## anonymous one year ago check my work

1. anonymous

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 $\frac{x^2}{r^2}.f''(r)-\frac{x^2}{r^3}.f'(r)+\frac{y^2}{r^2}.f''(r)-\frac{y^2}{r^3}.f'(r)+\frac{z^2}{r^2}.f''(r)-\frac{z^2}{r^3}.f'(r)+\frac{3}{r}.f'(r)$ $\frac{3}{r}.f'(r)+f''(r)(\frac{x^2+y^2+z^2}{r^2})-\frac{1}{r}.f'(r)(\frac{x^2+y^2+z^2}{r^2})$ $\implies \frac{3}{r}.f'(r)+f''(r)-\frac{1}{r}.f'(r)=f''(r)+\frac{2}{r}.f'(r)$

2. IrishBoy123

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

3. anonymous

idk, it's just a proof ques, am I right though??

4. anonymous

I haven't spherical coordinates before, I did with rectangular coordinate system

5. IrishBoy123

i'd need a cold towel to wade through that!!! i'll print it out and see if i can help....

6. anonymous

haven't used*

7. anonymous

@ganeshie8 gave me a medal so it must be good, no need to print it out :)

8. IrishBoy123

this looks shorter to me and my head now hurts

9. anonymous

Oh you used the quotient rule, I used triple product rule $\frac{d}{dx}(u(x).v(x).w(x))=\frac{du}{dx}.v.w+u.\frac{dv}{dx}.w+u.v.\frac{dw}{dx}$

10. anonymous

wow your method is indeed quite fast @IrishBoy123