Question

Hello all,
I am currently working on the worked example from Session 26 and I am lost at one part of the solution to part b. The concept of summations and nth degrees or derivatives has always eluded me, so I was wondering if anyone could help explain what the solution is telling me. I can work out most of this part on my own, but after the three vertical periods, they lose me. Thank you.

Answer 1

Khan's videos might be helpful. If you have time, see this
https://www.khanacademy.org/math/integral-calculus/sequences_series_approx_calc/maclaurin_taylor/v/maclauren-and-taylor-series-intuition
and the following ones. He is doing the same type of problem as your question, perhaps more clearly...
If after watching you have a specific question, please post it.

Answer 2

Okay, thanks for providing me with that link. Khan has some really helpful videos.

Answer 3

As for the answer to part (b), what we have is an example of poor communication.

They are (trying) to show the pattern when we take derivatives of x^n
Focusing just on the exponent, we see
n-1, n-2, ..., 2,1,0
in other words, each time we take a derivative, the exponent goes down by 1, and we get a descending sequence of integers.

They insisted on writing this in the form n-c:
n-1, n-2, ... n- c

To show the last few terms, for example for n-c = 2, (solving for c= n-2), they wrote
2 as n - (n-2) (which if you simplify you see becomes 2)
This is not very clear, and when we use them "out front" it becomes a bit indecipherable.

I would develop the pattern this way
\[ \frac{d}{dx} x^1 = 1 = 1! \\ \frac{d^2}{dx^2} x^2= 2 \frac{d}{dx} x^1= 2\cdot 1 = 2! \\
\frac{d^3}{dx^3} x^3= 3\cdot \frac{d^2}{dx^2} x^2 = 3\cdot 2 \frac{d}{dx} x^1= 3\cdot2\cdot 1 = 3!
\]
or in general
\[ \frac{d^n}{dx^n} x^n = n!\]
in words, the nth derivative of x^n is n! (a constant)
If we take one more derivative (the (n+1)th ) we get
\[ \frac{d}{dx} n! = 0 \]
because n! is a constant.
and notice any subsequent derivatives would be
\[ \frac{d}{dx}0= 0 \]
and we get 0 forever after.

Answer 4

Thus we have these "rules"
\[ \text{ for m>n, } \frac{d^{(m)}}{dx^{m}} x^n= 0 \]
\[ \text{ for m=n, } \frac{d^{(n)}}{dx^{n}} x^n= n!\]

Answer 5

Part b was actually what I was most confused with. I agree that the way they showed it was a bit confusing in itself, so thank you for taking the time to offer some clarification.