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Proof w/ natural log and Riemann Sum

  1. Jan 14, 2006 #1
    Problem states:

    (A) Use mathematical induction to prove that for [tex]x\geq0[/tex] and any positive integer [tex]n[/tex].
    [tex]e^x\geq1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}[/tex]

    (B) Use part (A) to show that [tex]e>2.7[/tex].
    (C) Use part (A) to show that
    [tex]\lim_{x\rightarrow\infty} \frac{e^x}{x^k} = \infty[/tex]
    for any positive integer [tex]k[/tex].

    I thought that I could easily show that e to the x power was greater than 1 and if I could show that it was greater than 1 plus the Riemann sum:
    [tex]\sum_{i=1}^n \frac{x^n}{n!}[/tex] then I would have my proof...
     
  2. jcsd
  3. Jan 14, 2006 #2
    Base case x = 0 then the LHS becomes [itex]e^0 = 1[/itex] and the RHS becomes [itex]1 + 0 + ... + 0 [/itex]. So the inequality holds. For x = k then you have [tex]e^k \geq 1 + k + \frac {k^2}{2} + ... + \frac {k^n}{n!}[/tex]. Now you do x = k+1 and you have [tex]e^k e = e^{k+1} \geq e + ek + e \frac {k^2}{2} + ... + \frac {ek^n}{n!}[/tex]. Clearly [tex]e>1,\ ek> k,\ e\frac {k^2}{2} > \frac {(k+1)^2}{2}[/tex] and so on. Therefore

    [tex]e^{k+1} \geq 1 + (k+1) + \frac {(k+1)^2}{2} + ... + \frac {(k+1)^n}{n!}[/tex]
     
    Last edited: Jan 14, 2006
  4. Jan 14, 2006 #3

    HallsofIvy

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    No! x is a real variable so you can't do induction on x! The "induction variable" is n. The base case is for n=0. Prove that for x>= 0, ex>= 1 (which is easy: ex is an increasing function.)

    Now, assume that ex>= 1+ x+ (1/2)x2+ ...+ (1/k!)xk for some k. You need to prove that ex>= 1+ x+ (1/2)x2+ ...+ (1/k!)xk+ (1/(k+1)!)xk+1.

    What happens if you assume that is not true, that ex< 1+ x+ (1/2)x2+ ...+ (1/k!)xk+ (1/(k+1)!)xk+1 and differentiate both sides?
     
    Last edited: Jan 14, 2006
  5. Jan 14, 2006 #4

    shmoe

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    Not much, differentiation might not preseve the inequality (I might not see what you have in mind here?)

    You can go from k to k+1 by a definite integral though (I'll leave it to Xcron to set up).
     
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