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Don't understand the fundamentals of this problem using MVT

  1. Nov 23, 2016 #1
    1. The problem statement, all variables and given/known data
    http://prntscr.com/daze68

    What I don't understand:

    1. "P be a polynomial with degree n"
    do these equations satisfy this description?:
    $$p(x) = (x^2 + x)^n$$
    $$p(x) = (5x^2 + 2x)^n$$
    etc.
    2. "C1 is a curve defined by y=p(x)"
    c1 is essentially just the curve of the polynomial?

    3. p(x) = e^x
    This is meant to find points of intersection(which are also called roots? Correct me if i'm wrong) right? That there are n solutions to this?

    4. h(x) = p(x) - e^x
    Where does this h(x) function come from? I thought this would equal to 0 if you rearrange the equation from
    3. Is it just a function introduced for the sake of answering this problem?

    5. "Since p is of degree n, the nth derivative is just a constant"
    What is this saying? That after differentiation n is still a constant? Since it's a polynomial it has to be some constant to begin wtih right?

    6. How do they go from h(x) = p(x) - e^x to h^(n)(x) = 0 and why does it have only one solution/root????
    If you differentiate then it becomes h'(x) = p'(x) - e^x right?

    I know that the MVT states( i think ) that the derivative will have n-1 roots. There exists a root between the interval of its original function but how does that apply here? Sorry if this is inappropiate for a question i just don't understand the problem at its core.
     
  2. jcsd
  3. Nov 23, 2016 #2

    fresh_42

    Staff: Mentor

    No. They are of degree ##2n##.
    Yes, or better: the graph ## \{ (x,p(x)) \, \vert \, x \in \mathbb{R}\}## of the polynomial.
    Yes. The term root means "solution of the equation". And at most ##n##. There might be less.
    Just a definition. It is a "help" function. Instead of looking for solutions of ##p(x)=e^x## we now may look for zeroes of ##h(x)##.
    So, the answer is yes. ##e^x## looks like a bent ##x-##axis. So the goal is to show, that what applies to the flat version also applies to the curved version.
    I don't understand what's your issue here. ##x^n## becomes ##const. \, x^1## after ##n-1## differentiations, ##const.## after ##n## and ##0## after ##n+1## differentiations. ##n## as the exponent is reduced by ##1## with every differentiation. But the (constant) coefficients of the highest term changes, however only from one constant to another, e.g. ##2x^n \rightarrow 2nx^{n-1} \rightarrow 2n(n-1)x^{n-2} \rightarrow \ldots ##
    Yes, and if you do it ##n## times, then it reads ##h^{(n)}(x)=const. \, - \, e^x## which has only one point, where ##h^{(n)}=0## is possible. (Why?)
    The trick is to go upwards again from ##h^{(n)}=0## since ##h^{n}(x) = h'\, (h^{(n-1)}(x) = 0## and climb back to ##h(x)## of which we want to find at most ##n## zeroes (roots). But I haven't read nor done the proof, so this is to be taken with a grain of salt.
     
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