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In the land of polynomials with 500 dimensions

\[\displaystyle \sum_{ i=1}^{500} \frac{1}{p_i+x_i}\]

Where \(p_i\) denotes the \(i^{th}\) prime number , for a polynomial mentioned below.

\(x^{500} + x^{300} - x^{3} + x - 1729\)

has roots \(x_{1} , x_{2} , ..... , x_{500}\)

Then evaluate the above summation.

See this

Note by Chinmay Sangawadekar
11 months, 3 weeks ago

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@Dev Sharma , @Swapnil Das , @Nihar Mahajan , @Harsh Shrivastava , @Shivam Jadhav @Aareyan Manzoor You might find this interesting. Chinmay Sangawadekar · 11 months, 3 weeks ago

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@Chinmay Sangawadekar does the summation mean \[\dfrac{1}{2+x_1}+\dfrac{1}{3+x_2}+....+\dfrac{1}{p_{100}+x_{100}}\] Aareyan Manzoor · 11 months, 3 weeks ago

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@Aareyan Manzoor I have been working on this problem as you said it cant be solved analytically yeah see for the totient I have mentioned it above Chinmay Sangawadekar · 11 months, 3 weeks ago

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@Chinmay Sangawadekar @Chinmay Sangawadekar, toitent here is \(\phi(x_i\)? Aareyan Manzoor · 11 months, 3 weeks ago

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@Aareyan Manzoor no toitent of I Chinmay Sangawadekar · 11 months, 2 weeks ago

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@Aareyan Manzoor I think this could be possible , take the summation as any constant and then compare it with 1/provided ... then apply vieta ? isnt it ? Chinmay Sangawadekar · 11 months, 2 weeks ago

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@Chinmay Sangawadekar what about p then? There are 100 values of p not any constant Dev Sharma · 11 months, 2 weeks ago

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@Dev Sharma no no take the whole summation as constant Chinmay Sangawadekar · 11 months, 2 weeks ago

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@Aareyan Manzoor But I think I can do the totient one Chinmay Sangawadekar · 11 months, 3 weeks ago

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

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@Aareyan Manzoor I think yes. Akshat Sharda · 11 months, 3 weeks ago

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@Akshat Sharda this can't be analytically solved i think.... it could have been if the polynomial at the negative of the primes attained the same value. Aareyan Manzoor · 11 months, 3 weeks ago

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I think that the question is not complete in it's statement. I mean, how do we define the pairing of a root with a prime number? What is \(x_1\)? Anupam Nayak · 11 months ago

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@Calvin Lin I have updated it accordingly. Chinmay Sangawadekar · 11 months, 2 weeks ago

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I am assuming that you want "where \(p_i \) denotes the ith prime number"?

If \(p\) is allowed to vary , then this problem is not well phrased because there isn't a (canonical) order to the roots of a polynomial. E.g. Assume that we only have 2 roots which are \(1+i \) and \( 1 - i \). What is the sum that we're interested in? Calvin Lin Staff · 11 months, 2 weeks ago

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@Calvin Lin sorry sir its pi ..denotes the ith prime no...I will update it Chinmay Sangawadekar · 11 months, 2 weeks ago

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@Chinmay Sangawadekar I understand. But, what is the order of the roots? Since we (most likely) have complex roots, there is no (canonical) order to the roots that we can set. In particular, which of the roots is \( x_ 1 \)? Calvin Lin Staff · 11 months, 2 weeks ago

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I have one more ,

Just replace \(p\) with totient(_i) Chinmay Sangawadekar · 11 months, 3 weeks ago

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@Chinmay Sangawadekar @Chinmay Sangawadekar, have you solved this, or just randomly picked some functions? Aareyan Manzoor · 11 months, 3 weeks ago

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@Aareyan Manzoor So I posted it on Brilliant Chinmay Sangawadekar · 11 months, 3 weeks ago

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@Aareyan Manzoor I jist had an idea abt it...then I tried , but no outcome Chinmay Sangawadekar · 11 months, 3 weeks ago

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@Aareyan Manzoor Can we find this for small degree polynomials Chinmay Sangawadekar · 11 months, 3 weeks ago

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@Aareyan Manzoor No I haven't solved this yet but can this be solved? Chinmay Sangawadekar · 11 months, 3 weeks ago

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i think not possible to find or may be i am wrong?? Dev Sharma · 11 months, 2 weeks ago

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@Dev Sharma u are correct, speaking analytically... Aareyan Manzoor · 11 months, 2 weeks ago

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@Aareyan Manzoor do you think totient one is possible to find? Dev Sharma · 11 months, 2 weeks ago

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@Dev Sharma hmm... i have not quite understand what toitent means here. eulers toitent function? Aareyan Manzoor · 11 months, 2 weeks ago

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@Aareyan Manzoor yeah........ Dev Sharma · 11 months, 2 weeks ago

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