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Help me prove this:
for some constant a[i cannot use the fact this is ln(x)]Last edited by {7/3} (2013-05-22 18:43:33)
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hi {7/3}
Are you wanting a proof from first principles? I usually start with the derivative of a^x, then e^x, then reverse these for the log.
If you may assume d(e^x)/dx = e^x then it will only take a few lines.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Proof from first principles will be better
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This is how I do powers and logs:
For the function
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
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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OR
I suspect there's a circular argument lurking here as power series probably depend on natural logs somewhere, but maybe it's ok.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi Bob
The Taylor series uses the derivatives, so you are still using the derivative of log there. Unless we get the Taylor series of log in a different manner.
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OR
This was posted by yeyui on
http://forums.xkcd.com/viewtopic.php?f=17&t=36281
I have kept the post but edited the variables to suit your problem.
First some change of variable magic:
Now apply this magic to the function of interest
So this function has the property f(ab)=f(a)+f(b) which means that it is some logarithm.
Now "just" evaluate it at any particular point to show that it is the right one.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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I really like that proof. Really elegant.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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That was an awesome proof,but i need one more favor,if f'(x)=f(x) and f(0)=1 than f(x)=a^x for some constant a,how do i prove this?
There are 10 kinds of people in the world,people who understand binary and people who don't.
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There are a whole set of functions that differentiate to give themselves. But they are all multiples of each other.
That is, if f'(x) = f(x) and g'(x) = g(x) then f = kg for some constant k.
Proof:
Consider
Show this is equal to zero which means that h(x) = constant.
I think that should enable you to do what you want.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thanks
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