HSC 2015 MX2 Marathon (archive) (1 Viewer)

Zlatman · Oct 20, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Next Question (relatively easy one):

$$ For $ x > 0, $ prove: $ x-\frac{x^3}{3} < \tan^{-1}{x} < x-\frac{x^3}{3} + \frac{x^5}{3}$

$x^4 $ could be used instead of $ \frac{5}{3} x^4 $, since $ \frac{x^5}{5} < \frac{x^5}{3} $ for $ x > 0 $.$$

$Also, the terms should be integrated from 0 to x, so that constants do not matter.$

glittergal96 · Oct 20, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Consider the sum$ \ P_n = \sum_{i=1}^n \frac{1}{p_i} \ $where$ \ p_i \ $is the$ \ i^{th} \ $prime number, so$ \\ p_1 = 2, \ p_2 = 3, \ p_3 = 5, \ $etc.$$

$\\ $i) Prove by mathematical induction that$ \ P_n \ $when simplified into a single fraction, is$ \ P_n = \frac{c_n}{d_n} \ $is such that$ \ c_n \ $is odd and$ \ d_n \ $is even. ($ c_n \ $and$ \ d_n \ $need not be in lowest terms)$$

$\\ $ii) Hence deduce that$ \ P_n \ $is never an integer$$

$\\ $iii) (advanced) Prove similarly or otherwise, that$ \ \sum_{i=1}^n \frac{1}{i} \ $is never an integer for$ \ n > 1$

An alternate solution to iii):

Let m be the largest integer such that 2^m =< n.

We must then have 2^m =< n < 2^{m+1}, which implies that 2^m > n/2.

Hence k.2^m > n for any positive integer k larger than 1.

Another way of saying this is that 2^m is the only integer in {1,2,...,n} that is divisible by 2^m.

So, assuming that H_n=N is an integer, we have:

$\sum_{k\leq n,k\neq 2^m}\frac{1}{k}=N-2^{-m}.$

Any sum of fractions can be written as a single fraction with denominator the LCM of the denominators of the individual fractions, and the largest power of 2 that occurs in the denominators on the LHS is 2^{m-1}. This implies that

$\frac{a}{2^{m-1}b}=N-\frac{1}{2^m}$

with b odd.

Multiplying both sides by $2^{m-1}b$ we get an integer on the LHS and a half integer (number of the form k +`1/2) on the RHS.

This contradiction completes the proof.

glittergal96 · Oct 20, 2015

Re: HSC 2015 4U Marathon

Paradoxica said:
$Hence, prove $H_n$ diverges.$

Hence? These facts are pretty unrelated, I would be surprised if there was a proof of divergence that used in an essential way that the H_n are non-integers.

An elementary proof of divergence can be obtained by writing:

1+1/2+1/3+...
=1+(1/2)+(1/3+1/4)+(1/5+1/6+1/7+1/8)+...
>1+1/2+2/4+4/8+...
=1+1/2+1/2+....

from which we see that the partial sums increase without bound.

Ekman · Oct 20, 2015

Re: HSC 2015 4U Marathon

Zlatman said:
$x^4 $ could be used instead of $ \frac{5}{3} x^4 $, since $ \frac{x^5}{5} > \frac{x^5}{3} $ for $ x > 0 $.$$

$Also, the terms should be integrated from 0 to x, so that constants do not matter.$

For your last statement written in latex, x^5 / 5 is not greater than x^5/3 . Its the other way around actually. But it doesn't really matter either way because it works since you can say: $1-x^2 + x^4 < 1 - x^2 +\frac{5x^4}{3}$ and so if $\frac{1}{x^2+1} < 1-x^2 + x^4$ the right hand side of the question is true.

Zlatman · Oct 20, 2015

Re: HSC 2015 4U Marathon

Ekman said:
For your last statement written in latex, x^5 / 5 is not greater than x^5/3 . Its the other way around actually. But it doesn't really matter either way because it works since you can say: $1-x^2 + x^4 < 1 - x^2 +\frac{5x^4}{3}$ and so if $\frac{1}{x^2+1} < 1-x^2 + x^4$ the right hand side of the question is true.

oh, lol, that was a typo, should've been the other way round.

Fixed, thanks!

Sy123 · Oct 20, 2015

Re: HSC 2015 4U Marathon

Good luck MX2 students!

HSC 2015 MX2 Marathon (archive) (1 Viewer)

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