Polynomial Proof (1 Viewer)

[Lukybear]

a)Suppose that the origin is the point of inflexion of f(x) = ax^3 + bx^2 + cx + d

We know that since its a odd function, passing through center with origin being pt of inflexion then, b=d=0

Prove that if l is a line through the origin crossing the curve again at A and B, then O is the midpoint of the interval AB

 

[Trebla]

The polynomial equation has the form y = ax³ + cx
The line l has the equation y = mx and is assumed to cut the cubic curve at three points

Upon solving simultaneously:
mx = ax³ + cx
x(ax² + c - m) = 0
Solutions are:


Since it is assumed there are three solutions then (m - c)/a > 0

The two non-zero solutions represent the x coordinates of A and B which clearly have the x-coordinate midpoint at x = 0.

i.e. Since the function is odd
f(- x) = - f(x)
=> [f(- x) + f(x)] / 2 = 0 and in this case



So the midpoint of the y-coordinates is at y = 0

Thus the midpoint must be the origin. I'm not sure if it can be shown anymore rigorously, because in my opinion the property is quite obvious lol.

 

[Lukybear]

thanks. What about this.



Suppose that n is even.
Show that En(x) has all stationary points lying above the x-axis.

 

[Trebla]

thanks. What about this.



Suppose that n is even.
Show that En(x) has all stationary points lying above the x-axis.

Do you mean this?


If so, then:
E_n'(x)= 1 + x/1! + x²/2! + .... + x^n-1/(n-1)!

For stationary points to exist at say some x = x₀ which is clearly non-zero:
1 + x₀/1! + x₀²/2! + .... + x₀^n-1/(n-1)! = 0

Sub this result into E_n(x) evaluated at x = x₀ giving:
E_n(x₀) = x₀ⁿ/n!

If n is even then x₀ⁿ > 0 which implies
E_n(x₀) > 0

Hence stationary points always yield positive y-values for E_n(x) when n is even

 

[Lukybear]

how then do we prove that e(x) >0 for all x

 

[untouchablecuz]

how then do we prove that e(x) >0 for all x

if all turning points are above the x axis, then the absolute minimum MUST be above the x axis

it thus follows that E_n(x) is greater than zero for all x

not really a proof. give us some rigour trebla? : )

 

[Lukybear]

but turning point can be concave downwards. Needs rigour from Trebla.

 

[vafa]

if all turning points are above the x axis, then the absolute minimum MUST be above the x axis

Can you provide a mathematical proof for this statement? I have got a function, staring off at -10, has a maximum above the x axis, and again a minimum above x axis, and then comes back to -10 again. Here absolute minimum is -10 and the the turning points are above x axis, thus your statement seems to be false.

 

[vafa]

but turning point can be concave downwards. Needs rigour Trebla.

I do not see any problem in Trebla's proof and I believe that it is correct. Could you please tell me what does showing E_n>0 have to do with turning point being concaving downwards? It is not important whether your turning points is concave downwards or upwards, all you want to show is that the y value of turning points is always a positive value.

 

[Lukybear]

Im referring to second question. Prove E(x)>0 for all real x. Un said if all t.p laid above y, then it is true. But thats not as you said. If only their E"(x)>0, then it would be the case.

 

[Cazic]

A differentiable function E_n(x) is greater than 0 for all x if all the turning points occur "above the x-axis" and



Since E_n(x) is a polynomial (and hence differentiable everywhere) and n is even, both limits tend to infinity.

 

[Iruka]

There is no way you can add a whole bunch of positive terms together and get zero.
Hence, if 1 + x₀/1! + x₀²/2! + .... + x₀^n-1/(n-1)! = 0,
then x₀<0.

En''(x)= 1 + x/1! + x²/2! + .... + x^n-2/(n-2)! =En'(x)- x^n-1/(n-1)!

So, E''(x₀) = En'(x₀)- x₀^n-1/(n-1)!
= - x₀^n-1/(n-1)!, which is positive.

 

[Cazic]

This is an otherwise method proof.

solution (tex3141592)

I have some problems with this.

In using the log function, which is the inverse of the exponential function (a fact that you use later on in your proof) and only defined for positive y, you're implicitly assuming the exponential function is positive before you start. But that is the fact you're trying to prove. This is why the last paragraph in your proof doesn't require n to be even.

There are other small problems throughout, like you say that "obviously as h -> 0, n -> oo". Well, no, only if we approach from the right.

Sorry

 

[addikaye03]

When I say n tentds to infinity, I mean it tends to plus and minus infinity (That is how I was taught, maybe nowadays, they write infinity for plus infinity but when I say infinity, I really mean plus and minus infinity.

I have not assumed that log functon is only defined for positive y value, in fact I could write |y| instead y, but I guess you might be right as there is no way to avoid this. But that should not make the proof, incorrect though.

Yeah that proof is a pretty well known for proving that statement

 

[Cazic]

When I say n tentds to infinity, I mean it tends to plus and minus infinity (That is how I was taught, maybe nowadays, they write infinity for plus infinity but when I say infinity, I really mean plus and minus infinity.

I have not assumed that log functon is only defined for positive y value, in fact I could write |y| instead y, but I guess you might be right as there is no way to avoid this. But that should not make the proof, incorrect though.

Let me clearer:

You want to show that the range of the one-to-one function f is positive. You start by considering the inverse of f, whose domain is positive. But this is equivalent to what you want to show, that is, you start off by assuming what you want to show. The comments about n at the end are meaningless.