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Dense Subspace: ONB This might be a duplicate. If so, then please let me know. Thanks!
Given a Hilbert space $\mathcal{H}$.
Consider a dense subspace $\overline{Z}=\mathcal{H}$.
Then it provides an ONB: $\mathcal{S}\subseteq Z$
(I guess it can be shown by slightly adjusting the usual proof via Zorn's lemma...)
|
I just realized that this cannot be true for the following reason:
Assume every dense subspace would provide an ONB for the Hilbert space:
$$\overline{\mathcal{S}}=\mathcal{H}\quad(\mathcal{S}\subseteq Z)$$
Then, it would serve as well as an ONB for the subspace:
$$\overline{\mathcal{S}}\supseteq Z$$
But there are preHilbert spaces which do not admit any ONB (see Bourbaki or Robert Isreal).
|
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|
Is the number 0.2343434343434.. rational? Consider the following number:
$$x=0.23434343434\dots$$
My question is whether this number is rational or irrational, and how can I make sure that a specific number is rational if it was written in decimal form.
Also, is $0.234$ rational or irrational?
|
We have $100x=23.434343\cdots$ and $10000x=2343.434343\cdots$ so
$$9900x=2320$$
hence $x$ is rational and
$$x=\frac{2320}{9900}=\frac{116}{495}$$
|
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|
Fallacy - where is the mistake? Could anyone help me to find the mistake in this fallacy? Because the actual result for $I$ is $\pi/2$
\begin{equation}
I = \int_{0}^{\pi} \cos^{2} x \; \textrm{d}x
\end{equation}
\begin{equation}
I = \int_{0}^{\pi} \cos x \cos x \; \textrm{d}x
\end{equation}
substitution:
\begin{equation}
\sin x = u
\end{equation}
\begin{equation}
\cos x \; \textrm{d}x = u
\end{equation}
\begin{equation}
\cos x = \cos (\arcsin x)
\end{equation}
and the limits:
\begin{align}
\begin{split}
x &= 0 \quad \Rightarrow \quad u=0 \\
x &= \pi \quad \Rightarrow \quad u=0
\end{split}
\end{align}
so
\begin{equation}
I = \int_{0}^{0} \cos (\arcsin u) \textrm{d}u
\end{equation}
\begin{equation}
I=0
\end{equation}
|
if $\sin(x)=u$ we get $x=\arcsin(u)$ and $dx=\frac{1}{\sqrt{1-u^2}}du$ thus we get
$\int \cos(x)^2dx=\int 1-\sin(x)^2dx=\int \sqrt{1-u^2}du$
|
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|
"Evaluated at" or "at" notation Normally a variable that is a function another variable would be represented as in the following fashion: $ V(t) $ (voltage as a function of time). However, my engineering professor (who also wrote the textbook) likes to use the vertical bar notation instead, so the flux at $ r = R_0 $ is $ N|_{r=R_0} $ instead of $ N(R_0) $ or $ N(r=R_0) $. Likewise, $ C|_z - C|_{z+\Delta z} $ and so on.
Personally, I would rather reserve the vertical bar for a more complex expression like evaluation of a derivative:
$$
\left.\frac{df}{dt} \right|_{t=4} = \left. t^2+3t\right|_{t=4}
$$
Which is the preferred notation? If it's just a matter of style, then what are the pros and cons of each option?
|
I've usually seen that notation mean "restricted to", which isn't too far from "evaluated at" in meaning, but it is more general. For instance,
$$
\left. f\right|_{[0,1]}
$$
could be interpreted to be a function with values that agrees with $f$'s values, but is only defined on the interval $[0,1]$.
One advantage is that you're not dependant on a function name in order to write down both the function and the evaluation point, as is witnessed by
$$
\left. t^2 +3t\right|_{t=4}
$$
Which with normal notation would be cumbersome to separate from, let's say
$$
\left. t^2 +12\right|_{t=4}
$$
One disadvantage is that you'd need to add brackets to really be able to tell where the expression begins.
Another one is clarity: it's not as much in use, and you would therefore have to explain it every time you use it. Composition of functions might also get messy if you get right down to it, with nested bars and subscripts. Compare $f(g(x))$ to
$$
f|_{g|_x}
$$
|
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|
Show a function for which $f(x + y) = f(x) + f(y) $ is continuous at zero if and only if it is continuous on $\mathbb R$
Suppose that $f: \mathbb R \to\mathbb R$ satisfies $f(x + y) = f(x) + f(y)$ for each real $x,y$.
Prove $f$ is continuous at $0$ if and only if $f$ is continuous on $\mathbb R$.
Proof: suppose $f$ is continuous at zero. Then let $R$ be an open interval containing zero. Then $f$ is continuous at zero if and only if $f(x) \to f(0)$ as $x \to 0$. Then $|f(x) - f(0)| < \epsilon$.
Can anyone please help me? I don't really know how to continue. Thank you.
|
First observe that $f(0) = f(0) + f(0)$, so that $f(0) = 0$.
Now suppose $f$ is continuous at $0$. Let $x \in \mathbb{R}, \epsilon > 0$. Let $\delta > 0$ be such that $|f(t)| < \epsilon$ whenever $|t| < \delta$.
If $|y - x| < \delta$, then setting $t = y -x$ we have $|f(y) - f(x)| = |f(t)| < \epsilon$, thus completing the proof.
Extra comments:
Of course we can prove much more than continuity if $f$ is continuous at zero: it must be of the form $f(x)= \alpha x$, for some $\alpha \in \mathbb{R}$.
There's a related fact (harder to prove, but not that hard): Suppose $f: \mathbb{R} \to \mathbb{R}$ satisfies $f(x+y) = f(x) + f(y)$, and suppose $f$ is not continuous. Then the graph of $f$, $\Gamma(f) = \{(x,f(x)) | x \in \mathbb{R}\}$, is dense in $\mathbb{R}^2$.
|
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|
How to show $P^1\times P^1$ (as projective variety by Segre embedding) is not isomorphic to $P^2$? I am a beginner.
This is an exercise from Hartshorne Chapter 1, 4.5. By his hint, it seems this can be argued that there are two curves in image of Segre embedding that do not intersect with each other while in $P^2$ any two curves intersect.
I feel this solution is very special. I would like to know more.
Is there any invariant to detect whether two birational equivalent varieties are iso or not?
Thanks!
|
The canonical bundles of these varieties both only have the zero section, so these sections cannot be used to prove the non-isomorphism of the varieties.
The anticanonical bundles $K^\ast=\Lambda^2T$ of these varieties however have spaces of global sections of different dimensions $h^0$ over the base field and prove that non-isomorphism:
$$h^0(\mathbb P^2, K^\ast_{\mathbb P^2})=h^0(\mathbb P^2,\mathcal O_{\mathbb P^2}(3))=10$$ whereas
$$h^0(\mathbb P^1\times \mathbb P^1, K^\ast_{{\mathbb P^1\times \mathbb P^1}})=h^0(\mathbb P^1\times \mathbb P^1,\mathcal O_{\mathbb P^1 }(2)\boxtimes \mathcal O_{\mathbb P^1 }(2))=9$$ (Whew, that was close: 10 and 9 !)
Edit: notation
Given the projections $p,q:\mathbb P^1\times \mathbb P^1\to \mathbb P^1$ of $\mathbb P^1\times \mathbb P^1$ onto its two factors, the tensor product $p^\ast\mathcal O_{\mathbb P^1}(a)\otimes_{\mathcal O_{\mathbb P^2}} q^\ast\mathcal O_{\mathbb P^1}(b)$ of the pull-backs $p^\ast\mathcal O_{\mathbb P^1}(a)$ and $q^\ast\mathcal O_{\mathbb P^1}(b)$ is denoted by $\mathcal O_{\mathbb P^1}(a)\boxtimes \mathcal O_{\mathbb P^1}(b)$ or simply by $\mathcal O_{\mathbb P^2}(a,b)$.
This convention is used in other answers too.
|
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|
Solving for n in the exponent. Well, it's another question I feel like I should know. I'm trying to model the number of successes before the first failure. The probability of successes is given as $p$, which makes the probability of failure $(1-p)$.
The probability mass function, as I've calculated it, turns out to be $p^{n-1}(1-p)$, since we will stop at the first failure.
I'm trying to solve the following equation for n, but I'm at a loss for how to get it out of the exponent.
$1=p^{n-1}(1-p)$
I would appreciate any help anyone can give me.
Thanks
|
If $X$ is the number of successes (not trials) until the first failure, then $\Pr(X=n)=p^n(1-p)$ for $n=0,1,2,\dots$.
This cannot ever be equal to $1$ if $p\ne 0$.
If you want to solve $p^n(1-p)=a$, given $0\lt p\lt 1$, rewrite the equation as
$p^n=\frac{a}{1-p}$.
and take the logarithm of both sides.
Remark: As was pointed out in the answer, except in the trivial case we cannot have $\Pr(X=n)=1$. Perhaps you want to show that
$$\sum_0^\infty p^n(1-p)=1.$$
Let $0\lt p\lt 1$. Then the above sum is an infinite geometric series, and by the usuual formula it does sum to $1$. We do not even have to compute, since $\sum_0^\infty \Pr(X=n)$ must be $1$.
|
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|
Convergence of the series $\sum\limits_{n=1}^{\infty}(-1)^{n}\frac{\ln(n)}{\sqrt{n}}$ I would like to see whether or not $$\sum\limits_{n=1}^{\infty}(-1)^{n}\dfrac{\ln(n)}{\sqrt{n}}$$ is a convergent series.
Root test and ratio test are both inconclusive. I tried the alternating series test after altering the form of the series:
$$\sum\limits_{n=1}^{\infty}(-1)^{n-1}\left[\dfrac{-\ln(n)}{\sqrt{n}}\right]\text{.}$$
After using L-Hospital, it's clear that $\lim\limits_{n \to \infty}\left[\dfrac{-\ln(n)}{\sqrt{n}}\right] = 0$. To show that it's decreasing led to me finding the derivative $\dfrac{\ln(n)}{2n^{3/2}}-\dfrac{1}{n^{3/2}}$, which I could set to be less than $0$, but a plot has shown that $n < e^{2}$ is not where $\dfrac{-\ln(n)}{\sqrt{n}}$ is decreasing.
So all that remains is a comparison test. I can't think of a clever comparison to use for this case. Any ideas?
|
$\dfrac{\ln(n)}{\sqrt{n}}$ is mono-tonic decreasing after $n=\lceil e^2 \rceil$ and remains bounded between $n=1$ to $\lfloor e^2 \rfloor$, so from alternating series test $\sum\limits_{n=1}^{\infty}(-1)^{n}\dfrac{\ln(n)}{\sqrt{n}}$, must converge.
|
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|
Group theory: subset of a finite group Given
*
*$G$ be a finite group
*$X$ is a subset of group $G$
*$|X| > \frac{|G|}{2}$
I noticed that any element in $G$ can be expressed as the product of 2 elements in $X$. Is there a valid way to prove this？
If the third condition was $|X| = \frac{|G|}{2}$ instead, does the above statement still hold?
Thank you.
|
Answer for the additional question: the statement need not be true if $|X|=|G|/2$.
For example, let $G$ be the cyclic group of order $2$ and $X$ consist of the non-identity element.
A (slightly) more general example: let $G$ be the cyclic group of order $2n$, let $g$ be a generator, and let
$$X=\{g^{2k}\mid k=0,1,\ldots,n-1\}\ .$$
And a bit more general again: if $X$ is a proper subgroup (not just a subset) of $G$ then the product of elements of $X$ is still in $X$, and hence not all elements of $G$ will be obtained.
|
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|
Change of variable in double integrals I need help to solve the following question(s).
a) Evaulate the integral
$$\iint_D (x-y) \, dx \, dy,$$
where $D$ is the triangle with vertices $(0,0)$, $(-1,1)$ och $(4,2)$.
b) Evaulate the integral
$$ \iint_D (y-x) dx \, dy,$$
where $D$ is the triangle with vertices $(0,0)$, $(4,1)$ och $(2,2)$.
My attempt
a)
Actually, I am not very sure where to correctly start. A suggestion would be to first find out that the lines (i.e. the lines of the triangle) can be described as
$y=-x \Leftrightarrow y-x=0$,
$y=x/5+6/5 \Leftrightarrow y-x/5=6/5$,
$y=x/2 \Leftrightarrow y - x/2 = 0$.
I suppose the change of variables should be found out in light of this. But I dont know how to continue. According to the (very short) solution, the substitution is supposed to be
$u = x+y$,
$v = x−2y$,
but this makes no sense to me.
|
So I may have found an method. Please comment if you have any suggestions of how it can be improved. So my remaining question is why the other change of variables (i.e. the change I mentioned in my first post) differ slightly from the following change of variables.
a) Evaulate the integral
$ \int \int_D (x-y) \, dx \, dy,$
where D is the triangle with vertices (0,0), (-1,1) och (4,2).
Solution
Let
$$
\left(\begin{array}{r}
x
\\
y
\end{array}\right)=
\left(\begin{array}{rr}
4 & -1
\\
2 & 1
\end{array}\right)
\left(\begin{array}{r}
u
\\
v
\end{array}\right) =
\left(\begin{array}{r}
4u-v
\\
2u+v
\end{array}\right),
$$
which is equivalent to
$$
\left(\begin{array}{r}
u
\\
v
\end{array}\right)=
\left(\begin{array}{r}
\frac{1}{6}(x+y)
\\
\frac{1}{3}(2y-x)
\end{array}\right).
$$
Plugging in the vertices gives us
$$(x,y)=(0,0) \Leftrightarrow (u,v)=(0,0), $$
$$(x,y)=(4,2) \Leftrightarrow (u,v)=(1,0), $$
$$(x,y)=(-1,1) \Leftrightarrow (u,v)=(0,1). $$
Jacobian $\ldots$
$$ \frac{d(x,y)}{d(u,v)} = 6 \Rightarrow dx \, dy = |6| \, du\, dv =6 \, du\, dv $$
Finally $\ldots$
$$\iint_D (x-y) \, dx \, dy = 6 \int^1_0 \int^{1-u}_0 (2v-2u) \, du\, dv = 0.$$
|
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Calculating the hitting probability using the strong markov property ** This problem is from Markov Chains by Norris, exercise 1.5.4.**
A random sequence of non-negative integers $(F)n)_{n\ge0}$ is obtained by setting $F_0=0$ and $F_1=1$ and, once $F_0,\ldots,F_n$ are known, taking $F_{n+1}$ to be either the sum or the difference of $F_{n-1}$ and $F_n$, each with the probability $1/2$. Is $(F_n)_{n\ge0}$ a Markov chain?
(a) By considering the Markov chain $X_n=(F_{n-1},F_n)$, find the probability that $(F_n)_{n\ge0}$ reaches $3$ before first returning to $0$.
(b) Draw enough of the flow diagram for $(X_n)_{n\ge0}$ to establish a general pattern. Hence, using the strong Markov property, show that the hitting probability for $(1,1)$, starting from $(1,2)$, is $(3-\sqrt{5})/2$.
(c) Deduce that $(X_n)_{n\ge0}$ is transient. Show that, moreover, with probability $1$, $F_n \rightarrow \infty$ as $n \rightarrow \infty$.
My attempt for (b): From $(1,2)$ the chain looks like $(1,2)\rightarrow (2,3)$ or $(1,2)\rightarrow (2,1)$ each with probability 1/2. From $(2,1)$ we can reach $(1,1)$. I want to calculate the probability generating function using the strong markov property $\phi(s)=\mathbb{E}_{(1,2)}(s^{H_{(1,2)}^{(1,1)}})$ where $H_{(1,2)}^{(1,1)}=\inf \{n\geq 0\colon X_n=(1,1) \text{ starting from } (1,2)\}$. I thought that if we start in $(2,3)$ and we want to reach $(1,1)$ we at least have to go trough $(1,2)$ again and then from $(1,2)$ to $(1,1)$. So I believe that $\mathbb{E}_{(2,3)}(s^{H_{(2,3)}^{(1,1)}})=\phi(s)^2$, but I am not sure if this true
I really need help with this exercise.
Thank you.
|
I also stumbled upon this problem. I did not solve (a) and (c) so far, but for you specific questions I have the answer.
If drawing the flow diagram of the Markov chain up to 4 steps from $(1,2)$ (i.e., about $2^4$ states can be reached), one notices, that one step into the wrong direction (e.g., from $(1,2)$ to $(2,3)$ requires 2 steps back.
An example: $(1,2) \rightarrow (2,3)$ requires 2 steps $(2,3) \rightarrow (3,1) \rightarrow (1,2)$ to get back to the initial position. This holds true for any state. With this observation in mind, we can write down the hitting probability of $(1,1)$ starting from $(1,2)$ denoted as $h_{(1,2)}^{(1,1)} := h_{(1,2)}$:
$$h_{(1,2)} = \frac{1}{2} h_{(2,3)} + \frac{1}{2} h_{(2,1)},$$
and
$$h_{(2,3)} = h_{(2,3)}^{(1,2)} h_{(1,2)} = h_{(1,2)} h_{(1,2)} = h_{(1,2)}^2,$$
because $h_{(2,3)}^{(1,2)}$ has the same distribution as $h_{(1,2)}$. A similar argument can be used to derive
$$h_{(2,1)} = \frac{1}{2-h_{(1,2)}}.$$
Putting this together, we end up at
$$0 = h_{(1,2)}^3 + 4h_{(1,2)}^2 - 4h_{(1,2)} + 1.$$
In this case, the only valid solution to this equation is $h_{(1,2)}=\frac{3-\sqrt{5}}{2}$.
|
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Solving Pythagoras Problem An aircraft hangar is semi-cylindrical, with diameter 40m and length 50 m. A helicopter places an inelastic rope across the top of the hangar and one end is pinned to a corner, a A. The rope is then pulled tight and pinned at the opposite corner, B. Determine the lenghth of the rope.
So, first I find the diagonal line from A straight to B.
c^2=50^2+40^2
The answer is 64.03124......
Then, I find out that there is a semi-circle shape. So I find the length of the arc.
r=32.05162.......
Length=2*Pi*32.05162.....
=201.16008m......
But the correct answer is 80.30m
Can anyone tell me where i did wrong?
|
This question is a simple Pythagoras problem.
First, we find the length of the arc whose diameter is 40m.
Then we assume it to be a side of the quadrilateral whose side is 50 m.
Next, we use Pythagoras theorem to find the diagonal.
The answer will be near to 80.34..
|
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Can you cancel out a term if equal to zero? quick question here:
In my proofs class we had a problem that after a little work we end up with:
$x(x-y)=(x+y)(x-y)$ where $ x = y $. Now, I know this is pretty basic, but my teacher said that for the next step, one cannot cancel out $(x-y)$ from both sides as $(x-y) = 0 $.
Can someone explain the logic and/or the reasoning behind this?
I'm pretty sure this falls under some obscure basic algebra rule that I've forgotten over the years but I cannot find anything about this on the internet.
Edit:
To clear up some confusion here, I am not looking for how to solve this problem, but rather the why this particular rule is so.
The problem I am working on gives a proof. I am supposed to mark the errors in the proof. For this problem, the error was that they cancelled out $(x-y)$ and I am trying to understand why that's an error.
|
There is no such thing as "cancelling out".
In this case you want to divide both sides of the equation by (x-y). But always remember before dividing: you can't divide by zero. But if x=y then in fact (x-y) = 0 and so you would in fact divide by zero which you are not allowed to. But only professors are allowed to divide by zero. :-) Just kidding.
|
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Twisted logarithm power series I recently encountered a power series similar to the one of the $\log(1-x)$ of the form
$$
F(x)= \sum_{n=1}^\infty \frac{\psi(n)x^n}{n},
$$
where $\psi$ is some Dirichlet character. Has anyone here seen a function like this? Here are some observations I have made:
1) The radius of convergence is the same as for $\log(1-x)$, so the power series converges for $|x| < 1$.
2) If $\psi(n)$ is the trivial characer mod $N$ then $F(x) = \sum_{n=1}^\infty \frac{x^n}{n}-\frac{x^{Nn}}{Nn}=-\log\left(\frac{1-x}{1-x^N}\right)$.
The next interesting case would be when $\psi$ is a quadratic character. I'd be happy about any reference or further observation on these functions.
|
Such a function is actually a combination of logarithms. Suppose $\psi$ is a Dirichlet character mod $N$, and let $\zeta$ be a primitive $N$th root of unity. The functions $a \mapsto \zeta^{ka}$, for $k=0, ..., N-1$, form a basis of $L^2(\mathbb Z/N\mathbb Z, \mathbb C)$, hence there exist numbers $b_k \in \mathbb C$ such that
$$\psi(n) = \sum_{k=0}^{N-1} b_k \zeta^{kn}$$
for all $n\in \mathbb Z/N\mathbb Z$. The numbers $b_k$ are essentially Gauss sums - the Fourier coefficients of $\psi$. Then your function is just
$$-\sum_{k=0}^{N-1} b_k \log(1-\zeta^kx).$$
Remark that we don't even need the fact that $\psi$ is a character - it could be any periodic function with a period of $N$, and the "twisted" logarithm would still decompose as above, with different $b_k$'s.
It is an interesting idea, nevertheless. For instance, the value at $x=1$ of your function is $L(\psi, 1)$, a number of significant arithmetic interest.
|
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closed but not exact I saw several times that $\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$ is closed but not exact. Closed, is obvious but I can't prove non exactness, can one please help me ?
My attempt, let $f\in \omega^{0}(U)$ and $df=\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y} dy$ and so $\frac{\partial f}{\partial x}(x,y)=\frac{-y}{x^2+y^2},\frac{\partial f}{\partial y}(x,y)=\frac{x}{x^2+y^2}$. Now let $g(\theta)=f(\cos \theta,\sin \theta)$ so $g'(\theta)=1 \implies g(\theta)=\theta+C$. Now i'm not getting any contradiction.
|
You know that if you integrate over a loop into which there is no singularity, as the form is exact the integral is 0.
Then let us try with a loop containing 0 inside: let us take the circle $C(0,1)$.
$$\int_{C(0,1)} -\frac y{x^2 + y^2}dx + \frac x{x^2 + y^2}dy
= \int_0^{2\pi} -\frac {r\sin\theta}{r^2} (-r \sin\theta d\theta)
+ {r\cos\theta}{r^2} (r \cos\theta d\theta)\\ = 2\pi \neq 0
$$
|
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|
Beautiful little geometry problem about sines Given triangles ABC and $A_1B_1C_1$ such that $\sin A = \cos A_1, \sin B = \cos B_1, \sin C = \cos C_1$. What are the possible values for the biggest of these 6 angles?
I tried some stuff like sine theorem but can't derive from it? How do we do this one? Assume the closest to the smallest angle?
|
Short answer: $\frac{3\pi}{4}$.
Detailed answer: $A_1,B_1,C_1<\frac\pi2$. The equality
$$\sin A=\cos A_1$$
gives
$$A=\frac\pi2-A_1\text{ or }A=\frac\pi2+A_1,$$
and similarly for $B$ and $C$. Working out the cases, the only possibility is
$$A=\frac\pi2+A_1,B=\frac\pi2-B_1,C=\frac\pi2-C_1.$$
Since $A+B+C=A_1+B_1+C_1=\pi$, $$A=A_1+\frac\pi2=B_1+C_1=\pi-A_1.$$
|
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Projective representaions of $(\mathbb{Z}/3\mathbb{Z})^2$ I have a very short question: is there a faithful projective representaion
$\rho: \mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/3\mathbb{Z}\to {\rm PGL}(4,\mathbb R)$?
Thanks!
|
Following Derek's hint, we first pick a faithful representation $\mathbb{Z}/3\mathbb{Z} \rightarrow GL(2, \mathbb{R})$.
Send $1$ to
$$\left(\begin{array}{cc}
1 & -1\\
1 & 0 \end{array}\right)$$
You can check that this is an element of order 3, so it gives us a representation. It's also obviously faithful.
Then, define a representation of $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ by sending the first generator to the above matrix in the first $2 \times 2$ block and the second generator to the above matrix in the second $2 \times 2$ block. This gives us a faithful representation of the group in which no element is mapped to a scalar matrix so it remains faithful after taking quotients.
|
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|
Inversion of n x n matrix A matrix F is given:
$$
F = [e^{i\frac{2\pi kl}{n}}]_{k,l=0}^{n-1}
$$
Find $$ F^{-1} $$
I know Gaussian method for inverting matrices but I suppose it doesn't apply to matrices with not given exact n value. Could you tell me what are the methods for inverting matrices like this?
|
HINT:
This is a famous matrix, $1/\sqrt{n}\cdot F$ is unitary. One can check this directly or look at
http://en.wikipedia.org/wiki/Discrete_Fourier_transform#Orthogonality
|
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|
Integrating algebraic functions The function $y = f(x)$, restricted on the domain $ 0 < x < 1$ and satisfying
$$y^{5}+y^{4} + x = 0,$$
seems to be well-defined and smooth. So how does one integrate this thing? That is, what is $\int_{0}^{1} f(x) dx$?
Of course, one can use Newton's method to approximate $f(x)$ for any given value of $x$ and numerically integrate the result. But this feels uninspiring. So I was wondering if there is a general "trick" or insight to integrating algebraic functions.
|
As MPW says, in general there is no reason to expect that the integral should be easy to compute. However, you can do the following. Differentiate the equation once to obtain $$0=5y'y^4+4y'y^3+1=y'y^3(y+4(y+1))+1. $$ Then use the equation to write $$ y+1 = -\frac{x}{y^4}$$ and plug this into the above to get $$-1= y'y^3(y+4(-\frac{x}{y^4}))=\frac{y'y^5-4xy'}{y},$$ or $$ y=4xy'-y'y^5.$$ Now you're in business: $$ \int_0^1 yd x=\int_0^1(4xy'-y'y^5)d x=4y(1)-4\int_0^1yd x -\frac{1}{6} y(1)^6.$$ Solving for the desired quantity you get $$ \int_0^1 yd x=-\frac{1}{30}y(1)^6+\frac{4}{5}y(1). $$
|
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|
Without actually calculating the value of cubes find the value of $(1)^3+(2)^3+2(4)^3+(-5)^3+(-6)^3$. Also write the identity used Without actually calculating the value of cubes find the value of $(1)^3+(2)^3+2(4)^3+(-5)^3+(-6)^3$.
Also write the identity used
|
You can use the identity that if $a + b + c=0$, then $a^3 + b^3 + c^3 = 3abc$. So we can say that:
*
*$2^3 + 4^3 + (-6)^3 = 3\cdot 2\cdot 4\cdot -6 = -144$.
*$1^3 + 4^3 + (-5)^3 = 3\cdot 1\cdot 4 \cdot -5 = -60$.
Adding both, we get:
$1^3 + 2^3 + 2 (4)^3 + (-5)^3 + (-6)^3 = -204$
|
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Real Analysis alpha Holder condition Holder condition
A function $f:(a,b)\rightarrow R$ satisfies a Holder condition of $\alpha$ order if $\alpha > 0$, and for some constant $H$ and for all $u,x \in (a,b)$,
$$|f(u)-f(x)| \leq H|u-x|^\alpha.$$
Prove that an $\alpha$ - Holder function defined on $(a,b)$ is uniformly continuous and infer that it extends uniquely to a continuous function defined on $[a,b]$. Is the extended function $\alpha$ - Holder?
I understand how to show uniform continuity. Just choose $\delta = ( \frac{\epsilon}{H})^ {1/ \alpha}$ and that works out nicely , but I don't understand extending uniquely to a continuous function. Any help with the second part? Thanks
|
Take a Cauchy sequence that converges to $a$, say $\{a_n\} \subset (a,b)$. By uniform continuity (or Holder continuity if you prefer), $\{f(a_n)\}$ is also a Cauchy sequence, hence it has a limit, say $f_a$.
It is easy to prove that this limit does not depend on the Cauchy sequence $\{a_n\}$, hence we can say that $\lim_{x \to a^+}f(x)$ exists and equals $f_a$.
Define your function on $[a,b)$ by setting $f(a) = f_a$. (This also implies that your function is continuous at $a$.)
Do the same for $b$ :)
|
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Fejer's theorem with Riemann integrable function
If $f$ is integrable and $f(x+), f(x-)$ exists for some $x$, then
$$
\lim_{N \rightarrow \infty} {\frac{1}{{2\pi }}\int_{ - \pi }^\pi {f\left( {x - t} \right){K_N}\left( t \right)dt} } = \frac{1}{2}[f(x+) + f(x-)]
$$ where $K_N(t)$ is Fejer kernel.
Here is my naive try:
First observe the difference
\begin{array}{l}
\left| {\frac{1}{{2\pi }}\int_{ - \pi }^\pi {f\left( {x - t} \right){K_N}\left( t \right)dt} - \frac{1}{2}[f(x + ) + f(x - )]} \right|\\
= \left| {\frac{1}{{2\pi }}\int_{ - \pi }^\pi {f\left( {x - t} \right){K_N}\left( t \right)dt} - \frac{1}{2}[f(x + ) + f(x - )]\frac{1}{{2\pi }}\int_{ - \pi }^\pi {{K_N}\left( t \right)dt} } \right|\\
= \frac{1}{{2\pi }}\left| {\int_{ - \pi }^\pi {\left[ {f\left( {x - t} \right) - \frac{1}{2}[f(x + ) + f(x - )]} \right]{K_N}\left( t \right)dt} } \right|\\
\le \frac{1}{{2\pi }}\int_{ - \pi }^\pi {\left| {\left[ {f\left( {x - t} \right) - \frac{1}{2}[f(x + ) + f(x - )]} \right]{K_N}\left( t \right)} \right|dt}
\end{array}
Then I stuck to say more words about my approach...
I think if I can rewrite $f(x+)$ and $f(x-)$ in terms of $f(x-t)$ maybe I can pursue further..
Thank you
|
The Fejer kernel $K_{N}(t)$ has several nice properties:
*
*$K_{N}(t) \ge 0$
*$K_{N}(t)=K_{N}(-t)$
*$K_{N}(t+2\pi)=K_{N}(t)$
*$\int_{0}^{\pi}K_{N}(t)\,dt = \pi$
*$K_{N}(t)$ tends uniformly to $0$ for $0 < \delta \le |t| \le \pi$ as $N\rightarrow\infty$.
Those are the only properties that you need. For example,
$$
\left|\frac{1}{2\pi}\int_{-\pi}^{0}K_{N}(t)f(x-t)\,dt
- \frac{1}{2}f(x+0)\right| \\
= \frac{1}{2\pi}\left|\int_{-\pi}^{0}K_{N}(t)(f(x-t)-f(x+0))\,dt\right| \\
\le \frac{1}{2\pi}\int_{-\pi}^{-\delta}K_{N}(t)|f(x-t)-f(x+0))|\,dt
+ \frac{1}{2\pi}\int_{-\delta}^{0}K_{N}(t)|f(x-t)-f(x+0)|\,dt
$$
The first term on the right is bounded by
$$ \frac{1}{2\pi}\left(\sup_{t\in[-\pi,-\delta]}K_{N}(t)\right)\left(\int_{-\pi}^{\pi}|f(t)|\,dt+\frac{1}{2}|f(x+0)|\right),
$$
a term which tends to $0$ as $N\rightarrow\infty$ because of property (5). The second term on the right is bounded by
$$
\frac{1}{2\pi}\int_{-\pi}^{0}K_{N}(t)\,dt \sup_{t\in[x,x+\delta]}|f(t)-f(x+0)| \le \frac{1}{2}\sup_{t\in[x,x+\delta]}|f(t)-f(x+0)|
$$
Therefore, for $\epsilon > 0$, there exists $\delta > 0$ such that the term on the right is bounded by $\epsilon/2$, assuming that $\lim_{t\downarrow 0}f(x+t)=f(x+0)$ exists. Then, for that fixed $\delta$, there exists $N_{0}$ large enough that $N \ge N_{0}$ guarantees that the previous term is bounded by $\epsilon/2$. Therefore, for $N \ge N_{0}$,
$$
\left|\frac{1}{2\pi}\int_{-\pi}^{0}K_{N}(t)f(x-t)\,dt
- \frac{1}{2}f(x+0)\right| < \epsilon.
$$
By definition of the limit,
$$
\lim_{N\rightarrow\infty}\frac{1}{2\pi}\int_{-\pi}^{0}K_{N}(t)f(x-t)\,dt=f(x+0).
$$
The other half is handled in a similar manner.
|
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|
Triangulation of 4 points why Delaunay maximizes the minimum? I have been going through the chapter on Delaunay triangulations from the book by DeBerg (http://www.cs.uu.nl/geobook/interpolation.pdf). In lemma 9.4, he simply says that "from Thales theorem" we can show that of the two possible triangulations of four points (depending on which diagonal you choose to join) the one where we cut through the angles that sum to less than 180 is "illegal" meaning the angles don't satisfy the maximin criterion (meaning of the six resulting angles of the triangulation, the minimum is not as high as it would have been if we had joined the other diagonal). I don't really see how one can get this from Thales theorem. Does any one have an outline for a proof as to why this might hold?
|
I guess that by “Thales' theorem” he's essentially referring to the inscribed angle theorem, probably in the form of Theorem 9.2 from that chapter.
Consider circles through $p_i$ and $p_k$. Concentrate on the arc above that line. Any point on such a circle will form the same angle with $p_i$ and $p_k$. The centers of all these circles will lie on the perpendicular bisector between $p_i$ and $p_k$. As the center of a circle moves up, the angle becomes smaller, while increasing angle corresponds to lower center. My point is this: in order to maximize an angle, you want to move the center of the corresponding circle as far down as you can.
So from all the circles I just described, there are two of special importance, namely the one through $p_j$ and through $p_l$. Of these two, the one through $p_l$ is the one which has the lower center position, so it corresponds to a larger angle.
$$\angle p_kp_lp_i > \angle p_kp_jp_i$$
So $\triangle p_kp_lp_i$ is the combination you want if you maximize the angle opposite side $p_ip_k$.
So far, this is comparing just two angles. Couldn't it be the case that switching from edge $p_ip_j$ to edge $p_kp_l$ decreases some other angle instead? The answer is no:
\begin{align*}
\angle p_kp_ip_l = \angle p_kp_ip_j + \angle p_jp_ip_l &> \angle p_kp_ip_j \\
\angle p_lp_jp_k = \angle p_lp_jp_i + \angle p_ip_jp_k &> \angle p_ip_jp_k
\end{align*}
So for both the other angles you're adding something, therefore they will increase not decrease. This is due to the quadrilateral being convex. The only angle which actually decreases is the angle at $p_k$, since that will get divided into two. To see that this is still an improvement, you'd do the whole consideration above, but this time for edge $p_lp_i$ instead of $p_ip_k$.
|
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feedback on my solution (improper integral) i have done this improper integral but i am not sure if i have followed the correct procedure or my answer is correct. Please help!
|
Your answer appears to be correct, though how you got there is unclear. Where you split the integral in two is technically correct, though it is unclear what purpose you felt that may have served. From then on, it's fine until you try to plug in, at which point you should have
$$5\times0^{1/5}-5\times(-1)^{1/5}+5\times32^{1/5}-5\times0^{1/5}$$
The first and last terms are both zero, though they would have both cancelled out regardless of where you split the integral. The remaining terms are equal to $5+10=15$. However, what you have written down is equal to $-5+10=5$. Again, it's unclear how you dropped the minus sign and still obtained positive $5$.
|
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Minimal $T_0$-topologies Let $X$ be an infinite set and let $\tau$ be a $T_0$-topology on $X$. Does $\tau$ contain a $T_0$-topology that is minimal with respect to $\subseteq$?
|
Larson’s example in the paper cited by Tomek Kania and its verification are simple enough to be worth giving here (in very slightly modified form) for easy reference.
Let $\tau$ be the cofinite topology on an uncountable set $X$, let $\tau_0\subseteq\tau$ be a $T_0$ topology on $X$, and let $\tau_0^*=\tau_0\setminus\{\varnothing\}$. For each $U\in\tau_0^*$ let $\tau_U=\{W\in\tau_0:U\subseteq W\text{ or }W\subseteq U\}$; clearly $\tau_U$ is a $T_0$ topology on $X$. If $\tau_0$ is minimal $T_0$, then $\tau_U=\tau_0$ for each $U\in\tau_0^*$, and $\langle\tau_0^*,\subseteq\rangle$ is a chain. But then $\langle\{X\setminus U:U\in\tau_0^*\},\subseteq\rangle$ is an uncountable chain of finite sets, which is absurd. Thus, $\tau$ contains no minimal $T_0$ topology.
|
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Solve $x^3 - x + 1 = 0$ Solve $x^3 - x + 1 = 0$, this cannot be done through elementary methods.
Although, this is way out of my capabilities, I would love to see a solution (closed form only).
Thanks!
|
Let $x=u+v$ and note that $(u+v)^3=3uv(u+v)+(u^3+v^3)$
This has the required form if $3uv=1$ and $u^3+v^3=-1$
So $u^3v^3=\frac 1{27}$, and $u^3, v^3$ are roots of $y^2+y+\frac 1{27}=0$
$u,v= \sqrt[3] {\frac {-1\pm \sqrt{1-\frac 4{27}}}2 }$
$x=u+v=\sqrt[3] {\frac {-1+ \sqrt{\frac {23}{27}}}2 }+\sqrt[3] {\frac {-1- \sqrt{\frac {23}{27}}}2 }$
|
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Question about $e^x$ Let $ p(x)=1+x+x^2/2!+x^3/3!+....+x^n/n!$ where $n$ is a large positive integer.Can it be concluded that $\lim_{x\rightarrow \infty }e^x/p(x)=1$?
|
No.
$$\frac{e^x}{p(x)}=1+\frac{\sum\limits_{k=n+1}^\infty \frac {x^k}{k!}}{p(x)}\stackrel{x\to\infty}\to\infty$$
|
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Solving a Extended Euclidean Algorithm related problem Alex has some (say, n) marbles (small glass balls) and he has going to buy some boxes to store them. The boxes are of two types:
Type 1: each box costs c1 Taka and can hold exactly n1 marbles
Type 2: each box costs c2 Taka and can hold exactly n2 marbles
He wants each of the used boxes to be filled to its capacity and also to minimize the total cost of buying them. Since he finds it difficult to figure out how to distribute his marbles among the boxes he seeks his friends' help.It is worthy of mention that the value of n,c1,n1,c2 and n2 are given.
His one friend solved this problem using extended euclidian algorithm and found out two values everytime .Thus his result for the input
1.n=43,c1=1,n1=3,c2=2,n2=4 the result will be 13 1
2.n=40,c1=5,n1=9,c2=5,n2=12,the result will be "no valid values".
Though I understand extended euclidian algorithm I can not catch how he solved this problem.Can anyone tell me how his friend sovled it using extended euclidian algorithm.Since I'm a novice learner of extended euclidian algorithm I need better explanation.
|
The following does not answer absolutely your question, but I think that it may help.
Lemma: Let $a$ and $b$ be coprime integers greater than $1$.
*
*The equation $ax+by=n$ has nonnegative integer solution for $n\geq ab-a-b+1$
*The equation $ax+by=ab-a-b$ has no nonnegative integer solution.
Proof: Suppose that $a>b$. First, note that $n-(b-1)a\geq ab-a-b+1-ab+a=1-b$
The set $\{n, n-a, n-2a\ldots, n-(b-1)a\}$ has $b$ elements that are
pairwise different mod $b$. Then, there is one of them, say $n-ax$,
that is a multiple of $b$. Then we can write that $ax+by=n$, but we
have to prove that $y$ is nonnegative. It suffices to show that
$n-ax\geq 0$.
If $x<b-1$ then
$$n-ax\geq n-a(b-2)\geq ab-a-b+1-ab+2a=a-b+1\geq 0$$
If $x=b-1$ then $n-ab+a$ is a multiple of $b$, that is, $n+a$ is also
a multiple of $b$. But $n+a\geq ab-b+1$ and the next multiple of $b$
is $ab$, so $n+a\geq ab$, that is $n\geq ab-a$. Then
$$n-ax=n-a(b-1)\geq ab-a-ab+a\geq 0$$
This proves 1.
To prove that $ax+by=ab-a-b$ has no nonnegative solution, just note
that $x=b-1$, $y=-1$ is an integer solution. Other solutions of this
diophantine equation are given by $$x=b-1+kb, y=-1-ka$$ where $k$ is
an integer. But if $x\geq 0$ then $b-1+kb\geq 0$ or $k\geq\frac1b-1$,
that is, $k\geq 0$ and $y\leq-1$. This proves that no solution has
both values nonnegative, as stated in $2$.
|
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Find series expansion of 1/cosx Find the series expansion of 1/cosx from basic series expansions.
I tried to find 1/cosx from the expansion of cosx but was unsure how to continue. When I found 1/cosx from the basic formula for finding series expansions I didn't get the same answer.
|
(going to the fifth term for an example purpose)
Using the basic expansions of cos(x) gives us
$$
\frac{1}{\cos(x)} = \frac{1}{1-\frac{x^2}{2}+\frac{x^4}{24} + \cdots}
$$
of the form $ \frac{1}{1-X} $ which has a known and easy expansion :
$$
1+X+X^2+X^3+X^4+X^5+\cdots
$$
where $ X = \frac{x^2}{2} + \frac{x^4}{24} $ (approching $0$ when $x$ approaches $0$). The smaller $x$ term is $x^2$, so we don't need to take more terms than $X^2$ in the above expansion (otherwise terms would exceed $x^5$ and be negligeable).
Hence,
$$
\frac{1}{\cos(x)} = 1+(\frac{x^2}{2}−\frac{x^4}{24})+(\frac{x^2}{2})^2+o_{x\to0}(x^5)
$$
$$
= 1+\frac{x^2}{2}+\frac{5x^4}{24}
$$
|
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Find this maximum of this $\frac{\int_{0}^{\pi}f(x) \, dx}{\int_{0}^{\pi} f(x)\sin x\,dx}$ Question:
Assmue that $\int_0^\pi f(x)\,dx$ and $\int_0^\pi f(x)\sin x\,dx$ is convergence,and $f(x)>0,\forall x\in(0,\pi)$ Find this maximum as possible for all function $f$
$$I=\dfrac{\int_0^\pi f(x)\,dx}{\int_0^\pi f(x)\sin{x}\,dx}$$
show that：
$$I\le\dfrac{4}{\pi}?$$
I think this problem is interesting,But I can't.
|
Hint: let
$$f_n(x):= n\chi_{(0,1/n)} + n^{-1}\chi_{[1/n,\pi]}, $$
with $\chi$ denoting the indicator (i.e. for any set $E$, $\chi_E=1$ on $E$, $\chi_E=0$ outside $E$).
You get
$$\int_0^\pi f_n(x)dx = 1+ (\pi-1/n)/n ,$$
while (for large $n$)
$$\int_0^\pi |f_n(x)\sin x| dx \leq n\int_0^{1/n} x dx + (\pi-1/n)/n ... $$
|
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Are there mathematical contexts where "finite" implicitly means "nonzero?" I recently gave my students in a discrete math class the following problem, a restatement of the heap paradox:
Let's say that zero rocks is not a lot of rocks (surely, 0 is not a lot of rocks) and that if you have a lot of rocks, removing one rock leaves behind a lot of rocks. Prove that no finite number of rocks is a lot of rocks.
A small number of students submitted proofs by induction with the base case starting at one rock rather than zero rocks. We deducted a point for this, saying that this left the case of zero rocks unaccounted for.
Some students replied back to us saying that zero is arguably not a finite number. Some students pointed out this dictionary definition of finite which explicitly excludes 0 as not finite.
My background is in discrete math, and I've never seen zero referred to as not finite. The empty set of zero elements is a finite set, for example. There are no finite groups of size zero, but that's a consequence of the group axioms rather than because 0 isn't finite.
Are there mathematical contexts in which zero is definitively considered to be not finite?
Thanks!
|
Maybe this is too applied for your question, but there is a context in which finite means "large" or possibly large. That is in elasticity theory, to distinguish the linear theory of small strain from the nonlinear theory of "finite" strain.
|
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|
Writing the integral $ \int_{t}^{\infty}(\frac{1}{4\pi s^{3}})^{1/2}\frac{r(|x|-r)}{|x|}e^{-\frac{(|x|-r)^{2}}{4s}}ds$ in simpler form? I was wondering if
$\int_{t}^{\infty}(\frac{1}{4\pi s^{3}})^{1/2}\frac{r(|x|-r)}{|x|}e^{-\frac{(|x|-r)^{2}}{4s}}ds$ can be written more simply, where $x,r\in \mathbb{R}$ ?
wolfram alpha doesn't give me anything.
Here are my attempts ( I will type as I go):
1)Taylor expansion
$\frac{r(|x|-r)}{|x|}\frac{1}{2\pi^{\frac{1}{2}}}\sum_{k=0}^{\infty}\frac{(|x|-r)^{2k}}{k!}(\frac{3}{2}+k+1)\frac{1}{t^{\frac{5}{2}+k}}$
I would like to get a big O estimate for it eg. $O(t^{\alpha})$. So does the above sum converge to some closed expression.
|
This can be simplified a lot. (I assume that $t \gt 0$ as well.) Take out the constants to get
$$\frac1{2 \sqrt{\pi}} \frac{r}{|x|} (|x|-r) \int_t^{\infty} ds \, s^{-3/2} e^{-(|x|-r)^2/(4 s)}$$
Sub $s=1/u^2$ to get that the integral is equal to
$$\frac1{\sqrt{\pi}} \frac{r}{|x|} (|x|-r) \int_0^{1/\sqrt{t}} du \, e^{-(|x|-r)^2 u^2} $$
This is easily expressed in terms of an error function:
$$ \frac{r}{2 |x|} \operatorname{erf}{\left ( \frac{|x|-r}{\sqrt{t}} \right )} $$
|
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For which of the following functions the $\sum_{x\in S(f)}\frac{1}{x}$ converges? For real valued function $f$ define
$$S(f)=\{x:x>0,f(x)=x\}$$
For which of the following functions the $\sum_{x\in S(f)}\frac{1}{x}$ converges?
$\tan x,\tan^2x,\tan{\sqrt{x}},\sqrt{\tan x},\tan 2x$
I do not have any idea to solve this problem,I think all these have infinitely many fixed points.I need help.
Thanks.
|
For $f(x)=\tan x$, there is a fixed point of $f$ in each interval $((n-\frac12)\pi,(n+\frac12)\pi)$, hence the series in question is essentially the harmonic series, diverging.
For $f(x)=\tan^2 x$ and $f(x)=\tan 2x$, the same argument applies.
$f(x)=\sqrt{\tan x}$ makes no difference compared to $\tan x$ (except that the irrelevant negative parts become undefined).
For $f(x)=\tan\sqrt x$, however, we find one fixed points in $((n-\frac12)^2\pi^2,(n+\frac12)^2\pi^2)$ instead, so the growth is quadratic and the series converges, just as $\sum \frac1{n^2}$ does.
|
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|
Is this formula satisfiable? I am confused whether or not my explanation for whether or not this formula is satisfiable is correct. Note that the question state it should be Brief and it should not
be necessary to write down a full truth table.
I asked someone and they said my explanation was not ok and one was only giving an example of the formula not being satisfiable. And that the formula was satisfiable...
Here is the formula: (p→q)→(¬p→¬q)
And here is my explanation why I think its not satisfiable
*
*A→B is satisfiable as long as you don't have both A= true and B= false.
*When p = false, and q = true
*A or (p→q)= true , and B or (¬p→¬q) = false
*Therefore is not satisfiable.
Could somone improve or confirm if my explanation is ok
|
Possibly you are mixed up between satisfiable and tautology.
A statement is a tautology if it is always true. You have shown that the statement may be false, so it is not a tautology. But this is not what was asked.
A statement is satisfiable if it is sometimes true. So you need to find one example of truth values for $p,q$ such that the statement is true. This should not be too hard by trial and error (or see other people's answers).
Also worth noting: a statement is satisfiable if its negation is sometimes false, that is, if its negation is not a tautology.
|
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|
No simple groups of order 9555: proof While reading through crazyproject, I came across the following proof that there were no simple groups of order $9555$. However, I do not understand the step that says:
"Moreover, since 7 does not divide 12 and 13 does not divide 48, $P_7P_{13}$ is abelian."
I don't quite follow this reasoning. If someone could explain this to me I'd be very grateful.
|
The idea is to show that elements of $P_7$ and $P_{13}$ commute. The proof seems to be using that $|\mathrm{Aut}(P_7)| = 48$ but this is wrong. Because $|P_7| = 49$ we know that $P_7$ isomorphic to one of $\mathbb{Z}_{49}$ or $\mathbb{Z}_{7} \times \mathbb{Z}_{7}$ and so $|\mathrm{Aut}(P_7)|$ is either $42$ or $48 \cdot 42$.
The idea of the proof can still be used. We know that $P_{13} \leq N_G(P_7)$ and $13$ does not divide $|\mathrm{Aut}(P_7)|$ so we must have $P_{13} \leq C_G(P_7)$. Therefore $P_7P_{13}$ is abelian, since both $P_7$ and $P_{13}$ are.
We could change the proof to avoid thinking about $P_7P_{13}$ as follows. Since $P_{13} \leq C_G(P_7)$ we also have that $P_7 \leq C_G(P_{13}) \leq N_G(P_{13})$ and then pick up the proof in the last sentence.
|
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|
Finding the sum of $\sin(0^\circ) + \sin(1^\circ) + \sin(2^\circ) + \cdots +\sin(180^\circ)$ I need help understanding the sum of $\sin(0^\circ) + \sin(1^\circ) + \sin(2^\circ) + \cdots +\sin(180^\circ)$ or $\displaystyle \sum_{i=0}^{180} \sin(i)$
This might be related to a formula to find the average voltage from a generator used to gauge waves: $V_\text{avg} = 0.637 \times V_\text{peak}$. I am currently learning about AC circuits in the military.
|
Wikipedia says
$$\sum_{k=0}^n \sin{(\gamma + k\alpha)} = \frac{\sin\tfrac{(n+1)\alpha}{2} + \sin{(\gamma + \tfrac{n\alpha}{2})}}{\sin\tfrac\alpha2}$$So we may write $\gamma = 0$, $n=180$ and $\alpha=\frac{\pi}{180}$ to get $$\sum_{k=0}^{180} \sin{\tfrac{k\pi}{180}} = \frac{\sin\tfrac{181\pi}{360} \sin\tfrac{\pi}{2}}{\sin\tfrac{\pi}{720}} = \frac{\sin\tfrac{181\pi}{360} }{\sin\tfrac{\pi}{720}}$$
A proof of the formula can be found here.
|
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|
Area and integration question, is this area under the curve? Find the exact area between $x$ and the graph $f(x)=(x-1)(x-2)(x-3)$.
$$f(x) = x^3-6x^2+11x-6$$
I found that this is an odd shaped positive polynomial with a maxima between 1 and 2 and minima between 2 and 3.
I am confused to what the question wants. I naturally want to integrate the expansion of $f(x)$ from 1 to 2 and add it with the absolute value of of $f(x)$ integrate from 2 to 3.
However I'm worried that the question wants the area under the curve which says not to include the area between 2 and 3 as it is below the $x$ axis.
|
$$A= \int_{1}^{2}x^3-6x^2+11x-6-\int_{2}^{3}x^3-6x^2+11x-6$$
$$A= \left[\frac{x^4}{4}-{2x^3}+\frac{11x^2}{2}-6x\right]_{1}^{2}-
\left[\frac{x^4}{4}-{2x^3}+\frac{11x^2}{2}-6x\right]_{2}^{3}=\frac{1}{2}$$
|
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|
Prove that g is differentiable Question: Suppose f, g, and h are defined on (a,b) and $a < x_0 < b$. Assume f and h are differentiable at $x_0$, $f(x_0) = h(x_0)$, and $f(x) \le g(x) \le h(x)$ for all x in (a,b). Prove that g is differentiable at $x_0$ and $f'(x_0) = g'(x_0) = h'(x_0)$.
I know the case where $g(x) = f(x)$ or $h(x)$ makes this problem very simple so we don't need to look at that one.
Other than that, all I really know is, based on the definition, that
$lim_x \to x_0$$(f(x)-f(x_0))/(x-x_0)$ and the same for $g(x)$ but I'm not really sure where to go from there.
Thanks for the help!
|
We have: $f(x_0) \leq g(x_o) \leq h(x_0)$, and $f(x_0) = h(x_0)$, this implies $f(x_0) = g(x_0) = h(x_0)$, you can continue. From this we have: $f'(x_0^{+}) \leq g'(x_0^{+}) \leq h'(x_0^{+})$, and also $f'(x_0^{-}) \geq g'(x_0^{-}) \geq h'(x_0^{-})$. But $f'(x_0^{+}) = f'(x_0^{-}) = f'(x_0) = h'(x_0) = h'(x_0^{+}) = h'(x_0^{-}) \to g'(x_0^{+}) = g'(x_0^{-}) = g'(x_0) = f'(x_0) = h'(x_0)$.
|
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|
Prove that $(\mathbb{N},\le)$ and $(\mathbb{Z}, \le)$ are not order isomorphic I want to prove that $(\mathbb{N},\le)$ and $(\mathbb{Z}, \le)$ are not order isomorphic. So what I want to show is that the following is not true:
$$x \le_\mathbb{N} y \iff f(x) \le_\mathbb{Z} f(y)\qquad\forall x,y \in \mathbb{N}$$
But I think it is true, isn't it? Or have I done a mistake in the definition of the orde isomorphism?
I could also want to show that $\mathbb{N}$ and $\mathbb{Z}$ are not bijections. Is this what I have to do rather than show that it contradicts the definition of orde isomorphisms? Do I have to show that $\mathbb{N}$ does not have a surjection $\phi$ to $\mathbb{Z}$?
|
$\mathbb{N}$ has a minimal element; $\mathbb{Z}$ doesn't. An order isomorphism preserves minimality of an individual element. Can you show this?
|
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|
Evaluate the limit $\displaystyle\lim_{x \to 0^{+}} x \cdot \ln(x)$ I have this assignment
Evaluate the limit:
5.$\displaystyle\lim_{x \to 0^{+}} x \cdot \ln(x)$
I don't think we are allowed to use L'Hopital, but I can't imagine how else.
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If you set $y=\ln x$, you can rewrite it as $\lim_{x\to-\infty}ye^y=-\lim_{y\to\infty}\frac{y}{e^y}$, and you probably know what that is. Of course that approach requires that you be allowed to use known limits.
|
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|
Positive definiteness of block matrices I have a $2 \times 2$ block matrix of the form $$M = \left[ {\begin{array}{*{20}{c}}
{\delta I}&A\\ {{A^T}}&kA \end{array}} \right]$$ where the matrix $A$ is positive definite and not symmetric, $I$ is the identity matrix, and $k > 0$ and $\delta > 0$.
*
*Can we choose $\delta > 0$ such that the matrix $M$ be positive definite?
*Is there any general formula for positive definiteness of block matrices? It seems that the Schur complement is only for symmetric matrices.
I really appreciate if anyone can help me regarding this problem.
|
There is always a $\delta$ large enought that turns $M$ positive definite.
First, since $A$ is positive definit, there is $\alpha>0$ such that
$$
x^TAx \ge \alpha\|x\|^2 \quad \forall x\in \mathbb R^n,
$$
where I used the vector norm $\|x\|^2 = x^Tx$.
Let $x = \pmatrix{x_1\\x_2}\in \mathbb R^{2n}$. Then
$$
x^TMx = \delta \|x_1\|^2 + 2 x_1^T Ax_2 + k x_2^TAx_2.
$$
By positive definiteness of $A$, $x_2^TAx_2 \ge \alpha \|x_2\|^2$.
Now we use Cauchy-Schwarz inequality and definition of matrix 2-norm to estimate
$$
x_1^T Ax_2 \le \|x_1\|\cdot \|A\|\cdot \|x_2 \|.
$$
Using the inequality $ab \le \frac\epsilon2 a^2 + \frac1{2\epsilon}b^2$ for all $a,b\ge 0$, we find
$$
2x_1^T Ax_2 \le 2\|x_1\|\cdot \|A\|\cdot \|x_2 \|
\le \frac{k\alpha}2\|x_2\|^2 + \frac{2\|A\|^2}{k\alpha}\|x_1\|^2
$$
Putting everything together, we find
$$
x^TMx = \delta \|x_1\|^2 + 2 x_1^T Ax_2 + k x_2^TAx_2 \ge (\delta - \frac{2\|A\|^2}{k\alpha}) \|x_1\|^2 + \frac{k\alpha}2\|x_2\|^2.
$$
Hence, $M$ is positive if $\delta > \frac{2\|A\|^2}{k\alpha}$.
|
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|
uniqueness of joint probability mass function given the marginals and the covariance Let X and Y be two nonnegative, integer-valued random variables. Is there a way to find the joint probability mass function, i.e.
$$
\mathbb{P}(X= k, Y= h)
$$
for some $k,h\geq 0$, given the marginals and the covariance?
Moreover, it is true that $\mathbb{E}[XY]=\sum_{k=0}^{\infty}\sum_{h=0}^{\infty}h\,k\,\mathbb{P}(X= k, Y= h)$.
I know that I can write
$$
\mathbb{E}[XY]=\sum_{k=0}^{\infty}\sum_{h=0}^{\infty}h\,k\,g_{h,k}
$$
Can I conclude that $\mathbb{P}(X= k, Y= h)=g_{h,k}$? Is there uniqueness?
Thanks in advance.
|
In general $E[X]=\sum_x xp(x)$; So, obviously, for $Z=XY$ we would have
\begin{align*}
E[Z]=&\sum_z zP(Z)\\
=&\sum_x \sum_y xy P(X,Y)\\
=&~a
\end{align*}
where $a$ is a constant. But your question is how many coefficient sets $\{c_{x,y}\}$ we can find so that
\begin{align*}
\sum_x \sum_y x~y~c_{x,y}=a
\end{align*}
For sure it is not unique. For example for an arbitrary pair ($x_1$ and $y_1$) if you change $c_{x_1,y_1}$ to $c_{x_1,y_1}+1$ then the total result would be increased by $x_1 y_1$. Now, for another arbitrary pair ($x_2$ and $y_2$), you can compensate this difference by decreasing $c_{x_2,y_2}$ value to become $(c_{x_2,y_2}-\frac{x_1 y_1}{x_2 y_2})$.
As you see there exist infinite sets of $\{c_{x,y}\}$ which all of them result in $a$.
|
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|
Compute the degree of the splitting field I need to compute the degree of the splitting field of the polynomial $X^{4}+X^{3}+X^{2}+X+1$ over the field $\mathbb{F}_{3}$. Quite honestly I don't really know where to begin, I know the polynomial is irreducible in this field. So I thought we could consider some element $\alpha \in E$ where E is some field extensions of $\mathbb{F}_{3}$ and try to find a relation between $\alpha$ and the other roots, but I am not 100% sure. Also I think the degree is $4$ but I am not sure why. Any hints would be apprecaited. Please beare in mind that I am only a few weeks into my galois theory course so it might take a while for me to follow.
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If $K$ is a finite field and $f \in K[X]$ is irreducible, then $K[X]/(f)$ is a splitting field of $f$. This follows from the fact that finite extensions of finite fields are always normal. If $\alpha$ is a root, the other roots are $\alpha^{p^n}$, $n \in \mathbb{N}$, where $p=\mathrm{char}(K)$. In particular, the degree is $\mathrm{deg}(f)$.
If $n$ is a natural number coprime to $p$, then it is a fact that the cyclotomic polynomial $\Phi_n$ is irreducible in $\mathbb{F}_p[x]$ if and only if $[p]$ generates $(\mathbb{Z}/n)^\times$.
Since $[3]$ generates $(\mathbb{Z}/5)^\times$, it follows that $\Phi_5=X^4+\dotsc+X+1$ is irreducible in $\mathbb{F}_3[X]$ and hence $\mathbb{F}_{3^4}$ is a splitting field. Explicitly, if $\alpha$ is a root of $\Phi_5$, then the other roots are $\alpha^2,\alpha^3,\alpha^4$.
|
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If $P$ and $Q$ are distinct primes, how to prove that $\sqrt{PQ}$ is irrational? $P$ and $Q$ are two distinct prime numbers. How can I prove that $\sqrt{PQ}$ is an irrational number?
|
Proof: Assume, to the contrary, that $\sqrt{pq}$ is rational. Then $\sqrt{pq}=\frac{x}{y}$ for two integers $x$ and $y$ and we further assume that $gcd(x,y)=1$. Observe that $pqy^2=(qy^2)p=x^2$. Since $qy^2$ is an integer, $p\mid x^2$ and by Euclid's Lemma, $p\mid x$. Thus $x=pd$ for some integer $d$ and so $pqy^2=x^2=(pd)^2$ and so $qy^2=p(d^2)$. Hence, $p\mid qy^2$. Since $p,q$ are distinct primes, $p \neq q$ and so $p \nmid q$, which means $p\mid y^2$. Thus, $p\mid y$. This contradicts our assumption that $gcd(x,y)=1$.
|
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Let $A,B,C,D$ be the vertices of a four sided polygon taken in anti clockwise. Given $|AB|=|BC|=3,|AD|=|CD|=4,|BD|=5$ , Find $|AC|$ Let $A,B,C,D$ be the vertices of a four sided polygon taken in anti clockwise.
Given $$|AB|=|BC|=3,|AD|=|CD|=4,|BD|=5$$
Find $|AC|$
My try:I have noticed trangles $ABD$ and $BCD$ are right triangles but I do not know how to solve it.
|
You know the angle $C$ and $A$ are right and $BD$ must be perpendicular to $AC$(since the length of BC and AB are the same). Suppose the intersection point is $M$. Hence you can $AM$ by using BCM is similar to BDC.
|
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Which theorem could be used? I want to write the $p-$adic expansion of $6!$ in $\mathbb{Q}_3$.
I have to solve the congruence $x \equiv 6! \pmod {3^n}$, right?
I found the following:
$$x_0 \equiv 6! \pmod 3 \Rightarrow x_0 \equiv 0 \pmod 3$$
$$x_1 \equiv 6! \pmod {3^2} \Rightarrow x_1 \equiv 0 \pmod {3^2}$$
$$x_2 \equiv 6! \pmod {3^3} \Rightarrow x_2 \equiv 18 \pmod {3^3}$$
$$x_3 \equiv 6! \pmod {3^4} \Rightarrow x_3 \equiv 72 \pmod {3^4}$$
Using the formula $x_n=\sum_{i=0}^n a_i 3^i$, I found: $a_0=0,a_1=0,a_2=2,a_3=2$.
Is there a theorem I could use, in order to find the coefficients of the infinite series?
A theorem that states, for example, that the solutions we find are periodic?
|
From the comments, it seems you understand now that $720$, just like any other integer, has a finite $3$-adic representation, i.e. one of the form $$720=a_0 3^0+a_1 3^1+\cdots +a_n 3^n,$$ and you want to find the digits $a_0,\cdots, a_n$, where $0\leq a_i\leq 2$.
We will do it step by step. How do we find $a_0$? If we reduce the expression above modulo $3$, we get $720 \equiv a_0\pmod{3}$. Do you understand this step? All positive powers of $3$ go away because we took the remainder modulo $3$.
Now, $720 \equiv 0 \pmod{3}$ so $a_0\equiv 0\pmod{3}$. But there is only one integer $a_0$ between $0$ and $2$ that is zero modulo $3$: the integer $a_0=0$. Thus we found the zeroth digit!
On to the first digit. We reduce the equation $720=a_0 3^0+a_1 3^1+\cdots +a_n 3^n$ modulo $3^2$. What do we get? All powers of $3$ above the first go away and we get $a_0+3a_1 \equiv 0 \pmod{3^2}$ (because $720$ is zero mod $9$). Hmmm, we did not quite get $a_1$, but no worry. We now know $a_0$ from the previous step, so how do we find $a_1$?
Since $a_0+3a_1$ is smaller than $9$ and zero mod nine, it has to equal zero in the integers. This is a crucial property: if a quantity $q$ reduced modulo $M$ is equivalent to $a$ with $0\leq a < M$ and we know that $0 \leq q <M$, then actually $q=a$ as equality of integers. So in our case we get $a_0+3a_1=0$.
Now subtract $a_0$ (a digit we already know) from the equation $a_0+3a_1=0$ and divide by three to get $a_1$. There, we found the first digit which (expectedly) equals zero.
On to the second digit: we reduce the equation $720=a_0 3^0+a_1 3^1+\cdots +a_n 3^n$ modulo $3^3$ to get $$a_0+3a_1 + 9a_2 \equiv 18 \pmod{3^3}.$$ Again, since $a_0+3a_1+9a_2< 27$ we conclude that $$a_0+3a_1 + 9a_2 = 18. $$ But from our previous divsion we had obtained $$a_0+3a_1=0.$$ Subtracting the equations and dividing by $9$, we find $a_2 = 2$. This gives us the second digit.
In general, this is the procedure: reduce your number modulo powers of $3$ consecutively. The first power will immediately give you the zeroth digit. To get the $k$-th digit, reduce $720$ modulo $3^{k+1}$ and modulo $3^k$; consider these remainders as integers, subtract them and divide by $3^k$. The integer you will get is your $k$-th digit.
So for instance you already computed $720 \equiv 234 \pmod{3^5}$ and $720 \equiv 72 \pmod{3^5}$. Subtract $234-72 = 162$ and divide that by $3^4$ to get $a_4=2$.
When do we stop? Your partial expressions $a_0+\cdots +a_k 3^k$ will be smaller than 720 but steadily increasing as you add more and more digits $a_k$, and eventually will become equal to 720. This is where you will stop. (If you continue, the previous process will keep giving you zero digits forever).
I hope this helped, but if you still have unclear points I think you should make a serious review of congruences and their laws before tackling $p$-adic numbers.
|
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How can I find the radius of a circle from a chord and a section of the radius? Draw a circle with center O.
Line AD is a chord that is 8cm long. The arc above is smaller than the one below.
B is the center of AD.
Line CB is a line that is 2cm long. It meets AD at 90°.
Diagram:
Given the facts above, is it possible to find CO (the radius of the circle)? If so, what is the radius, and how can I find it for any other circle?
|
Hint: Suppose $OB=x$, then $OA=OC=x+2$. Now apply Pythagoras theorem in triangle $AOB$
|
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|
Integration with square root in denominator I am honestly embarrassed to ask this because i feel like i should know how to do this but:
$
\int \frac{x}{\sqrt{2x-1}}dx
$
Try to use u-substitution please
|
Substitute $u=2x-1, du = 2dx$ so that you get
$$\int \dfrac{x}{\sqrt{2x-1}} \, dx = \frac12\int\frac{u+1}{2\sqrt{u}}\, du = \frac12\int\frac{\sqrt{u}}{2} + \frac{1}{2\sqrt{u}}\, du.$$
Do you see how to proceede?
|
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|
Decomposition of $_1F_2(1+n;1,2+n;x)$ I am looking for a way to decompose $_1F_2(1+n;1,2+n;z)$ for $n\in\mathbb{N}$ into either Bessel J functions or regularized confluent hypergeometric functions $_0\tilde F_1(b(n),z)$. Mathematica seems to be able to do the decomposition for $n=1,2,3,4$ but I am trying to get a closed form for any $n$. I have looked through Wolfram's functions site but have been unsuccessful in my search. This would be useful because numerical routines can deal with Bessels much faster than hypergeometrics.
|
I have found a somewhat simple relation based on the triangle of falling factorials, reading by rows (A068424). We have $$_1F_2(1+n;1,2+n;z)=\sum_{k=1}^{n+1}\begin{pmatrix}
n+1 \\
k
\end{pmatrix}k!(-1)^{k+1}\,_0\tilde F_1(;1+k;z),\,n\geq 0 \\
=\sum_{k=1}^{n+1}\begin{pmatrix}
n+1 \\
k
\end{pmatrix}k!(-1)^{k+1}(-z)^{-k/2}J_k(2\sqrt{-z}),\,n\geq 0.$$
Now we can use the recurrance relation to get it all in terms of 1 or 2 bessels/confluents in order to reduce computation time. By Wolfram's Functions we have $$_0\tilde F_1(;b;z)=z^{-m}(2-b)_{m-1}\left ( \, _0 \tilde F_1(;b-m-1;z)\sum_{j=0}^{m-1}\left [ \frac{(m-j-1)!(-z)^j}{j!(m-2j-1)!(2-b)_j(b-m)_j} \right ]\\+(1+m-b)\,_0\tilde F_1(;b-m;z)\sum_{j=0}^{m-1}\left[\frac{(m-j)!(-z)^j}{j!(m-2j)!(2-b)_j(b-m-1)_j} \right ] \right ) \\
=z^{-(b-2)}(2-b)_{b-3}\left ( \, _0 \tilde F_1(;1;z)\sum_{j=0}^{b-3}\left [ \frac{(b-j-3)!(-z)^j}{j!(j+1)!(b-2j-3)!(2-b)_j} \right ]\\-\,_0\tilde F_1(;2;z)\sum_{j=0}^{b-3}\left[\frac{(b-2-j)!(-z)^j}{(j!)^2(b-2-2j)!(2-b)_j} \right ] \right ),\,b\geq 3$$
|
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|
Prove that if p divides xy then p divides x or p divides y I am given that the following proposition is true. (Proved in class)
"Suppose that $x$, $y\in \Bbb Z$, not both zero. Then there exists $m$, $n\in\Bbb Z$ such that $$mx + ny = d$$ where $d$ is the greatest common factor of $x$ and $y$. Moreover, if $ax + by = f$ for some $a$, $b$, $f\in\Bbb Z$ then $d | f$."
Using the statement above, prove that if $p$ is prime and $p|xy$ then $p|x$ or $p|y$. (hint: if $p$ does not divide $x$, then $p$ and $x$ are relatively prime)(We just learned the definition of relatively prime, so that is probably why that hint is there.)
I just don't get why I need to use the statement above. The way I see it, if $p|xy$ then there are two possible cases:
*
*$p|x$
*$p$ does not divide $x$
And obviously, if $p|x$ then the original statement is true, and if $p$ does not divide $x$ then I just need to prove that $p|y$ in that case which I could do. That seems simpler than any other way, am I wrong? Working with prime numbers always throws me. I don't get why it matters if $p$ is prime, so if someone wants to explain that, it would be a nice addition.
|
Hint $gcd(p,x)$ is a divisor of $p$. Therefore it is either $p$ or $1$.
If the gcd is $p$ it means $p|x$ and you already treated that case.
Otherwise, by the lemma
$$mp+nx=1$$
Multiply this by $y$, and use that $p|xy$.
P.S. The key to the proof is that $p$ is prime, otherwise the gcd could be something else but 1 or p. And it was already shown by examples that in that case the statement is not true.
|
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|
Repeated substitution gone wrong It was an exam question.
$$ f(n)=
\begin{cases}
0 & \mbox{if } n \leq 1 \\
3 f(\lfloor n/5 \rfloor) + 1 & \mbox{if } n > 1 \\
\end{cases}$$
So by calculating some I have $f(5) = 1$, $f(10) = 4$, $f(50) = 13$.
I had to solve this recurrence. So to get rid of the floor operator I said lets $n$ be $5^k$ so a multiple of $5$. So started to solve this.
\begin{equation}
\begin{split}
f(5^k) & = 3\cdot f(5^{k-1}) + 1\\
& = 3\cdot[ 3f(5^{k-2}) + 1 ]+ 1\\
& = 3^2f(5^{k-2}) + 3 + 1\\
& = 3^2\cdot[ 3f(5^{k-3}) + 1 ]+ 3 + 1\\
& = 3^3f(5^{k-3}) + 9 + 3 + 1 \\
& = ...\\
& = 3^jf(5^{k-j}) + \sum_{j=0}^k 3^j\\
& = 3^jf(5^{k-j}) + \frac{3^{k+1} - 1}{2}\\
\text{let $j = k$}\\
& = 3^kf(5^0) + \frac{3^{k+1} - 1}{2}\\
& = 3^kf(1) + \frac{3^{k+1} - 1}{2}\\
& = 3^k\cdot 0 + \frac{3^{k+1} - 1}{2}\\
& = \frac{3^{k+1} - 1}{2}\\
\end{split}
\end{equation}
And now I'm stuck here trying to express my $3^{k+1}$ as a function of $n$.
I'm pretty I probably made some error up there, if I did I don't see them.
|
I didn't check your whole argument, but assuming it's correct, you can use the fact that if $n$ is $5^k$, then $k$ is $\log_5 n$ to express the result in terms of $n$.
(Side Question: are the $x$'s in the function definition supposed to be $n$'s?)
|
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|
Finding the points at which a surface has horizontal tangent planes Find the points at which the surface
$$
x^2 +2y^2+z^2 -2x -2z -2 = 0
$$ has horizontal tangent planes. Find the equation of these tangent planes.
I found that $$ \nabla f = (2x-2,4y) $$ I'm thinking that the gradient vector must
be equal to $(0,0)$ so $x = 1, y=0$ which implies $z = 3,-1$. So the points are $(1,0,3)$ and $(1,0,-1)$. How would we write the equation of these tangent planes?
|
Your answer is correct, but the logic is wrong. You should consider the above surface as a level surface of the function:
$$ f(x, y, z) = x^2 + 2y^2 + z^2 - 2x - 2z - 2$$
Then the gradient should be a vector with 3 components.
$$ \nabla f(x, y, z) = (2x - 2, 4y, 2z - 2) $$
If the tangent plane is horizontal, the gradient must point in the $z$-direction, therefore the $x$ and $y$ components are $0$. So it follows that $x = 1$ and $y = 0$
Finally, the equations are just $z = 3$ and $z = -1$ since horizontal planes have the same $z$ coordinates everywhere.
Another way is to treat $z$ as a function of $x$ and $y$, then set $\partial z / \partial x$ and $\partial z / \partial y$ equal to $0$. However, those expressions are not the same as what you wrote, since they have to be found through implicit differentiation. I'm not sure which method you were trying to use.
|
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|
Limit of $(1-2^{-x})^x$ I am observing that $(1-2^{-x})^x \to 1$ as $x \to \infty$, but am having trouble proving this. Why does the $-x$ "beat" the $x$?
I thought of maybe considering that $$1-(1-2^{-n})^n = 2^{-n}(1+2^{-n}+2^{-2n}+\cdots+2^{-n(n-1)})\le2^{-n} \frac{1}{1-2^{-n}} \to 0,$$
as $n \to \infty$ and then noting that $(1-2^{-x})^x$ is continuous to go from integers $n$ to any real $x$.
Is there a more elegant solution, or any better intuition? Moreover, it seems that $(1-2^{-ax})^{bx}\to 1$ holds as well for any $a,b>0$.
|
Look at the logarithm:
$$
\log(1-2^{-x})^x=x\log(1-2^{-x})\sim x\cdot(-2^{-x})\rightarrow 0.
$$
So
$$
(1-2^{-x})^x \rightarrow 1.
$$
|
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|
Why isn't the orthogonal vector to a direction vector of a plane not necessarily perpendicular to such plane? I had made a question, and the problem with my exercise was that I was trying to calculate a vector perpendicular to some plane in $\mathbb{R}^3$: given one line $L$ inside the plane, I grabbed the direction vector $(-2,0,3)$ of this line and then solve
$$(x,y,z) \cdot (-2,0,3) = 0$$
To get an orthogonal vector to the plane. But I was told that
Not all vector perpendicular to $(−2,0,3)$ gives you the normal vector
of the plane.
However I don't really see why. If $(-2,0,3)$ is a direction vector of the plane, why wouldn't a vector perpendicular to $(-2,0,3)$ necessarily be also perpendicular to said plane? In $\mathbb{R}^3$, at least, I can't really visualize why not.
|
The vector perpendicular to your line will be just that, perpendicular to your line; nothing more is guaranteed. To be perpendicular to the plane, you must be perpendicular to all vectors (or lines) in the plane at once; there is only one direction (and its opposite) which does that. The vector you found perpendicular to $(2,0,-3)$ could very well be inside the plane; there is certainly a direction within the plane that is perpendicular to your line. Or your perpendicular vector could happen to be a normal vector to the plane. Or anything in between.
One really cannot tell more just knowing that the vector is perpendicular to $(2,0,-3)$.
|
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|
Optimization problem: rowing across a lake
A woman at a point A on the shore of a circular lake with radius $r=3$ wants to arrive at the point $C$ diametrically opposite $A$ on the other side of the lake in the shortest possible time. She can walk at the rate of 10 mph and row a boat at 5 mph. What is the shortest amount of time it would take her to reach point $C$?
My Working
Let $C\hat{A}B=\theta$
$\begin{align*}\therefore AB+arcBC&=\sqrt{3^2+3^2-2\cdot3\cdot3\cdot\cos(\pi - 2\theta)}+3\cdot2\theta\\
&=3\sqrt{2-2\cos(\pi-2\theta)}+6\theta\end{align*}$
$\begin{align*}\therefore \text{time taken} =t=\frac{3\sqrt{2-2\cos(\pi-2\theta)}}{5}+\frac{6\theta}{10}\end{align*}$
To find the point $B$ at which $t$ is minimized, we make $\frac{\operatorname d t}{\operatorname d\theta}=0$:
$\begin{align*}\frac{\operatorname d t}{\operatorname d\theta}=\frac{3-4\sin(\pi-2\theta)}{10\sqrt{2-2\cos(\pi-2\theta)}}+\frac{3}{5}=0\end{align*}$
And the rest of my working gets messy from here. Is there a better method than this?
|
I would like to propose a calculus approach here. So the question is vague in the sense that, we don't really know if she has to walk first or row first. But it's actually symmetric in nature, if you consider walking (along the arc) first (say 'Q' miles) AND then row the remaining distance (say 'P' miles), it's literally symmetric to rowing the same previous distance ('P' miles) first followed by walking ('Q' miles).
Consider any case among these as it doesn't matter due to the above proof. I'll take latter one here. So, assuming $\angle{BAC}=\theta $, we can conclude few things, the 'P' meters becomes $ 2r \times cos(\theta)$ miles, and the 'Q' miles of arc length corresponds to $r \times 2 \theta$ miles. So, now plugging in the value for r as 3, we have the total time as $T_{total}= (\frac{2 \times \cos(\theta)}{5} + \frac{3 \times 2 \times \theta}{10}) \text{hrs}$.
Now we just have to find where this function has the least value. But we have to be sure of the domain we consider here, as we HAVE to reach the other side, we could start with $\theta=0$ or have a maximum angle of $\frac{\pi}{2}$ rad. (We cannot have $\theta > \pi$ because that would mean we're not wanting to reach the other side.)
Keeping these things in mind, we find that the critical point occurs at $\theta$ = 0.5235. Now this value if substituted in the total time equation, gives us the time as $\approx$ 1.667 hr. Now let's check how the function behaves at the end points in the domain (remember, we can include 0 and $\frac{\pi}{2}$ rads). $\text{T}_{\theta=0}= 1.2 hr$ and $\text{T}_{\theta=\frac{\pi}{2}} \approx 0.942$. So, turns out that if you just first row for 0 miles at an angle of $90^{o}$ and then walk for the remaining $(3 \times 2 \pi)$ miles, it takes the least amount of time.
|
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|
Find the sum of an infinite series of Fibonacci numbers divided by doubling numbers. How would I find the sum of an infinite number of fractions, where there are Fibonacci numbers as the numerators (increasing by one term each time) and numbers (starting at one) which double each time as the denominators?
I'm assuming this has something to do with limits.
|
Here's a reasonably efficient low-tech approach (i.e., not using Binet's formula or generating functions).
Prove by induction that
$$ \sum_{k=0}^n \frac{F_k}{2^k} = 2 - \frac{F_{n+3}}{2^n} \tag{1} $$
and that
$$ nF_n \le 2^n \tag{2} $$
(For (2) you might find it convenient to assume $n\ge 6$.) From (2) it follows that $\frac{F_{n+3}}{2^n} \le \frac8{n+3} \to 0$, so the partial sums in (1) converge.
|
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|
Rigorous proof that surjectivity implies injectivity for finite sets I'm trying to prove that, for a finite set $A$, if the map $f: A\rightarrow A$ is a surjection, then it's an injection. I've looked at this post: Surjectivity implies injectivity but the arguments there seem very round-about, proving that injectivity implies surjectivity first. I think I've seen other proofs, too, but they all sound kind of flimsy. They tend to go something like "If the function is surjective then no two arguments can go to the same image, since that would require some element of the range not being the image of any element." While that's true and I see what the argument has in mind, it's no more convincing than the statement of the claim that we're trying to prove! Without a more rigorous proof, both seem to rely on the intuition that everything has to go to one and only one thing, but we're supposed to be proving that.
But I've been trying to develop a more thorough and direct proof, and haven't been able to. I've tried induction:
The theorem holds trivially if $|A|=0$. Suppose the theorem holds up to $n=|X|$, and let $|A|=n+1$ and $f:A\rightarrow A$ be a surjection. Then let $x\in A$ and $f^{-1}(x)$ be the set of all elements $y\in A$ such that $f(y)=x$...
But at this stage I want to restrict my function, but that has the disadvantage that if $f(y)$ has more than one element in it, when I pluck it out of $A$ as the domain and pluck $x$ out of $A$ as the range, then I may not be dealing with a function mapping from a set to itself and so I don't get to use the inductive hypothesis.
Any other proofs you know about?
|
Let $f: A\to A$ be any function. For each $a\in A$, let $N(a)$ be the number of elements in $A$ that are mapped to $a$. Then $\sum_a N(a) = n = |A|$ because every element of $A$ is mapped to some element of $A$. (This is the statement that the fibers of $f$ partition $A$.)
If $f$ is surjective, then $N(a)\ge 1$ for all $a$. If $f$ is not injective, then $N(a_0)>1$ for some $a_0$. But then $\sum_a N(a) > n$, a contradiction.
This same technique works for the dual statement:
If $f$ is injective, then $N(a)\le 1$ for all $a$. If $f$ is not surjective, then $N(a_0)<1$ for some $a_0$. But then $\sum_a N(a) < n$, a contradiction.
|
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|
A question on a nonnegative quadratic form Denote $x,y,z$ as variables, and $a,b,c$ as coefficients. Suppose $a\leq b\leq 0\leq c$ and $a+b+c=0$.
Could anyone help me prove whether the following quadratic form positive semi-definite?
\begin{equation*}
\begin{split}
I(x,y,z)=&(a^2+4b^2+4c^2)a^2x^2+(4a^2+b^2+4c^2)b^2y^2\\
+&(4a^2+4b^2+c^2)c^2z^2\\
-&2ab(a^2+b^2+c^2+3ab)xy\\
+&2ac(a^2+b^2+c^2+3ac)xz\\
+&2bc(a^2+b^2+c^2+3bc)yz.
\end{split}
\end{equation*}
|
Well, I showed you how to try to attack these problems a couple of days ago, and you can use exactly the same strategy here.
A problem on positive semi-definite quadratic forms/matrices
Once again using MATLAB Toolbox YALMIP to compute a sum-of-squares certificate. Ordering of the variables is not required, but the equality appears to be necessary to exploit
sdpvar a b
c = -a-b
I=(a^2+4*b^2+4*c^2)*a^2*x^2+(4*a^2+b^2+4*c^2)*b^2*y^2+(4*a^2+4*b^2+c^2)*c^2*z^2-2*a*b* (a^2+b^2+c^2+3*a*b)*x*y+2*a*c*(a^2+b^2+c^2+3*a*c)*x*z+2*b*c*(a^2+b^2+c^2+3*b*c)*y*z;
% Floating-point decomposition
[~,v,Q] = solvesos(sos(I))
sdisplay(v{1})
Q{1}
% Integer decomposition
Q = intvar(7,7);
Match = coefficients(I-v{1}'*Q*v{1},[a b x y z])==0;
optimize([Match, Q >=0])
value(Q)
>> sdisplay(v{1})
ans =
'z*b^2'
'z*a*b'
'z*a^2'
'y*b^2'
'y*a*b'
'x*a*b'
'x*a^2'
>> value(Q)
ans =
5 6 1 1 1 -2 0
6 12 6 1 -1 -1 1
1 6 5 0 -2 1 1
1 1 0 5 4 -2 -2
1 -1 -2 4 8 -3 -2
-2 -1 1 -2 -3 8 4
0 1 1 -2 -2 4 5
|
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What is $\int_0^{\infty} x^2e^{\frac{(x-\mu)^2}{2 a^2}} dx$? How can we express the integral $\int_0^{\infty} x^2e^{-\frac{(x-\mu)^2}{2 a^2}} dx$ for example by means of the error function? The problem is of course, that the expectation value is shifted and we don't integrate from minus infinity to plus infinity. Thus, I doubt that it is possible to explicitely evaluate the integral.
|
Let $\Phi(t) = \int_{-\infty}^t e^{-x^2/2} \, dx$. Then
$$
\int_k^{\infty} x^2 e^{-x^2/2} \,dx = -\frac{d}{d\alpha}\Big|_{1/2}\int_k^{\infty}e^{-\alpha x^2}\,dx = - \frac{d}{d\alpha}\Big|_{1/2} \int_k^{\infty} e^{-(\sqrt{2 \alpha}x)^2/2}\,dx =\\ -\frac{d}{d\alpha}\Big|_{1/2} \Big( \frac{1}{\sqrt{2\alpha}} \int_{\sqrt{2 \alpha} k}^{\infty} e^{-x^2/2} \,dx \Big) = - \frac{d}{d\alpha}\Big|_{1/2}\Big[\frac{1}{\sqrt{2 \alpha}}\big( 1 - \Phi(\sqrt{2\alpha} \,k)\big)\Big] = \cdots
$$
Now
$$
x^2 = (x-\mu+\mu)^2 = (x-\mu)^2 +2 \mu(x-\mu) + \mu^2
$$
so
$$
\int_{0}^{\infty} x^2 e^{- \frac{(x-\mu)^2}{2a^2}}\,dx =\\ a^2\int_{0}^{\infty} \frac{(x-\mu)^2}{a^2} e^{- \frac{(x-\mu)^2}{2a^2}}\,dx + 2\mu \int_{0}^{\infty} (x-\mu) e^{- \frac{(x-\mu)^2}{2a^2}}\,dx + \mu^2 \int_{0}^{\infty} e^{- \frac{(x-\mu)^2}{2a^2}}\,dx\\
= a^2 \int_{-\mu/a}^{\infty}x^2 e^{-\frac{x^2}{2}} \,dx +\mu \int_{\mu^2}^{\infty}e^{-\frac{x}{2a^2}}\,dx + \mu^2 \int_{-\mu/a}^{\infty} e^{-\frac{x^2}{2}} \,dx
$$
The first term can now be found in terms of $\Phi$ using the derivation above, the third term is just $\mu^2 [1 - \Phi(-\mu/a)]$, and the middle term can be integrated easily.
|
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Prove that the number of automorphisms in $\mathbb Q[\alpha]$ equals $1$ $(|Aut\mathbb Q[\alpha]|)=1$ Please, help me to understand this problem:
Let $\alpha=\sqrt[3]{2}$ be a root of the polynomial $x^3-2$.
a) Prove that the number of automorphisms in $\mathbb Q[\alpha]$ equals $1$ $(|Aut\mathbb Q[\alpha]|)=1$;
b) If $L$ is the Galois Field of $x^3-2$ over $\mathbb Q$ ($Gal(x^3-2,\mathbb Q))$. Prove that the number of automorphisms in $L$ equals $6$ $(|Aut L|=6)$.
My solution:
a) I claim that there exists only one authomorphism in $\mathbb {Q}[\alpha]$: the identity function. In fact, an arbitrary element of $\mathbb{Q}[\alpha]$ is of the form $a_0+a_1\alpha+a_2\alpha^2$. Then , if $f$ is an automorphism, then $f(a_0+a_1\alpha+a_2\alpha^2)=a_0+a_1f(\alpha)+a_2f(\alpha^2)=a_0+a_1\alpha+a_2\alpha^2$.
Now, if I try to apply this method in "b", it fails. A generic element of $L$ is of the form $a_0+a_1\alpha+a_2\alpha^2+a_3u+a_4\alpha u+a_5\alpha^2u$, where "u" is the primitive 3th root of unity. Then $f(a_0+a_1\alpha+a_2 \alpha^2 +a_3u+a_4\alpha u+a_5\alpha^2 u)=a_0+a_1\alpha+a_2\alpha^2+a_3f(u)+a_4\alpha f(u)+a_5\alpha^2 f(u)$. There is no way to get 6 automorphisms from here. By the way: what is $f(u)$?
|
In a) how do you make the assumption that $f(\alpha)=\alpha$? In b) you are asked to compute the Galois group of $L$ over $\mathbb{Q}$.
Hint for both parts of the exercise: First show that any automorphism of the fields in question fixes $\mathbb{Q}$. Then try to find out where it could possibly map $\alpha$ and $u$; since $\alpha$ and $u$ satisfy a polynomial equation over $\mathbb{Q}$ the same is true for $\phi(\alpha)$ and $\phi(u)$.
|
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|
Show that ${n \choose k}{k \choose m} = {n \choose m}{n-m \choose k-m}$ Show that $\begin{align}{n \choose k}{k \choose m} = {n \choose m}{n-m \choose k-m}.\end{align}$
Not sure how to approach this exactly. I've tried to use the property $\begin{align}{n \choose k} = {n \choose n-k} \end{align}$, which seems like it could be useful, but doesn't gett me anywhere.
|
Hint: $$\binom{a}{b} = \frac{a!}{b!(a-b)!}.$$
|
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|
Trace of $L^p$ function For $U$ a bounded domain in $\mathbb{R}^n$, why is it that, in general, an $L^p$, $1\leq p<+\infty$, function does not have a trace on the boundary of $U$?
Thanks in advance.
|
Here is a counterexample for $n=1$, which can me easily modified to hold in higher dimensions:
Consider the sequence $f_n(x):=(1-nx)_+$ for $x\in (0,1)$. $(\cdot)_+$ denotes the positive part. By dominated convergence $f_n\to 0$ in $L^p(0,1)$. Assume there exists a contiunous and linear trace operator $T$ on $L^p(0,1)$ that evaluates the functions in $x=0$ (and of course provides the correct result for continuous functions). Then we must have $1=T(f_n)\to T(f)=0$ which is a contradiction. Note that the convergence of $f_n\to 0$ does not hold in $H^{1,p}(0,1)$.
For completeness: $p\in[1,\infty)$.
|
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|
Question regarding notation in algebraic topology My class has not been following a book and my professor's last bit of notation is a bit confusing to me.
This is the goal. We are given a path-connected space $Y$ and $H$ a subgroup of $\pi_1(Y,y)$. We want to find a cover $X$, of $Y$ such that $X$ is path-connected, $p(x)=y$, and the $im(p_*(\pi(X,x))=H$
So we ask what are the points of $X$ going to be
We look at all the points over $y$, the fibre $p^{-1}(y)$.
Since $X$ is path-connected, every point of $p^{-1}(y)$ can be connected by a path $\alpha$ starting at the given point $x$ and ending at $x'$
The $p \circ \alpha$ is a path in $Y$ starting and ending at $y$.
This says if $\textbf{this is where the notation confuses me}$
$[\gamma]=[p \circ \alpha] \in \pi_1(Y,y)$
$x' * [\gamma]=x$
How can $x' * [\gamma]=x$? Isn't this concatenation an entire path and not just one point??
|
That notation doesn't mean much to me, either. But to try to help:
I would guess that perhaps the prof has defined an action of $\pi_1(Y)$ on the fiber $p^{-1}(y)$, an action denoted by "*". So this says that the path $\gamma$, acting on $x'$, gives $x$. (The definition of the action would be "lift the path to one in X that starts at $x'$, and see where it ends; that endpoint is defined to be $x' * \gamma$"). [I make this guess because something like that is what people typically do in working with covering spaces.]
If you're looking for a book to peek at, you might want to check out "Algebraic Topology: An Introduction" by Massey, who does all this, and with more or less this notation.
|
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|
Sum of coefficients in binomial theory. While trying to get introduced to binomial theory at university's website, I learned about the sum of binomial coefficients, and they showed me some of the features, and one of them was the pyramid of coefficients sum which is:
$$1 = 1$$
$$1+1 = 2$$
$$1+2+1 = 4$$
$$1+3+3+1 = 8$$
$$1+4+6+4+1 = 16$$
and then they asked me to filled the blank with something that describes the sum:
$$\sum_{i=0}^n \binom{n}{i} = \left[ \phantom{\frac 1 1} \right]$$
If I understood right what they want, they how do I describe the sum ?
|
There are a variety of different ways to understand this. One of those is the combinatorial way of understanding it, which says that binomial coefficients count subsets of specified sizes (i.e. "combinations") and the sum of all the coefficients in one row counts all subsets of all sizes, of a set of a specified size.
That is worth understanding, but I'll give a quicker answer here. In constructing Pascal's triangle, each entry besides all the $1$s is the sum of the two entries above it. That means every number in each row gets added into the previous row twice. That means the sum of the entries in the next row will be twice as big. Every time you got down one row, the sum gets multiplied by exactly $2$.
|
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|
Show that $\boldsymbol{\mathrm{F}}$ is independent of path.
Consider a vector field $\boldsymbol{\mathrm{F}}(x,y) = \langle 2xy,
x^2 \rangle$ and three curves that start at $(1, 2)$ and end at
$(3,2)$. Explain why $$\int\limits_{C}\boldsymbol{\mathrm{F}}\cdot
\text{ d}\boldsymbol{\mathrm{r}}$$ has the same value for all three
curves, and what is this common value?
(There is a graph of three curves, but I'm pretty sure it's not necessary. For your reference, this is Stewart's Calculus, p. 1082, section 16.3 #11.)
My work: notice that
$$\begin{align}
&\dfrac{\partial}{\partial y}[2xy]=2x \\
&\dfrac{\partial}{\partial x}[x^2] = 2x
\end{align}$$
and thus, $\boldsymbol{\mathrm{F}}$ is conservative.
My understanding is that we need to show that $\boldsymbol{\mathrm{F}}$ is independent of path. Looking at my theorems provided doesn't help.
And if I do find such a theorem, I'm not sure what to use for $\boldsymbol{\mathrm{r}}$.
|
As you showed, the vector field is conservative, so it doesn't matter which path you take, the only thing you need are the starting and end point.
First, as $\mathrm{F}$ is conservative, you have to calculate a function $f$ such that $\nabla f=\mathrm{F}$. An easy way to do this is using this formula:
$$\displaystyle f(x,y) = \int_{0}^{x}\mathrm{F}_{1}(t,0)dt + \int_{0}^{y}\mathrm{F}_2(x,t)dt$$
Where $\mathrm{F}_1,\mathrm{F}_2$ are the first and second value of the vector field $\mathrm{F}$. Therefore $$\displaystyle \int_{C}\mathrm{F}\cdot dr = f(x,y)|^{r_1}_{r_0}$$
Where $r_1$ and $r_0$ are your end and starting points, respectively.
|
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|
Calculate limit $\lim_{n\rightarrow\infty}\frac{(4n-100)^{4n-100}n^n}{(3n)^{3n}(2n)^{2n}}?$ The limit $$\lim_{n\rightarrow\infty}\dfrac{(4n)^{4n}n^n}{(3n)^{3n}(2n)^{2n}}$$ can be calculated as $\lim_{n\rightarrow\infty}\dfrac{4^{4n}}{3^{3n}2^{2n}}=\lim_{n\rightarrow\infty}\left(\dfrac{4^4}{3^32^2}\right)^n=\infty$.
What about $$\lim_{n\rightarrow\infty}\dfrac{(4n-100)^{4n-100}n^n}{(3n)^{3n}(2n)^{2n}}?$$ Would it still be $\infty$?
|
Hints:
Can you see that $(4n-100)^{4n-100} \sim (4n)^{4n}$ as $n\to \infty$?
Then $$\lim_{n\rightarrow\infty}\dfrac{(4n-100)^{4n-100}n^n}{(3n)^{3n}(2n)^{2n}}=\lim_{n\rightarrow\infty}\dfrac{(4n)^{4n}n^n}{(3n)^{3n}(2n)^{2n}}$$
|
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|
Percentages not adding up I have a series of percentages:
132/220 (60%) and 88/220 (40%).
Now when I break them down into subcategories and then recalculate the percentages them come out 5% different.
81/140 (58%) and 59/140 (42%) (percentages ROUNDED).
21/40 (52.5%) and 19/40 (47.5%).
14/20 (70%) and 6/20 (30%).
16/20 (80%) and 4/20 (20%).
then we I do the averages I get:
65.125% and 34.875%
I have no idea what is happening???
|
If the samples/subcategories have different sizes, the average of the averages is not the same as the average of all numbers.
You have to make a weighted average: multiply each percentage by the size of the group, add them together and divide by the total size.
|
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|
An equation which has solution modulo every integer In the book Abstract Algebra by Dummit and Foote he remarks that there is an equation which has solutions modulo every integer but has no integer solutions. The equation he gives is $$3x^3+4y^3+5z^3=0$$My question is how do we prove that this has solutions modulo every integer and also that it has no integer solutions? If anyone can give hints it would be great.
|
Here is another
$$ x^2 + y^2 + z^9 = 216, $$ where we allow $z$ negative or positive or zero as desired. Same conclusion for
$$ x^2 + y^2 + z^9 = 216 p^3, $$ with prime $p \equiv 1 \pmod 4.$
See PDF
|
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|
Why doesn't $\arccos x = -\tfrac12\sqrt{3}$ have any solutions? I have this exercise with an unclear answer. The question is this:
$$\arccos x = -\frac{\sqrt3}{2}\,.$$
The answer is this:
$$\begin{gather*}
\varphi(x)= \arccos x\\
V_\varphi = [0,\pi]\\
-\frac{\sqrt3}{2}\notin V_\varphi\,.
\end{gather*}
$$
can someone this a bit more to me? How does this prove that there are no solutions?
|
As a couple of others have already pointed out, $-\sqrt{3}/2$ is simply not in the range of the arccosine. Here's an explanation as to why that's true.
Here's the graph of the cosine function over the interval $[-\pi,2\pi]$:
The issue is that this function is not one-to-one. As a result, we must restrict it to an appropriate domain where it is one-to-one in order to talk a about a restricted inverse function. Conventionally, the interval $[0,\pi]$ is chosen, which yields something like the following:
Now, the inverse of this restricted version of the cosine is what we know as the arccosine and its graph looks like so:
Of course, the domain and range have flipped - thus, $-\sqrt{3}/2$ (or any other negative number) is not in the range!
|
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|
When to use Integral Substitution? $e^x$$(1+e^x)^{1\over{2}}$
why can't use integration by part,
What is meant by in the form of f(g(x))g'(x)?
Can you give a few example?
Thank you
|
Integration by substitution says that
$$
\int f(g(x))g'(x) \; dx = \int f(u)\;du
$$
where $u = g(x)$. This is a tool that will let you compute the integral of all functions that look exactly like $f(g(x))g'(x)$. When you first see this, you might be thinking that this seems like a very specialised rule that only seem good for very special cases. But it turns out that this is very useful.
In your example, you have
$$
\int (1 + e^x)^{1/2}e^x\; dx.
$$
If you want to know whether or not you can use substitution you just have to determine if it is possible to write $(1 + e^x)^{1/2}e^x$ as $f(g(x))g'(x)$ for some functions $f$ and $g$. Mabe by trail and error you find that indeed if $f(x) = x^{1/2}$ and $g(x) = 1+e^x$, then $f(g(x))g'(x) = (1 + e^x)^{1/2}e^x$. And so with $u = g(x) = 1+e^x$, you get
$$
\int (1 + e^x)^{1/2}e^x\; dx = \int u^{1/2} \; du = \frac{2}{3}u^{3/2} + C = \frac{2}{3}(1+e^x)^{3/2} + C.
$$
You ask why you you can't use integration by parts. Integration by parts says that
$$
\int f(x)g'(x)\; dx = f(x)g(x) - \int f'(x)g(x)\; dx.
$$
So you would have to find $f$ and $g$ such that $f(x)g'(x) = e^x(1+e^e)^{1/2}$. You can of course choose $f(x) = (1 + e^x)^{1/2}$ and $g(x) = e^x$, so then you get
$$
(1 + e^x)^{1/2}e^x - \int \frac{1}{2}(1+e^x)^{1/2}e^xe^x\;dx.
$$
This is perfectly fine, but it doesn't do you any good. You wanted to find one complicated integral and now you just have another complicated integral. So you can use integration by parts, but, again, it doesn't help you much. It would be a good exercise to try other choices of $f$ and $g$ to convince yourself that this indeed doesn't help much.
|
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|
Square root and principal square root confusion A few months ago I asked a question about the $\pm$ symbol because I was confused about it... I still carry the same confusion (which really bugs me) but I think the real confusion has to do with the square root and principal square root. I hope I can finally grasp the concept with the following two questions...
Question 1:
$\sqrt{12x^2} = \sqrt{4\cdot3x^2} = 2|x|\sqrt{3}$
Why do we only have to keep in mind that $x$ could be negative, yet we just factorize $\sqrt{12}$ to $\sqrt{3\cdot4}$ and put the 2 right in front of the radical sign?
$\sqrt{12}$ could be factorized as $\sqrt{-3 \cdot -4}$ as well.
Question 2:
I've been told to take the square root of both sides in the following equation, but the square root doesn't have its own symbol right? Only the principal root does... So if you'd use the radical sign you're using the principal root and therefore you're missing out on a solution in the following scenario:
$(x + 8)² = 1 \iff \sqrt{(x + 8)^2} = \sqrt1 \iff x + 8 = 1 \iff x = -7$
Does that mean that whenever we want to take the square root of something (not the principal) we just use a $\pm$ symbol instead? For example:
$(x + 8)^2 = 1 \iff x + 8 = \pm 1 \iff x = -8\pm1$
|
Q1: If $a,b$ are nonnegative, then $\sqrt{ab}=\sqrt a\sqrt b$. If $a<0$ or $b<0$ (and you are not into complex numbers), you better not talk about the square roots on the right.
Q2:
In general $a=b\implies f(a)=f(b)$.
Since $(x+8)^2$ and $1$ are both nonnegative, we can ltake $\sqrt{\ }$ for the function $f$ and obtain
$$(x+1)^2=1\implies \sqrt{(x+1)^2}=\sqrt 1.$$
(Actually, since the square root function is injective on $[0,\infty)$, the use of "$\iff$" would be justifed, but itis not for general $f$).
Now recall that $$\tag0\sqrt {t^2}=|t|$$ for all real $t$ (wheras $\sqrt{t^2}=t$ only for nonnegative $t$, and we don't know if $x+1$ is nonnegative). Hence by simplificatiopn of expressions we find
$$\sqrt{(x+1)^2}=\sqrt 1\iff |x+1|=1$$
Now the appearence of $|\cdot|$ suggests that we investigate two cases:
$$\tag a x+a\ge 0\quad\land\quad x+1=1$$
$$\tag b x+a< 0\quad\land\quad -(x+1)=1$$
So the $\pm$ sign (or any other distinction of two cases, e.g. as I just did) ultimately comes from equation $(0)$.
An alternative way to proceed from the beginning is as follows, using the third binomial formula:
$$ (x+1)^2=1\iff (x+1)^2-1=0\iff((x++1)+1)((x+1)-1)=0$$
and now use that a product is zero if and only if at least one of the factors is zero.
|
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General solution of recurrence relation if two equal roots Consider the recurrence relation
$$
ax_{n+1}+bx_n+cx_{n-1}=0
$$
If the characteristic equation
$$
a\lambda^2+b\lambda+c=0
$$
has two equal roots, then the general solution is given by
$$
x_n=(A+nB)\alpha^n.
$$
Could you please explain that to me, I do not see that!
|
Let $D$ be defined as $(Dx)_n = x_{n+1}$. Assuming $a \neq 0$, we have
$x_{n+2}+{b \over a} x_{n+1} + {c \over a}x_n = 0$, or
$(D^2 +{b \over a} D + {c \over a}I)x = 0$. Suppose $\lambda$ is the repeated root of the quadratic $y^2+{b \over a} y + {c \over a} = 0$, then we can write the difference equation as
$(D-\lambda)^2 x = 0$.
We see that $(D-\lambda) x = 0$ iff $x_n = x_0 \lambda^n$ for some $x_0$ (with the convention that $0^n = \delta_{0n}$).
Also note that if $c$ is a constant, then $(D-1)x = c$ iff
$x_n = x_0 +nc$ for some $x_0$.
We can write the equation $(D-\lambda)^2 x = 0$ as the pair of equations $(D-\lambda)y = 0$, $(D-\lambda) x = y$.
The first gives $y_n = y_0 \lambda^n$ for some $y_0$.
If $\lambda = 0$, then solving $(D x)_n = y_0 \delta_{0n}$ gives $x_n = x_0 \delta_{0 n}+ y_0 \delta_{1 n}$ for some $x_0,y_0$ (that is, the sequence
$x=(x_0,y_0,0,...)$).
If $\lambda \neq 0$, then we have ${1 \over \lambda^n}((D-\lambda)x)_n = y_0$. Let $l$ denote the function $l_n = \lambda^n$, then we have
${ 1\over l} (D-\lambda) x = y_0$. Expanding gives
${1 \over l} Dx - {1 \over l} \lambda x = \lambda D {x \over l} - \lambda {x \over l} = \lambda (D-1) {x \over l} = y_0$, or $(D-1) {x \over l} = { y_0 \over \lambda}$.
Noting that $n \mapsto { y_0 \over \lambda}$ is a constant function, we see that this gives $({x \over l})_n = {x_n \over \lambda^n} = x_0 + n {y_0 \over \lambda}$, or $x_n = (x_0 + n {y_0 \over \lambda}) \lambda^n$, which is the desired form.
|
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|
Expected value with two uniformly distributed random variables A surveyor wishes to lay out a square region with each side having length L. However, because of a measurement error, he instead lays out a rectangle in which the north–south sides both have length X and the east–west sides both have length Y. Suppose that X and Y are independent and that each is uniformly distributed on the interval [L - A, L + A] (where 0 < A < L). What is the expected area of the resulting rectangle?
I tried using $E(x,y) = E(x)E(y) + Cov(x,y) = E(x)E(y)$ (since they are independent)
I ended up with $4A^2L^2$, but the answer should be just $L^2$
|
Hint:
If random variable $U$ is uniformly distributed over interval $[-1,1]$ then $U$ and $-U$ have the same distribution, so that $\mathbb{E}\left(U\right)=\mathbb{E}\left(-U\right)=-\mathbb{E}\left(U\right)$ hence: $$\mathbb{E}\left(U\right)=0$$
Note that $X=L+A.U$ is uniformly distributed over interval $[L-A,L+A]$ and find its expectation making use of the linearity of expectation.
|
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|
Why does induction procedure of Euler characteristic fail for non-convex polyhedra? What am I missing? Euler characteristic of convex polyhedra is always $V-E+F=2$.
Induction procedure reduces edges and vertices until we are down to one vertex whose $V-E+F=2$ and hence you are done.
The same induction procedure should also work for non-convex polyhedra and reduce to one vertex. So why does the induction procedure not work for non-convex polyhedra(giving $\neq 2$ values sometimes)?
Please use elementary language in your description (for example, a talk to freshmen).
|
The proof I know that $V-E+F=1$ for convex polyhedra starts by removing a face, then stereographically projecting the complement into the plane, to get a convex polygon in the plane subdivided into smaller polygons. Then one removes pairs of edges and vertices or edges and faces until you get down to a point.
The thing that fails in the nonconvex case is that after you remove a face, you can't necessarily project the rest into the plane. For example, if your polyhedron is shaped like a torus (so it has a hole in it), once you remove a face, the rest of the polyhedron can not be embedded in the plane.
|
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Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$ While doing some mathematical modelling of planetary orbits I have come up with two definite integrals $D_1$ and $D_2$ which appear to produce the same result R when tested with various values of $a$ ( where $0<a<1$).
$$
D_1
\, =\,
\int_0^{2\pi}f_1\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R
$$
and
$$D_2\, =\,\int_0^{2\pi}f_2\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R$$
The hypothesis: $D_1$ = $D_2$ has been proved in a separate question Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta = \int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta$ .
The remaining hypotheses $D_1$ = $R$ and $D_2$ = $R$ have not been proved. So the question is:-
Prove $D_1$ = $R$ or $D_2$ = $R$.
Only one proof is required because the other can then be obtained from $D_1$ = $D_2$.
For information
WolframAlpha computes expressions for the indefinite integrals $I_1,I_2$ as follows:-
$$I_1
\, =\,
\int\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
$$
$$constant1 + \frac
{a\,\sqrt{a^2-1}\sin\theta\,[-(2a^3+a)\cos^2\theta+3(a^2+1)cos\theta+a(2a^2-5)]}
{2(a^2-1)^{5/2}(a\cos\theta-1)^3}
$$
$$-\frac
{6a\,(a\cos\theta-1)^3\,\tanh^-1
\left(
\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}
\right)
}
{(2(a^2-1)^{5/2}\,(a\cos\theta-1)^3}
$$
and
$$I_2
\, =\,
\int\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
$$
$$constant2 -
\frac
{2a^2\sin\theta-sin\theta}
{2(a^2-1)^2(a\cos\theta-1)}
-\frac
{\sin\theta}
{2(a^2-1)(a\cos\theta-1)^2}
$$
$$
-\frac
{3a\tanh^-1\left(\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}\right)}
{(a^2-1)^{5/2}}
$$
Note that the final terms of each expression ( i.e. the terms involving $\tanh^{-1} $ and $\tan$ ) are equivalent to each other.
Also, note that
$$\int\frac{\cos\theta}{(1-a\cos\theta)^3}\,d\theta=
\frac{-\sin\theta}{(1-a\cos\theta)^3}
+\int \frac{3a\sin^2\theta}{(1-a\cos\theta)^4}\,d\theta.
$$
Written with StackEdit.
UPDATE 20141028
I have accepted TenaliRaman's answer. I don't yet understand all the steps but his helpful exposition gives me confidence that with time I can understand it because the methods cited (binomials, series) are ones I have learned (at high school).
The answer of M.Strochyk also appears to give a good proof. But the residue method is too advanced for me to understand at present.
UPDATE 20220713
I have now accepted Quanto's answer (because it is simple enough for me to understand). I have also added an answer based on Quanto's but with the intermediate steps written out.
|
With
$$
\int_{0}^{2\pi} \frac{1}{1-a \cos \theta} d \theta=\frac{2\pi}{\sqrt{1-a^2}}
$$
evaluate the following integrals successively
\begin{align}\int_{0}^{2\pi} \frac{1}{(1-a \cos \theta)^2} d\theta
=&\frac{d}{da} \int_{0}^{2\pi} \frac{a}{1-a \cos \theta} d \theta=\frac{2\pi}{(1-a^2)^{3/2}}\\
\int_{0}^{2\pi} \frac{\cos \theta}{(1-a \cos \theta)^3} d\theta
=&\frac{d}{da} \int_{0}^{2\pi} \frac{1}{2(1-a \cos \theta)^2 }d \theta=\frac{3a\pi}{(1-a^2)^{5/2}}\\
\int_0^{2\pi} \frac{3a\sin^2\theta}{(1-a\cos\theta)^4}\,d\theta
=&\int_0^{2\pi} \sin\theta \ d\left(-\frac{1}{(1-a\cos \theta)^3}\right)\\
\overset{ibp}=&\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}d\theta=\frac{3a\pi}{(1-a^2)^{5/2}}\\
\end{align}
|
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|
Multivariable limit .... no L'Hopital rule? I am looking a bit at limits for multivariable functions by myself, and I can't figure it out; my book only mentions them shortly, but now I am looking at an "assignments for those interested" and it says
$$\lim_{(x,y) \to (0,0)} \frac{\sin(xy)}{xy} \cos(x+y)$$
But that's a $0$ over $0$ expression... ?
Do we have a L'Hopital rule for these types of functions? How would I solve it?
|
Hint: $$\lim_{(x,y)\to(0,0)} \frac{\sin (xy)}{xy} = 1$$
And $\cos (x + y)$ is continuous.
|
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|
Rules of i ($\sqrt -1$) to a power $i^{2014}$ power =?
A. $i^{13}$
B. $ i ^{203}$
C. $i^{726}$
D. $i^{1993}$
E. $i^{2100}$
I don't understand the concept that powers of i repeat in fours and that "two powers of i are equal if their remainders are equal upon division by four". I especially don't understand the second part of the previous statement.
|
You know that $i^4=1$, because
$$
i^4=(i^2)^2=(-1)^2=1
$$
Now $i^5=i^4\cdot i=i$, $i^6=i^4\cdot i^2=-1$, $i^7=i^4\cdot i^3=-i$ and finally $i^8=i^4\cdot i^4=1$. You can go on forever, the powers of $i$ will repeat the same pattern
$$\dots\quad i\quad {-1} \quad {-i}\quad 1 \quad\dots$$
If $n=4q+r$ with $0\le r<4$, that is, $r$ is the remainder of the division of $n$ by $4$, you have
$$
i^n=i^{4q+r}=i^{4q}\cdot i^r=(i^4)^q\cdot i^r=1^q\cdot i^r=i^r
$$
In particular, if $m=4q_1+r$ and $n=4q_2+r$ have the same remainder, then
$$
i^m=i^r=i^n
$$
|
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|
Coprime numbers and equations Suppose $~m~$ and $~n~$ are coprime and both of them are greater than one. Is it right that equation $~mx + ny = (m-1)(n-1)~$ has solutions over non-negative integers?
For example $~ (x,y) = (6,0) ~$ satisfies equation $~3x + 10y = 18~$.
Thanks in advance.
|
Yes, this equation has a (unique) solution in non-negative integers.
Suppose $m > n$. Looking at it modulo $n$, if that equation holds, we have
$$mx \equiv (m-1)(n-1) \equiv -(m-1) \pmod{n}.$$
There is a unique integer $x_0 \in [0,n-1]$ with $mx_0 \equiv 1-m \pmod{n}$. Then $(m-1)(n-1) - mx_0$ is a multiple of $n$, and we need to see that it is not negative. Since $m(n-1) \equiv -m \not\equiv 1-m \pmod{n}$, we know $x_0 \neq n-1$, hence $0 \leqslant x_0 \leqslant n-2$ and
$$(m-1)(n-1) - mx_0 \geqslant (m-1)(n-1) - m(n-2) = m - (n-1) > 0.$$
|
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|
Distance between two skew lines I have 2 skew lines $L_A$ and $L_B$ and 2 parallel planes $H_A$ and $H_B$.
The line $L_A$ lies in $H_A$ and $L_B$ in $H_B$. If the equations of $H_A$ and $H_B$ are given like this:
$x+y+z = 0$ (for $H_A$)
$x+y+z = 5$ (for $H_B$)
Can I just simply say that the distance between two lines $L_A$ and $L_B$ is 5 since there the two planes they lies are separated apart by 5?
|
Yes,the shortest distance between the skew lines will equal the distance between the planes.
|
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|
Prove that $S$ is a subring of $\mathbb{Z}_{28}$ Question: $S=\{0,4,8,12,16,20,24\}.$ Prove that $S$ is a subring of $\mathbb{Z}_{28}$
Confusion 1: This might be a dumb question, but when we refer to $[4]$ in $S$, for example, is that the congruent class of $4$ modulo $28$ in our case? Because I assume $[4]=\{x\in\mathbb{Z}:x=4+nk\}$, and $n$ doesn't have to be $28$.
Confusion 2: In order to prove that set $S$ is closed under addition and multiplication, is there any other way than going case by case? I noticed that all elements in $S$ are multiples of $4$, but I don't really see how that would help me.
Thank you.
|
Since you are in $\mathbb{Z}_{28}$, everything is modulo $28$ i.e. $a=[a]=\{a+28k:k\in\mathbb{Z}\}$ for all $a\in\{0,2,\dots,27\}$.
Now you have noticed that all elements in $S$ are multiples of $4$, then it follows that if you add any two elements in $S$, the answer must still be a multiple of $4$, which must be an element of $S$. This shows that $S$ is closed under addition. Multiplication works the same way.
Edit: If the sum of two elements is a multiple of $4$, since $28$ is also a multiple of $4$, the sum modulo $28$ is again a multiple of $4$, which means it is in $S$.
|
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|
Recursion and Time Complexity Concept The question prompt is as follows:
Consider the function $f(n)$ defined as:
$$f(n) = \begin{cases}n(n-1)f(n-2) & n > 1\\1 & n=0,\; n=1\end{cases}$$
How may be $g(n)$ be defined to make $f(n) \in \mathcal O(g(n))$? (I'm supposed to choose one response from $1$ and a response from $2$)
*
*For $n>0$, define $g(n)$ as:
Option 1: $g(n) = g(n-1) + n$
Option 2: $g(n) = ng(n-1)$.
*Then, assign:
Option 1: $g(0) = 0$
Option 2: $g(0) = 100$.
Now, I solved the recursive equation and got the following:
$$f(n) = n!$$
If we define $g(n) = g(n-1) + n$:
$$g(n) = n(n+1)/2 + \{0\text{ or }100\}$$
If we define $g(n) = n\cdot g(n-1)$:
$$g(n) = n! \cdot \{0\text{ or }100\}$$
(The $\{0\text{ or }100\}$ because when $g(0)$ could be $0$ or $100$.)
However, I am not seeing how either definition of $g(n)$ yields $\lim_{n\to\infty} (f(n)/g(n)) = 0$.
|
Your problem is that you're seeking the wrong thing limit. In order for $f(n)$ to be $\mathcal O(g(n))$, you want:
$$\lim_{n\to\infty} \frac{f(n)}{g(n)} = c, \quad 0\leq c < \infty$$
Clearly, if we choose $g(n) = n!\cdot g(0)$ and $g(0) = 100$, the resulting value is $c=\frac{1}{100}$.
You probably were confused with the definition of little-oh:
$$f(n)\in o(g(n))\iff \lim_{n\to\infty} \frac{f(n)}{g(n)} = 0$$
To conclude: you did everything else right, except you messed up the limit definition of $\mathcal O(\cdot)$.
|
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|
Riemann Zeta Function at $s = 1 + 2 \pi i n / \ln 2$ I am aware that the function defined by
$$ \zeta(s) = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots $$ for $Re(s)>1$ can be extended to a function defined for $Re(s)>0$ by writing
$$ \zeta(s) = \frac{1}{1-2^{1-s}} \cdot \left( \frac{1}{1^s} - \frac{1}{2^s} + \frac{1}{3^s} - \frac{1}{4^s} + \cdots \right) $$
Using this representation, one sees that at $s=1$, since the sum evaluates to $\ln 2$, the Riemann Zeta Function has a simple pole of residue $1$. However, what happens at $$s= 1 + \frac{2\pi i n}{\ln 2} $$ where $n$ is a non-zero integer? The denominator is zero at these values of $s$, but it is well-known that the Riemann Zeta Function only has the one pole at $s=1$, so the the sum at these values must evaluate to $0$ to eliminate the poles which would otherwise form. How do we know though that that alternating sum has zeroes for these values?
|
This issue has been adressed by J. Sondow (2003) here: Zeros of the alternating zeta function on the line R(s)=1.
|
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|
How to prove the amounts of dominos with x+y=n+k = x+y=n-k? I've been trying to answer this question for hours with no luck at all.. The question is the following:
Question
*
*Imagine we have domino blocks of the following shape: [ x | y ] with x, y ∈ [0..n].
*Imagine 0 ≤ k ≤ n.
Show that the amount of blocks with
*
*x + y = n - k
is equal to the amount of blocks with
*
*x + y = n + k
The question also says:
In both cases, you will find (n - k + 1)/2, but you need to prove this.
Thanks a lot for the help!
|
For each $x$ from $0$ to $n-k$ let $y=n-k-x$. This makes $n-k+1$ blocks with $x+y=n-k$. But if $n-k+1$ is even, we must count only a half of them, because the block $[x|y]$ is the same as $[y|x]$. And if $x-k+1$ is odd, there is a block that has no pair, namely $[(x-k)/2, (x-k)/2]$.
For each block $[x|y]$ with $x+y=n-k$ we have exactly one block $[n-x|n-y]$, and $n-x+n-y=2n-(n-k)=n+k$, so the number of blocks with $x+y=n+k$ is the same as before.
|
{
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|
What is the generalized form of this identity and how to interpret it? I have learnt that for any inner product space of $\mathbb{C}$, we have
$$\langle f,g\rangle=\frac{1}{4}\Big[||f+g||^2-||f-g||^2+i\big(||f+ig||^2-||f-ig||^2\big) \Big]$$
I know how to prove it, but I have a hard time to remember the formula. I think maybe there are some general form or tricks which can help me to remember the formula, or there may be some geometric interpretation of the identity.
|
This was an answer to the original question, which specifically asked for help memorizing the identity.
You could try to remember it "in parts"
*
*$[f,g] = \|f+g\|^2-\|f-g\|^2$
*$\langle f,g \rangle = \frac{1}{4}([f,g]+i[f,ig])$
If someone can offer a geometric interpretation of $[f,g],$ that would certainly help.
|
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|
solve this simple equation:$ax^2+byx+c=0$ I need help solving the diophantine equation:$$ax^2+bxy+c=0$$
The quadratic formula does not seem to help much. Please help.
|
Usually we solve Diophantine equations for positive integer solutions. Therefore I think you need positive integer solutions.
Consider the equation $ax^2+bxy+c=0$ as a quadratic equation of $x.$ For integer values of $x$ discriminant of this should be a perfect square. $$△_x=(by)^2-4ac=z^2$$ for some integer $z.$ Since $(by+z)(by-z)=4ac,$ this can have only finitely many integer solution for $y$ and $z.$
For each of these values, you have two rational solution for $x$ given by $\dfrac{-by+z}{2a}$ and $\dfrac{-by-z}{2a}.$
Pick positive integer solutions.
|
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|
If $A$ is connected, is at least one of the sets $\mathrm{Int}A$ and $\mathrm{Bd}A$ connected? If $A$ is a connected subset of $X$, does it follow at least one of the sets $\mathrm{Int}A$ and $\mathrm{Bd}A$ are connected?
I have found counterexamples showing that they not both need to be connected, and was wondering whether this result can be strengthened or not.
|
Thanks to Daniel Fischer for his correction!
Consider the union $X$ of two full triangles $T,T'$ in the plane that meet at a vertex, and remove a small disk from the interior of $T$. Then $X$ is connected, and neither the interior nor the boundary of $X$ are connected.
|
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|
Why can't prime numbers satisfy the Pythagoras Theorem? That is, why can't a set of 3 prime numbers be a Pythagorean triplet? Suppose $a$, $b$ and $c$ are three prime numbers.
How to prove that $a^2 + b^2 \neq c^2$?
|
The sum of two odd numbers are even, so one of the numbers must be $2$.
If $a$ or $b$ are $2$ we have $a^2+4=c^2$ or $4=(c+a)(c-a)$ Since $c-a$ and $c+a$ have the same parity, this is impossible.
If $c=2$ we have $a^2+b^2=4$ but since $a$ and $b$ are positive, both must be $1$, but $1^2+1^2=2$.
|
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|
Show complex solutions exist Let A be a complex number and B a real number. Show that the equation
$\,\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B = 0\,$ has a solution iff $\,\lvert A^2\rvert \geq 4B$. If this is so, show that the solution set is a circle or a single point.
Well i am trying to do the first part first. So assuming the equation has a solution that would mean $z = x+iy$ satisfies the equation.
I was going to let $A = s+it$ for a complex number, but it is not working out for me. Wrong step?
|
We have
$$
0=\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B =
z\overline{z}+\frac{1}{2}(Az+\overline{Az})+\frac{1}{4}\lvert A\rvert^2-\frac{1}{4}\lvert A\rvert^2+B=\left\lvert z+\frac{1}{2}A\right\rvert^2-\frac{1}{4}\lvert A\rvert^2+B
$$
If $4B>\lvert A\rvert^2>0$, then $\lvert z^2\rvert+ \mathrm{Re}\, (Az) + B=\left\lvert z+\frac{1}{2}A\right\rvert^2-\frac{1}{4}\lvert A\rvert^2+B>0$, and hence no solutions.
If $4B>\lvert A\rvert^2\le 0$, set $C=\frac{1}{2}\sqrt{\lvert A\rvert^2-4B}$, and our equation is equivalent to
$$
\left\lvert z+\frac{1}{2}A\right\rvert^2=C^2,
$$
and hence equivalent to
$$
\left\lvert z+\frac{1}{2}A\right\rvert=C,
$$
the set of solutions of which is the circle centered at $-A/2$ with radius $C$.
|
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|
If I roll three dice at the same time, how many ways the sides can sum up to $13$? If I rolled $3$ dice how many combinations are there that result in sum of dots appeared on those dice be $13$?
|
At the lower division math level we can do the following easily given the low number of combinations:
1)list the number of potential combinations
116
265
355
364
454
2) Now we find out the numbers of way that we can arrange the listed numbers in which it is:
3
6
3
6
3 respectively
thus when we add the numbers we get 21
|
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|
How to find the value of a function where a relation is given If $g(x)$ is a polynomial satisfying
$$g(x)g(y)=g(x)+g(y)+g(xy)-2$$
For all real value of $x$ and $y$. And
$$ g(2)=5$$
then find
$$g(3)=?$$
|
We have that $\big(g(x)-1\big)\big(g(y)-1\big)=g(xy)-1$.
Set $f(x)=g(x)-1$, then
$$
f(xy)=f(x)f(y)\tag{1}
$$
Also, $f(x)\ge 0$, for all $x>0$, as $f(x)=(f(\sqrt{x}))^2$
Thus $f(x^n)=(f(x))^n$, as as $f(x)\ge 0$, then $(f(x^{1/n}))^n=f(x)\to f(x^{1/n})=(f(x))^{1/n}$. Altogether $f(x^{m/n})=(f(x))^{m/n}$, and as $f$ is continuous
$f(x^{a})=(f(x))^{a}$, for all $x>0$ and $a\in\mathbb R$.
If $g(2)=5$, then $f(2)=1=4$, and
$$
g(3)=f(3)+1=f(2^{\log_2 3 })+1=\big(f(2)\big)^{\log_2 3 }+1=4^{\log_2 3 }+1=2^{2\log_2 3}=9+1=10.
$$
Note. Alternatively, $(1)$ implies that $f(x)=x^n$, for some $n\in\mathbb N$, as $f$ is a polynomial. To do that first observe that if $f(x)=0$, then $x=0$, and hence $f(x)=ax^n$. Next, observe that $a>0$, and $f(2)=4$, and as $f(x)f(x)=f(x^2)$, then $a=1$. Thus $f(x)=x^n$. Finally, $f(2)=4$ implies that $n=2$ and that
$$
f(x)=x^2.
$$
|
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|
Find the power set $P(S)$ for $S=\{\emptyset, \{\emptyset\}, \{\emptyset \{\emptyset\}\}\}$ Find the power set $P(S)$ for $S=\{\emptyset, \{\emptyset\}, \{\emptyset \{\emptyset\}\}\}$
OK this problem confuses me for many reasons, but here is what I know. The cardinality of a set is $2^n$ where $n$ is the number of elements in the set. In this problem, however, how can the empty set be an element?
If I were just going to say this set has $2^n$ elements that would mean the set has $2^3$ or 8 elements, but I don't know what those elements would be other than empty sets.
Any help in understanding this problem is greatly appreciated!
|
Every set has the empty set as an element. However #(S) should be zero where #(S) is the cardinality of set S.
The power set of the empty set or sets of empty sets should be simply the empty set.
Remember however that the power set of a set is not a cardinality. It is a set constructor that builds a set from the elements of set S. So technically your answer should be the set which is the factorial combination of each element ie if set S = { a, b, c} then P(S) = {a, b, c, {a,b}, {a,c}, {b, c}, {a, b, c} }
IMHO of course !
|
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|
Integrate without substitution I'm wondering how I could integrate the following without substitution:$$\int \frac{4}{1 + e^{-x}}dx$$
I know we can factor out the constant so that $4 * \int \frac{1}{1 + e^{-x}}$ but I'm stumped as of what to do next. Could anyone help me out?
|
$$\int \frac{1}{1+\frac{1}{e^x}} dx = \int \frac{e^x}{1+ e^x} dx = \ln (1 + e^x) + C $$
|
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|
$f(x)=x^6(x-7)^3$ over the range $[-14, 10]$ Where does the function achieve its global minimum?
I've identified the global min as $14/3$ and was told this was incorrect. I tried its corresponding y value as well and received the same answer. Other critical points are $7$ and $0$ and they are also incorrect.
|
In $x=\frac{14}{3}$, $f$ achieve local minimun, note that
$f(-4)=(-4)^{6}(-4-7)^{3}= -5. 451\,8\times 10^{6}$
$f(14/3)=(14/3)^{6}(14/3-7)^{3}=-\frac{2582\,630\,848}{19\,683}=-1.312\,1\times 10^{5}$
|
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|
Usage of law of sines The vertex angle of an isosceles triangle is 35 degrees. The length of the base is 10 centimeters. How many centimeters are in the perimeter?
I understand the problem as there are two sides with length 10 and one side of unknown length.
I used laws of sines to find the side corresponding to the 35 degree angle.
$$
\frac{\sin35^\circ}{x} = \frac{\sin72.5^\circ}{10}
$$
I get 6 as the length of the side
The answer is 43.3. They get it by dropping an altitude from the vertex to the base and forming congruent right triangles. What am I doing wrong?
|
The other answer does not use the law of sines, as your title states.
You're given that the side of $10$ lies opposite the vertex angle of $35^\circ$.
First find the other two angles (the "base angles"). As the triangle is isoceles, they are equal. Denote one base angle as $\theta$.
By the angle sum of the triangle, $2\theta + 35^\circ = 180^\circ$, giving $\theta = 72.5^\circ$.
Now employ the law of sines. Denote one of the sides adjoining the vertex as $x$.
So $\displaystyle \frac{x}{\sin 72.5^\circ} = \frac{10}{\sin 35^\circ}$.
Solving that gives you $x \approx 16.63$cm.
Clearly there are two such sides with length $x$, giving a perimeter of $2x + 10 \approx 2(16.63) + 10 = 43.26$cm. I'm rounding off to $2$ decimal places.
|
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"url": "https://math.stackexchange.com/questions/992660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Prove that $\sum_{i=0}^n4^i$ = $1/3(4^{n+1} - 1)$ $\sum_{i=0}^n4^i$ = $1/3(4^{n+1} - 1)$
Attempt:
Let $n =0$ $4^0 = 1 \text{ and } (4^{0+1} -1)/3 = 1$
Assume true at $n = k \text{ so we have} \sum_{i=0}^k4^i = 1/3(4^{k+1} -1)$
The part I'm stuck at is the 3rd step. Can someone point me in the right direction after this?
$\sum_{i=0}^{k+1}4^i$ ?
|
Hint: $\displaystyle \sum_{i=0}^{k+1} 4^i = \displaystyle \sum_{i=0}^k 4^i + 4^{k+1}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/992746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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|
Inverse of an infinitely large matrix? This is probably a trivial problem for some people, but I've spent quite some time on it:
What is the inverse of the infinite matrix $$
\left[\begin{matrix}
0^0 & 0^1 & 0^2 & 0^3 & \ldots\\
1^0 & 1^1 & 1^2 & 1^3 & \ldots\\
2^0 & 2^1 & 2^2 & 2^3 & \ldots\\
3^0 & 3^1 & 3^2 & 3^3 & \ldots\\
\vdots & \vdots & \vdots & \vdots &\ddots
\end{matrix}\right]
$$
(Assume that $0^0=1$ for this problem).
I'm not sure if this problem has a solution or is well-defined, but if it has a solution, it would help greatly in a ton of stuff I'm doing (mostly related to generating functions and polynomial approximations). I began by taking the inverse of progressively larger square matricies, but I didn't see any clear pattern.
|
Numerically, we get some very interesting results for the matrix $M_{ij}=(i-1)^{j-1}$ if it is expressed in the Fourier basis with alternating row signs,
$$A := \begin{bmatrix}1 \\ & -1 \\ && 1 \\&&& -1 \\ &&&& \ddots \end{bmatrix}\mathcal{F}^{-1} \left[\begin{matrix}
0^0 & 0^1 & 0^2 & 0^3 & \ldots\\
1^0 & 1^1 & 1^2 & 1^3 & \ldots\\
2^0 & 2^1 & 2^2 & 2^3 & \ldots\\
3^0 & 3^1 & 3^2 & 3^3 & \ldots\\
\vdots & \vdots & \vdots & \vdots &\ddots
\end{matrix}\right] \mathcal{F}.$$
As the discretization size of $M$ goes up, $M$ becomes extremely ill-conditioned and the entries blow up, so numerical calculations using standard doubles (15 digits of accuracy) will fail when it becomes much larger than 10-by-10. However, there are toolboxes that let you do computation with greater precision, and I used one with 512 digits of accuracy to produce the following images of the real and complex entries of the matrices $A$ up to size 128-by-128:
The color of the $(i,j)$'th pixel in each plot represents $A_{ij}$, the value of the $(i,j)$'th entry of $A$. The first row is the real part of $A$, and the second row is the imaginary part. Red means large positive value, blue means large negative value, and the maximum real value in the plot is shown in the middle. The picture is big but scaled down for display on math.stackexchange - you can open it in a new tab to see it in more detail.
It looks like in this Fourier basis the normalized matrices are converging to an integral operator with a smooth kernel,
$$\frac{1}{N}A v \rightarrow \int_0^1 (K(x,y) + iJ(x,y)) v(y) dy,$$
where $K$ and $J$ are the smooth functions in the pictures, and $N$ is some renormalization factor.
Since the kernel is smooth, the action of $A$ will annihilate highly oscillatory functions. Recalling the definition $A = \text{diag}(1,-1,\dots) \mathcal{F}^{-1} M \mathcal{F}$, we see exactly how $M$ is ill-conditioned and what functions are in its numerical null space - functions that are the Fourier transform of a highly oscillatory functions.
Turning this around, there should exist an inverse $A^{-1}$ for the renormalized limit, acting on a space of functions that are sufficiently smooth. Indeed, the following is a plot of the spectrum of the renormalized $A$ in the 128-by-128 case (top right), and it's dominant singular vectors (left, real on top imaginary on bottom):
Here's the Matlab code I used:
%Using Advanpix multiprecision computing toolbox, http://www.advanpix.com/
mp.Digits(512+9);
mp.GuardDigits(9);
jjmin = 2;
jjmax = 7;
jjrange = jjmax - jjmin + 1;
for jj=jjmin:jjmax
N = 2^jj;
%Generate original matrix M, where M_nm = (n-1)^(m-1)
v = mp((0:N-1)');
M = mp(zeros(N,N));
for kk=0:(N-1)
M(:,kk+1)=v.^kk;
end
%Generate matrix D F^(-1) M F, where F is the fft, and
%D is the diagonal matrix with diagonal [1, -1, 1, -1, ...]
FMF = mp(zeros(N,N));
for k=1:N
ek = mp(zeros(N,1));
ek(k)=1;
FMF(:,k) = ifft(ifftshift(M*fft(fftshift(ek))));
end
for k=1:N
FMF(k,:) = (-1)^(k-1)*FMF(k,:);
end
%plot it
subplot(2, jjrange,jj - jjmin+1)
imagesc(real(double(FMF)))
title(['N=', num2str(N)]);
subplot(2, jjrange, jjrange + jj -jjmin+1)
imagesc(imag(double(FMF)))
format short
title(['max= ',num2str(max(real(FMF(:))),3)])
end
subplot(2,jjrange,1)
ylabel('real(D F^{-1} M F)')
subplot(2,jjrange,jjrange + 1)
ylabel('imag(D F^{-1} M F)')
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/992850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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|
How can i prove this interesting limit Was just playing around any thought it was interesting.
knowing now that $a>b>0$
$\lim\limits_{x\rightarrow \infty} (a^x-b^x)^{\frac{a+b}{x}} = a^{a+b}$
|
It can be done as follows:
$$
\lim_{x\to\infty} \left(a^x-b^x\right)^{\frac{a+b}{x}}=\lim_{x\to\infty} a^{a+b}\cdot\left(1-\left(\frac{b}{a}\right)^x\right)^{\frac{a+b}{x}}=
$$
Now we have $\lim_{x\to\infty} \left(\frac{b}{a}\right)^x=0$ and $\lim_{x\to\infty} \frac{a+b}{x}=0$ and therefore:
$$
\lim_{x\to\infty} a^{a+b}\cdot\left(1-\left(\frac{b}{a}\right)^x\right)^{\frac{a+b}{x}}=a^{a+b}\cdot\left(1-0\right)^0=a^{a+b}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/992945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
How to show that there is no odd $n$ bigger than 100 such that the group $U(n)$ will contain $2^n$ number of elements? Consider $U(n):=\{1\leq r\leq n: (r, n)=1\}$. Under multiplication modulo $n$ it forms an abelian group. Its order will be $\varphi(n)$ i.e. $\varphi(n)=|U(n)|$.
Assume that $n\geq 100$. Then my question is : does there exists any odd natural number $m_o$ such that the group $U(n)$ will have the order $2^n$ ? The reason I have chosen $n\geq 100$ is that : I have already checked the result below 100 and the answer I have obtained so far is NO if $33\leq n\leq 100$. No need to care about $n\leq 33$. I think the result is still NO if we go beyond 100.
But the question is : how can I show no such odd $m_o$ would exists ?
What I have approached is the following:
Since $m_o$ is odd so we can write $$m_o=\prod\limits_{i=1}^r p_i^{a_i}$$ where $p_i$ are odd primes and $a_i$ are positive integers. Then we must have
$$U(m_o)\simeq U(p_1^{a_1})\times \cdots \times U(p_r^{a_r})$$ where $\times$ denotes the external direct product of groups. I don't know if we can create some contradiction over here.
Also note that , the question is equivalent to say that if $m_o$ is one such odd number then there will be $2^n$ number of positive integers less than $m_o$ and relatively prime to $n$. We have to show no such $m_o$ is possible to find after 100.
in this way also no contradiction i am able to get.
Please help me. thanks in advance.
|
You want positive integers $n$ such that $\varphi(n)$ is a power of $2$. But if $n=p_1^{a_1}\cdots p_m^{a_m}$, then $\varphi(n)=\displaystyle\prod_{i=1}^m (p_i^{a_i}-p_1^{a_1-1})$.
Therefore, the only primes you can have are $2$ and those such that $p_i-1$ is a power of $2$. The only such primes which are known are $3, 5, 17, 257, 65537$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/992966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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|
Prove $\left|\sum_{i=1}^n x_i y_i \right| \le \dfrac{1}{a} \sum_{i=1}^n {x_i}^2 + \dfrac{a}{4}\sum_{i=1}^n {y_i}^2$ If $X,Y$ are vectors in $\mathbb{R}^n$ and $a>0$ show that:
$$\left|\sum_{i=1}^n x_i y_i \right| \le \dfrac{1}{a} \sum_{i=1}^n {x_i}^2 + \dfrac{a}{4}\sum_{i=1}^n {y_i}^2 (*)$$
I started with Cauchy–Schwarz inequality and got:
$$\left|\sum_{i=1}^n x_i y_i \right| \le {\sum_{i=1}^n {x_i}^2}^\frac{1}{2} \cdot{\sum_{i=1}^n {y_i}^2}^\frac{1}{2}(**)$$
So apparently we need to show that $(**) < (*)$
and I'm stuck. Don't really know what to do with $a$'s.
Please help!
|
Observe that the inequality follows first from Cauchy-Schwarz inequality and then by AM-GM inequality.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/993084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
How prove $\pi^2>2^\pi$ show that
$$\pi^2>2^\pi$$
I use computer found
$$\pi^2-2^\pi\approx 1.044\cdots,$$
can see this
I know
$$\Longleftrightarrow \dfrac{\ln{\pi}}{\pi}>\dfrac{\ln{2}}{2}$$
so let
$$f(x)=\dfrac{\ln{x}}{x}$$
so
$$f'(x)=\dfrac{1-\ln{x}}{x^2}=0,x=e$$
so $f(x)$ is Strictly increasing
on $(2,e)$, and is Strictly decreasing
on $(e,3) $
so I can't know $f(2)$ and $f(\pi)$ which is bigger?
maybe this problem exsit have easy methods by hand
|
Hint:
$$\frac{\ln 2}{2}=\frac{\ln 4}{4}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/993170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
}
|
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