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# 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

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

9 months, 1 week ago

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-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 $2016*2017=\boxed{4066272}$ · 9 months, 1 week 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. · 9 months, 1 week ago

i have said that "one root of this expression is -1...". at the first. · 9 months, 1 week ago

So when you add up $$\sum_{i}$$ you have 2017 terms. · 9 months, 1 week ago

i have said why we can neglect -1, reducing a root and giving us 2016 terms. · 9 months, 1 week ago

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. · 9 months, 1 week ago

true... the answer should be $$2017^2$$? · 9 months, 1 week ago

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 ;) · 9 months, 1 week ago

sorry for late response( electricity...), what a miracle indeed, i have edited it. · 9 months, 1 week ago

One more small correction: In your double sum, it is $$i$$ that goes to 2017, not $$k$$. · 9 months, 1 week ago

corrected once again... · 9 months, 1 week ago

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 $$j\leq 2016$$. · 9 months, 1 week ago

@Otto Bretscher , please try to solve using Vieta. · 9 months, 1 week 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) · 9 months, 1 week ago

yes,you are right @Otto Bretscher, please view my comment when you have time. · 9 months, 1 week 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. · 9 months, 1 week ago

hmm... i wrote that feature in my wiki "roots of unity" and proved it by the linked note. the same thing really. · 9 months, 1 week ago

This also would work · 9 months, 1 week ago

@Calvin Lin , what is your opinion for this ? What will be your approach ? · 9 months, 1 week ago

Cool problem .... the new problem which is based on our yesterday's conversation ... · 9 months, 1 week 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! · 9 months, 1 week ago

Bonus problem is complicated though @Otto Bretscher · 9 months, 1 week ago

No it's not! Not for a guy like you ;) · 9 months, 1 week 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 · 9 months, 1 week 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 ? · 9 months, 1 week ago

nope. · 9 months, 1 week ago

So am I right or wrong ? @Aareyan Manzoor · 9 months, 1 week ago

@Aareyan Manzoor , but why ? · 9 months, 1 week ago

i dont get "we seek a polynomial ...." · 9 months, 1 week ago

sorry we need a polynomial , so that we can find the summation using vieta , @Aareyan Manzoor · 9 months, 1 week ago

we really didnt use vietas. otto sir used roots of unity, i used a version of newtons sum. · 9 months, 1 week ago

But can this be solved by vieta ? according to me it could not @Aareyan Manzoor · 9 months, 1 week ago

i think not.... · 9 months, 1 week ago

Aareyan, are you not using Viete to make sure that the symmetric sums of the $$x_i$$ are zero? · 9 months, 1 week ago

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 · 9 months, 1 week ago

In my approach, I'm not using Viete and I certainly don't need such a polynomial. · 9 months, 1 week ago

I am just asking whether my approach is right ? (by vieta) @Otto Bretscher · 9 months, 1 week ago

@Otto Bretscher , but please also see the vieta approach ,,, is it right ? · 9 months, 1 week ago

Yes, Aareyan is applying Viete to the polynomial $$x^{2017}+1$$... that works. · 9 months, 1 week ago