80 reputation
7
bio website contraflo.ws
location Sydney, Australia
age 25
visits member for 2 years, 2 months
seen Mar 26 at 6:10

I am currently studying both Physics and Mathematics at the University of Sydney. My main interest areas are Differential Geometry and Particle Physics.


May
19
comment Show that $[l_1 \cdot l_2 \cdot l_3 ] = [l_1 + l_2 + l_3] \in H_1(X)$ The first Homology group of X
the $\cdot$ represents the usual path concatenation, whilst the addition $+$. I was hoping it was a conventional thing, as I guess that's where my confusion was coming from. I shall add additional parts of the question which may shed more light on the problem.
Oct
3
comment Proving $\frac{e^{z^2}}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^2}dt$ is bounded for $\Re(z) \leq 0$
You need a reputation of at least 15 and I'm afraid I'm new here and only have 8. I've tried already.
Oct
3
comment Proving $\frac{e^{z^2}}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^2}dt$ is bounded for $\Re(z) \leq 0$
I was just wanting clarification on the statement of the theorem. The limit calculation is fine.
Oct
3
comment Proving $\frac{e^{z^2}}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^2}dt$ is bounded for $\Re(z) \leq 0$
Thank you Giovanni, are you able to clarify what L'Hôpital's rule implies?
Oct
2
comment Proving $\frac{e^{z^2}}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^2}dt$ is bounded for $\Re(z) \leq 0$
Integration over the straight line contour ${t+z:-\infty \le t \le 0}$
Oct
2
comment Proving $\frac{e^{z^2}}{\sqrt{\pi}}\int_{-\infty}^{z}e^{-t^2}dt$ is bounded for $\Re(z) \leq 0$
@GiovanniDeGaetano, My claim came about when entering the function into wolfram alpha and seeing the exact result. I've proved this to myself and am quite sure of it. In regards to your upper bound, I feel that as there is still the exponential term out the front which will explode at -$\infty$, that it isn't as simple as this.