How to prove that eigenvectors from different eigenvalues are linearly independent How can I prove that if I have $n$ eigenvectors from different eigenvalues, they are all linearly independent?

I'll do it with two vectors. I'll leave it to you do it in general.

Suppose $\mathbf{v}_1$ and $\mathbf{v}_2$ correspond to distinct eigenvalues $\lambda_1$ and $\lambda_2$, respectively.

Take a linear combination that is equal to $0$, $\alpha_1\mathbf{v}_1+\alpha_2\mathbf{v}_2 = \mathbf{0}$. We need to show that $\alpha_1=\alpha_2=0$.

Applying $T$ to both sides, we get
$$\mathbf{0} = T(\mathbf{0}) = T(\alpha_1\mathbf{v}_1+\alpha_2\mathbf{v}_2) = \alpha_1\lambda_1\mathbf{v}_1 + \alpha_2\lambda_2\mathbf{v}_2.$$
Now, instead, multiply the original equation by $\lambda_1$:
$$\mathbf{0} = \lambda_1\alpha_1\mathbf{v}_1 + \lambda_1\alpha_2\mathbf{v}_2.$$
Now take the two equations,
$$\begin{align*}
\mathbf{0} &= \alpha_1\lambda_1\mathbf{v}_1 + \alpha_2\lambda_2\mathbf{v}_2\\
\mathbf{0} &= \alpha_1\lambda_1\mathbf{v}_1 + \alpha_2\lambda_1\mathbf{v}_2
\end{align*}$$
and taking the difference, we get:
$$\mathbf{0} = 0\mathbf{v}_1 + \alpha_2(\lambda_2-\lambda_1)\mathbf{v}_2 = \alpha_2(\lambda_2-\lambda_1)\mathbf{v}_2.$$

Since $\lambda_2-\lambda_1\neq 0$, and since $\mathbf{v}_2\neq\mathbf{0}$ (because $\mathbf{v}_2$ is an eigenvector), then $\alpha_2=0$. Using this on the original linear combination $\mathbf{0} = \alpha_1\mathbf{v}_1 + \alpha_2\mathbf{v}_2$, we conclude that $\alpha_1=0$ as well (since $\mathbf{v}_1\neq\mathbf{0}$).

So $\mathbf{v}_1$ and $\mathbf{v}_2$ are linearly independent.

Now try using induction on $n$ for the general case.

Suppose $\sum_k c_k v_k=0$ and $Av_k=\lambda v_k$. We only include in this sum eigenvectors corresponding to distinct eigenvalues, so that $\lambda_i\neq\lambda_j$ for $i\neq j$. We want to prove that this implies $c_k=0$.
Define the operators
$$A^{(i)}\equiv \prod_{k\neq i}(A-\lambda_k I).$$
Then $A^{(i)}\sum_k c_k v_k=d_i c_i v_i$ where
$d_i\equiv \prod_{k\neq i}(\lambda_i-\lambda_k)\neq0$.
It follows that $c_i=0$ for all $i$, and thus the vectors $\{v_i\}_i$ are linearly independent.


Obstruction to the splitness of short exact sequence in the category groups Let $1 \to K \to G \to H \to 1$ be a short exact sequence in the category of groups(interested in non-Abelian groups). My question is the following:

Where does the obstruction to the splitting of the above short exact sequence lie?

Any help will be appreciated.

Thank you so much.

The main problem arising when you deal with anabelian groups sequences is that the you don't have an unique notion of splitness (recall that groups don't form an abelian category).
There are two "canonically split" exact sequences: the direct product exact sequence:
$$1\longrightarrow G\longrightarrow G\times H\longrightarrow H\longrightarrow 1$$
and the semi-direct product exact sequence:
$$1\longrightarrow G\longrightarrow G\rtimes H\longrightarrow H\longrightarrow 1$$
Splitness is usually understood by means of semi-direct product, since this notion is more general and flexible. The obstructions look like the same ones in the abelian case (existence of a section or a retract) but there are some subleties because you need to distinguish when an exact sequence is of "direct product" type or "semi-direct product" type. Here the main result:
Theorem. Let $1\to H\to G\to K\to 1$ a short exact sequence. The following are equivalent:

there exists a morphism $G\to H$ which is a left inverse for $H\to G$;
there exists an isomorphism $G\to H\times K$ which fits in the following commutative diagram:

Similarly, the following are equivalent:

there exists a morphism $K\to G$ which is a right inverse for $G\to K$;
there exists an isomorphism $G\to H\rtimes K$ which fits in the following commutative diagram:

For more detailed informations, you can read this nice introduction: http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/splittinggp.pdf


Show algebraically that the graph of $y=x^2 + kx - 2$ will cut the $x$-axis twice for all values of $k$ A quadratics question.

  Show algebraically that the graph of $y=x^2 + kx - 2$ will cut the $x$-axis twice for all values of $k.$

I recently asked a similar question, but this problem seems broader in scope.

Use the quadratic function, with a= 1, b= k, and c= -2.

There's three different outcomes for a quadratic:

${b^2-4ac}\gt 0 \rightarrow$ we get two solutions, which means we have two x-intercepts.

${b^2-4ac}=0 \rightarrow$ we get one solution, which means we have one x-intercept

${b^2-4ac}\lt 0 \rightarrow$ no real solutions, which means no x-intercepts. You can think of the parabola of the axis being above the x-axis if it opens upwards, or below the x-axis if it opens downwards.

for graph Y = x^2 + Kx -2 to cut x - axis y coordinate must be zero thus,

x^2 + Kx -2 = 0

Now for this equation for solution to be finitly 2 , b^2 - 4ac > 0.

here b = K , a = 1 and c = -2.

for every value of K , b^2 - 4ac will be greater than zero.

for example k = 0 => (0)^2 - 4(1)(-2) = 8

for k = -1 => (-1)^2 - 4(1)(-2) = 9

hence for all value of K it will be greater than zero so now we can find the solution of an equation.


Write the generating function of a sequence as the quotient of two polynomials I want to write the generating function of the sequence
$$a_0=2, \;a_1=3, \; a_{n}=2a_{n-1}-2a_{n-2} \; \text{for }n\geq 2$$
as the quotient of two polynomials. But how can I do that? What is the best way?

I don't think this is possible. If you write the characteristic equation, you get $\lambda^2=2\lambda-2$, so $(\lambda-1)^2=-1$ i.e. $\lambda=1\pm i$ so the generic solution is of the form $a_n=c_1(1+i)^n+c_2(1-i)^n$ hence $c_1+c_2=2$ and $c_1(1+i)+c_2(1-i)=3$ which solves to
$$a_n=\left(1-\frac{i}{2}\right)(1+i)^n+\left(1+\frac{i}{2}\right)(1-i)^n$$
which is exponential (not a rational function) in $n$.


Prove the sequence is Cauchy and therefore Convergent Let $(x_n)_{n\in\mathbb{N}}$ be a sequence in $\mathbb{R}$. Let $r\in (0,1)$ and suppose that $|x_{n+1}-x_n|<r^n \forall n\in \mathbb{N}$. Show that $(x_n)_{n\in\mathbb{N}}$ converges.

I know that it is sufficient to show that the sequence is Cauchy. So, this is very similar to this past question (Please help prove that $(x_n)$ is a Cauchy sequence if $|x_{n+1} - x_n| \leq Cr^n$) but I don't completely understand the answers given so I'm hoping someone can help me out.

Let $k > 0$ be arbitrary. Then, using the triangle inequality once,
$$
|x_{n+k}-x_n| = |x_{n+k}-x_{n+k-1}+x_{n+k-1} - x_n|\le|x_{n+k}-x_{n+k-1}|+|x_{n+k-1} - x_n|.
$$
Applying it again on the last term gives
\begin{align*}
|x_{n+k}-x_n|
&\le|x_{n+k}-x_{n+k-1}|+|x_{n+k-1} - x_n|\\
&\le |x_{n+k}-x_{n+k-1}| + |x_{n+k-1} - x_{n+k-2}| + |x_{n+k-2}-x_n|.
\end{align*}
Using it over and over again, you end up with
$$
|x_{n+k}-x_n|\le\sum_{l=1}^k|x_{n+l}-x_{n+l-1}|.
$$
Now, use what you know:
\begin{align*}
|x_{n+k}-x_n|
&\le\sum_{l=1}^k|x_{n+l}-x_{n+l-1}|\le\sum_{l=1}^k r^{n+l-1} = r^n\sum_{l=1}^k r^{l-1}\\
&= r^n\sum_{l=0}^{k-1} r^{l}\le r^n\sum_{l=0}^\infty r^{l} = \frac{r^n}{1-r}.
\end{align*}
In the last equality we used the geometric series. As $k$ was arbitrary, we obtain $|x_m-x_n| < \frac{r^n}{1-r}$ for each $m\ge n$. Therefore, $(x_n)$ is Cauchy.


Invariant submanifolds Let $M$ be a smooth manifold, and let $N$ be a submanifold.  Let $V$ be a smooth vector field on $M$ which generates a flow $\Phi_t$ on $M$.  My intuition tells me (perhaps modulo some technical assumptions) that the following is true:

  If $V(p)$ is tangent to $N$ for all $p\in N$, then $N$ is an invariant submanifold of $\Phi_t$.

Is this true? What sorts of technical assumptions would I need to worry about to make the statement rigorous?  I imagine, for example, that there could be global topological issues so that perhaps the statement only holds locally.

Is there a good (basic) reference on invariant submanifolds?

I am a physicist, not a mathematician, so my answer may lack some of the rigor you expect.  Still, since no one else has taken a stab at this problem in the last year, I will shed what little light I can.

The reference you are looking for may be Sophus Lie's 1884 Differential Invariant Paper.  I was introduced to it in Oak Ridge in 1992 by Dr. Lawrence Dresner, an applied mathematician in Magnetics Division of the Y-12 lab.  Lie's work was being used to solve nonlinear PDE pertaining to superconductor stability problems in the Tokamak fusion reactor.

Sophus Lie was an 18th century Norwegian mathematician, but his style was rooted in the 17th century, so his work can be somewhat dense.  I recommend the translation by M Ackerman with the comments and additional material by Robert Hermann.  It was published in English in 1976 by MATH SCI PRESS, ISBN 0-915692-13-9.  The original paper is "On Differential Invariants", S. Lie, Math. Annallen, Vol. 24 (1884), 537-578.  I am just a student, and not really qualified to speak with authority on this issue, but as I understand it Lie's basic premise is precisely your statement.

Sophus Lie was trying to do for differential equations what Evariste Galois did for polynomials.  A Lie Group is a group that preserves the structure of the smooth manifold.  Lie Group stabilizers, or "differential invariants" as he called them, form an embedded, invariant submanifold.  Because of this, DEQ's may be rewritten in terms of group stabilizers.  Because a function of stabilizers is itself a stabilizer, this puts the DEQ into the kernel of the map with the Lie algebra of the group which is used to find a solution. As Hermann says in the preface, "The key idea is that one should study the structure of the orbit space of a symmetry group on the space of solutions."

Again, pardon the lack of rigor, but perhaps you might benefit from a practical example. If a projectile is fired into a fluid the force of friction is proportional to the square of the velocity, so deriving an equation for how far a projectile penetrates a fluid as a function of time can be a challenge.
$$ F=-\alpha v^2 \rightarrow m\ddot{y}=-\alpha \dot{y}^2$$Here y is the penetration distance, m is mass and $\alpha$ is the drag coefficient, a unitless constant dependent on projectile shape and the density and viscosity of the fluid. ($\dot{y}=\frac{dy}{dt}$ and $\ddot{y}=\frac{d^2y}{dt^2}$) Also, y(0)=0 and $\dot{y}(0)=v_o$, the initial velocity of the projectile.

This DEQ is invariant to the Lie group G(t,y)=$(\lambda t, \lambda^\beta y)\lambda_o=1$  Note that if $t'=\lambda t$ and $y'=\lambda^\beta y$, $\dot{y}'=\frac{dy'}{dt'}=\frac{\lambda^\beta dy}{\lambda dt}=\lambda^{\beta -1}\dot{y}$ and $\ddot{y}'=\lambda^{\beta -2}\ddot{y}$.  Applying these to the DEQ,
$$m\lambda^{\beta -2}\ddot{y}=-\alpha \lambda^{2\beta -2}\dot{y}^2$$
For invariance, $\beta =0$.  Now find the infinitestimal transformations of the primed variables and use the method of characteristics to find the stabilizers (differential invariants).

$$\bigg(\frac{dt'}{d\lambda}\bigg)_{\lambda _o=1}=t$$
$$\bigg(\frac{dy'}{d\lambda}\bigg)_{\lambda _o=1}=\beta y$$
$$\bigg(\frac{d\dot{y}'}{d\lambda}\bigg)_{\lambda _o=1}=(\beta -1)\dot{y}$$
$$\bigg(\frac{d\ddot{y}'}{d\lambda}\bigg)_{\lambda _o=1}=(\beta -2)\ddot{y}$$
$$d\lambda=\frac{dt}{t}=\frac{dy}{\beta y}=\frac{d\dot{y}}{(\beta -1)\dot{y}}=\frac{d\ddot{y}}{(\beta -2)\ddot{y}}$$
$$\frac{dt}{t}=\frac{dy}{\beta y}\rightarrow \beta ln t=ln y + \mu \rightarrow \mu=\frac{y}{t^\beta}\bigg|_{\beta =0}=y$$
$$\frac{dt}{t}=\frac{d\dot{y}}{(\beta -1)\dot{y}} \rightarrow \nu=\frac{\dot{y}}{t^{\beta -1}}\bigg|_{\beta =0}=t\dot{y}$$
$$\frac{dt}{t}=\frac{d\ddot{y}}{(\beta -2)\ddot{y}} \rightarrow \eta=\frac{\ddot{y}}{t^{\beta -2}}\bigg|_{\beta =0}=t^2\ddot{y}$$
These constants of integration, $\mu$, $\nu$, and $\eta$, are group stabilizers for the group of differential equations, a subset of the group of polynomials.  There are an infinite number of them but we only need the first three for our DEQ.  These differential invariants form an embedded submanifold in the solution space, which means the DEQ may be rewritten in terms of the invariants.  Multiplying the DEQ by $t^2$, we have
$$mt^2\ddot{y}=-\alpha(t\dot{y})^2 \rightarrow m\eta=-\alpha \nu^2$$
Since $$t\frac{d\nu}{dt}=\eta-(\beta-1)\nu$$ we can use this expression of the Lie algebra to solve our DEQ.  $$t\frac{d\nu}{dt}=\nu -\frac{\alpha}{m}\nu^2 \rightarrow \frac{d\nu}{\nu-\frac{\alpha}{m}\nu^2}=\frac{dt}{t}$$
This separation of variables is a direct result of the group invariance.  With a little fractional decomposition, some integration, substitution of $\nu=t\dot{y}$, more integration and application of initial conditions, you get
$$y=\frac{m}{\alpha}ln\bigg(1+\frac{\alpha}{m}v_o t\bigg)
$$This solution is easily verified and provides accurate results.

Again, the applied math may not be exactly what you are looking for, but hopefully it will spark some thoughts.  On page 102 of the afore-mentioned book is Lie's Theorem 4.4.1.  "Every infinite continuous group determines an infinite sequence of differential invariants, which can be defined as the solutions of complete systems."  In the following comments by Hermann, he states "As far as I can tell, Lie (in the spirit of the 18th century) assumes that everything is suitably 'general', and that these facts are self evident.  As far as I know, they are not proved to this day!"  What was true for Hermann in 1976 may still hold.  The method obviously works so the proof should exist, but since nobody has answered your post in a year it may be possible that such a valuable proof is still out there waiting to be discovered.


Cauchy-Schwartz Inequality problem Let $a,$ $b,$ $c,$ $d,$ $e,$ $f$ be nonnegative real numbers.
(a) Prove that
$$(a^2 + b^2)^2 (c^4 + d^4)(e^4 + f^4) \ge (ace + bdf)^4.$$
(b) Prove that
$$(a^2 + b^2)(c^2 + d^2)(e^2 + f^2) \ge (ace + bdf)^2.$$
I'm not sure how I should start approaching both problems. I believe I should use Cauchy-Schwartz, but I'm not sure.
Any help would be appreciated! Thanks in advance.

These are both CS inequality applications, you should try yourself.  Here is the first one:

$(c^4+d^4)(e^4+f^4) \geqslant (c^2e^2+d^2f^2)^2$
$(a^2+b^2)(c^2e^2+d^2f^2) \geqslant (ace+bdf)^2$

Now combine the two to get what you want.

For (a),
Just use Holder's Inequality.
For (b),
Expand $(a^2 + b^2) (c^2 + d^2) (e^2 + f^2) $. (It is equivalent to the LHS of my answer)
Expand $(ace + bdf)^2$. (It is equivalent to the RHS of my answer)
The problem is just proving
$$
a^2 c^2 e^2 + a^2 c^2 f^2 + a^2 d^2 e^2 + a^2 d^2 f^2 + b^2 c^2 e^2 + b^2 c^2 f^2 + b^2 d^2 e^2 + b^2 d^2 f^2 \ge a^2 c^2 e^2 + b^2 d^2 f^2 + 2abcdef
$$ or
$$
a^2 c^2 f^2 + a^2 d^2 e^2 + a^2 d^2 f^2 + b^2 c^2 e^2 + b^2 c^2 f^2 + b^2 d^2 e^2 \ge 2abcdef \quad \textrm{By cancelling 2 terms on the left and right side.}
$$
Now, that is just
$$
(acf - bde)^2 + a^2 d^2 e^2 + a^2 d^2 f^2 + b^2 c^2 e^2 + b^2 c^2 f^2 \ge 0
$$
We have proved by this that the inequality is valid for any real numbers, not only non-negative ones.


How to Show that $1+\left(\left\lceil\dfrac{x}{n}\right\rceil -1\right)n\leq x$? For $x$ integer in $\{1,\ldots,n^2\}$. I would like to show that
$$
1+\left(\left\lceil\dfrac{x}{n}\right\rceil-1\right)n \leq x.
$$
I start by the property of the ceiling function which gives me
$$
1+\left(\left\lceil\dfrac{x}{n}\right\rceil-1\right)n <x+1.
$$

By the property of the ceiling function we have:

$$
\left\lceil\dfrac{x}{n}\right\rceil <\dfrac{x}{n}+1,
$$
which gives me:
$$
1+\left(\left\lceil\dfrac{x}{n}\right\rceil-1\right)n <x+1.
$$

What I missed which seems some kind of obvious is that $\left(1+\left(\left\lceil\dfrac{x}{n}\right\rceil-1\right)n\right)$ is a positive integer and since it is less than $x+1$, it must be less than or equal to $x$. Thus,

$$
1+\left(\left\lceil\dfrac{x}{n}\right\rceil-1\right)n \le x.
$$

Okay divide both sides by $n$, then:
$$\left\lceil\dfrac{x}{n}\right\rceil-1 \le \frac{x-1}{n}$$ then we got:
$$\left\lceil\dfrac{x}{n}\right\rceil = \frac{x+k}{n}$$ where $k$ some integer number.
So we have inequality with $k$ and $n$ :
$$n \ge k-1$$ , now think about what set of valid $k$ and you will prove it.


$\limsup\left(\frac{a_1+a_{n+1}}{a_n}\right)^n\ge c$ Let $a_n>0,n\in\mathbb{N}$ be a sequence of positive real numbers. There exists a positive real number $c$ such that $\limsup\left(\frac{a_1+a_{n+1}}{a_n}\right)^n\ge c$ as $n\to\infty$ for all $\{a_n\}$. Find with proof the maximum possible value of $c$.

I will summarize Polya's solution, i.e., a proof that

$$\limsup_{n \rightarrow \infty} \left ( \frac{a_1+a_{n+1}}{a_n} \right ) \ge e$$

Assume the opposite, i.e.,

$$\limsup_{n \rightarrow \infty} \left ( \frac{a_1+a_{n+1}}{a_n} \right ) < \lim_{n \rightarrow \infty} \left ( 1 + \frac{1}{n} \right )^n = e$$

Then

$$\limsup_{n \rightarrow \infty} \left [ \frac{n(a_1+a_{n+1})}{(n+1) a_n} \right ]^n < 1$$

which implies that $\exists N \in \mathbb{N} : \forall \, n > N$

$$ \frac{n(a_1+a_{n+1})}{(n+1) a_n} < 1 \implies \frac{a_1 + a_{n+1}}{n+1} < \frac{a_n}{n}$$

or

$$\frac{a_{n+1}}{n+1} - \frac{a_n}{n} < -\frac{a_1}{n+1}$$

We may sum terms like this from, say, $n=N$ to some $K>N$ and get

$$\frac{a_{K+1}}{K+1} - \frac{a_N}{N} < -a_1 \sum_{k=N}^{K} \frac{1}{k+1}$$

Note that the sum on the right goes to $-\infty$ as $K \rightarrow \infty$.  Thus,

$$\lim_{K \rightarrow \infty} \frac{a_{K+1}}{K+1} = -\infty$$

which is a contradiction of the fact that $a_K >0$.  Thus, the conjecture is proven.

Let me complete the answer of Ron Gordon:

Ron Gordon proved that: For any sequence $(a_n)$ with positive terms:
$$\limsup_{n\to\infty} \Big(\frac{a_1+a_{n+1}}{a_n}\Big)^n\geq c,$$
with $c=e$.

We were also asked for the maximum value $c_{max}$ of such a constant $c$, thus it remains to prove that $e$ is the largest constant $c$ for which this inequality holds for any sequence $(a_n)$ with positive terms.

Indeed by choosing a sequence $(a_n)$ such that
$$a_n=\alpha+\beta (n-1),\quad\alpha\,,\beta>0$$
we get
$$\begin{align}
\big(\frac{a_1+a_{n+1}}{a_n}\big)^n&=\exp\big(n\ln(\frac{\alpha+\alpha+\beta n}{\alpha +\beta (n-1)})\big)\\
&=\exp\big(n\ln(1+\frac{\alpha+\beta}{\alpha+\beta (n-1)})\big)\\
&\leq\exp\big(n\frac{\alpha+\beta}{\alpha+\beta (n-1)}\big)\\
&\leq\exp(\frac{\alpha+\beta}{(\alpha-\beta)/n+\beta})\\
\end{align}$$
and thus $$\limsup_{n\to\infty} \Big(\frac{a_1+a_{n+1}}{a_n}\Big)^n \leq\exp(1+\alpha/\beta)\,,$$
and hence, for all $\alpha,\beta>0$,
$$c_{max}\leq\exp(1+\alpha /\beta)$$
which brings the inequality
$$c_{max}\leq e$$
since $\alpha/\beta$ can be chosen as small as we want.


How to show that $T(n) = T(n-1) + \Theta(n)$ is in $\Omega(n^2)$ In the class we have been shown the way to prove that
$T(n) = T(n-1) + \Theta(n)$ is in $O(n^2)$

$$
\begin{align}
T(n)&\le T(n-1) +cn &\\
    &\le c(n-1)^2+cn &\\
    &=cn^2-2cn+c+cn\\
    &=cn^2-cn+c\\
    &=c(n^2-n+1)\le cn^2 &\
\end{align}
$$

Then we were said that it is easy to see that $T(n) = T(n-1) + \Theta(n)$ is in $\Omega(n^2)$. But, I do not understand how to "see" this really.
As I understand we need to show that

$$
T(n-1) + \Theta(n)\ge cn^2
$$
Right? I have no idea how to do that. Need help. Thank you!

Let $c$ be the lower bound so that your $\Theta (n)$ is greater than or equal to $cn$. Then you have

$$T(n) \geq T(1) + c\sum_{j=1}^n j = T(1) + cn(n+1)/2 = \Omega(n^2)$$

The only tricky thing there is knowing the formula for $\sum_{j=1}^n j$. But that's a classical result, you can look up sum of consecutive integers for example.

The following definition is useful for your purpose  $f(n)=\Theta(g(n))$ means that

$$ c_1\, g(n) \leq  f(n) \leq c_2 \,g(n) $$

for some positive $k_1,k_2$.


$(A -3I)(A + 2I) = 0$ implies $A = 3I$ or $A = -2I$? I know that square matrices' multiplication $XY=0$ does not imply $X=0$ or $Y=0$.

However, what if $X = A - 3I $, $Y = A + 2I $, given $I$ the identity matrix.

$$(A - 3I)(A + 2I) = 0$$

If it goes with the same rule of $XY=0$, then $A$ can be one matrix other than $-3I$ and $2I$.
But, we have the restriction that we can find an $A$ such that  $A - 3I$ and $A + 2I$ are the multiplication that produce $0$.

Can I still find a matrix $A$ which is not equal to $3I$ or $-2I$?

To find all solutions in the $2 \times 2$-case, write $A= \begin{pmatrix}a & b\\ c & d\end{pmatrix}$. The condition $$(A-3I)(A+2I) = 0$$
leads to the system
$$\begin{cases}(a-3)(a+2) + bc = 0 \\ (a-3)b + b(d-2) = 0 \\ c(a+2) + c(d-3) = 0 \\ bc + (d-3)(d+2) = 0\end{cases}$$
In particular, $a= 3, b = 0, c = 0, d = -2$ satisfies this, so the matrix
$$A:=\begin{pmatrix} 3  & 0  \\ 0 & -2\end{pmatrix}$$
presents a counterexample.
There are even more solutions: see https://www.wolframalpha.com/input/?i=solve+for+a%2Cb%2Cc%2Cd%3A+%28a-3%29%28a%2B2%29%2Bbc+%3D+0%2C+%28a-3%29b+%2B+b%28d-2%29+%3D+0%2C+c%28a%2B2%29+%2B+c%28d-3%29+%3D+0%2C+bc+%2B+%28d-3%29%28d%2B2%29+%3D+0+

hint

By Cayley-Hamilton Theorem, if the charactetistic polynomial is

$$P(x)=(x-3)(x+2)=(3-x)(-2-x)$$ then matrix will satisfy $$(A-3I)(A+2I)=0$$


Differentiation of $\sin^{-1}(1/\sqrt{1+x})$ $$\arcsin \frac{1}{\sqrt{1+x}}.$$ I tried differentiating it but every time my answer came wrong. The answer in my book is $$- \frac{1}{2(x+1)\sqrt x}.$$

Rules I will use:

The derivative of $\arcsin x$ is $\frac{1}{\sqrt{1-x^2}}$
The derivative of $x^\alpha$ is $\alpha\cdot x^{\alpha-1}$
The chain rule, i.e. the derivative of $f(g(x))$ is $f'(g(x))\cdot g'(x)$.

First, take $f(x)=\arcsin(x)$ and $g(x)=\frac{1}{\sqrt{1+x}}$, and use the chain rule. The derivative of $f(g(x))$ is

$$\begin{align}f'(g(x))\cdot g'(x)&= \frac{1}{\sqrt{1-g(x)^2}}\cdot g'(x)\\ &= \frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{1+x}}\right)^2}}\cdot g'(x) \\&= \frac{1}{\sqrt{1-\frac{1}{1+x}}}\cdot g'(x)\\&=\frac{1}{\sqrt{\frac{1+x-1}{1+x}}}\cdot g'(x)\\&=\frac{\sqrt{1+x}}{\sqrt x}\cdot g'(x)\end{align}$$

I leave the computation of $g'(x)$ to you.

The function
$$
f(x) = \arcsin{\frac{1}{\sqrt{1+x}}}
$$
can be seen as a composite function $f(x) = u(s(x))$, where $u(x) = \arcsin{x}$ and $s(x)= \frac{1}{\sqrt{1+x}}$. Their derivatives are (I will assume these as given - I won't prove these)
$$
u'(x) = \frac{1}{\sqrt{1-x^2}}
$$
$$
s'(x) =  -\frac{1}{2(x+1)^{2/3}}
$$
Now, the derivative of a composite function is
$$
f'(x) = s'(x) u'(s(x)) = -\frac{1}{2(x+1)^{2/3}} \frac{1}{\sqrt{1-s^2(x)}}
$$
The latter terms is a bit tricky to evaluate, we should take it aside:
$$
\frac{1}{\sqrt{1-s^2(x)}} = \frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{1+x}} \right)^2}}
= \frac{1}{\sqrt{1-\left(\frac{1}{1+x} \right)}}
= \frac{1}{\sqrt{\frac{1+x}{1+x}-\frac{1}{1+x}}}
= \frac{1}{\sqrt{\frac{x}{1+x}}}
= \frac{(1+x)^{1/2}  }{  \sqrt{x} }
$$
therefore, the derivative can be simplified to
$$
f'(x) = -\frac{1}{2(x+1)^{2/3}} \frac{(1+x)^{1/2}  }{  \sqrt{x} }
 = -\frac{1}{2\sqrt{x}(x+1)}
$$


Solving the initial value problem for PDE Let consider the initial value problem given by $$xu_y-yu_x=u\quad\text{with}\ u(x,0)=h(x).$$ The answer that my teacher put on our exercises is $u=h(\sqrt{x^2+y^2})e^{\arctan(y/x)}$.

How I get the answer? I made some computations here and I find $$u=h\left(\frac{x^2-y^2}{x}\right)e^{y/x}.$$ Probably I'm wrong, but I can't see how to obtain the desired answer. Appreciate any help! :D
Remark: If someome wants to see my computations, I can put it here later.

General solution of pde is
$$u=F(\sqrt{x^2+y^2})e^{\arctan\frac{y}{x}}$$
Then solution of initial value problem is
$$u=h(\sqrt{x^2+y^2})e^{\arctan\frac{y}{x}}$$
In polar coordinates $$x=r\cos\phi,\quad y=r\sin\phi$$ pde is
$$\frac{\partial u(r,\phi)}{\partial\phi}=u(r,\phi)$$
with solution
$$u=F(r)e^\phi$$
If substitute $r=\sqrt{x^2+y^2}$, $\phi=\arctan\frac{y}{x}$ we get general solution.


What is the equation used to calculate a linear trendline? What are the equations to calculate a linear trendline over a set of points?

EDIT: "In excel it is done automatically but how to manually calculate a linear trendline over a set of points" was originally the question. At first I asked this question because I was simply doing it with excel all the time and couldn't figure out how it computed the result. People tend to focus on the excel part instead of the actual question so I just removed this mention.

I've been looking for this formula in this website so I've made it Q&A style :) Hope this helps.

Slope;

$$\alpha = {n\sum(xy) - \sum x \sum y \over n\sum x^2 - (\sum x)^2}$$

Offset:

$$\beta = {\sum y - \alpha \sum x \over n}$$

Trendline formula:

$$y = \alpha x + \beta $$

source : https://classroom.synonym.com/calculate-trendline-2709.html

When I have used the data analysis package for simple linear regression I always get a scatterplot of the data and have an option to add a trendline and also an option for the equation and R square to be printed on the graph.  these days this can easily be done with the chart wizard.


Every nonempty subset of the natural numbers has a least number Proposition: Every nonempty subset $A$ of $\mathbb{N}$ has a least element.

We assume the opposite: $$\exists \left( A \subseteq \mathbb{N} \wedge A \neq \varnothing \right): \forall s \in A: \exists a \in A: s > a$$

Be $A \subseteq \mathbb{N}$ and $A \neq \varnothing$.
Be $s \in A$ and $a_n \in A$ for every $n \in \mathbb{N}_{\ge 1}$ and be $s > a_1$ and $a_n > a_{n+1}$.

Proposition: $$s - a_n \ge n$$

Proof with mathematical induction:

The base case holds since $s > a_1 \implies s-a_1 \ge 1$.

Also: $a_n > a_{n+1} \implies a_n - a_{n+1} \ge 1$ and with this $s - a_n \ge n$ implies $s - a_{n+1} \ge n+1 \quad \square$

Since $s - a_n \ge n$ and $a_n \ge 0$ holds we get: $$s \ge n \quad \forall n \in \mathbb{N}_{\ge 1}$$
But it is $s < s+1 \in \mathbb{N}_{\ge 1} \text{ Contradiction!} \quad \square$

  First question: Is this proof valid?

  Second question: Do you know different proofs for this proposition?

Cheers

One standard argument is as follows:

Suppose $A\subseteq\Bbb N$ has no least element. If $0\in A$ then $A$ would have a least element.  So $0\notin A$.  Now suppose $0\notin A,1\notin A,\ldots, n-1\notin A$.  If $n\in A$ under these assumptions, then $A$ would have a least element.  So, by induction, $n\notin A$ for every $n\in\Bbb N$.  Therefore $A=\emptyset$.

Well, it might not be kosher to bring a baseball bat to a cricket game but:

$\mathbb N \subset \mathbb R$ and the reals have the least upper bound property.  A is bounded below by 0. so $x =\inf A$ exists.

It's easy to show x is an integer.  (Else there exists a $y$ such that $x < y < \lceil x\rceil$ so y is also a lower bound so $x$ is not inf.) And it's easy to show $x \in A$.  (Else $x + 1/10$ is also a lower bound and x is not inf.)

So $x = \inf A$ and $x \in A$ imply $x$ is the least element in A.

====

A bit more cricket:

Assume non-empty A has no minimum element.

We can do a downward induction to prove the following statement:  For any integer m, there is an element of A that is less than m.

Sub-base case:  Let a be in A.  then for all m > a the statement is true because a < m.

Base case: as A does not have a minimum element, a is not the minimum element.  So there is an element of A that is less than a.

Induction step: Assume the statement is true for m.  The element of A less than m must be less than or equal to m - 1.  If it isn't equal to m, than the there is an element of A less than m -1.  If it is equal to m, then m is an element of A but it isn't the minimum element as there is no minimum element.  So there is an element of A less than m - 1.

Hence this statement is true for all integers.

But it's not true for 0 or any negative integer.

So our assumption was wrong, and non-empty A has a minimum element.


Example of $I$-adic topology of submodule not matching subspace topology? I'm reading about the $I$-adic topology on $M$ for $R$ a commutative ring, $I$ an ideal of $R$ and $M$ an $R$-module. The references I'm reading don't provide examples, but they say that if $N$ is a submodule of $M$, it's not necessarily true that the subspace topology of $N$ in $M$ with the $I$-adic topology coincides with the $I$-adic topology on $M$ (of course, I know this holds for $R$ Noetherian and $M$ a finite $R$-module by Artin-Rees).

Does anyone know of any specific examples where these topologies don't coincide? Thank you very much!

Look this for instance: https://mathoverflow.net/questions/71699/counter-examples-to-krulls-intersection-theorem if it may help.


Is there a way to prove that the absurdity axiom is consistent? I'm currently looking for a way to prove that the Reductio ad absurdum axiom of classical logic is consistent.
That's what i've done :

You want to prove $A\implies B$

Your suppose
$\neg(A\implies B)$

You replace $(A\implies B)$ by it's equivalent $\neg (A) \vee B$, it gives you :
$\neg(\neg(A) \vee B)$, i can apply De Morgan law, which gives :
$A \wedge \neg(B)$
From that assertion i found a contradiction (something always false)
$(A \wedge \neg(B))\implies \perp$

I replace the implies by it's equivalent form
$\neg(A \wedge \neg(B)) \vee \perp$

$\perp$ is neutral for $\vee$, then, it's equivalent to
$\neg(A \wedge \neg(B))$

Now i can apply de Mogan law
$\neg (A) \vee B$
And i can replace $\neg(A) \vee B$ by it's equivalent form with implication :
$A\implies B$

So, is it a valid view of the Reductio ad absurdum axiom ?

Can you state that axiom as a well-formed formula (there's not necessarily just one axiom)?  Because, if you do, it probably ends up an easy matter to point out that the antecedent of the absurdity axiom can't unify with the axiom itself due to symbol clash.  Consequently, the only consequences of the absurdity axiom under detachment are it's substitution instances.  But, the rule of substitution is truth-preserving.  So, since the absurdity axiom is a tautology, all of it's consequences are tautologies also.  Therefore, the absurdity axiom is consistent.

But, you do have to show that symbol clash would occur or that no unification between the antecedent of the absurdity axiom and the absurdity axiom itself comes as possible.

Or did you mean that the axiom is consistent with some other axioms/rules of inference?

Edit: Suppose the absurdity axiom in Polish notation would get re-written, with - marks used informally to mark the antecedent, as C-CNpq-CCNpNqp.

Now, let's look at 1. CNab and 2. CCNpqCCNpNqp.  The first symbol of both is C, so unification might be possible so far.  The second symbol of CNab is N, and the second symbol of CCNpqCCNpNqp is C.  Thus, we have symbol clash, and there is no way to unify 2. with itself.  Consequently, nothing can get derived from 2. by detachment.  Only substitution instances can get derived, and since the rule of substitution preserves truth, and 2. is tautology, only tautologies can get derived.  Denote any member from the family of tautologies as $\tau$.  N$\tau$ is not a tautology, and thus not derivable.  Thus, for no formula derivable from 2. does $\vdash$$\tau$ and $\vdash$N$\tau$ hold.

Suppose the absurdity axiom, with 0 denoting falsity, has form CCCp0qCCCp0Cq0p and no definitions come as permissible.  Consider 1. CCa0b and 2. CCCp0qCCCp0Cq0p.  Let $\alpha$/$\beta$ denote substituting $\alpha$ with $\beta$.  So, in 1. and 2. let a/Cp0, q/0, b/CCCp0C00p.  Thus, they do have a most general unifier.  So, we'll draw out the consequences of the absurdity axiom as follows.

The absurdity axiom    1 CCCp0qCCCp0Cq0p
1 q/0                  2 CCCp00CCCp0C00p
1 p/Cp0 q/CCCp0C00p    3 CCCCp00CCCp0C00pCCCCp00CCCCp0C00p0Cp0
3 * C2-4               4 CCCCp00CCCCp0C00p0Cp0

Now there exist only two possibilities for detachable well-formed formulas.  Either 1 or a substitution instance of it gets used as the major premiss, or 4 or a substitution instance of it gets used as the major premiss.  We've already found the most general unifier of the antecedent of 1 above.  So, does the antecedent of 1 unify with 4, does the antecedent of 4 unify with 1, or does 4 unify with itself?

Considering the antecedent of 1 with 4, consider CCa0b and 4.  We can substitute a with CCp00, and we're alright.  But, now CCa0b becomes 1' CCCCp000b.  4 and 1' match up to their eighth symbol.  The eighth symbol of 1' is 0.  The eighth symbol of 4 is C.  Thus, we have symbol clash, and nothing can get inferred from some variant of 1 as the major premiss and 4 as the minor premiss.

Considering the antecedent of 4 with 1, consider CCCa00CCCCa0C00a0 and CCCp0qCCCp0Cq0p.  We substitute a/p and q/0 yielding 4' CCCp00CCCCp0C00p0 and 1' CCCp00CCCp0C00p.  But, now the 10th symbol of 4' is C and the 10th symbol of 1' is p.  The 10th through 12th symbols of 4' is Cp0. Thus, the smallest substitution for p in 1' would be Cp0.  But, then we have p substituted with some well-formed formula which has p in it also.  So, to unify p with Cp0 would require another substitution.  But any such substitution also has p in it.  So, we would require another substitution with p in it ad infinitum.  This is the occurs check.  Thus, the antecedent of 4 does not unify with 1.  Consequently, nothing can get detached with some variant of 4 as the major premiss and 1 as the minor premiss.

Lastly, let's consider 4-1 CCCCp00CCCCp0C00p0Cp0 and 4-2 CCCa00CCCCa0C00a0.  We'll substitute a/Cp0 yielding 4-3 CCCCp00CCCCCp00C00Cp00.  Now the first spot where 4-1 and 4-3 differ consists of their 12th symbol.  The smallest substitution would consist of substituting p with Cp0.  But, now the occurs check happens (which leads to a non-terminating recursive blow-up as suggested in the previous case).  Consequently, the antecedent of 4 does not unify with itself.

Since above cases qualify as exhaustive, the only theorems starting from 1 obtainable under detachment is 4 or one of it's substitution instances.  Now, both 1 and 4 are tautologies and so are any of their substitution instances.  Thus, from the absurdity axiom 1 we can only obtain tautologies and consequently no negation of tautology can get derived.  Therefore, this absurdity axiom is (self) consistent under detachment.


How to detect interdependence of functions Let $f,g:\mathbb{R}\to\mathbb{R}$ be two functions (assume them to be regular enough, let's say $L^2(\mathbb{R})$). It might be the case that these functions are not totally independent. For example we could have that the value of $f(x)$ depends in some way on the value of $g(x-t)$. How is it possible to detect such interdependences?

A first idea I had was to look at the Fourier transform of the two functions. Then if $f(x)$ depends somewhat strongly (e.g. linearly) on $g(x)$, the transforms should be basically the same except for multiplication by some factor (by the usual Fourier transform of a shifted function rule).

Another thing I found around is cross-correlation, but this does not work well for non-linear phenomena.

Is my first idea viable? And are there good references on the subject around?

Also, what are the correct tags for this question?

FunctionalDependence

page 17

Criterion for functional independence of $ u(x,y), v(x,y) $ is through the Jacobian

$$ J= \dfrac{\partial (u(x,y),v(x,y))}{\partial (x,y)} \ne 0 $$

In the event of vanishing J, function defining transformation $u$ to $v$ can be also found out.


Elimination of variables from a system of equation. Suppose $a,b,c,d,e,f,g,h,i$ are real numbers. Consider the system of equation
\begin{equation}
\begin{pmatrix}
1 \\ 1 \\ 1
\end{pmatrix}
= \begin{pmatrix}
a + bx  + cy \\ \frac{d}{x} + e + fz \\ \frac{g}{y} + \frac{h}{z} + i
\end{pmatrix}.
\end{equation}
We need to eliminate the variables $x,y$ and $z$ from the above (assuming that solution for the variables $x,y$ and $z$ exists) and get a single  equation in $a,b,c,d,e,f,g,h,i$. I do not know it is possible or not. I hope this is possible but I am unable to get the required. Any suggestion or further help is much appreciated. Thanks in advance.

Multiplying by $x,y,z$ we obtain three polynomial equations in $x,y,z$. Taking resultants we can solve it. The formulas, however, are rather complicated. To show an easy example, take the case of $b=0$. Then we obtain
$$
y=\frac{1-a}{c},\quad z=\frac{h( 1- a )}{ai - a - cg - i + 1}
$$
and
$$
x=\frac{d( - ai + a + cg + i - 1)}{aei - ae - afh - ai + a - ceg + cg -  ei + e + fh + i - 1}.
$$
This example also shows that the coefficients $a,b,\cdots ,i$ need not satisfy a nontrivial polynomial equation in general. In fact, all coefficients are arbitrary in the above formula except for the denominators being nonzero.


Probability of streaks I thought of this question lately, but I'm not satisfied with the answer I got:

If I flip a coin 100 times, what is the probability that I will get a streak of at least ten of the same side?

The way I thought of solving it was just: $\left(90 \times 0.5 ^ 9\right) = 0.0879$, but this has to be wrong because, for the probability of getting a streak of at least 4 in the same scenario, it yields: $\left(96 \times 0.5 ^ 3\right) = 12$, which obviously doesn't make sense.

According to this page, there is a closed form expression for just this problem.

  ...the probability, S, of getting K or more heads in a row in N
  independent attempts (where p is the probability of heads and q=1-p is
  the probability of tails) is:

  $$
S(N,K) = p^K\sum_{T=0}^\infty {N-(T+1)K\choose
T}(-qp^K)^T-\sum_{T=1}^\infty {N-TK\choose T}(-qp^K)^T
$$

With the unusual convention that ${A\choose B}= 0$ for $A < B$.  Numerical evaluation gives me 0.0441372 for the case of $p=1/2$, $N=100$, $K=10$.

Edit 1

Reworking it a bit to get rid of that weird convention just changes the upper limit.

$$
p^k \sum _{t=0}^{\frac{n-k}{k+1}} \binom{n-k (t+1)}{t} \left(-q p^k\right)^t-\sum _{t=1}^{\frac{n}{k+1}} \binom{n-k t}{t} \left(-q p^k\right)^t
$$

The following Mathematica code gives you numbers, just plug in p, n, k in the last substitution bit.

-\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(t = 1\),
FractionBox[\(n\), \(1 + k\)]]\(
\*SuperscriptBox[\((\(-
\*SuperscriptBox[\(p\), \(k\)]\)\ q)\), \(t\)]\ Binomial[n - k\ t,
       t]\)\) + p^k \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(t = 0\),
FractionBox[\(\(-k\) + n\), \(1 + k\)]]\(
\*SuperscriptBox[\((\(-
\*SuperscriptBox[\(p\), \(k\)]\)\ q)\), \(t\)]\ Binomial[
       n - k\ \((1 + t)\), t]\)\) //. {k -> 10, n -> 100, q -> 1 - p,
   p -> 1/2} // N

Edit 2

It has recently been pointed out by Mark L. Stone that the above is for a streak of heads, but not for the case of either streak occurring. I'd recommend reading his post below.

I think it would be easier to use this formula $\sum_{i=1}^n \binom{n}{i}p^i(1-p)^{n-i} $, where $n$ is the number of times the action occurred (in this case, the coin flip), and $p$ is the probability of the chosen event happening (in this case, $\frac{1}{1024}$ for a streak of ten).
Plugging that in would give us $\sum_{i=1}^{100} \binom{100}{i} (\frac{1}{1024})^i(1-\frac{1}{1024})^{100-i} \approx .0930826...$, or about 9.9%


Building Minimum warehouses A big international retailer is setting up shop in India and plans to open stores in N towns (3 ≤ N ≤ 1000), denoted by 1, 2, . . . , N.

There are direct routes connecting M pairs among these towns. The company decides to build warehouses to ensure that for any town X, there will be a warehouse located either in X or in an immediately neighboring town of X.

I need to find the minimum number of warehouses the company has to build.

Example : If we have N=10 towns and 7 routes between :

1 2
2 4
2 5
3 6
8 6
9 7
10 7

Here answer will be 3.
Help me device an algorithm to do this problem.

This problem is equivalent to finding the domination number of the graph, i.e., the smallest set $S$ of vertices such that every vertex of the graph lies in $S$ or its neighborhood $N(S)$.
Different graphs with the same $M$ (number of edges) may have different domination numbers.
I would thus try to find a domination number algorithm. The problem is NP hard, so it's not going to be fast for large graphs.
The brute force way to do this would be as follows.
Let $G$ be the graph whose vertices are towns, and whose edges are direct routes.
Run through every vertex set $S\subset V(G)$ of the graph from smallest to largest (so first check all single vertices, then all pairs, etc), and check if every vertex not in $S$ is adjacent to something in $S$ (so check if $V(G) \subseteq N(S)\cup S$).
As soon as you find a set $S$ that is adjacent to everything, stop the algorithm, and take the number of vertices in $S$ to be your answer.
It is important that you stop as soon as you find one, because this algorithm is already going to be painfully slow.

Use Algorithm to find the number of connected components in a graph. The number of components in the graph will show how many warehouses are needed.So, that every places is reachable.


Simplify the expression: $\sin x (1+\cot^2 x)$ Simplify the expression: $\sin x (1+\cot^2 x)$

Please help. I have tried all identities and still can't figure it out. Thanks.

Note the following identity:
$$\sin^2{x}+\cos^2{x}\equiv 1$$
If you divide both sides by $\sin^2{x}$, you can derive another identity:
$$1+\cot^2{x}\equiv \csc^2{x}$$
Can you continue?

I assume you mean the expression

$\sin{x} \left(1 + \cot^2{x} \right)  = \sin{x} \left(1 + \frac{1}{\tan^2{x}} \right)  = \sin{x} \left(1 + \frac{\cos^2{x}}{\sin^2{x}} \right)  = \sin{x} + \frac{\cos^2{x}}{\sin{x}} = \sin{x} + \cos{x}\cot{x}$.

However, it's not becoming much simpler. There are many forms that you can give for this expression. Ask Wolfram Alpha for more expressions. According to Wolfram, it can also be $\csc(x)$.


Let V be the plane with equation... Let $V$ be the plane with equation $x_1 - 4 x_2 - 2 x_3 =0$ in $\mathbb{R}^3$. Find the matrix $A$ of the linear transformation
$T(x)= \begin{bmatrix} 4& 2 & -8 \\ 2 & 1& -1 \\ -2& -1& -2\end{bmatrix} x$ with respect to the basis $\{\begin{bmatrix} 4 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix} 2 \\ 0 \\ 1\end{bmatrix}\}$

Note: $T(x)$ is a 3x3 matrix and the basis are 3x1 matrices

Figure out how $T$ acts on the basis vectors $v_1 = (4,1,0)$ and $v_2 = (2,0,1)$.
The vector $T(v_1)$ should be a linear combination of $v_1$ and $v_2$. Find $a$ and $b$ with $T(v_1) = av_1 + bv_2$.
Similarly, write $T(v_2) = cv_1 + dv_2$.
The matrix $A$ is:

$$\left(\begin{array}{cc} a & c \\ b & d \end{array}\right)$$

Edit: corrected order of variables in matrix


Exam Question - Convex Optimisation I had an exam question today and want to check if my answer is right or if there was a mistake...

Consider the following convex optimisation problem:

$$\begin{array}{ll} \text{minimize} & - \ln(1+x_1)- \ln(\frac{3}{2}+x_2)- \ln(4+x_3)\\ \text{subject to} & x_1+x_2+x_3=1\\ & x_1, x_2, x_3\geq0\end{array}$$

I got that the dual has optimal solution $\lambda =\frac{2}{5}$ and $x=(\frac{3}{2},1, -\frac{3}{2})$. However, obviously, the third variable violates its constraint. I'm thinking either I got the answer wrong or the question was wrong! Can someone help me here?

I am sorry to tell you this, but you did not get the correct answer (which I guess you already know since you have $x_3<0$ in your solution). The correct solution is

$$
x_1 = \frac{3}{4}\\
x_2 = \frac{1}{4}\\
x_3 = 0
$$

You can find this answer just by reasoning. How?  Well as you know, $\ln(x)$ has a negative second derivative. This means that an increase in $x$, say $\Delta x$, will have less impact on $\ln(x)$ for larger values of $x$ than for smaller. You can interpret the first constraint as a resource constraint. Thus, you have a total of $1$ to distribute over the three non-negative variables. If you give all your resources to $x_1$ it will still not "catch up" to $\ln(4)$. The same goes for $x_2$. So there is little point in giving any resources to $x_3$, thus $x_3=0$. Now the best way of distributing the resources between $x_1$ and $x_2$ is to make the two first terms equal (due to the negative second derivative). Thus solving
$$
\ln(1+x_1)=\ln(\frac{3}{2}+x_2) \text{     for $x_1+x_2=1$, and $x_1,x_2\geq0$}
$$
yielding
$$
x_1=\frac{3}{4}\\ x_2= \frac{1}{4}
$$
So the result can be derived just by realizing that the second derivative of $\ln{x}$ is negative. Hope this was helpful!

Question is correct but your answer is not!
If you put more detail here about your solution I can explain why you got wrong answer.
Anyway here it is my guess about your solution:

The problem is a convex optimization over compact constraint, and slater CQ holds therefore, $x^* $ is an optimal solution iff $x^*$ is KKT point. so if you set up KKT sys, you get a simple system of non linear equations and inequalities including the primal inequality i.e., $x=(x_1,x_2,x_3) \ge 0$. So if you have taken into account the later inequality you should have came up right answer which satisfies non-negativity condition! I bet you didn't consider this condition....


Derivative of inverse functions Show graphically that
$$
(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}. \\
$$
PS: I know that $f(x)$ and $f^{-1}(x)$ are symmetrical about the line $y=x$. It should be proved  by showing that multiplying the slopes of the tangents of points $x$ and $f(x)$ equals 1.

As you say, the graph of a function and its inverse are symmetrical about the line $y=x$ - or, one is a reflection about that line of the other.  This geometric fact corresponds to the algebraic fact that $y=f(x)$ if and only if $x=f^{-1}(y)$.  Again stated geometrically, the point $(x,y)$ is on the graph of $f$ if and only if the point $(y,x)$ is on the graph of $f^{-1}$.  To see why this interchange of $x$ and $y$ corresponds to reflection across the line $y=x$, refer to the following diagram:

If we move horizontally from the point $(x,y)$ to the line $y=x$, we don't change the $y$-coordinate and we arrive at a point whose $x$ coordinate is the same as its $y$ coordinate; thus we arrive at the point $(y,y)$.  Moving vertically, we don't change the $x$-coordinate; so moving vertically brings us to the point $(x,x)$ on the line $y=x$.  Reversing the process brings us to the point $(y,x)$. Thus, the point $(y,x)$ is on the opposite side of the line $y=x$ and along the diagonal of a square - exactly as the geometric reflection of $(x,y)$ would be.

We now ask - how does this interchange of $x$ and $y$ affect the slope of a line.  Given a line with points $(x_1,y_1)$ and $(x_2,y_2)$, it's slope is
$$\frac{\Delta y}{\Delta x} = \frac{y_2-y_1}{x_2-x_1}.$$
The corresponding line generated by interchanging $x$ and $y$ has slope
$$\frac{x_2-x_1}{y_2-y_1},$$
which is exactly the reciprocal.  We can again illustrate this to make it clear:

Finally, we apply these observations to the graph of an invertible function $f$ together with it's inverse function $f^{-1}$ and the tangent lines at $(x,y)$ and $(y,x)$.

What does mean "graphically"? It is not clear to me.

We know that $f(f^{-1}(x))=x$ by definition

Thus $[f(f^{-1}(x))]'=f'(f^{-1}(x))(f^{-1})'(x)=1$

And then $(f^{-1})'(x)=\dfrac{1}{f'(f^{-1}(x))}$

But this is not clearly "graphical"...


Prove that exist a matrix $A$ such that $T(X)=AX$ Let be F a field and T a lineal operator from $F^{n}$ to $F^{m}$. Prove that exist a matrix $A_{mxn}$ with entries in $F$ such that $T(X)=AX$ $\forall X \in F^{n}$.
I've done this:
Let be $A_{mxn}$ a $mxn$ matrix with entries in $F$ and let be $X \in F^{n}$.
The vector $X$ is also a $nx1$ matrix.
And we know that $AX$ must be a $mx1$ matrix, i.e., $AX \in F^{m}$
By other side, we have that $T:F^{n}\rightarrow F^{m}$, i.e., $T(X)=B$ with $X\in F^{n}$
And $B \in F^{m}$, and we've already seen that there exists elements in $F^{m}$ with this form: $AX$ with $A \in F^{mxn}$
So $B=AX \in F^{m}$. Therefore, we've finished.
Is it enough what I've done to prove it? I feel that my proof is very direct so I'm not sure if something is wrong in it.

Hint : Consider the $n$ elements $e_1=(1, 0, ..., 0)$, $e_2 =(0, 1, 0, ...,0)$, ..., $e_n=(0, ..., 0, 1)$ of $F^n$, and their images $f_1 = T(e_i)$, ..., $f_n=T(e_n)$ by $T$ (which are elements of $F^m$).
Try to prove that the matrix whose $i$th column is the vector $f_i$ is a solution to your problem.


Your readings are in mm/h.  Since $15$ minutes $= {1\over 4}$ hours, you should divide a reading by $4$ to get the total amount rain in mm that fell during that $15$ minute span.  Taking the sum of all readings and dividing by $4$ will thus give you the total rainfall for the day (in mm).

This is assuming that the readings given are averages over the time span.  If they are instantaneous readings taken $15$ minutes apart, this can give you a decent estimate, but to obtain a better one you could use a more advanced numerical integration method instead (this method amounts to the rectangle rule).

For instance, using  Simpson's rule you would take $$\frac 1 {6} \sum_{i=1}^{12}(r_{2i-1}+2r_{2i})$$where $r_i$ is the $i^{th}$ reading.

Firstly ,add all the readings.Then multiply it with 1/4.It gives the amount of rain fall in mm.Then divide it by 24.It gives the amount of rain fall for the particular day i.e;mm/day


comparing two integrals: a cancellation method Problem: whether proving $\int x\,f(x)\,\mathrm{d}x < \int x\,g(x)\,\mathrm{d}x$ is equivalent to proving $\int \, f(x)\,\mathrm{d}x < \int g(x)\,\mathrm{d}x$?
I have tried this and want to make sure this is right

Since the range is from $-\infty$ to $\infty$, the answer is no. Here is a counter example. Consider $f(x) = \frac{1}{1+x^4}$ and $g(x) = \frac{x}{1+x^4}$. Then
$$
\int_{-\infty}^{\infty}xf(x)dx = \int_{-\infty}^{\infty}\frac{x}{1+x^4}dx = 0 < \int_{-\infty}^{\infty}xg(x)dx = \int_{-\infty}^{\infty}\frac{x^2}{1+x^4}dx = \frac{\pi}{\sqrt{2}}
$$
And yet
$$
\int_{-\infty}^{\infty}f(x)dx = \int_{-\infty}^{\infty}\frac{1}{1+x^4}dx = \frac{\pi}{2} > \int_{-\infty}^{\infty}g(x)dx = \int_{-\infty}^{\infty}\frac{x}{1+x^4}dx = 0
$$
NOTE: I will not go through the steps to get these integrals. They are ugly. The general idea is just to have $f(x)$ be an even function and $g(x)$ be an odd one. Then, $xf(x)$ is odd and $xg(x)$ is even. The integral of an odd function (given this range) is 0.


Proof of sets A and B involving set theory, showing: $(B^c - AB)^c = B$ Use set algebra rules to show why the complement of $(B^c - AB)^c = B$

=> Let x be an object

Assume $x\in (B^c -AB)^c $ or $x \notin (B^c - AB)$

So then $x \in B $ but $x\notin AB$, therefore $x \in B$

<= Let x be an object

Assume $x \in B$ therefore $x \in B$ but  $x \notin AB$

In other words, $x \in (B^c)^c$ and $x \in (AB)^c$

or $x \in (B^c-AB)^c$

I feel like I'm on the right track but these can be cleaned up more

This looks completely fine to me.

You're doing the definition of subset, each way ( implies-subset relation ) and showing the result follows.

You haven't written it in the most easy to follow form, but it is correct.

I prefer stuff like:

  Assume $x\in(\ldots)^C$

  Then  $x\in\ldots$

  $\vdots$

  Thus $x\in(...)^C\implies x\in\ddots$ which by the implies-subset relation means $(\ldots)\subseteq\ddots$

But without the bullet points (cannot do indents here)


Proving the existence/non-existence of $\lim_{x\rightarrow0} x\tan\frac1x$ Find $$\lim_{x\rightarrow0} x\tan\frac1x$$

Now I tried to find the form of the limit ($0/0$ or $0\cdot \infty$ or $\infty/\infty$), but as $x\rightarrow 0$, $\tan(1/x)$ tends to $\tan \infty$, and since $\tan x$ is unbounded unlike $\sin x$ or $\cos x$, no particular value or range can be assumed for $\tan(1/x)$.
Then I tried to find LHL and RHL.
Let $\lim_{x\rightarrow0^+} x\tan{(1/x)}=L$.
Then $\lim_{x\rightarrow0^-} x\tan{(1/x)}=-L$, since $x$ is approaching from the negative side, the input $1/x$ of $\tan$ is the negative of the input in RHL, and $\tan (-x)=-\tan x$
Now if the limit exists, then $LHL=RHL$, thus $L=0$.
Thus I got that if the limit exists, then it must be equal to $0$. But this doesn't confirm that the limit exists (and it doesn't).
Please help me in proving that the limit doesn't exist, and also please point out the mistakes (if any) in the argument I presented above (sorry for I might be weak in limits and the basics of it)
EDIT: As pointed out by Shubham in the comments, I forgot to take the sign of $x$ too in the $LHL$, thus rendering the argument which proved $L=0$ moot.
THANK YOU

Let $f(x)=x\tan(1/x)$ when $x\neq 0.$
If $a_n=\frac{1}{2\pi n}$ then $f(a_n)=0$ for all $n.$ Thus $$\lim_{n\to\infty} f(a_n)=0.$$
On the other hand, let $$b_n=\frac1{2\pi n+\arctan(n)}$$
Then $$\frac{n}{(2n+1)\pi}<f(b_n)<\frac{n}{2n\pi}$$ so by the squeeze theorem, $$\lim_{n\to\infty} f(b_n)=\frac{1}{2\pi}$$
Finally, let $$c_n=\frac1{2\pi n+\arctan(n^2)}$$ and show $f(c_n)\to +\infty.$
But $a_n,b_n,c_n$ all converge to $0,$ so $$\lim_{x\to 0} f(x)$$ cannot exist.
(We only needed any two of these three sequences to disprove convergence.)

The limit does not exist.
Consider the sequence $xn=\{2/n\pi\} = \{2/\pi, 1/\pi, 2/3\pi, 1/2\pi,  \}$
$y_n = x_n *\tan(x_n)$ As $x_n\to0, \tan(x_n)$ does NOT exist as $\tan(x_n)$ varies from $-\infty$ to $+\infty$.
$x_n$ is well behaved, $\tan(x_n)$ is NOT. $y_n$ is not well behaved.


Reading Binomial Tables While reading a table of cumulative binomial probabilities, if I need to find the probability of, for example, exactly 4 successful events happening and all the rest failures occurring, how would I go about that ?

The table you show is for when $n=20$ and is cumulative.  If you are interested in finding the probability of at most $4$ successful events and the rest failures... you look at the row which begins with the bold 4, then look across until reaching the column which begins with the bold 0.xx where 0.xx is the probability of success.  Trace your finger across the row and down the column to see where they meet.  For example, at most $4$ successes out of $20$ where probability of success is $0.25$ occurs with probability approximately $0.415$.

Since you are interested in the value for exactly $4$ successes rather than at most $4$ successes, you can subtract the value right above it.  I.e. $Pr(X= 4) = Pr(X\leq 4) - Pr(X\leq 3)$, which for the case where $p=0.25$ yields approximately $0.415 - 0.225 = 0.19$.

Doing this directly yields $\binom{20}{4}0.25^40.75^{16} \approx 0.189685\cdots$ confirming the result above.

If you were interested in probabilities where $n$ were some number different than $20$ then you will need a different table.  If you are in a scenario where you have a calculator or calculator software available, I see no reason to bother with tables in the first place and would instead use the usual formulas directly.

Go into EXCEL, in the function field enter BINOMRASP and calculate. In this case, type =BINOMRASP(4;20;0.25;FALSE)


Solving the IVP $8y''+26y=0$, $y(0)=2$, and $y'(0)=7$ Im supposed to find $y$ as a function of $t$ given the equation:

$$8y''+26y=0$$

Initial values are $y(0)=2$ and $y'(0)=7$.

I found the function of $y$ is $y=c_1\cos (\frac{\sqrt{13}}{2}t)+c_2\sin(\frac{\sqrt{13}}{2}t)$. Then I take the derivative, $y'=-\frac{\sqrt{13}}{2}c_1 \sin (\frac{\sqrt{13}}{2}t)+\frac{\sqrt{13}}{2}c_2\cos (\frac{\sqrt{13}}{2}t)$. Then using the initial values i'd solve for both arbitrary constants $c_1$ and $c_2$ and I ended up getting $2$ and $\frac{14}{\sqrt{13}}$. However this answer isn't correct. There's gotta be a mistake somewhere but I've been struggling to find it.

The DE's characteristic equation has two distinct complex roots: $+\frac{\sqrt{13}}{2}i$ and $-\frac{\sqrt{13}}{2}i$, so the general solution (indefinite integral) is:

$$y=c_1\cos (\frac{\sqrt{13}}{2}t)+c_2\sin(\frac{\sqrt{13}}{2}t)$$

First initial condition, $y(0)=2$:

$$2=c_1\cos (\frac{\sqrt{13}}{2}0)+c_2\sin(\frac{\sqrt{13}}{2}0) \implies c_1=2$$

Second initial condition, $y'(0)=7$:

$$y'=-\frac{\sqrt{13}}{2}c_1 \sin (\frac{\sqrt{13}}{2}t)+\frac{\sqrt{13}}{2}c_2\cos (\frac{\sqrt{13}}{2}t)$$

$$7=-\frac{\sqrt{13}}{2}c_1 \sin (\frac{\sqrt{13}}{2}0)+\frac{\sqrt{13}}{2}c_2\cos (\frac{\sqrt{13}}{2}0) \implies c_2=\frac{14}{\sqrt{13}}$$

Your solution is correct.

With free Computer Algebra System Maxima 5.42.1:


Proof that a common brain teaser is wrong (Burning Rope) There is a common brain teaser that goes like this:

You are given two ropes and a lighter. This is the only equipment you can use. You are told that each of the two ropes has the following property: if you light one end of the rope, it will take exactly one hour to burn all the way to the other end. But it doesn't have to burn at a uniform rate. In other words, half the rope may burn in the first five minutes, and then the other half would take 55 minutes. The rate at which the two ropes burn is not necessarily the same, so the second rope will also take an hour to burn from one end to the other, but may do it at some varying rate, which is not necessarily the same as the one for the first rope. Now you are asked to measure a period of 45 minutes. How will you do it?

Now I usually love brain teasers but this one frustrated me for a while because I could not prove that if a rope of non-uniform density is burned at both ends it burns in time $T/2$. I think I have sketched a proof by induction that shows that it's not actually true.

Given a rope of uniform density the burn rate at either end is equal so clearly it burns in time $T/2$. Now, consider a rope of non-uniform density, the total time T for this rope to burn is the linear combination of the times of the uniform density "chunks" to burn, i.e. $T = T_1 + T_2 + \ldots + T_n$. So consider, $T/2 = T_1/2+ T_2/2 + \ldots + T_n/2$. If we look at each $T_i/2$ this is precisely the time it takes to burn the uniform segment $T_i$ if lit at both ends. Therefore, in order to arrive at a rope that burns in time $T/2$, one would need to light each uniform segment on both ends, not simply the end of both ends of the total rope. What am I doing wrong?

You say "Therefore, in order to arrive at a rope that burns in time $T/2$, one would need to light each uniform segment on both ends, not simply the end of both ends of the total rope. What am I doing wrong?"

If I understand what you mean here, the idea is that you cut the rope into short segments, each one of (approximately) uniform density. Then you light the first one at both ends. As soon as it's done burning, you light the next one at both ends, and so forth.

That's certainly one way to arrive at a rope that burns in time $T/2$.  Your mistake is thinking that's the only way -- i.e., instead of "one would need to," it's "one could."  But here's another way to do it:  Suppose you knew, or somehow guessed, exactly how far the rope would burn in time $T/2$ if lit from one end (say the left).  If you imagine cutting the rope at exactly that point, the other portion will also burn in time $T/2$.  But you don't actually have to make the cut in order to start burning the other portion -- you can light it from the other end.  So you don't need to know where to make the cut; you just light both ends.  When they're done burning, at time $T/2$, you'll know where the cut would have been.

Let's say yes half burns in 5 mins and half burns in 55 mins. We now lit fire in both the ends. Now whatever rate might be of burning, the total rope get burned out at 30 mins.

Now take the 2nd rope , lit it's two ends, and lit any two more portions of the rope. That makes the rope burn down in 15 mins. So now u have total of 30 + 15 = 45 mins


How to find radius of convergence with power series from differential equations So I have a question that says find the radius of convergence after I have found the power series solution of a given differential equation. I know to find the radius of convergence you take

$$ p=\lim_{n \rightarrow \infty} \left\lvert\frac{C_n}{C_{n+1}}\right\rvert $$

but, I don't understand where $C_n$ and $C_{n+1}$ come from. Could someone please explain how I would find $C_{n}$ and $C_{n+1}$?

An elementary (and not particularly smart) approach could be:

Let's say the problem is $$\begin{cases}y'(x)=f(x,y(x))\\ y(0)=y_0\end{cases}$$

Since $C_n=\dfrac{y^{(n)}(0)}{n!}$, you "only" need to compute $y^{(n)}(0)$.

You know that $y'(0)=f(0,y_0)$.

Deriving in $x$ the first equation you get $$y''(x)=\frac{\partial f}{\partial x}(x,y(x))+y'(x)\frac{\partial f}{\partial y}(x,y(x))$$

Whence $y''(0)=\dfrac{\partial f}{\partial x}(0,y_0)+y'(0)\dfrac{\partial f}{\partial y}(0,y_0)$. Notice that you have already calculated $y'(0)$ the step before.

Keep deriving
$$y^{(3)}(x)=\\=
\dfrac{\partial^2 f}{\partial x^2}(x,y(x))+2y'(x)\dfrac{\partial^2 f}{\partial x\partial y}(x,y(x))+y''(x)\dfrac{\partial f}{\partial y}(x,y(x))+(y'(x))^2\dfrac{\partial^2 f}{\partial y^2}(x,y(x))
$$

Again, you can evaluate everything in $x=0,\ y=y_0$ and get $y^{(3)}(0)$.

The formulas rapidly worsen the more you derive, but perhaps the specific instance of the problem simplifies the calculations.

$p(x) = \sum c_n x^n$

The ratio test: a series converges (power-series of otherwise) the ratio of all successive elements greater than some N are less than 1.

That is, if there exists an $N$ such that $n>N \implies |\frac{c_{n+1} x^{n+1}}{c_{n} x^{n}}|<1$ then the series converges.

$|\frac{c_{n+1} x^{n+1}}{c_{n} x^{n}}| = |\frac{c_{n+1}}{c_{n}}||x|$

$\lim_\limits{n\to\infty}$$|\frac{c_{n}}{c_{n+1}}| = r$ and $\frac {|x|}r < 1$

G. Sassatelli tells me that I might have misundrstood your problem.

How do you find the power series....

You say Suppose: $y = \sum_\limits{n=0}^{\infty} c_n x^n$

$y' = \sum_\limits{n=0}^{\infty} n c_n x^{n-1}$

And if necessary

$y'' = \sum_\limits{n=0}^{\infty} (n)(n-1) c_{n} n x^{n-2}$

Now you plug these into your original equation.
Set $c_0 = y(0)$
find and equation such that $c_{n+1} = f(c_n)$
And then try to find a general formula that for $c_n$

A simple example

$y' = y, y(0)= 1\\
y' = \sum_\limits{n=0}^{\infty} n c_n x^{n-1}\\
y' = \sum_\limits{n=1}^{\infty} (n+1)c_{n+1} x^{n}\\
\sum_\limits{n=1}^{\infty} (n+1)c_{n+1} x^{n} = \sum_\limits{n=0}^{\infty} c_{n} x^{n}\\
(n+1) c_{n+1} = c_n\\
c_n = \frac{c_0}{n!}\\
y = \sum_\limits{n=0}^{\infty} \frac{x^{n}}{n!}$

Now, in this case the radius of convergence is infinite, but that is not always true.


What is the difference between the two real numbers that satisfy this equation? What is the absolute difference between the two real numbers $x$ for which $(x+1)(x-1)(x-2) = (x+2)(x+3)(x-3)$? Express your answer in simplest radical form

I tried guessing solutions but seeing how there are no common zeroes to both the left- and right-hand sides I don't know what to do.

We have
$$(x^2-1)(x-2) = (x^2-9)(x+2) \implies x^3-2x^2-x+2 = x^3+2x^2-9x-18$$
which simplifies to
$$4x^2-8x-20 = 0 \implies x^2-2x-5 = 0 \implies (x-1)^2 = 6$$
Hence, the roots are $1\pm\sqrt6$, which means that the difference between the roots is $2\sqrt6$.


Direct sum of projective module Let $\left\{ P_{i}\right\} _{i\in I}$ be a family of $R$-module. I know that if each $P_{i}$ is projective then $\oplus_{i\in I}P_{i}$ is projective. Is the converse true, i.e if $\oplus_{i\in I}P_{i}$ is a projective module, is $P_{i}$ projective module for all $i\in I$?

Yes its true. $P$ is projective iff there is free module $F$ such that $F=P\oplus Q$. Now if $\bigoplus P_i$ is projective then there is free module $F$ such that $F=\bigoplus P_i\oplus Q=P_j\oplus(\bigoplus_{j\neq i}P_i\oplus Q)$. This means that $P_i$ is projective.

This is true. For injective modules,the corresponding results hold.


Why is there a preference to use the cumulative distribution function to characterise a random variable instead of the probability density function? Perhaps this is but a subtlety but I've noticed that in quite a few definitions in statistics and probability, definitions regarding the distribution of a variable or a sample of data choose to use the cumulative distribution function to characterise random variables as following a specific distribution as opposed to using the probability density function.

E.g with the Kolmogorov Smirnov test, we look at the difference between cdfs and not pdfs.

Is there a specific reason for this ?

Every random variable has a CDF. Not every random variable has a pdf (for instance, discrete or mixtures of discrete and continuous distributions). For instance, in the Kolmogorov-Smirnov test, you are comparing an empirical CDF, which is discrete, to a potentially continuous CDF. In addition, convergence in distribution is defined using CDFs.

A CDF behaves straightforwardly under transformation/substitution of variables while a PDF requires meddling with differential coefficients.


$f(x,y)=x^y$ is not continuous at (0,0) How to prove that the function $f(x,y)=x^y$ ($x>0$) is not continuous at the point $(0,0)$?

I tried $y=x^\alpha$ $(\alpha>0)$ but this does not work since for such $y$ we have $x^y \to 1$ as $x\to 0$.

Thank you in advance!

Try

$$y={-a\over\ln x}\qquad a\gt0$$

Then $y\gt0$ if $x\lt1$ and $y\to0$ as $x\to0$, but $x^y=e^{-a}$.

It is continous iff $$\lim_{(0,0)} y\log x$$ exists, but

$$\lim_{x\to 0}\lim_{y\to 0} y\log x = 0\\
\lim_{y\to 0^+}\lim_{x\to 0} y\log x = -\infty \\
$$
Hence, the limit does not exist.


Application of derivative - finding the point on curve Problem :

Find the point on the curve $ay^2=x^3$ in the fourth quadrant where normal to the curve makes equal non zero intercepts with axes.

Solution :

Let the point at which normal is drawn can be given by $(x_1,y_1)$ Then it must satisfy $ay^2=x^3$

i.e. $ay_1^2=x_1^3$

= $y_1 =\pm \sqrt{\frac{x_1^3}{a}}$

Now differentiating the  given equation we get :

$\frac{dy}{dx} =\frac{3x^3}{2ay}$ therefore $\frac{dy}{dx}$ at $(x_1,y_1) = \pm \frac{3}{2}\sqrt{\frac{x_1}{a}}$

Please guide further to clear the concepts.... thanks a lot..

HINT: The $\frac{dy}{dx}$ gives you slope of the tangent at that point [Differentiate correctly]. So find the slope of the normal at that point. Now you can write down the equation of the normal(point-slope form).
Further apply the condition on equal non zero intercepts.

Slope of tangent is $+1$ or $-1$,in $4^{\text{th}}$ quad, $x>0$, $y<0$ therefore find $dy/dx$ and put it equal to $\pm1$,  you get $x=4a/3$ and $y=-a(4/3)^{3/2}
$.


Sobolev space and equivalence of norms I'm considering the space $W\{n,p\}[0,1]$ of functions with $n-1$ continuous derivatives $f^{(n-1)}$ is absolutely continuous and $f^{(n)}$ is in $L^p[0,1]$. The usual norm is the sum of the $p$-norms of each derivative from $1$ to $n$ and the $p$ norm of the function.

Now just consider the $p$ norm of the function + $p$ norm of the $n^{th}$ derivative, i.e.$(\int |f(x)|^p)^{1/p}$ + $(\int |f^{(n)}|^p)^{1/p}$  I want to show that this is equivalent to the usual norm defined above.

This requires finding positive constants and sandwiching this new norm. In one direction it's obvious since the new norm is less than the usual norm for every $x$, say $\|x\|_2 \leq \|x\|_1$. I'm not sure how to show the other direction.

Here is a proof for $p>1$.
We want to show the existence of $c>0$ such that
$$
\|u \|_{W^{n,p}} \le c ( \|u \|_{L^p} + \|D^nu\|_{L^p}) \quad \forall u\in W^{n,p}(0,1).
$$
Assume this is not true. Then for every $k$ there is $u_k$ such that
$$
\|u_k\|_{W^{n,p}} > k ( \|u_k\|_{L^p} + \|D^nu_k\|_{L^p}).
$$
Clearly, $u_k\ne0$. We can rescale the $u_k$'s to achieve
$$
1 = \|u_k\|_{W^{n,p}} > k ( \|u_k\|_{L^p} + \|D^nu_k\|_{L^p}).
$$
So $(u_k)$ is bounded in $W^{n,p}$, moreover $u_k \to 0$ in $L^p$ and $D^nu\to 0$ in $L^p$. We can extract a weakly (weak-star for $p=\infty$) converging subsequence (still denoted by same index): $u_k \rightharpoonup^* u$ in $W^{n,p}$. Here, we need $p>1$. By compactness of the embedding, $u_k \to u$ in $W^{n-1,p}$. We already know that the limit is zero: $u=0$, so $\|u_k\|^p_{W^{n-1,p}} \to 0$.

Going back to the initial inequality (for $p<\infty$), proves
$$
1 = \|u_k\|^p_{W^{n,p}} = \|D^nu_k\|_{L^p}^p + \|u_k\|^p_{W^{n-1,p}} \to 0.
$$
Similarly, we get for $p=\infty$:
$$
1 = \|u_k\|_{W^{n,\infty}} =\max( \|D^nu_k\|_{L^\infty} , \|u_k\|_{W^{n-1,\infty})} \to 0.
$$
In both ways, we get a contradiction.

By the fundamental theorem of calculus, $f^{(k)}(x) = \int_0^x f^{(n+1)}(y) dy$. Thus
$$
\|f^{k}\|_{L^p} =\left [\int_0^1 \left(\int_0^x f^{(k+1)}(y) dy\right)^p\right]^{1/p}
$$
Minkowski's integral inequality tells us that
$$
\left[\int_0^1 \left(\int_0^x f^{(k+1)}(y) dy\right)^p\right]^{1/p} \leq \int_0^1 \left(\int_0^x |f^{(k+1)}(y)|^{p} dy\right)^{1/p}dx \leq \|f^{(k+1)}\|_{L^p}
$$
Now we see by induction that
$$
\sum_{k=1}^n \|f^{(k)}\|_{L^p} \leq (n-1)\|f^{(n)}\|_{L^p}
$$
So the norm equivalence easily follows. Note that the calculations here are much easier because the total measure space is one, as we're working on $[0,1]$.


Bar Notation Problem everyone! I came across a problem in math that dealt with bar notation. Does anyone know how, for instance, 1.234(with a bar notation over the 34) is expressed as a fraction? I know already how 1.22(with a bar notation over the 22) is expressed as 1 and 2/9. Any answer would be helpful. Thanks!


What is $\cot(\pi/2)$? Base on the unit circle, I know

$
\begin{align}
&\cot\left(\frac{\pi}{2}\right) \\
=&\frac{0}{1}\\
=&0
\end{align}
$

But it is also

$
\begin{align}
&\cot\left(\frac{\pi}{2}\right) \\
=&\frac{1}{\tan\left(\frac{\pi}{2}\right)}\\
=&\frac{1}{\frac{1}{0}}\\
=&undefined
\end{align}
$

And Google gives me this answer:

　$6.12303177 × 10^{-17}$

I am really confused now. Although I know it is $0$, I don't see why the other ones are wrong.

$$\cot x = \frac{1}{\tan x}$$ only when $\tan x \neq 0$ (i.e. $x \neq n\pi$ for any $n\in \mathbb {Z}$).

However, $\cot x$ is actually defined as

$$\cot x := \frac{\cos x}{\sin x}$$ so $\cot \left(\frac{\pi}{2}\right)=0$ is the correct answer.

$\tan \frac π2$ is actually infinity.

So if we divide 1 by $\tan \frac π2$ we will have $\frac 1{\infty}$ which leads to 0.

$$\cot \frac π2 = \frac 1{\tan \frac π2}=0$$

Furthermore:

$$\cot x=\frac 1{\tan x}=\frac{\cos x}{\sin x}$$

So:

$$\cot \frac π2=\frac{\cos \frac π2}{\sin \frac π2} = \frac 01 = 0$$


Show that the four points given below are the vertices of a rhombus. Show that the four points, $(5, 8), (7, 5), (3, 5)$ and $(5, 2)$ are the vertices of a rhombus.

I tried solving it, by finding out the distances by using the formula $\sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2}$

I found all of them but two sides are equal and other two are not. As far as I know, rhombus has all sides equal...but mine does not match! Help please :)

Hint: Here are the points on graph paper.

The sides are formed by joining the points as follows: $(5,8) - (7,5)$, $(7,5) - (5,2)$, $(5,2) - (3,5)$, and $(3,5)-(5,8)$. If you check these four lengths, you will find that they are all equal.


Question on fourier series expansion https://en.wikipedia.org/wiki/Basis_function#Fourier_basis
It is known that
form a basis for $L^2([0,1])$.
I want to see that $\{\sqrt2 \sin (2\pi nx+\pi x)\}, \{\sqrt 2 \cos (2\pi nx+\pi x)\}$, 1
also form a basis for $L^2([0,1])$. Assume that we already computed that these are orthonormal to each other.
(1) Is it true that those form a basis?
(2) How do I prove it? (heuristically or rigorously) Does it suffice to show that it is "complete"?
I'd appreciate any insight!

If I take the integers and add one to every element we will again obtain a copy of the integers.  With this in mind, consider the input to $\{\sqrt{2}sin(2\pi nx+\pi x)\}$.


RK4 method applied to $\frac{dy}{dt}=-\frac{y-t}{2}$ with $y(0)=1$ I tried to solve this question, I did it by Huen's method also, but I'm getting very different results than the answers sheet.

  Use the fourth-order Runge-Kutta method (RK4) to solve the following initial-value problem
  $$\frac{dy}{dt}=-\frac{y-t}{2} \text{ with } y(0) = 1$$
  on the interval $[0,3]$ by taking $\Delta t=0.125,\;0.25,\;0.5,$ and $1$ and compare the solutions. You may want to write a computer code to conduct the calculations for the time steps $\Delta t < 1$.

I'm sorry but I just feel it strange: why you take so much trouble to solve this equation in a numerical way since it is a very typical ODE that can be solved with the method of variation of constant?
Anyway I have got an explicit solution and I find it fits your ODE, which goes as follows:
$$y(t)=t+3e^{-\frac{1}{2}t}-2$$
Is this answer what you expect? I hope maybe you would find it helpful to you.


Finding the minimum of a function. I’m working on learning calculus with story problems. It’s easier to figure out when the question simply asks for specific answers, so I figured I needed to get better at solving problems based on the info given in the story problem.
Here’s an example that I’m trying out:

The delivery cost per ton of bananas, in thousands of dollars, when $x$ tons of bananas are shipped is given by $D = 3x + 100/x, x >0$. Find the value for $x$ for which the delivery cost per ton of bananas is a minimum, and find the value of the minimum delivery cost.

To begin with, we know this function is the delivery cost and $x$ is the number of tons of bananas.

To find the minimum of a function we use the first derivative, right? So,
$D = 3 - 100 x^{-2}$ ?
Then, I am unsure how to find the value. Do I find what $x$ is from the derivative and then plug it into the original function?

Let  $D(x) = 3x + \frac{100}{x}$ defined for $x>0$. Then, by differentiating, we yield :

$$D'(x) = 3 - \frac{100}{x^2}$$

To find an extreme point (minimum or maximum), one has simply to take the derivative equal to zero. Thus :

$$D'(x) = 0 \Rightarrow 3-\frac{100}{x^2} = 0 \Leftrightarrow x = \pm \sqrt{\frac{100}{3}}$$

But, note that the function is initially defined for $x>0$, so we ignore the negative value.

Thus, for $x = \frac{10\sqrt{3}}{3}$ the delivery cost is minimum. The value is yielded by plugging that $x$ into the cost equation :

$$D\bigg(\frac{10\sqrt{3}}{3}\bigg) = 10\sqrt{3} + \frac{100}{\frac{10\sqrt{3}}{3}} = 20\sqrt{3}$$

This is a minimum value and a minimum point and can be easily seen by checking the sign of the first derivative on corresponding intervals, or by implementing a second derivative test.

If you plot the function and the derivative of the function in the same figure you will see that the minimum of the function is exactly at the same x-value where the derivative intersects the x-axis.

So to find x-value for the coordinate of the minimum point you equate the derivative to zero.
$$D' = 3 -\frac{100}{x^2} = 0$$
We get $$3x^2 = 100$$
So the minimum is at $x = \sqrt{\frac{100}{3}}$ or $x= - \sqrt{\frac{100}{3}}$.
And because x>0 the minimum value is then $x = \sqrt{\frac{100}{3}}$

Also use https://www.desmos.com/calculator/qtzamwcdrk to get a better visual understanding of what is happening.


Sum of the roots of $x^2-6[x]+6=0$, where $[.]$ is GIF Sum of the roots of $x^2-6[x]+6=0$, where $[.]$ is GIF
I have done this ptoblem by inspection as $$\frac{x^2+6}{6}=[x] \implies x>0.$$
Let [x]=0, then $x$ is non real. Let $[x]=1$, then $x=0$ which contradicts. Let $[x]=2$, it gives $x=\sqrt{6}$, in agrrement. Similarly assuming $[x]=3,4$; we get correct roots as $\sqrt{12}$ and $\sqrt{18}$. But if er let $[x]=5$, it gives $x=\sqrt{24}$. which contradicts. So I get the sum of roots as $\sqrt{6}(1+\sqrt{2}+\sqrt{3}).$
The question is: Have I found all the real roots. In any case, what is(are) more appropriate method(s) of doing it.

$$\left\lfloor x \right\rfloor\le x$$
$$-6\left\lfloor x \right\rfloor\ge -6x$$
$$x^2-6\left\lfloor x \right\rfloor+6\ge x^2-6x+6$$
$$0\ge x^2-6x+6$$
$$x\in[3-\sqrt{3},3+\sqrt{3}] \ \ \ \& \ \ \left\lfloor x \right\rfloor\in\{1,2,3,4\}$$
so let's consider the cases:
$1^{\circ}$ $\left\lfloor x \right\rfloor=1\Rightarrow x=0$ which is not a root of $x^2-6\left\lfloor x \right\rfloor+6$
$2^{\circ}$ $\left\lfloor x \right\rfloor=2\Rightarrow x=\sqrt{6} $ which is a root of $x^2-6\left\lfloor x \right\rfloor+6$
$3^{\circ}$ $\left\lfloor x \right\rfloor=3\Rightarrow x=\sqrt{12} $ which is a root of $x^2-6\left\lfloor x \right\rfloor+6$
$4^{\circ}$ $\left\lfloor x \right\rfloor=4\Rightarrow x=\sqrt{18} $ which is a root of $x^2-6\left\lfloor x \right\rfloor+6$
$$\sum \text{roots}=\sqrt{6}+\sqrt{12}+\sqrt{18} $$

$x^2 = 6[x] - 6= 6([x]-1)$.  Forget about the greatest aspect of GIF and concentrate on the integer aspect of GIF. $x^2$ is multiple of $6$ and $x = \pm\sqrt {6k}$ where $k=[x]-1$.
$x^2 \ge 0$ so $[x]-1\ge 0$ so $x\ge [x] \ge 1$ so $k \ge 1$.
And if $k \ge 6$ then $x =\sqrt{6k} \le \sqrt {k^2} = k$ so $[x]\le x \le k =[x]-1$ is impossible.
We can pick them off one by one.  If $k=1$ and $x=\sqrt 6$ then $[x]=2$ and $6[x]-6=6=x^2$.  So that's one solution.  If $k=2$ and $x = \sqrt{12}$ then $[x]=3$ and $6[x]-6=12$ so that's another.  If $k=4$ then $x=\sqrt{24}$ and $[x]=4=k = [x]-1$.  Impossible.  If $k=5$ then $x=\sqrt{30}$ and $[x]=5$ and $k\ne [x]-1$ so that's impossible.
$x = \sqrt 6$ or $\sqrt{12}$ and the sum of the roots is $\sqrt 6 + \sqrt {12} = \sqrt 3(2+\sqrt 2)$.


Convergence of integral of multiplication of two positive functions I have two functions $f, g:\mathbb{R}\rightarrow \mathbb{R}_{\ge0}$, that are continuous. I know that $\int\limits_{-\infty}^\infty f(s) \, ds=C_1<\infty$, and $g(s)\le C_2$, with $C_1> 0$ and $C_2\ge0$, for all $s\in\mathbb{R}$.

For me, it is true that
$\int\limits_{-\infty}^\infty f(s)g(s) \, ds\le C_3<\infty$, with $C_3\ge0$. This statement would come from the fact that $f(s)g(s)\le C_2f(s)$, and then integrating on both sides.

This seems to be trivial, but I don't know if I should have additional conditions on the functions $f$ and $g$ so the previous statement is true. Any insights on this will be appreciated.

g(s) could be equal to a constant value per your statement then the integral would not be bounded.
To be true g(s) integral must be bounded also.


Convergence of $f_n = \sin(x+\frac{1}{n})$ as $n \rightarrow \infty$ Is the convergence of $$f_n = \sin \left(x+\frac{1}{n} \right)$$ as $n \rightarrow \infty$, uniform, where $x \in \mathbb{R}$?

It converges to $\sin x$ and $$||\sin (x+\frac{1}{n}) - \sin x ||_{\infty} \leq ||x + \frac{1}{n} - x||_{\infty} \rightarrow 0$$ as $n\rightarrow \infty$, so the convergence is uniform.

Is this an okay argument?

Your Argument is Okay as well

since

$$\sin a -\sin b  = 2\sin(\frac{a-b}{2})\cos(\frac{a+b}{2})$$
Together with $$|\sin x|\le |x|$$

One get

$$|\sin a -\sin b|  = |\sin(\frac{a-b}{2})\cos(\frac{a+b}{2})|\le |a-b|$$

Therefore

$$\sup_{x\in \Bbb R}\|\sin (x+\frac{1}{n}) - \sin x |\leq \sup_{x\in \Bbb R}|x + \frac{1}{n} - x|  =\frac1n\rightarrow 0$$

I don't think this argument is okay, because the second displayed equation is hardly different from a restatement of what you are trying to prove.  Why is it true, or rather, what about the $\sin$ function makes it true?

A good answer would make this clear, and would also make it clear if the same argument would work for any continuous periodic function, and not just the $\sin$ function.


Gift advice: present for high school graduate interested in math I am a PhD student in mathematics who recently found out that I will be attending my girlfriend's cousin's high school graduation party. I have never met the cousin, but hear that he is very interested in mathematics and is hoping to major in mathematics in college. He is taking Calculus BC (the equivalent of Calculus 2 at most colleges) now and is apparently doing quite well.

I am considering giving him a math book as a graduation present. The following texts immediately came to mind as decent candidates:

Elementary Number Theory by Underwood Dudley
Calculus by Michael Spivak
How to Prove it by Daniel Velleman

I think that they are all at about the right level. Further, these texts were instrumental in my early mathematical development. They provide for entertaining reads while still being substantive.

However, I would like some advice on the following matters,

Is a math text an appropriate graduation present?
What other math texts might I consider?
Would it be better to give a popular text such as Derbyshire's Prime Obsession?

I would argue that a math text is usually not an appropriate graduation present, even for somebody very interested in mathematics.
Just because he likes math, it doesn't mean that he likes reading books on it in his free time
I am an avid book reader myself, and have shifted heavily to non-fiction in later years. I have discussed books with friends frequently, and find out that they are much less inclined to read than I expect. And when they do read, they do it for entertainment, and if they start a non-fiction book, they seldom have the motivation to finish it. These people are mostly STEM graduates, quite a few of them have a Ph.D. too. Even the ones who do read non-fiction now (in their 30s and 40s) say that they developed the taste late.
As you can see here, many people do like reading this kind of book. But as far as I can tell, many more don't like it. And don't forget that math.stackexchange is not a place representative of all people who like maths, it is a place representative of people who like spending their free time on a web site about maths, which is probably a subset of the first ones. And I argue that it is a smallish subset.
Even if he likes math and likes reading books on it, it doesn't mean that he wants to get one from you.
First, book choice can be a rather personal matter. It is very hard to meet someone's taste, even if you know the general direction of what they like.
But there is this small problem with learning: All people love to learn, nobody likes being taught. (I don't remember whose aforism is this, maybe Churchill?). If you see curiosity and learning interest in a younger person, it is good to encourage them, help them discover the information they already thirst for, etc. But by just giving them a book because you think the world would be a better place if everybody has read it, you come across as condescending, the kind of 19th century educator who thinks it is his high moral duty to bring children on the path of rightness.
From your question, it seems that you fall at neither of these extremes, and are probably closer to the first case. But it still isn't close enough. You don't know him, don't know exactly what he is interested in. Maybe he isn't taking calculus because he loves calculus, but because he feels a high pressure to get into a good college. And he doesn't know you, and doesn't know your motives; it is easy for him to assume that you are the condescending kind even if you aren't. Even if he doesn't, choosing a book he might be interested in has a high chance to fail.
Even if he likes math and would read books on it, this is not necessarily the right occasion.
He just made it through school and wants to feel that he achieved something after making it through a pile of textbooks. He needs a feeling of closure, even as an intermediate step. Yes, there is college somewhere in his future, but he probably wants some time to stop, look back at what he did, and smile, before getting back on the treadmill. Having finished one level of something only to be presented with the next, higher level immediately is a sure way to demotivate somebody from ever achieving something. No, I have no guarantee that he feels this way, but you have no guarantee that he doesn't feel this way. And I have seen many people, including bright ones open for education, who are this way.
Besides, graduation is a social occasion. No matter how much of a nerd he might be, even if he is internally proud of being a nerd, he probably won't want his classmates to remember him as "the nerd who got textbooks for graduation day". Teenagers are very sensitive about their image in their class.
Conclusion
When you start with the fact that math belongs to his favorite school subjects and conclude that he will like receiving a math text as a graduation present, you are making a lot of assumptions, and there is a high risk that at least one isn't true. He might be one of the people who will truly enjoy a math book for graduation day, but the chance that he won't is quite high.
For your decision, you can try getting better intelligence. Let your girlfriend find out through the family grapewine how likely he is to appreciate this kind of gift. If you hear that he has earlier purchased math books for himself (which were not required for a course), then this is a very good indicator that your gift will be well received. Mentioning that you knew that he reads math books for fun will also clearly remove you from the "condescending old relative" zone.
If he hasn't done this, or you can't find conclusive information, consider some other, more accessible, gift. When you meet him, you can discretely evaluate his reading interest yourself, and if the result is positive, you can still get these books for him for Christmas.
If you don't go for a math book now, here are some other options which are math-themed:

a popular book, as you mentioned. There are a few good suggestions in this thread already. Beside things like Goedel Escher Bach, you can also get a puzzles book like one of Smullyan's. (Yes, it is still a book. But it is entertainment, and read in a very different context).
a puzzle, or set of puzzles. Most mathematicians I know love physical puzzles. Not the jigsaw ones, but the kind where you have to find out how to take a box apart, etc. Or just a beautiful chess set.
an entry for some event. This works a bit better with people interested in engineering, as there are more robots exhibitions than maths exhibitions :) but depending on your budget, you could find something.
and finally, you can of course make a gift which has to do with another hobby of his. One which is more of a hobby (= he has been seen spending free time on it without any benefit beside the joy of it) than an interest (= he perks up when he hears about the matter, and maybe prefers it in school, but does not necessarily allocate free time to it).

As a high school student in the same situation, I would recommend Excursions in Calculus by Robert M. Young. It contains almost all the fun of Spivak with none of the head bashing boringness (just my opinion).


LCM of irrational numbers So i read in a book that irrational and rational numbers do not have a common multiple and it said that lcm of irrational numbers is also only possible when both the irrational numbers have the same surd. I was wondering what this means.

Those are odd claims.

The first can, I think, be justified.  Let's say $\alpha$ is an irrational number and $\frac ab$ is rational (with $a,b\in \mathbb Z$).  Then it is certainly true that, for any non-zero integers $m,n$ we have $m\times \alpha$ is irrational and $n\times \frac ab$ is rational, so it is not possible for them to be equal.

But the second claim seems hard to follow, no matter what (standard) meaning you assign to "surd".

Originally, "surd" just mean "irrational".  These days, it more often means an expression in radicals, such as $\sqrt 2$ or $\sqrt[3] 3$.  However, numbers like $\pi$ and $2\pi$ clearly have common multiples so I'm not sure what meaning is intended.

https://www.quora.com/Does-the-L-C-M-of-two-irrational-numbers-exist
With $5*\sqrt{2}$ and $3*\sqrt{2}$ their least common multiple is $15*\sqrt{2}$, because it's the smallest number that's an integer multiple of both.

However, they don't always have an LCM. Take $\sqrt{2}$ and $\sqrt{3}$. There is no number $L$ such that $\frac {L}{\sqrt{2}} $ and $\frac {L}{\sqrt{3}}$ are integers, otherwise their quotient, $\sqrt{\frac {2}{3}} $, would be rational. And it isn't.


Two dynamical systems perpendicular I am given two dynamical systems:
$$x'=f(x)$$
$$x'=g(x)$$
where $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ and $g:\mathbb{R}^2\rightarrow \mathbb{R}^2$ are $C^1$ and perpendicular ($\langle f(x),g(x)\rangle=0$ for all $x's$). I am to show that if one of the systems has a (nontrivial) periodic orbit, then the other has a fixed point. Any help with this problem would be appreciated.

Suppose without loss of generality hat $f$ has a nontrivial periodic orbit. Consider the path $\gamma$ that describes the orbit. $\gamma$ is closed because the orbit is periodic; moreover, by the Existence and Uniqueness Theorem for ODEs, $\gamma$ must be simple. It is hence a Jordan curve and we may speak of the interior and exterior of $\gamma$.

  EDIT: If $g$ vanishes at some $x\in\gamma$, then $x$ is a trivial fixed point of the system induced by $g$. Assume then that $g$ is nowhere vanishing on $\gamma$.

Because $g$ is continuous, always orthogonal to $f$ and nowhere vanishing on $\gamma$, along $\gamma$ the field $g$ must face either always inwards (towards the interior) or always outwards (towards the exterior).

If $g$ faces always inwards, consider what happens as the time goes to $+\infty$. If $g$ faces always outwards, consider what happens as the time goes to $-\infty$. Do you think you can take it from here?

Hint: Choose a point $p\in\gamma$ and consider its orbit under $g$. Can the orbit intersect $\gamma$ at some point other than $p$? Use the existence and uniqueness theorem!


Prove the identity $\frac{d}{dt} |z(t)|= |z(t)|Re \frac{z'(t)}{z(t)}$ Let $t \mapsto z(t)$ be a complex-valued function of the real variable $t$. Assume further that $z'(t) \not= 0$ for every $t$. Prove the identity $\frac{d}{dt} |z(t)|= |z(t)|Re \frac{z'(t)}{z(t)}$. By the way what kind of identity is this I can not find anything about this problem.

you have $$|z|^2 = z \bar z \to 2|z|\frac{d|z|}{dt}=\frac{d}{dt}|z|^2 = z \frac{d\bar z}{dt} + \bar z\frac{dz}{dt} =2 Re\left(\bar z\frac{dz}{dt}\right) $$ dividing both side by $2 z\bar z$ gives you the desired result.


What makes elementary row operations "special"? This is probably a stupid question, but what makes the three magical elementary row operations, as taught in elementary linear algebra courses, special? In other words, in what way are they "natural" (as opposed to "arbitrary")?

It seems that they're always presented in a somewhat haphazard manner ("these are the three legendary elementary row operations, don't ask why, they just are"). From what I understand, they satisfy some nice properties, such as the inverse of each being an operation of the same type, etc. But is there something that characterizes them, i.e. is there some definition of what constitutes "elementary" that's only satisfied by the three types of elementary matrices, and no other matrix?

They're not special. They're just convenient. It's relatively easy to tell what happens to a matrix when you apply an elementary row operation to it, and this isn't quite as true for more complicated types of operations.

In the language of group theory, elementary matrices form a set of generators for the group of invertible square matrices. You could choose a different set of generators if you wanted to, but again, the elementary matrices are convenient.

Take a system of linear equations and apply the three elementary row operations to them, you will get an equivalent system of equations.


Investigate the convergence of $\sum a_n$ where $a_n = \int_0^1 \frac{x^n}{1-x}\sin(\pi x) \,dx$ Investigate the convergence of $\sum a_n$ where $a_n = \displaystyle\int_0^1 \dfrac{x^n}{1-x}\sin(\pi x) \,dx$.

We have thought about using the dominated convergence theorem to find $\lim a_n$, but that would result in something like $\lim a_n = \lim \displaystyle\int_0^1 \dfrac{x^n}{1-x}\sin(\pi x) \,dx = \displaystyle\int_0^1 \lim \dfrac{x^n}{1-x}\sin(\pi x) \,dx = \displaystyle\int_0^1 0 \,dx = 0$ which makes the convergence of $\sum a_n$ inconclusive.

Any tips on how to proceed? Thanks!

Using $\sin(\pi x) \geqslant \pi x(1-x)$ for $0\leqslant x \leqslant 1$
$$
   a_n = \int_0^1 \frac{x^n}{1-x} \sin(\pi x) \mathrm{d}x > \int_0^1 \pi x^{n+1} \mathrm{d}x = \frac{\pi}{n+2}
$$
it follows that $\sum_{n=0}^\infty a_n$ diverges.

$\newcommand{\+}{^{\dagger}}%
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$\ds{a_{n} = \int_{0}^{1}{x^{n} \over 1 - x}\,\sin\pars{\pi x}\,\dd x:\ {\large ?}}$.

\begin{align}
a_{n} &= \int_{0}^{1}\pars{1 - x}^{n}\,{\sin\pars{\pi x} \over x}\,\dd x
\\[3mm]
\verts{a_{n}} &< \pi\int_{0}^{1}\pars{1 - x}^{n}\,\dd x
= \left.\pi\,{\pars{1 - x}^{n + 1} \over -n - 1}\right\vert_{x = 0}^{x = 1}
= {\pi \over n + 1}
\end{align}

$$
\color{#0000ff}{\large\lim_{n \to \infty}a_{n} = 0}
$$


Solve these equations simultaneously (trig) Solve for $ x,y: $

\begin{equation}\cos x -\cos(x+y) = 0
\end{equation}

\begin{equation}\cos y -\cos(x+y) = 0
\end{equation}

The answers are $(0, 0), (\frac{2\pi}{3}, \frac{2\pi}{3})$.

I get $(0, 0)$, but how do you get the latter?

Sorry if this is a bit basic - I don't remember trig well...

tfa

Given

$$\cos x-\cos(x+y)=0\\
\cos y-\cos(x+y)=0$$

Subtracting the second equation from the first yields

$$\cos x=\cos y$$

Recall that $\cos x=\pm\sqrt{1-\sin^2x}$ and $\cos y=\pm\sqrt{1-\sin^2x}$.
This now becomes $\sqrt{1-\sin^2x}=\pm\sqrt{1-\sin^2y}$ yielding

$$\sin x=\pm\sin y$$

Recall that $\cos(x+y)=\cos x\cos y-\sin x\sin y$.
Thus, $\cos x-\cos(x+y)=0$ now becomes $\cos x-\cos^2x\pm\sin^2x=0$.
Solving $\cos x-\cos^2x-\sin^2x=0$ yields $\cos x=1$.
Solving $\cos x-\cos^2x+\sin^2x=0$ yields $\cos x-\cos^2x+1-\cos^2x=0$ which is equivalent to

$$(2\cos x+1)(-\cos x+1)=0$$

This yields two possibilities:  $\cos x=1$ or $\cos x=-\frac12$.  Can you do the rest?

I should begin dividing first equation by cos(x), then second by cos(y), ....