The Proof is Trivial!

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Solution 236

Here's a more systematic way:

u=\ln(x), du=\frac{dx}{x}, dx=e^udu

\int \sin( \ln(x))dx= \int \sin( u) e^u du= \Im (\int e^{u+iu})

=\Im \left (\frac{e^{u+iu}}{1+i} \right) =\Im \left ((1-i)\frac{e^{u+iu}}{2} \right)=\frac{e^u}{2} \left(\sin(u)- \cos(u) \right)=\frac{x}{2} \left(\sin(\ln(x))- \cos( \ln(x)) \right)

Ditto for

\cos( \ln(x))

Reply 1621

Doglamlico

Original post by Jkn

Edit: Another fun problem (not from IMO):

Problem 234*

Find all possible n-tuples of reals

x_1,x_2, \cdots ,x_n

such that

\prod_{i=1}^{n} x_i = 1

and

\prod_{i=1}^{k} x_n - \prod_{i=k+1}^{n} x_i = 1

for

0 \le k \le n-1

Perhaps add an explanation of the Pi notation and tuples, since they're not mentioned at A-level (or not on Edexcel, at least) - I know it's just notation, but when you don't recognise words/symbols it can put you off pretty quickly.

Reply 1622

bananarama2

Original post by The Polymath

Perhaps notation should get a note in the OP?

(edited 10 years ago)

Reply 1623

hecandothatfromran

Original post by The Polymath

Spoiler

Solution 238
I may be simplifying things too much here, but would it just be

\displaystyle \int_{0}^{x_a}\frac{f(x)}{x_a}dx

Yeah, that's it baby girl. How'd you solve it?

Reply 1624

Felix Felicis

Solution 237

Let

x = \dfrac{7}{\sqrt{5}} \tan \theta :

\displaystyle \int_{0}^{\infty} \frac{x^{29}}{(5x^{2} + 49)^{17}} dx = \frac{7^{30}}{49^{17} \cdot 5^{15}} \cdot \int_{0}^{\pi /2} \sin^{29} \theta \cos^{3} \theta \ d \theta = \frac{1}{7^{4} \cdot 5^{15} \cdot 480}

(edited 10 years ago)

Reply 1625

bananarama2

Original post by hecandothatfromran

Yeah, that's it baby girl. How'd you solve it?

Reply 1626

hecandothatfromran

Original post by bananarama2

Yeah, i know :rolleyes:

. That's how i talk. To me, it has an aura of eloquence about it, don't you think?

Reply 1627

jack.hadamard

Original post by The Polymath

...

Did you quote me? It seems so, but I can't find it.

Reply 1628

james22

Original post by Felix Felicis

Solution 237

Let

x = \dfrac{7}{\sqrt{5}} \tan \theta :

\displaystyle \int_{0}^{\infty} \frac{x^{29}}{(5x^{2} + 49)^{17}} dx = \frac{7^{30}}{49^{17} \cdot 5^{15}} \cdot \int_{0}^{\pi /2} \sin^{29} \theta \cos^{3} \theta d \theta = \frac{7^{30}}{49^{17} \cdot 5^{15} \cdot 480}

Yep, but it can be simplified further.

Reply 1629

jack.hadamard

Original post by Jkn

Had a nice afternoon avoiding STEP and indulging in some IMO problems!

Did you not like problem 6? Determine whether

f(5,y)

is a perfect cube for some

y

Reply 1630

Felix Felicis

Original post by james22

Yep, but it can be simplified further.

So it can

Reply 1631

Zakee

Original post by bananarama2

http://cdn.memegenerator.net/instances/300x300/38781038.jpg

Reply 1632

Jkn

Original post by The Polymath

Original post by bananarama2

Perhaps notation should get a note in the OP?

I think everyone knows at at this point, perhaps there should be some basic notation and even teaching resources in the OP, that would be great...

OMG I've just had a vision... like a TSR olympiad society where everyone learns random **** and then some of the older people find 6 or so problems and put them on and then we all have to do them in time conditions... that would be awesome! :O

I was a bit apprehensive in typing the problem up in that form though, without that notation, it is far too obvious to spot (I shall say no more :wink:

)

Also, I actually typed 10 problems up (rather than 4) and yet my browser froze and TSR only saved the first 4 :/ I'll probably put them up later...

Original post by jack.hadamard

Did you not like problem 6? Determine whether

f(5,y)

is a perfect cube for some

y

I gave it a minute or so but, as nothing jumped out at me, I decided not to try it (I don't have enough experience with functional equations to know whether or I'm on the right track...)

Spoiler

Reply 1633

Doglamlico

Original post by jack.hadamard

Did you quote me? It seems so, but I can't find it.

Yeah, a response to your question, but I realised that LotF had already given the answer so I deleted it. That quote notification is doomed to remain a the top of your "who quoted me" for the rest of time :colone:

(really annoying bug)

Reply 1634

jack.hadamard

Original post by The Polymath

(really annoying bug)

Okay, I was just puzzled. :tongue:

Original post by Jkn

I gave it a minute or so but, as nothing jumped out at me, I decided not to try it (I don't have enough experience with functional equations to know whether or I'm on the right track...)

Spoiler

This is a wise decision. Well, I have been out of practice (done only bookwork) for a while and feel atrophied already. :biggrin:

I did problem two easily, because I have been using Vandermonde-related identities recently. About problem three, I recognised Cassini's identity and got the right answer, but through a flawed proof. Problem six is actually a breeze.

Reply 1635

Jkn

Original post by jack.hadamard

Spoiler

Umm.. spoiler alert! :lol:

Spoiler

Reply 1636

Mladenov

Solution 232

Clearly the possible elements are from the set

\{1,2,\cdots, n-r+1\}

and, for each of those guys, we have

\dbinom{n-i}{r-1}

corresponding subsets. Hence,

\begin{aligned} \displaystyle \dbinom{n}{r}F(r,n)= 1\times \dbinom{n-1}{r-1}+\cdots+ (n-r+1) \times \dbinom{r-1}{r-1} = \sum_{i=r}^{n} \sum_{j=r-1}^{i-1} \dbinom{j}{r-1} = \dbinom{n+1}{r+1} \end{aligned}

.

Solution 233

We claim that all solutions to

n^{2}-mn-m^{2}

belong to the set

T

consisting only of consecutive Fibonacci numbers.
Suppose that there is a solution

(m,n)

, which is not in

T

. We can further suppose that

m+n

is minimal and that

n>m

. Thus

(n-m,m)

is a solution with smaller sum; whence, it is in

T

, implying that

(m,n)

is in

T

- contradiction.

Solution 234

Really?!

Solution 235

I denote the left handed side by

A_{n}

, for I am too lazy to type this sum. For

n=2

the inequality holds true. Suppose that it is true for all

i \in \{1,2, \cdots,n\}

.
Then,

\begin{aligned} \displaystyle A_{n+1}> A_{n}+ \frac{4}{n(n+1)(n+2)} = \frac{2n^{3}+6n^{2}+2n}{n(n+1)(n+2)} = \frac{2((n+1)^{2}+(n+1)-1)}{(n+1)(n+2)} \end{aligned}

.
It is quite weak.

(edited 10 years ago)

Reply 1637

bananarama2

Original post by ukdragon37

I can't believe it.... ten thousand words of category theory, done and submitted. I never thought I'd be able to :cry:

Now to get some sleep. :sleep:

Congrats btw :biggrin:

Reply 1638

jack.hadamard

Original post by Jkn

Umm.. spoiler alert! :lol:

...

Cassini's identity is regularly used in STEP. You are meant to recognise that and prove no other solutions.

Original post by Mladenov

...

You are on the team this year, then? It's approaching. :smile:

Reply 1639

Jkn

Original post by Mladenov

Solution 232

Clearly the possible elements are from the set

\{1,2,\cdots, n-r+1\}

and, for each of those guys, we have

\dbinom{n-i}{r-1}

corresponding subsets. Hence,

\begin{aligned} \displaystyle \dbinom{n}{r}F(r,n)= 1\times \dbinom{n-1}{r-1}+\cdots+ (n-r+1) \times \dbinom{r-1}{r-1} = \sum_{i=r}^{n} \sum_{j=r-1}^{n-1} \dbinom{j}{r-1} = \dbinom{n+1}{r+1} \end{aligned}

Seen as it is a problem that gives you a result to prove, you should probably elaborate on your last step :tongue:

(as a rule of thumb, if an IMO examiner would not award you the marks, you can probably assume that we will have trouble following :wink:

)

Solution 233

We claim that all solutions to

n^{2}-mn-m^{2}

belong to the set

T

consisting only of Fibonacci numbers.
Suppose that there is a solution

(m,n)

, which is not in

T

. We can further suppose that

m+n

is minimal and that

n>m

. Thus

(n-m,m)

is a solution with smaller sum; whence, it is in

T

, implying that

(m,n)

is in

T

- contradiction.

This is incorrect. You have assumed a statement without justification and then discarded the 'counterexample' in favour of the conjecture. You have also not fully justified how it is that you have a contradiction and you have also applied an algebraic manipulation that contradicts the generality of a result you later used. :tongue:

Solution 234

Really?!