Prof idea was to expose us to many things to get an idea of what is happening. However, this means we will lose a lot of idea about what is going on though
You'll understand more of the deRham stuff when you take algebraic topology soon. It would be good to explicitly compute some differential forms stuff. You might want to look at Chapter 8 of my book. I didn't say the word cohomology but I did a lot of stuff on this kind of stuff in there.
(Too many uses of the word "stuff" in those sentences. It's late and I played bridge all evening.)
Actually, when you have time, you might want to read or watch Lectures 44 and 45 of my course where I did a bunch of topology stuff with differential forms.
$\gamma$ : intesection of $z=x^{2}-y^{2}$ with $3x^{2}+4y^{2}=1$
Am supposed to calculate the line integral over the intersection.
$\int \bar{F}dr = \int \int Curl\bar{F}dS $
CurlF =$ (7y^{6},5z^{4},3x^{2})$
How to parametrize the surface here?
Thanks in advance
its a line integral in 3d , after using strokes theorem it will be transformed into suface integral in other region, then after doing that should be tranformed into double integral
my question is if its nessasy to transfor into double integral
Well the surface integral becomes a double integral after parametrisation (since integrating a surface integral has two parameters). The surface you are having with all the square terms suggest a parametrisation involving some variation of polar coordinates would be a good choice
calculate curl F (easiest in x,y coordinates), then use your parameterisation to convert that expression of x,y in terms of s,t. You can then dot product it with the normal vector N. By this point because of the parameterisation, your ds should become |J|dsdt where J is the jacobian of going from x,y variables to s,t variables. You should then end up with a double integral in s,t
exactly that is what i dont understand , seems like that jump from surface integral to double is not clear ,i did all the "work" allready on the surface , i dont know what to add to convert to double integral
allready have my normal it seems like to change the region and change dS with dA but that seems false
double integral over S : -14st^6 + 10 t(s^2-t^2)^4 +3s^2 dS
In that case, dS should just become dsdt, and then after you figuring out the limits, it becomes a double integral
This is because when you evaluate the normal, you already taken account of that jacobian in the dS term
(if you have calculated a unit normal instead (which is not what you have done here) then you will need to multiply the jacobian when you move from x,y to s,t coordinates)
In short, your double integral is: $$ \int_a^b\int_{c(s,t)}^{e(s,t)}-14st^6 + 10 t(s^2-t^2)^4 +3s^2 dsdt$$ where a,b,c,e are your limits corresponding to your parameters
@DHMO I don't understand something. When a problem says to "expand to fourth order in $\frac{y}{a}$", I know that it's looking for a taylor series expansion. But does that mean that my original function must have $\frac{y}{a}$ in it?
I was told to find an expression for the potential energy of a mass connected to 2 springs, fixed to only move in the y-direction. So I reasoned that my function should be $U=k|y|^2$.
This works because it gives the correct values for the potential energy at the minimum ($y=0$) and at the maximums $y=\pm a$.
Oh, forgot to mention: the other ends of the springs are fixed at $\pm a$
Forgot to tag: @DHMO
So to picture it, it just looks like a mass between 2 springs that moves up and down.
Thanks for your feedback. From my viewpoint, the question about the discontinuity of the series was right. Additonally I tried provide a reference where I was inspired (the paper only refers the calculations for which I am asking for a new example that I belive that is feasible to get). You are welcome and good weekend @ClementC.
Noticing elements multiplied to the left or right of the set S form an endomorphism on S to S. If the endomorphism is a permutation, then the proof of the + structure being relatively trivial follows shortly as once the additive identity is specified, the permutation ensures the + structure populates in a way isomorphic to the null semigroups
(NB one more example of a 3x3 associative zero term algebra on the next slide, not shown here. That one is also relatively trivial)
e=2791750015441845282449617808994392235718361609245357037995277819278848779957112605891676536208654961847315135289230917961030004642573453146672894161232247495786310160429472133265151079670176939601187261459706368113198201/1027027435571111514477741800526809128093952575118717306495725908654714438657261454220071898263977801963739801179019578071474848178499896970162708885651263114695666374690836204422930009139220332003160354967324851433288800 accurate to about 440 decimal places
@DHMO I took the liberty of removing "you know as well as I do" from your comment because I think that line might make p.m. defensive. See what I wrote just now.
Let's say I have a contour integral on a non-closed contour with starting point $z_0$ and ending point $z_1$. Am I allowed to do a substitution like this? And under what assumptions?
$$\displaystyle \int_{z_0}^{z_1} f (z) \, \mathrm d z = \int_{u^{-1}(z_0)}^{u^{-1}(z_1)} f (u (z))u'(z) \, \math...
In the mathematical field of complex analysis, contour integration is a method of evaluating certain integrals along paths in the complex plane.
Contour integration is closely related to the calculus of residues, a method of complex analysis.
One use for contour integrals is the evaluation of integrals along the real line that are not readily found by using only real variable methods.
Contour integration methods include
direct integration of a complex-valued function along a curve in the complex plane (a contour)
application of the Cauchy integral formula
application of the residue theorem
One...