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H: Maximal ideal in a quotient ring
Consider the ring $A=\mathbb{C}[x,y]/(y^2-x^4+5x-4)$, and consider the ideal $\mathfrak{m}=(y, x+2)$. Is $\mathfrak{m}$ maximal?
My sketched solution: consider an arbitrary element of $A$ and a representative $p(x,y)\in \mathbb{C}[x,y]$ for this element. Then $p(x,y)$ will be of the form $p(x,y)=y(\ldots)+g(x)$. Now by the Euclidean algorithm, there exist $q,r$ so that $g(x)=q(x)(x+2)+r(x)$ with $deg(r)<1$, i.e. $r(x)=c$ for some $c\in\mathbb{C}$. Summing up, $p(x,y)=y(\ldots)+(x+2)(\ldots)+c$ for some complex number $c$. Therefore there exist a surjective homorphism $\mathbb{C}\rightarrow A/\mathfrak{m}$, so $A/\mathfrak{m}\simeq\text{some quotient of $\mathbb{C}$}$. Clearly it must be $\mathbb{C}$ itself (since it's a field), so $A/\mathfrak{m}\simeq \mathbb{C}$ which implies that $\mathfrak{m}$ is maximal.
Is it a valid answer? Thanks in advance.
Edit: $A=\mathbb{C}[x,y]/(y^2-x^4+5x^2-4)$
AI: As Stefan Walter pointed out, you need to show that the quotient is nonzero. To see that, notice that we have
$$(y^2-x^4+5x^2-4,y,x+2)=(x^4-5x^2+4,y,x+2)=(y,x+2)$$
So $A/\mathfrak m\cong {\bf C}[x,y]/(x+2,y)\cong {\bf C}[x]/(x+2)\cong {\bf C}$. Since the quotient is a field, $\mathfrak m$ must be maximal. |
H: Equality constrained norm minimization
In these slides (5) the dual function for the norm minimization problem:
$$
\min_x \|x\| \quad \mbox{s.t.} \quad Ax =b
$$
is defined as:
$$
g(v) = \inf_x (\|x\| - ν^\intercal Ax + b^\intercal ν)
$$
what I don't understand is why the signs are reserved. The Lagrangian is according to the same author
$$
\|x\| + v^\intercal (Ax - b)
$$
so the dual function should have been
$$
g(v) = \inf_x (\|x\| + ν^\intercal Ax - b^\intercal ν)
$$
Is this correct?
AI: $$g^{'}(v) = \inf_x (\|x\| + ν^\intercal Ax - b^\intercal ν)$$ is equivalent to $$g(v) = \inf_x (\|x\| - ν^\intercal Ax + b^\intercal ν)$$
except for the final $v^T$ that you will obtain which leads to $\min ||{x}||$. It means $x_{min}$ which minimizes $||x||$ conditioned on $ Ax =b$ will be exactly the same if you use $g^{'}(v)$ or $g(v)$ |
H: Calculate "x % slower/faster"?
Ok, this might sound a bit stupid, but I find these kind of statements "x % slower/faster" pretty confusing.
Let's say I have algorithm A and algorithm B. Algorithm A takes 50 seconds to complete a task and algorithm B takes 25 seconds.
Now I could make the following statements:
"B is twice as fast as A"
"B is 100 % faster than A" (speedA - speedB)/speedB = 100%
"A is 50 % slower than B" (speedA - speedB)/speedA = 50%
Am I right about this?
AI: Yes, you're right about that.
In an ideal world, possibly, we would always use logarithms rather than ratios (percentages) when measuring relative differences. Then, supposing we standardized on the base-2 logarithm we could say
B is $1$ doubling faster than A.
A is $1$ doubling slower than B.
or if we used base-10 logarithms:
B is $0.3$ decades faster than A. ($\log_{10}\frac{50}{25}=0.3$)
A is $0.3$ decades slower than B. ($\log_{10}\frac{25}{50}=-0.3$)
The problem with that "ideal" world is that everyone who needed to speak about relative differences (a rather large fraction of the population) would have to know how logarithms work, which may or may not be a realistic goal. |
H: Finding constant speed
Lets say someone is flying at constant speed from place $X$ to place $Y$ and back.
Going to X takes 5 hours (with the wind) and coming back from X to Y takes 6 hours.
Lets assume that the wind is at a constant speed, then how long would it take for a piece of paper being propelled by the wind alone to travel from $X$ to $Y$.
AI: Heh, this problem is cute!
Suppose our (constant) speed is $v$, the (constant) speed of the wind is $w$, the time it takes for us to get from $Y$ to $X$ is $t_1$ and going back takes $t_2$. If the distance between $X$ and $Y$ is $d$, then because the speeds are all constant, we know
$$ (v+w)t_1 = d = (v-w)t_2,$$
where the plus vs. the minus sign is because the faster direction ($Y$ to $X$) is where we're travelling in the direction of the wind. Putting the times in, we get
$$ 5v + 5w = 6v - 6w,$$
and so $w = v/11$.
Now, the distance the piece of paper has to travel is that same as the distance we travel, so we have
$$ t_1(v+w) = 5(v+w) = d = wt,$$
where $t$ is the amount of time the piece of paper takes to travel. Solving for $t$, we get
$$ t = \frac{5(v + v/11)}{v/11} = \frac{5(12v/11)}{v/11} = 5\cdot 12 = 60.$$
So the piece of paper takes 60 hours to get from $Y$ to $X$. |
H: Why the contour integral of $f(z)=\frac{i-1}{z+i}$ is not zero although it should be because $f(z)$ is analytic?
Why the contour integral of $\,\displaystyle{f(z)=\frac{i-1}{z+i}}\,$ is not zero although it should be because $f(z)$ is analytic? I have used contour $z=\gamma(t)=2e^{it}$, where $0\leq t\leq\pi$.
AI: It makes no sense to say a function is analytic on its own - you must also specify an open subset of $\mathbb{C}$ that $f$ is analytic over. Here, $f$ is not analytic inside the specified contour - it has a simple pole at $z=-i$ which is contained inside. |
H: Existence of infinite discrete family of open sets in a non compact topological space
How can I prove if a Hausdorff topological space $X$ is not compact, then there exist a countably infinite discrete family of open sets in $X$.
AI: I don't think this is true. Note that any Hausdorff space $X$ which has an infinite discrete family of (nonempty) open sets cannot be countably compact, and there are certainly examples of Hausdorff countably compact non-compact spaces (e.g., the ordinal space $\omega_1$).
For more detail, suppose $\{ U_n \}_{n \in \omega}$ is a discrete family of nonempty open subsets of $\omega_1$. Without loss of generality, we may assume that each $U_n$ is of the form $( \alpha_n , \beta_n )$ for $\alpha_n < \beta_n$. Letting $\gamma = \sup_{n < \omega} \beta_n < \omega_1$, note that each $U_n$ is actually a subset of the ordinal space $[0,\gamma]$, which is compact. But this contradicts the assumed discreteness of the family in question (since no compact space can have such a family). |
H: Mathematical function for four corelated attributes
I have $4$ attributes $A,B,C,D$
each of them takes value between $[0,1]$
The more $A$ and $B$, the more the function value is.
The more $C$ and $D$, the less the function value is.
if C or D equals "one" the function value is one.
How can I model this function.
I tried:
$|AB-CD|$ but did not work.
UPDATE:
I want this function also to return value between [0,1]
AI: $$f(A,B,C,D)=AB(1-C)(1-D)+1$$ is increasing in $A$ and $B$ and decreasing in $C$ and $D$ as well as it satifies $f(A,B,C,D)=1\,\, for\,\, C=1\, or D=1$.
EDIT: Of course there are infinitely many such functions satisfying your conditions. Above is probably one of the simplest one.
One other thing is that one can not define $f(A,B,C,D)\in[0,1]$ satisfying your conditions and still continuous. Here is the reason:
$f$ is decreasing in $C$ and $D$. Let $f$ be somewhere in $[0,1]$ for given $A,B$. As $f\leq1$, $f$ starts decreasing in $C$ and $D$ at most from $1$ and for $C$, $D$, or both very close to $1$ we have $f<1+\Delta$ where $\Delta$ is not arbitrarily close to $0$ when $C\rightarrow 1$ or $D\rightarrow 1$ indicating that $0\leq f(x+\Delta)-f(x)\leq \Delta$ can not be bounded by arbitrary $\Delta$ $\rightarrow$ discontinuity at $C=1$, $D=1$ or $C=1,D=1$ |
H: Proof that $e^x$ is a transcendental function of $x$?
Let a function $f(x)$ be algebraic if it satisfies an equation of the form $$c_n(x)(f(x))^n + c_{n-1}(x)(f(x))^{n-1} + \cdots + c_0(x)=0,$$ for $c_k(x)$ rational functions of $x$, and let $f$ be called transcendental if it is not algebraic. Is it possible to use this definition directly to show that $e^x$ is transcendental?
One way I have been considering for any complex number is this:
Let $x_0\in\mathbb{C}$ and $x_n=x_0+2\pi i n$, where $n\in\mathbb{Z}$. Hence $x_n\neq x_m$ for all $n\neq m$, but we do have $e^{x_n}=e^{x_m}$ for all $n,m\in\mathbb{Z}$ (since $e^{2\pi i n} = 1$ for all $n\in\mathbb{Z}$). But since the Implicit Function Theorem suggests there exists an exact algebraic formula for $x$ using the above definition of an algebraic function, then $e^x$ can not be algebraic since there are an infinite number of representations $x_n$ of $x$.
AI: One could use the growth at infinity of the function $f:x\mapsto\mathrm e^x$.
Assume that $f$ is algebraic and choose a real number $x\geqslant0$. Then $|f(x)|\geqslant1$ and
$$
|c_n(x)|\,|f(x)|^n\leqslant b(x)|f(x)|^{n-1},\qquad b(x)=\sum\limits_{k=0}^{n-1}|c_k(x)|.
$$
Hence, for every real number $x\geqslant0$ such that $c_n(x)\ne0$, $|f(x)|\leqslant b(x)/|c_n(x)|$. But indeed, $c_n(x)\ne0$ for every real number $x$ large enough and $b(x)/|c_n(x)|$ can grow at most polynomially when the real number $x$ goes to $+\infty$ while $|f(x)|=\mathrm e^x$ grows... well, exponentially. This is a contradiction. |
H: How many numbers less than $x$ are co-prime to $x$
Is there a fast way , or a direct function to give the count of numbers less than $x$ and co-prime to $x$ .
for example if $x$ = 3 ; then $n = 2$ and if $x$ = 8 ; then $n = 4$.
AI: The relevant function is the Euler Totient function and the link contains several ways to compute it. |
H: reformulation of square residues for odd numbers?
If you look at the table of square residues:
http://en.wikipedia.org/wiki/Quadratic_residue#Table_of_quadratic_residues
you will find that for x^2 mod N where N is odd, we have N different residues but with simetry in half:
Example: N=25
mod 25
1,4,9,16,0,11,24,14,6,0,21,19 left side
19,21,0,6,14,24,11,0,16,9,4,1,0 right side
If we start from number 19 on the right side then difference between other right side numbers and 19 is
0, 2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, mod 25
http://oeis.org/A002378
formula for that sequence is n(n+1) or n(n-1)
If we go on with numbers we folow numbers of the left side. Residues of mod 25 is cycle of 25 numbers.
So if we want to find x^2 mod 25 = 14 is this the same as finding solution for
(y^2-y+19) mod 25 = 14 ?
example: from the above we get for 5 => (25-5+19) mod 25 = 14. Number five in that case means fifth number from the right sequence x^2 mod 25 which represents square number 289. Next square number with same residue is 16 places further(number 33^2)
AI: Yes, they are the same: $y^2 - y + 19$ is the same as $(y-13)^{2}$ (mod $25$), so the answers to one congruence are translates of the answers to the other. |
H: How to solve $100x +19 =0 \pmod{23}$
How to solve $100x +19 =0 \pmod{23}$, which is $100x=-19 \pmod{23}$ ?
In general I want to know how to solve $ax=b \pmod{c}$.
AI: Hint $\displaystyle\rm\,\ mod\ 23\!:\,\ x\,\equiv\, \frac{\color{brown}{-19}}{4\cdot \color{#0A0}{25}}
\,\equiv\,\frac{\color{brown}4}{4\cdot\color{#0A0} 2}
\,\equiv\, \frac{\color{blue}1}2
\,\equiv\, \frac{\color{blue}{24}}2
\,\equiv\, 12,\ \:$
by $\:\ \begin{array}{r}\color{brown}{{-}19\equiv 4}, & \color{blue}{1\equiv 24}\\ \color{#0A0}{25\equiv 2}\,\end{array} $ |
H: Finding all $x$ for $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$
I'm trying to find all $x$ for the inequality $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$. In order to do this, I want to factor one side so that I can find all values where $x$ determines the term to equal $0$.
$$\frac{2x - 13}{2x + 3} \lt \frac{15}{x} \iff \frac{x(2x - 13) - 15(2x+3)}{x(2x + 3)} \lt 0$$
$$\iff \frac{2x^2 - 13x - 30x - 45}{2x(x + \frac{3}{2})} \lt 0 \iff \frac{x^2 - \frac{43}{2}x - \frac{45}{2}x}{x(x + \frac{3}{2})} \lt 0 $$
$$\iff \frac{x^2 - \frac{43}{2}x - \frac{45}{2}x}{x(x + \frac{3}{2})} \lt 0 \iff \frac{(x - \frac{43}{4})^2 - \frac{1939}{4}}{x(x+\frac{3}{2})} \lt 0$$
I don't know how to get any further, and I'm starting to get too high values to handle.
The next step as I can see would be to find an $x$ that makes $(x - \frac{43}{4})^2 = \frac{1939}{4}$. This, alongside the obvious ones for $x$ and $x+\frac{3}{2}$ (creating division by $0$), would help me find the possible values for $x$.
But how do I get the last step? Or am I already dead wrong?
AI: Hint: From here:
$$\frac{2x^2 - 43x - 45}{2x(x + \frac{3}{2})} \lt 0$$
you could say that the fraction is less than zero if and only if either (1) the numerator is positive and the denominator is negative or (2) the numerator is negative and the denominator is positive. So for case (2), you solve:
$$
2x^2 - 43x - 45 <0 \quad \text{and}\quad 2x(x +\frac{3}{2}) > 0.
$$
You might already know this, but the way you solve each of these two inequalities is by finding possible roots. So for example for the $2x(x+ \frac{3}{2}) > 0$, the roots are $x = 0$ and $x = -\frac{3}{2}$. Now you just need to find the (constant) signs of the expression $2x(x+\frac{3}{2})$ on the intervals $(-\infty, -\frac{3}{2})$, $(-\frac{3}{2}, 0)$, and $(0, \infty)$. |
H: Complex analysis: Radius of convergence of power series
$$
\sum_{n=1}^\infty \frac{\cos n \theta}{(\sqrt{13})^{n+1}}x^n
$$
Find the radius of convergence for the above series. I have learnt to use the root test and ratio test but neither of them seem to work. I have problems manipulating.
Not sure if this is useful: $\cos z = \frac{1}{2} \left(e^{iz}+e^{-iz} \right)$
AI: Here is a hint. For any $N\in\mathbb{Z}^+$ we have that $\sup_{n>N}(\cos(n\theta))^{1/n} = 1$. Now try the root test. |
H: Radius of Convergence of this Series
This is a question from a GRE math subject test practice material.
$$ \sum^{\infty}_{n=1} \frac{n!x^{2n}}{n^n(1+x^{2n})} $$
The set of real numbers $x$ for which the series converges is: $\{0\}$, $\{-1 \leq x \leq 1\}$, $\{-1 < x < 1\}$, $\{-\sqrt{e} \leq x \leq \sqrt{e}\}$ or $\mathbb{R}$?
Attempt:
$\frac{n!}{n^n} = \frac{1}{n} \frac{2}{n} \frac{3(4)(5)...(n-1)(n)}{n^{n-2}} < (2/n^2)$
So by comparison test to $(2/n^2)$, $\frac{n!}{n^n}$ converges. Furthermore, $0 < \frac{x^{2n}}{(1+x^{2n})} < 1$ for all $x \in \mathbb{R}$. So this series converges for all real number $x$? The answer is $\mathbb{R}$?
Questions:
Is my logic and answer correct? Is there an easier way?
AI: Applying the root rest:
$$
\limsup_{n \to \infty} \left( \frac{n! x^{2n}}{n^n (1+x^{2n})} \right)^{1/n} = \limsup_{n \to \infty} \frac{x^2}{n \cdot \max(1,x^2)} \left( n! \right)^{1/n} = \mathrm{e}^{-1} \cdot \frac{x^2}{ \max(1,x^2)}
$$
The right-hand-side is bounded from about by $\mathrm{e}^{-1} < 1$ for all $x$, that is it converges for all $x \in \mathbb{R}$.
And if one really wants to use the ratio test, one can, though it certainly qualifies as doing things the hard way:
$$\begin{align*}
\left|\lim_{n\to\infty}\frac{\dfrac{(n+1)!x^{2n+2}}{(n+1)^{n+1}(1+x^{2n+2})}}{\dfrac{n!x^{2n}}{n^n(1+x^{2n})}}\right|&=\lim_{n\to\infty}\frac{n^n(n+1)!x^{2n+2}(1+x^{2n})}{(n+1)^{n+1}n!x^{2n}(1+x^{2n+2})}\\
&=\lim_{n\to\infty}\frac{n^n(n+1)x^2(1+x^{2n})}{(n+1)^{n+1}(1+x^{2n+2})}\\
&=\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n\frac{1+x^{2n}}{1+x^{2n+2}}\\
&=\lim_{n\to\infty}\left(1-\frac1{n+1}\right)^n\frac{1+x^{2n}}{1+x^{2n+2}}\\
&=\lim_{n\to\infty}\left(\left(1-\frac1{n+1}\right)^{n+1}\right)^{\frac{n}{n+1}}\frac{1+x^{2n}}{1+x^{2n+2}}\\
&=\frac1e\lim_{n\to\infty}\frac{1+x^{2n}}{1+x^{2n+2}}\\
&=\begin{cases}
\frac1e,&\text{if }|x|\le 1\\\\
0,&\text{if }|x|>1
\end{cases}\\
&<1\;.
\end{align*}$$ |
H: a non-monotonic function on [0,1] with infinitely many points of discontinuity such that the function is bounded & Riemann integrable on [0,1].
could any one Give an example of a non-monotonic function on $[0,1]$ with infinitely many points of discontinuity such that the function is bounded & Riemann integrable on $[0,1].$?
AI: Try $f=\sum\limits_{n=1}^{+\infty}(-1)^na_n\mathbf 1_{[0,x_n]}$ with $(a_n)_{n\geqslant1}$ and $(x_n)_{n\geqslant1}$ decreasing to $0$. |
H: Similarity of matrices over $\Bbb{Z}$
Let $A$ be a $2 \times 2$ matrix over $\Bbb{Z}$ with characteristic polynomial $x^2 + 1$. Determine whether $A$ is similar to
$$\pmatrix{0 & -1 \\ 1 & 0}$$
over $\Bbb{Z}$.
I'm so used to working with matrices over fields $\Bbb{Q}$ and $\Bbb{C}$ that I'm not sure how it differs when we switch to non-fields like $\Bbb{Z}$. Any advice would be much appreciated.
AI: Notice that any such $A$ is an element of order $4$ in $SL_2({\bf Z})$. But $SL_2({\bf Z})\cong {\bf Z}_4*_{{\bf Z}_2}{\bf Z}_6$.
In an amalgamated product, the only elements of finite order are those which are conjugate to the elements of amalgamated groups, so every such $A$ is conjugate to the $1\in {\bf Z}_4$ or $3\in{\bf Z}_4$ in $SL_2({\bf Z})$.
$1$ corresponds to the matrix you provided (with the usual identification of $SL_2({\bf Z})$ with the amalgamated product), while $3$ is its inverse. They are not conjugate in $SL_2({\bf Z})$, but they are in $GL_2({\bf Z})$ by $\begin{pmatrix}0&1\\1&0\end{pmatrix}$. Since any $A$ is conjugate to one of them, all the $A$ are similar. |
H: Relationship between different sequences generated through modulo arithmetic.
I am unsure the formal mathematical terminology/notation for dealing with sequences generated from integer modulo arithmetic. So first off, could someone recommend a book that focuses on the sequences generated from arithmetic operations on finite sets of integers? I bought an elementary number theory book and an abstract algebra book but neither ever discussed sequences.
Now the more explicit question. Consider:
$$f \left[ n \right] = a n \left( \text{mod} \; N \right) \\
g\left[n\right] = b n \left( \text{mod} \; N \right) $$
where $a,b \in \mathbb{Z}$, $n \in \{0, 1, 2, \ldots \}$, $N \in \{2, 3, 4, \ldots\}$ and $a \left( \text{mod} \; N \right)$ is the remainder of $a / N$. I want to prove that sometimes the sequence $g[n]$ "generated" (may not be correct term) by a choice for $b$ is a "permutation" (may not be correct term) of the sequence $f[n]$ generated by $a$ if $N$ is the same both. I also want to show that the new permutation can be generated by
$$g \left[ n \right] = f \left[ k n \right]$$
for some $k \in \mathbb{Z}$ whenever such a permutation exists. I have a suspicion that if $a$ and $b$ are relatively prime to $N$ then such a permutation exists from what little I know of congruence relations, but I can also think of cases where $a$ is relatively prime to $N$ and $b$ is not that this still works.
AI: As stated in the comments, if $a$ and $b$ are coprime to $N$, then each of the sequences has period $N$. In this case $g[n]=f[kn]$ with $k=a^{-1}b$, where $a^{-1}$ is the multiplicative inverse of $a\pmod N$. This also works if only $a$ is coprime to $N$ (but in this case $g[n]$ has a shorter period).
More generally, if $d:=\gcd(a,N)\mid b$, then with $a'=a/d$, $b'=b/d$ and $N'=N/d$ we have $g[n]=f[kn]$ with $k=a'^{-1}b'$, where the multiplicative inverse is taken $\bmod{N'}$.
If $d\nmid b$, then $g[n]$ takes values that $f[n]$ doesn't take, so there can be no $k$ with $g[n]=f[kn]$. |
H: about a not continuous function and its derivatives
I'll start with a function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ whose partial derivatives exists at a point, but is not continuous at that point.
Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be defined by
$f(x)=\begin{cases}1, &\mbox{if}& x=0 &\mbox{or if}& y=0\\ 0, &\mbox{otherwise}& \end{cases}$
because $f$ is constant on the $x$ and $y$ axes where it equals $1$,
$$\frac{\partial f}{\partial x}(0,0)=0$$
$$\frac{\partial f}{\partial y}(0,0)=0$$
But $f$ is not continuous at $(0,0)$ because $\displaystyle \lim_{(x,y)\rightarrow (0,0)}f(x,y)$ does not exist.
Now my issue:
if $f:\mathbb{R} \rightarrow \mathbb{R}$ it is defined by $f(x)=\begin{cases}1, \mbox{if}& x=0 \\ 0, &\mbox{otherwise} \end{cases}$
this function is not continuous but my question is: is there a derivative for this function?- I say no. For the function $f :\mathbb{R}^2\rightarrow \mathbb{R}$ it is possible because $x=y=0$ not in same time? Am I wrong?
Thanks:)
AI: Differentiability (when all variables are allowed to vary simultaneously) $\implies$ continuity. However, partial differentiability $\not\implies$ continuity. A multivariate function $\mathbf{f}:\mathbb{R}^m\rightarrow\mathbb{R}^n$ is differentiable if there exists a linear function $\mathbf{J}:\mathbb{R}^m\rightarrow\mathbb{R}^n$ such that:
\begin{align}
\lim_{\mathbf{h}\rightarrow \mathbf{0}} \frac{\mathbf{f}(\mathbf{x}+\mathbf{h}) - \mathbf{f}(\mathbf{x}) - \mathbf{J}(\mathbf{x})\mathbf{h}}{||\mathbf{h}||_2} &= 0
\end{align}
Note that the direction of approach $\mathbf{h}\rightarrow \mathbf{0}$ has not been specified, and the above limit should exist for all possible directions. In case $m = 1$, we have two directions in which $h\rightarrow 0$, corresponding to the left and right hand limits.
If a multi-variate function is differentiable, then the partial derivatives exist $\mathbf{J}$ is the Jacobian matrix. However, if the partial derivatives exist, the above limit may still not exist, and the function may not be differentiable. Your example illustrates this fact. However, in case of a uni-variate function, partial derivative and the derivative are the same, because you can talk about change in only one variable's direction. |
H: Little-o vs. asymptotic equivalence
As big- and little-o notation are a little too technical to me, I prefer an expression with asymptotic equivalence ($\sim$). However, how does one "translate" this expression to an asymptotic equivalence? It's about the primorials, as found on Wikipedia (http://en.wikipedia.org/wiki/Primorial).
$p_n\#=e^{(1+o(1))\cdot n\cdot\log{n}}$
Is it possible to express this using $\sim$-notation?
Help is greatly appreciated.
AI: You can't, at least not directly, since we have $e^{n\cdot\log{n}}\nsim e^{(1+1/n)\cdot n\cdot\log{n}}=e^{n\cdot\log{n}}\cdot n$.
You can say something like $\log(p_n\#)\sim n\cdot\log{n}$, though. |
H: What type of graph problem is this?
Lets say I have four group
A [ 0, 4, 9]
B [ 2, 6, 11]
C [ 3, 8, 13]
D [ 7, 12 ]
Now I need a number from each group(i.e a new group) E [num in A,num in B, num in C, num in D], such that the difference between the maximum num in E and minimum num in E should be possible lowest.What type of problem is this ? which graph algorithm will be better to solve this kind of problem ?
Thanks in advance.
AI: I can't think of any reasonable way to represent this problem using
graphs.
I suggest the following to deal with your problem:
1) keep the elements of each groups sorted .
denote $n$ as the total number of elements ($n=|A|+...+|D|)$, for
each $M\in\{A,B,C,D\}$ denote$U_{M}$ as the union of the other groups
with higher alphabetical order (for example $U_{B}=C\cup D)$.
2) since we have to choose one element from each group we have to
choose one from $M$, for every element in $M$ find an element in$U_{M}$
s.t the difference is minimal (since the elements are sorted you can
do this in $log(|U|)$ using binary search).
save this minimum in
a variable and update it when you find a new minimum (also save the
elements that gave you the minimum and the groups they came from so
you can know what elements in what groups give this minimum)
At the end of this procedure you will know what two elements in which
groups give you your minimum, of course you can take the other two
elements for $E$ in an arbitrary way without it affecting the minimum.
The time complexity is $O(nlog(n))$ since for every $M$ it holds
that $|M|\leq n$ and$|U_{M}|\leq n$. |
H: What is an irreducible character of a finite group?
Let $S_n$ be the group of permutations of $\{1, 2, \ldots, n\}$. A “character” for $S_n$ is a function $\chi\colon S_n \to \mathbb{C} \setminus \{0\}$ with $\chi(ab) = \chi(a)\chi(b)$ for all $a, b \in S_n$ [NOTE: this is not 100% accurate - see comments and answers below (OP)]. What is an irreducible character? I have searched the Web, and all the references I found were inadequate or told me more than I wanted to know.
AI: The homomorphisms $G\to \Bbb C^\times$ are actually not the whole story of character theory, but are a very tidy chapter in it. If $V$ is a vector space (over $\Bbb C$) and $G$ finite, the homomorphisms $G\to GL(V)$ from $G$ into the general linear group of invertible linear maps are called representations, which are essentially the ways to equip $V$ with a linear $G$ action. If $\rho$ is a representation, then the map given by $\chi_\rho:G\to \Bbb C:g\mapsto\mathrm{tr}\,\rho(g)$ (the trace of the linear map associated to $g$, which is independent of basis or coordinate choice for $V$) is called a character of $G$.
If $V$ is one-dimensional (in which case we call $\rho$ and $\chi_\rho$ one-dimensional as well) then $\rho=\chi_\rho$ and the characters are multiplicative. Note that $\mathrm{tr}\,\rho(e_G)=\dim\,V$ shows the dimension can be directly computed from the character, so there is no ambiguity with respect to what dimension a character may have. With a distinguished basis we have $V\cong \Bbb C^n$ in an obvious way, and so we can write $GL(V)$ as $GL_n(\Bbb C)$, in which case we are working with matrix representations specifically.
If there is a proper nontrivial subspace $W\subseteq V$ that is invariant under the linear $G$-action, that is if we have $\rho(g)W\subseteq W$ for each $g\in G$, then the restriction of $G$'s linear action to $W$ then forms a subrepresentation of $V$. If a representation $\rho:G\to GL(V)$ has no subrepresentations, it is called an irreducible representation, in which case $\chi_\rho$ is also called irreducible. This is well-defined because (over any field of characteristic zero) representations are uniquely determined by their characters, which again prevents any sort of ambiguity or conflicting descriptions from arising.
A one-dimensional space has no proper nontrivial subspaces, so one-dimensional representations and characters (the ones you are talking about) are all automatically irreducible. Using the very nifty Schur's lemma, one can see that representations of abelian groups decompose into a direct sum of one-dimensional representations and thus are irreducible precisely when they are one-dimensional (you may encounter Schur's and direct sums later).
I can recommend this excellent note on the representation theory of the symmetric groups and Young tableaux, where the irreducible representations are obtained through Specht modules. To understand this fully you will need relatively good familiarity with representation theory, algebra, combinatorics and permutations, so maybe come back to it later. (Also for other readers.)
Representation theory doesn't need to be done in $\Bbb C$, we can work in arbitrary fields. However, as I hinted at a moment ago, if the characteristic divides the order of the finite group $G$ then things get complicated and we are working in modular representation theory. Things also get complicated when $G$ is a topological group, in which case we need theory from abstract harmonic analysis. |
H: Solving the cubic polynomial equation $x^3+3x^2-5x-4=0$
How can I solve the cubic polynomial equation $$x^3+3x^2-5x-4=0$$
I simplified it to:
$$x(x^2+3x-5)=4$$
But I don't know where to go from here.
AI: The manipulation that you performed does not help. Use the rational root test: since the leading coefficient is $1$ and the constant term is $4$, the only possible rational roots of the cubic are fractions of the form $a/b$, where $a$ is a divisor of $4$ and $b$ is a divisor of $1$. In other words, if the cubic has any rational root at all, it must be one of the numbers $\pm1,\pm2$, and $\pm4$. Actually calculation shows that
$$(-4)^3+3(-4)^2-5(-4)-4=-64+48+20-4=0\;,$$
so $-4$ is a root of the cubic $x^3+3x^2-5x-4$. Now use the fact that $r$ is a root of a polynomial if and only if $x-r$ is a factor of that polynomial to conclude that $x^3+3x^2-5x-4$ is divisible by $x-(-4)=x+4$. When you perform this division $-$ either by polynomial long division or by synthetic division $-$ you’ll get a quadratic as the quotient. Let’s say that you get $x^2+bx+c$ as the quotient; then you know that
$$x^3+3x^2-5x-4=(x+4)(x^2+bx+c)\;.$$
This tells you that $x=-4$ is one solution to your original equation, and the others are the solutions to the quadratic equation $x^2+bx+c=0$, which you can find using the quadratic formula. |
H: Is this a sufficient condition for a subset of a topological space to be closed?
Let $X$ be a topological space, and let $\{U_i\}$ be an open cover. If $Y$ is subset of $X$ such that $Y\cap U_i$ is closed in $U_i$ (for each $i$), does this imply that $Y$ is closed in $X$?
AI: Note that $Y^c\cap U_i = U_i \setminus Y \cap U_i$ is open in $U_i$. Therefore it is open in $X$. Now, since $\bigcup U_i = X$, $Y^c = \bigcup (Y^c \cap U_i)$ is open in $X$. Hence $Y$ is closed. |
H: distance between centers of two touching circles
Lets say we have two circles with radii 2 and 3 respectively.
If we then put Circle radius 2 on a flat surface, then the other circle on the flat surface so they touch once, what is the distance between the two centers?
AI: A the point of contact the circles share a mutual external tangent and so the two radii are both perpendicular to that line. Therefore the center to center distance is simply $r_1 + r_2$, in this case $2 + 3 = 5$. |
H: What does $H(\kappa)$ mean?
As the typical references (Wikipedia, Mathworld, etc.) don't seem to address this satisfactorily, I figured this would be a good place to put a nice formal definition. Hence:
I've heard that if $\kappa$ is a strongly inaccessible cardinal, then $H(\kappa)$ (or sometimes $H_\kappa$) equals $V_\kappa$. What does $H(\kappa)$ mean in this instance and how is it defined?
AI: The elements of $H(\kappa)$ are the sets that are hereditarily of cardinality less than $\kappa$. If $x\in H(\kappa)$, then $|x|<\kappa$, $|y|<\kappa$ for every $y\in x$, $|z|<\kappa$ whenever there are $x$ and $y$, such that $z\in y\in x$, and so on.
This gives the intuitive idea, but it’s not really a definition. For that it’s easiest to start by defining the transitive closure of a set $x$:
$$\operatorname{tr cl}(x)=\bigcup_{n\in\omega}{\bigcup}^n(x)\;,$$
where
$${\bigcup}^n(x)=\begin{cases}
x,&\text{if }n=0\\\\
{\bigcup}\left({\bigcup}^{n-1}(x)\right),&\text{if }n>0\;.
\end{cases}$$
Then $$H(\kappa)=\{x:|\operatorname{tr cl}(x)|<\kappa\}\;.$$
(Although it doesn’t have the form required by the axiom schema of comprehension, this definition can be justified by showing that $H(\kappa)\subseteq V_\kappa$ for every infinite cardinal $\kappa$.)
Assuming AC, $H(\kappa)=V_\kappa$ iff either $\kappa=\omega$ or $\kappa$ is a beth fixed point; this is proved in the answer by François G. Dorais to this question. (Beth fixed points exist in ZFC.) |
H: Algebra over a Field
Let $\mathbb A$ be an algebra of dimension $k$ over the field $\mathbb F$. It is true that $\mathbb A$ is isomorphic to a subalgebra of the matrix algebra $M_k(\mathbb F)?$
AI: Yes, if the algebra $\mathbb{A}$ is associative then as Sean Eberhard points out the left-multiplication maps give linear transformations on the algebra $\mathbb{A}$ whose matrices form a subalgebra of $M_k(\mathcal{F})$ which is isomorphic to the given associative algebra.
Take $\mathbb{A} = \mathbb{F}^k$ which multiplication $*$. Let us examine where the associativity is necessary. Suppose $T: \mathbb{A} \rightarrow \mathbb{A}$ is left multiplication by $A \in \mathbb{A}$ this means $T_A(v) = A*v$. Now, suppose $T_B(v) = B*v$ is another such left-multiplication map. We should like $A \mapsto T_A$ to define an isomorphism. Consider $T_A \,{\scriptstyle \stackrel{\circ}{}}\, T_B$. Observe,
$$ (T_A \,{\scriptstyle \stackrel{\circ}{}}\, T_B)(v) = T_A(T_B(v)) = T_A(B*v)=A*(B*v) $$
Yet, $T_{A*B}(v) = (A*B)*v$. If $*$ is not an associative operation there is no reason that $T_A \,{\scriptstyle \stackrel{\circ}{}}\, T_B = T_{A *B}$ should hold true.
It may still be possible to find an isomorphism to some subset $S$ of $M_k(\mathcal{F})$ if we give the subset an operation other matrix multiplication. This means $S$ is not strictly speaking a subalgebra of $M_k(\mathcal{F})$ since the multiplication on $S$ is not inherited from the matrix multiplication operation on $M_k(\mathcal{F})$.
Some authors use different notation for the same point-set to indicate a different choice of multiplication. For example, $M(\mathbb{F})=\mathbb{F}^{n \times n}$ with matrix multiplication whereas $gl_n(\mathbb{F})= \mathbb{F}^{n \times n}$ with the commutator-bracket multiplication $[A,B] = AB-BA$. $gl_n(\mathbb{F})$ is not an associative algebra, the departure from associativity is quantified by the Jacobi identity $[A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0$. For a finite dimensional Lie algebra of dimension $k$ it turns out that you can find an isomorphic copy of the algebra in $gl_n(\mathbb{F})$ where $n \geq k$. Equality may not be possible. See Ado's Theorem; http://en.wikipedia.org/wiki/Ado%27s_theorem . This is also known for super Lie algebras.
I tend to think that if we have a finite dimensional algebra then it is possible to embedd it in matrices of sufficiently large order, even if it is nonassociative. But, the operation need not be simple matrix multiplication. |
H: nonlinearity of PDE's
(1) $u_x + u_y = 0$
(2) $u_x + yu_y = 0$
(3) $u_x + uu_y = 0$
(4) $u_{xx} + u_{yy} = 0$
(5) $u_{tt} − u_{xx} + u^3 = 0$
(6) $u_t + uu_x + u_{xxx} = 0$
(7) $u_{tt} + u_{xxxx} = 0$
(8) $u_t − iu_{xx} = 0$
Can someone provide an explanation from definition of linearity why (3), (5), and (6) are NOT linear PDE's?
I.e., I tried $L(cu) = cL(u)$ and $L(u+v) = Lu + Lv$ but can't get this definition to work out nicely for examples 3, 5, and 6 above.
Thanks
AI: Let's look at number 3 in full detail. Define
$$L(u)=u_x+u u_y$$
Then for arbitrary constant c, we have
$$L(c u)=(c u)_x+(c u )(c u)_y=c u_x+c u \left(c u_y\right)=c u_x+c^2 u u_y$$
Compare this with
$$c L(u)=c\left(u_x+u u_y\right)=c u_x+c u u_y$$
So
$$L(c u) - c L(u) = c u_x + c^2 u u_y -\left(c u_x+c u u_y\right)=\left(c^2-c\right)u u_y\neq 0$$
and therefore $L(u)$ is not linear. Similarly, examples 5 and 6 are also seen to have non-linear terms. |
H: Divisibility of consecutive natural numbers
I found this task followed by a hint,that I should try to apply Chinese remainder theorem to that:
Prove, that there exist 2012 consecutive natural numbers,which satisfy that every one of them is divisible by a cube of a natural number $\ge$ 2.
The problem is, I don't really see how to use the theorem above. Can anyone help?
AI: Look at the system of congruences
$x\equiv 0 \bmod{2^3}$, $\,x+1\equiv 0\bmod{3^3}$, $\,x+2\equiv 0\bmod{5^3}$, $x+3\equiv 0\bmod{7^3}$, $\,x+4\equiv 0\bmod{11^3}$, and so on. |
H: Why a quadratic equations always equals zero?
On evaluating quadratic equations, It always equals zero:
$$ax^2+bx+c=0$$
Why zero? Is it possible to use other number for another purpose?
AI: The value of c is a simple number with no variable. So you can move any value on the right side over to the left and it will just become part of c. Example:
$$x^2+x-6=6$$
$$x^2+x-12=0$$
Therefore, we can set the right hand side equal to any number we want. We usually set it equal to zero because this helps to solve later. Example:
$$(x+3)(x-2) = 6$$ vs
$$(x-3)(x+4) = 0$$
The second one is easier to solve because we know anything multiplied by 0 is 0. That means we can solve each part individually.
EDIT:
After we reach the factored form, we know the answer is in the form of something multiplied by something else equals a number. If that number is not 0 then we must take both parts into account. On the other hand, if it is 0 then we can simply ask what will make one of those parts zero? Then it doesn't matter what the other part is.
$$(x-3)(x+4) = 0$$
Here, we know that if $(x-3) = 0$ or $(x+4) = 0$ then the whole thing will equal zero, because anything multiplied by 0 is 0. So, we can just ask what value of $x$ will make $(x-3) = 0$ true?
Compare this to the version not set to zero:
$$(x+3)(x-2) = 6$$
Now, we can't make this any simpler. We must figure out what value of $x$ will make that entire thing true from the start. |
H: Dimension and basis of kernel and image of linear transformation
From a bank of past prelim exams:
Let $V$ be the real vector space of real polynomials of degree at most $5$, and define $T: V \to \Bbb{R}^3$ by $T(f) = (f(0), f(1), f(2))$. Find the dimensions of the kernel and image of $T$, and find a basis for each.
I've attempted a solution, working with the standard basis of $V$ and concluded that the kernel is the set of polynomials divisible by $x(x-1)(x-2)$. Thus, I concluded that the dimension of the kernel is $3$, and the dimension of the image is $6-3=3$ by the rank-nullity theorem.
I'm just unsure of how to write the basis of the kernel and image. Any help would be greatly appreciated.
AI: The nullspace is all polynomials of the form $$x(x-1)(x-2)(a + bx + cx^2) \text{ for some } a, b, c \in \Bbb{R}.$$ Expand and rewrite as:
$$ ax(x-1)(x-2) + bx^2(x-1)(x-2) + cx^3(x-1)(x-2) $$
i.e. all the kernel elements are scalar linear combinations of $3$ polynomials. Hence the kernel basis elements are
$$\{ x(x-1)(x-2), x^2(x-1)(x-2),x^3(x-1)(x-2)\} $$
Ops, I forgot the other part. The image is $\Bbb{R}^3$ whose basis is the canonical basis. Originally, I over-complicated things: consider every non-zero tuple $(r_0, r_1, r_2) \in \Bbb{R}^3$. We can show that we can construct $f(x) = ax^5 + bx^4 + cx^3 +dx^2 + ex + r_0$ such that $f(0) = r_0, f(1) = r_1, f(2) = r_2$. We can show that such $f$ always exists by an interpolation argument. Hence the image is all of $\Bbb{R}^3$.
Added: explicit construction for $f(x)$ such that $f(0) = r_0, f(1) = r_1, f(2) = r_2$: $$f(x) = \frac{r_2}{2}x(x-1) - r_1 x(x-2) + 2r_0(x-1)(x-2). $$ |
H: How to show that $\sum\limits_{n=1}^\infty \exp(i\,nz)$ converges?
How do I show that $$ f(z) = \exp(i\, z) + \exp(i\, 2z) + \ldots + \exp(i\, nz) + \ldots $$ converges? Problem is taken from a Yahoo! Answers question: "Find the infinite sum of sin(n)/n?".
AI: For the series to converge, you have the following condition,
$$ |{\rm e}^{iz}| < 1 \,.$$
Assuming $z=x+iy$, we have,
$$ \left|{\rm e}^{i(x+iy)} \right| = \left|{\rm e}^{ix-y)} \right| = {\rm e}^{-y} < 1 \Rightarrow -y < 0 \Rightarrow y > 0 \,.$$ |
H: does it come from cauchy integral formulae
Well I have to do the integral $$\int_{|z|=4}\frac{f(z) dz}{(z-2)(z-3)}$$
where $f(z)=\sin(\pi z^2)+\cos(\pi z^2)$
one breaks like: $$\int_{|z|=4}\frac{f(z) dz}{(z-2)(z-3)}$$ $$=\int_{|z-2|=1/2}\frac{f(z)/(z-3) dz}{(z-2)}$$ $$+\int_{|z-3|=1/2}\frac{f(z)/(z-2) dz}{(z-3)}$$
why it is true?Is there any easier method?
AI: If you "connect" the original path $\,|z|=4\,$ with those two paths $\,|z-2|=1/2\,,\,|z-3|=1/2\,$ via straight lines and you integrate forth and back those straight segments and around the new paths, the integral over each straight segment cancels out (both directions!) and you remain only with the integrals over the new paths, outside of which the integral equals zero as $\,\frac{1}{(z-2)(z-3)}\,$ is analytic.
Thus, the integral equals the sum of
$$\oint_{|z-2|=1/2}\frac{dz/(z-3)}{z-2}=2\pi i\left.\left(\frac{1}{z-3}\right)\right|_{z=2}=-2\pi i$$
$$\oint_{|z-3|=1/2}\frac{dz/(z-2)}{z-3}=2\pi i\left.\left(\frac{1}{z-2}\right)\right|_{z=3}=2\pi i$$
All in all, we get that
$$\oint_{|z|=4}\frac{dz}{(z-2)(z-3)}=-2\pi i+2\pi i =0$$ |
H: In how many ways a number $\gt 5000$ can be formed using given digits without repeating?
In how many ways one can form a number greater than $5000$ when allowed only to arrange digits taken from $2,3,4,5,8$ without repeating any digit?
I would think it would be $2\cdot4\cdot3\cdot2$.
Would this be correct?
AI: We need to think about all the ways to make a number bigger than $5000$,
any $5$ digit number we make will be bigger than $5000$, since none of our numbers are $0$.
so any arrangement of all $5$ does it, and there are $5!$ ways to arrange our $5$ numbers.
Then we consider $4$ digit numbers. For a $4$ digit number to be bigger than $5000$, its leading digit must be $\geq 5$ and so we have two possibilities for the first number (5 and 8).
For each choice of leading number we then have to order $3$ objects from a choice of $4$ objects. which means we have $\frac{4!}{(4-3)!} = 4!$ options for each leading digit.
So we have a total of $$5! + 4! + 4!$$
where the $5!$ is from all the $5$ digit numbers, and the two $4!$'s are from $4$ digit numbers with leading $5$ and $8$ respectively
EDIT: Throughout I've used the fact that the number of ways to order $m$ objects from a choice of $n$ objects is $$\frac{n!}{(n-m)!}$$
We can understand this easily enough:
For the first object we have $n$ choices, for the second object we have $(n - 1)$ choices and this continues until for the last object we have $(n - m + 1)$ choices.
So we have $$n(n-1)\cdots(n- m +1)$$ total possibilities.
Which can be expressed more succinctly as
$$\frac{n!}{(n-m)!}$$ |
H: Yet another differential equation
Hello,
I would appreciate any help solving the following equation:
$$\begin{align}
y''[t] + \dfrac{d}{m}y'[t] + \dfrac{k}{m}y[t] = G \\
\end{align}$$
subject to:
$$t[0] = t0$$
$$t'[0] = 0$$
This is not HW, this is for a website where I try to simulate object attached to a spring falling, as I presented in this question; I just forgot to add damping (d) force so that the object will stop its movement after a while.
Thank you!
AI: Multiply through by $m$ to give $my'' + dy' + ky = Gm.$ First, find the complementary function by solving the homogeneous equation $my'' + dy' + ky = 0.$ Use the trial function $y(t) = e^{\lambda t}$. We get $(m\lambda^2 + d\lambda +k)e^{\lambda t} = 0.$ Since $e^{\lambda t} > 0$ we need to solve $m\lambda^2 + d\lambda +k = 0$ for $\lambda.$ Whence:
$\lambda_{\pm} = \frac{1}{2m}(-d \pm \sqrt{d^2 - 4mk}) \, .$
The complimentary function is then $y(t) = ae^{\lambda_-} + be^{\lambda_+}$ for any $a$ and $b$. Next, we need to find the particular integral. Since the right hand side of $my'' + dy' + ky = Gm$ is constant, we try $y(t) = \alpha$ where $\alpha$ is a constant. Putting that into $my'' + dy' + ky = Gm$ gives $k\alpha = Gm$ and so $\alpha = Gm/k.$ The general solution is then:
$y(t) = ae^{\lambda_-} + be^{\lambda_+} + \frac{Gm}{k} \, .$
Finally, we impose the initial conditions that $y(0) = t_0$ and $y'(0) = 0.$ You will get two simultaneous equations in $a$ and $b$. Solve them gives:
$a = \frac{m\lambda_+(Gm-kt_0)}{k\sqrt{d^2-4mk}} \, , $
$b = \frac{m\lambda_-(Gm-kt_0)}{k(d^2-4mk)} \, . $
Provided I've not made a silly mistake, I get:
$y(t) = \frac{m}{k}\left[ G + \frac{(Gm-kt_0)}{\sqrt{d^2-4mk}}\left(\lambda_+e^{\lambda_-t} + \lambda_-e^{\lambda_+t}\right) \right] \, . $ |
H: Contradiction?: Partitions/Equivalence classes vs. Bijective homomorphism
I am reading through Michael Artin's Algebra, and the following passages seem contradictory to me.
On page 55: "The elements of the image correspond bijectively to the nonempty fibres, which are the equivalence classes."
On page 56: Proposition 2.7.15: "Let K be the kernel homomorphism $\phi: G \rightarrow G'$.
The fibre of $\phi$ that contains an element $a$ of $G$ is the coset $aK$ of $K$. These cosets partition the group $G$, and they correspond to elements of the image of $\phi$."
I would like to know what else is in the equivalence class of a? It could be alone, but it could also be b, and b~a, so $\phi(b) = \phi(a)$.
However, if the partition of $aK$ all equal $\phi(a)$, then how can $\phi$ be a bijective map?
Is it because everything in $aK$ is just the same thing? If this is the case, then groups are sets, and I was under the impression that in sets, repetition of elements is not allowed.
If anyone could clear my confusion, that would be very much appreciated. If there are simple examples that demonstrate what the textbook is trying to say, that would also be very helpful.
Many thanks.
AI: If $f:X\to Y$ is any map, the fibres of $f$ are the sets $f^{-1}[\{y\}]$ such that $y\in Y$; the particular set $f^{-1}[\{y\}]$ is called the fibre over $y$. The set $$\mathscr{F}=\Big\{f^{-1}[\{y\}]:y\in f[X]\Big\}$$ of all fibres of $f$ is a partition of $X$; it corresponds to the equivalence relation $x_1\sim x_2$ iff $f(x_1)=f(x_2)$. The correspondence $$y\leftrightarrow f^{-1}[\{y\}]\tag{1}$$ between $f[X]$ and $\mathscr{F}$ is clearly a bijection, and this has nothing to do with any extra structure on $X$ or $Y$. (It may not be a bijection between all of $Y$ and $\mathscr{F}$, because $f$ may not be a surjection: if $y\in Y\setminus f[X]$, then the fibre over $y$ is empty.) Note that the correspondence $(1)$ is not a bijection between $X$ and $f[X]$: it’s a bijection between $\mathscr{F}$, the set of non-empty fibres of $f$, and $f[X]$.
In the case of a group homomorphism $\varphi:G\to G'$, if $K=\ker\varphi$, then for $a,b\in G$ we have $\varphi(a)=\varphi(b)$ iff $a\in bK$ iff $b\in aK$. That is, $a$ and $b$ are in the same fibre of $\varphi$ iff they are in the same coset of $K$ in $G$. Thus, the non-empty fibres of $\varphi$ are precisely the cosets of $K$ in $G$. (Again we have to specify non-empty fibre, since $\varphi$ may not be a surjection.) |
H: How do I "describe the region" of $\Bbb R^3$ represented by an inequality?
I don't really understand what these two questions are asking. Nor do I know how to start it.
1) Describe in words the region of $\mathbb {R}^3$ represented by: $x^2 + z^2 \le 9$.
2) Write an inequality to describe: The solid upper hemisphere of the sphere of radius 2 centered at the origin.
Can some please explain what I am supposed to do here?
Thanks in advance.
AI: For question no. $1$, notice that if we plot $x^2+z^2 \le 9$ in $\mathbb R^2$, we get a filled circle of radius $3$. Add another axis, and what happens? For example, a plate is a circle, but what 3d shape is made when plates are stacked?
For question no. $2$, recall that an (empty, i.e. not solid) sphere of radius $r$ has the equation $x^2+y^2+z^2=r^2$. Noting that the radius is $3$, we have $x^2+y^2+z^2=9$. How do we now make this sphere filled? Look at what was done in question $1$ to make a filled circle. |
H: When does the product topology have a countable base?
Could any one tell me how to prove this one?
The product topology has a countable base if and only if the
topology of each coordinate space has a countable base and all
but a countable number of coordinate spaces are indiscrete
AI: The harder direction is to show that the product space has a countable base, then each of the factors also has a countable base, and all but countably many of them have the indiscrete topology.
Let $X=\prod_{i\in I}X_i$, and let $\mathscr{B}$ be a countable base for $X$. Fix $x=\langle x_i:i\in I\rangle\in X$. For $i\in I$ let $$Y_i=\Big\{\langle y_j:j\in I\rangle\in X:y_j=x_j\text{ for all }j\in I\setminus\{i\}\Big\}\;;$$ if $\pi_i:X\to X_i$ is the projection map, it’s a standard elementary result that $\pi_i\upharpoonright Y_i$ is a homeomorphism of $Y_i$ onto $X_i$. In particular, for each $B\in\mathscr{B}$ and $i\in I$, $\pi_i[B\cap Y_i]$ is open in $X_i$. For $i\in I$ let $\mathscr{B}_i=\{\pi_i[B\cap Y_i]:B\in\mathscr{B}\}$; clearly each $\mathscr{B}_i$ is countable, and I claim that $\mathscr{B}_i$ is a base for $X_i$.
Suppose that $p\in U\subseteq X_i$, where $U$ is open. Let $y$ be the unique point of $Y_i$ such that $\pi_i(y)=p$. Then $\pi_i^{-1}[U]$ is an open nbhd of $y$ in $X$, so there is some $B\in\mathscr{B}$ such that $y\in B\subseteq \pi_i^{-1}[U]$. It follows that $p\in\pi_i[B\cap Y_i]\subseteq U$, where $\pi_i[B\cap Y_i]\in\mathscr{B}_i$; this establishes the claim.
Now let $I_0$ be the set of $i\in I$ such that the topology on $X_i$ is not indiscrete, and suppose that $I_0$ is uncountable. For each $i\in I_0$ let $U_i$ be a non-empty proper open subset of $X_i$. Without loss of generality assume that the point $x$ fixed at the beginning of the proof is such that $x_i\in U_i$ for all $i\in I_0$. Then $x\in\pi_i^{-1}[U_i]$ for each $i\in I_0$, so for each $i\in I_0$ there is a $B(i)\in\mathscr{B}$ such that $x\in B(i)\subseteq\pi_i^{-1}[U_i]$. $\mathscr{B}$ is countable, and $I_0$ is uncountable, so there are a $B\in\mathscr{B}$ and an uncountable $I_1\subseteq I_0$ such that $B(i)=B$ for every $i\in I_1$. Clearly $x\in B\subseteq\prod_{i\in I_1}U_i\times\prod_{i\in I\setminus I_1}X_i$.
On the other hand, $B$ is open in the product topology on $X$, so so there are a finite $F\subseteq I$ and open sets $V_i$ in $X_i$ for $i\in F$ such that
$$x\in\prod_{i\in F}V_i\times\prod_{i\in I\setminus F}X_i\subseteq B\subseteq\prod_{i\in I_1}U_i\times\prod_{i\in I\setminus I_1}X_i\;.$$
Pick any $i\in I_1\setminus F$; then $X_i\subseteq\pi_i[B]\subseteq U_i\subsetneqq X_i$, which is absurd. Hence $I_0$ must be countable. |
H: Morse functions are dense in $\mathcal{C}^\infty(X,\mathbb{R})$.
In Shastri's Elements of Differential Topology, p.210-211, there is written:
Why do we get a Morse function $f_u$ on $X$? We know that for any $f\!\in\!\mathcal{C}^\infty(X,\mathbb{R})$, there is some $a\!\in\!\mathbb{R}^N$, such that $f_a(x)=f(x)\!+\!\langle x,a\rangle$ is a Morse function on $X$. Since $X$ is compact, the function $|\langle\_,a\rangle|$ attains its maximal value on $X$. Then, we define $$b := \frac{a\varepsilon}{\max_{x\in X}|\langle x,a\rangle|},$$ and we have $\sup_{x\in X}|f\!-\!f_b|=\sup_{x\in X}|\langle x,b\rangle|=\frac{\sup_{x\in X}|\langle x,a\rangle|}{\max_{x\in X}|\langle x,a\rangle|}\varepsilon=\varepsilon$. But why is this $f_b$ a Morse function on $X$? Its differential is $D(f_b)_p=D(f)_p+b$, so $p\!\in\!X$ is a critical point iff $D(f)_p\!=\!-b\!=\!-\frac{a}{\ldots}$. On the other hand, the critical points of $f_a$ are those for which $D(f)_p\!=\!-a$. I do not see how to make a conclusion here.
AI: By compactness of $X$, the function $\|x\|$ is bounded on $X$, and by the theorem, there exists
$$a\;\in\;\mathbb{B}^N\Big(0,\frac{\varepsilon}{\max_{x\in X}\|x\|}\Big),$$ for which $f_a$ is a Morse function. Then by Cauchy-Schwarz-Bunyakowsky,
$$\sup_{x\in X}|f(x)\!-\!f_a(x)|= \sup_{x\in X}|\langle x,a\rangle|\leq \sup_{x\in X}\|x\|\|a\|\leq \varepsilon.$$ |
H: find all the entire functions that satisfies a given condition
I have been struggling to find a solution for this problem:
Find all the entire analytic functions $f(z)$ (analytic in the complex plane) that satisfy the condition $|z^2f(z)-3+e^z|\leq3$ for all $z \in \mathbb{C}$.
Any ideas?
Thank you in advance.
AI: If $f(z)$ is entire, then $g(z)=z^2f(z)-3+e^z$ is entire. But $g$ is bounded and entire, so by Liouville's theorem it reduces to a constant. Solving for $f(z)$, we find it has to have a pole at $0$, so there are no solutions. |
H: how to show that power series is analytic inside the radius of convergence?
Let $f(z) = \sum a_n z^n$ be a power series with radius of convergence $R$. How do we show that $f$ is analytic in the circular region of radius $R$?
AI: Lemma
Let $R$ be the radius of convergence of $f(z) = \sum_{n=0}^{\infty} a_nz^n$.
Let $g(z) = \sum_{n=1}^{\infty} na_nz^{n-1}$.
Then the radius of convergence of $g(z)$ is $R$.
Proof:
Let $R'$ be radius of convergence of $g(z)$.
Since $|a_nz^n| \le |na_nz^n|$, $R' \le R$.
So it suffices to prove $R \le R'$.
Let $z$ be such that $|z| < R$.
Choose $r$ such that $|z| < r < R$.
Let $\rho = \frac{|z|}{r}$.
Since $\sum_{n=1}^{\infty} a_nr^{n-1}$ converges, there exists $M > 0$ such that
$|a_nr^{n-1}| \le M$ for all $n \ge 1$.
Then
$|na_nz^{n-1}| = n|a_n|r^{n-1}\rho^{n-1} \le nM\rho^{n-1}$
Since $0 \le \rho < 1$, $\sum_{n=1}^{\infty} nM \rho^{n-1}$ converges.
Hence $\sum_{n=1}^{\infty} na_nz^{n-1}$ converges.
Hence $R \le R'$.
QED
Proposition
Let $R$ be the radius of convergence of $f(z) = \sum_{n=0}^{\infty} a_nz^n$.
Let $g(z) = \sum_{n=1}^{\infty} na_nz^{n-1}$.
Then $f'(z) = g(z)$ for every $|z| < R$.
Proof:
Let $z$ be such that $|z| < R$.
Choose $r$ such that $|z| < r < R$.
Let $h$ be such that $0 < |h| \le r - |z|$
Consider $\frac{f(z+h) - f(z)}{h} = \sum_{n=0}^{\infty} a_n\frac{(z + h)^n - z^n}{h}$
By the formula $x^n - y^n = (x - y)(x^{n-1} + yx^{n-2} +\cdots+ y^{n-2}x + y^{n-1})$,
$|a_n\frac{(z + h)^n - z^n}{h}| = |a_n||(z + h)^{n-1} + z(z+h)^{n-2}+\cdots+ z^{n-2}(z+h) + z^{n-1})
\le n|a_n|r^{n-1}$
Define $\psi_n(h) = a_n\frac{(z + h)^n - z^n}{h}$ for $0 < |h| \le r - |z|$
Define $\psi_n(0) = n a_n z^{n-1}$.
Then $\psi_n(h)$ is continuous in $|h| \le r - |z|$.
Since $\sum_{n=1}^{\infty} n|a_n|r^{n-1}$ converges by the lemma,
$\Psi(h) = \sum_{n=1}^{\infty} \psi_n(h)$ converges uniformly in $|h| \le r - |z|$.
Hence $\Psi(h)$ is continuous in $|h| \le r - |z|$.
In particular $\lim_{h \to 0} \Psi(h) = \Psi(0)$.
This implies $f'(z) = g(z)$.
QED |
H: Leading coefficient of polynomial with more than one variable
What is the leading coefficient of a polynomial with more than one variable, when two or more terms have the same degree but different coefficients?
For example:
$3x^2y^2 + 5xy^3$.
The degree is 4. Is the leading coefficient 3, 5, both, none?
Thanks.
AI: There isn't a unique notion of leading coefficient in more than one variable. Which term you decide to be the leading term depends on what you want to do. The keyword here is monomial order. |
H: How many distinct degree 7 polynomials are there over the modular arithmeic modulo 7?
If it's infinite, is it countable or uncountable infinite?
I am a newbie to this topic... I don't know what modular arithmetic for polynomials means. Can someone please give me a link where I can learn?
AI: There are $8$ coefficients to be determined. The lead coefficient cannot be $0$. So the number is $(6)(7^7)$. |
H: How to find intersection of an ellipse and a line that passes through the foci
There are two lines, parallel to the $x$-axis, which pass through the foci and intersect the ellipse at four points. How can I find the points of intersection?
vertex: $(0,0)$
foci: $(0,10)$ and $(0,-10)$
co-vertices: $(20\sqrt2, 0)$ and $(-20\sqrt2, 0)$
AI: You do not even need the equation for the ellipse if you use the property that, for any point $(x,y)$ on the ellipse, the sum of the distances from $(x,y)$ to the foci is a constant. Since we know that $(20\sqrt2,0)$ is a point on the ellipse, we can find this constant:
$\lVert (20\sqrt 2, -10)\rVert+\lVert(20\sqrt 2, 10)\rVert=2\sqrt{ 800+100}=60$.
Now we know that the y-coordinate of the points of intersection for the line going through $(0,10)$ is $10$, so we just need the x-coordinate. So we have that $$\lVert(x,10)-(0,10)\rVert + \lVert(x,10)-(0,-10)\rVert = 60\\
\lvert x \vert + \sqrt {x^2+400}=60\\
x^2+400=60^2-120\lvert x\rvert+x^2\\
\lvert x \rvert=\frac {3200}{120}=\frac {80} 3$$
So the points of intersection are $(\frac {80} 3 , 10)$ and $(\frac{-80} 3,10)$. The same method will allow you to find the other two points, try it! (You could also appeal to symmetry, but it's good to work through the argument on your own as well.) |
H: What is the locus of $\mathrm{arg}(\frac{z-a}{z-b})=\theta$?
This isn't quite homework, but it might as well be. I'm working through Needham's Visual Complex Analysis and I'm finding myself stuck on the following problem:
Explain geometrically why the the locus of z such that
$\mathrm{arg}(\frac{z-a}{z-b})=\theta$ is an arc of a certain circle
passing through the fixed points a and b.
My first thought was to think about what $\mathrm{arg}(z)=\theta$ means, and I think it should be a line starting at the origin and going in the direction of $\theta$. I also understand that, supposing that we were able to find one z that satisfies this condition, we could find another z by moving the vector in a way that increases the argument of $z-a$ while decreasing the argument if $z-b$.
I'm looking for a hint. Can anyone subtly nudge me in the right direction? I'm having trouble seeing why this has to be the arc of a circle. Also, how can the arc pass through b? Would the fraction $\frac{z-a}{z-b}$ be undefined for z = b? How can it pass through a? Wouldn't the fraction be the zero vector for z = a ?
AI: The problem says that the circle passes through a and b, not necessarily the arc of the circle . I hope the image below jogs some memories from geometry class, or leads you to a new fun fact--as the case may be.
Good luck! |
H: Why do $r=\cos[2\theta]$ and $r=\frac{1}{2}$ have 8 intersections?
I am not sure if the question is a duplicate. The conclusion from the title is clear, for example by viewing the wolframalpha graph at here.
But on the other hand I feel intuitively there should be only 4, since a line intersect with $y=\cos[\theta]$ at 2 points, thus $y=1/2$ intersect with $y=\cos[2\theta]$ at 4 points in $[0,2\pi]$. Since $\cos[2\theta]=1/2$ implies a solution, the number of solutions of $$y=\cos[2x]-1/2,x\in [0,2\pi]$$ and $$r=\cos[2\theta],\theta\in \mathbb{S}^{1}$$ must be the same. So why there is such a difference?
AI: The problem is that there are two widely used conventions for interpreting $r=f(\theta)$ when $f(\theta)\lt 0$.
One convention is to say that if $f(\theta)\lt 0$, then the curve is undefined. Makes sense, $r$ is a distance.
Another convention is that if $f(\theta)\lt 0$, we plot $(|f(\theta)|,\theta)$ as usual, and then reflect the result across the origin, or equivalently rotate through $180^\circ$.
That second convention will give you $8$ points.
Remark: I slightly prefer the second convention, it gives prettier curves. |
H: How to approximate the value of $ \arctan(x)$ for $x> 1$ using Maclaurin's series?
The expansion of $f(x) = \arctan(x)$ at $x=0$ seems to have interval of convergence $[-1, 1]$
$$\arctan(x) = x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}+\mathcal{O}\left(x^{13}\right) $$
Does it mean that I cannot approximate $\arctan(2)$ using this series? Also I'm getting radius of convergence $|x| < 1$ using ratio test. How do I get $|x| \leqslant 1$?
AI: Use, for $x>0$
$$
\arctan(x) + \arctan\left(\frac{1}{x}\right) = \frac{\pi}{2}
$$
For $x<0$ use the parity.
Added: Use
$$
\tan\left(\frac{\pi}{2} - \phi\right) = \frac{\sin\left(\frac{\pi}{2}-\phi\right)}{\cos\left(\frac{\pi}{2}-\phi\right) } = \frac{\cos(\phi)}{\sin(\phi)} = \frac{1}{\tan(\phi)}
$$
Thus, for $x = \tan(\phi)$:
$$
\tan\left(\frac{\pi}{2} - \phi\right) = \frac{1}{x}, \quad \frac{\pi}{2} - \phi = \arctan\left(\frac{1}{x}\right), \quad
\frac{\pi}{2} - \arctan(x) = \arctan\left(\frac{1}{x}\right)
$$ |
H: Graph of the infinite sum $g(x)=\sum_{k=1}^\infty \sin^k(kx)$
For which real numbers $x$ does the infinite sum $\sin(x)+\sin^2(2x)+\sin^3(3x)+ \cdots+\sin^n(nx)+\cdots$ converge?
How does its graph look?
AI: Let $x=r\pi$ and consider the following cases:
$r=\frac{a}{2b}$: For infinitely many $n$ we have $n=(2m+1)b$ thus $\sin^n(nx)=\sin^n\left(\frac{(2m+1)a}{2}\pi\right)=\pm 1$ so the sequence does not go to $0$ hence the series does not converge.
$r=\frac{a}{2b+1}$: For any $m$ we have $\left|nx-\frac{2m+1}{2}\pi\right|=\pi\left|\frac{2na-(2m+1)(2b+1)}{4b+2}\right|\geq \frac{\pi}{4b+2}$ and so $|\sin(nx)|$ is bounded above by $\lambda=\sin\left(\frac{2b}{2b+1}\pi\right)<1$, hence we have $$\sum\limits_{n=1}^\infty |\sin^n(nx)|<\sum\limits_{n=1}^\infty \lambda^n=\frac{\lambda}{1-\lambda}$$ so the series converges absolutely.
$r$ is irrational: This is more difficult. Let $[r]$ denote the fractional part of $r$. The key is to examine terms which are nearly half-integers. Let $N=\lfloor1/[r]\rfloor$. Consider the points $[r],[2r],\ldots,[nr]$ in $[0,1)$ ordered as real numbers. It is a theorem that there exist $\alpha>\beta>\gamma$ such that the distance between any two consecutive points is one of these three, and that the number of pairs with distance equal to each of these values differs by at most $N$. In particular, there are at least $\frac{n-2N}{3}$ pairs a distance $\alpha$ apart, so $\alpha<\frac{3}{n-2N}$. Thus by the pigeonhole principle, we have some $m\leq n$ such that $\left|[mr]-\frac12\right|<\frac{3}{n-2N}$. For such $m$ we have $$|\sin(mx)|>\left|\sin\left(\frac{(2k+1)\pi}{2}-\frac{3}{n-2N}\right)\right|=\cos\left(\frac{3}{n-2N}\right)>1-\frac{9}{2(n-2N)^2}$$
thus we get
$$|\sin^m(mx)|>\left(1-\frac{9}{2(n-2N)^2}\right)^m>1-\frac{9m}{2(n-2N)^2}\geq 1-\frac{9n}{2(n-2N)^2}\to 1$$
and so the sequence does not converge to $0$, hence the series does not converge.
Since this does series not converge except on a set of measure $0$, I'm not sure how to meaningfully describe its graph. However, if you want to compute its value for $r=\frac{a}{2b+1}$ you can break the series up into sums over equivalence classes $\bmod 2b+1$ using absolute convergence, which gives
$$\begin{align}
\sum_{n=1}^\infty \sin^n(nx)&=\sum_{i=1}^{2b+1}\sum_{j=0}^\infty(-1)^j\sin^{j(2b+1)+i}\left(\frac{a+i}{2b+1}\pi\right)\\
&=\sum_{i=1}^{2b+1}\sin^i\left(\frac{a+i}{2b+1}\pi\right)\sum_{j=0}^\infty(-1)^j\sin^{j(2b+1)}\left(\frac{a+i}{2b+1}\pi\right)\\
&=\sum_{i=1}^{2b+1}\sin^i\left(\frac{a+i}{2b+1}\pi\right)\frac{1}{1+\sin^{(2b+1)}\left(\frac{a+i}{2b+1}\pi\right)}\\
\end{align}$$
which, while not the prettiest, is at least a finite sum. |
H: Solving simultaneous equations with imaginary numbers
Consider the following simultaneous equation:
$$\begin{cases} 5z-(3+i)w=7-i\\ (2-i)z+2iw = -1+i \end{cases} $$
What is the simplest way to manipulate one of the equations so that a variable can be eliminated and the equation solved?
AI: You do it the same way you do it over the reals. For example, you could solve the first equation for $z$, $z=(7-i+(3+i)w)/5$, substitute that into the 2nd equation, solve for $w$, then get $z$. Or, multiply the first equation by $2-i$, the second by 5, and subtract to eliminate $z$. |
H: $G/H$ is a finite group so $G\cong\mathbb Z$
Let $G$ is an abelian infinte group such that for all nontrivial subgroups $H$ $$\forall H\leq G, \left|\frac{G}{H}\right|<\infty$$ Prove that $G\cong\mathbb Z$.
What I have done:
Clearly, it is enough to show that $G$ is a cyclic group. Moreover, I see as $G/H$ is a finite group for all subgroups $H$, then $G$ cannot have any elements with the finite order. This means to me that if $H \leq G$ and $H$ is cyclic then $H$ cannot be finite. Help your friend for the rest. Thanks. :)
AI: Show that such a group must be finitely generated. Using the fundamental theorem of finitely generated abelian groups, you have that
$G \simeq \mathbb{Z}^r \times \mathbb{Z} / n_1 \mathbb{Z} \times ... \times \mathbb{Z} / n_k \mathbb{Z}$
where $n_i \in \mathbb{Z}$. There is an isomorphic copy of each factor of the direct product in $G$. Quotienting $G$ by $\mathbb{Z}^s$ for various $1 \leq s \leq r$ and the $\mathbb{Z} / n_i \mathbb{Z}$, you see that $r = 1$ ($r \geq 1$ for the group to be infinite and $r \leq 1$ for the finite quotient property) and $k = 0$ (for the finite quotient property). |
H: Calculate absolute values with unknown constant
I am to calculate all $x$ if $f(x) = g(x)$ and if
$$f(x)= |2x+2| + |3-2x|$$
$$g(x)= x + 14$$
How do I mix regular numbers with absolute values in such a sense? I thought I could calculate it like this:
$$|2x+2| + |3-2x| = 3x+2 + 3+2x = 3x+5$$
But then I realised that the end value of either of the absolute values is determined by $x$. It feels like a catch 22: how do I calculate this equation without knowing the value of $x$ until I've calculated it?
AI: HINT: You can assume three cases:
$x \geq \dfrac 32$
$-1<x< \dfrac 32$
$x \leq -1$ |
H: Subgroups of $\Bbb{R}^n$ that are closed and discrete
I am trying to prove that every closed discrete subgroup of $\Bbb{R}^n$ under addition is a free abelian group of finite rank. I have tried to do this by induction on the dimension $n$.
Base case: We claim that a subgroup $H \subset \Bbb{R}$ has a least positive element $\alpha$. The completeness axiom tells us that there is a least positive real number $\alpha$ among all positive reals in $H$, and being closed implies that $\alpha \in H$.
Now assume the inductive hypothesis and consider a closed, discrete subgroup $H$ of $\Bbb{R}^n$. Choose some element $x \in H$ of least positive distance to the origin. This can be done because otherwise we get a contradiction like the above. Now let $L$ be the subspace spanned by $x$ and write $\Bbb{R}^n = L \oplus W$ for some complement $W$. Consider the projection (as a linear map) $p : \Bbb{R}^n \to W$. Now this is also a group homomorphism and so we have
$$\textrm{Im} (H) \subset W$$
being a subgroup and so by the induction hypothesis is isomorphic to $\Bbb{Z}^l$ for some $1 \leq l \leq n-1$. However from here I am having some trouble concluding that $H$ itself must be isomorphic to $\Bbb{Z}^l \oplus \Bbb{Z}x$. If I know the existence of a map $l : \textrm{Im}(H) \to H$ such that
$$p \circ l = \textrm{id}_{\textrm{Im}(H)}$$
then by the splitting lemma I can conclude my problem. However, I don't have such a map so how do I conclude the problem? Please do not post complete solutions.
Thanks.
Edit: Perhaps I should add some context. I am trying to conclude that the kernel of the exponential map $\exp : \mathfrak{g} \to G$ where $G$ is a connected abelian matrix Lie group is a free abelian group of finite rank. We know that the assumptions on our matrix Lie group mean that $\exp$ is a group homomorphism, so that $\ker \exp$ is a subgroup of the Lie algebra $\mathfrak{g}$.
Edit: The problem is reduced to showing that the image of $H$ under the projection is indeed discrete, because otherwise we cannot apply the inductive hypothesis.
AI: I fully endorse Alex Becker's answer. I just want to address the question of discreteness of $p(H)$ in the original post.
Let $W$ be the orthogonal complement to $L$, so $p$ is the orthogonal projection onto $W$.
Lemma. For all elements $y\in \left(H\setminus\mathbb{Z}x\right)$ we have $||p(y)||\ge ||x||/2.$
Proof. Let $y\in \left(H\setminus\mathbb{Z}x\right)$ be arbitrary. Let us write
$$
y=rx+w,
$$
where $r\in\mathbb{R}$ and $w\in W$. Because $x$ generates $L\cap H$, we must have $y\notin L$, so $w\neq0$. There exists an integer $m$ such that $|m-r|\le1/2$. Consider the vector
$y'=y-mx=(r-m)x+w\in H$. Because $y\notin L$, $y'\neq0$. Therefore $||y'||\ge ||x||$
as $x$ was selected to be (one of) the shortest non-zero vectors in $H$. By construction
$||(r-m)x||\le ||x||/2$. By triangle inequality
$$
||p(y)||=||w||=||y'-(r-m)x||\ge ||y'||-||(r-m)x||\ge ||x||-\frac{||x||}2=\frac{||x||}2.
$$
Q.E.D.
The discreteness of the image follows immediately from the Lemma. If $p(y)$ and $p(y')$ are two distinct points in the group $p(H)$, then $y-y'\notin\mathbb{Z}x$, and therefore
$$
d(p(y),p(y'))=||p(y)-p(y')||=||p(y-y')||\ge\frac{||x||}2.
$$ |
H: Wilson's theorem related problem
prove $$18! \equiv -1 \pmod{437} $$
I do not want full solution to the above problem but if anybody can tell me how we can approach to it, I will really appreciate that.
AI: hints :
$437=19\cdot 23$ (as proposed by Sean)
Wilson's theorem :-)
$19\cdot 20\cdot 21 \cdot 22=(-4)(-3)(-2)(-1)\pmod{23}$ |
H: Proving $n^4 + 4 n^2 + 11$ is $16k\,$ for odd $n$
if $n$ is an odd integer, prove that $n^4 + 4 n^2 + 11$ is of the form
$16 k$.
And I went something like:
$$\begin{align*}
n^4 +4 n^2 +11
&= n^4 + 4 n^2 + 16 -5 \\
&= ( n^4 +4 n^2 -5) + 16 \\
&= ( n^2 +5 ) ( n^2-1) +16
\end{align*}$$
So, now we have to prove that the product of $( n^2 +5 )$ and $( n^2-1)$ is a multiple of 16.
But, how can we do this?
If anybody has any idea of how I can improve my solution, please share it here.
Edit updated to include the necessary hypothesis that $n$ is odd.
AI: $n=2k$:
$$n^4+4n^2+11\\=(n^2-1)(n^2+5)+16\\=(4k^2-1)(4k^2+5)+16\\=16k'^4+16k''^2+11\\=16k+11$$
Which is not $16k$.
$n=2k+1$:
$$n^4+4n^2+11\\=(n^2-1)(n^2+5)+16\\=(4k^2+4k)(4k^2+4k+6)+16\\=8\underbrace{k(k+1)}
_{2k}(2k^2+2k+3)+16$$
Which is $16k$. |
H: General question about solving equations involving a definite integral
Are there any well known techniques to solve a problem of the following form: $$\int_a^b f(x,\alpha) dx = g(\alpha),$$ where $a,b\in\mathbb{R}$ are fixed, $f$ and $g$ are known functions, $\alpha\in\mathbb{C}$ is the unknown variable, and the expression is not an identity. Put another way, given the above expression are there techniques available to find the values of $\alpha$ for which the expression holds true, assuming we know from empirical study that there do exist such $\alpha$ ?
AI: It's a question slightly strange, but under certain not-too-tight conditions, we have
$$\frac{d}{d\alpha}\left(\int_a^bf(x,\alpha)dx\right)=\int_a^b\frac{d}{d\alpha}\left(f(x,\alpha)\right)dx$$
So if you know both functions you could check whether $\,g'(\alpha)\,$ equals the above... |
H: Does a fundamental solution set exist for homogeneous first-order difference equations?
When $A$ is diagonalisable, $\vec{x}_{k+1}=A\vec{x}_k$ implies that
$\vec{x}_k = c_1\lambda_1^k\vec{v}_1 +...+c_n\lambda_n^k\vec{v}_n$
because an eigenbasis exists and any $\vec{x}$ can be decomposed into
a linear combination of $A$'s eigenvectors.
But when $A$ is not diagonalisable, is there a related general formula for (or a general technique to find) $\vec{x}_k$ in terms of $A$'s eigenvalues and eigenvectors?
I ask this because I have just learnt that for the differential system $\dot{\vec{x}}=A\vec{x}$, a fundamental solution set always exists so that even when $A$ is defective, $\vec{x}$ can still be expressed in terms of $A$'s eigenvectors and eigenvalues.
AI: It works pretty much the same way as for differential equations. Let $P^{-1}AP=J$ where $J$ is in Jordan form. Then $$x_k=A^kx_0=PJ^kP^{-1}x_0$$ and you just need to understand powers of a matrix in Jordan form, and the relation of $P$ to the eigenvectors and generalized eigenvectors.
EDIT: Here's the simplest example. Say $$A=\pmatrix{\lambda&1\cr0&\lambda\cr}$$ Then $$A^n=\pmatrix{\lambda^n&n\lambda^{n-1}\cr0&\lambda^n\cr}$$ Let $v_1$ be the eigenvector $(1,0)$, let $v_2$ be the generalized eigenvector $(0,1)$. Then $$A^nv_1=\lambda^nv_1\quad A^nv_2=n\lambda^{n-1}v_1+\lambda^nv_2$$ so
$$A^n(c_1v_1+c_2v_2)=(c_1\lambda^n+c_2n\lambda^{n-1})v_1+c_2\lambda^nv_2$$
But do check my arithmetic.
MORE EDIT: We can rewrite that last expression as $$x_n=c_1\lambda^nv_1+c_2\lambda^n((n/\lambda)v_1+v_2)$$ which looks a little more like the form you've given for the analogous difference equation. |
H: Spotting maximum, minimum, maximal and minimal elements in a poset
Let $(\mathbb N^* \times \mathbb N^*, \varphi)$ be a poset defined as follows:
$$\begin{aligned} (a,b)\varphi(c,d)\Leftrightarrow ab<cd \text{ or } (a,b) = (c,d)\end{aligned}$$
Check if $\varphi$ is a total order and determine maximum, minimum, maximal and minimal elements.
It's easy to prove that $\varphi$ is not total because as we consider $(a,b),(c,d) \in \mathbb N^* \times \mathbb N^* : a = d \text{ and } b = c$ then
$$\begin{aligned} ab \nless cd \text{, } cd \nless ab \text{ and } (a,b) \neq (c,d)\end{aligned}$$
What I am having a very hard time with is spotting maximum, minimum, maximal and minimal elements. In my opinion the poset doesn't have any maximum or maximal elements as $\mathbb N^* \times \mathbb N^*$ is infinite, but it does have a minimal element which is $(1,1)$. Given that $\nexists (\varepsilon, \varepsilon') : 1 = \varepsilon' \text{ and } 1 = \varepsilon$ and $(1,1) \neq (\varepsilon, \varepsilon')$ then $(1,1)$ is comparable to all elements in $\mathbb N^* \times \mathbb N^*$, does this make it also the minimum?
AI: I think your reasoning is sound enough.
If there is a maximal element, $(c,d)$ say, then $(a,b) \varphi (c,d)$ for all other elements $(a,b)$ in $\mathbb{N}^* \times \mathbb{N}^*$ which implies that:
\begin{equation}
ab < cd \text{ or } (a,b)=(c,d)
\end{equation}
But then $(c,d) \varphi (c+1,d)$ since $cd < (c+1)d$ contradicting the maximality of $(c,d)$.
Finally, $(1,1)$ is the minimal element as $1 \cdot 1=1$ is the smallest a product of two elements in $\mathbb{N}^*$ can be. That is, any other element $(e,f)$ has $e>1$ or $f>1$ In the first case, $(1,1) \varphi (e,f)$ since $1\cdot 1 < e \leq ef$. The second case is similar.
Also I like your profile: "Just a guy who hopes to live through his algebra exam." +1 for having realistic goals, haha. |
H: Proof for length of period in simple modulo $N$ sequence.
I am looking for a concise proof that the length of the smallest period of the sequence
$$f[n] = a n \pmod N $$
is $N$ if $(a,N) = 1$. From the Pigeonhole Principle, it is not hard to show that $f[n]$ is periodic with $N$, but how do I know that it is not periodic with a number $M < N$. Is a proof by contradiction the way to go?
I found a lemma in the proof of Fermat's Theorem here, that takes me most of the way by arguing that of $1a, 2a, 3a, \ldots, (N-1)a$ are not congruent modulo $N$ if $(a,N)=1$. This question seems so trivial and is used in discussions of the periodicity of sequences so I thought there might be several ways to show it to be true.
AI: If $an=f[n]=f[0]=0\pmod N$ for some period $n$, then $N$ divides $an$. Since $a$ is coprime to $N$, $N$ divides $n$. |
H: Find the largest prime factor
I just "solved" the third Project Euler problem:
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
With this on Mathematica:
Select[Divisors[600851475143], PrimeQ]
It will first give me a list with all the divisors of 600851475143 and then It's going to select the prime ones, giving me:
{71, 839, 1471, 6857}
And thus giving me the answer of 6857. I feel that my solution won't give me better insights into this (since I used almost null mental activity for it) and I would like to know how can I extend this problem, what other ways are accessible on solving it?
I initially thought that I should applying the idea of divisibility, some guys on the PE solved it using the sieve of Eratosthenes, can you suggest other ways that could solve this problem?
Thanks in advance.
AI: In general factoring problem is not easy, see:Theoretical Computer Science Q&A
There is a Pollard-Rho algorithm which you can find some information here: Wikipedia: Factoring integers
I solved that problem of Project Euler too! The output of MATLAB is:
71
839
1471
6857
Elapsed time is 0.002433 seconds.
>>
And the code in MATLAB:
clear;
clc;
tic;
p=600851475143 ;
m=3;
while(sqrt(m)<p)
j=rem(p,m);
if (j==0)
p=p/m;
disp(m);
m=m+2;
else
m=m+2;
end
end
toc;
By the way, there is factor command in MATLAB that directly outputs the factors of a number. But it has a 2^32 of input limitation size. I removed the limitation of the original code and it responded in 0.027501 seconds! |
H: Is the set $\{f: \int_{\mathbb{R}} f \leq h\}$ closed in $L^2(\mathbb{R})$?
If the sequence $\{f_n\}$ in $L^2(\mathbb{R})$ converges to $f$ in $L^2(\mathbb{R})$ and $\int_{\mathbb{R}}f_n(t)dt\leq h$ for some $h>0$ and all $n\in \mathbb{N}$; does $f$ satisfy $\int_{\mathbb{R}}f(t)dt\leq h$?
AI: Not necessarily. Try $f_n=2\,\mathbf 1_{(0,1)}-\frac1n\mathbf 1_{(1,n+1)}$, and $f=2\,\mathbf 1_{(0,1)}$. Then the integral of $(f_n-f)^2$ is $\frac1n$ hence $f_n\to f$ in $L^2$, but the integral of $f_n$ is $1$ and the integral of $f$ is $2$. |
H: Evaluating this determinant
I am asked to find the following determinant $$D = \begin{vmatrix} 1 & 2 & \cdots & n \\\ n+1 & n+2 & \cdots & 2n \\\ \vdots & \vdots & \vdots & \vdots \\\ n(n-1)+1 & n(n-1)+2 & \cdots & n^2 \end{vmatrix} \ $$ where $n\in\mathbb N$. Any help. Thank you.
AI: Step 1: Substract line 1 to line 2 and to line 3.
Step 2: Line 3 is now twice line 2, hence D=0.
Step 3: Check what happens when there is no line 3, that is, when n=1 or n=2. |
H: Number of elements of order $7$ in a group of order $28$
Given a group $G$ with order $28 = 2^2 \cdot 7$. Sylow-Theory implies that there is a exactly one $7$-Sylow-Subgroup of order $7$ in $G$, and $1$ or $7$; $2$-Sylow-Subgroups.
Where to go from here concerning the number of elements of order $7$?
AI: If $x$ is an element of order $7$ then the subgroup $\langle x\rangle$ generated by $x$ has order $7$... |
H: Integral of function over displaced sphere
Let $F$ be a vector field defined in $\mathbb{R}^{3}$ minus the origin.
$$F(r)=\frac{r}{|r|^{3}}$$ where $r=(x,y,z)$.
Let $S$ be the sphere of radius $1$ and center $(2,0,0)$.
Compute $\int_{S}{F\cdot n \ dS}$ where $n$ is the vector normal to the surface of $S$.
My initial reaction was to use the divergence theorem to get:
$$\int_{S}{F\cdot n \ dS} = \int{\operatorname{div}{F} \ dV} = -\int_{S}{\frac{2}{|r|^{4}} \ dV}$$
My multivariable calculus is a bit rusty, so I was having a bit of trouble writing out the triple integral.
The sphere is defined by the equation $(x-2)^{2}+y^{2}+z^{2}=1$ which amounts to $x^{2}+y^{2}+z^{2}=4x-3$ which means $r^2=4 r\sin(\phi)\sin(\theta)-3$.
This seems not too good... Did I make a mistake in my calculations?
AI: Yes, actually you made a calculation mistake. Indeed, careful calculation shows that
$$\nabla \cdot \mathbf{F} = 0$$
away from the origin. To be specific, let us rename the coordinate variables as $(x_1, x_2, x_3) = (x, y, z)$. Then
$$\begin{align*}\nabla \cdot \mathbf{F}
&= \sum_{i=1}^{3} \frac{\partial}{\partial x_i}\frac{x_i}{r^3}
= \sum_{i=1}^{3} \frac{\partial x_i}{\partial x_i}\frac{1}{r^3} + x_i \frac{\partial}{\partial x_i}\frac{1}{r^3} \\
&= \sum_{i=1}^{3} \frac{1}{r^3} + x_i \left(-\frac{3}{r^4}\right)\frac{\partial}{\partial x_i}\sqrt{x_1^2+x_2^2+x_3^2} \\
&= \frac{3}{r^3} - \frac{3}{r^4} \sum_{i=1}^{3} x_i \cdot \frac{x_i}{r} \\
&= \frac{3}{r^3} - \frac{3}{r^3}=0.
\end{align*}$$
Now you may apply the divergence theorem to obtain
$$\int_{S}\mathbf{F}\cdot\mathbf{n}\;da = \int_{V}\nabla \cdot \mathbf{F}\;dv = 0,$$
where $V$ is the region inside the sphere. This calculation is valid since the origin does not lie on and inside the sphere $S$.
If you made a bold and brave resolution to attack this problem by brutal integration, then you may do as follows: First, make the substitution $(x',y',z')=(x-2,y,z)$. This is equivalent to shifting the whole situation so that the center of $S$ coincides with the origin. (This makes the domain of integration simpler in expense of the simplicity of the integrand.) Then make a variant of spherical coordinate change as follows:
$$ \begin{align*}
x' &= r \cos\phi \\
y' &= r \sin\phi\cos\theta\\
z' &= r \sin\phi\sin\theta.
\end{align*}$$
Of course, $0 \leq \phi \leq \pi$ and $0 \leq \theta < 2\pi$. Then on $S$, with this new coordinate system, $S$ is characterizes as the locus of $r = 1$, and on $S$ we have $da = r^2\sin\phi\,d\phi d\theta = \sin\phi\,d\phi d\theta$. Thus our integral becomes
$$\begin{align*}\int_S \mathbf{F}\cdot\mathbf{n}\;da
&=\int_{0}^{2\pi}\int_{0}^{\pi} \frac{(x'+2,y',z')}{((x'+2)^2+y'^2+z'^2)^{3/2}} \cdot \frac{(x',y',z')}{(x'^2+y'^2+z'^2)^{1/2}}\;r^2 \sin\phi\,d\phi d\theta\\
&=\int_{0}^{2\pi}\int_{0}^{\pi} \frac{(x'+2,y',z') \cdot (x',y',z')}{(r^2 + 4r\cos\phi + 4)^{3/2}}\;r \sin\phi\,d\phi d\theta\\
&=\int_{0}^{2\pi}\int_{0}^{\pi} \frac{r (r^2 + 2r\cos\phi) \sin\phi}{(r^2 + 4r\cos\phi + 4)^{3/2} }\;d\phi d\theta\\
&=\int_{0}^{2\pi}\int_{0}^{\pi} \frac{(1 + 2\cos\phi) \sin\phi}{(5 + 4\cos\phi)^{3/2} }\;d\phi d\theta\\
&=2\pi \int_{0}^{\pi} \frac{(1 + 2\cos\phi) \sin\phi}{(5 + 4\cos\phi)^{3/2} }\;d\phi
\end{align*}$$
Finally, let $t = 4\cos\phi$. Then $dt = -4\sin\phi \, d\phi$ and therefore
$$\begin{align*} \int_S \mathbf{F}\cdot\mathbf{n}\;da
&= \frac{\pi}{4} \int_{-4}^{4} \frac{2 + t}{(5 + t)^{3/2} }\;dt \\
&= \frac{\pi}{4} \left[ -2 \cdot \frac{2 + t}{\sqrt{5 + t}}\right]_{-4}^{4} + \frac{\pi}{2} \int_{-4}^{4} \frac{1}{\sqrt{5-t}}\;dt \\
&= -2\pi + \frac{\pi}{2} \left[2\sqrt{5+t}\right]_{-4}^{4} = 0.
\end{align*}$$ |
H: How do I show there isn't an order isomorphism b/w the two sets $\{1, 2, 3,...\}$ and $\{1, 2, 3, ..., \omega \}$
That is, how can I prove there isn't a bijection $f$ from one set to the other such that
$f(x) < f(y)$ iff $x < y$?
AI: Suppose that $f:\{1,2,3,\dots\}\to\{1,2,3,\dots,\omega\}$ is an order-preserving bijection. There must be some $n\in\{1,2,3,\dots,\}$ such that $f(n)=\omega$. What can $f(n+1)$ be? |
H: When should I make a colon in a statement?
As an example, take a look at the Cauchy's convergence test:
$\forall_{\varepsilon>0} \exists_{n_0\in\mathbb{N}} \forall_{m,n \geq n_0} \colon \left|a_m-a_n \right|<\varepsilon $
or
$\forall_{\varepsilon>0} \colon \exists_{n_0\in\mathbb{N}} \colon \forall_{m,n \geq n_0} \colon \left|a_m-a_n \right|<\varepsilon $
or
$\forall_{\varepsilon>0} \exists_{n_0\in\mathbb{N}} \forall_{m,n \geq n_0} \left|a_m-a_n \right|<\varepsilon $
Is there a difference in the meaning of these three statements? (If yes, please explain the difference)
When should I make a colon in a mathematical statement and when not?
(I've found the first and the second in Wikipedia)
AI: It's just a matter of punctuation style. There is no mathematical meaning hidden in whether a colon is written or not.
Some authors prefer colons after every quantifier. Other omit them when the subformula itself starts with a quantifier. Yet other authors never use them.
Some authors use a dot instead of a colon. Some insist on parentheses around the $(\forall \epsilon > 0)$, or round the subformula, but rarely around both. Some will parenthesize the subformula except when it starts with a quantifier of its own, or depending on how tightly its top connective binds.
Some people (such as yours truly) do whichever things on alternate weeks, depending on the weather and the phase of the moon.
It's all just typography, and carries no logical content, as long as it's clear which parts of the formula each quantifier ranges over. (Beware that there are differing conventions about that, in the styles that don't insist on parentheses around the subformulas. So $\forall x:\phi\to\psi$ could mean either $(\forall x:\phi)\to\psi$ or $\forall x:(\phi\to\psi)$ depending on the exact conventions followed by the author).
In any case, note that writing the bound variable as a subscript to the quantifier, as you do in the question, is somewhat nonstandard. |
H: Irreducible polynomial over field of order p
Let $p$ be a prime and $F=\mathbb{Z}/p\mathbb{Z}$ and $f(t)\in F[t]$ be an irreducible polynomial of degree $d$.
I need to show that $f(t)$ divides $t^{(p^{n})}-t$ if and only if $d$ divides $n$.
AI: The first theorem of the following paper provides the proof:
http://www.jstor.org/stable/2316211
Or look at Theorem $7.6$ of the following article
http://www-groups.mcs.st-and.ac.uk/~neunhoef/Teaching/ff/ffchap3.pdf |
H: Making a logo with processing, can't figure out a coordinate...
I'm struggling to make a simple logo, I have no deep knowledge of Inkscape so I'm doing it with a little bit of processing.
The problem is I can't figure out how to determine one certain point's coordinate, considering I want this point to have certain "constraints" in my geometry construct.
Here is the construct:
Knowing AB (=AD), AC = 1, A = (0,0), C's coordinates are easily determined.
I'm having a hard time figuring out D's coordinates.
I tried CaRMetal, it figures it out, but I can't make a logo with it.
Is it solvable, or should I determine other lenghts instead of this ?
AI: Unless you know the radius of the inner circle, there are infinitely many points which $D$ could be. Once you know the radius of the inner circle you will be able to find two possible points for $D$.
Supposing you know the length of $\overline{AB}$ (or equivalently, $\overline{AD}$). Then, using vector notation, you know what $\mathbf D \cdot \mathbf D$ is. Now $\mathbf D$ and $\mathbf {D-C}$ are perpendicular, so $\mathbf D\cdot (\mathbf D - \mathbf C)=0\implies \mathbf D\cdot \mathbf D=\mathbf D\cdot\mathbf C$ so if we write $\mathbf D=(x,y)$, $\mathbf C=(c_1,c_2)$ and $\overline{AD}=k$ then the points of $\mathbf D$ are given by the intersection of a line and a circle:
$$x^2+y^2=k^2=xc_1+yc_2$$
This will have two points of intersection (unless $k=0$ or the line is tangent to the circle). Obviously if $k=0$ then $\mathbf D$ is at the origin as well. Now if $k\ne0$ and $c_1\ne0$, we can solve the linear equation as follows:
$$x=-\frac {c_2}{c_1} y+\frac {k^2} {c_1}$$
Now you can substitute this in the quadratic equation:
$$\left(-\frac{c_2}{c_1} y+\frac {k^2}{c_2}\right)^2+y^2=k^2$$
Simplifying this will yield a quadratic equation for $y$ without $x$, so you may use the quadratic formula to solve for $y$. This is where you will get two values for $y$, which you can then plug back in to the earlier linear equation to solve for $x$.
If $c_1=0$, the linear equation would already give a specific value for $y$, so you would just substitute that value into the quadratic equation to find two values for $x$. |
H: Limit of expression involving exponentials.
For $0 \le s < 1$, $t \ge 0$ let
$$G(s,t) := \frac{e^{-t} s}{\sqrt{1-(1-e^{-2t})s}}$$
For $\lambda > 0$ compute the limit of $G(e^{-2\lambda e^{-2t}},t)$ as $t \rightarrow \infty$.
AI: Well,
$$
G\left( \mathrm{e}^{-2 \lambda \exp(-2t)}, t\right) = \frac{ \exp(-t) \cdot \exp\left( -2 \lambda \exp(-2t) \right)}{\sqrt{1-(1-\exp(-2t) \exp\left( -2 \lambda \exp(-2t) \right) }}
$$
Let $u = \exp(-t)$. Then
$$
\lim_{t \to \infty} G\left( \mathrm{e}^{-2 \lambda \exp(-2t)}, t\right) = \lim_{u \downarrow 0} \frac{u \exp(-2 \lambda u^2)}{\sqrt{1-\left(1-u^2\right) \exp(-2 \lambda u^2) }}
$$
Use l'Hospital's rule, or Taylor series expansion of the exponential $$\exp(-2 \lambda u^2) = 1 - 2 \lambda u^2 + \mathcal{o}(u^2)$$
to get
$$
\lim_{t \to \infty} G\left( \mathrm{e}^{-2 \lambda \exp(-2t)}, t\right) = \lim_{u \downarrow 0} \frac{u \exp(-2\lambda u^2)}{\sqrt{1-(1-u^2)\exp(-2\lambda u^2)}} =
\lim_{u \downarrow 0} \frac{\color\green{\exp\left(- 2 \lambda u^2\right)}}{\sqrt{\color\red{\frac{1-\exp\left(- 2 \lambda u^2\right)}{u^2}} + \color\green{\exp\left(- 2 \lambda u^2\right)} }} = \frac{\color\green{1}}{\sqrt{\color\red{2 \lambda} + \color\green{1}}} = \frac{1}{\sqrt{1+2\lambda}}
$$
where the Taylor series can be used to see that $\lim_{u \downarrow 0} \frac{1-\exp\left(- 2 \lambda u^2\right)}{u^2} = 2 \lambda$. |
H: Calculate the edge length of a cube, which is inserted in a hemisphere
How can I calculate the edge length of a cube, which is inserted into a half-sphere with a radius $a$?
AI: This is an elaboration on @J.M.'s comment, and assumes we want the largest such cube (obviously we could put smaller cubes inside the half-sphere).
Imagine the half-sphere in $\Bbb R^3$, laying flat on the xy-plane. Let the side length of the cube be $r$.
The corners of the cube will touch the top of the half-sphere, so the length of a line from the origin to any top vertex of the cube will have length $a$, since this line is also the radius of the half-sphere. The line from the origin to a vertex on the xy-plane will be one-half the length of a diagonal on a face of the cube. Using the pythagorean theorem and letting the diagonal be $d$, we have that $d^2=r^2+r^2$ and so $d=\sqrt 2 r$, thus the line from the origin to the vertex on the xy-plane is $\frac 1 {\sqrt 2} r$. We have two sides of a triangle which has one vertex at the origin, one as a vertex of the cube on the xy-plane, and one as a top vertex of the cube. Using vertices of the cube which are on the same edge of the cube, we have that the last side of such a triangle has length $r$. Thus we just use the pythagorean theorem again:
$$r^2+\frac {r^2} 2 = a^2$$
Solving for $r$ in terms of $a$ gives the desired result. |
H: A cube is divided into two cuboids
A cube is divided into two cuboids. The surfaces of those cuboids are in the ratio $7: 5$. Calculate the ratio of the volumes.
How can I calculate this?
AI: Without loss of generality we can let the sides of the original cube be $1$.
The larger of the two cuboids has four of its sides equal to say $x$. (The other $8$ sides are $1$.) Then the smaller cuboid has four of its sides equal to $1-x$.
The surface area of the larger cuboid is $2+4x$. (Two $1\times 1$ faces, and four $x\times 1$ faces). Similarly, the surface area of the smaller cuboid is $2+4(1-x)=6-4x$.
We are told that
$$\frac{2+4x}{6-4x}=\frac{7}{5}.$$
Solve. We get $x=\frac{2}{3}$, making $1-x=\frac{1}{3}$. So the volumes are in the ratio $2:1$. |
H: Image of a strip under the map $e^z$
Please help me in finding the image of the strip $\lbrace (x,y) \in\mathbb{C} : mx-\pi<y<mx+\pi,-\infty<x<\infty\rbrace $ under the mapping $w=e^z $ ?
Where $m$ is any real number.
AI: $${\mathbb C}\setminus \{0\}$$
as for any complex $(a,b)\neq0$, you can find $\rho>0$ and $\theta\in[-\pi,\pi]$ such that
$(a,b)=\rho e^{i\theta}=e(\ln(\rho)+i\theta)$.
So take $x=\ln(\rho)$ and $y$ can be found in $[mx-\pi,mx+\pi]$ such that $y-\theta=2k\pi$ ($k\in\mathbb Z$)
Hence, $(x,y)$ is in your strip and $e^{(x+iy)}=(a,b)$
EDIT : except that your strip doesn't contain $y=mx\pm\pi$. So you will not have all points $\rho.e^{i.(m\rho+\pi)}$ that is some kind of spiral :
$${\mathbb C}\setminus \{\rho.e^{i(m\rho+\pi)}\;|\;\rho\ge0\}$$ |
H: Precedence of concatenation: Is $5/7y$ equal to $(5/7)\times y$ or $5/(7\times y)$?
Possible Duplicate:
Do values attached to integers have implicit parentheses?
What is 48÷2(9+3)?
What is “multiplication by juxtaposition”?
What is the precedence of the concatention operator when used for
multiplication?
If it's the same as multiplication, 5/7y would mean (5/7)*y.
However, it also seems reasonable to see it as 5/(7*y).
I realize I can avoid the issue by writing "5y/7" or using explicit
parentheses, but am curious about the "correct" answer.
Googling was surprisingly unhelpful: most results were about the
precedence of string concatention in programming languages.
AI: I think that you almost answered your own question:
I realize I can avoid the issue by writing "5y/7" or using explicit parentheses.
I don't know of any "correct" way to interpret $5/7y$, and I don't believe that there is any. There are lots of notations in math that are not that clear, and the only answer is to make them clear. So I would never write $5/7y$ on a blackboard if I meant $5/(7y)$.
If you by $5/7y$ mean $5/7*y$, then from what I understand, when programming a compiler would interpret this as $5*y/7$. |
H: How to derive distributivity from Boolean algebra laws
Let $(L,\le,\bot,\top)$ be a bounded lattice and $\neg: L \rightarrow L$ be a map that satisfies the following laws:
$a \wedge b = \bot \Leftrightarrow a \le \neg b$
$\neg\neg a =a$
I'd like to show distributivity, namely, $a\wedge(b\vee c) = (a \wedge b) \vee (a \wedge c)$ and $a \vee (b \wedge c) = (a \vee b) \wedge (a \vee c)$.
I've already shown the following laws:
$a \wedge \neg a = \bot$, $a \vee \neg a = \top$
The de Morgan's laws: $\neg (a \wedge b) = \neg a \vee \neg b$ and $
\neg (a \vee b) = \neg a \wedge \neg b$
I've also shown the distributive inequalities that hold for all lattices, and that the two forms of distributivity are equivalent. How should I use these laws to show what I'd like to?
AI: First, let's recall the notion of a Heyting implication $\to$: it is a binary operation on a lattice such that
$$a \le (b \to c) \text{ if and only if } (a \land b) \le c$$
A lattice equipped with a Heyting implication $\to$ is automatically distributive. Indeed, $a \land (b \lor c) \le d$ if and only if $(b \lor c) \le (a \to d)$, so if and only if $b \le (a \to d)$ and $c \le (a \to d)$, so if and only if $a \land b \le d$ and $a \land c \le d$. Hence, $a \land (b \lor c) \le (a \land b) \lor (a \land c)$ and $(a \land b) \lor (a \land c) \le a \land (b \lor c)$.
Now, a lattice equipped with a boolean negation automatically has a Heyting implication: take $(a \to b) = \lnot (b \land \lnot c)$, as usual. This works, because $a \land b \le c = \lnot \lnot c$ implies $a \land b \land \lnot c = \bot$, so $a \le \lnot (b \land \lnot c)$; why $a \le \lnot (b \land \lnot c)$ implies $a \land b \land \lnot c = \bot$, so $a \land b \le \lnot \lnot c = c$. |
H: Reducing simple trigonometric expression
I have an expression: $\sin(\pi/3) - \sin(\pi/6)$.
I have it reducing to $3 / \sqrt{2} - 1/2$,
but my text has $3 \sqrt{3} - 3$.
How does my eq become their eq?
Edited the eq
AI: Starting with their work $$\dfrac{\Delta y}{\Delta t} = \dfrac{\sin(\dfrac{\pi}{3}) - \sin(\dfrac{\pi}{6})}{\dfrac{\pi}{3} - \dfrac{\pi}{6}}$$
This then simplifies as $$\dfrac{\dfrac{\sqrt{3}}{2} - \dfrac{1}{2}}{\dfrac{\pi}{6}} = \dfrac{\sqrt{3}-1}{2} \cdot \dfrac{6}{\pi} = \dfrac{6(\sqrt{3}-1)}{2} \cdot \dfrac{1}{\pi} = \dfrac{3 \sqrt{3} - 3}{\pi}.$$ |
H: Periodic Function - Repeating Pattern Problem
For the following question:
A necklace is made by stringing N individual beads together in the repeating pattern of Red Bead , Green Bead , White , Blue and Yellow Bead. If necklace begins with a RED Bead and ends with a white bead , then N could be A)$16$ B)$32$ C)$41$ D)$54$ E)$68$
I think the answer should be 41 but my text says its 68. Is this a misprint?
Here is how I am solving it:
Let n= $1$ for Red and let n=$2$ for green and so on. Now for the answers if n=$16,32,54,68$ they are divisible by $2$ so they would end up in green but for n=$41$ we could write $40+1$ so its Green and then White.
Also one more question , say if we had n=40 that is divisible by both n=2(Green) and n=4(Blue) and also by n=5 (Yellow) then the 40th one which would it be Blue , Green or Yellow ?
AI: Ok so we need to find $N$. First off, there are 5 beads in the pattern. Red is bead number 1, and yellow is bead number 3. Since we are ending with bead number 3, and there are 5 beads in the pattern, then 5x=N - 3, where x must be an integer. What does this equation mean? You must subtract 3 from N, and the result must be divisible by 5. 41-3 = 38, which is not divisible by 5, 68-3 = 65 is divisible by 5.
Another way to look at this is see that bead 1 is red. Continuing down the cycle, bead 6 is also red. Every 5th bead onwards is read. Beads 11,16,21..36,41 are red. So bead 41 is also red, which would mean that there can't be 41 beads in the pattern. Bead 66 is also red. So bead 67 would be green, and 68 would be white.
As for your second question, you have to understand that it does not work like that. Suppose you have a pattern ABCDE of repeating items. You know that the pattern starts at A, and you have your N value. Let i = the index of any letter in that sequence. For A, i = 1; for B, i = 2; and so on. If you want to test if the letter at the i-th index corresponds to a certain N value, then you need to see if (N-i)/(i) is an integer number. For example, if you want to see if the 43rd bead is blue, where i = 4, then do (43-4)/4 = 39/4, which is not an integer answer, so the 43rd bead is not blue. |
H: Elementary Number Theory $(a+b,\frac{a^p+b^p}{a+b})=1 \text{ or } p$
Assume $(a,b)$=1, and let $p$ be an odd prime, prove that
$$(a+b,\frac{a^p+b^p}{a+b})=1 \text{ or } p$$.
I thought of letting $p=2k+1,k\in\mathbb{Z}$,
then use the identity
$$a^{2k+1}+b^{2k+1}=(a+b)(a^{2k}-a^{2k-1}b+a^{2k-2}b^2-\ldots+b^{2k})$$
So dividing throughout,
$$\frac{a^p+b^p}{a+b}=a^{2k}-a^{2k-1}b+a^{2k-2}b^2-\ldots+b^{2k}$$
However, after that, I am kind of stuck..
Sincere thanks for any help!
AI: Let me continue the hint I wrote above. We see that if $h$ divides $a +b,$ then $\frac{a^p +b^p}{a+b} \equiv pb^{p-1}$ (mod $h$). Hence, if we let $h = {\rm gcd}(a+b,\frac{a^p +b^p}{a+b}),$ we see that $h|pb^{p-1}.$ But $h$ is coprime to $b,$ since $h |(a+b)$ and ${\rm gcd}(a,b) = 1.$ Hence $h|p,$ as required.
Later edit: Although the question does not request a proof that both $1$ and $p$ can occur, I provide one, in response to the comment of Andr\'e. If $p$ divides $a+b,$ then taking $h =p$ above shows that $\frac{a^p +b^p}{a+b} \equiv pb^{p-1} \equiv 0$ (mod $p$), so that $p$ divides ${\rm gcd}(a+b,\frac{a^p +b^p}{a+b})$ in that case, and the gcd is equal to $p$ by what has been shown already. |
H: What do I need to know to understand the completion of the field of rational functions of a non-singular projective curve?
So the title gives the jist of my question. Specifically, let $X$ be a non-singular projective curve, $P$ a point on $X$, $v_P$ the discrete valuation associated to the ring $\mathcal{O}_P$. Then I have read that the completion of $k(X)$ with respect to the valuation $v_P$ is isomorphic to the field of formal Laurent series over $k$.
Stuff that might be relevant? I know some basic Galois theory, some very basic point set topology, and I'm just starting chapter 10 in An Introduction to Commutative Algebra by Atiyah and MacDonald.
I was hoping someone could tell me the material I will have to read to understand this along with good books that cover it. If there is an algebraic way of going about this I would prefer it as I'm really enjoying An Introduction to Commutative Algebra. Also if someone wanted to give me an overview of what is happening here that would be appreciated also.
Thanks for any help!
AI: You need some geometric facts because the statement is false for singular curves. You also need to assume $P$ is a rational point of $X$ (automatically true if the base field $k$ is algebraically closed).
This being said, the fact that $X$ is a non-singular curve implies that the maximal ideal of the local ring $O_P$ is generated by one element $t$. The hypothesis $P$ is rational means that $O_P/tO_P=k$. So
$$O_P=k+tO_P=k+t(k+tO_P)=...=k[t]+t^nO_P$$
for all $n\ge 1$. Passing to the limit (for the $t$-adic topology), we see that the completion of $O_P$ is $k[[t]]$. As $k(X)=\mathrm{Frac}(O_P)$, we get the desired result. |
H: Is $\text{rk}L=\text{rk}L^*L $ true for finite rank operators?
Let $L$ be a compact linear operator in an infinitedimensional space that has finite rank. Do the equations $$\text{rk}L=\text{rk}L^*L\ \text{and} \ \text{rk}L^*L=\text{rk}R,$$ where $R$ is the (unique) root of $L^*L$ have to hold ? If they do, why is this the case ?
The proof of this statement is easy in the finite dimensional case, but for finite rank operators I couldn't figure it out.
AI: Let $H$ and $K$ be Hilbert spaces. For given $\xi\in K$ and $\eta\in H$ we denote by $\xi\bigcirc\eta$ the rank one operatror sending $\zeta\in H$ to $\langle\zeta,\eta\rangle\xi$. Since $L:H\to K$ is a finite rank operator then by collorary of Hilbert-Schmidt representation theorem we have
$$
L=\sum\limits_{k=1}^n s_k (e_k'\bigcirc e_k'')\tag{1}
$$
for some orthonormal systems $\{e_k':k=1,n\}\subset K$ and $\{e_k'':k=1,n\}\subset H$ and positive numbers $\{s_k:k=1,n\}$ that called singular values. One may check that
$$
L^*L=\sum\limits_{k=1}^n s^2_k(e_k''\bigcirc e_k'')\qquad
R=(L^*L)^{1/2}=\sum\limits_{k=1}^n s_k(e_k''\bigcirc e_k'')\tag{2}
$$
From $(1)$ and $(2)$ it follows that
$$
\mathrm{rk}(L)=\mathrm{dim}\;\mathrm{Im}(L)=\mathrm{dim}(\mathrm{span}\{e_k':k=1,n\})=n
$$
$$
\mathrm{rk}(L^*L)=\mathrm{dim}\;\mathrm{Im}(L^*L)=\mathrm{dim}(\mathrm{span}\{e_k'':k=1,n\})=n
$$
$$
\mathrm{rk}(R)=\mathrm{dim}\;\mathrm{Im}(R)=\mathrm{dim}(\mathrm{span}\{e_k'':k=1,n\})=n
$$
So
$$
\mathrm{rk}(L)=\mathrm{rk}(L^*L)=\mathrm{rk}(R)
$$ |
H: Dedekind Cuts in Rudin' analysis - Step 4
This question refers to the construction of $\mathbb{R}$ from $\mathbb{Q}$ using Dedekind cuts, as presented in Rudin's "Principles of Mathematical Analysis" pp. 17-21.
More specifically, in the last paragraph of step 4, Rudin says that for $\alpha$ a fixed cut, and given $v \in 0^*$, setting $w=- v / 2$, there exists an integer $n$ such that $nw \in \alpha$ but $(n+1)w$ is not inside $\alpha$. Rudin says that this depends on the Archimedean property of the rationals, however he has not proved it.
Could somebody prove the existence of the integer $n$?
AI: By definition there is some rational $q$ that is not in $\alpha$. Let $p=|q|+1$; then $p>q$, so $p\notin\alpha$, and $p>0$. The rational $w$ is also positive, so by the Archimedean property of the rationals there is a positive integer $m$ such that $mw>p$, and therefore $mw\notin\alpha$.
We also know that there is some rational $r\in\alpha$. Let $s=-|r|-1$; then $s<r$, so $s\in\alpha$, and $-s>0$, so again by the Archimedean property there is a positive integer $k$ such that $kw>-s$. But then $(-k)w<s$, so $(-k)w\in\alpha$. We now have positive integers $k$ and $m$ such that $-kw\in\alpha$ and $mw\notin\alpha$.
Let $S=\{i\in\Bbb N:(i-k)w\notin\alpha\}$; $m+k\in S$, so $S\ne\varnothing$. Since $\Bbb N$ is well-ordered, $S$ has a smallest element, $j$. Note that $0\notin S$, since $-kw\in\alpha$, so $j>0$. Let $n=(j-1)-k$. Then $nw\in\alpha$, and $(n+1)w\notin\alpha$. |
H: How not to prove the Riemann hypothesis
I remember reading somewhere that there is a (probably a family of) quick false proof of the Riemann hypothesis that starts by using complex logarithms in a bad way, then does some elementary calculations and out pops the result. A perusal of the General Mathematics section of the arXiv also shows an abundance of presumably false proofs of the same. These tend to be somewhat... unclear and I'd prefer avoiding to look at them in any detail.
Question: Can anyone describe a quick false proof of the Riemann hypothesis?
Preferably I'd like one that has a very clear false step that misuses complex logarithms or $n$-th roots or some such gadget in an identifiable way. This is both because of my curiosity, and because I figure I could use it as an exercise problem for undergrads (if I ever end up teaching a course on complex analysis) to explain why they should be careful about taking logarithms.
AI: Matthew Watkins has a collection of 'proofs' here.
Describe them would make you loose part of their flavor...
(errors are sometimes commented and the oldest and 'less serious' ;-) are at the end...) |
H: Prove: Full Rank and a solution os linear system
I'm studying for my exam of linear algebra.. I want to prove the following corollary:
Given $A \in{R^{n\times n}}$, there is a solution $x$ to $Ax = y$ for all $y$, if and only if $A$
has rank $m$ (full row rank).
I know that the rank of a matrix is the maximum number of columns (rows respectively) that are linearly independent and is defined by:
$\operatorname{Img} (A) = \operatorname{Rg} (A):= y \in{C^m}:y = Ax, x \in{C^n}$
My problem is that I can not find a way to relate the two concepts in order to reach a formal proof. Any help?
AI: Write the system $Ax = y$ as
$$ \pmatrix{\mid & \mid & & \mid \\
a_1 & a_2 & \dots & a_n \\
\mid & \mid & & \mid }
\pmatrix{x_1 \\ x_2 \\ \vdots \\ x_n} = \mathbf{y} \tag{1} $$
or
$$ x_1 \pmatrix{\mid \\ a_1 \\ \mid } + x_2 \pmatrix{\mid \\ a_2 \\ \mid } + \cdots + x_n \pmatrix{\mid \\ a_n \\ \mid } = \mathbf{y} \tag{2} $$
For every $\mathbf{y} \in \Bbb{R}^n$, equation $(2)$ has a unique solution $x_1, x_2, \ldots, x_n$ iff the vectors $\{ a_1, a_2, \ldots, a_n \}$ form a basis of $\Bbb{R}^n$, which is true since $A$ is $n\times n$ and full rank. |
H: Free Associative Algebras
let $X=\{X_0, X_1,\ldots \}$ be a finite or countable set, $K$ a field, and $K\langle X\rangle$ the free associative algebra generated by $X$.
It is known that all associative algebras, generated by a finite or countable number of
elements, can be represented as the factor ring $K\langle X\rangle/I$, where $I$ is an ideal of $K\langle X\rangle$.
If we considere the matrix algebra $M_n(K)$, over the field $K$, what is the representation of this algebra in the form $K\langle X\rangle/I$?
AI: You need to be careful: there are two fields floating around here, $K$ and $\mathbb{F}$. The statement should be that all associative $K$-algebras are factor rings of some free $K$-algebra $K\langle X \rangle$; it is not true as you stated it.
Example: $\mathbb{F}_p$ is not a factor algebra of any free $\mathbb{Q}$-algebra
If $K=\mathbb{F}$, you may take the generating set $X$ to be $\{E_{ij}: 1 \leq i,j \leq n\}$ and the ideal $I$ to be generated by the elements of this free algebra coming from the usual relations amongst elementary matrices, namely $E_{ij}E_{kl} - \delta_{jk}E_{il}$ together with $1 - \sum_i E_{ii}$.
Here $\delta_{jk} = 0$ if $j \neq k$, otherwise it is $1$. |
H: A very simple question about a Harris' exercise
In the book of Harris Algebraic Geometry, a First Course, exercise 2.29, it is asked to show that the composition of $1\times \nu: \mathbb{P}^1\times\mathbb{P}^1 \to \mathbb{P}^1\times\mathbb{P}^2$ where $\nu$ is the Veronese embedding with the Segre embedding $\sigma:\mathbb{P}^1\times\mathbb{P}^2 \to \mathbb{P}^5$ can be represented (after identification of $\mathbb{P}^1\times\mathbb{P}^1$ with its image the quadric $Q: Z_0Z_3-Z_1Z_2=0$ in $\mathbb{P}^3$ by the Segre mapping), by the map $$\phi: [Z_0,Z_1,Z_2,Z_3]\to [F_0(Z),\cdots,F_1(Z)]$$ where $(Q,F_0,\cdots,F_5)$ is a basis of the 7 dimensional space of homogenous quadratic polynomials that contains the line $L: Z_0=Z_1=0$ of $Q$.
This seems strange to me because $\phi$ is not even defined on the line $L$. What did I miss ?
AI: The composition $\sigma\circ(1\times\nu)$ will map $((x:y),(z:w))\mapsto ((x:y),(z^2:zw:w^2))\mapsto (xz^2:xzw:xw^2:yz^2:yzw:yw^2).$
Identifying $\Bbb P^1\times\Bbb P^1\cong Q\subseteq\Bbb P^3$ takes $((x:y),(z:w))\mapsto (xz:xw:yz:yw)$ or conversely, $(a:b:c:d)\mapsto ((1:c/a),(1:b/a))=((a:c),(a:b))$ where we suppose that $a\neq 0.$ Thus, on the affine patch $a\neq 0,$ the composition is determined by
$$(a:b:c:d)\mapsto (a^3:a^2b:ab^2:ca^2:cab:cb^2)=(a^3:a^2b:ab^2:ca^2:a^2d:adb) =(a^2:ab:b^2:ac:ad:bd),$$ where we've used $ad=bc.$
Thus, we wish to show that $\langle a^2,ab,b^2,ac,ad,bd,ad-bc\rangle$ is a basis of the homogeneous quadratics containing $a=b=0.$ This part should be routine.
Edit: The identification on the patch $c\neq 0$ is given by $(a:b:c:d)\mapsto ((a/c:1),(1:d/c))=((a:c),(c:d)).$ On this patch the composition is determined by $$(a:b:c:d)\mapsto (ac^2:acd:ad^2:c^3:c^2d:cd^2)=(ac^2:bc^2:bcd:c^3:c^2d:cd^2)=(ac:bc:bd:c^2:cd:d^2).$$
On this patch we find a similar expression, but now our forms vanish on $c=d=0.$
I think the question is not saying that the first expression should be valid everywhere on $Q,$ since it does not make sense to evaluate it on $a=b=0,$ as you say. On $c\neq 0$ or $d\neq 0$ we have this alternate expression, which should coincide with the original one on the intersection $a,c\neq 0$ for example. This way we'll get a globally defined map on $Q.$ Unfortunately, the way the question seems to be stated doesn't make this precise at all. |
H: integrate two curves
I just started Calculus II and have been given the following problem:
Find the area inside the curves: $$y = 5 - 2x^2\\y = 3x^2 \\ x = -4 \\x = 4$$
I graphed the two curves on my calculator to get a rough idea of the problem but noticed that the x values aren't where the two curves cross.
I'm wondering if I should integrate on 4,-4 or if I should calculate where they cross and use those values as the minimum and maximum x values.
Thanks in advance
edit:
The beggining of the problem was just about sketching the graph and drawing approximating rectangles; however the part that I was confused on was worded as such:
y = 5 - 2x^2, y = 3x^2, x = -4, x = 4
find the area S of the region
AI: Draw the two parabolas (which meet at $x=\pm 1$), and the two lines. There are three finite regions in the picture. I think you are being asked to find their combined area. The issue is that in part of the picture, the parabola $y=5-2x^2$ is above the parabola $y=3x^2$, and in part of the picture, it is below $y=3x^2$. The simplest way to do the calculation is to treat these parts separately, then add up.
The "middle" region has $5-2x^2\ge 3x^2$. To find its area we calculate
$$\int_{-1}^1 ((5-2x^2)-3x^2)\,dx.$$
The region on the "right" has $3x^2\ge 5-2x^2$. To find its area we calculate
$$\int_1^4 (3x^2-(5-2x^2))\,dx.$$
The area of the region on the "left" can be calculated in a similar way. However, we might as well take advantage of symmetry. Now add up the three areas.
Remark: I would prefer to note the symmetry from the beginning, and observe that we can calculate the area to the right of the $y$-axis, then double. The area to the right of the $y$-axis is
$$\int_0^1 ((5-2x^2)-3x^2)\,dx+\int_1^4 (3x^2-(5-2x^2))\,dx. $$
We could express this as a "single" integral, by noting that the area to the right of the $y$-axis is
$$\int_0^4\left|(5-2x^2)-3x^2\right|\,dx.$$
However, the gain is illusory, since to deal with the absolute value sign, the best thing to do is to break up the integral at $x=1$. |
H: $H\vartriangleleft G$ and $|H|\not\equiv 1 (\mathrm{mod} \ p)$ then $H\cap C_{G}(P)\neq1$
Let $G$, a finite group, has $H$ as a proper normal subgroup and let $P$ be an arbitrary $p$-subgroup of $G$ ($p$ is a prime). Then $$|H|\not\equiv 1 (\mathrm{mod} \ p)\Longrightarrow H\cap C_{G}(P)\neq1$$
What I have done:
I can see the subsequent well-known theorem is an especial case of the above problem:
Let $G$ is a finite non trivial $p$-group and $H\vartriangleleft G$. Then if $H\neq1$ so $H\cap Z(G)\neq1$.
So I assume that $G$ acts on $H$ by conjugation and therefore $$|H|=1+\sum_{x\in H-\{1\}}|\mathrm{Orbit}_G(x)|$$ $|H|\not\equiv 1 (\mathrm{mod} \ p)$ means to me that there is $x_0\in H$ such that $p\nmid|\mathrm{Orbit}_G(x_0)|$. Am I doing right? Thanks.
This problem can be applied nicely in the following fact:
Let $p$ is an odd prime and $q$ is a prime such that $q^2\leqslant p$. Then $\mathrm{Sym}(p)$ cannot have a normal subgroup of order $q^2$.
AI: Instead of letting all of $G$ act on $H$, consider just the action of $P$ on $H$. Then everything you wrote above still holds, but since $P$ is a $p$-group, the size of the orbit of $x_0$ must be a power of $p$. Can you conclude from here? |
H: Maximum intersecting subsets
There are n elements. What is the maximum number of subsets chosen at any one time so that every pair of subsets from collection intersect?
AI: This problem is classical, and has a surprisingly trivial solution. Let $[n]=\lbrace 1,\ldots, n\rbrace$. Then $\mathcal{P}([n])$ (the set of all subsets of $[n]$) has size $2^n$. Say that $\mathcal{A}\subset\mathcal{P}([n])$ is intersecting if $A,B\in\mathcal{A}$ implies $A\cap B \ne \emptyset$.
Claim: Every maximal intersecting $\mathcal{A}\subset\mathcal{P}([n])$ contains exactly $2^{n-1}$ sets.
Proof: The upper bound is trivial: if $\mathcal{A}$ contains more than $2^{n-1}$ sets then it contains both $A$ and $A^c$ for some $A\subset[n]$. Now suppose that $\mathcal{A}$ is maximal intersecting, so that we can't add any set to $\mathcal{A}$ without destroying the intersecting property. Certainly $\mathcal{A}$ is an upset, i.e., if $A\in\mathcal{A}$ and $A\subset B$ then $B\in\mathcal{A}$. Suppose $A\notin\mathcal{A}$. Then by maximality there exists $B\in\mathcal{A}$ such that $A\cap B = \emptyset$, i.e., $B\subset A^c$, so $A^c\in\mathcal{A}$. |
H: How to find the time at which a function is maximized given two component functions
I haven't touched calculus in around a year and a half. However, this semester, I have to take Intro to Electrical Engineering as a requirement for my Computer Engineering major, and the first part of this class seems quite heavy on calculus. One of the problems I've been assigned is as follows:
The voltage and current at the terminals of the circuit element in Fig 1.5 are zero for $t < 0$. For $t \geq0$ they are:
v $= (16,000t + 20)e^{-800t}$ V
i $= (128t + 0.16)e^{-800t}$ A
At what instant of time is maximum power delivered to the element?
Now, I thought that all I had to do was multiply v times i and find the derivative of that equation, then find the maximum based on that, and then find the value of t for that. However, I tried that method on a practice problem and it didn't give me the correct answer. Additionally, the next part of this question is to actually find the maximum value for p, which makes it seem odd that I'd need to find the maximum value first, then the time at which it is at a maximum. So, please, help me figure out the actual method to solving this problem.
EDIT: Also, for clarity, Fig 1.5 simply shows an ideal circuit element, it isn't really necessary for solving the problem.
AI: Your approach is correct, multiply $vi$, take the derivative, set to zero. Without seeing your work, it is hard to say more. Having found the time, the value of the power is $vi$ at that moment. Usually I find the time first from the derivative and if you want the power you find it then. I plotted it in Alpha and it looks like the maximum is at $t=0$ as the equations are not valid before that. |
H: Convergence in $\sup$ norm $\Rightarrow$ Cauchy in $\sup$
Let $(f_n)$ be a sequence of bounded functions on a set $E \subseteq \mathbb R$ and suppose that $f$ is a bounded function such that $\|f_n - f\|_{\infty} \to 0$ as $n \to \infty$. Prove that $(f_n)$ is a Cauchy sequence in the $\sup$ norm.
My Thoughts
Method 1:
Since $(f_n) \to f$ in the $\sup$ norm, we have $$\tag{$*$} \lim_{n \to \infty} \|f_n - f \|_{\infty} = 0 $$ By a previous theorem, we have that $(*)$ is true iff for all $\epsilon >0$ there exists an $N$ so that for all $n \ge N$, $\|f_n - f\|_{\infty} \le \epsilon$.
Now, $$\tag{$\dagger$} \|f_n - f\|_{\infty} = \lim_{m \to \infty} \|f_n - f_m\|_{\infty}$$ My book brings a proof of the converse of this statement to this point and then essentially claims (in more formal language) "since the left hand side of $(\dagger)$ is equal to the right hand side, this implies $$\forall \epsilon > 0 \ \exists M\ \forall n, m > M \Big[\|f_n - f_m\|_{\infty} \le \epsilon\Big]$$
Is this faulty reasoning or not? If so, how else would I prove this?
AI: There's nothing special about the supremum norm here. In any metric space, a convergent sequence is a Cauchy sequence. It's just the triangle inequality: $d(f_n, f_m) \le d(f_n, f) + d(f, f_m)$. |
H: What is the meaning of $\mathbb R^+$?
For a function $f$ that maps set $A$ to $B$,
$f\colon\mathbb R^+\to\mathbb R^+$, $f(x) = x^2$ is injective.
$f\colon\mathbb R\to\mathbb R$, $f(x) = x^2$ is not injective since $(- x)^2 = x^2$.
what is the difference between $\mathbb R^+$ and $\mathbb R$?
Additionally, what is the difference between $\mathbb N$ and $\mathbb N^+$?
AI: $\mathbb R^+$ commonly denotes the set of positive real numbers, that is: $$\mathbb R^+ = \{x\in\mathbb R\mid x>0\}$$
It is also denoted by $\mathbb R^{>0},\mathbb R_+$ and so on.
For $\mathbb N$ and $\mathbb N^+$ the difference is similar, however it may be non-existent if you define $0\notin\mathbb N$. In many set theory books $0$ is a natural number, while in analysis it is often not considered a natural number. Your mileage may vary on $\mathbb N$ vs. $\mathbb N^+$. |
H: Is the statement "1/3 of the natural numbers are divisible by 3" true? Is anything similar to it true?
If we're talking about a finite set of the natural numbers, like those between 1 and 500 or 1 and a million, it seems to me that the fraction of numbers in that finite set that have a factor of 5 approaches $1/5$ as the set increases in size. Like roughly $1/2$ of all numbers in such a set have a factor of 2, roughly $1/3$ have a factor of 3, and so on; and this approximation grows less "rough" and more exact as the size of the set increases.
So, can we say that out of the entire set of the natural numbers, exactly $1/5$ are divisible by 5? Or perhaps that the limit of the fraction of the natural numbers less than or equal to a given n divisible by a given integer approaches 1/that integer as n approaches infinity?
(I would love to know how to ask this question with proper notation.)
AI: This can indeed be made formal. To formalize the statement "$x$ fraction of natural numbers satisfy the property $P$", we define the function
$$f(n)=\text{ number of natural numbers }\leq n\text{ which satisfy }P$$
and write $\lim\limits_{n\to \infty} \frac{f(n)}{n}=x$. In your first case, the function $f$ is given by $f(n)=\lfloor n/3\rfloor$ and the statement becomes
$$\lim\limits_{n\to\infty} \frac{\lfloor n/3\rfloor}{n}=\frac{1}{3}$$
which is easily seen to be true, since $\frac{1}{3}-\frac{1}{n}\leq \frac{\lfloor n/3\rfloor}{n}\leq \frac{1}{3}$. Similar results hold for any natural number in place of $k$. |
H: Solving this linear system based on the combustion of methane, has no constants
I recently discovered that I could solve a chemical reaction using a linear system. So I thought I would try something simple like the combustion of methane.
where x y z and w are the moles of each molecule
x $CH_4$ + y $O_2$ = z $H_2$O + w C$O_2$
the linear system for this would be:
x = w
2y = z + 2w
4x = 2z
I got as far as y = z and y = 2w but without any constants, I am stumped. Can anyone help me? I was assuming that elimination and substitution would suffice, but I must be wrong.
AI: You have three equations with four unknowns, which should lead you to expect a single undefined parameter. In this case you can multiply all your variables by any constant and still have a fine set of equations. The easiest cure is to set one of them to $1$. Maybe you choose $x=1$. Then you should be able to solve the rest easily. This would represent burning $1$ mole of methane. If you burned $2$ moles, you would have twice as much of each other reactant. |
H: Binomial/Tensor Identity
Let $k$ be a a field and consider the space $k[x] \otimes_k k[x]$. I would like to verify the equation
$$ \sum_{k=0}^{m+n} {m+n \choose k} x^k \otimes x^{(n+m)-k}= \sum_{i=0}^n \sum_{j=0}^m{n \choose i}{m \choose j} x^{i+j}\otimes x^{(n+m)-(i+j)}$$
however I am not getting very far with this. Could anyone give me some help? Thanks very much.
AI: This doesn't really have much to do with tensor products. If you group together the terms on the right hand side with a common value of $i+j$ (call the common value $k$), then the identity reduces to the binomial coefficient identity $$\sum_{i+j=k} \binom{n}{i} \binom{m}{j} = \binom{n+m}{k}.$$ This identity is proved by a combinatorial argument or by comparing the two sides of $(1+x)^{m+n} = (1+x)^m (1+x)^n$. |
H: Quadratic Equation Error
For the floating point system $(B, t, L, U) = (10,8,-50,50)$ and for the quadratic equation:
$ax^2 + bx + c$, I need to show error that arises in various cases and how to fix those.
$$a=10^{-30}$$
$$b= -10^{30}$$
$$c=10^{30}$$
I think there is cancellation error when plugging into quadratic formula. Can anyone help me verify that if multiply quadratic formula by conjugate, then I'll get $\frac{2c}{-b \pm \sqrt{b^2-4ac}}$ which will take away the error for the case that is subject to error?
Thanks!
AI: The basic idea is that if |$4ac| \ll b^2$, the square root is very close to $|b|$. Depending on the sign of $b$, one combination or the other will involve subtracting two nearly equal quantities.
To be specific, let us assume $b \gt 0$. Then one root is $\frac {-b+\sqrt{b^2-4ac}}{2a}$ and is the one subject to cancellation. If we multiply by the congugate:$$\frac {-b+\sqrt{b^2-4ac}}{2a}\frac {b+\sqrt{b^2-4ac}}{b+\sqrt{b^2-4ac}}=\frac{b^2-4ac-b^2}{2ab+2a\sqrt{b^2-4ac}}=\frac{-2c}{b+\sqrt{b^2-4ac}}$$ and the cancellation has disappeared. The case $b \lt 0$ is similar except you worry about the one with the minus sign. |
H: Numerical Analysis
I am trying to determine some numerical difficulties that arise from a couple problems, and a good way to re-write them to avoid those errors.
For instance, I have:
1) $\sqrt{x+\dfrac{1}{x}} - \sqrt{x-\dfrac{1}{x}}$ where $x\gg 1$
I think that since these two terms approximately equal each other, there will be cancellation error. So I multiplied the numerator and denominator by the conjugate yielding:
$\dfrac{\dfrac{2}{x}}{\sqrt{x+\dfrac{1}{x}}+\sqrt{x-\dfrac{1}{x}}}$
I think that this should get rid of the cancellation error, does anyone see anything wrong with this attempt?
If this looks right, then I will show my attempt on the second problem, but I hope to verify my method first.
2) $\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}}$ where $a\approx 0$ and $b\approx 1$
Thanks!
AI: For 1, you have successfully avoided the cancellation. If you want, you could go to $$\frac {\frac 2x}{\sqrt x (\sqrt{1+\frac 1{x^2} }+\sqrt{1-\frac 1{x^2} })}=\frac 2{x^{\frac 32} (\sqrt{1+\frac 1{x^2} }+\sqrt{1-\frac 1{x^2} })}\approx x^{-\frac 32}$$ but I am not sure that is an improvement.
For 2, you could have $\frac 1{a^2}$ overflow where $\frac 1a$ does not. To avoid this, you could rewrite it as $\frac 1a \sqrt {1+\frac {a^2}{b^2}}$. That still squares $a$, but if it underflows maybe it gets set to zero and you are OK. |
H: Why do we define the group law on elliptic curves only for Weierstrass forms and $O$ an inflexion point?
In almost all texts concerning the group law on an elliptic curve it is first proven that any nonsingular cubic can be given by a Weierstrass equation and then the group law using the point $O$ at infinity is described (which in this case is an inflexion point).
Since I seem to be able to geometrically define addition of points $P$ and $Q$ for any nonsingular cubic and any point $O$ on the curve chosen as the identity element, I wonder why this is not mentioned more often.
I see that $P+Q+R=O$ for any collinear points $P$, $Q$ and $R$ only if $O$ is an inflexion point. Of course it all gets much easier with this property at hand, but is it 'needed'?
Maybe I did miss some counterexample, but isn't it possible to have a nice geometric addition law for all sorts of elliptic curves, not only for the ones in Weierstrass form with $O$ the (inflection) point at infinity?
Thank you for your insights.
Edit: Probably the title was a bit misleading. The more basic thing to ask seems to be Why do we require $P+Q+R=O$ for collinear points?
AI: Actually, sometimes we don't. If you write an elliptic curve in Edwards normal form, the group law is not given by collinearity. Instead, it's given by is a polynomial function deforming the group law on a circle. Edwards normal form is used in cryptography, I think because certain operations are easier to perform with it.
(Curves in Edwards normal form can also be lifted in an appropriate sense to projective space in such a way that the group law is given in terms of coplanar rather than collinear points, where one of the four points is always the identity and can be arbitrary. I regret that I do not have any reference to give for this construction other than an old high school paper of mine; surely it is classical.)
In algebraic geometry textbooks, the form of the group law involving collinearity is a special case of a more general construction, namely that of the divisor class group or Picard group of an algebraic curve. This construction is an analogue of the construction of the class group of the ring of integers of a number field, and the role of collinear points is that they furnish principal divisors (analogous to principal ideals) which we quotient by to get the divisor class group. So one very good reason to present the group law this way is to aim towards this generalization. (The Picard group has the conceptual advantage of being defined without reference to a particular point on the curve.) |
H: machine learning project ideas
I am interested about playing with machine learning algorithm and time series analysis. Is there website/resource with a comprehensive list of sample projects/proposals one may be interested about?
AI: You may want to read this and look at the two terms of projects and additional projects.
Enjoy!
http://blog.smellthedata.com/2010/07/choosing-first-machine-learning-project.html |
H: Why are "irrational numbers" not named "nonrational numbers"?
Possible that I'm misunderstanding the concept of irrational numbers, but seems like the term nonrational would be much more clear. Why is "irrational" more clear than "nonrational"?
UPDATE: Just to be clear, it would be true to say the terms “irrational numbers” and “nonrational numbers” have the exact same meaning, and neither is something the other is not, correct?
AI: It is from Latin "irrationalis" ... so you have to blame those old Romans for this form. |
H: How to evaluate $ \int_c (\sin z)/ z^6$?
Evaluate:
$$ \int_c {\sin z\over z^6} \, dz$$
Where, $c$ is a circle of radius $2, |z| = 2$.
Don't understand why ${\pi i \over 5!} $ (Answer from book).
$$ {\sin z \over z^6} = {1 \over z^6}\left [ z-\frac{z^3}{6}+\frac{z^5}{120}-\frac{z^7}{5040}+\frac{z^9}{362880}-\frac{z^{11}}{39916800}+O(z)^{12} \right ]$$
What happens to $ \int_c {1 \over z^5}\,dz$ and $- {1\over 3!} \int_c {1 \over z^3} \, dz $?
AI: Hint: for a function $\,f(z)\,$ holomorphic within the domain inclosed by a closed smooth path $\,\gamma\,$, we have
$$f^{(n)}(a)=\frac{n!}{2\pi i}\oint_{\gamma}\frac{f(z)}{(z-a)^{n+1}}dz$$ |
H: The difference between convergence in $L^{\infty}$ and almost uniformly
I am reading these notes http://terrytao.wordpress.com/2010/10/02/245a-notes-4-modes-of-convergence/ by Terry Tao. I have a question about the difference between convergence in $L^{\infty}$ and convergence almost uniformly.
Is the difference that convergence almost uniformly guarantees that you can get uniform convergence outside a set of arbitrarily small but still positive measure, while convergence $L^{\infty}$ gets uniform convergence outside a set of exactly measure zero? Formal definitions follow to make ideas precise.
Let $(X, \mathcal{M}, \mu)$ be a measure space. Let $f, f_1, f_2, \ldots$ be a measurable functions.
We say that $f_n \to f$ in $L^{\infty}$ if for all $\varepsilon > 0$ there is an $N_{\varepsilon}$ such that $|f_n(x) - f(x)| \leq \varepsilon$ $\mu$--a.e. when $n \geq N_{\varepsilon}$.
We say that $f_n \to f$ almost uniformly if for all $\varepsilon > 0$ there is a set $E \in \mathcal{M}$ with $\mu(E) \leq \varepsilon$ such that $f_n \to f$ uniformly on $E^c$. I.e., for each $\delta > 0$ there is an $N_{\delta}$ such that $|f_n(x) - f(x)| \leq \delta$ for all $x \in E^c$ when $n \geq N_{\delta}$.
AI: Yes, that is right.
It is a good exercise to prove that your definition of $L^\infty$ convergence is equivalent to:
There is a set $E \subset \mathcal{M}$ with $\mu(E) = 0$ such that $f_n \to f$ uniformly on $E^c$.
The thing to notice is that your definition should be expanded as:
For all $\epsilon > 0$ there exists $N$ such that for all $n \ge N$ there exists $F \in \mathcal{M}$ such that $\mu(F) = 0$ and $|f_n(x) - f(x)| \le \epsilon$ for all $x \in F^c$.
That is, the set $F$ may depend on $\epsilon$ and $n$. In showing the equivalence with my statement, you have to find a single set $F$ that works for all $\epsilon,n$. |
H: Reed-Solomon Code calculation
I have a Reed-Solomon Code which can correct t=2 errors. The generator polynomial is $p(X) = X^3 + X + 1$ and $p(a) = a^3 + a + 1 = 0$ this means $a^3 = a + 1$
What is the degree of generator polynomial of this code?
Are the following code words valid, why or why not?
$(1, a+1, 1, 0, 0, a+1, a)$
$(a^2+a+1, a+1, a^2, a^2+a+1, a^2, 0, 0)$
The first question is easy to answer I think; the degree of the generator polynomial is 3.
However, the second question is way more challenging. I assume that the finite field $\operatorname{GF}(2^3)$ is used and therefore a code word consists of $n = 2^3 - 1 = 7$ symbols. This criteria is met by both code words mentioned in the second question.
How can I determine the validity of the 2 code words?
Probably I have to use the Discrete Fourier Transform (DFT) but because of $n=7$ this would generate a $7\times 7$ matrix which is to costly to do it by hand.
Is there a simple way to calculate all valid code words?
AI: OK, now that I have read your questions more carefully, here is some more
preliminary information. Your Reed-Solomon code has a generator polynomial
of degree $4$ which is not the same as what you call the generator
polynomial, viz., $p(x) = x^3 + x + 1$. In coding theory $p(x)$
would be called the irreducible polynomial used to construct the field.
In this instance, it would more likely be called the primitive
polynomial since its root $a$ is a primitive element of the field, meaning
that all $7$ nonzero elements of the field can be represented as powers of
$a$.
So, what is this generator polynomial (call it $g(x)$) of the Reed-Solomon code?
You might have been given this information as part of the problem, so
look for it. The generator polynomial of a $t$-error-correcting Reed-Solomon
code has the property that $2t$ successive powers of $a$ are roots of
the polynomial. Thus, we have
$$\begin{align*}
g(x) &= (x-a^i)(x-a^{i+1})(x-a^{i+2}\cdots(x-a^{i+2t-1})\\
&= g_{2t}x^{2t} + g_{2t-1}x^{2t-1} + \cdots + g_1 x + g_0
\end{align*}$$
where all the coefficients $g_j$ are necessarily nonzero. Popular
choices for $i$ are $0$ and $1$, but unless you know $g(x)$, it
is not possible to answer the question about which
vectors are valid codewords, except possibly in the general
form "This vector is not a valid codeword for any choice
of $i$, $0 \leq i \leq 6$" which would require a lot more
calculation of the form given below.
Valid codeword polynomials are divisible by the generator polynomial
$g(x)$. So, if you are given the second form of $g(x)$ as
described above, then divide the alleged codeword polynomial
$c(x)$ (which,
for the vector $(c_0, c_1, \ldots, c_6)$,
could be $c(x) = c_0 + c_1x + \cdots + c_6x^6$ or
$c(x) = c_0x^6 + c_1x^5 + \cdots + c_5x + c_6$
depending on the convention used in the document
you are reading) by the generator polynomial $g(x)$ which is given
to you. This is plain polynomial long division as you would
have learned in secondary school except
that you are working over a finite field instead of real numbers. If the remainder is zero, the vector is indeed a codeword; if the remainder is nonzero, it is not a codeword.
On the other hand, if you are given the $4$ roots of $g(x)$ but not the
coefficients $g_j$, then rather than
multiplying out the four factors to get the $g_j$ and then doing
the polynomial division, it is easier to simply
evaluate the alleged codeword polynomial $c(x)$
at the four roots of the generator polynomial (this is called computing the syndrome in
coding theory). If all four evaluations $c(a^i)$, $c(a^{i+1})$,
$c(a^{i+2})$, $c(a^{i+3})$
result in $0$, $c(x)$ is indeed a valid codeword polynomial;
if at at least one evaluation gives a nonzero result, $c(x)$
is not a valid codeword polynomial.
This way of checking can be adapted to determine whether the vector in question
is a codeword in any double-error-correcting Reed-Solomon code. Evaluate
the codeword polynomial at $a^i$ for all $i, 0 \leq i \leq 6$. This is
computing the finite-field discrete Fourier transform which you
mentioned. Then look for $4$ consecutive $0$ values in the sequence
$$c(1), c(a), c(a^2), c(a^3), c(a^4), c(a^5), c(a^6), c(1), c(a), c(a^2).$$
If there are $4$ consecutive $0$ values beginning with $c(a^i), 0 \leq i \leq 6$,
then $c(x)$ is a valid codeword polynomial in the double-error-correcting
Reed-Solomon code with generator polynomial
$$g(x) = (x-a^i)(x-a^{i+1})(x-a^{i+2})(x-a^{i+3}).$$
There may be more than one choice of $i$ for which this holds.
On the other hand, there might not be $4$ consecutive zeroes in which
case you have shown that
$c(x)$ is not a valid codeword polynomial in any double-error-correcting
Reed-Solomon code of length $7$ over the finite field of $8$ elements.
Finally, your question about creating a list of codewords has the
following answer. The codewords are all the multiples of $g(x)$.
So, write out all the $8^3$ polynomials $b(x)$ of degree $2$ or less
(remember that polynomials of degree $2$ have $3$ coefficients, and we
have $8$ choices for each coefficient. The codewords then are
the $8^3$ polynomials $b(x)g(x)$. An easier way is to create
three vectors of length $7$ corresponding to $g(x)$, $xg(x)$ and
$x^2g(x)$ respectively, and then take all possible linear
combinations $b_2(x^2g(x)) + b_1 (xg(x)) + b_0g(x)$. |
H: Convergence demonstration?
I got the following assigned as homework:
"Demonstrate that the following series are convergent:"
$$\sum_{k=0}^N a^k\\ \sum_{k=0}^\infty a^k \\ \sum_{k=0}^\infty ka^k \\\sum_{k=0}^\infty k(k-1)a^k\\ \sum_{k=0}^\infty k^2a^k$$
I know most of these converge when $|a| < 1$, but I'm not sure how I'm supposed to prove this, cause I don't think the ratio test applies with all the series? Especially the first one.
I apologize if this a really stupid question, cause it feels like one, but this is the first time I'm dealing with the convergence subject.
AI: The first sum converges because it contains finitely many finite terms. You can use the ratio test for proving the convergence of the others as follows:
\begin{align}
\lim_{k \rightarrow \infty} {\Big|}\frac{a^{k+1}}{a^k}{\Big|} &= \lim_{k \rightarrow \infty} |a| < 1 \\
\lim_{k \rightarrow \infty} {\Big|}\frac{(k+1)a^{k+1}}{ka^k}{\Big|} &= \lim_{k \rightarrow \infty} {\Big|}{\Big(}\frac{1}{k}+1{\Big)}a{\Big|} = |a| < 1 \\
\lim_{k \rightarrow \infty} {\Big|}\frac{(k+1)ka^{k+1}}{k(k-1)a^k}{\Big|} &= \lim_{k \rightarrow \infty} {\Big|}{\Big(}\frac{1+\frac{1}{k}}{1-\frac{1}{k}}{\Big)}a{\Big|} = |a| < 1 \\
\lim_{k \rightarrow \infty} {\Big|}\frac{(k+1)^2a^{k+1}}{k^2a^k}{\Big|} &= \lim_{k \rightarrow \infty} {\Big|}{\Big(}1+\frac{1}{k}{\Big)}^2a{\Big|} = |a| < 1 \\
\end{align}
In all the above cases, the ratio limit is $< 1$. Hence all series converge absolutely. |
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