Thoughts on 2015 CSSA MX2 Trials? (1 Viewer)

Ekman · Jul 30, 2015

Re: 4u trials CSSA

leehuan said:
$$It was $\sum _{ k=1 }^{ n }{ { \left( { a }_{ k } \right) }^{ 2 } }$

He typed the question slightly wrong.

I also feel stupid realising that I missed how to interpret my sums and products of roots.

Ok that makes sense since: $\sum _{ k=1 }^{ n }{ { \left( { a }_{ k } \right) }^{ 2 } } = - \sum _{ k=1 }^{ n }{ { \left( i{ a }_{ k } \right) }^{ 2 } }$

Drsoccerball · Jul 30, 2015

Re: 4u trials CSSA

Drsoccerball said:
$$Let $z \in \mathbb C$ and $n \in \mathbb Z $ such that $ \\ { \left( z-1 \right) }^{ n }+{ \left( z+1 \right) }^{ n }=0 \\ $(i) Show that $\left| z-1 \right| =\left| z+1 \right| $ and hence deduce that all the roots of the equation are in the form $z=i{ a }_{ k } \quad \forall \quad { a }_{ k }\in \mathbb R \quad \forall \quad k\in \mathbb N$

Easy 2 marker.

$\\ $(ii) Use the relationship between the roots and coefficients to show that$ \\ \sum _{ k=1 }^{ n }{ { \left( { a }_{ k } \right) }^{ 2 } } =n\left( n-1 \right)$

And that was out of 3.

Have a go if you want.

Click to expand...

i) $(z-1)^n=-(z+1)^n$

$|z-1|^n =|z+1|^n$

$\therefore |z-1|=|z+1|$

$$ Let $ z =x+iy$

$(x-1)^2+y^2 = (x+1)^2 +y^2$

$\therefore x=0 \therefore z=ia_k$

$z^2= -a_k ^2$

b) $P(x)= (z-1)^n +(z+1)^n$

$P(x)=2(z^n +\binom{n}{2}z^{n-2} +...+1)$

$-\frac{2 \cdot n! }{2! (n-2)!} = -\sum_{k=1}^n{a_k^2}$

$\therefore \sum _{ k=1 }^{ n }{ { \left( { a }_{ k } \right) }^{ 2 } } =n\left( n-1 \right)$

Hows this

Someone sent me the question in inbox
EDIT: If you square the result that means your halfing the roots doesn't that mean the sum should be up to $\frac{n}{2}$ That was my working out but i dont think its right
Actually makes sense because lets let u=z^2 and do sum of roots ! thats it fin

Drsoccerball · Jul 30, 2015

Re: 4u trials CSSA

Ekman said:
Edit: I noticed something fishy about roots being purely imaginary and that the double sum of roots being equal to a positive number, that could explain the reason why it would need to be positive n(n-1)?

Look at my working

InteGrand · Jul 31, 2015

Re: 4u trials CSSA

Drsoccerball said:
Someone sent me the question in inbox
EDIT: If you square the result that means your halfing the roots doesn't that mean the sum should be up to $\frac{n}{2}$ That was my working out but i dont think its right
Actually makes sense because lets let u=z^2 and do sum of roots ! thats it fin

It's just:

sum of squares of roots = (sum of roots)² - 2•(sum of roots taken two at a time) (just rearranging the expansion of (a₁ + a₂ + ... + a_n)²)

i.e.

$$\sum _{k=1}^n (-a_k ^2)=0-2\times \frac{2\binom{n}{2}}{2}$ (since the sum of roots is $0$ as the coefficient of $z^{n-1}$ is 0, and the sum of pairs of roots is $\frac{2\binom{n}{2}}{2}$, based on the coefficient of $z^{n-2}$ and the coefficient of $z^n$)$

$$\Rightarrow \sum _{k=1}^n a_k ^2=2\times \frac{2\binom{n}{2}}{2}=2\binom{n}{2}=2\times \frac{n(n-1)}{2}=n(n-1)$ (using $\binom{n}{2}=\frac{n(n-1)}{2}$).$

You should probably explain your steps a bit. And what do you mean let u = z². Why do we need to do that?

Edit: oh I see what you were doing, you were treating the polynomial as a polynomial in z² I think. But this argument only works for even n, because for odd n (e.g. imagine if the polynomial's first term was z⁵), substituting u = z² would not give us a polynomial in u (as the leading term's power of u would be a non-integer). And for even n, your sum on the RHS of third line of part b)'s working would indeed be twice the sum of roots of the polynomial in u (which your LHS is), but you would need to provide explanation for why this is the case (using the conjugate root theorem to explain it).

Drsoccerball · Jul 31, 2015

Re: 4u trials CSSA

InteGrand said:
It's just:

sum of squares of roots = (sum of roots)² - 2•(sum of roots taken two at a time) (just rearranging the expansion of (a₁ + a₂ + ... + a_n)²)

i.e.

$$\sum _{k=1}^n (-a_k ^2)=0-2\times \frac{2\binom{n}{2}}{2}$ (since the sum of roots is $0$ as the coefficient of $z^{n-1}$ is 0, and the sum of pairs of roots is $\frac{2\binom{n}{2}}{2}$, based on the coefficient of $z^{n-2}$ and the coefficient of $z^n$)$

$$\Rightarrow \sum _{k=1}^n a_k ^2=2\times \frac{2\binom{n}{2}}{2}==2\binom{n}{2}=2\times \frac{n(n-1)}{2}=n(n-1)$ (using $\binom{n}{2}=\frac{n(n-1)}{2}$).$

You should probably explain your steps a bit. And what do you mean let u = z². Why do we need to do that?

Edit: oh I see what you were doing, you were treating the polynomial as a polynomial in z² I think. But this argument only works for even n, because for odd n (e.g. imagine if the polynomial's first term was z⁵), substituting u = z² would not give us a polynomial in u (as the leading term's power of u would be a non-integer). And for even n, your sum on the RHS of third line of part b)'s working would indeed be twice the sum of roots of the polynomial in u (which your LHS is), but you would need to provide explanation for why this is the case (using the conjugate root theorem to explain it).

So id lose one mark since i didnt prove for odd n?

Thoughts on 2015 CSSA MX2 Trials? (1 Viewer)

Ekman

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Drsoccerball

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InteGrand

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Drsoccerball

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