Question

Prove the Power Rule using "n" factorization. I don't get how "n" factorization works.
Prove the Power Rule using "n" factorization. I don't get how "n" factorization works.
@Mathematics

Answer 1

What is "n factorization" and what is the power rule?

Answer 2

(a + b)^n
The Power Rule is nx^n-1

Answer 3

Ok to make it easier, I understand what factorials are. I know that 2! means 2*1 = 2.

Answer 4

But I have no idea how to apply any of that factorial for "n" stuff to power rule.

Answer 5

If you can explain, that will be awesome. This is the one thing that is throwing me off.

Answer 6

I mis-read the question completely. Factorials would be nice for the expansion of (a+b)^n but I don't get what exactly needs to be proved.

Answer 7

Ok wait one minute. Let me just copy and paste to you what my prof said. Don't worry, its brief.

Answer 8

Quote Professor:
Yes you do need it. I did not use the binomial theorem to prove the power rule. I used the factorization of A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+A^{n-3}B^2+.... +A B^{n-2}+B^{n-1}).



You need to know this factorization.

Answer 9

"it" is to know how to prove the Power rule.

Answer 10

When I looked at that factorization, I was like "WTH..".

Answer 11

Embarrassingly, I've read this proof recently but my mind has gone blank :(

Answer 12

Yes. This identity is very useful and this is the one he's using:
\[ a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + ... + a^2b^{n-3} + ab^{n-2} + b^{n-1}) \]

Answer 13

Ok how the heck does this work. I am focused, and I'll try really hard to understand.

Answer 14

Look at an example to make this clearer (hopefully!):

For instance a^5 - b^5:

\[ (a-b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4) \] \[= a (a^4 + a^3b + a^2b^2 + ab^3 + b^4) \] \[ - b (a^4 + a^3b + a^2b^2 + ab^3 + b^4) \]
\[ = a^5 + a^4b+a^3b^2 + a^2b^3 + ab^4 \]
\[ - a^4b - a^3b^2 - a^2b^3 - ab^4 - b^5 \]
Now see that nearly all the terms cancel out, leaving us with
\[ = a^5 - b^5 \]

Answer 15

One quick question: How did you get the long brackets?

Answer 16

So hopefully that gives you comfort that this identity for general n works.

Answer 17

One beside (a-b)

Answer 18

The long bracket beside (a-b), how did you get that?

Answer 19

I'm proving the identity, starting in this case with the right-hand side above and showing it is equal to the left-hand side

Answer 20

Alright continue on please.

Answer 21

Oh okay, so what you basically did was take (a-b)*(a-b)^4 to show (a-b)^5?

Answer 22

No. I didn't do that at all.

Answer 23

No induction use.

Answer 24

Just explicit calculation

Answer 25

Look at what I did for n = 5. Write out for yourself the identity for n = 6 so you feel comfortable with it.

Answer 26

Then where did you get (a^4 + a^3b + a^2b^2 + ab^3 + b^4) from?

Answer 27

This is what I was saying about the bracket beside (a-b)

Answer 28

It is the term we obtain when we evaluate
\[ \frac{a^5 - b^5}{a-b} \]

Answer 29

So either you prove that it works by doing what I did above; or try the long division.

Answer 30

Ok, so you are telling me to do: (a^6 - b^6)/(a-b)?

Answer 31

Better: show that
a^6 - b^6 = (a - b)(a^5 + a^4b + .... + b^5)

Answer 32

Good job JamesJ. Sometimes it is hard for students to make the jump from a specific example to an abstract one.
Students needs to realize proofs are abstract and apply to all cases.

Answer 33

OHH Ok, I get it so how do I use this approach with Proving the Power Rule? I understand now that that is how you get that bracket^(whatever no.).

Answer 34

Now then. To calculate the derivative of f(x) = x^n we need to evaluate the limit of the following difference ratio:

\[ \frac{ f(x) - f(a)}{x-a} \]
\[ = \frac{x^n - a^n}{x-a} \]

Answer 35

Why not:
f(x+h) - f(x)
---
h?

Answer 36

Same difference. It's actually slightly notationally easier with the notation I'm using, but they're otherwise identical. From my notation to yours: x --> x + h , a --> x

Answer 37

but we'll use the form you're using. Write g = x + h. Then we want

\[ \frac{ g^n - x^n}{g-x} \]

Answer 38

ok

Answer 39

Now \[g^n - x^n = (g-x)(g^{n-1} + .... + x^{n-1}) \]
hence the ratio is just
\[ g^{n-1} + .... + x^{n-1} \]

Answer 40

Notice that this sum has n terms where the power of g and the power of x sum to n-1.

Answer 41

and all the coefficients of these monomials g^j x^(n-1-j) are 1.

Answer 42

Now the limit of each term as h --> 0 is x^j . x^(n-1-j) = x^(n-1)

Answer 43

therefore the sum of the n of them is

\[ nx^{n-1} \]

Answer 44

Why do you have those dots in b/ween "...."

Answer 45

Because there are n of them in general, but I can't write down n terms in general explicitly without using some complicated notation. I.e., is it n = 2, or n = 17, or n = 256809?

Answer 46

ok

Answer 47

But if you want to see it explicitly, here it is:

\[ a^n - b^n \ = \ (a-b) \ \sum_{j=0}^{n-1} a^{n-1-j}b^j \]

Answer 48

I get that. So you are doing n choose j,

Answer 49

No

Answer 50

Make the leap QRAwarrior from specific to abstract.
JamesJ has done a great job of explaining it to you. :)

Answer 51

There are no binomial coefficients

Answer 52

So, formally,
if \( f(x) = x^n \) then

\[ f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{(x+h) - x} \]

Answer 53

\[ = \lim_{h \rightarrow 0} \frac{ ((x+h) - x) \sum_{j=0}^{n-1} (x+h)^{n-1-j} x^j }{(x+h) - h} \]

Answer 54

\[ = \lim_{ h \rightarrow 0} \ \ \sum_{j=0}^{n-1} (x+h)^{n-1-j} x^j \]
\[ = \sum_{j=0}^{n-1} x^{n-1-j} x^j \]
\[ = \sum_{j=0}^{n-1} x^{n-1} \]
\[ = x^{n-1} \ \sum_{j=0}^{n-1} \ 1 \]
\[ = n x^{n-1} \]