×

# 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

9 months, 1 week ago

Sort by:

@Dev Sharma , @Swapnil Das , @Nihar Mahajan , @Harsh Shrivastava , @Shivam Jadhav @Aareyan Manzoor You might find this interesting. · 9 months, 1 week ago

does the summation mean $\dfrac{1}{2+x_1}+\dfrac{1}{3+x_2}+....+\dfrac{1}{p_{100}+x_{100}}$ · 9 months, 1 week ago

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 · 9 months, 1 week ago

@Chinmay Sangawadekar, toitent here is $$\phi(x_i$$? · 9 months, 1 week ago

no toitent of I · 9 months, 1 week ago

I think this could be possible , take the summation as any constant and then compare it with 1/provided ... then apply vieta ? isnt it ? · 9 months, 1 week ago

what about p then? There are 100 values of p not any constant · 9 months, 1 week ago

no no take the whole summation as constant · 9 months, 1 week ago

But I think I can do the totient one · 9 months, 1 week ago

yes · 9 months, 1 week ago

I think yes. · 9 months, 1 week ago

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. · 9 months, 1 week ago

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 $$x_1$$? · 8 months, 3 weeks ago

@Calvin Lin I have updated it accordingly. · 9 months, 1 week ago

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? Staff · 9 months, 1 week ago

sorry sir its pi ..denotes the ith prime no...I will update it · 9 months, 1 week ago

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 $$x_ 1$$? Staff · 9 months ago

I have one more ,

Just replace $$p$$ with totient(_i) · 9 months, 1 week ago

@Chinmay Sangawadekar, have you solved this, or just randomly picked some functions? · 9 months, 1 week ago

So I posted it on Brilliant · 9 months, 1 week ago

I jist had an idea abt it...then I tried , but no outcome · 9 months, 1 week ago

Can we find this for small degree polynomials · 9 months, 1 week ago

No I haven't solved this yet but can this be solved? · 9 months, 1 week ago

i think not possible to find or may be i am wrong?? · 9 months, 1 week ago

u are correct, speaking analytically... · 9 months, 1 week ago

do you think totient one is possible to find? · 9 months, 1 week ago

hmm... i have not quite understand what toitent means here. eulers toitent function? · 9 months, 1 week ago