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May 27

Differentiability

If $\lim _{(h,k)\to (0,0)}{\frac {f(a+h,b+k)-f(a,b)-hf_{x}(a,b)-kf_{y}(a,b)}{\sqrt {h^{2}+k^{2}}}}=0$ then f is differentiable at (a,b). If this limit does not exist, can you conclude anything about the differentiability of f at (a,b)? Widener (talk) 00:56, 27 May 2011 (UTC)[reply]

If the limit doesn't exist, f is not differentiable at (a,b). Sławomir Biały (talk) 14:10, 28 May 2011 (UTC)[reply]

Latex Bracket Question

I want to show that three equations, when taken together, imply another equation.

I want to do this by having the three equations aligned, then having a curly bracket on the right to take them together, and then an arrow coming out of the curly bracket pointing to the new equation that combines their information.

I cannot find out how to do this. Trying to put them in an array means that I cannot align the three equations, as the "&" sign is taken as the border of the array cell.

Can anybody tell me how to do this?

130.102.78.164 (talk) 01:20, 27 May 2011 (UTC)[reply]

\left.{\begin{array}{l}a=b,\quad b=c\\c=d\\d=e\end{array}}\right\}\Rightarrow a=e

Is that what you want?(Igny (talk) 02:05, 27 May 2011 (UTC))[reply]

Almost. I would like the a=b, c=d and d=e to be aligned along their equals signs (in your example they are almost aligned anyway, but my equations have left-hand-sides of slightly different lengths).

I can't align them using "&=" because in the array environment a "&" is taken as a cell division. — Preceding unsigned comment added by 130.102.78.164 (talk) 01:10, 28 May 2011 (UTC)[reply]

Do you know about the align environment? --Trovatore (talk) 03:47, 28 May 2011 (UTC)[reply]

Yes I am using the align environment. The array environment goes inside the align environment. In the align environment equations are aligned using "&". But in the array environment, "&" denotes the sides of cells in the array. — Preceding unsigned comment added by 130.102.78.164 (talk) 08:20, 28 May 2011 (UTC)[reply]

How about this?

\left.{\begin{array}{rcl}abc&=&b\\b&=&cd\\cd&=&efg\end{array}}\right\}\Rightarrow abc=efg

or same with negative space

\left.{\begin{array}{rcl}abc&\!\!=\!\!&b\\b&\!\!=\!\!&cd\\cd&\!\!=\!\!&efg\end{array}}\right\}\Rightarrow abc=efg

?(Igny (talk) 04:48, 29 May 2011 (UTC))[reply]

Perfect! Thank you. — Preceding unsigned comment added by 130.102.158.15 (talk) 05:48, 30 May 2011 (UTC)[reply]

What about \implies to give

\implies

instead of \Rightarrow to give

\Rightarrow

? — Fly by Night (talk) 20:48, 30 May 2011 (UTC)[reply]

Rank of TC(x)

Is it true that the rank if the Transitive Closure of an set x is (rank of x) - 1? Where rank is defined in the set-theoretic sense. I think this is what I have calculated but I am not sure whether the 'right-subtraction' of -1 is well-defined and I may have made a mistake anyway.

Thanks very much 131.111.185.74 (talk) 09:47, 27 May 2011 (UTC)[reply]

I think maybe you have some unusual convention in mind for what transitive closure means? To me, the transitive closure of x is the set of all elements of x, elements of elements of x, and so on. So x itself is always a subset of its transitive closure, which means its rank can't go down. Can't go up, either, because the extra elements you're throwing in are always of lower rank. So the rank of the transitive closure of x equals the rank of x itself. --Trovatore (talk) 04:22, 28 May 2011 (UTC)[reply]

No, I think my definition is standard, I just probably made a mistake somewhere! It was all calculated using the fact that the rank is the supremum of the rank of the elements + 1, but I think maybe I just started off with 1 too many "elements of" and missed out including x itself. Your argument is a lot nicer - thanks! 131.111.185.74 (talk) 06:54, 28 May 2011 (UTC)[reply]

Inverse Laplace Transform

As a caveat, I am currently revising for a variety of exams and am asking the following questions because these are the ones that I have so far found almost completely intractable.

I want to compute, via the Bromwich contour, the inverse Laplace Transform of ${\frac {1}{\sqrt {s}}}$ . I have attempted this in two ways.

The first is by closing the contour with a single arc, creating a semicircle, that encloses the origin. My problem with this is that I cannot for the life of me work out how to evaluate the residue of the function ${\frac {e^{st}}{\sqrt {s}}}$ at the origin.

My second method, by construction, avoided this problem by making an indentation at the origin (accompanied by the arcs of two quarter circles and two line segments, one just above the negative real axia and one just below, connecting the contour encircling the origin to the two quarter circles). This was a mild improvement since the sum of all of the contour integrals is zero but I lost as much as I gained in the sense of not being able show that the sum of the integrals over the quarter circles is equal to zero in the limit that the radius of each circle goes to infinity.

Could someone please tell me what is going wrong and tell me how I can proceed to the correct answer (if possible, an explanation of how to resolve the issues with both methods would be appreciated but I realise that people may not have the time when resolution of one of the methods will suffice). Thank you. meromorphic [talk to me] 19:04, 27 May 2011 (UTC)[reply]

Try a substitution to get the Gaussian integral. A&S gives the answer as

{\frac {1}{\sqrt {\pi t}}}

if that helps.--RDBury (talk) 03:14, 28 May 2011 (UTC)[reply]

OK but which one of the methods does this work with because surely I'll still have the problems I mentioned above? meromorphic [talk to me] 08:45, 28 May 2011 (UTC)[reply]

There's a singularity at 0 but the improper integral converges if it's included on the contour. The real problem is that when you carry out the substitution the contour changes to two diagonal rays starting at the origin, not close the real axis like I thought it would be. The upshot is you get something more like a Fresnel integral. You still might be able to make it equal the Gaussian integral by estimating the integral on an arc from the real axis to the diagonal and showing this goes to 0 as the radius goes to infinity but it's getting more complicated that I thought so maybe it's just the wrong approach. Are you sure you have to use Bromwich? If not then you can just verify the answer from A&S, which really does reduce to the Gaussian integral.--RDBury (talk) 15:21, 28 May 2011 (UTC)[reply]

No, I definitely have to use Bromwich. Looking at a similar Laplace inversion, I think I have to choose the second contour, excluding the origin, but then I still don't know how to bound the integrals and show that they tend to zero. Any ideas? meromorphic [talk to me] 12:10, 29 May 2011 (UTC)[reply]

You need to use a theorem whose name I forgot. But you can easily prove this as follows. Naive estimation of the integral along the circles |I| < Maximumum of integrand times pi/2 does not work, because as a function of the radius this estimate increases as sqrt(R). But you can be more careful and estimate the integral over the polar angle more accurately. You write the exponential as

|exp(t z)| = |exp[t R cos(theta)]|.

You can overestimate the cosine by replacing that by a linear function. You can then compute the integral, you see that you get a factor 1/R, so the estimate decreases like

1/sqrt(R). Count Iblis (talk) 21:48, 29 May 2011 (UTC)[reply]

You should then find that the answer is minus the sum of the two integrals along the negative real axis, which you can write as:

{\frac {1}{\pi }}\int _{0}^{\infty }{\frac {\exp(-xt)}{\sqrt {x}}}dx

So, you find that the inverse Laplace transform is the Laplace transform divided by pi. Now, you could now, of course, simply calculate the Laplace transform by writing this as a Gaussian integral to obtain the answer, but that would defeat this whole exercise. Because the Laplace transform is proportional to the original function, you can directly obtain the answer from that without doing the contour integration (suppose that you did not verify that the integrals along the two quarter circles tend to zero and you computed the two integrals along the negative real axis first by rewriting them as a Gaussian integral, ou would have obtained the answer before completing the contour integration exercise).

You therefore have to refrain from trying to evaluate the Laplace transform. Instead, you substitute x = u/t to write it as:

\int _{0}^{\infty }{\frac {\exp(-xt)}{\sqrt {x}}}dx={\frac {c}{\sqrt {t}}}

(1)

where the constant c is given as:

c=\int _{0}^{\infty }{\frac {\exp(-u)}{\sqrt {u}}}du

The contour integration result can be written as:

L^{-1}\left[{\frac {1}{\sqrt {s}}}\right]={\frac {1}{\pi }}L\left[{\frac {1}{\sqrt {x}}}\right]

where on the r.h.s. the forward transform depends on the same variable t as the backward transform on the l.h.s. If you take the Laplace transform on both sides of the equation, you get:

{\frac {1}{\sqrt {s}}}={\frac {1}{\pi }}L^{2}\left[{\frac {1}{\sqrt {x}}}\right]

where the second Laplace transform takes a function depending on t and makes it depending on s. Using eq. (1) you can write the r.h.s. as

{\frac {c^{2}}{\pi }}{\frac {1}{\sqrt {s}}}

So, we see that

c={\sqrt {\pi }}

from which the answer follows. Count Iblis (talk) 00:18, 30 May 2011 (UTC)[reply]

Mapping from one region to another

Another question in complex analysis. After showing that the map $w=-i\log(z)$ , where I have used the branch of the logarithm function given by $-\pi <\arg(z)<\pi$ maps the region H, defined by the complex plan excluding the negative real axis, to the infinite strip S, defined by {z=x+iy: $-\pi <x<\pi$ }, I have to find an analytic map from the domain G, defined by $|z|<1$ , to S, as defined previously.

I have very little to go on, as I cannot see what motivates this result. I suspect that the answer will demand that I map G to H and then use the result from the first part of the question to map H to S but that's as much as I can think of. Any suggestions? Thanks. meromorphic [talk to me] 19:17, 27 May 2011 (UTC)[reply]

{\frac {1-z}{1+z}}

maps G to a 45° wedge. Then apply z⁸ to expand this to something that looks very similar to H. (I removed the math template since it was hiding the definition of S on my machine.)

I guess the intuition is gained by working with Möbius transformations and various elementary functions to see their effect on various simple regions in the plane. Use periodic functions to simplify strips and half strips, use Möbius transformations to turn circles into lines, and use power functions to expand angles.--RDBury (talk) 06:33, 28 May 2011 (UTC)[reply]

Conformal Mapping

In similar fashion to the above question, I have to find a conformal mapping from the the region $z:-{\frac {\pi }{2}}<x<{\frac {\pi }{2}}$ }, to the unit disc.

Again, I do not see an intuitive way of approaching this problem. I have tried a map involving the exponential function but I cannot come up with one that maps a point in the given domain to the origin. Help please? Thanks. meromorphic [talk to me] 19:26, 27 May 2011 (UTC)[reply]

Try starting with −cos z to map the given region to the upper half plane. You can then map the upper half plane to the unit circle by a Möbius transformation.--RDBury (talk) 03:40, 28 May 2011 (UTC)[reply]

Electromagnetism

For the first part of the question, I have to use Gauss' Law, in the integral form, to find the electirc field inside a uniformly charged sphere of radius R and total charge Q. I worked this out to be $\mathbf {E} ={\frac {Qr}{4{\pi }{R^{3}}\varepsilon _{0}}}\mathbf {e_{r}}$ where $\mathbf {e_{r}}$ is the basis vector in the radial direction.

I am now meant to use the relation $d\mathbf {F} =\mathbf {E} dq$ to find the total force exerted on the top half of the sphere by the bottom half. Again, I simply don't know what integral/equation/theorem I want to use. Can someone please point me in the right direction? Thanks. meromorphic [talk to me] 19:40, 27 May 2011 (UTC)[reply]

Take the z-direction to be orthogonal to the plane cutting the sphere in two halves. Then the force obviously only has a z-component. You can write dq = Q/(4/3 pi R^3) r^2 sin(theta)dr dtheta dphi. Multiplying by the z-component of E gives:

E_z dq = 3 Q^2/(16 pi^2 R^6) r^3 cos(theta)sin(theta) dr dtheta dphi

You need to integrate this over the half sphere, so from r = 0 to R, theta = 0 to pi/2 and phi = 0 to 2 pi. Count Iblis (talk) 23:44, 29 May 2011 (UTC)[reply]

Cayley′s theorem

I want a list of all finite groups G which are not isomorphic to a subgroup of Sym(|G|-1). --84.62.198.162 (talk) 21:27, 27 May 2011 (UTC)[reply]

You can start with groups of prime order. Then cyclic groups of prime power order and the quaternion group as well, any group that has a minimal non-trivial subgroup. The Klein 4-group is one though it doesn't have this property, not sure if there are others.--RDBury (talk) 02:59, 28 May 2011 (UTC)[reply]