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

**bob000** · July 15th 2011, 08:24 PM

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"

**Prove It** · July 15th 2011, 08:52 PM

The derivative of a constant is 0. 0 is a constant. So the derivative of 0 is 0.

**Soroban** · July 16th 2011, 02:49 AM

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

**Prove It** · July 16th 2011, 03:05 AM

Originally Posted by Soroban

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...

**FernandoRevilla** · July 16th 2011, 03:12 AM

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 $x\in\mathbb{R}$ .

**Prove It** · July 16th 2011, 03:50 AM

Originally Posted by FernandoRevilla

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...

**FernandoRevilla** · July 16th 2011, 04:02 AM

Originally Posted by Prove It

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 $e^x$ ?

**Prove It** · July 16th 2011, 04:04 AM

Originally Posted by FernandoRevilla

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.

**FernandoRevilla** · July 16th 2011, 04:09 AM

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

**Prove It** · July 16th 2011, 04:15 AM

Originally Posted by FernandoRevilla

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.

**FernandoRevilla** · July 16th 2011, 04:40 AM

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 $(x^n)'=nx^{n-1}$ .

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

**HallsofIvy** · July 16th 2011, 04:57 AM

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 $de^x/dx= e^x$ .

**FernandoRevilla** · July 16th 2011, 05:08 AM

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 $f$ satisfying $f'(x)=f(x)$ .

**chisigma** · July 16th 2011, 11:43 AM

Originally Posted by FernandoRevilla

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

$\chi$ $\sigma$

**FernandoRevilla** · July 16th 2011, 11:57 AM

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' $f(x)=0$ ...] 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.

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