Video: Finding a Function Given the Maclaurin Series Expansion

A function 𝑓 has a Maclaurin series given by π‘₯Β³ βˆ’ (π‘₯⁢/2) + (π‘₯⁹/3) + ... + (βˆ’1)^𝑛 (π‘₯^(3𝑛)/𝑛) + .... Which of the following is an expression for 𝑓(π‘₯)? [A] sin 3π‘₯ [B] ln (1 + π‘₯Β³) [C] 𝑒^(3π‘₯) βˆ’ cos π‘₯ [D] 𝑒^(3π‘₯) + cos π‘₯

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Video Transcript

A function 𝑓 has a Maclaurin series given by π‘₯ cubed minus π‘₯ to the sixth power over two plus π‘₯ to the ninth power over three plus all the way up to negative one to the 𝑛th power times π‘₯ to the power of three 𝑛 over 𝑛. Which of the following is an expression for 𝑓 of π‘₯? Is it (a) sin of three π‘₯, (b) the natural log of one plus π‘₯ cubed, (c) 𝑒 to the power of three π‘₯ minus cos of π‘₯, or (d) 𝑒 to the power of three π‘₯ plus cos of π‘₯?

Remember, the Maclaurin series is a special case of the Taylor series. It’s the Taylor series of a function 𝑓 centered or about zero. The Taylor series becomes 𝑓 of π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of the 𝑛th derivative of 𝑓 evaluated at zero over 𝑛 factorial times π‘₯ to the 𝑛th power. And this can be written as 𝑓 of zero plus 𝑓 prime of zero over one factorial times π‘₯ plus 𝑓 double prime of zero over two factorial times π‘₯ squared and so on. So one thing we could do is simply evaluate the Maclaurin series for each of our functions. Let’s begin by letting 𝑓 of π‘₯ be equal to sin of three π‘₯. We’re going to need to differentiate this function a few times. So let’s begin by finding the first derivative of sin of three π‘₯.

We know that when we differentiate sin of π‘₯, we get cos of π‘₯. And in fact, the derivative of sin of three π‘₯ is three cos of three π‘₯. We differentiate again. Now the derivative of cos of π‘₯ is negative sin of π‘₯. Since we’re differentiating three cos of three π‘₯, we get three times negative three sin of three π‘₯, which is negative nine sin of three π‘₯. We differentiate again, and we get three times negative nine cos of three π‘₯, which simplifies to negative 27 cos of three π‘₯.

Now, according to the formula for our Maclaurin series, we need to evaluate each of these derivatives at π‘₯ equals zero. And so by letting π‘₯ equal zero in the function 𝑓 of π‘₯, we get sin of three times zero, which is sin of zero, which is zero. Then, 𝑓 prime of zero is three cos of three times zero. That’s three times cos of zero or three times one, which is three. 𝑓 double prime of zero is negative nine times sin of three times zero, which is once again zero. And then, the third derivative 𝑓 triple prime of zero is negative 27 times cos of three times zero, which is negative 27.

Substituting these into our formula for the Maclaurin expansion and we get 𝑓 of π‘₯ equals zero plus three π‘₯ over one factorial plus zero π‘₯ squared over two factorial plus negative 27π‘₯ cubed over three factorial and so on. Continuing this process and we find that the first three terms are three π‘₯ minus nine over two π‘₯ cubed plus 81 over 40π‘₯ to the fifth power and so on. This is quite clearly not equal to the series in our question. So sin of three π‘₯ is not the correct answer.

Now, let’s look at the natural log of one plus π‘₯ cubed. Well, in this case, we can quote the result for the expansion of the natural log of one plus π‘₯. Its Maclaurin series expansion is π‘₯ minus π‘₯ squared over two plus π‘₯ cubed over three minus π‘₯ to the fourth power over four and so on. To find the Maclaurin series expansion for the natural log of one plus π‘₯ cubed, we simply replace π‘₯ with π‘₯ cubed. And we get π‘₯ cubed minus π‘₯ cubed squared over two plus π‘₯ cubed to the third power over three minus π‘₯ cubed to the fourth power over four and so on. This simplifies to π‘₯ cubed minus π‘₯ to the sixth power over two plus π‘₯ to the ninth power over three and so on.

If we compare this to the Maclaurin series in our question, we notice that they’re the same. So actually, this is an expression for the natural log of one plus π‘₯ cubed. And in fact, had we spotted originally that the Maclaurin series we were given looks a lot like the expansion for the natural log of one plus π‘₯, that could’ve saved us some time. And we could check our result by finding the Maclaurin series for 𝑒 to the power of three π‘₯ minus cos π‘₯ and 𝑒 to the power of three π‘₯ plus cos π‘₯. And when we do, we see that they are patently not the same as the one in our question.

And so either by fully finding the Maclaurin series or by recognizing the general form for the Maclaurin series expansion of the natural log of one plus π‘₯, we find the answer to be (b). It’s the natural log of one plus π‘₯ cubed.

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