Holy crap hahahaha you're rightLol sledgehammer .
Good solution.
Rearrangement gives required result.
You can prove that the sequence is monotonically increasing by defining and then proving that(I'm sure there is a more rigourous way to prove this inequality, I think it appeared in a HSC question before)
I think he meant for it to be proven algebraically, similarly to the classic "Prove e^pi > pi^e" problem.Um ... evaluate.
Interesting, is there a direct way of proving this without computing pi/4 whatsoever?Regarding the approximation of pi by an alternating series:
Note that the series zig-zags by progressively smaller amounts. In fact we have every partial sum with an odd number of terms as an upper bound and every partial sum with an even number of terms as a lower bound. Moreover, the more terms we use the tighter these bounds are.
Yep, by proving that all partial sums taking more terms are greater/less than our partial sum, do this by pairing adjacent terms and noting that the sequence is decreasing in size.Interesting, is there a direct way of proving this without computing pi/4 whatsoever?
I can find upper bounds for pi and e using primitive methods but they are too weak, and I end up getting something lower than 1/3 :/
I was able to get e < 3 using trapezium and integral of e^x from 0 to 1.
And pi < 4 using a circle inscribed in a hexagon. But these approximations are too weak.
The approximations need to be quite strong as well since:
So the approximations need to be such that we sort for an error of 10^{-2} ....
So yeah I think we would have to rely on the formal series for pi and e to get a really accurate approximation, because I tried things like integration and even geometry but they aren't good enough.
Also, is there a series for e which we can use to get an upper bound for it? Because the lower bound won't work when we flip it.
Ah ok I see.Yep, by proving that all partial sums taking more terms are greater/less than our partial sum, do this by pairing adjacent terms and noting that the sequence is decreasing in size.
So you need series approximations with bounds on error.
For e, we have:
this is pretty easy to prove by induction, if we define e by the infinite series you mentioned.
Unfortunately, the Gregory series for pi is terrible, it takes ages to get anywhere near accurate. You would be better off by circumscribing and inscribing regular k-gons by the unit circle (k=12 might be nice if k=6 isn't good enough) and comparing areas to get surdic estimates, which are much easier to compare to 1/3.
No, in fact the only 'exact' forms of pi and e are exactly: - they are merely symbolic representations - pi the ratio between the circumference and diameter, and e the base of the natural logarithm (among other interpretations). ANY form of the decimal approximations MUST be truncated at some point since the number is irrational and will have an infinite number of decimals. Your calculator uses series to approximate these constants: for e it uses , truncating (stopping at a certain number) at n. And for , again truncating somewhere to give you enough accuracy.nvm i thought pi and e on the calculator were exact...
oh right thanks!No, in fact the only 'exact' forms of pi and e are exactly: - they are merely symbolic representations - pi the ratio between the circumference and diameter, and e the base of the natural logarithm (among other interpretations). ANY form of the decimal approximations MUST be truncated at some point since the number is irrational and will have an infinite number of decimals. Your calculator uses series to approximate these constants: for e it uses , truncating (stopping at a certain number) at n. And for , again truncating somewhere to give you enough accuracy.
Great questionHere is a form of Taylor's theorem that can be easily proven by MX2 methods:
where
denotes the k-th derivative of f, and n is any non-negative integer.
Here is a form of Taylor's theorem that can be easily proven by MX2 methods:
where
denotes the k-th derivative of f, and n is any non-negative integer.
Yep, its one of my favourite elementary applications of IBP, Taylor's theorem is very important in many fields of mathematics.Great question
Consider the integral:
Integrating by parts:
Taking the sum from k=1 to n on both sides, the left sum telescopes: