- #71
I like Serena
Homework Helper
MHB
- 16,336
- 258
Dr. Seafood said:Only semi-circular! HAHA LOLlkasfksdlkvcslamc
LOL! And I've just made it only curvy!
Dr. Seafood said:Only semi-circular! HAHA LOLlkasfksdlkvcslamc
micromass said:Seems ok
\begin{Annoying-mathematician}
Of course, this only shows that the length of a quarter of a circle is independent of pi. Nothing is said about the entire circle.
\end{Annoying-mathematician}
I like Serena said:LMAO.
Well, what about the argument that 4 times the length of a quarter of a circle is equal to the length of an entire circle?
To make it Annoying-mathematician'ally clad, I guess we should define 4 integrals for each quarter of the circle.
But as my prof liked to say in his proofs: this is left as an exercise to the reader.
micromass said:But the function is not bounded. And the function can never be Riemann integrable since it has poles in 1 and -1. So you're working with improper Riemann integrals...
Now, improper Riemann integrals have been created to extend Riemann integration to some unbounded functions, but it is another interesting fact about the generalized Riemann integral that any function having an improper integral must already be integrable in the sense described in Definition 8.1.6.
micromass said:Now you see why my profs were always annoyed to have me as a student
My approach was as follows. Step 1: Prove that L(C)/r is independent of a,b and r. Step 2: Define [itex]\pi=L(C)/(2r)[/itex]. (Without step 1, step 2 doesn't make sense). However, as HallsofIvy and micromass has already mentioned, it's not obvious that my way of doing step 1 makes sense. To be more specific, it's not obvious that the function [itex]x\mapsto\frac{1}{\sqrt{1-x^2}}[/itex] is integrable on [-1,1]. It is, but that must be proved separately. If we can prove that, my calculation would then show that [tex]\int_{-1}^1\frac{1}{\sqrt{1-x^2}}\,dx=\pi.[/tex] This means that one of many possible ways to find approximations of the value of [itex]\pi[/itex] would be to do that integral numerically.Kindayr said:So would you just define [itex]\pi = \int_{-1}^1 \frac{1}{\sqrt{1-s^2}} ds[/itex] with [itex]s=\frac{t}{r} [/itex] and leave the approximation to someone else?
The function [itex]x\mapsto\frac{1}{\sqrt{1-x^2}}[/itex] defined on [-1,1] doesn't satisfy that requirement. I'm guessing that the null set you had in mind was {-1,1}, but the restriction of the function to the open interval (-1,1) isn't bounded.I like Serena said:What about:
"A function on a compact interval [a,b] is Riemann integrable if and only if it is bounded and continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure). "
Fredrik said:The function [itex]x\mapsto\frac{1}{\sqrt{1-x^2}}[/itex] defined on [-1,1] doesn't satisfy that requirement. I'm guessing that the null set you had in mind was {-1,1}, but the restriction of the function to the open interval (-1,1) isn't bounded.
I do believe that's true. However, Archimedes did make two significant assumptions (in addition to the assumption the perimeter exists):disregardthat said:The archimedes way (incribing and circumscribing by polygons) can be made logically sound (no circularity) without assuming any analytical properties of sin(x).
Hurkyl said:I do believe that's true. However, Archimedes did make two significant assumptions (in addition to the assumption the perimeter exists):
- The length of the circular arc truly is somewhere between the perimeters of the inscribed and circumscribed polygons
- The perimeters of the circumscribed polygons converges to the same number as the perimeters of the inscribed polygons
Which may or may not be difficult; I'm not sure.
This part, IIRC, isn't hard to prove, but it only works for the inscribed polygon. Some other method is required to show the circle's perimeter is less than that of the circumscribed polygon.disregardthat said:a straight line is the shortest path from one point to another
Fredrik said:That a rotation by 360 degrees is equivalent to no rotation at all? (Of course, that's just the definition of "degree").