Question Video: Integrating a Power Series Representation of a Function Mathematics • Higher Education

For the given function π(π₯) = tanβ»ΒΉ (2π₯), find a power series representation for π by integrating the power series for πβ².

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Video Transcript

For the given function π of π₯ is equal to the inverse tan of two π₯, find a power series representation of π by integrating the power series for π prime.

The question is asking us to integrate the power series for the function π prime so that we can find the power series representation of our function π of π₯ is equal to the inverse tan of two π₯. So letβs start by calculating what our function π prime of π₯ is. We know that the derivative with respect to π₯ of the inverse tan of ππ₯ is just equal to π divided by one plus π squared π₯ squared. We can use this to calculate that the function π prime of π₯ is equal to two divided by one plus two squared π₯ squared. And then, we can just simplify two squared π₯ squared to just be four π₯ squared.

Next, the question tells us that we should find the power series for our function π prime of π₯. And we can do this by using what we know about geometric series. We know that if the absolute value of our ratio π is less than one, then the sum from π equals zero to β of π multiplied by π to the πth power is just equal to π divided by one minus π. Next, we can notice that our function π prime of π₯ is already in the form π divided by one minus π if we set π to equal two and π to equal negative four π₯ squared.

So if we substitute this into what we know about geometric series, we get that if the absolute value of negative four π₯ squared is less than one, then we know that the sum from π equals zero to β of two multiplied by negative four π₯ squared all raised the πth power is equal to two divided by one minus negative four π₯ squared, where we can notice that our denominator, one minus negative four π₯ squared, is just one plus four π₯ squared. So this is equal to two divided by one plus four π₯ squared, which is equal to π prime of π₯.

So weβve now found the power series for our function π prime. That is, itβs the sum from π equals zero to β of two multiplied by negative four π₯ squared raised to the πth power. So we can write the power series for our function π prime of π₯ into our top equation. Now, weβre ready to simplify this power series. Letβs start by distributing the exponent π over our parentheses. This gives us negative four to the πth power multiplied by π₯ squared to the πth power.

Next, we can rewrite negative four as negative one multiplied by two squared. We can then distribute the exponent π over our first set of parentheses to get negative one to the πth power multiplied by two squared to the πth power. Finally, we can evaluate the exponents in our second and third set of parentheses, giving us a final answer of negative one to the πth power multiplied by two to the power of two π multiplied by π₯ to the power of two π. This gives us a new sum from π equals zero to β of two multiplied by negative one to the πth power multiplied by two to the power of two π multiplied by π₯ to the power of two π.

Now, we can simplify this further by noticing that two multiplied by two to the power of two π is just equal to two to the power of two π plus one, giving us the sum from π equals zero to β of two to the power of two π plus one multiplied by negative one to the πth power multiplied by π₯ to the power of two π. We now recall that the integral of our derivative function π prime of π₯ with respect to π₯ is equal to the function π of π₯ up to a constant of integration πΆ.

At this point, we substitute our power series representation for our derivative function π prime of π₯. At this point, weβre going to move the integral symbol inside our sum. And we know we can do this because weβre integrating a power series. At this point, we can notice that two to the power of two π plus one and negative one to the πth power are a constant with respect to π₯. So integrating the term inside our summand is equivalent to just integrating π multiplied by π₯ to the power of two π with respect to π₯, which we know is equal to π multiplied by π₯ to the power of two π plus one divided by two π plus one plus our constant of integration πΆ.

Instead of adding a constant of integration for every single term in our sum, weβre just going to add one right at the end. Evaluating the integral inside our sum gives us πΆ plus the sum from π equals zero to β of two to the power of two π plus one multiplied by negative one to the πth power multiplied by π₯ to the power of two π plus one divided by two π plus one.

We now have a power series representation for our function π of π₯. However, it still has an unknown variable which weβve labelled πΆ. To eliminate this unknown variable, we recall that this is a power series for our function π of π₯, which is equal to the inverse tan of two π₯. Therefore, if we were to substitute, say, π₯ is equal to zero into our function, we would get the inverse tan of zero is equal to π evaluated at zero is equal to our power series where we substitute π₯ equals zero.

So πΆ plus the sum from π equals zero to β of two to the power of two π plus one multiplied by negative one to the πth power multiplied by zero to the power of two π plus one divided by two π plus one. And we know we can justify using the substitution since itβs true if the size of negative four π₯ squared is less than one, which is true for a substitution π₯ is equal to zero. We see that, in this infinite sum, every term is being multiplied by zero. So this is just a sum of zeros and we can conclude is equal to zero.

Therefore, we can conclude that the inverse tan of zero is equal to πΆ. And of course, we know the inverse tan of zero evaluates to just equal to zero. Therefore, what we have shown is a power series representation of the inverse tan of two π₯ is equal to the sum from π equals zero to β of two to the power of two π plus one multiplied by negative one to the πth power multiplied by π₯ to the power of two π plus one divided by two π plus one.