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Calculus 1 proof

  1. Nov 5, 2009 #1
    given F(x)=[tex]\sum[/tex]akxk (0<k<n) a polynomial function of factor n while n is an odd number and an<0 , prove that :

    F(1/x) gets negative values for a small enough positive x
     
  2. jcsd
  3. Nov 5, 2009 #2

    Mark44

    Staff: Mentor

    [tex]F(x)~=~\sum_{k = 0}^n a_k x^k[/tex]

    Expanded, the sum looks like this:
    F(x) = a0 + a1x + a1x2 + ... + anxn.
    With the given conditions that n is an odd integer and an < 0, what do you get for F(1/x)?
     
  4. Nov 5, 2009 #3
    F(x)=[tex]\sum[/tex][tex]\frac{ak}{xk}[/tex]

    =a0+[tex]\frac{a1}{x}[/tex]+[tex]\frac{a2}{x2}[/tex]+...+[tex]\frac{an}{xn}[/tex]

    but i dont see anything here, all i see is that i am dividing by something which gets exponentially smaller, but i dont know what a1....an is, what i think from this is that since i am subtracting bigger numbers each time, it will become negative - but this is hardly a proof,
     
  5. Nov 5, 2009 #4

    Mark44

    Staff: Mentor

    You need to use your conditions that an < 0, n is odd, and x is a small positive number.
     
  6. Nov 5, 2009 #5
    how about this

    F(x)= xn*[a0/xn + a1x/xn + a2x2/xn......+an]


    F(1/x)= 1/xn*[a0*xn + a1x*xn + a2x2*xn......+an]


    lim F(1/x)= 1/xn*[0 + 0 +....an)
    n->inf
    x->0

    =an/xn
    and since an is negative F(1/x) is negative,


    is this correct?
     
  7. Nov 6, 2009 #6

    Mark44

    Staff: Mentor

    No. When you took the limit, you have two things changing - n and x - but you still ended up with x^n. Doesn't work that way. Also, you didn't use the given information that n is an odd integer.

    Think about what the graph of F(x) looks like. It's an odd-degree polynomial, with the coefficient of the highest-degree term being negative. So it must be true that
    [tex]\lim_{x \rightarrow -\infty} F(x) = \infty[/tex]
    and
    [tex]\lim_{x \rightarrow \infty} F(x) = -\infty[/tex]

    In other words, when x is very negative, F(x) will be very large and positive, and when x is very positive, F(x) will be very negative. This is the behavior of all odd-degree polynomials when the leading coefficient is negative. Another way to say this is that the graph starts out high in the 2nd quadrant, and ends up low in the 4th quadrant, with possibly some squiggles in between that could represent local maxima and local minima.

    Since F is a polynomial, it is continuous, and by the Intermediate Value Theorem, there must be at least one number, say x0, for which F(x0) = 0. There could be more values, but all we need is one.

    Now, sketch a graph of y = F(1/x), and think about what happens to this graph in comparison to the other graph, y = F(x).
     
  8. Nov 6, 2009 #7
    What difference does it make whether n is odd or even? F(x) will be negative for large x anyway.
     
  9. Nov 6, 2009 #8

    Mark44

    Staff: Mentor

    True, but an even-degree polynomial doesn't necessarily have any zeros. For example, f(x) = x^2 + 1 doesn't have any (real) zeroes.
     
  10. Nov 6, 2009 #9
    i sort of get what you are sating, the graph of (1/x) have an asymptote at x=0 with

    lim =infinity
    x->0-

    lim =-infinity
    x->0+

    lim = ao
    x->inf

    therefore there will obviously be some value for x that f(1/x) will be negative since its limit at 0+ is negative,
    but is this enough for a proof?
     
  11. Nov 6, 2009 #10
    But wouldn't not having any zeroes make the proof much simpler since then the function would just be negative everywhere? =)


    It really depends alot on what kind of course you are taking. Atleast you need to argue that F is continuous and that after certain point x_0, for all x>x_0, F(x_0) will be negative.
     
  12. Nov 6, 2009 #11
    im doing calculus1 at university for a computer science degree,

    so i would argue that since F(x) is continuous and reaches +/- infinity at its ends, there must be some point Xo that F(Xo)=0,
    then i sya that F(1/x) around +/-0 behaves like F(x) around +/-inf

    if F(Xo)=0 then F(1/xo) must do what? is it also 0? how can i prove that?


    i know that lim(x->inf)(f(1/x)=ao, then even if ao is positve, there must be an xo such that 0<x<xo returns negative values
     
  13. Nov 6, 2009 #12

    Mark44

    Staff: Mentor

    There is no evidence that there will be a zero at x = 0. There will be a zero somewhere, though. You can assume that for some number x0, F(x0) = 0.

    This means that for the graph of y = F(t) = F(1/x), there is a vertical asymptote at t = 1/x0. The left- and right-sided limits are more-or-less as you say above except they are as t --> 1/x0. I don't know what you mean with this, though:
    lim = ao
    x->inf
     
  14. Nov 6, 2009 #13

    Mark44

    Staff: Mentor

    Maybe or maybe not, but the point is that it is given that F is an odd-degree polynomial.
     
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