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Derivatives

 
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PostPosted: Wed Sep 26, 2007 12:49 am   Post maybe stupid    Post subject: Derivatives Reply to topic Reply with quote
Long time no see. sa_tongue.gif

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. icon_confused.gif
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Blocks
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PostPosted: Wed Sep 26, 2007 12:58 am   Post maybe stupid    Post subject: it's Reply to topic Reply with quote
0

As for the proof, iono, use induction or something.
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Dr Brain
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PostPosted: Wed Sep 26, 2007 8:51 am   Post maybe stupid    Post subject: Reply to topic Reply with quote
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.
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tcsoccerman
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PostPosted: Wed Sep 26, 2007 4:09 pm   Post maybe stupid    Post subject: Reply to topic Reply with quote
simple enough
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BDwinsAlt
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PostPosted: Wed Sep 26, 2007 6:09 pm   Post maybe stupid    Post subject: Reply to topic Reply with quote
I hate functions. Useless... icon_rolleyes.gif
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Samapico
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PostPosted: Wed Sep 26, 2007 8:00 pm   Post maybe stupid    Post subject: Reply to topic Reply with quote
......functions........useless????? you're a retard or what?
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PostPosted: Wed Sep 26, 2007 8:16 pm   Post maybe stupid    Post subject: Reply to topic Reply with quote
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. sa_tongue.gif
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CypherJF
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PostPosted: Wed Sep 26, 2007 8:35 pm   Post maybe stupid    Post subject: Reply to topic Reply with quote
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. tongue.gif
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Blocks
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PostPosted: Thu Sep 27, 2007 2:50 am   Post maybe stupid    Post subject: Reply to topic Reply with quote
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.
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Animate Dreams
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PostPosted: Fri Oct 12, 2007 11:37 am   Post maybe stupid    Post subject: Reply to topic Reply with quote
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. tongue.gif


Yeah, I suggest skipping your Math classes and just go straight to Ancient Greek. Much easier, and much more fun.
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Blocks
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PostPosted: Fri Oct 12, 2007 2:17 pm   Post maybe stupid    Post subject: Reply to topic Reply with quote
Hey Purge, how did this all turn out?
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SamHughes
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PostPosted: Wed Oct 17, 2007 1:40 pm   Post maybe stupid    Post subject: Reply to topic Reply with quote
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.
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PostPosted: Sun Oct 21, 2007 9:24 pm   Post maybe stupid    Post subject: Reply to topic Reply with quote
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. sa_tongue.gif

Thanks for the help, by the way.
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