Waste less time on Facebook — follow Brilliant.
×

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
1 year, 11 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} \( \boxed{123} \)

Comments

Sort by:

Top Newest

@Dev Sharma , @Swapnil Das , @Nihar Mahajan , @Harsh Shrivastava , @Shivam Jadhav @Aareyan Manzoor You might find this interesting.

Chinmay Sangawadekar - 1 year, 11 months ago

Log in to reply

does the summation mean \[\dfrac{1}{2+x_1}+\dfrac{1}{3+x_2}+....+\dfrac{1}{p_{100}+x_{100}}\]

Aareyan Manzoor - 1 year, 11 months ago

Log in to reply

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 - 1 year, 11 months ago

Log in to reply

@Chinmay Sangawadekar @Chinmay Sangawadekar, toitent here is \(\phi(x_i\)?

Aareyan Manzoor - 1 year, 11 months ago

Log in to reply

@Aareyan Manzoor no toitent of I

Chinmay Sangawadekar - 1 year, 11 months ago

Log in to reply

@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 - 1 year, 11 months ago

Log in to reply

@Chinmay Sangawadekar what about p then? There are 100 values of p not any constant

Dev Sharma - 1 year, 11 months ago

Log in to reply

@Dev Sharma no no take the whole summation as constant

Chinmay Sangawadekar - 1 year, 11 months ago

Log in to reply

But I think I can do the totient one

Chinmay Sangawadekar - 1 year, 11 months ago

Log in to reply

yes

Chinmay Sangawadekar - 1 year, 11 months ago

Log in to reply

I think yes.

Akshat Sharda - 1 year, 11 months ago

Log in to reply

@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 - 1 year, 11 months ago

Log in to reply

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 - 1 year, 10 months ago

Log in to reply

@Calvin Lin I have updated it accordingly.

Chinmay Sangawadekar - 1 year, 11 months ago

Log in to reply

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 - 1 year, 11 months ago

Log in to reply

sorry sir its pi ..denotes the ith prime no...I will update it

Chinmay Sangawadekar - 1 year, 11 months ago

Log in to reply

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 - 1 year, 11 months ago

Log in to reply

I have one more ,

Just replace \(p\) with totient(_i)

Chinmay Sangawadekar - 1 year, 11 months ago

Log in to reply

@Chinmay Sangawadekar, have you solved this, or just randomly picked some functions?

Aareyan Manzoor - 1 year, 11 months ago

Log in to reply

So I posted it on Brilliant

Chinmay Sangawadekar - 1 year, 11 months ago

Log in to reply

I jist had an idea abt it...then I tried , but no outcome

Chinmay Sangawadekar - 1 year, 11 months ago

Log in to reply

Can we find this for small degree polynomials

Chinmay Sangawadekar - 1 year, 11 months ago

Log in to reply

No I haven't solved this yet but can this be solved?

Chinmay Sangawadekar - 1 year, 11 months ago

Log in to reply

i think not possible to find or may be i am wrong??

Dev Sharma - 1 year, 11 months ago

Log in to reply

u are correct, speaking analytically...

Aareyan Manzoor - 1 year, 11 months ago

Log in to reply

do you think totient one is possible to find?

Dev Sharma - 1 year, 11 months ago

Log in to reply

@Dev Sharma hmm... i have not quite understand what toitent means here. eulers toitent function?

Aareyan Manzoor - 1 year, 11 months ago

Log in to reply

@Aareyan Manzoor yeah........

Dev Sharma - 1 year, 11 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...