nth Derivative Proof

[DJBruce]

As I worked on teaching my self calculus this summer I begin to wonder how to write a generic formula for the nth derivative of a equation of the form:

 

[math]y=x^{a}[/math] [math]where: a\geq 1[/math]

 

After thinking about it the pattern became obvious, but the proof took me a while to work out, in the end I came up with an inductive proof. Below is the proof I came up with.

 

[math]Basis:[/math]

[math]\frac{d^{n}y}{dx^{n}}=\left(\frac{a!}{(a-n)!}\right)x^{(a-n)}[/math]

 

[math]Let: n=1[/math]

 

[math]\frac{d^{1}y}{dx^{1}}=\left(\frac{a!}{(a-1)!}\right)x^{(a-1)}=ax^{(a-1)}[/math]

 

This step is verified because I came to the form proven by the Power Rule.

 

[math]Inductive: Let: n=(n+1)[/math]

 

[math]\bullet \left(\frac{d^{(n+1)}y}{dx^{(n+1)}}\right)= \frac{dy}{dx}\left(\frac{d^{n}y}{dx^{n}}\right)=[/math]

 

[math]\bullet \frac{dy}{dx}\left(\left(\frac{a!}{(a-n)!}\right)x^{(a-n)}\right)= [/math]

 

Substitution

 

[math]\bullet \left(\left(\frac{(a!)(a-n)}{(a-n)!}\right)x^{(a-n-1)}\right)= [/math]

 

Power Rule

 

[math]\bullet \left(\left(\frac{(a!)(a-n)}{\left((a-n)(a-n-1)(a-n-2)...(1)\right)}\right)x^{(a-n-1)}\right)= [/math]

 

Definition of Factorial

 

[math]\bullet \left(\left(\frac{(a!)}{\left((a-n-1)(a-n-2)...(1)\right)}\right)x^{(a-n-1)}\right)=[/math]

 

Cancellation

 

[math]\bullet \left(\left(\frac{(a!)}{((a-(n+1))!}\right)x^{(a-n-1)}\right)=[/math]

 

Definition of Factorial and Factoring

 

[math]\bullet \left(\left(\frac{(a!)}{\left((a-(n+1))!\right)}\right)x^{(a-(n+1))}\right) [/math]

 

Factoring

 

[math]So[/math]

[math]\bullet \frac{d^{n}y}{dx^{n}}=\left(\frac{a!}{(a-n)!}\right)x^{(a-n)}[/math]

 

I was wondering if my proof is correct and if it would be considered rigorous. Any correction or suggestions would be greatly appreciated.

Edited August 10, 2009 by DJBruce

[Xittenn]

When n is greater than a the resulting factorial is undefined!

[DJBruce]

When n is greater than a the resulting factorial is undefined!

 

Yes, you pointed this out to me last night so I will qualify the proof by stating: [math]n \leq a [/math], but if [math] n=a [/math] then you are going to get:

 

[math] \frac{a!}{0!}x^{0}=a! [/math]

so if get a constant this means the next derivative should be zero, so I don't think it is that large of a flaw.

[timo]

- Perhaps that's exactly what you wanted to avoid, but I find [math] \frac{a!}{(a-n)!} [/math] a rather obscuring way to write [math]\prod_{k=a-n+1}^{a}k[/math] or simply "product of all integers from a down to a-n+1". Also, the factorials don't work for negative exponents.

 

- Way too many parentheses for my taste.

 

- Actually, you can see the problems with n>=a directly from your proof if you properly catch the division by zero in the step you called "cancellation".

 

- The correct notation (your steps were fine just what you wrote down is not according to standard syntax) for your first step is [math] \frac{d^{n+1}y}{dx^{n+1}}= \frac{d}{dx}\left(\frac{d^{n}y}{dx^{n}}\right)[/math]. What you wrote would mean [math] y' \cdot \left(\frac{d^{n}y}{dx^{n}}\right)[/math]

 

- Apart from the division by zero and the notation mistake that's a nice, simple, correct and nicely presented (perhaps a few too many steps for my personal taste) proof by induction.

[DJBruce]

- Perhaps that's exactly what you wanted to avoid, but I find [math] \frac{a!}{(a-n)!} [/math] a rather obscuring way to write [math]\prod_{k=a-n+1}^{a}k[/math] or simply "product of all integers from a down to a-n+1". Also, the factorials don't work for negative exponents.

 

I have never actually seen that notation, before what exactly is that called? I realized the factorials do not work with negative exponents so that's why I put the restriction that the exponent must be greater than or equal to one.

 

- Actually, you can see the problems with n>=a directly from your proof if you properly catch the division by zero in the step you called "cancellation".

 

So what your saying is because when n=a you get:

[math]\left(\frac{(a!)(0)}{(0!)(0)}\right)x^{0}=\frac{0}{0}[/math]

that I should actually restrict this to [math]n< a[/math].

 

Thank you for taking the time to read my proof, your corrections and opinions are greatly appreciated.

[timo]

I have never actually seen that notation, before what exactly is that called?

Dunno. I call it "product". It's the multiplication version of the sum sign, so

[math] \sum_{X=a}^b X = a + (a+1) + \dots + b [/math]

and

[math] \prod_{X=a}^b X= a \cdot (a+1) \cdot \dots \cdot b [/math]

X, a and b can of course be any integer-valued expression, i.e. a number, a variable or a term.

 

And as always, someone wrote something in Wikipedia: http://en.wikipedia.org/wiki/Multiplication#Capital_pi_notation

 

So what your saying is because when n=a you get:

[math]\left(\frac{(a!)(0)}{(0!)(0)}\right)x^{0}=\frac{0}{0}[/math]

that I should actually restrict this to [math]n< a[/math].

Something in that direction, yes (I do not fully understand your question). But looking at it a 2nd time you'd have to catch the error one step further ahead in the "definition of the factorial" step. When a=n, then (a-n)! = (a-n)*(a-n-1)* ... * (1) does not hold true anymore. 0! is 1, not 0*...*1.

 

For your proof that would mean the following:

- You have proven that for n=1, your expression works.

- You have proven that if your expression works for n=x, then it also works for n=x+1 - if x is smaller than a.

=> So rather than the usual conclusion that your expression is correct for all n>1 you now can say that your expression is correct for all n from 1 to a.

[Xittenn]

misread statement delete post.......thank you!

 

[Edit]

 

here I'll make myself useful.......

 

Empty Product

Edited August 11, 2009 by buttacup