Mathematicians the world over

April 1, 2015 Philip Waring 1 Comment

Mathematics as much as music does bring us together, as evidenced by the fifteen or so responses to the Quora question below, coming I’d like to believe from nearly as many different lands across the globe (but to read their names, I wonder if they don’t all come from just a few countries, India, Russia, China certainly. Also, given that Americans come from probably every country in the world they could all be Americans. Music of course may reach us all, that is billions of us, but mathematics many fewer, millions at the most. Why is that?

I imagine a world where music and mathematics have somehow replaced religion in our lives. Would the people in such a world ever go to war with one another? Mathematicians, like musicians, speak the same language, and obey the same rules.

Would mathematicians ever fight for the rightness of their equations over those of other mathematicians, even those of other lands and other languages? Would musicians ever have to defend their particular music against someone else’s?

Whereas it does seem that believers in this or that God have always gone to war with believers in another God, or simply with the non-believers barbarians beyond their respective kingdoms and control.

Here is the Quora question put to the world online with a, actually my selection of some of the answers. I’m not enough of a mathematician myself to be able to tell you the best of them, and why. I’m working on it, you know, life-long learning.

Solutions by squaring both sides and solving for x

Srinath Srinivasa, philosophically curious about mathematics

Let $x = sqrt{2sqrt{2sqrt{2 dots}}}$

We have $x = sqrt{2x}$ or $x^2 = 2x$

which resolves to $x = 0$ or $x = 2$

But since the expression was clearly greater than 0, we can reject the $x = 0$ possibility, giving us $x = 2$ .

Upvote

Vincen Mathai, Drinking the wine of life

1 upvote by Bernd Gmelin.

This method used for finding the solution can get you the answer within a couple of steps , thus helping you crack such questions within a minute during the various exams such as GRE,GMAT, CAT etc.

We need to create the equation that can help us solve the an find values for the problem.

First we need to consider a suitable variable that can represent the problem

$y=$ $sqrt{2sqrt{2sqrt{2sqrt{cdots}}}}$

Next, square both sides and arrange the the resultant equation in a quadratic form .

$y^2= 2y$
$y^2- 2y=0$
$y(y-2)=0$

Solving the equation gives you two possible solution for y.

$y= 2$ or $y=0$

Since it is clear that variable y is positive from the problem given , the final answer is $y= 2$

Dan Tudorache

1 upvote by Mohammad Kanan.

Let x = sqrt(2*sqrt(2*sqrt(2*sqrt(2*sqrt(2…

Since the number of square roots is infinite, it does not matter if we “drop” the first “sqrt(2*”. Since we have an infinite number of square roots, and infinity minus one is still infinity, the value of x will be unchanged if we drop the first “sqrt(2*”.

Therefore we can write:
x = sqrt(2*x)

From the equation above, we have x^2 = 2*x. The solutions to this equation are 0 and 2.

Zero is obviously not equal to sqrt(2*sqrt(2*sqrt(2*sqrt(2*sqrt(2….

So the only solution is x = 2.

Since, many of you were demanding me to prove that 0 is not a solution of this sequence of nested roots,

For , X $_{0}$ >0, X $_{n}$ →2
To see this, note that X $_{n+1}$ =y(X $_{n}$ ), where the function y:x↦√2x is such that x<y(x)<2 for every x in (0,2), y(2)=2, and 2<y(x)<x if x>2. Thus (X $_{n}$ ) is increasing if 0<X $_{0}$ <2and decreasing if X $_{0}$ >2 and converges to 2 in both cases.

Rahul Verma

2 upvotes by Anusha Sridharan and Mahantesh Kumbar.

Let us assume that the solution is S

$S = sqrt{2sqrt{2sqrt{2sqrt{.....}}}}$

As the series goes up to infinite, we can re-write the equation

$S = sqrt{2cdot S}$

$S^2 = 2cdot S$

$S^2-2S=0$

$S(S-2)=0$

$S=0,or, 2$

$S = sqrt{2sqrt{2sqrt{2sqrt{.....}}}} = 2$

Pummy Kalsi

10 upvotes by Jakob Nebeling, Malcolm Harper, P. Michael Hutchins, (more)

The common proof for this is squaring both sides and solving the quadratic, as many have done here. However, when dealing with infinite series or expressions, we have to be very careful with the assumptions we are making. There is a clear assumption in this method, which is that this infinite expression is convergent and has a specific value.To be more rigorous, the proof using limits or even the one with the powers of 2 forming the well known convergent series is probably better.

Solutions using infinite series and convergence

Ahmed Besbes, Software Engineering Intern

2 upvotes by Okechukwu Ochia and Changqi Cai.

$X = sqrt{2sqrt{2sqrt{2sqrt{2sqrt{...}}}}}$
You may notice that squaring X gives
$X^{2} =$ $2sqrt{2sqrt{2sqrt{2sqrt{...}}}}$ = $2 X$
Therefore :
$X^{2} = 2 X$
$X$ being a non-zero value, by dividing by $X$ on both sides, we get : $X = 2$

Ok, now more formally :

Let’s consider the function f : $x mapsto sqrt{2x}$

$f circ f (x) = sqrt{2sqrt{2x}}$

$f circ f circ f (x) = sqrt{2sqrt{2sqrt{2x}}}$

$f^{(n)}(x) = sqrt{2sqrt{2sqrt{2sqrt{2sqrt{...sqrt{2x}}}}}}$

Let’s consider $U_{n} = f ^{(n)}(1)$

$U_{n}$ can be expressed by :

$U_{n} = 2^{frac{1}{2} +frac{1}{4} + frac{1}{8} + ... + frac{1}{2^{n}} } = 2^{1 - frac{1}{2^{n}}}$ ,

By taking the limit : $U_{n} rightarrow 2$

Ivanov Dmitry, CS/Math student. With rubber chicken.

Upvote37

37 upvotes by Sergio Torassa, Alex Petit, Constantin Dumitrascu, (more)

Aurélien Emmanuel

About the question “can 0 be an answer?”
It solely depends on how do we pose the problem.

If you define it as a limit (of (sqrt(2), sqrt(2sqrt(2)), …), then, 2 is the only solution.

If you define it all at once, then the square root only means that it can disappear if you square it, leading to a quadratic equation.

Similarly, 1+ 2 + 3 + 4 + 5 + 6 + … is obviously +infinity, if you define it to be the limit of a finite sum, but it also can be proven to be -1/12, and makes sense in some theories I sadly don’t know much. An easier example could be :

1/2 + 1/4 + 1/8 + 1/16 + …
Call it A
A – 1/2 = A/2
The only sulution is 1. Is it? no +- infinity works as well. I don’t know much the sense of it, but I guess there is no much difference between + and – infinity when we start doing that kind of stuff. Indeed, We can expect that
2^(+infinity) = infinity while 2^(-infinity) = 0, but if we don’t think in term of limit, 2^infinity is a number that satisfys:

2*x = x
0 and Infinity are then two possible solutions.

Włodzimierz Holsztyński

I’ve introduced the sequence $D_0 D_1 D_2 ldots$ in my previous answer (but I cannot find it!). Thus, induction, $D_0:=1$ and $D_n := sqrt{2cdot D_{n-1}}ge sqrt{2cdot 1},$ i.e. $D_ngesqrt 2$ for every positive integer $n.$

It follows, still for arbitrary positive integer $n,$ that:

$(2-D_n)cdot(2+D_n) = 2cdot(2-D_{n-1})ge 0$

again by induction., and furthermore

$2-D_n = frac 2{2+D_n}$ $cdot (2-D_{n-1}) le Acdot(2-D_{n-1})$

for $0 < A:=frac 2{2+sqrt 2}<1.$ Thus, again by induction,

$forall_{n=0 1 2 ldots} 0<2-D_n<A^n$

This shows that the limit $D$ is $2$ , and that the error terms are limited by the geometric progression $A^n$ .

Uday K Kumar, Thinker, Critic, Weekend coder

After struggling to edit with math tags, this is what I have, did any one mention this yet ?
a good old $log / anti log$ logic
$sqrt{2sqrt{2sqrt{2sqrt{cdots}}}} = x$
$log sqrt{2sqrt{2sqrt{2sqrt{cdots}}}} = log x$
$(1/2) log {2sqrt{2sqrt{2sqrt{cdots}}}} = log x$
$log {2sqrt{2sqrt{2sqrt{cdots}}}} = 2 log x$
$log 2 + log { sqrt{2sqrt{2sqrt{cdots}}}} = 2 log x$
$log 2 + log { sqrt{2sqrt{2sqrt{cdots}}}} = 2 log x$
$log 2 + log x = 2 log x$
$log x = log 2$
$x = 2$

$sqrt{2sqrt{2sqrt{2sqrt{cdots}}}} = 2$

Reg Dodds

a solution from the computer

from math import sqrt root = sqrt(2)
product = root for iteration in range(15):
print(iteration, root, product)
root = sqrt(root)
product *= root

Output:

1.4142135623730951 1.4142135623730951
1.189207115002721 1.6817928305074292
1.0905077326652577 1.8340080864093427
1.0442737824274138 1.9152065613971474
1.0218971486541166 1.9571441241754002
1.0108892860517005 1.978456026387951
1.0054299011128027 1.9891988469672661
1.0027112750502025 1.99459211217094
1.0013547198921082 1.9972942257819402
1.0006771306930664 1.9986466550053015
1.0003385080526823 1.9993232129924874
1.0001692397053021 1.999661577863858
1.0000846162726942 1.9998307817732266
1.0000423072413958 1.9999153890968617
1.0000105766425498 1.9999788469386284

This convinces me without any proof.

Simon Dwork

Wow, so many of the answers here are plainly incorrect. The way that the question is stated, the answer is that the question does not define one number. It in fact defines two! The CORRECT answer is: either 0 or 2.

It makes sense that you get more than one answer, because you’re not looking at a limit of number here. It would be a DIFFERENT question to ask what the following converges to: sqrt(2), sqrt(2*sqrt(2)), etc. It’s a different question, because then you would have the “…” on the left, not the right!

One thought on “Mathematicians the world over”

NGV says:

April 2, 2015 at 8:58 am

a very intriguing question/idea:
“Would mathematicians ever fight for the rightness of their equations over those of other mathematicians, even those of other lands and other languages? Would musicians ever have to defend their particular music against someone else’s?”
I’d like to think that people engaged in such demanding, full-filling, creative, all encompassing pursuits that require real thought and feeling, (not just musicians and mathematicians) would have the confidence in themselves and their world, to not ‘worry’ about making everyone else like or agree with them. Real dedication, real art, real passion, that goes a long way for building self-confidence and respect for others.

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