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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Given a scalar point function \[f(r)\] where \[r=\sqrt{x^2+y^2+z^2}\] Prove that \[\nabla^{2}f(r)=f''(r)+\frac{2}{r}f'(r)\] \[\nabla^{2}f(r)=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=\frac{\partial^2f}{\partial z^2}\] \[\therefore \frac{\partial}{\partial x}(\frac{\partial f}{\partial r}\frac{\partial r}{\partial x})+\frac{\partial}{\partial y}(\frac{\partial f}{\partial r}\frac{\partial r}{\partial y})+\frac{\partial}{\partial z}(\frac{\partial f}{\partial r}\frac{\partial r}{\partial z})\] Now \[r^2=x^2+y^2+z^2\] \[\implies 2r\frac{\partial r}{\partial x}=2x\]\[\implies \frac{\partial r}{\partial x}=\frac{x}{r}\] \[\implies \frac{\partial r}{\partial y}=\frac{y}{r}\]\[\implies \frac{\partial y}{\partial z}=\frac{z}{r}\] Now LHS again, \[\frac{\partial}{\partial x}(f'(r)\frac{x}{r})+\frac{\partial}{\partial y}(f'(r)\frac{y}{r})+\frac{\partial }{\partial z}(f'(r)\frac{z}{r})\] \[\frac{x}{r}.\frac{\partial f'}{\partial x}+\frac{1}{r}.f'(r)+xf'(r).\frac{\partial}{\partial x}(r^{-1})\] \[+\frac{y}{r}.\frac{\partial f'}{\partial y}+\frac{1}{r}.f'(r)+yf'(r)\frac{\partial}{\partial y}(r^{-1})\] \[+\frac{z}{r}\frac{\partial f'}{\partial z}+\frac{1}{r}.f'(r)+yf'(r).\frac{\partial }{\partial z}(r^{-1})\] Applying chain rule \[\frac{x}{r}.\frac{\partial f'}{\partial r}.\frac{\partial r}{\partial x}+xf'(r).\frac{\partial }{\partial r}(r^{-1}).\frac{\partial r}{\partial x}\] \[+\frac{y}{r}.\frac{\partial f'}{\partial r}.\frac{\partial r}{\partial y}+yf'(r).\frac{\partial }{\partial r}(r^{-1}).\frac{\partial r}{\partial y}\] \[+\frac{z}{r}.\frac{\partial f'}{\partial r}.\frac{\partial r}{\partial z}+zf'(r).\frac{\partial }{\partial r}(r^{-1}).\frac{\partial r}{\partial z}\] \[+\frac{3}{r}.f'(r)\] Now we have \[\frac{x^2}{r^2}.f''(r)-\frac{x^2}{r^3}.f'(r)+\frac{y^2}{r^2}.f''(r)-\frac{y^2}{r^3}.f'(r)+\frac{z^2}{r^2}.f''(r)-\frac{z^2}{r^3}.f'(r)+\frac{3}{r}.f'(r)\] \[\frac{3}{r}.f'(r)+f''(r)(\frac{x^2+y^2+z^2}{r^2})-\frac{1}{r}.f'(r)(\frac{x^2+y^2+z^2}{r^2})\] \[\implies \frac{3}{r}.f'(r)+f''(r)-\frac{1}{r}.f'(r)=f''(r)+\frac{2}{r}.f'(r)\]
are you allowed spherical? https://gyazo.com/abacaf4b950f6e57251b8fdd887162b8 if so, it comes down to expanding out the first term on the RHS of this as everything else = 0
idk, it's just a proof ques, am I right though??

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I haven't spherical coordinates before, I did with rectangular coordinate system
i'd need a cold towel to wade through that!!! i'll print it out and see if i can help....
haven't used*
@ganeshie8 gave me a medal so it must be good, no need to print it out :)
this looks shorter to me and my head now hurts
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Oh you used the quotient rule, I used triple product rule \[\frac{d}{dx}(u(x).v(x).w(x))=\frac{du}{dx}.v.w+u.\frac{dv}{dx}.w+u.v.\frac{dw}{dx}\]
wow your method is indeed quite fast @IrishBoy123

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