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The Proof is Trivial!

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Original post by Felix Felicis
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Original post by Zakee
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Thanks guys. :lol:

Now back to STEP.
Original post by Lord of the Flies

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Lol, my solution involves inverse fourier transforming everything in sight. Needless to say, I am not an elegant mathematician.
Reply 922
Avatar for shamika
shamika
16
Original post by Lord of the Flies
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:smile:

To top it off, you presented the answers really well. Très bien!
Original post by ben-smith
Needless to say, I am not an elegant mathematician.


Pf, nonsense Ben! Your solution to problem 69 was very neat for instance.

Original post by shamika
To top it off, you presented the answers really well. Très bien!


If only I could present things as neatly on paper... Mais cimer! :biggrin:
Problem 145**

Let P2(x)\mathcal{P}_2(x) be the set of all polynomials of degree at most 2.

Find q(x)∈P2(x)q(x) \in \mathcal{P}_2(x) such that ∫01p(x)q(x) dx=p(12)  \displaystyle\int_0^1 p(x)q(x)\ dx = p(\frac{1}{2}) \ for all p(x)∈P2(x)p(x) \in \mathcal{P}_2(x)
Reply 925
Avatar for Jkn
Jkn
11
I'm back with an awesome problem... ENJOY! :colone:

Problem 146**

Evaluate
Unparseable latex formula:

\displaystyle\sum_{p\in\mathmm{P}}^{\infty}\frac{1}{p}

Original post by Jkn
I'm back with an awesome problem... ENJOY! :colone:

Problem 146**

Evaluate
Unparseable latex formula:

\displaystyle\sum_{p\in\mathmm{P}}^{\infty}\frac{1}{p}



So.. you mean the sum of the reciprocals of all the primes? 'cause if so..

Spoiler

Original post by FireGarden
Problem 145**


I like the problem, so...

Let P≤n(x)P_{\leq n}(x) denote the set of all univariate polynomials, in xx, of degree atmost nn.

Define Φq(p)=∫01p(x)q(x) dx\displaystyle \Phi_q(p) = \int_{0}^{1} p(x) q(x)\ dx for p,q∈P≤n(x)p,q \in P_{\leq n}(x).

The goal is to find qq such that Φq(p)=p(a)\Phi_q(p) = p(a) for all p∈P≤n(x)p \in P_{\leq n}(x), where a∈Ra \in \mathbb{R}.

Notice that Φq(xn)=an\Phi_q(x^n) = a^n by assumption. In addition, we have:

i) Φaq=aΦq\Phi_{a q} = a\Phi_{q} for a constant aa and polynomial qq.
ii) Φq+p=Φq+Φp\Phi_{q + p} = \Phi_{q} + \Phi_{p} for polynomials qq and pp.

and Φxa(xb)=1a+b+1\displaystyle \Phi_{x^a}(x^b) = \frac{1}{a + b + 1}.

Spoiler

Problem 147 / ***

Assume ss is a set containing numbers. Let σ(s)\sigma(s) and π(s)\pi(s) denote the sum and product, respectively, of all the elements in the set ss.

Let SS be the set {1,2,...,n}\{1, 2, ..., n\}.

i) Find an expression for ∑s∈P(S)1π(s)\displaystyle \sum_{s \in \mathcal{P}(S)} \frac{1}{\pi(s)}.

ii) Prove that ∑s∈P(S),s≠∅σ(s)π(s)=n(n+2)−(n+1)∑k=1n1k\displaystyle \sum_{s \in \mathcal{P}(S), s \not= \varnothing } \frac{\sigma(s)}{\pi(s)} = n(n+2) - (n+1)\sum_{k=1}^{n} \frac{1}{k}.

Note that π(∅)=1\pi(\varnothing) = 1.
Reply 929
Avatar for Jkn
Jkn
11
Original post by FireGarden
So.. you mean the sum of the reciprocals of all the primes? 'cause if so..

Spoiler


Spoiler

Original post by Jkn

Spoiler



Spoiler

Reply 931
Avatar for Jkn
Jkn
11
Original post by bananarama2

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Spoiler

Reply 932
Avatar for Mladenov
Mladenov
1
Solution 147

Part i). Consider (1+1)×⋯×(1+1n)=n+1\displaystyle \left(1+1 \right)\times \cdots \times \left(1+\frac{1}{n} \right) = n+1. Hence, ∑s∈P(S)1π(s)=n+1\displaystyle \sum_{s \in \mathcal{P(S)}} \frac{1}{\pi(s)} = n+1. I include
Unparseable latex formula:

s = \O

.

Part ii). We use induction. The base case is obvious. Suppose that our result is true for all i∈{1,2,⋯ ,n}i \in \{1,2, \cdots, n\}. We prove it for n+1n+1.
Split our sum into two parts:
Unparseable latex formula:

\begin{aligned} \displaystyle \sum_{s \in \mathcal{P}(S_{n+1}), s \not= \O} \frac{\sigma(s)}{\pi(s)} & = \sum_{s \in \mathcal{P}(S_{n}), s \not= \O} \frac{\sigma(s)}{\pi(s)} + \sum_{s \in \mathcal{P}(S_{n} )} \frac{n+1 + \sigma(s)}{(n+1)\pi(s)} \\&= n(n+2) - (n+1)H_{n} + 1+n + \frac{1}{n+1}(n(n+2) - (n+1)H_{n}) \\& = n(n+2) - (n+1)H_{n} + 2n+2 - \frac{1}{n+1} -H_{n} \\&= (n+1)(n+3) - (n+2)H_{n+1} \end{aligned}

.

Remark: I denoted Sn={1,2,⋯ ,n}S_{n} = \{1,2,\cdots, n\} in part ii).

Solution 146

The kkth prime is less than 2kln⁡k+22k\ln k +2 for k≥2k \ge 2. Hence the series diverges.
(edited 10 years ago)
Reply 933
Avatar for Jkn
Jkn
11
Original post by Mladenov

Solution 146

The kkth prime is greater than k(ln⁡k+ln⁡(ln⁡k−1))k(\ln k + \ln(\ln k-1)) for k≥2k \ge 2. Hence the series diverges.

Reminds me of this:
And_then_a_miracle_happens_cartoon.jpg

Spoiler



Here's an easier one. Undergrads will find it child's play but it's a really nice result so I'll post it anyway:

Problem 148**

Evaluate ∫−∞∞sin2xx2dx\displaystyle\int_{-\infty}^{\infty}\frac{sin^2x}{x^2}dx
I shall use the same method as when I did ∫−∞∞sin⁡xx dx\displaystyle \int_{-\infty}^{\infty} \frac{\sin x}{x}\,dx in a previous solution.

Solution 148

f(t)=∫0∞(sin⁡txx)2 dx⇒L{f(t)}=∫0∞1x2∫0∞e−tssin⁡2tx dt dx=∫0∞1x2(2x2s(s2+4x2)) dx=π2s2      ⇒  L−1{π2s2}=πt2  ⇒  ∫−∞∞(sin⁡xx)2 dx=π\begin{aligned} f(t)=\displaystyle \int_{0}^{\infty} \left(\frac{\sin tx}{x}\right)^2\,dx\Rightarrow \mathcal{L}\{ f(t)\} &=\int_0^{\infty}\frac{1}{x^2} \int_0^{\infty}e^{-ts} \sin^2 tx\,dt\,dx \\ & =\int_0^{\infty}\frac{1}{x^2} \left( \frac{2x^2}{s(s^2+4x^2)}\right) \,dx\\&=\frac{\pi}{2s^2}\;\;\; \Rightarrow\; \mathcal{L}^{-1}\left\{\dfrac{\pi}{2s^2} \right\}=\dfrac{\pi t}{2}\;\Rightarrow\;\int_{-\infty}^{\infty} \left( \frac{\sin x}{x}\right)^2\,dx=\pi\end{aligned}

Spoiler


It would be interesting if we could figure out a closed form for ∫−∞∞(sin⁡xx)n dx\displaystyle \int_{-\infty}^{\infty} \left(\frac{\sin x}{x}\right)^n \,dx
(edited 10 years ago)
I really must learn about Laplace transforms.
Original post by bananarama2
I really must learn about Laplace transforms.


Have a look at this. It's a bit slow at times but it is clear and he's quite a fun lecturer!
Reply 937
Avatar for Mladenov
Mladenov
1
Original post by Jkn

Spoiler



I am too busy these days, and I considered the problem superficially.
I need an upper bound.
Here is one sufficient upper bound - for k>1k>1, we have pk<2kln⁡k+2p_{k} < 2k \ln k +2. Now, it is obvious that ∑p∈P1p>∑k≥212kln⁡k+2\displaystyle \sum_{p \in \mathbb{P}} \frac{1}{p} > \sum_{k \ge 2} \frac{1}{2k \ln k +2}; the last series diverges.
Original post by Lord of the Flies
Have a look at this. It's a bit slow at times but it is clear and he's quite a fun lecturer!


Indeed, that was quite good! I skipped bits of working and examples though :colone:
Original post by bananarama2
Indeed, that was quite good! I skipped bits of working and examples though :colone:


Ha, yes that's what I meant by "slow" :biggrin:

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