Question Number 204468 by MathedUp last updated on 18/Feb/24
$$\mathrm{How}\:\mathrm{Can}\:\mathrm{we}\:\mathrm{prove}\:\underset{{h}=−\infty} {\overset{\infty} {\sum}}\:{J}_{{h}} \left({z}\right)=\mathrm{1} \\ $$
Answered by Peace last updated on 19/Feb/24
$${J}_{{n}−\mathrm{1}} \left({x}\right)+{j}_{{n}+\mathrm{1}} \left({x}\right)=\frac{\mathrm{2}{n}}{{x}}{j}_{{n}} \left({x}\right)…..\left(\mathrm{1}\right) \\ $$$${let}\:{f}_{{x}} \left({y}\right)=\underset{−\infty} {\overset{\infty} {\sum}}{j}_{{n}} \left({x}\right){y}^{{n}} \\ $$$$\Rightarrow{yf}_{{x}} \left({y}\right)+\frac{{f}\left({y}\right)}{{y}}=\frac{\mathrm{2}}{{xy}}\Sigma{nj}_{{n}} \left({x}\right){y}^{{n}−\mathrm{1}} =\frac{\mathrm{2}{y}}{{x}}\left({f}_{{x}} '\left({y}\right)\right) \\ $$$$\Rightarrow{f}'_{{x}} \left({y}\right)={f}_{{x}} \left({y}\right)\left(\frac{{x}}{\mathrm{2}}+\frac{{x}}{\mathrm{2}{y}^{\mathrm{2}} }\right) \\ $$$${f}_{{x}} \left({y}\right)={e}^{\frac{{x}}{\mathrm{2}}\left({y}−\frac{\mathrm{1}}{{y}}\right)} =\underset{−\infty} {\overset{\infty} {\sum}}{j}_{{n}} \left({x}\right){y}^{{n}} ;{y}=\mathrm{1}\Rightarrow\mathrm{1}=\underset{−\infty} {\overset{+\infty} {\sum}}{j}_{{n}} \left({x}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by MathedUp last updated on 19/Feb/24
$${thx}! \\ $$