What is the Proof of Differentiation for nx^n-1?

[roger]

hi

I'm in a bit of a hurry but I'm looking for the proof or derivation of the formula for differentiation :
nx^n-1

I don't mean to show it works for all (most) functions but where it comes from
Thanks

Roger

 

[Curious3141]

You can prove it for positive integral n using the Binomial Theorem.

Hint : consider (x + \delta x)^n - x^n

 

[roger]

I am looking for the rigorous proof, not the binomial theory.

Perhaps as a series ?

Roger

 

[hypermorphism]

I am looking for the rigorous proof, not the binomial theory.

Perhaps as a series ?

Roger

Use the natural logarithm to turn exponents into products.

 

[Curious3141]

Use the natural logarithm to turn exponents into products.

This would mean assuming a less "obvious" result. I suppose the way this could be done would go something like this :

f(x) = x^n = e^{n\ln x}

f'(x) = (e^{n\ln x})(\frac{n}{x}) = nx^{n-1}

To me this is using a more advanced result to prove a less advanced one.

Instead, consider using the general Binomial expansion (which holds for all complex exponents) to prove the result. Please see the expansion in my post below to work out the details.

 

[cepheid]

You can prove it for positive integral n using the Binomial Theorem.

Hint : consider (x + \delta x)^n - x^n

So, as I understand it, you'd do this expansion inside the limit (from the definition of a derivative) in order to be as rigorous as possible, right?

 

[Curious3141]

So, as I understand it, you'd do this expansion inside the limit (from the definition of a derivative) in order to be as rigorous as possible, right?

I just edited my post, please read it. It took me a while to work out the LaTex, that's really harder than it should be.

 

[dextercioby]

To me this is using a more advanced result to prove a less advanced one.

Instead, consider using the general Binomial expansion (which holds for all complex exponents) to prove the result. This is very easy and is a general result.

And what would that be?Is it the one which includes the Gamma Euler function??Please,enlighten us...

Would u say Gamma Euler is simpler than natural logarithms and exponentials??

Daniel.

 

[mathwonk]

the derivative of x^n means the coefficient of the linear part of the change in y, compared to the change in x. so we write x, as it changes, as x+dx, and then raise it to the nth power, getting (x+dx)^n = x^n + n x^(n-1) dx
+ (n)(n-1)/2 x^(n-2)(dx)^2 +...

so we see that the linear part of this change in x, i.e. linear in dx, is n x^(n-1)dx. Thus the coefficient is n x^(n-1).

much of the discussion here so far has seemed ludicrous to me. so maybe this will seem so to you.

 

[Curious3141]

And what would that be?Is it the one which includes the Gamma Euler function??Please,enlighten us...

Would u say Gamma Euler is simpler than natural logarithms and exponentials??

Daniel.

Actually, it's not as difficult as you make out.

The binomial expansion of (1 + x)^n for any complex n and |x| < 1 is :

(1 + x)^n = 1 + x + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ...

is it not ?

Doesn't involve gamma functions now does it ? Where's the difficulty ?

It's fairly simple to work the rest out.

 

[dextercioby]

You're right.My formula is general.Yours does not cover real and complex exponents.

(1+x)^{n}= \sum_{k=0}^{+\infty} \frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}x^{k}

,for all complex "n",with:
|x|&lt;1

Daniel.

 

[Curious3141]

You're right.My formula is general.Yours does not cover real and complex exponents.

(1+x)^{n}= \sum_{k=0} \frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}x^{n-k}

Daniel.

Actually it does. When you simplify the gamma functions, it comes out to the exact same thing. Try it.

Hint : Let n = N + a, where N is an integer and a is the nonintegral portion between 0 and 1. Can you see where this is going ?

That would cover all real exponents. Complex exponents, I haven't tried yet.

 

[dextercioby]

Compare my formula with yours for "k=3" and "n=i+5".

I edited my previous post.

Daniel.

 

[Curious3141]

Compare my formula with yours for "k=3" and "n=i+5".

I edited my previous post.

Daniel.

I know for a fact that my formula covers all complex exponents. I can quite easily prove it for all reals at least, I just scribbled it out, do you want the proof ?

 

[dextercioby]

I asked you to see whether my formula and yours give the same values for "k=3" and "n=i+5".I didn't ask you for any proof...

Daniel.

P.S.There's no point in doing the calculation,because they don't...

 

[Curious3141]

I asked you to see whether my formula and yours give the same values for "k=3" and "n=i+5".I didn't ask you for any proof...

Daniel.

P.S.There's no point in doing the calculation,because they don't...

You are quite, quite wrong.

My way : coefficient of term with k = 3 is

\frac{(i + 5)(i + 4)(i + 3)}{3!}

Your way :

\frac{\Gamma (i + 6)}{(\Gamma(4))(\Gamma(i + 3))} = \frac{(i + 5)(i + 4)(i + 3)(i + 2)(i + 1)(i)\Gamma(i)}{3!(i + 2)(i + 1)(i)(\Gamma(i))} = \frac{(i + 5)(i + 4)(i + 3)}{3!}

Correct ?

 

[dextercioby]

Yeah,you're right,just because those gamma-s get symplified.But "area of a circle"?

Daniel.

 

[Curious3141]

You can use the same principle of cancellation to prove it for all reals (with the method I suggested). I honestly don't know enough about the gamma functions of complex arguments to try and tackle the proof for all complex exponents.

 

[dextercioby]

U don't have to know anything about the complex variable Gamma Euler.
\Gamma(z+1)=z\Gamma(z)

would be enough to prove that the Gamma ratio can be written:
\frac{\Gamma(z+1)}{\Gamma[(z-k)+1]}=\frac{\Gamma(z+1)}{\Gamma[(z+1)-k]}=(z+1-k)_{k}=(z+1-k)[(z+1-k)+1][(z+1-k)+2]...[(z+1-k)+(k-1)]

,where i made use of a substitution and the definiton of the Pochhammer symbol.

Daniel.

P.S.It's better if u didn't put the Pochhammer symbol in the binomial expansion.Just leave it with the 3 gammas Euler.

 

[Curious3141]

Yes, that would work.

 

[Curious3141]

Alternatively,

\frac{\Gamma(z+1)}{\Gamma[(z-k)+1]} = \frac{z(z-1)(z-2)...(z-k+1)(z-k)\Gamma(z-k)}{(z-k)\Gamma(z-k)}

which would give the same result.

 

[cepheid]

Oh man...and the poor guy just wanted a proof of the power rule for differentiation. I think what Curious and Mathwonk said is good enough to answer that question.

 

[Curious3141]

Oh man...and the poor guy just wanted a proof of the power rule for differentiation. I think what Curious and Mathwonk said is good enough to answer that question.

Well, he wanted a rigorous proof, all the preceding helps to establish that rigour.

 

[dextercioby]

Oh man...and the poor guy just wanted a proof of the power rule for differentiation. I think what Curious and Mathwonk said is good enough to answer that question.

He should say "thank you" because,beside the proof for what he wanted,he got a pretty decent and rigurous treatment of the binomial formula.Sometimes it's useful to diverge from the original question.


Daniel.

 

[matt grime]

I am looking for the rigorous proof, not the binomial theory.

That is the rigorous proof.

 

[roger]

thanks alot, everyone !

Roger

 

[HallsofIvy]

I am looking for the rigorous proof, not the binomial theory.

Perhaps as a series ?

Roger

Why isn't using the binomial theorem rigorous? Since you didn't say anything about n, it would be natural to assume it is intended to be a positive integer.

Somewhat simpler than using the binomial theorem is to use the product rule and proof by induction:
if n= 1, xⁿ= x and its derivative is 1= 1*x^1-1 so the formula is correct for n= 1.
Assume that (x^N)'= N x^N-1 for specific N.

x^N+1= x*x^N so, by the product rule,
(x^N+1)'= 1*x^N+ x*Nx^N-1= x^N+Nx^N= (N+1)x^N

By induction then, (xⁿ)'= nx^n-1 for all n.
If you mean the derivative of x^α where α can be any real number (other than -1), then logarithms.

 

[roger]

Somewhat simpler than using the binomial theorem is to use the product rule and proof by induction:
if n= 1, xⁿ= x and its derivative is 1= 1*x^1-1 so the formula is correct for n= 1.
Assume that (x^N)'= N x^N-1 for specific N.

x^N+1= x*x^N so, by the product rule,
(x^N+1)'= 1*x^N+ x*Nx^N-1= x^N+Nx^N= (N+1)x^N

By induction then, (xⁿ)'= nx^n-1 for all n.
If you mean the derivative of x^α where α can be any real number (other than -1), then logarithms.

Why isn't using the binomial theorem rigorous? Since you didn't say anything about n, it would be natural to assume it is intended to be a positive integer.

Yes, I was assuming n to be a positive integer.


Why isn't the biomial theory rigorous ? because now I feel compelled to ask where the binomial theory derives from ?


Thanks for any input on the above.
Roger

 

[HallsofIvy]

The binomial theorem,
(x+y)^n=\Sigma_{i=0}^n _nC_ix^iy^{n-i}
is pretty basic algebra. I would expect anyone taking calculus to be familiar with it.

 

[dextercioby]

Yes, I was assuming n to be a positive integer.
Why isn't the biomial theory rigorous ? because now I feel compelled to ask where the binomial theory derives from ?

It is rigurous,trust me...As soon as u use gamma Euler functions,it's as rigurous as any other mathematical result...

Archimedes and Newton did it.However,the generalization to complex numbers and exponents was done much later...

Daniel.