Polynomial induction

  • Thread starter quincyboy7
  • Start date
  • #1
32
0

Homework Statement



Suppose an/n+1 +...+a0/1=0.
Prove f(x) =anxn +...+a0 has a root between zero and one.


Homework Equations



I'm pretty sure this is induction, but I'm not completely sure.
Mean Value Theorem probably

The Attempt at a Solution



Well f(0)=a0 and f(1)=an + ... + a0.
If it is induction, it is easy to show that when n=0, f(x)=0 for all x and thus it is obviously constant and has infinitely many roots between 0 and 1.

Assuming a root between 0 and 1 for n=k, the case for n=k+1 just doesn't seem very manipulable to me to create any kind of cancellation or truth arising from n=k. I'm pretty sure I have to manipulate the an/n+1 +...+a0=0 deal, but how to do that I have no clue.

Also, should I aim to prove that f(0) and f(1) have opposite signs? This gives me the end result immediately from the Mean Value Theorem. How would I accomplish that? What if they are both zero?

Is induction the right method here, or am I barking up the wrong tree?

Any help would be much appreciated.
 

Answers and Replies

  • #2
160
2
You have to find a way to turn each term [itex]x^k[/itex] into [itex]1/(k+1)[/itex]. Can you think of an operation that does something like that?
 
  • #3
Dick
Science Advisor
Homework Helper
26,258
619
Consider the polynomial g(x)=a_n*x^(n+1)/(n+1)+a_(n-1)*x^n/n+...+a0*x.
 
  • #4
32
0
So I can integrate f and get

h(x)=anxn+1/n+1 + ... + a0*x + C.

I can factor out an x and get x(anxn/n+1 +...+a0)+C.
I'm still left riddled with x-terms and I don't see how the integral can help me find many roots...
 
  • #5
Dick
Science Advisor
Homework Helper
26,258
619
So I can integrate f and get

h(x)=anxn+1/n+1 + ... + a0*x + C.

I can factor out an x and get x(anxn/n+1 +...+a0)+C.
I'm still left riddled with x-terms and I don't see how the integral can help me find many roots...
Leave off the +C. What are h(0) and h(1)?
 
  • #6
32
0
Why can I leave off the +C?

h(0)=0 (or C)
h(1)=0 (or C) as well...so no area is accumulated by f from 0 to 1. Thus the areas "cancel out" and either f is constant at 0 or f(0) has an sign opposite that of f(1)...that makes sense, but how can I write that up rigorously?
 
  • #7
Dick
Science Advisor
Homework Helper
26,258
619
You can leave off the +C because you don't care what it is. The derivative of h(x) is still f(x). Why not make it zero? If you want to make the conclusion rigorous, why don't you use the Mean Value Theorem, like you said you thought you should do?
 
  • #8
32
0
Ok, so either the function accumulates zero area everywhere (i.e. is constant at zero) and I'm done, or it accumulates some positive area and the same amount of negative area.

Question, though. Does the accumulation of positive area implies there exists an x on the interval such that f(x)>0? I mean, that's incredibly intuitive, but how would you derive that directly from, say, the definition of an integral as INT f(x)=inf{U (f, P)}=sup{L (f, P)}, where U(f,P)=the sum of the supremums of the value of f(x) on a particular interval in partititon P times the length of the interval and L(f,P)=the sum of the infimums of the value of f(x) on a particular interval in P times the length of the interval
 
  • #9
Dick
Science Advisor
Homework Helper
26,258
619
Ok, so either the function accumulates zero area everywhere (i.e. is constant at zero) and I'm done, or it accumulates some positive area and the same amount of negative area.

Question, though. Does the accumulation of positive area implies there exists an x on the interval such that f(x)>0? I mean, that's incredibly intuitive, but how would you derive that directly from, say, the definition of an integral as INT f(x)=inf{U (f, P)}=sup{L (f, P)}, where U(f,P)=the sum of the supremums of the value of f(x) on a particular interval in partititon P times the length of the interval and L(f,P)=the sum of the infimums of the value of f(x) on a particular interval in P times the length of the interval
You've already said h(0)=h(1). Why aren't you using the Mean Value Theorem instead of philosophising about 'area accumulation'?
 
  • #10
32
0
Oh! I was always assuming I should use mean value on f, not h. Alright, mean value on h implies that there is a value c on (0,1) s.t. h'(c)=0 i.e. f(c)=0.

I'm stupid.
 
  • #11
Dick
Science Advisor
Homework Helper
26,258
619
I'm stupid.
No, you're not. You just look the wrong way sometimes and keep fixated on it instead of looking around.
 
  • #12
160
2
Incidentally, you can do it using the integration method. What you need is the following theorem:

Mean value theorem for integrals:
Let [itex]f:[a,b]\to\mathbb{R}[/itex] be continuous. Then there exists [itex]x\in (a,b)[/itex] such that
[tex]\int_a^b f(t)\mathrm{d}t=(b-a)f(x)[/tex]

Proof:
Let [itex]g(x)=\int_a^x f(t)\mathrm{d}t[/itex] for [itex]x\in[a,b][/itex]. Then by the fundamental theorem of calculus, g is differentiable in (a,b), with g'(x)=f(x), and continuous in [a,b]. By the mean value theorem, there is an x in (a,b) with g'(x)=(g(b)-g(a))/(b-a)=f(x), which is what we wanted.
 

Related Threads on Polynomial induction

  • Last Post
Replies
0
Views
1K
Replies
3
Views
4K
Replies
2
Views
2K
Replies
5
Views
5K
Replies
0
Views
1K
Replies
2
Views
4K
Replies
5
Views
1K
Replies
1
Views
4K
Replies
3
Views
3K
Replies
6
Views
1K
Top