Question Number 2207 by Yozzi last updated on 08/Nov/15
$${Find}\:{the}\:{sum}\:{to}\:{n}\:{terms}\:{of}\:{the}\:{series} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{S}\left({x}\right)=\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{e}^{−{x}^{{r}} } . \\ $$
Commented by 123456 last updated on 09/Nov/15
$$\mathrm{S}\left({x}\right)=\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{e}^{−{x}^{{r}} } =\frac{\mathrm{1}}{{e}}+\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{e}^{−{x}^{{r}} } \:\left({n}>\mathrm{0}\right) \\ $$$$=\frac{\mathrm{1}+\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{e}^{\mathrm{1}−{x}^{{r}} } }{{e}}\:\:\left({n}>\mathrm{0}\right) \\ $$$$=\frac{\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{e}^{\mathrm{1}−{x}^{{r}} } }{{e}} \\ $$
Answered by Rasheed Soomro last updated on 16/Nov/15
$${S}\left({x}\right)=\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}{e}^{−{x}^{{r}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}^{{x}} =\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+…+\frac{{x}^{{n}} }{{n}!} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$\:\:\:\:{e}^{−{x}^{\mathrm{0}} } ={e}^{−\mathrm{1}} =\mathrm{1}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+…\pm\frac{\mathrm{1}}{{n}!}\:\:\:\left[+\:{for}\:{even}\:{n},\:−\:{for}\:{odd}\:{n}\right] \\ $$$$+{e}^{−{x}^{\mathrm{1}} } ={e}^{−{x}} =\mathrm{1}−{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+…\pm\frac{{x}^{{n}} }{{n}!} \\ $$$$+{e}^{−{x}^{\mathrm{2}} } =\:\:\:\:\:\:\:=\mathrm{1}−{x}^{\mathrm{2}} +\frac{{x}^{\mathrm{4}} }{\mathrm{2}!}−\frac{{x}^{\mathrm{6}} }{\mathrm{3}!}+…\pm\frac{{x}^{\mathrm{2}{n}} }{{n}!} \\ $$$$+{e}^{−{x}^{\mathrm{3}} } =\:\:\:\:\:\:\:=\mathrm{1}−{x}^{\mathrm{3}} +\frac{{x}^{\mathrm{6}} }{\mathrm{2}!}−\frac{{x}^{\mathrm{9}} }{\mathrm{3}!}+….\pm\frac{{x}^{\mathrm{3}{n}} }{{n}!} \\ $$$$+{e}^{−{x}^{\mathrm{4}} } =\:\:\:\:\:\:\:=\mathrm{1}−{x}^{\mathrm{4}} +\frac{{x}^{\mathrm{8}} }{\mathrm{2}!}−\frac{{x}^{\mathrm{12}} }{\mathrm{3}!}+…\pm\frac{{x}^{\mathrm{4}{n}} }{{n}!} \\ $$$$+….\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\:\:\:\:\:\:…..\:\:\:\:\:….\:\:\:\:…..\:\:\:……. \\ $$$$+….\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\:\:\:\:\:\:……\:\:\:\:….\:\:\:…..\:\:\:……. \\ $$$$+{e}^{−{x}^{{n}} } \\ $$$${S}\left({x}\right)\:\:\:{is}\:\:\:{sum}\:{of}\:{n}+\mathrm{1}\:\:{geometric}\:{series} \\ $$$${S}\left({x}\right)=\left(\mathrm{1}+\mathrm{1}+…{n}+\mathrm{1}\:{terms}\right)−\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…{n}+\mathrm{1}\:{terms}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{2}!}\left(\mathrm{1}+{x}^{\mathrm{2}} +{x}^{\mathrm{4}} +…{n}+\mathrm{1}\:{terms}\right)−\frac{\mathrm{1}}{\mathrm{3}!}\left(\mathrm{1}+{x}^{\mathrm{3}} +{x}^{\mathrm{6}} +…{n}+\mathrm{1}\:{terms}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…+\frac{\mathrm{1}}{{n}!}\left(\mathrm{1}\pm{x}^{{n}} +{x}^{\mathrm{2}{n}} ….{n}+\mathrm{1}\:{terms}\right)\:\:\left[+{for}\:{even}\:{n},\:−{for}\:{odd}\right] \\ $$$$\:\:\:\:=\mathrm{1}\left({n}+\mathrm{1}\right)−\frac{\mathrm{1}\left(\mathrm{1}−{x}^{{n}+\mathrm{1}} \right)}{\mathrm{1}−{x}}+\frac{\mathrm{1}}{\mathrm{2}!}\left(\frac{\mathrm{1}\left(\mathrm{1}−\left({x}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} \right.}{\mathrm{1}−{x}^{\mathrm{2}} }\right)…. \\ $$$$\:\:\:=\left({n}+\mathrm{1}\right)−\frac{\left(\mathrm{1}−{x}^{{n}+\mathrm{1}} \right)}{\mathrm{1}−{x}}+\frac{\mathrm{1}}{\mathrm{2}!}\left(\frac{\mathrm{1}−\left({x}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{2}} }\right)−… \\ $$$${Continue} \\ $$