Learning Feynman's Trick for Integrals
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Learning Feynman's Trick for Integrals
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Re: Learning Feynman's Trick for Integrals
#2It starts off with a pretty major error.
I'(t)=\int_0^1 \partial/(\partial t)((x^t - 1)/(ln x))dx = \int_0^1 x^t dx=1/(t+1), when it is actually equal to \int_0^1 x^{t-1}/ln(x)dx.
These two are definitely not always equal to each other.
Re: Learning Feynman's Trick for Integrals
#3[deleted]
Re: Learning Feynman's Trick for Integrals
#4My issue with both this and u-substitution is that you don't know what expression to use. There are a LOT of expressions that plausibly simplify the integral. But you have to do a bunch of algebra for each one (and not screw it up!), without really knowing whether it actually helps.
OTOH, if I'm given the expression, it's just mechanical and unrewarding.
Re: Learning Feynman's Trick for Integrals
#5It starts off with a pretty major error. I'(t)=\int_0^1 \partial/(\partial t)((x^t - 1)/(ln x))dx = \int_0^1 x^t dx=1/(t+1), when it is actually equal to \int_0^1 x^{t-1}/ln(x)dx. These two are definitely not always equal to each other.
No, it is correct. The integral is with respect to x, and the ordinary/partial derivatives are with respect to t. Written out fully, the derivative computation is
d/dt (x^t - 1)/ln(x) = d/dt [exp(ln(x)t) - 1]/ln(x) = ln(x)exp(ln(x)t)/ln(x) = exp(ln(x)t) = x^t.
Edit: d/dt exp(ln(x)t) = ln(x)exp(ln(x)t) by the chain rule, while d/dt (1/ln(x)) = 0 since the expression is constant with respect to t.
There are convergence considerations that were not discussed in the blog post, but the computations seem to be correct.
Re: Learning Feynman's Trick for Integrals
#6I just finished Mathematica by David Bessis and I wish this information was presented in the way he talks about math: using words and imagery to explain what is happening, and only using the equations to prove the words are true.
I just haven’t had to use integral calculus in so many years, I don’t recall what the symbols mean and I certainly don’t care about them. That doesn’t mean I wouldn’t find the problem domain interesting, if it was expressed as such. Instead, though, I get a strong dose of mathematical formalism disconnected from anything I can meaningfully reason about. Too bad.
Re: Learning Feynman's Trick for Integrals
#7I just finished Mathematica by David Bessis and I wish this information was presented in the way he talks about math: using words and imagery to explain what is happening, and only using the equations to prove the words are true. I just haven’t had to use integral calculus in so many years, I don’t recall what the symbols mean and I certainly don’t care about them. That doesn’t mean I wouldn’t find the problem domain…
That's one of the things I like best about https://betterexplained.com -- it focuses on ways to gain intuition about a given math concept, using visuals and metaphors as appropriate. If only math education were always presented like that....
Re: Learning Feynman's Trick for Integrals
#8My issue with both this and u-substitution is that you don't know what expression to use. There are a LOT of expressions that plausibly simplify the integral. But you have to do a bunch of algebra for each one (and not screw it up!), without really knowing whether it actually helps. OTOH, if I'm given the expression, it's just mechanical and unrewarding.
That’s how most of math works past high school. It requires a lot of practice and intuition.
Re: Learning Feynman's Trick for Integrals
#9It’s interesting he mentions he doesn’t like contour integration since many integrals can be done either way.
Feynman’s trick is equivalent to extending it into a double integral and then switching the order of integration.
Re: Learning Feynman's Trick for Integrals
#10My issue with both this and u-substitution is that you don't know what expression to use. There are a LOT of expressions that plausibly simplify the integral. But you have to do a bunch of algebra for each one (and not screw it up!), without really knowing whether it actually helps. OTOH, if I'm given the expression, it's just mechanical and unrewarding.
That’s how most of math works past high school. It requires a lot of practice and intuition.
I don't know about this particular case though, I get the feeling there's a system to it that can be exploited by eg Wolfram. It's just that you're in the dark for a long time before you find the switch.