Video: Using Differentiation to Find the Maximum Height a Launched Rocket Attains

Consider the series 𝑓(π‘₯) = π‘₯/(1 βˆ’ π‘₯Β²) = βˆ‘_(𝑛 = 0)^(∞) π‘₯^(2𝑛 + 1). Differentiate the given series expansion of 𝑓 term by term to find the corresponding series expansion for the derivative of 𝑓.

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Video Transcript

Consider the series 𝑓 of π‘₯ equals π‘₯ over one minus π‘₯ squared, which is equal to the sum of π‘₯ to the power of two 𝑛 plus one for values of 𝑛 between zero and ∞. Differentiate the given series expansion of 𝑓 term by term to find the corresponding series expansion for the derivative of 𝑓.

Let’s begin by using the power series to write out the first few terms. The first term is when 𝑛 is equal to zero. So it’s π‘₯ to the power of two times zero plus one, which is one. We then let 𝑛 be equal to one. And we get two times one plus one, which is three. Next, we let 𝑛 be equal to two. And we find the exponent to be equal to five. And we can continue in this manner. We find the first few terms to be π‘₯ to the power of one plus π‘₯ cubed plus π‘₯ to the fifth power plus π‘₯ to the seventh power, and so on.

We’re told to differentiate the series expansion of 𝑓 term by term. And so we’re going to do just that. The derivative of π‘₯ with respect to π‘₯ is just one. And we recall that, for any polynomial term, we differentiate by multiplying by the exponent and then reducing that exponent by one. So the derivative of π‘₯ cubed is three π‘₯ squared. The derivative of π‘₯ to the fifth power is five π‘₯ to the fourth power. Differentiating π‘₯ to the seventh power, and we get seven π‘₯ to the sixth power, and so on.

We need to find a way to write this as a sum. So let’s have a look at what’s happened. Each time, we’ve multiplied by the exponent. And we saw that that exponent was generated by using the expression two 𝑛 plus one. Then we reduced each power by one, while the original exponent was two 𝑛 plus one. So our new exponent is two 𝑛 plus one minus one. The values for our sum remain the same. They’re from 𝑛 equals zero to ∞. And of course two 𝑛 plus one minus one is simply two 𝑛. And so by differentiating term by term, we found the corresponding series expansion for the derivative of 𝑓. It’s the sum of two 𝑛 plus one times π‘₯ to the power of two 𝑛 for values of 𝑛 between zero and ∞.

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