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B A function such that f(x)=f'(x)? (not e^x)

  1. Sep 19, 2017 #1
    Hello all,

    I was just experimenting around with some derivatives and ended up coming across a function whose derivative is the same as the original function. I know that the derivative of e^x is e^x, but I didn't know it was possible for other functions to also follow this. I'm just wondering if this is a rare occurrence or if it's not really big deal.
     
  2. jcsd
  3. Sep 19, 2017 #2

    MarkFL

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    To find the family of functions equal to their own first derivative, we can solve the ODE:

    ##\dfrac{dy}{dx}=y##

    For which the solution is:

    ##y(x)=Ce^x##
     
  4. Sep 19, 2017 #3

    Charles Link

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    I do think you likely made an error in your calculation. Since ## dy/dx=y ## this can be written as ## dy/y =dx ## so that ## ln|y|=x+C ##. Taking an expontial of both sides of the equation gives ## y=A e^x ##. I do think this solution is unique. If you have ## dy/dx=y ##, the function would be expressible in this form.
     
  5. Sep 19, 2017 #4

    Mark44

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    ##y = Ae^x## represents an entire family of functions, including the one for which A = 0 (##y \equiv 0##).
     
  6. Sep 19, 2017 #5

    MarkFL

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    Yes, the uniqueness is guaranteed by Picard-Lindelöf.
     
  7. Sep 19, 2017 #6

    Charles Link

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    I would like to see the OP's function. I do think he must have miscalculated.
     
  8. Sep 19, 2017 #7

    MarkFL

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    I agree, on both counts. :biggrin:
     
  9. Sep 19, 2017 #8

    Stephen Tashi

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    The OP doesn't say he found two distinct families of functions that satisfy f'(x) = f(x). It only says he found two distinct functions with that property.
     
  10. Sep 19, 2017 #9

    Charles Link

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    Perhaps he found that ## y=\cosh(x)+\sinh(x) ## has ## dy/dx=y ##. It's just a guess, but a possibility. LOL :)
     
  11. Sep 19, 2017 #10
    The function I found is

    ##\frac{-\pi e^x\cos \left(\frac{\pi e^x}{2}\right)}{2\sqrt{1-\sin ^2\left(\frac{\pi e^x}{2}\right)}}##

    If you tell WolframAlpha to solve for this function equaling it's derivative it outputs the result "True" (you need to leave the page to load for a few seconds though). Here's a link to it: https://www.wolframalpha.com/input/...os((e^x+π)/2))/(2+sqrt(1+-+sin^2((e^x+π)/2)))

    At first I thought this may be a big deal until I simplified the function and found that it actually has a general form of ##f\left(x\right)=e^x\cdot \frac{h\left(x\right)}{\left|h\left(x\right)\right|}##. So basically I discovered that:

    If: ##f\left(x\right)=e^x\cdot \frac{h\left(x\right)}{\left|h\left(x\right)\right|}##

    Then: ##f'\left(x\right)=e^x\cdot \frac{h\left(x\right)}{\left|h\left(x\right)\right|}##

    I'm not sure if this is of any significance or not.
     
  12. Sep 19, 2017 #11

    Mark44

    Staff: Mentor

    ##\frac{\cos(\pi/2 \cdot e^x)}{\sqrt{1 - \sin^2(\pi/2 \cdot e^x)}}## is just a complicated way of writing 1, so your function f(x) is essentially a constant (##-\frac \pi 2##) times ##e^x##.
     
    Last edited: Sep 20, 2017
  13. Sep 20, 2017 #12
    ##\frac{\cos \left(\frac{\pi }{2}e^x\right)}{\sqrt{1-\sin ^2\left(\frac{\pi }{2}e^x\right)}}## is only equal to 1 when ##\cos \left(\frac{\pi }{2}e^x\right)>0##. When ##\cos \left(\frac{\pi }{2}e^x\right)<0## it equals -1. Referring back to ##\frac{-\pi e^x\cos \left(\frac{\pi }{2}e^x\right)}{2\sqrt{1-\sin ^2\left(\frac{\pi }{2}e^x\right)}}##, the graph generated from this has values which alternate between being positive and negative depending on whether the solution is positive or negative. This is what it looks like graphically:

    [tex]f\left(x\right)=\frac{-\pi e^x\cos \left(\frac{\pi }{2}e^x\right)}{2\sqrt{1-\sin ^2\left(\frac{\pi }{2}e^x\right)}}[/tex]
    upload_2017-9-20_17-24-44.png
     
    Last edited by a moderator: Sep 20, 2017
  14. Sep 20, 2017 #13

    TeethWhitener

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    Differentiating
    $$f(x) = e^x\frac{h(x)}{|h(x)|}$$
    by the chain rule gives
    $$f'(x) = e^x\frac{h(x)}{|h(x)|}+ e^x\frac{d}{dx}\left(\frac{h(x)}{|h(x)|}\right)$$
    But the function
    $$\frac{h(x)}{|h(x)|}$$
    simply gives the sign of ##h(x)## at ##x##. This is a constant in the areas where ##h(x)\neq 0##, so its derivative ##dh/dx## is equal to zero in those areas. Where ##h=0##, the derivative is undefined.
     
  15. Sep 20, 2017 #14
    Exactly, this means that ##e^x\cdot \frac{d}{dx}\left(\frac{h\left(x\right)}{\left|h\left(x\right)\right|}\right)=0## for all values of ##x## except for ##h(x)=0##.

    Therefore ##f'\left(x\right)=e^x\cdot \frac{h\left(x\right)}{\left|h\left(x\right)\right|},\ h(x)\neq0## which is same as the original function ##f(x)##.
     
  16. Sep 20, 2017 #15

    TeethWhitener

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    Right. ##h(x)/|h(x)|## is piecewise constant, so if we multiply it by any function ##f(x)##:
    $$g(x) = f(x)\frac{h(x)}{|h(x)|}$$
    and differentiate, we get:
    $$g'(x) = f'(x)\frac{h(x)}{|h(x)|}$$
    where ##h(x) \neq 0##, and the derivative is undefined where ##h(x) = 0##.
     
  17. Sep 20, 2017 #16

    I like Serena

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    It's actually one of the (equivalent) definitions of the exponential function.
    See Exponential function on wiki, which says:
     
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