# Question Video: Finding the Taylor Series Expansion of a Function from a Formula Mathematics

For a function π: π(β4) = 6, πβ²(β4) = β6 and π^(π) (β4) = (β1/π) π^(π β 1) (β4) for π β₯ 2. Find the Taylor series expansion of π at π₯ = β4.

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### Video Transcript

For a function π, π evaluated at negative four is equal to six, π prime of negative four is equal to negative six, and the πth derivative of π of π₯ with respect to π₯ at negative four is equal to negative one divided by π multiplied by the previous derivative of π of π₯ with respect to π₯ at negative four for π greater than or equal to two. Find the Taylor Series expansion of π at π₯ is equal to negative four.

Weβre given some information about a function π. Weβre told that π of negative four is six; π prime of negative four is negative six. And then weβre given a formula to help us find the πth derivative of π of π₯ with respect to π₯ at π₯ is equal to negative four, where this formula is only valid for values of π greater than or equal to two. We need to use this information to find our Taylor series expansion of the function π at π₯ is equal to negative four. This means our Taylor series will be centered at π₯ is equal to negative four.

To do this, letβs start by recalling what we mean by the Taylor series expansion of a function π at π₯ is equal to π. This means the power series given by the sum from π equals zero to β of the πth derivative of π of π₯ with respect to π₯ evaluated at π divided by π factorial multiplied by π₯ minus π raised to the πth power. We call π the center of our Taylor series. In this question, weβre told to use π is equal to negative four. So letβs substitute π is equal to negative four into our Taylor series expansion definition.

To start, we can see we can simplify π₯ minus negative four to be π₯ plus four. So to find terms in this power series, weβre going to need to find the πth derivative of π of π₯ with respect to π₯ evaluated at negative four. And itβs worth pointing out at this point when we say the zeroth derivative of a function, we just mean that function itself. And weβre given some of these derivatives in our question and a formula to find the rest of these derivatives. So letβs try writing these out one by one and see if we can notice a pattern.

Letβs start with the zeroth derivative of π of π₯ with respect to π₯ evaluated at negative four. Remember, this is just π evaluated at negative four. And weβre told this in the question. Weβre told that this is equal to six. Letβs now find the first derivative of π of π₯ with respect to π₯ evaluated at π₯ is equal to four. We couldβve alternatively written this as π prime of negative four. And once again, weβre told this in the question. Weβre told this is equal to negative six.

To find the next term in our Taylor series expansion, we need to know the second derivative of π of π₯ with respect to π₯ evaluated at negative four. But weβre not told this directly. Instead, weβre given a formula to help us calculate this. Since weβre finding the second derivative, our value of π is equal to two. So we need to substitute π is equal to two into this expression. Substituting π is equal to two gives us the following expression. And we can simplify this since we know two minus one is equal to one. So this gives us negative one-half multiplied by the first derivative of π of π₯ with respect to π₯ evaluated at negative four.

In other words, this is negative one-half multiplied by our previous result. So weβll write this down to get the next term in our sequence. We multiplied by negative one-half, and we should calculate what this is. Itβs negative one-half multiplied by negative six, which we know is equal to three. Letβs now find another term to see if we can spot a pattern. To find the next term in our Taylor series expansion, we would need to know the third derivative of π of π₯ with respect to π₯ evaluated at negative four.

This time, our value of π is equal to three. So weβll substitute π is equal to three into the formula given to us in the question. This gives us negative one-third times the second derivative of π of π₯ with respect to π₯ evaluated at negative four. And now, weβre starting to see our pattern. This is just negative one-third multiplied by the previous result.

So it seems to find the next term in our sequence, weβre just multiplying by negative one over π. Letβs evaluate this expression just to make sure. We get negative one-third multiplied by three, which if we evaluate, we get negative one. And if we look at our formula, this seems to make sense. We take the previous derivative evaluated at negative four and then multiply this by negative one over π.

By using this information, we can actually simplify this expression. First, we notice to get from six to negative six, we multiplied by negative one. Or alternatively, we could write this as negative one divided by one. Now that we can more easily see this sequence, letβs see if we can find an easier expression for the πth derivative of π of π₯ with respect to π₯ at π₯ is equal to negative four. Well, our sequence starts at six. Next, we multiply six by negative one divided by one. Then we multiply this by negative one-half.

Then, we showed you need to multiply this by negative one-third. And in fact, we showed that to get the πth term, youβre going to need to multiply by all the way up to negative one over π. So we get the πth derivative of π of π₯ with respect to π₯ at negative four is equal to six times negative one over one times negative one-half. And we multiply all the way up to negative one over π. Well, at the moment, this formula wonβt be valid when π is equal to zero. But we can simplify this formula. First, we have π factors of negative one. So we can start by writing this as six times negative one to the πth power.

But then, if we were to multiply all of these fractions together, we would get one divided by a denominator which was π times π minus one multiplied by all the way down to one. In other words, we just get one divided by π factorial. And since zero factorial is now equal to one and negative one raised to the power of zero is also equal to one, we can see this formula is also valid when π is equal to zero. So this formula works for all of our integers greater than or equal to zero.

Now we can substitute this formula into our Taylor series expansion. This gives us that π of π₯ is equal to the sum from π equals zero to β of six times negative one to the πth power multiplied by one over π factorial all divided by π factorial. And then we multiply all of this by π₯ plus four all raised to the power of π. And we can simplify this. Instead of multiplying by one over π factorial, we can just divide by π factorial. But then, in our denominator, we now have π factorial multiplied by π factorial. So we could just write this as π factorial all squared. And then with a small bit of rearranging, we get our final answer.

Therefore, we were able to show if a function π has the following properties β π of negative four is six, π prime of negative four is negative six β and we can find the πth derivative of π of π₯ with respect to π₯ at negative four by taking negative one over π times our previous derivative at negative four for π greater than or equal to two, then the Taylor series expansion of π centered at π₯ is equal to negative four is given by the sum from π equals zero to β of negative one to the πth power times six divided by π factorial all squared all multiplied by π₯ plus four all raised to the πth power.