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 ?
2026-04-25 22:28:37.1777156117
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What would be the roots of the derivative of this polynomial
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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 $$f'=f\sum_{i=1}^{101}{\frac{1}{x-i}}.$$
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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\}\} $$
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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 $\displaystyle= \frac{100a\cdot\frac12\cdot101\cdot 102}{101a}=5100$
Original Answer
To answer the question in the title, I've asked Mathematica:
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: