# Are “most” sets in $\mathbb R$ neither open nor closed?

It seems intuitive to believe that most subsets of $\mathbb R$ are neither open nor closed.

For instance, if we consider the collection of all (open, closed, half-closed/open) intervals, then one can probably make precise the notion that "half" of all intervals in this collection are neither open nor closed. (Whether this will amount to a reasonable definition of what it means for most subsets to be neither open nor closed might be up for debate.)

If this intuition is correct, is there a way to formalise it? If not, how would we formalise its being wrong?

To be clear, I am happy for a fairly broad interpretation of the term "most". Natural interpretations include but are not limited to:

1. Measure-theoretic (e.g. is there a natural measure on (a $\sigma$-algebra on) the power set of $\mathbb R$ that assigns negligible measure to $\tau$?)
2. Topological (e.g. is there a natural topology on the power set of $\mathbb R$ where $\tau$ is meagre, or even nowhere dense?)
3. Set-theoretic (e.g. does the power set of $\mathbb R$ have larger cardinality than $\tau$?)

Here, $\tau$ is (obviously) the Euclidean topology.

Actually, that last version of the question in parentheses might have the easiest answer: Let $\mathcal B$ be the Borel sets on $\mathbb R$. We have that $|\tau| \le | \mathcal B | = | \mathbb R | < \left| 2^{\mathbb R} \right|$. (For details on the equality, see here. For a much simpler proof, see this answer.)

Are there alternative ways to make this precise?

• A trivial answer to 1. and 2. would be to define a topology on $P(\Bbb R),$ the power-set of $\Bbb R$, as $\{\emptyset, P(\Bbb R)\setminus \tau, P(\Bbb R)\}$ and a measure $m$ with $m(P(\Bbb R))=1$ and $m(\tau)=0.$ But I don't think this is quite what you're hoping for. – DanielWainfleet Sep 16 '18 at 1:12
• @DanielWainfleet Indeed. I was hoping the word "natural" would be enough to rule out such answers. Unless there are other reasons I'm not seeing that would make such a definition natural? – Theoretical Economist Sep 17 '18 at 22:43

Since each open non-empty subset of $\mathbb R$ can be written has a countable union of open intervals and since the set of all open intervals has the same cardinal as $\mathbb R$, the set of all open subsets of $\mathbb R$ has the same cardinal as $\mathbb R$. And since there is a bijection between the open subsets of $\mathbb R$ and the closed ones, the set of all closed subsets of $\mathbb R$ also has the same cardinal as $\mathbb R$. So, in the set-theoretical sense, most subsets of $\mathbb R$ are neither closed nor open.
• To the proposer: The set of bounded open real intervals with rational end-points is a countable base (basis) for $\tau.$ For any topology $T$ with a countable base $B$ we have $|T|=|\{\cup C: C\in P(B)\}|\leq |P(B)|\leq 2^{\aleph_0}=|\Bbb R|.$.... So $|\tau|\leq |\Bbb R|.$ And we also have $|\tau|\geq |\{(r,r+1):r\in \Bbb R\}|=|\Bbb R|.$ – DanielWainfleet Sep 16 '18 at 1:27


# Is the Trace of the Transposed Matrix the Same as the Trace of the Matrix?

## Problem 633

Let $A$ be an $n \times n$ matrix.

Is it true that $\tr ( A^\trans ) = \tr(A)$? If it is true, prove it. If not, give a counterexample.

## Solution.

The answer is true. Recall that the transpose of a matrix is the sum of its diagonal entries. Also, note that the diagonal entries of the transposed matrix are the same as the original matrix.

Putting together these observations yields the equality $\tr ( A^\trans ) = \tr(A)$.

Here is the more formal proof.

For $A = (a_{i j})_{1 \leq i, j \leq n}$, the transpose $A^{\trans}= (b_{i j})_{1 \leq i, j \leq n}$ is defined by $b_{i j} = a_{j i}$.

In particular, notice that $b_{i i} = a_{i i}$ for $1 \leq i \leq n$. And so,
$\tr(A^{\trans}) = \sum_{i=1}^n b_{i i} = \sum_{i=1}^n a_{i i} = \tr(A) .$

### More from my site

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• Prove that the Dot Product is Commutative: $\mathbf{v}\cdot \mathbf{w}= \mathbf{w} \cdot \mathbf{v}$ Let $\mathbf{v}$ and $\mathbf{w}$ be two $n \times 1$ column vectors. (a) Prove that $\mathbf{v}^\trans \mathbf{w} = \mathbf{w}^\trans \mathbf{v}$. (b) Provide an example to show that $\mathbf{v} \mathbf{w}^\trans$ is not always equal to $\mathbf{w} […] • If 2 by 2 Matrices Satisfy$A=AB-BA$, then$A^2$is Zero Matrix Let$A, B$be complex$2\times 2$matrices satisfying the relation $A=AB-BA.$ Prove that$A^2=O$, where$O$is the$2\times 2$zero matrix. Hint. Find the trace of$A$. Use the Cayley-Hamilton theorem Proof. We first calculate the […] • Determine Whether Given Matrices are Similar (a) Is the matrix$A=\begin{bmatrix} 1 & 2\\ 0& 3 \end{bmatrix}$similar to the matrix$B=\begin{bmatrix} 3 & 0\\ 1& 2 \end{bmatrix}$? (b) Is the matrix$A=\begin{bmatrix} 0 & 1\\ 5& 3 \end{bmatrix}$similar to the matrix […] • If Two Matrices are Similar, then their Determinants are the Same Prove that if$A$and$B$are similar matrices, then their determinants are the same. Proof. Suppose that$A$and$B$are similar. Then there exists a nonsingular matrix$S$such that $S^{-1}AS=B$ by definition. Then we […] • Matrix Operations with Transpose Calculate the following expressions, using the following matrices: $A = \begin{bmatrix} 2 & 3 \\ -5 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 0 & -1 \\ 1 & -1 \end{bmatrix}, \qquad \mathbf{v} = \begin{bmatrix} 2 \\ -4 \end{bmatrix}$ (a)$A B^\trans + \mathbf{v} […]

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# This note has been used to help create the Componendo and Dividendo wiki

See the complete wiki page here.

The method of Componendo et Dividendo allows a quick way to do some calculations, and can simplify the amount of expansion needed.

If $a, b, c$ and $d$ are numbers such that $b, d$ are non-zero and $\frac{a}{b} = \frac{c}{d}$, then

$\begin{array} {l r l } \text{1. Componendo:} & \frac{ a+b}{b} & = \frac{ c+d}{d}. \\ \text{2. Dividendo: } & \frac{ a-b}{b} & = \frac{ c-d} {d}. \\ \text{ Componendo et Dividendo: } & \\ \text{3. For } k \neq \frac{a}{b},& \frac{ a+kb}{a-kb} & = \frac{ c+kd}{c-kd} .\\ \text{4. For } k \neq \frac{-b}{d}, & \frac{ a}{b} & = \frac{ a + kc } { b + kd }. \\ \end{array}$

This can be proven directly by observing that

$\begin{array} {l r l } \text{ 1.} \frac{ a+b}{b} = \frac{ \frac{a}{b} + 1} {1} = \frac{ \frac{c}{d} + 1} {1} = \frac{ c+d}{d} . \\ \text{ 2.} \frac{ a-b}{b} = \frac{ \frac{a}{b} - 1} {1} = \frac{ \frac{c}{d} - 1} {1} = \frac{ c-d}{d} . \\ \text{ 3.} \frac{ a+kb}{a-kb} = \frac{ \frac{a}{b} + k } { \frac{a}{b} - k} = \frac{ \frac{c}{d} + k } { \frac{ c}{d} -k} = \frac{ c+kd} { c-kd} . \\ \text{ 4.} \frac{ a + kc} { b+ kd} = \frac{ a}{b} \times \frac{ 1 + k \frac{c}{a} } { 1 + k \frac{d}{b} } = \frac{ a}{b} . \end{array}$

## Worked examples

### 1. Show the converse, namely that if $a, b, c$ and $d$ are numbers such that $b, d, a-b, c-d$ are non-zero and $\frac{ a+b}{a-b} = \frac{c+d} { c-d}$, then $\frac{ a}{b} = \frac{c}{d}$.

Solution: We apply Componendo et Dividendo with $k=1$ (which is valid since $\frac{a+b}{a-b} \neq 1$ ), and get that $\frac{ 2a } { 2b} = \frac{ (a+b) + (a-b) } { (a+b) - (a-b) } = \frac{ (c+d) + (c-d) } { (c+d) - (c-d) } = \frac{ 2c} { 2d}.$

Note: The converse of Componendo and Dividendo also holds, and we can prove it by applying Dividendo and Componendo respectively.

### 2. Solve for $x$: $\frac{ x^3+1} { x+ 1} = \frac{ x^3-1} { x-1}$.

Solution: For the fractions to make sense, we must have $x \neq 1, -1$.

Cross multiplying, we get $\frac{ x^3+1}{x^3-1} = \frac{ x+1}{x-1}.$

Apply Componendo et Dividendo with $k=1$ (which is valid since $\frac{x+1}{x-1} \neq 1$ ), we get that $\frac{ 2x^3}{2} = \frac{ 2x}{2} \Rightarrow x(x^2-1) = 0$. However since $x \neq 1, -1$, we have $x=0$ as the only solution.

Note: We also need to check the condition that the denominators are non-zero, but this is obvious.

Note by Calvin Lin
5 years, 5 months ago

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if a/b =c/d what will be the result by componendo dividendo

- 5 years, 2 months ago

See the statements contained in the first box.

Staff - 5 years, 2 months ago

Sorry I'm slightly confused, could you clarify what Componendo and Dividendo integrate to?

- 3 years, 5 months ago

Check out the examples on the componendo and dividendo wiki page.

Staff - 3 years, 5 months ago

I found out (somewhat accidentally) that

$\frac{a+mb}{b+na} = \frac{c+md}{d+nc}$

is also true. It's quite easy to prove; it can also be derived from the 4th case stated above. It seems different enough, though, to be worth a mention, yet I never see it anywhere. Or is it perhaps that there are many other such corollaries and only the most basic ones are usually listed?

- 1 year, 7 months ago

Ohhh, that's a really nice identity. It is a generalization of Worked Example 1. Can you add it to the Componendo and Dividendo wiki under Problem Solving?

Like you said, it is essentially / can be derived from the 4th case, where we have $\frac{ a + mb } { c + md} = \frac{a}{c} = \frac{b}{d} = \frac{ b+na}{d + nc }$ (as long as the denominators are non-zero).

Staff - 1 year, 7 months ago

Have finally had a chance to add the identity to the Wiki; please take a look when you can and let me know if I need to make any changes. (I accidentally first added it to the Theorem section before I remembered that you said Problem Solving, I did move it to the correct section after that, hope it didn't cause any problems.) I also added an example applying it a little further down, in the section that introduced using C&D with non-linear terms. The Wiki mentioned the terms could be polynomial or exponential; my example uses trig functions, I believe it's still valid but please take a look and let me know if there are any issues. Thanks.

- 1 year, 7 months ago

That's a great writeup. I like how you highlighted Method 1, which "by right" should be how people understand this identity, and "by left" it is placed in this wiki because of the form it took.

Yes, your example with a trigo substitution is valid. Good one. Thanks!

Staff - 1 year, 7 months ago


# Thread: Parametric equations of plane problem

1. ## Parametric equations of plane problem

This problem was confusing me so any help would be appreciated!

Write both the parametric and symmetric equations of the line of intersection of the planes with the equations 2x - y + z = 5 and x + y - z = 1.

2. Originally Posted by clockingly
This problem was confusing me so any help would be appreciated!

Write both the parametric and symmetric equations of the line of intersection of the planes with the equations 2x - y + z = 5 and x + y - z = 1.
If $z=0$ then:
$2x-y=5$
$x+y=1$
Thus, $x=2$ and $y=-1$

If $z=1$ then:
$2x-y=4$
$x+y=2$
Thus, $x=2$ and $y=0$

Thus, we have that this line must contains the points $(2,-1,0)\mbox{ and }(2,0,1)$

Thus, the Symettric Equation is:
$\frac{x-2}{0} = \frac{y+1}{-1} = \frac{z-0}{-1}$

Note, this is a zero in the denominator of the first fraction, it is a useful notation I saw in a Russian Textbook* which simply means $x=2$ constantly, so it is just a shorthand notation.

*)If you are interested the Textbook was written by Perelman! Not Perelman you think it is but his grandfather!

3. Originally Posted by ThePerfectHacker
Thus, the Symettric Equation is:
$\frac{x-2}{0} = \frac{y+1}{-1} = \frac{z-0}{-1}$

Note, this is a zero in the denominator of the first fraction, it is a useful notation I saw in a Russian Textbook* which simply means $x=2$ constantly, so it is just a shorthand notation.

*)If you are interested the Textbook was written by Perelman! Not Perelman you think it is but his grandfather!
Thanks for that TPH. i did this question the same as you did, but i had no idea what to do with that 0 in the denominator, it kind of freaked me out, so i decided not to post

4. Originally Posted by Jhevon
Thanks for that TPH. i did this question the same as you did, but i had no idea what to do with that 0 in the denominator, it kind of freaked me out, so i decided not to post
The way it is taught in American schools is if that every happens then the fuction does not have a "symettric form" to it. It only has parametric. But Grandpa's Perelman notation is really useful like I said.

5. Hello, clockingly!

Write both the parametric and symmetric equations of the line of intersection
of the planes with the equations: . $\begin{array}{ccc}2x - y + z &= &5 \\ x + y - z &= &1\end{array}$
This is how I was taught to find the intersection of two planes.

We have: . $\begin{array}{cccc}2x - y + z & = & 5 & {\color{blue}[1]}\\ x + y -z & = & 1 & {\color{blue}[2]}\end{array}$

Add [1] and [2]: . $3x \,=\,6\quad\Rightarrow\quad x\,=\,2$

Substitute into [2]: . $2 + y - z \:=\:1\quad\Rightarrow\quad y \:=\:z-1$

We have $x,\,y,\,z$ as a function of $z\!:\;\;\begin{Bmatrix}x & = & 2 \\ y & = & z-1 \\ z & = & z\end{Bmatrix}$

On the right side, replace $z$ with a parameter $t$.

. . . . $\begin{Bmatrix}x & = & 2 \\ y & = & t - 1 \\ z & = & t\end{Bmatrix}$ . . . . There!

6. Originally Posted by Soroban
Hello, clockingly!

This is how I was taught to find the intersection of two planes.

We have: . $\begin{array}{cccc}2x - y + z & = & 5 & {\color{blue}[1]}\\ x + y -z & = & 1 & {\color{blue}[2]}\end{array}$

Add [1] and [2]: . $3x \,=\,6\quad\Rightarrow\quad x\,=\,2$

Substitute into [2]: . $2 + y - z \:=\:1\quad\Rightarrow\quad y \:=\:z-1$

We have $x,\,y,\,z$ as a function of $z\!:\;\;\begin{Bmatrix}x & = & 2 \\ y & = & z-1 \\ z & = & z\end{Bmatrix}$

On the right side, replace $z$ with a parameter $t$.

. . . . $\begin{Bmatrix}x & = & 2 \\ y & = & t - 1 \\ z & = & t\end{Bmatrix}$ . . . . There!

nice method

7. The customary notation used in textbooks in North America for the symmetric form in which one direction number is zero is to use a semicolon: $\frac{{y + 1}}{{ - 1}} = \frac{z}{{ - 1}}\, ;\,x = 2$.


# Proving $\frac{n^n}{e^{n-1}}<n!<\frac{(n+1)^{n+1}}{e^{n}}$ by induction for all $n> 2$.

I am trying to prove

$$\frac{n^n}{e^{n-1}}<n!<\frac{(n+1)^{n+1}}{e^{n}} \text{ for all }n > 2.$$

Here is the original source (Problem 1B, on page 12 of PDF)

Can this be proved by induction?

The base step $n=3$ is proved: $\frac {27}{e^2} < 6 < \frac{256}{e^3}$ (since $e^2 > 5$ and $e^3 < 27$, respectively).

I can assume the case for $n=k$ is true: $\frac{k^k}{e^{k-1}}<k!<\frac{(k+1)^{k+1}}{e^{k}}$.

For $n=k+1$, I am having trouble:

\begin{align} (k+1)!&=(k+1)k!\\&>(k+1)\frac{k^k}{e^{k-1}}\\&=e(k+1)\frac{k^k}{e^{k}} \end{align} Now, by graphing on a calculator, I found it true that $ek^k >(k+1)^k$ (which would complete the proof for the left inequality), but is there some way to prove this relation?

And for the other side of the inequality, I am also having some trouble: \begin{align} (k+1)!&=(k+1)k!\\&<(k+1)\frac{(k+1)^{k+1}}{e^{k}}\\&=\frac{(k+1)^{k+2}}{e^k}\\&<\frac{(k+2)^{k+2}}{e^k}. \end{align} I can't seem to obtain the $e^{k+1}$ in the denominator, needed to complete the induction proof.

• just curious, what level this exam is -to enter PhD program or some qual? – Alex Feb 15 '15 at 13:47
• @Alex According to the link provided in my question, it's UC Berkeley's "preliminary exam", which apparently their students must pass before they can continue their Ph.D program and eventually take their oral qualifying exam. – Cookie Feb 15 '15 at 17:52
• Also, there is a solution here math.berkeley.edu/sites/default/files/pages/… (page 12 of PDF), but the solution uses integration. I understand that method works, but I wanted to also prove this by induction. Induction must be possible to use here... – Cookie Feb 15 '15 at 17:55
• so it's a prelim to qual exam? – Alex Feb 16 '15 at 0:05
• It's a prelim required to stay in the Ph.D program. The student has two chances (or maybe three based on appeal) to pass the exam, or else the student is dismissed. – Cookie Feb 16 '15 at 0:27

Let's try an inductive proof from the original inquality for $n$, let's prove for $n+1$

$$\frac{n^n}{e^{n-1}}<n!<\frac{(n+1)^{n+1}}{e^{n}}$$

Okay, multiply both sides by $n+1$. At least the middle is correct

$$(n+1)\frac{n^n}{e^{n-1}}<(n+1)!<\frac{(n+1)^{n+2}}{e^{n}}$$

and we have to make the left and right sides look more appropriate

$$\left(\color{red}{\frac{n}{n+1}} \right)^\color{red}{n}\frac{(n+1)^{n+1}}{e^{n-1}}<(n+1)!<\frac{(n+2)^{n+2}}{e^{n}} \left(\color{blue}{\frac{n+1}{n+2}}\right)^{\color{blue}{n+2}}$$

Our induction is complete if we can prove two more inequalities:

$$\frac{1}{e} < \left(\frac{n}{n+1} \right)^n \text{ and } \left(\frac{n+1}{n+2}\right)^{n+2}< \frac{1}{e}$$

These two inequalities can be combined into one and we can take reciprocals. At least it is well-known.

$$\bigg(1 + \frac{1}{m}\bigg)^{m+1}> \mathbf{e} > \bigg(1 + \frac{1}{n} \bigg)^n$$

This is true for any $m, n \in \mathbb{N}$.

You shave off a little bit too much when you said $(k+1)^{k+2} < (k+2)^{k+2}$. Instead, you needed the more delicate:

$$(k+1)^{k+2} < \frac{1}{e} (k+2)^{k+2}$$

If you know1 that: $$a_n=\left(1+\frac{1}{n}\right)^n,\qquad b_n = \left(1+\frac{1}{n}\right)^{n+1}$$ give to sequences converging towards $e$, where $\{a_n\}$ is increasing while $\{b_n\}$ is decreasing, consider that:

$$n = \prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right),\tag{1}$$ so: $$n! = \prod_{m=2}^{n} m = \prod_{m=2}^{n}\prod_{k=1}^{m-1}\left(1+\frac{1}{k}\right) = \prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{n-k}=\frac{n^n}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k}\tag{2}$$ and your inequality trivially follows.

1) If you are not aware of such a classical result, then prove it by induction. It is rather easy.

• I think that it is not as easy as you think. – marty cohen Feb 15 '15 at 4:09
• @martycohen: it is also possible to use AM-GM in order to prove that $\{a_n\}$ is increasing and $\{b_n\}$ is decreasing. I think there are many questions/answers on MSE devoted to such well-known fact. – Jack D'Aurizio Feb 15 '15 at 4:12
• Here, for example: math.stackexchange.com/questions/389793/… – marty cohen Feb 15 '15 at 4:18

Proof: We will prove the inequality by induction. Since $e^2 > 5$ and $e^3 < 27$, we have $$\frac {27}{e^2} < 6 < \frac{256}{e^3}.$$ Thus, the statement for $n=3$ is true. The base step is complete.

For the induction step, we assume the statement is true for $n=k$. That is, assume $$\frac{k^k}{e^{k-1}}<k!<\frac{(k+1)^{k+1}}{e^{k}}.$$

We want to prove that the statement is true for $n=k+1$. It is straightforward to see for all $k > 2$ that $\left(1+\frac 1k \right)^k < e < \left(1+\frac 1k \right)^{k+1}$; this algebraically implies $$\left(\frac k{k+1} \right)^{k+1} < \frac 1e < \left(\frac k{k+1} \right)^k. \tag{*}$$ A separate induction proof for the left inequality of $(*)$ establishes $\left(\frac{k+1}{k+2} \right)^{k+2}<\frac 1e$. We now have \begin{align} (k+1)! &= (k+1)k! \\ &< (k+1) \frac{(k+1)^{k+1}}{e^k} \\ &= \frac{(k+1)^{k+2}}{e^k} \\ &= \frac{(k+1)^{k+2}}{e^k} \left( \frac{k+2}{k+2} \right)^{k+2} \\ &= \frac{(k+2)^{k+2}}{e^k} \left( \frac{k+1}{k+2} \right)^{k+2} \\ &< \frac{(k+2)^{k+2}}{e^k} \frac 1e \\ &= \frac{(k+2)^{k+2}}{e^{k+1}} \end{align} and \begin{align} (k+1)! &= (k+1)k! \\ &> (k+1) \frac{k^k}{e^{k-1}} \\ &= (k+1) \frac{k^k}{e^{k-1}} \left(\frac{k+1}{k+1} \right)^k \\ &= \frac{(k+1)^{k+1}}{e^{k-1}} \left( \frac k{k+1} \right)^k \\ &> \frac{(k+1)^{k+1}}{e^{k-1}} \frac 1e \\ &= \frac{(k+1)^{k+1}}{e^k}. \end{align} We have established that the statement $$\frac{(k+1)^{k+1}}{e^k}<(k+1)!<\frac{(k+2)^{k+2}}{e^{k+1}}$$ for $n=k+1$ is true. This completes the proof.

Some times it is easier for such an induction if we shift the sequence by one step, so the basis of the expressions is nicer for algebraic manipulations.
Let's rewrite your sequence of inequalities as
$$\displaystyle {n! \over e}<{(n+1)^{n+1} \over e^{n+1}} < {(n+1)! \over e} <{(n+2)^{n+2} \over e^{n+2}}< {(n+2)! \over e} \tag 1$$
and for simpler references below as $$a_0 \quad < \quad b_0 \quad <\quad a_1 \quad <\quad b_1 \quad < \quad a_2 \tag 2$$

Then we ask: does from $a_0<b_0<a_1$ follow that $a_1<b_1<a_2$ ?

Of course $a_1 = (n+1) \cdot a_0$ and so it might be useful to define $b_0$ as a fraction in the interval of $a_0$ and $a_1$: $$b_0 = (n+1)q_1 \cdot a_0 \text{ where } q_1<1 \tag {3.1 }$$. Consequently, define $$b_1 = (n+2)q_2 \cdot a_1 \text{ where also } q_2<1 \tag {3.2 }$$ Here the inequality $q_2 < 1$ is not known but expected and if this can be shown by induction from $q_1$ this would solve the problem .

So we start with $$q_1 = {b_0 \over a_0 (n+1)} = {(n+1)^{n} \over e^n n! } \tag {4.1 }$$ and by the beginning of the induction we know, that this is indeed smaller than 1 so $$q_1 < 1 \tag {4.2 }$$

Now we have simply $$q_2 = {b_1 \over a_1 (n+2)} = {(n+2)^{n+1} \over e^{n+1} (n+1)! } \tag {4.3 }$$ Next we consider the systematic progression in the sequence of $q_1,q_2,q_3,...$. To begin we determine the ratio $r_2={q_2 \over q_1}$ . We find $$\begin{eqnarray} r_2&=&{q_2 \over q_1}& =&{ {(n+2)^{n+1} \over e^{n+1} (n+1)! } \over {(n+1)^{n} \over e^{n} (n)! } } \\ &&&=& {(n+2)^{n+1} e^{n} (n)! \over e^{n+1} (n+1)! (n+1)^{n}} \\ &&&=& {(n+2)^{n+1} \over e (n+1)^{n+1}} \\ &&&=& \left({n+2 \over n+1}\right)^{n+1} \cdot \frac 1e \\ r_2&=&{q_2 \over q_1}& =& \left( 1 + {1 \over n+1} \right)^{n+1} \cdot \frac 1e \end{eqnarray} \tag {5.1 }$$

It is now needed to recognize/remember from the definition of $e$, that the last expression is smaller than 1 and that for $n \to \infty$ approximates monotonically 1.
Also we see by the expansion of the binomial-series $(1+1/x)^x=1+1+1/2+...$ in the general case that for $x>2$ this is greater than $2$ so $${2 \over e} \approx 0.73 < r_2 < 1 \tag{5.2}$$

From this we know now, that $q_2$ is not only smaller than $1$ but also smaller than $q_1$ but$q_{n+1} \to q_n$ for increasing $n$.

So we have $$\begin{eqnarray} &q_2 &=& q_1 \cdot r_2 < q_1 < 1 \\ \to& b_1 &<& (n+2) a_1 = a_2 \\ \to &a_1 &<& b_1 <a_2 \end{eqnarray} \tag {6 }$$ which we wanted to show.

Since several answers have already proven the result via induction, I see no harm in recording an additional proof that does not use induction.

We claim for all $n\ge 1$, $$\int_{0}^{1} (x\log(x))^n \, dx = \frac{(-1)^n n!}{(n+1)^{n+1}}$$ This can be shown using differentiation under the integral sign; set $k=n$ below: $$f(n) = \int_{0}^{1} x^n \, dx = \frac{1}{n+1} \implies \frac{d^k}{dn^k} f(n) = \int_{0}^{1} x^n (\log(x))^k \, dx = \frac{(-1)^{k} k!}{(n+1)^{k+1}}$$

Since $x \log(x)$ does not change sign, we have $$\int_{0}^{1} |x\log(x)|^n \, dx = \frac{n!}{(n+1)^{n+1}}$$ From this expression, we see that the given inequality is equivalent to $$\frac{1}{(n+1)e^n} < \int_{0}^{1} |x\log(x)|^n \, dx < \frac{1}{e^n}$$

A simple computation shows $|x\log(x)|^n$ attains a maximum of $e^{-n}$ at $x=e^{-1}$, so the upper bound is evident. To see the lower bound, note $|x\log(x)|^n$ is a concave function, so its graph encloses the triangle with vertices $(0,0)$, $(1,0)$, and $(e^{-n}, e^{-1})$. This triangle has area $e^{-n}/2 > e^{-n}/(n+1)$ since $n>2$.


# Formula of nth term of a pattern

1. Aug 27, 2014

### songoku

1. The problem statement, all variables and given/known data
Find the formula for nth term of:
.... , -450 , -270 , -180 , -90 , 90 , 180 , 270 , 450 , ...

2. Relevant equations
Arithmetic and geometric series
Recursive

3. The attempt at a solution
Usually when finding the formula of Un, n starts from 1. But since the pattern goes infinite in both ways, how can we set which one is the first term?

As for recursive, it usually involves the preceding term, Un-1. Again, how can we determine the preceding term for the above case?

Please give me idea how to start solving this. Thank you very much

2. Aug 27, 2014

### jackarms

For the general formula (in terms of n), you're right that the sequence going both ways complicates things. The answer is that you have to arbitrarily designate a term to be $a_{1}$, or perhaps $a_{0}$. You would write the whole formula based around that starting term, in a form like this:

$a_{n} = ....$, where $a_{0} = 3$

For the recursive definition, you'll just use a term and the preceding term. Since it's recursive, you don't need to know the particular values for either of those terms -- the recursive form just means you describe a term in terms of its preceding term. So it works with any two (adjacent) terms in the sequence, and so it isn't necessary to designate a starting term. The form for that would look like:

$a_{n} = a_{n-1}....$

So $a_{n}$ can refer to any term in the sequence, because this rule works for any pair of terms in the sequence, no matter how long it is and whether it goes infinitely in just one direction or two.

Last edited: Aug 27, 2014
3. Aug 27, 2014

### Ray Vickson

Just fix $n = 0$ anywhere you want; for example, you can take $U_0 = -450.$ All the $U_n$ for $n < 0$ are the ones you do not see to the left of what you have written and all the $U_n$ for $n > 7$ are the ones you don't see to the right of what you have written. Now just start taking differences and see what you get---there is a nice pattern.

4. Aug 28, 2014

### songoku

I still can't find the pattern. Do we have to make separate formula for n < 0 and n > 0?

I tried to set U0 = 0, at the middle of the positive and negative terms. What I got is 270 = 180 + 90 ; 450 = 270 + 180. Maybe it is related to Un = Un-1 + Un-2, but I don't know about 90 and 180.

Thanks

5. Aug 30, 2014

### SammyS

Staff Emeritus
The pattern is evident in the sequence as it's given, before you assign any starting point.

Hint: Make a difference table with 1st & 2nd differences.

6. Aug 31, 2014

### Ray Vickson

For the sequence
$$U = \{U_i,i=0,1,2, \ldots \} = \{-450 , -270 , -180 , -90 , 90 , 180 , 270 , 450 , \ldots\}$$
the sequence of first differences is
$$dU = \{U_{i+1}-U_{i}, i = 0,1,2,\ldots \} = \{ -270-(-450),-180-(-270), \ldots \} = \{ 180,90,90,180,90,90,\ldots \} \\ = 90 + 90 \{1,0,0,1,0,0,1,0,0,\ldots \}$$
If you can find a nice formula for the sequence $\{1,0,0,1,0,0,1,0,0,\ldots \}$ you are almost finished. If you know about complex variables and the three (complex) roots of unity, you can finish up with a finite sum of some geometric series, to get a fairly nice formula for the nth term of your original sequence.

7. Aug 31, 2014

### PeroK

Alternatively, you should be able to do something with n (mod3) to formulate the sequence of 1, 0, 0, 1, 0, 0 ...

8. Sep 2, 2014

### andrewkirk

If we are to be rigorous, we need to take care with our definitions in defining the pattern. Recursive definitions mostly use the Recursion Theorem, which is only defined on the positive integers $Z_+$. For more general well-ordered sets we can use the Theorem of Transfinite Recursive Definition, but that's not needed here.

We can't just use the Recursion Theorem here in a single step, because the base set is the full set of integers, which is not well-ordered. The following trick gets around this problem.

Define $a_1=90, b_1=-90$
Define rules to give $a_{n+1}$ and $b_{n+1}$ in terms of $a_n$ and $b_n$ as follows, for $n\in Z_+$:

$a_{n+1}=a_n+180$ if $n+1$ is divisible by 3, else $a_{n+1}=a_n+90$
$b_{n+1}=b_n-180$ if $n+1$ is divisible by 3, else $b_{n+1}=b_n-90$

Then the recursion theorem tells us that these definitions prescribe unique sequences $a_n$ and $b_n$.

Now we can just define $c:Z\to Z$ by $c(n)=a_{n+1}$ if $n\geq 0$, otherwise $c(n)=b_{-n}$.

Last edited: Sep 2, 2014
9. Sep 14, 2014

### songoku

I think I get the idea. I'll try it first. Thanks a lot for the help

10. Sep 25, 2014

### Simon Bridge

I'm always puzzled by these things - i.e. notice how the sequence is symmetrical about the -90,90 terms?
Why not exploit that? Put $a_1=90$, then $a_n = (n/|n|)a_{|n|}$ ... the sequence for $a_{|n|}$ should be easy ... though there are only four terms so you are spoiled for choice. {90, 180, 270, 450...} the differences than go {1, 1, 2, ...} in units of $a_1$, which is suggestive.

What approach you choose seems to depend on what assumptions you make about the pattern in question.


# Prove upper bound for recurrence

I am working on problem set 8 problem 3 from MIT's Fall 2010 OCW class 6.042J. This is covered in chapter 10 which is about recurrences.

Here is the problem:

$$A_0 = 2$$ $$A_{n+1} = A_n/2 + 1/A_n, \forall n \ge 1$$

Prove

$$A_n \le \sqrt2 + 1/2^n, \forall n \ge 0$$

I have graphed the recurrence and the upper bound and they seem to both converge on $\sqrt2$.

Also, if you ignore the boundary condition $A_0 = 2$ then you find that $\sqrt2$ is a solution to the main part of the recurrence. i.e. $\sqrt2 = \sqrt2/2 + 1/\sqrt2$.

The chapter and videos on recurrences have a lot to say about a kind of cookbook solution to divide and conquer recurrences which they call the Akra-Bazzi Theorem. But this recurrence does not seem to be in the right form for that theorem. If it were in the form $A_{n+1} = A_n/2 + g(n)$ then the theorem would give you an asymptotic bound. But $1/A_n$ is not a simple function of $n$ like a polynomial. Instead it is part of the recurrence.

Also, the chapter has a variety of things to say about how to guess the right solution and plug it into an inductive proof, but I haven't had much success. I have tried possible solutions of various forms like $a_n = \sqrt2+a/b^n$ and tried solving for the constants $a$ and $b$ but to no avail.

So, if someone can point me in the right direction that would be great. I always assume that the problem sets are based on something taught in the videos and in the text of the book but I am having trouble tracking this one down.

Bobby

We show by induction that $$\sqrt{2}\lt A_n\le \sqrt{2}+\frac{1}{2^n}.\tag{1}$$ Suppose the result holds at $n=k$. We show the result holds at $n=k+1$.

For the inequality on the right of (1), we need to show that $$\frac{A_k}{2}+\frac{1}{A_k}\le \sqrt{2}+\frac{1}{2^{k+1}}.$$ By the induction hypothesis, we have $$\frac{A_k}{2}+\frac{1}{A_k}\le \frac{\sqrt{2}}{2}+\frac{1}{2^{k+1}}+\frac{1}{\sqrt{2}}=\sqrt{2}+\frac{1}{2^{k+1}},$$ which takes care of the induction step for the inequality on the right of (1).

We still need to show that $\sqrt{2}\lt A_{k+1}$. Let $A_k=\sqrt{2}+\epsilon$, where $\epsilon$ is positive. Then $$A_{k+1}=\frac{\sqrt{2}+\epsilon}{2}+\frac{1}{\sqrt{2}+\epsilon}=\frac{4+2\sqrt{2}\epsilon+\epsilon^2}{2(\sqrt{2}+\epsilon)}\gt \frac{4+2\sqrt{2}\epsilon}{2(\sqrt{2}+\epsilon)}=\sqrt{2}.$$ This completes the induction step for the inequality on the left of (1).

Remark: The inequality (1) and squeezing show that $A_n$ indeed has limit $\sqrt{2}$.

• It didn't occur to me to prove that $A_n \ge \sqrt2$ and then substitute $\sqrt2$ in the $1/A_n$ term. Much easier that way. Thanks. – Bobby Durrett May 28 '16 at 2:36
• @BobbyDurrett: You are welcome. The inequality $A_n\gt \sqrt{2}$ is not mentioned explicitly in what you are asked to show, but when one tries to push the induction through, it becomes clear that it is necessary for the proof. – André Nicolas May 28 '16 at 2:40
• Nice induction for the $>\sqrt{2}$ part. I didn't manage it immediately, that's why I showed it studying the map $x\mapsto \frac{x}2+\frac{1}{x}$. – Daniel Robert-Nicoud May 28 '16 at 11:09

Let $$B_k = \frac{A_k}{\sqrt{2}}$$ Then $B_0 = \sqrt{2}$ and $$B_{n+1} = \frac12\left(B_n+\frac{1}{B_n} \right)$$ Thus $b_n$ is the $n$-th guess if you perform Newton's algorithm to try to find $\sqrt{1}$ starting with a guess of $\sqrt{2}$.

This recursion can be solved in closed form using the formula for the hyperbolic tangent of $2x$ in terms of $\tanh x$: $$\tanh(2x) = \frac{2 \tanh x}{1+\tanh^2{x}}$$ . The result looks something like $$B_n = \tanh\left( 2^n \theta\right)$$ where $\theta = \tanh^{-1} B_0$; the answer I have given is off in that every other term needs to be the reciprocal of what I wrote, but the general idea will work.

Once you have that, you can know exactly what $B_n$ is and prove the relation; or better yet, use the error analysis for Newton's method to get an estimate of how close to 1 you would be.

We want to apply induction, but we need also a lower bound on $A_n$ for the $\frac{1}{n}$ term. We can show that $A_n\ge\sqrt{2}$ as follows:

Let $f(x) = \frac{x}{2} + \frac{1}{x}$. Then $f'(x) = \frac{1}{2} - \frac{1}{x^2}$ which has a zero at $x = \sqrt{2}$ where $f(\sqrt{2}) = \sqrt{2}$, and is positive for $x>\sqrt{2}$. This shows that $f(x)\ge\sqrt{2}$ whenever $x\ge\sqrt{2}$, and as $A_{n+1} = f(A_n)$, we have that $A_n\ge\sqrt{2}$ for all $n\ge0$.

Then to conclude:

The case $n=0$ is trivially true.

For $n\ge0$ we have that $$A_{n+1} = \frac{A_n}{2} + \frac{1}{A_n}.$$ By induction hypothesis and what we have shown above, this has as upper bound $$A_{n+1}\le \frac{\sqrt{2} + 2^{-n-1}}{2} + \frac{1}{\sqrt{2}} = \sqrt{2} + 2^{-n-1}.$$

Three steps: Use the definition of $A_n$ and algebra to establish that $$A_{n+1}-\sqrt 2 = {(A_n-\sqrt 2)^2\over 2A_n}.\tag1$$ Next, use (1) to prove by induction that $$A_n\ge\sqrt 2\ \text{for every n.}\tag2$$ Finally, use (1) and (2) to prove by induction that $$A_n-\sqrt2 \le \frac1{2^{n}}\ \text{for every n.}\tag2$$

Your sequence is $$A_{n+1} = \frac{1}{2} A_n + \frac{1}{A_n}$$ where the last term features a division which reminds of the Newton-Raphson iteration.

Newton Raphson iteration takes the root of the tangent as next step for estimating a root of $f$: $$0 = T'(x_{n+1}) = f(x_n) + f'(x_n) (x_{n+1} - x_n) \iff \\ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ by comparison we have $$-\frac{1}{2}A_n + \frac{1}{A_n} = - \frac{A_n^2-2}{2A_n} = - \frac{f(A_n)}{f'(A_n)}$$ and see $f(x) = x^2 - 2$, the function used to iterate against the root $\sqrt{2}$.


# Finding the domain and range of a composite function

So I have two functions. $$f(x) = e^{-x^2+1}$$ and $$g(x)=\sqrt{x^2-4x+3}$$. I am then asked to determine the domain and range of

$$a)f∘g,$$

$$b)g∘f$$

I already did part $$a)$$ and the domain for part $$b)$$.

For part $$a)$$, the domain was $$(-\infty,1)\cup(3,\infty)$$ and the range was $$(0,e^2)$$.

For part $$b$$, I figured out that the domain was $$(-\infty,-1]\cup[1,\infty)$$. I am not sure how to find the range though. Normally, I would take the inverse of g∘f and find the domain of that, and although I can do it, I don't think I did it correctly.

Currently, I did figure out that $$g∘f$$ is $$\sqrt{e^{-2x^2+2}-4e^{-x^2+1}+3}$$. How do I find the range of this mess though? I attempted to take the inverse which I believe is:

$$y=\pm\sqrt{1-\ln(2\pm\sqrt{1+y^2})}$$

Although I know that Wolfram Alpha is not the arbitrator of what correct is, it's generally been right and my answer disagrees with what Wolfram alpha has obtained (As seen here). In addition, the range is something that I am not sure how Wolfram obtained (As seen here). This also looks REALLY messy.

Can anyone guide me as to how this was obtained? That would be much appreciated!

• I see how you got $(-\infty,1)\cup(3,\infty)$ (except maybe the choice of open/closed) but not how you got $e^2.$ When you were computing the range, did you forget the gap in the domain? I think that part is easier to solve if you do not work out the formula for $f\circ g.$ – David K Sep 11 '19 at 11:58
• I took the inverse of the function and found the domain of that to get the range. – Future Math person Sep 11 '19 at 18:23
• That seems like an unnecessarily difficult way to do it. Also error-prone, since (I think) it gives the wrong answer. What value of $x$ satisfies $f(g(x))=e^{3/2},$ for example? – David K Sep 11 '19 at 18:42
• Really? I didn't think it was too bad. I ended up getting $f(g(x))=-2 \pm \sqrt{2-\ln(x)}$. This would mean $2-\ln(x) \geq 0$ and thus $e^2 \geq x$ – Future Math person Sep 11 '19 at 18:44
• Also I get $g(1)=g(3)=0,$ and $0$ is in the domain of $f,$ so I would not exclude $1$ and $3$ from the domain in part a). I think you may need to be a lot more careful about checking what happens at your boundaries. (This is relevant to the range as well.) – David K Sep 11 '19 at 18:47

To find the range you want, namely of $$g(f(x))=F(x),$$ note that $$F(x)$$ is never negative. Also, it is defined and continuous at all real $$x.$$ Then it is an even function of $$x.$$ Thus, it suffices to consider only the range for positive $$x,$$ say.

Note that $$F(0)>0.$$ Also, as $$x\to +\infty,$$ we have $$F(x)\to \sqrt 3.$$ Thus, the range is at least $$[F(0),\sqrt 3),$$ by IVT. It only remains to see if the function attains $$\sqrt 3$$ at any point, or if it ever falls below $$F(0).$$ The first is easily answered in the negative (set $$F(x)=\sqrt 3$$ to deduce a contradiction), so the range is half-open. So does the function ever go below $$e^2-4e+3$$? In particular we may ask whether the function ever attains the minimum possible here, namely $$0.$$ Thus, setting $$F(x)=0,$$ we see that we want to see if there are real solutions to the quadratic equation $$p^2-4p+3=0,$$ where $$p=e^{-x^2+1}.$$ This has solutions. In particular, we get that $$e^{-x^2+1}=1,$$ or in other words that $$-x^2+1=0.$$ Thus $$F(x)$$ vanishes in at least two points.

It therefore is the case that the range is $$[0,\sqrt 3).$$

Note that the range of $$f$$ on $$(-\infty,-1]\cup[1,\infty)$$ is $$(0,1]$$. The range of $$g$$ on $$(0,1]$$ is $$[0,\sqrt{3})$$.

Let $$f,g$$ be given by $$\begin{cases} f(x)=e^{1-x^2}\\[4pt] g(x)=\sqrt{x^2-4x+3}\\ \end{cases}$$ $$\bullet\;\,$$Part $$(\mathbf{a}){\,:}\;\,$$Find the domain and range of $$f\circ g$$.

Since the domain of $$f$$ is $$\mathbb{R}$$, the domain of $$f\circ g$$ is the same as the domain of $$g$$.

Hence the domain of $$f\circ g$$ is $$(-\infty,1]\cup [3,\infty)$$.

Since $$f$$ is an even function, the range of $$f$$ is the same as the range of $$f$$ on the restricted domain $$[0,\infty)$$.

On the interval $$[0,\infty)$$, $$f$$ is strictly decreasing, and $$f$$ approaches zero from above as $$x$$ approaches infinity.

Since $$f$$ is continuous, it follows that the range of $$f$$ is $$(0,e]$$.

For $$x\ge 3$$, $$g$$ realizes all values in $$[0,\infty)$$, so the range of $$f\circ g$$ is the same as the range of $$f$$.

Hence the range of $$f\circ g$$ is $$(0,e]$$.

$$\bullet\;\,$$Part $$(\mathbf{b}){\,:}\;\,$$Find the domain and range of $$g\circ f$$.

The domain of $$g\circ f$$ is the set of all real $$x$$ such that $$f(x)$$ is in the domain of $$g$$, which is the set of all real $$x$$ such that $$f(x)\le 1$$ or $$f(x)\ge 3$$.

But we can't have $$f(x)\ge 3$$, since the maximum value of $$f$$ is $$e$$.

For the condition $$f(x)\le 1$$, we get $$f(x)\le1\iff e^{1-x^2}\le 1\iff 1-x^2\le 0\iff |x| \ge 1$$ hence the domain of $$g\circ f$$ is $$(-\infty,-1]\cup [1,\infty)$$.

Restricted to the domain $$(-\infty,-1]\cup [1,\infty)$$, the range of $$f$$ is $$(0,1]$$, hence the range of $$g\circ f$$ is the range of $$g$$ on the restricted domain $$(0,1]$$.

It follows that the range of $$g\circ f$$ is $$[0,\sqrt{3})$$.

$$\bullet\;\,$$To summarize the results:

• The domain of $$f\circ g$$ is:$$\;\,(-\infty,1]\cup [3,\infty)$$.
• The range of $$f\circ g$$ is:$$\;\,(0,e]$$.$$\\[6pt]$$
• The domain of $$g\circ f$$ is:$$\;\,(-\infty,-1]\cup [1,\infty)$$.
• The range of $$g\circ f$$ is:$$\;\,[0,\sqrt{3})$$.


Eigenvalue Inequalities for Hermitian Matrices

The eigenvalues of Hermitian matrices satisfy a wide variety of inequalities. We present some of the most useful and explain their implications. Proofs are omitted, but as Parlett (1998) notes, the proofs of the Courant–Fischer, Weyl, and Cauchy results are all consequences of the elementary fact that if the sum of the dimensions of two subspaces of $\mathbb{C}^n$ exceeds $n$ then the subspaces have a nontrivial intersection.

The eigenvalues of a Hermitian matrix $A\in\mathbb{C}^{n\times n}$ are real and we order them $\lambda_n\le \lambda_{n-1} \le \cdots \le \lambda_1$. Note that in some references, such as Horn and Johnson (2013), the reverse ordering is used, with $\lambda_n$ the largest eigenvalue. When it is necessary to specify what matrix $\lambda_k$ is an eigenvalue of we write $\lambda_k(A)$: the $k$th largest eigenvalue of $A$. All the following results also hold for symmetric matrices over $\mathbb{R}^{n\times n}$.

The function $f(x) = x^*Ax/x^*x$ is the quadratic form $x^*Ax$ for $A$ evaluated on the unit sphere, since $f(x) = f(x/\|x\|_2)$. As $A$ is Hermitian it has a spectral decomposition $A = Q\Lambda Q^*$, where $Q$ is unitary and $\Lambda = \mathrm{diag}(\lambda_i)$. Then

$f(x) = \displaystyle\frac{x^*Q\Lambda Q^*x}{x^*x} = \displaystyle\frac{y^*\Lambda y}{y^*y} = \displaystyle\frac{\sum_{i=1}^{n}\lambda_i y_i^2} {\sum_{i=1}^{n}y_i^2} \quad (y = Q^*x),$

from which is it clear that

$\notag \lambda_n = \displaystyle\min_{x\ne0} \displaystyle\frac{x^*Ax}{x^*x}, \quad \lambda_1 = \displaystyle\max_{x\ne0} \displaystyle\frac{x^*Ax}{x^*x}, \qquad(*)$

with equality when $x$ is an eigenvector corresponding to $\lambda_n$ and $\lambda_1$, respectively, This characterization of the extremal eigenvalues of $A$ as the extrema of $f$ is due to Lord Rayleigh (John William Strutt), and $f(x)$ is called a Rayleigh quotient. The intermediate eigenvalues correspond to saddle points of $f$.

Courant–Fischer Theorem

The Courant–Fischer theorem (1905) states that every eigenvalue of a Hermitian matrix $A\in\mathbb{C}^{n\times n}$ is the solution of both a min-max problem and a max-min problem over suitable subspaces of $\mathbb{C}^n$.

Theorem (Courant–Fischer).

For a Hermitian $A\in\mathbb{C}^{n\times n}$,

\notag \begin{aligned} \lambda_k &= \min_{\dim(S)=n-k+1} \, \max_{0\ne x\in S} \frac{x^*Ax}{x^*x}\\ &= \max_{\dim(S)= k} \, \min_{0\ne x\in S} \frac{x^*Ax}{x^*x}, \quad k=1\colon n. \end{aligned}

Note that the equalities $(*)$ are special cases of these characterizations.

In general there is no useful formula for the eigenvalues of a sum $A+B$ of Hermitian matrices. However, the Courant–Fischer theorem yields the upper and lower bounds

$\notag \lambda_k(A) + \lambda_n(B) \le \lambda_k(A+B) \le \lambda_k(A) + \lambda_1(B), \qquad (1)$

from which it follows that

$\notag \max_k|\lambda_k(A+B)-\lambda_k(A)| \le \max(|\lambda_n(B)|,|\lambda_1(B)|) = \|B\|_2.$

This inequality shows that the eigenvalues of a Hermitian matrix are well conditioned under perturbation. We can rewrite the inequality in the symmetric form

$\notag \max_k |\lambda_k(A)-\lambda_k(B)| \le \|A-B\|_2.$

If $B$ is positive semidefinite then (1) gives

$\notag \lambda_k(A) \le \lambda_k(A + B), \quad k = 1\colon n, \qquad (2)$

while if $B$ is positive definite then strict inequality holds for all $i$. These bounds are known as the Weyl monotonicity theorem.

Weyl’s Inequalities

Weyl’s inequalities (1912) bound the eigenvalues of $A+B$ in terms of those of $A$ and $B$.

Theorem (Weyl).

For Hermitian $A,B\in\mathbb{C}^{n\times n}$ and $i,j = 1\colon n$,

\notag \begin{aligned} \lambda_{i+j-1}(A+B) &\le \lambda_i(A) + \lambda_j(B), \quad i+j \le n+1, \qquad (3)\\ \lambda_i(A) + \lambda_j(B) &\le \lambda_{i+j-n}(A+B). \quad i+j \ge n+1, \qquad (4) \end{aligned}

The Weyl inequalities yield much information about the effect of low rank perturbations. Consider a positive semidefinite rank-$1$ perturbation $B = zz^*$. Inequality (3) with $j = 1$ gives

$\notag \lambda_i(A+B) \le \lambda_i(A) + z^*z, \quad i = 1\colon n$

(which also follows from (1)). Inequality (3) with $j = 2$, combined with (2), gives

$\notag \lambda_{i+1}(A) \le \lambda_{i+1}(A + zz^*) \le \lambda_i(A), \quad i = 1\colon n-1. \qquad (5)$

These inequalities confine each eigenvalue of $A + zz^*$ to the interval between two adjacent eigenvalues of $A$; the eigenvalues of $A + zz^*$ are said to interlace those of $A$. The following figure illustrates the case $n = 4$, showing a possible configuration of the eigenvalues $\lambda_i$ of $A$ and $\mu_i$ of $A + zz^*$.

A specific example, in MATLAB, is

>> n = 4; eig_orig = 5:5+n-1
>> D = diag(eig_orig); eig_pert = eig(D + ones(n))'
eig_orig =
5     6     7     8
eig_pert =
5.2961e+00   6.3923e+00   7.5077e+00   1.0804e+01

Since $\mathrm{trace}(A + zz^*) = \mathrm{trace}(A) + z^*z$ and the trace is the sum of the eigenvalues, we can write

$\notag \lambda_i(A + zz^*) = \lambda_i(A) + \theta_i z^*z,$

where the $\theta_i$ are nonnegative and sum to $1$. If we greatly increase $z^*z$, the norm of the perturbation, then most of the increase in the eigenvalues is concentrated in the largest, since (5) bounds how much the smaller eigenvalues can change:

>> eig_pert = eig(D + 100*ones(n))'
eig_pert =
5.3810e+00   6.4989e+00   7.6170e+00   4.0650e+02

More generally, if $B$ has $p$ positive eigenvalues and $q$ negative eigenvalues then (3) with $j = p+1$ gives

$\notag \lambda_{i+p}(A+B) \le \lambda_i(A), \quad i = 1\colon n-p,$

while (4) with $j = n-q$ gives

$\notag \lambda_i(A) \le \lambda_{i-q}(A + B), \quad i = q+1\colon n.$

So the inertia of $B$ (the number of negative, zero, and positive eigenvalues) determines how far the eigenvalues can move as measured relative to the indexes of the eigenvalues of $A$.

An important implication of the last two inequalities is for the case $A = I$, for which we have

\notag \begin{aligned} \lambda_{i+p}(I+B) &\le 1, \quad i = 1 \colon n-p, \\ \lambda_{i-q}(I+B) &\ge 1, \quad i = q+1 \colon n. \end{aligned}

Exactly $p+q$ eigenvalues appear in one of these inequalities and $n-(p+q)$ appear in both. Therefore $n - (p+q)$ of the eigenvalues are equal to $1$ and so only $\mathrm{rank}(B) = p+q$ eigenvalues can differ from $1$. So perturbing the identity matrix by a Hermitian matrix of rank $r$ changes at most $r$ of the eigenvalues. (In fact, it changes exactly $r$ eigenvalues, as can be seen from a spectral decomposition.)

Finally, if $B$ has rank $r$ then $\lambda_{r+1}(B) \le 0$ and $\lambda_{n-r}(B) \ge 0$ and so taking $j = r+1$ in (3) and $j = n-r$ in (4) gives

\notag \begin{aligned} \lambda_{i+r}(A+B) &\le \lambda_i(A), ~~\qquad\qquad i = 1\colon n-r, \\ \lambda_i(A) &\le \lambda_{i-r}(A + B), ~~\quad i = r+1\colon n. \end{aligned}

Cauchy Interlace Theorem

The Cauchy interlace theorem relates the eigenvalues of successive leading principal submatrices of a Hermitian matrix. We denote the leading principal submatrix of $A$ of order $k$ by $A_k = A(1\colon k, 1\colon k)$.

Theorem (Cauchy).

For a Hermitian $A\in\mathbb{C}^{n\times n}$,

$\notag \lambda_{i+1}(A_{k+1}) \le \lambda_i(A_k) \le \lambda_i(A_{k+1}), \quad i = 1\colon k, \quad k=1\colon n-1.$

The theorem says that the eigenvalues of $A_k$ interlace those of $A_{k+1}$ for all $k$. Two immediate implications are that (a) if $A$ is Hermitian positive definite then so are all its leading principal submatrices and (b) appending a row and a column to a Hermitian matrix does not decrease the largest eigenvalue or increase the smallest eigenvalue.

Since eigenvalues are unchanged under symmetric permutations of the matrix, the theorem can be reformulated to say that the eigenvalues of any principal submatrix of order $n-1$ interlace those of $A$. A generalization to principal submatrices of order $n-\ell$ is given in the next result.

Theorem.

If $B$ is a principal submatrix of order $n-\ell$ of a Hermitian $A\in\mathbb{C}^{n\times n}$ then

$\notag \lambda_{i+\ell}(A) \le \lambda_i(B) \le \lambda_i(A), \quad i=1\colon n-\ell.$

Majorization Results

It follows by taking $x$ to be a unit vector $e_i$ in the formula $\lambda_1 = \max_{x\ne0} x^*Ax/(x^*x)$ that $\lambda_1 \ge a_{ii}$ for all $i$. And of course the trace of $A$ is the sum of the eigenvalues: $\sum_{i=1}^n a_{ii} = \sum_{i=1}^n \lambda_i$. These relations are the first and last in a sequence of inequalities relating sums of eigenvalues to sums of diagonal elements obtained by Schur in 1923.

Theorem (Schur).

For a Hermitian $A\in\mathbb{C}^{n\times n}$,

$\notag \displaystyle\sum_{i=1}^k \lambda_i \ge \displaystyle\sum_{i=1}^k \widetilde{a}_{ii}, \quad k=1\colon n,$

where $\{\widetilde{a}_{ii}\}$ is the set of diagonal elements of $A$ arranged in decreasing order: $\widetilde{a}_{11} \ge \cdots \ge \widetilde{a}_{nn}$.

These inequalities say that the vector $[\lambda_1,\dots,\lambda_n]$ of eigenvalues majorizes the ordered vector $[\widetilde{a}_{11},\dots,\widetilde{a}_{nn}]$ of diagonal elements.

An interesting special case is a correlation matrix, a symmetric positive semidefinite matrix with unit diagonal, for which the inequalities are

$\notag \lambda_1 \ge 1, \quad \lambda_1+ \lambda_2\ge 2, \quad \dots, \quad \lambda_1+ \lambda_2 + \cdots + \lambda_{n-1} \ge n-1,$

and $\lambda_1+ \lambda_2 + \cdots + \lambda_n = n$. Here is an illustration in MATLAB.

>> n = 5; rng(1); A = gallery('randcorr',n);
>> e = sort(eig(A)','descend'), partial_sums = cumsum(e)
e =
2.2701e+00   1.3142e+00   9.5280e-01   4.6250e-01   3.6045e-04
partial_sums =
2.2701e+00   3.5843e+00   4.5371e+00   4.9996e+00   5.0000e+00

Ky Fan (1949) proved a majorization relation between the eigenvalues of $A$, $B$, and $A+B$:

$\notag \displaystyle\sum_{i=1}^k \lambda_i(A+B) \le \displaystyle\sum_{i=1}^k \lambda_i(A) + \displaystyle\sum_{i=1}^k \lambda_i(B), \quad k = 1\colon n.$

For $k = 1$, the inequality is the same as the upper bound of (1), and for $k = n$ it is an equality: $\mathrm{trace}(A+B) = \mathrm{trace}(A) + \mathrm{trace}(B)$.

Ostrowski’s Theorem

For a Hermitian $A$ and a nonsingular $X$, the transformation $A\to X^*AX$ is a congruence transformation. Sylvester’s law of inertia says that congruence transformations preserve the inertia. A result of Ostrowski (1959) goes further by providing bounds on the ratios of the eigenvalues of the original and transformed matrices.

Theorem (Ostrowski).

For a Hermitian $A\in \mathbb{C}^{n\times n}$ and $X\in\mathbb{C}^{n\times n}$,

$\lambda_k(X^*AX) = \theta_k \lambda_k(A), \quad k=1\colon n,$

where $\lambda_n(X^*X) \le \theta_k \le \lambda_1(X^*X)$.

If $X$ is unitary then $X^*X = I$ and so Ostrowski’s theorem reduces to the fact that a congruence with a unitary matrix is a similarity transformation and so preserves eigenvalues. The theorem shows that the further $X$ is from being unitary the greater the potential change in the eigenvalues.

Ostrowski’s theorem can be generalized to the situation where $X$ is rectangular (Higham and Cheng, 1998).

Interrelations

The results we have described are strongly interrelated. For example, the Courant–Fischer theorem and the Cauchy interlacing theorem can be derived from each other, and Ostrowski’s theorem can be proved using the Courant–Fischer Theorem.

2 thoughts on “Eigenvalue Inequalities for Hermitian Matrices”

1. Daniel Kressner says:

This is a very nice overview; thanks! Tangential to what Parlett is stating, most if not even all of these results can be concluded from the Eckart–Young–Mirsky theorem in the spectral norm.


# The Comparison Test for Series

I'm doing a homework problem - it asks us to show whether the series: $$\sum_{n=1}^{\infty}\frac{n^n}{n!}$$ converges or diverges. I looked at a graph of the sequence component $\big(\frac{n^n}{n!} \big)$ and saw it continued to increase. I then considered the sequence: $$\sum_{n=1}^{\infty}\frac{1}{n}$$ which diverges by the P test.

But, $$\frac{1}{n}\leq \frac{n^n}{n!},~ \forall~n\geq1$$ which would then mean that the first series I showed diverges by the comparison test.

My problem is that this seems too simple? Can I compare one series to any series or does the comparison series have to meet some certain requirements (besides those I've addressed).

Cheers.

• Looks fine to me. – Jacky Chong Oct 12 '16 at 6:15
• You could also do comparison test with the divergent series $\sum_{n=1}^\infty 1$. – angryavian Oct 12 '16 at 6:16
• That seems so simple, so I do not need to compare a series with changing exponents and factorials to a similar one? – Wharf Rat Oct 12 '16 at 6:17
• Is the necessary condition for convergence ($n^n/n!\to 0$) satisfied here? – A.Γ. Oct 12 '16 at 6:20

Your reasoning is fine. In fact, this can be shown to diverge with the simplest / most intuitive test there is: If $\ \displaystyle \sum_{n=1}^\infty a_n$ is a convergent series, it is a necessary condition that $\displaystyle \lim_{n \rightarrow \infty} a_n = 0$. In this scenario, $n^n \geq n!$ for all $n \geq 1$, implying $\displaystyle \frac{n^n}{n!} \geq 1$ for all such $n$, so this necessary condition does not hold.

I looked at a graph of the sequence component $\big(\frac{n^n}{n!} \big)$ and saw it continued to increase.
If there series converges then $n^n/n! \to 0$, but if you use Stirling's approximation: $n! \approx \sqrt{2\pi n} \cdot \left(\frac{n}{e}\right)^n$
$$\lim_{n} \frac{n^n}{n!} = \lim_n \frac{n^n}{\sqrt{2 \pi n} \cdot\dfrac{n^n}{e^n}} = \lim_n \frac{e^n}{\sqrt{2 \pi n}} = \frac{1}{\sqrt{2\pi}} \lim_n \frac{e^n}{\sqrt{n}} = \frac{1}{\sqrt{2\pi }} \lim_n \left(\frac{e^{2n}}{n}\right)^{\frac{1}{2}} \to \infty$$
• No need for Stirling here, we have $$\frac{n^n}{n!}=\frac{n\cdot n\cdots n}{1\cdot2\cdots(n-1)\cdot n}>\frac{n\cdot 2\cdots n}{1\cdot2\cdots(n-1)\cdot n}=n$$ – AD. Oct 12 '16 at 8:04


# Integrating two equations that equal, what happens to the constant on one of the sides?

In class, we were talking about Newton's 3rd law and how to integrate.

$\int(g)dt = \int(y''(t))dt \implies g(t) + C = y'(t)$

I am confused about why the right hand side of the equation doesn't get a constant. After asking the professor, he said that it was because the two constants would cancel each other out. But I still don't understand why that should prevent us from writing a constant on the right side.

• Recall, you can add constants together into a single constant (as was done with $C$). Also, you can define a $C_1$ and $C_2$ - one to each side. You can also show as being on the RHS of the equation. All will produce the same result. The choice is typically one of convenience to make solving easiest. – Amzoti Jan 16 '14 at 18:13
• As an example, take $y' = x y$. Solve it with a constant on one side after separation and integration. Then, have a constant on each side. What do you notice when you solve for $y$ in both approaches? $y = c e^{x^2/2}$. – Amzoti Jan 16 '14 at 18:21
• I think I understand, thank you! – user121860 Jan 16 '14 at 18:29
• I haven't taken calculus any calculus courses for about two years now and I've forgotten some of the rules, such as combining the constants. The professor told me that when we integrated both constants would be C1, ie. the same, and as such would end up canceling each other out by way of subtraction if we wanted to integrate more. So he wrote C1 on the left hand side, but it seems to make sense if he actually wanted to write C (the inclusion of both constants). – user121860 Jan 16 '14 at 18:33
• Though if you don't mind, examples are always helpful. Thank you for your time and help! – user121860 Jan 16 '14 at 18:33

Recall, you can add constants together into a single constant (as was done with C).

Also, you can define a $C_1$ and $C_2$ - one to each side. You can also show the constant as being on the RHS of the equation. All will produce the same result. The choice is typically one of convenience to make solving easiest.

Example: Consider the separable equation:

$$y' = x y$$

After separation, we can integrate both sides as:

$$\int \dfrac{1}{y}~dy = \int x~dx$$

Approach 1: Single constant (we could have also put $C$ on the LHS - try this)

$$\ln y = \dfrac{x^2}{2} + C$$

We take the exponential of each side and have:

$$y = e^{x^2/2 + C} = e^{x^2/2}~e^C = w~e^{x^2/2}$$

Note: $w = e^C$, which is just some constant (totally arbitrary).

Approach 2: Constant on each side

$$\ln y + c_1 = \dfrac{x^2}{2} + c_2$$

Taking exponential of both sides:

$$e^{\ln y + c_1} = e^{x^2/2 + c_2}$$

The RHS is as above and the LHS, we have ($q$ is just any arbitrary constant):

$$e^{\ln y}~e^{c_1} = y~q$$

Now we write:

$$q y = c_2~e^{x^2/2}$$

Dividing constants and just calling it $w$, we have:

$$y(x) = w~e^{x^2/2}$$

In both approaches, we just get some constant $w$.

Now, if you were provided with an initial condition like $y(0) = 2$, you would plug in $x=0$ and see that $w = 2$.


# Maximise population coverage subject to budget constraint

Let $t_i =$

$1$ if transmitter i is to be constructed

and $0$ otherwise,

$c_j =$

$1$ if community j is covered

and $0$ otherwise.

Obj func:

Max

$$z = [10, 15, ..., 10] \cdot c$$

s.t.

1. the budget constraint

$$[3.6, 2.3, ..., 3.10] \cdot t \le 15$$

1. If community $j$ is covered, it is done so by at least one constructed transmitter $i$:

eg if $c_1$ is covered, then $t_1$ or $t_3$ is constructed:

$$c_1 \to (t_1 \bigvee t_3)$$

$$\iff \neg c_1 \bigvee (t_1 \bigvee t_3)$$

$$\iff 1 - c_1 + t_1 + t_3 \ge 1$$

$$\iff c_1 \le t_1 + t_3$$

Similarly, we have:

$$c_2 \le t_1 + t_2$$

$$\vdots$$

$$c_{15} \le t_7$$

1. If a transmitter $i$ is constructed, at least one community $j$ is covered:

eg if $t_1$ is constructed, then $c_1$ and $c_2$ are covered:

$$t_1 \to (c_1 \bigwedge c_2)$$

$$\iff \neg t_1 \bigvee (c_1 \bigwedge c_2)$$

$$\iff (\neg t_1 \bigvee c_1) \bigwedge (\neg t_1 \bigvee c_2)$$

$$\iff 1 - t_1 + c_1 \ge 1 \ \text{and} \ 1 - t_1 + c_2 \ge 1$$

$$\iff c_1 \ge t_1 \ \text{and} \ c_2 \ge t_1$$

Similarly, we have:

$$c_2, c_3, c_5 \ge t_2$$

$$c_1, c_7, c_9, c_{10} \ge t_3$$

$$\vdots$$

$$c_{12}, c_{13}, c_{14}, c_{15} \ge t_7$$

Is that right?

From Chapter 3 here.

• Note if you choose transmitters $1, 2$, then there is overlap of communities, and your function will double count the population of community $2$. – Macavity Mar 15 '16 at 5:05
• @Macavity Oh thanks. What to do then? $z'=z-y \cdot x$ where $y_i$ corrects for double or triple counting? – BCLC Mar 15 '16 at 5:10
• @Macavity I think Kuifje found away around the double count. Do you agree? – BCLC Mar 17 '16 at 17:54

Here is a way to get the constraints right. Define $x_i$ exactly as you have done. Define $p_i$ to be an indicator $1/0$ depending on whether community $i$ has been covered or not.

The budget constraint remains as you have set, and the objective function is now of form $\max z = 10p_1+15p_2 + \cdots + 10p_{15}$

Now how do we ensure that $p_i$ is set correctly, for any given choice of $\{x_i\}$? One way is to note that the objective function gives positive weight to $p_i$, so define a constraint for each population based on transmitters which could cover it - e.g. for the first one : $p_1 \le x_1+x_3$, for the second $p_2 \le x_1 + x_2$ etc. This will force the population indicator to turn $0$ if none of the relevant transmitters are selected.

• Thanks Macavity. Edited. How is it? – BCLC Mar 15 '16 at 6:08
• Your $p_i$ needs to be replaced by $c_i$ in the objective function and $x_i$ has no role to play in the objective. – Macavity Mar 15 '16 at 6:10
• Thanks. We answered in class, and we sort of got different answers. Prof is still going to think about it. How do you find sol? – BCLC Mar 15 '16 at 11:31
• Using an $M$ is also a good way, though not needed here. Try out the Integer Programs and see how it works. – Macavity Mar 15 '16 at 11:43
• @BCLC Yes, your try 4 seems what I meant, and it works. – Macavity Mar 20 '16 at 18:31

This is a classical set covering problem.

Let $y_i$ be a binary variable that equals $1$ if and only if transmitter $i=1,\cdots,7$ is built, and $x_j$ another binary that equals $1$ if community $j$ is covered. Let $p_j$ be the population of community $j=1,\cdots,15$.

You want to maximize the total coverage: $$\mbox{Maximize }Z=\sum_{j=1 }^{15}p_jx_j$$ subject to budget constraints: $$\sum_{i=1}^7c_iy_i\le 15 \\$$ To link the variables, proceed like Macavity suggests: $$x_j\le \sum_{i\;|\;i \;covers\; j}y_i\quad \forall j=1,\cdots,15\\ y_i,x_j\in \{0,1\}\quad \forall i=1,\cdots,7\;\forall j=1,\cdots,15$$

• Sorry I was editing while you asked the question. Variables $x_j$ will make sure there is no double counting, and are activated as soon as one of the variables $y_i\; |\;i$ covers $j$ is. – Kuifje Mar 17 '16 at 18:03
• No apologies needed. I'm the one asking a favour. It's really 'Try 2' and not 'Try 1' ? My prof seems to disagree with Macavity – BCLC Mar 17 '16 at 18:14
• What do you mean by 'Try 1' and 'Try 2'? I think Macavity is right. He basically explained the nature of my second constraint. What specific point does your professor disagree with? – Kuifje Mar 17 '16 at 18:25
• I disagree with "Try1". Try 1 (first equation) implies that if communities $1$ and $2$ are covered, than transmitter $1$ is built. This is not necessarily true, as they can be covered by transmitters $3$ and $2$ for example. It is the other way around: if community $1$ is covered, than at least one transmitter among those that cover the community must be built. – Kuifje Mar 18 '16 at 0:34
• yes indeed, for the same reasons. – Kuifje Mar 18 '16 at 14:30


# Why does factoring eliminate a hole in the limit?

$$\lim _{x\rightarrow 5}\frac{x^2-25}{x-5} = \lim_{x\rightarrow 5} (x+5)$$

I understand that to evaluate a limit that has a zero ("hole") in the denominator we have to factor and cancel terms, and that the original limit is equal to the new and simplified limit. I understand how to do this procedurally, but I'd like to know why this works. I've only been told the methodology of expanding the $x^2-25$ into $(x-5)(x+5)$, but I don't just want to understand the methodology which my teacher tells me to "just memorize", I really want to know what's going on. I've read about factoring in abstract algebra, and about irreducible polynomials (just an example...), and I'd like to get a bigger picture of the abstract algebra in order to see why we factor the limit and why the simplified is equal to the original if it's missing the $(x-5)$, which has been cancelled. I don't want to just memorize things, I would really like to understand, but I've been told that this is "just how we do it" and that I should "practice to just memorize the procedure."
I really want to understand this in abstract algebra terms, please elaborate. Thank you very much.

• Very laudable that you want to understand instead of "just memorize". Aug 7, 2013 at 19:21
• The expression $\frac{(x+5)(x-5)}{x-5}=x+5$, just so long as $x\neq 5$, since $\frac{(x+5)(x-5)}{x-5}$ has the bad fortune of not being defined at $x=5$. Aug 7, 2013 at 19:29
• I don't understand what your question is. Are you asking why the function $\frac{x^2-25}{x-5}$ is undefined at $x=5$? or why we're allowed to cancel $(x-5)$ from the top and bottom? or why the limits before and after cancellation are equal? Aug 7, 2013 at 20:09
• @BlueRaja-DannyPflughoeft yes, all of those questions I'm asking :) Aug 7, 2013 at 22:35
• To answer the question - I've only been told the methodology of expanding the x2−25 into (x−5)(x+5). It goes like this: (in reverse) (x-5)(x+5) = x(x+5)-5(x+5) = x^2+5x-5x-25 = x^2-25 So, working above the other way takes x^2-25 back to (x-5)(x+5) Aug 8, 2013 at 22:52

First, and by definition, when dealing with

$$\lim_{x\to x_0}f(x)$$

we must assume $\,f\,$ is defined in some neighborhood of $\,x_0\,$ except , perhaps, on $\,x_0\,$ itself, and from here that in the process of taking the limit we have the right and the duty to assume $\,x\,$ approaches $\,x_0\,$ in any possible way but it is never equal to it.

Thus, and since in our case we always have $\,x\ne x_0=5\,$ during the limit process , we can algebraically cancel for the whole process.

$$\frac{x^2-25}{x-5}=\frac{(x+5)\color{red}{(x-5)}}{\color{red}{x-5}}=x+5\xrightarrow[x\to 5]{}10$$

The above process shows that the original function behaves exactly as the straight line $\,y=x+5\,$ except at the point $\,x=5\,$ , where there exists "a hole", as you mention.

• said hole can, of course be filled in by plugging it with the limit ... Aug 13, 2013 at 22:33
• Some definitions of the limit do not require that we consider points that are not equal to $x_0$. How do you handle these cases then? The definition goes as $\lim f(x) = L$ provided for each $\epsilon\gt 0$ there is some $\delta \gt 0$ such that $\vert x - x_0\vert \lt \delta\ldots$ and so on. Sep 9, 2016 at 15:50
• @AlexOrtiz The epsilon delta definition of the limit for general metric spaces has $0 < d(x,c) < \delta$. Note the first inequality, which implies $x \neq c$. Nov 21, 2020 at 0:19

This image of mine seems apropos:

In the case of $\lim_{x\to 5} \frac{x^2-25}{x-5}$, the message here is: Away from $x=5$, the function $\frac{x^2-25}{x-5}$ is completely identical to $x+5$; thus, what we expect to find as we approach $x=5$ is the value $5+5$. This anticipated value is what a limit computes.

The fact that the original function isn't defined at $x=5$ is immaterial. Walley World may be closed for repairs when you arrive, but that doesn't mean you and your dysfunctional family didn't spend an entire cross-country road trip anticipating all the fun you'd have there.

• Extremely helpful answer as always, Blue. You're incredible at understanding what people are really asking and providing strong intuition. Thank you for your contribution to this site! Jun 1, 2015 at 15:55
• @Blue Your answer made me think something: Can we always find another function which - away from a certain value - is completely identical to the former function? I mean, what is done in this example is the following: We have an algebraic expression and we can find another algebraic expression that have that property. Jul 6, 2017 at 1:21
• BTW: I read your image as the jingles in this video. Jul 6, 2017 at 1:21
• Super answer...the craziest I ever read in here Jul 13, 2019 at 22:17
• @Blue: Shameless plug: A downloadable poster version of the "Journey/Destination" image is available at my Etsy shop, and a t-shirt is available from Spring. (I'll delete this comment if there are objections.)
– Blue
Oct 12, 2021 at 10:00

Let's consider a simpler example first. Consider the function $f(x) = \frac{2x}x$. This says you take some number $x$, multiply by 2, then divide by the original number $x$. Obviously the answer is always 2, right? Except that when $x$ is zero, the division is forbidden and there is no answer at all. But for every $x$ except 0, we have $\frac{2x}x = 2$. In particular, for values of $x$ close to, but not equal to 0, we have $\frac{2x}x = 2$.

The function $\frac{x^2-25}{x-5}$ is similar, just a little more complicated. Calculating $x^2-25$ always gives you the same as $(x-5)(x+5)$. That is, if you take $x$, square it, and subtract 25, you always get the same number as if you take $x$, add 5 and subtract 5, and then multiply the two results. So we can replace $x^2-25$ with $(x+5)(x-5)$ because they always give the same number regardless of what you start with; they are two ways of getting to the same place. And then we see that $$\frac{x^2-25}{x-5} = \frac{(x+5)(x-5)}{x-5} = x+5$$

except that if $x-5$ happens to be zero (that is, if $x=5$) the division by zero is forbidden and we get nothing at all. But for any other $x$ the result of $\frac{x^2-25}{x-5}$ is always exactly equal to $x+5$. In particular, for values of $x$ close to, but not equal to 5, we have $\frac{x^2-25}{x-5} = x+5$.

The limit $$\lim_{x\to 5} \ldots$$ asks what happens to some function when $x$ close to, but not exactly equal to 5. And while this function is undefined for $x=5$, because to calculate it you would have to divide by zero, it is perfectly well-behaved for other values of $x$, and in particular for values of $x$ close to 5. For values of $x$ close to 5 it is equal to $x+5$, and so for values of $x$ close to 5 it is close to 10. And that is exactly what the limit is calculating.

I think that what confuses you is the difference between "solving the algebraic expression", and "finding the limit". Given:

$$f_1=\frac{x^2-25}{x-5} \quad f_2 = (x+5)$$

Then, $f_1$ and $f_2$ are most definitely NOT the same function. This is because they have different domains: 5 is not a member of the domain of $f_1$, but it is in the domain of $f_2$.

However, when we go from: $$\lim _{x\rightarrow 5}\frac{x^2-25}{x-5} \quad to \quad \lim _{x\rightarrow 5}\frac{(x-5)(x+5)}{x-5} \quad to \quad \lim_{x\rightarrow 5} (x+5)$$

We are not saying that the expressions inside the limits are equal; maybe they are, maybe they are not. What we are saying that they have the same limit. Totally different statement.

Above, the transformation of the second expression to the third one allows us to find a different function for which a) we know that the limit is the same, and b) we know how to trivially calculate that limit.

The big question, then: what transformations can I make to the function $f_1$ so that the limit stays the same? I think this is usually poorly explained in introductory courses -- a lot of hand-waving going on.

Obviously you can do any algebraic manipulation that leaves $f_1$ unchanged. You can also make any manipulation that removes and/or introduces discontinuities (points for which the function does not exist), as long as the new function stays continuous for an arbitrarily small neighborhood around $a$ (except possibly at $a$ itself). Your example is a case of such a transformation.

Here I'm myself cheating because I'm not defining 'continuity' for you. I'm sorry; please use an intuitition of what continuous means ("no holes, no jumps"), until you are presented with a formal one.

More complex transformations exist, but they have to be justified individually. You'll get to them eventually.

• you clarified a lot, thanks Aug 8, 2013 at 20:08
• Sixth line from the end - "You" will be "Your".
– MrAP
Jul 5, 2017 at 19:18
• @MrAP, thanks. You should feel free to post an edit directly Jul 6, 2017 at 1:04

Here's the basic idea: You're given a rational function, $f\colon \Bbb R \setminus \{5\}\to \Bbb R$, which is continuous everywhere in its domain.

You want to find the limit of that function at $5$.

One way to do that is to construct a continuous extension of that function, $g\colon \Bbb R \to \Bbb R$, such that $g(x) = f(x)$ whenever $x$ is in the domain of $f$. Then $$\lim_{x\to 5}f(x) = \lim_{x\to 5} g(x) = g(5).$$

In this case, factoring and cancelling accomplishes that objective.

• i really like this idea of a "continuous extension", what is the formal name of this rule/theorem? I'd like to read more about it. Thanks for the answer @dfeuer Aug 8, 2013 at 20:07
• @user4150: Try this google search: "removable discontinuity" "continuous extension" Aug 8, 2013 at 21:24
• @dfeuer Typo? It should be $$\lim_{x\to 5}f(x) = \lim_{x\to 5} g(x) = g(\color{royalblue}5).$$ Oct 1, 2014 at 8:06
• @Hakim, right you are. Fixed. Oct 1, 2014 at 9:55

The chill pill you are looking for is

If $f_1$ = $f_2$ except at $a$ then $\lim _{x{\rightarrow}a}f_1(x)=\lim_{x{\rightarrow}a}f_{2}(x)$

• Yes! You made my day! Apr 14, 2016 at 16:42
• Assuming both function to be continuous at $x=a$. May 26, 2017 at 11:45

Intuitively, we can start the other way round, by simply considering $\lim_{x\rightarrow 5} (x+5)$ which we're all agreed we understand. Now consider, independently,

$$\frac{x-5}{x-5}$$

this is obviously 1 everywhere, except where it's undefined at $x = 5$. So, what would you expect to be the effect of multiplying the two? It's just multiplying by 1, except that it also introduces a hole at $x = 5$

To understand this more broadly, it is convenient to check L'Hôpital's rule, which basically boils down to this:

Given a function $f(x)=\frac{g(x)}{h(x)}$, where for some $x=x_0$ both $g(x_0)=0$ and $h(x_0)=0$, the actual value $f(x_0)$ can be obtained as*

$$\lim_{x\to x_0}f(x) = \lim_{x\to x_0}\frac{g'(x)}{h'(x)}$$

(where the prime denotes derivation by $x$) Note that if $g'(x_0)=0=h'(x_0)$, you can re-apply the same rule again and again until either numerator or denominator are not 0. In your example, we get

$$\lim_{x\to 5}\frac{x^2-25}{x-5} = \lim_{x\to 5}\frac{2x}{1}=10=\lim_{x\to5}\,(x+5)$$

To get an even deeper understanding, the rule's proof might be an interesting read.

*: This is only one case, you can also have $|g(x_0)|=|h(x_0)|=\infty$, but not $g(x_0)=0$ and $f(x_0)=\pm\infty$

• I don't think it's appropriate to bring up L'Hôpital in a question like this. To justify the rule, the reader needs to be familiar with derivatives, differentiability, the sandwich theorem and Cauchy's mean value theorem just to get started. The OP is asking for understanding, not higher-level magic hand-waving. L'Hôpital's rule is a great practical tool, but it's also wielded too often as a substitute for thinking. Aug 8, 2013 at 11:53
• @EuroMicelli Woah there, I didn't intend to hand-wave magically, I just wanted to provide a different answer that, if one is interested enough, can not only explain the removable singularities in polynomial fractions but also in more complicated cases. Though I agree one shouldn't blindly apply Hôpital to just anything if there is another maybe more elegant way Aug 8, 2013 at 13:04
• Also you don't obtain the $f(x_0)$ value - you get $f(x_0)$ value provided it is defined and $f$ is continous. The OP function is not defined at $5$ but it's limit is. Aug 9, 2013 at 7:05
• @MaciejPiechotka True, stricly speaking the example function is equivalent to $$f(x) = \begin{cases} x+5 & x\neq 5 \\ \text{undefined} & x = 5\end{cases}$$ but the singularity at $x=5$ is removable as in $$f(x)= \begin{cases}\tfrac{x^2-25}{x-5} & x\neq 5 \\ 10 & x=5 \end{cases} \Rightarrow f(x) \equiv x+5$$ But you're right that one should always keep removed singularities in mind as they may have severe influence to e.g. applications in Physics (though I can't tell one ad hoc) Aug 9, 2013 at 7:10

One of definitions of $\lim_{x \to A} f(x) = B$ is:

$$\forall_{\varepsilon > 0}\exists_{\delta > 0}\forall_{0 < \left|x - A\right| < \delta}\left|f(x) - B\right| < \varepsilon$$

The intuition is that we can achieve arbitrary 'precision' (put in bounds on y axis) provided we get close enough (so we get the bounds on x axis). However the definition does not say anything about the value at the point $f(A)$ which can be undefined or have arbitrary value.

One of method of proving the limit is to find the directly $\delta(\varepsilon)$. Hence we have following formula (well defined as $x\neq 5$):

$$\forall_{0 < \left|x - 5\right| < \delta}\left|\frac{x^2-25}{x-5} - 10\right| < \epsilon$$

As $x\neq 5$ (in such case $\left|x - 5\right| = 0$) we can factor the expression out

$$\forall_{0 < \left|x - 5\right| < \delta} \left|x + 5 - 10\right| < \varepsilon$$ $$\forall_{0 < \left|x - 5\right| < \delta} \left|x - 5 \right| < \varepsilon$$

Taking $\delta(\varepsilon) = \varepsilon$ we find that:

$$\forall_{\varepsilon > 0}\exists_{\delta > 0}\forall_{0 < \left|x - 5\right| < \delta}\left|\frac{x^2-25}{x-5} - 10\right| < \varepsilon$$

The key thing is that we don't care about value at the limit.

Look at this example:

$$\lim _{\text{thing} \to zero}(\dfrac{\text{Thing}}{\text{Thing}}) \cdot \text{Another thing}$$

Here our $\text{'Thing'}$ is tending to zero, but not zero, it is something real, call it a real $\text{Thing}$, which can be cancelled merrily.

• What do you have against using letters as variables? Mar 8, 2014 at 18:54

The hole doesn't always disappear but when it does it is because one expression is equivalent to another except right at that hole. For example $$\frac{x^2 - 25}{x - 5}= \frac{(x+5)(x-5)}{x-5}= (x + 5)$$ everywhere except at x = 5. However, if we are very close to 5 and we had infinite place arithmetic (as we do theoretically) the calculation of the first expression $\frac{x^2 - 25} {x - 5}$ would be very close to the last expression $(x+5)$, that is the limit as x approaches 5 of the first expression is the same limit as the limit as x approaches 5 of last expression. Since that latter is a 'nice number', we can define the value at 5 for the first expression and remove the "everywhere except at $x = 5$" by just putting 10 as the value and not calculating it.

The trick is to think of limit a bit differently: A limit is the value a function would have at a point were it continuous at that point, with nothing else being different.

You can see this by comparing the epsilon-delta definitions of the two concepts. We say $$f$$ is continuous at $$c$$, if $$f$$ is defined at $$c$$ and for every $$\epsilon > 0$$ there is a $$\delta > 0$$ such that

$$|f(x) – f(c)| < \epsilon$$ whenever $$0 < |x – c| < \delta$$.

Likewise, $$f$$ has limit $$L$$ at $$c$$ if for every $$\epsilon > 0$$ there is a $$\delta > 0$$ such that

$$|f(x) – L| < \epsilon$$ whenever $$0 < |x – c| < \delta$$.

Thus, if we have a function with isolated gaps and there is a different function $$f^{*}$$ such that

1. $$f^{*}(x) = f(x)$$ wherever $$f(x)$$ is defined and continuous,
2. $$f^{*}$$ is defined and continuous everywhere, i.e. has domain $$\mathbb{R}$$ (versus the original function having domain $$\mathbb{R}$$ minus a set of isolated points),

then we must have that the value this $$f^{*}$$ takes at the points where $$f$$ was not defined or continuous must be the limiting values at those points.

And that's what you find by cancellation: by rewriting the expression describing the function in a form that involves only addition and multiplication with no division, you have created a formula that describes a function that must agree where the division part is defined, but since no division is involved, is defined everywhere. Moreover, since addition and multiplication (and powers, though you can consider them repeated multiplication) are continuous, then it follows that this function must be continuous as well. Hence the above are met, and the holes fill in.

The idea is that if two functions $f$ and $g$ differ only at $a$ but are identical otherwise, we have $\displaystyle \lim_{x \to a} f(x) = \lim_{x \to a} g(x)$. In this case $f(x) = \dfrac {x^2-25}{x-5}$ and $g(x) = x+5$ are identical everywhere except $x=5$, so $\displaystyle \lim_{x \to 5} \dfrac {x^2-25}{x-5} = \lim_{x \to 5} x+5 = 10.$

In limits the value of x is always tending and not exact

x→a means the value of x tends to a. Similarly x→5 means tending to 5. So basically the value of x is not 5, but a little more or less than 5.

Hence you cannot take the denominator 0 in any case as the denominator will be, again, tending to zero. That is why we try to factorize the numerator first in order to check if there are any terms common both on the numerator and the denominator.

$$L:=\lim_{x\to x_0}f(x)$$ is the value that would make $$f$$ continuous at $$x_0$$.

If you known another function $$g$$ which is continuous and coincides with $$f$$ around $$x_0$$, then perforce

$$L=g(x_0).$$


# HP Forums

You're currently viewing a stripped down version of our content. View the full version with proper formatting.
.
InvH:

« .577215664902 - InvPsi 1. -
»

InvPsi:

« DUPDUP DUP InvP InvP DUP InvP InvP DUPDUP InvP InvP DUP InvP DUP EXP .5 + Psi OVER - - EXP .5 +
»

InvP:

« DUP EXP .5 + Psi OVER - - EXP .5 + Psi OVER - -
»

(*) On the HP 49G, replace Psi with 0 PSI.

Examples:

4.012007 InvH --> 30.523595226 (in less than half a second); this is a very belated solution to one of the famous Valentin's challenges (#3 here).

.577215664902 InvH --> 0.46163214497 ; InvH(Euler-Mascheroni constant) = xmin (local minimum of the continuous factorial function).

4 pi 2 / - 3 ENTER 2 LN * - InvH --> 0.24999999999 ; one of the special values for fractional arguments examples here.
1.5 InvH --> 2. ; H2

1 Psi --> -0.577215664902 InvPsi --> 1.00000000009
2 Psi --> 0.422784335098 InvPsi --> 2.

Background:

Let H(x) be the continuous function associated with Harmonic Numbers. Then,

$H(x)=\gamma +\psi \left ( x+1 \right )$

$\psi \left ( x+1 \right )=H(x)-\gamma$

$x+1 =\psi^{-1} \left ( H(x)-\gamma \right )$

$x+1 =\psi^{-1} \left ( H(x)-\gamma \right )$

$x =\psi^{-1} \left ( H(x)-\gamma \right )-1$

or

$H^{-1}(x) =\psi^{-1} \left ( x-\gamma \right )-1$

That is, in order to obtain the inverse of H(x) we only need the Inverse Digamma Function and the Euler-Mascheroni constant. No problem with the constant, but the Inverse Digamma might be a problem since Digamma is not easily invertible.

A rough approximation is

$\psi^{-1} \left ( x\right )=e^{x}+\frac{1}{2}$

The equivalent HP 50g program is

P1:

« EXP .5 +
»

However, this is good only for x >= 10, not good enough for our purposes:

10 P1 --> 22026.9657948 Psi --> 10.0000000001.

But,

9 P1 --> 8103.58392758 Psi --> 9.00000000064
8 P1 --> 2981.45798704 Psi --> 8.00000000469
7 P1 --> 1097.13315843 Psi --> 7.00000003465

So, let's try to improve the accuracy a bit:

P2:

« DUP P1 Psi OVER - - P1
»

10 P2 --> 22026.9657926 Psi --> 9.99999999999
9 P2 --> 8103.58392239 Psi --> 8.99999999999
8 P2 --> 2981.45797306 Psi --> 8.
7 P2 --> 1097.13312043 Psi --> 7.
6 P2 --> 403.928690211 Psi --> 6.
5 P2 --> 148.912878357 Psi --> 5.00000000001

But,

4 P2 --> 55.0973869316 Psi --> 4.00000000039
3 P2 --> 20.5834634677 Psi --> 3.00000002131

Proceeding likewise, we get

P3:

« DUP P2 Psi OVER - - P2
»

This is good for x as low as 2, but not for x = 1:

2 P3 --> 7.88342863120 Psi --> 2.
1 P3 --> 3.20317150637 Psi --> 1.0000000139

A couple more steps suffice for x around -0.6 and greater, which is good enough for our purposes. That's what the InvPsi program above does, albeit in an inelegant way. Also, this is just an intuitive and somewhat inneficient approach. Better methods suggestions are welcome.

Edited to fix a couple of typos.

Edited again to include a printout of my current directory:

InvPsi will accept arguments greater or equal -1. About 0.05 s, 0.10 sor 0.50 s, depending on the arguments.
Cool, thanks! I was just trying to figure out an inverse harmonic number function the other day and getting nowhere. Neither Mathworld nor Wikipedia seemed to have much to say on the subject.

John
(04-27-2016 04:57 PM)John Keith Wrote: [ -> ]Cool, thanks! I was just trying to figure out an inverse harmonic number function the other day and getting nowhere. Neither Mathworld nor Wikipedia seemed to have much to say on the subject.

If the arguments are always plain harmonic numbers, that is, the ones obtained from the discrete function, then the inverse function can be implemented simply as

InvHn:

« .577215664902 - EXP IP »

Examples:

137 ENTER 60 / InvHn --> 5. ; 137/60 = H(5)

10 Hx --> 2.92896825397 ; H(10)
InvHn --> 10.

2E10 --> 24.2962137754 ; H(2x10^10)
InvHn --> 20000000000.

where Hx is the program for the continuous function:

Hx:

« 1. + Psi .577215664902 + »

Regards,

Gerson.
Reference URL's
• HP Forums: https://www.hpmuseum.org/forum/index.php
• :


Why do small angle approximations have to be in radians?

Why do small angle approximations only hold in radians? All the books I have say this is so but don't explain why.

• @EmilioPisanty perhaps you're correct, but anonymous disparagement is hardly the best way to greet a new user. Anyway, user28435, try Googling one-line answers before asking in future (this must have been asked hundreds of times on Math SE).
– Meow
Aug 19, 2013 at 21:36
• Don't think of sine as being defined in terms of angles - rather, arcs length of a circle. (The approximations come from the fact that a circle is approximately a straight line, especially near the point of tangency.) Oct 21, 2014 at 10:55

The reason why we use radians for $\sin x$ (or other trigonometric functions) in calculus is explained here and here in Wikipedia.

Having known that, notice that small angle approximation is just the Taylor expansion of $\sin x$ and $\cos x$ around $x=0$:

$$\sin x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}\tag{1}$$ $$\cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}\tag{2}$$

If you scale $x$ by some constant $\omega$, then you must replace $x$ with $\omega x$ in $(1)$ and $(2)$. So, working in degrees $($ $\omega=\frac{\pi}{180}$ $)$, the approximation will become: $$\sin \theta\approx \frac{\pi}{180}\theta\tag{\theta in degrees}$$

There are several correct ways to answer this that illuminate different aspects of what is going on (and I wouldn't be surprised if the answer is present on this site somewhere EDIT: indeed the other responses to this question are different ways of looking at it). Here is a geometrical answer.

The basic formula is

$$l = r \theta$$

where $l$ is the arc length of a segment of a circle with radius $r$ subtending an angle $\theta$. For this formula to be true, $\theta$ needs to be in radians. (Just try it out, is the arc length of the whole circle equal to $360 r$ or $2\pi r$?) In fact, that formula is really how you define what you mean by radians.

Now consider a straight line segment connecting the two endpoints of the arc subtended by $\theta$. Call it's length $a$. You can show by fiddling with triangles that $$a=2r\sin\left(\frac{\theta}{2}\right)$$

In the limit that the angle is small (so only a small piece of the circle is subtended), you should be able to convince yourself that $a\approx l$. This is the core of the small angle approximation.

Using the two relationships above we have

$$r\theta \approx 2r \sin\left(\frac{\theta}{2}\right)$$ or $$\sin\left(\frac{\theta}{2}\right)\approx\frac{\theta}{2}$$ You can see that using radians was crucial here because it allowed us to use $l=r\theta$.

A 'small angle' is equally small whatever system you use to measure it. Thus if an angle is, say, much smaller than 0.1 rad, it will be much smaller than the equivalent in degrees. More typically, saying 'small angle approximation' typically means $\theta\ll1$, where $\theta$ is in radians; this can be rephrased in degrees as $\theta\ll 57^\circ$.

(Switching uses between radians and degrees becomes much simpler if one formally identifies the degree symbol $^\circ$ with the number $\pi/180$, which is what you get from the equation $180^\circ=\pi$. If you're differentiating with respect to the number in degrees, then, you get an ugly constant, as you should: $\frac{d}{dx}\sin(x^\circ)={}^\circ \cos(x^\circ)$.)

In real life, though, you wouldn't usually say 'this angle should be small' without saying what it should be smaller than. If the latter is in degrees then the former should also be in degrees.

That said, though: always work in radians! Physicists tend to use degrees quite often, but there is always the underlying understanding that the angle itself is a quantity in radians and that degrees are just convenient units. Trigonometric functions, in particular, always take their arguments in radians, so that all the math will work well. Always differentiate in radians, always work analytically in radians. And at the end you can plug in the degrees.

It's because this relationship $$\lim_{x\rightarrow0}\frac{\sin(x)}{x}=1$$ (i.e. $\sin(x) \approx x$) only holds if $x$ is in radians, as, using L'Hoptial's rule, $$\lim_{x\rightarrow0}\frac{\sin(x)}{x}=\lim_{x\rightarrow0}\frac{\frac{d}{dx}\sin(x)}{\frac{d}{dx}x}=\lim_{x\rightarrow0}\frac{d}{dx}\sin(x)=\begin{cases} 1 & \text{ if } x\text{ is in radians}\\ \frac{\pi}{180} & \text{ if } x \text{ is in degrees} \end{cases}.$$ That is, $\frac{d}{dx}(\sin(x))=\cos(x)$ only if $x$ is in radians. The reason can be divined from this proof without words by Stephen Gubkin:

• Sure but WHY does this only hold in radians. That's the bit I can't understand. Everyone says it holds in radians but no one explains why lol. Thanks anyway. Aug 19, 2013 at 21:03
• @user28435: because in radians the sin of an angle basically equals the angle itself, at small enough angles, so it very much simplifies the algebra if you don't have to keep converting degrees.
– Mike Dunlavey
Aug 20, 2013 at 0:12

You need to clarify what you are really asking: in physics the small angle approx. is typically used to approximate a non-linear differential equation by a linear one - which is much easier to solve - by allowing us to approximate the sin(x) function by x. The resulting equation is only a good one for small angles. I.e., its a physics thing, not a math thing.

because the sine of an angle is dimensionless [i.e. just a number], the angle has to be also dimensionless [i.e. the radian is dimensionless] ... units on both sides of an equation have to match


# Determining the remainder of a small integer n raised to a high power when divided by m

I recently took a test that asked me to determine the value of $3^{82} \mod 5$.

I was unable to figure out how to do it on the test, but when I got home I noticed that there is a pattern to the remainders of $3^n$,

• $3^{1} \mod 5 = 3$
• $3^{2} \mod 5 = 4$
• $3^{3} \mod 5 = 2$
• $3^{4} \mod 5 = 1$
• $3^{5} \mod 5 = 3$
• $3^{6} \mod 5 = 4$
• $3^{7} \mod 5 = 2$
• $3^{8} \mod 5 = 1$
• $3^{9} \mod 5 = 3$

Is the best strategy to follow this pattern up to the $82^{\text{nd}}$ remainder (obviously in an intelligent way) or is there some much more obvious trick to this question that I missed?

-

As you have observed $$3^4 \equiv 1 \pmod 5$$ This means that $$3^{4k} \equiv 1 \pmod 5$$ Hence, $$3^{80} \equiv 1 \pmod 5 \implies 3^{82} \equiv 3^2 \pmod{5} = 4 \pmod 5$$ In general, $$3^{4k + r} \equiv 3^r \pmod{5}$$
Thank you for your answer. It's not entirely clear to me how $3^4 \mod 1 \equiv 1$ tells us that $3^{4k} \mod 1 \equiv 1$. Could you explain this step? –  crazedgremlin Nov 7 '12 at 1:07
@crazedgremlin $\rm\: \color{#0A0}{3^4}\equiv\color{#C00} 1\:\Rightarrow\: 3^{4k} = (\color{#0A0}{3^4})^k\equiv \color{#C00}1^k\equiv 1\:$ by the Congruence Product Rule. It is essential to learn the arithmetic of congruences, as opposed to less flexible mod operations. –  Bill Dubuque Nov 7 '12 at 1:23


# Thread: Limits / Series question

1. ## Limits / Series question

hi

Let
P = [(2^3-1)/(2^3+1)][(3^3-1)/(3^3+1)]........[(n^3-1)/(n^3+1)]
n=2,3,4,......

Find Limit P when N tends to infinity.

2. Originally Posted by champrock
hi

Let
P = [(2^3-1)/(2^3+1)][(3^3-1)/(3^3+1)]........[(n^3-1)/(n^3+1)]
n=2,3,4,......

Find Limit P when N tends to infinity.
Factorise $\frac{n^3-1}{n^3+1}$ as $\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}$, and notice that $n^2+n+1 = (n+1)^2 - (n+1) + 1$. You will then find that you have a telescoping product.

3. what should i do with n^2 - n + 1 ?

I was able to cancel the (n-1)/(n+1) terms. but dont know what to do with n^2 - n + 1 and n^2 + n + 1

4. You easily undestand if you write the sequence of terms...

$P= \prod_{n=2}^{\infty} \frac{(n-1)\cdot (n^{2} + n + 1)}{(n+1)\cdot (n^{2} - n + 1)}= \frac {1\cdot 7}{3\cdot 3}\cdot \frac {2\cdot 13}{4\cdot 7}\cdot \frac {3\cdot 21}{5\cdot 13}\cdot \frac {4\cdot 31}{6\cdot 21}\cdot \dots$

... and it is evident you can simplify both in numerator and denominator [3 with 3, 4 with 4, ... , 7 with 7, 13 with 13, 21 with 21,...]

Finally is...

$P=\frac{2}{3}$

Very nice!...

Kind regards

$\chi$ $\sigma$

5. Originally Posted by champrock
what should i do with n^2 - n + 1 ?

I was able to cancel the (n-1)/(n+1) terms. but dont know what to do with n^2 - n + 1 and n^2 + n + 1
Use the hint I gave before: $n^2+n+1 = (n+1)^2 - (n+1) + 1$. That tells you that the term $n^2+n+1$ in the numerator of each fraction cancels with the term $(n+1)^2 - (n+1) + 1$ in the denominator of the next one.

6. I think Only 1/3 is left. All the other terms cancel out. How are you getting 2/3

7. The term of order n is...

$a_{n}= \frac{(n-1)\cdot (n^{2} +n + 1)}{(n+1)\cdot (n^{2} - n + 1)}$

For $n=3$ is...

$a_{3}= \frac{2\cdot 13 }{4\cdot 7}$

Kind regards

$\chi$ $\sigma$