Mathematics

11:57 AM

Can anyone please explain how I can prove that $\phi(f^{i} g^{j})=f^{2i}g^{j}$ is a homomorphism on $S_{3}$? ($\phi$ is a mapping from $S_{3}$ to $S_{3}$). I'm having no luck with it, any help will be appreciated.

If i take two different elements in $S_{3}$ should their powers be the same or different?

@Paradox101 First, you need to specify what $f$ and $g$ are. I assume a 3-cycle and a transposition?

12:13 PM

@TobiasKildetoft yes so if $S_{3}= {e,f,f^2,g,fg,f^2 g}$ then how should I take two elements from the group to prove homomorphism?

@Paradox101 You should take two arbitrary elements (i.e. not necessarily with the same powers) and check that the map sends their product to the product of their images

@TobiasKildetoft but it has to be two elements from $S_3$ right? So if I take $fg$ and $f^2 g$ i get $\phi((fg) (f^2 g))=(fg)^{2} f^{2}g$. Is this correct so far?

Keith

Hi I'm in senior high and math is just kicking my butt. Any tips?

@Paradox101 You need to first write the product in the given form to be able to apply the map

@Keith Kick back

Keith

Hah , can't see it's butt so :)

12:21 PM

@TobiasKildetoft can you clarify? In what given form?

@Paradox101 in the form $f^ig^j$ so you need to switch the order of $g$ and $f$ in your product, using the relations in the group

@Paradox101 I need to go now though.

@TobiasKildetoft like $\phi((f^2)(f^1)=(f)^{4} f^{2}g$?

oh ok

1:05 PM

morning chat

Secret

1:23 PM

It is well known that not all integrals are elementary (i.e. have closed forms in terms of elementary functions. But is there any study on the criteria need to be met in order to have an integral that has no closed form even in terms of special functions?

depends on the scope of what you mean by "special functions"

i mean, while mathematica isn't able to come up with an expression for that indefinite integral, that doesn't mean that none exists in the mathematical literature.

it could be in terms of something awful, like generalized confluent hypergeometric functions. but whether you admit that as a 'special function' is a matter of definition.

Secret

Is there really a convention in the literature on defining what expressions will form a special functions and not?

For example, the Riemman zeta is defined to be a series, and gamma functions are integrals

Keith

@skillpatrol yes , amongst other topics I'm taking calculus and have trouble getting a intuition of the topic. Any help would be great!x

i'd challenge the first of those. the series definition of the Riemann zeta function is useful along the positive real axis, but what you're usually interested in is its analytic continuation to the entire complex plane

skill patrol

Have to tried the khan academy @Keith

Secret

1:34 PM

Perhaps another way to ask that is: Why are elementary functions elementary. What property that they have that special functions don't have?

We know that whether an integral is involved does not necessary makes a function elementary, because ln(x) is defined as the integral of 1/x yet it is elementary