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Question Number 167054 by shikaridwan last updated on 05/Mar/22

f(x)= { ((e^(−x^2 )   0<x<∞)),((e^(−x^4 )  −∞<x<0    )) :}     Find geometric mean of f(x)

$${f}\left({x}\right)=\begin{cases}{{e}^{−{x}^{\mathrm{2}} } \:\:\mathrm{0}<{x}<\infty}\\{{e}^{−{x}^{\mathrm{4}} } \:−\infty<{x}<\mathrm{0}\:\:\:\:}\end{cases} \\ $$ $$ \\ $$ $$\:{Find}\:{geometric}\:{mean}\:{of}\:{f}\left({x}\right) \\ $$

Answered by aleks041103 last updated on 05/Mar/22

g=lim_(a→∞) lim_(Δx→0) ((Π_(i=0) ^(((2a)/(Δx))−1) f(−a+iΔx)))^(1/(2a/Δx))   ⇒ln(g)=lim_(a→∞) lim_(Δx→∞) ((Δx)/(2a))(Σ_(i=0) ^(((2a)/(Δx))−1) ln(f(−a+iΔx)))=  =lim_(a→∞) (1/(2a))∫_(−a) ^a ln(f(x))dx=ln(g)  now  ln(f(x))= { ((−x^2 , 0<x)),((−x^4 ,x<0)) :}  ⇒g=lim_(a→∞) ((−1)/(2a))(∫_(−a) ^0 x^4 dx+∫_0 ^a x^2 dx)=  =lim_(a→∞) ((−1)/(2a))((a^5 /5)+(a^3 /3))→−∞  ⇒ln(g)→−∞  ⇒g=0

$${g}=\underset{{a}\rightarrow\infty} {\mathrm{lim}}\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\sqrt[{\mathrm{2}{a}/\Delta{x}}]{\underset{{i}=\mathrm{0}} {\overset{\frac{\mathrm{2}{a}}{\Delta{x}}−\mathrm{1}} {\prod}}{f}\left(−{a}+{i}\Delta{x}\right)} \\ $$ $$\Rightarrow{ln}\left({g}\right)=\underset{{a}\rightarrow\infty} {\mathrm{lim}}\underset{\Delta{x}\rightarrow\infty} {\mathrm{lim}}\frac{\Delta{x}}{\mathrm{2}{a}}\left(\underset{{i}=\mathrm{0}} {\overset{\frac{\mathrm{2}{a}}{\Delta{x}}−\mathrm{1}} {\sum}}{ln}\left({f}\left(−{a}+{i}\Delta{x}\right)\right)\right)= \\ $$ $$=\underset{{a}\rightarrow\infty} {{lim}}\frac{\mathrm{1}}{\mathrm{2}{a}}\underset{−{a}} {\overset{{a}} {\int}}{ln}\left({f}\left({x}\right)\right){dx}={ln}\left({g}\right) \\ $$ $${now} \\ $$ $${ln}\left({f}\left({x}\right)\right)=\begin{cases}{−{x}^{\mathrm{2}} ,\:\mathrm{0}<{x}}\\{−{x}^{\mathrm{4}} ,{x}<\mathrm{0}}\end{cases} \\ $$ $$\Rightarrow{g}=\underset{{a}\rightarrow\infty} {{lim}}\frac{−\mathrm{1}}{\mathrm{2}{a}}\left(\underset{−{a}} {\overset{\mathrm{0}} {\int}}{x}^{\mathrm{4}} {dx}+\int_{\mathrm{0}} ^{{a}} {x}^{\mathrm{2}} {dx}\right)= \\ $$ $$=\underset{{a}\rightarrow\infty} {{lim}}\frac{−\mathrm{1}}{\mathrm{2}{a}}\left(\frac{{a}^{\mathrm{5}} }{\mathrm{5}}+\frac{{a}^{\mathrm{3}} }{\mathrm{3}}\right)\rightarrow−\infty \\ $$ $$\Rightarrow{ln}\left({g}\right)\rightarrow−\infty \\ $$ $$\Rightarrow{g}=\mathrm{0} \\ $$

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