Page 1 of 2 12 LastLast
Results 1 to 15 of 16

Math Help - Proof of the existence of a function f such the f(x)=f'(x)

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    35
    Thanks
    1

    Proof of the existence of a function f such the f(x)=f'(x)

    A lot of the proofs I've found so far are subtly circular because they use properties of the exponential or logarithm function that are derived from the fact (e^x)'=e^x

    One good proof I've found defined lnx as the integral of 1/x and worked backwards via partial derivatives before using the proof to derive further properties of e and ln. Is there a more direct method to prove existence than this "I pulled an equation out of my hat (the inverse of the integral of 1/x) and am going to show that it's a solution to my DE thus proving existence"
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293

    Re: Proof of the existence of a function f such the f(x)=f'(x)

    The derivative of a constant is 0. 0 is a constant. So the derivative of 0 is 0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,682
    Thanks
    614

    Re: Proof of the existence of a function f such the f(x)=f'(x)

    Hello, bob000!

    I don't see any difficulty.


    We have: . f'(x) \,=\,f(x) \quad\Rightarrow\quad \frac{dy}{dx} \,=\,y

    Separate variables: . \frac{dy}{y} \,=\,dx

    Integrate: . \int \frac{dy}{y} \:=\:\int dx \quad\Rightarrow\quad \ln y \:=\:x + c

    Exponentiate: . y \:=\:e^{x+c} \:=\:e^x\cdot e^c \:=\:e^x\cdot C


    Therefore: . y \:=\:Ce^x

    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293

    Re: Proof of the existence of a function f such the f(x)=f'(x)

    Quote Originally Posted by Soroban View Post
    Hello, bob000!

    I don't see any difficulty.


    We have: . f'(x) \,=\,f(x) \quad\Rightarrow\quad \frac{dy}{dx} \,=\,y

    Separate variables: . \frac{dy}{y} \,=\,dx

    Integrate: . \int \frac{dy}{y} \:=\:\int dx \quad\Rightarrow\quad \ln y \:=\:x + c

    Exponentiate: . y \:=\:e^{x+c} \:=\:e^x\cdot e^c \:=\:e^x\cdot C


    Therefore: . y \:=\:Ce^x

    Assuming of course that you have proven what the integral of \displaystyle \frac{1}{x} is...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45

    Re: Proof of the existence of a function f such the f(x)=f'(x)

    Using the theory of functional series: \sum_{n=0}^{+\infty}\frac{x^n}{n!} has ratio of convergence \rho=+\infty. Using the uniform convergence and denoting f(x):=\sum_{n=0}^{+\infty}\frac{x^n}{n!} we have f'(x)=\sum_{n=0}^{+\infty}\frac{x^n}{n!}=f(x) for all x\in\mathbb{R} .
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293

    Re: Proof of the existence of a function f such the f(x)=f'(x)

    Quote Originally Posted by FernandoRevilla View Post
    Using the theory of functional series: \sum_{n=0}^{+\infty}\frac{x^n}{n!} has ratio of convergence \rho=+\infty. Using the uniform convergence and denoting f(x):=\sum_{n=0}^{+\infty}\frac{x^n}{n!} we have f'(x)=\sum_{n=0}^{+\infty}\frac{x^n}{n!}=f(x) for all x\in\mathbb{R} .
    Sorry but this is circular reasoning, as to get this series, you need to know what the derivative of e^x is...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45

    Re: Proof of the existence of a function f such the f(x)=f'(x)

    Quote Originally Posted by Prove It View Post
    Sorry but this is circular reasoning,
    Obviously it isn't.

    as to get this series, you need to know what the derivative of e^x is...
    Could you explain where I have mentioned e^x ?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293

    Re: Proof of the existence of a function f such the f(x)=f'(x)

    Quote Originally Posted by FernandoRevilla View Post
    Obviously it isn't.



    Could you explain where I have mentioned e^x ?
    Considering this is the Taylor series for e^x, which can only be found from knowing the derivative of e^x, then it is circular reasoning.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45

    Re: Proof of the existence of a function f such the f(x)=f'(x)

    Quote Originally Posted by Prove It View Post
    Considering this is the Taylor series for e^x, which can only be found from knowing the derivative of e^x, then it is circular reasoning.
    Are you saying that I can't choose the series \sum_{n=0}^{+\infty}x^n/n! without knowing what the function e^x is ?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293

    Re: Proof of the existence of a function f such the f(x)=f'(x)

    Quote Originally Posted by FernandoRevilla View Post
    Are you saying that I can't choose the series \sum_{n=0}^{+\infty}x^n/n! without knowing what the function e^x is ?
    I am saying that there must be a reason you would choose that particular series. If it's for any other reason than "I know it's the series for e^x" then it would not be circular reasoning.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45

    Re: Proof of the existence of a function f such the f(x)=f'(x)

    Quote Originally Posted by Prove It View Post
    I am saying that there must be a reason you would choose that particular series. If it's for any other reason than "I know it's the series for e^x" then it would not be circular reasoning.
    (a) In Mathematics, if we find an "object" and we prove that satisfies a determined property, the origin or way we have found the object it is irrelevant for the validity of the proof.

    (b) Although the above consideration is sufficient to show that the argument is not circular it would be very easy to find a power series such that its derivative series is the same series using only the fact (x^n)'=nx^{n-1} .

    P.S. It is not necessary to consider (b) if you don't want.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,373
    Thanks
    1314

    Re: Proof of the existence of a function f such the f(x)=f'(x)

    I agree with Fernando Revilla- it is perfectly valid to define E(x) to be that series (once you have shown that it converges for all x). Of course, before you can write it as " e^x", you would have to show that it is a power of some number and that is not too difficult.

    Another way is to first define, as bob000 says, L(x)= \int_1^x \frac{1}{t}dt, show that this function has all properties of the logarithm: L(xy)= L(x)+ L(y), L(x^y)= yL(x) and that L maps the positive real numbers one-to-one onto the real numbers. Once you have done that, define E(x) to be the inverse function to L(x).
    Of course, if y= E(x), then x= L(y) so that, for x not 0, [tex]1= (1/x)L(y)= L(y^{1/x})[tex] so that y^{1/x}= E(1) and then y= E(1)^x. That is, defining e to be E(1), we have shown that E(x) is e^x.

    (bob000, I would not call this "pulling a formula out of our hat". We know that the integral of x^n is \frac{1}{n+1}x^{n+1} for n any integer except n= -1. It makes perfect sense to define a function to be that integral and if it turns out to be the natural logarithm, so much the better!)

    And, since, by the fundamental theorem of calculus, dL(x)/dx= 1/x, with E(x) defined as its inverse, if y= E(x) then x= L(y) so that dx/dy= 1/y. From the chain rule, then, dy/dx= y= E(x).

    But there is really nothing "circular" about the way the derivative of e^x is handled in most texts. Look at the difference quotient: for any positive number a, \frac{a^{x+h}- a^x}{h}= \frac{a^xa^h- a^x}{h}= a^x\frac{a^h- 1}{h} Then \frac{da^x}{x}= \lim_{h\to 0}\frac{a^{x+h}- a^x}{h}= a^x\lim_{h\to 0}\frac{a^h- 1}{h} is just a matter of showing that \lim_{h\to 0}\frac{a^h- 1}{h} exists. Admittedly, most texts don't do a very rigorous job of that, but once you have that it exists, it follows that the derivative of a^x is a constant, depending continuously on a, time a^x. It is clear that for some numbers, such as a= 2, that constant is less than 1 while for others, such as a= 3, that constant is larger than 1. It follows that there exist a unique value of a such that the constant is 1.

    We define "e" to be that number and so de^x/dx= e^x.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45

    Re: Proof of the existence of a function f such the f(x)=f'(x)

    Quote Originally Posted by HallsofIvy View Post
    I agree with Fernando Revilla- it is perfectly valid to define E(x) to be that series (once you have shown that it converges for all x).
    That is the way many authors (for example Serge Lang) prove the existence of a function f satisfying f'(x)=f(x) .
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5

    Re: Proof of the existence of a function f such the f(x)=f'(x)

    Quote Originally Posted by FernandoRevilla View Post
    That is the way many authors (for example Serge Lang) prove the existence of a function f satisfying f'(x)=f(x) .
    In my opinion is difficult to believe that Mr Serge Lang didn't realize that a very large family of functions [among them also the 'null function' f(x)=0...] do satisfy the condition y^{'} = y... may be that somebody has forgotten to say that the exponential function y(x)= e^{x} satisfies the DE y^{'}=y with the 'initial condition' y(0)=1...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  15. #15
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45

    Re: Proof of the existence of a function f such the f(x)=f'(x)

    Quote Originally Posted by chisigma View Post
    In my opinion is difficult to believe that Mr Serge Lang didn't realize that a very large family of functions [among them also the 'null function' f(x)=0...] do satisfy the condition y^{'} = y... may be that somebody has forgotten to say that the exponential function y(x)= e^{x} satisfies the DE y^{'}=y with the 'initial condition' y(0)=1...
    Bah!, don't trust on marginalized functions.
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: October 19th 2010, 10:50 AM
  2. [SOLVED] Existence proof
    Posted in the Number Theory Forum
    Replies: 6
    Last Post: September 23rd 2010, 09:18 AM
  3. existence proof, how to start?
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 12th 2010, 09:34 PM
  4. proof of existence of z in x < z < y
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: May 24th 2009, 10:54 AM
  5. existence of proof in connection to x^2 = y^2 mod n
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: November 13th 2008, 04:24 PM

Search Tags


/mathhelpforum @mathhelpforum