Recent content by kduna

The easiest way would be to just write down all of the quadratics over this field and check whether or not each one has a root. If your question is only about monic polynomials, then there are only 16 such polynomials. It is possible to write down a formula that counts the number of monic...

T is a function that takes a vector ##v## in ##V## to a vector ##w## in ##W##. We are very used to the idea of function that take numbers as inputs. For instance, if ##f(x) = x^2 + 1##, then ##f## takes ##1## to ##2##. We denote this by ##f(1) = 2##. So ##T: V \rightarrow W## means a function...

You can also prove this without induction. Proposition: ##x^n - 1 = (x - 1)(1 + x + ... + x^{n-1})##. The above is easily shown by distributing the entire ##(1 + x + ... + x^{n-1})## on the right hand side to yield: ##(x + x^2 + ... + x^n) - (1 + x + ... + x^{n-1})##. Now if you plug in...

You don't need to write f[x] and g[x] in the plot. Plot[{x^(5) - x, 2}, {x, -100, 100}] will work.

You can in fact make #5 more general. x ~ y iff x = y + n*k for some k. There are n equivalence classes.

I only have a minute, but I'll see what I can get through: 1. First let's pick which digits are going to be used. There are 10 choices for the first paired digits, 9 choices for the second paired digits, and 8 choices for the lone digit. So there are ##10*9*8## ways to choose these digits...

You can use good old row reduction to do this. Set up an augmented matrix whose first three columns are ##u, v, w##. Then make the last column ##x##. Then row reduce this. Note: You could also do it by inspection, although that is harder and may require a lot of playing around/ getting lucky.

Notice that:## \frac{1}{u!(v-u)!} = \frac{1}{v!} {v \choose u} ##. When finding the marginal, you should be summing to infinity, which actually just means summing through v (since v choose u will be zero thereafter). Therefore your sum is: ## \frac{1}{v!} \sum_{u=0}^{v} {v \choose u}...

That looks great. In this case, not only is ##U^T = U^{-1}##. But you have that both of those are ##U## itself.

It might be easier to prove the contrapositive: If ##A \neq B##, then ##P(A) \neq P(B)##. Although now that I think about it, it isn't too hard to show directly. Suppose ##a \in A##. Try to argue that ##a \in B##.

If by ##C^∞## you mean the space of all infinitely differentiable functions, then there are a lot more than polynomials around. Let ##f \in C^∞##. Look at the power series: ##f(x) = ∑_{i=0}^{∞} a_i x^i##. If the fourth derivative of ##f## is 0, then you have that ##a_i = 0## for ##i \geq 4##...

Once a basis has been specified, each linear transformation has a unique matrix representation. Think about the vectors that they gave you ##L## acting on. Do they form a basis? If so, how would they be linked to the form of the matrix?

I'll assume you already identified a subgroup ##H## that is isomorphic to ##S_{41}##. Think about the cosets of ##H##. How many are there? (If this is a hard question, try looking over a proof of Lagrange's Theorem).

Also, in group theory, people tend to stay away from using the word commutative. Instead you should use abelian. The reason for this is that it is nice to have separate words for commuting in multiplication and commuting in addition when you get to ring theory.

Let ##e_1, e_2## denote the identities of ##A(p)## and ##A'## respectively. If ##(a_1, a_2) \in A##, ##(a_1, a_2) = (e_1, e_2)## iff ##a_1 = e_1## and ##a_2 = e_2##. Suppose ##(a_1, a_2)## has order ##p^k## for some ##k > 0##. Then it must be that ##a_1^{p^k} = e_1## and ##a_2^{p^k} =...