Find that $\lim_{a\to \infty}a^3 f(a)=?$ I would appreciate if somebody could help me with the following problem:

  Let $f(a)$ area of S, $A(a,a^2)$, $B(b,b^2)$ and $\overline{AB}=1$, given that:

Find that $$\lim_{a\to \infty}a^3 f(a)=?$$

I tried  but couldn't get it that way.

It can be shown that the area above an upward-facing parabola, defined by $y=q(x)$, that is cut off by the secant line between $x=a$ and $x=b$ is equal to
$$
(b-a) \bigg( \frac13 q(a) - \frac23 q\bigg(\frac{a+b}2\bigg) + \frac13 q(b) \bigg).
$$
(Indeed, this calculation is what goes into Simpson's Rule!)

In this problem, $b\approx a+\frac1{2a}$ when $a$ is large (this makes the "rise" of the segment tend to $1$ and the "run" tend to $0$). Therefore we want to calculate
$$
\lim_{a\to\infty} a^3 \cdot \frac1{2a}\bigg( \frac13 a^2 - \frac23 \bigg( a+\frac1{4a} \bigg) ^2 + \frac13 \bigg( a+\frac1{2a} \bigg) ^2 \bigg) = \frac1{48}.
$$
(To make this calculation rigorous, one would use a statement like $b=a+\frac1{2a}+O(a^{-2})$.)

$$\sqrt{(b-a)^2+(b^2-a^2)^2}=1\implies \sqrt{(b-a)^2(1+(b+a)^2)}=1\\
\implies (b-a)^2(1+(b+a)^2)=1$$
$$\lim_{a\to\infty}a^3f(a)=\lim_{a\to\infty}a^3\int_a^{b}\left(\frac{b^2-a^2}{b-a}(x-a)-x^2\right){\rm d}x$$
These two relations can get you started.


What does $e^{a*ln(x)}$ equal in terms of $a$ and $x$, and how is this found? I saw somewhere that it would be $x^a$, but I'm not sure why.

Only two logarithm rules are needed:
$$\ln(a^b) = b\cdot \ln(a)$$
$$e^{\ln(x)} = x$$
And thus
$$e^{a\cdot \ln(x)} = e^{\ln(x^a)} = x^a$$


In which order do I graph transformations of functions? In which order do I graph transformations of functions?

The 6 function transformations are:

Vertical Shifts
Horizontal Shifts
Reflection about the x-axis
Reflection about the y-axis
Vertical shifting or stretching
Horizontal shifting or stretching

Tell me if I'm wrong, but I believe that in any function, you have to do the stretching or the shrinking before the shifting. But where do the reflections fall in this process?

For $Af (Bx+C)+D$ perform the operations in order: C, B , $A $, $D $. For the reflection, say $-A $, it does not matter if you stretch or shrink by $A $ and then reflect. Try an example with a simple function like $-3x^2$.

Scale (Stretch or shrink) Vertical or horizontal order do not matter.
Reflect Vertical or horizontal order do not matter.
Shift Vertical or horizontal order do not matter.


Question about limit! How can I find this limit,

$$\lim_{x\rightarrow1}\frac{x^{1/3}-1}{x^{1/2}-1}.$$

I know how to find it using the L'Hospital theorem. I would like to find it without it.

Divide by x-1 in numerator and denominator then give limit saperately
lim x→1 (x^(1/3)) - 1)(x-1)/lim x→1 ((x^(1/2) - 1)/(x-1) = (1/3)/(1/2) = 2/3


Using Taylor's inequality for ln(1+x) I am using the Taylor series for $f(x) = ln(1+x)$, $a = 0$, to try and estimate $ln(0.5)$. I am trying to find how big $n$ must be in order to estimate $ln(0.5)$ to within $10^{-5}$. The problem I am running into is that when I compute $M$ to use in Taylor's Inequality, I get $M$ = $n!\cdot2^{n+1}$, which when I plug this into Taylor's inequality for $M$, I get $R_n(-1/2) = \frac{n!\cdot2^{n+1}}{(n+1)!\cdot2^{n+1}}$. which simplifies to $\frac{1}{n+1}$. But I can't imagine this is correct since this would mean $n$ would have to be $9999$ to estimate $ln(0.5)$ within $10^{-5}$. Any help you can give me is greatly appreciated.

The integral form of the remainder is:
$$R_n(x) = \int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n\;dt$$
this is exact, not an approximation, not a bound. If we now bound $|f^{(n+1)}(t)|$ on $[a,x]$ by some $M$ then

\begin{align}|R_n(x)|
&= \left|\int_a^x \frac{f^{(n+1)}(t)}{n!}(x-t)^n\;dt\right|
= \frac{1}{n!}\left|\int_a^x f^{(n+1)}(t)(x-t)^n\;dt\right|\\
&\overset{(*)}{\leq} \frac{1}{n!}\left|\int_a^x M|x-t|^n\;dt\right| \\
&= \frac{M}{n!}\frac{|x-a|^{n+1}}{n+1}
= \frac{M}{(n+1)!}|x-a|^{n+1}
\end{align}
This is of the form of the Taylor inequality. Taking the smallest such $M$ we get the typical inequality, $|R_n(x)| \leq U_n(x) \equiv \frac{\max_{t\in[x,a]}|f^{(n+1)}(t)|}{(n+1)!} |x-a|^{n+1}$.

My reason for starring $(*)$ is to emphasize the bluntness of this approximation. If $M|x-t|^n$ is a sufficiently loose upperbound for $|f^{(n+1)}(t)(x-t)^n|$ then the bound on the remainder could be way off.

In the specific case of $f(x) = \ln(1+x)$ we have $f^{(n+1)}(x) = (-1)^n \frac{n!}{(1+x)^{n+1}}$. As you correctly noted $|f^{(n+1)}(x)| \leq 2^{n+1} n!$ on $[-\tfrac12,0]$. The true remainder is:

\begin{align}
R_n(-\tfrac12) &= \frac{1}{n!}\int_0^{-\tfrac12} f^{(n+1)}(t)(-\tfrac{1}{2} - t)^n\;dt
= \frac{(-1)^n}{n!} \int_0^{-\tfrac12} (-1)^n\frac{n!}{(1+t)^{n+1}} (\tfrac12 + t)^n \;dt\\
&= \int_0^{-\tfrac12} \frac{1}{1+t} \left(\frac{\tfrac12 + t}{1+t}\right)^n  \;dt = \frac{1}{2^n}\int_0^{-\tfrac12} \frac{1}{1+t} \underbrace{\left(2 - \frac{1}{1+t}\right)^n}_{(**)}  \;dt
\end{align}
Using wolfram alpha to compute this integral for specific $n$ we can find $|R_{11}(-\tfrac12)| \approx 3.79\times 10^{-5}$ and $|R_{12}(-\tfrac12)| \approx 1.76\times 10^{-5}$ and $|R_{13}(-\tfrac12)| \approx 8.2\times 10^{-6}$. So the minimum $n$ is actually $13$. Indeed these match up with a direct calculation.

On the other hand $U_n(x) = \frac{2^{n+1}}{n+1} |x|^{n+1}$. This gives, as you found, $U_n(-\tfrac12) = \frac{1}{n+1}$ which is indeed a pretty dreadful upperbound on $|R_n(-\tfrac12)|$. In fact even to just get $U_n(-\tfrac12) < 10^{-1}$ we need $n>9$ terms, but at that point the true remainder is already $|R_9(-\tfrac12)|<10^{-3}$.

Note that the poorness of the approximation is due to $(**)$ in the integrand. For large $n$ it is only near $1$ very close to $0^-$ and is basically $0$ everywhere else on $[-\tfrac12,0]$. In other words the only significant contribution of $\frac{1}{1+t}$ to the integral is restricted to a small neighborhood near $0^-$. It should therefore be obvious that using a bound of the $n+1$st order derivative, far away from contributing portions of the integrand, could lead to a bad bound $U_n(x)$.

A bit beyond
With a little bit more work we can get a much better asymptotic bound on the true remainder. Starting from the $(**)$ step and substituting $y = \frac{1}{1+t} - 1$,

$$R_n(-\tfrac12)
= \frac{1}{2^n} \int_0^1 \underbrace{(y+1)}_{\frac{1}{1+t}} \cdot\underbrace{(1-y)^n}_{ \left(2 - \frac{1}{1+t}\right)^n} \cdot \underbrace{-\frac{dy}{(1+y)^2}}_{dt}
= -\frac{1}{2^n}\int_0^1 \frac{(1-y)^n}{1+y}\;dy
$$

Note that the integrand is positive in $(0,1)$, so the integral is positive, hence $R_n(-\tfrac12) < 0$. This makes sense as the terms of the Taylor approximations $-\sum_{n=1}^N \frac{1}{n\cdot 2^n}$ is decreasing in $N$, so it converges to $\ln(0.5)$ from above.

Using integration by parts with $u = \frac{1}{1+y}$ and $dv = (1-y)^n\;dy$ we get

\begin{align}
|R_n(-\tfrac12)|
&= \frac{1}{2^n}\left( \bigg[\underbrace{\frac{1}{1+y}}_{u}\cdot \underbrace{- \frac{(1-y)^{n+1}}{n+1}}_{v}\bigg]_0^1 - \int_0^1 \underbrace{\frac{(1-y)^{n+1}}{n+1}}_{v} \cdot \underbrace{-\frac{dy}{(1+y)^2}}_{du} \right)\\
&= \frac{1}{2^n} \frac{1}{n+1} - \frac{1}{2^n}\frac{1}{n+1}\int_0^1 \frac{(1-y)^{n+1}}{(1+y)^2} \;dy \\
&\leq \frac{1}{2^n(n+1)} - \frac{1}{2^n(n+1)}\int_0^1 \frac{(1-y)^{n+1}}{(1+(1))^2} \;dy \\
&= \frac{1}{2^n(n+1)} - \frac{1}{2^n(n+1)}\frac{1}{4} \frac{1}{n+2} \\
&= \frac{1}{2^n(n+1)} - \frac{1}{2^{n+2}(n+1)(n+2)} \\
\end{align}

To see just how much better this is than Taylor's approximation for $\ln(0.5)$, consider $n=100$. Taylor's gives the bound $U_{100}(-\tfrac12) = \frac{1}{100+1} \approx 10^{-2}$, whereas using the first term above gives us

$$
|R_{100}(-\tfrac12)|
\leq \frac{1}{2^{100}\cdot (100 + 1)}
= \frac {1}{\left(2^{10}\right)^{10} \cdot 101}
\leq \frac{1}{\left(10^3\right)^{10} \cdot 10^2}
= 10^{-32}
$$

To summarize, $$R_n(-\tfrac12) = -\frac{1}{2^n(n+1)} + \Theta\left(\frac{1}{2^n \cdot n^2}\right)$$

$\ln (1-x) = -\sum_\limits{n=1}^{\infty} \frac {x^n}{n}$

This sequence has a radius of convergence of $|x| < 1$

If you want to find the Taylor series of $\ln (1+x)$ just substitute $-x$ for $x$ above.

To find $\ln 0.5$ we are setting $x = 0.5$

This means that in our Taylor series error.

It should be on the order of $\frac {0.5^{n+1}}{n+1}$

And to get precision of $10^{-5}$ I think we can ignore the $\frac {1}{n+1}$ factor for now.

$\log 0.5^{(n+1)} = \log 10^{-5}\\
(n+1) = \frac {-5}{-\log 2}\\
\frac {5}{0.301} \sim 15\\
n\sim 14$

the factor of $\frac {1}{15}$ means we can do one less iteration.

$n = 13$

Still a fair amount of calculating but not impossible.


How to Prove that this Function is not $1-1$ I'm trying to show that the function

$$f(x)= \dfrac{x}{4}+x^2\sin\left(\dfrac{1}{x}\right)$$
is not $1-1$ for any neighborhood of $0$. I know that what I have to do is find two different points that yield the same value, but these are not so obvious.

Thanks for any help.

The derivative of $f(x)$ is

$$f'(x)=\frac 14+2x\sin\left(\frac 1x\right)-\cos\left(\frac 1x\right)$$

In any neighborhood of zero, you can find a very small positive $a$ where $\cos\left(\frac 1a\right)=1$. Then the second term of my expression for $f'(a)$ is near zero, thus $f'(a)$ is near $-\frac 34$, thus $f'(a)$ is negative, thus $f(x)$ is decreasing in a neighborhood of $a$ that is inside our neighborhood of zero.

In that same neighborhood of zero, you can find a very small positive $b$ where $\cos\left(\frac 1b\right)=-1$. Then the second term of my expression for $f'(b)$ is near zero, thus $f'(b)$ is near $\frac 54$, thus $f'(b)$ is positive, thus $f(x)$ is decreasing in a neighborhood of $b$ that is inside our neighborhood of zero.

Both $a$ and $b$ are positive, so $f(x)$ is continuous on the interval between $a$ and $b$. You can easily show that any function continuous on an interval that increases in a subinterval and also decreases in a subinterval is not injective on that interval. (Use the Intermediate Value Theorem on the four cases $a<b$ or $b<a$, $f(a)<f(b)$ or $f(b)<f(a)$). This is a useful theorem to use in many other problems, as well.

Thus your function $f(x)$ is not injective in any neighborhood of zero, in the positive part of that interval alone.

When $-\frac 1n<x<\frac 1n$ the argument of $\sin (\frac 1x)$ takes all the values of the interval $]n,\infty[$ so that $\sin (\frac 1x)$ takes infinitely many times all the values of the rank $[-1,1]$ of the function $\sin x$ (If you want to draw the graph of $\sin (\frac 1x)$, you have a shaded rectangle in the neighborhood of $0$).

This is the reason for which the function $g(x)=x^2\sin (\frac 1x)$ can't be 1−1 for any neighborhood of $0$ and so is for $f(x)=\frac x4 +x^2\sin(\frac 1x)$.


How many new bridges do we need to build in Königsberg to achive an Euler circuit? We know that an euler circuit in a graph is that the starting point and the ending point is the same and that we need to visit all bridge once how do i solve this problem?

*Note that I am talking about the Königsberg of Eulers time and not the modern Kaliningrad.

Here's a solution from wikipedia:
https://en.wikipedia.org/wiki/Seven_Bridges_of_K%C3%B6nigsberg#Solutions

I guess this should be the answer to the problem.

Notice all the degrees are even.

So This is an Eulerian path.


Condition for 3 distinct normals from a point $(h,k)$ to the parabola $y^2=4ax$ What are the conditions in terms of $h,k$ to be able to draw three
  distinct normals to the parabola $y^2= 4ax$ ?

Normal to the parabola from $(h,k)$ is given by: $am^3 +(2a-h)m+k=0$. This equation can yield three distinct slopes $m_1,m_2,m_3$ if $\Delta>0$ (source).

The $\Delta$(discriminant) of the equation is given by $-4a (2a-h)^3-27a^2k^2$.

For it to be greater than $0$ when $a>0$, I derived $h>2a$ and $k<-2\sqrt{2}a$. How do I derive the other constraints on $h,k$ i.e. $k>4\sqrt{2}{a}$ and $h>8a$
?

You have almost given the general condition yourself.

The Δ(discriminant) of the equation is given by $-4a (2a-h)^3-27a^2k^2$.

If discriminant of the cubic equation is positive, then the cubic equation has 3 real roots.
The discriminant of the cubic normal equation must be positive to have 3 real and distinct slopes, that means, three distinct normals from that particular point (h, k).
So the required condition for three normals to be drawn from $(h,k)$ to the
parabola $y^2 =4ax$ is just
$$ -4a (2a-h)^3-27a^2k^2 > 0 $$
On rearranging, we get a compact representation :
$$ \biggl(\frac{h-2a}{3}\biggr)^3> a\biggl(\frac{k}{2}\biggr)^2 $$
You need to know one of the coordinates $h$ or $k$, or conditions on one of the coordinate to obtain constraint for the other coordinate.
I have referred this site: https://brilliant.org/wiki/cubic-discriminant/

Apply the condition for normal $y=mx+am^3-2am$ for the given parabola.
Let $m_1$, $m_2$ and $m_3$ be the slopes of the required normals to the parabola.
For three distinct normals the slopes should satisfy the following condition: $m_1+m_2+m_3=0$.


Prove that $\lnot (p \implies q)$ is equivalent to $p \land \lnot q$? By equivalent I mean the biconditional, as in
$$\lnot (p \implies  q) \iff p \land \lnot q$$

Given the definition of implication, I understand why this is true, but I need a bit of help showing this with a formal proof using rules like $\lor-\text{Elim}$  and $\bot-\text{Intro}$.

Here is the pure natural deduction proof (Fitch-Style).

If $p \land \neg q$:

  $p$;

  $\neg q$.

  If $p \to q$:

    ...

    $\bot$.

  $\neg( p \to q )$.

If $\neg( p \to q )$:

  If $\neg p$:

    If $p$:

      $\bot$.

      $q$.

    $p \to q$.

    $\bot$.

  $\neg \neg p$.

  $p$.

  If $q$:

    ...

    $p \to q$.

    $\bot$.

  $\neg q$.

  $p \land \neg q$.

$p \land \neg q \leftrightarrow \neg( p \to q )$.

I'll leave you to fill in the blanks, which should be easy. See this post for a brief description of the general method to discovering such a proof. Sometimes it can be convenient to use LEM (law of excluded middle) to split cases in a clever manner to shorten the proof, but that way requires some experience.

As for the interesting side question of whether or not one can prove the equivalence in intuitionistic logic, which has the same inference rules except without DNE (double negation elimination), and hence LEM does not hold, the answer is that indeed it cannot be done. Note that the above proof only uses DNE once. Feel free to ignore the rest of this post if you are not interested in intuitionistic logic.

In intuitionistic logic, "$\neg p$" is simply a short-form for "$p \to \bot$". To prove that some sentence is not provable in intuitionistic logic, it suffices to construct a Kripke frame that does not satisfy it (see this post). Consider the Kripke frame $0 \to 1$ where $p$ is known only at $1$ and $q$ is not known at both $0,1$. Then $\neg( p \to q )$, which denotes $( p \to q ) \to \bot$, is known everywhere, but $p \land \neg q$ is not known at $0$.

Alternatively, note that if it were provable in intuitionistic logic, it would hold when $q = \bot$, but that gives $\neg( p \to \bot ) \to p \land ( \neg \bot )$, which is equivalent (by definition) to $\neg \neg p \to p$, which is not valid in intuitionistic logic.

The OP is looking for the following:

  Given the definition of implication, I understand why this is true, but I need a bit of help showing this with a formal proof using rules like ∨−Elim and ⊥−Intro.

The following proof uses disjunction elimination ("∨−Elim" or "vE") on line 20 and negation introduction ("⊥−Intro" or "⊥I") on lines 6 and 21.

For a description of the justification used see the text forallx and the associated proof checker referenced below.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/


Proof without using induction How to prove that
$$1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$
without using induction.

If we don't know the right side of this expression, how to get right expression. I tried with partial sums and binomial formula but can't get it.

So the problem is:
$$1^2+2^2+...+n^2=?$$

Thanks for replies.

With linear algebra:

You can do it with linear algebra, by noticing that the map $\Sigma$ which maps a sequence $u_n$ to the sequence $u_0 + \dots + u_n$ has, as a left inverse, the map $\Delta$, mapping the sequence $u_n$ to the sequence $u_n - u_{n-1}$; since, for any sequence $u_n = f(n)$ where $f$ is a polynomial of degree $d$, $\Delta u$ is polynomial of degree $d-1$, we see that $\Sigma$ maps polynomials of degree $d$ to polynomials of degree $d+1$, and we can find which ones by linear algebra.

With combinatorics:

An even simpler way is to use the standard basis $e_d$ where $e_d$ is the sequence $e_{d,n} = \binom{n+d}{d}$, for then we have
$$(\Delta e_d)_n = e_{d,n} - e_{d,n-1} = \binom{n+d}{d} - \binom{n+d-1}{d} = \binom{n+d-1}{d-1} = (e_{d-1})_{n}.$$
Then we only need to write $n^2$ as a linear combination of $e_{0,n} = 1$, $e_{1,n} = n$ and $e_{2,n} = n(n-1)/2$.

With integrals:

A third way is to use integrals: namely, with simple linear algebra we find that
$$ n^2 = \int_{n-1}^{n} (x^2+x+1/6) dx,$$
so that
$$ 0+1+\dots+n^2 = \int_0^{n} (x^2+x+1/6) dx = \frac{1}{6} (2n^3+3n^2+n).$$

With geometry:

A fourth way is the usual geometric proof, where you can tile together six pyramids in a parallelepiped.

Any statement on natural numbers can be proved only using induction. It is possible that the induction is hidden in the argument. So this question doesn't make sense.


Calculate the derivative of $g(x)=x^2.cos(1/x^2)$. Does the derivative of this function exist at $x=0$? Calculate the derivative of $g(x)=x^2.cos(1/x^2)$.  Does the derivative of this function exist at $x=0$?

I calculated the derivative to be $2.cos(1/x^2)x+2(sin(1/x^2))/x$.

I'm tempted to say that the derivative does not exist at zero because it is a vertical line, but I'm not sure this is the case.

Note that your function $g(x)$ is not defined at $x=0$, therefore neither is its derivative. A necessary condition for the derivative to exist at a point is to the function be defined and to be continuos at that point, which is not the case.

$g'$ does not have limit in $0$, but it does not mean $g$ not differentiable in $0$ :
$$\frac{g(x)-g(0)}{x}=x\cos\frac{1}{x^2}\xrightarrow[x\to0]{}0$$
so $g'(0)=0$.

All you can say is that $g'$ is not continuous in $0$.


How to prove a map is a local diffeomorphism Let $F:U \subset R^2 \rightarrow R^3$ by given by $F(u,v) = (u\sin{(\alpha)}\cos{(v)},u\sin{(\alpha)}\sin{(v)},u\cos{(\alpha}))$,
$(u,v) \in U = \{(u,v) \in R^2 \space; u > 0\}$ with $\alpha$  a constant.
Show F is a local diffeomorphism of U onto a cone C with the vertex at the origin and $2\alpha$ as the angle of the vertex.

From Do Carmo, problem 4.2.1

I was thinking of using the Inverse function theorem. $F$ is differentiable. We can compute the jacobian I suppose and show it's non-zero. I'm not sure how to construct this so it's a map onto a cone with the above properties.

Recall that the Inverse function theorem says if we have a function from $R^n \rightarrow R^n$ that is continuously differentiable on some open set containing a point $p$, and that the determinate of $Jf(a) \ne 0$ then there is some open set $W$ containing $f(a)$ such that $f:V \rightarrow W$ has a continuous inverse $f^{-1}:W \rightarrow V$ which is differentiable for all $y \in W$.    Basically it says $f$ is a diffeomorphism from $V$ to $W$ given it satisfies the above hypothesis.

Now above it wants us to prove it's a local diffeomorphism, not globally a diffeomorphism. That would just mean given any point $p$ in the domain of F, there exists an open set $U \subset \mathrm{dom}(F)$ containing $p$ such that $f:U \rightarrow f(U)$ is a diffeomorphism.

1) The Jacobian $Jac(F)(u,v)$ has rank $2$ at every point $(u,v)\in U$, so that $F$ is an immersion, but certainly not an embedding: see  point 3) below.
The image $F(U)$ is exactly the subset $z\gt0, x^2+y^2=z^2\tan ^2(\alpha)$ of $\mathbb R^3$.

2) This subset is indeed the intersection $C\subset \mathbb R^3$ of the upper half-space $z\gt 0$ with the cone of half-angle $\alpha$ with vertex at the origin and axis along the $z$-axis.
That (blunted !) half-cone $C$ is   a locally closed submanifold of $\mathbb R^3$.

3) The coinduced morphism of differential manifolds $F_0:U\to F(U)=C$ is the universal covering map of $C$, and its  restriction to any  vertical line $\{c\}\times \mathbb R\subset U \;(c\gt 0)$ is the universal  covering map of the  horizontal circle $C\cap\{z=c\cos \alpha\} $ of $C$ .

An opinion
All in all, this exercise  could have been written in a slightly more explicit way ...

You have to do two things:

Show that the image of $F$ is the required cone
Show that $F$ is a local diffeomorphism

As for point 1, you can see that $F(u,v)=u (\sin \alpha \cos v , \sin \alpha \sin v, \cos \alpha)$, so it is clearly a cone (if you fix the variable $v$, and let $u \in \mathbb{R}$ you get a straight line through $0$, while if you fix the variable $u$ you get circles). To see that the angle of the vertex is $2 \alpha$, you can use the fact that
$$F(u,v) \cdot (0,0,1) = ||F(u,v)|| \ ||(0,0,1)|| \cos( \mbox{angle between $F(u,v)$ and $(0,0,1)$ } ) $$
where $\cdot$ denotes the scalar product. LHS gives you $u \cos \alpha$, while RHS gives you $u \cos( \mbox{angle between $F(u,v)$ and $(0,0,1)$ } )$, so (supposing that $\alpha \in \left( 0,\pi \right)$)
$$\mbox{angle between $F(u,v)$ and $(0,0,1)$ } = \alpha$$

As for point 2, compute the Jacobian
$$J(u,v) = \left( \begin{matrix}
\sin \alpha \cos v & \sin \alpha \sin v & \cos \alpha \\
-u \sin \alpha \sin v & u\sin \alpha \cos v & 0
\end{matrix} \right)
$$
Since $u >0$ (this hypothesis is important otherwise you would have the second row vanishing), you can see that this Jacobian has always rank $2$, because
$$
\left| \begin{matrix}
\sin \alpha \cos v & \sin \alpha \sin v  \\
-u \sin \alpha \sin v & u\sin \alpha \cos v
\end{matrix} \right| = u \sin \alpha \neq 0
$$
($J(u,v)$ has an invertible $2 \times 2$ submatrix, so its rank is $\geq 2$). Since $2$ is the maximal rank $J(u,v)$ can achieve, "some version of the inverse function theorem" tells you that $F$ is a local diffeomorphism between $\mathbb{R}^2$ and the cone.

Clearly, all of this makes sense if $\sin \alpha \neq 0$, otherwise your map would be $F(u,v)=(0,0,u \cos \alpha)$, which gives you a line.


What is the difference between independent and mutually exclusive events? Two events are mutually exclusive if they can't both happen.

Independent events are events where knowledge of the probability of one doesn't change the probability of the other.

Are these definitions correct? If possible, please give more than one example and counterexample.

Yes, that's fine.

Events are mutually exclusive if the occurrence of one event excludes the occurrence of the other(s).  Mutually exclusive events cannot happen at the same time. For example: when tossing a coin, the result can either be heads or tails but cannot be both.

$$\left.\begin{align}P(A\cap B) &= 0 \\ P(A\cup B) &= P(A)+P(B)\\ P(A\mid B)&=0 \\ P(A\mid \neg B) &= \frac{P(A)}{1-P(B)}\end{align}\right\}\text{ mutually exclusive }A,B$$

Events are independent if the occurrence of one event does not influence (and is not influenced by) the occurrence of the other(s).  For example: when tossing two coins, the result of one flip does not affect the result of the other.

$$\left.\begin{align}P(A\cap B) &= P(A)P(B) \\ P(A\cup B) &= P(A)+P(B)-P(A)P(B)\\ P(A\mid B)&=P(A) \\ P(A\mid \neg B) &= P(A)\end{align}\right\}\text{ independent }A,B$$

This of course means mutually exclusive events are not independent, and independent events cannot be mutually exclusive.  (Events of measure zero excepted.)

Events are Independent when happening of one does not influence happening of other. Eruption of volcano on Earth and orbit of Mars do not influence each other, so are independent events.

Growth of human population and preservation of many other species are mutually exclusive, as the one can only happen if the other does not happen.

Strictly speaking, mutually exclusive does not imply that one of them must happen. If there is a large asteroid impact on Earth, then neither human population grows nor endangered species are preserved.


Expected value uniform decreasing function We are given a function $f(n,k)$ as

for(i=0;i < k;i++)
  n = rand(n);
return n;

rand is defined as a random number generator that uniformly generates values in the range $[0,n)$. It returns a value strictly less than $n$; also $\operatorname{rand}(0)=0$.

What is the expected value of our function $f(n,k)$ given $n$ and $k$?

I will make a guess that rand(n) returns uniform reals on the interval [0,n).
Since rand(n) gives the same distribution as n*rand(1). Thus for given $n$ and $k$ the random variable produced by the algorithm is
$$
   X = n \prod_{m=1}^k U_m
$$
where $U_m$ are independent identically distributed continuous uniform random variables on the unit interval. Thus
$$
   \mathbb{E}\left(X\right) = n \prod_{m=1}^k \underbrace{\mathbb{E}(U_m)}_{=\frac{1}{2}} = n 2^{-k}
$$

Responding to OP's request to consider the case when rand(n) returns unifrom random integers $[0,n]$. Let $Y_m \sim \mathcal{DU}\left([0,Y_{m-1}\right)$ be such a uniform random integer drawn on $m$-th iteration. Assume $Y_0 = n$. Then we seek to find
$$
    \mathbb{E}\left(Y_k\right) = \mathbb{E}\left( \mathbb{E}\left(Y_k \mid Y_{k-1}\right)\right) = \mathbb{E}\left(\frac{1}{2}Y_{k-1}\right) = \ldots =\frac{1}{2^k} Y_0 = \frac{n}{2^k}
$$
Thus the expectation is the same as in the case of continuous uniforms.

Now, assuming rand(n) generates uniform integers on $[0,n)$, instead of $[0,n]$. In that case $Y_k|Y_{k-1} \sim \mathcal{DU}\left( \left[0, \max(Y_{k-1}-1,0)\right]\right)$. Obtaining closed form is not feasible for arbitrary $k$, but with the help of Mathematica I was able to get expectations for low values of $k$:
$$ \mathbb{E}\left(Y_1\right) = \begin{cases}
 \frac{n-1}{2} & n > 1 \\[5pt]
 0 & \text{otherwise}
\end{cases}
$$
$$
\mathbb{E}\left(Y_2\right) = \begin{cases}
 \frac{(n-1)(n-2)}{4 n} & n>2 \\[5pt]
 0 & \text{otherwise}
\end{cases}
$$
$$
\mathbb{E}\left(Y_3\right) = \begin{cases}
 \frac{1}{8 n} \left(4 H_{n-1}+(n-1)(n-6)\right) & n>3 \\[5pt]
 0 & \text{otherwise}
\end{cases}
$$
$$
   \mathbb{E}\left(Y_4\right) = \begin{cases}
 \frac{1}{16 n} \left( 4 \left(H_{n-1}\right){}^2+12 H_{n-1}-4 H_{n-1}^{(2)}+(n-1)(n-14) \right) & n>4 \\[5pt]
 0 & \text{otherwise}
\end{cases}
$$

This is the reproducing code:

Block[{z, yc, ypr},
  Rest@NestList[(Piecewise[{{Expectation[(#1 /. z -> yc),
            Distributed[yc,
             DiscreteUniformDistribution[{0, Max[ypr - 1, 0]}]],
            Assumptions -> Element[ypr, Integers] && ypr >= 1],
           ypr >= 1}}, # /. z -> 0] /. ypr -> z) &, z, 4] /. z -> n] //
  Simplify[#, Element[n, Integers] && n >= 0] &

The rational part in the expectation appears to be $\frac{(n-1)(n+2-2^k)}{2^k n}$, thus in the large $n$ limit, expectation would agree with the continuous case. The expression involving harmonic numbers is of order $\mathcal{O}\left(\log(n)^{k-2}\right)$, and thus small compared to $n$.

With the guess-work, I was also able to find $\mathbb{E}(Y_5)$:
$$
  \mathbb{E}(Y_5) = [n > 5 ] \left( \frac{(n-1)(n-30)}{32 n} + \frac{1}{32n} \ell_n \right)
$$
where
$$
  \ell_n = -8 H_{n-1} H_{n-1}^{(2)}+\frac{8}{3} \left(H_{n-1}\right){}^3+12 \left(H_{n-1}\right){}^2+28 H_{n-1}-12
   H_{n-1}^{(2)}+\frac{16 }{3} H_{n-1}^{(3)}
$$
Here is confirmation with simulations:

In[153]:=
With[{n = 17},
    (28 HarmonicNumber[n-1] + 12 HarmonicNumber[n-1]^2 +
     8/3 HarmonicNumber[n-1]^3 - 12 HarmonicNumber[n-1,2] -
     8 HarmonicNumber[n-1] HarmonicNumber[n-1,2] +
     16/3 HarmonicNumber[n-1,3] + (n-1)(n-30))/(32 n)] // N

Out[153]= 0.131232

In[157]:=
Table[Nest[RandomInteger[{0,Max[#1-1,0]}]&, 17, 5], {10^7}] // N // Mean

Out[157]= 0.13157

If the random numbers are not restricted to integers but are uniformly generated on the entire interval $[0,n)$, the expected value is halved in each iteration, and thus by linearity of expectation the expected value of $f(n,k)$ is $2^{-k}n$.

If $n$ and the random numbers are restricted to integers, the expected value decreases from $a_i$ to $(a_i-1)/2$ in each iteration, so we need to shift by $1$ to obtain the recurrence $a_{i+1}+1=(a_i+1)/2$, so in this case the expected value of $f(n,k)$ is $2^{-k}(n+1)-1$.


Is it possible to solve any Euclidean geometry problem using a computer? By "problem", I mean a high-school type geometry problem.

If no, is there other set of axioms that allows that?
If yes, are there any software that does that?

I did a search, but was not able to find a single program that allows that. It is strange because even if it is impossible to solve any problem, most of the natural problems should be solvable.

There is a method to decide (true or not) any theorem in Euclid's Elements, by translating it into analytic geometry, or multivariate polynomials over the reals.

Of the philosophical controversies discussed by the Greeks that inspired Euclid to set up his axioms, common notions and definitions (like what an angle is, what really is the action of a 'compass'), most of them are de facto resolved by how operations and variables work in real 2D analytic geometry with the Pythagorean theorem (which has been proven to be proof-power equivalent to Euclid's fifth postulate).

The conversion works as follows:

An undefined point is given two variables, an $x$ and $y$ coordinate  $(x_1,y_1)$. Any possible distinct point should be named by some other pair $(x_2,y_2)$ (a proof that these two points coincide would show that $x_1=x_2$ and $y_1=y_2$).
a circle is defined by center $(x_3,y_3)$ and distance $d_1$ such that $(x-x_3)^2 + (y-y_3)^2 = d_1^2$. If you want a point on that circle, you instantiate this equation with the appropriate $x$ and $y$.
a line is defined by, well, two points. Just by stating two points, you have a line.
Frankly, I'm not sure how to do angles, but my memory tells me that it is possible to 'deal' with them.

So now you have a set of polynomial equations in many variables. And all you need is a procedure to find if there is a solution (some set of satisfying numbers or really a nontrivial set of equations) for them.

Suppose you only had points and lines. Then all you'd need is Gaussian elimination. But with circles, you can have arbitrary degree. Groebner basis completion (via Buchberger's algorithm from the mid 70's) is what you can use. Essentially, it does just what you'd expect (but didn't realize you could) on higher degree multivariate polynomials. In analogy with eliminating columns in a matrix, you try to eliminate 'maximal' terms one by one. It makes it easier if you multiply out everything so your equations are all of the form 'a sum of terms' = 0, where a term is a coefficient with a number of variables (possible none) multiplied, e.g. $17x_1^3y_{41}^2d_1^{1729}$. The algorithm assumes an order on these terms, and then follows an analog of the numerical GCD algorithm (coincidentally or not in Euclid).

So the algorithm tries to 'reduce' as much as possible (eliminating leading terms where it can). If it eliminates so much to the point where it has an equation $0=1$, then you know you have a contradiction, and then your equations have no way of being satisfied, so the polynomials that describe the theorem are inconsistent, so the theorem itself does not hold. Otherwise (and this is the beauty of it), the theorem -does- hold.

In some sense you could do this by hand for some very small trivial problems (It allows you to find the intersection of a given circle and line for example). Anyway, the details of the algorithm make it a decision procedure (will stop with an answer of 'yes' or 'no').

As a practical matter, you probably don't want to do this by hand. Groebner basis completion is (depending on what you consider the exact problem) PSPACE-complete, so mostly likely exponential (in the number of variables).

As to Tarski's result that elementary plane geometry is decidable, yes, that is a classic result (from the 50's). The theorem is the the theory of the reals is decidable, which includes the '$\lt$' relation. This includes Euclidean geometry, and is a bit more general. Tarski's algorithm is ingenious in that it solved a long outstanding problem, but is not particularly efficient. Collins' 'cylindrical algebraic decomposition' algorithm also from the mid 70's) is much more efficient (if you consider an increase in efficiency going from a stack of exponentials, to simply maybe just doubly exponential; the second is only merely astronomical, the first terribly so). It is of course still less efficient than Buchberger's algorithm. Because of the different fields these came out of, I'm not sure of any research that discusses any explicit comparison of the two.

Whatever the efficiency is, yes, there are a number of actual theorems proved mechanically in this style. A very nontrivial example is Morley's theorem (the trisectors of the angles of a triangle meet in an equilateral triangle) which Zeilberger has given an automated proof of using Maple. And Wu (of the Wu method mentioned, which is similar to Buchberger's algorithm) and his acolytes have more comprehensively proven whole sets of Euclidean theorems.

An interesting problem would be how to prove Euclidean theorems using Euclid's axioms as is (instead of translating). There has been some recent research by Avigad into that. And I've heard that one can do a translation of Euclid into Clifford algebras that is a bit more 'like' the original Euclid than the analytic approach.

(for anything unexplained, you might try googling or wikipedia with varying degrees of success)

So maybe that's not satisfying for whatever reason. Maybe there are other axiom systems for which one can natively (that is, synthetically, uninterpreted) have a decision procedure (which means write a computer program to do it. Hilbert's axioms are really a 'fix' of Euclid's; that is, it adds betweenness and some other axioms that Euclid, even in his arduous skepticism, left out. So I consider it essentially the same. Then there's Tarski's axioms. These are only mildly different, but still translate to the analytic version. (I am of the opinion that it wasn't a deliberate connection, and maybe an intellectual coincidence, between these axioms and Tarski's decision procedure for real closed fields.

As to computer programs, well there's Maple or Mathematica (or pretty much any computer algebra package that implements Groebner basis calculation); you still have to do the translation yourself into polynomials.

On quite the other hand there are a number of geometry 'editors', that allow you to 'draw' a theorem, allowing you to drag and drop the free parameters as objects, i.e. showing that the bisectors of a triangle meet at a point, allowing you to move all the vertices, showing you the intersection of the pairs of bisectors all meet at the same point. Some of these will allow you to 'prove' your construction (that some points always coincide for instance (Geometer's Sketchpad?). As far as I know, these 'proofs' are not symbolic (like GB calculation) but rather numeric (they notice that the difference among computed values is always less than some very small epsilon).

This reply is too long, so I can not add it as a comment. I hope someone will see it.

After reading some more on the subject (referenced from answers here), I must say that I am a little disappointed from the approach being used. I had imagined that any high school student can input his homework to a program, and it will output a solution. As I see, this is currently not the case.

I am only an egg, but I imagined that the axioms, assumptions and what is needed to be proved would be described in first order logic, and then some search algorithm would apply some rules of inference and search for the desired goal. This is similar to the way humans do math.

The problems with the polynomial approaches I see (you are welcomed to explain why I am wrong) are:

Previously generated mathematical knowledge is not being used.
Searches are all "brute force", a human mathematician usually use intuition and past experience to try different approaches and directions.
The algorithm can answer if something is true or false, but it can not produce a step by step proof (Zeilberger did that in his "geometry web book", but was his method different?).
Algorithm requires non trivial tuning for each different theorem.

Using the search algorithm, previously generated knowledge may be represented as more assumptions and rules of inference. Intuition would be represented by a heuristic function.


I don't understand the implication statement in a quadratic epsilon delta proof. If I have $\lim\limits_{x\to2} x^2=4$ and then  I use the $\epsilon-\delta$ proof, why is $|x+2| \leq 5$ , even though that is not possible because that would imply that $-5 \leq x+2 \leq 5$ but it is not that, it is $3 \leq x+ 2 \leq 5$? I meant to write less than not less than or equal to

In general, if you make some statement $P,$ and then you say "therefore $Q,$" you are not saying that $P$ and $Q$ are completely equivalent. There might be plenty of ways to satisfy $Q$ while $P$ is still false. What matters is that there is no way to make $P$ true and $Q$ false at the same time. In the examples you've given, we start with $P,$ a statement that is true for $x$ within some interval, and then make a statement $Q$ that is true for $x$ in a larger interval. No problem as long as the smaller interval fits completely inside the larger one.

Now, if someone were to say $3 < x+2 < 5,$ and then say that therefore
$\lvert x+2\rvert < 4,$ then you would have a faulty "proof," because part of the interval that makes $3 < x+2 < 5$ also makes $\lvert x+2\rvert < 4$ false, for example if $x=2.$

You claim that $\lim_{x\to 2}x^2=4$ with a $\epsilon-\delta$ argument. Fix $\varepsilon \in (0,1)$. Then it has to exist $\delta=\delta(\varepsilon)>0$ such that
$$
|x-2|<\delta \implies |x^2-4|<\varepsilon,
$$
i.e.,
$$
2-\delta<x<2+\delta \implies 4-\varepsilon<x^2<4+\varepsilon.
$$
To this aim, fix $\delta=\varepsilon/5$ (which is $<1$). Then
$$
x^2<(2+\delta)^2=4+\delta^2+4\delta \le 4+5\delta = 4+\varepsilon
$$
and
$$
x^2>(2-\delta)^2=4+\delta^2-4\delta>4-4\delta>4-5\delta=4-\varepsilon.
$$


If $|a|= \infty$, then $$ is infinite subgroup I'm trying to understand the proof of:

  Let $G$ be group and $a \in G$.
  If $|a|= \infty$, then $<a>$ is infinite subgroup.

The proof I have here goes like:

  Let $|a|=\infty$. Then $a^i ≠ a^j$ for all $i≠j$, because otherwise
  $|a|≤ |i-j|.$ Which proves the statement.

I don't understand why must $a^i ≠ a^j$ and how this relates to the order of the element (or the subgroup).

Suppose that for some $i\not= j$, you have $a^{i}=a^{j}$. Assume without loss of generality that $i<j$. Then, multiplying $a^{-i}$ to both sides gives $e=a^{j-i}$. Since $j-i>0$, this gives that $|a|\leq j-i$, contradicting that $|a|=\infty$. Therefore, it must be that for all $i\not= j$, $a^{i}\not= a^{j}$. Hence, $<a>=\{a^{i}\mid i\geq 0\}$ is an infinite set.

It is known that if $G$ is finite cyclic group and has order $n$ then the subgroup $H$ order will be number divisible by $n$, so let the subgroup $H$ order be $r$ then $r.x=n$ for positive $x$, now in your task assume that $H$ is finite then $n$=infinity must divide $r$ which is not true, so the subgroup $H$ is finite.


Finding the image of $f(x)=\frac{1}{1+x^{2}}$ $f(x)=\frac{1}{1+x^{2}}$ and $x\geq0$

To find the image:

$y=f(x)$

$y=\frac{1}{1+x^{2}}$

$x=y^{-1}$

$x=\sqrt{y^{-1}-1}$

$y\geq1$

Then the image of $f(x)=\frac{1}{1+x^{2}}$ is $y\geq1$

Is this correct?  If not, what information is missing?

For $\sqrt{y^{-1}-1}$ to be defined, $y^{-1}\ge 1$, i.e. $\frac{1}{y}\ge 1$ or $y\le 1$.  The range is indeed bounded above by 1; it is bounded below by zero since $1+x^2\ge 1>0$.  $0$ is not in the range, since the fraction is never zero; however we can get arbitrarily close.

Hence the range is: $$(0,1]$$

Your answer is correct, y should be greater than or equal to 1.


Open Ball in a Metric Space vs. Open Set in a Topological Space I'm having trouble understanding the notion of an open set when applied to a space without a metric defined on it. I have read that all metric spaces are naturally a topological space, but the converse is not true. Topological spaces definitions I have read use the idea of open sets, and I can't understand this abstract idea of openness or closedness of a set without having a notion of distance.

I can understand the notion of open and closed sets in a metric space from the definitions I have read using the idea of distance and open balls. So for example given a metric space $M$ with the metric $d$, we can say a set $U \subset M$ is open if $\forall x \in U, \exists B(x,r) \subset U$, which says that we can choose any point within the set $U$ and there will always be some sufficiently small distance $r$ that we can move in any direction to another point $y$ that is also contained within $U$. This has an intuitive conceptual meaning in my head regardless of space $M$ or metric $d$.

Now take the example of a space $F$, a fruit bowl with $3$ apples, $3$ oranges and $2$ bananas. There is no metric defined on the space $F$ to determine a distance between its elements, the fruit. Can we define an open set on this space? In order to be a topological space we need to be able to define open sets right?

Maybe an analogy will help. Supposing you start with an understanding of Real Analysis, you can generalize this understanding to Metric Spaces:

In Real Analysis, you have this fixed notion of distance between two points in $\mathbb{R}$ defined as $|x-y|$. Then using this definition you build your collections of open sets that you use to study $\mathbb{R}$. These sets look like open intervals $(x-r, x+r)$ centered at $x$ with radius $r$, and using these open sets you can study the space (continuity, connectedness, derivatives, etc.)
Then you jump to into a general a Metric Space and are told "Use whatever definition of distance that you like (as long as it follows a few rules to keep things sensible) and build your open sets with that!" Then you just proceed to do the same thing you did in Real Analysis only you are using open balls that look like $B(x,r)$, centered at $x$ with radius $r$.

Now we can see a related jump from understanding Metric Spaces to understanding General Topology:

In Metric Spaces, you have this fixed notion of distance between two points in your space $X$ defined as $d(x,y)$. Then using this definition you build your collections of open sets that you use to study $X$. These sets look like open balls $B(x,r)$ centered at $x$ with radius $r$, and using these open sets you can study the space (continuity, connectedness, derivatives, etc.)
Then you jump to into General Topology and are told "You don't need to bother  building the collection of open sets at all. Instead, I'm just going to hand you this collection of open sets! (and these open sets will follow a few rules to keep things sensible)" Then you just proceed to do the same thing you did with Metric Spaces only you are using these open sets that you were handed to you, and there might not be a easy way to describe what they look like.

Take a set $X$. Let $\lambda \in X$ , we say that a ball of radius epsilon ($\epsilon$) around the point $\lambda$ will be an open set iff $\ni$ a point $k\notin X$ such that a neighborhood of radius delta ($\delta$) around the point $k$ is not in the ball of radius $\epsilon$ around $\lambda$.

In mathematical terms , $B_\epsilon(\lambda)$ is open when,$$N_\delta(k) \cap B_\epsilon(\lambda)=\emptyset.$$

If you want to think in terms of a topological space:
We say that let $Y\subset X$ , then given a set $A$ in $Y$ we say that it is open if
$A=Y\cap U$ where $U$ is open in $X$


Seven points in the plane such that, among any three, two are a distance $1$ apart Is there a set of seven points in the plane such that, among any three of these points, there are two, $P, R$, which are distance $1$ apart?

Yes. Normally I like to give some more motivation for an answer, but in this case it's probably hard to say anything suggestive that the illustration below does not, besides perhaps that after experimenting one might guess that the four-point "diamond" configurations are helpful (and probably necessary) in constructing such an arrangement.

(I remember seeing this problem, by the way, during a mail-in high school mathematics competition circa 1999; probably the widespread availability of the Internet makes it impractical to hold this sort of competition today.)

Edit I've posed a follow-up question, asking whether this is the unique configuration up to Euclidean motions. Servaes wrote an excellent, detailed answer showing that it is.

EDIT: This answer (which starts with a bold "No.") is wrong as can be seen from Travis's answer (which starts with a justified "Yes."). In principle, I should delete it, but - for the moment - I'll leave it here because somebody may find it helpful in other ways, such as seeing how to arrive at some neccessary conditions (cf. some comments)

No.

Assume $d(A,B)$, $d(A,C)$, $d(A,D)$, $d(A,E)$ are all $\ne 1$. Then all distances amoing $B,C,D,E$ must be $=1$.
However, because the tetrahedron cannod be "flattened",

  Among any four points there must be two of distance $\ne 1$.

We conclude that $A$ can have at most three points at distance $\ne 1$, hence

  Every point hast at least three points at distance $=1$.

Hence if we pick any two points at distance $\ne1$, then one of the remaining five points must be one of the intersection points of the $1$-circles around them, i.e.,

  For any two points $A,B$ with $d(A,B)\ne 1$ there exists $C$ with $d(A,C)=d(B,C)=1$.

Can you take it from here?


How to explain the perpendicularity of two lines to a High School student? Today I was teaching my friend from High School about linear functions. One of the exercises we had to do was finding equations of perpendicular and parallel lines. Explaining parallel equations was quite easy, if we have the equation $y = ax + b$ it's not hard to show with a couple of examples that changing the parameter $b$ only "moves" the line up or down but doesn't change the angle, thus lines $k$ and $\ell$ are parallel iff $a_k = a_{\ell}$.

However, I couldn't find a clear way to explain why those lines are perpendicular iff $a_k \times a_{\ell}= -1$. Of course, it's obvious if we use the fact that $a = \tan (\alpha)$ with $\alpha$ being the angle at which line intersects the X axis and that $\tan (\alpha) = - \cot (\frac{\pi}{2} + \alpha)$. But this forces us to introduce trigonometry and rises oh so many questions about the origin of the equation above. Does anyone know a good, simple explanation that's easy to remember?

Draw a line through the origin, $y=mx$, and mark the point $(1,m)$. Rotate the drawing by a quarter turn and observe that the point is now at $(-m,1)$.

Old slope: $m$, new slope: $-\frac1m$.

Let $r$ a line through the origin with slope m. The direction of this line is given by the vector $u = (1, m_r)$. If $s$ is the perpendicular line $r$ Without loss of generality, we can assume that $s$ also passes through the origin. If $v = (a, b)$ is its direction vector, and $r$ is perpendicular a $r$, it follows that $u\cdot v = 0 \ => \ (1,m_r)(a, b) = 0$. Thus,
$$a + bm_r = 0 \ => \ m_r = -\dfrac{a}{b} = -\dfrac{1}{\dfrac{b}{a}} = -\dfrac{1}{m_s} \ => \ m_r\cdot m_s = -1$$


Find the four digit number? Find a four digit number which is an exact square such that the first two digits are the same and also its last two digits are also the same.

HINT:

So, we have $$1000a+100a+10b+b=11(100a+b)$$

$\implies 100a+b$ must be divisible by $11\implies 11|(a+b)$ as $100\equiv1\pmod{99}$

As $0\le a,b\le 9, 0\le a+b\le 18\implies a+b=11$

$$\implies11(100a+b)=11(100a+11-a)=11^2(9a+1)$$

So, $9a+1$ must be perfect square

We know that the only square that has last two digits as same is $12^2=144$. For a four digit perfect square it should either be $(50-12)^2$ or $(50+12)^2$ or $(100-12)^2$

$38^2= 1444 $
$62^2= 3844$
$88^2= 7744 $

As per the question we know $88^2$ is the perfect square that has first 2 digits & last 2 digits as the same.


Find closed formula $f(n)$ from generating function I'm asked to find a closed formula for

$f(n)=6f(n-1)-9f(n-2)$ for $n>1$ with $f(0)=-1. f(1)=0$,

using the ordinary generating function $F(X)$.

I found $F(X)=-1/(1-3x)^2$ but from there I don't manage to get a satisfactory formula for $f(n)$. Can anyone give me a hint ?

Thanks

Write the recurrence as:
$$
f(n + 2) = 6 f(n + 1) - 9 f(n) \qquad f(0) = - 1, f(1) = 0
$$
Multiply by $z^n$ and sum over $n \ge 0$. Recognize:
\begin{align}
\sum_{n \ge 0} f(n + 1) z^n &= \frac{F(z) - f(0)}{z} \\
\sum_{n \ge 0} f(n + 2) z^n &= \frac{F(z) - f(0) - f(1) z}{z^2}
\end{align}
and get:
$$
\frac{F(z) + 1}{z^2} = 6 \frac{F(z) + 1}{z} - 9 F(z)
$$
Solving for $F(z)$, and splitting into partial fractions:
$$
F(z) = \frac{1 - 6 z}{1 - 6 z + 9 z^2} = \frac{1}{(1 - 3 z)^2} - \frac{2}{1 - 3 z}
$$
This tells us:
$$
f(n) = \binom{-2}{n} (-1)^n 3^n - 2 \cdot 3^n
     = \left( \binom{n + 1}{1} - 2 \right) \cdot 3^n
     = \frac{n - 3}{2} \cdot 3^n
$$


Approximation formula for Higher-order derivative I have this assigment

  Derive by using Taylor approximation of $f$ a formula for approximation of $f′(x_0),f′′(x_0),f′′′(x_0),f′′′′(x_0)$ with an error term of order $h^4$.

I have already done $f′(x_0)$ and $f′′(x_0)$ but I'm stuck trying to get $f'''(x)$. I've done the Taylor expansion for $f(x+h),f(x-h),f(x+2h),f(x-2h),f(x+3h),f(x-3h)$ and don't know what else to do.

Can someone give me some advice o help me to get $f'''(x)$?

PD Excuse my bad English.

EDIT:

What the assigment want me to get is something like

$f'''(x)= \frac{-f(x+3h)+8f(x+2h)-13f(x+h)+13f(x-h)-8f(x-2h)+f(x-3h)}{8h^3}+O(h^4)$


Equations modulo $143$ Let $x=11$ and $y=13$, and $z=xy=143$.

(i) Show that $1$, $x+1$, $−1$ and $−(x+1)$ are the $4$ solutions of $n^2 \equiv 1\pmod z$.

(ii) Find the coset of $U_z(2)$ consisting of solutions to $n^2 \equiv 9\pmod z$.

I do not know at all how to do this problem. I am very iffy about this section.

Hints: For question (i), you should realise that $n^2\equiv1\pmod{11}$ and $n^2\equiv1\pmod{13}$. Can you see why this is? Find $n$ in each of these two settings, and then use (for example) the Chinese Remainder Theorem to prove the result. To find $n$ in these settings, you can (if you want) use the fact that $n^2-1=(n-1)(n+1)$. Note: We could not use this equality upstairs because of zero divisors. For example, $11\times 13\equiv0\pmod{z}$. This does not happen when we are working modulo a prime.

For question (ii), either proceed as above to find all solutions to the given equation, or multiply your solutions to part (i) to find all solutions to the given equation. I phrase this an an either/or because it is not obvious that the second method works (it is not obvious that you get all the solutions). So if you pick the second method (which is quicker) then you should convince yourself that it works (if I was marking your exam and you picked that method I would expect a few lines of working to explain why it worked). Then, suppose the solution set is $\{n_1, n_2, \ldots, n_q\}$. This means that your coset is written $2\{n_1, 2n_2, \ldots, 2n_q\}$, which is the set...what? You tell me. :-)


Inversible Antisymmetric Matrix Let $A$ be a $n \times n$ antisymmetric matrix. Show that if $A$ is invertible then $n$ is even.

I guess that this means that $\det(A)$ is zero when $n$ is odd and non-zero when it is even but I dont know how to show this.

Thanks

An antisymmetric matrix $A$ is a matrix with

$$A^T = -A.$$

Taking the determinant on both sides yields

$$\det (A^T) = \det (-A).$$

How could you simplify the RHS?

We have

det$(A) = $det$(-A^T) = (-1)^n$det$(A^T) = (-1)^n$det$(A)\Rightarrow (-1)^n =1 \Rightarrow n$ is even.


Determine if two straight lines given by parametric equations intersect Does
$[x,y,z] = [4,-3,2] + t[1,8,-3]$

intersect with

$[x,y,z] = [1,0,3] + v[4,-5,-9] ?$

Attempt
To find out if they intersect or not, should i find if the direction vector are scalar multiples? Clearly they are not, so that means they are not parallel and should intersect right? But the correct answer is that they do not intersect. How do you do this? Thanks!

In 3 dimensions, two lines need not intersect. The two lines intersect if and only if there are real numbers $a$, $b$ such that $[4,-3,2] + a[1,8,-3] = [1,0,3] + b[4,-5,-9]$. We have the system of equations:

$$
\begin{aligned}
4+a &= 1+4b &(1) \\
-3+8a &= -5b &(2) \\
2-3a &= 3-9b &(3)
\end{aligned}
$$

$-(2)+(1)+(3)$ gives
$$
9-4a=4 \\
\Downarrow \\
a=5/4
$$

$(2)$ then gives

$$7 =-5b \\
\Downarrow \\
b = -7/5
$$

But in $(1)$ we then have

$$4+5/4=1-28/5$$

which is false. So no solution exists, and the lines do not intersect.

I think they are not on the same surface (plane).  Two vectors can be: (1) in the same surface in this case they can either (1.1) intersect (1.2) parallel (1.3) the same vector; and (2) not in the same surface.


A conjecture on the closeness of twin primes Let $p_1$ and $p_2$ be twin primes, and let $p_1-1=a_1\times b_1$ and $p_2+1=a_2\times b_2$ be such that $|b_1-a_1|$ and $|b_2-a_2|$ are minimised. Similarly,  let $p_1+1=p_2-1=a\times b$ be such that $|b-a|$ is minimised.
Now assume the twin prime conjecture, and let $$\mathcal P_n=\frac{\#\text{twin primes} \le n:\min\{|b_1-a_1|,|b-a|,|b_2-a_2|\}\neq|b-a|}{\#\text{twin primes} \le n}$$ What is the value of $\mathcal P_\infty$?
Note that by convention the pair $(2,3)$ are not twin primes.

For twin primes less than $100$, we have that $p_1=5,17$. Hence I think that if there are infinitely many such primes then they will be exceedingly sparse.
Here is a table showing $\mathcal P_n$ for increasing values of $n$. Credits to Enzo Creti for running the program. $$\small\begin{array}{c|c}\log_{10}n&1&2&3&4&5&6&7&8&9&10&11\\\hline \mathcal P_n&\frac12&\frac14&\frac{17}{35}&\frac{93}{205}&\frac{600}{1224}&\frac{4326}{8169}&\frac{31939}{58980}&\frac{243876}{440312}&\frac{1928700}{3424506}&\frac{15661079}{27412679}&\frac{129632703}{224376048}\\\hline\text{decimal}&\small 0.5&\small0.25&\small0.4857&\small0.4537&\small0.4902&\small0.5296&\small0.5415&\small0.5539&\small0.5632&\small0.5713&\small0.5777\end{array}$$
This can be seen more clearly in the following plot; more interesting is the behaviour after $n=10^4$.

There is a slight dip at $n=10^2,10^4$ but otherwise it looks like there is a converging increase in $\mathcal P_n$. It is unlikely, but if there reaches a point where $\mathcal P_k=\mathcal P_{k+1}$ then the twin prime conjecture will be disproved.

Twin primes other than $3,5$ are of the form $6k\pm1$, which makes $p_1-1 = 6k-2$, $p_2 + 1 = 6k+2$, and $p_1+1 = p_2-1 = 6k$. Define $f(x) = \min\{|a-b|\}$ where $a b = x$, and you are asking are there infinitely many $k$ where $6k\pm1$ are both primes, and $f(6k-2) < f(6k)$ or $f(6k+2) < f(6k)$.

Experimentally, about $54\%$ of the twin primes are "close", not the $33\frac13\%$ that one would naively expect. Still, most likely infinitely many.


Where is $x^x$ continuous? The idea of continuity of a function is something I come across quite regularly, but I've never really understood it well. I'm trying to fix that by looking at some interesting functions.

What happens continuity wise for $x<0$ in this function:

$$f(x)=x^x$$

I can't find a quick answer to this. Apologies if this is a duplicate.
I'm looking for a detailed explanation.

As mentioned in another answer, writing $x^x$ is the same as writing $e^{x\log x}$. But complex analysis tells us that there are many possible branches of $\log x$ that you could choose. Namely, you can define it on any simply connected region in $\mathbb C$ not containing zero, and you can add $2\pi i$ as many times as you like.

Since you're asking about real negative values of $x$, you'd need a branch of the logarithm that avoids this half line in the complex plane. For example, you could choose the branch
$$\log z=\log|z|+i\theta$$
where $-\frac{\pi}{2}<\theta<\frac{3\pi}{2}$. In any such branch, the complex logarithm is analytic and therefore continuous on the negative real half-line.

To conclude, the answer to your question is that $x^x$ is always continuous for $x<0$ provided you've picked a well-defined meaning of the function.

I apologize that this explanation requires that you've seen complex analysis, but this question cannot be adequately answered otherwise.

The quickest and easiest way to make a statement on this function's continuity is to take a derivative. This requires logarithmic differentiation. The derivative is:
$$f'(x) = x^x(\ln(x) + 1)$$

Note that $\ln(x)$ is only valid for positive entries. So, this tells us that $f$ is continuous for positive numbers.


Commutator formula in infinite dimensions The commutator formula states that for $A,B$ elements of a Lie algebra,

$$ \lim_{n\to \infty}\left\{ \exp\left(-A\tfrac{t}{n}\right)\exp\left(-B\tfrac{t}{n}\right)\exp\left(A\tfrac{t}{n}\right)\exp\left(B\tfrac{t}{n}\right)\right\}^{n^2}=\exp\left(t^2[A,B]\right)$$

I am interested in the case where $A=iH_1$ and $B=iH_2$ with $H_i$  self-adjoint. For finite dimensions the above certainly holds, but what happens in infinite dimensions? Under which conditions? Bounded/unbouded operators? I know that Trotter's formula has some complications in infinite dimensions, I'd be very thankful for any hints here.

This is answered in "Topics in Dynamics 1: Flows" by Edward Nelson (Theorem 7, Chapter 8): if $A$ and $B$ are skew-adjoint operators on a Hilbert space $\mathcal{H}$ and the restriction of $[A,B]$ to $\mathcal{D}=\mathcal{D}(AB)\cap\mathcal{D}(BA)\cap\mathcal{D}(A^2)\cap\mathcal{D}(B^{2})$ is essentially skew-adjoint, then

$\lim_{n\rightarrow\infty}[e^{-\sqrt{\frac{t}{n}}A}e^{-\sqrt{\frac{t}{n}}B}e^{\sqrt{\frac{t}{n}}A}e^{\sqrt{\frac{t}{n}}B}]^nu=e^{\overline{[A,B]}t}u$ for all $u\in\mathcal{H}$, uniformly for $t$ in any compact subset in $[0,\infty)$.


Have I discovered a new significance to a previously discovered constant? I've been interested in infinite sums for a while, though I have no formal education of them. I was messing around with repeated division and addition (e.g. 1 + (1 / (1 + (1 /...)))) I then plugged fibonacci numbers into the above pattern, and as I calculated more and more layers, the result converged to around 1.39418655, which I just now found out was a constant called Madachy's constant.  However, from the research that I did on the constant (I found very little), it doesn't seem to be related to infinite series or Fibonacci numbers at all. Have I found a new way to calculate this number? Does it give it any more significance?

Did you see this? fq.math.ca/Scanned/6-6/madachy.pdf It looks like it is exactly your calculation.

Friendly edit: This is in Fibonacci Quarterly, Volume 6 No. 6 page 385, as seen at the cover of that issue.

The continued fraction you are calculating will converge to the golden ratio. It is most certainly a significant constant having history spanning several millennia.


How can I show $\lim_{t\to 0^{+}}\|S(t)-I\|\neq 0$ where $S(t)f(x)=e^{-t^2-2tx}f(x+t)$.. I need some help with the following:

For every  $t\in [0, \infty)$ let $S(t):C_0([0, \infty))\rightarrow C_0([0, \infty))$ be the bounded linear operator given by, $$S(t)f(x)=e^{-t^2-2tx}f(x+t)),$$ where $\displaystyle C_0([0, \infty))=\{f\in C^\infty([0, \infty), \mathbb R): \lim_{x\to \infty}f(x)=0\}$. How to show $$\lim_{t\to 0^{+}}\|S(t)-I\|\neq 0.$$  Recall the norm of a bounded linear operator is $\displaystyle \|A\|=\sup_{\|x\|=1}\|Ax\|$.

Notice that if we choose, for each $t$, $f(x)=e^{-x/t}$.

$S(t)f(0)=e^{-t^2-1}$, so $S(t)f(0)- f(0) = e^{-t^2-1} -1 $

Hence, by definition of sup, $||S(t)-I|| \geq 1 - e^{-t^2-1}$. Then we are done.

This is a an example of pointwise convergence not implying uniform convergence.


decresing sequence implies $\lim{n u_n} $ exists? if $(u_n)$ is a nonincreasing sequence of positive real
numbers; how to prove or disprove  that the sequence $(nu_n)$ admits a limit?
We know  two results
1- if $(u_n)$ is a  sequence of positive real
numbers such that $\sum_n a_n$ converges, and if $\lim\limits_{n \to \infty} n u_n$ exists, then $\lim\limits_{n \to \infty} n u_n = 0$
2- if $(u_n)$ is a nonincreasing sequence of positive real
numbers such that $\sum_n u_n$ converges, then $\lim_{n \rightarrow \infty} n u_n = 0$.
PS :  I corrected the usual notation for a sequence

Kenta S gave two valid examples in their comment: both $u_n = \frac1{\sqrt n}$ and $u_n=1$ have the property that $\lim_{n\to\infty} nu_n = \infty$. It's important to note that, despite the use of the equals sign in that notation, it indicates a specific way that $\lim_{n\to\infty} nu_n$ fails to exist. (Infinity is not a number.)
But other divergent behaviour is also possible. Consider for example
$$
u_n = 2^{-\lfloor \log_2 n\rfloor} = \min\bigg\{ \frac1{2^k}\colon \frac1{2^k} > \frac1n \bigg\},
$$
so that $(u_n) = (1,\frac12,\frac12,\frac14,\frac14,\frac14,\frac14,\frac18,\dots)$. It's not hard to show that $\limsup_{n\to\infty} nu_n = 2$ but $\liminf_{n\to\infty} nu_n = 1$, and so $\lim_{n\to\infty} nu_n$ does not exist.

Let $y_n = n u_n$. Then $L = \lim_{n \rightarrow \infty} y_n$ exists iff $\forall \varepsilon \exists m_0 \forall m \geq m_0 \mbox{ s.t. }|y_m - L| < \varepsilon$, i.e. for any small value, the $y_n$ beyond some point are all closer to $L$ than that value.
This means that $|n u_n - L| < \varepsilon$, so $|u_n - \frac{L}{n}| < \frac{\varepsilon}{n}$, or to rearrange it (and taking advantage of the fact that $u_n > 0$), $\frac{L - \varepsilon}{n} < u_n < \frac{L + \varepsilon}{n}$.
This means that the limit can only exist if $u_n = o(n^{-1})$ , i.e. $u_n$ shrinks faster than $\frac{1}{n}$.


Finding the range and domain of $f(x)=\tan (x)$ I am attempting to find the range and domain of $f(x)=\tan(x)$ and show why this is the case. I can seem to find the domain relatively well, however I run into problems with the range. Here's what I have done so far.

  Finding the domain of $f(x)=\tan(x)$

Consider $f(x)=\tan(x)$ is defined as $f(x)=\tan(x)=\frac{\sin(x)}{\cos(x)}$, it is clear the domain of $f(x)$ is undefined when $\cos(x)=0$. $\cos(x)=0$ whenever $x=\frac{\pi}{2}+\pi k$ for integers $k$, so the domain of $f(x) =\tan (x)$ can be stated as $x\in\mathbb{R}, x \ne \frac{\pi}{2}+\pi k\text{  for integers k}$

  Finding the range of $f(x)=\tan(x)$

To find the range of $f(x)=\tan(x)$ we must refer to the definition of $f(x)=\tan(x)=\frac{\sin(x)}{\cos(x)}$. From this we can see that $f(x)$ is undefined when $\cos(x)=0$, as the interval of $\cos(x)$ is $[-1,1]$ we will now need to split this into two cases: $-1\leqslant\cos(x)<0$ and $0<\cos(x)\leqslant1$.

Considering the first interval; $-1\leqslant\cos(x)<0$, as $\cos(x)\to0^-$: $\sin(x)\to1$ and $\tan(x)=\frac{\sin(x)}{\cos(x)}\approx\frac{1}{\text{very small(negative)}}\approx\text{very big(negative)}$.

In other words as $\cos(x)\to0^-$, $\tan(x)\to-\infty$

Considering the second interval; $0<\cos(x)\leqslant1$, as $\cos(x)\to0^+$: $\sin(x)\to1$ and $\tan(x)=\frac{\sin(x)}{\cos(x)}\approx\frac{1}{\text{very small(positive)}}\approx\text{very big(positive)}$.

In other words as $\cos(x)\to0^+$, $\tan(x)\to+\infty$.

This is where I am up to. What I want to know is how I can show definitively that $tan(x)$ can take all the values within the interval $[-\infty,\infty]$. Some people would call it a day here, and say that this shows that the range  of $f(x)$ is $-\infty<\tan(x)<\infty$. However I nearly ran into a similar error when I was finding the range of $\sec(x)$, only to discover that although it does tend to positive and negative infinity it doesnt take any values in the interval $(-1,1)$. Where do I proceed from here?

EDIT: I am not looking for answers that use differentiation. Answers should be pre-calculus level.

Firstly, I'd like to applaud you on your persistence in trying to get to a rigorous solution.

I'll give you a hint for a geometric solution: if you use the triangle definition of a tangent, can you think of a way of constructing a triangle giving any desired value of tangent?

Edit: some more detail

Lets say I want to find a value of $x$ such that $\tan(x) = y$ for some $y$. Then I can construct a right angled triangle with sides of lengths $1, y$ and $\sqrt{1+y^2}$. Then if $x$ is the appropriate angle, then $\tan(x) = y$

When $\theta$ is in $dom(\tan)={\mathbb R}\setminus (\frac{\pi}{2}+\pi{\mathbb Z})$, then  $tan(\theta)$ is the slope of the line through the origin and the point $(\cos(\theta),\sin(\theta))$. Since we can make this line as close as we want to the $y$-axis (with positive as well with negative slope), then the slope can take any real value. Hence $range(\tan)={\mathbb R}$.


In a 52-deck card, draw 6. Probability of getting 3 different suits, an 8 and a face card? A standard deck of cards has 52 cards. If order does not matter, what are the chances of getting 3 cards of different suits, an 8 and a face card?

I thought like this:

P(first card being a different suit) = $1/1 $
P(second card being a different suit) = $39/51 $
P(third card being a different suit) = $26/50$
P(fourth card being an 8) = $4/49 $
P(fifth card being a face card) = $12/48$

My answer : $\frac{1}{1} \cdot \frac{39}{51} \cdot \frac{26}{50} \cdot \frac{4}{49} \cdot \frac{12}{48} = 0.81\%$

Total no of cases are ${52 \choose 6}$ , i am gonna write ${n \choose r }$ as nCr.
(I am assuming that we have to take in consideration that which suit are of the 8 and face cards are of).

Now as we need 6 cards of three 3 different suites, the possible sums can be:-
$a)1+1+4\ \ b)1+2+3\ \  c)1+1+1+3\ \ d)1+1+2+2\ \ e)2+2+2$

a)1+1+4 : rather than counting an 8 appears and an face cars appers it's easier to count total - either one of them does not appears

= Total - 8 does not appear - face card does not appear + both 8 and face card does not appear.

(13C1.13C1.13C4  -  12C1.12C1.12C4 - 10C1.10C1.10C4 + 9C1.9C19C4)$\times$(number of ways of selection of suits)

Selection of suits -> 4C2(for selecting 2 suits from 1 cards come in) .2C(1) (from where 4 cards come)=12

So it comes out 465132

Same way b)->2135520 ,c)->757672 ,d)->1820448,
e)->570348

Final answer $\frac{465132+2135520+757672+1820448 +570348}{{52 \choose 6}}$

=$\frac{5749120}{20358520}$=0.28239380858

When order doesn't matter (i.e., the number of unordered samples/$\textit{combinations}$ is what we are interested in), then $n\choose{k}$ methods are usually the safest way to go.

With that said, when all outcomes of an experiment are equally likely, then the probability of an event $A$ is

$$
\mathbb{P}(A) = \frac{|A|}{|\Omega|},
$$
where $|\cdot|$ denotes the number of elements in the set, and $\Omega$ denotes the sample space.

For example, the probability of obtaining 2 aces, 1 King and a card not ace or King from a deal of 4 cards is
$$
\frac{{4\choose{2}} {4\choose{1}} {44\choose{1}}}{52\choose 4}.
$$
Think of it as "from the 4 aces available, we want 2" and so on. The handy thing about using $n\choose{k}$ approach is that once you've worked out what the correct numbers are, you don't have to worry about extra combinations on top of that.

So back to our example (I assumed from your working that we do not care what suit the 8 and the face card belong to). The number of ways 3 cards of different suits can be dealt from a pack is

$$
\text{(Number of ways of choosing 3 suits from 4)}\times\text{(Number of ways of choosing 1 card from each suit 3 times)}\\
= {4\choose{3}}{13\choose{1}}^3.
$$
With these 3 cards chosen, the number of ways of choosing one 8 from the four available is ${4\choose{1}}$; same for the King. This leaves us with
$$
\frac{|A|}{|\Omega|} = \frac{{4\choose{3}}{13\choose{1}}^3{4\choose{1}}{4\choose{1}}}{{52\choose{5}}}
\approx  0.054102 = 5.4102\%.
$$
I was concerned about the assumptions of the question, since it's not clear that the suit draws have to account for drawing an 8 or a King, but it may be the case that a priori we don't take this into account, since the cards are all being drawn simultaneously. I'm not sure. In any case, this would be my approach and I hope it at least sheds some light for you on how these questions can be tackled.


$F=y^{2}\mathbf{i}+x^{3}\mathbf{j}$ curve C is counterclockwise path around triangle with vertices $(0,0)$, $(1,0)$, and $(0,4)$ Find the work done by $F$ over the curve in the direction of increasing t.

$$
F=y^{2}\mathbf{i}+x^{3}\mathbf{j};
$$
curve $C$ is counterclockwise path around triangle with vertices $(0,0)$, $(1,0)$, and $(0,4)$ Thanks!

Hints

Along the segment $(0,0)$ to $(1,0)$, $y=0$.

Along the segment $(0,0)$ to $(0,4)$, $x=0$.

The hypotenuse can be parameterize as
$$
 t(0,1)+(t-1)(4,0), \quad 0 \le t \le 1.
$$

Solution

A diagram of the contour's three components against the vector field:

Quick check: is this vector field $F=P(x,y)\mathbf{i} + Q(x,y)\mathbf{j}$ conservative? Does $P_{y}=Q_{x}$ If yes, the closed contour is $0$:

$$
P_{y} = 2y, \quad Q_{x} = 3x^{2} \qquad \Rightarrow \qquad P_{y} \color{red}{\ne} Q_{x}
$$

Evaluate the pieces.

Leg 1: $y = 0$

$$\int_{0}^{1} f(x,y) dx = \int_{0}^{1} \left( 0, x^{3} \right) dx = \left( 0, \frac{1}{4} \right)$$

Leg 2: $y=4-x$
$$
\int_{0}^{1} f(x,y) dx = \int_{0}^{1} \left( \left( 4-4x \right)^{2}, x^{3} \right) dx =
\left(\frac{16}{3},\frac{1}{4}\right)
$$

Leg 3: $x = 0$
$$
\int_{0}^{1} f(x,y) dy = \int_{0}^{1} \left( y^{2}, 0 \right) dx =
\left( \frac{1}{3}, 0 \right)
$$

$$
I_{C} = I_{1} + I_{2} + I_{3} =
\left( 0, \frac{1}{4} \right) +
\left(\frac{16}{3},\frac{1}{4}\right) +
\left( \frac{1}{3}, 0 \right) =
\mathbf{\left( \frac{17}{3}, \frac{1}{2} \right)}
$$


Question on proof of $1+2+\dots+n=\frac{n(n+1)}{2}$ by induction. I saw some video where it needs to prove $1+2+\dots+n=\frac{n(n+1)}{2}$ inductively. So it has to be true if $k=1$ and $k+1$ are true.

So, for $k=1$:

$$1=\frac{1(1+1)}{2}=\frac{1(2)}{2}=\frac{2}{2}=1$$

it is valid.

For $k+1$ here is the proof he does:

$$
\begin{align}
1+2+\dots+k+k+1&=\frac{(k+1)(k+2)}{2}    &(1)\\
\frac{k(k+1)}{2}+k+1&=\frac{(k+1)(k+2)}{2}  &(2)\\
&=\frac{k^2+2k+k+2}{2}&(3)\\
&=\frac{k^2+3k+2}{2}&(4)\\
&=\frac{(k+1)(k+2)}{2}&\text{factoring (4)}
\end{align}$$

Therefore this formula is valid for $k+1$.

But is this true? I think not. He is just undoing what he have just done. To prove it I think I need to do this:

$$
\begin{align}
1+2+\dots+k+k+1&=\frac{(k+1)(k+2)}{2}\\
\frac{k(k+1)}{2}+k+1&=\frac{(k+1)(k+2)}{2}\\
\frac{k(k+1)+2(k+1)}{2}&=\frac{(k+1)(k+2)}{2}\\
\frac{(k+1)(k+2)}{2}&=\frac{(k+1)(k+2)}{2}\\
(k+1)(k+2)&=(k+1)(k+2)\\
k+1&=k+1\\
k&=k\\
0&=0\text{ or }1=1
\end{align}$$

Therefore, $k+1$ is valid in this formula.

Is this right or am I making a mistake somewhere?

The second way is correct but I would be more carefull and use like this:

$$1+2+...+k+(k+1)=\frac{k(k+1)}{2}+k+1=\frac{k^2+3k+2}{2}=\frac{(k+1)(k+2)}{2}$$

P.s: Not use the equality in the first place. The problem using the equality is that you need guarantee equivalence in every step what is not so easy in many problems.

Sua ultima afirmaçao esta errada. Parece que voce nao entendeu bem a "hipotese de induçao". "Se vale para k+1 entao valera para k".


Calculating the definite integral $\int^a_{-a} e^{-x^2} \, dx$ After I learnt about this proof that $$\int^\infty_{-\infty} e^{-x^2} dx=\sqrt\pi$$I wondered if it could be applied to calculate the more general case $$\int^a_{-a} e^{-x^2} dx$$ for real $a\ge0$. Following the proof, we have $$\left(\int^a_{-a} e^{-x^2} dx\right)^2=\int^a_{-a}e^{-x^2}dx\int^a_{-a}e^{-y^2}dy$$$$=\int^a_{-a}\int^a_{-a}e^{-x^2+y^2}dxdy$$$$2\pi\int^a_0re^{-r^2}dr, r=x^2+y^2$$ $$=-\pi\int^{r=a}_{r=0}e^sds,s=-r^2$$ $$=-\pi\left|e^{-r^2}\right|^{a}_0$$ $$=\pi\left(1-e^{-a^2}\right)$$ Then, $$\int^a_{-a} e^{-x^2} dx=\sqrt{\pi\left(1-e^{-a^2}\right)}$$

But when I tried to plot $\sqrt{\pi\left(1-e^{-a^2}\right)}$ and $\int^a_{-a} e^{-x^2} dx$, there are some discrepancies:

$\sqrt{\pi\left(1-e^{-a^2}\right)}$ and $\int^a_{-a} e^{-x^2} dx$">

As you can see, $\sqrt{\pi\left(1-e^{-a^2}\right)}=\int^a_{-a} e^{-x^2} dx$ only when $x=0$ and (asymtopically) $x=\infty$. Why do the two functions deviate from each other?

This approach doesn't work. The set $\{re^{i\theta} : 0\leq r \leq a, \theta \in [0,2\pi]\}$ is not the same as $\{x+iy, -a\leq x \leq a, -a\leq y \leq a\}$. The first one a disk and the second one a rectangle.

$$\left(\int^a_{-a} e^{-x^2} dx\right)^2=2\pi\int^a_0re^{-r^2}dr$$

No, this is not true.

In general, because the function $x\mapsto e^{-x^2}$ is even, we know that  $$\int_{-a}^ae^{-x^2}dx = 2\cdot \int_0^a e^{-x^2} dx$$

which means that finding a closed form for one expression is equivalent to finding a closed form for the other expression.

It is also a well known result in mathematics that the integral of the function $x\mapsto e^{-x^2}$ is not elementary, which means that it cannot be written in closed form.


Proof concerning outer measure Assume that $X:=[a,b]$ is a fixed interval in $\mathbb R$ and let $m^*$ be the outer measure on $X$. Suppose that $A \subset X$ is a null set, i.e. $m^*(A)=0.$ Show that for every $B\subset X$,

$m^*(B \cup A)=m^*(B\setminus A)=m^*(B)$

I'm studying measure theory and looking for some direction on this proof. Thanks!

Take my hint from above if you would like just that. Below is a more complete solution.

$m^*(B\setminus A)\leq m^*(B)$ by monotonicity. $m^*(B)\leq m^*(B\cup A)$ by the same. $m^*(B\cup A)\leq m^*(B\setminus A)+m^*(A)$ by subadditivity.  Hence:
$$
m^*(B\setminus A)\leq m^*(B)\leq m^*(B\cup A)\leq m^*(B\setminus A)+m^*(A)=m^*(B\setminus A).
$$
So all the inequalities collapse into equalities.

$B \cup A = (B \setminus A) \cup A $, that is, it is the union of two disjoint sets so
$$m^*(B \cup A) = m^* (B \setminus A) + m^*(A)= m^*(B \setminus A). $$

Now since $A$ is a null set, $ B \setminus A= B$, hence $m^* (B \cup A)= m^*(B)$.


If $ \vert z \vert = a $ then what is the real part of $ ( \frac{z-a}{z+a} ) $ I am beginner to complex numbers.
I am having no idea to solve this:
Let $ z $ be any non - zero complex number.
If $ \vert z \vert = a $ , then what is the real part of $ ( \frac{z-a}{z+a} ) $ ?

"a" is greater than 0.

Use $Re(w) = \frac {w + \bar{w}} 2, w \in \mathbb{C}$ for $w=\frac{z-a}{z+a}$ then make the calculus.

You'll get something like: $\frac 1 2 (\frac{z-a}{z+a} + \frac{\bar{z}-a}{\bar{z}+a}) = \frac 1 2 (\frac {a^2 + za-a\bar{z} - a^2 + a^2 - za+a\bar{z} - a^2}{(z+a)(\bar{z} + a)}) = 0$

WLOG $z=ae^{2it}$ where $t$ is real

$$\dfrac{z-a}{z+a}=\dfrac{e^{2it}-1}{e^{2it}+1}=\dfrac{e^{it}-e^{-it}}{e^{it}+e^{-it}}=\dfrac{2i\sin t}{\cos t}=?$$


Minimum for this function I thought of writing this question Minimum for this function in a different way, if it helps.

I want to minimize
 $$\sum_{i=1}^n a_ix_i  + \nu \sum_{i=1}^n b_i 2^{x_i} ,$$
where $a_i \in [0,1]$, $b_i \in (0,\infty)$, and $x_i \in [x_{\min},0)$, and
$$ \sum_{i=1}^n 2^{x_i} = 1 .$$

$x_{\min}$, $\nu$, $a_i$  and  $b_i$ are constants.

I guess the tricky part is to minimize the function while ensuring all $2^{x_i}$ sum up to $1$ for these $n$ variables.

Thank you for your patience and help.

First let $x = (x_1, \dots, x_n)$ and write $f(x) = \sum_i^n a_i x_i + \nu\sum_i^n b_i 2^{x_i}$ and $g(x) = \sum_i^n 2^{x_i} - 1$. Now the method of Lagrange's multipliers tells you that you have a solution $x'$ iff it is an extremal point of the function $f(x) + \lambda g(x)$ with $\lambda$ a parameter to be determined. So you get equations $a_i + (\nu b_i + \lambda) 2^{x'_i} \log{2} = 0$ for all $1 \leq i \leq n$. It's now easy to express $x'_i$ as a function of lambda and plugging this into the remaining constraint $g(x') = 0$ to find the lambda and complete the solution. Hope this helps.

I don't think rewriting this way helps. In the first place, how would you show that if $y$ is the optimal solution of $\min_y f(y)$, then $x = 2^y$ is the optimal solution to $\min_x f(\log x)$?


How many $3$ digit numbers with digits $a$,$b$ and $c$ have $a=b+c$ My question is simple to state but (seemingly) hard to answer. How many $3$ digit numbers exist such that $1$ digit is the sum of the other $2$. I have no idea how to calculate this number, but I hope there is a simple way to calculate it. Thank you in advance.

EDIT: The first digit should not be $0$

Parameterize those numbers by $\overline {abc}$.

We have three cases, one when the first digit equals the sum of the other two. The second, when the second digit equals the sum of the other two, and so on...

Case I

$\overline {abc}$, $a=b+c$

when $b$ is $0$, $c$ can be $\{1,2,...,9\}$

when   $b$ is $1$, $c$ can be $\{0,1,2,3,...,8\}$ and so on...

$9+9+8+...+1=55-1=54$

Case II

$\overline {bac}$, $a=b+c$

when $b$ is $1$, $c$ can be $\{0,1,...,8\}$

when $b$ is $2$, $c$ can be $\{0,1,...,7\}$ and so on...

Now, we've added some numbers here that we've also added in the previous case. Namely, the ones for which $b=a+c$ is true. Since we've added them in this case $b=a-c$ is also true, so $c$ must be $0$. $c$ is $0$ only when $b$ is $\{1,2,...,9\}$. So the final answer for this case is $45-9=36$.

Case III

$\overline {bca}$, $a=b+c$

when $b$ is $1$, $c$ can be $\{0,1,...,8\}$

when $b$ is $2$, $c$ can be $\{0,1,...,7\}$ and so on..

Here we've added again some numbers, namely, for which $b$ is equal $a+c$, in step one. Note that we did not add any numbers that we added in case $2$ since, for those numbers we would have $c=b+a$(from case $2$) and also $c=a-b$(from step $3$). That would make $b$ $0$.(and we did not count that).

So, the final answer is $45-9=36$, for this case.

In total, we have $126$ numbers with that property.

Those numbers are:
$$\begin{array}{|c|}
\hline
101&
110&
112&
121&
123&
132&
134&
143&
145&
154&
156&
165&
167&
176&
178&
187&\\\hline
189&
198&
202&
211&
213&
220&
224&
231&
235&
242&
246&
253&
257&
264&
268&
275&\\\hline
279&
286&
297&
303&
312&
314&
321&
325&
330&
336&
341&
347&
352&
358&
363&
369&\\\hline
374&
385&
396&
404&
413&
415&
422&
426&
431&
437&
440&
448&
451&
459&
462&
473&\\\hline
484&
495&
505&
514&
516&
523&
527&
532&
538&
541&
549&
550&
561&
572&
583&
594&\\\hline
606&
615&
617&
624&
628&
633&
639&
642&
651&
660&
671&
682&
693&
707&
716&
718&\\\hline
725&
729&
734&
743&
752&
761&
770&
781&
792&
808&
817&
819&
826&
835&
844&
853&\\\hline
862&
871&
880&
891&
909&
918&
927&
936&
945&
954&
963&
972&
981&
990&\\\hline
\end{array}
$$

For verification, the code is:

 #include<iostream>

 using namespace std;

 int main() {
         int q=0;
         for(int a=1;a<=9;a++)
             for(int b=0;b<=9;b++)
                 for(int c=0;c<=9;c++)
                         if(a==b+c || b==a+c || c==a+b)

{q++;cout<<a<<b<<c<<"\n";}

 cout<<q; return 0; }

This, in case you do not allow $2-digit$ or $1-digit$ numbers.

For some, this might not be the most intuitive way, but it is a way to solve this problem.

If $a=b+c$, then if a is an even number it gives us 2 exceptions, and if a is odd it gives 1 exception. Will keep exceptions in a seperate bucket.

The maximum sum that can be reached is 9.
So if $(b,c,a)$ represent the set of possible number then,
$(1,8,9),(2,7,9),(3,6,9),(4,5,9)$ and then $b,c$ change place. That's $4\times 2 =8$ numbers. And since places are not fixed, therefore its $8\times3=24$ numbers + 1 exception $(909)$

Now, For 8  $(1,7,8),(2,6,8),(3,5,8)$ that's $3\times2\times3=18$ numbers +2 exceptions $(448,808)$

For 7  $(1,6,7),(2,5,7),(3,4,7)$ that's $3\times2\times3=18$ numbers +1 exception $(707)$

For 6  $(1,5,6),(2,4,6)$ that's $2\times2\times3=12$ numbers +2 exception $(606,336)$

For 5  $(1,4,5),(2,3,5)$ that's $2\times2\times3=12$ numbers +1 exception $(505)$

For 4  $(1,3,4)$ that's $2\times1\times3=6$ numbers +2 exception $(404,224)$

For 3  $(1,2,3)$ that's $2\times1\times3=6$ numbers +1 exception $(303)$

So far, 96 numbers. Now, lets look at the combinations of number put in exception and will count and add seperately.
$(909,990,448,844,484,808,880,707,770,606,660,336,363,633,112,121,211,202,220,101,110)$

Total count of number :$117$


Permutation and combination with overcounting A lady gives a dinner party to six quests. The number of ways in which they may be selected from among ten friends, if two of the friends will not attend the party together is
(a) $112$     (b) $140$
(c) $164$      (d) None of these

what i tried:
 Since two friend will not attend party together i considered them as one, so now there 9 friends.
So the number of ways $6$ people are selected from $9$ people=$\frac{9\times8\times7\times6\times5\times4}{6\times5\times4\times3\times2\times1}$=$84$

So the number of ways $6$ people are selected from $9$ people=$84$
but we have considered two people as one because either will be attending party

So the actual number of arrangements with two friends not attending party together=$84$$\times$$2$=$168$
So is the Answer $168$ i.e None of these

First off, for the arrangements that do not include the rival, you should see how many parties you can invite from the 8 remaining.

Then you need to see how many ways you can match up Rival #1 with 5 people from the remaining 8.  Then how many ways can you match up Rival #2 with 5 people from the remaining 8.

ETA: Probably easier method would be ${10\choose 6}-{8\choose 4}$ which is total parties of 6 minus parties containing both rivals.

If neither of the two unamicable friends don't attend the party, we choose $\binom{8}{6} = 28$.

Now if either of them is selected for the party (but not the other), we have $ \binom{2}{1}\cdot \binom{8}{5} = 112$.

Thus the answer will be $112+28= 140$.


Why does Stolz- Cesaro fail to evaluate the limit of $\dfrac{n + n^2 + n^3 + n^4 + \ldots + n^n}{1^n + 2^n + 3^n + 4^n + \ldots +n^n}$, I need to find the limit of the sequence

$\dfrac{n + n^2 + n^3 + n^4  + \ldots + n^n}{1^n + 2^n + 3^n + 4^n + \ldots +n^n}$,

My strategy is to use Stolz's Cesaro theorem for this sequence.

Now, the numerator is given by :

$x_r = n^1+ n^2 +n^3 + \ldots +n^r$, so $x_{n+1} - x_{n} = n^{n+1}$

Similarly for denominator
$y_r = 1^n + 2^n + 3^n +\ldots +r^n$, so $y_{n+1}- y_{n} = (n+1)^n$

Using Stolz Cesaro, this limit is equivalent to

$\displaystyle \lim \dfrac{n^{n+1}}{(n + 1)^n}$, which diverges to $ +\infty$,

However ans given to me is $\dfrac{e-1}{e}$, Can anyone tell where is the error in my solution ?

Thanks.

Note that, as mentioned in the comments below, your computation of the ratio is incorrect. Regardless, the hypothesis of Stolz-Cesaro assumes that the limit $\lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n}$ exists. If it doesn't exist, it does not imply that the original limit does not exist.

A better way to approach is to write it as follows:
$$
\frac{n + n^2 + \cdots + n^n}{1^n + 2^n + \cdots +n^n} = \frac{n^{-(n-1)} + n^{-(n-2)} + \cdots + n^{-1} + 1}{\left(\frac{1}{n}\right)^n + \left(\frac{2}{n}\right)^n  + \cdots + \left(\frac{n-1}{n}\right)^n + 1}
$$
As $n \to \infty$, clearly the numerator $\to 1$. For the denominator, see this.

If you approximate the sums with the corresponding integrals it converges to 0
$$
L = \frac{\int_{1}^{n}n^x dx}{\int_{1}^{n} x^n dx} = \frac{n^{n+1} + n^n }{n^{n+1}\log n - \log n} = 0
$$


Using central limit theorem for normal distribution So i have the following question:

The times that patients spend in a doctor’s surgery have mean 5 minutes, and standard
deviation 2 minutes. On one particular day, the doctor sees 30 patients during his surgery
which starts at 4.30pm. Find an approximate probability that she finishes with her last
patient before 6.50pm. State clearly any assumptions you need to make.

I have so far got to the following:
Let X be the time (in minutes) spent at the surgery. From the question we have $\text{E}(X) = 5$
and $\text{Var}(X) = 4$.
Using a random samply 30 from $X_1, X_2, \ldots , X_{30}$.

Let $$Y =
\sum_{i=1}^{30}
X_i$$
By the Central Limit Theorem $Y ∼ N(\mu = 150, \sigma^2 = 120)$ approximately.
As surgery starts at 4 : 30, and the doctor desires to finish surgery at 6 : 50, we desire
$P(Y < 140)$

However I am not sure where to go from here? In the answer it says $P(Y < 140)=0.1807$ could someone explain how this was calculated? Thanks

Perhaps a statistician might be able to offer some insight, but I believe your application of the CLT as stated in the question is wrong. This question defines the CLT as follows:

  If $X_{1}, ... , X_{n}$ are i.i.d continuous random variables with
  mean $\mu$ and variance $\sigma^2$, then as $n \rightarrow \infty$,
  $$ \sqrt{n} \frac{\bar{X} - \mu}{\sigma} $$ will have the standard
  normal distribution

It seems like you're off by a factor of $\sqrt{n}$ somewhere. Let's break it down. First, the empirical mean of your sample, $\bar{X}$, is defined as:

$$\bar{X} = \frac{\sum_{i=1}^{30}
X_i}{30}$$

Applying the CLT, as above:

$$\sqrt{n} \frac{\bar{X} - \mu}{\sigma} = \sqrt{30} \frac{\frac{\sum_{i=1}^{30}
X_i}{30} - 5}{2} \approx N(\mu = 0, \sigma^2 = 1)$$

Now, we're interested in $P(Y<140)$, where $Y$ is defined as:

$Y = \sum_{i=1}^{30}{X_i} = 30\bar{X}$

So:
$$P(Y<140) = P(\bar{X}<\frac{14}{3})$$

Now, rearranging our normal approximation identity from above, with $n = 30$

$$\sqrt{30}(\bar{X} - \mu) \approx 2 N(\mu = 0, \sigma^2 = 1)$$
$$(\bar{X} - \mu) \approx \dfrac{2 N(\mu = 0, \sigma^2 = 1)}{\sqrt{30}}$$
$$\bar{X} \approx \dfrac{2 N(\mu = 0, \sigma^2 = 1)}{\sqrt{30}} + 5$$

So:

$$P(\bar{X}<\frac{14}{3}) \approx P(\dfrac{2 N(\mu = 0, \sigma^2 = 1)}{\sqrt{30}} + 5 < \frac{14}{3})$$

Rearranging again we have:

$$P(N(\mu = 0, \sigma^2 = 1) < \frac{\sqrt{30}(\frac{14}{3} - 5))}{2})$$

Crunching out the RHS, this boils down to:

$$P(N(\mu = 0, \sigma^2 = 1) < -0.91287092917)$$

Now that the LHS is your standard normal distribution, you can plug the RHS number into any bog-standard cumulative probability calculator such as https://stattrek.com/online-calculator/normal.aspx and you get a cumulative probability of $0.181$, which is the same as that quoted in the question up to a rounding error.

Assumptions

The Q asks to state assumptions. The assumptions we are making are:

In order to apply the CLT as above, we have to assume that the variables $X_i$ are indeed all independent and are indeed all identically distributed. This is a big assumption. Because in reality, if one patient takes longer, a doctor might rush the next one, for example.
We assume that a sample of $30$ patients is sufficiently large to apply the CLT, but in theory the CLT only guarantees us a normal distribution in the infinite limit. We are assuming that our underlying distributions for patient wait time are sufficiently "well behaved" that the convergence guaranteed by the CLT is enough to give us a good approximation with sample size $30$. However, in general, we can't always make that assumption. Some distributions are so skewed that we need many more than $30$ terms in our sum in order for the resulting distribution to start looking normal.

Common Sense Check

Does that answer make sense? Well, yeah. If everyone on average takes 5 mins, and you have 30 people, you'd expect them to take $150$ mins. $140$ mins is ten mins quicker than that, and this calculation gives that an $18\%$ chance of that happening, which seems pretty reasonable to me.

Explanation of Strategy

In general in these types of problems, you want applly the CLT to allow you to restate the problem in terms of the standard normal distribution, which you can then use to get your probabilities easily using standard tables or a calculator like I did.


Is there any commonality between Math induction and Logic induction? Logic induction is reasoning by probability.

Math induction seems to be related to just Natural numbers and used to prove a statement for every natural number.

From these definitions there is no relationship between the two concepts or are there details within each of these concepts that are common ?

Update :

Ref : https://www.youtube.com/watch?v=0F8IDj6I-tM&list=PLS8vfA_ckeuZ9UjAHhA1q-ROZGuE_h21V&index=3 at time 4:42

screenshot :

From the viewpoint of modern logic, no, they are just different things.  There is an historical connection though.  Historically, for some logicians, "induction" was inference from special cases to a general rule.  This is related to the definition you found, in that some logicians would say that special cases can make the general rule probable though they can not prove it deductively.

On the other hand, mathematical induction was so-called by Augustus de Morgan because he saw it as getting a general rule from successive cases.  See Gerald Edgar's answer to https://mathoverflow.net/questions/24102/historically-first-uses-of-mathematical-induction

Contrary to what a lot of people think, mathematical induction is not an inference made from infinitely many cases.  It is not the idea that given $P(0)$ and $P(1)$ and $P(2)$ and so on through $P(n)$ for every specific natural number $n$, you can infer "for all natural numbers $n$, $P(n)$.'' In fact that principle is called Hilbert's Omega rule, and it does not hold in  Peano Arithmetic, or in ZFC, or indeed in any consistent recursively axiomatized theory.

But mathematical induction does hold in those theories.  It is a principle about what you can infer given two things: $P(0)$ and a proof for an indeterminate $n$ that "If $P(n)$ then $P(n+1)$.''  Not infinitely many different concrete numbers $n$, but a single indeterminate $n$, logically speaking a variable.  Peano Arithmetic and ZFC both let you infer "for all natural numbers $n$, $P(n)$'' from those.

To see this intuitively think of  G\"odel's second incompleteness theorem.  If PA is consistent, then for every concretely given natural number $n$ it has to prove "$n$ does not code a proof in PA of $0=1$.''  That is because each concrete case can be checked concretely (given enough time).  But the different cases have little to do with each other.  The assumption that $n$ does not code such a proof gives you no leg up on showing that $n+1$ does not.  You cannot organize the reasoning to work for an actually indeterminate $n$.  And so PA does not prove "for all $n$, $n$ does not code a proof of $0=1$.

To clarify the usage: "mathematical induction" is normally understood the way the OP uses it.  It is reasoning  on the natural numbers, by proving a base case of $n=0$ and then an induction step where you assume the conclusion for an indeterminate $n$ and prove it for $n+1$. It is taken as an axiom scheme in formalized arithmetic, and (in one form or another, depending on details) is a part of the definition of the set of natural numbers in set theory.

This is far from the only kind of induction used in mathematics.  Other prominent notions of induction are transfinite induction used on transfinite ordinals in set theory, epsilon-induction used in Zermelo-Fraenkel set theory, and Noetherian induction used on ideals of Noetherian rings (or on open subsets of Noetherian spaces).  There is even a general theory of "well-founded relations," which in brief means relations that support a generalized concept of inductive proof.  These are well explained many places on line.

None of those senses of induction involve any notion of probability.  Nor can standard uses of "inductive reasoning" in probability be justified by somehow reducing them to these.  But there is the historical link described above.

I'll first indicate that mathematical induction does appear outside the natural numbers.  For instance, common proofs of the deduction meta-theorem use mathematical induction.

Both mathematical induction and logic induction start with at least one thing satisfying a property.

To use the example of the video, if I reason using logic induction one might argue as follows:

Base cases: The Sun rose the day before yesterday, yesterday, and today.

Therefore, the Sun will rise tommorow.

If I use mathematical induction I might write a proof like the following:

Base Cases: 0 is a natural number.  And 0 = 0.  1 = 0, and 2 = 0.

Induction case: The induction step involves us proving that for a natural number called "num", if num = 0, then (num + 1) = 0.  So, let's assume that num = 0.  Since num = 0, (num + 1) = (0 + 1).  And (0 + 1) = 1.  Thus, (num + 1) = 1.  By the second base case above, that is 1 = 0, it follows that (num + 1) = 0.  Consequently, we've proved that the inductive schema holds.

Therefore, the successor of any natural number equals 0.  Since every natural number equals the successor of some other natural number, all natural numbers equal 0.

Of course that isn't correct, because it isn't the case that 1 = 0 or 2 = 0.  But I'll also point out that the Sun doesn't rise.. the Earth turns inward towards the Sun!


A problem of limit: $\lim\limits_{n\to \infty}\left\{\frac{(n+1)^{n}}{n!} \right\}^{(1/n)}$ I need to find the limit of $$\lim\limits_{n\to \infty}\left\{\left(\frac{2}1\right) \left(\frac{3}2\right)^{2} \left(\frac{4}3\right)^{3} ... \left(\frac{n+1}n\right)^{n} \right\}^{(1/n)}$$

Cancelling out the same quantities in numerators and denominators, I have reached the step: $$\lim\limits_{n\to \infty}\left\{\frac{(n+1)^{n}}{n!} \right\}^{\frac1n}$$

Now I am stuck here. I have searched google for help, but found only the result for $\lim\limits_{n\to\infty}\left\{\frac{1}{n!} \right\}^{(\frac1n)}$.

Which formulae/properties should I use now to proceed from this stage?

Edit: The options for the answer are: a) $e$, b) $1/e$, c) $\pi$, d) $1/\pi$

Taking logarithm on both the sides,
we get $$\log(y_n)=\log((\frac{(n+1)^n}{n!})^{\frac{1}{n}})$$
$$=\frac{1}{n}\{\log\frac{2}{1}+2\log\frac{3}{2}+...+n\log\frac{n+1}{n}\}.$$
Using Cauchy's First theorem on limits
$$\lim_{n\to\infty}\frac{1}{n}\{\log\frac{2}{1}+2\log\frac{3}{2}+...+n\log\frac{n+1}{n}\}=\lim_{n\to\infty}n\log\frac{n+1}{n}=1$$

Well, $e$ is the limit of $\left(\frac{(n+1)^n}{n!}\right)^{1/n}$ as $n$ goes to infinity. Indeed:

By Stirling's Approximation, $n!$ is $(1+o(1)\left(\frac{n}{e}\right)^n\sqrt{2\pi n}$. So $\left(\frac{(n+1)^n}{n!}\right)^{1/n}$ is

$[\sqrt{2\pi n}]^{1/n} \left(\left(\frac{n}{e(n+1)}\right)^n\right)^{1/n}$ which is $e^{-1}$.

Was that the question that was meant to be asked?


Find the last two digit of $3^{3^{100}}$. Find the last two digit of $3^{3^{100}}$.
I know how to calculate if I have $3^{100}$. That I will use euler's theorem. which gives me $3^{40}\equiv 1 \pmod{100}$. And so on... but if I have $3^{3^{100}}$ what should I do?
I Tried:
Infact, i need $3^{3^{100}}\equiv x\pmod{ 100}$
For this, I need $3^{100}\equiv y\pmod{\phi{(100)}}$
So i got $y=1$ by using eulers thorem for the abouve cogruence.
That is $(3,40)=1\implies 3^{16}\equiv 1\pmod{40}$
So, i got $(3^{16})^63^4\equiv 1\pmod{40}$
Using this in first congruence i got $3^1\equiv 3\pmod{100}$.
So, the answer is $03$. Is it correct?
I did't use chinese remainder theorem and all. Is there any mistake in my arguments?

If we can find $3^{3^{100}}$ in mods 4 and 25 we're done by the Chinese Remainder Theorem.

Mod 4: Note that $3\equiv -1\pmod 4$, hence $3^{3^{100}}\equiv (-1)^{3^{100}}\equiv (-1)^{2k+1}\equiv -1\equiv 3\pmod 4$.

Mod 25: Since $\varphi(25)=20$ we have $3^{3^{100}}\equiv 3^{3^{100}\pmod {20}}\pmod {25}$. Similarly, $3^{100}\equiv 3^{100\pmod 8}\pmod {20}$, and $100\equiv 4\pmod 8$. Hence, working backwards we get:

$3^{100}\equiv 3^{100\pmod 8}\equiv 3^4\equiv 1\pmod {20}$,

$3^{3^{100}}\equiv 3^{3^{100}\pmod {20}}\equiv 3^1\equiv 3\pmod {25}$.

Thus, threading the congruences together we have:

$3^{3^{100}}=4a+3=25b+3$, or $4a=25b$. So since $\gcd(4,25)=1$ we need $a=25k$. Therefore $3^{3^{100}}=4(25k)+3=100k+3$, so the answer is $\boxed{03}$.


Prove that $\lim _{x \to \infty} x\sin x$ doesn't exist (using delta epsilon). Question:

Prove that $\lim _{x \to \infty} x\sin x$ doesn't exist (using delta epsilon).

What I did:
I've been struggling with this one for a long time. Really tried digging up the net for explanations, but couldn't find a really good source to explain how this proof should be done(some said proof by contradiction like I did, and some say without contradiction). I got too confused by now, I also read this Proving limit doesn't exist using the $\epsilon$-$\delta$ definition
which gave some guidance but I thought I'd use the floor function to make sure M is natural, otherwise the idea of using these values wouldn't have worked.
Anyway I'd really love someone to verify/correct this solution before I hand it in.
Thanks a lot.

We assume there exists a limit, therefore:
$\forall \varepsilon>0 \exists M>0: x>M \Rightarrow |f(x)-L|<\varepsilon$
and specifically, if we pick some $x_1,x_2>M$: $|f(x_1)-L|<\frac \varepsilon2,
|f(x_2)-L|<\frac \varepsilon2$
Also, this applies for all $\varepsilon$, and specifically for  $\varepsilon=0.5$
(1) Therefore using the triangle inequality: $|f(x_1)-f(x_2)| = |f(x_1)-L+L-f(x_2)| \le |f(x_1)-L |  + |f(x_2)-L| \le \frac \varepsilon2 +  \frac \varepsilon2 = \varepsilon$
This applies for all $x>M$ so specifically for: $x_1=\frac {\pi \lfloor M+1 \rfloor}{2} > M , x_2=\pi \lfloor M+1 \rfloor$
We put these values in (1): $|f(x_1)-f(x_2)|=|\frac {\pi \lfloor M+1 \rfloor}{2}\sin\frac {\pi \lfloor M+1 \rfloor}{2}-\pi \lfloor M+1 \rfloor \sin (\pi \lfloor M+1 \rfloor)|= \begin{cases} 1 & \lfloor M+1 \rfloor   odd  \\ 0 &\lfloor M+1 \rfloor  even\end {cases} $
So we have found an $\varepsilon $ that doesn't always satisfy the requirements of the limit definition so this is a contradiction. Therefore the limit doesn't exist.

Solution by means of sequences:

$$x_n:=\pi n\implies x_n\sin x_n=0\xrightarrow[n\to\infty]{}0$$

$$y_n:=\frac{(4n-3)\pi}2\implies y_n\sin y_n=\frac{(4n-3)\pi}2\xrightarrow[n\to\infty]{}\infty$$

you can always choose $x=k\pi$, then you can show by the $\epsilon-\delta$ language that if the limit exists, it must be $0$. But then, you choose $x=2k\pi+\frac\pi2$, which shows that the limit might not be $0$.
Ok, I have used the uniqueness of limits in $R$, but somehow you can always choose some special $x$, which makes $sinx$ become a constant to solve this question.