4 ms·

> I don't need any of that stuff to prove these identities. I just need a pocket calculator. But numerical results don't establish identities, and identities d

by lutusp 12y ago

> I don't need any of that stuff to prove these identities. I just need a pocket calculator. But numerical results don't establish identities, and identities don't depend on numerical results. After all, with only numerical results for guidance, all sorts of illusory outcomes appear to mean more (or less) than they do. For example, this integral is equal to Pi: http://i.imgur.com/tJS1ZaA.png http://i.imgur.com/tJS1ZaA.png If one approximates the above integral by summing a bunch of numerical results on the interval between -1 and 1, one can approximate Pi, but the above integral equals Pi -- it's an identity. It has all sorts of advantages over a numerical result, not least of which is the fact that one cannot express Pi using a finite number of decimal places. The article's examples weren't really identities, they were just coincidences (for the most part). > Finally, is .999 really "extremely" close to an integer? This is why mathematics has it all over words for expressing certain ideas.

nilkn 12y ago
These identities are numerical results. Your integral example is not really the same sort of identity. For that reason, I agree with you that "identity" is a bit of a misnomer here.
- lutusp 12y ago
  > These identities are numerical results. You missed my point -- the examples in the linked article aren't identities, not as mathematicians use that word. http://en.wikipedia.org/wiki/Identity_(mathematics) http://en.wikipedia.org/wiki/Identity_(mathematics)
  - nilkn 12y ago
    I responded to that part of your post as well. I am fully aware of what "identity" normally means in mathematics. I didn't feel the need to make a big deal out of the word in my original post since it's somewhat tangential and I thought it'd be clear what I was referring to (since the author of the post uses the same terminology). Your point that true identities can't be proven with numerical calculations is true but not relevant, because as we're both aware these identities are not the pure sort of identity you had in mind when you wrote that post.
nitrogen 12y ago
For example, this integral is equal to Pi: http://i.imgur.com/tJS1ZaA.png http://i.imgur.com/tJS1ZaA.png It's been years since I've done any calculus. Do I understand that integral correctly as finding the area of a circle of radius 1, which is π?
- Double_Cast 12y ago
  tl;dr yes. I just spent the past few minutes trying to reverse engineer it. I'm no expert, but here's how I figure things. For the left side of the equation: The formula for a circle's area is a = PI r^2. For a unit circle, the area is simply equal to PI. Now for the right side of the equation: r^2 = x^2 + y^2 is the Pythagorean Theorem and also the equation of a circle. If we solve for y, we have y = sqrt(r^2 - x^2). In other words, the height of any circle (centered at the origin) is sqrt(r^2 - x^2) given any particular place along the x-axis. This is the same expression which is integrated. For a unit circle, we assume r = 1 and ignore everything outside of -1 and 1 on the x-axis. This is why the integral is bound between -1 and 1. So now we have integrate(-1, 1){ sqrt(1^2 - x^2) }. By integrating we get the area of the top half of the unit circle. But we also need the bottom half. So we just double everything and that's what the extra 2 is for. We now have the area of a unit circle on the right side too. QED
  - lutusp 12y ago
    All good. :)
- lutusp 12y ago
  Yes -- well, 1/2 the area of a unit circle, which is why I multiply by 2. I used Sage (http://www.sagemath.org/ http://www.sagemath.org/) to create these quick results: f(x) = sqrt(1-x^2) 2 * integrate(f(x),x,-1,1) Sage replies: pi (Sage is smart enough to knows this result is equal to pi, so it doesn't provide a numerical approximation, but simply prints "pi"). plot(f(x),x,-1,1,gridlines=True,aspect_ratio=1) Sage replies: http://i.imgur.com/GZdnT8z.png http://i.imgur.com/GZdnT8z.png The function by itself produces a line equal to 1/2 the distance across a unit circle at the coordinate provided by x. Therefore a definite integral of the function on the interval -1 to 1 should produce an area equal to 1/2 that of a unit circle, or pi/2. My Calculus tutorial: http://www.arachnoid.com/calculus/ http://www.arachnoid.com/calculus/

by lutusp 12y ago

nilkn 12y ago

These identities are numerical results. Your integral example is not really the same sort of identity. For that reason, I agree with you that "identity" is a bit of a misnomer here.

lutusp 12y ago
> These identities are numerical results. You missed my point -- the examples in the linked article aren't identities, not as mathematicians use that word. http://en.wikipedia.org/wiki/Identity_(mathematics) http://en.wikipedia.org/wiki/Identity_(mathematics)
- nilkn 12y ago
  I responded to that part of your post as well. I am fully aware of what "identity" normally means in mathematics. I didn't feel the need to make a big deal out of the word in my original post since it's somewhat tangential and I thought it'd be clear what I was referring to (since the author of the post uses the same terminology). Your point that true identities can't be proven with numerical calculations is true but not relevant, because as we're both aware these identities are not the pure sort of identity you had in mind when you wrote that post.

nitrogen 12y ago

For example, this integral is equal to Pi: http://i.imgur.com/tJS1ZaA.png http://i.imgur.com/tJS1ZaA.png It's been years since I've done any calculus. Do I understand that integral correctly as finding the area of a circle of radius 1, which is π?

Double_Cast 12y ago
tl;dr yes. I just spent the past few minutes trying to reverse engineer it. I'm no expert, but here's how I figure things. For the left side of the equation: The formula for a circle's area is a = PI r^2. For a unit circle, the area is simply equal to PI. Now for the right side of the equation: r^2 = x^2 + y^2 is the Pythagorean Theorem and also the equation of a circle. If we solve for y, we have y = sqrt(r^2 - x^2). In other words, the height of any circle (centered at the origin) is sqrt(r^2 - x^2) given any particular place along the x-axis. This is the same expression which is integrated. For a unit circle, we assume r = 1 and ignore everything outside of -1 and 1 on the x-axis. This is why the integral is bound between -1 and 1. So now we have integrate(-1, 1){ sqrt(1^2 - x^2) }. By integrating we get the area of the top half of the unit circle. But we also need the bottom half. So we just double everything and that's what the extra 2 is for. We now have the area of a unit circle on the right side too. QED
- lutusp 12y ago
  All good. :)
lutusp 12y ago
Yes -- well, 1/2 the area of a unit circle, which is why I multiply by 2. I used Sage (http://www.sagemath.org/ http://www.sagemath.org/) to create these quick results: f(x) = sqrt(1-x^2) 2 * integrate(f(x),x,-1,1) Sage replies: pi (Sage is smart enough to knows this result is equal to pi, so it doesn't provide a numerical approximation, but simply prints "pi"). plot(f(x),x,-1,1,gridlines=True,aspect_ratio=1) Sage replies: http://i.imgur.com/GZdnT8z.png http://i.imgur.com/GZdnT8z.png The function by itself produces a line equal to 1/2 the distance across a unit circle at the coordinate provided by x. Therefore a definite integral of the function on the interval -1 to 1 should produce an area equal to 1/2 that of a unit circle, or pi/2. My Calculus tutorial: http://www.arachnoid.com/calculus/ http://www.arachnoid.com/calculus/