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$\sum^n_{i = n+1} 1$ is $0$ or underfined or identity element for the operation it's in? Let's say you have the $\sum^n_{i = n+1} 1$. The things to add don't exist, because $n+1 > n$. What do you do then?
Do you count it as $0$? Because $0$ is the identity element for addition? What if it was multiplying from $n+1$ to $n$? Would that be $1$?
Should you take the identity element for the operation that takes no arguments? Or should you take the identity element for the operation that the sum/product is used in? Example:
$$3\cdot \sum^n_{i = n+1} 1 = 3\cdot 0,$$
because the identity of additon is $0$ or
$$3\cdot \sum^n_{i = n+1} 1 = 3\cdot 1,$$
because the identity of multiplication is $1$?
Is there such a thing as undefined in math? How is the result of operations on $0$ elements that require more defined?
Thanks!
|
An 'empty sum' such as the one you're talking about is defined as 0, because as you say 0 is the additive identity. This is just a convention, though a useful one.
The fact that you then multiply this empty sum by something doesn't change this. The sum still evaluates to 0, which if you multiply by 3 you get 0.
|
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|
A non-nilpotent matrix $A\in \mathbb C^{2 \times2}$ has a square root Is there any quick argument to show that every non-nilpotent matrix $A\in \mathbb C^{2 \times2}$ has a square root? Just the existence without computing it.
Knowing that $A\in \mathbb C^{2 \times2}$ is non-nilpotent basically tells us that at least one eigenvalue is non-zero. But it's far from being diagonalizable.
My text haven't introduced factorisations based on orthogonality yet. So I'm expecting the proof to be based on Jordan form.
|
The result is trivial when $A$ is diagonalisable. So, we only need to consider the case where $A$ is non-diagonalisable and non-nilpotent. Hence $A$ has two equal but nonzero eigenvalues $\lambda$.
By Cayley-Hamilton theorem, $A^2=\operatorname{tr}(A)-\det(A)I=2\lambda A-\lambda^2I$. Therefore $A=\frac1{4\lambda}(A+\lambda I)^2$.
Alternatively, as $A$ is invertible, it has a matrix logarithm. Therefore $A=(e^{(\log A)/2})^2$.
|
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|
Integral: $\int \frac{dx}{\sqrt{x^{2}-x+1}}$ How do I integrate this? $$\int \frac{dx}{\sqrt{x^{2}-x+1}}$$
I tried solving it, and I came up with $\ln\left | \frac{2\sqrt{x^{2}-x+1}+2x-1}{\sqrt{3}} \right |+C$. But the answer key says that the answer should be $\sinh^{-1}\left ( \frac{2x-1}{\sqrt{3}} \right )+C$. Any answer is very appreciated.
|
Notice
$$ x^2 - x + 1 = \left(x-\frac{1}{2}\right)^2 + 1 - \frac{1}{4} = \left(x-\frac{1}{2}\right)^2 + \frac{3}{4}$$
|
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|
Knot theory: Genus of a surface Use Euler characteristic to determine the genus of the surface in Figure 4.24 in picture below.
I am stuck with this question 4.10 from Colin Adams, the Knot Book.
|
I know it isn't using the Euler characteristic, but I couldn't help it. Consider the following 'proof by picture':
Start by rounding everything out, so that your original picture looks like a ball with holes drilled out of it. Then just follow the pictures.
This is an ambient isotopy of the manifold, so the genus is preserved, and it follows that the genus of your original surface is 2.
|
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A 3-minute algebra problem I have just taken advance math test level 2 and there are several problems that have been bugging me. This is the first question:
If $x,y>0$, then determine the value of $x$ that satisfies the system
of equations: \begin{align} x^2+y^2-xy&=3\\ x^2-y^2+\sqrt{6}y&=3\\
\end{align}
I can answer this problem using a 'standard' algebra but it takes time more than 3 minutes. For the sake of curiosity, is there a way to answer this problem less than 3 minutes?
|
Take G.Bach's hint so that $2y=x+\sqrt 6$
Now add the two equations to obtain $$2x^2+(\sqrt 6-x)y=6$$ Substitute for $y$: $$2x^2+\frac 12(\sqrt 6-x)(\sqrt 6+x)=6=\frac 32 x^2+3$$
So that $x^2=2$
|
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Compute $\sum_{n=1}^\infty\frac{1}{(1-z^n)(1-z^{n+1})}z^{n-1}$ and show its uniform convergence Given the power series $$P:=\displaystyle\sum_{n=1}^\infty\frac{1}{(1-z^n)(1-z^{n+1})}z^{n-1}$$ I want to show that $P$ converges uniformly in $\mathbb{C}$ and compute its limit.
I've tried to multiply $P$ with $(1-z)z$ which yields $$\tilde{P}=\displaystyle\sum_{n=1}^\infty\left(\frac{1}{z^{n+1}-1}+\frac{1}{z^n-1}\right)$$ However, I'm unable to continue from here. What's the trick now?
|
Hint: Note that
$$
\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})}
=\frac1{1-z}\left(\frac{z^{n-1}}{1-z^n}-\frac{z^n}{1-z^{n+1}}\right)
$$
Therefore,
$$
\begin{align}
\sum_{n=1}^N\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})}
&=\frac1{1-z}\left(\frac1{1-z}-\frac{z^N}{1-z^{N+1}}\right)\\
&=1-\frac1{1-z}\frac{z^N}{1-z^{N+1}}
\end{align}
$$
You should be able to use this on compact subsets of $|z|\lt1$.
|
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|
How to evaluate $\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)}$ How to evaluate $$\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n-1}{2}\right)}$$, where n is integer > 0?
I know the gamma function formula will give
$$ \frac{(\frac{n-2}{2})!}{(\frac{n-3}{2})!}$$ How to simplify it?
|
This is an elaboration of the hint by Raymond Manzoni.
One of $n/2$ and $(n-1)/2$ is an integer, and the other is a half integer, and while there is a nice expression of the gamma function on integers as a factorial, evaluating the gamma function on half integers is more complicated. However, they can be evaluated using the following facts:
*
*$\Gamma(1/2)=\sqrt{\pi}$
*$\Gamma(x)=(x-1)\Gamma(x-1)$
We then have $\Gamma(n+1/2)=(n-1/2)\Gamma(n-1/2)=(n-1/2)(n-3/2)\Gamma(n-3/2)=\cdots$
Continuing the pattern, we can write
$$\Gamma(n+1/2)=\Gamma(1/2)\left(\frac{1}{2}\frac{3}{2}\cdots \frac{2n-1}{2} \right)=\frac{\sqrt{\pi}(2n-1)!!}{2^n}.$$
Here, we use the double factorial notation $n!!=n(n-2)(n-4)\cdots (1\text{ or } 2)$ However, we can simplify $(2n-1)!!$ and write it in terms of normal factorials and powers of $2$. Namely, we have
$$(2n-1)!!=1 \cdot 3 \cdots (2n-1)=\frac{(2n-1)!}{2\cdot 4 \cdots 2(n-1)}=\frac{(2n-1)!}{2^{n-1}(n-1)!}.$$
|
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|
Distance point on ellipse to centre I'm trying to calculate the distance of a certain point of an ellipse to the centre of that ellipse:
The blue things are known: The lengths of the horizontal major radius and vertical minor radius and the angle of the red line and the x-axis. The red distance is the desired result. It is not given where on the ellipse the point is. It can be anywhere on the ellipse. Is this problem possible? If so, in which can this be solved? Thanks in advance!
After reading Kaj Hansen's comment and trying a bit this is what I did, it still won't work though.
In a triangle, $tan(\theta)=\frac{\text{opposite side}}{\text{adjecent side}}$. The slope of a line is $\frac{\Delta y}{\Delta x}$. Therefor the slope of the red line is $\tan(\theta)$; the formula of the line is $y=\tan(\theta)\cdot x$.
The formula of the ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. When I put the two formulas together I get
$$\frac{x^{2}}{a^{2}}+\frac{(\tan(\theta)\cdot x)^{2}}{b^{2}}=1$$
After a bit of rearranging:
$$x=\pm \sqrt{\frac{a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}}$$
$$$$
$$y=\tan(\theta)\cdot x$$
$$y=\pm \tan(\theta) \cdot \sqrt{\frac{a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}} $$
Now with the help of Pythagoras' theorem $c=\sqrt{a^{2}+b^{2}}$ the red line should be
$$\sqrt{\left ( \sqrt{\frac{a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}}\right ) ^{2}+\left ( \tan(\theta) \cdot \sqrt{\frac{a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}} \right )^{2}}$$
which can be simplified:
$$\sqrt{\frac{a^{2}\cdot b^{2} \cdot (\tan(\theta))^{2}+a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}}$$
This, however, does not give the right answer. Let's try something:
$a=2$; $b=1$; $\theta=\frac{1}{2}\cdot \pi$ (The point is the point where the ellipse intersects with the minor radius)
$$\sqrt{\frac{2^{2}\cdot 1^{2} \cdot (\tan(\frac{1}{2}\cdot \pi))^{2}+2^{2}\cdot 1^{2}}{2^{2}\cdot (\tan(\frac{1}{2}\cdot \pi))^{2}+1^{2}}}$$
$$\sqrt{\frac{4 \cdot (\tan(\frac{1}{2}\cdot \pi))^{2}+4}{4\cdot (\tan(\frac{1}{2} \pi))^{2}+1}}$$
But wait, $\tan(\frac{1}{2}\cdot \pi)$ is undefined. The formula cannot be filled in completely, which is a requirement. I need a formula that can be filled in for every value of $\theta$ on the domain $[0,\frac{1}{2}\cdot \pi]$
|
We know that the Parametric equation of an Ellipse - not centered, and not parallel to the Axises - is:
$$
x(\alpha) = R_x \cos(\alpha) \cos(\theta) - R_y \sin(\alpha) \sin(\theta) + C_x \\
y(\alpha) = R_x \cos(\alpha) \sin(\theta) + R_y \sin(\alpha) \cos(\theta) + C_y
$$
Where:
- $C_x$ is center X.
- $C_y$ is center Y.
- $R_x$ is the major radius.
- $R_y$ is the minor radius.
- $\alpha$ is the parameter, which ranges from 0 to 2π radians.
- $\theta$ is the Ellipse rotation angle.
You can see this equation explanation here: What is the parametric equation of a rotated Ellipse (given the angle of rotation)
So all you need to do is enter your parameters to this equation to find the point (X, Y) and calculate the distance fron the Ellipse center to point (X, Y) on the Ellipse edge.
Distance equation:
$$
distance = \sqrt{(x_1 - x_2) ^ 2 + (y_1 - y_2) ^ 2}
$$
|
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Use Euclid's Algorithm to find the multiplicative inverse Use Euclid's Algorithm to find the multiplicative inverse of $13$ in $\mathbf{Z}_{35}$
Can someone talk me through the steps how to do this? I am really lost on this one.
Thanks
|
Basically you want to find $a$ such that $13a \equiv 1 \mod 35$ which is the same as: $$13a + 35k = 1, \qquad \text{For some }k\in \mathbb Z$$
Use Euclid's Algorithm on $13$ and $35$, the same way as for finding $\gcd(13,35)$.
So start with $35 = 2\cdot13 + 9$ and so on... Then substitute your answers in the line above (called reversing Euclid's Algorithm) and you'll find $a$ and $k$.
Your answer is $a$.
|
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|
Finding volume of the solid of revolution? Can anyone help me with finding the volume of a solid of revolution of f(x) about the x axis for the interval [1,6]. It's supposed to be able to be done without needing calculus but I am having trouble figuring it out.
$f(x) =
\begin{cases}
1 & 1 \leq x< 2\\
1/2 & 2 \leq x< 3\\
. & .\\
. & .\\
1/n & n\leq x< n+1\\
\end{cases}$
I know the volume would be found like this $\pi$ $\int_{1}^{6}(f(x))^2dx$ but I am unsure about how to go about it with this function.
Any help is appreciated.
Thanks
|
$f(x)$ is constant in $n$ intervals. Hence, $\int^b_a (f(x))^2dx=(f(x))^2(b-a)$. So, the volume is simply
$$\pi\left(1^2(2-1)+2^2(3-2)+\cdots+\dfrac{1}{n^2}(n+1-n)\right)=\pi\left(1^2+2^2+\cdots+\dfrac{1}{n^2}\right)$$
Though I believe that the $2$ should actually be $\dfrac{1}{2}$.
|
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If $\int_0^1 f(x) e^{nx} dx = 0$ for every n, then f=0 $f$ be a continuous function [0,1] to $R$.
$\int_0^1 f(x)e^{nx} dx = 0$ for all $n \in N\cup\{0\}$
how to prove $f(x)= 0$ in $[0,1]$ for all $x\in[0,1]$?
I solved "$\int_0^1 f(x)x^n dx = 0$ for all $n \in N\cup\{0\}$" with weierstrass theorem
|
You haved showed that if $\int_a^b f(x)x^n dx = 0$, then $f=0$. well, we will use it.
Let $e^x=y$,
$$\int_0^1 f(x)e^{nx}\, \mathrm dx=\int_1^e f(\ln y)y^{n-1}\,\mathrm dy$$
so $\int_1^e f(\ln x)x^{n-1}\, \mathrm dx=0$ for all $n \gt 0$. Hence
$$f(\ln x)=0, x\in [1,e]$$
that is $f(x)=0, x\in [0,1]$
|
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Is the particle in a ring a regular Sturm-Liouville problem? The problem of a particle in a ring is a well-known eigenvalue problem $$\frac{d^2}{d\theta^2} \psi(\theta) + V_0 \psi(\theta) = \lambda \psi(\theta)$$in physics and the Schrödinger equation has a Sturm-Liouville like form. The problem seems to be that the boundary condition $\psi(\theta + 2\pi) = \psi(\theta)$ does not seem to be very Sturm-Liouville like. Therefore, I am wondering: To which type of eigenvalue problem does this correspond?
|
Typical periodic Sturm-Liouville requires two conditions: $\psi(0)=\psi(2\pi)$ and $\psi'(0)=\psi'(2\pi)$.
|
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|
Series: Let $S=\sum\limits_{n=1}^\infty a_n$ be an infinite series such that $S_N=4-\frac{2}{N^2}$. Let $S=\sum\limits_{n=1}^\infty a_n$ be an infinite series such that $S_N=4-\frac{2}{N^2}$.
(a) Find a general formula for $a_n$.
(b) Find the sum $\sum\limits_{n=1}^\infty a_n$.
Can you explain to me how I can convert the partial sum to the general equation?
(a) What are the values of
$\sum\limits_{n=1}^{10} a_n$ and $\sum\limits_{n=4}^{16} a_n$?
$\sum\limits_{n=1}^{10} a_n=23433271/635040$
$\sum\limits_{n=4}^{16} a_n= 15799025474051/259718659200 $
(b) What is the value of $a_3$?
$167/18$
Why aren't these values correct as well?
|
Note that $S_{N}-S_{N-1}=a_N$, so that shall give you $a_n$
|
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|
Exponential growth precalc population The population of City A increases by 8% every 10 years. The population of City B triples every 120 years. The two cities had equal populations of 10,000 residents each in the year 2000. In what year will city B have twice as many residents as city A?
|
Let $t$ the time in years with $t=0$ correspondig to year $2000$, $P_A(t)$ and $P_B(t)$ can be written by
\begin{align}
P_A(t)&=(1.08)^{t/10}P_0& \text{and} & & P_B(t)&=3^{t/120}P_0
\end{align}
where $P_0=10000$, $P_B(t)=2P_A(t)\iff (1.08)^{t/10}=2\cdot3^{t/120}$, taking logs we have
\begin{align}
\frac{t}{120}\log3&=\log 2+\frac{t}{10}\log(1.08)\\
\left(\frac{1}{120}\log3-\frac{1}{10}\log(1.08)\right)t&=\log2\\
t&=\frac{120\log2}{\log3-12\log(1.08)}\\
&\approx 475.
\end{align}
Therefore, in year 2475 the population of City B will be twice that of the City A.
|
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Lipschitz and uniform continuity Show that f(x)=$\sqrt{x}$ is uniformly continuous, but not Lipschitz continuous.
I can prove that it's uniformly continuous. But why is it not Lipschits? How do I check the definition?
|
Given any $M>0$, choose $c=0$ and $ 0<x<\frac{1}{m^{2}}$, so that $\frac{1}{\sqrt{x}}>M$. then we have:
$$ \frac{|f(x)-f(c)|}{|x-c|}=\frac{|\sqrt{x}|}{|x|}=\frac{1}{\sqrt{x}}>M $$
Since $M$ was arbitrary, this shows that $F$ is not Lipschitz continuous.
|
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Calculating Eigenvalues is only Assume that the following is used:
$$
A = \begin{pmatrix}
0& 1&\\
2& 3&\\
4& 5&\\
6& 7&\\
8& 9&
\end{pmatrix}
$$
Then calculating the Coveriance matrix, which, gives me:
$$
C = \begin{pmatrix}
40& 40&\\
40& 40&\\
\end{pmatrix}
$$
Then using the following:
$$
det = (a+b) \cdot (a+b)-4 \cdot(a \cdot b - c \cdot c),
$$
where in this case, $a = 40, b = 40, c = 40$ gives the answer:
$$
\lambda_{1} = 80, \\
\lambda_{2} = 0,
$$
These are therefore the correct Eigen values. However, using this formula, if I have the following:
$$
A = \begin{pmatrix}
-4& -2&\\
-1& -3&\\
4& 5&\\
6& 7&\\
8& 9&
\end{pmatrix},
$$
where the Covariance matrix is given:
$$
C = \begin{pmatrix}
99.2& 103.4&\\
103.4& 116.8&\\
\end{pmatrix},
$$
gives the Eigenvalues as:
$$
\lambda_{1} = 218.119 \\
\lambda_{2} = -15.5189
$$
When the actual values are:
$$
\lambda_{1} =211.774 \\
\lambda_{2} = 4.226
$$
Could anyone tell me where I am calculating this wrong please?
EDIT:
For
$\lambda_{1} = (a + b + det)/2 \\
\lambda_{2} = (a + b - det)/2
$
|
Have you double-checked your calculations? The eigenvalues of any $2\times2$ real symmetric matrix $\pmatrix{a&c\\ c&b}$ are given by
$$\frac{a+b\pm\sqrt{(a+b)^2-4(ab-c^2)}}2.$$
Plug in the entries of $C$, I don't find any discrepancies.
|
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Is this possible physically? The length $x$ of a rectangle is decreasing at the rate of $5 \, cm/min$ and the width $y$ is increasing at the rate of $4\, cm/min$. When $x=8\, cm$ and $y=6 \,cm$, find the rates of change of
(a) the perimeter and (b) the area of rectangle.
$$ \frac{dx}{dt} =-5\,\,cm/min, \, \frac{dy}{dt} =4\,\, cm/min$$
(a) $\frac{dP}{dt} =-2\, cm/min$ (perimeter decreases at the rate of $2\, cm/sec$)
(b) $\frac{dA}{dt} = 2 \, cm^2/sec$ (area of the rectangle is increasing at the rate of $2 \,cm^2 /min$)
Now I ask that is this possible in the physical sense?
|
So the question is:
Can the perimeter decrease and the area increase?
Absolutely. Consider a rectangle with sides 3 and 4. It has area 12 and perimeter 14.
Now make the short side a bit longer and the long side a bit shorter, we get a new rectangle with sides (for example) 3.1 and 3.89. It has a larger area $$3.1 \cdot 3.89 = 12.059 > 12$$but a shorter perimeter $$2 \cdot 3.1 + 2 \cdot 3.89 = 13.98 < 14$$
|
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|
A function such that $f(f(n)) = -n$? This question from somebody's job interview made me puzzled:
Design a function f, such that:
$f(f(n)) = -n$ ,
where n is a 32 bit signed integer; you can't use complex numbers arithmetic. If you can't design such a function for the whole range of numbers, design it for the largest range possible.
Do you have any idea how to tackle it?
EDIT: I removed "Real number" from the title, since it just confused readers, but it was meant just to stress that complex numbers are not allowed.
|
This answer to a related question supplies the relevant analysis.
Assuming that a "signed 32-bit integer" means that you are considering the domain and range of $f$ to be the integers in the interval $[-2^{31}, 2^{31})$, a simple counting argument shows that the largest domain on which the identity $f(f(x))=-x$ can hold must be missing at least three points.
(aside: $-2^{31}$ must be excluded, because it is outside of the domain of unary negation. The interesting part is that we must exclude two other points too)
The sample function from that answer can be used:
$$ f(x) = \begin{cases} 0 & x == 0
\\ x+1 & x>0 \wedge \text{$x$ is odd}
\\ 1-x & x>0 \wedge \text{$x$ is even}
\\ x-1 & x<0 \wedge \text{$x$ is odd}
\\ -1-x & x<0 \wedge \text{$x$ is even}
\end{cases} $$
which has the desired property on $[-2^{31} + 2, 2^{31} - 2]$.
I won't discuss implementing a solution with bit twiddling: such a topic is beyond the scope of this forum.
EDIT: if we instead want the domain of $f$ to be the entire interval, we can make the identity hold for one more point. If we take a function $f$ given as above, and let the three points that are excluded be $a$, $-a$, and $2^{-31}$, then we can extend $f$ by setting
*
*$f(a) = 2^{-31}$
*$f(2^{-31}) = -a$
*$f(-a) = $ anything
and thus with this extension, $f(f(a)) = -a$ as well.
|
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|
Solving bernoulli differential equation How to solve $$t \frac{dy}{dt} + y = t^4 y^3$$
First I divided by $t$ to get $$\frac{dy}{dt} + \frac{y}{t} = t^3 y^3$$
Then I multiplied through by $y^{-3}$ to get $$y^{-3} \frac{dy}{dt} + \frac1{ty^2} = t^3$$
Then I used the subsitution $w = y^{-2}$ and $w'=-2y^{-3}\frac{dy}{dt}$
Subbing that in the equation becomes $$\frac1{y^3} \frac{dy}{dt} + \frac1{ty^2} = t^3$$
How do I solve it from here as I still have a $\frac1{ty^2}$ instead of just $y$ making it linear?
Also are all the rest of my workings correct?
|
You almost find the solution. You've made a good substitution by letting $w=\frac{1}{y^2}$ and $w'=-\frac{2y'}{y^3}$, then
\begin{align}
\frac{1}{y^3}\frac{dy}{dt}+\frac{1}{ty^2}&=t^3\qquad\rightarrow\qquad-\frac{1}{2}w'+\frac{w}{t}=t^3\qquad\rightarrow\qquad w'-\frac{2}{t}w=-2t^3
\end{align}
The first-order nonlinear ODE becomes a first-order linear ODE. Take the integrating factor
\begin{align}
e^{-\Large\int\;\frac{2}{t}\,dt}=e^{-2\ln t+C}=e^{\ln t^{-2}}e^C=At^{-2}
\end{align}
then multiply both sides of the first-order linear ODE by $At^{-2}$.
\begin{align}
At^{-2}w'-At^{-2}\frac{2}{t}w&=-2t^3At^{-2}\\
At^{-2}w'-2At^{-3}w&=-2At\\
\frac{d}{dt}\left(At^{-2}w\right)&=-2At\\
d\left(At^{-2}w\right)&=-2At\;dt
\end{align}
The last step is integrating both sides.
|
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Mathematicians ahead of their time? It is said that in every field there’s that person who was years ahead of their time. For instance, Paul Morphy (born 1837) is said to have retired from chess because he found no one to match his technique that very much resembled modern chess theory.
So, who was the Paul Morphy of mathematics?
|
Simon Stevin (1548 – 1620) had the unbelievable insight of thinking of an arbitrary number in terms of its unending decimal expansion. He was ahead of his time in the sense that the full significance of what he did was not appreciated until the 1870s when more abstract versions of the construction of the reals were given by Charles Méray and others. Unending decimals inspired Newton to introduce more general power series. The rest is history.
Another amazing contribution of Stevin was his argument immediately recognizable as a proof of the intermediate value theorem, which was not "officially" proved until the 1820s by Cauchy. See this article. Here he was ahead of his time by at least 200 years.
|
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|
Determine if $\displaystyle\sum_{k=1}^\infty\sin(kx)\sin\left(\cfrac{1}{kx}\right)$ is convergent Determine if the following series converges absolutely:
$$\sum_{k=1}^\infty \sin(kx)$$
and
$$\sum_{k=1}^\infty\sin(kx)\sin\left(\cfrac{1}{kx}\right)$$
I know how to deal with whether they converge. First one diverge by n-term test, second one converge by Dirichlet's Test. However, I do not know how to deal with whether they converge absolutely. Can someone teach me how to determine if both series converge absolutely. Thanks.
|
The series
$$\tag1 \sum_{k=1}^\infty \sin(kx)$$
converges absolutely for $x\in\pi \mathbb Z$. For all other cases, it doesn' even converge:
Note that
$$ \sin((k+1)x)=\sin(kx)\cos x+\cos(kx)\sin x$$
hence whenever $\sin(kx)\approx 0$ and hence $|\cos(kx)|\approx1$, then $|\sin((k+1)x)|\approx|\sin x|$.
More precisely, if $\sin x\ne0$, we can find $a>0$ with
$$ \sqrt{1-a^2}\cdot |\sin x|>2a.$$
Then $|\sin(kx)|<a$ implies $|\cos(kx)|>\sqrt{1-a^2}$ and hence
$$ |\sin((k+1)x)\ge |\cos(kx)\sin x|-|\sin(kx)\cos x|> \sqrt{1-a^2}|\sin x|-a>a$$
We conclude that $\sin(kx)\not\to 0$ unless $\sin x=0$.
In summary, the series $(1)$
*
*converges absolutely (trivially) for $x\in\pi\mathbb Z$
*diverges for $x\notin\pi\mathbb Z$.
We have seen above that for $x\notin\pi\mathbb Z$, at least one of two successive summands is $>a$ in absolute value.
Apply this to
$$\tag2 \sum_{k=1}^\infty \left|\sin(kx)\sin(\tfrac1{kx})\right|$$
For $k$ sufficiently large we have $\sin\frac1{kx}\approx\frac1{kx}$, say $|\sin\frac1{kx}|\ge\frac1{2k|x|}$.
Then for such $k$ we have $$\left|\sin(kx)\sin(\tfrac1{kx})\right|+\left|\sin((k+1)x)\sin(\tfrac1{(k+1)x})\right| \ge \frac{a}{2(k+1)|x|}.$$
We conclude by comparision with the harmonic series that $(2)$ diverges for $x\notin\pi\mathbb Z$. In summary, the series
$$ \sum_{k=1}^\infty\sin(kx)\sin\frac1{kx}$$
*
*is not defined for $x=0$
*converges absolutely (trivially) for $x\in\pi\mathbb Z\setminus\{0\}$
*converges (as you have shown, by the Dirichlet test), but not absolutely, for $x\in\mathbb R\setminus\pi\mathbb Z$.
|
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Understanding why $f(z) \cdot (z - p_1)^{n_1} \cdot \ldots \cdot (z - p_k)^{n_k}$ is analytic under a condition Setting: Let $f$ be meromorphic on $\mathbb{\hat{C}} = \mathbb{C} \cup \{\infty\}$. Let $\{p_i\}$ be the $k$ number of poles of $f$. Let $n_i$ denote the orders of each of the $p_i$.
Question: Why is it that
$$
g(z) = f(z) \cdot (z - p_1)^{n_1} \cdot \ldots \cdot (z - p_k)^{n_k}
$$
is an analytic function on $\mathbb{C}$? It seems to me that that given $f(z)$ is still a factor of $g(z)$, that $g(p_i)$ is still going to be undefined for all $i \in \{1, \ldots , k\}$.
Note: The motivation for this post is in trying to understand a statement made in this discussion.
|
If $f$ had a pole $z_1$ with order $a$. Then $g(z)=f(z)(z-z_1)^a$ is continuous in $z_1$ and holomorphic in a region of $z_1$. You can see then that it is holomorphic to this region .
|
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Conditions on certain entries of a matrix to ensure one Jordan block per eigenvalue In preparation for a future exam, I found the following problem:
Let $$A = \begin{pmatrix} 1 & 0 & a & b \\ 0 & 1 & 0 & 0 \\ 0 & c & 3 & -2 \\ 0 & d & 2 & -1 \end{pmatrix}$$ Determine conditions on $a,b,c,d$ so that there is only one Jordan block for each eigenvalue of $A$ (in the Jordan canonical form of $A$).
I found, readily enough, that the only eigenvalue of the $A$ is $\lambda = 1$. So the problem is in fact to find conditions of $a,b,c,d$ such that $A$ is similar to a single-block Jordan matrix. Thus we want the dimension of the eigenspace $E_1$ to be 1, or equivalently, the rank of $A-I$ to be 3.
$$A-I = \begin{pmatrix} 0 & 0 & a & b \\ 0 & 0 & 0 & 0 \\ 0 & c & 2 & -2 \\ 0 & d & 2 & -2 \end{pmatrix}$$
So would the conditions $a,b$ not both zero, $a \neq -b$ and $c \neq d$ be sufficient? More to the point, what is the most economical way of stating the conditions on $a,b,c,d$ to ensure the desired outcome?
|
Another approach: There is only one Jordan Block for each eigenvalue of a matrix when the minimum polynomial is the same as the characteristic polynomial. You have the characteristic equation: $(x-1)^4$, so you just need the minimum polynomial to be the same. This is equivalent to $(A-I)^3 \neq 0$. Now \begin{equation} (A-I)^3 = \begin{bmatrix} 0 & a (2 c - 2 d) + b (2 c - 2 d) &0 &0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}. \end{equation} So if we don't want this matrix to be the zero matrix, we must have $a (2 c - 2 d) + b (2 c - 2 d)\neq0$, which is the same conclusion arrived at by Brian Fitzpatrick...so this just shows a different way to get there...
|
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Solutions to differential equation Let $\left\{a,\lambda\right\}\subset\mathbb{R}$.
Let the following differential equation for a function $x\left(t\right)\in\mathbb{R}^{\mathbb{R}}$ be given:
$$ \boxed {\ddot{x}\left(t\right)=4\lambda\left(x\left(t\right)^{2}-a^{2}\right)x\left(t\right) } $$
I am trying to find all solutions to this equation which obey the following boundary condition: $x\left(-\infty\right)=-a$ and $x\left(\infty\right)=a$.
I have found one such family of solutions, indexed by $\tau\in\mathbb{R}$: $$\boxed{x\left(t\right)=a\,\tanh\left[\frac{\omega}{2}\left(t-\tau\right)\right]}$$ It is easy to verify this is indeed a solution.
My question is:
*
*Is it the only set of solutions for these boundary conditions? If yes, how to prove no others exist? If not, what are all the other solutions?
*I suspect that there is another set of (at least approximate) solutions: $$ \boxed{x\left(t\right) = a\prod_{j=1}^{n} \tanh\left[\frac{\omega}{2}\left(t-\tau_j\right)\right] }$$ where $n\in2\mathbb{N}+1$, and $\left\{\tau_j\right\}_{j=1}^n\subset\mathbb{R}$ are such that $\tau_j < \tau_{j+1} \forall j\in\left\{1,\dots,n-1\right\}$. If these are solutions, how do you prove that? (I tried induction and failed) If they are not solutions, in what way are they approximate solutions? (what is the margin of error?)
This problem comes from trying to find instanton solutions to the double well potential in quantum mechanics (see Coleman "Aspects of Symmetry" page 272).
|
These are indeed all solutions. Multiplying the equation by $2\dot{x}$, one easily finds a first integral (the energy):
$$\dot{x}^2=2\lambda \left(x^2-a^2\right)^2+E,\tag{1}$$
and then boundary conditions imply $E=0$.
Now write $x=a\tanh u$, then (1) with $E=0$ gives the equation $\dot{u}^2=2\lambda a^2$, with the only solutions given by linear functions.
|
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Convergence/Divergence of $\sum_{n=2}^{\infty} \frac{2^n}{3^n+4^n}$ Using Comparison Test Use the comparison test to show if the series converges/diverges?
$\sum_{n=2}^{\infty} \frac{2^n}{3^n+4^n}$
|
Hint:
$\dfrac{2^n}{3^n+4^n}\leq\dfrac{2^n}{3^n}$
Does $\dfrac{2^n}{3^n}$ converge?
|
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Why are roots (x-intercept) called the solutions of a quadratic equation? As I understand it, all the points in the hyperbole are the solutions. Although I see that it's easier to draw the curve after you discovered the roots, I don't see why to call them 'the solutions.' It has infinite solutions.
|
It is by definition. The x's that make the entire equation equal 0 are called the "solutions" to the equation. In other words, the x coordinates of the points where the graph intersects the x axis would be the solutions or otherwise called zeros.
|
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Tricky Triangle Area Problem This was from a recent math competition that I was in. So, a triangle has sides $2$ , $5$, and $\sqrt{33}$. How can I derive the area? I can't use a calculator, and (the form of) Heron's formula (that I had memorized) is impossible with the$\sqrt{33}$ in it. How could I have done this? The answer was $2\sqrt{6}$ if it helps.
Edited to add that it was a multiple choice question, with possible answers:
a. $2\sqrt{6}$
b. $5$
c. $3\sqrt{6}$
d. $5\sqrt{6}$
|
From the law of cosines ($C^2 = A^2 + B^2 - 2AB\cos \theta$), we get that $(\sqrt{33})^2 = 2^2 + 5^2 - 2 \cdot 2 \cdot 5 \cos \theta$.
Simplifying this, we get $33 = 29 - 20 \cos \theta$, which means that $\displaystyle \cos \theta = -\frac{1}{5}$
Because $\cos^2 \theta + \sin^2 \theta = 1$, we get that $\displaystyle \sin^2 \theta = \frac{24}{25}$. This means that $\displaystyle \sin \theta = \frac{2\sqrt{6}}{5}$ (note that, because $0 \le \theta \le \pi$, $\sin \theta \ge 0$).
The area of the triangle is $\displaystyle \frac{1}{2} A B \sin \theta = \frac{1}{2} \cdot 2 \cdot 5 \cdot \frac{2\sqrt{6}}{5} = 2\sqrt{6}$.
|
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Point separating function I'm trying to prove that the set $P = \{p: [0,1] \times [0,1] \to R \; \mid \; \text{p is a polynomial}\}$ is dense in $C( [0,1] \times [0,1], R)$. I'm stuck trying to find a points separating function. (obviously using the Stone Wierestrass approx).
Here is what I got so far: I thing the function that is the graph of $x$ and $\sin(x)$ should be points separating right?
Thanks!!
|
I'm sorry I misunderstood your notation initially.
$x \mapsto \sin(x)$ isn't a polynomial function, so we'll have to try something else.
Here's something. Pick a point $(x_0,y_0) \in [0,1]^2$. Then the square of the distance between $(x_0,y_0)$ and another point $(x,y)$ is a familiar polynomial in $x$ and $y$, and is only $0$ if $(x,y) = (x_0,y_0)$.
Note that this is actually way more than we need; this function works for our initial point $(x_0,y_0)$ and any other point.
|
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Why a linear numerator for fractions with irreducible denominators? For example: (2x^3+5x+1)/((x^2+4)(x^2+x+2)) breaks down to (ax+b/(x^2+4))+(cx+d/(x^2+x+2)). I have been told that since the denominators are irreducible, the numerators will be either linear or constant. Now my question is for something like (2x^3+5x+1/(x^2-4)) you would make it equal (A/(x+2))+(B/(x-2)), why do assume that the numerators are constant? Why couldn't the numerators be linear like the irreducible one?
|
With a fraction like $ax+b\over cx+d$ it is always possible to remove $a\over c$ copies of the denominator from the numerator, leaving a constant in the numerator:
$${ax+b\over cx+d}={ax+b-ax-\frac ac d\over cx+d}+\frac ac={b-\frac{ad}c\over cx+d}+\frac ac$$
In other words, we have the choice of leaving the fraction in a "linear over linear" form, or changing it to be in a "constant over linear plus constant" form. Note that a function like $\dfrac a{bx+c}+d$ is easier to graph than one like $\dfrac{ax+b}{cx+d}$ because of the constant offsets compared with the affine transformation involved in the second case.
|
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How to prove that this function is continuous? If $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ is continuous on the rectangle $R=[a,b] \times [c,d]$, prove that the function
$g(x) := \int\limits_{c}^{d} f(x,y) dy$
is continuous on $[a,b]$.
Thanks in advance!
|
Here is how you advance. We need to prove that
$\forall \epsilon >0 $ thers exists $\delta$ such that
$$ |h|<\delta \implies |g(x+h) - g(x)| < \epsilon $$
We advance as
$$ g(x+h) - g(x)= \int\limits_{c}^{d} f(x+h,y) dy - \int\limits_{c}^{d} f(x,y) dy $$
$$ = \int\limits_{c}^{d} (f(x+h,y) - f(x,y)) dy $$
$$ \implies |g(x+h) - g(x)| =\Bigg| \int\limits_{c}^{d} (f(x+h,y) - f(x,y)) dy \Bigg| $$
$$\implies |g(x+h) - g(x)| \leq \int\limits_{c}^{d} |f(x+h,y) - f(x,y)| dy. $$
You need to use the fact that the function $f(x,y)$ is continuous on $[a,b]\times [c,d]$ to finish the problem.
|
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Why is max($\frac{2}{||w||}$)= min($\frac{1}{2}$)($||w||^2$)? I was watching a video on machine learning.
The instructor says that maximizing ($\frac{2}{||w||}$)is difficult (why?) so instead we prefer to minimize $\frac{1}{2}||w||^2$. $w$ is a vector.
How are these two functions equivalent?
****ADDENDUM ****
I plotted both statements. So I can see that the max of ($\frac{2}{||w||}$) is at 0 (it reaches infinity) and the min of $\frac{1}{2}||w||^2$ is 0. Am I thinking of this correctly?
Isn't it obvious to begin with? I couldve figured out that the max of ($\frac{2}{||w||}$) is 0 without doing any fancy conversion.
|
Let $x = ||w||$, then $\dfrac{2}{||w||} = \dfrac{2}{x}$. So if $\dfrac{2}{x} \leq M$, then:$x \geq \dfrac{2}{M}$, and it follows that: $x^2 \geq \dfrac{4}{M^2}$. This means that instead of finding the max of $\dfrac{2}{x}$ which can be cumbersome, you equivalently solve for the min of $x^2$ which is much simpler because you will deal painlessly with a much simpler object: a polynomial of second degree instead of a painful rational expression !
|
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The sign representation of the Symmetric Group I am currently trying to learn some of the basics of Representation Theory through Sagan's The Symmetric Group: Representations, Combinatorial Algorithms, and Symmetric Functions.
On page 11, after briefly describing the concept of a submodule, he provides an example with $G:=\{g_1,g_2,\ldots,g_n\}$ and $V:=\mathbb{C}[G]$. Furthermore, he proves that $W:=\mathbb{C}[g_1+\dotsb+g_n]$ is a submodule of $V$ on which $G$ acts trivially.
The reader should verify that if $G= S_n$, then the sign representation can also be recovered by using the submodule $$W:=\mathbb{C}\left[\sum_{\pi\in S_n}sgn(\pi)\pi\right].$$
I'm not completely sure what he means by that. I first proved that $W$ is a submodule of $V:=\mathbb{C}[S_n]$ as
$$\sum_{\pi\in S_n}sgn(\pi)\sigma\pi =\pm\sum_{\pi\in S_n}sgn(\pi)\pi$$
holds for every $\sigma\in G$.
What would be my next step in order to show that the sign representation can be recovered from this module?
|
It suffices to prove that $$\sigma\cdot w=sgn(\sigma)w$$ for all $w\in W$. Indeed, in this case, any vector space isomorphism $W\rightarrow\mathbb{C}$ can be seen to be an isomorphism of $W$ with the sign representation.
Note that $$\sigma\left(\sum_{\pi\in S_n}sgn(\pi)\pi\right)=\sum_{\pi\in S_n}sgn(\pi)\sigma\pi$$ $$=\sum_{\pi\in S_n}sgn(\sigma)sgn(\sigma\pi)\sigma\pi$$ $$=sgn(\sigma)\sum_{\pi\in S_n}sgn(\pi)\pi,$$ as desired.
|
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Trigonometric identity involving sum of "Dirichlet kernel like" fractions Computing the eigenvalues of a matrix related to fast Fourier transform, we stumbled upon the following identities.
Let $k$ and $N$ be natural numbers with $k<N$, then:
$$\sum\limits_{j=1}^N (-1)^{j+1}\frac
{\sin \left( \frac{2\pi j k}{2N+1}\right)}
{\sin\left( \frac{\pi j}{2N+1}\right)}=k$$
and
$$\sum \limits_{j=1}^{N}(-1)^{j+1}
\sin \left( \frac{\pi j k}{N}\right)
\cot \left( \frac{\pi j}{2N} \right)=k.$$
How can these be proven? (This is not a homework question. I have seen lots of similar formulae but not exactly these ones. These can be reformulated in various ways, e.g., the first is simultaneously an identity for Chebyshev polynomials of second kind.)
|
(In the spirit of Dirichlet kernel &c) let's rewrite the sum as
$$
\sum_{j=1}^N(-1)^{j+1}2\left[\cos\left(\frac{\pi j}{2N+1}\right)+\cos\left(\frac{3\pi j}{2N+1}\right)+\dots+\cos\left(\frac{(2k+1)\pi j}{2N+1}\right)\right]
$$
and change the order of summation — since
$$
2\sum_{j=1}^N(-1)^{j+1}\cos\left(\frac{(2l+1)\pi j}{2N+1}\right)=
\sum_{j=1}^{2N}(-1)^{j+1}\cos\left(\frac{(2l+1)\pi j}{2N+1}\right)=1
$$
we are done.
|
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Calculate $ \frac{\partial f}{\partial x} (x,y) $ Calculate $ \frac{\partial f}{\partial x} (x,y) $ of
$$ f(x,y) = \int_{x^2}^{y^2} e^{-t^2}\, dt$$
|
Let us call $E=E(t)$ a fixed antiderivative of $e^{-t^2}$. Then $$f(x,y)=E(y^2)-E(x^2).$$ Now, $$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( E(y^2)-E(x^2) \right) = -2x E'(x^2)=-2x e^{-x^4}.$$
|
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$n$ is a square and a cube $a^2 = n = b^3\Rightarrow n\equiv 0,1\pmod{7}$
Verify that if an integer is simultaneously a square and a cube, then it must be either of the form ${7k}$ or ${7k +1}$.
I have no idea on how to proceed.
|
If an integer $a$ is simultaneously a square and a cube, it must be the sixth power of an integer $b$
Now as $7$ is prime either $7|b$ or $(7,b)=1$
If $7|b, 7|b^n$ for integer $n\ge1$
Else by Fermat's Little Theorem , $7|(b^{7-1}-1)$
|
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|
Factorising $D^{2n} - I$ and $D^n - I$ $\rm(ii)$ Let $\mathcal P_{11}(\Bbb R)$ be the vector space of polynomials of degree $\leqslant11$ over the field $\Bbb R$ and let $D:\mathcal P_{11}(\Bbb R)\to\mathcal P_{11}(\Bbb R)$ be the linear map given by differentiation. Write down the least positive integer $n$ for which $D^{2n}=0$ on $\mathcal P_{11}(\Bbb R)$. By factorisation of the formal expressions $D^{2n}-I$, $D^n-1$ or otherwise, show that the mapping $$D^9-D^6+D^3-I:\mathcal P_{11}(\Bbb R)\to\mathcal P_{11}(\Bbb R)$$
I can't seem to get a factor of $D^9-D^6 + D^3 - I$,could someone show me how to get the factor they require? I'm assuming it's in the form of $D^{12}-I$, but all I have is $D^{12} - I = (D^4-I)(D^8+D^4+I)$ by guess work
|
Hint: Use polynomial division.
|
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|
$\mathbb{C}\otimes_\mathbb{C} \mathbb{C} \cong \mathbb{R}\otimes _\mathbb{R} \mathbb{C}$ I am trying to show that $\mathbb{C}\otimes_\mathbb{C} \mathbb{C} \cong \mathbb{R}\otimes _\mathbb{R} \mathbb{C}$ as abelian groups.
I've tried to come up with various maps but gotten nowhere.
Thanks for any help
|
The groups $\mathbb C\otimes_{\mathbb C}\mathbb C$ and $\mathbb R\otimes_{\mathbb R}\mathbb C$ are both isomorphic to $\mathbb C$. More general, for every $\mathbb C$-vector space $V$ you have $\mathbb C\otimes_{\mathbb C}V\cong V\cong\mathbb R\otimes_{\mathbb R}V$ because for any unitary (commutative) ring $A$ and any $A$-module $M$ you have $A\otimes_AM\cong M$.
A proof of this isomorphism.
The function $A\times M\to M:(a,x)\mapsto ax$ is $A$ bilinear, hence it induces a group homomorphism $h:A\otimes_A M\to M:a\otimes x\to ax$. Since $1\in A$ this is surjective, while injectivity follows from $0=h(\sum_ia_i\otimes x_i)=\sum_i a_ix_i$ implies $\sum_ia_i\otimes x_i=1\otimes(\sum_ia_i x_i)=0$.
|
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1-dim. linear finite elements, von neumann boundary condition. We have the inhomogeneous poisson equation $-\Delta u = f$ on the unit interval and a non-uniform mesh $0=x_0 \ldots x_N=1$. When we write down the stiffness matrix we get a linear system of dimension $N+1$.
Enforcing Dirichlet boundary conditions, the dimension reduces by 1 for each side of the boundary.
What about von Neumann boundary conditions? For example, if I use a Dirichlet boundary condition $u(1)=b$ and a von neumann boundary condition $u'(0)=a$ how do the dimensions change?
First of all the Dirichlet side will cut the dimension by 1 again. It seems to me that the dimension should stay at $N$ even with von neumann at the other side but I don't know how to show it.
|
A Neumann boundary condition simply gets into the system of equations through the Green's first identity when applied to the Laplacian term, i.e.:
$$ \int_L u_{xx} \, \omega \, dx = u_x(1) \omega(1) - u_x(0) \omega(0) - \int_L u_x \omega_x \, dx, $$ where $\omega$ is the test function used in your FEM approach. Since you have $u(1) = b$, you may enforce $\omega(1) = 0$, but on the other hand, for the VN b.c. just substitute it back in the previous equation, yielding:
$$ \int_L u_{xx} \, \omega \, dx = -a \, \omega(1) - \int_L u_x \omega_x \, dx, $$ which will be further simplified if you conveniently choose $\omega(x)$ such that $\omega(0) = 1$, i.e., for example choosing $\omega(x)$ as the family of piecewise linear polynomials in $[0,1]$, known as hat functions. Therefore, we would have:
$$u(x) = \sum_{i=0}^N u_n \omega_n (x) = u_N \omega_N(x) + \sum_{i=0}^{N-1} u_n \omega_n(x), $$ where $u_n$ are the unknowns. The Dirichlet boundary condition tells us:
$$u(1) = u_N \cdot 1 + 0 = b \Rightarrow u_N = b, $$ and we have reduced the order of the system by 1, as you pointed out before. Notice now that substituting the VN b.c. would tell us nothing and therefore $u_N$ remains unknown.
I hope this helps.
Cheers!
|
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|
Real Analysis convergence help I want to show that $\sum \frac{x_{k}}{(1+ x_{k})}$ diverges if $\sum x_{k}$ diverges. I am asked to consider separately where the sequence is bounded and when it is not.
For the bounded case I believe that I have been able to show converges by the Comparison Test.
However, I am stuck at the unbounded case. So far, all I know is that the sequence is less than one, and each term is close to but less than one so it shouldn't converge.
Thanks!
Thanks everyone for your answers! I get it now!
|
In the unbounded case, $x_k/(1+x_k)$ becomes close to $1$ and hence bounded away from zero, for infinitely many $k$. Thus the series cannot converge because the terms do not go to zero. The hard case is when $\{x_k\}$ is bounded.
|
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|
Finding the Polar Equation of a line
Hello, this problem is giving me some trouble. I'm confused as to how I find the equation of the line. I know that x^2+y^2 = r^2, x=r$\cos\theta$, y=r$\sin\theta$, and that tan$\theta$= y/x. I'm just unsure as to how I can derive the polar equation.
|
The equation of the circle is $$x^2+y^2=370$$ Differentiating implicitly, we find $$2x+2y\frac{dy}{dx}=0$$ so $$\frac{dy}{dx}=-\frac{x}{y}$$So the slope of your line is $-\frac{19}{3}$. Therefore, in Cartesian coordinates, the slope of the line is $$y=-\frac{19}{3}(x+19)-3$$Now substitute $x=r\cos\theta$ and $y=r\sin\theta$ and solve for $r$.
Alternatively, if you don't want to use calculus, you can use a little elementary geometry: tangents are perpendicular to radii drawn to the point of tangency. Here, the slope of the line containing the radius from $(-19,-3)$ is obviously $\frac{3}{19}$, and perpendicular lines have slopes that are negative reciprocals, so the slope of the tangent line is $-\frac{19}{3}$.
|
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|
Limit as x approches infinity - Trouble with calculus I have this problem on a practice exam:
$\displaystyle \lim_{x\to\infty} 3x - \sqrt{9x^2+2x+1}$
We are dealing with L'hospitals rule, so when you plug $\infty$ in for $x$ you get
$\infty - \infty$. I multiplied by the conjugate to get:
$\displaystyle \frac{-2x-1}{3x + \sqrt{9x^2+2x+1}}$
After applying L'hospitals a few times I was lost so I peeked at the solutions sheet and my teacher went from the above step straight to:
$\displaystyle \frac{-2}{3+3} = -\frac{1}{3}$
I'm kind of lost as to how she came to that answer. Can anyone explain to me how that works?
|
Let $L$ be the given expression, then $L = \dfrac{-2x - 1}{3x + \sqrt{9x^2 + 2x + 1}} = \dfrac{-2 -\dfrac{1}{x}}{3 + \sqrt{9 + \dfrac{2}{x} + \dfrac{1}{x^2}}} \to \dfrac{-2}{6} = \dfrac{-1}{3}$ when $n \to \infty$
|
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|
Difficult time finding critical points using Lagrange The function is $f(x,y,z) = xyz$ on $x^2 + y^2 + z^2 = 1$. So I have:
$yz = 2x \lambda \\
xz = 2y \lambda \\
xy = 2z \lambda \\
x^2 + y^2 + z^2 = 1$
I guessed $x = \pm 1, y = 0, z = 0, \lambda = 0$ but apparently this isn't a critical point. There's many more that I can't seem to find using algebra either. I can't divide anything out because I can't divide by $0$, and I also can't factor anything. How can I find all the critical points?
|
Perhaps something like this:
$$ yz = 2x \lambda |\times x\\
xz = 2y \lambda |\times y \\
xy = 2z \lambda |\times z\\ $$
$$ xyz = 2x^2 \lambda\\
xyz = 2y^2 \lambda \\
xyz = 2z^2 \lambda \\ $$
Now we can something like this for example divide firs two equations and get:
$$ x^2=y^2=z^2 $$
Now solve this step by step - consider octants for each case and you'll get a bunch of solutions I guess. Symmetrical equations always include something bit more than the ordinary ones.
|
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|
How to find the values of a and b? If the polynomial 6x4 + 8x3 - 5x2 + ax + b is exactly divisible by the polynomial 2x2 - 5, then find the values of a and b.
|
The roots of $2x^2-5$ must be roots of our polynomial:
$$\left\lbrace
\begin{array}{c}
6\cdot\frac{25}4+15\sqrt{\frac52}-\frac{25}2+\sqrt{\frac52}a+b=0\\
\phantom{1}\\
6\cdot\frac{25}4-15\sqrt{\frac52}-\frac{25}2-\sqrt{\frac52}a+b=0
\end{array}
\right.$$
Adding up the equations you get
$$50+2b=0$$
And substracting them:
$$30\sqrt{\frac52}+2\sqrt{\frac52}a=0$$
So $a=-15$, $b=-25$.
|
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Quadratic Equality Constraints via SDP I want to know if it is possible to solve a QCQP problem with quadratic equality constraints in SDP. I know it is possible to convert a QCQP to an SDP by using the Shur complement. The following worked for me thus far:
$$
\begin{equation}
\begin{array}{cccccc}
\underset{x}{min} & x^{T}Q_{0}x+q_{0}^{T}x+c_{0} & & \underset{t,x}{min} & t\\
s.t & x^{T}Q_{i}x+q_{i}^{T}x+c_{i}\leq0 & \Longleftrightarrow & s.t & \left(\begin{array}{cc}
I & M_{0}x\\
x^{T}M_{0}^{T} & -c_{0}-q_{0}^{T}x+t
\end{array}\right) & \succeq0\\
& & & & \left(\begin{array}{cc}
I & M_{i}x\\
x^{T}M_{i}^{T} & -c_{i}-q_{i}^{T}x
\end{array}\right) & \succeq0\\
& i=1,2,...,m & & & i=1,2,...,m &
\end{array}
\end{equation}
$$
Where $M_{j}M_{j}^{T}=Q_{j}$ (Eigen Decomposition can also be used, thanks Alt)
However it seems this only applies to quadratic inequality constraints.
I considered converting the desired inequality constraints to equalities by using slack variables but I think that technique only works with linear constraints.
I also considered having two constraints like:
$$
\begin{equation}
\begin{array}{c}
x^{T}Qx+q^{T}x+c\leq0\\
-(x^{T}Qx+q^{T}x+c)\leq0
\end{array}
\end{equation}
$$
To force an equality but the second constraint is not convex so it wont work.
So is there another way to do it?
Edit: turns out quadratic equality constraints are non-convex thus cant be solved directly with this approach
|
Hint:
You don't need the constraints $M_{j}M_{j}^{T}=Q_{j}$. For example if $UDU^T$ is the eigen decomposition of $Q_j$, Then $M_j=UD^{\frac{1}{2}}$ (if $Q_j$ is symmetric)
Note that if you introduce new variable $M_j$, then the bilinear forms in the constraints make your problem non-convex, (Besides the fact that equality constraints with quadratic form $M_{j}M_{j}^{T}=Q_{j}$that you have to add also makes the problem non-convex.)
|
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|
Calculate $\lim_{n\rightarrow\infty}\frac{1}{n}\left(\prod_{k=1}^{n}\left(n+3k-1\right)\right)^{\frac{1}{n}}$ I'm need of some assistance regarding a homework question:
$$
\mbox{"Calculate the following:}\quad \lim_{n \to \infty}
\frac{1}{n}\left[%
\prod_{k = 1}^{n}\left(n + 3k -1\right)\right]^{1/n}\
\mbox{"}
$$
Alright so since this question is in the chapter for definite integrals ( and because it is similar to other questions I have answered ) I assumed that I should play a little with the expression inside the limit and change the product to some Riemann sum of a known function.
$0$K so I've tried that but with no major breakthroughs$\ldots$
Any hints and help is appreciated, thanks $!$.
|
The product $P_n$ may be expressed as follows:
$$P_n = \left [ \prod_{k=1}^n \left (1+\frac{3 k-1}{n}\right ) \right ]^{1/n} $$
so that
$$\log{P_n} = \frac1{n} \sum_{k=1}^n \log{\left (1+\frac{3 k-1}{n}\right )}$$
as $n \to \infty$, $P_n \to P$ and we have
$$\log{P} = \lim_{n \to \infty} \frac1{n} \sum_{k=1}^n \log{\left (1+\frac{3 k-1}{n}\right )} = \lim_{n \to \infty} \frac1{n} \sum_{k=1}^n \log{\left (1+\frac{3 k}{n}\right )}$$
which is a Riemann sum for the integral
$$\log{P} = \int_0^1 dx \, \log{(1+3 x)} = \frac13 \int_1^4 du \, \log{u} = \frac13 [u \log{u}-u]_1^4 = \frac{8}{3} \log{2}-1$$
Therefore,
$$P = \frac{2^{8/3}}{e} $$
|
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How come if $\ i\ $ not of the following form, then $12i + 5$ must be prime? I know if $\ i\ $ of the following form $\ 3x^2 + (6y-3)x - y\ $ or $\ 3x^2 + (6y-3)x + y - 1, \ \ x,y \in \mathbb{Z}^{+},i \in \mathbb Z_{\ge 0}$,
then $\ 12i + 5\ $ must be composite number, because:
$12(3x^2 + (6y-3)x + y - 1) + 5 = 36x^2 + (72y - 36)x + (12y - 7) = (6x + 12y - 7)(6x + 1)$
How come if $\ i\ $ not of the following form $\ 3x^2 + (6y-3)x - y\ $ and $\ 3x^2 + (6y-3)x + y - 1\ $,
then $12i + 5$ must be prime? $\ x,y \in \mathbb{Z}^{+},i \in \mathbb Z_{\ge 0}$.
For example: $\ 5 = \ 3*1^2 + (6*1-3)*1 - 1\ $ ,when $\ x = y = 1\ $, as proved before,$\ 5*12+5\ $ must be composite number;
$\ 0, \ 1. \ 2.\ 3.\ 4\ $ can't of the following form $\ 3x^2 + (6y-3)x - y\ $ and $\ 3x^2 + (6y-3)x + y - 1\ $,
$\ 12*0 + 5 = 5\ $, $\ 5 $ is prime,$\ 12*1 + 5 = 17\ $, $\ 17 $ is prime, and so on,
|
This is rather elementary.
Suppose that $12i+5$ is composite, say $12i+5 = ab$.
Then looking modulo $6$, we get that one of the elements $a,b$ is $1 \pmod{6}$ and the other is $-1 \pmod{6}$; assume w.l.o.g. that $a$ is $1 \pmod{6}$, and write $a = 6r+1$ and $b = 6s-1$.
Observe that $r$ and $s$ must have different parity. Indeed, if they are both even or both odd, then computing modulo $12$, we would get that $ab \equiv -1 \pmod{12}$, which is false.
If $s>r$, then we set $x = r$, and $s = r + (2y-1)$ for some positive number $y$. Then compute $ab$, to get $i = 3x^2 + (6y-3)x + y - 1$.
If $s<r$, then we set $x = s$, and $r = s + (2y-1)$ for some positive number $y$. Then compute $ab$, to get $i = 3x^2 + (6y-3)x - y$.
|
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Topology of space of symmetric matrices with fixed number of positive and negative eigenvalues Let $M$ be real non-singular symmetric $n \times n$ matrix with $p$ positive and $n-p$ negative eigenvalues. What is the topology of the space of such matrices?
For a trivial case $n=1$ the matrix is an ordinary number (which is also its eigenvalue), there are two 1-dimensional spaces with a trivial topology, one for negative and one for positive eigenvalue.
For $n=2$, if eigenvalues have the same sign, space of such matrices is 3-dimensional space of trivial topology, but if they have different signs, then space of such matrices is isomorphic to $\mathbb R^2 \times S^1$.
How to find the topology for a general case of $p$ positive and $n-p$ negative eigenvalues?
|
Here are my comments as an answer (with few more details):
Suppose you fix the eigenvalues and set $q=n−p$. Then, in view of the orthogonal diagonalization of quadratic forms, what you get is the homogeneous space $$O(n)/(O(n)\cap O(p,q))=O(n)/(O(p)\times O(q))$$
which is the Grassmannian $G_p(R^n)$ of $p$-dimensional subspaces in $R^n$. There is a huge literature describing topology of Grassmannians. If you do not want to fix the eigenvalues, only their sign, you get a certain $R^n$-bundle over the above Grassmannian. The homotopy type, of course, will be the same.
See Books on topology and geometry of Grassmannians for references on topology of Grassmannians. In addition to the answers there, Hatcher has a good discussion of Grassmannians in his "Algebraic Topology".
|
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|
If $a_n$ is a strictly increasing unbounded sequence, does $\sum_n \frac{a_{n+1} - a_n}{a_n}$ diverge? So I've been thinking through some test cases. If $a_n = n$ then $\sum_n \frac{a_{n+1} - a_n}{a_n}$ is the harmonic series which diverges. And if $a_n = \sum_{k=1}^n 1/k$ then $\sum_n \frac{a_{n+1} - a_n}{a_n}$ diverges like $\sum_n 1/(n \log n)$. So that got me thinking, if $a_n$ is a strictly increasing unbounded sequence, does $\sum_n \frac{a_{n+1} - a_n}{a_n}$ necessarily diverge?
|
Yes.
Let $b_n=\frac{a_{n+1}-a_n}{a_n}> 0$. Suppose to the contrary that $\sum_n b_n<A$. We show that
$$
a_{n+1}=a_n (1+b_n)=a_1\Pi_{1\le k\le n}(1+b_k)
$$
converges, contradicting the premise.
In fact
$$
\Pi_{1\le k\le n}(1+b_k)\le (\frac{\sum 1+b_k}{n})^n\le (1+\frac{A}{n})^n\to e^A.
$$
|
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|
Is there a generating function for $\sqrt{n}$? I tried to come up with a closed form for the ordinary generating function for the sequence $\{\sqrt{n}\}_0^{\infty}$ but I could not. Is there a way to derive it using the recurrence relation $$a_{n+1} = \sqrt{a_n^2+1}. $$
Because if there is, it is no obvious to me how to do so.
I noticed that the task is trivial if we use a Dirichelt series generating function, namely $\zeta(s-\frac{1}{2})$ but this seems less in interesting to me than having a closed form for the ogf or perhaps even the exponential generating function.
|
It does exist, defined as $g(z) = \sum_{n \ge 0} \sqrt{n} z^n$. It is even a nice function, in that it is analytic in a region around the origin (apply your favorite test). It doesn't have a representation in terms of elementary functions, however.
|
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|
is 121 divides the my pattern for base 2? Is $121|2^{120}-1$? If yes, is there any online free calculation to check these type of values?
Advanced thanks to one and all!
-Richard Sieman
|
If $p$ is prime, then you have $p$ divides $2^{p-1} - 1$ by Fermat's little theorem, which would show that your result holds if $121$ is prime. However you have that $121 = 11^2$ is not prime, so it's more complicated. A simple calculation shows that the congruence indeed does not hold.
|
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A closed form for the infinite series $\sum_{n=1}^\infty (-1)^{n+1}\arctan \left( \frac 1 n \right)$ It is known that $$\sum_{n=1}^{\infty} \arctan \left(\frac{1}{n^{2}} \right) = \frac{\pi}{4}-\tan^{-1}\left(\frac{\tanh(\frac{\pi}{\sqrt{2}})}{\tan(\frac{\pi}{\sqrt{2}})}\right). $$
Can we also find a closed form for the value of $$\sum_{n=1}^{\infty} (-1)^{n+1} \arctan \left(\frac{1}{n} \right)? $$
Unlike the other infinite series, this infinite series only converges conditionally.
|
This is just a possible way to proceed, and doesn't provide in any way a complete solution to the problem, thus I will put it as community wiki, and pray you to contribute if you have any insights.
The series converges (check!), inserting the Taylor series for $\arctan(x)$:
$$\begin{array}{ll}\sum_{n=1}^\infty(-1)^{n+1}\arctan\left(\frac{1}{n}\right)&=\sum_{n=1}^\infty(-1)^{n+1}\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\frac{1}{n^{2k+1}}\\&=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{n^{2k+1}}\\&=\sum_{k=0}^\infty\frac{(-1)^k(1-2^{-2k})}{2k+1}\zeta(2k+1)\end{array}$$
where in the second equality we have used the fact that everything converges, and in the third we noticed that the second series in the second line is the Dirichlet eta function.
I don't know if this leads anywhere, but since the zeta function has been studied pretty heavily maybe there's some results somewhere that we can use to finish from here.
|
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|
$r=(x,y,z)$ prove that $\mathrm{curl}\; r = 0$
Example $\bf 84\,\,\,$ Let $\,\mathbf r=(x,y,z)$ and $r=|\!\,\mathbf r|=\sqrt{x^2+y^2+z^2}$. Then $$\operatorname{div}\mathbf r=
\dfrac{\partial x}{\partial x}
+
\dfrac{\partial y}{\partial y}
+
\dfrac{\partial z}{\partial z}
=3; \\
\operatorname{curl}\mathbf r=
\left|\begin{matrix}
\mathbf i & \mathbf j & \mathbf k \\
\tfrac{\partial}{\partial x} & \tfrac{\partial}{\partial y} & \tfrac\partial{\partial z}\\
x & y & z
\end{matrix}\right|=\rlap{\rlap{0}\rlap00}0.$$
I fail to see how this equals zero
Is $\tfrac{dz}{dy} - \tfrac{dy}{dz} = 0$ and same for other terms too?
Thanks
|
The reason the curl is $0$ is because $\mathbf{r}$ has continuous second-order partial derivatives. It's a known theorem.
You should also note that this immediately implies $\mathbf{r}$ is a conservative field.
|
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|
Does a series have to start from 1 to be able to apply convergence tests? In the ratio test, for example, we know that the test applies for a sum from n=1 to n=infinity.
Can I directly apply this test to sums from n=x to n=infinity (for example, a sum from n=0 -> infinity)?
|
You can always write your sum as a sum starting with the index $1$ by performing an index shift:
$$
\sum_{k=k_0}^\infty a_k = \sum_{k=1}^\infty a_{k_0+k-1}.
$$
|
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On the hypothesis of the Additive Cousin Problem The Additive Cousin Problem is the following:
Assume that $D$ is a region (open, connected) of $\mathbb{C}^n$. Assume that the Dolbeault Cohomology Group of $D$, $H^1_{\bar{\partial}}(D)$ is equal to zero, and that we have an open covering of $D$ given by $\{ U_j:j\in J\}$. Suppose that we have meromorphic functions $m_j\in\mathcal{M}(U_j)$ for every $j$ such that, whenever $U_i\cap U_j\neq \emptyset$, $$m_j-m_i\in\mathcal{O}(U_i\cap U_j).$$ Given this, we ask if there is $m\in\mathcal{M}(D)$ such that, in $U_j$, we have $$ m-m_j\in\mathcal{O}(U_j).$$
Finding such $m$ is called the Additive Cousin Problem.
Can the existence of the $m_j$ meromorphic functions be a consequence of the hypothesis $H^1_{\bar{\partial}}(D)=0$ ? Range's book seems to be implying this, but I do not know how this can be possible.
|
No, the problem really assumes you have meromorphic functions satisfying certain conditions (perhaps with certain poles). You could always take all the $m_j$ to be constant and then there's no hypotheses at all needed for them to exist.
Perhaps you already understand how this goes, but if $H^{0,1}_{\bar\partial}(D)=0$, then by the Dolbeault isomorphism we have $H^1(D,\mathcal O) = 0$. Thus, the $\check{\rm C}$ech $1$-cocycle $f_{ij} = m_i-m_j\in Z^1(\scr U,\mathcal O)$ is a coboundary, so we have holomorphic functions $f_i\in \mathcal O(U_i)$ with $f_i-f_j=m_i-m_j$ on $U_i\cap U_j$. Therefore, the functions $f_i-m_i$ glue to give a global meromorphic function $m$.
|
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is there a higher dimensional analogue of the first isogonic center? I'm curious to know if, given four points $a, b, c, d$, you can always find a point $p$ such that last lines $pa, pb, pc, pd$ form equal angles pairwise.
I'd also appreciate resources on 3d geometry especially if there is an analogue of inscribed angles for a circle.
|
No, there is not. Take $ABC$ and $BCD$ as two equilateral triangles, let $M$ be the midpoint of $BC$, $\pi_A,\pi_D,\pi_M$ be the planes where $ABC,BCD$ and $ADM$ lie. The locus $l_A$ of points $P$ such that $P$ "see" all the sides of $ABC$ under the same angle is a line through the center of $ABC$ orthogonal to $\pi_A$. By symmetry, $l_A$ and $l_D$ meet on $\pi_M$, above $AD$, while the loci associated with the $ABD$ and $ACD$ faces meet on $\pi_M$, too, but below $AD$, so there is no 3d-equivalent of the Steiner-Torricelli-Fermat point in this configuration.
This is even more evident if you take $ABCD$ as a degenerate tetrahedron, in which $A,B,C,D$ all lie on the same plane and $ABC,BCD$ are equilateral triangles, with $A$ and $D$ symmetric with respect to the midpoint of $BC$.
|
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How to find a counter example for non convexity? Consider a simple function $f(x,y)=\frac{x}{y}, x,y \in (0,1]$, the Hessian is not positive semi definite and hence it is a non convex function. However, when we plot the function using Matlab/Maxima, it "appears" convex. For the sake of clarity we want to find points which violate the definition of convexity, in other words identify the region (in the plot) where the function is non-convex. Can you suggest a procedure to do the same ?
|
If you write down the Hessian matrix you'll see that for $x,y\in (0,1]$ the Hessian is indeed positive semi-definite. This is call conditional positive semi-definiteness. You can see the Hessian as a matrix parametrized by $x$ and $y$: it is positive semi-definite depending on their values. Try to compute the eigen-values as a function of $x$ and $y$.
|
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|
Behaviour of Solutions to $x^2y'' + \alpha xy'+ \beta y = 0$ as $x \to 0$ and $x \to \infty$
Consider the Euler equation $x^2y'' + \alpha xy' + \beta y = 0$.
Find conditions on $\alpha$ and $\beta$ so that:
*
*All solutions approach zero as $x \rightarrow 0$
*All solutions are bounded as $x \rightarrow 0$
*All solutions approach zero as $x \rightarrow \infty$
*All solutions are bounded as $x \rightarrow \infty$
*All solutions are bounded both as $x \rightarrow 0$ and $x \rightarrow \infty$
I am having trouble solving the problem. It must obviously have to do with the solutions to the indicial equation $ r^2 + (\alpha - 1)r + \beta= 0 \implies r = \dfrac{1 - \alpha
\pm \sqrt {(\alpha - 1)^2 - 4\beta^2}}{2} $. When these roots are identical or complex the solutions involve an $\ln x$ term the limit of which when $x \to 0$ is undefined I think? So my feeble argument is that the solutions for $r$ should be distinct to begin with? Please help.
|
A logarithmic term, if present, only changes the behaviour as $x \to 0$ or $x \to \infty$ when $\text{Re}(r) = 0$. That is, if $\text{Re}(r) \ne 0$ then both
$x^r$ and $x^r \log(x)$ have the same limit or lack of limit as $x \to \infty$ and as $x \to 0$. I don't know what that "or complex" is doing there: in the Euler equation a logarithmic term only appears when the roots are equal.
|
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|
Are rank of $T$ and $T^*$ equal? Let $H$ be a infinite dimensional Hilbert space and $T:H\to H$ be an operator on $H$.Is it true that $\operatorname{rank}T=\operatorname{rank}T^*$. We know that this is true for finite dimension. Is it true for infinite dimensional Hilbert space or is there any example of $T$ and $T^*$ that their rank are different?
Thanks!
|
The operator $Tv \mapsto T^*v$ is actually a partial isometry. The closure of the ranges are Murray-von Neumann equivalent.
|
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|
A subsequence of a convergent sequence converges to the same limit. Questions on proof. (Abbott p 57 2.5.1) Solutions to Homework 3 doesn`t duplicate. We have to prove that if $(a_{n})$ is a sequence in $\mathbb{R}$ with $\displaystyle \lim_{n\rightarrow\infty} a_n =a$, and if $(a_{n_{k}})_{k\in \mathbb{N}+}$ is a subsequence of $(a_{n})$ , then $\displaystyle \lim_{k\rightarrow\infty}a_{n_{k}}=a$.
We first need to know that $n_{k}\geq k$ for all $k\in Z_{>0}$. This is proved by induction on $k$. I omit this.
Let $e >0$. $\displaystyle \lim_{n\rightarrow\infty}a_{n}=a$ is posited, so there's $\color{violet}{N}\in \mathbb{N}$ such that for all $n\in\mathbb{N}$, $n \ge N \implies |a_n-a| < e$.
Let $k\in \mathbb{N}$ with $k\geq N.$ Then ${n_k}\geq k\geq N$. Therefore $|a_{n_{k}}-a|<e$.
I know we must find $ N\in\mathbb{N}$ such that $\color{red}{n_k} \ge N\implies |a_\color{red}{n_k}-a| < e \quad (♫)$.
As the proof overhead shows, this $N$ is the same as the posited $\color{violet}{N}$. But what engenders $(♫)$ ?
Is proof saying $n \ge N \implies |a_n-a| < e$
implies $\color{red}{k} \ge N \implies |a_\color{red}{k}-a| < e$
implies $(♫)$, because $n_k \ge k \ge N$?
If yes, then I don't understand how $k$ can be replaced by $n_k$?
|
I'm going to post a separate answer because I think I can be clearer than the accepted answer.
I know we must find $ N\in\mathbb{N}$ such that $\color{red}{n_k} \ge N\implies |a_\color{red}{n_k}-a| < e \quad (♫)$.
No, this is not what you need to show! You need to show that there exists $N$, such that for all $\boldsymbol{k} \ge N, |a_{n_k} - a| < \epsilon$.
This is because $\boldsymbol{a_{n_k}}$ is a sequence in $\boldsymbol{k}$, not a sequence in $\boldsymbol{n}$. It takes in an integer $\boldsymbol{k}$ and gives you an integer in return.
Is proof saying $n \ge N \implies |a_n-a| < e$
implies $\color{red}{k} \ge N \implies |a_\color{red}{k}-a| < e$
implies $(♫)$, because $n_k \ge k \ge N$?
Here is what the proof is saying. Fix any $k \ge N$.
Then $n_k \ge k \ge N$.
Thus $|a_{n_k} - a| < \epsilon$, Q.E.D.
If you have had trouble understanding any of the English in this answer, please let me know and I will try to clarify.
|
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How to solve this limit, hint only $$\lim_{n\to\infty}\bigg(\frac{1}{\sqrt{9n^2-1^2}}+\frac{1}{\sqrt{9n^2-2^2}}+ \dots +\frac{1}{\sqrt{9n^2-n^2}}\bigg)$$ I need a hint. I see that maybe compute with integral. But what the integrable function?
|
$$\text{As }\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$
Here the $r$( where $1\le r\le n$)th term $=\displaystyle\frac n{\sqrt{9n^2-r^2}}=\frac1{\sqrt{9-\left(\frac rn\right)^2}}$
|
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|
Orthogonal Projector Let $(H,\langle\,|\,\rangle)$ be a separable Hilbert space on $\mathbb{C}$.
$P_{\psi}:=\langle\psi,\,\rangle\psi$, where $\psi\in H$ is such that $\|\psi\|=1$.
I have to prove that $P_{\psi}$ is an orthogonal projector on $H$.
I can prove $P_{\psi}P_{\psi}=P_{\psi}$, but I can't prove $P_{\psi}^*=P_{\psi}$.
|
For $\xi,\eta\in H$ we have
$$\langle P_\psi(\xi),\,\eta\rangle\ =\ \big\langle \langle \psi,\xi\rangle\cdot\psi\,,\ \eta\big\rangle\ =\ \langle\psi,\xi\rangle\cdot\langle \psi,\eta\rangle$$
(assumed that conjugation happens in second variable).
And, we get the same for $\langle \xi,\,P_\psi(\eta)\rangle\,$.
|
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|
Green's Function Solutions So, I'm considering a PDE and trying to find its Green's function first. To this end, I solve the following helmholtz equation:
$$\frac{d^2g}{dx^2}+\frac{d^2g}{dy^2}+\frac{d^2g}{dz^2}-\alpha^2g=\delta(x-\xi)\delta{(y-\eta)}\delta{(z-\rho)}$$
Well, I can solve this PDE for $g$ , but what happens to the solution when $x=\xi$,$y=\eta$, etc? In that case, would the Green's function fails to solve the PDE? So, it fails to solve the homogeneous case at the "separation point"? Just trying to understand what happens at this odd location.
|
Yes, the Green's function (a.k.a. "fundamental solution") does not solve the homogeneous equation "at" the separation point.
There are at least two ways to think about this. One is to approximate (weakly) the Dirac deltas by spikes that are nevertheless smooth pointwise-valued functions. For elliptic operators and such, solving such an equation provably approximates the genuine solution. This heuristic has certain limitations, but is the obvious starting point.
A philosophically slightly different attitude is to let go a little of the "pointwise" notion of "function", to allow thinking of distributions (such as Dirac delta) as "generalized functions", despite not necessarily having pointwise values everywhere. That is, instead of using the lack of pointwise sense at "the separation point" to "prove" that Dirac delta "is not a function", interpret that difficulty as an argument in favor of relaxing the notion of function to not-necessarily require pointwise values, etc.
|
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How can I define $e^x$ as the value of infinite series? I understand the definition of $e^x$ by limit. But I do not know how to come up with:
$$e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$$
without using Taylor series. more explicitly without using calculus. how do we know if a function can be expressed as series or not ?
|
One definition of the exponential function is the limit
$$
\lim_{n\to\infty} \Big(1 + \frac{x}{n}\Big)^n=e^x.
$$
Let $P_n(x)$ denote the polynomial $(1+x/n)^n$, so that $e^x=\lim_{n\to\infty} P_n(x)$; I will show that
$$
\lim_{n\to\infty} P_n(x) = \sum_{n=0}^\infty \frac{x^n}{n!}.
$$
If you expand out $P_n(x)$ using the binomial theorem, you'll find that
$$
P_n(x) = \sum_{k=0}^n \binom{n}{k}\frac{x^k}{n^k}.
$$
Now
\begin{align}
\frac{1}{n^k}\binom{n}{k}
&= \frac{1}{n^k}\cdot\frac{n}{k}\cdot\frac{n-1}{k-1}\cdots\frac{n-k+1}{1} \\
&= \frac{1}{k}\cdot\frac{1-1/n}{k-1}\cdots\frac{1-(k-1)/n}{1}.
\end{align}
Since $k$ is fixed, $\lim_{n\to\infty}\binom{n}{k}/n^k=\frac{1}{k!}$. It follows that
$$
\lim_{n\to\infty} P_n(x) = \sum_{k=0}^\infty\lim_{n\to\infty}\binom{n}{k}\frac{x^k}{n^k}
= \sum_{k=0}^\infty\frac{x^k}{k!}.
$$
You should look at the Wikipedia article about characterizations of the exponential function; it has five.
As for your question about whether a function can be expressed as a series or not, to answer it I believe you need to say something about calculus. What I mean is that if a "nice" function $f(x)$ has a series representation at a point $a$ then the series is given by
$$
f(x)=\sum_{k=0}\frac{f^{(k)}(a)}{k!}(x-a)^k,
$$
where $f^{(k)}(a)$ is the $k$th derivative of $f$ at $a$.
|
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Gram-Schmidt process Question:Apply the Gram-Schmidt process to find an orthonormal basis for the subspace.
$S= \mathrm{span} [{(1,2,-4,-1),(-3,0,5,-2),(0,7,2,-6)}]$
The span is suppose to look like a matrix but I couldn't get it to look right here so I wrote it that way.
I was able to do the process but I wanted to check something. When I was working on my 3rd vector I didn't get a cancellation. Is it wrong if something doesn't cancel out?
I got my answers to be $u_1= \frac{1}{\sqrt{22}} \left(\begin{smallmatrix}1\\ 2\\-4\\-1\end{smallmatrix}\right)$, $u_2= \sqrt{\frac{22}{395}} \left(\begin{smallmatrix}-45/22\\ 21/11\\13/11\\-65/22\end{smallmatrix}\right)$, and $u_3= \sqrt{\frac{395}{7947}} \left(\begin{smallmatrix}258/79\\ 929/395\\782/395\\ 4/79\end{smallmatrix}\right)$
|
Wow what a nasty question...well I get exactly those $u_1,u_2,u_3$, and I cross-checked on Mathematica - it is correct. I am not sure exactly what you thought should have cancelled out, but it seems as if it is not "wrong" the fact that it didn't cancel out.
But I think it might be worthwhile to try and find a simpler basis for that subspace before applying Gram-Schmidt?
|
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Use logarithmic differentiation to find the derivative of $y = (1 +\frac1x)^{2x}$ Can someone guide me through solving this problem?
$$\dfrac{\mathrm d}{\mathrm dx}\left(1 +\dfrac1x\right)^{2x}$$
|
Let
$$
y=\left(1 +\dfrac1x\right)^{2x}
$$
then
$$
\ln y=2x\ln\left(1 +\dfrac1x\right).
$$
Thus
$$
\begin{align}
\frac{d}{dx}\ln y&=\frac{d}{dx}2x\ln\left(1 +\frac1x\right)\\
\frac1y\frac{dy}{dx}&=2\ln\left(1 +\frac1x\right)+\frac{2x}{1 +\frac1x}\cdot\left(-\frac1{x^2}\right)\\
&=2\ln\left(1 +\frac1x\right)-\frac{2}{x +1}\\
\frac{dy}{dx}&=y\left(2\ln\left(1 +\frac1x\right)-\frac{2}{x +1}\right)\\
&=\left(1 +\dfrac1x\right)^{2x}\left(2\ln\left(1 +\frac1x\right)-\frac{2}{x +1}\right).\\
\end{align}
$$
|
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Show that $\int_0^de^{-tx}g(x)dx\sim_{t\rightarrow\infty} \frac{g(0)}{t}$ Let $d>0$. Let $g\in C^0([0,d])$ such that $g(0)\ne0$.
Show that $$\int_0^de^{-tx}g(x)dx\sim_{t\rightarrow\infty} \frac{g(0)}{t}$$
How can I prove that ?
It's the first time I see this kind of exercise
Thank you in advance for your time
|
Here is a start. Making the change of variables $y=tx$ yields
$$ \int_0^de^{-tx}g(x)dx = \frac{1}{t}\int_{0}^{td} g\left(\frac{y}{t}\right)e^{-y} dy. $$
I think you can advance now.
Note:
$$ \int_{0}^{\infty} e^{-y}dy = 1. $$
|
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Prove that if m is prime and m|kl then either m|k or m|l. Proofs homework question, here's what I've figured out thus far.
Suppose m doesn't divide k. We need to then prove that m|l. If m doesn't divide k and m is a prime then we know m and k are co-prime - hcf (m,k) = 1. Which means 1 = ks + mt (for some integers; s,t).
I'm not really sure where to go from here to prove that m|l based on these facts, help?
|
The Bezout-based proof is a bit more intuitive interpreted modulo $\,m.\,$ Notice that the Bezout identity $\,\color{#c00}{1 = \gcd(k,m)} = a k\! +\! b m\,$ becomes $ $ mod $\,m\!:\ 1\equiv ak,\,$ hence $\,\color{#c00}{k^{-1}} \equiv a$ exists mod $\,m.\,$ Hence $\, m\mid k\,\ell\,\Rightarrow\,k\,\ell\equiv 0\!\!\!\!\! \overset{\large\quad \times\,\color{#c00}{k^{-1}}}\Rightarrow\!\!\ell\equiv 0\,\Rightarrow\,m\mid \ell.\ $ Therefore, mod $\,m,\,$ the proof amounts to the simple fact that elements $\,k\,$ coprime to the modulus are, by Bezout, invertible, so cancellable.
Remark $\ $ If you know the gcd Distributive Law $\rm\color{#0a0}{(DL)}$ you can write the proof very concisely
$$m\mid k\ell,m\ell\ \Rightarrow\ m\mid (k\ell,m\ell)\overset{\rm\color{#0a0}{(DL)}} = \color{#c00}{(k,m)}\ell = \ell\qquad\qquad$$
|
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|
Triple Integral Troubles I'm having trouble calculating this integral. I can do the first one just fine, but it's in simplifying and calculating the third integral where I get stuck.
$16\int_0^\frac{\pi}{4}\int_0^1\int_0^{\sqrt{1-r^2cos^2(\theta)}}rdzdrd\theta$
$16\int_0^\frac{\pi}{4}\int_0^1r\sqrt{1-r^2cos^2(\theta)}drd\theta$
with u-substitution:
$16\int_0^\frac{\pi}{4}\frac{-1}{3}(\frac{(1-cos^2(\theta))^(\frac{3}{2})}{cos^2(\theta)})d\theta$
I know the answer should be $16-8\sqrt{2}$
|
You're forgetting a "$-1$" term. Let $u = 1 - r^2\cos^2\theta$ so that $du = -2r\cos^2\theta \, dr$. Then observe that:
\begin{align*}
\int_0^1 r \sqrt{1 - r^2\cos^2\theta} \, dr
&= \frac{1}{-2\cos^2\theta}\int_1^{1-\cos^2\theta} \sqrt{u} \, du \\
&= \frac{1}{-2\cos^2\theta}\left[\frac{u^{3/2}}{3/2}\right]_1^{1-\cos^2\theta} \\
&= \frac{1}{-3\cos^2\theta}\left[(1-\cos^2\theta)^{3/2} - 1\right] \\
\end{align*}
Hence, we obtain:
$$
\frac{-16}{3}\int_0^{\pi/4} \frac{(1-\cos^2\theta)^{3/2} - 1}{\cos^2\theta} \, d\theta
$$
Using Wolfram|Alpha, this integral does indeed equal $16 - 8\sqrt2$, as desired.
If you want to actually compute the antiderivative by hand, then here are some trig identities that will magically simplify things:
\begin{align*}
\frac{(1-\cos^2\theta)^{3/2} - 1}{\cos^2\theta}
&= \frac{(\sin^2\theta)^{3/2} - 1}{\cos^2\theta} \\
&= \frac{\sin^3\theta - 1}{\cos^2\theta} \\
&= \frac{\sin\theta(1 - \cos^2\theta) - 1}{\cos^2\theta} \\
&= \frac{\sin\theta - \sin\theta\cos^2\theta - 1}{\cos^2\theta} \\
&= \frac{1}{\cos\theta} \cdot \frac{\sin\theta}{\cos\theta} - \sin\theta - \frac{1}{\cos^2\theta} \\
&= \sec\theta\tan\theta - \sin\theta - \sec^2\theta
\end{align*}
Thus, the antiderivative is:
$$
\sec\theta + \cos\theta - \tan\theta
$$
|
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Why is calculating the area under a curve required or rather what usage it would provide I understand Integration and Differentiation and see a lot of Physics / Electrical Theory using them.
Take for example a sine wave. So area for me means the space any object would occupy. So what's usage it comes to find the area of a sine curve?
There are so a many formulas that calculates the area by Integration - but why calculation is required - I mean what information we can get (isn't it just space occupied) or rather what data we can find by calculating area of a curve via Integration?
|
A function, differentiate it represent how fast it changes, integrate it represent how much change had accumulated.
If we know the derivative of the function (how fast it changes), by integrate it, we find function itself. If we know how much change had accumulated, by differentiate it, we can find of the function of this thing.
|
{
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|
a function that is in $L^2$ (the right version) Sorry I made a mistake when posting the last question. Actually my question is: can you give a $f(x) $ such that $ f \in L^2 ( \mathbb R)$ but $ x^{-\frac{1}{2}} f \notin L^1 ( \mathbb R ) $. Thanks!
|
Take $$f(x)=\frac 1{\sqrt x\log x}\mathbf{\chi}_{[2,\infty)}(x).$$
Then $x\mapsto f^2(x)=(x\log^2x)^{-1}\mathbf{\chi}_{[2,\infty)}(x)$ is integrable and $x\mapsto x^{-1/2}f(x)=(x\log x)^{-1}\mathbf{\chi}_{[2,\infty)}(x)$ is not.
|
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|
$\ln|x|$ vs.$\ln(x)$? When is the $\ln$ antiderivative marked as an absolute value? One of the answers to the problems I'm doing had straight lines:
$$ \ln|y^2-25|$$
versus another problem's just now:
$$ \ln(1+e^r) $$
I know this is probably to do with the absolute value. Is the absolute value marking necessary because #1 was the antiderivative of a squared variable expression that could be either positive or negative (and had to be positive because, well, natural log) and the second was positive by default?
Sorry if this is me asking and answering my own question; I'd just love to get confirmation in case I'm wrong.
|
logarithm is a function only defined on domain $(0,\infty)$, so it make no sense to input negative value.
|
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|
Show that if $p$ is a prime of the form $p=4n+1$, then we can solve $x^2\equiv -1\mod p$(with $x$ an integer).
Show that if $p$ is a prime of the form $p=4n+1$, then we can solve $x^2\equiv -1\mod p$(with $x$ an integer).
My attempt:If $p$ is a prime, then $U_p=${$[x]|1\leq x<p$} is cyclic.
|
The following is an alternative proof that goes back to Dirichlet. Note that for every $x$ in the interval $1$ to $p=1$, there is a unique $y$ in that interval such that $xy\equiv 1\pmod{p}$. Let $p=4k+1$. Suppose that $x^2\equiv -1\pmod{p}$ has no solution. We will show that leads to a contradiction.
If $x^2\equiv -1\pmod{p}$ has no solution, then the unique $y$ described above is not congruent to $x$ modulo $p$.
Thus the numbers from $1$ to $p-1$ can be divided into "couples" with product congruent to $-1$ modulo $p$. It follows that $(p-1)!\equiv (-1)^{2k}\equiv 1\pmod{p}$. This contradicts Wilson's Theorem.
|
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|
Find the value of the real parameters a,b, if there exists a P(X) binomial and the following is true Find the value of the real parameters a,b, if there exists a P(X) binomial and the following is true
$(X^3-aX^2-bX+1) : P(X) = X^2-X+1 $
I tried to divide and to equalize the remainder to zero, but I think i'm missing something.
|
you can make a polynomial division like this: $(X^3-aX^2-bX+1):(X^2-X+1)=P(X)$
It´s just a transformation of the origininal equation. Then compare the values of the parameters.
greetings,
calculus.
|
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|
Combinatorics (combinations problem) How many ways are there to pick a group of $4$ people from $10$ people (each of a different height) and then pick a second group of $3$ other people such that all the people in the first group are taller than all the people in the second group? (Hint: Consider two ways)
I try to illustrate it to sort the shortest to the tallest, then I found out four different cases. But my lecturers mean to use 2 ways, can anyone help me??
|
Just pick $7$ people from $10$, and let the $3$ shortest ones be called the second group. This can be done in $\binom{10}{7}$ ways.
Remark: The cases approach of the post does some double-counting. One can adjust it, by organizing by "shortest in the group of $4$," If the shortest in that group is to be say Person $6$, then we need to choose three from the $5$ tallest.
|
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|
no. of positive integral solutions of ||x - 1| - 2| + x = 3 What are the no. of positive integral solutions of ||x - 1| - 2| + x = 3 ?
My effort
||x - 1| - 2| = 3 - x
|x - 1| - 2 = 3 - x OR |x - 1| - 2 = x - 3
|x - 1| = 5 - x OR |x - 1| = x - 1
x - 1 = 5 - x OR x - 1 = x - 5 OR x - 1 $\geq$ 0
2x = 6 OR x $\geq$ 1
Therefore, x $\geq$ 1
But the any value of x greater than 1 except 3 does not satisfy the equation. Where have I gone wrong?
|
$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\Li}[1]{\,{\rm Li}_{#1}}
\newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}&\color{#66f}{\large\sum_{x\ =\ 0}^{\infty}
\delta_{\verts{\vphantom{\large A}\verts{x\ -\ 1}\ -\ 2}\ +\ x,3}}
=\sum_{x\ =\ 0}^{\infty}\oint_{\verts{z}\ =\ 1}
{1 \over z^{-\verts{\verts{\vphantom{\Large A}x\ -\ 1}\ -\ 2}\ -\ x\ +\ 4}}
\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ 1}{1 \over z^{4}}
\sum_{x\ =\ 0}^{\infty}z^{\verts{\verts{\vphantom{\Large A}x\ -\ 1}\ -\ 2}\ +\ x}
\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}{1 \over z^{4}}\pars{%
z + z^{3} + \sum_{x\ =\ 2}^{\infty}z^{\verts{x\ -\ 3}\ +\ x}}
\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ 1}{1 \over z^{4}}\pars{%
z + z^{3} + z^{3} + z^{3} + \sum_{x\ =\ 4}^{\infty}z^{2x - 3}}
\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ 1}\pars{%
{1 \over z^{3}} + {3 \over z} + \sum_{x\ =\ 4}^{\infty}{1 \over z^{7 - 2x}}}
\,{\dd z \over 2\pi\ic}
\\[5mm]&=\overbrace{\oint_{\verts{z}\ =\ 1}{1 \over z^{3}}\,{\dd z \over 2\pi\ic}}
^{\ds{=\ \dsc{0}}}\ +\
\overbrace{\oint_{\verts{z}\ =\ 1}{3 \over z}\,{\dd z \over 2\pi\ic}}
^{\ds{=\ \dsc{3}}}\ +\
\sum_{x\ =\ 4}^{\infty}\overbrace{%
\oint_{\verts{z}\ =\ 1}{1 \over z^{7 - 2x}}\,{\dd z \over 2\pi\ic}}
^{\ds{=\ \dsc{\delta_{x,3}}}}\ =\
\color{#66f}{\LARGE 3}
\end{align}
|
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|
Text similar to chapter 9 of Topology from James Munkres I'm self-studying chapter 9 of Topology from James Munkres.
I like to read different books about the same topic at the same time. Can someone recommend some text/book that is about the same subjects as found in chapter 9?
This chapter is about the fundamental group. It is from part 2 of the book, which is called algebraic topology. The sections are called homotopy of paths, the fundamental group, covering spaces, the fundamental group of the circle, retractions and fixed points, the fundamental theorem of algebra, the Borsuk-Ulam theorem, deformation retracts and homotopy type, the fundamental group of S* and fundamental groups of some surfaces.
Thanks in advance !
|
Algebraic Topology by F.H Croom will be a good choice for a beginner in algebraic topology
|
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|
key generation in RSA cryptosystem: why it can be performed in polynomial time? Suppose that I want to generate the keys of the RSA cryptosystem: the public key will be the couple $(n,e)$ where $n$ is the product of two primes $p$ and $q$ and gcd$(\phi(n),e)=1$.The private key will be the integer $d$ such that $ed=1$ (mod $\phi(n)$). Suppose that the length of $n$ (represented as bit string) is $k$, therefore the following is an algorithm that outputs $n$, $e$ and $d$:
*
*Choose two very large random prime integers:
$p$ and $q$ such that $|pq|_2=k$
*Compute $n$ and $φ(n)$:
$n = pq$ and $φ(n) = (p-1)(q-1)$
*Choose an integer $e$, $1 < e < φ(n)$ such that:
gcd$(e, φ(n)) = 1$
*Compute $d$, $1 < d < φ(n)$ such that:
$ed ≡ 1$ (mod $φ(n)$)
I have showed that points 1,2,4 can be computed in polynomial time of $k$, but I can't find any algorithm that can compute the point 3 in polynomial time.
Do you have any idea about it?
Edit: I'm wrong about the fact that the point $1$ can be performed in polynomial time. I had in my mind a wrong brute force argument. In an answer above, the user @TonyK says that this is an open problem.
|
Step 1 is the hard one! I think that it still hasn't been proven to be polynomial, although it is believed to be.
Step 3 is easy, if all you need is an $e$ such that $(e,\phi(n)) = 1$ (although a real-life public key protocol will require more than this). Just let $e$ be the smallest prime that doesn't divide $\phi(n)$. This is clearly polynomial in $\log_2(n)$.
|
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|
Writing answers to trigonometric equation I wonder how to write answers to trigonometric equations in more elegant form. For instance if we have $ \displaystyle \sin x = \frac{\sqrt{2}}{2} \vee \sin x=-\frac{\sqrt{2}}{2}$ then I write four cases instead of just one where $\displaystyle x=\frac{\pi}{4}+\frac{k\pi}{2}$
Can anyone explain how to obtain such forms ?
|
$$\sin x=-\frac{\sqrt2}2=-\frac1{\sqrt2}=\sin\left(-\frac\pi4\right)$$
$$\implies x=n\pi+(-1)^n\left(-\frac\pi4\right)$$ where $n$ is any integer
for $\displaystyle n=2m\implies x=2m\pi-\frac\pi4$
for $\displaystyle n=2m+1\implies x=(2m+1)\pi+\frac\pi4=2m\pi+\frac{5\pi}4$
Similarly, $\displaystyle\sin x=\frac{\sqrt2}2\implies $
for $\displaystyle n=2m\implies x=2m\pi+\frac\pi4$
for $\displaystyle n=2m+1\implies x=(2m+1)\pi-\frac\pi4=2m\pi+\frac{3\pi}4$
Observe that the values can be merged into $$n\cdot\frac\pi2+\frac\pi4$$ i.e., each of the four Quadrant has the offset $\dfrac\pi4$
|
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|
Equivalence Relation Proof for modular arithmatic Given this modular relation:
$x^3 \equiv y \pmod{3}$
how would you go about proving the transitivity of the system? I have proven the reflexivity, and symmetry pretty easily but the transitivity is giving me many problems, and I feel like im not setting up the problem correctly.
|
By little Fermat (or directly) note $\,{\rm mod}\ 3\!:\ x^3\equiv x\,$ for all integers $\,x,\,$ thus $\,x^3\equiv y\iff x\equiv y.\,$ Thus your relation is the same as the standard congruence relation '$\equiv$', which is transitive since
$\quad x\equiv y,\ y\equiv z\pmod 3\,\Rightarrow\, 3\mid x\!-\!y,y\!-\!x\,\Rightarrow\, 3\mid x\!-\!y+y\!-\!z = x\!-\!z\,\Rightarrow\,x\equiv z\pmod 3$
Remark $\ $ Note that the inference in the first line already uses the fact that '$\equiv$' is an equivalence relation (transitive, etc). Thus if you don't already know that, then you will need to prove that too (if so, be careful to prove things in the correct order to avoid a circular proof).
|
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|
Find maximum of a complex function $f(z)$ I am trying to find the following maximum, whose existence is justified by the compactness of the close ball $\Delta$ of $\mathbb C$ and continuity of the function
$$f(z)=\sum_{k=1}^\infty (-1)^k \frac{2z}{k^2 \pi^2-z^2}\cos kt$$
from $\mathbb C$ to $\mathbb C$
$$\large { \displaystyle \max_{z \in {\mathbb C},|z| \leq \frac{\pi}{4}} \left|f(z)\right|}$$
any suggestions please? I try to use Maximum Modulus Principle:
$$\large { \displaystyle \max_{z \in {\mathbb C},|z| \leq r} \left|\sum_{k=1}^\infty (-1)^k \frac{2z}{k^2 \pi^2-z^2}\cos kt\right|}=\large { \displaystyle \max_{z \in {\mathbb C},|z| = r} \left|\sum_{k=1}^\infty (-1)^k \frac{2z}{k^2 \pi^2-z^2}\cos kt\right|}=\large { \displaystyle \max_{\theta\in \mathbb R} \left|\sum_{k=1}^\infty (-1)^k \frac{2re^{i\theta}}{k^2 \pi^2-r^2e^{i2\theta}}\cos kt\right|}$$. How can I go on?
|
Hint: holomorphic functions on some bounded open area attaint their maximum on the boundary of the area.
|
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|
Fibonacci's proof that $x^3+2x^2+10x=20$ has no solution in radicals? I read on a poster today that Fibonacci showed that $x^3+2x^2+10x=20$ has no solution expressible in radicals, way back when.
I couldn't find the proof anywhere. Does anyone know where I can find it?
|
He proved that the solution cannot be one of Euclid's irrationals. All Euclid's irrationals are strictly contained in the set of numbers of the form
$$
\sqrt[4]{p}\pm\sqrt[4]{q},
\qquad p,q\in\mathbb{Q}.
$$
The proof would be similar to (but of course more complicated than) how you prove $\sqrt2$ is not rational.
|
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|
Using random numbers to evaluate $\sum_{k=1}^{10000}e^{\frac{k}{10000}}$ I tried using the Monte Carlo Method to approximate the sum $\sum_{1}^{10000}e^{\frac{k}{10000}}$. First I genarating 100 random numbers in (1, 10000). Then by the strong law of large numbers:
$$\sum_{i=1}^{100}\frac{f(U_i)}{k} \to E[f(U)] = \theta$$
where:
$$\theta = \int_{1}^{10000}e^{\frac{x}{10000}}dx$$
The exercise says to use only 100 random numbers. So dont know if that thought is right, because this method requires k→∞
|
You are confusing a few different concepts:
*
*You write $\frac{1}{k}\sum_{i=1}^{100}f(U_i)\to\mathbb{E}[f(U)]$ when you clearly mean $\frac{1}{100}\sum_{i=1}^{100}f(U_i)\approx\mathbb{E}[f(U)]$.
*You talk about random numbers in $\{1,\dots,10000\}$ (implying that you choose them uniformly). This is therefore a discrete random variable. To evaluate its expectation, you then integrate between 1 and 10000 as if it were continuous (and you don't even normalize the density).
Let us go back to the basics. You know that for any continuous function $f$ on $[0,1]$,
$$\frac{1}{N}\sum_{k=0}^Nf\left(\frac{k}{N}\right)\xrightarrow[N\to+\infty]{}\int_0^1f(x)\,\mathrm{d}x.$$
By identification, do you see what function $f$ you should choose to approximate $\sum_{k=1}^{10000}e^{k/1000}$?
Let us assume that you have found such an $f$, and $\sum_{k=1}^{10000}e^{k/1000}\approx\int_0^1f(x)\,\mathrm{d}x$. You may write
$$\int_0^1f(x)\,\mathrm{d}x=\mathbb{E}[f(U)],$$
where $U$ is a uniform random variable on $[0,1]$. Then, do you see how to approximate $\mathbb{E}[f(U)]$ by using the Monte Carlo techniques you mention in your post?
|
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|
2 Trains and Fly Problem. Find the number of trips made by the fly back and forth. Question: A Train A is approaching at a speed of 10m/sec, another Train B moving in the opposite direction at a speed of 20m/sec. A fly whose absolute speed is 50m/sec goes repeatedly from A to B and back, without loosing any time at any of the trains. The repeated moving back and forth stops when the 2 trains crash into each other.
The initial distance between the 2 trains is 300m.
1>Find the distance traveled by the fly
2>Find the number of trips made by the fly back and forth
My Question: I get that the answer to the first question is 500m, since it takes 10 seconds for the 2 trains to crash, so in 10 seconds at a speed of 50m/sec the fly covers 50m/sec*10sec = 500m.
But how do I find the number of trips(back and forth) the fly makes until the trains crash?
|
The problem can be solved much more simpler by using graphs. A plot of d(t) vs t for the trains A and B will be two straight line who intersect at the time and position where the two trains crash. Furthermore, plotting the d(t) of the bird on the same plane will show clearly where the fly keeps going to and fro between the two train plots….The graphical solution is simple and easy to understand
|
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|
Non-Hamiltonian $k$-connected $k$-regular graphs ($k>3$)? The Petersen graph provides us with an example of an non-Hamiltonian 3-connected 3-regular graph. Are any 4-connected 4-regular graphs known to be non-Hamiltonian? What about generic $k$-connected $k$-regular graphs?
|
According to this paper (Regular n-Valent n-Connected NonHamiltonian Non-n-Edge-Colorable Graphs by G. H. J. Meredith) the answer is yes (by construction) for $k=4$ and $k \ge 8$ but the other cases are not shown there. You can see a nice picture of the construction (which goes via the Petersen graph) for $k=4$ on page 5.
|
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$\sum {a_n}$ be a convergent series of complex numbers but let $\sum |{a_n}|$ be divergent.. I am stuck on the following problem that says:
Let $\sum {a_n}$ be a convergent series of complex numbers but let $\sum |{a_n}|$
be divergent.
Then it follows that
a. $a_n \to 0$ but $|{a_n}|$ does not converge to $0$.
b. the sequence $\{a_n\}$ does not converge to $0$.
c. only finitely many $a_n$’s are $0$.
d. infinitely many $a_n$’s are positive and infinitely many are negative.
e. none of the above.
I am not sure which of the aforementioned options is correct . I think option (a) is the right choice but am unable to prove it. Wiki has been of little help.
Can someone explain? Thanks and regards.
|
1) Is not true. For example, take $a_n = \dfrac{(-1)^n}{n}$, $n > 0$.
2) Also not true. Take the same sequence.
3) Not true. Take $a_n = \dfrac{(-1)^n}{n}$ if $n$ is odd, and $0$ if $n$ is even. Then, $a_n$ converges and $|a_n|$ does not, but there are not finitely many $a_n$'s that are $0$.
4) True. If finitely many $a_n$'s are negative or positive, we can take them out and the series would be convergent, and since all of the would be positive or negative, the absolute value of the sequence would converge too.
For the fourth one, you could prove it more rigorously, but I let that for you.
|
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|
Find a continuous function on the reals where $f(x) >0$ and $f'(x) < 0$ and $f''(x) < 0$ We need to find a function $f(x)$ where $f(x) >0 $and $f'(x) < 0$ and $f''(x) < 0$ where $f$ is continuous for all real numbers.
We have tried $ f(x) = \sqrt{-x}$ however this is not defined for $x>0$ and therefore is only continuous where $x<0$.
Is there even such a function? Because $f$ is positive but decreasing (...increasingly)
Any help is appreciated.
Thanks.
|
Such function does not exist!
If not, there is such $f$ concave, so
$$f(x)\leq f'(0)x +f(0), \qquad \forall x \in \Bbb R$$
Since $f'(0)\lt0$, so if $x \gt -\dfrac{f(0)}{f'(0)}$, then
$$f(x)\leq f'(0)x +f(0)\lt0$$
|
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|
easy question about conservative vector fields Suppose I have a vector field $f: A \subset R^2 \to R^2 $. Write $f = (M,N) $. Does it follow this?
$$ \frac{ \partial M}{\partial y} = \frac{ \partial N}{ \partial x } \quad \iff \quad f \text{ is conservative}$$
|
No. It does not. If $A$ is the punctured plane then the vector field with $M = -y/(x^2+y^2)$ and $N = x/(x^2+y^2)$ is not conservative. This can be seen by integrating around any closed contour which contains the origin. It will give a nontrivial result depending on how many times the curve winds around the origin. On the other hand, if $A$ is simply connected then there is an equivalence between satisfying the criteria $M_y=N_x$ and $(M,N)$ being conservative.
|
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|
Using properties of determinants, show that Using Properties of determinants, show that:
$$
\begin{vmatrix}
a & a+b & a+2b\\
a+2b & a & a+b\\
a+b & a+2b & a
\end{vmatrix}
= 9b^2 (a+b)
$$
I've tried it but not getting $9b^2$
|
Add all the three columns to get
$$\left \vert \begin{bmatrix} 3a+3b & a+b & a+2b\\ 3a+3b & a & a+b\\ 3a+3b & a+2b & a\end{bmatrix}\right \vert = 3(a+b)\left \vert \begin{bmatrix} 1 & a+b & a+2b\\ 1 & a & a+b\\ 1 & a+2b & a\end{bmatrix}\right \vert$$
Subtract first row from row and first from third to get
$$3(a+b)\left \vert \begin{bmatrix} 1 & a+b & a+2b\\ 1 & a & a+b\\ 1 & a+2b & a\end{bmatrix}\right \vert = 3(a+b)\left \vert \begin{bmatrix} 1 & a+b & a+2b\\ 0 & -b & -b\\ 0 & b & -2b\end{bmatrix}\right \vert = 3(a+b)(2b^2+b^2) = 9b^2(a+b)$$
|
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|
Prove the identity $\frac{\cos B}{1-\tan B} + \frac{\sin B}{1-\cot B}=\sin B+\cos B$ I have worked on this identity from both sides of the equation and can't seem to get it to equal the other side no matter what I try.
$\displaystyle\frac{\cos B}{1-\tan B} + \frac{\sin B}{1-\cot B} =\sin B+\cos B$
|
HINT:
$\displaystyle\frac{\cos B}{1-\tan B}=\frac{\cos B}{1-\dfrac{\sin B}{\cos B}}=\frac{\cos^2B}{\cos B-\sin B}$
$\displaystyle\frac{\sin B}{1-\cot B}=\frac{\sin B}{1-\dfrac{\cos B}{\sin B}}=\frac{\sin^2B}{\sin B-\cos B}=-\frac{\sin^2B}{\cos B-\sin B}$
|
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2 questions about counting and permutation I have a test on monday and I couldn't solve these 2 questions, I'd be grateful if you help me
1-) How many ways are there to distribute 18 balls among 6 dierent persons if
a) each ball is dierent and each person should get 3 balls
b) all balls are identical ?
2-)How many permutations of the letters a a a b b b c c c d d d are there with no three consecutive letters the same?
|
Question 2: We use the Principle of Inclusion/Exclusion. There are $\frac{12!}{3!3!3!3!}$ permutations (words). We now count the bad words, in which there are $3$ consecutive occurrences of a or of b or of c or of d.
We count the words with $3$ consecutive a. Group the three a into a superletter, which we will call A. Then we have $10$ "letters," which we can arrange in $\frac{10!}{3!3!3!1!}$ ways.
The same is true for $3$ consecutive b, c, or d. But if we add these $4$ numbers (or equivalently multiply the answer for aaa by $4$, the number $4\cdot \frac{10!}{3!3!3!1!}$ double-counts the words that have both aaa and bbb, and the same for all $\binom{4}{2}$ ways of choosing $2$ letters. By replacing aaa by the "letter" A, and bbb by B, we can see that there are $\frac{8!}{3!3!1!1!}$ words that have a a triple a or a triple b. So our new estimate of the number of bad words is
$4\cdot \frac{10!}{3!3!3!1!}-\binom{4}{2}\cdot \frac{8!}{3!3!1!1!}$.
However, we have subtracted too much, for we have subtracted once too many times the number of words that have three triple letters. Our new estimate of the number of bad words is $4\cdot \frac{10!}{3!3!3!1!}-\binom{4}{2}\cdot \frac{8!}{3!3!1!1!}+\binom{4}{3}\cdot \frac{6!}{3!1!1!1!}$.
However, we have added back once too often the bad words with all $4$ letters occurring in groups of $3$. So we need to subtract $\binom{4}{4}\cdot\frac{4!}{1!1!1!1!}$.
Now put things together. The number of good words is
$$\binom{4}{0}\cdot \frac{12!}{3!3!3!3!}-\binom{4}{1}\cdot \frac{10!}{3!3!3!1!}+\binom{4}{2}\cdot \frac{8!}{3!3!1!1!}-\binom{4}{3}\cdot \frac{6!}{3!1!1!1!}+\binom{4}{4}\cdot\frac{4!}{1!1!1!1!}.$$
Remark: If $k_1+k_2+\cdots +k_t=n$, then $\frac{n!}{k_1!k_2!\cdots k_t!}$ is often denoted by the multinomial symbol $\binom{n}{k_1,k_2,\dots,k_t}$. That notation would make our formulas look better.
We do not wish to write out solutions for Question 1. But the answer to (a) will turn out to simplify to $\frac{18!}{3!3!3!3!3!3!}$. Part (b) is ambiguous, it is not clear whether we still want $3$ to each. But then the problem is trivial. If there is no restriction on the numbers given to each person, then a Stars and Bars argument gives the answer $\binom{23}{5}$.
|
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|
Proof using complex numbers Prove that $\left|\dfrac{z-w}{1-\bar{z}w}\right| = 1$ where $\bar{z}$ is conjugate of $z$ and $\bar{z}w\ne 1$ if either $|z| = 1$ or $|w| = 1$.
I used $|c_1/c_2| = |c_1|/|c_2|$ and multiply out with $z = x + iy$ and $ = a+ib$ but I am getting stuck near finish.
|
Suppose $|z|=1$: then
$$
\left|\dfrac{z-w}{1-\bar{z}w}\right|=
\frac{1}{|z|}\left|\dfrac{z-w}{1-\bar{z}w}\right|=
\left|\dfrac{z-w}{z(1-\bar{z}w)}\right|=
\left|\dfrac{z-w}{z-z\bar{z}w}\right|=
\left|\dfrac{z-w}{z-w}\right|=1
$$
If $|w|=1$, consider that $|1-\bar{z}w|=|1-\bar{w}z|$ because they are conjugate.
|
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|
Prove that the union of two disjoint countable sets is countable This is a question from my proofs course review list that I have had trouble understanding.
I understand the concept of disjoint sets. I'm not sure what they mean by countable. How would one prove a set is countable and furthermore, that the union of two countable sets is countable.
Please help me get started with the question by understanding this concept.
|
(In the following, countable means what is often referred to as countably infinite, i.e. countable and not finite)
A set $A$ is called countable if there exists a bijection $f$ between $A$ and the set of natural numbers $\mathbb{N}$. In other words, $A$ is countable if there is some mapping $f \,:\, A \to \mathbb{N}$ such that $f$ hits every element of $\mathbb{N}$ exactly once. Note that any such $f$ has a uniquely defined inverse $f^{-1}$.
It follows that there exists a bijection $f_{A,B}$ between any two countable sets $A$ and $B$, because if $f_A$ is a bijection from $A$ to $\mathbb{N}$, and $f_b$ a bijection from $B$ to $\mathbb{N}$, then $x \mapsto f_B^{-1}(f_A(x))$ is a bijection from $A$ to $B$.
Therefore, to show that the union of two arbitrary disjoint countable sets is countable, it suffices to show that the union of two specific disjoint countable sets $A,B$ is countable. This works because if $X,Y$ are disjoint and countable, by the above there are bijections $f_X \,:\, X \to A$, $f_Y \,:\, Y \to B$, and a bijection $g \,:\, A \cup B \to \mathbb{N}$. We can therefore define a biection from $f \,:\, X \cup Y$ to $\mathbb{N}$ by setting $$
f(z) = \begin{cases}
g(f_X(z)) &\text{if $z \in X$,} \\
g(f_Y(z)) &\text{if $z \in Y$.}
\end{cases}
$$
(I leave the proof that this actually is a bijection to you)
Since we can view $\mathbb{Z}$ as the disjoint union of $\mathbb{N} = \{0,1,\ldots\}$ and $\mathbb{Z} \setminus \mathbb{N} = \{\ldots,-2,-1\}$, it therefore suffices to
*
*show that $\mathbb{Z} \setminus \mathbb{N}$ is countable, i.e. to find a bijection between $\{\ldots,-2,-1\}$ and $\{0,1,\ldots\}$, and
*show that $\mathbb{Z}$ is countable.
|
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|
Calculating the residues of $\frac{\zeta^{\prime}{(s) x^{s}}}{\zeta(s)\cdot s}$ Calculating the poles of $\frac{\zeta^{\prime}{(s) x^{s}}}{\zeta(s)\cdot s}$, where x is a fixed real number.
I am trying to calculate the poles of this function at the trivial zeros of $\zeta$, namely the even negative integers.
To do so I image one looks at
$\lim_{s \to \ -2k} \frac{(s+2k)\zeta^{\prime}{(s) x^{s}}}{\zeta(s)\cdot s}$
but I have trouble when evaluating the $\zeta^{\prime}$ at $-2k$.
Is my strategy bad or am I able to calculate $\zeta^{\prime}(-2k)$ for every k? I suspect the value to be 1.
This is almost exactly the same question, but no satisfying answer has been given.
|
For a fixed $x\in \mathbb{R}\setminus \{0\}$, $s\mapsto x^s$ is an entire function (you can choose different branches of that function by choosing different logarithms of $x$, but they all are entire), so that factor contributes no poles. In simple poles of the function, it modifies the residue just by multiplication with the value of $x^s$ in the pole.
The factor $\frac{1}{s}$ contributes a simple pole in $0$ (since $\zeta(0)\neq 0$), and the residue
$$\operatorname{Res}\left(\frac{\zeta'(s)\cdot x^s}{\zeta(s)\cdot s}; 0\right) = \frac{\zeta'(0)}{\zeta(0)} = \frac{-\frac12\log (2\pi)}{-\frac12} = \log (2\pi)$$
is easy to find.
So the poles arising from $\frac{\zeta'(s)}{\zeta(s)}$ remain. Generally, if $f$ is holomorphic in a punctured neighbourhood of $z_0$ and has a pole or zero in $z_0$, we have a representation $f(z) = (z-z_0)^k\cdot h(z)$ in that punctured neighbourhood, where $h$ is holomorphic with $h(z_0) \neq 0$. Then the logarithmic derivative of $f$ in that punctured neighbourhood is
$$\frac{f'(z)}{f(z)} = \frac{k(z-z_0)^{k-1}\cdot h(z) + (z-z_0)^k\cdot h'(z)}{(z-z_0)^k\cdot h(z)} = \frac{k}{z-z_0} + \frac{h'(z)}{h(z)}.$$
Since $\frac{h'}{h}$ is holomorphic in a neighbourhood of $z_0$, we see that $\frac{f'}{f}$ has a simple pole in $z_0$ with residue $k$ [so the residue is the multiplicity of the zero if $f(z_0) = 0$, and the negative of the order of the pole if $f$ has a pole in $z_0$].
Since the trivial zeros of $\zeta$ are all simple (which is easy to see from the functional equation), it follows that
$$\frac{\zeta'(s)}{\zeta(s)}$$
has simple poles with residue $1$ there, so
$$\operatorname{Res}\left(\frac{\zeta'(s)\cdot x^s}{\zeta(s)\cdot s}; -2n\right) = \frac{x^{-2n}}{-2n} = - \frac{1}{2n x^{2n}}$$
for $n$ a positive integer. Generally, if $\rho$ is a zero of $\zeta$ with multiplicity $\mu(\rho)$, we have
$$\operatorname{Res}\left(\frac{\zeta'(s)\cdot x^s}{\zeta(s)\cdot s};\rho\right) = \mu(\rho)\frac{x^\rho}{\rho}.$$
The pole of $\zeta$ at $1$ adds a simple pole of $\frac{\zeta'(s)\cdot x^s}{\zeta(s)\cdot s}$ with residue $-x$.
|
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|
Cartier divisors and global sections I have a brief question - I seem to have a vague recollection that if we have a Cartier divisor $D$ on a scheme $X$ , then we can determine whether $D$ is effective by saying whether $\mathcal{O}_X(D)$ has a global section or not. I have tried to prove this fact, but can't seem to do it (is it true?) . If anyone could confirm / deny this I would be very happy, and I would also be glad if someone could give me a reference or a proof of the fact!
Sincerely
Tedar
|
Effective cartier divisors are just closed subschemes which locally are cut out by the vanishing of a non-zero divisor.
I don't know what you mean by $\mathcal{O}_X(D)$, but if you mean $\mathcal{O}_D(D)$, then certainly it need not have a nonconstant global section. Take for example a family of elliptic curves over $\mathbb{A}^1$. Each elliptic curve is a cartier divisor, but being a projective curve, doesn't have any nonconstant global sections.
See, for example, section 9 of http://stacks.math.columbia.edu/download/divisors.pdf
|
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|
How to show these inequalities? To show that $\forall n \in \mathbb{N}$
$$n \log{n}-n+1 \leq \log{(n!)} \leq (n+1) \log{n}-n+1$$
do I have to use induction? Once at the one inequality and then at the other?
Or is there an other way to show this?
|
$$\log n!=\log 1+\log 2+\cdots +\log n$$ so $$\log n!\le \int_1^{n+1}\ln x\,\mathrm{d}x=(n+1)\ln(n+1)-(n+1)+1=(n+1)\ln(n+1)-n$$ and perhaps you can toy with that to get the right side.
For the other direction, $$\log n!=\log 2+\cdots +\log n\ge \int_1^n \ln x\,\mathrm{d}x=n\ln n-n+1$$
In general for an increasing function, we have $$\int_{a-1}^b f(x)\,\mathrm{d}x\le f(a)+f(a+1)+\cdots +f(b)\le \int_a^{b+1}f(x)\,\mathrm{d}x$$
|
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|
What is the reason for this geometric answer?
$$\frac{AB}{CD} = \frac{2}{2.6} = 0.77\ldots$$
$$\frac{AC}{AD} = \frac{2}{2.6} = 0.77\ldots$$
$$\frac{BC}{AC} = \frac{2}{2.6} = 0.77\ldots$$
Therefore $\triangle ABC$ and $\triangle ACD$ are similar.
I know from the answer sheet that $y$ is $47^\circ$ and $x$ is $109^\circ$. I don't know the reasons why though. The diagram does not look like a parallelogram because both opposite sides are not parallel.
|
Continuing,
since the triangles are similar,
$x = 109$
and
$y
= CAD
=180-(109+47)
=180-156
=24
$.
|
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|
Problem understanding the Axiom of Foundation I am just beginning to learn the ZF axioms of set theory, and I am having trouble understanding the Axiom of Foundation. What exactly does it mean that "every non-empty set $x$ contains a member $y$ such that $x$ and $y$ are disjoint sets." In particular, how can $y$ be an element of a set $x$, and also be disjoint from it (I have seen this be called the epsilon-minimal element). My intuitive understanding of sets (which is obviously wrong) tells me that if $y$ is an element of $x$, then the intersection of $\{y\}$ and $x$ should be $\{y\}$. How can it be $\varnothing$?
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The axiom isn't saying that $x\cap \{y\}=\emptyset$. It's saying $x\cap y=\emptyset$. Keep in mind that in ZFC everything is a set, including the elements of other sets.
|
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|
How to prove the inequalities $\int_{0}^{1}\sin{(x^n)}dx\ge\int_{0}^{1}(\sin x)^ndx\ge 0$
Show that:
$$\int_{0}^{1}\sin{(x^n)}dx\ge\int_{0}^{1}(\sin x)^ndx\ge 0$$
My idea:maybe $\sin{(x^n)}\ge (\sin{x})^n？$
|
Let $n\geqslant1$ and $u(x)=\sin(x^n)-(\sin x)^n$ then $$u'(x)=n\cdot\left(x^{n-1}\cos(x^n)-(\sin x)^{n-1}\cos x\right).$$ For every $x$ in $(0,1)$, $x^n\leqslant x$ hence $\cos(x^n)\geqslant\cos x$, and $x\geqslant\sin x$ hence $x^{n-1}\geqslant(\sin x)^{n-1}$. Thus, $u'(x)\geqslant0$ for every $x$ in $(0,1)$. Since $u(0)=0$, this proves that, for every $x$ in $(0,1)$, $$\sin(x^n)\geqslant(\sin x)^n,$$ which yields the comparison of the integrals for $n\geqslant1$ (the fact that both integrals are nonnegative is direct).
|
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|
Learning Advanced Mathematics I'm a 12th grade student and I've recently developed a passion for mathematics . Currently my knowledge in this particular area is comprised by : single-variable calculus , trigonometry , geometry , basic notions of linear algebra and set theory .
I'm particularly interested in calculus and I need some advice as I intend on coupling my future math skills with the study of Quantum Physics .
Where should I go from here in order to understand multivariable calculus ?Can anyone recommend an interesting advanced calculus textbook ?
(Hope the prerequisites needed for understanding the textbooks you suggest match my current mathematical skills and knowledge)
|
As DonAntonio commented a good idea is to review single variable calculus emphasizing the theorems and their proofs. A good book to go beyond just the praxis of solving integrals and into the demonstrations is Tom Apostol's "Calculus and Linear Algebra" volumes 1 and 2. The first is primarily concerned with single variable calculus and analytic geometry, the second volume with proper linear algebra and multivariable calculus, with nice add-ons like differential equations, probability and some numerical analysis. I do think both volumes address a "starters set" in both math and physics.
Regarding quantum mechanics, although linear algebra, differential equations and Fourier analysis are usually enough to grasp the methods I strongly suggest you invest in learning the classical physics first, as in classical mechanics, thermodynamics and electromagnetism. An option is to try Feynman's Lectures that are now fully available online in high resolution here http://www.feynmanlectures.caltech.edu/
|
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How to find PV $\int_0^\infty \frac{\log \cos^2 \alpha x}{\beta^2-x^2} \, \mathrm dx=\alpha \pi$ $$
I:=PV\int_0^\infty \frac{\log\left(\cos^2\left(\alpha x\right)\right)}{\beta^2-x^2} \, \mathrm dx=\alpha \pi,\qquad \alpha>0,\ \beta\in \mathbb{R}.$$
I am trying to solve this integral, I edited and added in Principle value to clarify the convergence issue that the community pointed out. I tried to use $2\cos^2(\alpha x)=1+\cos 2\alpha x\,$ and obtained
$$
I=-\log 2 \int_0^\infty \frac{\mathrm dx}{\beta^2-x^2}+\int_0^\infty \frac{\log (1+\cos 2 \alpha x)}{\beta^2-x^2}\mathrm dx,
$$
simplifying
$$
I=\frac{ \pi \log 2 }{2\beta}+\int_0^\infty \frac{\log (1+\cos 2 \alpha x)}{\beta^2-x^2}\mathrm dx
$$
but stuck here. Note the result of the integral is independent of the parameter $\beta$. Thank you
Also for $\alpha=1$, is there a geometrical interpretation of this integral and why it is $\pi$?
Note this integral
$$
\int_0^\infty \frac{\log \sin^2 \alpha x}{\beta^2-x^2} \,\mathrm dx=\alpha \pi-\frac{\pi^2}{2\beta},\qquad \alpha>0,\beta>0
$$
is also FASCINATING, note the constraint $\beta>0$ for this one. I am not looking for a solution to this too obviously on the same post, it is just to interest people with another friendly integral.
|
Consider the function $$ f(z) = \frac{\log(1+e^{2i \alpha z})}{z^{2}-\beta^{2}} \ , \ (\alpha,\beta >0)$$
which is well-defined on the complex plane if we omit the real axis and restrict $z$ to the upper half-plane while defining $\log (1+e^{2iaz})$ to be $\log(2)$ just above the origin.
Notice that $$\text{Re} \big( f(x) \big) = \frac{1}{2} \frac{\log(2+2 \cos 2 \alpha x)}{x^{2}-\beta^{2}} = \frac{1}{2} \frac{\log \big(4 \cos^{2} ( \alpha x) \big)}{x^{2}-\beta^{2}}.$$
Now integrate around a contour that consists of the line segment just above the line segment $[-R,R]$ (with half-circle indentations of radius $r$ around the branch points at $z= \frac{(2n+1)\pi}{2 \alpha}$ and the simple poles at $z = \pm \beta$) and the upper half of the circle $|z|=R$.
Along the upper half of $|z|=R$, $\log(1+e^{2iaz}) \to 0$ as $R \to \infty$. So the integral clearly vanishes along there as $R \to \infty$.
And since $\lim_{z \to \frac{(2n+1)\pi}{2 \alpha}}\left(z- \frac{(2n+1)\pi}{2 \alpha} \right) f(z) = 0$, the contributions from the indentations around the branch points vanish as $r \to 0$.
So we have
$$\begin{align} \text{PV} \int_{-\infty}^{\infty} \frac{\log \left(4 \cos^{2}( \alpha x)\right)}{\beta^{2}-x^{2}} \ dx &= -2 \ \text{Re} \ \text{PV} \int_{-\infty}^{\infty} \frac{\log(1+e^{2i \alpha x})}{x^{2}-\beta^{2}} \ dx \\ &= -2 \ \text{Re} \Big( i \pi \ \text{Res}[f(z),\beta)] + i \pi \ \text{Res}[f(z),-\beta)]\Big) \\ &= - 2 \ \text{Re} \ i \pi \Big(\frac{\log(1+e^{2i \alpha \beta})}{2 \beta} + \frac{\log(1+e^{-2i \alpha \beta})}{-2 \beta} \Big) \\ &= \frac{2\pi}{\beta} \left[\arctan\Big(\frac{\sin 2 \alpha \beta}{1+\cos 2 \alpha \beta}\Big) \right] \\ &= \frac{2 \pi}{\beta} \arctan \left(\tan ( \alpha \beta) \right) , \end{align}$$
which implies
$$ \log(4) \ \text{PV} \int_{-\infty}^{\infty} \frac{1}{\beta^{2}-x^{2}} \ dx + \text{PV} \int_{-\infty}^{\infty} \frac{\log \cos^{2}(\alpha x)}{\beta^{2}-x^{2}} \ dx = \frac{2 \pi}{\beta} \arctan \left( \tan (\alpha \beta) \right).$$
But $$\text{PV} \int_{-\infty}^{\infty} \frac{1}{\beta^{2}-x^{2}} \ dx =0$$
since the residues at $\pm \beta$ cancel each other.
Therefore,
$$ \text{PV} \int_{0}^{\infty} \frac{\log \cos^{2}(\alpha x)}{\beta^{2}-x^{2}} \ dx = \frac{\pi}{\beta} \arctan \left(\tan (\alpha \beta) \right).$$
And if $\alpha \beta < \frac{\pi}{2}$, $$ \text{PV} \int_{0}^{\infty} \frac{\log \cos^{2}(\alpha x)}{\beta^{2}-x^{2}} \ dx = \frac{\pi}{\beta} \left(\alpha \beta \right) =\pi \alpha.$$
|
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