Problems with the Riemann tensor in general relativity

Ineedhelpimbadatphys
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Homework Statement
In the picture.
Relevant Equations
Also in pictures.
IMG_2750.jpeg
IMG_2752.jpeg

After Taylor expansion and using equations (2), I have no problem getting to equation (1). Now obviously I have to somehow use (3.71) ,which I do know how, to derive to express the second order derivative.
On the internet I found equation (3), and I have tried to understand where this comes from (4).
Is equation (3) correct, if yes, how am i supposed to contineu from what I have in (4). It should equal 3 times the secobd order derivative?

If equation (3) is not correct, any tips how to continue from what I have?
 
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The sentence after (3.71) was a little unclear. I meant that I do know how to derive (3.71) using equations (2).
 

Thread 'Please tell me where I am going wrong in this integral'

##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
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Thread 'Number of quantum accessible states of a particle given T, N?'

It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...
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