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Question Video: Finding the First Four Terms of a Taylor Series of a Product of Functions Mathematics • Higher Education

Write the first four terms for the Taylor series expansion of 𝑓(π‘₯) =11𝑒^(2π‘₯) in ascending powers of (π‘₯ βˆ’ 2).

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Video Transcript

Write the first four terms for the Taylor series expansion of 𝑓 of π‘₯ is equal to 11 times 𝑒 to the power of two π‘₯ in ascending powers of π‘₯ minus two.

We need to find the first four terms of the Taylor series expansion for the function 𝑓 of π‘₯ is equal to 11 times 𝑒 to the power of two π‘₯ in ascending powers of π‘₯ minus two. Remember, ascending powers means the exponents of π‘₯ minus two will be increasing. And the fact that we’re told to use ascending powers of π‘₯ minus two tells us that we need to center our Taylor expansion at two.

Let’s recall what we mean by a Taylor series expansion for a function 𝑓 of π‘₯. The Taylor series expansion for a function 𝑓 of π‘₯ centered at π‘Ž is given by the following power series: the sum from 𝑛 equals zero to ∞ of the 𝑛th derivative of 𝑓 of π‘₯ with respect to π‘₯ evaluated at π‘Ž divided by 𝑛 factorial multiplied by π‘₯ minus π‘Ž raised to the 𝑛th power.

We call π‘Ž the center of our Taylor series. In our case, we’re told to use π‘Ž is equal to two since then our series will be given in terms of powers of π‘₯ minus two. So to find the first four terms of this series, we’re going to need to start differentiating our function 𝑓 of π‘₯.

Our series starts from 𝑛 is equal to zero. So let’s find the first term in this series. That’s when 𝑛 is equal to zero. To do this, we’re going to need to find the zeroth derivative of 𝑓 of π‘₯ with respect to π‘₯ evaluated at two. And we need to remember what this means. The zeroth derivative of a function is just equal to that function itself. In other words, we just need to find 𝑓 evaluated at two. So we just substitute π‘₯ is equal to two into our function 𝑓 of π‘₯. This gives us 11 times 𝑒 to the power of two times two, which we can evaluate is equal to 11𝑒 to the fourth power.

To find the next term in our series, we’re going to want to find the first derivative of 𝑓 of π‘₯ with respect to π‘₯ evaluated at two. But we can’t do this directly. So instead, we’re going to need to differentiate our function 𝑓 of π‘₯. That’s the derivative of 11𝑒 to the power of two π‘₯ with respect to π‘₯.

And to differentiate this, we need to recall our rules for differentiating exponential functions. We know for any constant π‘˜, the derivative of 𝑒 to the power of π‘˜π‘₯ with respect to π‘₯ is equal to π‘˜ times 𝑒 to the power of π‘˜π‘₯. In our case, our value of π‘˜ is equal to two. So differentiating 𝑓 of π‘₯, we get 11 times two 𝑒 to the power of two π‘₯, which of course we can simplify to give us 22𝑒 to the power of two π‘₯.

We’re now ready to find our first derivative of 𝑓 of π‘₯ with respect to π‘₯ evaluated at two. We just substitute π‘₯ is equal to two into our expression for 𝑓 prime of π‘₯. We get 22𝑒 to the power of two times two. And we can just simplify this to get 22𝑒 to the fourth power.

To find the third term in our Taylor series expansion, we’re going to need to find the second derivative of 𝑓 of π‘₯ with respect to π‘₯ evaluated at two. Again, we can’t do this directly. So we’re going to need to differentiate our function 𝑓 prime of π‘₯ with respect to π‘₯. Our second derivative of 𝑓 of π‘₯ with respect to π‘₯ is equal to the derivative of 22𝑒 to the power of two π‘₯ with respect to π‘₯.

Again, we can do this by using our derivative rule for exponential functions. This time, our value of π‘˜ is also equal to two. So by using this rule, we get the second derivative of 𝑓 of π‘₯ with respect to π‘₯ is equal to 22 times two 𝑒 to the power of two π‘₯. And of course, we can simplify this to get 44𝑒 to the power of two π‘₯.

We can now use this to find the second derivative of 𝑓 of π‘₯ with respect to π‘₯ evaluated at two. We just substitute π‘₯ is equal to two into 𝑓 double prime of π‘₯. We get 44𝑒 to the power of two times two, which we can simplify to give us 44𝑒 to the fourth power.

We just need to do this one more time to find the fourth term in our Taylor series expansion. We need to find the third derivative of 𝑓 of π‘₯ with respect to π‘₯ evaluated at two. Just as we did before, we can’t do this directly. So we need to differentiate our second derivative of 𝑓 of π‘₯ with respect to π‘₯. Just as we did before, we’ll differentiate this by using our rules for differentiating exponential functions. And our value of π‘˜ is equal to two.

This gives us our third derivative of 𝑓 of π‘₯ with respect to π‘₯ is equal to 44 times two 𝑒 to the power of two π‘₯. And we can simplify this to get 88𝑒 to the power of two π‘₯.

And now we can use this expression to find the third derivative of 𝑓 of π‘₯ with respect to π‘₯ evaluated at two. We just substitute π‘₯ is equal to two into this expression. We get 88𝑒 to the power of two times two, which simplifies to give us 88𝑒 to the fourth power.

We now need to substitute these four values into our Maclaurin series. And remember, we’re only writing out the first four terms of this series. Our first term when 𝑛 is equal to zero is 11𝑒 to the fourth power divided by zero factorial multiplied by π‘₯ minus two to the zeroth power. Of course, we can simplify this. We know zero factorial is just equal to one. And π‘₯ minus two all raised to the zeroth power is also equal to one. So our first term just simplifies to give us 11𝑒 to the fourth power.

Let’s now find the second term in our series. Remember, we substitute 𝑛 is equal to one and π‘Ž is equal to two into the summand in our Taylor series. And we already showed that the first derivative of 𝑓 of π‘₯ with respect to π‘₯ at two is 22𝑒 to the fourth power. So using this, our second term is 22𝑒 to the fourth power divided by one factorial multiplied by π‘₯ minus two raised to the first power. And once again, we can simplify this.

First, π‘₯ minus two raised to the first power is just π‘₯ minus two. Next, one factorial is just equal to one. And dividing by one doesn’t change our value. So we just get 22𝑒 to the fourth power times π‘₯ minus two.

The third term in our series will be when 𝑛 is equal to two. And we already showed the second derivative of 𝑓 of π‘₯ with respect to π‘₯ at two is 44𝑒 to the fourth power. So substituting these values and π‘Ž is equal to two into the summand in our Taylor series, we get the third term is 44𝑒 to the fourth power divided by two factorial multiplied by π‘₯ minus two squared.

And once again, we can simplify. First, two factorial is just equal to two. And then 44 divided by two can be simplified to just give us 22. So our third term simplifies to give us 22𝑒 to the fourth power times π‘₯ minus two all squared.

We just need to do this one more time to find the fourth term in our series. That’s when 𝑛 is equal to three. This time, by substituting these values into the summand of our Taylor series, we get 88𝑒 to the fourth power divided by three factorial times π‘₯ minus two all cubed. And we can simplify this term. First, three factorial is equal to three times two times one. And now we can cancel the shared factor of two in our numerator and our denominator. This gives us 44 divided by three. So we were able to simplify our fourth term to 44𝑒 to the fourth power divided by three times π‘₯ minus two all cubed. And this is our final answer.

Therefore, we were able to show the first four terms in the Taylor series expansion of 𝑓 of π‘₯ is equal to 11𝑒 to the power of two π‘₯ in ascending powers of π‘₯ minus two is given by 11𝑒 to the fourth power plus 22𝑒 to the fourth power times π‘₯ minus two plus 22𝑒 to the fourth power multiplied by π‘₯ minus two all squared plus 44𝑒 to the fourth power over three multiplied by π‘₯ minus two all cubed.

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