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"

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.

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

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

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

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

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

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

Quote:

Originally Posted by

**Soroban** Hello, bob000!

I don't see any difficulty.

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

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

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

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

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

Assuming of course that you have proven what the integral of $\displaystyle \displaystyle \frac{1}{x}$ is...

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

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

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

Quote:

Originally Posted by

**FernandoRevilla** Using the theory of functional series: $\displaystyle \sum_{n=0}^{+\infty}\frac{x^n}{n!}$ has ratio of convergence $\displaystyle \rho=+\infty$. Using the uniform convergence and denoting $\displaystyle f(x):=\sum_{n=0}^{+\infty}\frac{x^n}{n!}$ we have $\displaystyle f'(x)=\sum_{n=0}^{+\infty}\frac{x^n}{n!}=f(x)$ for all $\displaystyle x\in\mathbb{R}$ .

Sorry but this is circular reasoning, as to get this series, you need to know what the derivative of e^x is...

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

Quote:

Originally Posted by

**Prove It** Sorry but this is circular reasoning,

Obviously it isn't.

Quote:

as to get this series, you need to know what the derivative of e^x is...

Could you explain where I have mentioned $\displaystyle e^x$ ?

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

Quote:

Originally Posted by

**FernandoRevilla** Obviously it isn't.

Could you explain where I have mentioned $\displaystyle e^x$ ?

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.

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

Quote:

Originally Posted by

**Prove It** 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 $\displaystyle \sum_{n=0}^{+\infty}x^n/n!$ without knowing what the function $\displaystyle e^x$ is ?

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

Quote:

Originally Posted by

**FernandoRevilla** Are you saying that I can't choose the series $\displaystyle \sum_{n=0}^{+\infty}x^n/n!$ without knowing what the function $\displaystyle e^x$ 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.

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

Quote:

Originally Posted by

**Prove It** 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 $\displaystyle (x^n)'=nx^{n-1}$ .

P.S. It is not necessary to consider (b) if you don't want.

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 "$\displaystyle e^x$", 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, $\displaystyle L(x)= \int_1^x \frac{1}{t}dt$, show that this function has all properties of the logarithm: L(xy)= L(x)+ L(y), $\displaystyle L(x^y)= yL(x)$ 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 $\displaystyle y^{1/x}= E(1)$ and then$\displaystyle y= E(1)^x$. That is, defining e to be E(1), we have shown that E(x) **is** $\displaystyle e^x$.

(bob000, I would not call this "pulling a formula out of our hat". We know that the integral of $\displaystyle x^n$ is $\displaystyle \frac{1}{n+1}x^{n+1}$ 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 $\displaystyle e^x$ is handled in most texts. Look at the difference quotient: for any positive number a, $\displaystyle \frac{a^{x+h}- a^x}{h}= \frac{a^xa^h- a^x}{h}= a^x\frac{a^h- 1}{h}$ Then $\displaystyle \frac{da^x}{x}= \lim_{h\to 0}\frac{a^{x+h}- a^x}{h}= a^x\lim_{h\to 0}\frac{a^h- 1}{h}$ is just a matter of showing that $\displaystyle \lim_{h\to 0}\frac{a^h- 1}{h}$ **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 $\displaystyle a^x$ is a constant, depending continuously on a, time $\displaystyle a^x$. 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 $\displaystyle de^x/dx= e^x$.

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

Quote:

Originally Posted by

**HallsofIvy** 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 $\displaystyle f$ satisfying $\displaystyle f'(x)=f(x)$ .

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

Quote:

Originally Posted by

**FernandoRevilla** That is the way many authors (for example Serge Lang) prove the existence of a function $\displaystyle f$ satisfying $\displaystyle f'(x)=f(x)$ .

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' $\displaystyle f(x)=0$...] do satisfy the condition $\displaystyle y^{'} = y$... may be that somebody has forgotten to say that the exponential function $\displaystyle y(x)= e^{x}$ satisfies the DE $\displaystyle y^{'}=y $ with the 'initial condition' $\displaystyle y(0)=1$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

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

Quote:

Originally Posted by

**chisigma** 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' $\displaystyle f(x)=0$...] do satisfy the condition $\displaystyle y^{'} = y$... may be that somebody has forgotten to say that the exponential function $\displaystyle y(x)= e^{x}$ satisfies the DE $\displaystyle y^{'}=y $ with the 'initial condition' $\displaystyle y(0)=1$...

Bah!, don't trust on marginalized functions.