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For the given function π of π₯ equals the natural log of one plus two π₯, find a power series representation for π by integrating the power series for π prime.
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We donβt have a nice power series expansion for the function π of π₯ equals the natural log of one plus two π₯.
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But we should notice that the derivative of π, π prime of π₯, is two over one plus two π₯.
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So weβre going to start with an equation that weβve seen before.
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That is, one over one minus π₯ is equal to one plus π₯ plus π₯ squared plus π₯ cubed, and so on.
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And we write this as the sum of π₯ to the πth power for values of π between zero and β.
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We write our derivative as two times one over one plus two π₯.
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And then weβre going to use the equation weβve seen before, replacing π₯ with negative two π₯.
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And this is because we want it in the form one minus something.
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And one minus negative two π₯ gives us the one plus two π₯ weβre after.
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We can therefore use this equation to write one over one plus two π₯ as one plus negative two π₯ plus negative two π₯ squared plus negative two π₯ cubed, and so on.
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Which can of course be written as the sum of negative two π₯ to the πth power for values of π between zero and β.
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π prime of π₯ is two times one over one plus two π₯.
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So thatβs two times the sum of negative two π₯ to the πth power for values of π between zero and β.
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Two is independent of π.
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So weβre going to take it inside the sum.
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And weβre going to rewrite negative two π₯ as two times negative one times π₯.
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We then distribute this exponent across each term.
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And we see that we have the sum of two times two to the πth power times negative one to the πth power times π₯ to the πth power.
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Two times two to the πth power can be written as two to the power of π plus one.
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And now we have our expression for π prime of π₯.
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Weβre trying to find a power series representation for π.
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So we recall that we can achieve this by integrating our expression for π prime of π₯.
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And we can do so by integrating each individual term in the series.
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Thatβs called term-by-term integration.
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Here thatβs the integral of the sum of two to the power of π plus one times negative one to the πth power times π₯ to the πth power between π equals zero and β with respect to π₯.
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And then of course since weβre dealing with a power series, we can write this as the sum of the integrals.
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So how are we going to integrate two to the power of π plus one times negative one to the πth power times π₯ to the πth power?
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Well, no matter the value of π, two to the power of π plus one times negative one to the πth power is a constant.
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This means we can take it outside of the integral and focus on integrating π₯ to the πth power itself.
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Now when we integrate π₯ to the πth power, we know that π is positive.
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So we simply add one to the exponent and then divide by that new number.
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So we get π₯ to the power of π plus one over π plus one.
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And of course we had that constant of integration πΆ, which is outside the summation.
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So how do we find that constant of integration πΆ?
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Well, letβs go back to π of π₯.
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Weβre told that itβs equal to the natural log of one plus two π₯.
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And we know that if we let π₯ be equal to zero, we get quite a nice value for π of zero.
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Itβs the natural log of one plus two times zero, which is the natural log of one, which is of course zero.
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By replacing π of π₯ with zero and π₯ with zero, we see that zero is equal to the sum of two to the power of π plus one times negative one to the πth power plus zero to the power of π plus one over zero plus one plus πΆ.
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Now zero to the power of π plus one over zero plus one is always zero.
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And so we have the sum of zeros, which is zero.
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And we find the constant of integration itself is zero.
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And so by integrating the power series for π prime, we found a power series representation for π.
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Itβs the sum of two to the power of π plus one times negative one to the πth power times π₯ to the power of π plus one over π plus one.