Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ ¹ ² ³ °
 

You are not logged in. #2 20130523 21:14:11
Re: Help me prove thishi {7/3} You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #4 20130523 23:26:37
Re: Help me prove thisThis is how I do powers and logs: at (0,1), the derivative is: Even though I don't know what that is, it will have a value; let's say k. Now the derivative at other points So all graphs in the family have the property that the gradient function at x is a^x times the gradient at (1,0) In the family there will be one value of a for which k = 1 Call that one a = e then so Now suppose Taking logs base e for the first expression: Differentiating wrt x which means we now know the value of k ... and so [still working on this last bit but I think I'll post before I lose it all] No good. I seem to be stuck here because if k  ln a this becomes ln x and I was trying hard to avoid that. I seem to have gone too far and proved the log is base e. I'll come back to it later after a think. Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #6 20130524 02:10:32
Re: Help me prove thisHi Bob The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #7 20130524 05:35:27
Re: Help me prove thisOR
Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #8 20130524 06:07:33
Re: Help me prove thisI really like that proof. Really elegant. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #10 20130524 16:40:18
Re: Help me prove thisThere are a whole set of functions that differentiate to give themselves. But they are all multiples of each other. Show this is equal to zero which means that h(x) = constant. I think that should enable you to do what you want. Bob You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei 