Question Video: Transcendental Functions as Power Series | Nagwa Question Video: Transcendental Functions as Power Series | Nagwa

Question Video: Transcendental Functions as Power Series Mathematics • Higher Education

Consider π(π₯) = π^(π₯). Find the Maclaurin series of π(π₯). Use the first three terms of this series to find an approximate value of π^(0.4) to 2 decimal places.

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Video Transcript

Consider π of π₯ equals π to the π₯ power. Thereβs two parts to this question. The first part says, find the Maclaurin series of π of π₯. And the second part says, use the first three terms of this series to find an approximate value of π to the power of 0.4 to two decimal places.

Letβs start by writing out the Maclaurin series expansion for a general function π of π₯. So, letβs begin by finding some of the derivatives of our function π of π₯. π of π₯ equals π to the π₯ power, so π of zero equals π to the zero power, which is one. We differentiate π to the π₯ to get the first derivative. So, we recall the fact that the derivative of π to the power of π₯ with respect to π₯ is just π to the power of π₯. So, π prime of π₯ equals π to the π₯ power. And so, it follows that π prime of zero is one. And in fact, we see that all the derivatives of π of π₯ are π to the power of π₯.

So, now letβs make the substitutions. As our function is defined as π of π₯, we can replace π with π and begin to make some substitutions. We found that π of zero is one. And we found that π prime of zero is one. And we found that π double prime of zero is one. And π triple prime of zero is one. And this will continue in this way for all the future terms. We can bring the π₯, the π₯ squared, and the π₯ cubed, and so on on to the top of the fraction. And then, we can spot that we can actually write this as the sum from π equals zero to β of π₯ to the πth power over π factorial. So, this gives us the Maclaurin series expansion of π to the π₯ power.

The second part of this question asks us to use the first three terms of this series to find an approximate value of π to the 0.4 power to two decimal places. Here are the first three terms of our series. And we want to approximate π to the π₯ when π₯ equals 0.4. So, we replace π₯ with 0.4, and we find that this is one plus 0.4 over one factorial. But we know that one factorial is just one. So, this is 0.4. And 0.4 squared is 0.16. And thatβs over two factorial, which is two multiplied by one, which is two. But 0.16 over two is just 0.08. So, adding these up, we find that our approximation is 1.48. We could then use a calculator to check how good our approximation is. Using a calculator gives us the π to the power of 0.4 is 1.4918 to four decimal places. So, we can see that even though we only use the first three terms of our series, our approximation is actually quite good.

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