HSC 2013 MX2 Marathon (archive) (11 Viewers)

seanieg89 · Jun 20, 2013

Re: HSC 2013 4U Marathon

Lol sledgehammer

.

Sy123 · Jun 20, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Lol sledgehammer .

Holy crap hahahaha you're right

Wow I can't believe I went through all that and not notice that simple transformation...

asianese · Jun 21, 2013

Re: HSC 2013 4U Marathon

Interesting approach with the logarithms.

asianese · Jun 21, 2013

Re: HSC 2013 4U Marathon

$Prove that$

$e^{1/e}+e^{1/\pi} \geq 2e^{1/3}.$

braintic · Jun 21, 2013

Re: HSC 2013 4U Marathon

asianese said:
$Prove that$

$e^{1/e}+e^{1/\pi} \geq 2e^{1/3}.$

Um ... evaluate.

Carrotsticks · Jun 21, 2013

Re: HSC 2013 4U Marathon

braintic said:
$\begin{align*} \dfrac {(m+n)!}{m!.n!} . m^m . n^n &= {}^{m+n}C_m . m^m . n^n \\&= \text{term in } m^m . n^n \text{ in expansion of } (n+m)^{m+n}\\&< (n+m)^{m+n} \end{align*}$

Rearrangement gives required result.

Good solution.

I noticed that in your code, you used {}^{m+n}C_m to denote nCr.

May I suggest that you can also use \binom{n}{r} to have the same effect?

Sy123 said:
(I'm sure there is a more rigourous way to prove this inequality, I think it appeared in a HSC question before)

You can prove that the sequence is monotonically increasing by defining $a_k = \left ( 1+\frac{1}{k} \right )^k$ and then proving that $a_k < a_{k+1} \qquad , \forall k\geq1$

braintic said:
Um ... evaluate.

I think he meant for it to be proven algebraically, similarly to the classic "Prove e^pi > pi^e" problem.

Sy123 · Jun 21, 2013

Re: HSC 2013 4U Marathon

asianese said:
$Prove that$

$e^{1/e}+e^{1/\pi} \geq 2e^{1/3}.$

$a+b \geq 2\sqrt{ab}$

$\therefore e^{1/e} + e^{1/\pi} \geq 2e^{\frac{1}{2e + \frac{1}{2\pi}}}$

Now, lets establish some approximations for e and pi.

$We will assume that$

$e = \sum_{k=0}^{\infty}\frac{1}{k!}$

Since this sum is increasing, therefore:

$e > \sum_{k=0}^{3} \frac{1}{k!} = \frac{8}{3}$

The only series for Pi that I know that could be used to approximate Pi is an alternating series which I will have a hard time proving that Pi is greater than it.
So: Consider a regular 12 gon inscribed within a circle of radius 1. It is possible to find the exact form of the area of the 12 gon which I know is less than the area of the circle:

Finding area of hte triangles and multiplying by 12, we get: $\pi > 3$

$\therefore \frac{1}{2} \left(\frac{1}{e} + \frac{1}{\pi} \right )> \frac{1}{2} \left(\frac{3}{8} + \frac{1}{3} \right ) = \frac{17}{48} > \frac{17}{51} = \frac{1}{3}$

$\therefore \ \frac{1}{2} \left(\frac{1}{e} + \frac{1}{\pi} \right ) > \frac{1}{3}$

$\therefore \ e^{1/e} + e^{1/\pi} > 2e^{1/3}$

EDIT: Nope nope nope I have to find Pi and e as the lower bound. Redoing.

seanieg89 · Jun 21, 2013

Re: HSC 2013 4U Marathon

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.

Sy123 · Jun 21, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
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.

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: $1/2(1/e+ 1/\pi) \doteq 0.34$

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.

HeroicPandas · Jun 21, 2013

Re: HSC 2013 4U Marathon

nvm i thought pi and e on the calculator were exact...

seanieg89 · Jun 22, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
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: $1/2(1/e+ 1/\pi) \doteq 0.34$

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.

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:

$\frac{1}{(n+1)!}<e-\sum_{k=0}^n \frac{1}{k!}<\frac{1}{n\cdot n!}$

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.

Sy123 · Jun 22, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
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:

$\frac{1}{(n+1)!}<e-\sum_{k=0}^n \frac{1}{k!}<\frac{1}{n\cdot n!}$

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.

Ah ok I see.

So, we can establish a fairly accurate bound for e:

$e < \frac{87}{32}$ using the inequality you made above.

Then I guess we could use:

$\int_0^1 \frac{x^4(1-x)^4}{1+x^2} = \frac{22}{7} - \pi$

Since that function is always positive then:

$\pi < \frac{22}{7}$

$\therefore \frac{1}{2e} + \frac{1}{2\pi} > \frac{32}{87 \times 2} + \frac{7}{22 \times 2} = \frac{1313}{3828} > \frac{1313}{1313 \times 3} = \frac{1}{3}$

The use of 22/7 is a little cheaty, but the use of k-gons won't give a good enough approximation despite the one for e giving a great one.
In order to get an upper bound for pi, I need to consider inscribing a circle into the k-gon, and the only way I can do that and still have nice angles is k=6, since I need to split the 6 quadrilaterals in half in order to find the area.
Proof is then complete.

Sy123 · Jun 23, 2013

Re: HSC 2013 4U Marathon

$$Consider for some positive real values$ \ \ a, \ \ b, \ \ k$

$$Define the sequence$ \ \ \{S_n \}$

$S_n = \frac{a}{kn+b} + \frac{a}{kn+2b} + \frac{a}{kn+3b} + \dots + \frac{a}{n(k+b)}$

$i) Prove that$

$0 < S_n < \frac{a}{k} \ \ \ \fbox{2}$

$ii) Prove that$

$\frac{b}{t+b} < \int_t^{t+b} \frac{dx}{x} < \frac{b}{t} \ \ \ \fbox{1}$

$iii) Hence prove that$

$\frac{a}{b} \ln \left(\frac{n(k+b)+b}{kn+b} \right) < S_n < \frac{a}{b} \ln \left(\frac{b+k}{k} \right)$

$$Hence find$ \ \ \lim_{n \to \infty}S_n \ \ \ \ \fbox{4}$

asianese · Jun 23, 2013

Re: HSC 2013 4U Marathon

HeroicPandas said:
nvm i thought pi and e on the calculator were exact...

No, in fact the only 'exact' forms of pi and e are exactly: $\pi $ and $ \mathrm{e}$ - 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 $\mathrm{e}=\sum_{k=0}^\infty \frac{1}{k!} \approx \sum_{k=0}^n \frac{1}{k!}$ , truncating (stopping at a certain number) at n. And for $\pi=4\sum_{k=0}^\infty \dfrac{(-1)^k}{2k+1}\approx 4\sum_{k=0}^n \dfrac{(-1)^k}{2k+1}$ , again truncating somewhere to give you enough accuracy.

HeroicPandas · Jun 23, 2013

Re: HSC 2013 4U Marathon

asianese said:
No, in fact the only 'exact' forms of pi and e are exactly: $\pi $ and $ \mathrm{e}$ - 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 $\mathrm{e}=\sum_{k=0}^\infty \frac{1}{k!} \approx \sum_{k=0}^n \frac{1}{k!}$ , truncating (stopping at a certain number) at n. And for $\pi=4\sum_{k=0}^\infty \dfrac{(-1)^k}{2k+1}\approx 4\sum_{k=0}^n \dfrac{(-1)^k}{2k+1}$ , again truncating somewhere to give you enough accuracy.

oh right thanks!

seanieg89 · Jun 23, 2013

Re: HSC 2013 4U Marathon

Here is a form of Taylor's theorem that can be easily proven by MX2 methods:

$Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a function that is smooth (can be differentiated any number of times). Prove that:\\ \\ $f(x)=\sum_{k=0}^n \frac{x^kf^{(k)}(0)}{k!}+\int_0^x \frac{t^nf^{(n+1)}(x-t)}{n!}\, dt$$

where

$f^{(k)}$

denotes the k-th derivative of f, and n is any non-negative integer.

Sy123 · Jun 23, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Here is a form of Taylor's theorem that can be easily proven by MX2 methods:

$Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a function that is smooth (can be differentiated any number of times). Prove that:\\ \\ $f(x)=\sum_{k=0}^n \frac{x^kf^{(k)}(0)}{k!}+\int_0^x \frac{t^nf^{(n+1)}(x-t)}{n!}\, dt$$

where

$f^{(k)}$

denotes the k-th derivative of f, and n is any non-negative integer.

Great question

Consider the integral:

$I_k = \int_0^{x} \frac{t^{k-1}}{(k-1)!} f^{(k)} (x-t) \ dt$

Integrating by parts:

$u = f^{(k)} (x-t)$

$du = -f^{(k+1)}(x-t) \ dt$

$dv= \frac{t^{k-1}}{(k-1)!} dt$

$v = \frac{t^k}{k!}$

$I_{k} = \frac{t^k}{k!} f^{(k)} (x-t) |_0^x + I_{k+1}$

$I_k - I_{k+1} = \frac{x^k}{k!} f^{(k)} (0)$

Taking the sum from k=1 to n on both sides, the left sum telescopes:

$I_1 - I_{n+1} = \sum_{k=1}^{n} \frac{x^k}{k} f^{(k)} (0)$

$I_1 = \int_0^{x} f'(x-t) \ dt = -f(x-t) |_0^x = -f(0) + f(x)$

$\therefore \ -f(0) + f(x) - I_{n+1} = \sum_{k=1}^{n} \frac{x^k}{k!} f^{(k)} (0)$

$\therefore \ f(x)=\sum_{k=0}^n \frac{x^kf^{(k)}(0)}{k!}+\int_0^x \frac{t^nf^{(n+1)}(x-t)}{n!}\, dt$

Sy123 · Jun 23, 2013

Re: HSC 2013 4U Marathon

If you don't mind me expanding on this

seanieg89 said:
Here is a form of Taylor's theorem that can be easily proven by MX2 methods:

$i) Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a function that is smooth (can be differentiated any number of times). Prove that:\\ \\ $f(x)=\sum_{k=0}^n \frac{x^kf^{(k)}(0)}{k!}+\int_0^x \frac{t^nf^{(n+1)}(x-t)}{n!}\, dt$$

where

$f^{(k)}$

denotes the k-th derivative of f, and n is any non-negative integer.

$ii) Using part (i), prove that$

$e = \sum_{k=0}^{n} \frac{1}{k!} + \frac{(-1)^{n+1}}{n!}e \int_0^1 t^n e^{-t} \ dt \ \ \ \fbox{3}$

$Let$

$J_n = \int_0^{1} t^n e^{-t} \ dt$

$iii) Show that$

$0 < J_n < \frac{1}{n+1} \ \ \ \fbox{1}$

$iv) Hence prove that$

$e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots \ \ \ \fbox{1}$

$v) Prove by mathematical induction$

$\frac{1}{(n+1)!} < e - \sum_{k=0}^n \frac{1}{k!} < \frac{1}{n \cdot n!} \ \ \ \fbox{2}$

$Assume that there exists some integers$ \ \ p \ \ $and$ \ \ q \ \ $ with no common factor such that$

$e= \frac{p}{q}$

$vi) Prove that$

$\frac{1}{n+1} < \frac{p}{q} n! - N < \frac{1}{n} \ \ \ \fbox{1}$

$$For some integer$ \ \ N$

$$vii) Hence prove that$ \ \ e \ \ $is irrational$ \ \ \ \ \fbox{2}$

seanieg89 · Jun 23, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Great question

Consider the integral:

$I_k = \int_0^{x} \frac{t^{k-1}}{(k-1)!} f^{(k)} (x-t) \ dt$

Integrating by parts:

$u = f^{(k)} (x-t)$

$du = -f^{(k+1)}(x-t) \ dt$

$dv= \frac{t^{k-1}}{(k-1)!} dt$

$v = \frac{t^k}{k!}$

$I_{k} = \frac{t^k}{k!} f^{(k)} (x-t) |_0^x + I_{k+1}$

$I_k - I_{k+1} = \frac{x^k}{k!} f^{(k)} (0)$

Taking the sum from k=1 to n on both sides, the left sum telescopes:

$I_1 - I_{n+1} = \sum_{k=1}^{n} \frac{x^k}{k} f^{(k)} (0)$

$I_1 = \int_0^{x} f'(x-t) \ dt = -f(x-t) |_0^x = -f(0) + f(x)$

$\therefore \ -f(0) + f(x) - I_{n+1} = \sum_{k=1}^{n} \frac{x^k}{k!} f^{(k)} (0)$

$\therefore \ f(x)=\sum_{k=0}^n \frac{x^kf^{(k)}(0)}{k!}+\int_0^x \frac{t^nf^{(n+1)}(x-t)}{n!}\, dt$

Yep, its one of my favourite elementary applications of IBP, Taylor's theorem is very important in many fields of mathematics.

The moral is that we can approximate smooth (or sufficiently differentiable) functions by polynomials, and the size of the error decays at the rate of a polynomial of degree one greater. (To see this note that each of the derivatives of f will be bounded in a small interval about 0, and bound the integral expression for the error using this).

One nice application of this is that if the difference between two sufficiently smooth functions is bounded by a constant times x^{n+1} (at least close to 0), then their first n derivatives at zero must be equal (try to see why this must be true). We say such a pair of functions are "equal up to order n". Eg sin(x) is equal to x up to order 1 about 0.

Another is that this gives us a quite intuitive way of proving the 0/0 form of L'Hospital's rule (try to see this also).

Sy123 · Jun 23, 2013

Re: HSC 2013 4U Marathon

$$For some positive real values$ \ \ \{x_k \}$

$Prove that$

$\frac{1}{n} \sum_{k=1}^{n} x_k \geq \left(\frac{1}{n} \sum_{k=1}^n x_k^{-1} \right)^{-1}$

$State the condition for equality$

(don't feel the obligation to answer the questions I'm posting in order, answer whichever ones you want to)

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