HSC 2015 MX2 Marathon (archive) (5 Viewers)

Sy123 · Jul 24, 2015

Re: HSC 2015 4U Marathon

$\\ $Let$ \ S_m(n) = 1^m +2^m + \dots +n^m \ $for integers$ \ m \geq 0$

$\\ $i) By considering the expansion of$ \ (k+1)^{m+1} - k^{m+1} \ $show that$ \\ \\ 1 + \sum_{k=0}^m S_k(n) \binom{m+1}{k} = (n+1)^{m+1}$

$\\ $ii) Hence or otherwise show that$ \ S_m(n) \ $ (with$ \ m \geq 1) \ $are polynomials in$ \ n \ $of degree$ \ m+1 \ $and by considering them as polynomials, have$ \ n=0 \ $and$ \ n=-1 \ $as roots$$

shervos · Jul 25, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Let$ \ S_m(n) = 1^m +2^m + \dots +n^m \ $for integers$ \ m \geq 0$

$\\ $i) By considering the expansion of$ \ (k+1)^{m+1} - k^{m+1} \ $show that$ \\ \\ 1 + \sum_{k=0}^m S_k(n) \binom{m+1}{k} = (n+1)^{m+1}$

$\\ $ii) Hence or otherwise show that$ \ S_m(n) \ $are polynomials in$ \ n \ $of degree$ \ m+1 \ $and by considering them as polynomials, have$ \ n=0 \ $and$ \ n=-1 \ $as roots$$

not sure whether i understand the problem, what does Sm(-1) look like? In fact, what does S_m (-n) look like? Is it (-1)^m+(-2)^m+....+(-n)^m?

InteGrand · Jul 25, 2015

Re: HSC 2015 4U Marathon

shervos said:
not sure whether i understand the problem, what does Sm(-1) look like?

He meant show that the polynomials that the power series sum (taken as polynomials defined for any real n) to have -1 and 0 as roots.

shervos · Jul 25, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
He meant show that the polynomials that the power series sum (taken as polynomials defined for any real n) to have -1 and 0 as roots.

I could be misunderstanding, but from the definition of S_m (n) (in the first line of the question), does this mean that S_m (n) holds positive integer values only?

InteGrand · Jul 25, 2015

Re: HSC 2015 4U Marathon

shervos said:
I could be misunderstanding, but from the definition of S_m (n) (in the first line of the question), does this mean that S_m (n) holds positive integer values only?

$Yes, but basically forget about the $S_m(n)$ definition for the last part. E.g. $S_1 (n) = 1+2+3+\cdots + n = \frac{n(n+1)}{2}$, and calling $P_1 (x):=\frac{x(x+1)}{2}$, we can see this has $x=0$ and $x=-1$ as roots. So if it helps notationally, call the polynomials $P_m (x)$ (rather than $S_m$), replacing $n$'s with $x$ (just to emphasise it's a polynomial for any real input), defined for all real $x$, and show that for each positive integer $m$, $P_m$ has $-1$ and $0$ as roots.$

shervos · Jul 25, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Let$ \ S_m(n) = 1^m +2^m + \dots +n^m \ $for integers$ \ m \geq 0$

$\\ $i) By considering the expansion of$ \ (k+1)^{m+1} - k^{m+1} \ $show that$ \\ \\ 1 + \sum_{k=0}^m S_k(n) \binom{m+1}{k} = (n+1)^{m+1}$

$\\ $ii) Hence or otherwise show that$ \ S_m(n) \ $are polynomials in$ \ n \ $of degree$ \ m+1 \ $and by considering them as polynomials, have$ \ n=0 \ $and$ \ n=-1 \ $as roots$$

subbing in n=-1 gives 1+ sigmaS_k(-1)*{m+1)Ck=(-1+1)^m+1=0??

the LHS is 1 if S(-1)=0 for all k?

Sy123 · Jul 25, 2015

Re: HSC 2015 4U Marathon

shervos said:
subbing in n=-1 gives 1+ sigmaS_k(-1)*{m+1)Ck=(-1+1)^m+1=0??

the LHS is 1 if S(-1)=0 for all k?

The first argument doesn't work, just because the sum of all of them is zero doesn't mean all of them are zero (for instance 1 + (-1) +2 + (-2) = 0 but they aren't all zero).
Try induction!

(Also yea the property "S_m(n) has roots n=0 and n=-1" is only true for m >= 1, I'll edit that in)

shervos · Jul 25, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
The first argument doesn't work, just because the sum of all of them is zero doesn't mean all of them are zero (for instance 1 + (-1) +2 + (-2) = 0 but they aren't all zero).
Try induction!

(Also yea the property "S_m(n) has roots n=0 and n=-1" is only true for m >= 1, I'll edit that in)

? Not sure what you mean here? Since S_k (-1) =0 as you claimed for all integers k, doesn't this mean that sigmaS_k(-1)*{m+1)Ck equals 0? (Since all the terms are zero, their sum is zero....am i wrong here?)

simpleetal · Jul 25, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Let$ \ S_m(n) = 1^m +2^m + \dots +n^m \ $for integers$ \ m \geq 0$

$\\ $i) By considering the expansion of$ \ (k+1)^{m+1} - k^{m+1} \ $show that$ \\ \\ 1 + \sum_{k=0}^m S_k(n) \binom{m+1}{k} = (n+1)^{m+1}$

$\\ $ii) Hence or otherwise show that$ \ S_m(n) \ $ (with$ \ m \geq 1) \ $are polynomials in$ \ n \ $of degree$ \ m+1 \ $and by considering them as polynomials, have$ \ n=0 \ $and$ \ n=-1 \ $as roots$$

Breaking my Bored of Studies fast to help a fellow Boser in need.
$\\ i) $Obviously statement holds true for n=1$. \ $Assume true for some n=j. We proceed by induction.$ 1+ \sum_{k=0}^{m}S_{k}(j+1)\binom{m+1}{k+1} = 1+\sum_{k=0}^{m}S_{k}(j)\binom{m+1}{k+1}+\sum_{k=0}^{m}(j+1)^k \binom{m+1}{k+1}= (j+1)^{m+1}+(j+2)^{m+1}-(j+1)^{m+1}=(j+2)^{m+1} \\ ii)$Another induction proof. (Strong). Base case of m=1 gives the polynomial $S_{1}(n)=n(n+1)/2. $Suppose that $S_{k}(n)$ are polynomials in n of degree k+1 for all integers k= 1,2....m-1 . Then,$ \binom{m+1}{m+1}S_{m}(n)=S_{m}(n)= (n+1)^{m+1}- \sum_{k=0}^{m-1}\binom{m+1}{k+1}S_{k}(n)-1 $Where the first factor on the RHS is dgree m+1, and the highest power on the second factor is m, hence the LHS is a polynomial of degree m+1. Now, suppose that for all integers k=1... m-1,$ S_{m-1}(0)=0. $Then, subbing into equation gives that$ S_{m}(0)=0$ as well. Notice that $S_0(n)=n$, and so using strong induction again ($S_{k}(-1)=0$ for k=1,2,....m-1), we get that RHS= 0-(-1)+1=0. Done$

Sy123 · Jul 25, 2015

Re: HSC 2015 4U Marathon

shervos said:
? Not sure what you mean here? Since S_k (-1) =0 as you claimed for all integers k, doesn't this mean that sigmaS_k(-1)*{m+1)Ck equals 0? (Since all the terms are zero, their sum is zero....am i wrong here?)

The second part of my post was saying that it's not true that S_0(-1) = 0, so that's why I edited in the question to make it "prove for m>=1 that S_m(-1) = 0" (which is true).

So, when you sub in n = -1 into the expression in part (i), we get $1 + S_0(-1) \binom{m+1}{0} + \sum_{k=1}^m S_k(-1) \binom{m+1}{k} = 0$

Which is true as expected, since $S_0(n) = 1^0 + 2^0 + \dots + n^0 = n$ , meaning S_0(-1) = -1, and if it were true that the rest of S_k(-1) were zero, then the equation indeed holds true

simpleetal · Jul 25, 2015

Re: HSC 2015 4U Marathon

Fixing up my notation big time, please wait

Sy123 · Jul 25, 2015

Re: HSC 2015 4U Marathon

simpleetal said:
Breaking my Bored of Studies fast to help a fellow Boser in need.

$\\ $i) Obviously statement holds true for n=1$. \ $Assume true for some n=k. We proceed by induction.$ \\ \\ 1+ \sum_{k=0}^{n+1}S_{k}(n+1)\binom{k+1}{n+1} = 1+ \sum_{k=0}^{m+1}S_{k}(n)\binom{k+1}{n+1}+\sum_{k=0}^{m}(j+1)^k \binom{k+1}{m+1}= (n+1)^{m+1}+(n+2)^{m+1}-(n+1)^{m+1}=(n+2)^{m+1} \\ \\ $ii) Another induction proof. (Strong). Suppose they're polynomials for all integers upto and include m-1. Then,$ \\\\ S_{m}(n)= (n+1)^{m+1}- \sum_{k=0}^{m-1}\binom{m+1}{k+1}S_{k}-1 \\\\ $Where the first factor on the RHS is dgree m+1, and the highest power on the second factor is m, hence the LHS is a polynomial of degree m+1. Now, suppose that for all integers m>=1 upto and including m-1,$ S_{m-1}(0)=0. $Then, subbing into equation gives that$ S_{m}(0)=0$ as well. Notice that $S_0(n)=n, and so using strong induction again ($S_{m-1}(-1)=0$ for 1,2,....m-1), we get that RHS= 0-(-1)+1. Thus, $S_m (-1)=0$ as well (LHS)$$

Fixed that for you

(and yes that is the approach I was going for)

Though for part (i), using the expansion we simply do:

$\\ \sum_{k=1}^n (k+1)^{m+1} - k^{m+1} = 2^{m+1} - 1^{m+1} + 3^{m+1} - 2^{m+1} + \dots + (n+1)^{m+1} - n^{m+1} = (n+1)^{m+1} - 1 \ (*)$

$(k+1)^{m+1} - k^{m+1} = \sum_{i=0}^{m} \binom{m+1}{i} k^i$

$\\ \sum_{k=1}^n (k+1)^{m+1} - k^{m+1} = \sum_{k=1}^n \sum_{i=0}^m \binom{m+1}{i} k^i = \sum_{i=0}^m \left(\binom{m+1}{i} \left(\sum_{k=1}^n k^i\right)\right) = \sum_{i=0}^{m} S_i(n) \binom{m+1}{i} \ (**)$

$\\ $Using$ \ (*) \ $and$ \ (**) \ $we have the result$$

Sy123 · Jul 25, 2015

Re: HSC 2015 4U Marathon

$\\ $Show that$ \ \sin^{-1} \left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x \ $for$ \ - 1 \leq x \leq 1$

Drsoccerball · Jul 25, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Show that$ \ \sin^{-1} \left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x \ $for$ \ - 1 \leq x \leq 1$

$\sin(2\tan^{-1}x)$

$$ Let $ tan^{-1}x = \alpha$

$2\sin \alpha \cos \alpha = \frac{2x}{x^2+1}$

$\therefore \frac{2x}{1+x^2}= sin(2\tan^{-1}x)$

$\therefore sin^{-1}{\frac{2x}{1+x^2}} = 2tan^{-1}x$

Sy123 · Jul 25, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$\sin(2\tan^{-1}x)$

$$ Let $ tan^{-1}x = \alpha$

$2\sin \alpha \cos \alpha = \frac{2x}{x^2+1}$

$\therefore \frac{2x}{1+x^2}= sin(2\tan^{-1}x)$

$\therefore sin^{-1}{\frac{2x}{1+x^2}} = 2tan^{-1}x$

You should mention when the domain is restricted!

integral95 · Jul 26, 2015

Re: HSC 2015 4U Marathon

$$Show that any root z of$ \ \ \ z^4+z+3 = 0 \ \ \ $satisfies$ \ \ \ |z| >1 \\ \\ $and that any root z of$ \ \ \ 4z^4+z+1 = 0 \Rightarrow \ \ |z| \leq 1$

glittergal96 · Jul 27, 2015

Re: HSC 2015 4U Marathon

integral95 said:
$$Show that any root z of$ \ \ \ z^4+z+3 = 0 \ \ \ $satisfies$ \ \ \ |z| >1 \\ \\ $and that any root z of$ \ \ \ 4z^4+z+1 = 0 \Rightarrow \ \ |z| \leq 1$

Key is the triangle inequality.

For |z| <= 1, we have

|p(z)| >= 3- |z+z^4| >= 3-|z|-|z|^4 >= 1 > 0.

and

|z^4q(1/z)| = |z^4+z^3+4| >= 4-|z|^3 - |z|^4 >=2 > 0.

So p has no roots inside the closed unit disk and q has no roots outside the open unit disk (and so certainly has no roots outside the closed unit disk).

Drsoccerball · Jul 27, 2015

Re: HSC 2015 4U Marathon

glittergal96 said:
Key is the triangle inequality.

For |z| <= 1, we have

|p(z)| >= 3- |z+z^4| >= 3-|z|-|z|^4 >= 1 > 0.

and

|z^4q(1/z)| = |z^4+z^3+4| >= 4-|z|^3 - |z|^4 >=2 > 0.

So p has no roots inside the closed unit disk and q has no roots outside the open unit disk (and so certainly has no roots outside the closed unit disk).

Do we have to use the triangle inequality? I thought by differentiating to find the min stationary point and subbing back into P(x) we obtain what we need?

InteGrand · Jul 27, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Do we have to use the triangle inequality? I thought by differentiating to find the min stationary point and subbing back into P(x) we obtain what we need?

$What do you mean? In this case, $P(z)$ takes on non-real complex values since the input can be any complex number, so the concept of a ``minimum'' is unclear since we don't order complex numbers.$

Librah · Jul 27, 2015

Re: HSC 2015 4U Marathon

Simplify/sum the following 1+2^2+3^2+4^2+5^2.....n^2.

HSC 2015 MX2 Marathon (archive) (5 Viewers)

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