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[BokaBB]

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[wzkchem5]

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A few mathematical puzzles.
(1) (easy) Suppose we arrange all rational numbers between [0,1) into an array. Now construct a number x such that the i'th digit after the decimal point of x is identical to the i'th digit after the decimal point of the i'th rational number, i=1,2,3,.... Prove: x is irrational.
(2) (medium) We all know that the sum of 1/n^2, n=1,2,3,... is pi^2/6. But what about the sum of 1/(n^2+1)?
(3) (hard) Find an everywhere smooth function f(x) such that f(f(x))=x^2+x, x∈R.
(4) (medium) Define d(n) (n∈N+) as the number of decimal digits of n. Define a(n)=n*d(n)*d(d(n))*d(d(d(n)))*... (obviously the infinite product converges for all positive integers n, since after finitely many terms the d(d(...d(n)...)) will be 1). Is the infinite sum of 1/a(n), n=1,2,3,... convergent or divergent?
(5) (easy) The probability distribution of a random variable is a normal distribution with mean 0. A research group wants to prove that the mean of the distribution is not 0. They repeatedly perform independent measurements of the variable, and as soon as it can be concluded from available data that the mean of the distribution is not zero at 95% confidence level, they stop measuring and publish the result. What is the probability that they eventually publish this (wrong) result? (Obviously, the probability is larger than 5%.)

[Moosey]

(1) (easy) Suppose we arrange all rational numbers between [0,1) into an array. Now construct a number x such that the i'th digit after the decimal point of x is identical to the i'th digit after the decimal point of the i'th rational number, i=1,2,3,.... Prove: x is irrational.

uh what, that's not correct

there's no ordering described, so you can just take {0,0.3,0.03,0.003,0.0003...} as the first aleph-naught rational numbers, and you get 1/3 -- a rational number

[wzkchem5]

(1) (easy) Suppose we arrange all rational numbers between [0,1) into an array. Now construct a number x such that the i'th digit after the decimal point of x is identical to the i'th digit after the decimal point of the i'th rational number, i=1,2,3,.... Prove: x is irrational.

uh what, that's not correct

there's no ordering described, so you can just take {0,0.3,0.03,0.003,0.0003...} as the first aleph-naught rational numbers, and you get 1/3 -- a rational number

I said "all rational numbers", which means that every rational number must have a finite index.

[Moosey]

I said "all rational numbers", which means that every rational number must have a finite index.

Fair enough (I assumed that there was no need for finite indices because it was not explicitly stated)

[wzkchem5]

I said "all rational numbers", which means that every rational number must have a finite index.

Fair enough (I assumed that there was no need for finite indices because it was not explicitly stated)

My fault, I should have stated that clearly.

[toroidalet]

(1) (easy) Suppose we arrange all rational numbers between [0,1) into an array. Now construct a number x such that the i'th digit after the decimal point of x is identical to the i'th digit after the decimal point of the i'th rational number, i=1,2,3,.... Prove: x is irrational.

Consider a number whose digits never agree with x (like 1-x or x with every digit +1). If x were rational, then this other number would have some place i in the array. But by definition, its i'th digit is different from x's i'th digit, and we have a contradiction.

(4) (medium) Define d(n) (n∈N+) as the number of decimal digits of n. Define a(n)=n*d(n)*d(d(n))*d(d(d(n)))*... (obviously the infinite product converges for all positive integers n, since after finitely many terms the d(d(...d(n)...)) will be 1). Is the infinite sum of 1/a(n), n=1,2,3,... convergent or divergent?

Up until 10^^k, this sum is O(log^(k-1)(n)), which diverges very slowly. (this is because the integral of 1/x*1/xln(x)*1/xln(ln(x))...*1/xln^n(x) is ln^(n+1)(x))

[wzkchem5]

(1) (easy) Suppose we arrange all rational numbers between [0,1) into an array. Now construct a number x such that the i'th digit after the decimal point of x is identical to the i'th digit after the decimal point of the i'th rational number, i=1,2,3,.... Prove: x is irrational.

Consider a number whose digits never agree with x (like 1-x or x with every digit +1). If x were rational, then this other number would have some place i in the array. But by definition, its i'th digit is different from x's i'th digit, and we have a contradiction.

(4) (medium) Define d(n) (n∈N+) as the number of decimal digits of n. Define a(n)=n*d(n)*d(d(n))*d(d(d(n)))*... (obviously the infinite product converges for all positive integers n, since after finitely many terms the d(d(...d(n)...)) will be 1). Is the infinite sum of 1/a(n), n=1,2,3,... convergent or divergent?

Up until 10^^k, this sum is O(log^(k-1)(n)), which diverges very slowly. (this is because the integral of 1/x*1/xln(x)*1/xln(ln(x))...*1/xln^n(x) is ln^(n+1)(x))

Correct. Congratulations!

[MathAndCode]

I said "all rational numbers", which means that every rational number must have a finite index.

Let's suppose the opposite, i.e. there is some rational x and some ordering satisfying the condition. Because x is rational, its decimal expansion must start repeating at some point. If the repeated sequence is n digits long, then there are 9ⁿ possible repeating decimal expansions that don't have any decimal digits in common in the same place, as well as 9²ⁿ-9ⁿ possibilities for rational numbers whose repeating decimal sequence is twice as long as x's repeating decimal sequence, 9³ⁿ-9ⁿ possibilities for rational numbers whose repeating decimal sequence is three times as long as x's repeating decimal sequence, 9⁴ⁿ-9²ⁿ possibilities for rational numbers whose repeating decimal sequence is four times as long as x's repeating decimal sequence…, and infinitely many possibilities for the digits before the repeating part. If x has k digits before it starts repeating, then only k of those numbers can be on the list, but there are infinitely many more, thus creating a contradiction.

[BokaBB]

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