finance - compound interest with geometric series

Were studying geometric sequences in maths and this came up as one of the questions:

A mortgage is taken out for 150000 and is repaid annually with 20000 installments.
Interest is charged on the outstanding debt at 10%, calculated annually.
If the first repayment us made one year after the mortgage is taken out, find the number of years it takes for the mortgage to be repaid.

When i looked up the solutions online to this, they didn't use a geometric series formula to solve they just did it manually and they eventually got to an answer of 15 years. My maths teacher told us to first find the Tn formula (ar the power of n-1) and then try and use that to form the series formula . I don't know how to write out the series formula here so can anyone help me ?

modular arithmetic - How to find remainder modulo $n$, when $n$ is a large number

I am doing RSA questions and I really could use help! Can someone show me a simple way to find $25^9 \pmod{33}$?

linear algebra - Sylvester rank inequality: $operatorname{rank} A + operatorname{rank}B leq operatorname{rank} AB + n$

If $A$ and $B$ are two matrices of the same order $n$, then
$$ \operatorname{rank} A + \operatorname{rank}B \leq \operatorname{rank} AB + n. $$

I don't know how to start proving this inequality. I would be very pleased if someone helps me. Thanks!

Edit I. Rank of $A$ is the same of the equivalent matrix $A' =\begin{pmatrix}I_r & 0 \\ 0 & 0\end{pmatrix}$. Analogously for $B$, ranks of $A$ and $B$ are $r,s\leq n$. Hence, since $\operatorname{rank}AB = \min\{r,s\}$, then $r+s\leq \min\{r,s\} + n$. (This is not correct since $\operatorname{rank} AB \leq \min\{r,s\}$.

Edit II. A discussion on the rank of a product of $H_f(A)$ and $H_c(B)$ would correct this, but I don't know how to formalize that $\operatorname{rank}H_f(A) +\operatorname{rank}H_c(B) - n \leq \operatorname{rank}[H_f(A)H_c(B)]$.

elementary number theory - Let $p$ be prime and $(frac{-3}p)=1$. Prove that $p$ is of the form $p=a^2+3b^2$

Let $p$ be prime and $(\frac{-3}p)=1$, where $(\frac{-3}p)$ is Legendre symbol. Prove that $p$ is of the form $p=a^2+3b^2$.

My progress:

$(\frac{-3}p)=1 \Rightarrow$ $(\frac{-3}p)=(\frac{-1}p)(\frac{3}p)=(-1)^{\frac{p-1}2}(-1)^{\lfloor\frac{p+1}6\rfloor}=1 \Rightarrow$ $\frac{p-1}2+\lfloor\frac{p+1}6\rfloor=2k$
I'm stuck here. This is probably not the way to prove that.

Also tried this way:
$(\frac{-3}p)=1$, thus $-3\equiv x^2\pmod{p} \Rightarrow$ $p|x^2+3 \Rightarrow$ $x^2+3=p\cdot k$
stuck here too.

Any help would be appreciated.

Answer

First part:
$$\left(\frac{-3}{p}\right)=1 \text{ if and only if }\; p\equiv{1}\!\!\!\!\pmod{3}.\tag{1}$$
This can be achieved through the Gauss quadratic reciprocity theorem in the most general form, or through the following lines. If $p=3k+1$, by the Cauchy theorem for groups there is an order-3 element in $\mathbb{F}_p^*$, say $\omega$; from $\omega^3=1$ follows $\omega^2+\omega+1\equiv 0\pmod{p}$, hence:
$$(2\omega+1)^2 = 4\omega^2+4\omega+1 = 4(\omega^2+\omega+1)-3 = -3,$$
and $-3$ is a quadratic residue $\pmod{p}$. On the other hand, if $-3$ is the square of something $\pmod{p}$, say $-3\equiv a^2\pmod{p}$, then:
$$\left(\frac{a-1}{2}\right)^3\equiv\frac{1}{8}(a^3-3a^2+3a-1)\equiv\frac{1}{8}\cdot 8\equiv{1},$$
and $\frac{a-1}{2}$ is an order-3 element in $\mathbb{F}_{p}^*$. From the Lagrange theorem for groups it follows that $3|(p-1)$.

---

Second part:
$$\text{If }p\equiv 1\pmod{3},\qquad p=a^2+3b^2.\tag{2}$$
Since by the first part we know that $-3$ is a quadratic residue $\pmod{p}$, there exists an integer number $c\in[0,p/2]$ such that:
$$ c^2+3\cdot 1^2 = k\cdot p.\tag{3}$$
The trick is now to set a "finite descent" in order to have $k=1$. Let $d$ the least positive integer such that $c\equiv d\pmod{k}$. Regarding $(3)$ mod $k$, we have:
$$ d^2+3\cdot 1^2 = k\cdot k_1.\tag{4}$$
Since the generalized Lagrange identity states:
$$(A^2+3B^2)(C^2+3D^2)=(AC+3BD)^2 + 3(BC-AD)^2,\tag{5}$$

by multiplying $(3)$ and $(4)$ we get:
$$ (cd+3)^2 + 3(c-d)^2 = k^2 pk_1.$$
Since $cd+3\equiv c^2+3\equiv 0\pmod{k}$ and $c\equiv d\pmod{k}$, we can rewrite the last line in the following form:
$$ \left(\frac{cd+3}{k}\right)^2+3\left(\frac{c-d}{k}\right)^2 = k_1\cdot p.\tag{6}$$
Now a careful analysis of the steps involved in the algorithm reveals that $k_1$$ p = a^2 + 3b^2$$
as wanted.

real analysis - $|x_n| to infty implies |f(x_n)| to infty$

Let $f:\mathbb{R} \to \mathbb{R}$ a continuous function.

Show that

1) $\lim \limits_{x\to +\infty} |f(x)|=\lim \limits_{x\to -\infty}|f(x)|=+\infty$

implies

2) $|x_n|\to \infty \implies |f(x_n)|\to \infty$.

I know that $f$ is continuous iff $x_n \to a \implies f(x_n)\to f(a)$.

But, can I use

$\lim|f(x_n)|= |f(\lim(x_n))|=\lim\limits_{x\to \infty} |f(x)|=+\infty$

for infinite cases so directly?

If the solution is not this way, could you help me by giving a hint how to do it?

Answer

The first condition means

$$\forall M \quad \exists \bar x : \forall x\, |x|>\bar x \quad |f(x)|>M$$

and from here the second follows, indeed we have

$$\forall \bar x \quad \exists \bar n : \forall n>\bar n \quad |x_n|> \bar x$$

that is

$$\forall M \quad \exists \bar n : \forall n>\bar n \quad |f(x_n)|>M$$

calculus - Evaluating $lim_{nto infty } , left(sum _{k=1}^{infty } frac{1}{n}right)$

$$\lim_{n\to \infty } \, \left(\sum _{k=1}^{\infty } \frac{1}{n}\right)$$

Intuitively, it seems that you are adding infinite of $\frac{1}{n}$ and then taking the limit as n goes to infinity, which would seem to give zero.

Further, manipulating it as follows:

$$\sum _{k=1}^{\infty } \lim_{n\to \infty } \, \frac{1}{n}$$

gives that you are adding up an infinite number of zeroes, which would support the idea that the limit is zero.

Is this right?

Answer

The answer is not defined, under the usual definitions of real analysis, or any definition of limits I know. (However, there may exist other definitions of limits that I'm unaware of, in settings larger than calculus or real analysis, and I'd guess that under definitions, the limit, if defined at all, would be $\infty$, whatever that means.)

In real analysis, we have definitions for $\lim_{n \to \infty} a_n$, where $a_n$ is a sequence of real numbers. Sometimes we use the symbol $\infty$ to denote that a sequence of real numbers grows arbitrarily large, and that therefore (in particular) the sequence has no limit in the real numbers.

However, what we have here is not the limit of a sequence of real numbers:
$$\lim_{n \to \infty} (\sum_{k=1}^{\infty} \frac1n) = \lim_{n \to \infty} a_n$$ where $a_n = \sum_{k=1}^{\infty} \frac1n$. Here, the expression $a_n$, itself an infinite sum (and therefore being a limit of finite sums, under the usual definitions of infinite sums) happens in this case to not be a real number:
$$a_n = \sum_{k=1}^{\infty} \frac1n = \lim_{m\to \infty} \sum_{k=1}^{m} \frac1n = \lim_{m\to\infty} \frac{m}n = \infty$$ (here I have used "$= \infty$" in the last step as notation to say that it is unbounded: does not exist in the real numbers).

So you are trying to find $\lim_{n\to\infty} a_n$, where $a_n$ is a limit that does not exist. If you want to allow "$\infty$" as an expression for manipulation, you could write $$\lim_{n \to \infty} \infty,$$ which is still not covered by the usual definitions.

But if you extend the definitions in any reasonable way to cover cases like this, I guess you'd define $\lim_{n \to \infty} \infty = \infty$. Note however that this is not standard.

linear algebra - Jordan Canonical Form - Similar matrices and same minimal polynomials

I have to prove the following result:

Let $A, B$ be two $n \times n$ matrices over the field $\sf F$ and $A,B$ have the same characteristic and minimal polynomials. If no eigenvalue has algebraic multiplicity greater than $3,$ then $A$ and $B$ are similar.

I have to use the following result:

If $A, B$ are two $3 \times 3$ nilpotent matrices, then $A, B$ are similar if and only if they have same minimal polynomial.

Please suggest how to proceed.

Answer

Here's a sketch...

If $A$ and $B$ are similar, then $A=PBP^{-1}$ and so $A^k=(PBP^{-1})^k=PB^kP^{-1}$ and hence $f(A)=0$ if and only if $f(B)=0$. Thus similar matrices share the same minimal polynomial.

For the other direction. Suppose $A$ and $B$ share the same minimal polynomial and let $\lambda$ be one their (common) eigenvalues. Since the algebraic multiplicity of $\lambda$ is no more than $3$, there are only a few possibilities for the corresponding Jordan block(s) (in the Jordan form of $A$ or $B$). If the multiplicity is exact $3$ the possibilities are:

$$ J \qquad = \qquad \begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix}, \qquad \begin{bmatrix} \lambda & 1 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix}, \qquad \mbox{or} \qquad \begin{bmatrix} \lambda & 1 & 0 \\ 0 & \lambda & 1 \\ 0 & 0 & \lambda \end{bmatrix}$$

Notice that the first block is annihilated by $x-\lambda$, the second by $(x-\lambda)^2$, and the third by $(x-\lambda)^3$. Thus the minimal polynomial informs you as to which type of blocks appear. Thus since the matrices share the same Jordan form, they must be similar.

Notice the necessity of the hypothesis "algebraic multiplicity of $3$ or less". If you allow multiplicity $4$ you could have:

$$ \begin{bmatrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda \end{bmatrix} \qquad \mbox{and} \qquad \begin{bmatrix} \lambda & 1 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \end{bmatrix} $$

Both of the above are annihilated by $(x-\lambda)^2$ but they are not similar (one has 2 blocks and so has 2 indep. eigenvectors and the other has 3 blocks and so has 3 indep. eigenvectors).