The Proof is Trivial!

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Reply 440

Slowbro93

Original post by bananarama2

Sorry

Although I think I may have done you a favor, it took and hour and a half to type that. :pierre:

It's okay

the problem that I have with these questions isn't the theory behind it, it's actually computing it (the amount of mistakes I made first time around was scary :zomg:

)

Reply 441

bananarama2

Original post by cpdavis

It's okay

the problem that I have with these questions isn't the theory behind it, it's actually computing it (the amount of mistakes I made first time around was scary :zomg:

)

Agreed. I missed of loads of terms and was on the verge of giving up, it's distinctly dodgy approximation too.

Reply 442

dknt

Well done guys! :awesome:

Reply 443

Anony1234

Sorry

(edited 11 years ago)

Reply 444

Lord of the Flies

Solution 53

The proof is trivial for

k<3

. Now suppose that for some

k\geq 3

there exists

m

such that

m^2\equiv -7\pod{2^k}

.
If

2^{-k}(m^2+7)=2p\Rightarrow m^2\equiv -7\pod{2^{k+1}}

. On the other hand, if

2^{-k}(m^2+7)=2p-1\Rightarrow m^2+2^k\equiv -7\pod{2^{k+1}}

hence by letting

m'=m+2^{k-1}\Rightarrow m'^2\equiv -7\pod{2^{k+1}}

. Noting that

1^2\equiv -7\pod{2^3}

it follows inductively that there is always some

q_0

such that

q_0^2\equiv -7\pod{2^k}

Define

q_i=q_0+i2^k\Rightarrow q_i^2\equiv -7\pod{2^k}\quad (\Leftrightarrow q_i^2=n_i2^k-7)

. The proposition follows.

Reply 445

und

Problem 60**/***

Evaluate the integral

\displaystyle I=\int^{\infty}_{-\infty}sin{\left({\pi}^{4}x^{2}+\frac{1}{x^2}\right)}dx

(edited 11 years ago)

Reply 446

Mladenov

Solution 48

I have spent about one day on this functional equation, despite the fact that I had done hundreds of functional equations.

We see

f(x^{3})=f(x)^{2}

. Further plugging

x=-1,0,1

, we get

y^{2}-y=0

has three solutions, namely

f(-1),f(0),f(1)

. Hence two of these have to be equal. Suppose, for example,

f(-1)=f(1)

. Then, since

w(f(x))=x^{3}

, we have

-1=1

. Similarly, by checking the cases

f(-1)=f(0)

and

f(0)=f(1)

, we conclude that there are no functions

f,w

such that

f\big(w(x)\big)=x^2

and

w\big(f(x)\big)=x^3

.

Now, define

f(x)= \exp(\sqrt{\ln(x)})

w(x)=\exp(4\ln(x)^{2})

, for all

x>1

, for

x=1

f(1)=w(1)=1

, for

0<x<1

f(x)=\exp(\sqrt{-\ln(x)})

w(x)=\exp(-4\ln(x)^{2})

f(0)=w(0)=0

, and

f(x)=f(-x)

w(x)=w(-x)

for

x<0

.

More functional equations?

Problem 61**

Find all functions

f: \mathbb{R^{+}} \to \mathbb{R^{+}}

satisfying

f(x^{y})=f(x)^{f(y)}

Reply 447

Hasufel

Original post by Star-girl

Problem 14**

Prove that

2^{60}-1

is divisible by 61.

don`t know if anyone has solved this and, i`ve been unable to find a direct proof - which really annoys me. Here`s my best attempt (forgive me please, i`ve left it here or i`ll go nuts):

2^{60}-1

can be factorised using the formula for a^n-1^n.

One of the factors is:

2^{16}+2^{14}-2^{10}-2^{8}-2^{6}+5=1321 \times 61

i know! i know! but, i can now unclench my hair and stop grinding my teeth and move on!

Reply 448

Star-girl

Original post by Hasufel

don`t know if anyone has solved this and, i`ve been unable to find a direct proof - which really annoys me. Here`s my best attempt (forgive me please, i`ve left it here or i`ll go nuts):

2^{60}-1

can be factorised using the formula for a^n-1^n.

One of the factors is:

2^{16}+2^{14}-2^{10}-2^{8}-2^{6}+5=1321 \times 61

i know! i know! but, i can now unclench my hair and stop grinding my teeth and move on!

There has been a solution that's different to yours but both are equally valid. :smile:

See the first post in the thread, there is a link to it there (yours will probably be added in soon as well).

The question is a lot easier with a handy little tool called modular arithmetic and even easier of you know a result called Fermat's Little Theorem. You have provided a more accessible solution. :smile:

(edited 11 years ago)

Reply 449

Hasufel

Original post by Star-girl

There has been a solution that's different to yours but both are equally valid. :smile:

thanks! Didn`t see the other solution - will check it out - thank again for the comment!

Reply 450

Star-girl

Original post by Hasufel

thanks! Didn`t see the other solution - will check it out - thank again for the comment!

No problem. :hat2:

Reply 451

bananarama2

Original post by Star-girl

You have provided a more accessible solution. :smile:

* Better. None of the number theory nonsense :tongue:

Reply 452

Star-girl

Original post by bananarama2

* Better. None of the number theory nonsense :tongue:

This is a thread full of mathematicians... careful what you say. :wink:

Also, number theory is pretty awesome. :pierre:

Reply 453

bananarama2

Original post by Star-girl

This is a thread full of mathematicians... careful what you say. :wink:

Also, number theory is pretty awesome. :pierre:

I like to live dangerously.

(\square + \mu ^2 ) \psi =0

Look at that beauty.

Edit: It won't let me put put the D'alembert symbol.

(edited 11 years ago)

Reply 454

Star-girl

I haven't seen very many inequality questions around, so I'll throw some in.

Problem 62*

Show that if

p>m>0,

then

\dfrac{p-m}{p+m} \leq \dfrac{x^2-2mx+p^2}{x^2+2mx+p^2} \leq \dfrac{p+m}{p-m} ,\ \forall x \in \mathbb{R}

Problem 63*

Prove

\forall x,y,z,w, \ x^2+y^2+z^2+w^2+1 \geq x+y+z+w

Reply 455

bananarama2

Solution 63

\forall x,y,z,w, \ x^2+y^2+z^2+w^2+1 \geq x+y+z+w

\iff \displaystyle x^2-x+y^2-y+z^2-z+w^2-w \geq -1

\iff (x-\frac{1}{2})^2 -\frac{1}{4} + (y-\frac{1}{2})^2 -\frac{1}{4} + (z-\frac{1}{2})^2 -\frac{1}{4} + (w-\frac{1}{2})^2 -\frac{1}{4} \geq -1

\iff (x-\frac{1}{2})^2 + (y-\frac{1}{2})^2 + (z-\frac{1}{2})^2 + (w-\frac{1}{2})^2 \geq 0

Which is true since a square is equal greater than one.

(edited 11 years ago)

Reply 456

Dark Lord of Mordor

Original post by bananarama2

I like to live dangerously.

Unparseable latex formula:

(\box + \mu ^2 ) \psi =0

Look at that beauty.

Edit: It won't let me put put the D'alembert symbol.

Why not use \square

Reply 457

bananarama2

Original post by Dark Lord of Mordor

Why not use \square

One does not simply use a square.

Spoiler

(edited 11 years ago)

Reply 458

metaltron

Original post by Star-girl

There has been a solution that's different to yours but both are equally valid. :smile:

I have another solution that I'm quite happy with so I thought I'd post it on here:

2^{60} - 1 = \frac{1}{2}(2^{61} - 2) = \frac{1}{2}(\displaystyle \binom{61}{1} + \displaystyle \binom{61}{2} + \displaystyle \binom{61}{3} + ... + \displaystyle \binom{61}{60} )

The bracket is divisible by 61, as 61 is prime, and divisible by 2 since only 61C1 and 61C60 are odd, due to the otherwise everpresent factor of 60. Hence: