Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Constant e

  1. Nov 17, 2013 #1
    Find a number "k" such that exist only one intersection between the curve exponential k^x e and the straight x·k. This number is the constant e!

    Have you noticed this? This relationship must have many important implications and indirect that we already know...
  2. jcsd
  3. Nov 17, 2013 #2
    The [itex]e[/itex] after [itex]x^k[/itex] is unclear. Is it [itex]x^{ke}[/itex] or [itex]x^k \cdot e[/itex]?
  4. Nov 17, 2013 #3

    "between the curve exponential k^x and the straight x·k"

    I press "e" not wanting... sorry!
  5. Nov 17, 2013 #4
    Interesting. Do you know the proof? If you don't then I can post it if you are interested.
  6. Nov 17, 2013 #5
    I don't know, I just observed... Yes, I'm interesting!
  7. Nov 17, 2013 #6
    Actually, there is also only one intersection for [itex]0 < k \leq 1[/itex]. Anyway, here's the proof:

    Let [itex]k > 0[/itex]. For [itex]0 < k \leq 1[/itex], we have [itex]k^x[/itex] constant or decreasing while [itex]kx[/itex] is increasing. Hence it is easy to show that there is only one intersection, in fact at [itex]x = 1[/itex].

    Now assume [itex]k > 1[/itex] and let [itex]f(x) = k^x - kx[/itex] so that [itex]f(x) = 0[/itex] whenever [itex]k^x = kx[/itex]. Now we can notice a few things.

    Firstly, as [itex]x \to -\infty[/itex], we have [itex]k^x \to 0[/itex] and [itex]kx \to -\infty[/itex], and so [itex]f(x) \to \infty[/itex].

    Secondly, as [itex]x \to \infty[/itex], we have [itex]\frac{k^x}{kx} \to \infty[/itex] since [itex]k > 1[/itex] and so [itex]f(x) \to \infty[/itex].

    Lastly, we can see that [itex]f(1) = k^1 - 1 \cdot k = 0[/itex] for any [itex]k[/itex].

    Considering the limits shown above, if [itex]f(x) < 0[/itex] for some [itex]x[/itex], then there must be at least two zeros of [itex]f[/itex] by the intermediate value theorem. Thus, if there is only one intersection, we must have [itex]f(x) \geq 0[/itex] for all [itex]x[/itex]. So [itex]f[/itex] is a minimum at its only zero (which must be [itex]1[/itex] by our last observation above). Since [itex]f[/itex] differentiable, this means we must have [itex]f'(1) = 0[/itex]. Since [itex]f'(x) = \log(k)k^x - k[/itex] so the criteria [itex]f'(1) = 0[/itex] implies [itex]\log(k)k - k = 0[/itex]. But [itex]k > 0[/itex] and so [itex]\log(k) = 1 \implies k = e[/itex] which you observed.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook