for all real values of x?
Well, now you do.
That is all.
Did you know that
for all real values of x? Well, now you do. That is all. | |
May 5 2006, 11:53 pm
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Garthor
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Why not factor out a 1/4 from both sides?
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You could make hundreds of equations like that with both sides being equal no matter what number you use for x. 1x/4+1=.25x(8)/8+1
I know that equation is retarded, but if you took the time you could make hundreds of those. The only reason the one you have is mildly interesting is because somebody actually took the time to sit there and figure that out. | |
Math fails me. :(
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Why not factor out a 1/4 from both sides? I probably should have done that, yes, but the point is that those are both correct answers to the same integral (namely the integral of 1/(x*sqrt(25*x^2+16)) with respect to x), and they look completely different. One's an arcsinh and one's a natural log. Just thought it was odd that they came out looking so different, yet they're both correct. I just saw the two different answers, realised they were equivalent, and thought "woah, that's weird". Of course it becomes less interesting when you realise that arcsinh is typically defined in terms of natural logs, but it was a slight "WTF" moment and I wanted to share it. :-) | |
arcsinh() gives you natural logs because of how hyperbolic sine is constructed. Specifically, sinh(x)=(e^x-e^(-x))/2, so if you solve for the arcsinh:
x=sinh(y)=(e^y-e^(-y))/2 x=sinh(y)=(e^y-e^(-y))/2 2xe^y=e^(2y)-1 e^(2y)-2xe^y-1=0 e^y=x±sqrt(x^2+1) And since sqrt(x^2+1)>x, one of the values for e^y is negative and therefore only exists for complex numbers. The only real result is e^y=x+sqrt(x^2+1), which gives us: arcsinh(x)=ln(x+sqrt(x^2+1)) Now, to apply that to arcsinh(4/(5x)): arcsinh(4/(5|x|)) = ln(4/(5|x|)+sqrt(16/(25x^2)+1)) = ln(4+sqrt(25x^2+16))-ln(5|x|) Since (sqrt(25x^2+16)+4)(sqrt(25x^2+16)-4)=(25x^2+16)-16=25x^2, 2ln(5|x|)=ln(sqrt(25x^2+16)+4)+ln(sqrt(25x^2+16)-4). Notice that equation uses all x^2 terms, meaning it's ignoring the sign. This is why |x| is required; the substitutions we're using here aren't valid for negative numbers. Cut that in half and you get ln(5|x|)=ln(sqrt(25x^2+16)+4)/2+ln(sqrt(25x^2+16)-4)/2. Then substitute this in for ln(5|x|) in the above equation--which you can only do using |x|, not x, because in this case you're specifically ignoring signs--and you get: arcsinh(4/(5|x|)) = [ln(sqrt(25x^2+16)+4)-ln(sqrt(25x^2+16)-4)]/2 Which, when multipled by -1/4, provides your equation. The substitution leaves you with just x^2 terms on the right side, which again is why this equation requires |x| instead of x. | |
I don't know sinh.
Our calc teacher figured it wouldn't be important enough, so I didn't learn it. Meh | |