Video: Find the Maclaurin Series of a Composition between the Exponential Function and the Square Function

Use the Maclaurin series of 𝑒^π‘₯ to express ∫ 𝑒^(π‘₯Β²) dπ‘₯ as an infinite series.

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Video Transcript

Use the Maclaurin series of 𝑒 to the power of π‘₯ to express the integral of 𝑒 to the power of π‘₯ squared with respect to π‘₯ as an infinite series.

The question wants us to use the Maclaurin series of 𝑒 to the power of π‘₯ to represent this integral as an infinite series. We recall the Maclaurin series of 𝑒 to the power of π‘₯ tells us that 𝑒 to the power of π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of π‘₯ to the 𝑛th power divided by 𝑛 factorial. And this is valid for all real numbers π‘₯. We want to use this Maclaurin series for 𝑒 to the power of π‘₯ to represent the integral of 𝑒 to the power of π‘₯ squared with respect to π‘₯.

To do this, we’re going to use the fact that if we have a Maclaurin series for a function 𝑓 of π‘₯, which is valid on the radius of convergence 𝑅. Then we can generate a new power series using this Maclaurin series, which is valid on the absolute value of 𝑔 of π‘₯ is less than 𝑅. Since the question involves 𝑒 to the power of π‘₯ squared, we’ll set our function 𝑔 of π‘₯ to be equal to π‘₯ squared.

So we’ll start with our Maclaurin series for 𝑒 to the power of π‘₯, which we know is true for all real numbers 𝑅, which is equivalent to saying the radius of convergence is ∞. And we’re going to change all of our values of π‘₯ to π‘₯ squared. This gives us that 𝑒 to the power of π‘₯ squared is equal to the sum from 𝑛 equals zero to ∞ of π‘₯ squared raised to the 𝑛th power divided by 𝑛 factorial when the absolute value of π‘₯ squared is less than the radius of convergence for our Maclaurin series of 𝑒 to the π‘₯.

However, we’ve already shown that our Maclaurin series for 𝑒 to the power of π‘₯ has an infinite radius of convergence. So our power series will be true for all real numbers π‘₯. We can simplify the numerator of our summand by noticing that π‘₯ squared all raised to the 𝑛th power is just equal to π‘₯ to the power of two 𝑛. Next, the question wants us to find an expression for the integral of 𝑒 to the power of π‘₯ squared with respect to π‘₯. So we’re going to integrate both sides of our equation with respect to π‘₯. This gives us the integral of 𝑒 to the power of π‘₯ squared with respect to π‘₯ is equal to the integral of the sum from 𝑛 equals zero to ∞ of π‘₯ to the power of two 𝑛 divided by 𝑛 factorial with respect to π‘₯.

And since we’re integrating the infinite sum of a power series, we can switch the order of our integral and our sum. This gives us the sum from 𝑛 equals zero to ∞ of the integral of π‘₯ to the power of two 𝑛 divided by 𝑛 factorial with respect to π‘₯. Next, we’re going to use the fact that if 𝑛 is not equal to negative one and π‘Ž is a constant, then to integrate π‘Ž multiplied by π‘₯ to the 𝑛th power with respect to π‘₯, we add one to our exponent of π‘₯ and then divide by this new exponent. Then, we add our constant of integration 𝑐. Since our sum starts from 𝑛 equals zero, two 𝑛 is never equal to negative one. So our exponent of π‘₯ is never equal to negative one. And 𝑛 is independent of our variable π‘₯. This means we can consider it a constant with respect to π‘₯.

So we can now use our integral rule to integrate each term of our sum. We add one to our exponents giving us π‘₯ to the power of two 𝑛 plus one and then we divide by two 𝑛 plus one. This gives us the sum from 𝑛 equals zero to ∞ of π‘₯ to the power of two 𝑛 plus one divided by 𝑛 factorial multiplied by two 𝑛 plus one. And instead of adding a constant of integration for all of our terms, we’ll just add one constant of integration outside of our sum, which we will call 𝑐.

Therefore, we’ve shown that we can represent the integral of 𝑒 to the power of π‘₯ squared with respect to π‘₯ as the sum from 𝑛 equals zero to ∞ of π‘₯ to the power of two 𝑛 plus one divided by 𝑛 factorial multiplied by two 𝑛 plus one plus a constant of integration 𝑐.

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