Server Help

Trash Talk - Derivatives

Purge - Wed Sep 26, 2007 12:49 am
Post subject: Derivatives

Long time no see.

Anyway, I'm having trouble with one of my Calculus problems (it's not really a problem, just theory).

If any of you can recall derivatives of functions, you probably know that e^x is its own derivative. It is also known that only e^x is its own derivative. However, according to my teacher, there is another function NOT involving any exponential functions that is its own derivative. I just need to know this other function and how it can be proved.

I personally think he's mistaken, but this is an actual assignment... so I'm looking for a logical answer.

Blocks - Wed Sep 26, 2007 12:58 am
Post subject: it's

0

As for the proof, iono, use induction or something.

Dr Brain - Wed Sep 26, 2007 8:51 am
Post subject:

f(x) = 0 is it's own derivative, yes.

It can be proved using the linearity property, or just go straight to:

df/dt = (f(x) - f(x+dt))/t

plug in zero, and get out zero.

tcsoccerman - Wed Sep 26, 2007 4:09 pm
Post subject:

simple enough

BDwinsAlt - Wed Sep 26, 2007 6:09 pm
Post subject:

I hate functions. Useless...

Samapico - Wed Sep 26, 2007 8:00 pm
Post subject:

......functions........useless????? you're a retard or what?

Purge - Wed Sep 26, 2007 8:16 pm
Post subject:

It's not f(x) = 0, according to my teacher.

He didn't tell the solution since it's due on Friday, so more thinking for me.

CypherJF - Wed Sep 26, 2007 8:35 pm
Post subject:

actually proving calculus functions begins around abstract algebra level; which is way past the point where you run out of alpha-letters and you have to use greek.

Blocks - Thu Sep 27, 2007 2:50 am
Post subject:

I believe the only class of functions that are their on derivatives are of the form f(x) = c*e^x, where c is an arbitrary constant. If you substitute the Taylor series expansion of e^x, you technically not involving exponentials, but if you teacher means that, he is a wonk.

Animate Dreams - Fri Oct 12, 2007 11:37 am
Post subject:

CypherJF wrote:

actually proving calculus functions begins around abstract algebra level; which is way past the point where you run out of alpha-letters and you have to use greek.

Yeah, I suggest skipping your Math classes and just go straight to Ancient Greek. Much easier, and much more fun.

Blocks - Fri Oct 12, 2007 2:17 pm
Post subject:

Hey Purge, how did this all turn out?

SamHughes - Wed Oct 17, 2007 1:40 pm
Post subject:

Blocks wrote:

I believe the only class of functions that are their on derivatives are of the form f(x) = c*e^x, where c is an arbitrary constant.

Yep, you discover that by solving the differential equation f' = f and applying the uniqueness theorem.

Purge - Sun Oct 21, 2007 9:24 pm
Post subject:

Blocks wrote:

Hey Purge, how did this all turn out?

Well, I turned in my paper which was the proof of f(x) = 0 to be its own derivative, and I got full credit for it... and he said that it wasn't f(x) = 0.

Oh well, I'm not complaining.

Thanks for the help, by the way.