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Is the integral of 1/x
ln x og log x?
ln x og log x?
That is not a proof at all. The natural number e is found with the definition of ln, the integral of 1/x. There is no "proof", it's a definition to start with. The only thing that needs to be proven is that ln behaves as a logarithmic function.Proof:
y=ln(x)
d/dx ln(x)= dy/dx
x=e^y
dx/dy=x
so dy/dx=1/x
Whatever dude.That is not a proof at all. The natural number e is found with the definition of ln, the integral of 1/x. There is no "proof", it's a definition to start with. The only thing that needs to be proven is that ln behaves as a logarithmic function.
That is eqivalent to the definition that e is the unique number that fulfills [tex]\lim_{h\to 0} \frac{e^h -1}{h} = 1 [/tex].Whatever dude.
It's a proof given the defn. d/dx e^x=e^x and defining ln to be the inverse operation to e^x.
I remember mathwonk saying things are much easier and more rigourous if the ln function is defined as
[tex]\int_1^x \frac{1}{t} dt = \ln x[/tex]. From this definition we can define its base, and we Euler decided to call it e. So e can be defined to be the number which fulfills the condition that
[tex]\int_1^e \frac{1}{x} dx = 1[/tex]
[tex]\lim_{h\to 0} \frac{e^{x+h} - e^x}{h}...it avoids having to prove that
[tex]\lim_{x\rightarrow 0}\frac{a^x- a}{x}[/tex]
exists.