Question Number 216800 by depressiveshrek last updated on 21/Feb/25
Answered by MrGaster last updated on 21/Feb/25
![Prove:f(x)=a(x−r_1 )^m_1 (x−r_2 )^m_2 …(x−r_k )^m_k where a>0,r_1 ,r_2 ,…,r_k are the distinct real roots of f(x),and m_1 ,m_2 ,…m_k are their respective multiplicities (1) The derivatives of f(x)are: f^((j)) (x)=aΣ_(i=1) ^k ((m_i !)/((m_i −j)!))(x−r_i )^(m_i −j) Π_(l≠i) (x−r_i )^m_l for j=1,2,…,n. g(x): g(x)=f(x)+f′(x)…+f^((n)) (x) Substituting the expressionsfor thes derivative have: g(x)=aΣ_(j=0) ^n Σ_(i=1) ^k ((m_i !)/((m_i −j)))(x−r_i )^(m_i −j) Π_(l≠j) (x−r_l )^m_l (2) since f(x)≥0 for all x,each term in the sum f(x)+f′(x)+…+f^((n)) is nonnegative. Specificallyr fo each root r_i ,the terms involving (x−r_i )^(m−j) for j=0,1,…m_i are nonnegative because m_i −j≥0 Therefore the sum of theser tems is also nonnegative.c Hene,g(x)is a sum of nonnegatives term which implies: determinant (((g(x)≥0))) [Q.E.D]](https://www.tinkutara.com/question/Q216801.png)
$$\mathrm{Prove}:{f}\left({x}\right)={a}\left({x}−{r}_{\mathrm{1}} \right)^{{m}_{\mathrm{1}} } \left({x}−{r}_{\mathrm{2}} \right)^{{m}_{\mathrm{2}} } \ldots\left({x}−{r}_{{k}} \right)^{{m}_{{k}} } \\ $$$$\mathrm{where}\:{a}>\mathrm{0},{r}_{\mathrm{1}} ,{r}_{\mathrm{2}} ,\ldots,{r}_{{k}} \mathrm{are}\:\mathrm{the}\:\mathrm{distinct} \\ $$$$\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:{f}\left({x}\right),\mathrm{and}\:{m}_{\mathrm{1}} ,{m}_{\mathrm{2}} ,\ldots{m}_{{k}} \mathrm{are} \\ $$$$\mathrm{their}\:\mathrm{respective}\:\mathrm{multiplicities}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{The}\:\mathrm{derivatives}\:\mathrm{of}\:{f}\left({x}\right)\mathrm{are}: \\ $$$${f}^{\left({j}\right)} \left({x}\right)={a}\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{{m}_{{i}} !}{\left({m}_{{i}} −{j}\right)!}\left({x}−{r}_{{i}} \right)^{{m}_{{i}} −{j}} \underset{{l}\neq{i}} {\prod}\left({x}−{r}_{{i}} \right)^{{m}_{{l}} } \\ $$$$\mathrm{for}\:{j}=\mathrm{1},\mathrm{2},\ldots,{n}. \\ $$$${g}\left({x}\right): \\ $$$${g}\left({x}\right)={f}\left({x}\right)+{f}'\left({x}\right)\ldots+{f}^{\left({n}\right)} \left({x}\right) \\ $$$$\mathrm{Substituting}\:\mathrm{the}\:\mathrm{expressionsfor}\:\mathrm{thes} \\ $$$$\mathrm{derivative}\:\:\mathrm{have}: \\ $$$${g}\left({x}\right)={a}\underset{{j}=\mathrm{0}} {\overset{{n}} {\sum}}\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{{m}_{{i}} !}{\left({m}_{{i}} −{j}\right)}\left({x}−{r}_{{i}} \right)^{{m}_{{i}} −{j}} \underset{{l}\neq{j}} {\prod}\left({x}−{r}_{{l}} \right)^{{m}_{{l}} } \left(\mathrm{2}\right) \\ $$$$\mathrm{since}\:{f}\left({x}\right)\geq\mathrm{0}\:\mathrm{for}\:\mathrm{all}\:{x},\mathrm{each}\:\mathrm{term}\:\mathrm{in}\:\mathrm{the}\:\mathrm{sum}\:{f}\left({x}\right)+{f}'\left({x}\right)+\ldots+{f}^{\left({n}\right)} \mathrm{is}\:\mathrm{nonnegative}.\:\mathrm{Specificallyr} \\ $$$$\mathrm{fo}\:\mathrm{each}\:\mathrm{root}\:{r}_{{i}} ,\mathrm{the}\:\mathrm{terms}\:\mathrm{involving}\:\left({x}−{r}_{{i}} \right)^{{m}−{j}} \mathrm{for}\:{j}=\mathrm{0},\mathrm{1},\ldots{m}_{{i}} \mathrm{are}\:\mathrm{nonnegative}\:\mathrm{because}\:{m}_{{i}} −{j}\geq\mathrm{0} \\ $$$$\mathrm{Therefore}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{theser} \\ $$$$\mathrm{tems}\:\mathrm{is}\:\mathrm{also}\:\mathrm{nonnegative}.\mathrm{c} \\ $$$$\mathrm{Hene},{g}\left({x}\right)\mathrm{is}\:\mathrm{a}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{nonnegatives} \\ $$$$\mathrm{term}\:\mathrm{which}\:\mathrm{implies}: \\ $$$$\begin{array}{|c|}{{g}\left({x}\right)\geq\mathrm{0}}\\\hline\end{array} \\ $$$$\left[{Q}.\mathrm{E}.\mathrm{D}\right] \\ $$