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Homework Help: Polynomial Proof

  1. Aug 2, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that if a0+a1x+a2x2+a3x3+....+anxn=0 then a0=0, a1=0 ... an=0

    2. Relevant equations

    3. 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)


    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:

    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.
  2. jcsd
  3. Aug 2, 2010 #2


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    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?
  4. Aug 2, 2010 #3
    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 [tex] a_{n+1} [/tex], and then plug back in and use induction hypothesis. That should prove it holds for all n pretty easily.
  5. Aug 3, 2010 #4
    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+....+anxn 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 nth. (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, ..., nth derivatives are all zero we just need a generalized expression for the mth 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+....+anxn
    m=1 -> f'(x)=a1+2a2x+3a3x2+....+nanxn-1
    m=2 -> f''(x)=2a2+6a3x+24a4x2+....+n(n-1)anxn-2
    general case -> fm(x)=m!am+(m+1)!am+1x+(m+2)!am+2x2+....+n!/(m-n)!anxn-m

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


    Again, please let me know how I might be able to improve this and thanks again for your help.
  6. Aug 3, 2010 #5
    After a little more thought I realized my formula for fm(x) could be better obtained by the formulation for repeated use of the power rule, (di/dxi) xk=k!/(k-i)! xk-i where i [tex]\leq[/tex] 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.
  7. Aug 3, 2010 #6


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    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.
  8. Aug 3, 2010 #7
    If you don't think its needed I won't bother with it. Thanks again for all your help.
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