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//This question was solved as by the hints given in the comments section, thank you to all!

Here's the exact question:

If $p(x)$ denotes a polynomial of degree $n$, such that $p(k) = 1/k$ for $k = 1, 2 , 3 , ..., n + 1$, determine $p(2019)$ for $n = 2017$

This question may not be similar to Suppose that $P(x)$ is a polynomial of... regarding the final solution

So this question should also not be a duplicate.....

Here was my initial approach:

Take $f(x) = p(x) - 1/x$

Let the leading coefficient of $f(x)$ be $m$

Then, $f(x) = m(x - 2)(x - 3)(x - 4)......(x - 2017)(x - 2018)$ will be zero for $x = k$

I tried to plug in $0$ , but $1/0$ won't help at anything...

Then I tried $-1$ , but that didn't help either

And then, I tried $1$ , and it only suggests that $m$ should be $0$ which is not probable for a $n$th degree polynomial

Can anyone please help me at this, any suggestions/edits would greatly help! Thank you

Cuckoo Beats
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    $f$ is not a polynomial, so it doesn't have a leading coefficient. – aschepler Mar 27 '23 at 03:11
  • @dxiv I am really sorry to find the approach similar to that of the linked question. But I couldn't help myself with the uploaded answers, and the solution seems a bit different too. But I will try to not repeat questions with similar approaches

    Thank you for your hint, I am trying

     – Cuckoo Beats Mar 27 '23 at 03:13
  • Alternatively, take $P(x)$ from the linked problem. Then your $p(x) = 1 - P(x-1)$. (i.e. It's not just solved in the same manner as your own linked problem - it's solved by the linked problem) – Brian Moehring Mar 27 '23 at 03:13
  • @BrianMoehring I could not get what you meant, can you please be a bit more open to your hint, thank you! – Cuckoo Beats Mar 27 '23 at 03:15
  • @aschepler is that really true? I donot think the same way though :( – Cuckoo Beats Mar 27 '23 at 03:16
  • @CuckooBeats Correction to the hint: $,xp(x) - 1 = c(x-1)(x-2)\dots(x-n)(x-n-1),$. Determine $c$ from the condition at $x=0$. This is entirely similar to how the linked problem is solved. – dxiv Mar 27 '23 at 03:17
  • @dxiv thank you so much! I solved the question now :) – Cuckoo Beats Mar 27 '23 at 03:18
  • @CuckooBeats Glad it helped. It is perfectly fine, even encouraged, to post a self-answer. – dxiv Mar 27 '23 at 03:19
  • @dxiv okay, i will do it! – Cuckoo Beats Mar 27 '23 at 03:20

1 Answers1

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So this is the complete solution, as discussed in the comments:

According to the question, $p(x)$ is a $n$th degree polynomial so it should have $n$ roots

So something like $x(p(x))$ should be a $(n+1)$th degree polynomial having $(n+1)$ roots

So, we modify the $f(x)$ in my question into $x(p(x) - 1/x) = xp(x) - 1$

Then, plug in $x = 0$

Which equals, $0 - 1 = m(0 - 1)(0 - 2).....(0 - 2017)(0 - 2018)$

From this, we get $m = -1/2018!$

So $2019p(2019) - 1 = (-1/2018!)(2019 - 1)(2019 - 2)....(2019 - 2016)$

Which equals, $2019p(2019) - 1 = (-1/2018!)(2018!) = -1$

Hence, $2019p(2019) = 0$ or $p(2019) = 0$

Cuckoo Beats
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  • Proof looks good, only you need to justify the first step. Something like "$xp(x)-1$ is a polynomial of degree $n+1$ and each of $x=1,2,\dots,n+1$ is a root, so $xp(x)-1=m(x-1)(x-2)\dots(x-n-1)$ for some constant $m$". – dxiv Mar 27 '23 at 03:39
  • @dxiv oh okay I will do it – Cuckoo Beats Mar 27 '23 at 03:46