Question Number 55685 by tm888 last updated on 02/Mar/19
$${proof}\:{that}\: \\ $$ $$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{a}_{{i}} }{{a}_{{i}} −{x}}=\mathrm{2015}\:{has}\:{exactly}\:{n}\:{real}\: \\ $$ $${roots}.{o}<{a}_{\mathrm{1}} ....<{a}_{{n}} \\ $$
Answered by mr W last updated on 03/Mar/19
![let y=f(x)=Σ_(i=1) ^n (a_i /(a_i −x)) y′=Σ_(i=1) ^n (a_i /((a_i −x)^2 ))>0 ⇒the function is strickly increasing. lim_(x→a_k ^((−)) ) f(x)=Σ_(i=1) ^n [lim_(x→a_k ^((−)) ) (a_i /(a_i −x))] =Σ_(i=1,≠k) ^n (a_i /(a_i −a_k ))+lim_(x→a_k ^((−)) ) (a_k /(a_k −x))=+∞ lim_(x→a_k ^((+)) ) f(x)=Σ_(i=1) ^n [lim_(x→a_k ^((+)) ) (a_i /(a_i −x))] =Σ_(i=1,≠k) ^n (a_i /(a_i −a_k ))+lim_(x→a_k ^((+)) ) (a_k /(a_k −x))=−∞ that means for x∈(a_k ,a_(k+1) ) with k=1,2,...n−1 f(x) is strickly increasing lim_(x→a_k ) f(x)=lim_(x→a_k ^((+)) ) f(x)=−∞ lim_(x→a_(k+1) ) f(x)=lim_(x→a_(k+1) ^((−)) ) f(x)=+∞ i.e. f(x)=c has one and only root in (a_k ,a_(k+1) ) with c=any real number, e.g. 2015. lim_(x→−∞) f(x)=Σ_(i=1) ^n [lim_(x→−∞) (a_i /(a_i −x))]=+0 that means for x∈(−∞,a_1 ) f(x) is strickly increasing lim_(x→−∞) f(x)=+0 lim_(x→a_1 ) f(x)=lim_(x→a_1 ^((−)) ) f(x)=+∞ i.e. with c>0, e.g. c=2015 f(x)=c has one and only root in (−∞,a_1 ) lim_(x→+∞) f(x)=Σ_(i=1) ^n [lim_(x→+∞) (a_i /(a_i −x))]=−0 that means for x∈(a_n ,+∞) f(x) is strickly increasing lim_(x→a_n ) f(x)=lim_(x→a_1 ^((+)) ) f(x)=−∞ lim_(x→+∞) f(x)=−0 i.e. with c>0, e.g. c=2015 f(x)=c has no root in (a_n ,+∞) summary: f(x)=c>0,e.g. c=2015 has one and only one root in (−∞,a_1 ) one and only one root in (a_k ,a_(k+1) ) with k=1,2,...,n−1 no root in (a_n ,+∞) totally f(x)=c has exactly n roots.](Q55721.png)
$${let}\:{y}={f}\left({x}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{a}_{{i}} }{{a}_{{i}} −{x}} \\ $$ $${y}'=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{a}_{{i}} }{\left({a}_{{i}} −{x}\right)^{\mathrm{2}} }>\mathrm{0} \\ $$ $$\Rightarrow{the}\:{function}\:{is}\:{strickly}\:{increasing}. \\ $$ $$ \\ $$ $$\underset{{x}\rightarrow{a}_{{k}} ^{\left(−\right)} } {\mathrm{lim}}{f}\left({x}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\underset{{x}\rightarrow{a}_{{k}} ^{\left(−\right)} } {\mathrm{lim}}\frac{{a}_{{i}} }{{a}_{{i}} −{x}}\right] \\ $$ $$=\underset{{i}=\mathrm{1},\neq{k}} {\overset{{n}} {\sum}}\frac{{a}_{{i}} }{{a}_{{i}} −{a}_{{k}} }+\underset{{x}\rightarrow{a}_{{k}} ^{\left(−\right)} } {\mathrm{lim}}\frac{{a}_{{k}} }{{a}_{{k}} −{x}}=+\infty \\ $$ $$\underset{{x}\rightarrow{a}_{{k}} ^{\left(+\right)} } {\mathrm{lim}}{f}\left({x}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\underset{{x}\rightarrow{a}_{{k}} ^{\left(+\right)} } {\mathrm{lim}}\frac{{a}_{{i}} }{{a}_{{i}} −{x}}\right] \\ $$ $$=\underset{{i}=\mathrm{1},\neq{k}} {\overset{{n}} {\sum}}\frac{{a}_{{i}} }{{a}_{{i}} −{a}_{{k}} }+\underset{{x}\rightarrow{a}_{{k}} ^{\left(+\right)} } {\mathrm{lim}}\frac{{a}_{{k}} }{{a}_{{k}} −{x}}=−\infty \\ $$ $${that}\:{means}\:{for}\:{x}\in\left({a}_{{k}} ,{a}_{{k}+\mathrm{1}} \right)\:{with}\:{k}=\mathrm{1},\mathrm{2},...{n}−\mathrm{1} \\ $$ $${f}\left({x}\right)\:{is}\:{strickly}\:{increasing} \\ $$ $$\underset{{x}\rightarrow{a}_{{k}} } {\mathrm{lim}}\:{f}\left({x}\right)=\underset{{x}\rightarrow{a}_{{k}} ^{\left(+\right)} } {\mathrm{lim}}{f}\left({x}\right)=−\infty \\ $$ $$\underset{{x}\rightarrow{a}_{{k}+\mathrm{1}} } {\mathrm{lim}}\:{f}\left({x}\right)=\underset{{x}\rightarrow{a}_{{k}+\mathrm{1}} ^{\left(−\right)} } {\mathrm{lim}}{f}\left({x}\right)=+\infty \\ $$ $${i}.{e}.\:\:{f}\left({x}\right)={c}\:{has}\:{one}\:{and}\:{only}\:{root}\:{in}\:\left({a}_{{k}} ,{a}_{{k}+\mathrm{1}} \right) \\ $$ $${with}\:{c}={any}\:{real}\:{number},\:{e}.{g}.\:\mathrm{2015}. \\ $$ $$ \\ $$ $$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}{f}\left({x}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{{a}_{{i}} }{{a}_{{i}} −{x}}\right]=+\mathrm{0} \\ $$ $${that}\:{means}\:{for}\:{x}\in\left(−\infty,{a}_{\mathrm{1}} \right) \\ $$ $${f}\left({x}\right)\:{is}\:{strickly}\:{increasing} \\ $$ $$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:{f}\left({x}\right)=+\mathrm{0} \\ $$ $$\underset{{x}\rightarrow{a}_{\mathrm{1}} } {\mathrm{lim}}\:{f}\left({x}\right)=\underset{{x}\rightarrow{a}_{\mathrm{1}} ^{\left(−\right)} } {\mathrm{lim}}{f}\left({x}\right)=+\infty \\ $$ $${i}.{e}.\:{with}\:{c}>\mathrm{0},\:{e}.{g}.\:{c}=\mathrm{2015} \\ $$ $${f}\left({x}\right)={c}\:{has}\:{one}\:{and}\:{only}\:{root}\:{in}\:\left(−\infty,{a}_{\mathrm{1}} \right) \\ $$ $$ \\ $$ $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}{f}\left({x}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{a}_{{i}} }{{a}_{{i}} −{x}}\right]=−\mathrm{0} \\ $$ $${that}\:{means}\:{for}\:{x}\in\left({a}_{{n}} ,+\infty\right) \\ $$ $${f}\left({x}\right)\:{is}\:{strickly}\:{increasing} \\ $$ $$\underset{{x}\rightarrow{a}_{{n}} } {\mathrm{lim}}\:{f}\left({x}\right)=\underset{{x}\rightarrow{a}_{\mathrm{1}} ^{\left(+\right)} } {\mathrm{lim}}{f}\left({x}\right)=−\infty \\ $$ $$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:{f}\left({x}\right)=−\mathrm{0} \\ $$ $${i}.{e}.\:{with}\:{c}>\mathrm{0},\:{e}.{g}.\:{c}=\mathrm{2015} \\ $$ $${f}\left({x}\right)={c}\:{has}\:{no}\:{root}\:{in}\:\left({a}_{{n}} ,+\infty\right) \\ $$ $$ \\ $$ $${summary}: \\ $$ $${f}\left({x}\right)={c}>\mathrm{0},{e}.{g}.\:{c}=\mathrm{2015}\:{has} \\ $$ $${one}\:{and}\:{only}\:{one}\:{root}\:{in}\:\left(−\infty,{a}_{\mathrm{1}} \right) \\ $$ $${one}\:{and}\:{only}\:{one}\:{root}\:{in}\:\left({a}_{{k}} ,{a}_{{k}+\mathrm{1}} \right)\:{with}\:{k}=\mathrm{1},\mathrm{2},...,{n}−\mathrm{1} \\ $$ $${no}\:{root}\:{in}\:\left({a}_{{n}} ,+\infty\right) \\ $$ $${totally}\:{f}\left({x}\right)={c}\:{has}\:{exactly}\:{n}\:{roots}. \\ $$
Commented byotchereabdullai@gmail.com last updated on 03/Mar/19
$$\mathrm{The}\:\mathrm{great}\:\mathrm{and}\:\mathrm{ideal}\:\mathrm{professor}\:\mathrm{W} \\ $$
Commented bymr W last updated on 03/Mar/19
Commented bymr W last updated on 03/Mar/19
$${we}\:{can}\:{see} \\ $$ $${f}\left({x}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{a}_{{i}} }{{a}_{{i}} −{x}}=\mathrm{0}\:{has}\:{exactly}\:{n}−\mathrm{1}\:{roots}\:{and} \\ $$ $${f}\left({x}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{a}_{{i}} }{{a}_{{i}} −{x}}={c}\neq\mathrm{0}\:{has}\:{exactly}\:{n}\:{roots}. \\ $$