249 reputation
17
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location United Kingdom
age
visits member for 3 years, 2 months
seen 19 hours ago

I live in Essex, England.


Jul
2
awarded  Curious
Sep
27
awarded  Popular Question
Jun
27
awarded  Critic
Jun
5
comment British maths style guide
Thanks for that. I did feel a bit deflated when my question was so quickly closed for being off topic. I'm not sure about the correct terminology but, to me, my example looks like two independent clauses joined by an 'and'. My grammar textbook tells me in that case there should be a comma before the 'and'. I wondered if the same rule applied to maths. With hindsight I guess there's no reason why, in this case, there should be any difference between UK or US punctuation.
May
31
asked British maths style guide
Mar
26
accepted Equation of a plane passing through a line and a separate point.
Mar
26
comment Equation of a plane passing through a line and a separate point.
Of course. Thanks
Mar
26
asked Equation of a plane passing through a line and a separate point.
Jan
16
comment General solution of $\frac{d^{2}u}{d\rho^{2}}=\frac{l\left(l+1\right)}{\rho^{2}}u$
By the way, although someone has tagged this as Homework, it isn't for me. In my sixtieth decade, I'm simply trying to struggle through Griffiths' book.
Jan
16
comment General solution of $\frac{d^{2}u}{d\rho^{2}}=\frac{l\left(l+1\right)}{\rho^{2}}u$
Afraid I don't understand that. But I got there eventually.
Jan
16
accepted General solution of $\frac{d^{2}u}{d\rho^{2}}=\frac{l\left(l+1\right)}{\rho^{2}}u$
Jan
16
comment General solution of $\frac{d^{2}u}{d\rho^{2}}=\frac{l\left(l+1\right)}{\rho^{2}}u$
Thanks. I couldn't see the wood for the trees.
Jan
16
comment General solution of $\frac{d^{2}u}{d\rho^{2}}=\frac{l\left(l+1\right)}{\rho^{2}}u$
Yes, I just couldn't see it. Now it seems embarrassingly obvious.
Jan
16
comment General solution of $\frac{d^{2}u}{d\rho^{2}}=\frac{l\left(l+1\right)}{\rho^{2}}u$
Obviously, what's obvious to you is not obvious to me.
Jan
16
awarded  Yearling
Jan
16
asked General solution of $\frac{d^{2}u}{d\rho^{2}}=\frac{l\left(l+1\right)}{\rho^{2}}u$
Mar
13
accepted Query about Cartesian vector notation.
Mar
13
comment Query about Cartesian vector notation.
Yes, I do. I think that's why it puzzled me why textbooks use $\mathbf{a}=a_{x}\hat{e}_{x}+a_{y}\hat{e}_{y}+a_{z}\hat{e}_{z}$ instead of $\mathbf{V}=V^{x}\hat{e}_{x}+V^{y}\hat{e}_{y}+V^{z}\hat{e}_{z}=V^\mu \hat{e}_{\mu}$ which, to my eyes, looks more balanced with upper and lower indices.
Mar
12
comment Query about Cartesian vector notation.
One further question. The commonest notation for ordinary Cartesian vectors seems to be (I might well be wrong here)$\mathbf{a}=a_{x}\hat{e}_{x}+a_{y}\hat{e}_{y}+a_{z}\hat{e}_{z}$ with downstairs indices on both the components and basis vectors. What are the pros and cons of this notation and your $\mathbf{W}=W_{x}\hat{e}^{x}+W_{y}\hat{e}^{y}+W_{z}\hat{e}^{z}=W_\mu \hat{e}^{\mu}$ notation? Sorry if I'm making heavy weather about this.
Mar
12
comment Query about Cartesian vector notation.
Thanks. So if I'm talking about ordinary Cartesian vectors I use the notation $\mathbf{W}=W_{x}\hat{e}^{x}+W_{y}\hat{e}^{y}+W_{z}\hat{e}^{z}=W_\mu \hat{e}^{\mu}$ with a downstairs index for the components and upstairs for the basis vectors? Afraid you lost me after that but no matter. Is it possible to explain in fairly easy language what the problem is with using V as the vector symbol?