Integration of polynomials from first principles

[TheLivingMartyr]

Hello all,

 

I can now differentiate or integrate most functions using the rules that I have learnt, so I set myself the challenge of trying to derive the known formulae:

 

[math]

\frac{d}{dx} kx^m = mkx^{m-1}

[/math]

[math]

\int_{0}^{a} kx^m dx = \frac{ka^{m+1}}{m+1}

[/math]

 

from first principles (And I don't mean just doing the reverse of the derivative for the integral, I mean by actually summing rectangles of width tending to zero). By this I mean without any external help except for binomial expansions and one summation identity

 

Anyway, so I managed to do it fine for differentiation, and I did it for integration for the special case kx², but this relied on me being given the identity:

 

[math]

\sum_{r=0}^{n-1}r^2 = \tfrac{1}{6}(n-1)n(2n-1)

[/math]

 

I can't find anywhere the general rule for the summation:

 

[math]

\sum_{r=0}^{n-1}r^m

[/math]

 

If anyone knows the general summation, it would be a great help XD I can manage all the rest fine, it's just I don't think i'll have the willpower to inductively work out that summation.

 

thankyou!!!!!!

[mathematic]

As far as I can tell there is no general formula, although you can easily find formulas for m up to about 8.

[questionposter]

I think for some wierd reason you can't generate forumlas past a certain point, so what you'd have to do is break the polynomial up into its constituent squares and cubes.

After thinking about it, I think its no different than not being able to have a formula to get the angles of any octagon or any nonagon or etc. There's just too many parameters that can change, so instead you break it up into triangles and solve those.

Edited January 17, 2012 by questionposter

[TheLivingMartyr]

so you mean to prove for the general rule kx^m I would have to use a taylor power series rather than a summation of n^m? I'm just trying to figure out how else to go about proving it, because that's the route I followed to prove kx² Thanks though, I would have been on a wild goose chase otherwise XD

[mathematic]

so you mean to prove for the general rule kx^m I would have to use a taylor power series rather than a summation of n^m? I'm just trying to figure out how else to go about proving it, because that's the route I followed to prove kx² Thanks though, I would have been on a wild goose chase otherwise XD

If you have the energy you can always calculate the sum formula for n^m for any given m. It will be a polynomial of degree m+1, so you need to calculate by hand the sum for n= 1 to k, for k up to m+1, and solve the resultant set of m+1 linear equations.

[questionposter]

so you mean to prove for the general rule kx^m I would have to use a taylor power series rather than a summation of n^m? I'm just trying to figure out how else to go about proving it, because that's the route I followed to prove kx² Thanks though, I would have been on a wild goose chase otherwise XD

 

http://en.wikipedia....Ruffini_theorem

 

I think somewhere there it states mathematically why you can't generalize all polynomials to get exact answers, but you can really only approximate.

 

" This clearly excludes the possibility of having any formula that expresses the solutions of an arbitrary equation of degree 5 or higher in terms of its coefficients, using only those operations, or even of having different formulas for different roots or for different classes of polynomials, in such a way as to cover all cases."

Edited January 18, 2012 by questionposter

[Cap'n Refsmmat]

http://en.wikipedia....Ruffini_theorem

 

I think somewhere there it states mathematically why you can't generalize all polynomials to get exact answers, but you can really only approximate.

 

" This clearly excludes the possibility of having any formula that expresses the solutions of an arbitrary equation of degree 5 or higher in terms of its coefficients, using only those operations, or even of having different formulas for different roots or for different classes of polynomials, in such a way as to cover all cases."

That's not relevant to the problem at hand; the Abel-Ruffini theorem shows it's impossible to create a general equation to find the roots of those polynomials, but says nothing about a general equation to find their antiderivatives.

[TheLivingMartyr]

I've worked out a less messy method to prove that the antiderivative (ie. following the differentiation process backwards) will yield the area when bounds are applied. It relies on having first proved that dy/dx is the change y with respect to x, but allow me that assumption. It would also be nice if i could have a diagram to go with this, but try and imagine it XD. here goes

 

Let A be the sum of the signed areas between a fixed point on y = 0 a, and a variable point x. The corresponding y coordinate of x will therefore be f(x)

 

Now increment x by a small amount dx. We can say that A has consequently increased by a small change dA. We can approximate this dA by using a rectangle, f(x)*dx such that,

 

[math]

dA \approx f(x)dx

[/math]

 

It therefore follows to say that,

 

[math]

\frac{dA}{dx} \approx f(x)

[/math]

 

It can be observed that as dx decreases, the rectangle f(x)dx becomes closer to the true value of dA. We therefore say that,

 

[math]

\lim_{dx\to0} \frac{dA}{dx} = f(x)

[/math]

 

Since it can be shown that,

 

[math]

\lim_{dx\to0} \frac{dy}{dx}

[/math]

 

expresses the change in y with respect to x, ie. is the derivative of y with respect to x; this means that dA/dx is the derivative of the area with respect to x. If the derivative of the area is equal to the original function f(x), then it follows to say that if the inverse process of differentiation is applied to f(x), the equation for the area will be yielded.

 

[math]

\int dA = \int f(x)dx

[/math]

[math]

A = \int f(x)dx

[/math]

 

so, for example, following the opposite steps to differentiation for a polynomial gives,

 

[math]

f(x) = kx^n

[/math]

[math]

f'(x) = nkx^{n-1}

[/math]

[math]

F(x) = \frac{kx^{n+1}}{n+1}

[/math]

 

Well, i think that proof covers pretty much everything, it's a bit messy at the moment, but I think it follows a good logical route.

If there's anything I've missed please tell me