What would be the roots of the derivative of this polynomial

Question

I got this $$ f=a(x-1)(x-2)...(x-101) $$ knowing a and that $1 ,2 \ldots ,101$ are it's roots ... I need the roots of $f'$ , how should I do it ? More specifc how can I derivate a polynomial like that, is there a formula or something ?

yeah mb , i know the value of a but only 1 , 2 ,3 ... are it's roots — VlAd TbK, Jul 16 '19 at 07:47
How to compute the derivative? Well, you see that $f$ is a product of several simpler terms. You can use the chain rule. — Matti P., Jul 16 '19 at 07:47
Apply the product rule over and over again, like the triple product rule. Start from $f'(x) = a\left((x-1)(x-2) \cdots (x-100))' (x-101))+ ((x-1)(x-2) \cdots (x-100)) \cdot (x-101)' \right) $. — Toby Mak, Jul 16 '19 at 07:47
dosen't realy leads to something usefull ... how can i extract the roots of that — VlAd TbK, Jul 16 '19 at 08:06
From one of your comments to other posts, it appears you may only want the sum of the roots of the derivative. If so, you need to first correct your question. It is best to ask exactly what you want for others to help, you can always share what you have done in case that’s something others can complete. — Macavity, Jul 16 '19 at 14:05

score 3 · Answer 1 · answered Jul 16 '19 at 08:20

3

You can get the derivative thus.

Taking logs, we get $$\log f=\log a+\sum_{i=1}^{101}{\log(x-i)},$$ so that differentiating gives $$f'=f\sum_{i=1}^{101}{\frac{1}{x-i}}.$$

answered Jul 16 '19 at 08:20

Allawonder

13,327

Actually you can get exactly the same result without the log-trick, by derivating directly the product. Yet it doesn't yield the roots of the derivative in an obvious way. – Evargalo Jul 16 '19 at 09:06
@Evargalo: this way to evaluate the derivative is much more efficient. It yields a formula with a product of 101 factors by the sum of 101 terms, vs. the sum of 101 terms each made of 100 factors, i.e. expression length $O(n)$ vs. $O(n^2)$. I would recommend it to evaluate the derivative, except in a close neighborhood of a root of $f$. (To find the roots of $f'$, you can ignore the factor $f$.) – Jul 16 '19 at 09:48
1

and in the end ... how am i supposed to do the sum of the roots ? the answer says it is 5100 but i realy don't understand how you get there on paper – VlAd TbK Jul 16 '19 at 10:14
1

@VlAdTbK Do you need the roots or the sum of the roots? You talked about the derivative and its roots only in OP. – Allawonder Jul 16 '19 at 11:48
The sum of the roots would be much easier. I expect it is exactly $5100$. – Evargalo Jul 16 '19 at 13:00
@YvesDaoust : but both formulas (factorised or expended) can be reached by either method since you can pass easily from one to the other. – Evargalo Jul 16 '19 at 13:03
@Evargalo: you miss the point. The amount of work is much larger by the derivative-of-product formula. By a factor 50. – Jul 16 '19 at 13:17
@YvesDaoust Indeed I had missed the point of too much complexity because I had reached the result in three lines... A simple recurrence (as example, it is not that the recurrence is needed) reduces the 100 steps to only 2, so the amount of work is not that overwhelming ! – Evargalo Jul 16 '19 at 13:28
@Evargalo: computationally, $O(n^2)$ vs. $O(n)$. – Jul 16 '19 at 13:31

score 2 · Answer 2 · answered Jul 16 '19 at 14:06

2

From one of your comments to other posts, it appears you may only want the sum of the roots of the derivative. If so, you need to first correct your question. Assuming this is the case, all you need is Vieta’s rule and the first two terms of $f’$.
i.e. $f = a\,x^{101}-a(1+2+3+\cdots+101)\,x^{100}+\cdots$
$\implies f’ = 101a\,x^{100} + 100a \cdot(\frac12\cdot101\cdot 102)\, x^{99}+\cdots$
$\implies $ sum of roots $\displaystyle= \frac{100a\cdot\frac12\cdot101\cdot 102}{101a}=5100$

answered Jul 16 '19 at 14:06

Macavity

46,381

yeah that's what i needed , but how did you done it ? I stil don't rly understand how you got to it – VlAd TbK Jul 17 '19 at 13:23
Well, edit the question, then tell me which step i need to explain. – Macavity Jul 17 '19 at 14:41
i found the answer , thank you though – VlAd TbK Jul 17 '19 at 17:05

score -1 · Answer 3 · answered Jul 16 '19 at 09:06

As it was mentioned in previous comments, there will be a root of $f'$ on each interval $]k,k+1[, \quad k=1, \cdots , 100$. If you want a numerical approximation you can simply use Newton's method on each of these intervals , taking for instance as initial approximation $x^{(0)}_k = k + \frac{1}{2}$. if you use Mathematica you can get all the solution at once by issuing the commands

$$f[x\_]:=Product[x-i,\{i,1,101\}]$$ $$NSolve[f'[x]==0,x, WorkingPrecision \to 20]$$

$$ \{\{x\to 1.1808291122087612164\},\{x\to 2.2100807699421299589\},\{x\to 3.2293475049454603245\},\{x\to 4.2443987137725789046\},\{x\to 5.2570499827658189815\},\{x\to 6.2681262040946746698\},\{x\to 7.2780782056966400068\},\{x\to 8.2871819479287411490\},\{x\to 9.2956198975910239024\},\{x\to 10.303519661327208567\},\{x\to 11.310974371094841091\},\{x\to 12.318054322065932283\},\{x\to 13.324814035448533182\},\{x\to 14.331296759703496401\},\{x\to 15.337537456561865561\},\{x\to 16.343564849324371068\},\{x\to 17.349402868386108329\},\{x\to 18.355071696567098598\},\{x\to 19.360588541240691774\},\{x\to 20.365968215372014879\},\{x\to 21.371223582016873610\},\{x\to 22.376365899396812647\},\{x\to 23.381405092345185275\},\{x\to 24.386349968394921691\},\{x\to 25.391208391672422480\},\{x\to 26.395987424230801866\},\{x\to 27.400693441971712575\},\{x\to 28.405332230530290552\},\{x\to 29.409909065211691911\},\{x\to 30.414428778123500341\},\{x\to 31.418895814946718692\},\{x\to 32.423314283260973905\},\{x\to 33.427687993939492301\},\{x\to 34.432020496822844042\},\{x\to 35.436315111643480438\},\{x\to 36.440574954988405275\},\{x\to 37.444802963942305840\},\{x\to 38.449001916938810424\},\{x\to 39.453174452256307664\},\{x\to 40.457323084521758049\},\{x\to 41.461450219527214269\},\{x\to 42.465558167616346035\},\{x\to 43.469649155859836112\},\{x\to 44.473725339207302760\},\{x\to 45.477788810778033494\},\{x\to 46.481841611432215602\},\{x\to 47.485885738747685453\},\{x\to 48.489923155513841196\},\{x\to 49.493955797843767137\},\{x\to 50.497985582997415037\},\{x\to 51.502014417002584963\},\{x\to 52.506044202156232863\},\{x\to 53.510076844486158804\},\{x\to 54.514114261252314547\},\{x\to 55.518158388567784398\},\{x\to 56.522211189221966506\},\{x\to 57.526274660792697240\},\{x\to 58.530350844140163888\},\{x\to 59.534441832383653965\},\{x\to 60.538549780472785731\},\{x\to 61.542676915478241951\},\{x\to 62.546825547743692336\},\{x\to 63.550998083061189576\},\{x\to 64.555197036057694160\},\{x\to 65.559425045011594725\},\{x\to 66.563684888356519562\},\{x\to 67.567979503177155958\},\{x\to 68.572312006060507699\},\{x\to 69.576685716739026095\},\{x\to 70.581104185053281308\},\{x\to 71.585571221876499659\},\{x\to 72.590090934788308089\},\{x\to 73.594667769469709448\},\{x\to 74.599306558028287425\},\{x\to 75.604012575769198134\},\{x\to 76.608791608327577520\},\{x\to 77.613650031605078309\},\{x\to 78.618594907654814725\},\{x\to 79.623634100603187353\},\{x\to 80.628776417983126390\},\{x\to 81.634031784627985121\},\{x\to 82.639411458759308226\},\{x\to 83.644928303432901402\},\{x\to 84.650597131613891671\},\{x\to 85.656435150675628932\},\{x\to 86.662462543438134439\},\{x\to 87.668703240296503599\},\{x\to 88.675185964551466818\},\{x\to 89.681945677934067717\},\{x\to 90.689025628905158909\},\{x\to 91.696480338672791433\},\{x\to 92.704380102408976098\},\{x\to 93.712818052071258851\},\{x\to 94.721921794303359993\},\{x\to 95.731873795905325330\},\{x\to 96.742950017234181018\},\{x\to 97.755601286227421095\},\{x\to 98.770652495054539676\},\{x\to 99.789919230057870041\},\{x\to 100.81917088779123878\}\} $$

Any thoughts on why the downvote? – PierreCarre Sep 02 '19 at 13:29 — PierreCarre, Sep 02 '19 at 13:29

infinitezero · Answer 4 · 2019-07-16T09:02:36.750

-2

Original Answer

To answer the question in the title, I've asked Mathematica:

NSolve[D[f[x, 101], x] == 0, x]

where

$$f_n(x)=\prod \limits_{k=1}^n (x-k)$$

~~Turns out the result $x_1=x_2=...=x_{100}=51$. I'm still thinking about how you can do this with some brain power or pen and pencil, but I'm sure there are brighter minds than me around here.~~

Update

After fixing the Mathematica code, these are the (numerically found) roots:

edited Jul 16 '19 at 09:02

answered Jul 16 '19 at 08:10

infinitezero

774

so the roots are the same as from the first polynomial , except the last one ? I don't realy undesrstand tbh – VlAd TbK Jul 16 '19 at 08:12
No, you have 100 roots and they are all the same. $x_1=51, x_2=51, ... x_{100}=51$ – infinitezero Jul 16 '19 at 08:13
i guess it works this way too , i knew the sum of them was 5100 but i was thinking it was from gauss sum , haveing the roots from 1 to 100 – VlAd TbK Jul 16 '19 at 08:17
Are you sure of that ? – Claude Leibovici Jul 16 '19 at 08:22
5

This result is plain wrong, there is no root of multiplicity $100$. The derivative must have a root in every $[k,k+1]$ interval. The OP is wrong to accept blindly. – Jul 16 '19 at 08:24
I was confused about this as well. I might as well ask on mathematica.stackexchange how Mathematica figured this one out. – infinitezero Jul 16 '19 at 08:28
Could you provide the Mathematica syntax you used for the defintion of f[x,n] ? – Claude Leibovici Jul 16 '19 at 08:35
See here https://mathematica.stackexchange.com/questions/202166/nsolve-gives-weird-result-for-n-101-with-nsolve-fracddx-prod-limits-k – infinitezero Jul 16 '19 at 08:36
@infinitezero It is a matter of numerical precision. If you add the option WorkingPrecision->20 to your NSolve command you will get the correct answer. This is why we can't leave everything to machines! – PierreCarre Jul 16 '19 at 08:56
Yes, thank you. I blindly thought, though it seemed weird, I must agree, that it was in the spirit of the question that for $n=101$ this somehow produces a nice result. – infinitezero Jul 16 '19 at 09:03

What would be the roots of the derivative of this polynomial

4 Answers4

Update

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