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

[Saracen Rue]

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.

 

[MarkFL]

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:

$y(x)=Ce^x$

 

[Charles Link]

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.

 

[Mark44]

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 $e^{}$ of both sides 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.

$y = Ae^x$ represents an entire family of functions, including the one for which A = 0 ($y \equiv 0$).

 

[MarkFL]

...I do think this solution is unique...

Yes, the uniqueness is guaranteed by Picard-Lindelöf.

 

[Charles Link]

Yes, the uniqueness is guaranteed by Picard-Lindelöf.

I would like to see the OP's function. I do think he must have miscalculated.

 

[MarkFL]

I would like to see the OP's function. I do think he must have miscalculated.

I agree, on both counts.

 

[Stephen Tashi]

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.

 

[Charles Link]

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.

Perhaps he found that $y=\cosh(x)+\sinh(x)$ has $dy/dx=y$. It's just a guess, but a possibility. LOL :)

 

[Saracen Rue]

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.

 

[Mark44]

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/?i=d/dx+-(e^x+π+cos((e^x+π)/2))/(2+sqrt(1+-+sin^2((e^x+π)/2)))+=+-(e^x+π+cos((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.

$\frac{\cos(\pi/2 \cdot e^x)}{\sqrt{1 - \sin^2(\pi/2 \cdot e^x)}}$ is just a complicated way of writing 1, so your function f(x) is essentially a constant ($-\frac \pi 2$) times $e^x$.

 

[Saracen Rue]

$\frac{\cos(\pi/2 \cdot e^x)}{\sqrt{1 - \sin^2(\pi/2 \cdot e^x)}}$ is just a complicated way of writing 1, so your function f(x) is essentially a constant ($-\frac \pi 2$) times $e^x$.

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

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

 

[TeethWhitener]

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.

 

[Saracen Rue]

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.

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 $f(x)$.

 

[TeethWhitener]

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 $h(x) = 0$.

 

[I like Serena]

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.

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

The exponential function is often defined as the unique solution of the differential equation $y'=y$ such that y(0) = 1.