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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

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

Note by Chinmay Sangawadekar
11 months, 3 weeks 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 \[2016*2017=\boxed{4066272}\] Aareyan Manzoor · 11 months, 3 weeks ago

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@Aareyan Manzoor 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 · 11 months, 3 weeks ago

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@Otto Bretscher i have said that "one root of this expression is -1...". at the first. Aareyan Manzoor · 11 months, 3 weeks ago

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@Aareyan Manzoor So when you add up \(\sum_{i}\) you have 2017 terms. Otto Bretscher · 11 months, 3 weeks ago

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@Otto Bretscher i have said why we can neglect -1, reducing a root and giving us 2016 terms. Aareyan Manzoor · 11 months, 3 weeks ago

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@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 · 11 months, 3 weeks ago

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@Otto Bretscher true... the answer should be \(2017^2\)? Aareyan Manzoor · 11 months, 3 weeks ago

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@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 · 11 months, 3 weeks ago

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@Otto Bretscher sorry for late response( electricity...), what a miracle indeed, i have edited it. Aareyan Manzoor · 11 months, 3 weeks ago

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@Aareyan Manzoor One more small correction: In your double sum, it is \(i\) that goes to 2017, not \(k\). Otto Bretscher · 11 months, 3 weeks ago

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@Otto Bretscher corrected once again... Aareyan Manzoor · 11 months, 3 weeks ago

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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 · 11 months, 3 weeks ago

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@Otto Bretscher @Otto Bretscher , please try to solve using Vieta. Chinmay Sangawadekar · 11 months, 3 weeks ago

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@Chinmay Sangawadekar 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 · 11 months, 3 weeks ago

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@Otto Bretscher yes,you are right @Otto Bretscher, please view my comment when you have time. Aareyan Manzoor · 11 months, 3 weeks ago

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@Aareyan Manzoor 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 · 11 months, 3 weeks ago

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@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 · 11 months, 3 weeks ago

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@Otto Bretscher This also would work Chinmay Sangawadekar · 11 months, 3 weeks ago

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@Calvin Lin , what is your opinion for this ? What will be your approach ? Chinmay Sangawadekar · 11 months, 2 weeks ago

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@Otto Bretscher ,

Cool problem .... the new problem which is based on our yesterday's conversation ... Chinmay Sangawadekar · 11 months, 3 weeks ago

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@Chinmay Sangawadekar 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 · 11 months, 3 weeks ago

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@Otto Bretscher Bonus problem is complicated though @Otto Bretscher Chinmay Sangawadekar · 11 months, 3 weeks ago

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@Chinmay Sangawadekar No it's not! Not for a guy like you ;) Otto Bretscher · 11 months, 3 weeks ago

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@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 Chinmay Sangawadekar · 11 months, 3 weeks ago

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@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 ? Chinmay Sangawadekar · 11 months, 3 weeks ago

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@Chinmay Sangawadekar nope. Aareyan Manzoor · 11 months, 3 weeks ago

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@Aareyan Manzoor So am I right or wrong ? @Aareyan Manzoor Chinmay Sangawadekar · 11 months, 3 weeks ago

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@Aareyan Manzoor @Aareyan Manzoor , but why ? Chinmay Sangawadekar · 11 months, 3 weeks ago

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@Chinmay Sangawadekar i dont get "we seek a polynomial ...." Aareyan Manzoor · 11 months, 3 weeks ago

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@Aareyan Manzoor sorry we need a polynomial , so that we can find the summation using vieta , @Aareyan Manzoor Chinmay Sangawadekar · 11 months, 3 weeks ago

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@Chinmay Sangawadekar we really didnt use vietas. otto sir used roots of unity, i used a version of newtons sum. Aareyan Manzoor · 11 months, 3 weeks ago

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@Aareyan Manzoor But can this be solved by vieta ? according to me it could not @Aareyan Manzoor Chinmay Sangawadekar · 11 months, 3 weeks ago

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@Chinmay Sangawadekar i think not.... Aareyan Manzoor · 11 months, 3 weeks ago

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@Aareyan Manzoor Aareyan, are you not using Viete to make sure that the symmetric sums of the \(x_i\) are zero? Otto Bretscher · 11 months, 3 weeks ago

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@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 Chinmay Sangawadekar · 11 months, 3 weeks ago

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@Chinmay Sangawadekar In my approach, I'm not using Viete and I certainly don't need such a polynomial. Otto Bretscher · 11 months, 3 weeks ago

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@Otto Bretscher I am just asking whether my approach is right ? (by vieta) @Otto Bretscher Chinmay Sangawadekar · 11 months, 3 weeks ago

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@Otto Bretscher @Otto Bretscher , but please also see the vieta approach ,,, is it right ? Chinmay Sangawadekar · 11 months, 3 weeks ago

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@Chinmay Sangawadekar Yes, Aareyan is applying Viete to the polynomial \(x^{2017}+1\)... that works. Otto Bretscher · 11 months, 3 weeks ago

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