Compare: Which is bigger? $3^{\sqrt2}$ or $2^{\sqrt 3}$ Compare: Which is bigger?
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad$ $\Large{3^{\sqrt2}}\quad$ or  $\quad \Large{2^{\sqrt 3}}$

My attempts:
Let,  $\quad$ $a=3^{\sqrt 2}$ and $b=2^{\sqrt 3}$
We have

$$a^\sqrt{3}=3^\sqrt{6}\\ b^\sqrt{3}=8<3^{\sqrt{4}}<3^\sqrt{6}=a^\sqrt{3} $$
$$a^\sqrt{3}>b^\sqrt{3} \Longrightarrow a>b$$

Question-1:

Is my solution correct?

Question-2:

I saw the following solution was approved on another site.  But, the last line seems wrong to me. Am I right?

The handwritten one is simply the wrong approach.  If you raise $a$ and $b$ to different powers you simply will not (usually) be able to conclude anything by comparing the results.  How can you?  You raised them to different powers so whatever inequality you get doesn't relate to the original values in any way!

Everything it came up with is true (but not helpful)  up to $a^{\sqrt 3} > b^{\sqrt 2}$ this has been shown[1].  But the conclusion just doesn't make sense.

$a^{\sqrt 3} > b^{\sqrt 2} \not \implies a =b$(!!!!)

Why would it?  $\sqrt 2 \ne \sqrt 3$.

This is like claiming $3^3 > 4^2 \implies 3 > 4$.

(Or simply if $a = 2.99$ and $b = 3.01$ then  $2.99^{\sqrt 3} \approx  6.6663278104352795010160893782624$ while $3.01^{\sqrt 2}\approx 4.7511115647726528859943327888079$.

(So $a^{\sqrt 3} > b^{\sqrt 2} $ but $a < b$.)

$a^{bigger number} > b^{smallernumber}$ doesn't tell us anything.[2]

We could try

$a^{\sqrt 3} > b^{\sqrt 2} \implies a > b^{\frac {\sqrt 2}{\sqrt 3}}$

or

$a^{\sqrt 3} > b^{\sqrt 2} \implies a^{\frac {\sqrt 3}{\sqrt 2}} > b$

but we can't get anywhere from there.

....

But yours is fine.

=======

[1](It is nothing more than claiming $3^{\sqrt 6} > 2^{\sqrt 6}$.

[2] However if we got the other result $a^{smaller power} > b^{biggerpower}$ (and if we know $a, b > 1$) we can conclude $a > b$ because

a) $a^{biggerpower} > a^{smallerpower} > b^{bigger power}$

b) $a^{smallerpower} > b^{biggerpower} > b^{smallerpower}$.

But if the powers and the inequality are in disagreement, we don't know which "overpowers" the other.

This is assuming $a, b > 1$.  If $a,b < 1$ the exact OPPOSITE applies.

And if you don't know if $a,b$ are bigger smaller than $1$ then you can't compare different powers at all because you don't know anything.  That's another reason the handwritten one was the wrong approach.

Yes because $(x,y)\mapsto x^y$ is strictly increasing in both variables on $(1,\infty)\times (0,\infty)$.
No: $a^{\sqrt3}>b^{\sqrt 2}$ does not imply $a>b$. See $a=b=1.5$ for comparison.


Deriving the exponential distribution from a shift property of its expectation (equivalent to memorylessness). Suppose $X$ is a continuous, nonnegative random variable with distribution function $F$ and probability density function $f$. If for $a>0,\ E(X|X>a)=a+E(X)$, find the distribution $F$ of $X$.

About the necessary hypotheses (and in relation to a discussion somewhat buried in the comments to @bgins's answer), here is a solution which does not assume that the distribution of $X$ has a density, but only that $X$ is integrable and unbounded (otherwise, the identity in the post makes no sense).

A useful tool here is the complementary PDF $G$ of $X$, defined by $G(a)=\mathrm P(X\gt a)$. For every $a\geqslant0$, let $m=\mathrm E(X)$. The identity in the post is equivalent to $\mathrm E(X-a\mid X\gt a)=m$, which is itself equivalent to $\mathrm E((X-a)^+)=m\mathrm P(X\gt a)=mG(a)$. Note that $m\gt0$ by hypothesis. Now, for every $x$ and $a$,
$$
(x-a)^+=\int_a^{+\infty}[x\gt z]\,\mathrm dz.
$$
Integrating this with respect to the distribution of $X$ yields
$$
\mathrm E((X-a)^+)=\int_a^{+\infty}\mathrm P(X\gt z)\,\mathrm dz,
$$
hence, for every $a\gt0$,
$$
mG(a)=\int_a^{+\infty}G(z)\,\mathrm dz.
$$
This proves ${}^{(\ast)}$ that $G$ is infinitely differentiable on $(0,+\infty)$ and that $mG'(a)=-G(a)$ for every $a\gt0$. Since the derivative of the function $a\mapsto G(a)\mathrm e^{ma}$ is zero on $a\gt0$ and $G$ is continuous from the right on $(0,+\infty)$, one gets $G(a)=G(0)\mathrm e^{-ma}$ for every $a\geqslant0$.

Two cases arise: either $G(0)=1$, then the distribution of $X$ is exponential with parameter $1/m$; or $G(0)\lt1$, then the distribution of $X$ is a barycenter of a Dirac mass at $0$ and an exponential distribution. If the distribution of $X$ is continuous, the former case occurs.

${}^{(\ast)}$ By the usual seesaw technique: the RHS converges hence the RHS is a continuous function of $a$, hence the LHS is also a continuous function of $a$, hence the RHS integrates a continuous function of $a$, hence the RHS is a $C^1$ function of $a$, hence the LHS is also a $C^1$ function of $a$... and so on.

Yes, the exponential distribution is the ONLY continuous distribution satisfying the property, because the above differential equation has a unique solution. For the proof of this fact you can also have a look at the page on the exponential distribution at Statlect (the rate parameter and its interpretation - proof).


When are these two definitions equivalent? I can't remember how to show this but I feel like it must be true: if I have a continuous function $f$, then how do I show that

$$\lim_{n\rightarrow \infty}\frac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}}= c \quad\text{implies}\quad \lim_{\delta\rightarrow 0}\frac{f(x+\delta)-f(x)}{\delta}=c$$

EDIT: Sorry for the confusion, I did indeed mean $n\in \mathbb{N}$.

This is not true. Consider

$$f(x) = \begin{cases} x\cos\left(\frac {2\pi} x\right) &\text{if }x\neq 0,\\ 0 & \text{if }x=0.\end{cases}$$

Then $f$ is continuous on $\mathbb R$,

$$\frac{f\left(\frac 1n\right) - f(0)}{\frac 1n} = 1$$

for all $n\in \mathbb N$, but $f$ is not differentiable at $0$.

New Answer

I wrote this under the impression that $n$ and $\delta$ were both real, before the question was clarified. It is critical to the question that $n\in\mathbb{N}$, not $\mathbb{R}$, because then we can take advantage of periodic functions like sine or cosine to make the left-side limit exist. It is also critical that ${1\over\delta}\in\mathbb{R}$, not $\mathbb{N}$, because we can't take advantage of periodic functions unless they're discrete (which violates the "continuous function" requirement in the question).

Essentially, the left equation is no longer the definition of a derivative, though it would still serve just fine in most cases, because we're taking the limit discretely instead of continuously. But the right equation is the true derivative, with the limit taken continuously. So they aren't identical, and there are therefore ways to show implication is false, as shown by John Ma.

Old Answer

Basically, we can ignore most of the equation. The difference between the left side and the right side is that on the left we're taking the limit as ${1\over n}\to{1\over\infty}$ while on the right we're taking the limit as $\delta\to 0$.

So let's look at the limits directly.

$lim_{n\to\infty}{1\over n}$ can be solved by looking at a table or graph, and noting the value gets arbitrarily close to 0 as $n\to\infty$. There might be a more formal way to do it, but that's how we "proved" it in my calculus class.

$lim_{\delta\to 0}{\delta}$ can be solved by just plugging 0 into $\delta$ to get 0.

So we can let $\delta={1\over n}$ and note that as $n\to\infty$, ${1\over n}\to 0$ and $\delta\to 0$.

So once we've convinced ourselves that $\delta\to0$ is the same as ${1\over n}$, $n\to\infty$, the original implication makes sense. Start by noting that $n\to\infty\iff{1\over n}\to 0$:

$lim_{n\to\infty}{f(x+{1\over n})−f(x)\over{1\over n}}=c\iff$ $lim_{{1\over n}\to 0}{f(x+{1\over n})−f(x)\over{1\over n}}=c$

Now make the substitution, $\delta={1\over n}$:

$lim_{{1\over n}\to 0}{f(x+{1\over n})−f(x)\over{1\over n}}=c$ $\iff lim_{\delta\to 0}{f(x+\delta)−f(x)\over\delta}=c$

As John Ma's answer points out, it's possible to have a continuous function where the above limit doesn't exist for one or more $x$ values. It's actually possible to make a continuous function where the above limit exists nowhere. That's why we say a derivative only exists where the function is continuous and "smooth" or "well-behaved" near the point in question.

However, the implication doesn't say "the limit exists for all $x$". The implication says "if the limit of the left equation exists for a given $x$ and $f(x)$, then the limit of the right equation for the same $x$ and $f(x)$ not only exists, but has the same value, $c$".

We can go a step further by making it if and only if, as I've done above. If the limit exists for one side, it must exist for the other side. If the limit doesn't exist for one side, it can't exist for the other side. Because ultimately the two equations say the same thing.


If $0<x<\frac{\pi}{4}$ then prove that $\left(\sin x\right)^{\sin x}<\left(\cos x\right)^{\cos x}$. If $0<x<\frac{\pi}{4}$ then prove that $\left(\sin x\right)^{\sin x}<\left(\cos x\right)^{\cos x}$.

My attempt:

If $0<x<\frac{\pi}{4}$

then $\sin x <\cos x$

which means  $\left(\sin x\right)^{\sin x}<\left(\cos x\right)^{\sin x}<\left(\cos x\right)^{\cos x}$.

Am I correct. Can a more rigorous proof be given

I saw this solution in a textbook($103$ Problems in Trigonometry by Titu Andreescu) but i wanted a second opinion. The way the solution is it can only be answered by someone who has created the question and the solution.I will quote Titu:

Observe that the logarithmic function is concave down.

We apply Jensen's inequality to the numbers $\sin x<\cos x<\sin x+\cos x$

with weights $\lambda_{1}=\tan x$ and $\lambda_{2}=1-\tan x$(because for $0<x<\frac{\pi}{4}$ we have $\lambda_{1}=\tan x>0$ and $\lambda_{2}=1-\tan x>0$) to obtain

$\log(\cos x)$

$=\log\left(\tan x\sin x+(1-\tan x)(\sin x+\cos x)\right)$

$>\tan x\log(\sin x)+(1-\tan x)\log(\sin x+\cos x)$

Since $\sin x+\cos x=\sqrt{2}\sin(x+\frac{\pi}{4})>1$ and $\tan x<1$

We can say that $\log(\cos x)>\tan x\log(\sin x)$

Multiplying both sides by $\cos x$ and then exponientating we obtain the required result.

Please suggest some other way.


Lattice which is not bounded lattice I want to find an example of a lattice which is not a bounded lattice . Diagrams would be good with an explanation .

All finite lattices are bounded. $(\Bbb N,|)$ is an example of an unbounded lattice. Or you can add the Hasse diagram of $\Bbb Z$ above the maximum element of  the Hasse diagram of any bounded lattice to get an unbounded lattice.

The given poset is a lattice because any given pair of elements do have a lowest upper bound (LUB), but it tends to positive infinite for very large integers. By a similar reasoning, the GLB tends to negative infinite for very large negative integers. Therefore, LUB and GLB, even though they exist, are not bounded. Thus, the given poset is an unbounded lattice. This concept would be clearer if you were to draw the same conclusions from the Hasse diagram.


Maximum of the difference What is the maximum value of
$f(… f(f(f(x_{1} – x_{2}) – x_{3})-x_{4}) … – x_{2012})$
where $x_{1}, x_{2}, … , x_{2012}$ are distinct integers in the set ${1, 2, 3, …, 2012}$ and $f$ is the absolute value function?

I am solving with respect to the general case that is $$\max_{\substack{
   {x_i\in\{1,2,...,n\}\\ x_i\ne x_j,\ i\ne j}}} f(… f(f(f(x_{1} – x_{2}) – x_{3})-x_{4}) … – x_{n}), \quad f(x)=|x|$$

Since we are looking for a maximum solution, then it is sufficient to distribute the numbers $$1,2,3,...,n$$ in an order so that we get a maximum value for any value of $n$ and that order is what I explained below.

Consider the case where $\forall k=1,2,\ ...,\ n-1$ we have $x_k=n-k$ and $x_n=n$ i.e $x_1=n-1$, $x_2=n-2$, $x_3=n-3$ $ , ..., $ $x_k=n-k$ $, ...,$ $ x_{n-1}=1$  but $x_n=n$.

So that for $n=8$ for example, we compute the value of $f(f(f(f(f(f(f(7-6)-5)-4)-3)-2)-1)-8)$

Let $f_1=x_1$, $f_2=(x_{1} – x_{2})$, $f_3=(f(x_{1} – x_{2}) – x_{3})$, $...,$ $f_{n }=f(… f(f(f(x_{1} – x_{2}) – x_{3})-x_{4}) … – x_{n})$

Using the following notation where $0\le n_4,k_4\le3$ $$\quad\quad \quad\quad\quad n\equiv n_4 \mod 4 \quad\quad\quad\text { and }$$ $$k \equiv k_4 \mod 4$$ we have the following pattern depending on the value of $n$ (it's preferable to work out some cases to verify what I wrote below).
$$\boxed{\begin{array}{ll} f_1=n-1\\
f_2=1 \\
f_3=n- 4 \\
f_4=0 \end{array}} $$ $$ \boxed{f_{n-k}=\left\{\begin{array}{ll}1 \quad \quad \quad \quad \quad \quad\text { if } (n_4,k_4)=\{(0,2),(1,3),(2,0),(3,1)\} \\0  \quad \quad \quad \quad \quad \quad\text { if } n_4=k_4\\f_{n-k-4}+4 \end{array}\right.}$$ $$ \boxed{\begin{array}{ll}f_{n- 4 }=\left\{\begin{array}{ll}0 \quad\text   { if } n_4= 0 \quad      \\4   \quad\text   { if } n_4= 1 \quad     \\1 \quad\text   { if } n_4= 2 \quad      \\3 \quad\text   { if } n_4= 3 \quad      \end{array}\right.\\f_{n-3}=\left\{\begin{array}{ll}3 \quad\text   { if } n_4= 0 \quad      \\1 \quad\text   { if } n_4= 1 \quad     \\2 \quad\text   { if } n_4= 2 \quad     \\0 \quad\text   { if } n_4= 3 \quad      \end{array}\right.\\ f_{n-2}=\left\{\begin{array}{ll}1 \quad\text   { if } n_4= 0,1       \\0 \quad\text   { if } n_4= 2 \quad     \\2 \quad\text   { if } n_4= 3 \quad      \end{array}\right.\\ f_{n-1}=\left\{\begin{array}{ll}0 \quad\text   { if } n_4= 0,1       \\1 \quad\text   { if } n_4= 2,3       \end{array}\right.\end{array}}$$

And lastly $$f_n=|f_{n-1}-x_n| = |f_{n-1}-n|=\left\{\begin{array}{ll}n&\text   { if } n_4= 0,1       \\n-1&\text   { if } n_4= 2,3       \end{array}\right.$$

$$2012\equiv 0 \mod 4 \quad \implies \quad f_{2012}=2012$$

I don’t know the answer, but I wrote some code that might help. My code is too slow to calculate the actual answer, with $2012$ in the problem. But if you replace $2012$ with the numbers $1$ through $10$, the answers to those 10 problems are [1, 1, 2, 4, 5, 5, 6, 8, 9, 9]. Maybe you can see a pattern in those solutions.

This CoffeeScript code calculates the answer with brute-force, trying all possible permutations and recording the maximum value. One function it provides is maximumValueForSetSize, which would return the answer if called with 2012 as the argument. But it runs in $O(n!)$ time, far too slowly to practically provide the answer to your problem. However, another function, maximumValuesForSetsUpTo, can provide data points for the results. It also runs in $O(n!)$ time, but this is reasonable because the function is still useful for small values of $n$. I used that function to calculate the 10 answers that I included above.

Here is the code. You can run it from this JS Bin. Scroll to the “useful calculations” section at the bottom and change the numbers if you want to experiment.

# environment setup

# this is CoffeeScript (http://coffeescript.org/)
# and I'm using the Underscore.js library (http://underscorejs.org/)

log = console.log

testFunctionUsingTestCases = (fun, funName, testCases) ->
  for testCase in testCases
    expectedResult = _(testCase).first()
    callArguments = _(testCase).rest()
    actualResult = fun.apply(undefined, callArguments)

    if ! _.isEqual(actualResult, expectedResult)
      log("test failed for function "+funName+" - arguments, expected, and actual:", callArguments, expectedResult, actualResult)
      return
  log("all tests passed for function "+funName)

# defining evaluatePermutation

absOfDifference = (num1, num2) ->
  Math.abs(num1 - num2)

# Big O: O(n) for n = permutation size
evaluatePermutation = (permutation) ->
  current = _(permutation).first()
  for number in _(permutation).rest()
    current = absOfDifference(current, number)
  return current
evaluatePermutationTestCases = [
  [1, [1]]
  [1, [1, 2]]
  [2, [1, 2, 3]]
  [2, [2, 1, 3]]
  [0, [3, 1, 2]]
  [0, [3, 2, 1]]
  [2, [1, 2, 3, 4]]
  [4, [1, 3, 2, 4]]
  [2, [2, 1, 3, 4]]
]
testFunctionUsingTestCases(evaluatePermutation, "evaluatePermutation", evaluatePermutationTestCases)

permute = `function(input) {
    var permArr = [],
    usedChars = [];
    function main(input){
        var i, ch;
        for (i = 0; i < input.length; i++) {
            ch = input.splice(i, 1)[0];
            usedChars.push(ch);
            if (input.length == 0) {
                permArr.push(usedChars.slice());
            }
            main(input);
            input.splice(i, 0, ch);
            usedChars.pop();
        }
        return permArr
    }
    return main(input);
};` # from http://stackoverflow.com/a/11509565/578288

# Big O: O(n)
setWithSize = (size) ->
  [1..size]

# Big O: O(n!)
permutationsOfSetWithSize = (size) ->
  return _.compose(permute, setWithSize)(size)
permutationsOfSetWithSizeTestCases = [
  [[ [1] ], 1]
  [[ [1, 2],[2, 1] ], 2]
  [[ [1, 2, 3],[1, 3, 2],[2, 1, 3],[2, 3, 1],[3, 1, 2],[3, 2, 1] ], 3]
]
testFunctionUsingTestCases(permutationsOfSetWithSize, "permutationsOfSetWithSize", permutationsOfSetWithSizeTestCases)

# defining maximumValueForSetSize

# Big O: O(n*m) for n = num of permutations, m = permutation size
# evaluatePermutation is O(m)
# map is O(n)
# max is O(n)
maximumValueUsingPermutations = (permutations) ->
  return _.max( _(permutations).map(evaluatePermutation) )

# Big O: O(n!)
# maximumValueUsingPermutations is O(n^2)
# permutationsOfSetWithSize is O(n!)
maximumValueForSetSize = (size) ->
  return _.compose(maximumValueUsingPermutations, permutationsOfSetWithSize)(size)
maximumValueForSetSizeTestCases = [
  [1, 1]
  [1, 2]
  [2, 3]
]
testFunctionUsingTestCases(maximumValueForSetSize, "maximumValueForSetSize", maximumValueForSetSizeTestCases)

# defining maximumValuesForSetsUpTo

# Big O: O(n!)
maximumValuesForSetsUpTo = (maxSize) ->
  for size in [1..maxSize]
    maximumValueForSetSize(size)
maximumValuesForSetsUpToTestCases = [
  [[1, 1], 2]
  [[1, 1, 2], 3]
]
testFunctionUsingTestCases(maximumValuesForSetsUpTo, "maximumValuesForSetsUpTo", maximumValuesForSetsUpToTestCases)

# useful calculations

maxSize = 5
log("maximum values for sets of sizes up to #{maxSize}:", maximumValuesForSetsUpTo(maxSize) )

# this takes too long to run - estimated over a year
#idealSize = 2012
#log("maximum value for set of size #{idealSize}:",  maximumValueForSetSize(idealSize) )


Prove a real valued function is increasing Prove $$-\frac{x(1-x^c)\ln x}{(1-x)^2}$$
 $c>0$ is an increasing function on $[0,1]$.

I have a relatively cumbersome proof sketched below. I would like to see other ideas, particularly simpler ones.

Define $$g(x,c) = -\frac{x(1-x^c)\ln x}{c(1-x)^2}.$$ Then
$$\frac{g(x,c_2)}{g(x,c_1)}=\frac{c_1(1-x^{c_2})}{c_2(1-x^{c_1})},$$
$0\le c_1<c2$ is increasing for $x\in[0,1]$. Thus, if $g(x,c_1)$ increases then $g(x,c_2)$ also increases.

Take $c_1=0$, $g(x,0)=\frac{x(\ln x)^2}{(1-x)^2}$. We can show $g(x,0)$ is increasing in $x$ by substituting $x = e^{-2t}$. Therefore, $g(x,c)$ is increasing in $x$ for $c>0$.

$$-\frac{x(1-x^c)\ln x}{(1-x)^2} \;\; = \;\; x \cdot (x^c - 1) \cdot \ln x \cdot \frac{1}{(1-x)^2}$$

Note that this is a product of $4$ functions, each of which is increasing on the open interval defined by $0 < x < 1.$

NOTE: This is incorrect reasoning, as explained in the comments. I'll leave it up so that others can see why the kind of simplistic approach I used doesn't work, and in case I think of something reasonably simple that does work.


A hammer and a nail cost $1.10, and the hammer costs one dollar more than the nail. How much does the nail cost? The answer is 0.05. I used algebra. But my friends say, why not 0.10, and they also say, it can be that the hammer is 1.04 and the nail 0.06. How do I tell them that 0.05 is the definite  answer, nothing else; I need mathematical evidence and proof to convince folks.

Show your friends that their proposed alternate solutions don't fit the bill.

If the nail is 10 cents, then the hammer is a dollar more: $\$1.10$, and then the total is $\$1.20$.

If the nail is 6 cents and the hammer is $\$1.04$, then that is not a dollar more for the hammer; it's 98 cents more.

Given that H+N = 1.1 and H = N+1 we can solve for N for both. H = 1.1-N AND H = 1+N so 1.1-N = 1+N, then 1.1 = 1+2N, so 0.1 = 2N, therefore .1/2 = N, thus .05 = N. Substitute .05 for N either original equation and H = 1.05.


Is every Polish space the quotient of some subset of Baire space? Question is as in the title:

Is every Polish space the quotient space of some subset of Baire space $\omega^\omega$ ?

Yes. Actually, there are two cases. Take an arbitrary Polish space $X$. If $X=\emptyset$, then the subset of $\omega^\omega$ must be $\emptyset$. If $X$ is nonempty, then it is a quotient (a continuous image) of all of $\omega^\omega$.

You can find a proof in Kechris' Classical Descriptive Set Theory, Sect. 7.E. The main idea is that for $X\neq\emptyset$, you can construct a family $(F_s)_s$ of $F_\sigma$ subsets of $X$ indexed by finite sequences of natural numbers such that

$F_\emptyset= X$;
$F_s=\bigcup_i F_{s\smallfrown i}=\bigcup_i \overline{F_{s\smallfrown i}}$; and
the diameter $F_s$ is less than $2^{-|s|}$,

where $|s|$ is the length of $s$ and $(s_0,\dots,s_k)\smallfrown i = (s_0,\dots,s_k,i)$. For every $y\in \omega^\omega$ such that $\bigcap_m F_{y\restriction m}\neq\emptyset$, this intersection is a singleton and therefore it induces a function between the set $D$ of those $y$s and $X$. This map is continuous, and $D$ is closed. After that, you must use the result that every closed subset of the Baire space is a retract.

Remark:
Prof Terraf's answer proves that every Polish space is the continuous surjective image of $w^w$, but it doesn't prove that it is the $quotient$ $space$.


Show that $\int_C A\cdot dr=\frac{5}{3}$, where $A=(x^2-3y^2)i + (y^2-2x^2)j$ and $C$ is a closed curve in the $xy$-plane Show that $\int_C \mathbf{A}\cdot d\mathbf{r}=\frac{5}{3}$, where $\mathbf{A}=(x^2-3y^2)\mathbf{i}+(y^2-2x^2)\mathbf{j}$ and $C$ is a closed curve in the $xy$-plane such that
$$x=3\cos t, ~~ y=2\sin t, ~~~ 0\leq t\leq 2\pi.$$

\begin{align*}
\int_C \mathbf{A}\cdot d\mathbf{r}&=\int_C(x^2-3y^2) dx + (y^2-2x^2) dy\\&=\int_C \left[-3\, \left( 9\,  \cos^{2} \left( t \right)   -12\,
  \sin^{2} \left( t \right)  \right)   \sin \left( t
 \right) +2\, \left( -18\, \cos^{2} \left( t \right)   +
4\,  \sin^{2} \left( t \right) \right) \cos \left( t
 \right)\right] dt\\&=\int_0^{2\pi}(-63\,\sin \left( t \right)  \cos^{2} \left( t \right)
-44\, \cos^{3} \left( t \right)   +36\,\sin \left( t
 \right) +8\,\cos \left( t \right))dt\\&= 0
\end{align*}

Don't know where is the mistake. Please help me to solve.

Hint: Use Green's theorem to get the solution easily.


Multivariable calculus - find total derivative I want to find the total derivative of the function $f: \mathbb R^n \to \mathbb R^n$, $f(x)=\frac{x}{|x|}$

If I was to copy what the teacher taught, I should find the limit of $\lim_{t \to 0} \frac{f(a+th)-f(a)}{t}$ but I don't know how to find that limit.

$$\lim_{t \to 0} \frac{f(a+th)-f(a)}{t}=\lim_{t \to 0} \frac{\frac{a+th}{|a+th|}-\frac{a}{|a|}}{t}$$

what now?

For all $x\neq 0\in\mathbb R^n$ the function $f$ is differentiable at  $x$, i.e.
$$f(x+th)-f(x)=\langle \nabla f(x),th\rangle+O(t^2), $$
for all $t\in \mathbb R$ and $h\in \mathbb R^n$ s.t. $x+th\neq 0$, as the partial derivatives $\partial_i f(x)$ exist and are continuous in suitable open neighborhoods $U$ of $x$; this is show by   their explicit formulae, i.e.
$$\partial_i f(x):=\frac{\partial f}{\partial x_i}(x)=\frac{e_i}{\|x\|}-\frac{x_i}{\|x\|^3}x. $$
We denote by $e_i=(0,\dots,0,i,0,\dots,0)$ the $i$-th versor in $\mathbb R^n$.

Then
$$f(x+th)-f(x)=\frac{t}{\|x\|}\langle e_i,h\rangle-
\frac{t x_i}{\|x\|^3}\langle x,h\rangle +O(t^2), $$
which, in the limit $t\rightarrow 0$ gives

$$D_hf(x):=\lim_{t\rightarrow 0}\frac{f(x+th)-f(x)}{t}=\frac{h_i}{\|x\|}-
\frac{x_i\langle x,h\rangle}{\|x\|^3}. $$

Hint: I find it easier to break up problems that have absolute values in them into positive and negatives parts:
$$
  f(x) = \begin{cases}
          \cfrac{x}{x} & \quad \text{for} \quad x \ge 0 \\
          \cfrac{x}{-x} & \quad \text{for} \quad x < 0
         \end{cases}
$$
For n-dimensional vectors you have to define what the norm means.  If it is a $L_2$ norm,
$$
  f(\mathbf{x}) = \frac{\mathbf{x}}{\sqrt{\mathbf{x}\cdot\mathbf{x}}}
$$
The derivative is directional and is defined in the direction $\mathbf{y}$ as
$$
  \frac{\partial f}{\partial \mathbf{x}}\cdot\mathbf{y} = \left[\frac{d}{d\alpha} f(\mathbf{x} + \alpha \mathbf{y})\right]_{\alpha = 0}
$$
Update: For the purposes of calculation it is easier if we move into index notation
$$
  \frac{\partial f}{\partial \mathbf{x}} \equiv \frac{\partial}{\partial x_j}\left(\frac{x_i}{\sqrt{x_k\,x_k}}\right)
$$
After working out the algebra we get
$$
  \frac{\partial f}{\partial \mathbf{x}} = \frac{\mathbf{I}}{||\mathbf{x}||} - \frac{\mathbf{x}\otimes\mathbf{x}}{||\mathbf{x}||^3} \,.
$$
Of course, these are all partial derivatives but, presumably, the total derivative can be found if these partial derivatives are known.


Why did mathematicians introduce the concept of uniform continuity? I have solved many problems regarding uniform continuity, but still I can't understand the following:

Is there any practical application of this concept, or it is just a theoretical concept? Is there any wide application of this concept in any theorem or problem? In short, how did the concept of uniform continuity arise?

A Parable (that makes no claims of historical accuracy): Suppose your mathematical universe contains only the rational numbers. You study rational-valued functions of rational numbers, and discover the $\varepsilon$-$\delta$ definition of continuity.

Then you discover the real numbers, and the fact that the rationals are dense in the reals. It's natural to ask: If $f$ is a continuous function (on the rationals), does $f$ extend to a continuous function on the reals? Perhaps surprisingly, "no". For example, $f(x) = 1/(x^{2} - 2)$ is a continuous (!), rational-valued function on the set of rational numbers that is easily seen not to extend to a continuous function on the reals.

There are many ways to interpret "what went wrong" in this example, but the explanation serving the preceding parable comes down to the following:

Question 1: Suppose $f$ is a continuous (rational- or real-valued) function on some domain $D$, and suppose $(x_{k})$ is a Cauchy sequence in $D$. Is the image sequence $\bigl(f(x_{k})\bigr)$ Cauchy?

If you don't know the answer, it's worth making a serious effort at a proof before reading further. There's an obvious strategy: Fix $\varepsilon > 0$ arbitrarily. Since $f$ is continuous, we can make $|f(y) - f(x)| < \varepsilon$ by taking $y$ sufficiently close to $x$. And a sequence is Cauchy if "its terms can be made as close as we like". It all seems to work at first glance....

   The problem is, the answer is "no". An easy example is to take $f(x) = 1/x$ on the set of positive numbers, and to take $x_{k} = 1/k$. Or, if $f(x) = 1/(x^{2} - 2)$, use the recursive rational sequence $x_{1} = 2$, $x_{k+1} = \frac{1}{2}(x_{k} + \frac{2}{x_{k}})$.

The attempted proof above is a cautionary lesson about why mathematicians become so pedantic when teaching analysis: It's necessary to formulate and use precise definitions to avoid going logically awry. In the attempt to answer Question 1, the proof hits a snag because a Cauchy sequence in $D$ need not converge to a point $\ell$ of $D$, so we can't pick a $\delta > 0$ such that if $|x - \ell| < \delta$, then $|f(x) - f(\ell)| < \varepsilon$. And then, after careful examination, we discover a second, fatal snag: If $\delta_{k} > 0$ satisfies the definition of continuity at $x_{k}$ (i.e., if $|y - x_{k}| < \delta_{k}$ implies $|f(y) - f(x_{k})| < \varepsilon$), nothing ensures the sequence $(\delta_{k})$ has a positive lower bound.

Well, what if we require more of $f$, not merely continuity, but a condition that for every $\varepsilon > 0$, there exists a single $\delta > 0$ that "works throughout $D$" in the sense that if $x$ and $y$ are points of $D$ such that $|x - y| < \delta$, then $|f(x) - f(y)| < \varepsilon$.

Ah: We've just discovered uniform continuity on the set $D$.

Theorem: If $f$ is uniformly continuous on $D$, and if $(x_{k})$ is a Cauchy sequence in $D$, then $\bigl(f(x_{k})\bigr)$ is a Cauchy sequence.

Proof: See above, except now there are no snags.

All right, a (locally) uniformly continuous function maps Cauchy sequence to Cauchy sequences; so what?

Well, now we can have some mathematical fun. First, the definitions above extend with no effort to an arbitrary metric space $(X, d)$, and the proofs go through with obvious modifications (i.e., writing "$d(x, y)$" instead of "$|x - y|$").

Every metric space $(X, d)$ embeds isometrically into a complete space $(\overline{X}, d)$. One construction of $\overline{X}$ is to use equivalence classes of Cauchy sequences in $X$. A bit of analytic yoga shows that if $f:X \to Y$ is a uniformly continuous mapping into a complete metric space $(Y, d')$, there exists a unique (uniformly) continuous mapping $\overline{f}:\overline{X} \to Y$ that extends $f$.

Here are two "interesting" applications:

If $D$ is a bounded set of real numbers and $f$ is a real-valued function on $D$, then $f$ is uniformly continuous on $D$ (if and) only if $f$ extends continuously to the closure of $D$. This gives an essentially visual criterion for detecting uniform continuity (in the same sense that continuity is detectable by "the graph having no breaks", or differentiability is detectable by "the graph having a tangent line at each point"). For example, if $D$ denotes the set of non-zero real numbers, the functions
$$
f(x) = \sin\frac{1}{x},\qquad
g(x) = \frac{x}{|x|}
$$
are continuous in $D$, but (by a glance at their graphs) not uniformly continuous.
The space $X$ of continuous, real-valued functions on $[0, 1]$ is a vector space with distance function
$$
d(f, g) = \int_{0}^{1} |f(t) - g(t)|\, dt.
$$
The integration operator $I(f) = \int_{0}^{1} f(t)\, dt$ is (more-or-less obviously) uniformly continuous on $(X, d)$. Now, $(X, d)$ is not complete, but if $(f_{k})$ is a Cauchy sequence in $(X, d)$, then $\bigl(I(f_{k})\bigr)$ is a Cauchy sequence of real numbers, and therefore converges. This allows us to extend the Riemann integral (uniquely by continuity) to the completion of $(X, d)$.

The completion turns out to be $L^{1}[0, 1]$, the space of Lebesgue-integrable functions on $[0, 1]$, and the Lebesgue integral is the unique continuous extension of the Riemann integral.

It is a theorem in analysis that all functions that are continuous on a closed interval (or more generally, a compact set) are also uniformly continuous on that closed interval (resp. compact set).

As mentioned in the other answers, uniform continuity is much stronger than other notions of continuity. One of the reasons it is useful is that if we are working with a continuous function on a closed interval, we get uniform continuity for free.


What facts are known about the hypothetical smallest divergent integer in the collatz conjecture? If there is a divergent integer in the collatz conjecture then there must be a smallest divergent number by the WOP. We can observe some properties of this number such as it must be odd because if it was even then the next number in the series $n/2$ is smaller which is a contradiction. Are there any other known facts about the smallest number?

Such a number, call it $k$, cannot be congruent to $2 \;(\bmod\; 3)$, because otherwise it is possible to create a new counterexample.

For example, consider $\frac{2k-1}{3}$.

$$\frac{2k-1}{3}<k$$

For $k > -1$.

As $\frac{2k-1}{3}$ is odd, we iterate to $2k$. Being even, $2k$ goes to $k$.
Therefore, not only is $k$ a counterexample, so is $\frac{2k-1}{3}$. As it is lesser, this contradicts, $k$ as the least counter-example.

Therefore, $\frac{2k-1}{3}$ cannot exist. This happens when $k \not\equiv 2 \;(\bmod\; n)$.

This logic passes the "$3n-1$" test, as if we take $7$, a known counter-example, we get $5=\frac{2\times7+1}{3}$, and $5$ is also a known counterexample.

Such a number must be of the form $4n-1$ since if it was of the form $4n+1$ you would lead to $3n+1$ which is smaller.


Existence of an analytic function under some given conditions Which of the followings is(/are) correct?

There exists an entire function $f:\mathbb C \to \mathbb C$ which takes only real values & is such that $f(0)=0$ & $f(1)=1$.
There exists an entire function $f:\mathbb C \to \mathbb C$ such that $f\bigl(n+\dfrac{1}{n}\bigr)=0$ , $\forall n\in \mathbb N$.
There exists an entire function $f:\mathbb C \to \mathbb C$ which is onto & which is such that, $f(\frac{1}{n})=0$ , $\forall n\in \mathbb N.$

I tried through the Uniqueness Theorem for options (2) & (3) & apply consequence of Schwarz-Pick lemma & I get (1) & (3) are correct.

Am I right? I think I am wrong, since it contradicts the answer of the book?

Where my fallacy ?

1. If $f$ would be such a function, then $g(z)=e^{if(z)}$ would be entire, bounded, and non-constant.

2. $f(z) =0$...

3. The functions $f$ and $0$ are entire and equal on a Set which has an accumulation point....

False as range(f) cannot have dimension one { Consequence of Open mapping theorem} unless f is a constant.
True as f(n+1/n)=0 for all n implies f assumes value 0 in the nbd. of each n€N. By Identity theorem f must be identically zero.
False as f(1/n)=0 for all n implies f assumes value 0 in the nbd. of zero. Use Identity theorem to get f is identically zero hence it cannot be onto.


Uniform distribution with random support I have a uniform distribution $ X \sim U(A,B) $ where the limits themselves are random: $A \sim N(\mu_A,\sigma_A^2)$ and $B \sim N(\mu_B,\sigma_B^2)$. Hence the support of $X$ is random.

$A$, $B$ are independent. Also consider $\mu_B-\mu_A \gg \sigma_A+\sigma_B$

Any clues on the probability density function of this one?

The pdfs of A and B are known since they are normal distributions

The pdf of X given A and B is
$$
f_{X|A,B}(x|a,b)=1_{x \ge a} 1_{x \le b} \frac{1}{|a-b|}
$$

The joint pdf of X, A and B is
$$
f_{X,A,B}(x,a,b)=f_{X|A,B}(x|a,b) f_A(a) f_B(b)
$$

And the pdf of X is given as a marginal distribution from the latter
$$
f_X(x)= \int_{D_B} \int_{D_A} \! f_{X,A,B}(x,a,b) \, \mathrm{d}a \mathrm{d}b.
$$
where $D_A$ and $D_B$ the domains of A and B respectively.

So, let's be a bit more specific. The idea is that you first draw the samples for A and B. Then, you draw the sample for X.

I am assuming that if it turns out to be A>B, then X is set to an undefined value. In that case $\int_{\mathbb{R}} \mathbb{P}(X=x) dx < 1 $, ie. you are losing some probability mass, but since this event is rare (under your assumption), I would say this is ok. Also, A and B are independent.

However, X can't be independent of A,B or I am missing something (here comes my comment on that you should start stating your question by first defining what the underlying $\sigma$-algebras are).

So, using total probability:
$ \mathbb{P}(X=x) = \int_{\mathbb{R}} \int_{\mathbb{R}} \mathbb{P}(X=x|A=a,B=b) \cdot \mathbb{P}(A=a) \cdot\mathbb{P}(B=b) \cdot da \cdot db $

where:
$\mathbb{P}(X=x|A=a,B=b) = 1_{x\geq a} 1_{x\leq b} \frac{1}{|a-b|} $

$\mathbb{P}(A=a) = \frac{1}{\sigma_{A}\sqrt{2\pi}} exp\left(-\frac{(a-\mu_{A})^{2}}{\sigma_{A}^{2}}\right) $

$\mathbb{P}(B=b) = \frac{1}{\sigma_{B}\sqrt{2\pi}} exp\left(-\frac{(b-\mu_{B})^{2}}{\sigma_{B}^{2}}\right) $

I realize that computing the integral might be non-trivial.

I hope this sheds some light, or, again, I might miss something on the statement.


How to prove that limit of arctan(x) as x tends to infinity, is $\pi/2$? While working on some probability question, I had to evaluate $\lim_{x \to \infty} \arctan(x)$. I knew the answer intuitively as $\pi/2$, yet I cannot figure out how to prove it by elementary means (without resorting to $\epsilon-\delta$ arguments). How does one prove it (preferably, without resorting to L'Hopital's rule)?

The $\arctan$ function is the inverse function of $$\tan:\left(-\frac{\pi}2,\frac{\pi}2\right)\rightarrow\Bbb R$$
and since this function is monotonically increasing then
$$\lim_{x\to\frac\pi 2}\tan x=+\infty\iff \lim_{x\to+\infty}\arctan x=\frac\pi2$$

$$\lim_{x \to +\infty} \arctan x = \frac{π}{2}$$
and
$$\lim_{x \to -\infty} \arctan x = -\frac{π}{2},$$
then in total
$$\lim_{x \to \infty} \arctan x = \frac{π}{2} \mathrm{sgn}(x).$$


Show that this polynomial is positive Consider the following polynomial in two variables :

$$
Q(k,x)=27x^6 - 144kx^4 + 80k^2x^3 + 240k^2x^2 - 192k^3x + (64k^4 - 128k^3)
$$

Then for any integer $k \geq 5$, the polynomial $Q(k,.)$ (in one variable $x$)
seems to be always positive (i.e, $Q(k,x) >0$ for any real number $x$).
  Prove or disprove.

$x\mapsto Q(5, x)$ is positive. There is numerous ways to prove that, more or less algorithmically.
Assume that there exists a $k$ such that $x\mapsto Q(k,x)$ is not positive, and consider $L$ be the infimum of all real numbers $k \geqslant 5$ such that $x\mapsto Q(k, x)$ is not positive.
I claim that $Q(L,x)$ has a double real root, this is equivalent to say that $L$ is a root of the discriminant of $Q$ w.r.t. the variable $x$.
The discriminant of $Q$ with respect to $x$ is (computed with Maple) :
$$ \operatorname{disc}_x Q = 92162779488452608 k^{16} (k-4)^4 $$
The real roots of this discriminant are 0 and 4, thus $L$ cannot be finite, and hence the claim.

Proofs

Point 1 can be proved using Sturm sequences, and you can compute these sequences using Maple :

sturm(subs(k=5, Q),x,-infinity, infinity);
      0

This means that $Q(5, x)$ has no real roots. Since the leading coefficient is positive, the polynomial $Q(5, x)$ is positive whenever $x\in \mathbb R$.

To prove point 2 consider $x_0$ a real numbers such that $Q(L, x)$ is minimal. We have $Q(L, x_0) = 0$, because if $Q(L, x_0) > 0$ then $x\mapsto Q(L+\epsilon, x)$ is positive for $\epsilon > 0$ small enough, which contradict the definition of $L$. If $Q(L, x_0) < 0$, then $Q(L-\epsilon, x_0)$ is stille negative for $\epsilon > 0$ small enough, which is again a contradiction. So $x_0$ is a root of $Q(L,x)$ and of $\partial_x Q(L,x)$, thus it is a double root. The discriminant thing is usual.
The point 3 is Maple :

factor(discrim(Q, x));

You can try to prove your assumption using induction :

$1.$ Show that :

$Q(5,x)=27\cdot x^6-720 \cdot x^4+2000 \cdot x^3+6000\cdot x^2-24000\cdot x+24000 > 0$  ,for all $x$

$2.$ Suppose that :

$Q(k,x)=27x^6 - 144kx^4 + 80k^2x^3 + 240k^2x^2 - 192k^3x + (64k^4 - 128k^3) >0$ ,for all $x$

$3.$ Try to prove that : $Q(k+1,x) >0 $  , for all $x$ using assumption from step $2$


How much is the minimum of $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$$

We have three unknowns if they were two it were easy but I have no idea when it becomes three unknowns any hints?

note:There isn't any information about value of $a$,$b$ and $c$

Using AM-GM inequality, we have

  $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq 3 \cdot \sqrt[3]{\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a}}=3$$

The equality is indeed attained when

  $$\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\tag{1}$$

From $(1)$, we must have $a=b=c$ to attain equality.

Note:

The AM-GM inequality holds true in this case, only when $a,b,c>0$ or $a,b,c<0$. This can be seen by substituting $x=\frac{a}{b},y=\frac{b}{c},
z=\frac{c}{a}$ and setting $x,y,z >0$
A simple counter-example to show that the sign of $a,b,c$ must be same is $a=b=-1$ and $c=1$.(This can also be proved using contradiction in the previous argument.)
If $a,b,c \in \mathbb{R}$ and no other condition is given, then as pointed out by @Behrouz Maleki, $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \rightarrow -\infty.$

Well, the best I can think of, [that you stated the minimum, not the least absolute value]:

if $A$ is some arbitrarily large negative, i.e. $-10^{100}$
and $B$ is some arbitrarily small positive, i.e. $10^{-100}$ (so therefore $A/B= -10^{200}$)
and $C$ is some arbitrarily non-zero number

As $B$ decreases, $B/C$ approaches $0$, and as $A$ approaches $-\infty$,  $C/A$ approaches $0$ as well, but $A/B$ approaches $-\infty$.

As for minimizing the absolute value, that would take much more complex [or just different] maths. To summarize, $A=-\infty$ and $B=1/\infty$ and $C\ne0$

I am almost certain you meant minimizing the absolute value... but just in case, here's what I could think of.


Prove that $H_n=\{\sigma \in S_n \mid \sigma (i) \equiv i \pmod 3 \}$ with $n≥2$ is a subgroup of $S_n$. Prove that $H_n=\{\sigma \in S_n \mid \sigma (i) \equiv i \pmod 3 \}$ with $n≥2$ is a subgroup of $S_n$.

I'm doing this problem and I don't know if my approach is correctly done.

First of all, we see that $id\in H_n$, because $id(i)=i\equiv i\! \pmod 3$, and $H_n$ is not empty.
We have to see now that if $\sigma$, $\tau \in H_n$, then $\sigma\tau^{-1}\in H_n$.
First, we see that if $\tau \in H_n$, then $\tau (i) \equiv i$ (mod $3$). Furthermore, $\tau^{-1} (i) \equiv i \pmod 3$. So  $$\sigma\tau^{-1}(i)=\sigma(\tau^{-1}(i))=\sigma(i+3k)\equiv i+3k \!\!\!\pmod 3 \equiv i\!\!\! \pmod 3 \in H_n.$$
So $H_n$ is a subgroup of $S_n$.

I know that this is an easy exercise, but I'm having some troubles with the permutation groups, so I would like someone to say if I made any mistakes
Thank you.

There seems to be absolutely no problem with your proof. You don't seem to have any issues with symmetric groups at all. What I sometimes find useful is just to draw a picture. Something like this is what I'm talking about. It helps you comprehend how elements are "mapped" to each other which is very helpful in giving you intuition and insight before coming up with a proof. For an example if you drew the picture with respect to this problem you will see that any element must be mapped to itself or can be displaced by three elements at a time either way. You  can also discover that you can draw out all the elements that are congruent to each other modulo $3$ and see that elements in this set must be mapped to itself.

Another advice would be to delve deeper into exercises you find easy and look for other hidden patterns. For an example, given this problem it would be a fabulous exercise to think of how many elements actually are in $H_n$. That is given $S_n $ find out the number of mappings $\sigma \in S_n$ such that $\sigma(i) \equiv i \pmod 3$. I think for $n = 3(k + 1)$ there are $ 3 k! $, for $n = 3k + 2$ there are $ 2k! + (k- 1)! $ and for $n = 3k + 1$ there are $k! + 2(k-1)!$ number of such mappings. You wanna get better at manipulating symmetric groups? Then check if my claim is right.


Could someone please check my result for this complex logarithm exercise? I solved the following exercise and would appreciate it if someone would check my answer:

Exercise: Find the principal value of the logarithm of the following numbers:

$(i(i-1))^i$ and $i^i (i-1)^i$

My answer:

$$ \log((i(i-1))^i) = i \log (i (i-1)) = i(\log \sqrt{2} + i \arg e^{i \pi  5 \over 4}) = i \log \sqrt{2} + {3 \over 4 }\pi$$

and

$$ \log i^i(i-1)^i = \log (e^{i \pi \over 2})^i(e^{-i \pi \over 4})^i = \log e^{-{\pi \over 2}} e^{\pi \over 4} = {\pi \over 4 } - {\pi \over 2} = - {\pi \over 4}$$

$i(i-1)=-1-i$ so both expressions are equal to $(-1-i)^i = (\sqrt 2 e^{i\frac{5\pi}{4}})^i$

$log (-1-i)^i = i \;log(\sqrt 2 e^{i\frac{5\pi}{4}})= -\frac{5 \pi}{4}+i\;log\sqrt2$  is correct for both expressions.


the existence of sequences that approach infimum and supremum without knowing monotonicity Let $f: [a,b] \rightarrow$ be a continuous function. Then $f$ is bounded, which implies that $\sup f$ and $\inf f$ exist. I don't understand the following statement: there exists sequences, $f([a,b])$ that approach supremum and infimum. That is, there exist $\{f(x_n)\}$ and $\{f(y_n)\}$ such that $\lim \{f(x_n)\} = \inf f([a,b])$ and $\lim \{f(y_n)\} = \sup f([a,b])$. As far as I know, the sequence converges if the sequence is bounded and monotone, but here, we don't know the sequence is monotone. Then, how do we know the existence of sequences that approach infimum and supremum?

In case of continuous functions on closed intervals, the supremum is in fact the maximum and the infimum is the minimum.

Both of these values are attained.

You may generate  monotonic sequences  $f(x_n)$ and $f(y_n)$ to approach these values  using the definition of supremum and infimum along with the continuity of $f(x).$

Note that the monotonicity is not a requirement here, but an arbitrary choice as long as the limits are what you want them to be.

Let $M = \sup f([a, b])$. Then by definition of a supremum, for all $\epsilon > 0$ there must exists $y \in f([a, b])$ such that $M - \epsilon < y \leq M$ (else $M$ would not be a least upper bound). Also $y \in f([a, b])$ therefore $y = f(x)$ for some $x \in [a, b]$.

Try using this with $\epsilon = \frac{1}{n}$ for $n = 1, 2, \ldots$ to get a sequence $x_n$ such that $f(x_n) \to M$. A similar method is used to deal with the infinum.


ring homorphism from $\mathbb{Z+Z}$ to $\mathbb{Z+Z}$ Determine all the ring homorphisms from $\mathbb{Z+Z}$ to $\mathbb{Z+Z}$

This homorphism can be determined by $f((0,1))$ and $f((1,0))$ then how to proceed further?
This question is discussed here Determine all ring homomorphism from$ \mathbb{Z}\oplus \mathbb{Z}$into $\mathbb{Z}\oplus \mathbb{Z}$
But I am not understanding.

Note that $(0,1)$ and $(1,0)$ both square to themselves: $$(1,0)(1,0)=(1,0),\qquad (0,1)(0,1)=(0,1).$$  Thus they must map to elements of the form: $(x,y)$ with $$(x,y)=(x,y)(x,y)=(x^2,y^2).$$  Thus $x^2=x$ and $y^2=y$ so $x,y\in\{0,1\}.$
That gives $16$ possibilities, with each of $(1,0)$ and $(0,1)$ mapping to one of $(0,1),(1,0), (1,1), (0,0)$.
However a ring homomorphism must map $(1,1)$ to $(1,1)$.  Thus you are left with $4$ ring homomorphisms:
$$\begin{array}{ccccccc}(1,0)\mapsto (0,0), && (1,0)\mapsto (0,1),&&(1,0)\mapsto (1,0),&&(1,0)\mapsto (1,1).\\
(0,1)\mapsto (1,1) && (0,1)\mapsto (1,0)&&(0,1)\mapsto (0,1)&&(0,1)\mapsto (0,0)
\end{array}$$
Note these correspond to the identity, swapping the factors, and projecting the first or second factor onto the diagonal.


A Problem on Inverse Trigonometric Functions and Differential Equation. $f(x)=\left(\sin \left(\tan ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right)^{2}-1,\ |x|>1$
If $\displaystyle\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} x}\left(\sin ^{-1}(f(x))\right)$ and
$y(\sqrt{3})=\frac{\pi}{6}$, then $y(-\sqrt{3})$ is equal to :
Options:
$1. \quad-\frac{\pi}{6}\\
2. \qquad \frac{2 \pi}{3}\\
3. \qquad \frac{5 \pi}{6}\\
4. \qquad\frac{\pi}{3}$

It can be easily shown that $f(x) = [\sin(\tan^{-1}x) + \sin(\cot^{-1}x  )]^2 -1 = \frac{2x}{1+x^2}$.
$\frac{d}{dx} (\sin^{-1}\frac{2x}{1+x^2}) = \frac{-2}{1+x^2}$
Now $\frac{dy}{dx} = \frac{1}{2} \frac{ d(\sin^{-1}f(x))}{ dx} = \frac{-1}{1+x^2}$ from which we get $y = - \tan^{-1}x  + C. $
Now $y (\sqrt 3) = \frac{\pi}{6}$. So $C= \frac{\pi}{2}$.
So $y = \frac{\pi}{2} -  \tan^{-1}x $.
$y(-\sqrt 3) = \frac{5\pi}{6}$
Have I gone wrong any where? Can anyone please check my solution?

$$\dfrac{d}{dx}(\sin^{-1}f)= \dfrac{1}{\sqrt{1-f^2}}\cdot \dfrac{df}{dx}=P\cdot Q \;;$$
$$P=\dfrac{1}{\sqrt{1-(2x/(1+x^2))^2}} = \dfrac{1+x^2}{1-x^2} $$
Differentiating by Quotient rule
$$Q=\dfrac{2}{(1+x)^2}; \quad PQ=\dfrac{2}{1-x^2};\; $$
EDIT1:
Stopped too early by mistake.  Continuing,
$$\dfrac{dy}{dx}= \dfrac{1}{1-x^2}$$
Integrates to
$$y=\log_e\sqrt {\dfrac{1+x}{1-x}} +c$$
given
$$ y(\sqrt3)=\dfrac{\pi}{6}$$
$y$ is not real, constant of integration cannot be evaluated. For this result no option available.


Where do the higher order terms in Taylor series come from? I can see the first order approximation for a Taylor series: if we want to approximate $f(x)$ near $x_0$, then it's close to the line with slope $f'(x_0)$ that intersects $f(x_0)$, giving

$f(x) \approx f(x_0) + f'(x_0)(x-x_0)$

However, I don't see a good way to understand the higher order terms $f^{(n)}(x_0)/n! (x-x_0)$. I understand the proof of Taylor's theorem, but I don't have any real intuition for it. How were Taylor expansions discovered (i.e. what would lead one to guess the form of Taylor expansions)?

To simplify, let's discuss first what happens near $x_0 = 0$.

The way the local linear approximation works is that we are looking for the line that "best approximates" the function near $0$. This means that we want a function $y=a+bx$ with the property that the error between $y=a+bx$ and $y=f(x)$ goes to $0$ faster than $x$ approaches $0$. The first That is, we want
$$\lim_{x\to 0}\frac{f(x) - (a+bx)}{x} = 0.$$
A bit of work shows that the only $a$ and $b$ that can work are $a=f(0)$ and $b=f'(0)$:
$$\begin{align*}
\lim_{x\to 0}\frac{f(x)-(a+bx)}{x} &= \lim_{x\to 0}\frac{f(x)-f(0)+f(0)-(a+bx)}{x}\\
&= \lim_{x\to 0}\frac{f(x)-f(0)}{x-0} + \lim_{x\to 0}\frac{f(0)-(a+bx)}{x}\\
&= \lim_{x\to 0}\frac{f(x)-f(0)}{x-0} + \lim_{x\to 0}\frac{f(0)-a}{x} - \lim_{x\to 0}\frac{bx}{x}.
\end{align*}$$
If the function is differentiable, then the first limit is $f'(0)$; the last limit is $b$; and if we want the limit in the middle to exist, we need $f(0)=a$. So the only way this limit is equal to $0$ is if $b=f'(0)$ and $a=f(0)$.

This tells us that the line of "best approximation" to $f(x)$ near zero is $y = f(0) + f'(0)x$, which is precisely the local linear approximation.

(The fraction $\frac{f(x)-(a+bx)}{x}$ is the "relative error", because the numerator measures the error, but by dividing it by $x$ we are trying to take into account how large the quantity is; if I tell you that I measured a distance and I was off by 2 miles, you don't know if it was a very good approximation or a very bad one; it would be very good if I were measuring the distance to the moon; it would be very bad if I was measuring the length of my desk; the denominator tries to "normalize" the measurement so that it tells us how big the error is relative to how big the thing we are measuring is)

Now, the advantage of the local linear approximation is that it is very simple; the disadvantage is that it can get bad pretty quickly; for example, the local linear approximation of $y=\cos x$ at $0$ is $y=1$, which gets "bad" very quickly.

So we may want to approximate with something else which, while still easy, has a better chance of being a good approximation.

One possibility is to go from linear to quadratic: let's try to approximate $y=f(x)$ with a quadratic function, $y = a+bx+cx^2$; again, in order for the approximation to be very good, we want the error to go to zero faster than $x$ goes to $0$. And since now we have a parabola, we can require that the curvature of the parabola at $0$ be the same as the curvature of $y=f(x)$ at $0$ (this is what messes us up with $y=\cos x$: it has big curvature, but the local linear approximation is flat).

Now, for this parabola to approximate $y=f(x)$ well near $0$, we are going to want their local linear approximations to be the same (if they have different local linear approximations, then we can't expect $y=a+bx+cx^2$ to be "close to" $f(x)$). The local linear approximation to $y=a+bx+cx^2$ at $x=0$ is give by $a+bx$, so we want $a+bx = f'(0)x+f(0)$; so $a=f(0)$, and $b=f'(0)$.

To get the same curvature, we want $f''(0) = y''(0)$. Since $y''(0) = 2c$, we want $c = \frac{1}{2}f''(0)$. So our $y$ will be
$$y = f(0) + f'(0)x + \frac{f''(0)}{2}x^2.$$

In fact, this is indeed a very good approximation: it has the same value as $f(x)$ at $x=0$; the relative error goes to zero, since
$$\begin{align*}
\lim_{x\to 0}\frac{f(x) - (f(0)+f'(0)x + \frac{1}{2}f''(0)x^2)}{x} &=
\lim_{x\to 0}\frac{f(x) - (f(0)+f'(0)x)}{x} - \lim_{x\to 0}\frac{f''(0)x^2}{2x}\\
&= 0 - \lim_{x\to 0}\frac{1}{2}f''(0)x = 0.
\end{align*}$$
But more than that: the relative error between the derivatives goes to $0$:
$$\begin{align*}
\lim_{x\to 0}\frac{f'(x) - (f'(0)-f''(0)x)}{x} &= \lim_{x\to 0}\frac{f'(x)-f'(0)}{x}- f''(0)\\
&= f''(0)-f''(0) = 0.
\end{align*}$$

So not only does $y=f(0) + f'(0)x + \frac{1}{2}f''(0)x^2$ have the same value, and have the best possible relative error, the derivative is the best possible approximation to the derivative of $f(x)$; so the graph will have very similar shape near $0$.

If we then go on to a degree $3$ approximation, $y=a+bx+cx^2+dx^3$; if we want the relative error to go $0$, we are going to need $a=f(0)$, $b=f'(0)$. If we want the relative error in the derivative to go to $0$, we are going to need $c=\frac{1}{2}f''(0)$ as before. What if we want the relative error of the second derivative to go to $0$ as well? The second derivative of $y=a+bx+cx^2+dx^3$ is $2c+6dx = f''(0) + 6dx$. So we have:
$$\begin{align*}
\lim_{x\to 0}\frac{f''(x) - (f''(0)+6dx)}{x} &= \lim_{x\to 0}\frac{f''(x)-f''(0)}{x} - \lim_{x\to 0}6d\\
&= f'''(0) - 6d.
\end{align*}$$
For this to be equal to $0$, we need $6d = f'''(0)$, or $d = \frac{1}{6}f'''(0)$.

If we then go to a degree $4$ approximation and ask that the relative error, the relative error of the derivatives, the relative error of the second derivatives, and the relative errors of the third derivatives all go to $0$, then we find that we need the function $y= f(0) +f'(0)x + \frac{1}{2}f''(0)x^2 + \frac{1}{6}f'''(0)x^3 + ex^4$, where
$f^{(4)}(0) = 12e$, leading to $e = \frac{1}{12}f^{(4)}(0)$.

It is not hard to then see, inductively, that the coefficient we get for $x^n$ if we proceed this way will always be $\frac{f^{(n)}(0)}{n!}$.

You can repeat the same idea around any point $x_0$, but if you do it directly it turns out that you cannot easily use the coefficients you found for, say, the linear approximation, in the quadratic approximation; you end up with a system of equations instead of simply repeating the old coefficients. The simple way of "fixing" this is to shift everything to $0$, solve the problem at $0$, and then shift it back.

So if you want to solve the problem at $x_0$, we consider instead $g(x) = f(x+x_0)$, because then approximating $f$ near $x_0$ is the same as approximating $g$ at $0$. Moreover, $g'(x) = f'(x+x_0)(x+x_0)' = f'(x+x_0)$, $g''(x) = f''(x+x_0)$, etc. So, from the work above, we see that the degree $n$ approximation to $g$ near $0$ that has best possible relative error, best possible relative error between derivatives, best possible relative error between 2nd derivatives, etc. is
$$g(0) + g'(0)x + \frac{1}{2}g''(0)x^2 + \cdots + \frac{1}{n!}g^{(n)}(0)x^n.$$
But form the above, this is the same as
$$f(x_0) + f'(x_0)x + \frac{1}{2}f'(x_0)x^2  +\cdots + \frac{1}{n!}f^{(n)}(x_0)x^n.$$
This is the local approximation to $f(x+x_0)$ near $x=0$. Replacing $x$ with $x-x_0$, we get $f(x-x_0+x_0) = f(x)$, the "old $x$" was near $0$, so the new $x$ needs to be near $x_0$, and we get that for $x$ near $x_0$, we have
$$f(x)\approx f(x_0) + f'(x_0)(x-x_0) + \frac{1}{2}f'(x_0)(x-x_0)^2 + \cdots + \frac{1}{n!}f^{(n)}(x_0)(x-x_0)^n$$
exactly the formula for the $n$th Taylor polynomial approximation to $f(x)$ near $x_0$.

Sam L. raises the fair question of why we would want to assume that our approximations are polynomials. If you are going via the "make the relative error go to $0$; make the relative error of the relative error go to $0$", etc., you will discover that you are naturally led to polynomials.

Here's another motivation: when we encounter a function which we want to integrate and for which we are unable to find an antiderivative, we end up with two possible approaches:

Attempt to approximate the value of the integral via Riemann sums; in essence, the integral is a limit, so we can approximate the limit by using terms of the sequence whose limit we are trying to compute. This leads to left, right, midpoint, trapezoid, Simpson approximations, and other methods of numerical approximations to the integral using the function $f$.
Find a function $\mathcal{F}$ which approximates $f$ but is easy to integrate. (In fact, this is in part what is behind the Simpson rule approximation, in that it approximates the function $f$ in each subinterval by a quadratic function). Since polynomials are generally easy to integrate, trying to find polynomials that are good approximations to $f$ seems like a good idea; that is, the class of polynomials are a good proving ground to try to find "good approximations to $f$", because they are easy to integrate.

I borrow this GIF from Wikipedia. As you can see, a higher order approximation is more accurate:


How and what do I assume for my induction? Hi I wrote a formula to know the 100th term for the following sequence:
$$3, 10, 17, 24, 31$$
Note that each following number in the sequence will be 7 more than the previous number.
And here is the formula:
$$7N - 4$$
$$
\begin{array}{c|c}
Term   & 1 & 2  & 3  & 4  & 5  & Nth       & 100 \\ \hline
Number & 3 & 10 & 17 & 24 & 31 & 7\times N - 4 & 696 \\
\end{array}
$$
I proved it with the base case $$5 \times 7 - 4 = 31$$
which is correct.
And I proved it with$$100\times7 - 4 = 696,$$
which is correct.
Also I was looking at this induction example:

So that I can apply on my own induction. But the problem is I can't really apply it, because I don't know what to assume/claim to prove it. This is where I'm stuck:
$$
\mathrm{Induction \enspace step;  \enspace Assume  \enspace true  \enspace for  \enspace n=k,  \enspace show \enspace true \enspace n=k+1}
$$
$$
\mathrm{Assume: 3,10,17,24,...,[WHAT \enspace DO \enspace I \enspace NEED \enspace TO \enspace WRITE \enspace HERE?] = 7N-4}
$$
Does someone has any idea what I need to write in the above assumption? Thanks in advance.

The induction step will be trivial.
Assuming the $k$th term is given by $7k+4$, the $(k+1)$th term is $(7k+4) + 7 = 7(k+1)+4$.
This completes the induction.

Let $a_n$ be the $n$'th term in an arithmetic sequence, where:
$$
a_1 = 3
$$
$$
d = 7
$$
We find the $n+1$ term as:
$$
a_{n+1} = 3 + 7n
$$
If the the sequence is arithmetic, then :
$$
a_{n} = a_{n+1} - 7
$$
$$
\downarrow
$$
$$
a_{n}+7 = a_{n+1}
$$
Now we can see from the third row and the last row:
$$
a_{n+1}= a_n + 7
$$
$$
a_{n+1} = 3 + 7n
$$
$$
\downarrow
$$
$$
a_n + 7 = 3 + 7n
$$
$$
a_n = 7n-4
$$


Problem with solving the ODE equation $ty'=3y+t^5y^\frac{1}{3}$. I have problem with solving the following equation:

  $$ty'=3y+t^5y^\frac{1}{3}$$

I know it's easy without the $y^\frac{1}{3}$ term, but I'm confused now.

Any help would be appreciated.

It's a Bernoulli's ODE, first $y=0 $ is a solution , now by dividing by $y^{1/3} $ we find $ \frac{y'}{y^{1/3}} - 3 \frac{ty}{y^{1/3}} = t^4  $ , Now we put $z= \frac{1}{y^{\frac{-2}{3}}}$ we will find by replacing in the ODE $ \frac{2}{3}z'-\frac{3}{t}z=4t^4 $ by solving it you will find easily $z$ then $y$.

for your Control the solution is given by $$\left( y \left( t \right)  \right) ^{2/3}-2/9\,{t}^{5}-{t}^{2}{\it
\_C1}=0
$$


Finding the limit of $ \frac{\tan^2x-x^2}{x^2\tan^2x} $ How to find the limit of $ \frac{\tan^2x - x^2}{x^2\tan^2x}$ as $x$ approaches $0$.

I don't know how to approach and starting differentiating the numerator and denominator many times which after the 4th time gave me the result $\frac{2}{3}$. I want to know if there is any good simplification which can lead me to a simple and quick solution. Please give some ideas.

Thank you.

Note that as $x \to 0$, we have $\tan(x) = x + \dfrac{x^3}{3} + O(x^5)$ (this is really a 2 line calculation). So,
\begin{align}
\tan^2(x) &= x^2 + \dfrac{2x^4}{3} + O(x^6)
\end{align}
Hence,
\begin{align}
\dfrac{\tan^2x - x^2}{x^2 \tan^2x} &= \dfrac{\frac{2x^4}{3} + O(x^6)}{x^4 + O(x^6)} \\
&= \dfrac{\frac{2}{3} + O(x^2)}{1 + O(x^2)},
\end{align}
where I divided top and bottom by $x^4$. Thus, if we take the limit $x \to 0$, the answer is $\dfrac{2}{3}$.

By the way, observe that it is no coincidence you had to differentiate $4$ times, because as you can see in the above derivation, the Taylor polynomials of the numerator and denominator begin at the $x^4$ term, so it is precisely the ratio of these coefficients which you're calculating if you use something like L'Hopital's rule.

Recall that $\lim_{x\to 0}\dfrac{\sin(x)}{x}=1$. The proof of this involves the definition of trigonometric functions by unit circle, which I will omit here. An immediate consequence of this is $\lim_{x\to 0}\dfrac{\tan(x)}{x}=1$. Hence dividing both the numerator and the denominator by $x^2$, we see that the desired limit is indeed 0.


Can't determine if given relation is equivalence relation Definition of relation  ~

$(a,b)$ ~ $(c,d)$ $\iff$ $bc^2=da^2$,
where $(a,b),(c,d)$ are from $\mathbb{R}\times\mathbb{R}$ and $(a,b),(c,d)$ are different from $(0,0)$

First of all, I wonder if R can be equivalence relation if it is not defined on $\mathbb{R} \times \mathbb{R}$($(0,0)$ is not included ), because I will not be able to find classes that divide the whole $\mathbb{R} \times \mathbb{R}$.

If I can observe $\mathbb{R} \times \mathbb{R}$ without $(0,0)$, then I should see if this relation is reflexive, symmetric and transitive.
Clearly, this relation is reflexive, but I have troubles with determining if it is symmetric and transitive.
( Because of elements like $(a,0)$ or $(0,b)$ and dividing with zero )
Also, if it is, I can't imagine what classes are.

Anything you can think of will be helpful.
Thank you in advance.

Reflexive:
We need to prove that $(a,b)R(a,b)$, let us check if this is true. We have $$ba^2 = ba^2 \implies (a,b)R(a,b).$$

So our relation is reflexive.

Symmetric: We need to prove that if we have $(a,b)R(c,d)$ then this implies $(c,d)R(a,b)$. We have $$bc^2 = da^2 \implies da^2 = bc^2 \equiv (c,d)R(a,b).$$

So our relation is symmetric.

Transitive: We need to show that if we have $(a,b)R(c,d)$ and $(c,d)R(e,f)$ then we have $(a,b)R(e,f)$. We do so below: We need to show $be^2 = da^2$

$$bc^2 = da^2 \text{ and } de^2 = fc^2 \implies c^2 = \frac{de^2}{f}$$

Substituting the latter into the first equation, we get $$b \cdot \frac{de^2}{f}  = da^2 \iff be^2 = fa^2 \equiv (a,b)R(e,f).$$

To complete the proof of transitivity, you need to consider the case $f=0$, look in the comment section for how to do so.

Since our relation is reflexive, symmetric and transitive, it must be an equivalence relation and as you can see, there was no need to divide by $0$ or any other funky stuff. You can split the set $\mathbb{R} \times \mathbb{R}$ not including $(0,0)$ into equivalence classes.

Note: I used $R$ instead of ~ for the relation.

Unless otherwise stated, assume that all $x$-coordinates are nonzero. Also note that this answer is mostly for fun; I wanted to visualize the equivalence classes!

If we say that $(x_1, y_1) \sim (x_2, y_2)$ if and only if $y_1x_2^2 =  y_2x_1^2$, then (assuming neither $x_1$ nor $x_2$ is $0$) dividing both sides by $x_1^2x_2^2$ we have

$$\frac{y_1}{x_1^2} = \frac{y_2}{x_2^2}.$$

Now, choose a fixed $x_1$ and $y_1$, and let $k =\frac{y_1}{x_1^2}$. Any pair $(x, y)$ related to $(x_1, y_1)$ satisfies $\frac{y}{x^2} = k = \frac{y_1}{x_1^2}$, so that $y = kx^2$. In other words, pairs $(x_1, y_1)$ and $(x_2, y_2)$ with nonzero $x$-coordinates are related if their $y$-coordinates are proportional to the square of their $x$-coordinates, with the same constant of proportionality.

For a fixed constant of proportionality $k$, we have then that points satisfying the equation $y = kx^2$ are related. For points with nonzero $x$-coordinates, that the relation is an equivalence relation follows from the properties of equality (namely, that equality is an equivalence relation).

Note that all points whose only nonzero coordinate is $y$ are related (as they correspond to $k = 0$). I'll let you think about what to do with points whose $x$-coordinate is $0$, and what all they're related to.

Thus, (most of) the equivalence classes can be visualized as the graphs of parabolas $$y = kx^2,$$ with one equivalence class for each choice of a real number $k$. Since each of these parabolas go through $(0, 0)$, it should be clear why we needed to remove it from the set under consideration: Every other pair would have been related to it!


solve equation with sum $\ln x-\ln(\sqrt[2]{ x})+\ln(\sqrt[4]{ x})-\ln(\sqrt[8]{x})+...=2$ How to solve this? Any advice?
$$\ln x-\ln(\sqrt[2]{ x})+\ln(\sqrt[4]{ x})-\ln(\sqrt[8]{x})+...=2$$
Next step I do this
$\sum\limits_{n=0}^\mathbb{\infty}(-1)^n\ln(x^{\frac{1}{2^n}}) = 2$
But I don't know next step. Thanks.

For the property of logarithms you can rewrite your sum as:

$$\sum_{n=0}^{\infty}\left(-\frac 12\right)^n\ln x=\ln x-\frac 12\ln x+\frac 14\ln x-\dots=2$$

Now remember the formula for the geometric series:

$$\frac 1{1-z}=\sum_{n=0}^{\infty}z^n=1+z+z^2+z^3+...$$

You can easily see from there that:

$$1-\frac 12+\frac 14-\frac 18+\dots=\frac 23$$

So you have that:

$$\sum_{n=0}^{\infty}\left(-\frac 12\right)^n\ln x=\ln x\sum_{n=0}^{\infty}\left(-\frac 12\right)^n=\frac 23\ln x$$

Now your equation is:

$$\frac 23\ln x=2$$

$$\ln x=3$$

$$x=e^3$$


transformations for graphs - precalculus The problem is for homework,  but I will change the problem a bit so as to try to be able to do it by myself from what someone or some people will answer here.  Ok.

Let f be a function whose domain is [0,3] and range is [-3,0]. For each function g below, describe the transformations necessary to obtain the graph of g from the graph of f :

(a) g(x) = 3f(x-4) + 3   /////
(b) g(x) = -3(f(-x) - 3)

Im not sure of a couple things,  in (b) f(-x) what does it become,  and the 3f in (a) what do I do with that multiply it by the range?

Bijan

You should brush up on the general principles, like say here: http://www.purplemath.com/modules/fcntrans.htm

In (a), multiplying $f(x-4)$ by 3 will stretch the graph vertically by a factor of 3. It's true that the endpoints of the range of $f(x-4)$ (which is not the same as the range of $f(x)$) will be multiplied by 3; and so will all the y-values of the output.

In (b), the graph of $f(-x)$ is the same as the graph of $f(x)$ having been reflected across the y-axis. If $h(x) = f(-x)$, and for example $f(2) = 0$ (so $f$ has a point at $(2,0)$), then $h(-2) = 0$ (so $h$ has a point at $(-2, 0)$), and so forth.


Can I have a "remark" after a "lemma" before its "proof"? I've developed a lemma which is accompanied by its proof, naturally. But the lemma is a little bit complicated and I'd rather present some more explanations about it, before digging into the proof.

I'm an engineering student with pretty trivial experience in formal argumentation. So, I'm wondering whether I can put a remark section after the lemma to enlighten the reader about its meaning, then presenting the proof.

Actually, as I checked some proofs, authors often consider extra explanations after the proof. But I believe that such explanations are necessary before the proof, in this case.

Sure, you can do that -- many math textbooks have cases where they state a theorem, then proceed to give some intuition about what it means, applications of it, until finally the proof appears. (In some applied fields it is even the preferred publication style to postpone proofs that don't directly illuminate the main message of the article, all the way to an appendix).

Just make sure it is clear to the reader that a proof will follow later, so he doesn't think the proof is supposed to be so obvious that you don't even deign to give it.

In other words, don't just put "Remark." at the beginning of a paragraph and then start blathering. Write something like

  Before we prove the lemma, let's observe that it doesn't actually contradict the results of the previous section ...

(or whatever).

In my opinion, yes, since etymologically it is a proposition which is taken for granted, and the proof thereof is postponed (λῆμμα = ‘what is taken’ or ‘what is given’).


Question regarding an inequality How to prove that $$ \frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+\cdots+\frac{x_n}{1+x_1^2+\cdots+x_n^2}<\sqrt{n} $$ knowing that $(x_n)$ is a positive sequence ? I looked up all kinds of inequalities such AM-GM, Chebyshev, Cauchy-Schwarz, but I couldn't manage to obtain anything useful..
Can anyone help ?

I think the proof can be divided to two parts: when all elements are equal, and then when they are not equal, try to prove that the sum would be less than sum of all equal elements (either max or min of the sequence).

First step: any x can be replaced by 1/x to increase the sum, if x > 1.

Second step: For x < y <= 1 and C >= 1, we have x / (C + x^2) < y / (C + y^2)

Third step: we replace all items with x(k) = max of sequence and have a more convenient formula to prove.

Fourth step: when all elements are equal, sum is less than 1/2 + 1/3 + 1/4 ... Which is the exact case when all elements are equal to 1, and this is a known sequence, as I remember


Calculate Rotation Matrix to align Vector A to Vector B in 3d? I have one triangle in 3d space that I am tracking in a simulation. Between time steps I have the the previous normal of the triangle and the current normal of the triangle along with both the current and previous 3d vertex positions of the triangles.

Using the normals of the triangular plane I would like to determine a rotation matrix that would align the normals of the triangles thereby setting the two triangles parallel to each other. I would then like to use a translation matrix to map the previous onto the current, however this is not my main concern right now.

I have found this website http://forums.cgsociety.org/archive/index.php/t-741227.html
that says I must

determine the cross product of these two vectors (to determine a rotation axis)
determine the dot product ( to find rotation angle)
build quaternion (not sure what this means)
the transformation matrix is the quaternion as a 3 by 3 ( not sure)

Any help on how I can solve this problem would be appreciated.

Suppose you want to find a rotation matrix $R$ that rotates unit vector $a$ onto unit vector $b$.

Proceed as follows:

Let $v = a \times b$

Let $s = \|v\|$ (sine of angle)

Let $c = a \cdot b$ (cosine of angle)

Then the rotation matrix R is given by:
$$R = I + [v]_{\times} + [v]_{\times}^2\frac{1-c}{s^2},$$

where $[v]_{\times}$ is the skew-symmetric cross-product matrix of $v$,
$$[v]_{\times} \stackrel{\rm def}{=} \begin{bmatrix}
\,\,0 & \!-v_3 & \,\,\,v_2\\
\,\,\,v_3 & 0 & \!-v_1\\
\!-v_2 & \,\,v_1 &\,\,0
\end{bmatrix}.$$

The last part of the formula can be simplified to
$$
\frac{1-c}{s^2} = \frac{1-c}{1-c^2} = \frac{1}{1+c},
$$
revealing that it is not applicable only for $\cos(\angle(a, b)) = -1$, i.e., if $a$ and $b$ point into exactly opposite directions.

I have a simpler method comes from Erigen's "Mechanics of Continua". R is rotational matrix that rotate vector "a" align with vector "b" Matlab Code:

%%%%%% Rotate vector a align with vector b%%%%%%%%%%

syms ax ay az bx by bz k real

a=[ax ay az]'
au=a./sqrt(ax^2+ay^2+az^2)

b=[bx by bz]'
bu=b./sqrt(bx^2+by^2+bz^2)

R=[bu(1)*au(1) bu(1)*au(2) bu(1)*au(3);

   bu(2)*au(1) bu(2)*au(2) bu(2)*au(3);

   bu(3)*au(1) bu(3)*au(2) bu(3)*au(3)]

To verify:

c=R*a
cu=c./sqrt(c(1)^2+c(2)^2+c(3)^2)
simple(bu-cu)

A zero result means that $c$ (rotated $a$) and $b$ are aligned with each other.

simple(sqrt(c(1)^2+c(2)^2+c(3)^2)-sqrt(c(1)^2+c(2)^2+c(3)^2))

A zero result means that $c$ (rotated $a$) and $a$ are of the same length.


Calculus: $\lim_{x \to 1} \frac{x^3 - 1}{x-1}$ Is the limit $0$ or $3$?

$x^3 -1$ can be $(x-1)(x^2 +x +1)$, with the transformation, the limit will be $3$?
Why cannot we just say $x$ is to be $1$, $x^3 -1$ is going to be $0$ so the limit is $0$?

You have shown that you can see why the limit of the expression should be $3$. Regarding why it shouldn't be $0$, consider how $\displaystyle \frac{x^3-1}{x-1}=\left(x^2+x+1\right)\frac{x-1}{x-1}$ would be evaluated at different $x$-values. At $x=3$, we have $\require{cancel}\displaystyle \left(3^2+3+1\right)\frac{\cancel{2}}{\cancel{2}}$ and likewise, at $x=0.5$, we would have $\require{cancel}\displaystyle \left(0.5^2+0.5+1\right)\frac{\cancel{0.5}}{\cancel{0.5}}$. The pattern should be clear; for any $x\ne1$, the terms in the fraction are nonzero and can cancel with no problem. But this includes all $x$ in the vicinity of $1$, which is exactly what we pass through as we approach $1$. Since a limit is not interested in what happens at $x=1$ but in what happens near $x=1$, and we never need to hit $1$ exactly, we never need to evaluate $\cfrac00$ and it does not affect the answer.

Since x can not be 1，and x^3-1=(x-1)(x^2+x+1) ,we can transform the former one to x^2 + x + 1. The limit of it is 1+1+1=3.
For why it can’t be 0. The limit of (x-1) is also 0. We can’t make a conclusion that how much is the limit of 0/0.


What should my $u$ be for substitution? $$\int \limits_{\frac 16}^{\frac 12} \csc \left( \frac{\pi}{t} \right) \cot \left(\frac{\pi}{t}\right) \, dt$$

Having trouble deciding on a substitution for this integral .

Your integral diverges. On the other hand, $$\int_{1/2}^{1/6} \csc(\pi t)\cot (\pi t) dt$$ converges and has a nice closed form. I'll assume a typo and show this.

Rewriting $\csc$ and $\cot$ in terms of $\sin$ and $\cos$ we obtain $$\int_{1/2}^{1/6}\frac{\cos \pi t}{\sin^2 \pi t}dt=\int_{1/2}^{1/6}\frac{\cos \pi t}{1-\cos^2 \pi t}dt.$$ Now the substitution we need is $u=\cos \pi t \leftrightarrow dt=-1/\pi \displaystyle\frac{du}{\sqrt{1-u^2}},$ which produces $$-\frac1\pi\int_{\sqrt{3}/2}^{0}\frac{u}{(1-u^2)^{3/2}}du=\frac1\pi\left\lvert\frac{1}{\sqrt{1-u^2}}\right\rvert_0^{\sqrt3/2}=\frac2\pi-\frac1\pi=\bbox[5px,border:2px solid #C0A000]{\frac1\pi.} $$

Firstly, you need to write that expression in terms of sine and cosine then you will know how to proceed further from there.


need help with a systems of equations question I have these systems of equations:

$$\frac{3}{4}x-\frac{1}{2}y=12$$

$$ax-by=9$$

If $a$ and $b$ are constants, what is $\frac{a}{b}$?

I know that no solutions means that LHS $=0$ basically. so i tried making the RHS equal to 36 by multiplying the top equation by 3 and bottom by 4. that way when i subtract the two, i have the LHS $=0$. however this didn't work. any other way i can do this? the answer is $\frac{3}{2}$

Hint: Plugging $$y=\frac{3}{2}x-24$$in the second equation we get
$$ax-b\left(\frac{3}{2}x-24\right)=9$$
Can you finish?

Unless you've left something out, this isn't solvable. For example, letting $y=\frac32 x-24$ and subbing in to the second equation, we get $$ax-\frac32 bx+24b=9,$$ so $$\left(a-\frac32b\right)x=9-24b.$$ From this, we can conclude either that $a-\frac32b=0$ and $9-24b=0$ (so that $a=\frac32b$ and $b=\frac38,$ so that $a=\frac9{16}$ and, indeed, $\frac ab=\frac32$) or that $$x=\frac{9-24b}{a-\frac32}=\frac{18-48b}{2a-3}.$$ In the latter case, we can then conclude that $$y=\frac{27-72b}{2a-3}-24=\frac{99-72b-48a}{2a-3}.$$ However, this tells us nothing about $a$ and $b$! As long as $a\ne-\frac32$ and $b\ne 0,$ $\frac{a}{b}$ could be literally anything. In fact, since $b$ can be $0,$ then $\frac{a}{b}$ need not even be defined!!

If you know (for example) that both equations are supposed to be for the same line, then you can solve it. The easiest way to do so is to note that the lines should have the same slope, since $\frac ab$ will be the slope; all you have to do, then, is convert them both to slope-intercept form, and you're done.


Discount in Infinity I fully understand that $0.9999...$ mathematically will equal 1.0 exactly as the repeating decimal continues infinitely. This would be contrary to conventional logic where such a fraction is considered approximately 1.0 and not exactly 1.0. Here, of course, the conventional logic fails to understand what it mathematically means when the decimal is repeated in infinity.

However can this principle be translated to an infinite discount as well?

Say a number is reduced by 10% constantly (n-10%-10%-10%...). Will this too mathematically equal zero (as 0.0), or will it only ever get near zero?

Again, conventional logic will say that this will never actually be zero, it will only ever get closer to zero. Is conventional logic mathematically correct in such a case?

Thank you.

The phrase "is reduced by e.g. 10% in infinity" is perhaps ambiguous, but let me try to make it more precise:

If you reduce a quantity $A$ by $10\%$ (i.e., a proportion of $\frac{1}{10}$, then the remaining value is
$$\left(1 - \frac{1}{10}\right) A = \frac{9}{10} A,$$
and, inductively, if you do so $n$ times, then the remaining values is
$$\frac{9}{10}\left( \cdots\frac{9}{10}\left(\frac{9}{10} A\right) \cdots \right) = \left(\frac{9}{10}\right)^n A.$$
But for any small number $\epsilon > 0$, we can make this quantity small than $\epsilon$. (In fact, a little algebra shows that taking $n \geq \left\lceil \frac{\log \epsilon - \log A}{\log r}\right\rceil$ will do.) On the other hand, for any $n$, we have $\left(\frac{9}{10}\right)^n A \geq 0$, and so the as $n \to \infty$, the remaining quantity $\left(\frac{9}{10}\right)^n A$ goes to zero.

In a sense, it's equal to zero in the same way your example is equal to 1. Both converge on to their target, but only after an infinite number of digits/reductions.


zero matrix to the power of 0 Why $0^0=I$?

I'd tried prove that considering $N^0$ where N is a Nilpotent matrix and then using the Cayley -Hamilton theorem

Thanks in advance.

The same caveats as for the exponential of real numbers apply here. If exponentiation means repeated multiplication, then $A^0 = I$ is the base case for all $A$. Exponentiation by a continuous real parameter, on the other hand, should insist that $\mathbf{0}^0$ is undefined.

0^0 must be undefined because here in case of matrices if we take matrix's power than it implies the product of that matrix that power times and it does not make sense for the zero element's power zero I.e. Zeros' power zero , which is undefined.


Sketching the graph of trigonometric functions involving absolute value function How do I sketch the graph of $\tan |x|?$

I know that the modulus over $|x|$ can be thought of as the angle (in radians) will always be positive which implies that the angle is always measured anti-clockwise. But I seem to get no idea about sketching it correctly.

I know how a graph changes when we put absolute value function.
Also I don't have any texts that explains such graphs.

Any reference from the web to learn such graphs are welcome.

Kindly help.

First, draw the graph for $x>0$. This will be the same graph than $\tan x$. Then, since $\tan |-x|=\tan |x|$, the graph of the function will be symmetric w.r.t the axis $x=0$. Hence, you can draw the part $x<0$ just by using the symmetry of the function.

Draw or plot together on same graph sheet

$$  y=  \tan x,(x>0) ,\,  y= - \tan x, (x<0)  $$

The graph lines have symmetry about y-axis. It is an even  function, even or mirrored about y-axis with a slope discontinuity  at the origin. In fact if you hold up a mirror on y-axis and look at the real graph and virtual image you can see the full graph.


Probability of retirement event This is elementary, but not clear to me.

Suppose I know that the mean age of retirement is $\mu$ and the standard deviation $\sigma$.
What is the probability that someone of age $x$, who has not yet retired,
will retire sometime in the next year,
i.e., between $x$ and $x+1$?

Clearly for small $x$, far below the mean, the probability is near zero, and for large $x$, it approaches $1$.  So it is a type of cumulative distribution...

Thanks for your help!

\begin{align}
\mathbb{P}(X \in [x,x+1) \vert X \geq x) & = \dfrac{\mathbb{P}(X \in [x,x+1))}{\mathbb{P}(X \geq x)} = \dfrac{F_X(x+1) - F_X(x)}{1-F_X(x)}\\
& = 1 - \dfrac{1-F_X(x+1)}{1-F_X(x)}
\end{align}
Since you have not clarified if you are looking for a bound on the probability, this is what we get.

Integrate the Gaussian probability density function from x to x+1


Showing that a Lebesgue integral tends to zero as $n\to\infty$ I want to show that $\displaystyle\int_{0}^{\infty} \frac{\sin(e^x)}{ 1+nx^2}dx \to 0 $ as $n\to\infty$ from the point of view of Lebesgue integration. Is this as simply as bounding the numerator and saying the denominator tends to zero. Any help would be appreciated.

The integrand satisfies $\displaystyle \left| \frac{\sin (e^x)}{1 + nx^2} \right| \le \frac{1}{1+x^2}$ for all $n \ge 1$. Since $\displaystyle \int_0^\infty \frac{1}{1+x^2} \, dx < \infty$ you may apply LDCT: $$\lim_{n \to \infty} \int_0^\infty \frac{\sin (e^x)}{1 + nx^2} \, dx = \int_0^\infty \lim_{n \to \infty} \frac{\sin (e^x)}{1 + nx^2} \, dx = \int_0^\infty 0 \, dx =  0.$$

The direct route would be

$$
\left| \int_0^\infty \frac{\sin e^x}{1+nx^2}\,dx \right| \leq \int_0^\infty \frac{dx}{1+n x^2} = \frac{1}{\sqrt{n}} \int_0^\infty \frac{dt}{1+t^2}\,dt.
$$


What allows us divide/multiply dx in calculus? I've read nearly all of the threads on this topic but none seem to answer my question or lead me in the best direction.
When performing U-substituion or even in it's most basic form:
$y = 2x$, $dy=2\,dx$.
What allows us to do this and what/where does in come in a calculus textbook? None of my teachers have been able to explain it to me. It makes it difficult for me to proceed when even the basics don't seem intuitive.
I've read that the idea of differentials varies but is there a single understanding regarding them to which I can pertain? I feel I'm missing something by just doing it in my calculations and not understanding why. I've genuinely read all posts regarding this, it's been something on my mind for a long time so please don't close this.
Cheers

The chain rule says $\dfrac d {dx} f(g(x)) = f'(g(x)) g'(x)$.  Therefore

$$
\underbrace{\int f'(g(x)) g'(x)\,dx = \int \left( \frac d {dx} f(g(x)) \right)\,dx}_\text{chain rule} = f(g(x)) + C.
$$

If one abbreviates $g(x)$ as $u$ and $g'(x)\,dx$ as $du$, one writes this as
$$
\int f'(u)\,du = f(u) + C.
$$

That usage is consistent with writing $\dfrac {du} {dx} = g'(x)$, as you learned to do earlier.

If your instructor couldn't tell you that this is about the chain rule (just as integration by parts is about the product rule) then either your instructor either doesn't know the material well or didn't understand what you were asking.

You can understand calculus very well without ever manipulating differentials at all.  In fact, in a rigorous treatment of calculus (such as in Spivak's book Calculus), differentials like $dx$ have been banished (except for lingering in the notation).  To understand $u$-substitution, just use the chain rule in reverse.  $\int f(g(x)) g'(x) \, dx = F(g(x)) + C$, where $F$ is an antiderivative of $f$.  You can check this by taking the derivative of the right hand side, using the chain rule.


Real and distinct roots of a cubic equation The real values of $a$ for which the equation $x^3-3x+a=0$ has three real and distinct roots is

The derivative cancels at $x=1$ and $x=-1$. To these correspond a maximum value of $a+2$ and a minimum value of $a-2$. In order to have three real roots, you need three $x$ intercepts; this means that you must have $a+2>0$ and $a-2<0$. So, the condition is $|a|<2$.

Here is a graphical approach. Note that '+ a' shifts the graph up or down.

1) Disregard '+ a'.

2) Graph $ f(x) = x^3 - 3x $

http://www.wolframalpha.com/input/?i=plot%28x%5E3+-+3x%29#

3) from the graph we see that we must find the y values at the local extrema. You did that already getting y = 2 and -2

4) It should be obvious now by inspection that a must be between -2 and 2

Final answer ,

-2 < a < 2


Definition of quotient ring I have a question about the definition of quotient ring (I mean a commutative ring  case) .

Let $(a+I)$ be a coset of $R$ and $(b+I)$ be another left coset of $R$.

Then the multiplication in the quotient ring is defined as

$(a+I)(b+I)=(ab+I)$

However, if we consider the setwise product of $(a+I)$ and $(b+I)$ we know that their product is just a subset of a fixed coset of $I$. Then why should I believe the multiplication in the quotient ring is well-defined?

I suppose you mean that you are considering cosets of an ideal $I$ in a ring $R$, then $$(a+I)(b+I) = ab +aI+bI +I*I = ab + I + I + I= ab + I$$. Because $aI = I$, $I*I = I$ and $I + I=I$ as consequences of the definition of an ideal.


what is the value of angle A The triangle ABC is random. The line $AD$ is twice big as the line $DC$ ($AD=2*DC$). We know only the two angles that are shown in the picture. What's the value of angle $A$?

If you only want the measure of the angle (and after deduce with a geometric/trigonometric procedure) Geogebra gives the following:

I think the correct is answer equal to $75°$

I did it by construction

First of all I made triangle $BDC$ with ur provided information and doing some calculations ie ($\measuredangle DBC=15°$, $\measuredangle BDC=120^{\circ}$ ,$\measuredangle BCD=45°$) than I continued line $CD$ till point $A$ with ur information ($2CD=AD$) as I got point $A$ , I joined it with $B$ and calculated $\measuredangle BAD$ which comes up to be $75°$


If a point has no dimension and no area how can there be space? Everyone knows that a point on any plane has zero magnitude, height, width, or volume. A line is a collection of points, and a plane is a collection of lines, while a cube is a collection of planes and so forth.

How is it then that any line, surface or cube can be made of points that have no magnitude area etc.

In order to calculate the "sum" of lengths/areas/volumes etc. for a collection of subsets, you need the collection of subsets to be countable. But if you take a curve or surface etc. and look at all the points on the set, it is uncountable. So you can't reason that the total length/area/volume etc. is $0 + 0 + \ldots = 0$. If anything, you have to consider that e.g. for length of $S$ along the line you get $dx + dx+ \ldots$, "summed" over all points in $S$, which is indeed zero for a countable point set $S$ but for all points on a line segment you get an integral (not countable sum) of $dx$, which gives the length of the line segment.

I think the idea is that lines are sets of points. The length of a line is equal to the cardinality of the set of the points that comprise it.

Said otherwise, let line segment $P_{1}P_{3}$ designate the set $\langle P_1,P_2,P_3 \rangle$. The line segment is 3 units long.

In this way, assigning actual measures, such as inches, to units isn't any different than assigning actual things to numbers. For example, $2$ apples.

Math is abstract. Abstraction involves leaving things out (alternatively, concretion involves adding things in; however you prefer). Leaving out physical things allows us to generalize, which is a reason that math is so broadly applicable.


Finding all submodules of G-modules Let V; W be irreducible G-modules that are not isomorphic to each other.
How to prove that the only G-submodules of M:= $V \oplus W$, other than $0$ and M itself, are $V =
V \oplus  0$
and $W =
0 \oplus W.$

Let $R$ be any ring, and let $M$ be any semisimple left $R$-module.  Let $\{S_i\}_{i \in I}$ be a set of representatives for the isomorphism classes of simple left $R$-modules.  For each $i \in I$, let $M_i$ be the direct sum of all simple submodules isomorphic to $S_i$, the $\bf{S_i}$-isotypic component of M.  It is immediate that $M = \bigoplus_{i \in I} M_i$: this is the isotypic decomposition of $M$.  Note that no choices have been made: this decomposition is unique.

(What is not completely obvious is that the number of independent copies of $S_i$ in $M_i$ is a well-defined invariant of $M$.  If $M$ is finitely generated this follows from the Jordan-Holder Theorem.  In the general case see $\S$ 2.3 of these notes.  But this is not needed in what follows.)

Now let $N$ be a submodule of $M$.  Then $N_i$, the $S_i$-isotypic component of $N$, is $N \cap M_i$: both are the direct sum of all simple submodules of $N$ isomorphic to $S_i$.

In the case that $M = S_1 \oplus S_2$ is a direct sum of two nonisomorphic simple modules, we get immediately that $N = (S_1 \cap N) \oplus (S_2 \cap N)$.  Since $S_1$ and $S_2$ each have exactly two submodules, this shows that the obvious submodules $0$, $S_1 \oplus 0$, $S_2 \oplus 0$, $S_1 \oplus S_2$ are indeed the only four submodules of $M$.

There is an evident generalization to semisimple modules each of whose isotypic components is simple.

Assume that V,W are irreducible G-modules.  Define M to be the direct sum of V and W.  A submodule of M has the form of a submodule of V direct sum a submodule of W.  By the assumptions that V and W are irreducible there are only two choices for each of V and W for what those submodules can be.


Lyapunov stability question from Arnold's trivium V.I. Arnold put the following question in his Mathematical Trivium:

  Can an asymptotically stable equilibrium position become unstable in the Lyapunov sense under linearization?

It puzzled me for a while, since my experience doesn't include such a pathological case - but I'm still not sure about this one.

For the system $dx/dt=y-x^3$, $dy/dt=-y^3$, the origin is an
asymptotically stable equilibrium. (It is even globally attracting,
as can be seen by sketching the phase portrait with the help of the
nullclines $y=x^3$ and $y=0$.)

The linearized system at the origin is $dx/dt=y$, $dy/dt=0$,
which is not Lyapunov stable: the solution starting at $(x_0,y_0)=(0,\epsilon)$
is $(x(t),y(t)) = (\epsilon t, \epsilon)$,
which goes to infinity if $\epsilon \neq 0$.

The matrix corresponding to the linearized system is $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$,
which has zero as a double eigenvalue.
If both eigenvalues have nonzero real part, then the linearized system determines the stability
of the original system. If the eigenvalues are $\pm c i$ (for some real $c \neq 0$), then
the linearized system is a neutral center (hence Lyapunov stable).
So the phenomenon in question can only occur if at least one eigenvalue is zero.
(I haven't really thought about what can or cannot happen in the case with one
zero eigenvalue and one real nonzero eigenvalue.)

(without proof)

Linearized system has an eigenvalue at the imaginary axis => no
conclusion can be made.
Linearized system only eigenvalues at open left complex halfplane =>
nonlinear system asympt. stable.
Linearized system at least one eigenvalue in the open right complex
half plane => nonlinear system unstable

therefore you can conclude: an asymptotical nonlin. sys. after linearisation can be unstable if you get eigenvalues at the imaginary axis with algebraic multiplicity > geometric multiplicity


Difference Between Simpsons Rule and 3/8 rule So, I was studying Numerical Integration from Hildebrand where I came across Simpson's Rule and 3/8 rule. Now, Simpson's rule has an error of order 5 and degree of precision 3, which is same as 3/8 rule. The difference lies in the Lagrange polynomials we use to interpolate.

In 3/8 rule we need an extra computation point, hence I feel it's computationally insufficient as compared to Simpson's rule. Or if we restate Simpson's rule is better. My question lies, does 3/8 rule have any advantage?

Simpson's Rule: $\int_a^b f(x) dx\approx \frac{b-a}{6}\Big{[}f(a)+4f(\frac{a+b}{2})+f(b)\Big{]}$ with error as $\frac{1}{90}\Big{(}\frac{b-a}{2}\Big{)}^5|f^{4}(z)|$

3/8 Rule: $\int_a^b f(x)dx\approx \frac{(b-a)}{8}\Big{[}f(a)+3f(\frac{2a+b}{3})+3f(\frac{a+2b}{3})+f(b)\Big{]}$ with error as $\Big{|}\frac{(b-a)^5}{6480}f^4(z)\Big{|}$

Reference: Wikipedia for Formula

There are two advantages of the 3/8ths rule: First, the error term is smaller than Simpson's rule.

The second more important use of the 3/8ths rule is for uniformly sampled function integration. Suppose you have a function known at equally spaced points. If the number of points is odd, then the composite Simpson's rule works just fine. If the number of points is even, then you have a problem at the end. One solution is to use the 3/8ths rule. For example, if the user passed 6 samples, then you use Simpson's for the first three points, and 3/8ths for the last 4 (the middle point is common to both). This preserves the order of accuracy without putting an arbitrary constraint on the number of samples.

For Newton-Cotes formulas, when 'n' is an even integer the degree of precision is n+1 and when n is odd degree of precision is'n'(Ref. Numerical Analysis by Burden). Could someone explain how?


Using Logic Laws to prove $p \leftrightarrow q \equiv (p\lor q)\to(p \land q)$ I am trying to prove that

  $p \leftrightarrow q \equiv (p\lor q)\to(p \land q)$

and am really lost in the steps to solve this.

So far I have:

  $p \leftrightarrow q \equiv (p\to q)\land(q\to p) \qquad$|equivalence
   $p \leftrightarrow q\equiv (\neg p\lor q)\land(\neg q\lor p) \qquad$|implication

and I am not sure how to proceed from here. Any advice would be greatly appreciated!

edit:

Thanks to lord farin i now have

  ≡ (~p ∨ q) ∧ (q → p)  Implication Law
  ≡ (q → p) ∧ (~p ∨ q)  Commutative Law
  ≡ (q → p ∧ ~p ) ∨ (q → p ∧ q) Distributive Law
  ≡ (~p ∧ q → p) ∨ (q ∧ q → p) Commutative Law

but i am unsure of how to get there still.

The trick is to expand only one of the two implications materially at first:

$$\begin{align*}
(p\to q)\land(q \to p) &\equiv (\neg p \lor q)\land (q\to p)\\
&\equiv \underbrace{(\neg p \land q \to p)}_{\downarrow} \lor \underbrace{(q \land q\to p)}_\downarrow& & \text{distribution}\\
&\equiv \underbrace{(\neg p \land \neg q) \hspace{1.5em} \lor\quad(p \land q)}_\downarrow\\
&\equiv \hspace{2em}(p \lor q) \to (p \land q)
\end{align*}$$

I'll leave it up to you to take care of the transitions represented by the arrows.

Additional details: For the first two, you can rewrite the conditionals, then use distribution of the $\land$. For e.g. the first, this will leave $(\neg p \land \neg q) \lor (\neg p \land p)$, the second term of which is always false. Thus the whole expression is equivalent to $\neg p \land \neg q$.

A similar approach works for $q \land q \to p$.

You could write equivalence as $(p \land q) \lor (\lnot p \land \lnot q)$, which gives us, using commutation and implication law, the result you desire.


If a system is complete, must it be inconsistent? Godel's incompleteness theorem basically says that a set of axioms cannot prove everything. But you can add those unprovable truths to your set of axioms to expand it. Suppose you keep expanding your set of axioms until it contained all truths (which I think isn't possible in finite time). Or suppose somehow, you had a complete set of axioms. Would it necessarily be inconsistent, because if it wasn't, then it would be consistent and compete?

Goedel's first incompleteness theorem states that every theory that is strong enough (containing the peano axioms) is either inconsistent or incomplete. If it is is inconsistent, it can prove everything ! If it is consistent, there are necessarily statements that cannoe be proven within this theory. This does not mean that it can prove nothing.

Goedel's second incompleteness theorem states that no system that is strong enough (containing the peano axioms) can prove its own consistency.

Goedel's completeness theorem states that a statement is provable within a theory if and only if it is true under every interpretation within this theory.

Yes, you could have a system $S$ in which all propositions that are true in $S$ are axioms of $S$. The system $S$ is then, by definition, complete.

If $S$ is powerful enough to represent all the truths of the arithmetic of natural numbers then we know there are an infinite number of such true statements, so $S$ must have an infinite number of axioms.

But then you cannot be sure that $S$ is consistent because there is no effective decision procedure (a procedure that always halts eventually) that can show that $p$ and $\lnot p$ are not both axioms of $S$.