Question Video: Find the Taylor Series from the Derivatives

For the function 𝑓: 𝑓(3) = 2, 𝑓′(3) = 7 and 𝑓^(𝑛) (3) = βˆ’(1/2)𝑛𝑓^(𝑛 βˆ’ 1)(3) for 𝑛 β‰₯ 2. Find the first five terms of the Taylor series representation of 𝑓 at π‘₯ = 3.

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Video Transcript

For the function 𝑓, 𝑓 evaluated at three is equal to two. The first derivative of 𝑓 evaluated at three is equal to seven. And the 𝑛th derivative of 𝑓 evaluated at three is equal to negative a half multiplied by 𝑛 multiplied by the 𝑛 minus oneth derivative of 𝑓 evaluated at three for 𝑛 is greater than or equal to two. Find the first five terms of the Taylor series representation of 𝑓 at π‘₯ is equal to three.

We recall that we call the sum from 𝑛 equals zero to infinity of the 𝑛th derivative of 𝑓 evaluated at π‘Ž divided by 𝑛 factorial multiplied by π‘₯ minus π‘Ž raised to the 𝑛th power the Taylor series representation of the function 𝑓 at π‘Ž. The question asks us for the first five terms of this Taylor series representation. And we’re told that this is at π‘₯ is equal to three. So we set π‘Ž equal to three.

So we’ll substitute π‘Ž equals three into our Taylor series representation. And we’ll only do a partial sum up to four to find the first five terms. For our first term, when 𝑛 is equal to zero, we get the zeroth derivative of 𝑓 evaluated at three divided by zero factorial multiplied by π‘₯ minus three to the zeroth power.

We can simplify this term by noticing that zero factorial is equal to one and π‘₯ minus three raised to the zeroth power is also equal to one. Then we know that the zeroth derivative of 𝑓 is just equal to our function 𝑓. So our entire first term just simplifies to 𝑓 evaluated at three.

Now let’s do the same of our second term when 𝑛 is equal to one. This gives us the first derivative of 𝑓 evaluated at three divided by one factorial multiplied by π‘₯ minus three to the first power. We have that one factorial is just equal to one, and π‘₯ minus three to the first power is just equal to itself, π‘₯ minus three. And we can rewrite the first derivative of 𝑓 to just be equal to 𝑓 prime. So this gives us that our second term can be written as 𝑓 prime evaluated at three multiplied by π‘₯ minus three.

We’ll then just write out our third, fourth, and fifth term by substituting in 𝑛 is equal to two, three, and four, respectively. The question tells us that 𝑓 evaluated at three is equal to two. So we can change our first term to just be equal to two. The question also tells us that 𝑓 prime evaluated at three is equal to seven. So we can change the second term in our series to be equal to seven multiplied by π‘₯ minus three.

The later terms in our series all use higher-order derivatives of our function 𝑓. Luckily, the question gives us the formula for calculating the 𝑛th derivative of 𝑓 evaluated at three when 𝑛 is greater than or equal to two. Substituting 𝑛 is equal to two into this formula gives us the second derivative of 𝑓 evaluated at three is equal to negative a half multiplied by two multiplied by 𝑓. But we take two minus one derivatives and evaluate this at three.

We can simplify this by noticing that negative a half multiplied by two is equal to negative one. And taking two minus one derivatives of the function 𝑓 is the same as just taking one derivative. This gives us that the second derivative of 𝑓 evaluated at three is equal to negative 𝑓 prime evaluated at three, which is equal to negative seven.

We can now substitute 𝑛 equals three into this formula to find that the third derivative of 𝑓 evaluated at three is equal to negative a half multiplied by three multiplied by 𝑓. But we take three minus one derivatives and evaluate at three. Taking three minus one derivatives of 𝑓 and evaluating at three is evaluating the second derivative of 𝑓 at three, which we calculated above to be negative seven. So the third derivative of 𝑓 evaluated at three is equal to negative a half multiplied by three multiplied by negative seven, which is equal to 21 divided by two.

Finally, we can do the same by substituting 𝑛 equals four into our formula. We can get an expression for the fourth derivative of 𝑓 evaluated at three. Using our answer from before that the third derivative of 𝑓 evaluated at three is 21 divided by two. We get that the fourth derivative of 𝑓 evaluated at three is equal to negative a half multiplied by four multiplied by 21 divided by two, which is equal to negative 21.

We can now substitute these values in to find the coefficients of terms three, four, and five in our Taylor series representation. Substituting these values in and evaluating our factorials gives us the coefficient of the third term as negative seven divided by two, the coefficient of the fourth term as 21 divided by two all divided by six, and the coefficient of the fifth term as negative 21 divided by 24. We can evaluate negative seven divided by two to be equal to negative seven over two.

If we calculate 21 divided by two all divided by six, we’ll get seven divided by four. And we can calculate negative 21 divided by 24 to be equal to negative seven divided by eight. Giving us that the first five terms of the Taylor series representation of the function 𝑓 to find in the question about π‘₯ is equal to three are. Two plus seven multiplied by π‘₯ minus three minus seven over two multiplied by π‘₯ minus three squared plus seven over four multiplied by π‘₯ minus three cubed minus seven over eight multiplied by π‘₯ minus three to the fourth power.

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