Help with Polynomial, please, (1 Viewer)

Jason Xie · May 31, 2012

The first two questions i got it, but not the last one, can anyone do it, show working please, Thank you so much!!!

Jason Xie · May 31, 2012

I think, the last one is much harder, guys, try ur best to solve it!!, thank you!!

qrpw · May 31, 2012

Hmm, weird question, just by looking at it, it's pretty easy to prove without induction.

deswa1 · May 31, 2012

I haven't tried it yet but I think you need to use the stronger form of induction i.e. assume true for n=k and all integers up to k (1,2,...,k) and then prove for k+1. I'll do it tomorrow if no one has done it by then.

Carrotsticks · May 31, 2012

Jason Xie said:
The first two questions i got it, but not the last one, can anyone do it, show working please, Thank you so much!!!

View attachment 25346

From where did you get this question? I'm inclined to think that the question has a typo because the Auxiliary equation (T_n = T_{n-1} - T_{n-2}) does not match the quadratic polynomial given.

Verifying it now.

Carrotsticks · May 31, 2012

Yep confirmed typo in Question 2. Should be:

$T_n = -T_{n-1} - T_{n-2}$

Will post solution shortly.

Nooblet94 · May 31, 2012

Carrotsticks said:
Yep confirmed typo in Question 2. Should be:

$T_n = -T_{n-1} - T_{n-2}$

Will post solution shortly.

Ahh fuck, just wasted 15 minutes wondering what the hell I was doing wrong

bleakarcher · May 31, 2012

Carrotsticks said:
Yep confirmed typo in Question 2. Should be:

$T_n = -T_{n-1} - T_{n-2}$

Will post solution shortly.

Carrotsticks, can you please put the solution to this one?

Carrotsticks · May 31, 2012

Here's the first two for the mean time.

Doing the third one now, keep making silly mistakes and getting ridiculous solutions.

jeffwu95 · May 31, 2012

jun u noob

thorax94 · May 31, 2012

Carrotsticks said:
From where did you get this question? I'm inclined to think that the question has a typo because the Auxiliary equation (T_n = T_{n-1} - T_{n-2}) does not match the quadratic polynomial given.

Verifying it now.

Hi Carrotsticks, I am Jason's classmate and this question came from some homework that our 4u teacher gave us. There is indeed a typo.

Carrotsticks · May 31, 2012

Rest of (my) solution. May have made a silly mistake somewhere, a bit tired now.

EDIT: Tried to make image a bit larger for you to see.

jeffwu95 · May 31, 2012

Carrotsticks · May 31, 2012

jeffwu95 said:
View attachment 25348

As beautiful as your solution may/may not have been, I'm afraid I can't see it =/.

Btw for my one, you can download the image here:

And then zoom in.

Trebla · May 31, 2012

Third part requires strong induction.

$$When $n=1\\\\$LHS $= T_1 = -1\\\\$RHS $= 2\cos \frac{2\pi}{3} = -1 = $ LHS$\\\\$When $n=2\\\\$LHS $= T_2 = -1\\\\$RHS $= 2\cos \frac{4\pi}{3} = -1 = $ LHS$\\\\$Assume the statement holds for integers $n \leq k\\\\T_k = 2\cos\frac{2k\pi}{3}\\\\T_{k-1} = 2\cos\frac{2(k-1)\pi}{3}$

$$We need to prove that $\\\\T_{k+1} = 2\cos\frac{2(k+1)\pi}{3}\\\\$LHS $\\\\= T_{k+1}\\\\= T_{k} - T_{k-1}\\\\= 2\cos\frac{2k\pi}{3} - 2\cos\frac{2(k-1)\pi}{3}\\\\= 2\cos\frac{2k\pi}{3} - 2\cos\frac{2k\pi}{3}\cos\frac{2\pi}{3} - 2\sin\frac{2k\pi}{3}\sin\frac{2\pi}{3}\\\\$But $\cos\frac{2\pi}{3} = - \frac{1}{2}\\\\= - 2\cos\frac{2k\pi}{3} + \cos\frac{2k\pi}{3} - 2\sin\frac{2k\pi}{3}\sin\frac{2\pi}{3} \\\\ = - \cos\frac{2k\pi}{3} - 2\sin\frac{2k\pi}{3}\sin\frac{2\pi}{3} \\\\ = 2\left(-\cos\frac{2k\pi}{3}\times \frac{1}{2} - \sin\frac{2k\pi}{3}\sin\frac{2\pi}{3}\right)\\\\$But $\cos\frac{2\pi}{3} = - \frac{1}{2}\\\\= 2\left(\cos\frac{2k\pi}{3}\cos\frac{2\pi}{3} - \sin\frac{2k\pi}{3}\sin\frac{2\pi}{3}\right)\\\\=2\cos \dfrac{2(k+1)}{3}\\\\=$ RHS$\\\\$The statement holds for $n=k+1$ if it holds for integers $n\leq k$ and since it holds for $n=1$ and $n=2$, then the statement is true by induction.$

Just in case you can't see jeffwu95's solution:

(i)
By sum of roots:

$T_1\\\\ = a+b \\\\= -1$

Using sum and product of roots:

$T_2\\\\= a^2+b^2\\\\= (a+b)^2 - 2ab \\\\ = (-1)^2 - 2\\\\= -1$

(ii)
If a and b are the roots of x² + x + 1 = 0 then

$a^2 + a + 1 = 0 \\\\$Multiply both sides by $a^{n-2} \\\\\Rightarow a^n + a^{n-1} + a^{n-2} = 0\\\\$Similarly $b^n + b^{n-1} + b^{n-2} = 0\\\\$Adding the two equations $\\\\ (a^n+b^n) + (a^{n-1}+b^{n-1}) + (a^{n-2}+b^{n-2}) = 0 \\\\\Rightarrow T_n + T_{n-1} + T_{n-2} = 0\\\\\therefore T_n = - T_{n-1} - T_{n-2}$

jeffwu95 · May 31, 2012

there ya go

Carrotsticks · May 31, 2012

jeffwu95 said:
View attachment 25350 there ya go

Wonderful!

seanieg89 · Jun 1, 2012

Seems silly to require it to be done by induction when de moivre's is a course theorem...

Help with Polynomial, please, (1 Viewer)

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