Video: Finding Ther Fucntion a Power Series Converges To

For π‘₯ > 0, which of the following does the power series π‘₯ βˆ’ (π‘₯Β³/2!) + (π‘₯⁡/4!) βˆ’ (π‘₯⁷/6!) + ... + (βˆ’1)^(𝑛) (π‘₯^(2𝑛 + 1)/(2𝑛!)) + ... converge to? [A] sin π‘₯ [B] π‘₯ sin π‘₯ [C] π‘₯ cos π‘₯ [D] (sin π‘₯)/π‘₯

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Video Transcript

For π‘₯ is greater than zero, which of the following does the power series π‘₯ minus π‘₯ cubed over two factorial plus π‘₯ to the fifth power over four factorial minus π‘₯ to the seventh power divided by six factorial and then we add an infinite number of these terms together where the 𝑛th term is negative one to the 𝑛th power times π‘₯ to the power of two 𝑛 plus one divided by two 𝑛 factorial converge to? Option (a) the sin of π‘₯, option (b) π‘₯ times the sin of π‘₯, option (c) π‘₯ times the cos of π‘₯, or option (d) the sin of π‘₯ divided by π‘₯.

The question gives us a power series. And we’re told that this power series will converge for any value of π‘₯ greater than zero. We need to determine which function this power series converges to. We’re given four options. If we look at the power series we’re given in the question, we can see each term only includes π‘₯. This should remind us of our Maclaurin series or alternatively our Taylor series centered at zero. And we can see that our answers only include the sin of π‘₯, the cos of π‘₯, and π‘₯. So let’s start by recalling the Maclaurin series for the sin of π‘₯ and the cos of π‘₯.

First, the Maclaurin series for the sin of π‘₯ is π‘₯ minus π‘₯ cubed over three factorial plus π‘₯ to the fifth power over five factorial and we add infinitely many terms of this form where the 𝑛th term is negative one to the 𝑛th power times π‘₯ to the power of two 𝑛 plus one divided by two 𝑛 plus one factorial. Second, the Maclaurin series for the cos of π‘₯ is one minus π‘₯ squared over two factorial plus π‘₯ to the fourth power over four factorial and we add infinitely many terms of this form where the 𝑛th term is negative one to the 𝑛th power times π‘₯ to the power of two 𝑛 divided by two 𝑛 factorial. And we know both of these Maclaurin series are valid for all real values of π‘₯.

We can use these two Maclaurin series to find Maclaurin series for all four of the answers given to us in the question. For example, in option (a), we want to find the Maclaurin series for the sin of π‘₯. We’ve already done this. So we can move on to option (b) π‘₯ times the sin of π‘₯. To find the Maclaurin series for π‘₯ times the sin of π‘₯, we’re going to use our Maclaurin series for the sin of π‘₯.

So we’ll substitute this into our expression. This gives us π‘₯ times the sin of π‘₯ is equal to the following expression. Remember, we know the Maclaurin series for the sin of π‘₯ is valid for all real values of π‘₯. So we can just multiply each term by π‘₯. It will still be valid. To multiply each term by π‘₯, we just need to increase the exponent of π‘₯ by one. This gives us the following expression for π‘₯ times the sin of π‘₯. And we can see this is not equal to the power series given to us in the question. For example, our power series here has a term for π‘₯ squared. However, the power series given to us in the question does not have a term for π‘₯ squared.

And although not strictly necessary to answer this question, it’s worth pointing out that the power series of a function centered at π‘₯ is equal to π‘Ž is unique. It must be equal to the Taylor series of that function centered at π‘₯ is equal to π‘Ž. So far, we’ve found power series for option (a) and option (b). Let’s now find one for option (c) π‘₯ times the cos of π‘₯.

We’ll do this in the same way we did for π‘₯ times the sin of π‘₯. We’ll substitute our Maclaurin series for the cos of π‘₯ into this expression. Substituting in our Maclaurin series for the cos of π‘₯, we get the following expression for π‘₯ times the cos of π‘₯. Again, we know the Maclaurin series for the cos of π‘₯ is convergent for all real values of π‘₯. So to multiply this by π‘₯, we’ll just multiply each term by π‘₯. And to do this, we just need to add one to our exponent of π‘₯ on each term. This gives us π‘₯ minus π‘₯ cubed over two factorial plus π‘₯ to the fifth power over four factorial minus π‘₯ to the seventh power over six factorial and we add an infinite number of terms of this form where the 𝑛th term is negative one to the 𝑛th power times π‘₯ to the power of two 𝑛 plus one divided by two 𝑛 factorial.

And we can see this is exactly the same as the power series given to us in the question. So the power series given to us in the question must converge to the function π‘₯ times the cos of π‘₯, which was our option (c).

It’s also worth noting we could’ve done the same for option (d). We would’ve got the sin of π‘₯ divided by π‘₯ is equal to the sum from 𝑛 equals zero to ∞ of negative one to the 𝑛th power times π‘₯ to the power of two 𝑛 divided by two 𝑛 plus one factorial. Of course, this wouldn’t be valid when π‘₯ was equal to zero because the sin of π‘₯ divided by π‘₯ is not defined when π‘₯ is equal to zero. But this wasn’t the correct answer either. Therefore, we were able to show for values of π‘₯ greater than zero, the power series given to us in the question converges to π‘₯ times the cos of π‘₯, which was option (c).

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