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Question Number 59682 by aliesam last updated on 13/May/19
if H=X^2 +Y^2 +Z^2 prove (∂^2 H/∂X^2 )+(∂^2 H/∂Y^2 )+(∂^2 H/∂Z^2 )=(2/H)
$${if} \\ $$$$ \\ $$$${H}={X}^{\mathrm{2}} +{Y}^{\mathrm{2}} +{Z}^{\mathrm{2}} \\ $$$$ \\ $$$${prove} \\ $$$$ \\ $$$$\frac{\partial^{\mathrm{2}} {H}}{\partial{X}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {H}}{\partial{Y}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {H}}{\partial{Z}^{\mathrm{2}} }=\frac{\mathrm{2}}{{H}} \\ $$
Commented by mr W last updated on 13/May/19
H^2 =X^2 +Y^2 +Z^2 2H(∂H/∂X)=2X H(∂H/∂X)=X⇒(∂H/∂X)=(X/H) (∂H/∂X)×(∂H/∂X)+H(∂^2 H/∂X^2 )=1 (X^2 /H^2 )+H(∂^2 H/∂X^2 )=1 ((X^2 +Y^2 +Z^2 )/H^2 )+H((∂^2 H/∂X^2 )+(∂^2 H/∂Y^2 )+(∂^2 H/∂Z^2 ))=3 1+H((∂^2 H/∂X^2 )+(∂^2 H/∂Y^2 )+(∂^2 H/∂Z^2 ))=3 H((∂^2 H/∂X^2 )+(∂^2 H/∂Y^2 )+(∂^2 H/∂Z^2 ))=2 ⇒(∂^2 H/∂X^2 )+(∂^2 H/∂Y^2 )+(∂^2 H/∂Z^2 )=(2/H)
$${H}^{\mathrm{2}} ={X}^{\mathrm{2}} +{Y}^{\mathrm{2}} +{Z}^{\mathrm{2}} \\ $$$$\mathrm{2}{H}\frac{\partial{H}}{\partial{X}}=\mathrm{2}{X} \\ $$$${H}\frac{\partial{H}}{\partial{X}}={X}\Rightarrow\frac{\partial{H}}{\partial{X}}=\frac{{X}}{{H}} \\ $$$$\frac{\partial{H}}{\partial{X}}×\frac{\partial{H}}{\partial{X}}+{H}\frac{\partial^{\mathrm{2}} {H}}{\partial{X}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{X}^{\mathrm{2}} }{{H}^{\mathrm{2}} }+{H}\frac{\partial^{\mathrm{2}} {H}}{\partial{X}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{X}^{\mathrm{2}} +{Y}^{\mathrm{2}} +{Z}^{\mathrm{2}} }{{H}^{\mathrm{2}} }+{H}\left(\frac{\partial^{\mathrm{2}} {H}}{\partial{X}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {H}}{\partial{Y}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {H}}{\partial{Z}^{\mathrm{2}} }\right)=\mathrm{3} \\ $$$$\mathrm{1}+{H}\left(\frac{\partial^{\mathrm{2}} {H}}{\partial{X}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {H}}{\partial{Y}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {H}}{\partial{Z}^{\mathrm{2}} }\right)=\mathrm{3} \\ $$$${H}\left(\frac{\partial^{\mathrm{2}} {H}}{\partial{X}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {H}}{\partial{Y}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {H}}{\partial{Z}^{\mathrm{2}} }\right)=\mathrm{2} \\ $$$$\Rightarrow\frac{\partial^{\mathrm{2}} {H}}{\partial{X}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {H}}{\partial{Y}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {H}}{\partial{Z}^{\mathrm{2}} }=\frac{\mathrm{2}}{{H}} \\ $$
Commented by aliesam last updated on 13/May/19
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by tanmay last updated on 13/May/19
pls recheck question...
$${pls}\:{recheck}\:{question}… \\ $$
Commented by aliesam last updated on 13/May/19
H^2
$${H}^{\mathrm{2}} \\ $$
Commented by aliesam last updated on 13/May/19
=X^2 +Y^2 +Z^2
$$={X}^{\mathrm{2}} +{Y}^{\mathrm{2}} +{Z}^{\mathrm{2}} \\ $$
Answered by tanmay last updated on 13/May/19
x^2 +y^2 +z^2 =H^2 H=(√(x^2 +y^2 +z^2 )) (∂H/∂x)=(1/(2(√(x^2 +y^2 +z^2 ))))×2x (∂^2 H/∂x^2 )=(((√(x^2 +y^2 +z^2 )) ×(∂x/∂x)−x×(∂/∂x)((√(x^2 +y^2 +z^2 )) )/((x^2 +y^2 +z^2 ))) (∂^2 H/∂x^2 )=(((√(x^2 +y^2 +z^2 )) −(x/(2(√(x^2 +y^2 +z^2 ))))×2x)/((x^2 +y^2 +z^2 ))) (∂^2 H/∂x^2 )=((y^2 +z^2 )/((x^2 +y^2 +z^2 )^(3/2) )) so (∂^2 H/∂x^2 )+(∂^2 H/∂y^2 )+(∂^2 H/∂z^2 ) =(((y^2 +z^2 )+(x^2 +z^2 )+(x^2 +z^2 ))/((x^2 +y^2 +z^2 )^(3/2) )) =((2(x^2 +y^2 +z^2 ))/((x^2 +y^2 +z^2 )^(3/2) )) .((2H^2 )/((H^2 )^(3/2) )) =(2/H)
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={H}^{\mathrm{2}} \\ $$$${H}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }\: \\ $$$$\frac{\partial{H}}{\partial{x}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}×\mathrm{2}{x} \\ $$$$\frac{\partial^{\mathrm{2}} {H}}{\partial{x}^{\mathrm{2}} }=\frac{\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }\:×\frac{\partial{x}}{\partial{x}}−{x}×\frac{\partial}{\partial{x}}\left(\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }\:\right.}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)} \\ $$$$\frac{\partial^{\mathrm{2}} {H}}{\partial{x}^{\mathrm{2}} }=\frac{\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }\:−\frac{{x}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }}×\mathrm{2}{x}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)} \\ $$$$\frac{\partial^{\mathrm{2}} {H}}{\partial{x}^{\mathrm{2}} }=\frac{{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$${so}\:\frac{\partial^{\mathrm{2}} {H}}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {H}}{\partial{y}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {H}}{\partial{z}^{\mathrm{2}} } \\ $$$$=\frac{\left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)+\left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)+\left({x}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$=\frac{\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$.\frac{\mathrm{2}{H}^{\mathrm{2}} }{\left({H}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$=\frac{\mathrm{2}}{{H}} \\ $$
Commented by aliesam last updated on 13/May/19
genius
$${genius} \\ $$

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