Do You Believe In Magic. Pretty Wierd little find
- Thread starter Ben-Nice
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digitalwanderer
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Hey; if half a chicken laid half an egg in half a day how much would a pound of butter weigh?
OpenGL guy
Veteran
This isn't strictly correct. The proper form is:K.I.L.E.R said:
d/dx e^x = e^x
Now, the question is, can you take the differential equation:OpenGL guy said:
d/dx y(x) = y(x)
...with the initial condition that y(0) = 1, and prove that the answer is e^x, without using anything you know about the definition of the function e^x? That is, can you prove that the function that satisfies the above differential equation is some number to the power of x, with that number being forced to be the number we define as e?
Edit: This isn't just to OpenGL guy, I don't want to put him on the spot
Just an open question to anybody out there.OpenGL guy
Veteran
Chalnoth said:Now, the question is, can you take the differential equation:
d/dx y(x) = y(x)
...with the initial condition that y(0) = 1, and prove that the answer is e^x, without using anything you know about the definition of the function e^x? That is, can you prove that the function that satisfies the above differential equation is some number to the power of x, with that number being forced to be the number we define as e?
Edit: This isn't just to OpenGL guy, I don't want to put him on the spot
Code:
y = f(x)
y = y'
1 = y'/y
antiderivative (1) = antiderivative(y'/y)
x + c = ln(y)
y = e^(x+c)
Since f(0) = 1, c = 0
y = e^xdigitalwanderer
Legend
Yup, but you would be surprised at how many people you can ask that to in a bar/pub and have them answer "half a pound".Chalnoth said:
There were a few more dumb ones like that, I'll try and remember them.
OpenGL guy
Veteran
Here's a goofy idea then:Chalnoth said:Yeah, but you're still using what you know about e^x in that derivation, I think. You may not like that, but I think it's a bit fun to consider trying to do this without using logarithms or exponentials. It's a challenge, I think, but quite possible
Code:
y = f(x)
y = y'
f(0) = 1 and, hence, f^(n)(0) = 1 (nth derivative)
Take the Taylor series at 0:
y = f(0) + f'(0)x/1! + f''(0)x^2/2! + ...
y = 1 + x/1! + x^2/2! + x^3/3! + ...
Note that this is the Taylor series for e^x.Anyway, I'll go ahead and answer my own question. Here's the method I'll use:
First, let's take the differential equation and expand it in terms of the definition of a derivative:
dy/dx = (y(x+dx) - y(x))/dx
Now, since dy/dx = y(x), we can rearrange the above equation as:
y(x+dx) = y(x)(1+dx)
...so now we have a way to get from a position y(x) to a position y(x+dx): multiply by (1+dx). So far, the above equation is only valid in the limit as dx approaches zero, but let's see if we can't find the value of y at some significant distance from x. For this, I'll define:
dx = 1/n (with the limit that n->infinity)
...and claim that we want to move a total of deltax. In order to move by 1, we would have to multiply by (1+1/n) n times. In order to move by deltax, we would have to multiply by (1+1/n) (n*deltax) times. So, we have:
y(x + deltax) = y(x) (1+1/n)^(n*deltax)
...or:
y(x + deltax) = y(x) [(1+1/n)^n]^deltax
We can use the initial condition y(0) = 1 to write:
y(x) = [(1+1/n)^n]^x
Now, in the limit that n->infinity, the expression [(1+1/n)^n] is just a number, a number that happens to be equal to e
Anyway, I like to ask this kind of question because I'd like to know if I could do calculus with special functions if the only thing that I knew was basic algebra.
First, let's take the differential equation and expand it in terms of the definition of a derivative:
dy/dx = (y(x+dx) - y(x))/dx
Now, since dy/dx = y(x), we can rearrange the above equation as:
y(x+dx) = y(x)(1+dx)
...so now we have a way to get from a position y(x) to a position y(x+dx): multiply by (1+dx). So far, the above equation is only valid in the limit as dx approaches zero, but let's see if we can't find the value of y at some significant distance from x. For this, I'll define:
dx = 1/n (with the limit that n->infinity)
...and claim that we want to move a total of deltax. In order to move by 1, we would have to multiply by (1+1/n) n times. In order to move by deltax, we would have to multiply by (1+1/n) (n*deltax) times. So, we have:
y(x + deltax) = y(x) (1+1/n)^(n*deltax)
...or:
y(x + deltax) = y(x) [(1+1/n)^n]^deltax
We can use the initial condition y(0) = 1 to write:
y(x) = [(1+1/n)^n]^x
Now, in the limit that n->infinity, the expression [(1+1/n)^n] is just a number, a number that happens to be equal to e
Anyway, I like to ask this kind of question because I'd like to know if I could do calculus with special functions if the only thing that I knew was basic algebra.
... I thought butter was made out of milk, not eggs...?
