ππ for ππ.
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When we do, we find that ππ is equal to the πth derivative of π evaluated at π over π factorial.
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Itβs useful to remember that this formula does remain valid even if π is equal to zero, since zero factorial is just one and the zeroth derivative of π is simply π.
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So we obtain our first theorem.
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If π has a power series representation at π β β that is, if π of π₯ is equal to the sum of ππ times π₯ minus π to the πth power for values of π between zero and infinity β β where the modules of π₯ minus π is less than sum capital π
.
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Then its coefficients are given by the formula ππ equals the πth derivative of π evaluated at π over π factorial.
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Weβre going to substitute this back into our original series.
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We find that if π has a power series expansion at π, then the Taylor series of the function π at π or about π is the sum of the πth derivative of π evaluated at π over π factorial times π₯ minus π to the πth power for values of π between zero and infinity.
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And this can alternatively be written as π of π plus π prime of π over one factorial times π₯ minus π plus π double prime of π over two factorial times π₯ minus π all squared and so on.
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Itβs useful to know that if π is equal to zero, we do have a special case that has its own name.
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Itβs called the Maclaurin series.
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And we wonβt be investigating its properties in this video.
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Letβs now have a look at a couple of examples of how to apply the formula for the Taylor series.
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What are the first four terms of the Taylor series of the function π of π₯ equals the square root of π₯ about π₯ equals four?
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We recall that the Taylor series of a function π about π is given by the sum of the πth derivative of π evaluated at π of π factorial times π₯ minus π to the πth power for values of π between zero and infinity.
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So what do we know about our function?
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Itβs given as π of π₯ equals the square root of π₯, which we might alternatively say is π₯ to the power of one-half.
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We want to find the Taylor series about π₯ equals four, so weβre going to let π be equal to four.
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Using the second form of the Taylor series, we find that π of π₯ is equal to π of four plus π prime of four over one factorial times π₯ minus four plus π double prime of four over two factorial times π₯ minus four squared plus π triple prime of four over three factorial times π₯ minus four cubed.
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We can work out π of four fairly easily.
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Weβll substitute four into our original function.
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But to be able to work out π prime of four, π double prime of four, and π triple prime of four, weβre going to need to differentiate our function π of π₯ with respect to π₯ three times.
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We said that π of π₯ was equal to π₯ to the power of one-half and we recall that to differentiate a polynomial term, we multiply the entire term by the exponent and then reduce that exponent by one.
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This means π prime of π₯, the first derivative of π is a half π₯ to the power of negative one-half.
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π double prime of π₯ is negative a half times a half π₯ to the power of negative three over two, which simplifies to negative one-quarter π₯ to the power of negative three over two.
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Finally, π triple prime of π₯ is negative three over two times negative a quarter π₯ to the power of negative five over two, which is three-eighths π₯ to the power of negative five over two.
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Weβre now going to substitute π₯ equals four into our functions for π of π₯, π prime of π₯, π double prime of π₯, and π triple prime of π₯.
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π of four is the square root of four; itβs simply two.
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π prime of four is a half times four to the power of negative one-half which is a quarter.
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π double prime of four is negative one-quarter times four to the power of negative three over two, which is negative one over 32.
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And π triple prime of four is three-eighths times four to the power of negative five over two, which is three over 256.
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Our final job is simply to substitute these into our expansion.
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When we do, we find the first four terms of the Taylor series of the function π of π₯ equals the square root of π₯ about π₯ equals four to be two plus a quarter times π₯ minus four minus one over 64 times π₯ minus four squared plus one over 512 times π₯ minus four cubed.
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Weβll have a look at another example of this type.
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Consider the function π of π₯ equals cos of π₯.
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Find the Taylor series expansion of π of π₯ equals cos of π₯ at π₯ equals π and write the Taylor series expansion of π of π₯ in sigma notation.
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And we recall that the Taylor series expansion of a function about π₯ equals π is given by π of π₯ equals π of π plus π prime of π over one factorial times π₯ minus π plus π double prime of π over two factorial times π₯ minus π all squared and so on.
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In this question, our function is cos of π₯.
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And weβre looking to find the Taylor series expansion at π₯ equals π.
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So weβre going to let π be equal to π.
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And so we see that π of π₯ here is equal to π of π plus π prime of π over one factorial times π₯ minus π and so on.
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Now, we can quite easily evaluate π of π.
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We would substitute π into the function cos of π₯.
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But what about π prime of π, π double prime of π, and so on?
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Well, weβre going to differentiate our function with respect to π₯.
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We might recall the first derivative of cos of π₯ to be negative sin of π₯.
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Then to get the second derivative, we differentiate negative sin π₯ with respect to π₯ and we get negative cos π₯.
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Differentiating once more, we find that π triple prime of π₯ is sin π₯.
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And in fact, weβll repeat this process one more time because when we differentiate sin π₯, we get cos π₯ again.
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And you might notice, we have a cycle.
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The fifth derivative of π will be negative sin π₯, the sixth derivative will be negative cos π₯, and so on.
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And letβs use all of these to evaluate π of π, π prime of π, π double prime of π, and so on and look for a pattern.
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π of π is cos π, which is negative one.
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π prime of π is negative sin π, which is zero.
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π double prime of π is negative cos π, which is one.
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And π triple prime of π is sin π, which is once again zero.
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It, of course, follows that the fourth derivative of π will be cos of π again, which is negative one.
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Letβs substitute all of these back into our Taylor series expansion.
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Now of course, every other term is going to be equal to zero as weβve seen.
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And so we find the Taylor series expansion to be negative one plus a half times π₯ minus π all squared minus one over 24 times π₯ minus π to the fourth power plus one over 720 times π₯ minus π to the sixth power and so on.
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The second part of this question asks us to write the Taylor series expansion of π of π₯ in sigma notation.
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And so letβs see if thereβs a way we can find a pattern.
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We saw that π prime of π was equal to zero, while π triple prime of π is equal to zero.
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And extending the pattern, we would see that the fifth derivative, the seventh derivative, and so on would also be equal to zero.
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And so our derivative alternate between negative one, one, negative one, and so on.
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Similarly, our denominators are factorials of ascending even numbers each time this even number is the same as the exponent.
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So letβs define that even number as two π.
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Then the denominator is two π factorial and the exponent of π₯ minus π is two π.
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To achieve alternating powers of negative one, we write negative one times π plus one.
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That means when π is zero, negative one to the power of one is negative one, when π is one, negative one to the power of two gives us one, and so on.
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And so we find that π of π₯ equals the sum of negative one to the power of π plus one times π₯ minus π to the power of two π over two π factorial for values of π between zero and infinity.
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Weβre also able to find the radius of convergence of a Taylor series.
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We say that there exists the number capital π
such that the power series will converge for absolute values of π₯ minus π less than π
and will diverge for absolute values of π₯ minus π greater than π
.
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And this number capital π
is called the radius of convergence of the series.
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And note that the series may or may not converge, if the absolute value of π₯ minus π is equal to π
.
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We then say that the interval of convergence is an interval containing all the values of π₯, for which the series converges.
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Now, this may or may not include the end points of the interval and even be the whole set of real numbers.
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And we say that if we know the radius of convergence of a power series is π
, then for values of π₯ greater than π minus π
and less than π plus π
, the power series converges.
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For values of π₯ less than π minus π
and greater than π plus π
, the power series diverges.
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It also follows that the power series will always converge for π₯ is equal to π.
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Letβs have a look at an application of this theory.
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Consider the function π of π₯ equals π to the power of two π₯.
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Find the Taylor series representation of π about π₯ equals three.
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And in fact, there are two further parts of this question that weβll consider in a moment.
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We begin by recalling that the Taylor series of a function π about π is given by the sum of the πth derivative of π evaluated at π over π factorial times π₯ minus π to πth power values of π between zero and infinity.
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And so what do we know about our function?
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Well, itβs given by π of π₯ equals π to the power of two π₯.
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We want to find the Taylor series about π₯ equals three.
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So we let π be equal to three.
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Then we see that π of π₯ is equal to the sum of the πth derivative of π evaluated at three of π factorial times π₯ minus three to the πth power for values of π between zero and infinity.
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This is, of course, equal to π of three plus π prime of three over one factorial times π₯ minus three plus π double prime of three over two factorial times π₯ minus three all squared and so on.
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Itβs quite clear that weβre going to need to work out π of three, π prime of three, π double prime of three, and possibly look for a pattern.
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π of three is quite straightforward.
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Itβs π to the power of two times three which is, of course, π to the power of six.
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So letβs next look for the derivative of π of π₯.
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The derivative of π to the power of two π₯ is two π to the power of two π₯.
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So π prime of three must be two π to the power of two times three, which is two π to the power of six.
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Weβll differentiate once again.
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This time we get four π to the power of two π₯.
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And π double prime of three is for π to the power of two times three, which is four π to the sixth power.
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Now actually, we can see that the πth derivative of π of π₯ is equal to two to the πth power times π to the power of two π₯.
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And this means the πth derivative of π evaluated at three is two to the πth power times π to the sixth power.
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Our Taylor series expansion using sigma notation is, therefore, the sum of two to the πth power times π to the sixth power times π₯ minus three to the πth power all over π factorial for values of π between zero and infinity.
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The second part of this question asks us to find the interval of convergence of the Taylor series representation of π about π₯ equals three.
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We say that the intervals of all π₯s for which the power series converges is called the interval of convergence of the series.
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Now actually, we know that the power series will converge for π₯ equals π.
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So here π₯ equals three.
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But thatβs all we know right now.
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Weβll use the ratio test to determine the rest of the π₯s for which the power series will converge.
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This says suppose that we have some series, the sum of ππ, and we defined π to be equal to the limits as π approaches infinity of ππ plus one over ππ.
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If π is less than one, the series is absolutely convergent and hence convergent.
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This is the scenario weβre interested in.
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So our π must be the limit as π approaches infinity of the absolute value of two to the power of π plus one times π to the sixth power times π₯ minus three to the power of π plus one over π plus one factorial divided by two to the πth power times π to the sixth power times π₯ minus three to the πth power over π factorial.
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Now of course, when we divide by a fraction, itβs the same as multiplying by the reciprocal of that fraction.
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So we can rewrite our limit as shown.
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Then we notice we can perform some simplification.
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We can cancel out π to the power of six and two to the power of π.
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Similarly, we can cancel π₯ minus three to the power of π.
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And if we, finally, recall that π plus one factorial is the same as π plus one times π factorial, we see that we can cancel by dividing through by π factorial.
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Well, this all simplifies really nicely.
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We see that π is equal to the limit as π approaches infinity of the absolute value of two times π₯ minus three over π plus one.
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Now in fact, the expression two times π₯ minus three is independent of π.
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So we can factor this, remembering that we must write the absolute value of two times π₯ minus three.
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And so π is equal to the absolute value of two times π₯ minus three times the limit as π approaches infinity of the absolute value of one over π plus one.
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Now, as π gets larger, one over π plus one gets smaller.
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So the limit approaches zero.
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This means π is equal to zero.
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This is of course less than one.
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And it doesnβt change no matter the value of π₯.
00:15:14.220 --> 00:15:19.270
And so we say our series is absolutely convergent for all values of π₯.
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Its interval of convergence is negative infinity to infinity.
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The third part of this question asks, what is the radius of convergence of the Taylor series representation of π about π₯ equals three?
00:15:31.130 --> 00:15:43.610
Recall that there exist a number capital π
, which we call the radius of convergence, so that the power series will converge for absolute values of π₯ minus π less than π
and will diverge for absolute values of π₯ minus π greater than π
.
00:15:43.900 --> 00:15:47.350
Well here, weβve seen that our power series is absolutely convergent.
00:15:47.490 --> 00:15:50.170
And thereβs a special case for the radius of convergence.
00:15:50.250 --> 00:15:54.450
In cases such as these, we say that the radius of convergence is infinity.
00:15:54.640 --> 00:15:57.300
Capital π
is, therefore, equal to infinity.
00:15:58.960 --> 00:16:15.470
In this video, weβve seen that if π has a power series expansion at π, then the Taylor series of the function π at π or about π is the sum of the πth derivative of π evaluated at π over π factorial times π₯ minus π to the πth power for values of π between zero and infinity.
00:16:15.610 --> 00:16:26.830
We saw that capital π
represents the radius of convergence such that the power series will converge for absolute values of π₯ minus π less than π
and will diverge for absolute values of π₯ minus π greater than π
.
00:16:27.080 --> 00:16:32.230
And finally, we learned that we can use the ratio test or other suitable tests to find the intervals of convergence.
00:16:32.310 --> 00:16:37.710
The power series converges for values of π₯ greater than π minus π
and less than π plus π
.
00:16:37.760 --> 00:16:41.330
And it follows that the power series will always converge when π₯ is equal to π.