Probability that $3$ hands of a clock in the same semi-circle? I am wondering how to calculate the probability that $3$ hands of a clock are in the same semi-circle ? I know a similar question is that if we randomly choose n points in a circle, then the probability that all of them in the same semi-circle will be $n/2^{(n-1)}$ with $n = 3$.
But when it comes to the dial hands, where I guess there are correlations between them, I am not sure whether my question is equivalent to the previous one.
If they are not equivalent, and this question will be hard to calculate, then whether we can determinate this probability will be greater or smaller than $3/4$?
Anyone can help? Thanks.

The hands in fact move deterministically; it is just the time when we view them that can be random. The movement repeats ever twelve hours, so assume we view the hands at some moment uniformly distributed between noon and midnight. Then the probability they will be within a semicircle is the number of hours they spend in a semicircle, divided by $12.$
Starting at noon, the hands are within one semicircle until the second hand is opposite the hour hand. The second hand makes $12\times 60 = 720$ revolutions in $12$ hours while the hour hand makes one revolution in the same time,
so the time until the second and hour hands first align after noon is $12/719$ hour,
which is $720/719$ minutes;
but the time from noon until the hands are exactly opposite is half that time,
$360/719$ minute.
For a short time after the first $360/719$ minute, the hands are not within a semicircle. They are in a semicircle again starting at the instant when the second hand is opposite the minute hand.
The second hand aligns with the minute hand for the first time after noon at
$60/59$ minutes after noon, so the first time they are opposite is at
$30/59$ minute after noon.
Let $t_h = 360/719$ and $t_m = 30/59,$
and let's take the convention that "at time $t$" means at $t$ minutes past noon.
So the first time interval when the hands are not in a semicircle
starts at $t_h$ and ends at $t_m.$ The length of this interval is $t_m - t_h$ minutes;
specifically, the time during which the hands are continuously not in a semicircle,
in minutes, is
$$
\frac{30}{59} - \frac{360}{719} = \frac{330}{42421}.
$$
The second such time interval starts at $3t_h$ and ends at $3t_m$ for
$3(t_m - t_h)$ minutes not in a semicircle,
the third starts at $5t_h$ and ends at $5t_m$, and so forth as long as the minute hand remains less than $180$ degrees clockwise from the hour hand.
This is a promising pattern: after $n$ intervals like this, we will have spent a total of
$n^2(t_m - t_h)$ minutes during which the hands do not fit in a semicircle.
Alas, the full solution is not that simple.
The intervals increase in length until they are almost half a minute long,
and then this simple pattern is broken.
The minute hand first becomes exactly opposite the hour hand at $6/11$ hour after noon,
that is, at $360/11$ minutes after noon.
Shortly before that, at $65t_h$ minutes, the second hand passes opposite to the hour hand, making the hands not within a semicircle;
but at $360/11$ minutes after noon the second hand is suddenly finds itself between the minute and hour hands so that the hands are all within a semicircle again,
even though $65t_m$ minutes have not yet passed.
After this abbreviated interval of not-in-a-semicircle time,
the next interval is almost half a minute long, and the next several intervals decrease in length. But once they have decreased enough, they start increasing again, and the pattern repeats but with different intervals each time as the relative position of the second hand is different each time the minute hand overtakes the hour hand or comes opposite to it.
At about this point in the analysis I decided I would rather have a computer count the time for me. The events where the hands start fitting in a semicircle or stop fitting in a semicircle all occur at exact integer multiples of
either $1/11,$ $1/59,$ or $1/719$ minute after noon.
Since those fractions have a common denominator of
$11\times 59\times 719 = 466631,$
every event of interest (where the hands start or stop being within a semicircle) occurs at some multiple of $1/466631$ minute after noon.
By taking $720\times 466631$ equal timesteps of that length we cover twelve hours.
At the start of each timestep we know from the relative positions of the three hands whether the hands will fit in a semicircle or not during that timestep. (It is one way for the entire duration of the timestep.)
I wrote the analysis in the python script shown below.
I had the script count the first six hours, take a checkpoint, and then do the last six hours. Everything after $6$ o'clock is a mirror image of what happened before, in reverse order, so the same sequence of lengths of intervals occurs in reverse order after $6$ o'clock. I had the script record the start, end, and duration of each interval to verify this. Note that since each interval starts about a minute after the previous one, there are almost $360$ intervals in six hours, but not quite $360$ due to the way the hands "catch up" with each other.
According to my script, it takes
$720\times 11\times 59\times 719 = 335974320$
timesteps to complete twelve hours, and during $83982960$ of those timesteps the three hands do not fit within a semicircle. The proportion of time not in a semicircle is therefore
$$
\frac{83982960}{335974320} = \frac{1977}{7909} \approx 0.2499683904412694.
$$
So the probability that the hands are not within a semicircle is almost, but not quite, $1/4$;
the probability that they are within a semicircle is therefore slightly over $3/4.$

To relate these calculations to G Cab's method, let $x$ be the number of units of $12/11$ hour since noon (that is, $x$ is $0$ at noon and $11$ at midnight). If $k = \lfloor x\rfloor$ (an integer),
then for each $k$ (that is, for each period of $12/11$ hour) there are two sequences of intervals during which the three hands are not within a semicircle.
During the first half of this period, while the angle from the hour hand to the minute hand is between $0$ and $180$ degrees clockwise, the three hands are not in a semicircle whenever
$$
m + \frac12 \leq \frac{15(2k)}{22} + \frac{719}{11} \tau < m + \frac12 + \tau
$$
for some integer $m,$ where $\tau = x - k,$ and during the second half of this period, while the angle from the minute hand to the hour hand is between $0$ and $180$ degrees clockwise, the three hands are not in a semicircle when
$$
n + \tau \leq \frac{15(2k + 1)}{22} + \frac{719}{11} \tau < n + \frac12
$$
for some integer $n,$ where $\tau = x - k - \frac12.$
In either case we also have the restriction that $0 \leq \tau < \frac12.$
In the typical case, the beginning or end of an interval occurs when one of the inequalities becomes an equation; the exceptions occur when the equation is true only for $\tau < 0$ or $\tau > \frac12$.
Solving these equations for $\tau,$ we find
$$\tau = \frac{11}{719} \left(m + \frac12 - \frac{15(2k)}{22}\right)\tag1$$
at the start and
$$\tau = \frac{11}{708} \left(m + \frac12 - \frac{15(2k)}{22}\right)\tag2$$
at the end of a typical interval in the first sequence, but
$$\tau = \frac{11}{708} \left(n - \frac{15(2k+1)}{22}\right)\tag3$$
at the start and
$$\tau = \frac{11}{719} \left(n + \frac12 - \frac{15(2k+1)}{22}\right)\tag4$$
at the end of one in the second sequence.
The length of the first interval (in the same units as $\tau$)
is therefore
\begin{align}
\Delta\tau_1
&= \frac{11}{708} \left(m + \frac12 - \frac{15(2k)}{22}\right)
   - \frac{11}{719} \left(m + \frac12 - \frac{15(2k)}{22}\right) \\
&= \frac{121}{708(719)} \left(m - \frac{30k - 11}{22}\right)
\end{align}
and the length of the second interval is
\begin{align}
\Delta\tau_2
&= \frac{11}{719} \left(n + \frac12 - \frac{15(2k+1)}{22}\right)
   - \frac{11}{708} \left(n - \frac{15(2k+1)}{22}\right) \\
&= \frac{121}{708(719)} \left(\frac{30k + 723}{22} - n\right)
\end{align}
The combined length of two intervals, one from each sequence, is
\begin{align}
\Delta\tau_1 + \Delta\tau_2
&= \frac{121}{708(719)} \left(m - \frac{30k - 11}{22}\right)
 + \frac{121}{708(719)} \left(\frac{30k + 723}{22} - n\right) \\
&= \frac{11}{708(719)} \left(367 + 11(m - n)\right) \\
\end{align}
If we hold $m - n$ constant, $\Delta\tau_1 + \Delta\tau_2$
is independent of the value of $m$ as long as all of the $\tau$ values from which we computed $\Delta\tau_1$ and $\Delta\tau_2$ are between $0$ and $\frac12.$
For example, if we set $n = m$, we get a pairing of intervals in which each pair of intervals has total length $\frac{11(367)}{708(719)}$ except in some cases where an interval starts at $\tau=0$ or ends at $\tau=\frac12$ or where an interval has no corresponding interval in the other sequence.
Trial and error shows that if we set $n=m+1$ we minimize the number of intervals that are unmatched or that are in a pair where one interval ends at $\tau=0$ or $\tau=\frac12$ instead of where the equations above say it will
(there are $14$ such intervals);
each matched pair then has total length
$\frac{11(367-11)}{708(719)} = \frac{11(89)}{3(59)(719)}$
in the units of $\tau.$
This is equal to $\frac{356}{59(719)}$ hour,
which is $234960$ timesteps of the brute-force method.
According to my spreadsheet calculations, if $n = m+1$ there are $32$ matched pairs of intervals for each value of $k.$ since there are $11$ values of $k,$ the total time in intervals that belong to typical matched pairs is
$352 \times 234960$ timesteps. It then remains to account for the $14$ intervals that are not in typical matched pairs.
Those intervals can be identified for each $k$ by finding the maximum $m$ such that right-hand side of Equation $(1)$ is negative
and the minimum $n$ such that the right-hand side of Equation $(4)$
is greater than $\frac12.$
A relatively quick hand calculation should therefore be possible, though it would still be longer than I would like to write in detail in this answer.
Update: The calculations are not as long as I feared after all, as shown in G Cab's edited answer.

Here is the script for the brute-force method:
# Program to figure out what proportion of the time the three hands
# of a clock are not within a semicircle.

# Assume there are 59*719*720 equally-spaced positions around the clock face.
# Note: 59*719 == 42421; 42421*11 = 466631.

# Each timestep of this program is 1/42421 of a minute.
# In one timestep, the second hand moves 720 positions, the minute hand
# moves 12 positions, and the hour hand moves 1 position.
# In 42421 timesteps, one minute has passed and the second hand returns
# to the starting position,

numPositions = 11*59*719*720

# Returns the new position after having moved the given amount forward
# from the given position. That is, the return value is the sum of
# the two input values, modulo numPositions.
# It is assumed the input values are in the range 0 to numPositions - 1.
def newPosition(fromPosition, movement):
    toPosition = fromPosition + movement
    if toPosition >= numPositions:
        toPosition -= numPositions
    return toPosition

# Returns the number of steps clockwise to get from position1 to position2.
# positions measured in number of steps from the start.
def stepsClockwise(position1, position2):
    distance = position2 - position1
    if distance < 0:
        distance += numPositions
    return distance

# Returns 1 if the hands are at the start of a timestep in which the
# smallest angle encompassing all three hands will be greater than 180 degrees,
# 0 if the angle will be less than 180 degrees.
def isLargeAngle(hoursPosition, minutesPosition, secondsPosition):
    # A 180 degree angle
    halfCircle = numPositions // 2
    # A position exactly opposite the seconds hand
    antiSecondsPosition = newPosition(secondsPosition, halfCircle)
    minutesOffset = stepsClockwise(hoursPosition, minutesPosition)
    if minutesOffset < halfCircle:
        # The minutes hand is < 180 degrees clockwise from the hour hand
        antiOffset = stepsClockwise(hoursPosition, antiSecondsPosition)
        if (antiOffset >= 0 and antiOffset < minutesOffset):
            # The anti-seconds hand will be between the hours and minutes
            # during this timestep, so the seconds hand will be more than
            # 180 degrees past the hour hand but less than 180 degrees
            # before the minute hand.
            return 1
        else:
            return 0
    else:
        # The hour hand is <= 180 degrees clockwise from the minute hand.
        # During the next timestep the short arc between the hands will be
        # clockwise from the minute hand to the hour hand.
        hoursOffset = stepsClockwise(minutesPosition, hoursPosition)
        antiOffset = stepsClockwise(minutesPosition, antiSecondsPosition)
        if (antiOffset >= 0 and antiOffset < hoursOffset):
            # The anti-seconds hand will be between the minutes and hours
            # during this timestep, so the seconds hand will be more than
            # 180 degrees past the minute hand but less than 180 degrees
            # before the hour hand.
            return 1
        else:
            return 0

# Runs the given number of timesteps from the given starting position
# and returns the number of times the hands spanned more than 180 degrees.
def countLargeAngles(hoursPosition,
                     minutesPosition,
                     secondsPosition,
                     startingTimestep,
                     numTimesteps):
    numLargeAngles = 0
    wasLargeAngle = 0
    largeAngleStarted = 0
    numIntervals = 0
    for i in range(numTimesteps):
        if isLargeAngle(hoursPosition, minutesPosition, secondsPosition):
            numLargeAngles += 1
            if wasLargeAngle == 0:
                largeAngleStarted = startingTimestep + i
                wasLargeAngle = 1
        else:
            if wasLargeAngle == 1:
                largeAngleStopped = startingTimestep + i
                numIntervals += 1
                print(">180 degrees interval ", numIntervals,
                      " from ", largeAngleStarted,
                      " to ", largeAngleStopped, " for ",
                      (largeAngleStopped - largeAngleStarted), " timesteps")
                wasLargeAngle = 0
        hoursPosition   = newPosition(hoursPosition, 1)
        minutesPosition = newPosition(minutesPosition, 12)
        secondsPosition = newPosition(secondsPosition, 720)
    print("After ", numTimesteps, " timesteps")
    print("hour hand at ", hoursPosition)
    print("minute hand at ", minutesPosition)
    print("second hand at ", secondsPosition)
    print(numLargeAngles, " positions with > 180 degrees")
    return numLargeAngles, hoursPosition, minutesPosition, secondsPosition

# Run 360*466631 timesteps to simulate 6 hours,
# then another 360*466631 timesteps to simulate the next 6 hours.
halfTotalTimesteps = numPositions // 2
hoursPosition = 0
minutesPosition = 0
secondsPosition = 0

(numLargeAngles,
 hoursPosition,
 minutesPosition,
 secondsPosition) = countLargeAngles(hoursPosition,
                                     minutesPosition,
                                     secondsPosition,
                                     0,
                                     halfTotalTimesteps)
(additionalLargeAngles,
 hoursPosition,
 minutesPosition,
 secondsPosition) = countLargeAngles(hoursPosition,
                                     minutesPosition,
                                     secondsPosition,
                                     halfTotalTimesteps,
                                     halfTotalTimesteps)
numLargeAngles += additionalLargeAngles

print("Total number of timesteps was ", numPositions)
print("Total number of > 180 degrees was ", numLargeAngles)

Take two of the hands. On average they will be a quarter of a circle apart. Distance varies uniformly from zero to half a circle apart.
Turn the dial so that one hand is at 12 and the other hand at 3.
Now when the third hand is in the interval from 9 to 6 all three hands are on the same semicircle.
When the third hand is between 6 and 9 they are not on the same semicircle.
So on average the three hands will be on the same semicircle 3/4 of the time.


Why does implicit differentiation fail here? To differentiate an implicit function $y(x)$, defined by an equation $R(x, y) = 0$ one can totally differentiate $R(x, y) = 0$ with respect to $x$ and $y$ and then solve the resulting linear equation for $\frac{dy}{dx}$ to explicitly get the derivative in terms of $x$ and $y$.

Consider the following example: Let $y(x)$ be defined by the following relation:

$$(x^2-y^2)^{1/2}+\arccos\frac{x}{y}=0. \,(y\neq 0.)$$

Clearly, the equation defines $y$ as a function of $x$. In fact, it is easy to see that $y=x$. However, when I apply the method of implicit differentiation to $(x^2-y^2)^{1/2}+\arccos\frac{x}{y}=0$, I failed to get the desired result $\frac{dy}{dx}=1$ (since $y=x$). Why does implicit differentiation fail here?

Edit: I did not do the implicit differentiation by hand as it is too tedious; instead I trusted the result on WolframAlpha:

With a bit of work, you can show that $R(x,y)$ is only defined for $y=x$ and $y=-x$ excluding the point $(0,0)$. Hence your function does not satisfy the conditions of the implicit function theorem, which states that your function should be at least continuously differentable to apply the method of implicit differentiation.

Note, you could extend your problem to complex numbers, in that case the domain of applicability of $R(x,y)$ would be extended. However, even in that case, it would still not be the entire plane because no definition of the square root or the inverse cosine is analytic on the entire complex plane. In particular, if you choose the most "natural" extension to the complex plane of the inverse cosine, it is not defined on the branch points $-1$ and $1$. Those correspond exactly to the cases $y=x$ and $y=-x$. In other words, naïve implicit differentiation does not work in that case either. You would therefore need analytic extensions of the square root and the inverse cosine such that they are valid in the case $y=x$. But, those extensions will not correspond to the definitions you adopted in the real case.

$ dR(x,y)=\dfrac{\partial R}{\partial y}dy+\dfrac{\partial R}{\partial x}dx=0 $

Since $y(x)$ it follows that $dy=\frac{dy}{dx}dx$

$dR(x,y)=\dfrac{\partial R}{\partial y}dy+\dfrac{\partial R}{\partial x}dx=\dfrac{\partial R}{\partial y}\dfrac{dy}{dx}dx+\dfrac{\partial R}{\partial x}dx=\big(\dfrac{\partial R}{\partial y}\dfrac{dy}{dx}+\dfrac{\partial R}{\partial x} \big)dx=0$

$\rightarrow \dfrac{dy}{dx}=-\dfrac{\dfrac{\partial R}{\partial x}}{\dfrac{\partial R}{\partial y}}$

If you try it on Wolfram Alpha with

-d/dx(√(x^2-y^2)+arccos(x/y))/d/dy(√(x^2-y^2)+arccos(x/y))

it provides the required answer


the real sequence $\{p_n\}$ is defined as $p_1=2, p_{n+1}=( 2+p_n)/(2p_{n}+1)$ please show the procedure of this question, I am not sure how to start.

Note that if we let $f(z) = \frac{z+2}{2z +1}$ then $p_n = f^{n-1}(p_1)$. $f$ is a Moebius transform, so we can determine its powers by determining the powers of its associated matrix, $$ A_f=\begin{pmatrix} 1 & 2 \\ 2 & 1\end{pmatrix}.$$

You can work out the powers of this matrix in a few ways, the Jordan decomposition being the most usual, but a more direct approach, particularly useful when dealing with two by two symmetric matrices, is to write $$A_f = 3 \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & 1/2\end{pmatrix} - \begin{pmatrix} 1/2 & -1/2 \\ -1/2 & 1/2\end{pmatrix} = 3R -S$$ and note that $R^2 = R$, $S^2=S$ and $RS = SR = 0$. Therefore $$A_{f^n} = A_f^n = 3^n R + (-1)^nS = \frac{1}{2}\begin{pmatrix} 3^n +(-1)^n & 3^n -(-1)^n \\ 3^n -(-1)^n & 3^n +(-1)^n\end{pmatrix}.$$ Therefore, $$p_n = f^{n-1}(2) = \frac{ (3^{n-1} + (-1)^{n-1})\cdot 2 + (3^{n-1} - (-1)^{n-1}) }{ (3^{n-1} - (-1)^{n-1})\cdot 2 + (3^{n-1} + (-1)^{n-1}) } = \frac{3^n + (-1)^{n-1}}{3^n - (-1)^{n-1} },$$ and the limit is easy to calculate from there.

This is almost certainly not how this question was meant to be answered.

Observe that $p_n > \dfrac{1}{2}, \forall n \geq 1$. If $L$ is the limit, then $L = \dfrac{2+L}{2L+1}\to 2L^2 = 2 \to L = -1,1$. Since $L > 0$, we have $L = 1$. This led us to define: $|p_{n+1} - 1| = \left|\dfrac{2+p_n}{2p_n+1} - 1\right| = \left|\dfrac{2+p_n - 2p_n - 1}{2p_n+1}\right|= \left|\dfrac{1-p_n}{2p_n+1}\right|<\dfrac{|1-p_n|}{2}< ...< \dfrac{|1-p_1|}{2^n}= \dfrac{1}{2^n} \to 0$. Thus $p_{n+1} \to 1$ as claimed.


Confused with proof that all Cauchy sequences of real numbers converge. First the textbook proves that all Cauchy sequences are bounded, and so have a convergent subsequence, $\{a_{n_{k}}\}$ that converges to a limit, say $L$. Now we use this to prove that all Cauchy sequences are convergent.

So an $N_1$ exists such that $$\left|a_{n_{k}}-L\right|<\frac{\epsilon}{2}$$ for all $ k > N_1$,
and an $N_2$ exists such that $$\left|a_m-a_n\right|<\frac{\epsilon}{2}$$ for all $n,m > N_2 $.

Pick any $k > N_1$ such that $n_k > N_2$. Then for every $n > N_2$, $$\left| a_n - L \right | \leq \left| a_n - a_{n_{k}} \right| + \left| a_{n_{k}} - L \right| < \epsilon/2 + \epsilon/2 = \epsilon$$.

So $$\lim_{n \rightarrow \infty}a_n = L$$

I'm fine with this proof until the last part - I'm confused as to why we can pick an abitrary $k$ like we do? Should the limit not depend only on $n$? Now it appears like it also depends on $k$ and if we pick a $k < N_1$, the inequality isn't true. Can someone clarify this for me please?

At that point in the proof you’re trying to show that if $n>N_2$, then $|a_n-L|<\epsilon$. If you can find a real number $x$ such that $|a_n-x|<\frac{\epsilon}2$ and $|x-L|<\frac{\epsilon}2$, the triangle inequality will give you the desired result, so the proof boils down to finding such an $x$.

What things do we know are close to $L$? Terms $a_{n_k}$ of the subsequence, provided that $k$ is sufficiently large.
What things do we know are close to $a_n$? Terms $a_m$ of the original sequence, provided that $m$ and $n$ are sufficiently large.

We take care of (1) first: there is an $N_1$ such that $|a_{n_k}-L|<\frac{\epsilon}2$ whenever $k>N_1$. This means that we can take our $x$ to be any $a_{n_k}$ with $k>N_1$, and we’ll have $|x-L|<\frac{\epsilon}2$.

Then we take care of (2): there is an $N_2$ such that $|a_n-a_m|<\frac{\epsilon}2$ whenever $m,n>N_2$. This means that since we’ve already specified that $n>N_2$, we can take our $x$ to be any $a_m$ with $m>N_2$, and we’ll have $|a_n-x|<\frac{\epsilon}2$.

Can we combine the two requirements? Is there an $a_{n_k}$ with $k>N_1$ that is also an $a_m$ with $m>N_2$?
Equivalently, is there an $a_{n_k}$ with $k>N_1$ such that $n_k>N_2$? Sure: the sequence $\langle n_k:k\in\Bbb N\rangle$ is unbounded, so its tail $\langle n_k:k>N_1\rangle$ is also unbounded and contains a term $n_k>N_2$. Thus, we can set $x=a_{n_k}$ and satisfy both requirements, so that we have

$$|a_n-L|\le|a_n-x|+|x-L|<\frac{\epsilon}2+\frac{\epsilon}2=\epsilon$$

as desired. Note that this calculation doesn’t actually depend on a specific value of $k$: we could set $x=a_{n_\ell}$ for any $\ell\ge k$, the the calculation would be the same. As I said at the beginning, we’re really just trying to find one number $x$ that we can use to ‘tie’ $a_n$ to $L$ to within $\epsilon$; it turns out that there are lots of them, and it doesn’t matter which one we use.

The real problem is to understand the logical construction of the real numbers from Dedekind cuts. This is where the LUBP(least upper bound property) comes from, i.e. $\mathbb{R}$ is order-complete.

The LUBP easily implies that every monotonic bounded sequence converges.

From this $\mathbb{R}$ has the finite-intersection property, i.e. If $[a_n,b_n]_{n\in {\mathbb N}$ is a nested sequence of closed bounded intervals then its intersection is a closed bounded interval or a single point.

From any Cauchy sequence $(x_n)$ you can systematically define a nested sequence of closed bounded intervals $(I_n)$ with $x_n \in I_n$ and such that the lengths of the intervals goes to zero. Thus, the intersection of all such intervals is $\{x\}$ for some $x$ by the finite intersection property and this $x$ must be the limit of the Cauchy sequence.

If you don't go through an axiomatic justification of our number systems in a Discrete math class and don't see it in real analysis, your math program is fundamentally flawed. When you work hard to understand these formal logical underpinnings, all of analysis becomes much easier.

The above approach works for any metric space, i.e. a metric space is complete if and only if it has the appropriately defined version of the finite-intersection property which again is just that every Cauchy sequence is convergent. It just happens that R has a natural linear order but this is not important for metric completeness but requires a little help from general topology to guarantee that arbitrary intersections of closed sets are closed because of the general definition of an open set in a topological space.


Riemann Integrability of a simple function $$ f(x) =\begin{cases}\frac{x^2}{2}+4 & x \geq 0\\-\frac{x^2}{2}+2 & x < 0\end{cases} $$

Is $f(x)$ Riemann integrable in the interval $[-1, 2] ?$

Does there exist a function $g(x)$ such that $g'(x)=f(x)$ ?

Solution: I can see that the function is Riemann Integrable as there is only one point of discontinuity in the interval mentioned.

Also by First fundamental theorem of calculus there exists a function $g(t)$ such that

$\displaystyle g(t)=\int_{-1}^t f(x) \, dx$ where $t$ is in the interval $[-1, 2]$

But $f$ is not continuous at $x=0$. Thatwould mean $g'(c) \neq f(c)$ at all points in the interval.

What does it mean? Does there exist $g(x)$ as asked in the problem?

Suppose such a function $g$ exists.  Try finding the one-sided derivatives of $g$ at $0$:

$$
\lim_{h\,\downarrow\,0} \frac{g(0+h)-g(0)}{h} = \lim_{h\,\downarrow\,0} \frac{\int_0^h f(x)\,dx}{h}= \lim_{h\,\downarrow\,0} \frac 1 h \left[ \frac{x^3}{6}+4x \right]_{x:=0}^{x:=h} = \lim_{h\,\downarrow\,0} \left(\frac{h^2}{6} + 4\right) = 4.
$$

Then find the other one-sided derivative, involving $\displaystyle\lim_{h\,\uparrow\,0}$.  If they're not equal, then there can be no function $g$ satisfying the requirements.

By the first fundamental theorem of calculus, the function $$g(t)=\int_a^t f(x) dx$$ has a derivative in all points where $f$ is continuous. In those points, $g'(t) = f(t)$.


Binomial expansion for any n I teach A-Level maths, and in the second year we do the general binomial expansion, which is even provided for the students in the formula book.

For values of $n$ that are not positive integers: (I realise this form does also work for postive integers as well.)

$$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2+\ldots$$

Every textbook I have read says that if you then need to expand $(a+x)^n$, then you have to do $a^n(1+\frac{x}{a})^n$, expand the bracket and then multiply the constant back in. You have to have it in the form $(1+\mathrm{f}(x))^n$. You can then give the validity of the expansion $\left|\frac{x}{a}\right|<1 \Rightarrow |x| < |a|$.

We discuss the convergence issue, that in order for the infinite sequence to converge we need the '$x^n$' terms to converge to zero, and hence the validity statement.

So is there any reason why we cannot miss out that step and just do the following, which will result in exactly the same expansion?

$$(a+x)^n = a^n + na^{n-1}x + \frac{n(n-1)}{2!}a^{n-2}x^2+\ldots+\frac{n(n-1)\ldots(n-r+1)}{r!}a^{n-r}x^r+\ldots$$

This combines the second year learning, with the version they are given in the first year for positive integer values only:

$$(a+b)^n = a^n + ^nC_1a^{n-1}b + ^nC_2a^{n-2}b^2 + \ldots + ^nC_ra^{n-r}b^r +\ldots + b^n$$

Is it always provided using the first method just in order to simplify the learning for students? While I think I'll keep teaching it the way it is presented in their textbooks, I don't want to tell students they can't do it like that if they're already using that approach.

As a pedagogical point, the series expansion

$$ (1+x)^n = 1 + x + \frac{n(n-1)}{2} x^2 + \ldots $$

is somewhat simpler. This has some pedagogical value:

When doing a "complicated" calculation, it's best to make the things you're calculating with as simple as possible
If you're going to memorize something, memorizing the simpler one above is arguably easier than trying to memorize $(a+x)^n$
It's a nice example in the art of reducing the general case to a special particular case
The actual calculus work is purely single-variable calculus, rather than multi-variable calculus with all but one variable held constant

As an aside, you may like to take this opportunity to introduce generalized binomial coefficients $\binom{n}{r}$ in the case where $n$ is no longer restricted to be a nonnegative integer.

The reason why you need to do it the first way is because combinatorics are only defined within the domain of the natural numbers.  Therefore, the notation you are using for the "simplified" approach is garbleglorp and will confuse your students.  The first approach is the "generalized binomial theorem" and cannot be proven using combinatorics.  You, therefore, need to prove it by showing the series is convergent for negative and fractional rational exponents as well.  This is proven to be true for all rational numbers n iff the equation is (1 + x)n and |x| is small.

EDIT: I misread your question.


Deductions from Carmichael's theorem 1.) May I seek your advice on how to prove the following assertion(without recourse to (2)):

If $ 2014 \equiv 14 \ (\text{mod}  \ 2000), $ then $2014^{2014} \equiv 14^{14} \ (\text{mod 2000}).$

2.) Also, how we prove from Carmichael's theorem that, "If $a \equiv b \ (\text{mod m})$ and $c \equiv d \ (\text{mod} \ \lambda(m) ),$ then $a^c \equiv b^d \ (\text{mod m})$ ", where $\lambda (m)$ is defined as such: http://en.wikipedia.org/wiki/Carmichael_function

Thank you.

$\lambda(n)$ is the smallest number such that for any k we have $k^{\lambda(n)}\equiv 1 \bmod n$ for any $k$. suppose we have $b\equiv c \bmod \lambda n$. Then $b=c+x\lambda(n)$ for some integer x. Then we have $k^{c}=k^{b+x\lambda(n)}=k^b*(k^{\lambda(n)})^x\equiv k^b*1^x\equiv k^b \bmod n$


Looking for an example of a bijective continuous function $f:\mathbb{Q} \to \mathbb{Q}$ such that $f(-1)=0$, $f(0)=1$ and $f(1)=-1$? Clearly such a function does not exist from $\mathbb{R}$ to itself, but apparently it does in $\mathbb{Q}$ and I don't see how it could... Can you give me an example and explain to me how you thought of this one?

Thank you very much!

I first simply thaugh of putting the values of $f$ at the three special values of $x$ and making it $f(x)=x$ everywhere else, or constant, but it doesn't work...

Define $f:\Bbb{Q} \to \Bbb{Q}$ by

$$
f(x)=
\begin{cases}
x+1 & \sqrt{2}-3 < x < \sqrt{2}-1 \\
x-2 & \sqrt{2}-1 < x < \sqrt{2}\\
x & \text{otherwise}
\end{cases}
$$

Then it's easily checked that $f$ assumes the desired values. Also, $f$ has an obvious extension to $\Bbb{R}$ whose only points of discontinuity are irrational; thus it is continuous on $\Bbb{Q}$.

To see that it's bijective, note that the function
$$
g(x)=
\begin{cases}
x+2 & \sqrt{2}-3 < x < \sqrt{2}-2\\
x-1 & \sqrt{2}-2 < x < \sqrt{2} \\
x & \text{otherwise}
\end{cases}
$$
is its inverse.

Intuitively, $f$ is a function that swaps the order of some intervals in $\Bbb{Q}$. This sounds like a discontinuous operation, but by making the interval endpoints irrational, we can ensure that $\Bbb{Q}$ doesn't notice.


Prove that stochastic prcess trajectories are continous. Consider gaussian process $\{X_{t}, t \in [0,1]\}$ with zero mean and covariance function $R(X_{s}, X_{t}) = \min (s,t) -st$.

We want to know does this process has continous trajectories, i.e. $\lim_{t \to t_0} X_t = X_{t_0}$ a.e.

My attempt :

First of all let's prove that this process continous in $L_{2}$. Consider $\mathbb{E}(X_{t+h} - X_{t})^{2} = h \to 0$ (the same result may be proved by proving continuity of covariance function).

Next we have that it's continous in measure (because of convergence in average). But we know that probability of $\mathbb{P}(\{ \lim X_t \to X_{t_0} \})$ is continious function, so we may use Riesz theorem which give us subsequence converging almsot sure. Am I right? Hard problem for me actually, I don't know how it can be proven only by definitions.

Assume that such a process $X_t$ exists and is continuous (i.e. $X_t$ is a Brownian bridge).

Take $U$ an independent uniform variable over $[0,1]$ and let $Y_t=X_t+1_{U.t \in \mathbb Q}$.

Then for all $t$, $\mathbb P(X_t=Y_t)=\mathbb P(U.t \not \in \mathbb Q)=1$ so $Y_t$ is a gaussian process with the same mean and covariance as $X_t$, however almost surely it is continuous nowhere.

This shows in general that mean and variance of a stochastic process never guarantee its continuity.


Use the definition of a limit to prove this limit Use the definition of a limit to show that $\frac{3x-2}{x^2+1} \to \frac{4}{5}$ as $x\to 2$.

I am having trouble isolating the $(x-2)$ from the fraction to use as a delta please help!

$$\left|\frac{3x-2}{x^2+1}-\frac45\right|=\left|\frac{15x-10-4x^2-4}{5(x^2+1)}\right|=\left|\frac{4x^2-15x+14}{5(x^2+1)}\right|=\frac{\left|4(x-2)\left(x-\frac74\right)\right|}{5(x^2+1)}=$$
(taking $\,x\,$ very close to $\,2\,$, say $\,x\in\left(2-0.001,\,2+0.001\right)\,$ , we can assure that $\,x-\frac74>0\,$ , so)

$$=\frac{4x-7}{5(x^2+1)}|x-2|\;\;\color{red}{(**)}$$

and now we can estimmate, for $\;x\;$ as above (no need to be specially gentle or subtle with this...):

$$\frac{4x-7}{5(x^2+1)}<\frac2{5\cdot(1+1)}=\frac1{5}\;\;\color{blue}{(*)}$$

so given any $\epsilon>0\;$ we can choose $\;\delta=5\epsilon\;$ , and then:

$$\color{green}{(*)}\;\;|x-2|<\delta\implies\left|\frac{3x-2}{x^2+1}-\frac45\right|\stackrel{\color{red}{(**)}}=\frac{4x-7}{5(x^2+1)}|x-2|\stackrel{\color{green}{(*)+\color{blue}{(*)}}}<\frac15\cdot5\epsilon=\epsilon$$

You can simply set x=2 and solve the equation. You can do this because the equation is defined at x=2. If it were not then you would have to determine what value the function is approaching (usually infinity or zero, but can be anything).


If $A$ and $B$ are independent events, show that $P(A | B \cap C) = P(A|C)$ If $A$ and $B$ are independent events, show that $P(A | B \cap C) = P(A|C)$.

Here $B$ and $C$ are actually dependent events but I suppose that doesn't matter.

It's false. Here is a counterexample.

Consider $3$ independent tosses of a fair coin.

Let $A$ be the event: the first and second tosses come up the same.

Let $B$ be the event: the first and third tosses come up the same.

Let $C$ be the event: the second and third tosses come up the same.

The three events $A,B,C$ are pairwise independent: $P(A)=P(B)=P(C)=\frac12$, $P(A\cap B)=P(A\cap C)=P(B\cap C)=\frac14=P(A)P(B)=P(A)P(C)=P(B)P(C)$.

$P(A)=P(A|B)=P(A|C)=\frac12$, but $P(A|B\cap C)=1$.

$ P(A \vert B \cap C) = \frac{P(A \cap B \cap C)}{P(B \cap C)} $

And,$ P(A \cap B \cap C)$ can be written as $P(A \cap B \vert C)P(C)$.

But since A and B are independent, we can separate them to get :

$ P(A \cap B \vert C)P(C) = P(A \vert C)P(B \vert C)P(C)$ and substituting $P(B\vert C) = \frac{P(B\cap C)}{P(C)}$ gives the answer.


How to calculate PDF given CDF? (and related problems) I know how to find the CDF given the PDF. And I know that I must differentiate for this problem above, but I'm not sure how to go about solving these problems

(a) What is a CDF?   You should have a list of properties that identify a function as a valid cumulative distribution function.   Which of these is affected by the value of term $a$?   What value of $a$ satisfies that property?

(b) Yes, indeed, differentiating the CDF will give the pdf except if there is a step discontinuity.   Check to see if that is the case, and take steps to deal with having a mixed distribution.

The CDF, since it is the integral of the PDF, is always continuous.  Here that means that $\lim_{x\to 0^+} \frac{a- e^{-x}}{2}= \lim_{x\to 0^-} \frac{1}{2}x+ \frac{1}{2}$.  What does a have to be so those two limits are the same?


roots of functions involving several sines Is it possible to find exact solutions (in $\mathbb{R}$) of equations of the type

$$\alpha_1\sin(\beta_1 t)+\alpha_2\sin(\beta_2 t)+1=0$$
for $\alpha_i,\beta_i\in\mathbb{R}$?

In a comment to this question, Yulia V gave an explanation why there is probably none: "it is like a polynom, but more general, and there are no general solutions for polynoms for the degree greater than 4." But maybe in this particular case (two sines and one constant), somebody could infirm/confirm that intuition -- which is probably also right here -- with some even better arguments?

Depending on the four given constants we have a set of real roots and complex roots. It is a rotation of two vectors with different speeds with a constant matching y-projection.

However, the derivative can be solved in closed form by letting
 $$ \cos T = \dfrac {\alpha_1 \beta_1 }{\sqrt{(\alpha_1 \beta_1)^2 + (\alpha_2 \beta_2)^2} }  $$ etc. which can be integrated. (Sorry no, my error !)

EDIT1:
Amplitude and frequency modulation of signals are quite different this way.

Let
$$x(t)= \alpha_1\sin(\beta_1 t)+\alpha_2\sin(\beta_2 t)+1$$

Differentiating
$$x^{'}(t)= \alpha_1 \beta_1\cos(\beta_1 t)+\alpha_2 \beta_2\cos(\beta_2 t) $$

Letting
 $$ \cos T = \dfrac {\alpha_1 \beta_1 }{\sqrt{(\alpha_1 \beta_1)^2 + (\alpha_2 \beta_2)^2} },\, \sin T = \dfrac {\alpha_2 \beta_2 }{\sqrt{(\alpha_1 \beta_1)^2 + (\alpha_2 \beta_2)^2} }   $$

Then
$$x^{'}(t)=  const \cdot [ \cos T \cos(\beta_1 t) + \sin T \cos(\beta_2 t) ] $$

which cannot be simplified due to three frequencies. Read under signal frequency modulation, you may still be lucky.

EDIT2:
Second order differentiation

$$ - x^{''}(t)= \alpha_1 {\beta_1}^2 \sin(\beta_1 t)+\alpha_2 {\beta_2}^2\sin(\beta_2 t) $$

letting each signal expressible in terms of derivatives.


"Advice to young mathematicians" I have been suggested to read the Advice to a Young Mathematician section  of the Princeton Companion to Mathematics, the short paper Ten Lessons I wish I had been Taught by Gian-Carlo Rota, and the Career Advice section of Terence Tao's blog, and I am amazed by the intelligence of the pieces of advice given in these pages.

Now, I ask to the many accomplished mathematicians who are active on this website if they would mind adding some of their own contributions to these already rich set of advice to novice mathematicians.

I realize that this question may be seen as extremely opinion-based. However, I  hope that it will be well-received (and well-answered) because, as Timothy Gowers put it,

  "The most important thing that a young mathematician needs to learn is
  of course mathematics. However, it can also be very valuable to learn
  from the experiences of other mathematicians. The ﬁve contributors to
  this article were asked to draw on their experiences of mathematical
  life and research, and to offer advice that they might have liked to
  receive when they were just setting out on their careers."

My advice would be:
$\bullet $ Do many calculations
$\bullet \bullet$ Ask yourself  concrete questions whose answer is a number.
 $\bullet \bullet \bullet$ Learn a reasonable number of formulas by heart. (Yes, I know this is not fashionable advice!)
$\bullet \bullet \bullet \bullet$  Beware the illusion that nice general theorems are the ultimate goal in your subject.

I have answered many  questions tagged algebraic geometry on this site and I was struck by the contrast between the excellent quality of the beginners in that field and the nature of their questions: they  would know and really understand abstract results (like, say, the equivalence between the category of commutative rings and that of affine schemes) but would have difficulties answering more down-to-earth questions like:    "how many lines cut   four skew lines in three-dimensional projective space ?" or "give an example of a curve of genus $17$".

In summary  the point of view of  some quantum physicists toward the philosophy of their subject
Shut up and calculate !    contains more than a grain of truth for mathematicians too (although it could be formulated more gently...)

Nota Bene
The  above exhortation is probably due to David Mermin, although it is generally misattributed to Richard Feynman.

Edit
Since @Mark Fantini asks for more advice in his comment below, here are some more (maybe too personal!) thoughts:
$\bigstar$ Learn mathematics pen in hand but after that go for a stroll and think about what you have just learned. This  helps  classifying new material in  the brain, just as sleep is well known to do.
$\bigstar \bigstar$ Go to a tea-room with a mathematician friend and scribble mathematics for a few hours in a relaxed atmosphere.
I am very lucky to have had such a friend since he and I were beginners and we have been working together in public places ( also in our shared office, of course) ever since.
$\bigstar \bigstar \bigstar$ If you don't understand something, teach it!
I had wanted to learn scheme theory for quite a time but I backed down because I feared the subject.
One semester  I agreed to teach it to graduate students and since I had burned my vessels I really had to learn the subject in detail and invent simple examples to see what was going on.
 My students did not realize that I was only one or two courses ahead of them and  my teaching was maybe better in that  the material taught was as new and difficult for me as it was for them.
$\bigstar \bigstar \bigstar \bigstar$ Last not least: use this site!
Not everybody has a teaching position, but all of us can answer here.
I find using this site and MathOverflow  the most efficient way of learning or reviewing mathematics . The problems posed are often quite ingenious, incredibly varied and the best source for questions necessitating explicit calculations (see points $\bullet$ and $\bullet \bullet$ above).

New Edit (December 9th)
Here are a few questions posted in the last 12 days which I find are in the spirit of what I recommend in my post: a), b), c), d), e), f), g), h).

Newer Edit(December 17th)
Here is a fantastic question, brilliantly illustrating how to aggressively tackle mathematics, asked a few hours ago by  Clara: very concrete, low-tech and naïve  but quite disconcerting.
This question also seems to me absolutely original : I challenge everybody to find it in any book  or any on-line document !

If I could give a younger me advice, I would tell myself to lower my pride and accept as much help as I could in the times it was available to me, and shed my fear of humiliation and inferiority that kept me from even considering talking to others entirely let alone other mathematicians.

While you should strive to be more receptive to constructive criticism every day, I also think it's important to disregard criticism when it's source is from an individual that considers their authority in the field in question to be a conclusive presumption that must be held by yourself and all others.


Why “characteristic zero” and not “infinite characteristic”? The characteristic of a ring (with unity, say) is the smallest positive number $n$ such that $$\underbrace{1 + 1 + \cdots + 1}_{n \text{ times}} = 0,$$ provided such an $n$ exists. Otherwise, we define it to be $0$.

But why characteristic zero? Why do we not define it to be $\infty$ instead? Under this alternative definition, the characteristic of a ring is simply the “order” of the additive cyclic group generated by the unit element $1$.

My feeling is that there is a precise and convincing explanation for the common convention, but none comes to mind.  I couldn't find the answer in the Wikipedia article either.

There are two orderings of the set $\mathbb N = \{0,1,\dots\}$:

magnitude $a \leq b$
divisibility $a\mid b$ (i.e. $\exists c. b = a c$)

They are mostly compatible - usually when $a \mid b$, it holds $a \leq b$.

Some definitions are phrased using "greater than" ordering, while in fact the "divisibility" ordering is the real essence.

For example, the greatest common divisor of $a$ and $b$ might be defined as the greatest number which is a common divisor of both $a$ and $b$. Characteristic of a ring $R$ might be defined as smallest number $n>0$ which satisfies $n \cdot 1 = 0$.

Under such commonly taught definitions, it seems natural that $\operatorname{gcd}(0,0)=\infty$ and $\operatorname{char} \mathbb Z = \infty$.

However, those definitions implicitly rely on ideals, and are better phrased using divisibility order. The incompatibility is then more visible: $0$ is the largest element in divisibility order, while it is smallest in magnitude order. Magnitude has no largest element, and often $\infty$ is added to cover this case.

So let's formulate the definitions again, but this time using divisibility ordering.

The greatest common divisor of two numbers $a,b$ is greatest number (in sense of $\mid$) that is a divisor of $a$ and $b$ (i.e. is smaller than $a$ and $b$ in divisibility ordering). This is prettier - $\operatorname{gcd}$ is now the $\wedge$ operator in lattice $(\mathbb N, \mid)$; it also forms a monoid, with $0$ as identity element. Additionally, the definition can be adapted to any ring.
The characteristic of a ring $R$ is the smallest number $n$ (in sense of $\mid$) that satisfies $n \cdot 1 =0$. As a bonus, compared to previous definition, we can remove the $n>0$ restriction: zero is always a valid "annihilator" but it is often not the smallest one. Now we get $\operatorname{char} \mathbb Z = 0$.

Characteristic is a "multiplicative" notion, like gcd. If you have a homomorphism of rings $f: A \to B$, it must hold $\operatorname{char} B \mid \operatorname{char} A$. For example, you cannot map ${\mathbb Z}_2$ to ${\mathbb Z}_4$ - in a sense, ${\mathbb Z}_2$ is "smaller" than ${\mathbb Z}_4$. "Bigger" rings have "more divisible" characteristic, their characteristics are greater in the sense of divisibility. And the "most divisible" number is 0. Another example is $\operatorname{char} A \times B = \operatorname{lcm}(\operatorname{char} A, \operatorname{char} B)$.

In a bit more abstract language: given any ideal $I \subseteq \mathbb Z$, we associate to it the smallest nonnegative element, under the divisibility order. By properties of $\mathbb Z$, every other element of $I$ is a multiple of it. Let's call this number $\operatorname{min}(I)$.

We can now define $\operatorname{gcd}(a,b)=\operatorname{min} ((a) + (b))$, and $\operatorname{char} R = \min (\ker f)$, where $f \colon \mathbb Z \to R$ is the canonical map.

The definition of $\operatorname{min}(I)$ works for any PID, it does not require magnitude order. In any PID, $I = (\operatorname{min}(I))$.

(I dislike saying the ideal $\{0\}$ is "generated" by $0$; although this is true, it also generated by empty set. We do not say that $(2)$ is generated by $0$ and $2$.)

Not only does adding 1 infinitely times in a field not make sense, but also if adding 1 a positive number of times never yields zero, then certainly adding it zero times yields zero, thus it only seems natural to refer to such a field as "characteristic zero".


No proper dense open subsets of $\mathbb{R}^n$? Let $U$ be a dense open subset of $\mathbb{R}^n$, a positive integer, and suppose that $X:=\mathbb{R}^n-U$ is $0$-dimensional; in the sense of the Lebesgue covering dimension.  Then is $U$ homeomorphic to $\mathbb{R}^n$ if and only if $U=\mathbb{R}^n$?

What about $$\mathbb{R}^2 \setminus \lbrace (x,0), x \geq 0 \rbrace \quad \quad ?$$

You are right.
Clearly, if $U=\mathbb{R}^n$, $U$ is a dense open subset of itself.
For the other implication, let us assume $\exists U\subsetneq \mathbb{R}^n$, a dense open subset of $\mathbb{R}^n$, with $V:=\mathbb{R}-U$ $0$-dimensional. Since $V$ is $0$-dimensional, for any $\mathbf{y}\in V$, $B^n_{\epsilon}(\mathbf{y})\cap U\neq \emptyset,\;\forall \epsilon>0$ (Where $B^n_\epsilon(\mathbf{y})$ is $n$-dimensional ball, centered in $\mathbf{y}$ and radius $\epsilon$,).
Let, $$x_0\in \bigcap_{\epsilon >0} B^n_{\epsilon}(\mathbf{y})\cap U,$$
Then, since $U$ is an open set, the  $\exists\delta >0\;|\;B_{\delta}^n(\mathbf{x}_0)$ is contained in $U$. But, by construction, $\mathbf{y}\in B_{\delta}^n(\mathbf{x}_0)$, for any $\delta >0$. This is a contradiction, since we assumed $\mathbf{y}\notin U$.


A non constant complex valued function with zero derivative I need an example of a non constant complex valued function whose derivative is zero at every point in in the complex plane.

A complex valued function on a connected set with derivative $0$ is a constant. (Refer Stein - Shakarachi Complex analysis Chapter 1 for proof.)
If the function is defined on disconnected regions then its derivative is zero need not imply function is constant.
For example take $f$ to be $1$ on $|z|<1$ and $2$ on $|z-3|<1$. Then $f$ is non-constant but $f'(z)=0$.

A function  with zero derivative (anywhere) is always constant.

Let $f:\mathbb{C}\to\mathbb{C}$ a function with zero derivative at any $z_0\in\mathbb{C}$.

well write $f(z)=f(x+iy)=u(x,y)+iv(x,y)$ and $z_0=x_0+iy_0$.

So the derivative exists and $f'(z_0)=lim_{n\to\infty}\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z_0}$, and from its existence we conclude $f'(z_0)=lim_{\Delta z\to0}\frac{f(z_0+\Delta x)-f(z_0)}{\Delta x}=\frac{\partial u}{\partial x}(x_0,y_0)+i\frac{\partial v}{\partial x}(x_0,y_0)=0$ and...

$f'(z_0)=lim_{\Delta z\to0}\frac{f(z_0+i\Delta y)-f(z_0)}{i\Delta y}=-i\frac{\partial u}{\partial y}(x_0,y_0)+\frac{\partial v}{\partial y}(x_0,y_0)=0$

Thus for any $x,y\in\mathbb{R}$,  $0=\frac{\partial u}{\partial x}=\frac{\partial u}{\partial y}$ so $u:\mathbb{R}^2\to\mathbb{R}$ is constant.

Moreover $0=\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y}$ so $v:\mathbb{R}^2\to\mathbb{R}$ is constant.

Hence $u+iv=f$ is constant.

I think it holds ...


Calculus question about the existence of antiderivative Let $f(x)$ be the function equal $0$ on $\mathbb R_{<0}$ and equal $1$ otherwise. Is clear to me that $f$ has derivative undefined at $x=0$. Is intuitive because the derivative would be $\infty$ as $f$ increases by $1$ in an infinitely small space.

Now I was wondering whether antiderivative exists at $0$. My thoughts: antiderivative corresponds to area under the function. Before zero the are is always zero and after zero the area is $1$ so that like in derivative case the area increases by $1$ in an infinitely small space. But I don't know if intuition can be used in this case.

When does antiderivatives of functions exists in general and when not?

For such function does not exist any antiderivative; that is because even if $F$ is derivable with derivative discontinuous $F'$ always respect the intermediate value property, so assuming $0$ and $1$ and never $\frac{1}{2}$, $f$ is not a derivative of any function.

What exists is its definite integral, in any derivable point of this integral the derivative is the value of $f$ but such integral is not derivable in $0$ only continuous.

The function F(x) = 0 for x < 0, F(x) = x for x ≥ 0 is an antiderivative for the function f that you describe.  Or if you prefer F(x) = c for x < 0 and F(x) = x + c for x ≥ 0, where c is any real number. Notice that F(x) is continuous on the entire real line, but not differentiable at x = 0 (which is as it should be).

Any piecewise continuous function on the interval (a,b) will have an antiderivative there, and most of the functions you are likely to run into will be at least piecewise continuous.  To avoid having an antiderivative, the function has to avoid being piecewise continuous, which is very discontinuous indeed, but can happen.  For example, the function g(x) = 0 where x is rational and 1 where x is irrational does not have an antiderivative on the interval (a,b).

In general integration is much more forgiving than differentiation.  That is because addition is additive, in fact is really an average, which tends to smooth things out. (That's why people use 4 week running averages to analyze trends-- they are smoother than the underlying daily data.)

Whereas, the existence of a f'(c)  (c any number) implies that f is locally linear in the neighborhood of c.  That is, f and the tangent line to f at c are very similar in a small enough interval of c.  Obviously a function with a discontinuity cannot be locally linear at that point, nor can a function like |x| (at zero).


Does there exists a onto continuos function $f\colon (0,1) \to[0,1]$ which is one to one . Iam trying to solve this problem , i can visualize graphically it is not possible. iam trying to do the proof by contradiction. if f:(0,1) to [0,1] is a continuos onto function  then i have to prove that f can never be 1-1 . so iam letting that if f is 1-1 then either f is strictly increasing or strictly decreasing function. so case (1) if f is strictly increasing function then there exits some t in (0,1) such that f(t)=0 , so for x lesser than t f(x)k , f(x) <0 which is also not possible .so f can't be 1-1 . My method is right ??? is there is any alternative approach then plz tell

This is merely an elaboration of the solution provided by Júlio Cesar under the original comment (all credit to him), which is in itself "only" making precise your original correct reasoning.

Assume $f:(0,1) \to [0,1]$ was continuous and bijective. Then we find unique $a := f^{-1}(0)$, $b := f^{-1}(1)$ and may assume wLog that $0 < a < b < 1$. Choose $x \in (0,a)$ and let $y := f(x) \in [0,1]$. By the intermediate value theorem, we find some $c \in (a,b)$ with $f(c) = y = f(x)$. Since $x < a < c$, $f$ is not injective, so we arrive at a contradiction.

$[0,1] $ is closed.

$f $ continuous and one to one implies $f^{-1}([0,1]) $ is closed.

this is not the case for $(0,1) $

Using sequential characterisation of the limit, it is easy to prove the continuity of $f^{-1} $.


Why is epsilon not a rational number? I was wondering why epsilon, the smallest positive number, isn't a rational number. I was watching a video a few days ago about surreal numbers, and I've learned that, in the field of surreal numbers, o.(9) is not equal to 1, in contrast to the field of the real numbers, where they represent the same number. In the field of surreal numbers, you would get epsilon by subtracting 0,(9) from 1. If you were to do this in the rationals, you would just get 0. But I think there is a method do get epsilon even in the rationals, you would just take the following limit:

$$\lim\limits_{n \to\infty} \sum_{i=0}^n {1\over 10^i}$$
Am I making a wrong mathematical assumption or...? Is there a reason for which epsilon couldn't be a rational number?

See Surreal number :

Consider the smallest positive number in $S_ω$:

$\varepsilon =\{S_{-}\cup S_{0}|S_{+}\}=\{0|1,{\tfrac {1}{2}},{\tfrac {1}{4}},{\tfrac {1}{8}},...\}=\{0|y\in S_{*}:y>0\}$.

This number is larger than zero but less than all positive dyadic fractions. It is therefore an infinitesimal number, often labeled $ε$.

Thus epsilon is, "by definition" less than (and so different from) all rational in the $(0,1)$ interval.

Okay.  I do not know much about the surreal numbers.  So I looked on wikipedia and these were the VERY FIRST two sentences.

"In mathematics, the surreal number system is an arithmetic continuum containing the real numbers as well as infinite and infinitesimal numbers, respectively larger or smaller in absolute value than any positive real number. The surreals share many properties with the reals, including a total order ≤ and the usual arithmetic operations (addition, subtraction, multiplication, and division); as such, they form an ordered field."

This answers why epsilon, the smallest positive (surreal) number, can not be rational.  The rational numbers and the real numbers have the archimedian property that between any two rationals and between any two reals there exist a third.  So it is impossible to have a smallest positive real or rational number.  So if a smallest positive number is possible (as apparently it is) it can not be real or rational otherwise it would be the smallest real or rational number which is impossible.

So, I interpret the OP question to be why doesn't his construction of $\lim_{n \rightarrow \infty} \prod_{i=1}^n\frac 1{10^i} $ produce a least rational number?

Because the lim = 0.  (It also needn't be rational {although it is}.  But it does need to be real-- and there is no least positive real either.)

=====

Okay, I wasn't aware that epsilon exists in the field of surreal numbers and does have a definition as the smallest positive surreal number (I guess).

But here, fundamentally, the surreals are not the reals.  The rationals are the rationals and as $0 < 1/(n+1) < 1/n$ there can be no smallest positive rationals.  Thus any system that allows a smallest positive number (which the reals do not), such a number can not be rational.

=====================

1) epsilon is not the smallest positive number.  epsilon is not a set number at all.  It is a variable to represent possible numbers.  It is no more or less a number than, x, n, or i are numbers for exactly the same reasons and in exactly the same way.

2) There is no smallest positive |real| number.  This is fundamental to mathematics real analysis. [ed: and so any smallest positive number can not be real.]

3) The limit of a sequence (or sum) of rational numbers need not be rational at all.  Indeed it is a fundamental principal of real [ed: emphasis added later] numbers that all real [ed: but not surreal] numbers  (rational and irrationals) are limits of sequences of rational numbers.

4) The limit of the infinite product you gave is 0.


Assume $f:[a,b] \to \mathbb R$ is continuous and for every $x \in \mathbb R :f(x) \ge 0$,if $M:=\sup\left\{f(x): x \in [a,b]\right\}$ Assume $f:[a,b] \to \mathbb R$  is continuous and for every $x \in \mathbb R :f(x) \ge 0$,if $M:=\sup\left\{f(x): x \in [a,b]\right\}$,show that the sequence $$\left\{\left(\int_{a}^{b}f^{n}\left(x\right)dx\right)^{\frac{1}{n}}\right\}_n$$ Is convergent to $M$.

If $M=\sup\left\{f(x): x \in [a,b]\right\}$ then exists $x \in [a,b]$ such that foe every $\epsilon>0:M - \epsilon < f(x) \le M$,so :
$$\left(M-\epsilon\right)^{n}\left(b-a\right)=\int_{a}^{b}\left(M-\epsilon\right)^{n}dx\le\int_{a}^{b}f^{n}\left(x\right)dx\le\int_{a}^{b}M^{n}dx=M^{n}\left(b-a\right)$$
Taking the $n$th root of the both sides yields:
$$\left(M-\epsilon\right)\left(b-a\right)^{\frac{1}{n}}\le\left(\int_{a}^{b}f^{n}\left(x\right)dx\right)^{\frac{1}{n}}\le M\left(b-a\right)^{\frac{1}{n}}$$
Taking the limit of both sides:
$$\left(M-\epsilon\right) \le \lim_{n \to \infty} \left(\int_{a}^{b}f^{n}\left(x\right)dx\right)^{\frac{1}{n}} \le M$$
I have some questions:

How do we know that $M-\epsilon$ is not negative?
We used that $(M - \epsilon )^n< f^n(x)\le M^n $ and from here $$\int_{a}^{b}\left(M-\epsilon\right)^{n}dx\le\int_{a}^{b}f^{n}\left(x\right)dx\le\int_{a}^{b}M^{n}dx$$

Why that's true??

There are two cases. If $f \equiv 0$ then the claim follows immediately and there's no need to consider the inequality. In all other cases, there exists some point in the interval in which $f$ is positive and so $M > 0$. Hence, for small enough $\varepsilon$, the expression $M - \varepsilon$ will be positive.
You are using the monotonicity of the integral. If $f(x) \leq g(x)$ for all $x \in [a,b]$ and they are integrable then $\int_{a}^b f(x) \, dx \leq \int_a^b g(x) \, dx$. This follows from the fact that $g(x) - f(x) \geq 0$ for all $x \in [a,b]$ and that the integral of a non-negative function is non-negative.

First question: if $f(x)=0$ for all $x \in [a,b]$, then you are done.
Hence we assume that $f(x)>0$ for some $x \in [a,b]$. Then we have that $M>0.$
Then consider $ \epsilon >0 $ such that $\epsilon <M.$
Second question: if $f_1$ and $f_2$ are integrable and $f_1 \le f_2$ on $[a,b]$ , then we have
$$\int_a^b f_1(x) dx \le \int_a^b f_2(x) dx.$$


Differential function problem Let $f: \mathbb{R} → \mathbb{R} $ be a function such that $f(x)$ is differentiable on all $\mathbb{R}$ and $\lim_{x\to \infty}(f(x)-f(-x))=0$.

Prove there exists $x_{0} \in \mathbb{R}$ such that $f'(x_{0})=0$

I tried proving it by contradiction using that the limits at infinity and minus infinity are equal I'm having hard time formalizing it.

Any help appreciated.

Either $f$ has a local extremum somewhere at $x_0\in\mathbb R$ (a maximum or minimum), either it is monotone. In the first case, then necessarily $f'(x_0)=0$, in the second case, the fact that $\lim_{x\to+\infty}(f(x)-f(-x))=0$ implies that $f$ is constant, and thus $f'=0$ everywhere.

Since $\lim_{x\to \infty}(f(x)-f(-x))=0$,

and $f(x)$ is differentiable and assumed continuous everywhere, by the Intermediate Value Theorem, there exists a $x_{0}$ such that:

$$f^{'}(x_{0})=\frac{\lim_{x\to \infty}f(x)-\lim_{x\to \infty}f(-x)}{x+x}=0.$$ (proved)


Are there arbitrarily large gaps between consecutive primes? I made a program to find out the number of primes within a certain range, for example between $1$ and $10000$ I found $1229$ primes, I then increased my range to $20000$ and then I found $2262$ primes, after doing it for $1$ to $30000$, I found $3245$ primes.

Now a curious thing to notice is that each time, The probability of finding a prime in between $2$ multiples of $10000$ is decreasing, i.e it was $$\frac{2262-1229}{10000}=0.1033$$ between $10000$ and $20000$, and $$\frac{3245-2262}{10000}=0.0983$$ between $20000$ and $30000$,

So from this can we infer that there will exist two numbers separated by a gap of $10000$ such that no number in between them is prime? If so how to determine the first two numbers with which this happens? Also I took $10000$ just as a reference here, what about if the gap between them in general is $x$, can we do something for this in generality?

Thanks!

Can we infer that there exist two numbers separated by a gap of $10000$, such that no number in between them is prime?

We can infer this regardless of what you wrote.

For every gap $n\in\mathbb{N}$ that you can think of, I can give you a sequence of $n-1$ consecutive numbers, none of which is prime.

There you go: $n!+2,n!+3,\dots,n!+n$.

So there is no finite bound on the gap between two consecutive primes.

There several factors determining the size of gaps.  As per you example 10,000 has a square root of 100 which give you 25 prime number divisors for the 10,000 and the 20,000 square root is 141 and 34 divisors, and the 30,000 square root is 173 and 40 divisors.  The number of divisors helps to determine the gaps.  The number of factors needed is based on the square root of your max number.  The square roots are decreasing percentage of the max number: example 100 / 10000 = 1.000%, 141 / 20000 = .70500% and 173 / 30000 = .57666%.  The number of factors increased against the square root but decreases against the max number.  Example:  106,749,747,000,000 the square root is 10,331,977 the square root % is .00000968% with 685,159 factors the max gap in my sample 712.  The large gaps seam to have a cycle pattern to them.


Isomorphism of an endomorphism ring, how can $R\cong R^2$? I'm aware that over a commutative ring, two free finitely generated modules of finite rank are isomorphic if and only if they have the same rank.

I'm trying to understand a curious example of why this does not hold over a noncommutative ring.

  Let's take $M=\oplus_{i\in\omega}\mathbb{Z}$. Then let $R=\text{End}(M)$, which is not commutative. Then $R\cong R^2$ as left $R$-modules.

It seems like one could possibly make a basis of $R$ using only endomorphisms which act on the even or odd coordinates of elements of $M$, which might suggest such an isomorphism. Is there a clear proof of why $R\cong R^2$?

One is tempted (as I was originally), to argue as follows: since $M\cong M\oplus M\cong M\times M$, we have
$$R = \mathrm{Hom}(M,M) \cong \mathrm{Hom}(M,M\times M) \cong \mathrm{Hom}(M,M)\times\mathrm{Hom}(M,M) = R\times R,$$
and stop there. The problem with the argument is that in some instances we are dealing with these objects as $\mathbb{Z}$-modules rather than as $R$-modules (specifically the isomorphisms are isomorphism from $\mathrm{Hom}(M,M)$ to $\mathrm{Hom}(M,M\times M)$ is, right now, just an isomorphism as $\mathbb{Z}$-modules), so some care needs to be exercised to make the argument actually work.

First, note that $M\cong M\times M$  as $\mathbb{Z}$-modules.
Define a homomorphism $\varphi\colon M\to M\times M$ by maping $(m_i)$ to $\bigl((m_{2i-1}), (m_{2i})\bigr)$. That is, $(m_1,m_2,m_3,\ldots)$ maps to $\bigl( (m_1,m_3,m_5,\ldots),(m_2,m_4,m_6,\ldots)\bigr)$.

The elements of $R$ can be thought of as infinite "column-finite matrices"; that is, each endomorphism $M\to M$ corresponds to a family of functions $(\mathbf{f}_i)_{i\in\omega}$, with $\mathbf{f}_i\colon\mathbb{Z}\to M$; hence $\mathbf{f}_i = \sum_{j\in\omega}f_{ji}e_j$, where $e_j$ is the element of $M$ that has $1$ in the $j$th coordinate and $0$s elsewhere, $f_{ji}\in\mathbb{Z}$, and $f_{ji}=0$ for almost all $j$.

If we take an element of $\mathbf{f}=(f_{ij})$ of $R$, and compose it with the isomorphism $M\to M\times M$, we obtain a homomorphism $M\to M\times M$, given by $\bigl( (f_{i,2j-1}), (f_{i,2j})\bigr)$. So the map $\mathrm{Hom}(M,M)\to\mathrm{Hom}(M,M)\times \mathrm{Hom}(M,M)$ maps $(f_{ij})$ to $\bigl( (f_{i,2j-1}),(f_{i,2j})\bigr)$.

If we compose maps $\mathbf{f}=(f_{ij})$ and $\mathbf{g}=(g_{ij})$, we get the map
$(h_{ij})$, where
$$h_{ij} = \sum_{k=1}^{\infty}g_{ik}f_{kj}.$$
Since $f_{kj}=0$ for almost all $k$, the sum is finite and $h_{ij}$ makes sense.

This gives the action of $R$ on $\mathrm{Hom}(M,M)$. The action of $R$ on $\mathrm{Hom}(M,M)\times\mathrm{Hom}(M,M)$ is then given by
$$
\mathbf{g}\bigl((f_{ij}),(f'_{ij})\bigr) = \bigl( \mathbf{g}(f_{ij}), \mathbf{g}(f'_{ij})\bigr)
= \bigl( (h_{ij}), (h'_{ij})\bigr)$$
where
$$h_{ij} = \sum_{k=1}^{\infty}g_{ik}f_{kj},\quad\text{and}\quad h'_{ij}=\sum_{k=1}^{\infty}g_{ik}f'_{kj}.$$
To see that the map from $\mathrm{Hom}(M,M)$ to $\mathrm{Hom}(M,M\times M)$ respects the action of $R$ (that is, that we get an $R$-module homomorphism, not merely a $\mathbb{Z}$-module homomorphism), let $(f_{ij})\in\mathrm{Hom}(M,M)$ and $\mathbf{g}\in R$. The element $\mathbf{g}(f_{ij})$ is mapped to
$$\left( \Bigl(\sum_{k=1}^{\infty}g_{ik}f_{k,2j-1}\Bigr),\Bigl( \sum_{k=1}^{\infty}g_{ik}f_{k,2j}\Bigr)\right)$$

On the other hand, if we let $\mathbf{g}$ act on $\bigl( (f_{i,2j-1}), (f_{i,2j})\bigr)$, we get
$$\mathbf{g}\left( (f_{i,2j-1}), (f_{i,2j})\right) =  \left(\Bigl( \sum_{k=1}^{\infty}g_{ik}f_{k,2j-1}\Bigr), \Bigl( \sum_{k=1}^{\infty}g_{ik}f_{k,2j}\Bigr)\right),$$
that is, the same as the image of $\mathbf{g}(f_{ij})$. So the homomorphism $\mathrm{Hom}(M,M)\to\mathrm{Hom}(M,M)\times\mathrm{Hom}(M,M)$ given by $(f_{ij})\longmapsto ((f_{i,2k-1}),(f_{i,2k}))$ is actually a homomorphism as $R$-modules. Since the original map as $\mathbb{Z}$-modules was a bijection, so is this one, hence $\mathrm{Hom}(M,M)$ and $\mathrm{Hom}(M,M)\times\mathrm{Hom}(M,M)$ are isomorphic not only as $\mathbb{Z}$-modules, but also as $R$-modules. So we obtain:
$$R = \mathrm{Hom}(M,M) \cong \mathrm{Hom}(M,M)\times\mathrm{Hom}(M,M) =R\times R,$$
with the isomorphism being an isomorphism of $R$-modules, as desired.

You probably mean that "two FREE finitely generated $R$-modules  $M_1$ and $M_2$ are isomorphic iff they have equal rank".

It's not so surprising this doesn't work in the infinite dimensional case. Just look at infinite-dimensional vector spaces.

You should be careful when you talk about modules over a non-commutative ring. You should indicate whether you're talking about left/right modules, etc.

Edit: Not sure if this is correct, but you can write down an explicit isomorphism once you choose a bijection from the set of integers to the set of even integers. (Or something like that.)


Condition for integrability on finite measure space Let $(X,\mathcal{F},\mu)$ be a finite measure space. If $f:X\rightarrow \mathbb{R}$ is a measurable real function, show that, $f\in L^1(\mu)$ iff $\sum\limits_{n=0}^{\infty}\mu(\{f\geq n\})<\infty$. Am a bit stuck on the $(\Leftarrow)$ direction so any help is appreciated, thanks.

Consider the sets $A_n=\{|f|\geqslant n\}$ and integrate the double inequality
$$
\sum_{n=1}^\infty\mathbf 1_{A_n}\leqslant |f|\leqslant\sum_{n=0}^\infty\mathbf 1_{A_n}.
$$
Thus, $f$ is integrable if and only if
the series $\sum\limits_n\mu(A_n)$ converges.

The hypothesis that $\mu$ is finite is there to ensure that the $n=0$ term $\mu(A_0)=\mu(X)$ is finite.

The double inequality above stems from the pointwise relations, valid for every nonnegative $t$,
$$
\lfloor t\rfloor=\sum_{n=1}^\infty\mathbf 1_{t\geqslant n}\leqslant t\lt1+\lfloor t\rfloor=\sum_{n=0}^\infty\mathbf 1_{t\geqslant n}.
$$

In general, the $\Leftarrow$ direction does not hold because if we
can find $g\geq1$ with $\int g=\infty$. Put $f=-g$, then $\sum_{n=0}^{\infty}\mu([f\geq n])=0$
but $f$ is not integrable.

To avoid typing the sign of absolute value, I modify $f$ to ''$f:X\rightarrow[0,\infty)$''.

For each $n=0,1,2,\ldots,$let $A_{n}=\{x\mid f(x)\in[n,n+1)\}$.
Let $B_{n}=[f\geq n]$. Observe that $A_{n}=B_{n}\setminus B_{n+1}$,
so $\mu(A_{n})=\mu(B_{n})-\mu(B_{n+1})$. Also, $A_{0},A_{1},\ldots$
are pairwisely disjoint with $\cup_{n}A_{n}=X$.

Suppose that $\int f<\infty$. By Monotone Convergence Theorem, $$\int f=\sum_{n=0}^{\infty}\int_{A_{n}}f\geq\sum_{n}n\mu(A_{n})=\lim_{N\rightarrow\infty}\sum_{n=0}^{N}n\left(\mu(B_{n})-\mu(B_{n+1})\right)\geq\sum_{n=0}^{N}n\left(\mu(B_{n})-\mu(B_{n+1})\right)$$
for any $N$. Observe that $\sum_{n=0}^{N}n\left(\mu(B_{n})-\mu(B_{n+1})\right)=\sum_{n=1}^{N}\mu(B_{n})-N\mu(B_{N+1})$.
Therefore, $\sum_{n=0}^{N}\mu(B_{n})\leq N\mu(B_{N+1})+\int f+\mu(B_{0})$.
Note that $N\mu(B_{N+1})\leq\int_{B_{N+1}}f\leq\int f.$ Hence, $\sum_{n=0}^{N}\mu(B_{n})\leq2\int f+\mu(B_{0})$.
Letting $N\rightarrow\infty$, we have $\sum_{n=0}^{\infty}\mu(B_{n})\leq2\int f+\mu(B_{0})<\infty$.

Conversely, suppose that $\sum_{n=0}^{\infty}\mu(B_{n})<\infty$.
Note that $\int f=\sum_{n=0}^{\infty}\int_{A_{n}}f\leq\sum_{n=0}^{\infty}(n+1)\mu(A_{n})$.
For each $N$, $\sum_{n=0}^{N}(n+1)\mu(A_{n})=\sum_{n=0}^{N}(n+1)\left(\mu(B_{n})-\mu(B_{n+1})\right)=\sum_{n=0}^{N}\mu(B_{n})-(N+1)\mu(B_{N+1})\leq\sum_{n=0}^{N}\mu(B_{n})$.
Letting $N\rightarrow\infty$, we have $\int f\leq\sum_{n=0}^{\infty}(n+1)\mu(A_{n})\leq\sum_{n=0}^{\infty}\mu(B_{n})<\infty$.


Union of the Intersection of Infinite Sets So, I'm trying to make sense of this question:

"Suppose we have an $A_n$ for each n $\in$ N (i.e. N is the Naturals. Each $A_n$ is an arbitrary set). Fill in the blanks with four words to get a true statement. Justify your answer.

x $\in$ $\bigcup_{k=1}^{\infty}\bigcap_{n=k}^{\infty}$ $A_n$ iff x $\in$ $A_n$ for [blank] [blank] [blank] [blank] n $\in$ N"

So far, I understand that for a given k, if x $\in$ ($\bigcap_{n=k}^{infinity}$ $A_n$ = $A_k$ $\cap$ $A_{k+1}$ $\cap$ $A_{k+2}$ $\cap$ ....) $\implies$ x $\in$ $A_n$ for all n$
\geqslant$ k

But I'm stuck on interpreting the next stage. If x $\in$ $\bigcup_{k=1}^{\infty}\bigcap_{n=k}^{\infty}$ $A_n$, then x $\in$ ({$A_1$ $\cap$ $A_{2}$ $\cap$ $A_{3}$ $\cap$...} $\bigcup$ {$A_2$ $\cap$ $A_{3}$ $\cap$ $A_{4}$ $\cap$...} $\bigcup$....). Which just means that $\exists$ some b $\in$ N s.t. x $\in$ $A_n$ for all $n\geqslant$ b.

Have I interpreted this all correctly, and if so how could I translate what I've got so far into the 4 words? If I've misinterpreted the situation, have you any thoughts as to what I'm lacking?

Many thanks, really appreciate it.

"all except finitely many" is what I would write there. That is the definition of [\liminf ]. By the way, remember the command for "infinite" is \infty.


Closed form for $\int_0^1 \frac{x^a dx}{(1+x^b)^c}$. By means of a substitution, the integral can be reduced to

$$\int_0^1 \frac{x^p dx}{(1+x)^q}$$
but no method that I've thusfar tried seems tenable.

This Beta-like integral cannot be reduced into the Beta function (i.e. $\int_0^\infty \frac{x^p dx}{(1+x)^q}$) by any immediately obvious method (the denominator ranges from $1$ to $2$, not $0$ to $1$, which no subsitution can resolve). Numerically the integral tends to be a fraction (suggesting that $\Gamma(x)$ may be involved), unless $c=1$, in which case the result is easy to work out.

As @OL points out the integral is a hypergeometric function:

$$\frac{\, _2F_1(p+1,q;p+2;-1)}{p+1}$$

You can get this by expanding the denominator in a series (by the binomial theorem), integrating term by term, and observing that you get a hypergeometric series. If there is a relationship of some sort between $p$ and $q,$ this might be simplifiable.

You can use the variable $t=1+x$ and use the binomial theorem (for extended binomials coefficient to real numbers)


Showing that solutions to an ODE does not escape to $\infty$ in finite time I have an ODE

$$x'=f(x)$$

and that there are constants $C$, $K$ such that $|f(x)| ≤ C|x| + K$ for all x.

How can I show that the solutions to this ODE does not escape to infinity in finite time?

I tried to write it as $\frac{|x'|-K}{|x|} ≤ C$ but don't think I could integrate this.

You can
 bound $|x(t)|$ by a solution of the linear equation $y' = C y + K$.

Hint: Use the mean value theorem: $\mid x(u)-x(v)\mid \leq \mid x'(c)(u-v)\mid =\mid f(c)(u-v)\mid \leq (C\mid c\mid+K)(u-v)$ where $c\in [u,v]$.


Limit of $\displaystyle \lim_{x\to 0+}\frac 1x-\frac{\cos x}{\sin x}=0$ How to best show that
$$\lim_{x\to 0+}\frac 1x-\frac{\cos x}{\sin x}=0.$$

I've tried to make a fraction out of it, but then I get 0/0, and so I take the derivative using L'Hopitals, and it just gets messier and messier.

Would just like a hint.

$$\begin{align}\lim_{x\to 0+}\frac 1x-\frac{\cos x}{\sin x}&=\lim_{x\to 0+}\frac{\sin x-x\cos x}{x\sin x}\\&=\lim_{x\to 0+}\frac{(\sin x-x\cos x)'}{(x\sin x)'}\\&=\lim_{x\to 0+}\frac{\cos x-(\cos x-x\sin x)}{\sin x+x \cos x}\\&=\lim_{x\to 0+}\frac{\sin x}{(\sin x/x)+\cos x}\\&=\frac{0}{1+1}\\&=0.\end{align}$$
Here, note that $\lim_{x\to 0}\frac{\sin x}{x}=1.$

$$\lim_{x \to 0}\frac{1}{x}-\frac{\cos (x)}{\sin(x)}=\lim_{x \to 0}\frac{1}{\sin (x)}-\frac{\cos (x)}{\sin(x)}=\frac{1-\cos (x)}{\sin(x)}=\lim_{x \to 0} \frac{\ \sin(x)}{\cos(x)}=\frac{0}{1}=0$$

Here we used L'Hopital (after the third $=$ sign) and we used the identity $\lim_{x \to 0} \frac{ x}{\sin (x)}=1$


Explanation circular permutation Well if one looks at the formula of circular permutations $Pc = (n-1)!$

But as we come to that formula, I need a concrete example and an explanation.

Another thing, I have seen that when working with a bracelet, the formula changes. A concrete example please along with an explanation

Finally how the formula is worked when there are repeated elements that are permuted.

Thank you

Circular permutations

Consider an arrangement of blue, cyan, green, yellow, red, and magenta beads in a circle.

For this particular arrangement of the six beads, there are six ways to list the arrangement of the beads in counterclockwise order, depending on whether we start the list with the blue, cyan, green, yellow, red, or magenta bead.  They correspond to the six linear arrangements shown in the rows below.

Conversely, each of these six linear arrangements can be transformed into the circular arrangement above by joining the ends of a row.

More generally, any circular arrangement of these six beads corresponds to six linear arrangements.  Since there are $6!$ linear arrangements of six distinct beads, the number of distinguishable circular arrangements is
$$\frac{6!}{6} = 5!$$

Unless other specified, only the relative order of the objects matters in a circular permutation.  Therefore, circular arrangements are considered to be rotationally invariant.

Given a circular arrangement of $n$ objects, they can be rotated $0, 1, 2, \ldots, n - 1$ places clockwise without changing the relative order of the objects.  Hence, the number of distinguishable arrangements of $n$ objects in a circle is the number of linear arrangements divided by $n$, which yields
$$\frac{n!}{n} = (n - 1)!$$

Alternatively, given $n$ objects, we measure the order relative to a given object.  Fix that object.  As we proceed counterclockwise around the circle, the remaining objects can be arranged in $(n - 1)!$ orders.

Bracelets

Now suppose we place these beads on a bracelet.

Observe that if you remove the bracelet at left from your wrist, twist it through a half-turn, then place it back on your wrist, it will look like the bracelet at right, where the beads are arranged in the opposite order as you proceed counterclockwise around the circle.  Thus, we can form the same bracelet by arranging the blue, cyan, green, yellow, red, and magenta in clockwise or counterclockwise order.  Hence, the number of bracelets we can form with the six beads given above is
$$\frac{5!}{2}$$

More generally, if a bracelet has no clasp or opening that allows us to distinguish a linear order, it is invariant with respect to both rotations and reflection.  Hence, the number of distinguishable arrangements of a bracelet with $n$ objects is
$$\frac{1}{2} \frac{n!}{n} = \frac{(n - 1)!}{2}$$
provided $n > 2$.  If $n = 1$, there is only one possible arrangement for the bracelet.  If $n = 2$, there is only one distinguishable arrangement for the bracelet.

Circular permutations with repetition

This is a much trickier problem.  To see why, consider an arrangement of nine blue and three red beads in a circle.  Two such arrangements are shown below.

The first time I saw such a problem, I attempted to solve it by choosing three of the $12$ positions for the red beads, then divide by $12$ to account for rotational invariance.
$$\frac{1}{12}\binom{12}{3} = \frac{1}{12} \cdot \frac{12!}{3!9!} = \frac{11!}{3!9!}$$
Alas, this is not an integer.  The reason it is not an integer is the arrangement at left.  While the circular arrangement at right corresponds to $12$ different linear arrangements, the one at left does not.  Given its symmetry, there are only four distinguishable linear arrangements corresponding to the twelve possible starting points of the linear arrangement, depending on whether the first red bead is in the first, second, third, or fourth position of the linear arrangement.  Therefore, we have counted this linear arrangement $1/3$ times.  Hence, the actual number of circular arrangements is
$$\frac{1}{12}\binom{12}{3} + \frac{2}{3}$$

While that observation solves this particular problem, in general, you will need to master the use of Burnside's lemma to handle these problems.

Both Linear and Circular permutations are quite same but the way I have understood it is you have to just remove the rotations of linear permutation to get circular permutation. And that's why we divide it by n


Calculus volume question The region enclosed by the circle defined by the equation $(x-2)^2+y^2 = 4$ is rotated about the y-axis to generate a solid of revolution. Find the volume of the solid.

The suggest answer is
$\int_0^4$ $2{\pi}x2{\sqrt{(4-(x-2)^2)}}$
and I want to know why it is $2{\sqrt{(4-(x-2)^2)}}$ instead of ${\sqrt{(4-(x-2)^2)}}$, where the 2 came from? Thanks.

The area will be given by

$$A=\int_0^42\pi rh\,dx$$

where $r=x$ and $h=2y=2\sqrt{4-(x-2)^2}$

thus

$$A=\int_0^44\pi x\sqrt{4-(x-2)^2}\,dx$$

Because the circle is symmetric. When you use shell method to calculate the volume, you need to consider the area below x axis,


easy explanation of Newton's method I was reading a tutorial of training a neural network using Newton's method and it says, "The maximum error reduction (of the error surface function) depends on the ratio of the gradient to the curvature. So, a good direction to move in is one with a high ratio of gradient to curvature, even if the gradient itself is small"

Can anybody give an intuitive explanation as to why the high ratio of gradient to curvature is a good direction to move?

I'm assuming by 'curvature' they mean '2nd derivative'. The reason for the formula is pretty simple to derive. Suppose that $y$ is the actual minimum of $f$ and $x$ is the current point. Then the 2nd order Taylor approximation of $f$ about $x$ gives

$$f(y) \approx f(x) + f'(x)(y-x) + \frac{1}{2}f''(x)(y-x)^2$$

If you just differentiate with respect to $y$ and use that $f'(y)=0$ then you get

$$y\approx x-\frac{f'(x)}{f''(x)}.$$

So it's the maximum decrease in the error according to a local approximation of the error by a quadratic function.

Newton's method only uses the first derivative of the operator to approximate its behavior and estimate how to adjust the inputs to reach the desired outputs.  If the operator is actually close to linear, this estimate will work well.  If the operator is highly nonlinear, i.e. it has a lot of curvature or large higher derivatives, then that curvature will have a large effect that the estimate doesn't take into account and the estimate will be bad.


Sufficient conditions for being a PID Let R be a commutative ring with identity. If every ideal generated by two elements of R is principal, then can we conclude that R is a PID? Also, if every finitely generated ideal of R is principal, can we conclude that R is a PID?

A domain with every two-generated  ideal principal $\rm\:(a,b) = (c)\:$ is called a Bezout domain. By induction this is equivalent to every finitely-generated ideal being principal. This does not imply that every ideal is principal (PID). Indeed, the ring of all algebraic integers is Bezout (Kaplansky, Commutative Rings, Theorem 102), but its ideal $\:(2^{1/2},\,2^{1/3},\ldots)\:$ is not principal.

A domain is a PID iff it's Bezout and satisfies ACCP (ascending chain condition on principal ideals). The direction $(\Rightarrow)$ is clear. For $(\Leftarrow)$ note that ACCP implies that the divisibility relation is well-founded, hence a nonzero ideal I is generated by an element c minimal wrt to divisibility, since if $\rm\:c\nmid i\in I\:$ then $\rm\:(c,i) = (d)\subset I\:$ and   $\rm\:d\:|\:c\:$ properly (else $\rm\,c\mid d\mid i\,),$ contra minimality of c. This is essentially an application of the Dedekind-Hasse Test for a PID.

If we are given a Bezout UFD $(\Rightarrow$ ACCP) then we can use the number of prime factors as a metric for the Dedekind-Hasse test. Then that the above $\rm\,c\in I\,$ is divisibility minimal translates to it being an element of $\,\rm I\,$ with least number of prime factors.

There is a theorem of Isaacs which says that if every prime ideal is principal then so is every ideal.


Cauchy's functional equation for involutions: $ f ( x + y ) = f ( x ) + f ( y ) $ and $ f \big( f ( x ) \big) = x $ It is well known that Cauchy's functional equation for $ f : \mathbb R \to \mathbb R $,
$$
f ( x + y ) = f ( x ) + f ( y ) \quad \forall x , y \in \mathbb R \text ,
$$
admits highly pathological solutions if no further conditions are given. Is the condition
$$
f \big( f ( x ) \big) = x \quad \forall x \in \mathbb R \text ,
$$
i.e. $ f $ being an involution, enough to ensure that no "ugly" function is a solution?

Nope.  For example, we can define $f$ by $f(x) = x$ for $x \in \Bbb Q$ and $f(x) = -x$ over all other cosets of $\Bbb Q$.

see Does this uniquely specify F(x)=x, on [0,1] to [0,1]; if F(x) is a function from [0,1] to [0,1]; F:[0,1]to [0,1], and satisfies F(F(x))=x is and is strictly monotonic increasing (or just monotone increasing) (possibly only increasing), then it appears that F(x)=x on that interval is the only solution; ven without any further regularity or continuity or cauchy's equation F(x+y)=F(x)+F(y), being explicitly presumed.

According to one one answer, then it appears that F(x)=x, under the presumption of only monotone increasing (not even strict increasing);see
Does this uniquely specify F(x)=x, on [0,1] to [0,1]. I presume that the reasoning therein is correct, but perhaps it should be checked. There is a another solution given by commentator that it works even under F being only monotone increasing (rather than necessarily strictly increasing).

In a ranked say probability system this (the monotonicity increasing (or strict) hard-codes this rank order constraint into a functional form)  this makes some kind of sense. AS it would be able to capture, transcendental values (it appears), in the same sense that it can capture any other value. An inequality would be violated (or strict monotonicity) would be violated; F(1/pi)neq1/pi for example; instantly, ie one singular, or finitary step.

This is because if, F(1/pi)neq1/pi then there is some m in [0,1[ s.t F(1/pi)=m m>1/pi, or m<[1/pi] equal in value to some m smaller or large, and F(m)=1/pi by involution, the rank (or monotone strict increasing would be violated at the first move if m<1/pi, F(m)m=F(1/pi) ie F(m)>F(1/pi) by involution equation

that an involution function (often a form of point reflection symmetry, could technically capture even transcendental values; F(x)~=x, the rank order is immediately violated, whether x, or F(x) is an algebraic irrational, transcendental  or algebraic number-


Grid with both squares and equilateral triangles Is it possible to have a grid that contains both squares and equilateral triangles?

By grid I mean any set of the form $M \mathbb Z^2$, with $M \in GL_2\mathbb R$.

I think this is impossible, because it would contain a subgrid which is invariant under rotations of $\pi / 2$ and a subgrid which is invariant under rotations of $\pi / 3$. And I know there's no grid which is invariant under both rotations. But I can't get a conclusion from this point.

We may view $M\Bbb Z^2$ as a subset of $\Bbb C$ and by applying a rotation and/or scaling, we may assume wlog. that $M(1,0)^T=(1,0)^T=1$.
Then $M\Bbb Z=\Bbb Z + \alpha\Bbb Z$ where $\alpha=M(0,1)^T$.

Assume the lattice contains a square with vertices (in counter-clockwise order) $a,b,c,d$. Then $d-a=i(b-a)$, i.e., $M\Bbb Z$ contains two (non-zero) elements differing by a factor $i$. In other words, there exist $u,v,x,y\in\Bbb Z$, not all zero, such that $x+y\alpha=iu+iv\alpha$. Then $(x-iu)(y+iv)=(y^2+v^2)\alpha$. As the left hand side cannot be zero, we conclude $\alpha\in\Bbb Q[i]$.

Assume the lattice contains a triangle with vertices (in counter-clockwise order) $a,b,c$. Then $c-a=\omega(b-a)$ where $\omega=\frac {1+i\sqrt 3}2$, i.e., $M\Bbb Z$ contains two (non-zero) elements differing by a factor $\omega$. In other words, there exist $u,v,x,y\in\Bbb Z$, not all zero, such that $x+y\alpha=\omega u+\omega v\alpha$. Then $(x-\omega u)(y+(1-\omega)v)=(y^2+yv+v^2)\alpha$. As the left hand side cannot be zero, we conclude $\alpha\in\Bbb Q[\omega]$.

The field $\Bbb Q[i]\cap\Bbb Q[\omega]$ is a proper subfield of $\Bbb Q[i]$, hence must equal $\Bbb Q$. We conclude $\alpha\in \Bbb Q$, which would allow neither squares nor triangles in $\Bbb Z+\alpha\Bbb Z$.

We can cover an infinite long/finitely wide band with all triangles and another with all squares and we can cover the plane with these bands.   Not sure there is any other way.


Using Euler derive the relationships between cosine and exponential function $$\cos\phi=\frac{1}{2}(e^{j\phi} + e^{-j\phi}).$$
Please i was told to do this assignment but cant prove if anyone can help me out

I would write out what is $\frac{1}{2}(e^{ix} + e^{-ix})$ using euler's formula, and see if anything cancels. hint: something cancels


Interval bounds for symmetric doubly-stochastic matrices (designed with Metropolis weights). I'm facing an unusual problem with doubly-stochastic matrices, in the context of some undirected graph. I assume that it is connected, but this is not so important for this problem.

Let me introduce the problem. I assume I have an undirected graph; each agent has degree $d_i$ and a neighbor set $\mathcal{N}_i$. Correspondingly, we define a symmetric doubly-stochastic matrix $W=[w_{ij}]$ by:

\begin{equation*}\label{eq:N}
w_{ij} :=
 \left\{
 \begin{array}{ll}
  \frac{1}{\max\{d_i, d_j\}+1} & \text{if} \: j \in \mathcal{N}_i, \\
  1-\sum_{j\in\mathcal{N}_i}w_{ij} & \text{if} \: j=i, \\
  0 & \text{otherwise}.
 \end{array}
 \right.
\end{equation*}
These are the so called Metropolis weights, and it is easy to check that $W$ is indeed symmetric doubly stochastic. It is generally known that the eigenvalues $\lambda$ are in the interval $(-1,1]$ for any doubly-stochastic matrix. But, here comes the thing, I want to show the following:

Prove that the eigenvalues $\lambda$ are in $[-\frac{1}{2},1]$, whether the graph is connected or not.

Now this is an unusual thing to show, I know that. Unfortunately, Gershgorin theorem won't work for me. You may wonder at this point if this may actually be true, well, I checked it with matlab for over $10\:000$ randomly computed graphs with $100$ nodes, either connected or not and with all kinds of amount of edges. There was no case where there was even some eigenvalue $\lambda<-0.4$ ...

A property that might be useful:
$$w_{ii}\geq \frac{1}{d_i}\geq \frac{1}{\max\{d_i, d_j\}+1}=w_{ij},$$
for any $j \in \mathcal{N}_i$. Then I hope to show that the matrix
$$Q:=W+cI$$
is nonsingular for any $c>0.5$. This would prove the statement. I attempted this by showing that the $LU$-decomposition $LU = Q$ gives a matrix $U$ with a positive diagonal. Something I also observed with the help of Matlab ofcourse.  This is exactly were I got stuck and I hope that somebody has some out of the box ideas that can bring me any further.

edit: $LU$-decomposition should be on $Q$ of course.

The Metropolis weights are analyzed in this paper http://web.stanford.edu/~boyd/papers/pdf/avg_metropolis.pdf. See Proof of Lemma 1 in Section 3.2


Simple Proof by Induction: Question About Formulation of the Inductive Step. Prove $n^2 > n+1, n\geq2$:

Basis: n = 2

$n^2 > n+1$

$2^2 > 2+1$

$4>3$

Inductive Hypothesis: Assume P(k) is true, i.e.,

$k^2 > k+1$

Inductive Step: Show P(k+1) is true, i.e.,

$(k+1)^2 > (k+1) + 1$

$k^2 + 2k + 1 > (k+1) + 1$

$k^2 + 2k + 1 > k + 2$

$k^2 + 2k > k+1$

$k(k+2) > k + 1$

I know that I need to try and get the expression to read $k^2 > k+1$. The steps above are the closest I could get. Can some explain if my format for inductive proofs is correct and point me in the right direction to completing the proof?

Format-wise, I think what you're doing is perfectly fine. I think some more wording would be appropriate though: for instance, the way you have your inductive step written, it almost seems like you're assuming the inequality holds at every step along the way. Usually, I would start at the left-hand side and do manipulations to verify the right-hand side does hold (or the reverse). Still, don't be afraid to use words: being able to explain what you're doing and why is very important in proof-writing!

So, on the premise $k^2 > k+1$, you wish you show $(k+1)^2 > k+2$. We see that

$$(k+1)^2 = k^2 + 2k + 1$$

Invoking the induction hypothesis ($k^2 > k+1$), we see that

$$k^2 + 2k + 1 > k+1 + 2k + 1 = 3k + 2 > k+2$$

(for which the inequality holds since $k$ is assumed positive). Thus, $(k+1)^2 > k+2$ as desired.


Proof that the number $\sqrt[3]{2}$ is irrational using Fermat's Last Theorem Suppose that $\sqrt[3]{2}=\frac{p}{q}$. Then $2q^3 = p^3$ i.e $q^3 + q^3 = p^3$, which is contradiction with Fermat's Last Theorem.

My question is whether this argument is a correct mathematical proof, since Fermat's Last Theorem is proven, or does it loop on itself somewhere along the proof of the Theorem?

In other words, does the proof of Fermat's Theorem somehow rely on the fact that $\sqrt[3]{2}$ is irrational?

UPD:
As pointed out in comments, this actually is a valid argument, no matter what was used in the proof of the Fermat's Last Theorem (which from now on will be referred to as the Proof). What really interests me, is whether the Proof uses on some step the fact that $\sqrt[3]{2}$ is irrational?

In this comment BCnrd argues that this proof is "essentially circular", because converting an FLT counterexample to a Frey curve with certain congruence conditions as in the Wiles proof requires an argument equivalent to establishing irrationality of $\sqrt[3]{2}$.

does the proof of Fermat's Theorem somehow rely on the fact that $\sqrt[3]{2}$ is irrational?

Edit Added: Even if $\sqrt[3]{2}$ is irrational was contained in FLT, it would have had to be proven by some means, so as long as FLT did not assume FLT then it doesn't matter that a specific instance of FLT was contained in proof of FLT

FLT being true implies $\sqrt[3]{2}$ is irrational , ($\sqrt[3]{2}$ is irrational is a specific case of FLT).

$\sqrt[3]{2}$ is irrational is not enough to imply FLT.

as $\sqrt[3]{2}$ is irrational is a specific instance of FLT, if it was not true then it would be in contradiction of a more general theorem containing it as a specific instance of the theorem.

in other words FLT implies all specific instances of itself, where as infinitely many instances of FLT do not imply FLT.


Combinatorics: Getting all of the chocolates in a square at minimum cost Consider a $101$ by $101$ square array. Each unit square is surrounded by a fence. However, if two squares share a common edge, there is a doorway between the fence along their common edge. It costs $\$1$ each time a person wants to pass through a doorway.

In each of $5101$ of the unit squares, one chocolate is placed. Jake wants to collect all the chocolates. He is permitted to select any square in the array to begin his journey and may only change squares by passing through a doorway. What is the minimum amount of cash Jake should carry with him in order to be guaranteed, that if he is smart enough, he will collect all of the chocolates.

I've tried to do a worst possible case type argument, but it's hard to work out a worst possible case given the added complexity of Jake being able to choose his starting location.

Is the fact that $101$ prime, relevant? Could we try small primes with $\frac{p+1}{2}$ chocolates and do some sort of induction?

Color the squares like a checkerboard.  If the corners are white, there are $5101$ white squares.  The worst case is that the chocolates are in all the white squares.  Jake will have to go through all the squares to get all the chocolate and will need $\$10200$ to do that.

This applies if every pass through a door costs.  If you can leave a door open and go back through it free, you can do better.  I think you can miss almost half the black squares then.

The greedy chocolate eater will have to go from one square to the next, so they can't do better than 5100 doors. Just starting on one corner and going down along the border, then the next line up, and so clearly doesn't do "too many" doors.


Incomparable languages $B \nleq_T A$ and $A \nleq_T B$ I came across a problem to prove that there are sets $A$ and $B$ termed 'Turing-Incomparable' languages $B \nleq_T A$ and $A \nleq_T B$. The only languages I could think of are if $A$ and $B$ are disjoint regular expressions such as $B=1^*$ and $A=0^*$. This way, one could not act as an 'oracle' for the other, but is there something missing ?

"Disjoint" is by no means sufficient. The first problem is that the two languages you suggest are both computable, so you can automatically compute one from the other (there's no obligation to actually make use of an oracle you're given).

Constructing two Turing-incomparable sets is a very complex problem. In particular, there are no known "natural" examples - that is, basically, the only known examples of Turing-incomparable sets are sets cooked up specifically for proving that these pairs exist.

The standard construction of a pair of Turing-incomparable sets is the Friedberg-Muchnik finite-injury argument (http://www1.maths.leeds.ac.uk/~pmt6sbc/3163/FMslides.pdf gives an OK description of this argument). It's not a difficult technique by the standards of research math, but it's not easy to explain in a few sentences either.

Your example of symbol-disjoint languages doesn't seem relevant as we can still have a Turing computable biijection between the two.

In this context, the point of Turing reducibility, comparability, jumps and the likes is to acknowledge the effects of Turing degrees on set recognition by different oracle machines.

Your problem is equivalent to proving the non-existence of a total order over the Turing degrees.
A general example is that of 2 non-computably-enumerable sets A and B such that there's no computably-enumerable function mapping one onto the other.


Is there a pair of prime $(p,q)$ such that $p >q>7$ and $2q-p=3$ and $p, q$ are successive? I'm failed to find at a least one pair of primes$(p,q)$  $p >q>7$ and $2q-p=3$ with $p, q$ are successive , I think tha's impossible because we do not know more about Gaps between prime and if there is those pair it should be large , then my question how i can find that pair or to disproof it's existence ?

Edit:  I have added the missed condition wich is $p, q$ are successives

The key requirement is that the primes $p,q$ be sequential (that is, that there should be no prime between them).

Your assumption implies that $$q=\frac {p+3}2$$

However, a strong form of Bertrand's postulate  tells us that, for $n>25$, there must be a prime between $n$ and $\frac 65 n$.  Applying this to $n=\frac {p+3}2$ shows that, except for very small $p$, there must be a prime between $\frac {p+3}2$ and $p$ and from there a simple finite search shows that your desired primes do not exist.

Since $p, q$ are successive then you can rewrite your equation as $$2p_n-p_{n+1}=3$$
This shows that $p_{n+1}$ is almost double $p_n$ which seems to be forbidden for sufficiently large primes. This is because between $x$ and $2x$ exist many primes.
Write your equation in the form $p_{n+1}=2p_n-3$. This means that the number of primes in the interval $(x, 2x-3]$ is one (for $x=p_n$).
It is proved here that for $x\ge25$ there is a prime between $x$ and $\frac{6}{5}x$.
So, we have one prime between $p_n$ and $\frac{6}{5}p_n$ and one more between $\frac{6}{5}p_n$ and $\frac{6}{5}\cdot \frac{6}{5}p_n=\frac{36}{25}p_n$ if $p_n\ge 25$.
Since you have only one prime between $p_n$ and $2p_n-3$ we must have $2p_n-3<\frac{36}{25}p_n$vwhich gives $14p_n<75$ or $p_n<6$. Hence $p=2, 3$ or $5$.
 The prime $5$ is the only good one to satisfy your equation.