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6h
comment Show that a solution is the general solution.
Vector space because if $(x_n)$ and $(y_n)$ are solutions, so is $(px_n+qy_n)$. Two dimensional because if $(a_n)$ is a solution that satisfies initial condition $a_0=1$, $a_1=0$, and $(b_n)$ one that satisifies $b_0=0$, $b_1=1$ then every solution is a linear combination of $(a_n)$ and $(b_n)$. This is basically because any solution $c_n$ which is $0$ at $0$ and $1$ is $0$ everywhere.
6h
comment Show that a solution is the general solution.
Yes, verification that $\alpha^n$ and $\beta^n$ are solutions is as you write. And one is not a multiple of the other, so they are linearly independent.
6h
answered Show that a solution is the general solution.
7h
revised Probability Question: Dividing Groups
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7h
answered Probability Question: Dividing Groups
7h
comment Probability in Combination of 5 colour in 9 spaces
There are $\binom{9}{5}$ ways to choose where the white goes. For each, there are $4!$ ways to arrange the other colours.
8h
comment Probability in Combination of 5 colour in 9 spaces
We assume that all colour sequences are equally likely. (This should be in some form part of the statement of the problem.) There are $5^9$ such sequences. Now count how many sequences there are with $5$ W and one each of the others.
15h
reviewed No Action Needed The number of the solutions of $‎ x^{10}=‎ ‎ ‎\begin{bmatrix}1&0\\‎ ‎0&1‎ ‎\end{bmatrix}‎$
15h
reviewed Looks OK Examples of automorphisms on structures
15h
reviewed Looks OK Countable union of countable sets
15h
reviewed Leave Open probability distribution function of two independent variables
15h
reviewed Leave Open Automorphisms of $(\mathbb{R}, +)$
15h
revised straightforward calculus problem
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15h
comment Is multiplication normally/binomially distributed?
The two curves have vaguely similar shapes, up and then symmetrically down. But looked at more closely, they are very different. The multiplication curve $y=x(a-x)$ has a broader flattish area in the middle, and decays less rapidly in the tails. It really does not look "bell-shaped."
16h
comment straightforward calculus problem
You are welcome. A picture from a graphing program can help to make things more obvious.
16h
comment How to show that $at^2+bt+c$ can be written as $\displaystyle \begin{equation} a \left( t+\frac{b}{2a}\right) ^2-\frac{1}{4a}(b^2-4ac)\end{equation}$?
For using the completing the square procedure, and generalizations, you will need to be familiar with the process in the other direction.
16h
comment How to show that $at^2+bt+c$ can be written as $\displaystyle \begin{equation} a \left( t+\frac{b}{2a}\right) ^2-\frac{1}{4a}(b^2-4ac)\end{equation}$?
Expanding is perfectly correct. Note that we need $a\ne 0$.
16h
comment Examples of automorphisms on structures
Very clean example.
16h
revised straightforward calculus problem
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16h
comment straightforward calculus problem
There are $4$ pieces of identical length, one in each quadrant, because our curve does not care about the sign of $x$ and $y$. The first quadrant part, by the symmetry between $x$ and $y$, is divided into two pieces of the same length at $a$. The $8$ is $(4)(2)$.