Welcoming 2016 with a Polynomial.

\[ x^{2016}-x^{2015}+x^{2014}-\cdots+x^{2} -x+1\ \]

has roots , $x_{1},x_{2},\ldots,x_{2016}$ . Evaluate

$\left ( \displaystyle \sum_{k=1}^{2016}\sum_{i=1}^{2016}(1+x_{i})^{k} \right )$

#Algebra

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Comments

Taking a quick look at this during our x-mas festivities, I would suspect that the inner sum is always 2017, so that the whole sum would be 2016*2017.

[added later] We observe that the $y_i=-x_i$ , together with $y_0=1$ , are the 2017th roots of unity. Now

$\sum_{k=1}^{2016}\sum_{i=0}^{2016}(1-y_i)^k=\sum_{k=1}^{2016}\sum_{i=0}^{2016}(1-ky_i+....\pm y_i^{k})=\sum_{k=1}^{2016}2017=2016*2017$

since the sum of the $j$ th powers of the 2017th roots of unity is 0 for $j\leq 2016$ .

Otto Bretscher - 5 years, 5 months ago

yes,you are right @Otto Bretscher, please view my comment when you have time.

Aareyan Manzoor - 5 years, 5 months ago

I thought about it in a slightly different but equivalent way.

We observe that the $y_i=-x_i$ , together with $y_0=1$ , are the 2017th roots of unity. Now $\sum_{i=0}^{2016}(1-y_i)^k$ $=\sum_{i=0}^{2016}(1-ky_i+....\pm y_i^{k})=2017$ since the sum of the $j$ th powers of the 2017th roots of unity is 0.

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – This also would work

A Former Brilliant Member - 5 years, 5 months ago

@Otto Bretscher – hmm... i wrote that feature in my wiki "roots of unity" and proved it by the linked note. the same thing really.

Aareyan Manzoor - 5 years, 5 months ago

@Otto Bretscher , please try to solve using Vieta.

A Former Brilliant Member - 5 years, 5 months ago

My solution is quite short and straightforward.... why should I look for another solution? ;) (I guess I could use Viète to prove that the sum of the $j$ th powers of the roots of unity is 0)

Otto Bretscher - 5 years, 5 months ago

-1 is obviously not a root. we multiply by (x+1) to find $f(x)=x^{2017}+1=0$ one root of this expression is -1, and if we put it in the sum,i.e. $(1+x_i)=(1-1)=0$ . so we can go on withput worries. we are just searching for: $\sum_{k=1}^{2016}\sum_{i=1}^{2017} (x_i+1)^k=\sum_{i=1}^{2017} (\dfrac{(x_i+1)^{2017}-1}{x_i+1-1}-1)=\sum_{i=1} (x_i^{2016}+\binom{2017}{1}x_i^{2015}+....+\binom{2017}{2016}-1)$ gp used here. view this.yes, the sum of all 2016th to 1st power is zero. the $\binom{2017}{2016}-1=2016$ is left. added 2017 times results in $2016*2017=\boxed{4066272}$

Aareyan Manzoor - 5 years, 5 months ago

There may be a small error in the count (or rather two errors that cancel out). Your sum actually involves 2017 $i$ 's (you have added -1) but you have to subtract 1 from the summand.

Otto Bretscher - 5 years, 5 months ago

i have said that "one root of this expression is -1...". at the first.

Aareyan Manzoor - 5 years, 5 months ago

@Aareyan Manzoor – So when you add up $\sum_{i}$ you have 2017 terms.

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – i have said why we can neglect -1, reducing a root and giving us 2016 terms.

Aareyan Manzoor - 5 years, 5 months ago

@Aareyan Manzoor – But if you use only 2016 terms then it is no longer true that the sum of the 2016th to first powers is zero. Think about it... you will see what I mean.

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – true... the answer should be $2017^2$ ?

Aareyan Manzoor - 5 years, 5 months ago

@Aareyan Manzoor – No, there is another small error that compensates for this one... you are losing a term $-1$ along the way... the constant term in your sum should be 2017-1=2016, so it all works out... a Christmas miracle ;)

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – sorry for late response( electricity...), what a miracle indeed, i have edited it.

Aareyan Manzoor - 5 years, 5 months ago

@Aareyan Manzoor – One more small correction: In your double sum, it is $i$ that goes to 2017, not $k$ .

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – corrected once again...

Aareyan Manzoor - 5 years, 5 months ago

@Otto Bretscher , @Aareyan Manzoor , I have slight doubt ,

the inner summation will be definitely 2017.

So we need a polynomial whose roots are , $2017^{1} , 2017^{2} , 2017^{3} , .... , 2017^{2016}$ , isnt it ?

A Former Brilliant Member - 5 years, 5 months ago

nope.

Aareyan Manzoor - 5 years, 5 months ago

@Aareyan Manzoor , but why ?

A Former Brilliant Member - 5 years, 5 months ago

@A Former Brilliant Member – i dont get "we seek a polynomial ...."

Aareyan Manzoor - 5 years, 5 months ago

@Aareyan Manzoor – sorry we need a polynomial , so that we can find the summation using vieta , @Aareyan Manzoor

A Former Brilliant Member - 5 years, 5 months ago

@A Former Brilliant Member – In my approach, I'm not using Viete and I certainly don't need such a polynomial.

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – @Otto Bretscher , but please also see the vieta approach ,,, is it right ?

A Former Brilliant Member - 5 years, 5 months ago

@A Former Brilliant Member – Yes, Aareyan is applying Viete to the polynomial $x^{2017}+1$ ... that works.

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – I am just asking whether my approach is right ? (by vieta) @Otto Bretscher

A Former Brilliant Member - 5 years, 5 months ago

@A Former Brilliant Member – we really didnt use vietas. otto sir used roots of unity, i used a version of newtons sum.

Aareyan Manzoor - 5 years, 5 months ago

@Aareyan Manzoor – But can this be solved by vieta ? according to me it could not @Aareyan Manzoor

A Former Brilliant Member - 5 years, 5 months ago

@A Former Brilliant Member – i think not....

Aareyan Manzoor - 5 years, 5 months ago

@Aareyan Manzoor – Lets try again write a solution on the work that you have done , on vieta approach , then I will post mine @Aareyan Manzoor , @Otto Bretscher

A Former Brilliant Member - 5 years, 5 months ago

@Aareyan Manzoor – Aareyan, are you not using Viete to make sure that the symmetric sums of the $x_i$ are zero?

Otto Bretscher - 5 years, 5 months ago

So am I right or wrong ? @Aareyan Manzoor

A Former Brilliant Member - 5 years, 5 months ago

@Otto Bretscher , do we need to find a polynomial whose roots are , $2017 , 2017^2 ,.... 2017^2016$ ? As the inner summation is 2017 @Aareyan Manzoor help

A Former Brilliant Member - 5 years, 5 months ago

@Otto Bretscher ,

Cool problem .... the new problem which is based on our yesterday's conversation ...

A Former Brilliant Member - 5 years, 5 months ago

Yes, thanks for that suggestion! For square-free numbers the count turns out to be pretty easy... the general case is a mess, though! Nobody has solved it yet... I'm counting on you!

Otto Bretscher - 5 years, 5 months ago

Bonus problem is complicated though @Otto Bretscher

A Former Brilliant Member - 5 years, 5 months ago

@A Former Brilliant Member – No it's not! Not for a guy like you ;)

Otto Bretscher - 5 years, 5 months ago

@Calvin Lin , what is your opinion for this ? What will be your approach ?

A Former Brilliant Member - 5 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$