Proof by induction of polynomial differentiability

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Homework Help Overview

The discussion revolves around proving the derivative of the polynomial function (ax^n) using mathematical induction. The original poster expresses difficulty in completing the proof after establishing the base case.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the necessity of using the definition of the derivative versus applying rules such as the product rule. There is an emphasis on the base case and the transition from P(k) to P(k+1) in the induction process.

Discussion Status

Some participants have provided guidance on the steps to take next, including clarifying the use of limits in the base case. There is an acknowledgment of differing assumptions regarding the requirements of the proof, which has led to further exploration of the topic.

Contextual Notes

The original poster is in an analysis class, which may impose specific requirements on the proof method, such as the use of limits in the definition of the derivative.

ssayan3
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Homework Statement


Prove that (ax^n)' = nax^n-1 using induction.

I am very weak with induction proof, and I haven't had much trouble proving the basis step, but I can't seem to finish it...

Homework Equations





The Attempt at a Solution


1. Prove (ax)' = a

(a(x+h) - a(x))/h = (ax + ah - ax)/h = (ah)/h = a

2. Prove (ax^n)' = nax^n-1
? I can't seem to get my algebra right...
 
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In proving your base case using the definition of the derivative, you need to do this as a limit. Are you required to use the definition of the derivative in this problem? If not it's much simpler to show that d/dx(ax) = a d/d(x) = a, using the constant multiple rule and the power rule.

The next step is to assume that d/dx(axn) = naxn+1. Then use that to prove that d/dx(axn+1) = (n+1)axn+2. If you're not required to use the definition of the derivative, you can do this using the product rule, keeping in mind that axn+1 = axn * x.
 
This is for an analysis class, so yes, I would think that I would have to use the definition of derivative in this one...
 
Just because it's an analysis class doesn't necessarily mean that you have to use the definition of the derivative. That's an assumption you are making that may or may not be justified.

If if turns out that you do have to use the definition, it shouldn't be that hard for this proof. If you use the definition, however, you need to include limits. What you showed for your base case is very sloppy, not using limits at all. It should look something like this.
\frac{d(ax)}{dx}~=~\lim_{h \rightarrow 0}\frac{a(x + h) -ax}{h}~=~\lim_{h \rightarrow 0}\frac{ax +ah - ax}{h}~=~\lim_{h \rightarrow 0}\frac{ah}{h}~=~a
 
Let P(n) be the statement that for the natural number n, (ax^n)' = nax^{n-1}. Since you've already proven that P(1) holds, assume that P(k) holds and complete the proof by showing that this implies that P(k+1) holds. So you should now start with,

(ax^{k+1})' = (ax^kx)'

Can you see the next step that you should take?
 
Haha, fantastic! Thanks to both of you. That makes things much easier to understand for me.
 

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