UKC

I was looking through these problems today.

http://nrich.maths.org/5986&part=

Number 7 caught my eye.

Simplify.
(n+(n+(n+(n+(n+(n+n^0.5)^0.5)^0.5)^0.5)^0.5)^0.5)^0.5)

I can't see anything I can do with it. Any ideas?

In reply to ablackett:

Well I've made a mess of my notepad and not achieved much

In reply to ablackett:

Nope

http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427e73...

In reply to zebidee:

I have just noticed a ... in the original statement of the problem, so it's an infinite sequence of nested ((n+n^0.5)^0.5)

That changes things a bit.

In reply to ablackett:
n^(1-(1/(2^7))

If you expand the brackets you get:

n^0.5 + n^0.25 + ... + n^(1/(2^7))

which is diminishing return on the power of 2. This means you are left with 1 - the sum of the powers of 2^7 as your exponent.


EDIT:

to account for the ...

n^(1-(1/(2^∞))) which coincidentally equals n

In reply to ablackett:

So I managed to come up with a probably answer by messing around numerically, then demonstrate that that answer is a fixed point of
x -> n + x^0.5.

But I haven't manage to get the answer directly by simplifying the original expression...

In reply to Ramblin dave:





Should that be

x -> x + x^0.5 ?

In reply to ablackett:

No - the limit in question is the limit of the sequence

n + n^0.5
n + (n + n^0.5)^0.5
n + (n + (n + n^0.5)^0.5^0.5
...

If you call that limit x, then by definition
n + x^0.5 = x
because applying another iteration of the "take the square root and then add n" process won't make any difference.

Subtract n from both sides, square both sides, and then solve the resulting quadratic for x and you've got the result.

In reply to ablackett:

42 innit.

In reply to ablackett:

Final section of this:
http://www.math.northwestern.edu/~mlerma/courses/math214-2-03f/notes/c2-seq...
talks in a bit more mathematical detail about the approach Ramblin Dave is taking (finding limit of a recursively-defined sequence).

In reply to ablackett:

Clearly that sums to > n, because n+sqrt(n) > n.

Equally clearly it sums to less than 4n, because for any n,

n+sqrt(4n) < 4n.

Therefore it converges. Say it converges on N. Then it must be the case that

n + sqrt(N) == N.

If x == sqrt(N), then

x^2 - x - n == 0,

therefore

sqrt(N) == (1 + sqrt(1+4n))/2

sqrt(N) == 0.5 + sqrt(n + 0.25)

N == n + sqrt(n+0.25) + 0.5

In reply to ablackett:
Why do you want to do anything with it?
I am totally ignorant of this form of maths, and wonder what it has to do with life? Please tell me that it has some purpose.
I'm going to cry

In reply to Skol:


It's the escape route from Zeno's paradox(s).
And other problems that present themselves as infinite recursions.

In reply to Skol:

Mathematical logic is one of the ways to reduce unnecessary tears in life!

In reply to Skol:

It is closely related to the golden ratio and Fibonacci series. Ubiquitous in nature, and the subject of a whole chapter in Ian Stewart's book The Mathematics of Life.

It isn't the same for everybody but some people see the same aesthetic beauty in a mathematical formula that others (or they themselves) see in the interlaced spiral patterns of seeds in a sunflower head.

In reply to Ramblin dave:

Well done. Now deduce a continued fraction for the same result

In reply to Ramblin dave:
OK from what Ramblin dave said at 17:14 Mon.

let X= the value our infinite sum converges to.

so

n+sqrt(X) = X because sqrt then + n doesn't change the value as it converges.

sqrt(X) = X - n ............square both sides
X = X^2 - 2nX + n^2 ............subtract X from both sides
0=X^2 - (2n+1)X + n^2 .............complete the square
0=(X-(n+0.5))^2 - (n+0.5)^2 + n^2 ............expand the middle term
0=(X - (n+0.5))^2 - (n^2 + n + 0.25) + n^2 ............simplify and solve
sqrt(n+0.25) + n + 0.5 = X

In reply to Nutkey:

Which is the result Nutkey got to, but I didn't follow all of his working.

In reply to ablackett:

I have also enjoyed 2 and 19 from the same site.

2. What is the algebraic expression in terms of x,y and z for the area of the triangle with vertices (x,0,0), (0,y,0), and (0,0,z)


19) An experiment is repeated many times: two real numbers A and B are randomly, uniformly and independently chosen between &#8722;1 and 1 and the smaller number taken. What is the average value of this smaller number? (NB: by 'smaller' we mean here that A is smaller than B if and only if A<B)

In reply to Paul Robertson:

I thought it was about running for a bus

In reply to ablackett:

2: vector cross product...

3: I make it -1/3, based on a method that starts by sketching out a 2D area (x axis is A and y axis is B), splitting it along the diagonal, and considering for each sector separately how to calculate what the expected value of the smaller number is.

In reply to jonny taylor:

Physics vs Maths.

I think physics wins on both of these. I hadn't identified 2 as a vector cross product, went quite a long way round using Pythagoras, cosine rule, sin^2 + cos^2 = 1, and A=0.5abSinC.

3) are you talking about a centre of mass type argument for expectation? If so, I like it. I went a long way round again digging up all my probability knowledge from 12 years ago. Setting W as a variable which is the minimum of X and Y, and calculating a cdf for W, then differentiating to find a pdf for W, then calculating the expectation by multiplying by w and integrating between -1 and 1 to get -1/3.

I must admit that I generated 1000 values in excel first to see what I was aiming for.

In reply to ablackett:


Fortunately I've been doing quite a bit of vector geometry and differential geometry lately, so it's in the forefront of my mind.


Yes, that's the general idea. I was quite pleased that it worked out, I started out just with a general suspicion that there was going to be a nice geometric solution. My instinct was that pi was going to be involved, but drawing it out made it clear it was more centre-of-mass-of-a-triangle. I definitely didn't fancy doing it probabilistically/statistically.


Likewise, for verification!


So are you up for climbing this weekend, or what?

In reply to ablackett:

Not much time, but can see the solutions to 4 (may be wrong), 5, 10, 5, 16.

I knew 12 before, so that does not count.

Will have a look at this page more often, nice distraction from the normal daily grind of cell biology.

Thanks,

CB