Proving Polynomial Equation: a0+a1x+a2x2+...+anxn=0

[Vaal]

Homework Statement

Prove that if a₀+a₁x+a₂x²+a₃x³+...+a_nxⁿ=0 then a₀=0, a₁=0 ... a_n=0

Homework Equations

none

The Attempt at a Solution

I think I can do this for n up to 2 in the following way (please tell me if you see any gaps in my logic here):

f(x)=a₀+a₁x+a₂x²=0 (from the statement above)

f(0)=a₀+a₁(0)+a₂(0)²=a₀=0

f(1)=(0)+a₁(1)+a₂(1)²=a₁+a₂=0
f(-1)=(0)+a₁(-1)+a₂(-1)²=-a₁+a₂=0
here it can be easily shown that a₁ must be the zero vector for reals, 0, using properties of the zero vector

if a₀=0 and a₁=0 the we have:
f(1)=(0)+(0)(1)+a₂(1)²=a₂=0


I am really not sure where to go from here. At least at first glance, I can't seem to make this same process work for n=3 and even if I could use this process on higher n I'm not sure how/if I could use it to prove the general case for all n. Any help/prodding in the right direction would be much appreciated. Thanks.

 

[Dick]

That's a good start. And you can do it that way for general n, but it's gets extremely tedious as n gets large. Try thinking of it this way. If f(x)=a0+a1*x+...+an*x^n=0 then all the derivatives of f(x) must be zero. Right? Are they?

 

[Raskolnikov]

Even for large n, it's not really tedious if you prove by induction. For the "n+1"-th case, differentiate n + 1 times to solve for a_{n+1}, and then plug back in and use induction hypothesis. That should prove it holds for all n pretty easily.

 

[Vaal]

Thank you both, that helped a lot. I actually thought about taking successive derivatives before but for some reason the fact that all the derivatives would be zero for all x didn't occur to me at the time. Here is what I have so far, I ended up not really using induction so please let me know what you think/if there is a way I can make this a little more rigorous.

We know from the original statement that the function f(x)=a₀+a₁x+a₂x²+a₃x³+...+a_nxⁿ is equal to zero for all values of x. Since zero is a constant this means that the function is not changing over x and therefore has a derivative of zero at all x. Since the first derivative is 0 for all x then using the same logic we can say the second derivative is zero for all x, therefore so is the third, fourth, fifth ... all the way up to the n^th. (while I feel this logic is sound, I would love a more concise and technical way of expressing it if anyone has any ideas)

Now that we know the first, second, third, ..., n^th derivatives are all zero we just need a generalized expression for the m^th derivative of f(x) where m is an integer less than or equal to n. We can look at the first few derivatives and extrapolate the pattern (if there is a better way to do this than "extrapolate the pattern" please let me know)

m=0 -> f(x)=a₀+a₁x+a₂x²+a₃x³+...+a_nxⁿ
m=1 -> f'(x)=a₁+2a₂x+3a₃x²+...+na_nx^n-1
m=2 -> f''(x)=2a₂+6a₃x+24a₄x²+...+n(n-1)a_nx^n-2
general case -> f^m(x)=m!a_m+(m+1)!a_m+1x+(m+2)!a_m+2x²+...+n!/(m-n)!a_nx^n-m

From the logic above (second paragraph) we know f^m(x)=0 for all x, so we evaluate at x=0 and obtain:

f^m(0)=m!a_m+(m+1)!a_m+1(0)+(m+2)!a_m+2(0)²+...+n!/(m-n)!a_n(0)^n-m=0
m!a_m+0+0+...+0=0
a_m=0/m!
a_m=0

Again, please let me know how I might be able to improve this and thanks again for your help.

 

[Vaal]

After a little more thought I realized my formula for f^m(x) could be better obtained by the formulation for repeated use of the power rule, (dⁱ/dxⁱ) x^k=k!/(k-i)! x^k-i where i \leq k, rather than "extrapolate the pattern".

I'm still at a loss for a better way to express the content of the second paragraph though.

 

[Dick]

I really don't think there's anything wrong with what you have. f^(m)(0)=m!*a_m about sums it up. And stating that the mth derivative of f(x)=0 is zero? I can't really say that I think that calls for a formal proof. You could prove it for m=1 and use induction if you really want to be picky about it.

 

[Vaal]

If you don't think its needed I won't bother with it. Thanks again for all your help.