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The Proof is Trivial!

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Reply 120
Avatar for shamika
shamika
16
Original post by Mladenov
Hint (Problem 23)

Spoiler



There should be a more concise solution.


There is :colone:

There is a way of doing this question where I'd peg the method as **/***, but it is definitely possible with very simple methods
(edited 11 years ago)
Reply 121
Avatar for Slumpy
Slumpy
16
Subbing post (and admission of spending a couple of hours here instead of doing revision...)
Reply 122
Avatar for Noble.
Noble.
17
Original post by shamika
There is :colone:

There is a way of doing this question where I'd peg the method as **/***, but it is definitely possible with very simple methods


It's an interesting problem, and it looks like:

∑(i,j,k)∈Sijk=(n+2)!(n−3)!5!\displaystyle\sum_{(i,j,k)\in S} ijk = \dfrac{(n+2)!}{(n-3)!5!}

and for:

i+j+k=n+1i+j+k = n+1

x+y+z=nx+y+z = n

∑(i,j,k)∈S1(ijk)−∑(x,y,z)∈S2(xyz)=(n+34)\displaystyle\sum_{(i,j,k)\in S_1} (ijk) - \displaystyle\sum_{(x,y,z)\in S_2} (xyz) = {n+3 \choose 4}

Where (n+34){n+3 \choose 4} is the nth pentatopic number.
(edited 11 years ago)
Reply 123
Avatar for shamika
shamika
16
Original post by Noble.
It's an interesting problem, and it looks like:

∑(i,j,k)∈Sijk=(n+2)!(n−3)!5!\displaystyle\sum_{(i,j,k)\in S} ijk = \dfrac{(n+2)!}{(n-3)!5!}

and for:

i+j+k=n+1i+j+k = n+1

x+y+z=nx+y+z = n

∑(i,j,k)∈S1(ijk)−∑(x,y,z)∈S2(xyz)=(n+34)\displaystyle\sum_{(i,j,k)\in S_1} (ijk) - \displaystyle\sum_{(x,y,z)\in S_2} (xyz) = {n+3 \choose 4}

Where (n+34){n+3 \choose 4} is the nth pentatopic number.


So for the original sum...

Spoiler

Can you prove it?
(edited 11 years ago)
Solution 29

By Van der Waerden's theorem (an)n≥1(a_n)_{n\geq 1} must contain arbitrarily long arithmetic progressions hence there are infinitely many finite sequences (pn)1≤n≤k(p_{n})_{1\leq n\leq k} for which apn=ap1+(n−1)ha_{p_{n}}=a_{p_{1}}+(n-1)h with k>ap1k>a_{p_1}. By fixing n−1=ap1n-1=a_{p_1} we have ap1∣apna_{p_1}|a_{p_n} as desired. There are infinitely many such sequences hence there are infinitely many ordered pairs (p,q): ap∣aq(p,q):\,a_{p}|a_{q}
(edited 11 years ago)
Problem 30*

More limits & integrals?

Evaluate lim⁡n→∞(∫01dx1+xn)n\displaystyle \lim_{n\to\infty} \left(\int_0^1 \frac{dx}{1+x^n}\right)^n
Original post by metaltron
...


You have integrated incorrectly, ddyeyn≠ney\dfrac{d}{dy}e^{yn}\neq ne^{y}

Spoiler

(edited 11 years ago)
Original post by Lord of the Flies
You have integrated incorrectly, ddyeyn≠ney\dfrac{d}{dy}e^{yn}\neq ne^{y}

Spoiler



Thanks. Got rid of it now as it was all completely wrong then! Might look at it again tomorrow when I'm more awake.

Spoiler

(edited 11 years ago)
Original post by Indeterminate
...


Check that expansion - it isn't valid on the interval!
Original post by Lord of the Flies
Check that expansion - it isn't valid on the interval!



Indeed :banghead:

I was wondering why it seemed so easy. I'll try something different :smile:
Reply 130
Avatar for und
und
OP
16
Solution 30

Using the binomial expansion and integrating, we find that:

Unparseable latex formula:

\displaystyle\lim_{n\to \infty} \left( \int^1_0 \frac{dx}{1+x^n} \right)^n=\lim_{n\to \infty}\left(1-\frac{1}{n+1}+\frac{1}{2n+1}-\frac{1}{3n+1}+...\right)^n\\& \displaystyle=\lim_{n\to \infty}\left(1-\left(\frac{1}{n}-\frac{1}{2n}+\frac{1}{3n}+... \right) \right)^n \\& =\displaystyle\lim_{n\to \infty}\left(1-\frac{1}{n}\int^1_0 \frac{dx}{1+x}\right)^n \\& =\displaystyle\lim_{n\to \infty}\left(1-\frac{\ln2}{n}\right)^n \\& =\displaystyle\frac{1}{2}

(edited 11 years ago)
Reply 131
Avatar for aznkid66
aznkid66
1
Original post by und
Solution 30

-snip-

Unparseable latex formula:

& =\displaystyle\lim_{n\to \infty}\left(1-\frac{\ln2}{n}\right)^n \\& =\displaystyle\frac{1}{2}



Lol, way to trip right in front of the finish line..

EDIT: Right, ignore this then.
(edited 11 years ago)
Original post by aznkid66
Lol, way to trip right in front of the finish line..


Blithering idiot.

Unparseable latex formula:

\displaystyle\lim_{n\to \infty}\left(1-\frac{\ln2}{n}\right)^n & =\displaystyle\frac{1}{2}



Lol, I got warning points for this.
(edited 11 years ago)
Reply 133
Avatar for und
und
OP
16
Problem 31*

Find all real numbers x, y and z which satisfy the simultaneous equations x2−4y+7=0x^2-4y+7=0, y2−6z+14=0y^2-6z+14=0 and z2−2x−7=0z^2-2x-7=0.
(edited 11 years ago)
Original post by und
Problem 31

Find all real numbers x, y and z which satisfy the simultaneous equations x2−4y+7=0x^2-4y+7=0, y2−6z+14=0y^2-6z+14=0 and z2−2x−7=0z^2-2x-7=0.


Solution 31

Adding the three equations:

x2−2x+y2−4y+z2−6z=−14x^2 - 2x + y^2 - 4y + z^2 - 6z = -14

Completing the square:

(x−1)2+(y−2)2+(z−3)2=0(x-1)^2 + (y-2)^2 + (z-3)^2 = 0

Hence x=1,y=2,z=3x = 1, y = 2, z = 3
(edited 11 years ago)
Reply 135
Avatar for und
und
OP
16
Problem 32*

Find all positive integers nn such that 12n−11912n-119 and 75n−53975n-539 are both perfect squares.
(edited 11 years ago)
Original post by und
...


Original post by Upper Echelons
...

I thought und answered his own question there for a second :lol:

Solution 32

32 Solution

(edited 11 years ago)
Reply 137
Avatar for Benjy100
Benjy100
5
Problem 33

Determine ∫ba1(a−x)(x−b)dx\int\limits_b^a {\frac{1}{{\sqrt {(a - x)(x - b)} }}} {\rm{ }}dx
Original post by Benjy100
Problem 33

Determine ∫ba1(a−x)(x−b)dx\int\limits_b^a {\frac{1}{{\sqrt {(a - x)(x - b)} }}} {\rm{ }}dx


Surely this has asymptotes in the interval you've given?
Solution 33

Method 1:

Letting x=bcos⁡2t+asin⁡2tx=b\cos^2 t+a\sin^2 t gives:

∫badx(a−x)(x−b)=∫0π/22dt=π\displaystyle \int_b^a \frac{dx}{\sqrt{(a-x)(x-b)}} = \int_0^{\pi/2} 2 dt =\pi

Method 2:

∫ba1(a−x)(x−b)  dx=i∫ab1(a−x)(b−x)  dx\displaystyle\int_b^a \frac{1}{\sqrt{(a-x)(x-b)}}\;dx=i\int_a^b \frac{1}{\sqrt{(a-x)(b-x)}}\;dx


=−2i∫ab(−12a−x+−12b−x)(1a−x+b−x)  dx=\displaystyle -2i\int_a^b\left(\frac{-1}{2\sqrt{a-x}}+\frac{-1}{2\sqrt{b-x}}\right)\left(\frac{1}{\sqrt{a-x}+\sqrt{b-x}}\right)\;dx


=−2iln⁡(a−x+b−x)∣ab=−iln⁡(a−bb−a)=π=-2i\ln(\sqrt{a-x}+\sqrt{b-x})|_a^b=-i\ln\left(\dfrac{a-b}{b-a}\right)=\pi

(taking the principal branch - I'm sure that can be justified)
(edited 11 years ago)

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