Video: Find a Coefficient in a Taylor Series Using a Known Taylor Series

The Taylor series for cos π‘₯ centered at π‘₯ = 0 begins as follows: 1 βˆ’ (π‘₯Β²/2!) + (π‘₯⁴/4!) βˆ’ (π‘₯⁢/6!) + .... What is the coefficient of π‘₯Β² in the Taylor series for π‘₯Β² cos π‘₯Β² centered at π‘₯ = 0?

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Video Transcript

The Taylor series for cos of π‘₯ centered at π‘₯ equals zero begins as follows. One minus π‘₯ squared over two factorial add π‘₯ to the fourth power over four factorial minus π‘₯ to the sixth power over six factorial and so on. What is the coefficient of π‘₯ squared in the Taylor series for π‘₯ squared cos of π‘₯ squared centered at π‘₯ equals zero?

So we need to find the Taylor series for this function, π‘₯ squared cos of π‘₯ squared. We of course know the general form of a Taylor Series where we can find the necessary derivatives and substitute them into this general form. However, we’ve been given the series for cos of π‘₯ centered at π‘₯ equals zero in the question. So, actually, we can use this instead. There’s a really useful technique to find the Taylor series of harder functions by using the Taylor series of common functions just like cos of π‘₯.

If we take this Taylor series and replace π‘₯ with π‘₯ squared, we’ve got the Taylor series for cos of π‘₯ squared centered at π‘₯ equals zero. It’s really useful when we’re doing this to keep whatever we’re substituting in brackets. So let’s simplify these exponents first. We remember the index law that tells us if we raise something to a power and then raise that to another power, we can just multiply the powers together. So that’s the Taylor series for cos of π‘₯ squared. But remember, we’re really looking for the Taylor series of π‘₯ squared multiplied by cos of π‘₯ squared.

So we can do this by multiplying each term by π‘₯ squared. This is because we’re multiplying the whole series by π‘₯ squared to get the Taylor series for π‘₯ squared cos of π‘₯ squared centered at π‘₯ equals zero. We can do this by multiplying each term individually by π‘₯ squared. We can then simplify this. Remembering that if we have π‘₯ to the π‘š power and we multiply it by π‘₯ to the 𝑛 power, this is the same as π‘₯ to the π‘š add 𝑛 power.

Remember that the question asked us what is the coefficient of π‘₯ squared. So we can see here that π‘₯ squared is on its own. So we can say that the coefficient of π‘₯ squared in the Taylor series for π‘₯ squared cos of π‘₯ squared is one.

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