- #1
Gerenuk
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What do you think is some interesting or/and sensible way to define functions like exp(), cos() provided basic algebra rules (including integration) are known?
I make a suggestion I came up with
[tex]\ln x:=\int_1^x \frac{\mathrm{d}t}{t}[/tex]
from where key properties follow directly.
For example
[tex]\ln (xy)=\int_1^{xy} \frac{\mathrm{d}t}{t}=\int_{1/x}^y \frac{\mathrm{d}t}{t}=\int_1^y \frac{\mathrm{d}t}{t}+\int_{1/x}^1 \frac{\mathrm{d}t}{t}=\int_1^y \frac{\mathrm{d}t}{t}+\int_1^x \frac{\mathrm{d}t}{t}=\ln x+\ln y[/tex]
If we define the inverse function to be
[tex]\exp:=\ln^{-1}[/tex]
then from the logarithm rule
[tex]\exp(\ln(xy))=\exp(\ln x+\ln y)[/tex]
[tex]\exp(\ln(\exp(a)\exp(b)))=\exp(\ln \exp a+\ln \exp b)[/tex]
[tex]\exp(a)\exp(b)=\exp(a+b)[/tex]
Also it follows easily that for [itex]t=\exp x[/itex]
[tex]\frac{\mathrm{d}}{\mathrm{d}x}\exp(x)=\frac{1}{\frac{\mathrm{d}}{\mathrm{d}t}\ln t}=t=\exp x[/tex]
And hence
[tex]\exp x=1+\int_0^x \exp t\mathrm{d}t=1+\int_0^x \left(1+\int_0^t \exp t'\mathrm{d}t'\right)\mathrm{d} t=1+x+\int_0^x \int_0^t \exp t'\mathrm{d}t'\mathrm{d} t=1+x+\frac{x^2}{2!}+\dotsb[/tex]
And as I mentioned in another post I strongly support
[tex]\cos x:=\Re(\exp \mathrm{i}x)[/tex]
[tex]\sin x:=\Im(\exp \mathrm{i}x)[/tex]
So much for playing around with functions late at night
I make a suggestion I came up with
[tex]\ln x:=\int_1^x \frac{\mathrm{d}t}{t}[/tex]
from where key properties follow directly.
For example
[tex]\ln (xy)=\int_1^{xy} \frac{\mathrm{d}t}{t}=\int_{1/x}^y \frac{\mathrm{d}t}{t}=\int_1^y \frac{\mathrm{d}t}{t}+\int_{1/x}^1 \frac{\mathrm{d}t}{t}=\int_1^y \frac{\mathrm{d}t}{t}+\int_1^x \frac{\mathrm{d}t}{t}=\ln x+\ln y[/tex]
If we define the inverse function to be
[tex]\exp:=\ln^{-1}[/tex]
then from the logarithm rule
[tex]\exp(\ln(xy))=\exp(\ln x+\ln y)[/tex]
[tex]\exp(\ln(\exp(a)\exp(b)))=\exp(\ln \exp a+\ln \exp b)[/tex]
[tex]\exp(a)\exp(b)=\exp(a+b)[/tex]
Also it follows easily that for [itex]t=\exp x[/itex]
[tex]\frac{\mathrm{d}}{\mathrm{d}x}\exp(x)=\frac{1}{\frac{\mathrm{d}}{\mathrm{d}t}\ln t}=t=\exp x[/tex]
And hence
[tex]\exp x=1+\int_0^x \exp t\mathrm{d}t=1+\int_0^x \left(1+\int_0^t \exp t'\mathrm{d}t'\right)\mathrm{d} t=1+x+\int_0^x \int_0^t \exp t'\mathrm{d}t'\mathrm{d} t=1+x+\frac{x^2}{2!}+\dotsb[/tex]
And as I mentioned in another post I strongly support
[tex]\cos x:=\Re(\exp \mathrm{i}x)[/tex]
[tex]\sin x:=\Im(\exp \mathrm{i}x)[/tex]
So much for playing around with functions late at night
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