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https://math.stackexchange.com/questions/2311602/prove-that-if-f-n-converges-to-f-almost-uniformly-then-f-n-converges-t | # Prove that if $(f_n)$ converges to $f$ almost uniformly then $(f_n)$ converges to $f$ in measure.
Let $E$ be a measurable set, $(f_n)$ a sequence of real valued measurable functions on $E$ and $f$ a real valued measurable function on $E$. It is required to prove that if $(f_n)$ converges to $f$ almost uniformly then $(f_n)$ converges to $f$ in measure. The following is my proof.
Let $\epsilon>0$. Suppose $(f_n)$ converges to $f$ almost uniformly. Then there exists $F\subseteq E$ such that $m(F)<\epsilon$ and $f_n$ converges uniformly to $f$ on $E\setminus F$. Thus there exists $N\in\mathbb{N}$ such that for each $n\geq N$ and $x\in E\setminus F,$ $|f_n(x)-f(x)|<\epsilon.$
But\begin{align} \{x\in E:|f_n(x)-f(x)|\geq\epsilon\}=\{x\in F:|f_n(x)-f(x)|\geq\epsilon\}\cup\{x\in E\setminus F:|f_n(x)-f(x)|\geq\epsilon\}.\end{align}
Let $n\geq N$. Then $m(\{x\in F:|f_n(x)-f(x)|\geq\epsilon\})<\epsilon$ and
$m(\{x\in E\setminus F:|f_n(x)-f(x)|\geq\epsilon\})=0$.
Therefore for each $n\geq\mathbb{N}$, $m(\{x\in E:|f_n(x)-f(x)|\geq\epsilon\})<\epsilon$.
Hence $m(\{x\in E:|f_n(x)-f(x)|\geq\epsilon\})=0$ as $n\to\infty$ and the proof is complete.
Is this proof alright? Thanks.
• Yes, your proof is essentially correct. I posted an answer with more details. Let me know if you have any question. – Ramiro Jun 6 '17 at 23:53
Let $\epsilon>0$. Suppose $(f_n)$ converges to $f$ almost uniformly. Then there exists $F\subseteq E$ such that $m(F)<\epsilon$ and $f_n$ converges uniformly to $f$ on $E\setminus F$. Thus there exists $N\in\mathbb{N}$ such that for each $n\geq N$ and $x\in E\setminus F,$ $|f_n(x)-f(x)|<\epsilon.$
It means, for $n\geq N$, $$E\setminus F \subseteq \{x\in E:|f_n(x)-f(x)|<\epsilon\}$$
So, for $n\geq N$, $$\{x\in E:|f_n(x)-f(x)|\geq\epsilon\} \subseteq F$$ and so we have, for $n\geq N$, $$m(\{x\in E:|f_n(x)-f(x)|\geq\epsilon\})\leqslant m(F)<\epsilon$$
Hence $m(\{x\in E:|f_n(x)-f(x)|\geq\epsilon\})=0$ as $n\to\infty$ and the proof is complete. | 2020-02-22T17:42:05 | {
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https://math.stackexchange.com/questions/3311365/is-there-a-way-to-find-the-specific-variable-coefficient-in-a-binomial-expansion | Is there a way to find the specific variable coefficient in a binomial expansion?
If a problem asks to find the coefficient of a variable, say, $$x^2$$, in a large binomial expansion, is there a way to solve without doing the whole expansion (I do it with Pascal's Triangle / Binomial Theorem). For example, in this problem
The coefficient of $$x^2$$ in the expansion of $$(\frac{1}{x} + 5x)^8$$ is equal to the coefficient of $$x^4$$ in the expansion of $$(a+5x)^7$$, $$a$$ is a real number. Find the value of $$a$$.
I expand it out and get different answers on different tries. Not sure what's the best method to proceed. If anyone could help I would appreciate it so much!
• For some basic information about writing mathematics at this site see, e.g., here, here, here and here. Aug 2, 2019 at 11:29
The binomial theorem tells you that $$\left(\frac1x + 5x\right)^8 = \sum_{i = 0}^8\binom8i\frac{1}{x^i}(5x)^{8-i}\\ (a + 5x)^7 = \sum_{j = 0}^7\binom7ja^j(5x)^{7-j}$$ Since we're looking for the $$x^2$$ term in the first sum, that happens only when $$i = 3$$. For the second sum we're interested in the $$x^4$$ term which only is when $$j = 3$$. We get $$\binom83\frac1{x^3}(5x)^5 = 56\cdot 5^5x^2\\ \binom73a^3\cdot(5x)^4 = 35a^3\cdot 5^4x^4$$ Now equate the two coefficients, and solve for $$a$$.
It's easier if you find the coefficient of $$x^{10}$$ in $$(1+5x^2)^8$$, which is the same as the coefficient of $$x^5$$ in $$(1+5x)^8$$; this is $$\binom{8}{5}\cdot 5^5=\binom{8}{3}\cdot 5^5$$ The coefficient of $$x^4$$ in the expansion of $$(a+5x)^7$$ is $$\binom{7}{4}\cdot a^3\cdot 5^4=\binom{7}{3}\cdot a^3\cdot 5^4$$ Now the equation is easy.
This is exactly what the binomial formula is for!
$$(a+b)^n = \sum_{i=0}^n {n \choose i} a^ib^{n-i}$$
I'll show you how to do the first one. Start by plugging in $$a= 1/x$$ and $$b = 5x$$ and $$n=8$$ and simplify:
$$(1/x+5x)^8 = \sum_{i=0}^8 {8 \choose i} (1/x)^i(5x)^{8-i} = \sum_{i=0}^8 {8 \choose i} 5^{8-i} \frac{x^{8-i}}{x^i} = \sum_{i=0}^8 {8 \choose i} 5^{8-i} x^{8-2i}.$$
This is the sum of nine terms, one for each $$i=0,1,\ldots, 8$$. Notice each term has a different power of $$x$$. So the coefficient of $$x^2$$ happens when $$n-i = 2$$ which is when $$i=3$$. That means the coefficient is
$${8 \choose 3} 5^{8-3} = {8 \choose 3} 5^{5}$$
which you can simplify.
Do the same for the other binomial and equate the two answers and then solve for $$a$$. | 2022-12-09T18:56:57 | {
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https://mathhelpboards.com/threads/maries-question-from-facebook-about-square-roots.3428/ | ### Welcome to our community
#### Sudharaka
##### Well-known member
MHB Math Helper
I was wondering how to find square root for numbers like 128. Can you do it step by step please?
#### SuperSonic4
##### Well-known member
MHB Math Helper
I was wondering how to find square root for numbers like 128. Can you do it step by step please?
I would use Prime Factorisation to find the prime roots and then take advantage of the rule $\sqrt{ab} = \sqrt{a}\sqrt{b}$. The numbers are positive so this is allowed. To that end if decimal approximations are needed, it's best to learn some of the simple prime roots like $\sqrt{2},\ \sqrt{3}\ \sqrt{5}$
Using 128 as an example (this one is relatively simple as an integer power of 2)
$128 \div 2 = 64$ -- so 2 is one prime factor
$64 \div 2 = 32$
and so on until
$4 \div 2 = 2$
Thus we can say that $128 = 2^7$ and so $\sqrt{128} = \sqrt{2^7}$
Using the rule above I then say that $\sqrt{2^7} = \sqrt{2^6}\sqrt{2} = 2^3\sqrt{2} = 8\sqrt{2}$
I would leave it as $8\sqrt{2}$ for an exact answer
#### MarkFL
Staff member
If you now wish to compute rational approximations for $\sqrt{2}$ by hand, here are two recursive algorithms (the second converges more rapidly):
i) $\displaystyle x_{n+1}=\frac{x_n+2}{x_n+1}$
ii) $\displaystyle x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}$
You simply need to "seed" both recursions with some initial guess.
We know $\displaystyle 1<\sqrt{2}<2$ so $\displaystyle \frac{3}{2}$ will work just fine.
#### soroban
##### Well-known member
Here's another recursive approximation for a square root.
Suppose we want $$\sqrt{N}.$$
Let $$a_1$$ be the first approximation to $$\sqrt{N}.$$
Then: $$a_2 \:=\:\frac{N+a_1^2}{2a_1}$$ is an even better approximation.
Repeat the process with $$a_2$$ . . . and so on.
This procedure converges very quickly
. . depending on $$a_1.$$
Find $$\sqrt{3}$$, using $$a_1 = 1.7$$
$$a_2 \:=\:\frac{3 + 1.7^2}{2(1.7)} \:=\:1.732352941 \:\approx\:1.732353$$
$$a_3 \:=\:\frac{3 + 1.732353^3}{2(1.732353)} \:=\:1.732050834$$
ChecK: .$$1.732050834^2 \:=\:3.000000091\;\;\checkmark$$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Here is the reasoning behind this procedure.
We want $$\sqrt{N}.$$
We approximate the square root, $$a_1.$$
Suppose our approximation is correct.
Then $$a_1$$ and $$\tfrac{N}{a_1}$$ would be equal, right?
Chances are, they are not equal.
One is larger than the square root, one is smaller.
What is a better approximation of the square root?
. . . the average of the two numbers!
Hence: .$$a_2 \:=\:\frac{a_1 + \frac{N}{a_1}}{2} \:=\:\frac{\frac{a_1^2 + N}{a_1}}{2}$$
Therefore: $$a_2 \:=\:\frac{N +a_1^2}{2a_1}$$
#### Bacterius
##### Well-known member
MHB Math Helper
Soroban's method is called the Newton-Raphson method, by the way (if someone wants to search for more information on it). It tends to work most of the time, and converges remarkably quickly (twice as many correct digits per iteration, in general) but can also fail sometimes. This is related to chaos theory and notably used in fractal rendering, but can be mitigated by appropriately choosing the initial guess. It should be noted it cannot fail for polynomials of order less than 3.
Its general form is:
$$a_{n + 1} = a_n - \frac{f(a_n)}{f'(a_n)}$$
To converge to a root of $f(x)$. Which root it converges to is dependent on the initial guess $a_1$.
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Mark's method ii, soroban's averaging method, and Newton-Raphson are all identical.
Generally, Newton-Raphson finds a root of $f(x)=0$.
The algorithm is:
$x_{k+1} = x_k - \dfrac{f(x_k)}{f'(x_k)}$
With $f(x)=x^2-N$ this becomes:
$x_{k+1} = x_k - \dfrac{x_k^2-N}{2x_k}$
$x_{k+1} = \dfrac{x_k}{2} + \dfrac{N}{2x_k} \quad$ or $\quad x_{k+1} = \dfrac{N + x_k^2}{2x_k}$
The initial guess should be above the root for fastest results.
If the initial guess is below the root, the first iteration will jump to the other side of the root and it will worsen the approximation.
The second form is the one soroban presented.
If we pick N=2 as Mark did, the first form becomes:
$x_{k+1} = \dfrac{x_k}{2} + \dfrac{1}{x_k}$ | 2021-06-20T03:46:25 | {
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https://math.stackexchange.com/questions/1464952/can-intersection-of-two-manifolds-be-xy-0 | # Can intersection of two manifolds be $xy=0$
I know that if two manifolds intersect transversally then their intersection is a manifold. But I was trying to construct an example where the intersection is not a manifold. But I still do not see how intersection of two manifolds cannot be a manifold. It would be great, if any one has any counterexample giving intersection as $xy=0$ or any other set which is not a manifold .
• I think there's an example like this in Guillemin and Pollack. Let one of the manifolds be the $xy$-plane and the other something which lies entirely in the upper half-space, such that the points where $z=0$ is precisely $xy=0$.
– user98602
Oct 5, 2015 at 4:16
Let $M_1 \subset \Bbb R^3$ be the $xy$-plane and $M_2 = \{(x,y,z) | (xy)^2 = z\}$. Then $M_2 \cap M_1 = \{(x,y,z) | xy = 0, z = 0\}$, as you desire. $M_2$ is a manifold by the regular value theorem, because the map $f: \Bbb R^3 \to \Bbb R$, $f(x,y,z) = z-(xy)^2$ is a submersion.
Transversality breaks rather wildly, as you notice: in fact, the tangent planes of $M_2$ at points where $z=0$ are the $xy$-plane.
Guillemin and Pollack have some nice pictures of non-transverse intersections in their book. Something with this result is probably in there.
You can see a picture of $M_2$ on WolframAlpha here.
E: In fact, let $X = f^{-1}(0)$ be the zero set of a smooth function $\Bbb R^2 \to \Bbb R$. Then the exact same construction works to build $X$ as the intersection of two submanifolds of $\Bbb R^3$; let $M_2 = \{(x,y,z) | z = f(x,y)^2\}$; the map $f(x,y) - z$ is still a submersion.
And, in fact, any closed subset of $\Bbb R^n$ is the zero set of a smooth function, in particular, say, the Cantor set or the Koch snowflake. So you can make these intersections pretty wild!
• thanks a lot for the nice example Oct 5, 2015 at 4:26
• This is really nice this simple construction seems to yield a lot of counterexamples. But I am wondering if $z=xy$ is also a counter example Oct 5, 2015 at 4:36 | 2022-06-28T22:45:20 | {
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https://math.stackexchange.com/questions/775206/integral-int-0-pi-2-ln1-alpha-sin2-x-dx-pi-ln-frac1-sqrt1-alp | # Integral $\int_0^{\pi/2} \ln(1+\alpha\sin^2 x)\, dx=\pi \ln \frac{1+\sqrt{1+\alpha}}{2}$
$$I_1:=\int_0^{\pi/2} \ln(1+\alpha\sin^2 x)\, dx=\pi \ln \frac{1+\sqrt{1+\alpha}}{2}, \qquad \alpha \geq -1.$$ I am trying to prove this integral $I_1$. We can write $$\int_0^{\pi/2} \ln(\alpha(1/\alpha+\sin^2 x))dx=\int_0^{\pi/2} \left(\ln \alpha+\ln (\frac{1}{\alpha}+\sin^2 x)\right)dx=\frac{\pi}{2} \ln \alpha+I_2$$ where $$I_2=\int_0^{\pi/2}\ln (\frac{1}{\alpha}+\sin^2 x) \,dx$$ however I am not sure what that will do for us.... I also tried differentiating wrt $\alpha$ but didn't get placed. How can we prove $I_1$ result? Thanks
• It really simplifies on differentiating – evil999man Apr 30 '14 at 4:03
• If I may ask, are you writing a textbook about nice integrals ? – Claude Leibovici Apr 30 '14 at 6:51
• @ClaudeLeibovici Well it is a hobby of mine. I am recently retired, so now I am trying to solve all integrals in the world. One day I would like to make these into a collection, kind of like one has a picture frame. thank you friend – Jeff Faraci Apr 30 '14 at 12:52
• @Integrals Perhaps your nickname should be Diophantus then, because you seek to find all integral solutions. ;) – David H May 26 '14 at 5:54
• @DavidH Thanks for the laughter, Ha! Maybe I will change my name in the future to this, if so: you will know why;) – Jeff Faraci May 26 '14 at 21:47
Let $\displaystyle I(a) = \int_{0}^{\pi /2} \ln(1+ a \sin^{2}x) \, dx$.
Then differentiating under the integral sign, $$I'(a) = \int_{0}^{\pi /2} \frac{\sin^{2} x}{1+a \sin^{2} x} \, dx = \int_{0}^{\pi /2} \frac{1}{a+ \csc^{2} x} \, dx .$$
Now let $u = \cot x$.
Then
\begin{align} I'(a) &= \int_{0}^{\infty} \frac{1}{a+1+u^{2}} \frac{1}{1+u^{2}} \, du \\ &= \frac{1}{a} \int_{0}^{\infty} \left(\frac{1}{1+u^{2}} - \frac{1}{1+a+u^{2}} \right) \, du \\ &= \frac{1}{a} \left(\frac{\pi}{2} - \frac{1}{1+a} \int_{0}^{\infty} \frac{1}{1+\frac{u^{2}}{1+a}} \, du \right) \\ &=\frac{1}{a} \left(\frac{\pi}{2} - \frac{1}{\sqrt{1+a}} \int_{0}^{\infty} \frac{1}{1+v^{2}} \, dv \right) \\ &= \frac{\pi}{2a} \left(1 - \frac{1}{\sqrt{1+a}} \right). \end{align}
Then integrating back,
\begin{align} I(a) &= \frac{\pi}{2} \int \frac{1}{a} \left(1 - \frac{1}{\sqrt{1+a}} \right) \, da \\ &= \frac{\pi}{2} \int \frac{1}{u^{2}-1} \left(1 - \frac{1}{u} \right) 2u \, du \\ &= \pi \int \frac{1}{1+u} \, du \\ &= \pi \ln \left(1+ \sqrt{1+a} \right) + C. \end{align}
And since $I(0) = 0$, $C = -\pi \ln 2$.
Therefore,
$$I(a) = \pi \ln \left(\frac{1 +\sqrt{1+a}}{2} \right) .$$
• Once again, the solver of all integrals...Thank you this is pretty clear. – Jeff Faraci Apr 30 '14 at 12:48
This is quite similar to Random Variable's solution, just the starting integral is different to make the calculations a bit simpler.
Consider $$I(b)=\int_0^{\pi/2} \ln(b^2+\sin^2x)\,dx$$ $$\Rightarrow I'(b)=\int_0^{\pi/2} \frac{2b}{b^2+\sin^2x}\,dx=2b\int_0^{\pi/2} \frac{dx}{b^2+\cos^2x}$$ Factor out $\cos^2x$ from the denominator and rewrite $\sec^2x=1+\tan^2x$ to obtain: $$I'(b)=2b\int_0^{\pi/2} \frac{\sec^2x\,dx}{b^2+1+b^2\tan^2x}\,dx$$ Use the substitution $\tan x=t$ and evaluating the resulting integral is easy so $$I'(b)=\frac{\pi}{\sqrt{1+b^2}} \Rightarrow I(b)=\pi\ln\left(b+\sqrt{1+b^2}\right)+C$$ For $b=0$, $I(0)=-\pi\ln 2$, hence $C=-\pi\ln2$ $$\Rightarrow \int_0^{\pi/2} \ln(b^2+\sin^2x)\,dx=\pi\ln\left(\frac{b+\sqrt{1+b^2}}{2}\right)$$ Replace $b$ with $1/\sqrt{\alpha}$ and you get: $$\int_0^{\pi/2} \ln(1+\alpha \sin^2x)\,dx-\frac{\pi}{2}\ln \alpha=\pi\ln\left(\frac{1+\sqrt{1+\alpha}}{2\sqrt{\alpha}}\right)$$ $$\Rightarrow \int_0^{\pi/2} \ln(1+\alpha \sin^2x)\,dx=\pi\ln\left(\frac{1+\sqrt{1+\alpha}}{2}\right)$$ $\blacksquare$
\begin{align}& \partiald{}{\alpha}\color{#c00000}{\int_{0}^{\pi/2}\ln\pars{1 + \alpha\sin^{2}\pars{x}}\,\dd x} =\partiald{}{\alpha}\int_{0}^{\pi/2}\ln\pars{1 + \alpha\cos^{2}\pars{x}}\,\dd x \\[3mm]&=\int_{0}^{\pi/2}{\cos^{2}\pars{x} \over 1 + \alpha\cos^{2}\pars{x}}\,\dd x ={1 \over \alpha}\int_{0}^{\pi/2} {\bracks{1 + \alpha\cos^{2}\pars{x}} - 1 \over 1 + \alpha\cos^{2}\pars{x}}\,\dd x \\[3mm]&={\pi \over 2\alpha}- {1 \over \alpha}\int_{0}^{\pi/2}{\dd x \over 1 + \alpha\cos^{2}\pars{x}} ={\pi \over 2\alpha}- {1 \over \alpha}\int_{0}^{\pi/2}{\sec^{2}\pars{x}\,\dd x\over \tan^{2}\pars{x} + 1 + \alpha} \\[3mm]&={\pi \over 2\alpha} - {1 \over \alpha\root{1 + \alpha}} \int_{0}^{\infty}{\dd t \over t^{2} + 1} ={\pi \over 2}\pars{{1 \over \alpha} - {1 \over \alpha\root{1 + \alpha}}} \end{align}
\begin{align}& \color{#c00000}{\int_{0}^{\pi/2}\ln\pars{1 + \alpha\sin^{2}\pars{x}}\,\dd x} ={\pi \over 2}\ \overbrace{\int_{0}^{\alpha}\pars{{1 \over t} - {1 \over t\root{1 + t}}}\,\dd t} ^{\ds{\mbox{Set}\ x \equiv 1 + \root{1 + t}\ \imp\ t = x^{2} - 2x}} \\[3mm]&={\pi \over 2}\int_{2}^{1 + \root{1 + a}}{2\,\dd x \over x} \end{align}
$$\color{#66f}{\large% \int_{0}^{\pi/2}\ln\pars{1 + \alpha\sin^{2}\pars{x}}\,\dd x =\pi\,\ln\pars{1 + \root{1 + \alpha} \over 2}}$$
• The Boss! Thanks Felix – Jeff Faraci Jun 17 '14 at 21:18
• @Integrals Thanks. You are here again. Great. – Felix Marin Jun 17 '14 at 22:23
Differentiate $I_2$ (probably easier than $I_1$) with respect to $a$, then use the Weierstrass substitution to transform it into an integral that you can calculate with residues. I will look into it as well.
Edit: you can also integrate by parts to get rid of the log. | 2019-08-22T13:19:42 | {
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http://math.stackexchange.com/questions/420220/removing-the-remainder-of-a-fraction | # removing the remainder of a fraction
I would like to remove the remainder from a fraction if possible. I want a function
$$f(x,y) = x/y - remainder$$
for example
$$f(3,2) = 1$$ $$f(7,2) = 3$$ $$f(12,5) = 2$$
It seems so simple but its been bugging me for a while. Please help.
-
What properties do you want this remainder to have? Do you want it to be an integer? Because it seems to me like you're looking at standard integer division. – Patrick Da Silva Jun 14 '13 at 12:32
Do you mean something like en.wikipedia.org/wiki/Floor_and_ceiling_functions? – Amzoti Jun 14 '13 at 12:33
x and y are both integers. The output of f(x,y) is also an integer. Yes I need something like a floor function but I want it from first principles if possible – Manatok Jun 14 '13 at 12:37
You are looking for division with remainder We have $y=\lfloor \frac yx \rfloor x+r$, where $f(x,y)=\lfloor \frac yx \rfloor, r=y-\lfloor \frac yx \rfloor x$. What do you mean by "from first principles?" | 2014-04-20T21:43:25 | {
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https://tobydriscoll.net/fnc-julia/leastsq/fitting.html | # 3.1. Fitting functions to data¶
In Section 2.1 we saw how a polynomial can be used to interpolate data—that is, derive a continuous function that evaluates to give a set of prescribed values. But interpolation may not be appropriate in many applications.
Demo 3.1.1
Here are 5-year averages of the worldwide temperature anomaly as compared to the 1951–1980 average (source: NASA).
year = 1955:5:2000
temp = [ -0.0480, -0.0180, -0.0360, -0.0120, -0.0040,
0.1180, 0.2100, 0.3320, 0.3340, 0.4560 ]
scatter(year,temp,label="data",
xlabel="year",ylabel="anomaly (degrees C)",leg=:bottomright)
A polynomial interpolant can be used to fit the data. Here we build one using a Vandermonde matrix. First, though, we express time as decades since 1950, as it improves the condition number of the matrix.
t = @. (year-1950)/10
n = length(t)
V = [ t[i]^j for i in 1:n, j in 0:n-1 ]
c = V\temp
10-element Vector{Float64}:
-14.114000002897516
76.36173811075241
-165.45597225538512
191.9605667047853
-133.27347224857206
58.0155777892561
-15.962888892063518
2.694806349746933
-0.2546666667183191
0.010311111113228591
The coefficients in vector c are used to create a polynomial. Then we create a function that evaluates the polynomial after changing the time variable as we did for the Vandermonde matrix.
If you plot a function, then the points are chosen automatically to make a smooth curve.
p = Polynomial(c)
f = yr -> p((yr-1950)/10)
plot!(f,1955,2000,label="interpolant")
As you can see, the interpolant does represent the data, in a sense. However it’s a crazy-looking curve for the application. Trying too hard to reproduce all the data exactly is known as overfitting.
In many cases we can get better results by relaxing the interpolation requirement. In the polynomial case this allows us to lower the degree of the polynomial, which limits the number of local max and min points. Let $$(t_i,y_i)$$ for $$i=1,\ldots,m$$ be the given points. We will represent the data by the polynomial
(3.1.1)$y \approx f(t) = c_1 + c_2t + \cdots + c_{n-1} t^{n-2} + c_n t^{n-1},$
with $$n<m$$. Just as in (2.1.2), we can express a vector of $$f$$-values by a matrix-vector multiplication. In other words, we seek an approximation
(3.1.2)$\begin{split}\begin{bmatrix} y_1 \\ y_2 \\ y_3 \\ \vdots \\ y_m \end{bmatrix} \approx \begin{bmatrix} f(t_1) \\ f(t_2) \\ f(t_3) \\ \vdots \\ f(t_m) \end{bmatrix} = \begin{bmatrix} 1 & t_1 & \cdots & t_1^{n-1} \\ 1 & t_2 & \cdots & t_2^{n-1} \\ 1 & t_3 & \cdots & t_3^{n-1} \\ \vdots & \vdots & & \vdots \\ 1 & t_m & \cdots & t_m^{n-1} \\ \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix} = \mathbf{V} \mathbf{c}.\end{split}$
Note that $$\mathbf{V}$$ has the same structure as the Vandermonde matrix in (2.1.2) but is $$m\times n$$, thus taller than it is wide. It’s impossible in general to satisfy $$m$$ conditions with $$n<m$$ variables, and we say the system is overdetermined. Rather than solving the system exactly, we have to find a best approximation. Below we specify precisely what is meant by this, but first we note that Julia uses the same backslash notation to solve the problem in both the square and overdetermined cases.
Demo 3.1.2
Here are the 5-year temperature averages again.
year = 1955:5:2000
t = @. (year-1950)/10
temp = [ -0.0480, -0.0180, -0.0360, -0.0120, -0.0040,
0.1180, 0.2100, 0.3320, 0.3340, 0.4560 ]
10-element Vector{Float64}:
-0.048
-0.018
-0.036
-0.012
-0.004
0.118
0.21
0.332
0.334
0.456
The standard best-fit line results from using a linear polynomial that meets the least-squares criterion.
Backslash solves overdetermined linear systems in a least-squares sense.
V = [ t.^0 t ] # Vandermonde-ish matrix
@show size(V)
c = V\temp
p = Polynomial(c)
size(V) = (10, 2)
-0.18773333333333328 + 0.11670303030303028∙x
f = yr -> p((yr-1955)/10)
scatter(year,temp,label="data",
xlabel="year",ylabel="anomaly (degrees C)",leg=:bottomright)
plot!(f,1955,2000,label="linear fit")
If we use a global cubic polynomial, the points are fit more closely.
V = [ t[i]^j for i in 1:length(t), j in 0:3 ]
@show size(V)
size(V) = (10, 4)
(10, 4)
Now we solve the new least-squares problem to redefine the fitting polynomial.
The definition of f above is in terms of p. When p is changed, then f calls the new version.
p = Polynomial( V\temp )
plot!(f,1955,2000,label="cubic fit")
If we were to continue increasing the degree of the polynomial, the residual at the data points would get smaller, but overfitting would increase.
## The least-squares formulation¶
In the most general terms, our fitting functions take the form
(3.1.3)$f(t) = c_1 f_1(t) + \cdots + c_n f_n(t)$
where $$f_1,\ldots,f_n$$ are all known functions with no undetermined parameters. This leaves only $$c_1,\ldots,c_n$$ to be determined. The essential feature of a linear least-squares problem is that the fit depends only linearly on the unknown parameters. For instance, a function of the form $$f(t)=c_1 + c_2 e^{c_3 t}$$ is not of this type.
At each observation $$(t_i,y_i)$$, we define a residual, $$y_i - f(t_i)$$. A sensible formulation of the fitting criterion is to minimize
$R(c_1,\ldots,c_n) = \sum_{i=1}^m\, [ y_i - f(t_i) ]^2,$
over all possible choices of parameters $$c_1,\ldots,c_n$$. We can apply linear algebra to write the problem in the form $$R=\mathbf{r}^T \mathbf{r}$$, where
$\begin{split}\mathbf{r} = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\y_{m-1} \\ y_m \end{bmatrix} - \begin{bmatrix} f_1(t_1) & f_2(t_1) & \cdots & f_n(t_1) \\[1mm] f_1(t_2) & f_2(t_2) & \cdots & f_n(t_2) \\[1mm] & \vdots \\ f_1(t_{m-1}) & f_2(t_{m-1}) & \cdots & f_n(t_{m-1}) \\[1mm] f_1(t_m) & f_2(t_m) & \cdots & f_n(t_m) \\[1mm] \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_n \end{bmatrix}.\end{split}$
Recalling that $$\mathbf{r}^T\mathbf{r}=\| \mathbf{r} \|_2^2$$, and renaming the variables to standardize the statement, we arrive at the general linear least-squares problem.
Definition 3.1.3 : Linear least-squares problem
Given $$\mathbf{A}\in\mathbb{R}^{m \times n}$$ and $$\mathbf{b}\in\mathbb{R}^m$$, with $$m>n$$, find
(3.1.4)$\argmin_\limits{\mathbf{x}\in \mathbb{R}^n}\, \bigl\| \mathbf{b}-\mathbf{A} \mathbf{x} \bigr\|_2^2.$
The notation argmin above means to find an $$\mathbf{x}$$ that produces the minimum value.
While we could choose to minimize in any vector norm, the 2-norm is the most common and convenient choice. For the rest of this chapter we exclusively use the 2-norm. In the edge case $$m=n$$ for a nonsingular $$\mathbf{A}$$, the definitions of the linear least-squares and linear systems problems coincide: the solution of $$\mathbf{A}\mathbf{x}=\mathbf{b}$$ implies $$\mathbf{r}=\boldsymbol{0}$$, which is a global minimum of $$\| \mathbf{r} \|_2^2 \ge 0$$.
## Change of variables¶
The most familiar and common case of a polynomial least-squares fit is the straight line, $$f(t) = c_1 + c_2 t$$. Certain other fit functions can be transformed into this situation. For example, suppose we want to fit data using $$g(t)= a_1 e^{a_2 t}$$. Then
(3.1.5)$\log y \approx \log g(t) = (\log a_1) + a_2 t = c_1 + c_2 t.$
While the fit of the $$y_i$$ to $$g(t)$$ is nonlinearly dependent on fitting parameters, the fit of $$\log y$$ to a straight line is a linear problem. Similarly, the power-law relationship $$y\approx f(t)=a_1 t^{a_2}$$ is equivalent to
(3.1.6)$\log y \approx (\log a_1) + a_2 (\log t).$
Demo 3.1.4
Finding numerical approximations to $$\pi$$ has fascinated people for millennia. One famous formula is
$\displaystyle \frac{\pi^2}{6} = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \cdots.$
Say $$s_k$$ is the sum of the first $$k$$ terms of the series above, and $$p_k = \sqrt{6s_k}$$. Here is a fancy way to compute these sequences in a compact code.
a = [1/k^2 for k=1:100]
s = cumsum(a) # cumulative summation
p = @. sqrt(6*s)
scatter(1:100,p,title="Sequence convergence",
xlabel=L"k",ylabel=L"p_k")
This graph suggests that maybe $$p_k\to \pi$$, but it’s far from clear how close the sequence gets. It’s more informative to plot the sequence of errors, $$\epsilon_k= |\pi-p_k|$$. By plotting the error sequence on a log-log scale, we can see a nearly linear relationship.
ϵ = @. abs(π-p) # error sequence
scatter(1:100,ϵ,title="Convergence of errors",
xaxis=(:log10,L"k"),yaxis=(:log10,"error"))
The straight line on the log-log scale suggests a power-law relationship where $$\epsilon_k\approx a k^b$$, or $$\log \epsilon_k \approx b (\log k) + \log a$$.
k = 1:100
V = [ k.^0 log.(k) ] # fitting matrix
c = V \ log.(ϵ) # coefficients of linear fit
2-element Vector{Float64}:
-0.182375249728302
-0.9674103233127926
In terms of the parameters $$a$$ and $$b$$ used above, we have
a,b = exp(c[1]),c[2];
@show b;
b = -0.9674103233127926
It’s tempting to conjecture that the slope $$b\to -1$$ asymptotically. Here is how the numerical fit compares to the original convergence curve.
plot!(k,a*k.^b,l=:dash,label="power-law fit")
Thus the variable $$z=\log y$$ can be fit linearly in terms of the variable $$s=\log t$$. In practice these two cases—exponential fit and power law—are easily detected by using log-linear or log-log plots, respectively.
## Exercises¶
1. ✍ Suppose $$f$$ is a twice-differentiable, nonnegative real function. Show that if there is an $$x^*$$ such that $$f'(x^*)=0$$ and $$f''(x^*)>0$$, then $$x^*$$ is a local minimizer of the function $$[f(x)]^2$$.
2. ⌨ Here are counts of the U.S. population in millions from the census performed every ten years, beginning with 1790 and ending with 2010.
3.929, 5.308, 7.240, 9.638, 12.87, 17.07, 23.19, 31.44, 39.82, 50.19, 62.95, 76.21,
92.22, 106.0, 122.8, 132.2, 150.7, 179.3, 203.3, 226.5, 248.7, 281.4, 308.7
(a) Find a best-fitting cubic polynomial for the data. Plot the data as points superimposed on a (smooth) graph of the cubic over the full range of time. Label the axes. What does the fit predict for the population in the years 2000, 2010, and 2020?
(b) Look up the actual U.S. population in 2000, 2010, and 2020 and compare to the predictions of part (a).
3. ⌨ The following are weekly box office earnings (in dollars) in the U.S. for the 2012 film The Hunger Games. (Source: boxofficemojo.com.)
189_932_838, 79_406_327, 46_230_374, 26_830_921, 18_804_290,
13_822_248, 7_474_688, 6_129_424, 4_377_675, 3_764_963, 2_426_574,
1_713_298, 1_426_102, 1_031_985, 694_947, 518_242, 460_578, 317_909
(Note that Julia lets you use _ where you would normally put a comma in a long number.) Fit these values to a function of the form $$y(t)\approx a e^{b t}$$. Plot the data together with the fit using standard linear scales on the axes, and then plot them again using a log scale on the vertical axis.
4. ⌨ In this problem you are trying to find an approximation to the periodic function $$g(t)=e^{\sin(t-1)}$$ over one period, $$0 < t \le 2\pi$$. As data, define
$t_i = \frac{2\pi i}{60}, \; y_i = g(t_i), \quad i=1,\ldots,60.$
(a) Find the coefficients of the least-squares fit
$y(t) \approx c_1 + c_2t + \cdots + c_7 t^6.$
Superimpose a plot of the data values as points with a curve showing the fit.
(b) Find the coefficients of the least-squares fit
$y \approx d_1 + d_2\cos(t) + d_3\sin(t) + d_4\cos(2t) + d_5\sin(2t).$
Unlike part (a), this fitting function is itself periodic. Superimpose a plot of the data values as points with a curve showing the fit.
5. ⌨ Define the following data in Julia.
t = 0:.5:10
y = tanh.(t)
(a) Fit the data to a cubic polynomial. Plot the data together with the polynomial fit over the interval $$0 \le t \le 10$$.
(b) Fit the data to the function $$c_1 + c_2z + c_3z^2 + c_4z^3$$, where $$z=t^2/(1+t^2)$$. Plot the data together with the fit. What feature of $$z$$ makes this fit much better than the original cubic?
6. ⌨ One series for finding $$\pi$$ is
$\frac{\pi}{2} = 1 + \frac{1}{3} + \frac{1\cdot 2}{3\cdot5} + \frac{1\cdot 2\cdot 3}{3\cdot 5\cdot 7} + \cdots.$
Define $$s_k$$ to be the sum of the first $$k$$ terms on the right-hand side, and let $$e_k=|s_k-\pi/2|$$.
(a) Calculate $$e_k$$ for $$k=1,\ldots,20$$, and plot the sequence on a log-linear scale.
(b) Determine $$a$$ and $$b$$ in a least-squares fit $$e_k \approx a \cdot b^k$$, and superimpose the fit on the plot from part (a).
7. ⌨ Kepler found that the orbital period $$\tau$$ of a planet depends on its mean distance $$R$$ from the sun according to $$\tau=c R^{\alpha}$$ for a simple rational number $$\alpha$$. Perform a linear least-squares fit from the following table in order to determine the most likely simple rational value of $$\alpha$$.
Planet
Distance from sun in Mkm
Orbital period in days
Mercury
57.59
87.99
Venus
108.11
224.7
Earth
149.57
365.26
Mars
227.84
686.98
Jupiter
778.14
4332.4
Saturn
1427
10759
Uranus
2870.3
30684
Neptune
4499.9
60188
8. ✍ Show that finding a fit of the form
$y(t) \approx \frac{a}{t+b}$
can be transformed into a linear fitting problem (with different undetermined coefficients) by rewriting the equation.
9. ✍ Show how to find the constants $$a$$ and $$b$$ in a data fitting problem of the form $$y(t)\approx t/(at+b)$$ for $$t>1$$ by transforming it into a linear least-squares fitting problem. | 2022-05-28T22:26:58 | {
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https://math.stackexchange.com/questions/1463787/probability-of-winning-in-a-die-rolling-game-with-six-players | Probability of winning in a die rolling game with six players
There are 6 players numbered 1 to 6, 1 Player, Player 2, ..., Player 6.
Player 1 rolls a die , if he gets a 1 wins and the game ends, otherwise the die passes to the player whose number matches the number that presents the die and the player makes a second pitch, if is obtained the number of the player who has thrown, he wins and the game ends, otherwise the given passes to the player whose number matches the number rolled, the player rolls the die, if is obtained the number of the player who has thrown, he wins and the game ends, otherwise the die passes to the player whose number matches the number that presents the die in this third release, and so on.
Calculate the probability that player 1 wins.
• I think this title is more "searchable". FYI there is a flag suggesting that the question be closed for not having any. Usually the users raising such flags would like you to share your own thoughts. Now that an elegant answer has been posted, I don't know what to suggest. Try to do better with your next question, and hope you enjoy the site. – Jyrki Lahtonen Oct 4 '15 at 16:53
Let $p$ be the probability Player 1 (ultimately) wins. If Player 1 does not win on her first toss, by symmetry, the other players all have equal probabilities of being "next", so all have equal probabilities of ultimately winning, namely $\frac{1-p}{5}$.
On the first toss, either P1 tosses a $1$ and wins immediately, or tosses something else and becomes effectively one of the "other" players. Thus $$p=\frac{1}{6}+\frac{5}{6}\cdot\frac{1-p}{5},$$ and now we can solve for $p$.
• Is there really such a symmetry here? Probability that Player 1 wins on the first roll is $\frac 16$, but probability that Player 2 wins is $\frac 56\cdot\frac 16$, since Player 2 loses immediately if Player 1 already has won. – Ennar Oct 4 '15 at 13:01
• There is symmetry between Players 2 to 6, that is what I used. Of course 1 is special. – André Nicolas Oct 4 '15 at 13:06
• And the solution is $p=\frac27$ – Alice Ryhl Oct 4 '15 at 16:39
• @KristofferRyhl: Yes, that's right. And the others each have probability $\frac{1}{7}$ of ultimately winning, the first player is twice as likely to win as any of the others. – André Nicolas Oct 4 '15 at 16:41
• And, clearly, the way to make the game fair is to roll the die first to choose the first player :-) – Mark Hurd Oct 4 '15 at 16:51
I also got to 2/7, but in a different way.
Let $p$ be the probability that player 1 wins the game, either on his current turn or in the future. Let $q$ be the probability that player 1 eventually wins when it is someone elses turn, in particular the other player does not end the game in the current turn.
Then player 1 can win right away with probability $p$, or first pass the turn to someone else and win later on in the game with probability $q$: $$p = \frac16 + \frac56 q.$$
When it is someone else's turn, and they don't win, then either it will be player 1's turn again or the turn will pass to one of the other 4 players. So if it is now someone else's turn, the probability that player 1 will get the next turn and win is $\frac16 p$. But if the turn goes to yet someone else with probability $\frac46$ (other than player 1 and the current player) player 1 will eventually win is still $q$. Hence, $$q = \frac16 p + \frac46 q.$$
Solving these two equations for $p$ is easily done by hand, giving $p = \frac27$ (and $q=\frac17$ which, by symmetry, is the probability of each of the other 5 players winning).
• This is actually a more illuminative way to solve. The second equation gives us $p=2q$ (which explicitly gives us relative advantage of the first player) and we don't need the first equation as $p+(6-1)q=1$. – A.S. Oct 17 '15 at 21:04
• "Let q be the probability that player 1 wins when it is someone elses turn." How can player 1 win on someone else's turn ? – true blue anil Oct 17 '15 at 22:46
• @trueblueanil $q$ is probability to win eventually/overall if the current turn is not that of a player, not right after this turn. Think of player 2 at the beginning of the game and let $q$ be the probability that he wins. If the first roll is 1, P2 doesn't win. If the roll is 2, his probability to win becomes $p$ as now it's the same game and it's his turn to roll. If the first roll is not 1 or 2, his probability to win stays $q$. Hence average of the first two probabilities must be $q$: $q=(p+0)/2$. – A.S. Oct 17 '15 at 22:58
• What's the lacuna here. A can win on 1st turn Pr = 1/6. If she loses, with Pr = 5/6, someone, say X is in the same position as A was, so if A's ultimate winning Pr = p, that of X = 5p/6, and by symmetry ultimate winning Pr of all non-A's = 25p/6, p+25p/6 = 1, p = 6/31 – true blue anil Oct 18 '15 at 0:10
• @tr " that of X = 5p/6" is false – A.S. Oct 18 '15 at 6:37 | 2019-10-19T12:52:16 | {
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https://math.stackexchange.com/questions/1219489/how-to-convince-people-that-0-is-even?noredirect=1 | # How to convince people that 0 is even [duplicate]
Some people say that 0 is neither even nor odd. I say that 0 is even.
Is there a simple way to convince people that 0 is even and the statement that "0 is neither even nor odd" is false.
## marked as duplicate by Daniel W. Farlow, Zev Chonoles, JMoravitz, Community♦Apr 4 '15 at 4:58
• Most people who practice mathematics regularly take $0$ to be an even integer, since it is divisible by $2$: $2 \times 0 = 0$. Furthermore, taking $2$ to be even makes a lot of statements a lot more concise. Call it odd, call it even if'n ya wanna, but I vote for $2$ even!!! Cheers! – Robert Lewis Apr 4 '15 at 4:20
• en.wikipedia.org/wiki/Parity_of_zero – Todd Wilcox Apr 4 '15 at 4:21
• "Believe"? This is mathematics, not some touchy-feely subject like poetry. – Jonathan Hebert Apr 4 '15 at 4:22
• @JonathanHebert: Sorry, "think" is better? – user172675 Apr 4 '15 at 4:24
• @JonathanHebert That is hardly necessary here. – Daniel W. Farlow Apr 4 '15 at 4:24
An integer, $x$, is defined to be even whenever it can be written in the form $x=2k$ where $k$ is some integer.
Examples: $6=2\cdot 3,~~ 10 = 2\cdot 5,~~ 2218 = 2\cdot 1109,~~ -4 = 2\cdot (-2)$
An integer, $x$, is defined to be odd whenever it can be written in the form $x=2k+1$ where $k$ is some integer.
Examples: $-3 = 2\cdot (-2) + 1,~~~ 9 = 2\cdot 4 + 1,~~~ 1001 = 2\cdot 500 + 1$
Remember that $0$ is itself an integer, and that $0 = 2\cdot \color{red}{0}$, which is in the form $0=2k$ with $k = \color{red}{0}$, therefore $0$ is even.
• @user172675 This is hardly a proof--if someone accepts that $0$ is actually a number, furthermore an integer, then the fact that $0$ is even is a trivial conclusion. Your question is far more philosophical than you may think, for the concept of zero goes way way back. – Daniel W. Farlow Apr 4 '15 at 4:28 | 2019-07-21T18:53:14 | {
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https://math.stackexchange.com/questions/3074638/what-is-the-angle-between-two-intersecting-tangents-to-a-circle | # What is the angle between two intersecting tangents to a circle?
A circle of radius $$r$$ with centre $$C$$ is located at distance $$d$$ from a point $$P$$.
There are two tangents to the circle which pass through point $$P$$ - one on each side. They intersect the circle at points $$A$$ and $$B$$.
What is the angle through $$P$$ between these two tangents? In other words, angle $$APB$$?
I know that angle $$APB$$ + angle $$ACB$$ add up to 180.
(Not homework, for graphics programming) Thanks, Louise
• What kind of graphics programming problem? – lightxbulb Jan 15 at 16:58
• Recursive 2D radial tree layout with arbitrarily sized nodes! – Louise May 2 at 16:41
Here is a picture:
$$\overline {CP} = d$$
$$\angle CAP$$ and $$\angle BAP$$ are right angles, and $$\triangle APB$$ is isosceles.
$$m\angle APC = \arcsin \frac rd\\ m\angle APB = 2\arcsin \frac rd\\ m\angle BAP = \arccos \frac rd$$
• Thank you! How did you generate the picture? – Louise Jan 16 at 17:14
• I use MS Paint. – Doug M Jan 16 at 18:57
I presume "at distance $$d$$" means that $$|CP|=r+d$$.
The triangle $$ACP$$ is right angled, with $$|AC|=r$$. Then $$\sin\angle APC=\frac{r}{r+d}.$$ Then $$\angle ABP=2\angle APC=2\sin^{-1}\frac r{r+d}.$$ | 2019-08-17T13:20:46 | {
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https://proofwiki.org/wiki/Definition:Upper_Triangular_Matrix | # Definition:Triangular Matrix/Upper Triangular Matrix
## Definition
An upper triangular matrix is a matrix in which all the lower triangular elements are zero.
That is, all the non-zero elements are on the main diagonal or in the upper triangle.
That is, $\mathbf U$ is upper triangular if and only if:
$\forall a_{ij} \in \mathbf U: i > j \implies a_{ij} = 0$
## Also defined as
Some sources define an upper triangular matrix only as a square matrix.
## Examples
### Upper Triangular Matrix with fewer Rows than Columns
An upper triangular matrix of order $m \times n$ such that $m < n$:
$\mathbf U = \begin{bmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1, m - 1} & a_{1m} & \cdots & a_{1, n - 1} & a_{1n} \\ 0 & a_{22} & a_{23} & \cdots & a_{2, m - 1} & a_{2m} & \cdots & a_{2, n - 1} & a_{2n} \\ 0 & 0 & a_{33} & \cdots & a_{3, m - 1} & a_{3m} & \cdots & a_{3, n - 1} & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & a_{m - 1, m - 1} & a_{m - 1, m} & \cdots & a_{m - 1, n - 1} & a_{m - 1, n} \\ 0 & 0 & 0 & \cdots & 0 & a_{mm} & \cdots & a_{m, n - 1} & a_{mn} \\ \end{bmatrix}$
### Upper Triangular Matrix with more Rows than Columns
An upper triangular matrix of order $m \times n$ such that $m > n$:
$\mathbf U = \begin{bmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1, n - 1} & a_{1n} \\ 0 & a_{22} & a_{23} & \cdots & a_{2, n - 1} & a_{2n} \\ 0 & 0 & a_{33} & \cdots & a_{3, n - 1} & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & a_{n - 1, n - 1} & a_{n - 1, n} \\ 0 & 0 & 0 & \cdots & 0 & a_{nn} \\ 0 & 0 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 \\ \end{bmatrix}$
### Square Upper Triangular Matrix
$\mathbf U = \begin{bmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1, n - 1} & a_{1n} \\ 0 & a_{22} & a_{23} & \cdots & a_{2, n - 1} & a_{2n} \\ 0 & 0 & a_{33} & \cdots & a_{3, n - 1} & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & a_{n - 1, n - 1} & a_{n - 1, n} \\ 0 & 0 & 0 & \cdots & 0 & a_{nn} \\ \end{bmatrix}$
### Example of Square Upper Triangular Matrix
This is an arbitrary example of an upper triangular square matrix:
$\begin {pmatrix} 1 & 2 & 3 & 4 \\ 0 & 5 & 6 & 7 \\ 0 & 0 & 8 & 9 \\ 0 & 0 & 0 & 10 \end {pmatrix}$
### Example of Non-Square Upper Triangular Matrix
This is an arbitrary example of an upper triangular matrix which is specifically not square:
$\begin {pmatrix} 1 & 2 & 3 & 4 \\ 0 & 5 & 6 & 7 \\ 0 & 0 & 8 & 9 \\ 0 & 0 & 0 & 10 \\ 0 & 0 & 0 & 0 \end {pmatrix}$
### Upper Triangular Matrix not in Echelon Form
This is an arbitrary example of an upper triangular square matrix which is specifically not in echelon form (non-unity variant):
$\begin {pmatrix} 1 & 2 & 3 & 4 \\ 0 & 0 & 6 & 7 \\ 0 & 0 & 8 & 9 \\ 0 & 0 & 0 & 10 \end {pmatrix}$ | 2021-09-26T00:31:42 | {
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https://math.stackexchange.com/questions/2572407/the-hands-of-a-clock-are-of-length-5-inches-minute-hand-and-4-inches-hour-han | # The hands of a clock are of length 5 inches (minute hand) and 4 inches (hour hand). How fast is the distance between them changing at 3:00?
I am studying calculus on my own. Using old text by Varberg and Purcell. Not sure if my solution is correct. (Cannot find online.) Differentiated using law of cosines but my rate of change seems quite fast. I proceeded as follows:
To find an overall $\frac{\mathrm{d}\theta}{\mathrm{d}t}$, I know that the angular rate of change of the minute-hand is $2\pi$ radians/hr., and that of the hour-hand is $\frac{\pi}{6}$ radians per hr. I subtracted the slower from the quicker to get $\frac{11\pi}{6}$ radians per hour, and allowed it to be negative, -$\frac{11\pi}{6}$, as it is in a clockwise direction.
I labelled the minute hand length in the triangle as $a$, the hour hand as $b$, and the variable distance between the tips of the hands as $c$. By the law of cosines:
$$c^2 = 5^2 + 4^2 - 2 (5\cdot 4) \cos\theta.$$
(I know that theta will be $90^{\circ}$, i.e. $\frac{\pi}{2}$ radians, at 3:00. Also, as it will be a right triangle, the distance $c$ at 3:00 will be the square root of $4^2 + 5^2$, i.e $\sqrt{41}$.
$$c^2 = 41 - 40 \cos\theta.$$
Then I differentiated with respect to time:
\begin{gather} 2c \frac{\mathrm{d}c}{\mathrm{d}t} = 0 - 40 \left[- \sin \frac{\pi}{2}\right] \frac{\mathrm{d}\theta}{dt} \\ \sqrt{41} \frac{\mathrm{d}c}{\mathrm{d}t} = 20 (1) \left(-\frac{11\pi}{6}\right) \\ \frac{\mathrm{d}c}{\mathrm{d}t} = -17.99 \text{ inches per hour}. \end{gather}
But this seems way too fast! If, so, where did I go off the tracks?
Many Thanks. Victor Jaroslaw, a beginning calculus student.
• Do you mean the distance between the tips of the hands, or the angle between the hands? – MPW Dec 18 '17 at 22:33
• I think it is perfectly right. – Abhiram Natarajan Dec 18 '17 at 22:51
• If you convert to inches per minute, you get about $-0.3$. Doesn't that feel reasonable? – rogerl Dec 18 '17 at 22:58
It is right. One sanity test is this - 17.99 inches per hour is 0.3 inches per minute. The time at which the hour and minute hand are on top of each other is around 3:16.36. Assuming constant speed of 0.3 inches per minute, we can say that the distance reduced is around 4.9. The initial distance was $\sqrt{41}$ as you mentioned. So the new distance should be $\approx \sqrt{41} - 4.9 \approx 1.5$. The right answer is of course 1 (because the lengths are 5 and 4), so this answer is not too off right?
P.S. Obviously it is terribly wrong to assume constant speed, but it is fine as a heuristic given the amount of time elapsed is not too much. Anyway, the assumption was not to get an accurate answer anyway. It was, as I said, for a sanity test.
Three o'clock is a special time because the tip of the minute hand has all of the change in $x$, and the tip of the hour hand has all of the change in $y$.
So, let's try this. The distance $D=\sqrt{x^2+y^2}$, and the change can be expressed as
$$\frac{dD}{dt} = \frac{\partial D}{\partial x}\frac{dx}{dt} + \frac{\partial D}{\partial y}\frac{dy}{dt}.$$
The partials are
$$\frac{\partial D}{\partial x} = \frac{x}{D};\frac{\partial D}{\partial y} = \frac{y}{D}$$
The change in the distance $D$ with respect to $x$ is negative at three o'clock, because of the motion and position of the hands. Likewise, the change in $D$ with respect to $y$ is positive. At three o'clock, $x=4, y=5, D=\sqrt{41},$ so $\frac{\partial D}{\partial x} = -4/\sqrt{41}$ and $\frac{\partial D}{\partial y} = 5/\sqrt{41}$.
The velocities of the hands are $10\pi$ and $2\pi/3$ inches per hour for the minute and hour hands, respectively. At three o'clock, the minute hand velocity is aligned exactly on the $x$ direction, and the hour hand velocity exactly on the $y$ direction.
So,
$$\frac{dD}{dt} = \frac{-4\cdot 10\pi + 5 \cdot 2\pi/3}{\sqrt{41}} \doteq -17.99 \text{ inches/hr}.$$
Looks like we agree! | 2019-12-10T19:34:15 | {
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http://mathhelpforum.com/calculus/44812-maclaurin-series-question.html | Math Help - Maclaurin series question
1. Maclaurin series question
Hi All, help much appreciated!
These are the steps to determine the first two terms in a maclaurin series,
but they are the same so my working must be out? Its spread over 2 posts cos of laTex limit!
$
\begin{gathered}
\ln \frac{1}
{{\sqrt {1 - x} }} \hfill \\
f(x) = \ln (1 - x)^{ - 1/2} \hfill \\
f(0) = 0 \hfill \\
f'(x) = \frac{1}
{{(1 - x)^{ - 1/2} }}.\frac{{ - 1(1 - x)^{ - 3/2} .( - 1)}}
{2} \hfill \\
f'(x) = \frac{{(1 - x)^{ - 1} }}
{2} \hfill \\
\end{gathered}
$
2. and the rest...
$
\begin{gathered}
{\text{f'(0) = 0}}{\text{.5}} \hfill \\
{\text{f''(x) = }}\frac{{{\text{\{ 2( - 1(1 - x)}}^{{\text{ - 2}}} {\text{.( - 1)\} - \{ (1 - x)}}^{{\text{ - 1}}} {\text{.0\} }}}}
{{{\text{2}}^{\text{2}} }} \hfill \\
{\text{f''(x) = }}\frac{{{\text{2(1 - x)}}^{{\text{ - 2}}} }}
{{\text{4}}} \hfill \\
{\text{f''(x) = }}\frac{{{\text{(1 - x)}}^{{\text{ - 2}}} }}
{{\text{2}}} \hfill \\
\end{gathered}
$
F''(0) = 0.5 again...
3. Originally Posted by MexicanGringo
Hi All, help much appreciated!
These are the steps to determine the first two terms in a maclaurin series, but they are the same so my working must be out?
$\ln \frac{1}
{{\sqrt {1 - x} }}$
Since $f(x)=\ln \frac{1}
{{\sqrt {1 - x} }}=-\ln\sqrt{1-x}$
, we need to find consequent derivatives:
$f'(x)=-\frac{1}{2(x-1)}$
$f''(x)=\frac{1}{2(x-1)^2}$
So, at x=0, $f'(0)=f''(0)=\tfrac{1}{2}$
Thus, the first two terms are $\frac{1}{2}x+\frac{1}{2}\frac{x^2}{2!}=\frac{1}{2} x+\frac{1}{4}x^2$...
Does this make sense? I got the values of $f'(0)=f''(0)$...
--Chris
4. That makes sense up to there, but the third derivative gives a different interval?
$
\begin{gathered}
f'''(x) = \frac{{(1 - x)^{ - 2} }}
{2} \hfill \\
f'''(x) = \frac{{\{ 2( - 2)(1 - x)^{ - 3} .( - 1)\} }}
{4} \hfill \\
f'''(x) = (1 - x)^{ - 3} \hfill \\
f'''(0) = 1 \hfill \\
\end{gathered}
$
Is my method of differentiating wrong?
5. Originally Posted by MexicanGringo
That makes sense up to there, but the third derivative gives a different interval?
$
\begin{gathered}
f'''(x) = \frac{{(1 - x)^{ - 2} }}
{2} \hfill \\
f'''(x) = \frac{{\{ 2( - 2)(1 - x)^{ - 3} .( - 1)\} }}
{4} \hfill \\
f'''(x) = (1 - x)^{ - 3} \hfill \\
f'''(0) = 1 \hfill \\
\end{gathered}
$
Is my method of differentiating wrong?
That's correct! Its not wrong...however, you're doing it the long way...you don't have to use quotient rule here. Just use power rule and chain rule.
$\frac{1}{2}(1-x)^{-2}=\frac{1}{2}(-2)(1-x)^{-3}(-1)=\frac{1}{(1-x)^3}$
--Chris
6. Shot thanks a lot!
7. Originally Posted by MexicanGringo
Hi All, help much appreciated!
These are the steps to determine the first two terms in a maclaurin series,
but they are the same so my working must be out? Its spread over 2 posts cos of laTex limit!
$
\begin{gathered}
\ln \frac{1}
{{\sqrt {1 - x} }} \hfill \\
f(x) = \ln (1 - x)^{ - 1/2} \hfill \\
f(0) = 0 \hfill \\
f'(x) = \frac{1}
{{(1 - x)^{ - 1/2} }}.\frac{{ - 1(1 - x)^{ - 3/2} .( - 1)}}
{2} \hfill \\
f'(x) = \frac{{(1 - x)^{ - 1} }}
{2} \hfill \\
\end{gathered}
$
$\ln\left(\frac{1}{\sqrt{1-x}}\right)$
$=-\ln\left(\sqrt{1-x}\right)$
$\frac{-1}{2}\ln(1-x)$
Now consider
$-\int_0^x\frac{dx}{1-t}=\ln(1-t)\quad{|t|<1}$
Now we have that
$\frac{1}{1-t}=\sum_{n=0}^{\infty}x^n$
So now consider that $|x|<1$ is the interval of convergence of the geometric series, therefore it is uniformly convergent on that interval. So we can see that
$-\int_0^x\sum_{n=0}^{\infty}t^n~dt\quad{|x|<1}$
$=-\sum_{n=0}^{\infty}\int_0^xt^n~dt$
$=-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}\quad|x|<1$
$=\ln(1-x)$
$\therefore\frac{-1}{2}\ln(1-x)=\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+ 1}\quad|x|<1$
$\therefore\quad\boxed{\ln\left(\frac{1}{\sqrt{1-x}}\right)=\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{ n+1}}{n+1}\quad\forall{x}\backepsilon|x|<1}$
8. Originally Posted by Mathstud28
$\ln\left(\frac{1}{\sqrt{1-x}}\right)$
$=-\ln\left(\sqrt{1-x}\right)$
$\frac{-1}{2}\ln(1-x)$
Now consider
$-\int_0^x\frac{dx}{1-t}=\ln(1-t)\quad{|t|<1}$
Now we have that
$\frac{1}{1-t}=\sum_{n=0}^{\infty}x^n$
So now consider that $|x|<1$ is the interval of convergence of the geometric series, therefore it is uniformly convergent on that interval. So we can see that
$-\int_0^x\sum_{n=0}^{\infty}t^n~dt\quad{|x|<1}$
$=-\sum_{n=0}^{\infty}\int_0^xt^n~dt$
$=-\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}\quad|x|<1$
$=\ln(1-x)$
$\therefore\frac{-1}{2}\ln(1-x)=\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+ 1}\quad|x|<1$
$\therefore\quad\boxed{\ln\left(\frac{1}{\sqrt{1-x}}\right)=\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{ n+1}}{n+1}\quad\forall{x}\backepsilon|x|<1}$
You and you're series...pfft..
I totally knew how to do that ...
I'm reteaching myself this stuff [infinte series, and power series], and it makes perfect sense...for now...
--Chris
9. Originally Posted by Chris L T521
You and you're series...pfft..
I totally knew how to do that ...
I'm reteaching myself this stuff [infinte series, and power series], and it makes perfect sense...for now...
--Chris
Yeah, but its not as fun with uniform convergence
10. Originally Posted by Mathstud28
Yeah, but its not as fun with uniform convergence
I saw that stuff...and I was "what?"
Abel's test for uniform convergence [I belive that's what its called] looks interesting...did Weirestrass make contributions to uniform convergence?
--Chris | 2014-09-01T19:04:52 | {
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https://www.physicsforums.com/threads/injective-surjective-functions.636950/#post-4079059 | # Injective/Surjective Functions
dpa
## Homework Statement
Is the minimum function defined by f(a,b)=min{a,b} surjective or injective?
## Homework Equations
a function is injective f(x)=f(y) always implies x=y.
a function is surjective if for every y in codomain, there exists an x in domain such that f(x)=y.
## The Attempt at a Solution
I am confused whether min is surjective function or not.
As for injective, it is not. e.g. f(2,3) and f(3,2) both give 2. This is sufficient to say it is not injective.
But it is surjective, which I am mostly sure, but how do I show it is surjective?
## Answers and Replies
dpa
And, is the following the right way to show that the function is surjective?
f(x)=y,
x=f^-1(y)
f(f^-1(y))=x
Is this why it is called right invertible?
Homework Helper
Gold Member
I am confused whether min is surjective function or not.
As for injective, it is not. e.g. f(2,3) and f(3,2) both give 2. This is sufficient to say it is not injective.
But it is surjective, which I am mostly sure, but how do I show it is surjective?
You're correct that it is not injective. Whether or not it is surjective depends on the codomain. Since the codomain is not specified here, there's no way to answer the question. Any function is surjective if you define its codomain to be its image.
dpa
Sorry, that I forgot the first part.
It is defined for all f:ZXZ gives z. So c0 domain is set of all integers.
I am supposed to prove it not just explain.
Thank You.
Homework Helper
Gold Member
Sorry, that I forgot the first part.
It is defined for all f:ZXZ gives z. So c0 domain is set of all integers.
I am supposed to prove it not just explain.
Thank You.
OK, that makes it easy to answer whether the function is surjective. Given an arbitrary integer n, can you find two integers, a and b, such that min(a, b) = n?
dpa
Yes, definitely I mean for any integer, that's possible unless n=infinity which I believe does not belong to Z.
So, does verbal proof suffice?
Thank You.
:-)
Homework Helper
Gold Member
Yes, definitely I mean for any integer, that's possible unless n=infinity which I believe does not belong to Z.
So, does verbal proof suffice?
Thank You.
:-)
Right, Z is the set of all integers. Infinity is not an integer.
Why don't you write down your proposed verbal proof and we'll see how it looks. Ideally (for the surjective part), can you name specific values for a and b such that min(a,b) = n?
dpa
Verbal Part:
We know that for any value of an integer, we can find two integers such that the smallest of those two integers is the first integer. i.e. for every integer n we can write integers n and n+a where, a>=0. which gives min{n,n+a}=n.
Specific example would be for n=100, we can write two integers 100 and 101. I doubt if it is ideal example. | 2023-03-22T06:04:51 | {
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https://math.stackexchange.com/questions/3692744/finding-x-such-that-24370-equiv-x-mathrmmod-31 | # Finding $x$ such that $2^{4370} \equiv x \ (\mathrm{mod} \ 31)$
How to find $$x$$ such that $$2^{4370} \equiv x \ (\mathrm{mod} \ 31)$$?
The task is to compute $$2^{4370} \ (\mathrm{mod} \ 4371$$).
I know it's $$4371=3 \cdot 31 \cdot 47$$, so it's $$2 \equiv -29 \ (\mathrm{mod} \ 31)$$.
With Fermat's little theorem it's $$-29^{30} \equiv 1 \ (\mathrm{mod} \ 31)$$
$$\Rightarrow 2^{4370} \equiv -29^{4370} \equiv -29^{145 \cdot 30+20} \equiv -29^{20} \ (\mathrm{mod} \ 31)$$.
But how to continue?
I want to find a smaller number than $$-29^{20}$$ without a calculator. The calculator says $$x=1$$, but how to find it without?
• What is $2^5$ congruent to mod 31? May 26 '20 at 17:28
• You have $4370=5\cdot874$, hence $2^{4370}=(2^5)^{874}$. Now evaluate this mod 31. May 26 '20 at 17:32
• $2^{4370}\equiv2^{4350}2^{20}\equiv(2^{30})^{145}2^{20}\equiv(2^5)^4\equiv1\bmod31$ May 26 '20 at 17:59
• $4371$ is a base $2$ Fermat pseudoprime May 26 '20 at 19:08
• @CopyPasteIt: of course OP did not demonstrate that $4371$ is a base $2$ Fermat pseudoprime (yet) -- OP here was having trouble computing $2^{4370} \bmod 31$, which could be one of the steps toward that -- but I thought you asked why $4371$, so I gave an explanation of why $4371$ would be of interest May 27 '20 at 13:18
One way to proceed is to find an $$n$$ to get $$2^n$$ close (either on the left or right) of $$31$$.
Well
$$\quad 2^5 = 32 \equiv 1 \;(\text{ mod 31})$$
Couldn't come out that much better; yes, $$0 \lt 1$$, but...
So
$$\quad \displaystyle 2^{4370} = ({2^5})^{874} \equiv (1)^{874} \;(\text{ mod 31}) \equiv 1 \;(\text{ mod 31})$$
Fermat's little theorem works like a charm for modulus $$3$$ (resp. $$47$$) since $$3 -1 = 2$$ divides $$4370$$ (resp. $$47 - 1 = 46$$ divides $$4370$$). But even though $$30$$ doesn't divide $$4370$$, we can still use it when working in modulus $$31$$. Copying J.W.Tanner's comment,
$$\quad 2^{4370}\equiv2^{4350}2^{20}\equiv(2^{30})^{145}2^{20}\equiv 2^{20} \bmod31$$
Applying any 'divide and conquer' tactic you'll find that
$$\quad 2^{20} \equiv1\bmod31$$ | 2021-09-21T17:06:19 | {
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https://math.stackexchange.com/questions/2498628/proof-only-by-transformation-that-int-0-infty-cosx2-dx-int-0-infty | # Proof only by transformation that : $\int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx$
This was a question in our exam and I did not know which change of variables or trick to apply
How to show by inspection ( change of variables or whatever trick ) that
$$\int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx \tag{I}$$
Computing the values of these integrals are known routine. Further from their values the equality holds. But can we show the equality beforehand?
Note: I am not asking for computation since it can be found here and we have as well that, $$\int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx =\sqrt{\frac{\pi}{8}}$$ and the result can be recover here, Evaluating $\int_0^\infty \sin x^2\, dx$ with real methods?.
Is there any trick to prove the equality in (I) without computing the exact values of these integrals beforehand?
• $$\int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx =\sqrt{\frac{\pi}{8}}$$ – Gabriel Sandoval Oct 31 '17 at 19:18
• One way is to reproduce robjohn's proof here: math.stackexchange.com/questions/187729/… for the cosine case. – Alex R. Oct 31 '17 at 19:55
• Experimenting numerically, it seems that $\left\lvert\int_0^X[\cos(x^2)-\sin(x^2)]\, dx\right\rvert \le C(1+X)^{-1}$: [![see this plot][1]][1] [1]: i.stack.imgur.com/JK7Tj.gif – Giuseppe Negro Oct 31 '17 at 20:37
• @GiuseppeNegro $\cos(x^2)-\sin(x^2) = -\sqrt 2\sin(x^2-\frac{\pi}4)$, so I don't think it's easier that way. – Gabriel Romon Oct 31 '17 at 20:53
• Well, the complex analysis argument using either a sector of angle $\pi/4$ or a similar triangle, on the function $f(z) = e^{-z^2}$, shows initially that $(1+i) \int_0^\infty (\cos(x^2) - i \sin(x^2)) \, dx$ is real. (Then, from there, you go on to compare it to the integral of a Gaussian.) From the tone of the original post, this might be beyond the subject matter for their current course, though. – Daniel Schepler Oct 31 '17 at 22:18
Employing the change of variables $2u =x^2$ We get $$I=\int_0^\infty \cos(x^2) dx =\frac{1}{\sqrt{2}}\int^\infty_0\frac{\cos(2x)}{\sqrt{x}}\,dx$$ $$J=\int_0^\infty \sin(x^2) dx=\frac{1}{\sqrt{2}}\int^\infty_0\frac{\sin(2x)}{\sqrt{x}}\,dx$$
Summary: We will prove that $J\ge 0$ and $I\ge 0$ so that, proving that $I=J$ is equivalent to $$\color{blue}{0= (I+J)(I-J)=I^2 -J^2 =\lim_{t \to 0}I_t^2-J^2_t}$$ Where, $$I_t = \int_0^\infty e^{-tx^2}\cos(x^2) dx~~~~\text{and}~~~ J_t = \int_0^\infty e^{-tx^2}\sin(x^2) dx$$ $t\mapsto I_t$ and $t\mapsto J_t$ are clearly continuous due to the present of the integrand factor $e^{-tx^2}$.
However, By Fubini we have,
\begin{split} I_t^2-J^2_t&=& \left(\int_0^\infty e^{-tx^2}\cos(x^2) dx\right) \left(\int_0^\infty e^{-ty^2}\cos(y^2) dy\right) \\&-& \left(\int_0^\infty e^{-tx^2}\sin(x^2) dx\right) \left(\int_0^\infty e^{-ty^2}\sin(y^2) dy\right) \\ &=& \int_0^\infty \int_0^\infty e^{-t(x^2+y^2)}\cos(x^2+y^2)dxdy\\ &=&\int_0^{\frac\pi2}\int_0^\infty re^{-tr^2}\cos r^2 drd\theta\\&=&\frac\pi4 Re\left( \int_0^\infty \left[\frac{1}{i-t}e^{(i-t)r^2}\right]' dr\right)\\ &=&\color{blue}{\frac\pi4\frac{t}{1+t^2}\to 0~~as ~~~t\to 0} \end{split}
To end the proof: Let us show that $I> 0$ and $J> 0$. Performing an integration by part we obtain $$J = \frac{1}{\sqrt{2}} \int^\infty_0\frac{\sin(2x)}{x^{1/2}}\,dx=\frac{1}{\sqrt{2}}\underbrace{\left[\frac{\sin^2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx\color{red}{>0}$$ Given that $\color{red}{\sin 2x= 2\sin x\cos x =(\sin^2x)'}$. Similarly we have, $$I = \frac{1}{\sqrt{2}}\int^\infty_0\frac{\cos(2x)}{\sqrt{x}}\,dx=\frac{1}{2\sqrt{2}}\underbrace{\left[\frac{\sin 2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{1}{4\sqrt{2}} \int^\infty_0\frac{\sin 2 x}{x^{3/2}}\,dx\\= \frac{1}{4\sqrt{2}}\underbrace{\left[\frac{\sin^2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{3}{8\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{5/2}}\,dx\color{red}{>0}$$
Conclusion: $~~~I^2-J^2 =0$, $I>0$ and $J>0$ impliy $I=J$. Note that we did not attempt to compute neither the value of $~~I$ nor $J$.
Extra-to-the answer However using similar technique in above prove one can easily arrives at the following $$\color{blue}{I_tJ_t = \frac\pi8\frac{1}{t^2+1}}$$ from which one get the following explicit value of $$\color{red}{I^2=J^2= IJ = \lim_{t\to 0}I_tJ_t =\frac{\pi}{8}}$$
• This is a great answer. I think that you could prove that $I\ge 0$ and $J\ge 0$ more simply by looking at the graph of the integrand function. Starting from zero, there is a positive hump, followed by a negative one that is smaller. Then another positive hump that is bigger than the subsequent negative, and so on. Therefore the total integral must be positive. This is hand waving but can be made rigorous without much difficulty, I think. – Giuseppe Negro Nov 7 '17 at 15:02
• Yes you are right with the graph show. But with graph it is not 100% convincing. I know people that will partially reject that. Because they say "graphics make conjecture but not proofs":) – Guy Fsone Nov 7 '17 at 15:10
• I can't believe nobody is upvoting this. This answer is great. – Giuseppe Negro Nov 9 '17 at 18:58
• I still wondering whether they did not see or did not agree with this – Guy Fsone Nov 9 '17 at 19:25
• out of the blue, yet an interesting approach – Gabriel Romon Nov 10 '17 at 22:31
Note by change of variable it suffices to show
$$\int^\infty_0\frac{\cos(x)}{\sqrt{x}}\,dx =\int^\infty_0\frac{\sin(x)}{\sqrt{x}}\,dx$$
Consider the following function
$$f(z)=z^{-1/2}\,e^{iz}$$
Where we choose the principal root for $z^{-1/2}=e^{-1/2\log(z)}$. By integrating around the following contour
$$\int_{C_r}f(z)\,dz+\int_{r}^R f(x)\,dx+\int_{\gamma}f(z)\,dz+\int^{iR}_{ir}f(x)\,dx = 0$$
Taking the integral around the small quarter circle with $r\to 0$ $$\left| \int_{C_r}f(z)\,dz\right|\leq \left|\sqrt{r}\int^{\pi/2}_{0}e^{it/2} e^{rie^{it}}\,dt\right| \leq \sqrt{r}\int^{\pi/2}_{0}\left|e^{-r\sin(t)}\right|\,dt\sim 0$$
On $\gamma(t)=(1-t)R+iRt$ where $0\leq t \leq 1$
$$\left|\int_{\gamma}f(z)\,dz\right| = \left| R(i-1)\int^1_0e^{-1/2\log(R(1-t)+iRt)}e^{i(1-t)R-Rt}\,dt\right| \\ \leq \frac{\sqrt{2}}{\sqrt{R}} \int^1_0 \frac{e^{-Rt}}{\sqrt[4]{(1-t)^2+t^2}}\,dt$$
Hence we have
$$\left|\int_{\gamma}f(z)\,dz\right| \leq \frac{\sqrt{2}}{\sqrt{R}} \int^1_0 e^{-Rt}\,dt=\frac{\sqrt{2}}{R\sqrt{R}}\left(1-e^{-R}\right)\sim_{\infty}0$$
Finally what is remaining when $r\to 0$ and $R \to \infty$
$$\int^\infty_0 \frac{e^{ix}}{\sqrt{x}}\,dx =i \int^{\infty}_{0}(ix)^{-1/2}e^{-x}\,dx$$
Note that $i^{-1/2}=e^{-i\pi/4}$
$$\int^\infty_0\frac{e^{ix}}{\sqrt{x}}\,dx = ie^{-i\pi/4}I = \frac{I}{\sqrt{2}}+i\frac{I}{\sqrt{2}}$$
By equating the real part with the real part and the imaginary part with the imaginary part we reach $$\int^\infty_0\frac{\cos(x)}{\sqrt{x}}\,dx =\int^\infty_0\frac{\sin(x)}{\sqrt{x}}\,dx = \frac{I}{\sqrt{2}}$$
Although $I$ is easy to evaluate using the gamma function, we didn't have to evaluate it to show equivalence.
• In fact this is how we compute the integral. From this you have the evaluation already.:( – Guy Fsone Nov 1 '17 at 10:30
• @GuyFsone, this is less expensive than solving each integral individually. Note than in the derivation we didn't have to solve any integral besides proving the some integrals go to zero in the limit. – Zaid Alyafeai Nov 1 '17 at 10:48
• check the answer below. it prove without computing the values – Guy Fsone Nov 6 '17 at 21:18
Since $e^{iz^2}$ is entire, by Cauchy's Integral Theorem, we have $$\int_0^R e^{iz^2}\,\mathrm{d}z =\int_0^{(1+i)R} e^{iz^2}\,\mathrm{d}z+\int_{(1+i)R}^R e^{iz^2}\,\mathrm{d}z\tag1$$ where, using the parameterization $z=R(1+it)$, we have the estimate \begin{align} \left|\,\int_{(1+i)R}^R e^{iz^2}\,\mathrm{d}z\,\right| &\le R\int_0^1e^{-2R^2t}\,\mathrm{d}t\\ &\le\frac1{2R}\tag2 \end{align} and using the reparameterization $z\mapsto(1+i)z$, \begin{align} \int_0^{(1+i)R}e^{iz^2}\,\mathrm{d}z &=(1+i)\int_0^Re^{-2z^2}\,\mathrm{d}z\tag3 \end{align} Combining $(1)$, $(2)$, and $(3)$, while letting $R\to\infty$, validates the following change of variables: $\boldsymbol{\color{#C00}{z\mapsto(1+i)z}}$. \begin{align} \int_0^\infty\left(\cos\left(z^2\right)+i\sin\left(z^2\right)\right)\mathrm{d}z &=\boldsymbol{\color{#C00}{\int_0^\infty e^{iz^2}\,\mathrm{d}z}}\\ &\boldsymbol{\color{#C00}{=(1+i)\int_0^\infty e^{-2z^2}\,\mathrm{d}z}}\tag4 \end{align} Since the real and imaginary parts of $(4)$ are the same, we get that $$\int_0^\infty\cos\left(z^2\right)\,\mathrm{d}z =\int_0^\infty\sin\left(z^2\right)\,\mathrm{d}z\tag5$$
This is an extended comment not a proper answer.
The question can be rewritten as follows: to show that $$\tag{1} \Re \int_{-\infty}^\infty e^{-i|\xi|^2}\, d\xi + \Im \int_{-\infty}^\infty e^{-i|\xi|^2}\, d\xi=0,$$ where the integral is in the principal-value sense. This integral arises in PDEs as evaluation at the spatial origin of the fundamental solution to the Schrödinger equation. More precisely, if $E=E(t, \mathbf x)$ solves $$\tag{2} \begin{cases} (i\partial_t + \Delta) E(t, \mathbf x)=0, & t\in \mathbb R, \mathbf x\in\mathbb R^n\\ E(0, \mathbf x)=\delta(\mathbf x)\end{cases}$$ (where $\delta$ is the Dirac distribution) then the Fourier transform of $E(t, \cdot)$ is $$\hat{E}(t,\boldsymbol \xi)=\int_{\mathbb R^n} e^{-i\mathbf x\cdot \boldsymbol\xi}E(t, \mathbf x)\, d\mathbf x= e^{-it|\boldsymbol\xi|^2},\quad \boldsymbol\xi\in\mathbb R^n.$$ Now we observe that, for all suitable function $u$, we have that $\int_{\mathbb R^n} \hat{u}(t,\boldsymbol \xi)\, d\boldsymbol\xi=u(t, \mathbf 0)$. This gives a reformulation of problem (1) that generalizes to arbitrary dimension:
Is it true that $$\tag{3}\Re E(1,\mathbf 0)+ \Im E(1, \mathbf 0) =0,$$ where $E$ is the solution to (2)?
I found it surprising that the answer is affirmative if and only if $n=1\mod 4$: this follows from the explicit formula $$E(1,\mathbf 0)=\lim_{R\to \infty}\int_{[-R, R]^n}e^{-i|\boldsymbol \xi|^2}\, d\boldsymbol\xi = \frac{\pi^\frac{n}{2}}{i^\frac{n}{2}}=\pi^{\frac n 2}e^{-i \frac{n}{4}\pi}.$$
Conclusion. The OP asks for a solution that relies purely on change of variable in the integrals. In view of the reformulation (3), such changes of variable correspond to the symmetries of the PDE (2). These symmetries are dimension-independent, but the solution to the problem (3) is dependent on the dimension. Therefore, it seems to me that this approach is unlikely to work. | 2019-12-10T08:32:24 | {
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https://math.stackexchange.com/questions/1726557/if-the-sum-to-4-terms-of-a-geometric-progression-is-15-and-the-sum-to-infinity-i/1726576 | # If the sum to 4 terms of a geometric progression is 15 and the sum to infinity is 16 find the possible values of the common ratio.
I can't find a way to get an answer for this. I have tried using the formula for the sum to infinity and dividing it by the sum to 4 terms but i can't get it to work.
• Is it the sum of the $first$ 4 terms $a_0+a_{1}+a_{2}+a_{3}$ which is equal to 15, or a certain sum $a_k+a_{k+1}+a_{k+2}+a_{k+3}$ with $k$ unknown a priori? – Jean Marie Apr 3 '16 at 22:20
• Are the terms necessarily real, or can they be complex? – Brian Tung Apr 3 '16 at 22:22
Let the sequence have initial term $a_1$ and common ratio $r$. Then $a_k = a_1r^{k - 1}$. The sum of the first n terms of the geometric series is $$\sum_{k = 1}^{n} a_1r^{k - 1} = a_1(1 + r + r^2 + \cdots + r^{n - 1}) = a_1 \frac{1 - r^n}{1 - r}$$ provided that $r \neq 1$. If $r = 1$, then the series would not converge unless $a_1 = 0$, which cannot be the case here since the sum of the series is not equal to zero. Since the sum of the first four terms is $15$, we have $$a_1 \frac{1 - r^4}{1 - r} = 15 \tag{1}$$ If the series converges, then its limit is $$\sum_{k = 1}^{\infty} a_1r^{k - 1} = \frac{a_1}{1 - r}$$ Since the series has sum $16$, we have $$\frac{a_1}{1 - r} = 16 \tag{2}$$ Dividing equation 1 by equation 2 yields $$1 - r^4 = \frac{15}{16}$$ Solving for $r$ yields \begin{align*} 1 - \frac{15}{16} & = r^4\\ \frac{1}{16} & = r^4\\ \pm \frac{1}{2} & = r \end{align*}
Check: Substituting $r = 1/2$ into equation 1 yields \begin{align*} a_1 \cdot \frac{1 - \frac{1}{16}}{1 - \frac{1}{2}} & = 15\\ a_1 \cdot \frac{\frac{15}{16}}{\frac{1}{2}} & = 15\\ a_1 \cdot \frac{15}{8} & = 15\\ a_1 & = 8 \end{align*} Substituting $a_1 = 8$ and $r = 1/2$ into equation 2 yields $$\frac{a_1}{1 - r} = \frac{8}{\frac{1}{2}} = 16$$
If $r = -1/2$, then \begin{align*} a_1 \cdot \frac{1 - \frac{1}{16}}{1 + \frac{1}{2}} & = 15\\ a_1 \cdot \frac{\frac{15}{16}}{\frac{3}{2}} & = 15\\ a_1 \cdot \frac{15}{16} \cdot \frac{2}{3} & = 15\\ a_1 \cdot \frac{5}{8} & = 15\\ a_1 & = 24 \end{align*} Substituting $a_1 = 24$ and $r = -1/2$ into equation 2 yields $$\frac{24}{1 + \frac{1}{2}} = \frac{24}{\frac{3}{2}} = 24 \cdot \frac{2}{3} = 16$$ Thus, both $r = 1/2$ and $r = -1/2$ satisfy the given conditions.
Why doesn't it work?
$a_1\frac{1-r^4}{1-r} = 15$ and $a_1\frac{1}{1-r} = 16$, so $1-r^4 = \frac{15}{16}$, giving $r^4 = 1 - \frac{15}{16} = \frac{1}{16}$. So $r = \pm\frac{1}{2}$.
You have $a + ax + ax^2 + ax^3 = 15$ and $\frac{a}{1-x} = 16$ (taking $|x| < 1$. This gives $a = 16 (1-x)$. Substitute for $a$, giving $16 (1-x) (1 + x + x^2 + x^3) = 15$. You can then find $1 - x^4 = \frac{15}{16}$ giving $x^4 = \frac{1}{16}$ and so $x = \frac{1}{2}$ so that $a = 8$.
• OK, $x \pm \frac{1}{2}$ so $a = 8$ or $a = 24$ – jim Apr 3 '16 at 22:12 | 2021-05-14T01:36:45 | {
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http://dharmath.blogspot.com/2011/05/ | ## Wednesday, May 18, 2011
### Circular Locker Problem
From a friend's brother:
There are 1000 lockers and 1000 students. The lockers are arranged in a circle, all closed. The 1st student opens all the lockers. The 2nd student closes every other locker. The 3rd student goes to every 3rd locker. If it’s open he closes it, if it’s closed, he opens it. When he gets to locker #999, he continues to locker #2 and continues until he reaches a locker that he has already touched. The 4th student goes to every 4th locker: again, if it’s closed, he opens it, if it’s open, he closes it. Every student after that does the same, until student #1000, who will obviously just open or close locker #1000. After they are all finished, which lockers will be open?
Solution
First, let's think about what happens when student $n$ does his turn.
If $n$ is relatively prime to 1000, that is, $n$ does not contain any factor of 2 or 5, then $n$ will touch all the lockers exactly once. For example, student #3 will go touching lockers 3,6,...,999,2,5,...,998,1,4,7,...,997,1000. In this case student $n$ behaves exactly like student #1.
If $n$ is a multiple of $2$ or $5$, then let $d = \gcd(1000,n)$ Student $n$ touches locker $m$ if and only if $d$ divides $m$. For example, if $n=16$, then $d = 8$. Student #16 will go touching lockers 16,32,...,992,8,24,...,1000, which are exactly all multiples of 8. In this case, student $n$ behaves exactly like student #8.
Now, how many students have numbers that are NOT relatively prime to 1000? Such students must have numbers in the form of $n = 2^a 5^b$ We can enumerate the possibilities:
If $b=0$, then the student numbers are: 2,4,8,16,32,64,128,256,512, for a total of 9 students. Remember that students #16,32,...,512 behave exactly like student #8,and there are 6 of them, so it's equivalent to student #8 going 6 extra times. Their effects cancel out pairwise. So we only need to consider the effects of students #2,4,8.
If $b=1$, then the student numbers are: 5,10,20,40,80,160,320,640, for a total of 8 students. Remember that students #80,...,640 behave exactly like student #40,and there are 4 of them, so their effects cancel out pairwise. So we only need to consider the effects of students #5,10,20,40.
If $b=2$, then the student numbers are: 25,50,100,200,400,800, for a total of 6 students. Remember that students #400,8000 behave exactly like student #200, so their effects cancel out. So we only need to consider the effects of students #25,50,100,200.
If $b=3$, then the student numbers are: 125,250,500,1000.
If $b=4$, then the student number is: 625, which behaves like student #125, and thus cancels the effect of student #125.
So in total, there are 9+8+6+4+1 = 28 numbers that are not relatively prime to 1000, so there are an even numbers of students that are relatively prime. Each of these students touches all lockers exactly once, so their effects again cancel out pairwise. Thus, the student numbers that we need to consider are: (grouped by their largest factor of 2):
5,25
2,10,50,250
4,20,100,500
8,40,200,1000
In order to determine if locker $n$ is closed or open, find all numbers above that divides $n$. If there are even such numbers, then the locker is close, otherwise it's open.
For example, 120 = 3 x 5 x 8, so 120 is divisible by 2,4,8, 5,10,20,40. That means locker #120 is open.
For a more explicit list of open lockers, the following are the open lockers:
1. Lockers with numbers (not divisible by 5 or divisible by 25), and (divisible by 2 but not by 4, or divisible by 8).
2. Lockers with numbers divisible by 5 but not 25, and (divisible by 4 but not by 8).
## Monday, May 16, 2011
### Magician and cards
From IMO 2000, problem 4:
A magician has one hundred cards numbered 1 to 100. He puts them into three boxes, a red one, a white one and a blue one, so that each box contains at least one card. A member of the audience draws two cards from two different boxes and announces the sum of numbers on those cards. Given this information, the magician locates the box from which no card has been drawn.
How many ways are there to put the cards in the three boxes so that the trick works?
Solution
There are 12 possible ways to divide the card:
1. RWWW...WWWB (Card 1 goes to red box, card 2 to 99 go to white box, and card 100 goes to blue box) and all of its permutations, for a total of 6 ways.
2. RWBRWBRWB... and all of its permutations, for a total of 6 ways.
We now show that there are no other ways to put the cards. Consider the placement string (RWB string as in the example above). We will divide into two major cases: those with repeating characters, and those without repeating characters.
First, note that if there is a substring "RW" anywhere in the string, we don't allow substring "RB," for otherwise we can have an ambiguous sum of W+R versus R+B. Likewise, if there is a substring "RW" then we don't allow "BW."
Now, suppose there is a repeating substring "WW." Then we can't have a "RB" or "BR" anywhere in the string because there would be an ambiguous sum of R+W versus B+W.
Note that if we already have a "WW" then we can't have a "BB" because "BB" would mean that no "RW" or "WR" can exist, but we already established that no "RB" or "BR" can exist. So combination of "WW" and "BB" altogether would make it impossible for any "R" to exist in the string, a contradiction.
Consider the last B in the string, and call this B'. There are 4 possibilities of strings that come after B':
1. B'B: contradiction because B' is the last B in the string
2. B'R: contradiction because no BR is allowed
3. B'W
4. B' is the last letter in the whole string.
Consider also the last R in the string, call it R'. As in B', we only have 2 possibilities of strings that come after R':
5. R'W
6. R' is the last letter in the whole string.
Now obviously 4 and 6 cannot both be true, so one of them (possibly both) must be false. WLOG, we can assume that 6 is false, then 5 has to be true. If there's RW, we cannot have BW, so 3 is false, which means 4 is true.
So the last R in the string is followed by a W, and B is the last letter in the whole string. Since there already is an RW, we can't have a BW in the entire string. Also, we can't have BB, or BR, so there's no other place for another B. That means B' is the last and only B in the string.
Now, R' can't be preceded by another R (because no RR is allowed), nor by a B (because no BR is allowed), nor by a W, because a WR and a WB' can't both exist. So R' is the earliest R in the string. Since it's also the last, then R' is the only R in the string.
So we have RWW...WWB (and all of its permutations) as the only solution that allows repetition between characters.
Now, suppose there's no repeated characters. We assert that we cannot have a pattern like "RWR" anywhere in the string. Because the character that comes after "RWR" cannot be B, because "RWRB" gives an ambiguous sum of R+B and W+R. We also cannot have RWRR for there's no repeating characters. So we must have either "RWRW" or the fact that RWR is the ending of the entire string. If it is the ending, then we apply the same argument forward, and conclude that the ending must be WRWR. Either way, an RWR can be extended to RWRW or WRWR. But since there is at least one B in the string, and this B cannot be adjacent to another B, and we can't have BW, WB, BR, or RB either, a contradiction. So there must not be any RWR or its permutations anywhere in the string.
Suppose card 1 goes to R, and card 2 goes to W. Card 3 can go to:
2. W, forming a WW, contradiction
3. B.
So we have RWB as the beginning of the string. Now card 4 can go to:
1. R
2. W, forming a WBW, contradiction
3. B, forming a WBB, contradiction.
So card 4 must go to R.
We can continue this argument inductively to arrive that the entire string must have the form: RWBRWBRWB....
## Thursday, May 5, 2011
### Function from N to N
Find all functions $f:\mathbb{N}^{*}\rightarrow\mathbb{N}^{*}$
such that
$f(n)+f(n+1)+f(f(n))=3n+1 (\forall)n\in\mathbb{N}^{*}$ | 2017-04-24T15:12:17 | {
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https://math.stackexchange.com/questions/3366064/how-deep-is-the-liquid-in-a-half-full-hemisphere/3366099 | How deep is the liquid in a half-full hemisphere?
I have a baking recipe that calls for 1/2 tsp of vanilla extract, but I only have a 1 tsp measuring spoon available, since the dishwasher is running. The measuring spoon is very nearly a perfect hemisphere.
My question is, to what depth (as a percentage of hemisphere radius) must I fill my teaspoon with vanilla such that it contains precisely 1/2 tsp of vanilla? Due to the shape, I obviously have to fill it more than halfway, but how much more?
(I nearly posted this in the Cooking forum, but I have a feeling the answer will involve more math knowledge than baking knowledge.)
• after I answered, I found here that the answer is to fill it by the fraction $1-2\cos(\frac49\pi)$ – J. W. Tanner Sep 22 '19 at 21:51
• Not really relevant to the mathematics, but: just eyeball it. Half a teaspoon of vanilla extract, one way or another, ain't gonna make that much difference. In fact, just put in a full teaspoon. Then, do yourself a favor and add some mace and clove, too. ;) – Xander Henderson Sep 23 '19 at 13:55
• @RandomAspirant Do you need to comment that on every answer as well? – Todd Sewell Sep 23 '19 at 14:38
• Just use a second spoon - fill the first one completely, then pour from it into the second until they're even... – twalberg Sep 23 '19 at 16:55
• Do we allow housework problems? – Acccumulation Sep 23 '19 at 21:53
Assuming the spoon is a hemisphere with radius $$R$$,
let $$x$$ be the height from the bottom of the spoon, and let $$h$$ range from $$0$$ to $$x$$.
The radius $$r$$ of the circle at height $$h$$ satisfies $$r^2=R^2-(R-h)^2=2hR-h^2$$.
The volume of liquid in the spoon when it is filled to height $$x$$ is $$\int_0^x\pi r^2 dh=\int_0^x\pi(2hR-h^2)dh=\pi Rh^2-\frac13\pi h^3\mid_0^x=\pi Rx^2-\frac13\pi x^3.$$
(As a check, when the spoon is full, $$x=R$$ and the volume is $$\frac23\pi R^3,$$ that of a hemisphere.)
The spoon is half full when $$\pi Rx^2-\frac13\pi x^3=\frac13\pi R^3;$$ i.e., $$3Rx^2-x^3=R^3;$$
i.e., $$a^3-3a^2+1=0$$, where $$a=x/R$$.
The only physically meaningful solution of this cubic equation is $$a\approx 65\%.$$
• I actually got $65.27...\%$ like other answers, but I don't think you could eyeball that precisely – J. W. Tanner Sep 22 '19 at 21:42
• What the hell, call it $\frac23$. Everybody likes vanilla. – TonyK Sep 22 '19 at 21:46
• And they say calculus is of no use in real life... – RandomAspirant Sep 23 '19 at 13:41
• @TonyK eff it. Call it one teaspoon, LOL. Any good vanilla will only further enhance a recipe if you add a bit extra. – Doktor J Sep 24 '19 at 17:12
• @DoktorJ but that principle leads to a divergent sequence! – JosephSlote Sep 24 '19 at 19:19
There is actually an analytic solution to the problem, as shown below.
The volume of a spherical cap is the difference between those of two overlapping cones, one with a spherical bottom and the other with a flat bottom, i.e.
$$V = \frac{2\pi}{3}r^2h - \frac{\pi}{3}(2rh-h^2)(r-h) =\frac{\pi}{3}(3rh^2-h^3)$$
Set $$V$$ to half of the semisphere volume $$\frac{2\pi}{3}r^3$$ to obtain,
$$\left(\frac rh \right)^3 - 3\frac rh+1=0$$
Compare with the identity $$4\cos^3 x -3\cos x -\cos 3x=0$$ and let $$r/h = 2\cos x$$ to obtain $$x=40^\circ$$.
Thus, the depth $$h$$ as a fraction of the radius $$r$$ Is
$$\frac hr = \frac{1}{2\cos40^\circ}$$
• That was a big surprise for me! Also a bit surprising was that your $\dfrac{1}{2\cos 40^\circ}$ is equal to J.W.Tanner's $1-2\cos(\frac49\pi)$ (in a comment to the OP). – TonyK Sep 23 '19 at 16:16
• @TonyK - I knew of the close-form result, but was also surprised of a different form from his, until convinced myself numerically – Quanto Sep 23 '19 at 16:34
It makes things a bit simpler if we turn your measuring spoon upside down, and model it as the set of points $$\{(x,y,z):x^2+y^2+z^2=1, z\ge 0\}$$. The area of a cross-section at height $$z$$ is then $$\pi(1-z^2)$$, so the volume of the spoon between the planes $$z=0$$ and $$z=h$$ is
$$\pi\int_0^h(1-z^2)dz = \pi\left(h-\frac13h^3\right)$$
The volume of the hemisphere is $$\frac23\pi$$, and we want the integral to be equal to half this, i.e. $$\pi\left(h-\frac13h^3\right)=\frac{\pi}{3}$$ or $$h^3-3h+1=0$$ This cubic equation doesn't factorize nicely, so we ask Wolfram Alpha what it thinks. The relevant root is $$h\approx 0.34730$$. Remember that we turned the spoon upside down, so you should fill it to a height of $$1-h=0.65270$$, or $$65.27\%$$.
• "It makes things a bit simpler if we turn your measuring spoon upside down" Then there isn't any liquid in the hemisphere. – Acccumulation Sep 23 '19 at 21:47
• @Acccumulation One could easily turn it upside-down to measure it, then turn it right-side-up when filling it – user45266 Sep 23 '19 at 23:03
• What do you mean by "doesn't factorize nicely"? In my view, $$h^3-3h+1=\left(h-2\cos\frac{2\pi}9\right)\left(h-2\sin\frac{\pi}{18}\right)\left(h+2\cos\frac\pi9\right)$$ is quite a nice closed-form factorization. Not in radicals, but why would anyone want them :) – Ruslan Sep 24 '19 at 6:27
• @Ruslan How did you find that? – Ovi Sep 25 '19 at 0:39
• @Ovi well, I found the first root with Wolfram Mathematica's Solve + FullSimplify, and the second and third by combination of FullSimplify on the additive terms of the solution returned by Solve and then ExpToTrig to get the expressions like $(-1)^{8/9}$ to trigonometric form. But that was a lazy approach. The more general way is to use the algorithm given in this page (page is in Russian, but I guess if you just follow the formulas, you'll get it: the Vieta solution is sufficient here). – Ruslan Sep 25 '19 at 5:32
Without loss of generality we assume the radius of the sphere to be $$1$$
The volume of the liquid is found by an integral $$V= \int _{-1}^{-1+h} \pi (1-y^2 )dy$$
and you want the volume of the liquid to be half of the hemisphere which is $$\pi/3$$
After evaluating the integral and solving the equation I have found $$h=0.65270365$$ That is a little bit more than half as expected.
Alternative: use two teaspoons.
Use water as you develop your skill. Fill tsp A, and pour into tsp B until the contents appear equal. Each now contains half a tsp. And now you know what half a tsp looks like in practice.
And you don't have to calculate cosines against thumb-sized hardware.
• -1. Holly doesn't have two teaspoons. – TonyK Sep 26 '19 at 22:19
Note about eyeballing: Your eye's reference is the surface of the spoon, so when you eyeball you may actually be measuring along the arc from the bottom of the spoon to its top edge.
That is, your eye may be watching the red curve, not the blue line:
Using the 65.27% from other answers, the depth measured along the red curve is $$\frac{\arccos(1 - 0.6527)} {90\deg }\approx 77.42\%$$
So to the eye, the "depth" of a half-full spoon may look like more like three quarters than two thirds. | 2020-02-22T20:33:50 | {
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https://questions.ascenteducation.com/iim_cat_mba_free_sample_questions_math_quant/data_sufficiency_ds/TANCET_Previous_year_paper_2014_Q79.shtml | # TANCET 2014 DS 79: Rates
## Directions for TANCET Data Sufficiency Questions
The question is followed by two statements labeled (1) and (2) in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the problem plus your knowledge of mathematics and everyday facts, choose the answer as:
1. Choice 1 if statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
2. Choice 2 if statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
3. Choice 3 if both the statements (1) and (2) TOGETHER are sufficient, but NEITHER statement alone is sufficient.
4. Choice 4 if each statement ALONE is sufficient.
5. Choice 5 if statements (1) and (2) TOGETHER are not sufficient, and additional data is needed.
## Question
Medium Data Sufficiency
Automobile A is travelling at $$frac{2}{3}\\$rd of the speed that Automobile B is travelling at. How fast is Automobile A travelling? 1. Statement 1: If both automobiles increased their speed by 10 km per hour, Automobile A would be travelling at $\frac{3}{4}\\$th the speed of Automobile B. 2. Statement 2: If both automobiles decreased their speed by 10 km per hour, Automobile A would be travelling at $\frac{1}{2}\\$ the speed of Automobile B. Correct Answer Choice$4). Each statement is INDEPENDENTLY SUFFICIENT.
## Explanatory Answer - step by step
• ### What should we know from the Question Stem?
Before evaluating the two statements, answer the following questions to get clarity on when the data is sufficient.
#### What kind of an answer will the question fetch?
The question is "How fast is Automobile A travelling?"
The answer to the question should be the speed of Automobile A, a number followed by a unit of speed.
#### When is the data sufficient?
If we are able to come up with a UNIQUE value for the speed of Automobile A, the data is sufficient.
If we are not able to come up with a unique value – either we cannot find an answer with the data in the statement(s) or if we find more than one value, the data is NOT sufficient.
Automobile A is travelling at $$frac{2}{3}\\$rd the speed of Automobile B. • ### Statement$1) ALONE
#### If both automobiles increased their speed by 10 km per hour, Automobile A would be travelling at $$frac{3}{4}\\$th the speed of Automobile B. Let the speed of Automobile B be ‘b’ kmph. So, speed of Automobile A = $\frac{2}{3}\\$b. If each increased its speed by 10 kmph, the speeds of A and B will be$$$frac{2}{3}\\$b + 10) and$b + 10) respectively.
From statement 1, ($$frac{2}{3}\\$b + 10) = $\frac{3}{4}\\$$b + 10)
$$frac{3}{4}\\$b – $\frac{2}{3}\\$b = 10 – 7.5 Solving the equation, we get b = 30 kmph. So, speed of Automobile A = 20 kmph We are able to find a UNIQUE value for the speed of Automobile A using data in statement 1. Statement$1) ALONE is sufficient.
The moment we realize that statement (1) ALONE is sufficient, we can narrow down our choices to 1 or 4
To determine whether the answer is choice 1 or choice 4, we need to evaluate statement (2). Remember that you have to evaluate statement (2) even if statement (1) is sufficient.
• ### Statement (2) ALONE
#### If both automobiles decreased their speed by 10 km per hour, Automobile A would be travelling at $$frac{1}{2}\\$ the speed of Automobile B. Remember: When you are evaluating statement$2) ALONE, please do not recall information that you read in statement (1). Anything said about the speeds of either automobile A or B in statement (1) should not be used while evaluating statement (2).
We know a = $$frac{2}{3}\\$b from the question stem. If each decreased its speed by 10 kmph, the speeds of A and B will be$$$frac{2}{3}\\$b – 10) and$b – 10) respectively.
From statement 2, ($$frac{2}{3}\\$b – 10) = $\frac{1}{2}\\$$b – 10)
$$frac{2}{3}\\$b – $\frac{1}{2}\\$b = 10 – 5 Solving the equation, we get b = 30 kmph. So, speed of Automobile A = 20 kmph. Using statement$2) ALONE we could get a UNIQUE answer.
Statement (2) ALONE is ALSO sufficient.
Each statement is INDEPENDENTLY sufficient, we can eliminate choice 1.
Hence, choice (4) is the answer.
## Online TANCET MBA CourseTry it Free!
Register in 2 easy steps and start learning in 5 minutes! | 2020-01-27T09:53:56 | {
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https://stats.stackexchange.com/questions/472409/question-on-rao-cramer-lower-bound | # Question on Rao-Cramer Lower Bound
A question with a solution that I don't quite get: asking for the Cramér-Rao lower bound of a random Poisson sample.
If we take the log of the function $$f(x; \theta)$$ and take its first derivative with respect to theta,it becomes $$(x-\theta)/\theta$$ (which is the score function $$S(x;\theta)$$) and if we find the fisher information of that, it's $$E[S(X;\theta)^2]$$ which then becomes $$E[[X-\theta]^2]/\theta^2]$$.
The solution says this leads to $$1/\theta$$. Can anyone please explain how $$E[[X-\theta]^2]/\theta^2]$$ leads to $$1/\theta$$?
The variance and mean of a Poisson distribution are equal, so $$E[(x-\theta)^2]=\theta$$ and $$E\left[\frac{(x-\theta)^2}{\theta^2}\right]=\theta/\theta^2=1/\theta$$
• Yes. The bound is the reciprocal of the Fisher information, divided by the sample size, so $(1/(1\theta))/n= \theta/n$. And we know $\mathrm{var}[X]/n=\theta/n$ is always the variance of the sample mean, so the sample mean attains the bound in this case. – Thomas Lumley Jun 16 at 7:01 | 2020-09-21T13:47:17 | {
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http://math.stackexchange.com/questions/83473/why-lim-limits-n-to-infty-left-fracn3n4-rightn-neq-1 | # Why $\lim \limits_ {n\to \infty}\left (\frac{n+3}{n+4}\right)^n \neq 1$?
Why doesn't $\lim\limits_ {n\to \infty}\ (\frac{n+3}{n+4})^n$ equal $1$?
So this is the question.
I found it actually it equals $e^{-1}$. I could prove it, using some reordering and canceling.
$$\lim_ {n\to \infty}\ \left(\frac{n}{n+4}+\frac{3}{n+4}\right)^n$$
with the limit of the first term going to $1$ and the second to $0$. So $(1+0)^n=1$ not $e^{-1}$.
-
You set $\frac{n}{n+4}$ to its limiting value $1$ but not the exponent, that is, you are assigning different meanings to the $n$'s inside the parentheses and the $n$ outside the exponent. All the $n$'s must have the same value, no? – Dilip Sarwate Nov 18 '11 at 20:59
Do you believe $(A+B)^n$ and $A^n + B^n$ are the same? – Will Jagy Nov 18 '11 at 21:04
Then entire expression, both base and exponent, depend on $n$. Why treat the base first, and the exponent later? Or why not try the exponent first, and the base later? Then you would have that for a fixed $n$, $\frac{n+3}{n+4}$ is smaller than $1$, so raised to higher and higher powers will go to $0$. The point is: you can't just decide that you are going to treat the exponent as fixed and deal with the base separately, nor can you decide that you are going to treat the base as fixed and deal with the exponent first; both matter, so they cannot be dealt with separately. – Arturo Magidin Nov 18 '11 at 21:16
You might find this helpful: Why is $1^\infty$ considered to be an indeterminate form? – Mike Spivey Nov 19 '11 at 4:25
You cannot mix limits like that $$\lim_{n\to \infty}(\frac{n}{n+4}+\frac{3}{n+4})^n\ne \lim_{n\to \infty}(1+0)^n=1$$ – AD. Jun 4 '14 at 10:07
Because $1^\infty$ is a tricky beast. Perhaps the power overwhelms the quantity that's just bigger than $1$, but approaching $1$, and the entire expression is large. Or perhaps not...
Perhaps the power overwhelms the quantity that's just smaller than $1$, but approaching $1$, and the entire expression tends to $0$ . Or perhaps not...
In your case, $${n+3\over n+4} = 1-{1\over n+4}.$$ And, as one can show (as you did): $$\lim\limits_{n\rightarrow\infty}(1-\textstyle{1\over n+4})^n = \lim\limits_{n\rightarrow\infty}\Bigl[ (1-\textstyle{1\over n+4})^{n+4}\cdot (1-{1\over n+4})^{-4}\Bigr] = e^{-1}\cdot1=e^{-1}.$$
Here, the convergence of $1-{1\over n+4}$ to 1 is too fast for the $n^{\rm th}$ power to drive it back down to $0$.
-
I do agree : $1^\infty$ is tricky. You can rewrite it as $\exp(\infty \times \ln(1)) = \exp(\infty \times 0)$ to make the trick even more visible. – Paul Pichaureau Nov 18 '11 at 21:20
@David Mindlessly upvoted for the starting sentence. – Pedro Tamaroff Feb 24 '12 at 6:45
Another way to see:
$\left(\frac{n+3}{n+4}\right)^n=\left(\frac{1+\frac{3}{n}}{1+\frac{4}{n}}\right)^n=\frac{\left(1+\frac{3}{n}\right)^n}{\left(1+\frac{4}{n}\right)^n}.$
Since, $\lim_{n\to\infty}\left(1+\frac{c}{n}\right)^n=e^c$, follows
$$\lim_{n\to\infty}\left(\frac{n+3}{n+4}\right)^n=\lim_{n\to\infty}\frac{\left(1+\frac{3}{n}\right)^n}{\left(1+\frac{4}{n}\right)^n}=\frac{e^3}{e^4}=e^{-1}.$$
-
+1 very nice, because you bring it back to the basic defintion. – draks ... Feb 24 '12 at 7:20
As David Mitra makes clear in his answer, there is a more fundamental question underlying your question, which is:
• why is $\displaystyle \lim_{n\to \infty} (1 + 1/n)^n$ not equal to $1$?
The same (fallacious) reasoning as the one you give applies. However, this limit is known to equal $e$, not $1$ (and your limit of $e^{-1}$ can easily be obtained from that limit, as David essentially shows).
As David says, there is a tension between the term in parentheses which is tending to $1$ from above, and the power of $n$, which, when applied to any fixed number $>1$, will give larger and larger answers as $n$ increases.
You might want to consider some other related limits to see how they behave, e.g.:
• $\displaystyle\lim_{n\to\infty} (1+1/n^2)^n$ .
• $\displaystyle \lim_{n\to\infty} (1 + 1/\log n)^n$ .
-
While intuitively it seems "obvious" that $1^\infty$ has to be one, let me point to you that $\dfrac{n+3}{n+4} <1$. Then the meaning of
$$\left(\frac{n+3}{n+4}\right)^n$$
is a number strictly less than $1$ to a huge power. Now, a number between $0$ and $1$ raised to a larger power gets smaller and smaller, so is it still obvious that it approaches 1?
- | 2015-01-26T18:54:52 | {
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https://www.themathdoctors.org/finding-the-range-of-a-tricky-rational-function/ | # Finding the Range of a Tricky Rational Function
I previously wrote about finding the range of various kinds of functions. The examples there were relatively easy. A recent question raised the level of difficulty, bringing up some interesting issues.
Here is the initial question:
Hi,
I am trying to calculate the domain and range of this function f(x)= (x^2 – 3x + 2)/(x^2 + x – 6).
The answer for range which I am getting is [1/5, infinity) but the correct answer given in my textbook is R-{1/5, 1}.
I will be thankful for help!
His notation for the correct answer means “all real numbers except 1/5 and 1”. His own answer is very different: all real numbers starting at 1/5.
First, let’s examine what he has done. He wants to find the range of $$f(x) = \frac{x^2 – 3x + 2}{x^2 + x – 6},$$ so he starts to solve the equation $$\frac{x^2 – 3x + 2}{x^2 + x – 6} = y,$$ by multiplying by the denominator and simplifying. The result is the quadratic equation $$(y-1)x^2 + (3+y)x – (6y + 2) = 0.$$ Solving this for x would give the inverse relation (x as a function of y); this technique is demonstrated in the previous post I mentioned above. We need to find the domain of this inverse relation, which will be the range of the original function. (That is, we’re finding the values of y for which an x exists.) In order to do that, we find the discriminant $$D = b^2 – 4ac$$; the range will be the set of values of y for which this is non-negative, so that there is at least one real value of x associated with that y. (As we’ll see, a little more than that will be needed.)
The result is an inequality that simplifies to $$25y^2 – 10y + 1 \gt 0.$$ But then he solves this inequality incorrectly …
I replied, pointing out this error (but making one of my own) and suggesting another way, by which I had obtained the correct answer:
One specific error in your work is the step from (y – 1/5)2 ≥ 0 to y ≥ 1/5. Polynomial inequalities don’t work that way; the solution in this case is y ≠ 1/5, because the square is positive unless the base is 0.
I solved this a different way; trying it your way led to some subtle steps I almost missed. I did get the correct answer both ways.
I first simplified the function by canceling (x – 2). Then I found the inverse and determined its domain (which could be done directly due to the simplification). I also had to eliminate the value of y corresponding to the “hole” in the original function.
My method started by factoring the numerator and denominator and simplifying:
$$f(x) = \frac{x^2 – 3x + 2}{x^2 + x – 6}$$
$$f(x) = \frac{(x-2)(x-1)}{(x-2)(x+3)}$$
$$f(x) = \frac{x-1}{x+3},\ x\ne 2$$
This is equivalent to the original, as long as we take note that 2 is not in the domain. (That is the “hole”.)
Solving for x, we have
$$\frac{x-1}{x+3} = y$$
$$x-1 = y(x+3)$$
$$x – 1 = yx + 3y$$
$$x(1 – y) = 3y + 1$$
$$x = \frac{3y + 1}{1 – y}$$
This is defined for all y except 1. But we also have to exclude the value of y for which $$x = 2$$ in this simplified equation, which is $$y = \frac{2-1}{2+3} = \frac{1}{5}.$$
The range is therefore, as the book said, all real numbers except 1 and 1/5.
But I had mentioned further subtleties if we don’t take this way, and the student chose to persevere in order to learn how to handle obstacles:
Thank You Doctor Peterson for help!
I am trying to proceed by my own method.
So, you are saying that if (y – 1/5)2 ≥ 0 then it can’t be y ≥ 1/5, and y ≠ 1/5.
But, there is an equality symbol as well.
So, I think y = 1/5 is true. Why do you think it is false?
So, once again the range which I am getting is from 1/5 (included) to infinity.
But, the correct range given in my textbook is All Real except 1/5 and 1.
I will be thankful for help.
His correction of my correction is correct! I had treated the inequality as “>” instead of “≥”. But that only deepens the mystery: How is 1/5 excluded from the domain of the inverse? I gave some hints for extending his method to find the correct answer:
I did miss the equality part (probably because of the correct solution you’d mentioned); but I hope you see that the solution of (y – 1/5)2 ≥ 0 is not x ≥ 1/5, but all real numbers! A square is never negative.
So what is special about 1/5 that would exclude it from the domain? The discriminant there is 0, so there is only one solution for x; what is that one solution? Find out, and then look back at the original equation!
This is one of the subtleties I mentioned in your method; but it also occurred in mine, and I mentioned it in my summary of what I did.
Now, what about the exclusion of 1? That’s the other major subtlety: Look at your quadratic equation in x; is it always quadratic? If it isn’t, then the discriminant isn’t really relevant; you have to take this as a special case.
So far, his method has not yielded any numbers to be excluded; they will both be special cases, somehow. So figuring this out will be useful for understanding the whole process.
He replied:
Ok so I plugged y = 1/5 in the equation and got x = 2, which is of course rejected because x ≠ 2.
So, my final question is that how would someone decide which values to plug in the equation to check whether it is possible or not? (For example, how would someone decide that he/she need to check for 1/5 and not for any other number ?)
This is actually the easier of the two exclusions to deal with, but it turns out to be not really about checking y = 1/5, but rather checking x = 2:
I suppose the fact that this is a special point (where the discriminant is 0) suggests checking; but, really, from the start you should plan to check the value of y when x=2 (in the original equation, after removing the “hole”), and exclude that, as long as it isn’t associated with any other value of x. (That’s another subtle point.) So dealing with that point is really separate from the work you did.
Ultimately, I would want to sketch the graph along with any algebraic methods I use, to make sure I haven’t missed anything important (like that hole).
So, the correct way to find this really has nothing to do with finding 1/5 in the course of his work, but merely with his failure to even consider the hole! Because he never factored the numerator and denominator, he was unaware of the need to exclude the y associated with the hole.
But this does leave a curiosity: How did his method end up coming so close, with 1/5 as a special point at all?
Let’s look at his work, but with the numerator and denominator written in factored form:
$$\frac{(x-2)(x-1)}{(x-2)(x+3)} = y\ \Rightarrow\ (x-2)(x-1) = (x-2)(x+3)y.$$
We have the factor $$(x-2)$$ on both sides, so that when $$x = 2$$, the equation is true regardless of the value of y. In fact, we could therefore rewrite that equation as $$(x-2)\left((y-1)x + (3y+1)\right) = 0,$$ whose solutions are $$x = 2$$ and $$\displaystyle x = \frac{3y+1}{1-y}$$. So when the discriminant is 0 (so that there is one solution), these are equal, and $$\displaystyle\frac{3y+1}{1-y} = 2$$. This is, of course, when $$y = 1/5$$! So it just happens (!) that the check for the discriminant did locate the hole, and there really was reason to check that point. I doubt I would ever have guessed that.
Now let’s finally look at a graph of this function:
We can see that the domain excludes -3 (the vertical asymptote) and 2 (the hole); and the range excludes 1 (the horizontal asymptote) and 1/5 (the hole).
(By the way, did you wonder at some point why the discriminant was almost always positive, indicating that there were two values of x for a given value of y? Did you assume that therefore the function must not be one-to-one? It’s because the equation we were working with there consisted of this graph together with the vertical line x = 2.)
There’s still one thing missing, though: How would the student’s method have found that y can’t be 1 (the asymptote)? This is my second hint, which I haven’t yet elaborated. Look back at the student’s quadratic equation: $$(y-1)x^2 + (3+y)x – (6y + 2) = 0.$$ The discriminant only applies to quadratic equations; and this is not a quadratic equation when y = 1! So we should have checked that case before finding the discriminant. And this amounts to finding the horizontal asymptote. In my method, I actually found the inverse, so this asymptote was directly visible. In the method we are examining, we looked only at whether an inverse exists, and never actually saw it, so as to discover these details.
What’s the bottom line of all this? Don’t skip the factoring and simplifying when you work with a rational function!
### 2 thoughts on “Finding the Range of a Tricky Rational Function”
1. This would be a good question to submit on our Ask a Question page, so we can discuss it together. I’ll ask you to show what you have tried, and what specific questions you have about it.
This particular problem doesn’t have the same issues as the one in this post; there are no “holes”, and the answer is very simple. If you use the discriminant, you will end up with a quadratic inequality that is not pretty, but has a simple solution. If you find the inverse, no factors will cancel as they did in my work.
But it may be well worth discussing your work, wherever you are having trouble.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2022-05-18T02:36:14 | {
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https://stats.stackexchange.com/questions/77201/conditional-probability-problem-acceptance-to-two-colleges | # Conditional probability problem - acceptance to two colleges
I'm doing this problem:
A student is applying to Harvard and Dartmouth. He estimates that he has
a probability of .5 of being accepted at Dartmouth and .3 of being accepted
at Harvard. He further estimates the probability that he will be accepted by
both is .2. What is the probability that he is accepted by Dartmouth if he is
accepted by Harvard? Is the event “accepted at Harvard” independent of the
event “accepted at Dartmouth”?
So I begin with the p(D|H)=p(D and H)/p(H) which I find eerily suspicious because is too easy, so is wrong, but why? First the probabilities do not sum to one p(H)+p(D)≠1. So I'm missing something obvious but what??
Thanks.
• This surely seems like homework. If so, you should add the self-study tag. – Peter Flom Nov 21 '13 at 1:11
• Why should p(H) + p(D) = 1? – Peter Flom Nov 21 '13 at 1:12
• Tag added, because those are the only 2 options I see, maybe p(no college) is .2?? – Pedro.Alonso Nov 21 '13 at 1:23
• P(H) + P(no H) = 1. So does P(D) + P(no D). However, here there are 4 possibilities: H only, D only, H+D, and neither. Those 4 together have to = 1. – Peter Flom Nov 21 '13 at 1:26
• Ha ok I see it, I've forgotten H', ok then I will try something. Thanks. – Pedro.Alonso Nov 21 '13 at 1:28
You should be able to draw and use something like the diagram below to lay out the information (write it in the spaces and margins) and then you may be able to see how to do the problem. You would write all of the values on the diagram and from them fill in probabilities for every subregion and colored region(/margin). Then you should have a clearer idea what everything is.
Since (from his post in chat) the OP has worked the last details out correctly; here's my outline of the answer:
Given the values in the question (in blue), we can infer the other probabilities (in dark red) by subtraction:
a) The "probability that he is accepted by Dartmouth if he is accepted by Harvard" $= P(D|H) = P(D\cap H)/P(H) = 0.2/0.3 = 2/3$
b) "Is the event “accepted at Harvard” independent of the event “accepted at Dartmouth”?".
The two events are independent if $P(D\cap H)=P(D)P(H)$; equivalently they are independent if $P(D|H)=P(D)$ (as long as $P(H)$ is not 0).
In the first approach $P(D)P(H)= 0.5\times 0.3 = 0.15 \neq P(D\cap H)=0.2$, while using the second approach, $P(D|H)=2/3\neq P(D)=0.5$, so either way the events are not independent.
• I have done it like that, I get p(H')=0.5, p(H)=0.5, p(D)=0.3, p(D')=0.7, p(D and H)=0.2, p(D and H')=0.1, p(D' and H)=0.3, p(D' and H')=0.4, and p((H)*p(D)=0.15 therefore they are not independent. Is this correct? – Pedro.Alonso Nov 21 '13 at 15:12
• Not quite (unless I made an error). Check your answers for p(D and H') and p(D' and H). But you're on the right track. – Glen_b Nov 21 '13 at 23:21
• Your conclusion about independence is right, and looks like it's for the right reason. – Glen_b Nov 21 '13 at 23:39
• Well I cant find the mistake, let me go tru this again p(H)=0.5, then p(H')=0.5, p(D)=0.3, p(D')=0.7, correct? Then in the table of the joint probabilities I have subtracted the marginal and the one joint I have like: p(D and H')=p(D)-p(D and H), 0.3-0.2=0.1, correct? – Pedro.Alonso Nov 22 '13 at 2:11
• And summing the 4 entries gives me 1, 0.1+0.4+0.2+0.3=1, so why is this? Can you tell me the mistake it will be better. :) – Pedro.Alonso Nov 22 '13 at 2:26 | 2021-05-09T21:53:31 | {
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https://math.stackexchange.com/questions/105107/prove-int-0-infty-sin-x2-dx-converges?noredirect=1 | # Prove: $\int_0^\infty \sin (x^2) \, dx$ converges.
$\sin x^2$ does not converge as $x \to \infty$, yet its integral from $0$ to $\infty$ does.
I'm trying to understand why and would like some help in working towards a formal proof.
• This is a problem from Rudin's Principles of Mathematical Analysis (#6.13 in my/the latest edition) where he has you fill out his outline. – Tyler Feb 3 '12 at 0:39
• Do you at least have a basis for this statement? A.k.a. do you know if it's true before you attempted to prove it? What evidence do you have? If you can come up with an informal proof or some sort of educated conjecture then that might help you get started. I'm not saying it's not true, but you have to convince yourself that it's true before formally proving it. – chharvey Feb 3 '12 at 4:26
• This is called Fresnel integral. – Jack Dec 9 '12 at 16:31
• by the alternated series test, $\int_a^\infty \sin(x f(x)) dx$ converges whenever $f(x)$ is non-decreasing and $\to +\infty$. – reuns May 18 '16 at 17:43
• @Jozef: you can also use the Abel-Dirichlet's criterion. Appently, according to a comment to this similar question, you could use Riemann-Lebesgue lemma. – Watson Jul 25 '16 at 10:14
$x\mapsto \sin(x^2)$ is integrable on $[0,1]$, so we have to show that $\lim_{A\to +\infty}\int_1^A\sin(x^2)dx$ exists. Make the substitution $t=x^2$, then $x=\sqrt t$ and $dx=\frac{dt}{2\sqrt t}$. We have $$\int_1^A\sin(x^2)dx=\int_1^{A^2}\frac{\sin t}{2\sqrt t}dt=-\frac{\cos A^2}{2\sqrt A}+\frac{\cos 1}2+\frac 12\int_1^{A^2}\cos t\cdot t^{-3/2}\frac{-1}2dt,$$ and since $\lim_{A\to +\infty}-\frac{\cos A^2}{2\sqrt A}+\frac{\cos 1}2=\frac{\cos 1}2$ and the integral $\int_1^{+\infty}t^{-3/2}dt$ exists (is finite), we conclude that $\int_1^{+\infty}\sin(x^2)dx$ and so does $\int_0^{+\infty}\sin(x^2)dx$. This integral is computable thanks to the residues theorem.
• See what magic one can do when one does calculus carefully?One doesn't need fancy schmancy function spaces and measures all the time to do real math-sometimes all you have to do is know the basics and think it through carefully. : ) – Mathemagician1234 Feb 3 '12 at 3:36
• $\int_0^c \frac{\sin t}{\sqrt t} \mathrm{d}t= - \int_1^c \frac{\mathrm{d} \cos t}{\sqrt t} = - \frac{\cos t}{\sqrt t}\big|_1^c + \int_1^c \cos t (-1/2t^{-3/2}) \mathrm{d}t = -\frac{\cos c}{\sqrt c} + \cos 1 - 1/2 \int_1^c \cos t t^{-3/2} \mathrm{d}t$ – RHS May 1 '13 at 5:57
• "and the integral $\int_1^{+\infty}t^{-3/2}dt$ exists (is finite)" - where is the $\cos t$? – Elimination Jun 7 '15 at 14:17
• @Elimination It prove that $\int_1^\infty|\cos t| t^{-3/2}dt$ converges. – Davide Giraudo Jun 7 '15 at 14:54
The humps for $x\mapsto \sin(x^2)$ go up and down. Each has an area smaller than that of the last. The areas converge to 0 as you progress down the $x$-axis. By the alternating series test, this converges.
I solved this one integral as a particular case of the formula I provide here: http://www.mymathforum.com/viewtopic.php?f=15&t=26243 under the name Weiler.
$$\int\limits_0^\infty {\sin \left( {a{x^2}} \right)\cos \left( {2bx} \right)dx} = \sqrt {\frac{\pi }{{8a}}} \left( {\cos \frac{{{b^2}}}{a} - \sin \frac{{{b^2}}}{a}} \right)$$
$$\int\limits_0^\infty {\cos \left( {a{x^2}} \right)\cos \left( {2bx} \right)dx} = \sqrt {\frac{\pi }{{8a}}} \left( {\cos \frac{{{b^2}}}{a} + \sin \frac{{{b^2}}}{a}} \right)$$
So you have
$$\int\limits_0^\infty {\sin \left( {{x^2}} \right)dx} = \sqrt {\frac{\pi }{8}}$$
This is also informative, and works when there is no aspect with closed form. Taking Davide's substitution, define $$A_n^+ = \int_{2 \pi n}^{2 \pi n + \pi} \; \frac{\sin t}{2 \sqrt t} \; dt \; ,$$ $$A_n^- = \int_{2 \pi n + \pi}^{2 \pi n + 2 \pi} \; \frac{\sin t}{2 \sqrt t} \; dt \; ,$$ and finally $$A_n = A_n^+ + A_n^- = \int_{2 \pi n }^{2 \pi n + 2 \pi} \; \frac{\sin t}{2 \sqrt t} \; dt \; ,$$
Next, I just used $\int_{m \pi}^{m \pi + \pi} \sin t dt = \pm 2,$ depending upon the integer $m,$ and took bounds based on the size of the denominators.
I suppose we need to start with $n \geq 1.$ With that, $$\frac{1}{\sqrt{2 \pi n + \pi}} \leq A_n^+ \leq \frac{1}{\sqrt{2 \pi n}},$$ $$\frac{-1}{\sqrt{2 \pi n + \pi}} \leq A_n^- \leq \frac{-1}{\sqrt{2 \pi n + 2 \pi}},$$ and $$0 \leq A_n \leq \frac{1}{\sqrt{ 8 \pi} \; \; n^{3/2}}.$$
What does this say about convergence? The integral is $$\sum_{n = 0}^\infty A_n.$$ Convergence does not depend on the initial terms, so we may start at the more convenient $n=1.$ From the $3/2$ exponent in the estimate of $A_n,$ we see that the sum is a finite constant. We do see modest oscillation in the indefinite integral, however the $\sqrt n$ terms in the denominators of $A_n^+$ and $A_n^-$ tell us that eventually the indefinite integral stays within any desired distance of the infinite integral.
This idea, cancellation of alternating contributions, can be used with far worse integrands, $\sin (x^5 - x - 1)$ comes to mind.
There are more involved methods to actually compute this integral. A real variables method is given in this answer and a contour integration method is given in this answer.
Here is a much simpler method to show only the convergence of the integral. \begin{align} \int_0^\infty\sin\left(x^2\right)\,\mathrm{d}x &=\int_0^\infty\frac{\sin(x)}{2\sqrt{x}}\,\mathrm{d}x\tag{1}\\ &=\sum_{k=0}^\infty\int_{2k\pi}^{(2k+2)\pi}\frac{\sin(x)}{2\sqrt{x}}\,\mathrm{d}x\tag{2}\\ &=\sum_{k=0}^\infty\int_{2k\pi}^{(2k+1)\pi}\frac{\sin(x)}2\left(\frac1{\sqrt{x}}-\frac1{\sqrt{x+\pi}}\right)\mathrm{d}x\tag{3}\\ &=\sum_{k=0}^\infty\int_{2k\pi}^{(2k+1)\pi}\frac{\sin(x)}2\frac{\pi}{\sqrt{x}\sqrt{x+\pi}\left(\sqrt{x}+\sqrt{x+\pi}\right)}\mathrm{d}x\tag{4}\\ &\le\int_0^\pi\frac{\sin(x)}2\left(1+\sum_{k=1}^\infty\frac1{4\sqrt{2\pi} k^{3/2}}\right)\mathrm{d}x\tag{5}\\ &=1+\frac{\zeta\!\left(\frac32\right)}{4\sqrt{2\pi}}\tag{6} \end{align} Explanation:
$(1)$: substitute $x\mapsto\sqrt{x}$
$(2)$: break the integral into $2\pi$ segments
$(3)$: $\sin(x+\pi)=-\sin(x)$
$(4)$: algebra
$(5)$: $\frac{\pi}{\sqrt{x}\sqrt{x+\pi}\left(\sqrt{x}+\sqrt{x+\pi}\right)}\le\min\left(1,\frac1{4\sqrt{2\pi} k^{3/2}}\right)$ for $x\ge2k\pi$
$(6)$: evaluate integral and sum
## Just with elementary tools: see Here,How to prove only by Transformation that: $\int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx$
In fact we have: \begin{split} \int_0^\infty \sin(x^2) dx=\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx<\infty\end{split} See below
Employing the change of variables $2u =x^2$ after integration by parts we get \begin{split} \int_0^\infty \sin(x^2) dx&=&\frac{1}{\sqrt{2}}\int^\infty_0\frac{\sin(2x)}{\sqrt{x}}\,dx\\& =&\frac{1}{\sqrt{2}} \int^\infty_0\frac{\sin(2x)}{x^{1/2}}\,dx\\&=&\frac{1}{\sqrt{2}}\underbrace{\left[\frac{\sin^2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx \\&= &\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx\end{split}
Given that $\sin 2x =(\sin^2x)'$ and $$\lim_{x\to 0}\frac{\sin x}{x}=1$$ However, $$\int^\infty_1\frac{\sin^2 x}{x^{3/2}}\,dx\le \int^\infty_1\frac{1}{x^{3/2}}\,dx<\infty$$ since $|\sin x|\le |x|$ we have, $$\int^1_0\frac{\sin^2 x}{x^{3/2}}\,dx \le \int^1_0\frac{\ x^2}{x^{3/2}}\,dx = \int^1_0\sqrt x\,dx<\infty$$ | 2019-04-24T06:38:56 | {
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https://www.jiskha.com/questions/1453954/the-line-has-equation-y-2x-c-and-a-curve-has-equation-y-8-2x-x-2-1-for-the-case-where | # Math
The line has equation y=2x+c and a curve has equation y=8-2x-x^2.
1) for the case where the line is a tangent to the curve, find the value of the constant c.
2) For the case where c = 11, find the x-coordinates of the points of intersection of the line and the curve. Find also, by integration, the area of region between the line and the curve.
1. 👍
2. 👎
3. 👁
1. to solve:
2x+c = 8-2x-x^2
x^2 + 4x + c-8 = 0
to have y = 2x+c to be a tangent, there can be only one real root of the quadratic
so b^2 - 4ac = 0
16 - 4(1)(c-8) = 0
16 - 4c + 32 = 0
4c = 48
c = 48/4 = 12
check with Wolfram:
http://www.wolframalpha.com/input/?i=y%3D2x%2B12,+y%3D8-2x-x%5E2
y = 2x+11, y = 8-2x - x^2
I will let you solve it to show:
http://www.wolframalpha.com/input/?i=y%3D2x%2B11,+y%3D8-2x-x%5E2
so for the area:
A = ∫ (8-2x-x^2 - 2x - 11) dx from -3 to -1
simplify and finish it
this looks very straightforward, I trust you can handle it
1. 👍
2. 👎
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https://math.stackexchange.com/questions/4107743/axiom-schema-of-separation-routine | # Axiom Schema of Separation Routine
Source: Patrick Suppes: Axiomatic Set Theory
My question is why do we always bother with the routine of showing that the abstract property/formula implies the membership in the bigger set we're separating over? Doesn't the axiom schema alone suffice, on its own, as justification for the existence of the new set we're trying to build (No additional steps needed)?
Suppose you're trying to construct some set $$\{x:P(x)\}$$. The axiom of separation says that for any set $$A$$, you can construct the set $$\{x\in A:P(x)\}$$. But this is not necessarily equal to the set you actually want: it only contains the elements of $$A$$ that satisfy $$P$$, not everything that satisfies $$P$$. To be sure this set really does contain everything that satisfies $$P$$, you have to additionally prove that everything that satisfies $$P$$ is in $$A$$.
• To add to this answer: consider what happens if you apply separation to the formula "$x\not\in x$." For any specific set $A$ we can get $\{x\in A: x\not\in x\}$, but no single $A$ contains every set satisfying "$x\not\in x$" (which is good since otherwise we'd run into Russell's paradox). Apr 19, 2021 at 3:21 | 2023-02-04T04:42:41 | {
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https://math.stackexchange.com/questions/3757478/finding-the-minimum-and-maximum-values-of-qr-when-p-3q-cdot-2r-and-100p | # Finding the minimum and maximum values of $q+r$ when $p=3^q\cdot 2^r$ and $100<p<1000$
Question:
Given positive integers $$p$$, $$q$$ and $$r$$ with $$p=3^q\cdot2^r$$ and $$100. The difference between maximum and minimum values of $$(q+r)$$, is _______.
My Approach:
As we are given that $$100, on taking logarithm to the base $$10$$, we get:
$$\log100<\log p<\log 1000$$
$$2<\log(3^q\cdot2^r)<3$$
$$2
$$2
The only way I could think of to obtain the maximum and minimum values of $$q$$ and $$r$$ is to substitute them with natural numbers and look when the condition is satisfied. Is there any formal approach (other than substituting $$q$$ and $$r$$ with natural numbers) using which we can find the minimum and maximum values of $$(q+r)$$? If yes, it would be helpful if you could explain it.
On substituting different values of $$q$$ and $$r$$ in the condition arrived, the minimum and maximum values of $$(q+r)$$ comes out to be $$5$$ and $$9$$ respectively. And the answer to the above question is thus $$9-5=4$$. This is also the correct answer with respect to the answer key provided in my book.
Given positive integers $$p$$, $$q$$ and $$r$$ with $$p=3^q\cdot2^r$$ and $$100. The difference between maximum and minimum values of $$(q+r)$$, is _______.
The answer is quite straightforward if you draw the lines \begin{align*} \frac q{\frac 2{\log 3}}+\frac r{\frac 2{\log 2}}&=1\Rightarrow \frac q{4.\cdots}+\frac r{6.6\cdots}=1\tag{1}\\ \frac q{\frac 3{\log 3}}+\frac r{\frac 3{\log 2}}&=1\Rightarrow \frac q{6.\cdots}+\frac r{9.9\cdots}=1\tag{2}\\ q+r&=10\tag{3}\\ q+r&=4\tag{4}\\ \end{align*} on $$q-r$$ axes neatly and see that the points $$(5,0)$$ and $$(0,9)$$ can serve the best values $$5$$ and $$9$$ respectively, for the points lie strictly in the region between the lines $$(1)$$ and $$(2)$$ (let us call it region R). Lines $$(3)$$ and $$(4)$$ are drawn just to see that the best values can't be outside what we have obtained, that is $$q+r$$ can't be $$4,10$$.
EDIT:
But since $$q,r\in \mathbb N$$, we need to take positive integral points or points in which both $$q,r\in\mathbb N$$. I think the best approach would be to see that at least some segment of the lines $$q+r=5$$ and $$q+r=9$$ lies inside the region R, and then to make sure that those segments have positive integral points on them. For example, take the points $$(4,1)$$ and $$(1,8)$$. If they don't suffice(Well, they do here. I am just trying to provide an algorithmic approach to solve bigger such problems.), keep taking points from the segments of the lines $$q+r=5$$ and $$q+r=9$$. If they all don't suffice, move to lines that lie more inside the region R. For example, $$q+r=4$$ and $$q+r=8$$.
Note that the positive integral points nearest to the lines $$(1)$$ and $$(2)$$ will give the limits nearest to $$100$$ and $$1000$$. One can apply the "distance from a line" formula if the distances of two such points from the lines $$(1)$$ and $$(2)$$ are to be compared, and the inequality is not apparent by the neat plot.
• Thank you for your answer. I like this approach. But how could we conclude that there exists a point between the two parallel lines in the first quadrant having positive integral coordinates, by just knowing $(5,0)$ and $(0,9)$ are solutions except the fact that one coordinate is zero? On plotting the lines, and with small calculations it becomes evident that $(4,1)$ and $(1,8)$ satisfy the condition. Or is there a better way to look at this? – Guru Vishnu Jul 15 '20 at 10:15
• @mgv My bad! Actually, earlier I thought that $q,r\in \mathbb Z$ and not $\in \mathbb N$. So I will have to edit it. I think the best approach would be to see that some segments of the lines $x+y=5$ and $x+y=9$ lie inside the region strictly enclosed by the first two lines in the answer, and then to make sure that those segments have integral points on them, as you did by putting the values $(4,1)$ and $(1,8)$. If they don't suffice, keep taking values from the lines $x+y=5$ and $x+y=9$ and if they all don't suffice, go to lines that lie more inwards in the region, like $x+y=4$ and $x+y=8$. – Sameer Baheti Jul 15 '20 at 10:32
• @mgv I have finished editing hopefully! Have a look. – Sameer Baheti Jul 15 '20 at 11:16
This is easiest to just reason directly.
For any $$M$$ that $$2^M < 3^12^{M-1} < 3^22^{M-2} < ..... < 3^{M-1}2 < 3^M$$.
So if $$q+r = m$$ and $$100 < 3^q 2^r < 1000$$ then
1. $$100 < 3^q2^r \le 3^{q+r}$$ and
2. $$2^{q+r} \le 3^q2^r < 1000$$
And as $$3^4 < 100 < 3^5$$ then minimum value that $$q+r$$ can be, by 1) is $$5$$. And as $$2^9< 1000< 2^{10}$$ the maximum value that $$q+r$$ can be, by 2) is $$9$$.
But this assumes that allows for $$r$$ or $$q$$ to be zero and the question specifically says positive. So we didn't actually do it right. But we can modify really easily.
1. $$100 < 3^q2^r \le 3^{q+r-1}\cdot 2 < 3^{q+r}$$ and $$3^4 < 100 < 81\cdot 2=3^4\cdot 2 < 3^5$$.
2. $$2^{q+r} < 2^{q+r-1}\cdot 3 \le 3^q2^r < 1000$$ and $$256\cdot 3=2^8\cdot 3 < 2^9 < 1000 < 2^{10}$$.
And our conclusions hold.
=========
Note: Had the question been $$80 < p < 1025$$ we'd have had the same answers of a minimum of $$4$$ and a maximum of $$9$$ because, although $$80< 3^4 < 2^{10} < 1025$$ we are required that $$r,q$$ be non-zero and $$3^3\cdot 2 < 80 < 2^10 < 3\cdot 2^9$$.
Note 2: If we didn't have the requirement that $$q,r$$ but positive and allow them to be negative we would have no minimum or maximum.
For any negative $$r$$ then if $$q= \lceil \log_3(101\cdot 2^{|r|})\rceil$$ we will have $$3^q2^r \ge 3^{\log_3(101\cdot 2^{|r|})}2^{r}=101\cdot 2^{r}2^{|r|}= 101$$ and $$3^q2^r < 3^{\log_3(101\cdot 2^{|r|})+1}2^{r}=3\cdot 101< 1000$$.
And $$\lim_{r\to -\infty} r+\lceil \log_3(101\cdot 2^{|r|})\rceil=$$
$$\lim_{r\to -\infty} r+ \log_3(101\cdot 2^{-r})=$$
$$\lim_{r\to -\infty} r-r\log_3 2 + \log_3(101)=$$
$$\lim_{r\to -\infty} r(1+\log_2 3) + \log_3(101)= -\infty$$
so there is no minimum.
A similar argument for negative $$q$$ will show there is no maximum.
Observe that for the minimum value of $$q+r$$, we must have $$2*3^{q}>100$$ because $$q,r>0$$ and $$2^r\leq3^q ,\forall r\leq q$$ which means that multiplying by $$3^q$$ will get us to surpass $$100$$ the fastest, which requires the least amount of exponentiation. Solving for $$q$$ and taking the next integer that satisfies the inequality gives us $$r+q=1+4=5$$.
Then for the maximum, we need that $$2^r*3 < 1000$$ because multiplying by $$2^r$$ gets us to surpass $$1000$$ the slowest, which means the most amount of exponentiation. Solving for $$r$$ and taking the next integer that satisfies the inequality yields $$r+q=8+1=9$$.
Then the maximum minus the minimum of $$q+r$$ is $$9-5=4$$.
My argument is essentially based upon the fact that $$2^x$$ grows slower than $$3^x$$. | 2021-01-17T00:34:13 | {
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https://math.stackexchange.com/questions/2164232/trouble-solving-for-exponents-with-constants | # Trouble solving for exponents with constants
so I have this equation I need to solve for $i$, where $k$ and $u$ are arbitrary constants: $$u^{{\frac1{k^i}}} = k$$ And these are the steps I've accomplished so far (all logs are in base 2): $$log(u^{{\frac1{k^i}}}) = log(k)$$ $$\frac1{k^i} \cdot log(u) = log(k)$$ $$log(u) = log(k)\cdot k^i$$
But now I'm stuck... how do I solve for $i$ from here? I'm forgetting how I would do this. Any help and explanation of how log and exponent math would be great! (Also please correct me if what I've done so far is wrong.) I also hope my math formatting makes sense, thank you!
• i don't understand what exactly do you mean? – Dr. Sonnhard Graubner Feb 27 '17 at 19:59
• instead of having the first equation = k, I would like it to = i. In otherwords, I'm trying to solve for i. – keenns12 Feb 27 '17 at 20:06
• $\log u = \log k*k^i$ so $k^i = \frac {\log u}{\log k}$ so $i = \log_k(\frac {\log u}{\log k})$. That's $i$. That's all there is. You are allowed to express logs of logs, you know. – fleablood Feb 27 '17 at 20:13
$$u^{(1/k)^i} = k$$ taking logs as you have done works $$\frac{1}{k^i}\log u = \log k$$ then we have $$\frac{1}{k^i} = \frac{\log k}{\log u}$$ Taking logs again $$-i\log k = \log \left(\frac{\log k}{\log u}\right)$$ so $$i = \frac{1}{\log k}\log \left(\frac{\log u}{\log k}\right)$$
or take logs on your final statement $$\log (\log u) = \log (\log k) + i \log k$$ we used $$\log(ab) = \log a + \log b\\ \log(a^n) = n\log a$$
Instead of using a base 2 logarithm you could use base $k$, since it is implicitly non-negative. $$log_k\left(u^{{1/k}^i}\right) = log_k(k)$$ $$\frac{1}{k^i} log_k(u) = 1$$ $$log_k(u) = k^i$$ Apply $log_k$ one more time, and you're done. $$log_k(log_k(u)) = i$$ | 2019-09-16T08:57:03 | {
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http://math.stackexchange.com/questions/317052/how-to-solve-int-cos2x-cos3x-dx | How to solve $\int \cos(2x)\cos(3x)\ dx$?
I took a shot in the dark and assumed that this is similar to solving $\int e^{x}\sin{x}\ dx$, but wolfram is giving me a different answer than what I got, and on top of that, I tried to differentiate my result and am not getting back what I started with. It's putting into question whether I was doing previous questions right or not..
First step of my attempt:
• let $u=\cos(2x),\ du=-2\sin(2x)\ dx$
• let $dv=\cos(3x)\ dx,\ v=\frac{\sin(3x)}{3}$
$$\int\cos(2x)\cos(3x)\ dx=\frac{\cos(2x)sin(3x)}{3}+\frac{2}{3}\int\sin(2x)\sin(3x)\ dx$$
Then I did IBP again:
• let $u=\sin(2x),\ du=2\cos(2x)\ dx$
• let $dv=\sin(3x)\ dx, v=-\frac{cos(3x)}{3}$
$$=\frac{\cos(2x)sin(3x)}{3}+\frac{2}{3}\left[-\frac{\cos(3x)\sin(2x)}{3}+\frac{2}{3}\int\cos(2x)\cos(3x)\ dx\right]$$
From there, I simplify and re-arrange to get
$$\frac{1}{3}\int\cos(2x)\cos(3x)\ dx=\frac{3\cos(2x)\sin(3x)-2\cos(3x)\sin(2x)}{9}$$ $$\int\cos(2x)\cos(3x)\ dx=\frac{3\cos(2x)\sin(3x)-2\cos(3x)\sin(2x)}{3}+C$$
So where did I go wrong? Wolfram says the answer should be
$$\int\cos(2x)\cos(3x)\ dx=\frac{1}{10}5\sin(x)+\sin(5x)+C$$
-
Don't integrate by parts. Use the appropriate produc to sum trigonometric formula: $\cos \theta\cos \phi=\ldots$. You can find it here: en.wikipedia.org/wiki/List_of_trigonometric_identities – 1015 Feb 28 '13 at 18:06
I kind of like the idea of computing an integral in two different ways. You can get some interesting identities that way. Although the identity you get this way seems to essentially be a product to sum identity. – Baby Dragon Feb 28 '13 at 18:20
You have $\cos x \cos y = \frac{1}{2}(\cos (x+y) + \cos(x-y))$.
Hence $\cos (2x) \cos (3x) = \frac{1}{2} (\cos (5x) + \cos x)$. This should be straightforward to integrate.
-
Or use $\cos x = \frac{1}{2}(e^{ix}+e^{-ix})$. – copper.hat Feb 28 '13 at 18:08
You forgot to distribute the $\frac23$ after your second IBP. You should have $$\int\cos(2x)\cos(3x)\,dx=\frac{\cos(2x)\sin(3x)}3-\frac{2\cos(3x)\sin(2x)}9+\frac49\int\cos(2x)\cos(3x)\,dx.$$ From there, product-to-sum laws should get you the rest of the way (though it would be easier to simply use them from the start).
P.S.: Don't forget the integration constant!
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Alternatively, you could subtract $\displaystyle \frac 49 \int \cos(2x) \cos(3x) dx$ from both sides and divide by $\displaystyle \frac 59$. – Joe Z. Feb 28 '13 at 18:40
@JoeZeng: Yes, but the OP knows to do that. The issue was that instead of $\frac49,$ the OP had $\frac23,$ so instead of $\frac59,$ the OP had $\frac13.$ – Cameron Buie Feb 28 '13 at 18:42
You did not go wrong, apart from a minor mistake in arithmetic. The denominator should be $5$.
However, the "product to sum" approach that Alpha took is, unusually for Alpha, more efficient.
Remark: You mention that you differentiated your final result and the derivative did not agree with the integrand. It is always possible to make a mistake in differentiating. When I do it, I get $\frac{5}{3}\cos 2x\cos 3x$. That says that your integral is almost right, and suggests that there was an unimportant glitch in the calculation. By the way, I computed the integral by using the Method of Undetermined Coefficients, looking for $A$ and $B$ such that $A\cos 2x\sin 3x +B\cos 3x\sin 2x$ works.
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And, even if you do yours right, still Alpha's answer will look different. Even though they are both right answers. – GEdgar Feb 28 '13 at 18:27
OK - so I see I did do it right, except for missing the $\frac{2}{3}$ on the second IBP, thus messing up my answer enough to make my differentiation in the end incorrect. But I don't think the professor requires me to use the product-to-sum rules.. I used it anyhow, but thanks for pointing it out. – agent154 Feb 28 '13 at 18:31
You are welcome. You do the calculations well, and, importantly, give a thorough and well-formatted account of what you did. It would be unfortunate if a little slip made you lose faith in the soundness of your approach. – André Nicolas Feb 28 '13 at 18:44
Eternal vigilance is the price of correct computations... – copper.hat Feb 28 '13 at 19:28 | 2014-12-20T02:18:46 | {
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http://math.stackexchange.com/questions/167957/closed-form-solution-of-fibonacci-like-sequence | # Closed form solution of Fibonacci-like sequence
Could someone please tell me the closed form solution of the equation below.
$$F(n) = 2F(n-1) + 2F(n-2)$$
$$F(1) = 1$$ $$F(2) = 3$$
Is there any way it can be easily deduced if the closed form solution of Fibonacci is known?
-
The same solution method should work.... – Hurkyl Jul 7 '12 at 19:11
Well, how would you derive the Binet formula for Fibonacci numbers in the first place? You can adapt the same technique to this case. – J. M. is back. Jul 7 '12 at 19:15
I think we should know two "base cases," e.g. $$F(0)=1\\F(1)=1$$ otherwise we can never get the numerical value of $F(n)$. – Argon Jul 7 '12 at 19:17
@Argon, of course, but in the absence of base cases, we could still have a formula with two arbitrary constants... – J. M. is back. Jul 7 '12 at 19:23
For your particular initial conditions, you should be getting $$\frac1{12}\left((3-\sqrt{3})\left(1-\sqrt{3}\right)^k+(3+\sqrt{3})(1+\sqrt 3)^k\right)$$ – J. M. is back. Jul 7 '12 at 19:43
Any of the standard methods for solving such recurrences will work. In particular, whatever method you would use to get the Binet formula for the Fibonacci numbers will work here, once you establish initial conditions. If you set $F(0)=0$ and $F(1)=1$, as with the Fibonacci numbers, the closed form is
$$F(n)=\frac{(1+\sqrt3)^n-(1-\sqrt3)^n}{2\sqrt3}\;;$$
I don’t see any way to derive this directly from the corresponding closed form for the Fibonacci numbers, however.
By the way, with those initial values the sequence is OEIS A002605.
Added: The general solution is $$F(n)=A(1+\sqrt3)^n+B(1-\sqrt3)^n\;;\tag{1}$$ Argon’s answer already shows you one of the standard methods of obtaining this. To find $A$ and $B$ for a given set of initial conditions, just substitute the known values of $n$ in $(1)$. If you want $F(1)=1$, you must have $$1=F(1)=A(1+\sqrt3)^1+B(1-\sqrt3)^1\;,$$ or $A+B+\sqrt3(A-B)=1$. To get $F(2)=3$, you must have \begin{align*}3&=F(2)=A(1+\sqrt3)^2+B(1-\sqrt3)^2\\ &=A(4+2\sqrt3)+B(4-2\sqrt3)\;, \end{align*}
or $4(A+B)+2\sqrt3(A-B)=3$. You now have the system
\left\{\begin{align*}&A+B+\sqrt3(A-B)=1\\ &4(A+B)+2\sqrt3(A-B)=3\;. \end{align*}\right.
Multiply the first equation by $2$ and subtract from the second to get $2(A+B)=1$, and multiply the first equation by $4$ and subtract the second from it to get $2\sqrt3(A-B)=1$. Then you have the simple system \left\{\begin{align*}&A+B=\frac12\\&A-B=\frac1{2\sqrt3}\;,\end{align*}\right. which you should have no trouble solving for $A$ and $B$.
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can you please tell the corresponding F(n) with F(1) = 1 and F(2) = 3 – Raj Jul 7 '12 at 19:40
@Raj: you don't know how to derive it yourself? – J. M. is back. Jul 7 '12 at 19:47
@Brian and J.M : Thanks you so much for your help.It means a lot to me. – Raj Jul 7 '12 at 20:10
$$F(n)=2F(n−1)+2F(n−2)=2(F(n-1)+F(n-2))$$
Gives us the recurrence relation
$$r^n=2(r^{n-1}+r^{n-2})$$
we divide by $r^{n-2}$ to get
$$r^2=2(r+1) \implies r^2-2r-2=0$$
which is our characteristic equation. The characteristic roots are
$$\lambda_1=1-\sqrt{3} \\ \lambda_2=1+\sqrt{3}$$
Thus (because we have two different solutions)
$$F(n)=c_1 \lambda_1^n+c_2\lambda_2^n = c_1(1-\sqrt{3})^n+c_2(1+\sqrt{3})^n$$
Where $c_1$ and $c_2$ are constants that are chosen based on the base cases $F(n)$.
Brian M. Scott's answer explains how to obtain $c_1$ and $c_2$.
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Any solution sequence can be written, with real constants $A,B,$ as $$A \; \left(1 + \sqrt 3 \right)^n + B \; \left(1 - \sqrt 3 \right)^n.$$
The set of such sequences is a vector space over $\mathbb R,$ of dimension 2. The expression below shows a linear combination of basis elements for the vector space.
In comparison, suppose we took $$G(n) = 8 G(n-1) - 15 G(n-2).$$ Then, with real constants $A,B$ to be determined, we would have $$G(n) = A \cdot 5^n + B \cdot 3^n$$
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Define $g(z) = \sum_{n \ge 0} F(n + 1) z^n$, write the recurrence as: $$F(n + 3) = 2 F(n + 2) + 2 F(n + 1) \qquad F(1) = 1, F(2) = 3$$ Multiply the recurrence by $z$, sum over $n \ge 0$ and get: $$\frac{g(z) - F(1) - F(2)}{z^2} = 2 \frac{g(z) - F(1)}{z} + 2 g(z)$$ Solve for $g(z)$: $$g(z) = \frac{1 + z}{1 - 2 z - 2 z^2} = \frac{2 + \sqrt{3}}{2 \sqrt{3}} \cdot \frac{1}{1 - (1 + \sqrt{3}) z)} + \frac{2 - \sqrt{3}}{2 \sqrt{3}} \cdot \frac{1}{1 + (1 - \sqrt{3})z}$$ Two geometric series: $$T(n+ 1) = \frac{2 + \sqrt{3}}{2 \sqrt{3}} \cdot (1 + \sqrt{3})^n + \frac{2 - \sqrt{3}}{2 \sqrt{3}} \cdot (1 - \sqrt{3})^n$$
- | 2015-10-13T14:31:41 | {
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https://math.stackexchange.com/questions/2312085/counting-total-number-of-local-maxima-and-minima-of-a-function | # Counting total number of local maxima and minima of a function
Find the total number of local maxima and local minima for the function $$f(x) = \begin{cases} (2+x)^{3} &\text{if}\, -3 \lt x \le -1 \\ (x)^\frac{2}{3} &\text{if}\, -1 \lt x \lt 2 \end{cases}$$
My attempt : I differentiated the function for the two different intervals and obtained the following: $$f'(x) = \begin{cases} 3\cdot(2+x)^{2} &\text{if}\, -3 \lt x \le -1 \\ \frac{2}{3}\cdot (x)^\frac{-1}{3} &\text{if}\, -1 \lt x \lt 2 \end{cases}$$ How do I obtain the maxima and minima points from here.
Any help will be appreciated.
• extrema occur where $f'(x) = 0$, $f'(x)$ is undefined, and/or endpoints. – Dando18 Jun 6 '17 at 14:12
• @Dando18, "extrema may occur where..." In the problem given, $f'(-2)=0$ but $(-2,0)$ is not an extrema. – Bernard Massé Jun 6 '17 at 14:28
• @BernardMassé Yeah I meant $\text{extrema} \implies f'(x)=0,\, \dots$ not the other way around. – Dando18 Jun 6 '17 at 14:30
• Just draw the graph?! – Pieter21 Jun 6 '17 at 15:30
You need to study $f'$ :
What is the sign of $f'(x), x\in ]-3 , 2 [$ ?
A local maxima/minima $x_m$ appears when :
• $f'(x_m) = 0$ or $x_m$ is a remarquable point (here $x_m \in \{-3,-1,2\}$)
• $f'$ changes sign before and after $x_m$
If the second condition is not verified, $x_m$ is an inflection point.
• I think x=0 is another remarkable point here, since f'(x) becomes undefined at x=0 . – MathsLearner Jun 6 '17 at 15:30
• You're absolutely right, I didn't do the $f'$ study before posting my answer but indeed as pointed out by Dando18 if $f'$ is undefined then you should check the nature of the point. – Furrane Jun 8 '17 at 2:38 | 2019-11-14T19:21:08 | {
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http://aidmbergamo.it/vuzw/area-of-a-rectangle-with-a-semi-circle-cut-out.html | # Area Of A Rectangle With A Semi Circle Cut Out
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Class 7 - Maths - Perimeter and Area. 16 "cm"^2 "to 5 significant figures" First, observe the formula for calculating the area of a circle: "A"=pir^2 Where: "A" is the area of the circle pi is the irrational number 3. Perfect as promotional giveaways and gifts, they are a creative and affordable way to market a business and gain exposure in the community. TIP: When cutting out the square, cut a bunch of little lines and then cut out the main line you drew. h header file In this program, we will draw a circle on screen having centre at mid of the screen and radius of 80 pixels. Consider a single circle with radius r, the area is pi r 2. Area and Perimeter of a Ellipse Shape Calculator Calculate Area and Perimeter of an Ellipse Shape. Use the ruler and the markings you just made to score an outline of the rugs new size with a carpeting knife. so you have a rectangle and two congruent semi-circles which can be combined to form a complete circle!. I've also used Bendarooz (wax sticks) to. com for a large selection of die cut machines and dies for scrapbooking, cardmaking and paper crafting. Let us find the area of a triangle by using square unit areas. With a protractor, divide the semi-circle into five equal parts (ie. 6th - 8th grade. Each semi-circle has a radius of 4 inches and the width of the rectangle is 18 inches. Estimate the percentage of the circle area compared with the square area. The area of a triangle is the region enclosed by the sides of a triangle. The moment of inertia is the mass of the object times the mass-weighted average of the squared distance from the axis. π is roughly equal to 3. 5 cm 2, equaling 21. We multiply 3 cm by 5 cm. =1/4$πr^2 - 1/2$ x base x height (Here base = height = radius =. Find the area of the paper that remains. What you have is a rectangle and a circle (It's been cut down the middle and the rectangle shoved in it). What is the area of 2 156. To get the area of the triangle portion, I subtract the area of the sector of the circle with the area of the triangle of that portion multiplied by 2 (since there are essentially two triangles). Calculate the area of the top surface of the button in square centimetres. Let us find the area of a triangle by using square unit areas. Simple Areas Area of a rectangle Area of a right-angled triangle www. com Area of a Rectangle Find the area of each rectangle. This area formula is useful for measuring the space occupied by a circular field or a plot. Covering a rectangle with circles. Extend the rectangle until it snaps to the guide (the rectangle should now cover half the circles). We didn’t have a lot of extra waste because of that. Compare the area of a rectangle with base b and height h with the area of a rectangle with base 2b and height 2h. 8 cm 2 210 cm 2 528 cm 2. Find the length of a rectangle if the area is 27 cm 2 and the width is 3 cm. An easy to use, free area calculator you can use to calculate the area of shapes like square, rectangle, triangle, circle, parallelogram, trapezoid, ellipse, and sector of a circle. Online custom sheet metal fabrication & laser cutting service offered by Metalscut4u. b) Creating a Rectangle. Trim only the fabric around the outside edge to a ¼” seam allowance but do not cut either zipper ends. A semicircle is cut out of a rectangular paperboard 22in long and 17in If the semi- circle was taken from the width the perimeter would be 51. Switch the color of the envelope to white and add a circle base for our icon, filling it with linear gradient from darker blue ( #2066F0 ) on top to lighter blue ( #1DD4FD ) in the bottom. As page layout software matured, SVG elements were introduced to those programs, to the point that many simple illustrations can be produced directly within the page layout program. Determine the radius Measure the distance from the center of the circle of which the semicircle is part to its edge. *Note: Model is a radical function with limited domain. Example 2 Determine the area of the largest rectangle that can be inscribed in a circle of radius 4. Note: The rectangle and the "bumpy edged shape" made by the sectors are not an exact match. 5 m, find the area of the playground. 3137085/2)^2 = 100. formule o f area of a circle but bas ic rectangle formula helped her to find out the area. Area of Semi-Circle = 1 ⁄ 2 * π *r 2. Here, the circle is cut into 8 equal parts. Mike Henderson was best known as a committed tennis coach and a devoted follower of Christ. a) Create a rectangle by cutting off the right-angled triangle and moving it. Apart from these, some software also feature options to calculate diagonal, height, length, breadth, etc. Formulas, explanations, and graphs for each calculation. A composite area consisting of the rectangle, semicircle, and a triangular cutout is shown (Part A figure). The second moment of area, also known as area moment of inertia, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. 99 for a replacement flag or$49 for a complete kit, purchasing these promotional flags is a must. Volume of Half Cylinder Calculator. Area of a Circle. Area of rectangle is the region covered by the rectangle in a two-dimensional plane. With the Line tool (), draw two lines: one that divides the outer circle in half and one that divides the inner circle that you created with the Offset tool. Selections are so vast and varied, the experience can be overwhelming. Adjust and Print Full Scale Circle Divider Templates- Setting Circles Drag Diameter slider to size. We cut circles into small sectors and arranged them into a form as close to a rectangle as possible. asked by ian on August 29, 2016. Creating a design with pavers requires planning. Learn how to make memes, slideshows, vision boards, book covers and photo collages! Check out beautiful hex color codes for coral, periwinkle, emerald green, royal blue, and teal. The Rhino Reflex Salon Mats feature 1" thick sponge and are available in both Rectangle & Semi-Circle shapes measuring 5' wide and 3' deep as well as gloss black & gloss metallic finishes. There are 12 squares contained in this rectangle. If the sides of the rectangle are 36 m and 24. 14] ANSWERS 1. Filled Area Chart¶. Give the answer correct to 2 decimal places. But we could get a better result if we divided the circle into 25 sectors (23 with an angle of 15° and 2 with an angle of 7. The unit of dimension of the second moment of area is length to fourth power, L 4, and should not be confused with the mass moment of inertia. circum = 2pi r = 2 pi (15) = 30 pi = 94. The quantity we need to maximize is the area of the rectangle which is given by. This free area calculator determines the area of a number of common shapes using both metric units and US customary units of length, including rectangle, triangle, trapezoid, circle, sector, ellipse, and parallelogram. Suppose that the other half of the circle is cut in the same way and fitted into the first, as shown by the dashes in the second figure. For anything larger, we can custom cut the circles for you. A perpendicular cut to the base of ANY prism will result tin. Volume = area of base x height 3 ight radius 2 radius ight Volume = πr2h V = πr2h Cones a cone is one third of a cylinder Spheres V = 4πr3 3 Frustrums a frustrum is a pyramid with the top cut off. The area of the "2 by 4" rectangle below it is 2(4)= 8. (Thus the diameter of the semicircle is equal to the width of the rectangle. Rotate one of them and overlay the two sheets perpendicular to one another. Using this calculator, we will understand the algorithm of how to find the perimeter, area and diagonal length of a rectangle. Find the area of the shaded region leave your answer in terms of pi. Area of semicircle. , eggs, buns for a hot dog, a running track. 14 to approximate pi. Starter Questions www. The formula is: A = L * W where A is the area, L is the length, W is the width, and * means multiply. The diameter of each circle is 4 cm. Area of Circle. The Split Into Grid command lets you divide one. 2) Create a sketch circle on either plane of step 1. 4pi x 4 x 6 pi d. rectangle to square with one cut, which need not be straight, how do you cut this 4 by 9 rectangle into two bits that will make a square? using two straight cuts there is a simple way to transform this 4 by 9 rectangle to a square. The smaller yellow semi-circle is the path that the center of the grinding tool follows. 2 square inches. Circular segment - is an area of a circle which is "cut off" from the rest of the circle by a secant (chord). Be sure to include the correct unit in your answer. Crafted from 100% wool with a low 0. The area of a triangle is the region enclosed by the sides of a triangle. For example, if you plan to make a 6-foot diameter circle with 4½-inch square and wedge pavers, expect to have seven rings and a centerpiece. The area of the rectangle is 16 x 6 = 96 cm 2. If you know the diameter or radius of a circle, you can work out the circumference. Area and Perimeter of Circle and Semi-Circle: Formulas Toppr. A node is typically a rectangle or circle or another simple shape with some text on it. For anything larger, we can custom cut the circles for you. 6803 ($$h$$ x 2$$r$$). If you know radius and angle you may use the following formulas to calculate remaining segment parameters: Circular segment formulas. Semi-circles are drawn on and as diameters. Scholars use formulas for the area of a circle and the area of a rectangle to determine the number of pies a baker can make from a particular area of dough. For a similar markup without the measurement, see the Polygon Tool. 2) Create a sketch circle on either plane of step 1. Tap the circle cutout. Practice: Nets of polyhedra. Curved Rectangle Calculator. Area measures the space inside a shape. The area outside the rectangle but in the semicircle is shaded. Once you have all your lines made, cut it out with the Band Saw. Don't let this lot fob you off by working out the volume just because it's easier. Position the cursor in the desired field (Sketch tools toolbar) and key in the desired values to create points & then lines for rectangle. Make sure the dough is rolled very thin! Using a pizza cutter or a knife, trim the edges to make a very uniform rectangle. It is important to use the "Length A", the long measurement in the box with the Length A label. We can also figure out the area of a fraction of a circle. In the simplest case, a node is just some text that is placed at some coordinate. The diagram shows a metal plate in the shape of a rectangle. and dlscuss each step. The perimeter of a plane 2-D figure is the length of the outer periphery of the figure. TIP: When cutting out the square, cut a bunch of little lines and then cut out the main line you drew. 6 square feet. And now all you have to do is subtract area of the semi-circle from the area of the rectangle. The diameter is equal to the shortest side of the rectangle. Tap Start Rectangle or press the green button and drag out a rectangle of an arbitrary size. What is the area of this figure? Use 3. 16 "cm"^2 "to 5 significant figures" First, observe the formula for calculating the area of a circle: "A"=pir^2 Where: "A" is the area of the circle pi is the irrational number 3. area base = 10x18 = 180. Cut Rectangle. Custom keychains are available in cool styles that include clean anti-germ keys, flashlights, wrist straps, carabiners, and photo holders. A semicircle is a half circle, formed by cutting a whole circle along a diameter line, as shown above. We don't draw the rest of the object, just the shape made when you cut through. Volume = area of base x height 3 ight radius 2 radius ight Volume = πr2h V = πr2h Cones a cone is one third of a cylinder Spheres V = 4πr3 3 Frustrums a frustrum is a pyramid with the top cut off. The top part of the rectangle is 1" and so is the bottom. If either wear down enough to expose the red indicator, you know it's time to buy a new harness. Diameter cuts the circle into two equal parts. 18400 + π × 20 2 = 18400 + 1256. A s = s 2 (2) where. 56 + 8 = 20. In this solving problems worksheet, students determine the area of a rectangle inscribed in a circle. Area of a rectangle formula. The area of the rectangle is 16 x 6 = 96 cm 2. Once you have all your lines made, cut it out with the Band Saw. 5 cm squared. Now we all know that the area of a rectangle is its length multiplied by its height. a) Create a rectangle by cutting off the right-angled triangle and moving it. Round Corner Calculator. This is the equivalent of adding all four sides, since opposite sides are of equal length by definition. The belay loop and upper tie-in point have red indicator strips underneath the outer nylon material. !Work out the area of the semi-circle. If you have only the width and the height, then you can easily find all four sides (two sides are each equal to the height and the other two sides are equal to the width). Make sure the end of the ruler at the corner point doesn't shift position. China Marble Medallion catalog of Custom Made Medallion Square Marble Floor Medallions for Entrance/Lobby Floor Decoration, Classic Flora Style Circle Marble Medallion/Inlay/Waterjet Floor Pattern for Foyer Floor Decor provided by China manufacturer - Sparkle Design & Decor Co. If you're not sure which sides to label, pick any corner. =1/4$πr^2 - 1/2$ x base x height (Here base = height = radius =. Now the length is increased by 10% and its breadth is decreased by 20% and the resulting area is 484 m 2. Photoshop fills the area with a checkerboard pattern, which is how Photoshop represents transparency:. 6803 ($$h$$ x 2$$r$$). please help Answer by Paul(988) (Show Source):. We multiply 3 cm by 5 cm. 15-04-2018. The semi-circle fits half the length of the square perfectly. A semicircle is simply half of a circle 🌗. com Area and Perimeter of Circle and Semi-Circle Perimeter. Cross sections are usually parallel to the base like above. Tap the circle cutout. Watch this tutorial to see how it's done!. Semi Circle Shape. 14159… ø = Circle diameter; Diameter of Circle. A square unit area is a square having sides of one unit which can either be centimetres or metres. Consider one of the rectangle's sides to be of length s. How to calculate the area of a square or rectangle. 99) Find great deals on the latest styles of Half circle area rugs. Calculations at a curved rectangle, a flat, four-sided shape with directly opposite, parallel and congruent sides, with circular arcs and straight lines als side pairs. Area of a triangle calculation using all different rules, side and height, SSS, ASA, SAS, SSA, etc. Use that to check your integration. center point: the area of part of a circle is _____. The formula is Area of ABC = 1/2 base x height where the base is the length of the bottom of the triangle (the length of the rectangle), and the height is the length of BD (the height of the rectangle). Visit https://maisonetmath. We can see from the diagram that we can fit 15 one-square-inch squares into the rectangle--thus, the rectangle has an area of 15 square inches. Partial Circle Area Calculator. The diameter of each circle is 4 cm. A semicircle is cut out of a rectangular paperboard 22in long and 17in If the semi- circle was taken from the width the perimeter would be 51. 57 cm Top of a table has a radius of 50 cm, work out the area. Another one is "diamond. Area of square = 4x4 =16 square cm Area of semi-circle = ½ x3. Much like the top of a water bottle. One is a square and the other is a rectangle 18 cm longer than the square but 9 cm less wide than the square. Solution:Given; Length of the garden = 90m, Width of the garden = 75m, Width of the path around the garden = 5m. 1415 \times 225 = 706. How to calculate the area of a square or rectangle. docx from MATH 15 at Delhi Public School, R. We have an isosceles triangle with vertices A (center of one of the circles), E, and D (points of intersection of the circles). The area of a rectangle is found by multiplying the base times the height of the rectangle. Any diameter of a circle cuts it into two equal semicircles. Dunne rugs are very luxurious and are around 2cm in thickness, each rug is hard-wearing and easy to clean. Area of semicircle = (1/2) Are of circle = (1/2)(πd²/4) = πd²/8. How to find area of shaded region involving polygons and circles, Find the Area of a Circle With Omitted Inscribed Triangle, Find the area of a shaded region between and inscribed circle and a square, Find the area of a shaded region between a square inscribed in a circle, examples with step by step solutions, How to Find the Area of a Rectangle within Another Rectangle, Grade 7. The Dunne Hand-Tufted Wool Blue Rug from 100% pure wool then hand-carved and finished with a cotton backing. The task is to calculate the area of the crescents. Pieces 1 through 5 are for the same fraction of a period, 1/10 th period. Let be the radius of the semicircle, one half of the base of the rectangle, and the height of the rectangle. Volume of Half Cylinder Calculator. c) Creating an Oriented Rectangle. π (pronounced "pie" and often written "Pi") is an infinite decimal with a common approximation of 3. A = Circle area; π = Pi = 3. 1: Still Irrigating the Field (5 minutes) Here is a picture that shows one side of a child's wooden block with a semicircle cut out at the bottom. All you need are two measurements and you can calculate its perimeter by hand, or by using our perimeter of a rectangle calculator above. Area of a Rectangle Learning Intention Success Criteria • To be able to state area formula for a rectangle. In this python program, we will find area of a circle using radius. Else create a dimension. If i had a rectangle of 40 meters x 20. In geometry, the area enclosed by a circle of radius r is π r 2. I took those pieces down and transferred the to my pattern paper. Now we all know that the area of a rectangle is its length multiplied by its height. The task is to calculate the area of the crescents. Switching the input values above changes the layout and gives. !Work out the area of the semi-circle. Semicircle Calculator. If we know the radius then we can calculate the area of a circle using formula: A=πr² (Here A is the area of the circle and r is radius). This is for circle. What you have is a rectangle and a circle (It's been cut down the middle and the rectangle shoved in it). A training field is formed by joining a rectangle and two semicircles, as shown below. Cut a 3” line vertically though the black lining fabric, this will be hidden later. Multiply the radius of the semicircle by itself. What is the distance travelled by the centre of the circle, when the circle has travelled once around the rectangle? 14. The Rhino Hide double stuff mat is 7/8" thickness producing incredibly soft & resilient rebound properties needed by salon professionals. The rectangle has dimensions of 16cm by 6cm. Hence the diameter of a circle with area equal to 40 square cm is found to be 7. Also find the area of the garden in hectare. Rd_sharma Solutions for Class 10 Math Chapter 15 Areas Related To Circles are provided here with simple step-by-step explanations. A cross section is the shape we get when cutting straight The cross section of this circular cylinder is a circle. Answer: Total area of large semi. So the area of a semicircle is when you cut a full circle into half you get the area of a semicircle. Find the area of the shaded region. Rectangle And Semi-Circle - Duration:. Angles Games. The area of. 1165 1225 112. You can also design in the four standard shapes for roll labels. 28 square cm. The rectangle is 95m long and 74m wide. the diagram shows a square with a semi-circle on its left and on its bottom. Read customer reviews and common Questions and Answers for Orren Ellis Part #: W001872817 on this page. A square will be cut from each corner of the cardboard and the sides will be turned up to form the box. And now all you have to do is subtract area of the semi-circle from the area of the rectangle. If two identical circles are cut out of the wood and the area of EACH circle is x2 – 2, find the area of the remaining piece of wood. I only want the sides of the circle. Hence the shaded area = Area of the square - The area of the circle = 144 - 113. Tie the ribbon to zipper pull. Perimeter, Area and Volume Short Problems This is part of our collection of Short Problems. Record the length (l) and width (w) of a square or rectangle. The rectangle has length 20 cm and width 12 cm. Exam Style Questions Ensure you have: Pencil, pen, ruler, protractor, pair of compasses and eraser 3. Tes Global Ltd is registered in England (Company No 02017289) with its registered office at 26 Red Lion Square London WC1R 4HQ. Practice: Shaded areas. Cut off the spent blooms to keep clear for foot traffic. How could you work out the AREA without counting every square?. To find the area of a rectangle, use the formula. Determine the radius Measure the distance from the center of the circle of which the semicircle is part to its edge. crop a circle in the image, is an online tool, used to crop round circle in your images. Use formula to work out area of different size circles, semi-circles and quarter circles. Measure the hole, find the area. Find the area of the paper that remains. 6803 ($$h$$ x 2$$r$$). 1 square cm. Possible responses of the rectangle and the half-circle but. Rectangle in Semicircle: Find the area of the largest rectangle that can be drawn inside of a semicircle with radius 5 cm. π is roughly equal to 3. Area of a rectangle=(length*breadth) Therefore length = (area/breadth) and breadth=(area/length) 2. how many pipes or wires fits in a larger pipe or conduit Smaller Rectangles within a Large Rectangle - The maximum number of smaller rectangles - or squares - within a larger rectangle (or square). A semicircle is simply half of a circle 🌗. Find the area of the training field. Rectangles have special properties that can be very useful in helping you solve a problem. The formula used to calculate circle area is: A = π x (ø/ 2) 2. Rotate one of them and overlay the two sheets perpendicular to one another. Position the cursor in the desired field (Sketch tools toolbar) and key in the desired values to create points & then lines for rectangle. A sector is a fraction of a circle and therefore its area is a fraction of the area of the whole circle. com Area and Perimeter of Circle and Semi-Circle Perimeter. 1) A circle has a radius of 5 cm. Area of rectangle = 4 × 8 = 32 m² A circle of radius 8 cm is cut into 6 parts of equal size, as shown in the diagram. A button is made in the shape of a circle, with two congruent rectangles cut out, as shown in the diagram. The area of a partial circle is the area of the circular segment equal to the area of the circular sector minus the area of the triangular portion formed from the. So cropping is quick, highly secured and consumes less bandwidth. The dimensions are m = 80. To find the perimeter of a rectangle, add the lengths of the rectangle's four sides. Cut a rectangle (more or less about 8 X 16 inches) of your 'swimsuit' stretch fabric. Tie the ribbon to zipper pull. The perimeter of a plane 2-D figure is the length of the outer periphery of the figure. **Many of our plants are listed as Not a Good Fit for Kid Traffic/Dog Traffic/Around Pools. Geometry calculator solving for arc length given central angle and circle radius. – Project-Based Learning. But we could get a better result if we divided the circle into 25 sectors (23 with an angle of 15° and 2 with an angle of 7. Semi-circular arcs are drawn with the sides of the rectangle as diameters. 99) Find great deals on the latest styles of Half circle area rugs. To find the total area, find the area of each part and add them together. The figure can be any regular or irregular polygon; it does not have to be a regular geometric figure. Angles Games. 68 cm 2 Example #2: Find the surface area of a cylinder with a radius of 4 cm, and a height of 3 cm. A filled rounded rectangle with b=0 (or a=0) is called a stadium. 14 for pie , and do not round your answer. cropping is much Faster, since we are not uploading your images to our server. The rectangle has a dimensions 5 length and 4 width, the semicircle has a radius of 2 using 3. Hence the diameter of a circle with area equal to 40 square cm is found to be 7. The round cut-out in the front is also a perfect semi-circle. Find the area of the remaining card board. The area of. This shape is designed using 3 semi. the volume of both similar shapes in frustrum problems. b h Parallelogram 2. Determine the diameter of the circle you intend to make. Therefore, a r e a o f c i r c l e = πr 2 = 22 7 × 49 = 154 cm 2. The diameter should be measured in feet (ft) for square footage calculations and if needed, converted to inches (in), yards (yd), centimetres (cm), millimetres (mm) and. Smaller Circles within a Larger Circle - Estimate the number of small circles that fits into an outer larger circle - ex. Then, multiply the base by the height of the rectangle to get the area. You may not know the area of a semicircle, but that's fine - we know the area of a circle, or pir^2. 28 × 2 SA = 25. For the clipping, every path inside the clipPath is inspected and evaluated together with its stroke properties and transformation. 18400 + π × 20 2 = 18400 + 1256. The hole can be cut with a 6-inch hole saw and an electric drill but be very careful, a hole saw of this size is prone to grab in the wood and twist the drill out of your hands. 14159 r^2 is the square of the radius of the circle. 5 cm squared) from the area of the square (100 cm squared) to determine the area outside the circle, but still within the square. Area of a triangle calculation using all different rules, side and height, SSS, ASA, SAS, SSA, etc. 14 for pie , and do not round your answer. Define a semi-rectangle to be a quadrilateral with opposite sides equal, and at least one right angle. how many pipes or wires fits in a larger pipe or conduit Smaller Rectangles within a Large Rectangle - The maximum number of smaller rectangles - or squares - within a larger rectangle (or square). Designing a patio can be quite exciting when you realize how much freedom you have. In this image, the largest yellow circle is the overall diameter of the hole ground by the rock abrasion tool and the largest yellow rectangular shape is the area of the grinding wheel bit. Brett and Sara Wilson grew up as close friends in the same local church and area high school in rural, southern Ohio. 14 times the radius squared (πr 2). The equation for the area, A, of a square or rectangle area can be written as: A = b * h. also a red ribbon is fixed around the sides of the heart shaped cake with the ends overlapping by 5cm. Visual on the figure below: π is, of course, the famous mathematical constant, equal to about 3. The area of a rectangle can also be calculated using the following. Circumference of a semi-circle = = πr and the perimeter of a semi-circular shape = (π + 2) r units. How to calculate the area of a square or rectangle. π (pronounced "pie" and often written "Pi") is an infinite decimal with a common approximation of 3. Find out why Close. write an expression for the area of the entire. A 3x8 rectangle is cut into two pieces then rearranged to form a right-angled triangle. Iron out any creases or wrinkles. Make sure the end of the ruler at the corner point doesn't shift position. the face of the house, I would find the area of the large rectangle … then I would subtract out the two small upstairs windows and the door and the larger downstairs window. Knowing their names can be very helpful when you actually get to the point of purchase. A = Circle area; π = Pi = 3. Radius and diameter refer to the original circle, which was bisected through its center. Here is a List of Best Free Geometry Calculator Software for Windows that feature a number of tools to solve geometry problems. Each part is called semi circle. A rectangle is one of the many fundamental shapes you'll see in math. Round your answer to the nearest hundredth. A square has sides of length 12 cm. To make the pies, roll the pastry on a floured surface into a large rectangle, about 12 inches by 24-plus inches. Learn how to calculate perimeter and area for various shapes. The distance around the outside of a semicircle is generally called its perimeter. It is evident that if we make a large number of cuts, the figure formed will approximate a rectangle whose length is equal to one-half of the circumference and whose width is equal to the radius. Fold the entire piece of fabric in half to create a large square. Area of a circle intuition. Requires knowledge of Conic Sections. make up Remember, by o shape. Creating a design with pavers requires planning. Repeat the same method the other side. where r is the radius of the circle. The word ‘area’ stands for the space occupied by a flat object or figure. To get the area of the triangle portion, I subtract the area of the sector of the circle with the area of the triangle of that portion multiplied by 2 (since there are essentially two triangles). Draw and cut out the wing shape from the flap and insert into the slit. 1165 1225 112. Calculations at a semicircle. 28 square cm. asked by ian on August 29, 2016. How to Find the Area of a Semicircle To find the area of a semi-circle, you need to know the formula for the area of a circle. Semi-circles are drawn on and as diameters. Determine the diameter of the circle you intend to make. 4 in L, Base 6in - 4694841. in Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969 Page 251 16. Practice: Area of a circle. In the article below, we provide the semicircle definition and explain how to find the perimeter and area of a semicircle. What is the objective function in terms of the base of the rectangle, x? (Type an expression. The Gimp provides the Rectangle, Ellipse and Lasso selection tools to help you cut out specific parts of a photo or illustration to keep, eliminate or edit. Area of circle Learning Intention Success Criteria. Find out why Close. This is for circle. Draw a half circle contur with very thick border and convert it to path if needed. These mats are made to order and are non-returnable and non-refundable. There are two ways to draw filled shapes: scatter traces and layout. The diagram shows a metal plate in the shape of a rectangle. Download PDF. The formula for finding the area of a full circle is πr 2, where "r" represents the radius of the circle. Use formula to work out area of different size circles, semi-circles and quarter circles. The GIMP open source graphics program includes a number of editing tools, including a cropping tool and selection tool. Area of a sector. With the Eraser tool (), erase the top half of the second circle and the face that represents the inside of the. (847) 223-7000 [email protected] ws Shapes Preschool Activities and Crafts. Rectangle glass table tops are ideal for wooden furniture. Apart from these, some software also feature options to calculate diagonal, height, length, breadth, etc. Consider one of the rectangle's sides to be of length s. This free circle calculator computes the values of typical circle parameters such as radius, diameter, circumference, and area, using various common units of measurement. and for each shape, cut the net out, fold it to see how it makes the 3D shape, but do NOT glue it together. We now have a semicircle without having to deal with arcs in path elements. – Project-Based Learning. Instead of two circles make a single, but thick circle shape. Use the edge of the top sheet as a guide to cut off the protruding section from the lower sheet. Hwy 83 Grayslake, IL 60030. And the more we divided the circle up, the closer we get to being exactly right. The area of the figure will be the sum. what percent of area is lost as trimming Percent of CIRCLE lost due to trimming: % You can do the check!! ===== If you need a complete and detailed solution, let me know!! Send comments, "thank-yous," and inquiries to "D" at [email protected] Hence the diameter of a circle with area equal to 40 square cm is found to be 7. Enter length and width of the rectangle, as well as the radius of the circle that makes the corners. Example: What is the area of this triangle? Height = h = 12. Learn more about pi, or explore hundreds of other calculators addressing finance, math, fitness, health, and more. Express the formulas for the area and perimeter of a square using s for the length of a side. This is the currently selected item. The shaded part has an area of $$\frac13 \pi r^2$$. calculate the perimeter of the shaded shape correct to 1 decimal place. circles circles. The dimensions are m = 80. Therefore. The equation for the area, A, of a square or rectangle area can be written as: A = b * h. 227 centimeters squared. See formulas for each shape below. Figure 1: Semi-circle is essentially half a circle. Because the ring is hollow, all of its mass has to sit at a distance R from the center; hence, you have =R 2 and I = MR 2. The area of a rectangle can also be calculated using the following. What is the distance travelled by the centre of the circle, when the circle has travelled once around the rectangle? 14. To work out what its area is, we just need to find out what fraction of the circle the sector is occupying. A filled rounded rectangle with b=0 (or a=0) is called a stadium. Python Program to find Area Of Circle using Radius. The quantity we need to maximize is the area of the rectangle which is given by. Instead of two circles make a single, but thick circle shape. Easy to clean, these mats are made in the USA and include a 5 year manufacturer's warranty. We multiply the length, 3 meters, times the width, so times 3 meters, to get 3 times 3 is 9-- 9 square meters. Draw a temporary rectangle that covers half of the circle, then with the top circle and the rectangle selected, click the Minus Front button from the Pathfinder panel to trim it into a semi-circle. Shop Scrapbook. Work out the shaded area. Square & Rectangle Cut Out At Craftparts. It needs to be large enough that it can be folded in half and still have room for your pattern to fit on it. Area of a sector. a rectangle of largest area is cut out from a circle. Cut along the two different dotted lines. Area of a Semicircle Calculator A semicircle is nothing but half of the circle. Use the shapes for crafts, math, and shapes-themed learning activities. Talking about the area of a semicircle. Jan 06, 2016 · You can achieve a transparent cut out circle with 2 different techniques : 1. Show all working and. A path 5 m wide is to be built outside and around it. View Answer. The area of the figure will be the sum. to that add the 2 sides of the rectangle (15+15). Area of Semi-Circle = 1 ⁄ 2 * π *r 2. Figure 1: Semi-circle is essentially half a circle. Technically, a semi circle always has a degrees of 180, hence the term semi, which means half of a circle. So all you need to do now is divide the answer by 4: Area. This is typically written as C = πd. The rectangular figure of greatest area within a circle is a square. Calculate the intersection area of two circles July 14th, 2016. The area of the circle = sum of the areas of the pie-shaped sections. of a semicircle we calculate the area of the whole circle and then half the answer. NOT TO SCALE. Step 3: Add the diameter. 53 in 2; Area of Object = 44. length How many squares are in one row? 6 So how many would be in two rows? 6 + 6 = 12. A teachıng experınce wıth a blınd student ’ s fıngers : the area of a cırcle. The circle is made with 2 arc commands: With the mask element :. So we can say that area of semi circle will be equals to the half of area of circle. Ans : [Board Term-2, 2012, Set (12)] Area of sector OAPB 36. How could you work out the AREA without counting every square?. Common sense would suggest that the optimal rectangle would be a square, but here is a rigorous proof. Cross Sections. An arc is part of the circumference of a circle. For example, if you are cutting out an image of a person with curly hair, the Pen tool is the best thing to use. The dimensions are m = 80. Step - 7: System. And, the bottom two points of the rectangle touch the diameter of the semicircle. Learn how to find the area of circle by activity method. Featured here are exercises offered as geometrical shapes and in word format in three distinct levels; grouped based on the shapes used like triangles, squares, rectangles, parallelograms in level 1, trapezoids and circles are added in level 2, kites and rhombus are included in level 3. Determine the area of the rectangle below. This becomes 100 cm 2 - 78. | 2020-11-30T07:48:03 | {
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https://www.physicsforums.com/threads/minimum-distance-of-a-point-on-a-line-from-the-origin.603909/ | # Minimum distance of a point on a line from the origin?
1. May 7, 2012
### hivesaeed4
Consider the graph of $${x^2 + 2 y^2 = 1}$$.
What is the minimum distance of a point on the graph to the origin?
When I calculated the point which has minimum distance from origin it came out to be (1,0) implying the min distance to be 1. The answer is 1/2. Help?
2. May 7, 2012
### JJacquelin
Hi !
Show your calcul in whole details. So, it will be possible to locate the mistake.
3. May 7, 2012
### hivesaeed4
Okay.
We're given x^2+2*y^2=1.
so x^2=1-2y^2
now using distance formula
d^2=x^2+y^2
since x^2=1-2y^2, substituting it in the distance formula we get:
d^2=1-2y^2+y^2=1-y^2;
differentiating and then setting the eq to 0 we get;
0=-4y
or y=0. now x^2=1-2y^2=1
so x=+-1
so point having min distance form origin is (+-1,0)
using the distance formula now
d^2=x^2+y^2
d=sqrt(1+0)=1
4. May 7, 2012
### JJacquelin
Your computation is correct, but there is a snag :
If the derivative of a function is 0, then the function is minimum or maximum.
The function 1-y² is decreassing, so for y=1 we have the maximum of d², not the minimum.
(1-y²) is the smallest for the heighest value of y². Since x²+2y²=1, the highest value is y²=1/2. Finally, the smallest d=sqrt(1/2) is obtained at (x=0, y=sqrt(1/2))
Less confusing, d² = x²+y² = x²+(1-x²)/2 = (1/2)+(x²/2)
The function of x is increassing. The derivative leads to x=0 and the function is minimum at the point (x=0, y=sqrt(1/2)).
5. May 7, 2012
### hivesaeed4
Okay I get the second part but still don't get the first part.
Let's talk about the second part first. You defined x in terms of y and then solved. I defined y in terms of x. Since your answer is smaller then mine so it's logical that your correct. Now does htat mean that everytime I have to do each question twice i.e. first x interms of y and then y interms of x. And if both your and mine methods are correct should'nt both answers be same.
Now could you explain the first part in greater detail. I could'nt understand the following line:
The function 1-y² is decreassing, so for y=1 we have the maximum of d², not the minimum.
If y=1; d^2=1-(1)^2=0
So how is that the max of d^2?
6. May 8, 2012
### JJacquelin
Think of this : no point exists on the curve x²+2y²=1 with y=1.
So you cannot write : d²=1-(1)²=0. The distance between the origin and a non-existing point of the curve is a nonsense.
Just draw the curve and see what is the range of y for really existing points.
Last edited: May 8, 2012
7. May 8, 2012
### hivesaeed4
Ok I get it now. Thanks.
8. May 9, 2012
### mathwonk
that is obviously an ellipse, with top and bottom at (0,±1/sqrt(2)) and right hand endpoint at (±1,0).
Thus there are 4 critical points for the distance function, namely the two points furthest away and the two points nearest the origin.
9. May 9, 2012
### Hawkeye18
Hi hivesaeed4
Your calculations are correct, but you missed one detail.
When looking for max/min of a function on an interval, you need to compare the values at the critical points (i.e. the points where derivative is 0 or does not exist) and at the endpoints.
In your computations you forget to check endpoints $y=\pm 1/\sqrt2$, and each endpoint give you $d^2=1/2<1$. So the minimal distance is $1/\sqrt2$.
10. May 9, 2012
### hivesaeed4
Okay. But I've noticed in mathwonk's post that 4 critical points are taken into account. Does that mean the error I made was checking only two endpoints, instead of all four for the min/max values of the function?
Does it also mean that for a given function whose min and/or max values are to be calculated, we should first find it's possible critical points and then use distance formula for each critical point?
11. May 9, 2012
### mathwonk
heres another point of view. the function x^2 + 2y^2 has gradient (2x, 4y). Look for points of the ellipse where the radius vector is parallel to this gradient.
I.e. look for solutions (x,y) of x^2 + 2y^2, where (x,y) is parallel to (2x,4y). But these are never parallel unless either x or y = 0. there are 4 such points. | 2017-10-18T08:23:31 | {
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https://linnaxu.com/rcfnlg/integration-of-acceleration-d00aff | integration of acceleration
That is, $\ds \int_{x_0}^xf(x)dx$ is bad notation, and can lead to errors and confusion.) and the acceleration is given by $a\left( t \right) = \frac{{dv}}{{dt}} = \frac{{{d^2}x}}{{d{t^2}}}.$ Using the integral calculus, we can calculate the velocity function from the acceleration function, and the position function from the velocity function. It is a vector quantity (having both magnitude and direction). Homework Equations Is this because of the constant of Integration? $\begingroup$ Actually I should not use those definitions because that is average acceleration with respect to displacement. In this video we will look at how to convert between displacement, velocity and acceleration using integration acceleration using the integral x(t).vi the waveform is offset. In physics, jerk or jolt is the rate at which an object's acceleration changes with respect to time. This negative answer tells you that the yo-yo is, on average, going down 3 inches per second.. What is the general form of acceleration with respect to position? a at time t. In this limit, the area under the a-t curve is the integral of a (which is in eral a function of t) from tl to t2• If VI is the velocity of the body at time tl and v2 is velocity at time t2, then The change in velocity V is the integral of acceleration a with respect to time. The acceleration function is linear in time so the integration involves simple polynomials. Homework Statement Acceleration is defined as the second derivative of position with respect to time: a = d 2 x/dt 2.Integrate this equation with respect to time to show that position can be expressed as x(t) = 0.5at 2 +v 0 t+x 0, where v 0 and x 0 are the initial position and velocity (i.e., the position and velocity at t=0). Maximum and minimum velocity of the yo-yo during the interval from 0 to 4 seconds are determined with the derivative of V(t): Set the derivative of V(t) — that’s A(t) — equal to zero and solve:. From Calculus I we know that given the position function of an object that the velocity of the object is the first derivative of the position function and the acceleration of the object is the second derivative of the position function. In general, it is not a good idea to use the same letter as a variable of integration and as a limit of integration. I want a more accurate function of acceleration with respect to displacement. Think about it. Acceleration tells you the rate of change or “slope” of velocity. Jerk is most commonly denoted by the symbol and expressed in m/s 3 or standard gravities per second (g/s). I require integration regardless of the phase, that gives me a positive negative velocity and displacement signal and I see that the svt integration.vi is used , according to your The double integration of acceleration gives the position of an object. Now, evaluate V(t) at the critical number, 2, and at the interval’s endpoints, 0 and 4: Direct double integration of acceleration as a single integration. $\endgroup$ – obliv Mar 9 '16 at 22:02 Section 6-11 : Velocity and Acceleration. The Fundamental Theorem of Calculus says that Formally, double integration of acceleration, a(t), to obtain displacement, s(t), can be written as (8) s(T)=∫ 0 T ∫ 0 t a(t ′) d t ′ d t where is assumed that the accelerometer is initially at rest with zero displacement. In this section we need to take a look at the velocity and acceleration of a moving object. In , we see that if we extend the solution beyond the point when the velocity is zero, the velocity becomes negative and the boat reverses direction. 4.3. If you have velocity then you take the time derivative you'll get acceleration. It is a vector quantity ( having both magnitude and direction ) what is the of... 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Integration involves simple polynomials as a single integration or “ slope ” velocity! Quantity ( having both magnitude and direction ) a look at the velocity and acceleration of moving. Is average acceleration with respect to position acceleration with respect to time is a vector quantity ( having magnitude! | 2022-08-15T00:30:37 | {
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https://math.stackexchange.com/questions/656918/prove-rm-tr-aat-rm-tr-ata-for-any-matrix-a/656972 | # Prove ${\rm tr}\ (AA^T)={\rm tr}\ (A^TA)$ for any Matrix $A$
Prove ${\rm tr}\ (AA^T)={\rm tr}\ (A^TA)$ for any Matrix $A$
I know that each are always well defined and I have proved that, but I am struggling to write up a solid proof to equate them. I know they're equal.
I tried to show that the main diagonal elements were the same but if I say that $A$ is $n\times m$ then $$(AA^T)_{ii} = (a_{11}^2+\dots +a_{1m}^2) + \dots + (a_{n1}^2+\dots +a_{nm}^2)$$ and $$(A^TA)_{ii} = (a_{11}^2+\dots +a_{n1}^2) + \dots + (a_{1m}^2+\dots +a_{nm}^2)$$
• please write up the pieces of your proof and some one will make it solid... good luck! – user87543 Jan 30 '14 at 6:35
• rearrange terms. Both are the same expression. – voldemort Jan 30 '14 at 7:03
• @b00n, re-read your formula. – Martín-Blas Pérez Pinilla Jan 30 '14 at 8:52
• Thank you @Martín-BlasPérezPinilla! How naive of mine – b00n heT Jan 30 '14 at 8:55
• Well, was true! :-) – Martín-Blas Pérez Pinilla Jan 30 '14 at 8:57
By $(AA^T)_{ii}$ you seem to mean the $i$-th term on the diagonal of $AA^T$. But instead what you've writen is already $\mathrm{tr}(AA^T)$. Which is ok. Now, you just have to realize that both sums are the same, up to the order of the addends -which doesn't matter.
For instance: you have $a_{11}^2$ on both sums, haven't you?
Also $a_{1m}^2$ appears on both sums. Also $a_{n1}^2$...
Write a few more terms on both sums. Or write all terms in the particular case $2\times 3$ and you'll see it.
• Thank you, I have working for hours on hours so I sort of derped out and over looked the obvious. – e2DAeyePi Jan 30 '14 at 7:10
• @e2DAeyePi To make it a little more concrete, try writing each expression as a double sum in $\Sigma$ notation. If you're consistent with how you form the sums, you should get the same for each expression except that the order of summation is swapped. – G. H. Faust Jan 30 '14 at 7:18
To give you an idea of how to properly write these sort of proofs down, here's the proof.
For a matrix $X$, let $[X]_{ij}$ denote the $(i,j)$ entry of $X$. Let $A$ be $m\times n$ and $B$ be $n\times m$. Then \begin{align*} \mathrm{tr}\,(AB) &= \sum_{i=1}^n[AB]_{ii} \\ &= \sum_{i=1}^n\sum_{k=1}^m[A]_{ik}\cdot[B]_{ki} \\ &= \sum_{k=1}^m\sum_{i=1}^n[B]_{ki}\cdot[A]_{ik} \\ &= \sum_{k=1}^m[BA]_{kk} \\ &= \mathrm{tr}\,(BA) \end{align*} Your question is a special version of this result with $B=A^\top$.
All approaches to the question made so far are pretty good. I just want to add a small observation that will make the proof solid.
trace($AA^T$) or trace($A^TA$) is sum of squares of all elements of the matrix $A$ (which is the same as the sum of squares of all elements of the matrix $A^T$). | 2019-10-22T14:15:49 | {
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https://math.stackexchange.com/questions/1687038/three-digit-numbers-whose-digits-and-digit-sum-are-all-prime | # Three-digit numbers whose digits and digit sum are all prime
How many 3-digit numbers are there such that each of the digits is prime, and the sum of the digits is prime?
Shouldn't it be $0$, because the only one digit primes are $2,3,5,7$, and so the possible combinations of those numbers are (not particularly in primes) $235, 237, 257, 357$? And not one single group's digits add up to any prime number. But then why'd $0$ be a wrong answer?
• What about $335$? – 5xum Mar 7 '16 at 14:26
• so... I came up with the answer 7, (223, 227, 353, 557, 757, 337, 773) did anyone else come up with the same answer? Just want to check if I got this right. – jjhh Mar 7 '16 at 14:57
• I came up with more than $7$. For example, $223$ is missing from your list. – 5xum Mar 7 '16 at 15:04
• um, no, because 223 is the first number in the list (in brackets) – jjhh Mar 9 '16 at 8:30
• Sorry, I meant $227$. That one is missing from your list – 5xum Mar 9 '16 at 8:41
Here's a small hint: $335$ is one such number.
So $0$ is a wrong answer indeed.
It is not said that you cannot repeat digits.
• ooops... read it over so many times but never realised that... thank you – jjhh Mar 7 '16 at 14:41
The smallest sum is $2+2+2=6$.
The largest sum is $7+7+7=21$.
So the only possible prime sums are $7,11,13,17,19$:
• A sum of $7$ can be generated from the $3$ permutations of $223$
• A sum of $11$ can be generated from the $6$ permutations of $227$ and $353$
• A sum of $13$ can be generated from the $6$ permutations of $337$ and $355$
• A sum of $17$ can be generated from the $6$ permutations of $377$ and $557$
• A sum of $19$ can be generated from the $3$ permutations of $577$
Hence there are $3+6+6+6+3=24$ such numbers.
A short Python script in order to confirm the above:
count = 0
for a in [2,3,5,7]:
for b in [2,3,5,7]:
for c in [2,3,5,7]:
if a+b+c in [7,11,13,17,19]:
count += 1
print count
No, one such number is $223$. Now, there are $4^3$ $3$-digit numbers with digits $2, 3, 5, 7$, but since the conditions are independent of the order of the digits, it's enough to check just the $20$ such numbers whose digits are nondecreasing. Moreover, the sum of the three digits of any number will be $> 2$ and so if it is prime, it will be odd, and in particular must have an even number of $2$'s, so we need only check $12$ numbers: $223, 225, 227, 333, 335, \ldots$.
For example, $2 + 2 + 3 = 7$ is prime, so the $3$-digit numbers $223, 232, 322$ all have the desired property. | 2019-06-26T17:48:23 | {
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https://www.physicsforums.com/threads/jacobian-in-spherical-coordinates.706930/ | # Jacobian in spherical coordinates?
1. Aug 24, 2013
### Uan
Hi,
Started to learn about Jacobians recently and found something I do not understand.
Say there is a vector field F(r, phi, theta), and I want to find the flux across the surface of a sphere. eg:
∫∫F⋅dA
Do I need to use the Jacobian if the function is already in spherical coordinates?
My notes has an similar example that show you do use the Jacobian but I do not understand why. My understanding is you only use the Jacobian when there is a change in coordinates, but the function F is already in the desired coordinate system.
Thanks!
Uan
Last edited: Aug 24, 2013
2. Aug 24, 2013
### tiny-tim
Hi Uan! Welcome to PF!
Suppose you want to find a volume (or area) by integrating, and everything is already in spherical coordinates …
you still need to use the jacobian (instead of just drdθdφ) because volume (or area) is defined in terms of cartesian (x,y,z) coordinates, so you have made a transformation!
Similarly, flux is defined in terms of cartesian coordinates.
3. Aug 24, 2013
### Uan
Ok that makes sense.
One other question...
How do they get the side lengths r*d(theta) and r*sin(theta)*d(phi) of element dA in the diagram below?
4. Aug 24, 2013
### tiny-tim
Hi Uan!
rdθ (the length of the side of A) is length of the side of that triangle with two sides r and angle dθ …
so it's 2rsin(dθ/2), = 2r(dθ/2), = rdθ
(alternatively, if you're happy using arc-length instead of "straight" length, then the arc-length is rdθ by definition)
and the other one is calculated the same way, except that the side of the triangle is rsinθ instead of r
5. Aug 24, 2013
### Uan
Ohh yeaahhh!! Small angle approximation duh! Thanks tiny-tim! I really appreciate your help!
By the way, I derived it as rsin(dθ), = rdθ, as when the angle dθ goes infinitely small in the triangle with sidelengths r, r and dθ (I've redrawn it), its like it has 2 right angles (180° ~= 90° + 90° + ~0°) so hence rsin(dθ), = rdθ
[Broken]
My way is correct yes?
How did you get your 2s in 2rsin(dθ/2)?
(Time for F1 break)
Last edited by a moderator: May 6, 2017
6. Aug 24, 2013
### tiny-tim
(what's "F1 break"?)
i used a triangle with two equal sides, and split it in two to make two right-angled triangles, so that my result of 2rsin(dθ/2) was precise
7. Aug 24, 2013
### Uan
Ah yes! That works out beautifully! Cheers! :thumbs:
(Formula 1 Qualifying at Spa! )
8. Aug 27, 2013
### vanhees71
Of course, for the spherical shell it's pretty easy to get the surface vector element $\mathrm{d}^2 \vec{A}$ in this geometrical way, but it can become difficult for more complicated shapes. Thus, here the general way to get it.
The surface-element vector is defined as a vector perpendicular to the surface and of the length of the area element. If you have given the surface $S$ in parameter form
$$S: \quad \vec{r}=\vec{r}(u,v),$$
where $u$ and $v$ are real parameters, then the surface-element vector is given by
$$\mathrm{d} \vec{A}=\mathrm{d} u \mathrm{d} v \frac{\partial \vec{r}}{\partial u} \times \frac{\partial \vec{r}}{\partial v}.$$
That's so, because obviously this vector is perpendicular to the two (by assumption linearly independent) tangent vectors $\partial_u \vec{r}$ and $\partial_v \vec{r}$ and thus to all tangent vectors of the surface in the point under consideration, and the cross product has the magnitude corresponding to the infinitesimal parallelogram spanned by the vectors $\mathrm{d} u \partial_u \vec{r}$ and $\mathrm{d} v \partial_v \vec{r}$. You only must be careful concerning the orientation of the surface element, beause obviously it switches sign when you change the order of the parameters, because the cross product in skew symmetric.
For the sphere you take
$$\vec{r}=R \begin{pmatrix} \cos \varphi \sin \vartheta \\ \sin \varphi \sin \vartheta \\ \cos \vartheta \end{pmatrix}$$
with $u=\vartheta$ and $v=\varphi$. This gives
$$\mathrm{d} \vec{A} =\mathrm{d} \vartheta \mathrm{d} \varphi R^2 \begin{pmatrix} \cos \varphi \cos \vartheta \\ \sin \varphi \cos \vartheta \\ -\sin \vartheta \end{pmatrix} \times \begin{pmatrix} -\sin \varphi \sin \vartheta \\ -\cos \varphi \sin \vartheta \\ 0 \end{pmatrix}=R^2 \mathrm{d} \vartheta \mathrm{d} \varphi \begin{pmatrix} \cos \varphi \sin \vartheta \\ \sin \varphi \sin \vartheta \\ \cos \vartheta \end{pmatrix}$$.
For the full sphere the parameters run over the ranges $\vartheta=(0,\pi)$ and $\varphi \in [0,2 \pi)$. | 2018-02-20T00:58:45 | {
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https://forum.allaboutcircuits.com/threads/capacitor-discharge-time-problem.181597/ | # Capacitor discharge time problem.
#### quest271
Joined Apr 1, 2016
23
I am studying for CET certification. I am hung up on a practice problem envolving capacitor discharge time. Here is the question. A 0.09 microfarad capacitor is charged to 220 Volts. How long in milliseconds will it discharged to a level of 110 V if the discharged resistor has a resistance of 20, 000 ohms? The answer is 1.25 milliseconds. I can up with 9 milliseconds. This is what I did. t=rc 20k times 0.09 micro farads then I multiplied by 5 time constants.
#### Ian0
Joined Aug 7, 2020
3,517
Why five time constants?
You need to start with the equation for the discharge of a capacitor
$$V=V_0e^{\frac{-t}{RC}}$$
#### dl324
Joined Mar 30, 2015
13,320
Welcome to AAC!
Why did you choose 5 time constants? We use that number for calculating complete discharge time.
#### quest271
Joined Apr 1, 2016
23
I am studying for CET certification. I am hung up on a practice problem envolving capacitor discharge time. Here is the question. A 0.09 microfarad capacitor is charged to 220 Volts. How long in milliseconds will it discharged to a level of 110 V if the discharged resistor has a resistance of 20, 000 ohms? The answer is 1.25 milliseconds. I can up with 9 milliseconds. This is what I did. t=rc 20k times 0.09 micro farads then I multiplied by 5 time constants.
I assumed five time constants were used.
#### quest271
Joined Apr 1, 2016
23
I am studying for CET certification. I am hung up on a practice problem envolving capacitor discharge time. Here is the question. A 0.09 microfarad capacitor is charged to 220 Volts. How long in milliseconds will it discharged to a level of 110 V if the discharged resistor has a resistance of 20, 000 ohms? The answer is 1.25 milliseconds. I can up with 9 milliseconds. This is what I did. t=rc 20k times 0.09 micro farads then I multiplied by 5 time constants.
Do you have to solve that equation for the t in the numerator of -t/rc ?
#### dl324
Joined Mar 30, 2015
13,320
Do you have to solve that equation for the t in the numerator of -t/rc ?
That's the only way that makes sense to me.
#### Papabravo
Joined Feb 24, 2006
17,028
OK so the algebra looks like this
$$110\;=\;220e^{\left(-\cfrac{t}{RC}\right)}$$
$$ln(0.5)\;=\;\cfrac{-t}{RC}$$
$$-0.693\;=\;-\cfrac{t}{RC}$$
$$t\;=\;0.693RC$$
On my calculator the right answer drops out. That is approximately 70% of ONE time constant
Last edited:
#### quest271
Joined Apr 1, 2016
23
#### quest271
Joined Apr 1, 2016
23
OK so the algebra looks like this
$$110\;=\;220e^{\left(-\cfrac{t}{RC}\right)}$$
$$ln(0.5)\;=\;\cfrac{-t}{RC}$$
$$-0.693\;=\;-\cfrac{t}{RC}$$
$$t\;=\;0.693RC$$
On my calculator the right answer drops out. That is approximately 70% of ONE time constant
Thanks for your help. It makes sense now.
#### quest271
Joined Apr 1, 2016
23
#### quest271
Joined Apr 1, 2016
23
I am a little rusty with some of this. Thanks again.
#### quest271
Joined Apr 1, 2016
23
#### dl324
Joined Mar 30, 2015
13,320
I am a little rusty with some of this.
Since everyone is giving you equations...
You know that:
$$V_f=V_ie^{\frac{-t}{RC}}$$
Solving for t, you get:
$$t=-RCln(\frac{V_f}{V_i})$$
#### Papabravo
Joined Feb 24, 2006
17,028
Since everyone is giving you equations...
...
Nothing else matters
#### quest271
Joined Apr 1, 2016
23
Since everyone is giving you equations...
You know that:
$$V_f=V_ie^{\frac{-t}{RC}}$$
Solving for t, you get:
$$t=-RCln(\frac{V_f}{V_i})$$
I know how to solve for t . I was unsure if I was on the right path to getting the answer. Thanks again.
#### MrAl
Joined Jun 17, 2014
8,504
I know how to solve for t . I was unsure if I was on the right path to getting the answer. Thanks again.
The only thing you did wrong was assumed that it took 5 full time constants to reach the voltage they gave as the final voltage (110v). After 5 time constants the voltage would reach a level much lower than that and the usual way of describing that is that after 5 time constants the voltage reaches a level below 1 percent of the starting voltage. 1 percent of 220v is 2.20v and that is much lower than 110v so it 'cant be 5 time constants it must be less. If you calculate the correct time you can express that in terms of the time itself in milliseconds or you could express it in terms of the time constants. 1 time constant takes the voltage down to about 37 percent of the starting voltage and since that would still be less than 110v then that would mean the time must be even less than 1 full time constant.
The above kind of reasoning is what we sometimes call a "sanity check" because it gives us a secondary way to calculate the answer in rough terms and that tells us if our more accurate calculation is within reason.
#### Papabravo
Joined Feb 24, 2006
17,028
The only thing you did wrong was assumed that it took 5 full time constants to reach the voltage they gave as the final voltage (110v). After 5 time constants the voltage would reach a level much lower than that and the usual way of describing that is that after 5 time constants the voltage reaches a level below 1 percent of the starting voltage. 1 percent of 220v is 2.20v and that is much lower than 110v so it 'cant be 5 time constants it must be less. If you calculate the correct time you can express that in terms of the time itself in milliseconds or you could express it in terms of the time constants. 1 time constant takes the voltage down to about 37 percent of the starting voltage and since that would still be less than 110v then that would mean the time must be even less than 1 full time constant.
The above kind of reasoning is what we sometimes call a "sanity check" because it gives us a secondary way to calculate the answer in rough terms and that tells us if our more accurate calculation is within reason.
It is hard to know those things for a sanity check if you have never solved an exponential equation. Useful for those members of the cognoscenti who have, useless otherwise.
#### MrAl
Joined Jun 17, 2014
8,504
It is hard to know those things for a sanity check if you have never solved an exponential equation. Useful for those members of the cognoscenti who have, useless otherwise.
Not anymore | 2021-10-26T08:30:02 | {
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http://math.stackexchange.com/questions/252360/taylor-polynomial-of-arctan-of-given-degree-and-error | # Taylor Polynomial of $\arctan$ of given Degree and Error
Replace the following function by its taylor polynomial of the given grade, and approximate the error in the given interval:
$$f(x) = \arctan(x) \textrm{ by } T_3(f,x,0) \textrm{ in } |x| \le\frac{1}{10}$$
My solution and thoughts
We only need the first three derivatives and the fourth one for the error
$$f'(x) = \frac{1}{1+x^2} \\ f''(x) = \frac{2x}{(x^2+1)^2} \\ f'''(x) = \frac{6x^2-2}{(x^2+1)^3} \\ f^{(iv)}(x) = \frac{24x(x^2-1)}{(x^2+1)^4}$$
And by definition we know that
$$T_3(f,x,0) = \sum\limits_{k=0}^3 \frac{f^{(k)}(0)}{k!}x^k$$
we get
$$T_3(f,x,0) = x - \frac{x^3}{3}$$
We know that
$$R_3(x) = \frac{f^{(iv)}(c)}{4!}x^4, |x| \le \frac{1}{10}$$
by plugging in the maximal value of $x = c = \frac{1}{10}$ we get for any $c$:
$$\left|f^{(iv)}(c)\right| = \left|\frac{24c(c^2-1)}{(c^2+1)^4}\right| \le \\ \le \frac{24\cdot 99 \cdot 100^4}{1000\cdot 101^4}$$
$$\Rightarrow \left|R_3(x)\right| \le \frac{24\cdot 99 \cdot 100^4}{1000\cdot 101^4 \cdot 4!}\cdot\left|x^4\right| \Leftrightarrow \\ \Leftrightarrow \left|R_3(x)\right| \le \frac{10}{101^4}$$
Is this solution correct? How can I make it more formally right? (I know I lack some mathematical formalism).
-
What do you think, should I approximate the fourth differential by $\frac{24c^3}{c^8}$? – Flavius Dec 6 '12 at 16:01
This should be considered a minor addition to the answer by Mike Spivey.
Note that the fourth derivative is $0$ at $0$. So the Taylor polynomial that we get by expanding until the third derivative is exactly the same as the Taylor polynomial we get by expanding until the fourth derivative!
Thus you can use the formula for the remainder that involves the fifth derivative. There is not a lot of gain for the extra work, but there is some. The fifth derivative, I think, is $\dfrac{24(5x^4-10x^2+1)}{(x^2+1)^5}$.
The analysis leads to an upper bound on the error which is about one-fifth of the error upper bound we get by using the Lagrange estimate without noting that stopping at $3$ gives the same polynomial as stopping at $4$.
Remark: You may have worked too hard in doing the arithmetic. Let's do exactly as you did, using the fourth derivative instead of the better fifth. At a certain stage, you reached $$\left|f^{(iv)}(c)\right| = \left|\frac{24c(c^2-1)}{(c^2+1)^4}\right|$$ Note that $c^2+1\gt 1$ and $|c^2-1|\lt 1$. So our fourth derivative has absolute value $\lt 24\cdot \dfrac{1}{10}$. We have harly given away anything in making this simple estimate. Now divide by $24$, multiply by $\left(\dfrac{1}{10}\right)^4$. We find that the absolute value of the error is $\lt 10^{-5}$.
-
Your work looks good to me, except for a mistake in the last step (probably a typo). Other than that, it's pretty much exactly the way I would hope one of my students would solve this problem.
The mistake is that you lost the $99$ in the final step, so that at the end you should have $$\left|R_3(x)\right| \le \frac{99 \cdot 10}{101^4}.$$
(Also, the final implication is $\Rightarrow$, not $\Leftrightarrow$. Since you're substituting a value for $x$ here the two inequalities are not equivalent.)
As a side note, there's an easier way to calculate the Taylor series about $0$ for $f(x) = \arctan x$.
Since $$\frac{d}{dx} \arctan x = \frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 \pm \ldots,$$ via the geometric series formula when $|x| < 1$, $$\arctan x = C + x - \frac{1}{3} x^3 + \frac{1}{5}x^5 - \frac{1}{7} x^7 \pm \ldots.$$ Since $\arctan 0 = 0$, we have $C = 0$. This gives you $$T_3(f,x,0) = x - \frac{1}{3}x^3$$ rather painlessly.
You do still need to find the fourth derivative of $\arctan x$ to use the remainder formula for Taylor polynomials, however.
-
Thanks for your encouragement and the additional information, great help! – Flavius Dec 6 '12 at 17:19
@Flavius: I added one more minor comment you might want to look at. And you're welcome! – Mike Spivey Dec 6 '12 at 17:21 | 2015-10-07T15:50:10 | {
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https://math.stackexchange.com/questions/2940138/urn-probability-problem-containing-algebraic-variables | # Urn Probability Problem containing algebraic variables
Urn A contains $$x$$ red marbles and $$y$$ white marbles, and Urn B contains $$z$$ red marbles and $$v$$ white marbles.
If a marble is drawn from Urn A and put into Urn B and then a marble is drawn from Urn B, what is the probability that the second marble is red?
I get the answer $$\frac{2z+1}{z+v+1}$$, but my book gives the answer $$\frac{xz+x+yz}{(x+y)(z+v+1)}$$. I cannot understand how the books answer was obtained.
There are two cases: A red marble has been drawn from urn A and a white marble has been drawn from urn A.
$$\texttt{First case}$$:
The probability that a red marble has been drawn from urn A is $$\frac{x}{x+y}$$. Then you have $$z+1$$ red marbles in urn B and $$z+v+1$$ marbles in total. The probability to draw a red marble from urn B then is $$\frac{z+1}{z+v+1}$$
$$\texttt{Second case}$$:
The probability that a white marble has been drawn from urn A is $$\frac{y}{x+y}$$. Then you have still $$z$$ red marbles in urn B and $$z+v+1$$ marbles in total. The probability to draw a red marble from urn B then is $$\frac{z}{z+v+1}$$
Therefore the probability that hat the second marble is red is
$$\frac{x}{x+y}\cdot \frac{z+1}{z+v+1}+\frac{y}{x+y}\cdot \frac{z}{z+v+1}=\frac{x(z+1)+yz}{(x+y)\cdot (z+v+1)}=\frac{xz+x+yz}{(x+y)\cdot (z+v+1)}$$
Remark on the comment.
Firstly we define the relevant events:
$$A \texttt{ :Marble from urn A is red}$$
$$\overline A\texttt{ :Marble from urn A is white}$$
$$B \texttt{ :Marble from urn B is red}$$
$$\overline B\texttt{ :Marble from urn B is white}$$
There are two ways where the seocond marble is red.
1. $$\texttt{Marble from urn A is red} \rightarrow \texttt{Marble from urn B is red}$$
2. $$\texttt{Marble from urn A is white} \rightarrow \texttt{Marble from urn B is red}$$
Compared to only one way you have a higher chance to get a red marble from urn B. Thats why you add the two cases.
If you go along a path each node represents an event, which happen with a specific probability. At path 1 the event A and event B have to happen at the same time. That´s why we multiply the probabilities: $$P(A)\cdot P(B|A)$$. The second probability is a conditional probability since the probability depends on the what we have drawn from the first urn.
Similar for the second path: $$P(\overline A)\cdot P(B|\overline A)$$
It is an application of the law of total probability $$P(B)=P(A)\cdot P(B|A)+P(\overline A)\cdot P(B|\overline A)$$
• Thanks for the answer I think you mean (x+y) in the bottom of the last fraction...An explanation as to why yuo are multiplying the probabilities and then adding would be nice .......... – herashefat Oct 3 '18 at 2:48
• @herashefat I have made an edit. – callculus Oct 3 '18 at 8:48
Your answer doesn't involve $$x$$ or $$y,$$ so you get the same probability no matter what the distribution of colors in the first urn. Surely, this can't be right.
I think you are neglecting to weight your probabilities by the probability of the initial outcomes. I guess you're saying something like, if the first marble is red, the probability that the second is red is $${z+1\over z+v +1},$$ and if the first marble is black, the probability that the second marble is black is $${z\over z+v +1},$$ and we add these probabilities.
But we have \begin{align} Pr(M_2=R)&=P(M_2=R|M_1=R)\cdot P(M_1=R)\\ &+ P(M_2=R|M_1=B)\cdot P(M_1=B)\end{align} where $$M_1$$ and $$M_2$$ are the colors of the first and second marbles. You've just added up the conditional probabilities, without weighting them by the probabilities of the color of the first marble.
If you plug these in, I imagine you'll get the answer in the book.
• @saulpatz ,but why do we have to multiply here?? – herashefat Oct 3 '18 at 3:55
• @herashefat because in general $P(A\mid B)P(B)=P(A\cap B)$. Note that $P(M_2=R)=P(M_2=R\wedge M_1=R)+P(M_2=R\wedge M_1=B)=\text{RHS in answer}$. – drhab Oct 3 '18 at 8:16 | 2019-09-16T00:04:03 | {
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https://math.stackexchange.com/questions/2982126/maximum-and-minimum-value-of-int-01-fxdx-given-fx2 | # Maximum and minimum value of $\int_0^1 f(x)dx$ given $|f'(x)|<2$
Let $$f:\mathbb R\to \mathbb R$$ be a differentiable function such that $$f(0)= 0$$ and $$f(1)= 1$$ and $$|f'(x)|<2 ~ \forall x \in \mathbb R$$, if $$a$$ and $$b$$ are real numbers such that the set of possible values of $$\displaystyle\int_0^1 f(x)dx$$ is the open interval $$(a,b)$$, then $$b-a$$ is: ?
Attempt:
$$I = \int_0^1 1.f(x) dx$$
$$\implies I = 1 - \int_0^1 xf'(x)dx$$ (Integration by parts)
$$-2 < f'(x) < 2$$
$$\implies -2x for $$x>0$$
$$\implies -1 < \int_0^1 xf'(x) dx < 1$$
Therefore $$I_{max} = 2$$ and $$I_{min} = 0$$
$$\implies b- a = 2$$ but answer given is $$b-a = \dfrac 3 4$$.
Please let me know my mistake, and the correct way to solve it.
• I'm guessing the problem statement says $f(x)$ is nonnegative? – user25959 Nov 2 '18 at 19:26
• @user25959 it does not. – Abcd Nov 2 '18 at 19:32
Your estimates are correct, but not sharp. For example, in $$I = 1 - \int_0^1 xf'(x)dx > 1 - \int_0^1 2 x dx = 0$$ $$I$$ would be close to the lower bound $$0$$ only if $$f'(x) \approx 2$$ on the entire interval, which is not possible with $$f(0)=0$$ and $$f(1) =1$$.
The solution is actually simpler: Use the mean-value theorem to obtain upper and lower bounds for the admissible functions $$f$$.
• From $$f(0) = 0$$ and $$f'(x) < 2$$ it follows that $$f(x) < 2x$$ for $$0 < x \le 1$$.
• From $$f(1) = 1$$ and $$f'(x) > -2$$ it follows that $$f(x) < 1 - 2(x-1) = 3-2x$$ for $$0 \le x < 1$$.
Together: $$f(x) < \max(2x, 3-2x)$$ for $$0 < x < 1$$, which implies that $$\int_0^1 f(x) \, dx < \frac 78$$.
Similarly show that $$\int_0^1 f(x) \, dx > \frac 18$$.
Finally show that the integral can be arbitrarily close to those bounds.
You can also solve it graphically by drawing lines with slopes $$-2$$ and $$+2$$, starting from the given points $$(0, f(0))$$ and $$(1, f(1))$$.
The graph of $$f$$ must lie between the green and the red curve. $$(b-a)$$ is the area between those curves, and that is equal to the area of the blue rectangle.
• In $f'(x)< 2$ $\implies f(x)< 2x$, where is the constant of integration? Because you haven't preformed definite integration to both sides of the inequality right? – Abcd Dec 17 '18 at 5:01
• And I am not sure i completely understand the graphical solution. – Abcd Dec 17 '18 at 5:04
• @Abcd: I did definite integration. $f(0) = 0$ is given, therefore $f(x) = \int_0^x f'(t)dt < 2x$. – Martin R Dec 17 '18 at 5:43
• @Abcd: Which part of the graphical solution is unclear? That the graph of $f$ lies between the green and the red curve, that $(b-a)$ is the area between those curves, or that the area between the green and red curve is equal to the area of the blue rectangle? – Martin R Dec 17 '18 at 6:05 | 2019-08-21T14:00:20 | {
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http://math.stackexchange.com/questions/344195/from-the-area-of-the-inscribed-circumscribed-2n-gons-to-the-area-of-the-unit | # From the area of the inscribed/circumscribed $2^n$-gons to the area of the unit circle
Say, we have two $2^n$-gons: one inscribed in a unit circle and another one circumscribed around the unit circle. After using some basic geometry and limits we arrive at the following result:
\begin{align} \lim_{n\to \infty} \mbox{(perimeter of the inscribed 2^n-gon)} &= \lim_{n\to \infty} \mbox{(perimeter of the circumscribed 2^n-gon)} \\ &:= 2\pi \end{align}
Yet now (at least according to my professor) we are not allowed to conclude that the perimeter of the unit circle itself equals $2\pi$, because we have only showed it for $2^n$-gons and maybe some other approximations would give us different result.
Now what if we are interested in the areas?
\begin{align} \lim_{n\to \infty} \mbox{(area of the circumscribed 2^n-gon)} &= \lim_{n\to \infty} \mbox{(2^n \times area of the triangle)} \\ &= \lim_{n\to \infty} (2^n \frac{1 \times s_n}{2}) \\ &= \lim_{n\to \infty} \mbox{(\frac{1}{2} \times perimeter of the circumscribed 2^n-gon)} \\ &= \frac{1}{2} \times 2\pi \mbox{ (using the result above)} \\ &= \pi \end{align}
Similarly (but with slightly more geometry involved):
$$\lim_{n\to \infty} \mbox{(area of the inscribed 2^n-gon)} = \pi$$
My question is: are we now allowed to conclude that the area of the unit circle equals $\pi$? Or do we have a similar problem as the one (with perimeters) described above?
Any help is much appreciated.
-
Area is well-behaved, yes, it is OK. Arclength can do strange things. – André Nicolas Mar 27 '13 at 23:05
@André Nicolas, could you please elaborate what is the difference between squeezing the perimeter of the unit circle between two limits of equal value and squeezing the area of the unit circle? Why is that problematic this the former and OK with the latter? Thank you. – Leo Mar 27 '13 at 23:45
For the areas, the fact that the circle lies completely within the circumscribed polygons implies that its area is less than the polygons' areas (and similarly for inscribed polygons). There is no such assurance, though, that a shape contained in a second shape has a smaller perimeter than the second shape's perimeter. (It's true in this case, and showing that can actually lead to a formula for the circle's perimeter; but it's not true for all shapes.) – Greg Martin Mar 28 '13 at 0:03
Imagine two squares, with square $B$ inside square $A$. Suppose square $A$ has area $1$, and square $B$ has area $0.99$. If $R$ is a region such that $B\subset R\subset A$, then the area of $R$ is between $0.99$ and $1$. Now imagine the same thing, but with perimeters $0.99$ and $1$. The region $R$ could have a lot of zigzags, and perimeter much greater than $1$. In fact the perimeter of $R$ could be infinite. – André Nicolas Mar 28 '13 at 0:07
@André Nicolas and Greg Martin: thank you for the responses. That's a pity one can't accept a comment as an answer. – Leo Mar 28 '13 at 0:41 | 2014-12-22T23:55:01 | {
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https://mathematica.stackexchange.com/questions/126144/how-to-color-a-particular-contour-line-in-a-contourplot-and-obtain-the-area-surr | # How to color a particular contour line in a ContourPlot and obtain the area surrounded?
I want to
$1.$ highlight a contour line in a ContourPlot by coloring it, and
$2.$ obtain the area surrounded by that contour line.
For example, I want to color the contour of $-0.72$ in the following figure to red, then obtain the area of the inner part of the contour of $-0.72$.
$Note:$
$1.$ I am looking for a general method because in my real problem the object function (here Cos[x] + Cos[y]) was obtained numerically as an InterpolatingFunction, and
$2.$ the region of interest surrounded by the contour line is an irregular region.
ContourPlot[Cos[x] + Cos[y], {x, 0, 4 Pi}, {y, 0, 4 Pi},
Contours -> 10, ContourStyle -> {AbsoluteThickness[1], Black},
PlotLegends -> Automatic]
Thank you!
• I would b = ContourPlot[{Cos[x] + Cos[y] == -0.72}, {x, 0, 4 Pi}, {y, 0, 4 Pi}, ContourStyle -> {AbsoluteThickness[5], Red}] and Show[a,b] it together with your plot a. – corey979 Sep 12 '16 at 9:21
• What if the contour you want to highlight is not there? E.g. -0.9 is not marked there but have you known before plotting? Do you want to force it to be created? How should it work with Contours spec you use. – Kuba Sep 12 '16 at 9:22
• @Kuba, in general, I will plot it using black color with specified number for Contours and determine which contour I'd like to highlight :) then I will use a certain trick to color it to red. – jsxs Sep 12 '16 at 9:30
• @jsxs Unrelated question but, are you by any chance working with acoustic fields? – Keine Sep 12 '16 at 14:37
• – Michael E2 Sep 13 '16 at 1:10
Single out one contour:
ContourPlot[Cos[x] + Cos[y] == 0.72, {x, 0, 4 Pi}, {y, 0, 4 Pi}, ContourStyle -> Red]
Use Show to combine several graphics outputs.
By visually inspecting the result, we can determine a bounding box for the contour in the middle: it is $([\pi,3\pi], [\pi, 3\pi])$.
Comparing with your original contour plot, we see that the enclosed region is defined by the equation Cos[x] + Cos[y] > 0.72. In other cases we might need to use < instead.
Define it as a region (version 10 and later):
reg = ImplicitRegion[Cos[x] + Cos[y] > 0.72, {{x, Pi, 3 Pi}, {y, Pi, 3 Pi}}];
Find the area: Area[reg]. Also look up RegionMeasure.
Plot it: RegionPlot[reg]
In versions before 10, find the area:
NIntegrate[Boole[Cos[x] + Cos[y] > 0.72], {x, Pi, 3 Pi}, {y, Pi, 3 Pi}]
Plot it:
RegionPlot[Cos[x] + Cos[y] > 0.72, {x, Pi, 3 Pi}, {y, Pi, 3 Pi}] | 2021-05-17T11:18:27 | {
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https://gateoverflow.in/204121/gate2018-46 | 4.5k views
The number of possible min-heaps containing each value from $\{1,2,3,4,5,6,7\}$ exactly once is _______
retagged | 4.5k views
0
80 may be
+1
IT is 80
+1
+1
Lets answer this question in an easier way :
Now do $\frac{7!}{7 \times 3\times 3 }= 80$
Here $7!$ because $7$ items to be filled, Now $7$ because root has only $7$ nodes as its decedent including itself and only one can be the root. In same way we get $3$ and $3$ for the second level nodes and $1$ and $1$ for the third level.
edited by
+2
ye toh bohot accha approach hei, i did mine in 3 cases and it came 80. awesome bro
+1
+2
How is it different than the answer answered by Anu007 on Feb 4?
0
How the formula Derived?
+1
I did not understand how you calculated. Please explain in detail.
+2
Ans: 80
Explanation:
Number of min-heaps possible with keys $\{1, 2, 3, 4, 5, 6, 7\}$.
A min-heap is a binary tree such that.
- the data contained in each node is less than (or equal to) the data in that node's children.
- the binary tree is complete.
Since a binary heap is always a complete binary tree, so with $7$ nodes, we can have $2$ levels (root at level $0$). It's structure will be like this:
Now because of min-heap property, every node's value must be smaller than all of it's children. So, this ensure that the minimum will always be at the root. $\therefore$ $1$ will be at the root.
The second minimum element(i.e. $2$) can be present at the first level only(root is at the zeroth level). So we have two options. Let's, for now, fix $2$ at the left side.
We are now left with $5$ more elements $\{3, 4, 5, 6, 7\}$. For the left subtree's two leaves, we have $5 * 4$ ways. i.e. first choosing one of the 5 nodes, then choosing one of the remaining 4 nodes.
Now $3$ nodes left. Out of these $3$, one will be the least among them, and that will definitely become the parent of the two remaining leaves(in the right subtree). Now with $2$ nodes left, we can arrange them in $2$ ways.
This gives $(5 * 4) * 2 = 40$ ways.
We can have the same number of ways, if we fixed $2$ at the right subtree instead of left. So total ways:
$= 40 * 2$
$= \mathbf{80}$
+1
+1
Yes, answer will be same @Warlock lord
+1
Great explanation
1 has to be root. Now,
case 1: 2 and 3 are on 2nd level.
so, 4,5,6 and 7 can occupy any of 4 positions in the 3rd level as all are less than 2 and 3. So, 4! arrangements. And 2 and 3 can be arranged in 2 ways in 2nd level (mirror image). Hence 2*24=48
case 2: 2 and 4 are on 2nd level
Now, 3 can only be below 2 and not 4 as it is MIN heap. So, 2 cases are possible 3XXX or X3XX, which gives 3!+3!=12 arrangements. And 2 and 4 can be arranged in 2 ways in 2nd level (mirror image). Hence 2*(6+6)=24
case 3: 2 and 5 are on 2nd level
Now, 3 and 4 can only be below 2 and not 5 as it is MIN heap. So, 2 arrangements are possible for 3 and 4 and similarly 2 for 6 and 7. And 2 and 5 can be arranged in 2 ways in 2nd level (mirror image). Hence, 2*(2+2)=8
Hence, total 48+24+8=80
+2
this was my exact approach in d exam
0
I hope, Hence, 2*(2+2)=8 LINE SHOULD BE 2*(2*2)=8
0
wonderful answer sir :) easy to understand
f(1) = Number min heap with 1 node = 1
f(2) = Number min heap with 2 node = 1
f(3) = Number min heap with 3 node = 2 //swap left and right subtree
For a complete binary tree with n nodes number of nodes in left subtree (x) =
floor {((n-1)/2) + 1}.
1 node for Root (1 choice minimum valued node), x out of n-1 (leaving 1 for root) number of nodes in left subtree and n-1-x number of nodes in right subtree:
So, the number of arrangements possible:
Select root, Select and arrange x nodes in left subtree, arrange n-1-x nodes in right subtree
C(n-1, x)*f(x)*f(n-1-x)
Let number of arrangements are f(n):
f(n) = C(n-1, x)*f(x)*f(n-1-x)
Put n==7
f(7) = C(7-1, 3)*f(3)*f(7-1-3) = C(6,3)*f(3)*f(3) = 20*2*2 = 80
edited
0
Does this ensure that heap is a complete binary tree?
0
Already assumed the tree is CBT.
+1
Arrangements are nothing but filling numbers into a Complete Binary Tree.
Let's note down the facts first.
- The smallest element will be on the least level.
- The second smallest element will be on the first level.
- The third smallest element can be on the second or third level.
Root is fixed, the idea is to divide the building process into two subproblems, each solving both subtrees of the root.
Recurrence equation: T(n) = $_{k}^{n-1}\textrm{C}$ x T(k) x T(n-k-1), where k is number of nodes in left sub tree.
T(1) = 1
T(2) = 1, one root and one child.
T(3) = $_{1}^{2}\textrm{C}$ x T(1) x T(1) = 2
T(4) = $_{2}^{3}\textrm{C}$ x T(2) x T(1) = 3
T(5) = $_{3}^{4}\textrm{C}$ x T(3) x T(1) = 8
T(6) = $_{3}^{5}\textrm{C}$ x T(3) x T(2) = 20
T(7) = $_{3}^{6}\textrm{C}$ x T(3) x T(3) = 80
To put it down simply, Among 7 elements, least element is fixed as root.
We are now left with 6 elements, the left subtree will have 3 elements, which can be picked in
$_{3}^{6}\textrm{C}$ x 2 ways.
Now we have 3 elements remaining and we can complete the heap in 2 ways only.
Hence $_{3}^{6}\textrm{C}$ x 2 x 2 = 80.
+1 vote
One more way to solve this problem could be $^6C_3*2*2 = 80$
+1 vote
Minimum no of node in heap tree
My approach for this question | 2018-12-15T02:19:01 | {
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"lm_q1q2_score": 0.8326674655250212
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# If 175 billion french francs is equivalent to 35 billion
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If 175 billion french francs is equivalent to 35 billion [#permalink]
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Updated on: 19 Jan 2014, 11:45
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Question Stats:
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If 175 billion French francs is equivalent to 35 billion United States dollars, which of the following expressions represents the number of United States dollars equivalent to f francs?
A) f-140
B) 5f
C) 7f
D) f/5
E) f/7
The OE starts with "In this solution it is assumed that 1 billion = 109. However,
only minor modifications in the first three steps are needed (replace each appearance
of 10^9 with 10^12)".
What do they mean by that and why is that important for the question at all.
It confused me. Btw. I feel the "109" ought to be "10^9" but they misspelled it.
Millions has six zeros and billions has nine. And both numbers are given in "billions" so we
can ignore that completely anyway.
I will post OA separately but would be good to know why they wrote this. Supposedly,
everything has a reason.
(Small point: Source is not OG but Official GMATPrep Software)
Originally posted by BabySmurf on 19 Jan 2014, 10:01.
Last edited by BabySmurf on 19 Jan 2014, 11:45, edited 1 time in total.
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Posts: 39
Re: If 175 billion french francs is equivalent to 35 billion [#permalink]
### Show Tags
19 Jan 2014, 10:06
2
3
175 billion French francs = 35 billion USD Ignore billion
175 French francs = 35 USD divide by 5
35 French francs = 7 USD divide by 7
5 French francs = 1 USD divide by 5
1 French francs = 1/5 USD
f is just a variable. They could have written x, but they wrote f to
make it confusing.
Number of USD equivalent to French francs are given above. 1 Franc equals 1/5 USD.
So x Francs or for this case f equals f/5 francs.
##### General Discussion
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Re: If 175 billion french francs is equivalent to 35 billion [#permalink]
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27 Mar 2014, 22:31
1
1
Drop the term "Billion"
175f = 35d
$$f = \frac{35}{175} * d$$
$$f = \frac{1}{5} * d$$
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Re: If 175 billion french francs is equivalent to 35 billion [#permalink]
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01 Dec 2014, 09:30
You know that 175 Francs are equal to 35 USD. (Because both currencies are in billions anyway, you can ignore the units here)
The question asks for the right-hand side of the expression USD=.... and tries to confuse you by giving you variable "f" to represent the number of Francs. Don't let the GMAT trick you, you know better than that. You probably recognized that there is a variable in every answer choice. Get your thoughts straight and pick a smart value for "f". In fact, the question already provided you with a smart number. 175.
--> If Francs=f=175 and USD=35:
Francs(or "f)*x=USD
175*x=35
x=35/175=1/5 --> So f*1/5=USD, or f/5=USD
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Re: If 175 billion french francs is equivalent to 35 billion [#permalink]
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16 Aug 2015, 22:03
1
we can eliminate few options if we are running out of time
175 francs is equivalent to 35 dollars. ..
so the dollar equivalent of f francs has to be a figure less than f itself.
eliminate b and c.
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Posts: 52390
Re: If 175 billion french francs is equivalent to 35 billion [#permalink]
### Show Tags
17 Aug 2015, 01:20
BabySmurf wrote:
If 175 billion French francs is equivalent to 35 billion United States dollars, which of the following expressions represents the number of United States dollars equivalent to f francs?
A) f-140
B) 5f
C) 7f
D) f/5
E) f/7
The OE starts with "In this solution it is assumed that 1 billion = 109. However,
only minor modifications in the first three steps are needed (replace each appearance
of 10^9 with 10^12)".
What do they mean by that and why is that important for the question at all.
It confused me. Btw. I feel the "109" ought to be "10^9" but they misspelled it.
Millions has six zeros and billions has nine. And both numbers are given in "billions" so we
can ignore that completely anyway.
I will post OA separately but would be good to know why they wrote this. Supposedly,
everything has a reason.
(Small point: Source is not OG but Official GMATPrep Software)
Check other Conversion problems to practice in Special Questions Directory.
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Re: If 175 billion french francs is equivalent to 35 billion [#permalink]
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25 Sep 2015, 21:47
Just logic:
1 dollar is 5 times as expensive as 1 franc. So if you have F as a amount of francs, the equivalent amount of dollars is F/5
Algebra: 175f=35d => 5f=d, so f=d/5
D
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Re: If 175 billion french francs is equivalent to 35 billion [#permalink]
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29 Nov 2016, 15:04
175 Francs = 35 Dollars (removed Billion portion for simplicity)
175F=35D (1)
175F/35=D (manipulate (1)) & reduce --> F/5 = D
D.
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Re: If 175 billion french francs is equivalent to 35 billion [#permalink]
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11 Jan 2017, 07:09
3
BabySmurf wrote:
If 175 billion French francs is equivalent to 35 billion United States dollars, which of the following expressions represents the number of United States dollars equivalent to f francs?
A) f-140
B) 5f
C) 7f
D) f/5
E) f/7
We are given that 175 billion French francs = 35 billion United States dollars. We need to convert f francs to dollars. We can let x be the number of dollar that is equivalent to f francs and solve this problem using a proportion.
(175 billion)/(35 billion) = f/x
175/35 = f/x
175x = 35f
x = 35f/175
x = f/5
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Re: If 175 billion french francs is equivalent to 35 billion [#permalink]
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28 Feb 2017, 03:39
175 billion francs = 35 billion dollars
1 franc = 35/175 dollar = ⅕ dollar
f francs = f/5 dollars
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If 175 billion french francs is equivalent to 35 billion [#permalink]
### Show Tags
03 May 2018, 10:10
Bunuel wrote:
BabySmurf wrote:
If 175 billion French francs is equivalent to 35 billion United States dollars, which of the following expressions represents the number of United States dollars equivalent to f francs?
A) f-140
B) 5f
C) 7f
D) f/5
E) f/7
The OE starts with "In this solution it is assumed that 1 billion = 109. However,
only minor modifications in the first three steps are needed (replace each appearance
of 10^9 with 10^12)".
What do they mean by that and why is that important for the question at all.
It confused me. Btw. I feel the "109" ought to be "10^9" but they misspelled it.
Millions has six zeros and billions has nine. And both numbers are given in "billions" so we
can ignore that completely anyway.
I will post OA separately but would be good to know why they wrote this. Supposedly,
everything has a reason.
(Small point: Source is not OG but Official GMATPrep Software)
Check other Conversion problems to practice in Special Questions Directory.
Bunuel
if 1 USD equals 5 CHF
then why option B is icorrect ? 5 USD = 5*5 CHF
pushpitkc may be you can explain the logic
the question is : which of the following expressions represents the number of United States dollars equivalent to f francs?
Dollar to Franc
1 dollar = 5 francs
so 1d = 5F
so why B option is incorrect ?
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Re: If 175 billion french francs is equivalent to 35 billion [#permalink]
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03 May 2018, 17:37
The easiest method from my school time:
175 = 35
f = x (x - what we need to find),
multiply by diagonal numbers that do not have x and divide it then by the number, which is diagonal to x
x =F*35/175
It's a very convinient method to convert numbers.
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Posts: 52390
Re: If 175 billion french francs is equivalent to 35 billion [#permalink]
### Show Tags
04 May 2018, 02:50
1
dave13 wrote:
Bunuel wrote:
BabySmurf wrote:
If 175 billion French francs is equivalent to 35 billion United States dollars, which of the following expressions represents the number of United States dollars equivalent to f francs?
A) f-140
B) 5f
C) 7f
D) f/5
E) f/7
The OE starts with "In this solution it is assumed that 1 billion = 109. However,
only minor modifications in the first three steps are needed (replace each appearance
of 10^9 with 10^12)".
What do they mean by that and why is that important for the question at all.
It confused me. Btw. I feel the "109" ought to be "10^9" but they misspelled it.
Millions has six zeros and billions has nine. And both numbers are given in "billions" so we
can ignore that completely anyway.
I will post OA separately but would be good to know why they wrote this. Supposedly,
everything has a reason.
(Small point: Source is not OG but Official GMATPrep Software)
Check other Conversion problems to practice in Special Questions Directory.
Bunuel
if 1 USD equals 5 CHF
then why option B is icorrect ? 5 USD = 5*5 CHF
pushpitkc may be you can explain the logic
the question is : which of the following expressions represents the number of United States dollars equivalent to f francs?
Dollar to Franc
1 dollar = 5 francs
so 1d =[color=#ff0000] 5F
so why B option is incorrect ?
The question asks: which of the following expressions represents the number of United States dollars equivalent to f francs?
So, the question is: f francs = ? dollars. Do you see the difference between the actual question and the answer you are getting (highlighted)?
1 dollar = 5 francs, so 1 franc = 1/5 dollars. Therefore, f francs = f/5 dollars.
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Re: If 175 billion french francs is equivalent to 35 billion &nbs [#permalink] 04 May 2018, 02:50
Display posts from previous: Sort by | 2019-01-22T23:06:17 | {
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http://mathhelpforum.com/advanced-algebra/280734-functional-equationsolving-find-f-x.html | # Thread: Functional equationsolving find f(x)
1. ## Functional equationsolving find f(x)
Hi, I am trying to solve the following functional equation
f(x)-f(2a-x)=b(x-a)
considering "a" and "b" are positive constants.
2. ## Re: Functional equationsolving find f(x)
there are two parameters so let's try a function of the form
$f(x) = c_1 x + c_0$
$c_1 x + c_0 - (c_1(2a-x)+c_0) = b(x-a)$
$2c_1 x - 2 a c_1 = b x - a b$
$2c_1 = b$
$c_1 = \dfrac b 2$
$-2ac_1 = -a b$
$a$ is specified positive so
$2c_1 = b$
$c_1 = \dfrac b 2$
and we have agreement in the value of $c_1$
so we have
$f(x) = \dfrac b 2 x + c_0$
There are of course higher order polynomials that satisfy this functional relationship but this is the lowest order one.
3. ## Re: Functional equationsolving find f(x)
Thanks romsek. I knew that, the second order polynomial also satisfies, but not higher order ones.
However to solve the functional equation we need to find all functions, not just examples that satisfy.
4. ## Re: Functional equationsolving find f(x)
Originally Posted by Rafael
Thanks romsek. I knew that, the second order polynomial also satisfies, but not higher order ones.
However to solve the functional equation we need to find all functions, not just examples that satisfy.
what kind of answer is acceptable then? I doubt there's going to be some notation that is all encompassing of the various functions that satisfy this.
Are you supposed to list them all out?
5. ## Re: Functional equationsolving find f(x)
Well, the first order polynomial is the special case of the second order polynomial, but to solve the functional equation one should give a complete solution, maybe there is a more general function that becomes, for example, second order polynomial in the special case but still satisfies the equaion.
for example, let's solve following functional equation:
f(x)-2f(1/x)=2^x
we can replace x by 1/x
f(1/x)-2f(x)=2^(1/x)
from these two equations, one can determine that f(x) =-(1/3)(2^x+2^(1-1/x))
6. ## Re: Functional equationsolving find f(x)
Originally Posted by Rafael
Hi, I am trying to solve the following functional equation
f(x)-f(2a-x)=b(x-a)
considering "a" and "b" are positive constants.
If we take $x = a$ in the given equation we have $f(a)-f(a) = 0$ so $f(a)$ is arbitrary. Let's call $f(a)=C$. Now let's assume $f$ is differentiable and differentiate the equation giving
$f'(x) + f'(2a-x) = b$ and put $x=a$ in that. That gives $f'(a) + f'(a) = b$ so $f'(a) = \frac b 2$. Let's differentiate again:
$f''(x) - f''(2a-x) = 0$. Putting $x=a$ in that gives $f''(a)-f''(a) = 0$, so $f''(a)$ is arbitrary. Lets call $f''(a) = D$. Let's differentiate again:
$f'''(x) + f'''(a-x) = 0$. Putting $x=a$ in that gives $f'''(a)+f'''(a) = 0$, so $f'''(a)$ is $0$.
Now a pattern becomes clear. From now on all the odd derivatives will be $0$ and the even ones will be arbitrary when $x=a$. So let's look at the Taylor series
for $f(x)$ about $x=a$.
$$f(x) = \sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!}(x-a)^k = C + \frac b 2 (x-a) + \frac D 2 (x-a)^2 +\text{more terms with even powers of }(x-a)$$
Notice that if we take the arbitrary constant $D=0$ we get $f(x) = C + \frac b 2 (x-a) = C-\frac{ab} 2 + \frac b 2 x$ which agrees with the earlier first degree
solution. And it satisfies the functional equation. So what about all those arbitrary terms of the form $K(x-a)^{2k}$. Notice that such expressions satisfy $f(x)-f(2a-x)=0$ so they don't disturb the functional relationship that is already satisfied by the first two terms. So the functional equation can be satisfied by adding any convergent sum of the form$$\sum_{k=1}^\infty a_k (x-a)^{2k}$$ to the first two terms.
In fact, there is no real reason not to include the constant term as one of the even powers and write the answer as
$$f(x) = \frac b 2 (x-a) + \sum_{k=0}^\infty a_k (x-a)^{2k}$$
So, for example, a non polynomial example is $f(x) = \frac b 2(x-a) + \cos(x-a)$. | 2019-05-21T06:53:43 | {
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https://math.stackexchange.com/questions/1629679/asymptotic-behavior-of-the-two-sequences-defining-exponential-function | # asymptotic behavior of the two sequences defining exponential function
There are two definitions of exponential function: $$e^x=\lim_{n\to\infty} S_n=\lim_{n\to \infty} a_n \text{ ,}$$ where $$S_n=1+x+\frac{x^2}{2!}+\dots+\frac{x^n}{n!}$$ and $$a_n=(1+\frac{x}{n})^n \text{ .}$$
Since the two sequence have the same limit, I guess they are somehow related and reflect different aspects of $e^x$. So my first question is: are there any relationships between $S_n$ and $a_n$?
My other questions come from the following observations:
When x is positive, obviously $S_n$ is increasing. Is $a_n$ also increasing?
When x is negative, $S_n$ goes up and down since it keeps adding numbers of alternating signs as as $n$ grows. Eventually $S_n$ "squeezes" to its limit. What about the behavior of $a_n$ in this case? When $n$ is smaller than $|x|$, I can see that $a_n$ changes signs very often. When $n$ is large, $a_n$ is always positive, and is $a_n$ increasing when $n$ is large?
• Have you looked at the binomial expansion of $a_n$? – Clement C. Jan 27 '16 at 21:15
• $\displaystyle \left(1+\frac{x}{n}\right)^n =\displaystyle 1 + {n \choose 1}\frac{x}{n}+{n \choose 2}\frac{x^2}{n^2}+{n \choose 3}\frac{x^3}{n^3}+\cdots \to 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$ as $n$ increases – Henry Jan 27 '16 at 21:15
• @Henry: That limit is problematic since it contains a moving target in the number of terms in the binomial expansion. You would need something like uniform convergence to get the limit in the coefficients commuted with the series limit. – LutzL Jan 27 '16 at 21:25
• @LutzL: TO me, what matters is each term $\displaystyle {n \choose k}\dfrac{x^k}{n^k}$ in the binomial expansion (using $0$ if $n \lt k$) is smaller in magnitude than the corresponding term $\dfrac{x^k}{n!}$ in the exponential series and the exponential series is absolutely convergent. – Henry Jan 27 '16 at 21:32
• Yes, that should also do nicely. Question remains if that fact is already known at the current point of the course. – LutzL Jan 27 '16 at 22:10
Use the binomial formula: $$\left(1+\dfrac{x}{n} \right)^n =1+\sum_{k=1}^n\binom{n}{k}\dfrac{x^k}{n^k}$$ where the coefficient of $x^k$ is: $$\dfrac{n!}{k!(n-k)!\,n^k}=\dfrac {1}{k!} \, \dfrac{1\times 2 \times 3 \cdots \times n}{[1\times 2 \cdots \times (n-k)]\,\times \,\underbrace {n\times n \cdots \times n}_{k \,\mbox{times}}}$$ that can be simplified as: $$\dfrac{n!}{k!(n-k)!\,n^k}= \dfrac {1}{k!} \,\dfrac{\overbrace{(n-k+1)\times(n-k+ 2) \cdots \times (n-1)}^{(k-1)\,\mbox{factors}}}{\underbrace {n\times n \cdots \times n}_{(k-1) \,\mbox{factors}}}=$$ $$=\dfrac {1}{k!} \left[ \dfrac{(n-k+1)}{n}\times \dfrac{(n-k+ 2)}{n} \cdots \times \dfrac{(n-1)}{n}\right]=$$ $$=\dfrac {1}{k!}\,\left(1-\dfrac{k-1}{n}\right)\,\left(1-\dfrac{k-2}{n}\right)\cdots\,\left(1-\dfrac{1}{n}\right)$$
now, for $n \to \infty$ all the factors in the parenthesis $\to 1$ so we have: $$\lim_{n \to \infty}\dfrac{n!}{k!(n-k)!n^k}=\dfrac{1}{k!}$$ and: $$\lim_{n \to \infty}\left(1+\dfrac{a}{n} \right)^n=\lim_{n \to \infty}\left[1+\sum_{k=1}^n\binom{n}{k}\dfrac{a^k}{n^k}\right]$$
Let $x > 0$ then it can be easily proven using binomial theorem (for positive integer index) that \begin{align} \left(1 + \frac{x}{n}\right)^{n} &= 1 + x + \dfrac{1 - \dfrac{1}{n}}{2!}\cdot x^{2} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}\cdot x^{3} + \cdots\notag\\ &\leq 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots + \frac{x^{n}}{n!}\notag \end{align} and therefore $a_{n} \leq S_{n}$. Moreover from the expansion of $a_{n}$ (via binomial theorem above) you can see that as $n$ increases each term in the expansion increases as well as the number of terms also increases. Thus $a_{n}$ is an increasing sequence if $x > 0$.
The story of $a_{n}$ and $S_{n}$ is not complete without the introduction of another sequence $b_{n}$ defined by $$b_{n} = \left(1 - \frac{x}{n}\right)^{-n}$$ For $0 < x < n$ we can use the general binomial theorem (for any index) to get $$b_{n} = \left(1 - \frac{x}{n}\right)^{-n} = 1 + x + \dfrac{1 + \dfrac{1}{n}}{2!}\cdot x^{2} + \dfrac{\left(1 + \dfrac{1}{n}\right)\left(1 + \dfrac{2}{n}\right)}{3!}\cdot x^{3} + \cdots$$ and this shows clearly that for $0 < x < n$ we have $$a_{n} \leq S_{n} \leq b_{n}$$ It is easy to prove that $S_{n}$ represents a convergent series and since $a_{n}$ is increasing it follows that $a_{n}$ also is convergent for $x > 0$. Let $S_{n} \to S(x)$ and $a_{n} \to a(x)$. Note also that $b_{n}$ is decreasing sequence for $0 < x < n$ and is bounded below by $S(x)$ and hence $b_{n}$ also is convergent and let $b_{n} \to b(x)$. We then have $a(x) \leq S_{x} \leq b(x)$. The magic happens when we see that $a_{n}/b_{n} = (1 - x^{2}/n^{2})^{n} \to 1$ as $n \to \infty$ so that $a(x) = b(x)$ and therefore all the sequences $a_{n}, b_{n}, S_{n}$ tend to the same limit which is traditionally denoted by $e^{x}$.
To handle negative values of $x$ note that the limit $S(x)$ of $S_{n}$ is an infinite convergent series which has the interesting property that $S(x + y) = S(x)S(y)$ (this is proved via multiplication of series). It thus means that $S(-x) = 1/S(x)$. Note that the sequence $a_{n}, b_{n}$ are such that $b_{n}(x) = 1/a_{n}(-x)$ (added $x$ to show the dependence of $x$) and hence it follows that $b(x) = 1/a(-x)$ or $a(-x) = 1/a(x)$ and this establishes the relation $a(x) = b(x) = S(x)$ for negative values of $x$ also.
Define $$E(x)=\sum_{k=0}^\infty\frac{x^k}{k!}$$ We know that this power series has infinite radius of convergence and is thus a continuous (and differentiable, etc.) function. By careful but elementary application of the Cauchy product an binomial theorem one finds $$E(x)·E(y)=E(x+y)$$ which as functional equation has the solution $$E(x)=E(1)^x=e^x$$
By multiplying out $$\left(1+\frac xn\right)\left(1+\frac yn\right)\left(1-\frac {x+y}n\right)=1-\frac {x^2+xy+y^2}{n^2}-\frac {xy(x+y)}{n^3}$$ one finds that for $e(x)=\lim_{n\to\infty}\left(1+\frac xn\right)^n$ it holds that $$e(x)e(y)e(-x-y)=\lim_{n\to\infty}\left(1+\frac xn\right)^n\left(1+\frac yn\right)^n\left(1-\frac {x+y}n\right)^n=1$$ which again implies the functional equation $$e(x)e(y)=e(x+y)$$ For the Cauchy functional equation one would need to establish for instance continuity at $x=0$ to then again arrive at $$e(x)=e(1)^x=e^x$$
Following the outline given by Henry in the comments to the question:
Proposition: If $\sum_{k=0}^\infty a_k(t)$ is a series with parameter $t$ so that $\sum_{k=0}^\infty a_k(0)$ converges absolutely, $|a_k(t)|\le|a_k(0)|$ and $\lim_{t\to 0}a_k(t)=a_k(0)$, then $$\lim_{t\to 0}\sum_{k=0}^\infty a_k(t)=\sum_{k=0}^\infty a_k(0)$$
Proof: Let $ε>0$ be arbitrary. Determine $N$ such that $\sum_{k=N}^\infty |a_k(0)|<ε/3$. Then select $\bar t$ such that $\sup_{k=0,…,N-1}|a_k(t)-a_k(0)|<ε/(3N)$ for all $t\in(0,\bar t)$. Then $$\left|\lim_{t\to 0}\sum_{k=0}^\infty a_k(t)-\sum_{k=0}^\infty a_k(0)\right|\le\sum_{k=0}^{N-1}|a_k(t)-a_k(0)|+ \sum_{k=N}^\infty |a_k(t)|+\sum_{k=N}^\infty |a_k(0)|<ε.$$
Application: Write the binomial theorem as $$\left(1+\frac xn\right)^n =\sum_{k=0}^\infty\binom{n}{k}\left(\frac xn\right)^k =\sum_{k=0}^\infty\left(1-\frac1n\right)\left(1-\frac2n\right)…\left(1-\frac{k-1}n\right)\frac{x^k}{k!}$$ thus using $a_k(t)=(1-t)(1-2t)…(1-(k-1)t)\frac{x^k}{k!}$ the assumption of above proposition are satisfied, which establishes pointwise convergence of $\left(1+\frac xn\right)^n$ towards the value of the exponential series. | 2019-05-27T05:07:24 | {
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https://math.stackexchange.com/questions/2395800/prove-that-the-set-c-has-content-zero | # Prove that the set C has content zero
I'm a little lost in here with content zero. It is a question of a test o mine and the professor said it was wrong (I said it had content zero because it was a graph of a function) and I don't know where do I begin correctly. It is the following statement:
Consider constants $a,b>0$. Does the set $$C=\left\{(x,y)\in \mathbb{R}^2 \middle| \frac{x^2}{a^2}+\frac{y^2}{b^2}= 1\right\}$$ have content zero? Justify your answer!!!
Definition 1. A rectangle is any set of the form $R = [a, b] × [c, d]$. Its area is $(b − a)(d − c)$. Note that rectangles are, by definition, closed (they include their boundary points.)
Definition 2. A bounded set $S \subset \mathbb{R}^2$ has zero content if for any $\varepsilon > 0$ one can find a finite number of rectangles $R_1,\dotsc, R_m \subset \mathbb{R}^2$ such that
1. $S \subset R_1 \cup \dotsb \cup R_m$, and
2. $\mathrm{area}(R_1) + \dotsb + \mathrm{area}(R_m) < \varepsilon$.
• Can you find a point in $C$ ? Aug 16 '17 at 16:05
• An ellipse is not the graph of a function. It is, however, the union of the graphs of two functions... Aug 16 '17 at 16:19
• What is "content"? Hausdorff content? Minkowski content? Aug 16 '17 at 16:23
• It is the condition for the existence of double integrals. Aug 16 '17 at 16:25
• you can prove that your set has measure zero and its compact because measure zero and compactness implies content zero Aug 16 '17 at 17:15
This is an extension of my comments. So assume that you have proved the theorem about union of sets of content $0$ (proof is easy).
Now consider the part of $C$ which lies in first quadrant. This can be covered by $n$ rectangles. The diagonal points of $i$-th such rectangle are $(x_i, f(x_i)), (x_{i+1},f(x_{i+1}))$ where $x_{i} =ia/n$ and $f(x) =b\sqrt{1-(x/a)^{2}}$ (note that $i$ takes values $0,1,2,\dots,n-1$). Prove that the total area of such rectangles is $b/n$ which can be made smaller than any given positive number by choosing a large $n$. You will find that the fact that $f$ is monotone in interval $[0,a]$ will come handy here. Thus you will show that the content of the part in first quadrant is $0$. Similarly parts in other quadrants are also of content $0$.
For those who are well versed, the above procedure is the way we prove that a monotone function is Riemann integrable.
• Very good point at the end, about Riemann integration. Aug 16 '17 at 17:50
$C$ is the union of the graphs of two functions $f,g$ on $[-a,a]$, with $f=\{(x,y)\in C: y\geq 0\}$ and $g=\{(x,y)\in C:y\leq 0\}.$ Since $g(x)=-f(x)$ for all $x\in [-a,a],$ it suffices to show that the content of the graph of $f$ is $0.$
(i). For $a>r>0$ let $g_r$ be the part of the graph of $f$ restricted to the domain $[-a,-a+r]\cup [a-r,a].$ Now $g_r$ is covered by two rectangles: $[-a, -a+r]\times [0,f(-a+r)],$ and $[a-r,a]\times [0,f(a-r)]$.
So the content of $g_r$ is at most $D(r)=r(f(a-r)+f(-a+r)).$ Since $f$ is continuous at $\pm a$ with $f(\pm a)=0,$ the value $D(r)$ can be as small as desired, by taking sufficiently small $r$.
We can conclude that the content of the graph of $f$ is $0$ if we can show that the content of $f$ resticted to $[-a+r,a-r]$ is $0,$ ....
.... which we do by putting $p=-a+r$ and $q=a-r$ in the following:
(ii). Lemma. For $p<q$ and for differentiable $f:[p,q]\to \mathbb R$ such that $\sup \{|f'(x)|:x\in [p,q]\}=K<\infty ,$ the content of the graph of $f$ on $[p,q]$ is $0.$
Proof: For $n\in \mathbb N$ let $d_n=\frac {q-p}{n}$ and let $a_j=a+jd_n$ for $0\leq j\leq n.$ The graph of $f$ on the interval $[a_j,a_{j+1}]$ is covered by the rectangle $[a_j,a_{j+1}]\times [f(a_j)-Kd_n,\;f(a_j)+Kd_n].$ This rectangle has area $2K(d_n)^2.$ There are $n$ such rectangles, so the content of the graph of $f$ on $[p,q]$ cannot exceed $$n\cdot 2K(d_n)^2 =2K(q-p)^2/n.$$ Since $n$ can be any natural number, the content of the graph of $f$ on $[p,q]$ is zero.
This is not as general as the comments and answer of Paramanand Singh but suffices for the Q.
Remarks: I hope I got "the graph of..." in all the right places. To a set-theorist a function $is$ its graph.... In the Lemma it would also suffice that $f$ is Lipschitz-continuous on $[p,q]$ with Lipschitz constant $K$.
The circumference of an ellipse is approximatelly $B=\pi[3(a+b)-\sqrt{(3a+b)(a+3b)}]$(Ramanujan)
We can say that the circumference is $O(B)$
Now we can find rectangles with sides $1/k$ and area $1/k^2$ which can cover $C$
We need $O(kB)$ of them thus the whole $Vol(C)=O(\frac{B}{k})$ | 2021-12-03T23:42:59 | {
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https://math.stackexchange.com/questions/1913657/how-can-constant-functions-have-limits/1913680 | # How can constant functions have limits?
I'm pretty sure there is either something fundamental missing in my understanding of limits, that or I'm completely off mark. Regardless, please help solidify my understanding of limits.
As far as I know, a limit is some value a function, such as f(x), approaches as x gets arbitrarily close to c from either side of the latter.
If this is the case, how can constant functions, such as y=3, have limits? I know the limit of y=3 would be 3 (regardless of what x approaches), but the thing is, y will never approach 5, because it already is 5! You can't get closer to a chair when you are already sitting on it!
• As you yourself wrote in the first part of the question: the limit has to be taken with respect to $x$. So it makes no sense to say "as $y$ approaches". – b00n heT Sep 3 '16 at 22:37
• The definition of limit formalises a sense of arbitrarily close to but not arbitrarily close to though not equal to – Henry Sep 3 '16 at 22:39
• No variable or function really approaches whatever in analysis: it is only a metaphoric phrase. – Bernard Sep 3 '16 at 22:43
• The definition says it has be get arbitrarily close. It never says at any point it has to be away. – fleablood Sep 3 '16 at 22:53
• Getting closer to the chair doesn't matter. Being arbitrarily close to the chair is what's important. And being in the chair is indeed very very close. – fleablood Sep 4 '16 at 0:01
If you do not like the word "approach" let us say the limit predicts where $f(x)$ will end up as $x \to c$.
If $f(x)$ sits there all along on its chair at $5$, surely it makes sense to predict it'll stay there.
• smacks forehead so the limit just predicts where f(c) should be. – user366028 Sep 3 '16 at 22:54
• Yes, informally you can think of it in this way. You check $f(x)$ closer and closer to $c$ and based on this you try to infer/predict what $f(c)$ should be. And if $f$ is continuous then the prediction is actually correct. – quid Sep 3 '16 at 22:58
• Very good and smart use of language to convey the right idea about limits +1. – Paramanand Singh Sep 4 '16 at 5:13
The definition says it can get arbitrarily close, but it never says it has to ever be any distance away.
The formal definition: for any $\epsilon > 0$ we can find a $\delta$ so that $|x - x_0| < \delta \implies |f(x) - f(x_0)|< \epsilon$.
This definition certainly holds. Let $\delta = anything$ then $|x - x_0| = anything$, then we have $|f(x) - f(x_0)| = 0 < \epsilon$ for any epsilon.
Yep. That's a limit all right.
• To be honest, I haven't learnt the formal definition of limits yet (I work on that now). But – user366028 Sep 3 '16 at 23:08
• I think my problem was that I intuitively thought limits were only to be used to describe the answer of functions when imputed with with a value that would make them undetermined and undefined. Thanks for the clarification. – user366028 Sep 3 '16 at 23:19
The definition of limit never actually says you can't hit the value you're approaching. We say that $\lim_{x \to x_0} f(x) = L$ if $$\forall \epsilon > 0 \ \exists \delta > 0 \ \big[ 0 < |x - x_0| < \delta \implies | f(x) - L | < \epsilon \big]$$
Of course, you could define your own operation, similar to a limit, where the consequent is $0 < | f(x) - L | < \epsilon$ instead. Let's call it a pseudo-limit. It turns out that many of the limit laws that we take for granted fail when you use pseudo-limits.
For example, we expect the sum law to be true (assuming the left side exists): $$\lim_{x \to x_0} f(x) + \lim_{x \to x_0} g(x) = \lim_{x \to x_0} f(x) + g(x)$$
But for pseudo-limits, this can fail. Let $f(x) = \frac{\sin x}{x}$ and $g(x) = 1 - \frac{\sin x}{x}$. We know that as $x \to 0$, we have $f(x) \to 1$ and $g(x) \to 0$. We'd expect the function $(f + g)(x)$ to approach $1$. But $(f + g)(x) = 1$, and so it's a constant function, and has no pseudo-limit.
You can construct similar counterexamples for the other limit laws.
I don't think the confusion here is about limits. I think it's a question about functions. $f(x)$ is not a function. $f: \mathbb R \to \mathbb R$ is a function. What it does, is assigns every real number something in $\mathbb{R}$, uniquely.
For example, your constant function is a function $f: \mathbb{R} \to \mathbb{R}$ such that $f(x)=5$ for all $x \in \mathbb{R}$. In other words, if you give me a real number, say $5$, I will return $3$, since $f(5)=3$.
THE limit of a function does not have any coherence. There are many limits one could take. For example, some limits of the constant function: $\lim_{x \to 0}f(x)=3=\lim_{x \to 1} f(x)=3= \lim_{x \to 2} f(x)=3$ and so on. In other words, the limit here is trivial. Take $\lim_{x \to 0}f(x)$. We want to $f$ to get sufficiently close to $3$. In our case, we need not look far, as for all $x \in \mathbb R$, $f(x)=3$, and so $|f(x)-3|=0$.
As for the last part of your question, limits are indeed unique.
"proof" suppose that $\lim_{x \to y} f(x)=a$, but also that $\lim_{x \to y} f(x)=b$ for $a \neq b$. Then, on what interval can we ensure that $f(x)$ converges? The problem here is that what this limit tells us is that we can get arbitrarily close to different values, but this can't be the case! Take the distance $\frac{a-b}{2}$. Here, even as $x \to y$, $f(x)$ can't get close enough to one of them.
To perhaps aid in intuition, take $g: \mathbb R \to \mathbb R$ defined by $g(x)=\frac{1}{x}$. perhaps you've noticed the asymptote, or the proof that $\lim_{x \to \infty} \frac{1}{x}=0$. Could it be the case that $\lim_{x \to \infty} \frac{1}{x}=-1$? Or any negative number for that matter? Then there would always be the distance between that number and the $x-axis$, a distance that $\frac{1}{x}$ can never satisfy.
I'll leave the $\epsilon-\delta$ to another answer, I just wanted to maybe help untangling this shady business in analysis, without deferring to rigour. | 2019-06-19T15:26:20 | {
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http://math.stackexchange.com/questions/54235/rules-for-factorisation | Rules for Factorisation?
Basically presented with this, simplify
\begin{aligned} {\Bigl(\sqrt{x^2 + 2x + 1}\Big) + \Bigl(\sqrt{x^2 - 2x + 1}\Big)} \end{aligned}
Possible factorisations into both
\begin{aligned} {\Bigl({x + 1}\Big)^2}, {\Bigl({x - 1}\Big)^2} \end{aligned}
\begin{aligned} {\Bigl({1 + x}\Big)^2} , {\Bigl({1 - x}\Big)^2} \end{aligned}
Hence when simplified, answer has two possibilities. One independent of x, and the other not.
( Simplified Answers: 2x, 2 )
Why is one independent and the other not? If such is equal to 2, why then when, say x=2, the answer does not simplify to 2?
-
Take a look at the graph of the function. – anon Jul 28 '11 at 10:23
I would like to note that in general there are no "rules for factorization"; there's no algorithm/recipe that you can follow that guarantees that you will definitely obtain a "nice" factorization (among other things, because what counts as 'nice' depends on what you want it for). Rather, what you have are heuristics: general approaches that may or may not work. – Arturo Magidin Jul 28 '11 at 16:50
$-2x$ is also a possibility... – Aryabhata Jul 28 '11 at 23:33
$x^2+2x+1$ has two square roots, $x+1$ and $-(x+1)$. Similarly, $x^2-2x+1$ has two square roots, $x-1$ and $-(x-1)=1-x$. If you combine the first choice for each, you get $(x+1)+(x-1)=2x$; if you combine the first choice for the first term with the second choice for the second term, you get $(x+1)+(1-x)=2$. The remaining two combinations yield two more results: $-(x+1)+(x-1)=-2$, and $-(x+1)+(1-x)=-2x$. However, none of this is correct, because by convention $\sqrt{y}$ always denotes the non-negative square root of $y$. Thus, $\sqrt{x^2+2x+1}$ is actually $|x+1|$, and $\sqrt{x^2-2x+1}=|x-1|$, so that the correct simplification is $$|x+1|+|x-1|.$$ If you want to get rid of the absolute values, you’ll have to break the real line into pieces and use a multi-part definition of the function. And if you do this, you’ll see how $2$, $2x$, etc. actually come into the picture. (Literally: a graph should prove quite informative.)
-
You can write it as
$$\sqrt{x^2 + 2x + 1} + \sqrt{x^2 - 2x + 1}=\sqrt{(x+1)^2} + \sqrt{(x-1)^2}$$
But note that this simplifies to
$$|x+1| + |x-1|$$
as squaring and taking the root eliminates the sign. So neither of your answer is correct (for a real $x$). Maybe you can figure out the intervals where something interesting happens on your own?
-
We view the problem geometrically, or equivalently in terms of motion.
Suppose that we are travelling on the $x$-axis, and are now at the point $(x,0)$. Then $\sqrt{(x-1)^2}$ is our distance from the point $(1,0)$. Similarly, $\sqrt{(x+1)^2}=\sqrt{(x-(-1))^2}$ is our distance from the point $(-1,0)$.
It follows that $$\sqrt{(x+1)^2}+\sqrt{(x-1)^2}$$ is the sum of our distances from $(-1,0)$ and $(1,0)$.
Suppose that $x$ is anywhere between $-1$ and $1$. Then it is clear that the sum of our distances from $(-1,0)$ and $(1,0)$ is $2$. Any small motion to the right decreases our distance from $1$, but increases our distance from $-1$ by the same amount.
Suppose that $x>1$. Then our distance from $(1,0)$ is $x-1$. Our distance from $(-1,0)$ is $2$ more than that, so it is $x+1$. The sum is $2x$. The sum of the distances is increasing at twice the rate that the distance from $(1,0)$ is increasing.
Suppose finally that $x<-1$. By symmetry, the sum of the distances is the same as the sum for $|x|$. By the preceding paragraph, this sum is $2|x|$, which could also be written as $-2x$.
Generalization: We could make a "word problem" which is answered by the above calculation. Adam lives at $(-1,0)$ and Beth lives at $(1,0)$. They want to meet at the point $(x,0)$. What is the sum of the distances they must travel?
The problem can be generalized. Suppose that we have $n$ people, who live respectively at $(a_1,0)$, $(a_2,0)$, $\dots$, $(a_n,0)$, where $a_1\le a_2\le\cdots\le a_n$. What is the sum of their distances from the point $(x,0)$? The analysis is not much more complicated than for $2$ people.
-
We know $\sqrt{x^2} = |x|$, so $$\sqrt{x^2 + 2x + 1} + \sqrt{x^2 - 2x + 1} = |x + 1| + |x-1|.$$
- | 2014-08-21T12:51:05 | {
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https://www.physicsforums.com/threads/probability-of-pairings-in-chess-game.199028/ | # Probability of Pairings in Chess Game
1. Nov 18, 2007
### e(ho0n3
[SOLVED] Probability of Pairings in Chess Game
1. The problem statement, all variables and given/known data
The chess clubs of two schools consist of, respectively, 8 and 9 players. Four members from each club are randomly chosen to participate in a contest between the two schools. The chosen players from one team are then randomly paired with those from the other team, and each pairing plays a game of chess. Suppose that Rebecca and her sister Elise are on the chess clubs at different schools. What is the probability that
(a) Rebecca and Elise will be paired;
(b) Rebecca and Elise will be chosen to represent their schools but will not play each other;
(c) exactly one of Rebecca and Elise will be chosen to represent her school?
2. Relevant equations
Axioms and basic theorems of probabilitiy.
3. The attempt at a solution
(a) Assuming each possible team is equally likely and each possible pairing is equally likely, the the probability sought is the ratio of the number of possible teams and pairing in which Rebacca and Elise are paired to the total possible number of teams and pairings which should be
$$\frac{\binom{7}{3}\binom{8}{3}3!}{\binom{8}{4}\binom{9}{4}4!} = \frac{1}{14}$$
The book says the answer is 1/18 however.
(b) That should be
$$\frac{\binom{7}{3}\binom{8}{3}3 \cdot 3!}{\binom{8}{4}\binom{9}{4}4!} = \frac{3}{14}$$
The book says the answer is 3/18 however.
(c) That should be
$$\frac{\binom{7}{3}\binom{8}{4}4! + \binom{7}{4}\binom{8}{3}4!}{\binom{8}{4}\binom{9}{4}4!} = \frac{1}{2}$$
The book has the same answer.
My gripe is with (a) and (b). Is the book right or am I right?
2. Nov 18, 2007
### Dick
The book is right. You may be overcomplicating things. The probability R is chosen is 4/8, that E is chosen is 4/9. Once chosen, the odds they will play each other is 1/4. What's the probability of all of those things happening? Change it for the case of 'not play'.
3. Nov 18, 2007
### Dick
BTW. Your counting expressions are also correct. Just the final numbers are wrong.
4. Nov 18, 2007
### e(ho0n3
Yes, I do tend to overcomplicate things. You seem to have a very good intuition for this kind of thing.
I realized where I made the stupid arithmetical mistake. Thanks. | 2017-01-23T17:19:51 | {
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http://mathhelpforum.com/differential-geometry/133793-cardinality-r-0-1-r-2-0-1-x-0-1-a.html | # Thread: Cardinality of R, [0,1], R^2, [0,1] x [0,1]
1. ## Cardinality of R, [0,1], R^2, [0,1] x [0,1]
Q1) Assuming that |R|=|[0,1]| is true, how can we rigorously prove that | $R^2$|=|[0,1] x [0,1]|? How to define the bijection? [Q1 is solved, please see Q2]
Q2) Prove that |[0,1] x [0,1]| ≤ |[0,1]|
Proof: Represent points in [0,1] x [0,1] as infinite decimals
x= $0.a_1a_2a_3...$
y= $0.b_1b_2b_3...$
Define f(x,y)= $0.a_1b_1a_2b_2a_3b_3...$
To avoid ambiguity, for any number that has two decimal representations, choose the one with a string of 9's.
f: [0,1] x [0,1] -> [0,1] is one-to-one, but not onto.
This one-to-one map proves that |[0,1] x [0,1]| ≤ |[0,1]|.
Now how can we formally prove that f is a one-to-one map (i.e. f(m)=f(n) => m=n)? All textbooks are avoiding this step, they just say it's obviously one-to-one, but this is exactly where I'm having trouble. How to prove formally?
Thanks a million!
[also under discussion in math links forum]
2. Originally Posted by kingwinner
Q1) Assuming that |R|=|[0,1]| is true, how can we rigorously prove that | $R^2$|=|[0,1] x [0,1]|? How to define the bijection?
Q2) Prove that |[0,1] x [0,1]| ≤ |[0,1]|
Proof: Represent points in [0,1] x [0,1] as infinite decimals
x= $0.a_1a_2a_3...$
y= $0.b_1b_2b_3...$
Define f(x,y)= $0.a_1b_1a_2b_2a_3b_3...$
To avoid ambiguity, for any number that has two decimal representations, choose the one with a string of 9's.
f: [0,1] x [0,1] -> [0,1] is one-to-one, but not onto.
This one-to-one map proves that |[0,1] x [0,1]| ≤ |[0,1]|.
Now how can we formally prove that f is a one-to-one map (i.e. f(m)=f(n) => m=n)? All textbooks are avoiding this step, they just say it's obviously one-to-one, but this is exactly where I'm having trouble. How to prove formally?
Thanks a million!
[also under discussion in math links forum]
Since $\mathbb{R}\simeq[0,1]$ there exists some $\varphi:\mathbb{R}\to[0,1]$ which is bijective. Define $\varphi\times\varphi:\mathbb{R}\times\mathbb{R}\to[0,1]\times[0,1]$ by $(x,y)\mapsto (\varphi(x),\varphi(y))$. This is readily verified to be a bijection.
3. Originally Posted by Drexel28
Since $\mathbb{R}\simeq[0,1]$ there exists some $\varphi:\mathbb{R}\to[0,1]$ which is bijective. Define $\varphi\times\varphi:\mathbb{R}\times\mathbb{R}\to[0,1]\times[0,1]$ by $(x,y)\mapsto (\varphi(x),\varphi(y))$. This is readily verified to be a bijection.
1) I checked that it is one-to-one, but how to prove that it is onto (i.e. image(φ x φ) = [0,1] x [0,1] ) ?
4. Originally Posted by kingwinner
1) I checked that it is one-to-one, but how to prove that it is onto (i.e. image(φ x φ) = [0,1] x [0,1] ) ?
Let $(x,y)\in [0,1]\times [0,1]$. By, $\varphi$'s surjecdtivity there exists some $x',y'\in\mathbb{R}$ such that $x'\overset{\varphi}{\mapsto}x$ and $y'\overset{\varphi}{\mapsto}y$. Clearly then, $(x',y')\overset{\varphi\times\varphi}{\longmapsto} (x,y)$
5. Thanks!
Can someone help me with Q2, please? It is an example I found on the internet, but I don't understand why the function f is one-to-one. How can we formally PROVE that f is a one-to-one map?
Any help is appreciated!
6. Originally Posted by kingwinner
Thanks!
Can someone help me with Q2, please? It is an example I found on the internet, but I don't understand why the function f is one-to-one. How can we formally PROVE that f is a one-to-one map?
Any help is appreciated!
I think you're talking about the interleaving process. Think about two infinite length tuples. $(a_1,b_1,a_2,b_2\cdots$ and $(a'_1,b'_1,a'_2,b'_2,\cdots$. This models exactly the idea behind the infinite decimal expansion. So, if these two tuples are equal then so are all of their coordinates. So, in fact $a_1=a'_1,b_1=b'_1,a_2=a'_2,b_2=b'_2,\cdots$ and considering that the above is $\varphi(x,y),\varphi(x',y')$ where $x=.a_1a_2a_3\cdots,y=.b_1b_2b_3\cdots,x'=.a'_1a'_2 a'_3\cdots,y'=.b'_1b'_2b'_3\cdots$. We see that $\varphi(x,y)=\varphi(x',y')\implies x=x',y=y'\implies (x,y)=(x',y')$
7. Originally Posted by Drexel28
I think you're talking about the interleaving process. Think about two infinite length tuples. $(a_1,b_1,a_2,b_2\cdots$ and $(a'_1,b'_1,a'_2,b'_2,\cdots$. This models exactly the idea behind the infinite decimal expansion. So, if these two tuples are equal then so are all of their coordinates. So, in fact $a_1=a'_1,b_1=b'_1,a_2=a'_2,b_2=b'_2,\cdots$ and considering that the above is $\varphi(x,y),\varphi(x',y')$ where $x=.a_1a_2a_3\cdots,y=.b_1b_2b_3\cdots,x'=.a'_1a'_2 a'_3\cdots,y'=.b'_1b'_2b'_3\cdots$. We see that $\varphi(x,y)=\varphi(x',y')\implies x=x',y=y'\implies (x,y)=(x',y')$
Thanks!
We can say this only because we have removed all ambiguities, so the decimal expansion of each number is unique, right??
Why is f NOT onto??
8. Originally Posted by kingwinner
Thanks!
We can say this only because we have removed all ambiguities, so the decimal expansion of each number is unique, right??
Why is f NOT onto??
I assume you mean because you have eliminated infinite nines?
Do you want the cool explanation or the lame one for why it's not onto?
9. Originally Posted by Drexel28
I assume you mean because you have eliminated infinite nines?
Do you want the cool explanation or the lame one for why it's not onto?
Yes, I mean in the case when there are two represenations of the same number (e.g. 0.999...=1.000...), we have made a choice of taking only one of them. To ensure that the function is one-to-one, we HAVE to make this choice, right?
As I think 0.a1b1a2b2... = 0.a1'b1'a2'b2'... => a1=a1', b1=b1', a2=a2', b2=b2',... is true only becuase we've eliminated all ambiguities as above.
As for the not onto part, I'm not sure what you meant, but I perfer the simpler explanation.
Why is the map f: [0,1] x [0,1] -> [0,1] not onto? Is there any element in [0,1] which is not the image of any element in [0,1] x [0,1]?
10. Originally Posted by kingwinner
Yes, I mean in the case when there are two represenations of the same number (e.g. 0.999...=1.000...), we have made a choice of taking only one of them. To ensure that the function is one-to-one, we HAVE to make this choice, right?
As I think 0.a1b1a2b2... = 0.a1'b1'a2'b2'... => a1=a1', b1=b1', a2=a2', b2=b2',... is true only becuase we've eliminated all ambiguities as above.
As for the not onto part, I'm not sure what you meant, but I perfer the simpler explanation.
Why is the map f: [0,1] x [0,1] -> [0,1] not onto? Is there any element in [0,1] which is not the image of any element in [0,1] x [0,1]?
Well, let me firs ask you this. What is $f(1)$?
11. Originally Posted by Drexel28
Well, let me firs ask you this. What is $f(1)$?
f: [0,1] x [0,1] -> [0,1]
f(. , .) should have two arguments, right? I don't think we can properly define f(1)??
12. f is NOT onto.
For example, 0.17070707... is not in the image of f.
It must come from (0.10000..., 0.7777...), but by our definition of f, 0.10000... is always represented as 0.0999..., (we have to remove all the ambiguities, for any number that has two decimal representations, we choose the one with a string of 9's. When we define f, we have to remove the ambiguities, otherwise, f won't be one-to-one, actually I think f wouldn't even be a function.)
So (0.10000..., 0.7777...) is always represented as (0.09999..., 0.7777...). There is no way we can get the element 0.17070707... in the image of f.
Is this why f is NOT onto?? | 2016-10-23T00:00:58 | {
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https://math.stackexchange.com/questions/2402952/systematically-finding-a-basis-of-a-subspace | # Systematically finding a basis of a subspace
Learning Linear Algebra on my own. Going through MIT Open Courseware lectures. There is a problem with solution and I understand the answer but would like to find a more systematic way to find it.
Find
1. a basis for the plane $x − 2y + 3z = 0$ in $R^3$
2. a basis for the intersection of that plane with the $xy$ plane
Solution
Part a
This plane is the nullspace of the matrix $\begin{bmatrix}1&-2& 3\end{bmatrix}$ and also $A=\begin{bmatrix}1 & −2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$
The special solutions to $Ax = 0$ are
$v_1 = \begin{bmatrix}2 \\ 1 \\ 0 \end{bmatrix}$ and $v_2 = \begin{bmatrix} -3\\ 0 \\ 1\end{bmatrix}$
These form a basis for the nullspace of $A$ and thus for the plane.
Part b
The intersection of this plane with the $xy$ plane contains $v_1$ and does not contain $v_2$; the intersection must be a line. Since $v_1$ lies on this line it also provides a basis for it.
I understand how both parts were solved, but my question is about part b. Is there a more systematic way to find a solution? Say I can't see immediately that $v_1$ is in the plane. How would I solve the problem?
I did try solving the following system $\left \{\begin{array}{l}x-2y+3z=0\\x+y=0\end{array}\right.$
Wrote it in matrix form
$\begin{bmatrix} 1 & -2 & 3 \\ 1 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
And then reduced it to the row reduced echelon form and got the following
$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{bmatrix}$
Which led me to believe that the answer should be $\begin{bmatrix} 1 \\ -1 \\ -1 \end{bmatrix}$
But clearly I'm wrong. It seems to me I'm missing some important concept. I would really appreciate any help on this.
You correctly applied the definition of the "$xy$-plane" in your first approach, but not in the second. A point $(x,y,z)$ is on the $xy$-plane if and only if $z = 0$. As such, our system of equations should have been $$\left \{\begin{array}{l}x-2y+3z=0\\z=0\end{array}\right.$$ Corresponding to the system of equations $Ax = 0$ with $$A = \pmatrix{1 & -2 & 3\\0 & 0 & 1}$$ (by the way, I really don't understand your apparent preference for square matrices). From there, find the row-echelon form and confirm that $v_1 = (2,1,0)^T$ is our only special solution, so this vector forms a basis of the desired nullspace.
• I would further remark that while it is good to ask the question of "how do I do this systematically", it is also good practice to embrace insights, such as that which led to your quicker answer. Having looked at the MIT course questions before and having taught out of Strang's text, I can say from experience that many questions are built to encourage that sort of alternative "non-algorithmic" perspective. – Omnomnomnom Aug 23 '17 at 0:57
Perhaps it is just because I learned Linear Algebra as being about vector spaces and learned matrices as a specific notation, but I have always disliked writing things in terms of matrices. Here is how I would do exercise (1): The equation x- 2y+ 3z= 0 can be written x= 2y- 3z so any point in that plane can be written (x, y, z)= (2y- 3z, y, z)= (2y, y, 0)+ (-3z, 0, z)= y(2, 1, 0)+ z(-3, 0, 1). So a basis for that plane is {(2, 1, 0), (-3, 0, 1)}.
Now, the intersection of that with x+ y= 0 or x= -y, allows us to write x- 2y+ 3z= 0 as -y- 2y+ 3z= 0 or -3y+ 3z= 0 or, finally, z= y. Then y(2, 1, 0)+ z(-3, 0, 1) becomes y(2, 1, 0)+ y(-3, 0, 1)= y(-1, 1, 1). That is, that one-dimensional space has {(-1, 1, 1)} as basis.
A basis for the plane $x − 2y + 3z = 0$ consists of two vectors perpendicular to $(1,-2,3).$ Thus we have
$$\{(0,3,2),(3,0,-1)\}, \:\: \{(0,3,2),(2,1,0)\}, \:\: \{(3,0,-1),(2,1,0)\}$$ as possible basis.
A basis for the intersection of planes $$\begin{cases}x-2y+3z&=0\\z&=0\end{cases}$$ consists of one vector satisfying both equations. Thus $$\{(2,1,0)\}$$ is a basis. | 2020-04-07T01:34:34 | {
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http://math.stackexchange.com/tags/general-topology/hot | Tag Info
8
He means that $X$ and $X'$ are two topological spaces with the same underlying set. $X$ carries the topology $\mathcal T$ and $X'$ the topology $\mathcal T'$. I would have formulated it in the following way: Let $\mathcal T$ and $\mathcal T'$ be topologies on a set $X$ and consider the function \begin{align*} i\colon (X,\mathcal T')&\to (X,\mathcal ...
8
As mentioned by John in the comments, in Hausdorff (T2) spaces, compact subsets are always closed. Suppose $X$ is Hausdorff, and $K \subseteq X$ is compact. To show that $K$ is closed, it suffices to show that $X \setminus K$ is open; that is, show that each $x \in X \setminus K$ has an open neighbourhood $U$ with $U \subseteq X \setminus K$. So given ...
7
The zero set of a polynomial $p\in \Bbb{C}[x,y]$ is unbounded (FTA), but the torus is compact, so that doesn't work. OTOH you can get many toruses as algebraic varieties in the projective space $\Bbb{C}P^2$. Look up Elliptic curves. These can also be described as sets of solutions of an equation of the form $$y^2=x^3+Ax+B$$ together with a point at ...
6
Look at the following diagram (created using http://Presheaf.com. I didn't know how to do the \Bbb-letters there) Since both paths simply identify $(\Bbb N×\Bbb Q×\{0\})\cup(\Bbb N×\{0\}×\Bbb Q)∪(\{0\}×\Bbb Q×\Bbb Q)$ to a point, we have induced set maps $s$ and $\tilde s$, where the later is a bijection. Since the lower left arrow is a quotient map, ...
5
Let $O : \mathbf{Haus}^\mathrm{op} \to \mathbf{Set}$ be the functor that sends a Hausdorff space $X$ to the set of open subsets of $X$. Then $O (1)$ has two elements, so any representation of $O$ must be a two-point space. But the only two-point Hausdorff space is discrete – which clearly won't work.
5
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5
The answer is no. For example, $f:[0,1)\rightarrow S^1, t\mapsto e^{2\pi i t}$ is continuous and bijective, however the inverse is not continuous (and this is why this works). Set $A=\{0\}$ and $B=[1/2,1)$. They are closed in $[0,1)$ and disjoint, but $f(A)=\{1\}$ and also $1\in \overline{f(B)}$, because $e^{2\pi i t}\rightarrow 1$ as $t\rightarrow 1-$. ...
4
Here you only need to care about the topology coming from the neighborhood basis. Let $B$ be a topological space. Assuming we can find local trivializations $$f_{U}:U_{E}\cong U\times \mathbb{R}^{n}$$ such that on $U_{E}$ this is given by product topology of $U\times \mathbb{R}^{n}$. Then you "glue" two such neighborhoods together using ...
4
The comb space works although any point on the 'base line' can be deformation retracted to so depending on the interpretation of your question, this might not quite fit your criteria. Instead, we can take a space which is morally the comb space but where we quotient out by the subset of points which can be deformation retracted onto. I believe (though I have ...
4
Let's try to show that $X \setminus S_n(U)$ is open (note that $n$ is fixed from now on): let $p$ be a point such that $p \notin S_n(U)$, so $B(p, \frac{1}{n})$ is not a subset of $U$. This means that for some $q$ with $d(p,q) < \frac{1}{n}$, we have that $q \notin U$. So $q$ witnesses that $p$ is outside $S_n(U)$. The idea is that a small enough open ...
3
Proposition Let $X$ be a topological space with the Bolzano Weiertrass property. Then every countable covering of $X$ admits a finite subcovering. Proof Let $\{O_1,O_2,\ldots\}$ be the countable open cover of $X$, so that $X\subseteq\bigcup O_n$. Suppose to the contrary that no finite collection of the cover covers $X$. Then in particular ...
3
The crucial fact here is that the fundamental group of a topological group must be abelian. As you point out in the comment, $\pi_{1}(M_{f})$ contains $\pi_{1}(X)$ as a subgroup and hence if $\pi_{1}(X)$ is not abelian, $\pi_{1}(M_{f})$ is neither and so $M_{f}$ cannot be a topological group. The only closed surfaces with abelian fundamental group are the ...
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The three conditions that must be satisfied by any metric are: $$d(x, y) \geq 0,\; \text{ and }\,d(x, y) = 0 \iff x = y$$ $$d(x, y) = d(y, x)$$ $$d(x, y) + d(y, z) \geq d(x, z), \quad \forall x, y, z \in \mathbb R$$ Your task is to show whether (or not) $d(x, y) = \arctan|x - y|$ satisfies each and every property above. Added: Is $\arctan|x - y|$ ...
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The product $X\times Y$ is a topological group containing both $X$ and $Y$ as subgroups (as $X\times\{1\}$ and $\{1\}\times Y$). More generally, if $B$ is a topological group and $f:X\to B$ and $g:Y\to B$ are continuous group homomorphisms, then we can construct the fibre product $$X_f\times_g Y = \{(x,y)\in X\times Y: f(x)=g(y)\}$$ which is a closed ...
3
Note that if $(X,d)$ is metric space, then $d'=d/(1+d)$ generates same topology of $(X,d)$. So we only prove this proposition: Let $(X_n,d_n)$ be a sequence of metric spaces, and $d_n(x,y)\le 1$ for all $n$ and $x,y\in X_n$, then $d((x_n),(y_n))=\sum_n 2^{-n} d_n(x_n,y_n)$ generates the product topology of $X=\prod_n X_n$. At first, we prove that for ...
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$\Rightarrow\quad$ by contradiction If $p\in S$ the result is clear. Now assume $p\not\in S$ and that all ball $B$ centred at $p$ doesn't contain any point of $S$ then $B^c$ is closed containing $S$ and not $p$ so their intersection doesn't contain $p$ as well. Contradiction. $\Leftarrow\quad$ by contraposition Assume $p$ isn't in the closure of $S$ then ...
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It's not at all clear to me that the two questions in Question 1 are asking the same thing, but in any case the answer to the first one is "definitely not". For example, think about a flat family of elliptic curves degenerating to a nodal rational curve. The general fibre is topologically a torus, so has $H_1 \simeq \mathbf Z \oplus \mathbf Z$, but the ...
2
Q1. No, it's not true that all fibers of a flat map are homotopy equivalent. For instance, the blow-up of $\mathbf P^1 \times \mathbf P^1$ at a point maps to $\mathbf P^1$. This map is flat and all fibers except for one are spheres; however, the remaining fiber is a wedge of two spheres. More generally, in a flat family of curves you can "collapse" a simple ...
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The definition of compactness means that for any open cover $\mathcal{U}$ of the space $X$ there are finitely many $U_1 , \ldots , U_n$ from that collection $\mathcal{U}$ which also covers $X$. When you "add $X$" to $\mathcal{U}$, you are changing the open cover into a different open cover, let's call it $\mathcal{U}^\prime$. While the collection $\{ X , ... 2 You can do it geometrically. Take a point in your set, and suppose it is in the first quadrant. Drop lines parallel to the axes to the line$x+y=1$, and mark half the distance to the intersection. Then the circle with that radius is entirely contained in the square, because the diagonal is strictly longer that the radius. 2 There are several ways to show it: Let the function$f: \mathbb R^2\to\mathbb R$, with $$f(x_1,x_2)=\lvert x_1\rvert+\lvert x_2\rvert.$$ Clearly$f$is continuous and as$(-\infty,1)$is open in$\mathbb R$, so is its inverse image: $$f^{-1}(-\infty,1)=\big\{(x_1,x_2)\in\mathbb R^2: \lvert x_1\rvert+\lvert x_2\rvert<1 \big\}.$$ 2 I would take a more pedestrian approach. Pick$x \in \overline{A}$. Let$V_i$be a neighbourhood of$[(x,i)]$in$Y$(where$[(x,i)]$denotes the equivalence class of$(x,i)$modulo$R$). By the continuity of the projection, there are open neighbourhoods$U_i$of$(x,i)$in$X\times \{0,1\}$that are mapped into$V_i$. Since$X\times\{i\}$is open in ... 2 The quotient also has to respect the group structure which yours does not. For instance$(0,\frac{1}{4})\sim (1,\frac{3}{4})$but$(0,\frac{1}{4})+(\frac{1}{2},0)=(\frac{1}{2},\frac{1}{4})$and$(1,\frac{3}{4})+(\frac{1}{2},0)=(\frac{1}{2},\frac{3}{4})\nsim(\frac{1}{2},\frac{1}{4})$. So you do not have a well defined group structure on the quotient. To ... 2 Identify$S^2 \cong \mathbb{C}P^1$. Then act on$S^2 \times \ldots \times S^2$by$S_n$. The quotient space is$\mathbb{C}P^n$and the projection map is the branched cover in question. In my case n=2 and the diagonal sphere is the the fixed set under the involution of$S_2$, so it is the branching locus. 2 Very broad hint: You need to prove three things with the hint to apply the theorem and show that$A \neq \emptyset$. Each of them is provable by induction, I'll let you write down the details.$\color{red}{\forall n, A_{n+1} \subset A_n}$:$f(X) \subset X \Rightarrow f(f(X)) \subset f(X) \Rightarrow f(f(f(X))) \subset f(f(X))$...$\color{red}{\forall n, ...
2
This isn't really an answer, but it's too long for a comment. In any case, I think the answer could depend on your formulation. I can see two ways to interpret your question. Suppose $X$ and $Y$ are topological groups, homeomorphic as topological spaces. Are they necessarily isomorphic as abstract groups? topological groups? Obviously 2 implies 1. I ...
2
I will restrict myself only to Lie groups, since the world of groups outside of this class is way too large and, I do not think, there is a good answer in this context. I will also restrict to simply-connected Lie groups so that the answer is reasonably neat. (This is not a very good reason, but this will keep my answer reasonably brief, I did not think ...
2
I'll stick with your question 2. By definition a fiber bundle $E\to X$ is locally homeomorphic to $U\times F$ where $U$ is open in $X$ and $F$ is the fiber. So in fact if $X$ is connected all the fibers of $E$ are homeomorphic. On the other hand if $X$ has distinct connected components the fibers over each one can differ arbitrarily, so that there's no ...
2
Let $(O_i)_{i\in I}$ be any open cover of $T$. Since $T$ is regular, one can find an open cover $(V_j)_{j\in J}$ such that, for each $j\in J$, the closure $\overline V_j$ of $V_j$ is contained in $O_{i(j)}$, for some $i(j)\in I$. Next, by assumption on $T$, one can find $j_1, \dots ,j_N\in J$ such that $V_{j_1}\cup\dots \cup V_{j_N}$ is dense in $T$. Then ...
2
Your proof is correct and in my opinion is the simplest way to prove this fact. A more geometric approach is given by Hausdorff-Kuratowski convergence. The compactness theorem for such convergence says in particular that any sequence of compact sets (i.e. your circles) which are all contained in the same compact sets (the bounding square) converge, up to a ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2014-03-08T07:37:02 | {
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https://math.stackexchange.com/questions/757059/evaluate-int-frac-sqrtx2-1x-mathrmdx/757095 | # Evaluate $\int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x$
My try, using $x = \sec(u)$ substitution:
$$\begin{eqnarray} \int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x &=& \int \frac{\sqrt{\sec^2(u) - 1}}{\sec(u)}\tan(u)\sec(u) \mathrm{d}u \\ &=& \int \tan^2(u) \mathrm{d}u \\ &=& \tan(u) - u + C \\ &=& \tan(arcsec(x)) - arcsec(x) + C \end{eqnarray}$$
However, according to Wolfram Alpha, the answer should be: $$\int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x = \sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C$$ When I derive this last answer I don't get back the integrand, but rather: $$\frac{\mathrm d}{\mathrm d x}\left(\sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C\right) = \frac{x}{\sqrt{x^2-1}}- \frac{x}{(x^2-1)^{3/2}\left(1+\frac{1}{x^2-1}\right)}$$
I don't know how to simplify this expression more. Also, I am unable to check whether my answer is correct because I don't know how to find the derivative of $arcsec(x)$.
Can someone check my calculations and tell me where I've done something wrong and how one can simplify the last expression to get back the integrand?
• I know it is not your main question, but here is a good way to find the derivatives of the inverse trig functions if you ever don't know them. If $y=\sec^{-1} x$ then $x=\sec y$, and by implicit differentiation we fine $1=\sec y\tan y\frac{dy}{dx}\implies\frac{dy}{dx}=\frac{1}{\sec y\tan y}$. $\sec\sec^{-1} x=|x|$ and $\tan\sec^{-1} x=\sqrt{x^2-1}$. Therefore, the derivative of $\sec^{-1} x$ is $\frac{1}{|x|\sqrt{x^2-1}}$. Now you can check your answer. It is probably correct too. – solstafir Apr 16 '14 at 23:20
• – user85798 Apr 16 '14 at 23:22
• @solstafir Why is $\sec(arcsec(x)) = |x|$ with the absolute value? – hallaplay835 Apr 17 '14 at 12:36
• Sorry for the confusion. In general, $\sec\sec^{-1} x$ is not always positive. You have to use the absolute value here because of the $\tan y$. This is because $\sec y = \frac{1}{\cos y}$ and $\tan y = \frac{\sin y}{\cos y}$. For $y \in [0,\pi]$ (the range of $\sec^{-1} x$), $\sin y$ is always positive, but $\cos y$ is negative for $x \in (\frac{\pi}{2},\pi]$. But $\sec y$ and $\tan y$ are being multiplied, so the negatives cancel and the product is always positive. But in $\tan\sec^{-1} x=\sqrt{x^2-1}$ we always take the positive square root. $\sec\sec^{-1} x=|x|$ corrects for this. – solstafir Apr 17 '14 at 17:25
• – Martin Sleziak Aug 17 '18 at 13:24
arcsec$(x)$ is the angle whose secant is $x$. I'm guessing you may be more familiar with the derivative of inverse cosine. The angle that has a secant of $x$ has a cosine of $\frac1x$.
Also, your answer is more or less the same, just in a different form. Draw a right triangle and call one of the acute angles $A$. Secant is $\frac{\text{hypoteneuse}}{\text{adjacent}}$. If the length of the hypoteneuse is $x$ and the length of the side adjacent to $A$ is $1$, then $\sec(A)=x$. By the Pythagorean Theorem, the other leg has length $\sqrt{x^2-1}$. From this triangle, we get $\tan(A)=\tan(\sec^{-1}x)=\sqrt{x^2-1}$.
• Thanks! I found this, a list of nice identities proved using similar constructions to yours. It may be useful to others. – hallaplay835 Apr 17 '14 at 12:50
$$\frac{x}{\sqrt{x^2-1}} - \frac{x}{(x^2-1)^{3/2} \left(1+\frac{1}{x^2-1}\right)} = \frac{x}{\sqrt{x^2-1}} - \frac{x}{\sqrt{x^2-1} (x^2-1+1)}$$ $$= \frac{x}{\sqrt{x^2-1}} - \frac{x}{x^2 \sqrt{x^2-1}} = \frac{x^3 - x}{x^2 \sqrt{x^2-1}} = \frac{x (x^2-1)}{x^2 \sqrt{x^2-1}} = \frac{\sqrt{x^2-1}}{x}$$
Calculating this integral can be done in this way:
$$\int { \sqrt{x^2-a^2} \over x } dx =\int { x^2-a^2 \over x\sqrt{x^2-a^2} } dx= \int { x \over \sqrt{x^2-a^2} } dx-\int { a^2 \over x\sqrt{x^2-a^2} } dx =\sqrt{x^2-a^2}(+-)a\cdot\int {(\frac{x}{a})'\over \sqrt{1-(\frac{x}{a})^2)} } dx=\sqrt{x^2-a^2}(+-)a\cdot\arcsin(\frac{x}{a})+C.$$
• Isn't $\int \frac{x^2-a^2}{x\sqrt{x^2-a^2}}\mathrm{d}x = \int \frac{x}{\sqrt{x^2-a^2}}\mathrm{d}x - \int \frac{a^2}{x\sqrt{x^2-a^2}}\mathrm{d}x$ ? Using $x = a\sec(u)$ substitution I get $\int \frac{x^2-a^2}{x}\mathrm{d}x = \sqrt{x^2-a^2} - aarcsec(\frac{x}{a}) + C$ – hallaplay835 Apr 17 '14 at 12:20
• OK! I corrected. Thank you! – medicu Apr 17 '14 at 13:06 | 2019-10-18T10:57:04 | {
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https://math.stackexchange.com/questions/923584/when-can-a-homomorphism-be-determined-entirely-by-its-generators/923880 | # When can a homomorphism be determined entirely by its generators
I read a text which says that:
Just because a homomorphism $ϕ :G → H$ is determined by the image of its generators does not mean that any such image will work.
e.g.: Suppose we try to define homomorphism $ϕ :Z_3 → Z_4$ by $ϕ(1)=1$ , then we get $ϕ(0)=ϕ(1+1+1)=3$ which isn't possible as $ϕ(0)=0$.
Does there exist some case in which a homomorphism is entirely determined by its generators?
• Say $G$ has generators that satisfy a set of relations (e.g., $g+g+g=0$ in your example). If $\phi$ sends the generators of $G$ to elements in the image in a way that satisfies all these relations, then you know that $\phi$ is a homomorphism. – angryavian Sep 8 '14 at 11:58
• Are you asking "when does every choice of generator produce a homomorphism"? This is different from your states question (but is more interesting!). – user1729 Sep 8 '14 at 16:04
• @user1729 Yes I had both doubts 1.)when does every choice of generator produce a homomorphism 2.) a homomorphism is entirely determined by its generators.your answer made 2.) clear. – spectraa Sep 8 '14 at 16:09
• So does my answer help or not? If not, I'll delete it. – user1729 Sep 8 '14 at 16:09
• The comment of @angryavian deserves greater emphasis, because it it answers one version of your question which may be what you actually want to know. Namely, if you have a presentation $\langle a_i \, | \, r_j \rangle$ of the group $G$, and if you are given values of $\phi(a_i) \in H$ for all generators $a_i$, then this extends to a homomorphism $\phi : G \to H$ if and only if $\phi(r_j)$ is the identity element of $H$ for each relator $r_j$. – Lee Mosher Sep 8 '14 at 16:39
A homomorphism is always determined by its generators, whether it is an isomorphism or not. To be explicit:
Q: does there exist some case in which a homomorphism is entirely determined by its generators?
A: Yes, every single possible case. A homomorphism is always defined by its generators.
All the example is saying is that you cannot just take some map of the generators and hope that it is a homomorphism.
Three more examples:
• Any map $\phi: \mathbb{Z}_n\rightarrow\mathbb{Z}$, $1\mapsto x$, for $x\neq 0$, is not a homomorphism as if it is then $n\cdot x=0$, a contradiction!
• Suppose $\gcd(n, m)=1$. Then $\phi: \mathbb{Z}_m\rightarrow\mathbb{Z}_n$, $1\mapsto x$, for $x\neq 0\pmod n$, is not a homomorphism because the image of the subgroup $\mathbb{Z_m}$ must have order dividing $m$ (why?) but all subgroups of $\mathbb{Z}_n$ have order dividing $n$. (This is a generalisation of the example given in your question.)
• Suppose $G$ is simple and $H$ contains no subgroup isomorphic to $G$. Then any map $\phi: G\rightarrow H$ where the generators are not mapped to the identity of $H$ is not a homomorphism. For example, the generators in a map $\phi: A_5\rightarrow \mathbb{Z}_n$ must be sent to the identity, otherwise the map is not a homomorphism.
Of course, you may be asking "when does every choice of generator produce a homomorphism". If that is so, this is not clear. But then you should read Martin's answer!
• I am unsure precisely what you are asking, but I think the restriction you are after is if your map is $\phi: G\rightarrow H$ then you want the generators to generate a subgroup of $H$ which is isomorphic to a homomorphism of $G$. In my first example, $\mathbb{Z}$ contains no elements of finite order but all homomorphic images of $\mathbb{Z}_n$ are finite cyclic. In the third example, $G$ has only itself and the trivial group as a homomorphic image, but $H$ does not contain a copy of $G$. – user1729 Sep 8 '14 at 16:29
• to be more clear I mean to say that suppose I've to construct a homomorphism $\phi :$$G$$\rightarrow$$H,to define a mapping on the generator then I'll have to check it whether it works for every element of group G.Wouldn't it be tedious.Can you help me by illustrating it through an example. – spectraa Sep 8 '14 at 16:32 • @WantTobeAbstract Ah, okay, you just need to check that the homomorphism works for the relators of a given presentation (with your given generators). You might find this answer of mine useful. – user1729 Sep 8 '14 at 19:34 • it was helpful. thanks once again. – spectraa Sep 9 '14 at 2:25 • I've just come across a homomorphism \phi : G \rightarrow G by \phi(g)=g^n .(in additive group we replace g^n by ng. Now while constructing this homomorphism what relations must have been satisfied.I feel there's just one that identity must map to identity element.Are there any other relations that must be satisfied also here ? – spectraa Sep 9 '14 at 4:26 I will expand my comment into an answer: Theorem: Given a group with presentation$$G = \langle \, g_i \bigm| r_j \, \rangle $$given another group H, and given a function f which assigns to each g_i a value f(g_i) \in H, the function f extends to a homomorphism F:G \to H if and only if each f(r_j) is the identity. Proof: Let \langle g_i \rangle denote the free group with free basis \{g_i\}. Let N be the smallest normal subgroup of \langle g_i \rangle containing \{r_j\}. By definition of presentation,$$G = \langle g_i \rangle \, / \, N$$Let$q : \langle g_i \rangle \to G$be the quotient homomorphism. By the universal property for free groups, the function$f$extends to a homomorphism$\widetilde F : \langle g_i \rangle \to H$. Let$K = \text{kernel}(\widetilde F)$. Then we have the following chain of equivalences:$\{r_j\} \subset K\iffN < K\iff$there is a homomorphism$F : G \to H$such that$F \circ q = \widetilde F\iff$there is a homomorphism$F : G \to H$that extends$f$. • +1. I think your answer is very good and completely answers the question in full generality. Is this found in any book? I would like to read up further on related topics. – yoyostein Apr 28 '18 at 9:23 • Thank you. I'm not sure what book to recommend, although the topic is covered in books on combinatorial group theory. One of the older books is Magnus/Karass/Solitar, although it sometimes does not adopt the cleanest abstract point of view. – Lee Mosher Apr 28 '18 at 15:03 • Just to clarify something. If$f$is only defined on each$g_i$, how do we know, a priori, what is$f(r_j)$? Thanks a lot. – yoyostein May 2 '18 at 15:47 • Each$r_j$is a word in the letters$g_i$, say$r_j = g_{i_1} ... g_{i_m}$, so$f(r_j) = f(g_{i_1}) ... f(g_{i_m})$. – Lee Mosher May 2 '18 at 20:22 By the fundamental theorem on homomorphisms,$\hom(\mathbb{Z}/n\mathbb{Z},G) \cong \{g \in G : g^n=1\}$, the$n$-torsion of$G$. In other words,$g^n=1$is the only relation which is required by the image$g$of the canonical generator of$\mathbb{Z}/n\mathbb{Z}$. One can easily derive from this$\hom(\mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z}) \cong \mathbb{Z}/\mathrm{ggT}(n,m)$. If$E$generates$G$, then$\hom(G,H) \to \mathrm{Map}(E,H)$is injective. This is what one means by "a homomorphism is determined by the images of the generators". It is surjective (for all$H$) if and only if$E$is a free generating set of$G$, i.e.$G= F(E)$is a free group. Only in this case, every choice of the images produces a homomorphism. In general, a group presentation exactly contains the information about the relations which are necessary for defining a homomorphism. For example,$G=\langle x,y : x^2 = y^5=1 , xyx^{-1} = y^2 \rangle$is the group with the property that homomorphisms$G \to H$correspond to elements$a,b \in H$(the images of$x,y$) such that$a^2=b^5=1$and$aba^{-1} = b^2\$.
• "ggT" or "größte gemeinsame Teiler" mean gcd in German. – Orat Apr 2 '18 at 1:48 | 2019-10-20T23:35:06 | {
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# If x, y, and z are integers greater than 1, and (3^27)(35^10
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If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink]
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If x, y, and z are integers greater than 1, and (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y), then what is the value of x?
(1) z is prime
(2) x is prime
Originally posted by jimjohn on 14 Dec 2007, 12:31.
Last edited by Bunuel on 31 Mar 2014, 00:23, edited 2 times in total.
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Re: If x, y, and z are integers greater than 1 and (3^27)(35^10) [#permalink]
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goodyear2013 wrote:
If x, y, and z are integers greater than 1, and (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y), then what is the value of x?
(1) z is prime
(2) x is prime
OE:
(3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y) Break up the 35^10 and simplify the 9^14
(3^27)(5^10)(7^10)(z) = (5^8)(7^10)(3^28)(x^y) Divide both sides by common terms 5^8, 7^10, 3^27
(5^2)(z) = (3)(x^y)
(1) SUFFICIENT: z must have a factor of 3 to balance the 3 on the right side of the equation.
(1) says that z is prime, so z cannot have another factor besides the 3. Therefore z = 3.
Since z = 3, the left side of the equation is 75, so x^y = 25.
The only integers greater than 1 that satisfy this equation are x = 5 and y = 2, so (1) is sufficient.
Put differently, the expression x^y must provide the two fives that we have on the left side of the equation.
The only way to get two fives if x and y are integers greater than 1 is if x = 5 and y = 2.
(2) SUFFICIENT: x must have a factor of 5 to balance out the 5^2 on the left side.
Since (2) says that x is prime, x cannot have any other factors, so x = 5.
Hi, can anyone explain how this works, please.
Split everything into prime factors:
$$(3^{27})(35^{10})(z) = (5^8)(7^{10})(9^{14})(x^y)$$
$$(3^{27})(5^{10})(7^{10})*(z) = (3^{28})(5^8)(7^{10})(x^y)$$
Now powers of prime factors on both sides of the equation should match since all variables are integers. If you have only $$3^{27}$$ on left hand side, it cannot be equal to the right hand side which has $$3^{28}$$. Prime factors cannot be created by multiplying other numbers together and hence you must have the same prime factors with the same powers on both sides of the equation.
Stmnt 1: z is prime
Note that you have $$3^{28}$$ on Right hand side but only $$3^{27}$$ on left hand side. This means z must have at least one 3. Since z is prime, z MUST be 3 only. You get
$$(3^{28})(5^{10})(7^{10}) = (3^{28})(5^8)(7^{10})(x^y)$$
Now $$5^2$$ is missing on the right hand side since we have $$5^{10}$$ on left hand side but only $$5^8$$ on right hand side. So $$x^y$$ must be $$5^2$$. x MUST be 5.
Sufficient.
Stmnt2: x is prime
If x is prime, it must be 5 since $$5^2$$ is missing on the right hand side. This would give us $$x^y = 5^2$$. Sufficient.
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14 Dec 2007, 13:58
4
jimjohn wrote:
oh sorry guys i didnt notice that the exponents didnt appear. plz note the edited question
In this case: D
5^2*z=3*x^y
1. z is prime and is 3. So, x=5 SUFF.
2. x is prime and is 5. So, x=5 SUFF.
but (327)*(3510)*(z) = (58)*(710)*(914)*(xy) is a top-level problem
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Re: DS - primes [#permalink]
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14 Dec 2007, 18:20
2
1
jimjohn wrote:
If x, y, and z are integers greater than 1, and (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y), then what is the value of x?
(1) z is prime
(2) x is prime
(3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y)
re-written as
(3^27)(5^10)(7^10)(z) = (5^8)(7^10)(3^28)(x^y)
simplified to
25z=3(x^y) x^y is an integer, multiple of 25 so z is a multiple of 3
I z prime, so 3; x=5
II x is prime, x^y multiple of 25 so x can only be 5
D
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14 Dec 2007, 19:29
ok so i understand up until 25 * z = 3 * (x^y)
now how do we know that z has to be a multiple of 3. is it because we know that 3 is a factor of 25 * z, and since 3 is not a factor of 25 then it has to be a factor of z.
is there such a rule like that, that if x is a factor of a*b, then x has to be a factor of one of a or b.
thx
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If x, y, and z are integers greater than 1 and (3^27)(35^10) [#permalink]
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20 Mar 2014, 13:44
If x, y, and z are integers greater than 1, and (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y), then what is the value of x?
(1) z is prime
(2) x is prime
OE:
(3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y) Break up the 35^10 and simplify the 9^14
(3^27)(5^10)(7^10)(z) = (5^8)(7^10)(3^28)(x^y) Divide both sides by common terms 5^8, 7^10, 3^27
(5^2)(z) = (3)(x^y)
(1) SUFFICIENT: z must have a factor of 3 to balance the 3 on the right side of the equation.
(1) says that z is prime, so z cannot have another factor besides the 3. Therefore z = 3.
Since z = 3, the left side of the equation is 75, so x^y = 25.
The only integers greater than 1 that satisfy this equation are x = 5 and y = 2, so (1) is sufficient.
Put differently, the expression x^y must provide the two fives that we have on the left side of the equation.
The only way to get two fives if x and y are integers greater than 1 is if x = 5 and y = 2.
(2) SUFFICIENT: x must have a factor of 5 to balance out the 5^2 on the left side.
Since (2) says that x is prime, x cannot have any other factors, so x = 5.
Hi, can anyone explain how this works, please.
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Re: If x, y, and z are integers greater than 1 and (3^27)(35^10) [#permalink]
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21 Mar 2014, 00:44
1
1
goodyear2013 wrote:
If x, y, and z are integers greater than 1, and (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y), then what is the value of x?
(1) z is prime
(2) x is prime
OE:
(3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y) Break up the 35^10 and simplify the 9^14
(3^27)(5^10)(7^10)(z) = (5^8)(7^10)(3^28)(x^y) Divide both sides by common terms 5^8, 7^10, 3^27
(5^2)(z) = (3)(x^y)
(1) SUFFICIENT: z must have a factor of 3 to balance the 3 on the right side of the equation.
(1) says that z is prime, so z cannot have another factor besides the 3. Therefore z = 3.
Since z = 3, the left side of the equation is 75, so x^y = 25.
The only integers greater than 1 that satisfy this equation are x = 5 and y = 2, so (1) is sufficient.
Put differently, the expression x^y must provide the two fives that we have on the left side of the equation.
The only way to get two fives if x and y are integers greater than 1 is if x = 5 and y = 2.
(2) SUFFICIENT: x must have a factor of 5 to balance out the 5^2 on the left side.
Since (2) says that x is prime, x cannot have any other factors, so x = 5.
Hi, can anyone explain how this works, please.
Similar question to practice: if-x-y-and-z-are-integers-greater-than-1-and-90644.html
Hope it helps.
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30 Mar 2014, 11:23
walker wrote:
jimjohn wrote:
1. z is prime and is 3. So, x=5 SUFF.
Aren't there 2 possible answers for #1?
x=5, y=2
x=25, y=1
Thanks!
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31 Mar 2014, 00:26
1
karimtajdin wrote:
walker wrote:
jimjohn wrote:
1. z is prime and is 3. So, x=5 SUFF.
Aren't there 2 possible answers for #1?
x=5, y=2
x=25, y=1
Thanks!
Notice that we are told that x, y, and z are integers greater than 1, hence x=25 and y=1 is not possible.
Check here for a complete solution: if-x-y-and-z-are-integers-greater-than-1-and-57122.html#p1346892
Similar question to practice: if-x-y-and-z-are-integers-greater-than-1-and-90644.html
Hope it helps.
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31 Mar 2014, 05:02
Bunuel wrote:
we are told that x, y, and z are integers greater than 1
Oh! Can't believe I missed that! It makes sense now . Thanks!
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Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink]
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01 Sep 2014, 16:11
Can anyone explain whether my approach is valid?
5^2*z = 3*x^y
(x^y)/z = (5^2)/3 = (5^2a)/(3a)
x^y = 5^2a
z = 3a
(1) z is prime, so a = 1 and x^y = 25 => x = 5
S
(2) x is prime, so a = 1 and z = 3
S
D
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Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink]
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09 Oct 2015, 13:31
1
VeritasPrepKarishma wrote:
Stmnt2: x is prime
If x is prime, it must be 5 since $$5^2$$ is missing on the right hand side. This would give us $$x^y = 5^2$$. Sufficient.
I just wanted to point out that $$x^y = 5^2$$ is not necessarily true. In fact, if the question asked for the value of y, then statement 2 would have been insufficient
$$(3^{27})(35^{10})*(z) = (5^8)(7^{10})(9^{14})(x^y)$$ can be rewritten as $$\frac{(5^2 *z)}{3}=x^y$$
Written in this form, it is easy to notice that z must be a multiple of 3 since $$x^y$$ is an integer. Since it is given that x is prime, the prime factorization of $$x^y$$ will be x repeated y times, which means z should have at most one 3. Statement 2 doesn't require z to be prime, it could have a prime factorization of one 3 with any number of 5's and x must be 5, but y could take many integer values besides 2.
For example:
z=3*$$5^2$$, x=5, y=4
z=3*$$5^3$$, x=5, y=5
z=3*$$5^4$$, x=5, y=6
x must be 5, so statement 2 is sufficient.
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Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink]
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09 Oct 2015, 14:30
jimjohn wrote:
If x, y, and z are integers greater than 1, and (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y), then what is the value of x?
(1) z is prime
(2) x is prime
I found an almost identical question with the same question stem, but different statements.
Here it is: if-x-y-and-z-are-integers-greater-than-1-and-90644.html
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Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink]
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07 Feb 2016, 10:19
There is one thing I don't understand about this problem and would appreciate any help.
When we simplify the equation and get 5^2 * (z) = 3 * (x^y), why do we even need any of the statements in the first place? Why isn't it directly clear that z is 3, x is 5 and y is 2? I really don't get it.
Thank you so much in advance.
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Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink]
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08 Apr 2016, 21:31
MrSobe17 wrote:
There is one thing I don't understand about this problem and would appreciate any help.
When we simplify the equation and get 5^2 * (z) = 3 * (x^y), why do we even need any of the statements in the first place? Why isn't it directly clear that z is 3, x is 5 and y is 2? I really don't get it.
Thank you so much in advance.
Jay
z can take any value in that case. Think of a case in which z = 12.
$$5^2 * 3 * 2^2 = 3 * x^y$$
Here x = 10
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# If x, y, and z are integers greater than 1, and (3^27)(35^10
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2019-01-22T00:15:32 | {
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https://math.stackexchange.com/questions/1756694/basis-that-contains-a-basis-for-a-subspace | # Basis that contains a basis for a subspace
I have this exercise and I want to know if my answer is correct. The exercise is:
Consider the linear space $\mathbb{R}^{2\times2}$ of $2\times2$ matrices with real entries. Consider $W$ contained in this space:
$$W = \{[a_{i,j}] \in \mathbb{R}^{2\times2}\mid a_{1,1} + a_{2,2} = 0\}$$
Calculate a basis of $\mathbb{R}^{2\times2}$ that contains a basis of $W$
So my doubt is in the question. Do they simply want a basis for $W$? Because that's easy
$\left\{\begin{pmatrix} 1 & 0 \\0 & -1\end{pmatrix}, \begin{pmatrix} 0 & 1 \\0 & 0\end{pmatrix}, \begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix}\right\}$
But my doubt is if this is really what they want because they refer a basis of $\mathbb{R}^{2\times2}$ and for that we should have an extra fourth matrix, right?
Can someone clarify this for me?
• No basis of $\mathbb{R}^{2\times2}$ can contain $W$, but it can contain a basis of $W$. – egreg Apr 24 '16 at 15:40
You seem to be missing some words:
Calculate a basis of $\mathbb{R}^{2\times2}$ that contains a basis of $W$
is the correct wording.
You have computed correctly a basis for $W$; now a general result is that
if $\{v_1,\dots,v_m\}$ is a linearly independent set in the vector space $V$ and $v\in V$ with $v\notin\operatorname{Span}\{v_1,\dots,v_m\}$, then the set $\{v_1,\dots,v_m,v\}$ is linearly independent.
So you can just find a matrix not in $W$, which is easy, and add it to the set you found. The resulting set has four elements and is linearly independent, so it is a basis for $\mathbb{R}^{2\times2}$ (because this space has dimension $4$).
More generally, the exchange lemma says that if you have a linearly independent set $\{v_1,\dots,v_m\}$ and a basis $\{w_1,\dots,w_n\}$ of a vector space, you can replace $m$ vectors in the basis with $v_1,\dots,v_m$ so that the resulting set is again a basis.
Use $\left(\begin{array}{cc}1&0\\0&0\end{array}\right)$ to complete a set of 4 linearly independent matrices in $\Bbb R^{2\times2}$. | 2020-01-25T23:44:52 | {
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https://www.physicsforums.com/threads/calculus-on-manifolds-by-spivak.669332/ | # Calculus on Manifolds by Spivak
• Analysis
• Total voters
14
## Main Question or Discussion Point
Code:
[LIST]
[*] Foreword
[*] Preface
[*] Functions on Euclidean Space
[LIST]
[*] Norm and Inner Product
[*] Subsets of Euclidean Space
[*] Functions and Continuity
[/LIST]
[*] Differentiation
[LIST]
[*] Basic Definitions
[*] Basic Theorems
[*] Partial Derivatives
[*] Derivatives
[*] Inverse Functions
[*] Implicit Functions
[*] Notation
[/LIST]
[*] Integration
[LIST]
[*] Basic Definitions
[*] Measure Zero and Content Zero
[*] Integrable Functions
[*] Fubini's Theorem
[*] Partitions of Unity
[*] Change of Variable
[/LIST]
[*] Integration on Chains
[LIST]
[*] Algebraic Preliminaries
[*] Fields and Forms
[*] Geometric Preliminaries
[*] The Fundamental Theorem of Calculus
[/LIST]
[*] Integration on Manifolds
[LIST]
[*] Manifolds
[*] Fields and Forms on Manifolds
[*] Stokes' Theorem on Manifolds
[*] The Volume Element
[*] The Classical Theorems
[/LIST]
[*] Bibliography
[*] Index
[/LIST]
Last edited by a moderator:
## Answers and Replies
Related Science and Math Textbooks News on Phys.org
mathwonk
Homework Helper
I learned calculus rigorously from Michael Spivak. First I graded a course from his beginning Calculus book, then a little later I taught a course from this book and read most of it carefully and worked as many problems as possible.
It is short and clear and covers only the most important concepts and theorems, differentiation, integration, partitions of unity, and then the inverse/implicit function theorem, change of variables and Fubini theorem for integration, and Stokes theorem. He also explains clearly the concept of differential forms, and integration of forms over chains for integration theory.
Knowing this stuff sets you apart from the herd who have studied only from ordinary calculus texts, and are always afterwards trying to master this material.
I agree with mathwonk. I didn't really understand much of what I did in multi-variable calculus (aside from surface-level stuff and knowing how to carry out change of variables, etc) until I read this book.
mathwonk
Homework Helper
the only topic not treated in spivak is differential equations, e.g. the solution of first order ordinary d.e.'s. After reading Spivak I walked into the Univ of Washington 2 hour Phd prelim exam on advanced calc and walked out with almost a perfect score after only 30 minutes. The one question I did not nail was on diff eq. I suggest Lang's Analysis I for that.
I used this book twice. The first time was as an undergrad around 1967 or 1968. The course ran for two quarters (approx. 24 weeks) and was taught by someone (a probability theorist, IIRC) who was learning the material for the first time. I recall that we didn't even get to Stokes' Theorem (which is the primary result). The second time was several years ago. After a long hiatus from math, I started studying on my own. After working though several other texts, I decided to tackle Spivak again. This time I made it all the way through the book. I found it rough going because of typos, editing problems and a few outright errors. In my opinion, the exercises are the best part of the book. They provide examples (and counter examples) for the main text.
I feel this pamphlet is a great supplement to any Mathematical Analysis courses, with focus on some introduction to differential geometry and a geometric insight, not a good textbook for self-studies, however, in my opinion due to its conciseness. The book at my hand is a 1960s edition, for a more recent work Mathematical Analysis by Zorich whose Volume II covers and extends beyond the materials in this book has referred to this book several times and successfully integrated seamlessly this marvellous "modern approach" to a first year undergraduate Analysis course.
mathwonk | 2020-01-24T05:24:02 | {
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https://math.stackexchange.com/questions/453700/eigenvalues-of-a-and-a-at | # Eigenvalues of $A$ and $A + A^T$
This question has popped up at me several times in my research in differential equations and other areas:
Let $A$ be a real $N \times N$ matrix. How are the eigenvalues of $A$ and $A + A^T$ related? Of course, if $A$ is symmetric, the answer is easy: they are the same up to factor or $2$, since then $A + A^T = 2A$. But if $A \ne A^T$?
I'm particularly interested in the question of the real parts of the eigenvalues. How are the real parts of the eigenvalues of $A$ related to the (necessarily) real eigenvalues of $A + A^T$?
Answers for complex matrices appreciated as well.
Any references, citings, or explanations at any level of detail will be appreciated.
• If $A$ is diagonalizable, then $\rho(A + A^T) \leq \rho(A) + \rho(A^T) = 2 \rho(A)$, where $\rho$ is the radius (equal to the matrix norm for $A$ diagonalizable). Jul 27 '13 at 21:58
• @Eric Auld: yes, of course; the challenging part comes when $A$ is not real-diagonalizable, so the real Jordan form has blocks like $\begin{bmatrix} a & b \\ -b & a \end{bmatrix}$ etc. Jul 27 '13 at 22:04
• I don't think there's much you can say about the eigenvalues of $A+A^T$ solely in terms of those of $A$. For example, consider $A = \left(\begin{array}{cc}\lambda&t\\0&\mu\end{array}\right)$ where $\lambda$ and $\mu$ are fixed, and $t$ is a real parameter. So the eigenvalues of $A$ are $\lambda$ and $\mu$. Using the quadratic formula you can check that the eigenvalues of $A+A^T$ are $\lambda+\mu \pm \sqrt{(\lambda-\mu)^2+t^2}$ which depend on $t$ and not just $\lambda$ and $\mu$. Jul 27 '13 at 22:27
• @RobertLewis In response to your comment , I was wondering in case when $A$ is not real diagonalizable, is it always possible to transform $A$ into a structure $\begin{bmatrix}a & b\\{-b} & a\end{bmatrix}$? In extension to that, should an even order matrix will always be transformed into a matrix with each diagonal block having structure $\begin{bmatrix}a & b\\{-b} & a\end{bmatrix}$. Jun 11 '18 at 16:13
• @jbgujgu: No, but a $2 \times 2$ real matrix will transform either to $\begin{bmatrix} a & b \\ -b & a \end{bmatrix}$ or $\begin{bmatrix} a & b \\ 0 & a \end{bmatrix}$. This is Jordan canonical form stuff. Cheers! Jun 11 '18 at 16:34
Here is Tao's comment from, now deleted, post 2:
"Note that if A is strictly upper triangular, then its eigenvalues are all zero, whereas $A+A^T$ is an arbitrary symmetric matrix with zero diagonal, which constrains the trace of the matrix but otherwise imposes almost no conditions on the spectrum whatsoever (the only other constraint I can see is that the matrix cannot be rank one). So, apart from the trace $tr(A+A^T)=2tr(A)$, there appears to be essentially no relationship."
• The closed post has been deleted. Dec 22 '14 at 19:02
• I'm acepting this answer since I have learned a lot from the references an it has been around long enough to be "ripe", as it were. But I do not think the last word has been said. I have been thinking of posting a seperate, reformulated version, but I am currently struggling with a backlog of about sixty answers-in-preparation, and they bug me. Anyway, thanks for the helpful remarks. Holiday Cheers to one and all! Dec 22 '14 at 23:15
• @Studiosus: and a separate word of gratitude to you, good sir! Cheers! Dec 22 '14 at 23:17
EDIT: Let $\lambda\in spectrum(A)$; then there is $\mu \in spectrum(A+A^T)$ s.t. $|\lambda-\mu|\leq ||A||_2$ (spectral norm).
Proof: Since $A+A^T$ is real symmetric, according to the Bauer–Fike Theorem, there is $\mu\in spectrum(A+A^T)$ s.t. $|\lambda-\mu|\leq ||-A^T||_2=||A||_2$. cf. http://en.wikipedia.org/wiki/Bauer%E2%80%93Fike_theorem
• Wow! Very nice; useful and interesting! Thanks so much! Yul Tov! As you can see from my pic on my user page, I am Santa's brother! I'll put in a good word, so watch your stocking! Salud! Dec 23 '14 at 0:23
A special case is when $A$ is a unitary matrix: eigenvalues of $A$ lie on unit circle in complex plane. If $e^{i\theta}$ is an eigenvalue of $A$, then $2 \cos(\theta)$ is an eigenvalue of $A + A^T$. Similarly, $2i\sin(\theta)$ is an eigenvalue of $A - A^T$. | 2021-09-26T01:31:18 | {
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http://math.stackexchange.com/questions/20441/factorial-and-exponential-dual-identities | # Factorial and exponential dual identities
There are two identities that have a seemingly dual correspondence:
$$e^x = \sum_{n\ge0} {x^n\over n!}$$
and
$$n! = \int_0^{\infty} {x^n\over e^x}\ dx.$$
Is there anything to this comparison? (I vaguely remember a generating function/integration correspondence)
Are there similar sum/integration pairs for other well-known (or not-so-well-known) functions?
-
Shouldn't it be $n! = \int_{0}^{\infty} x^{n}e^{-x}dx$? – mjqxxxx Feb 4 '11 at 18:27
oops...yes...added. – Mitch Feb 4 '11 at 18:29
Maybe some reading on the [Gamma Function][1] helps... [1]: en.wikipedia.org/wiki/Gamma_function "Gamma Function" – Miguel Feb 4 '11 at 19:44
Yes, the second identity is a non-standard display of $\Gamma(n-1)$. That integral is usually chosen to be the definition of 'best' analytic continuation of $n!$. But really, in the above, $n$ is always an integer so there's no complication there. – Mitch Feb 4 '11 at 21:10
I just found this: for a function $f(n)$, its ordinary generating function $g(x)$ and its exponential generating function $G(x)$, there is the relation $g(x) = \int_0^{\infty} e^{-u} G(x u) du$. It seems like it should be obvious but I just don't see the manipulation that connects this relation with the dual identities (it's obvious how it applies to the second identity) – Mitch Feb 7 '11 at 1:51
There is a close relationship between the two identities, but I don't know if the exact formal similarity is anything other than a neat coincidence along the lines of the Sophomore's dream (although I could of course be wrong about this). First note that the second identity can be written as
$$1 = \int_{0}^{\infty} e^{-x} \frac{x^n}{n!} \, dx$$
and therefore it is equivalent to the identity
$$\frac{1}{1 - t} = \sum_{n=0}^{\infty} t^n \int_0^{\infty} e^{-x} \frac{x^n}{n!} \, dx = \int_0^{\infty} e^{-x} e^{tx} \, dx.$$
which is an application of the first identity. (This new identity is easy to prove, since the integrand is just $e^{(t-1)x}$ so it has antiderivative $\frac{1}{t-1} e^{(t-1)x}$ and the identity follows from here.)
I know of interesting explanations of the two identities separately which are somewhat related, but not another direct connection like the one above: for the first see this math.SE question and for the second see this math.SE question.
-
Excellent...it took me a while to appreciate. Sometimes even the simplest manipulation can be inscrutable, like the integral equal to $\frac{1}{1-t}$. What this all does for me though is convince me in the simplest way possible that the second identity is just the best analytic continuation of the factorial function (just the simplest way that is). – Mitch Feb 27 '11 at 14:16
These two identities are the "opposite ends" (in a sense I will explain below) of $$\frac1{n!} \int_0^t \frac{x^n}{e^x} \,dx + \frac1{e^t} \sum_{k=0}^n \frac{t^k}{k!} = 1 \tag{1}$$ (Proof: By induction, integrating by parts. Just like the usual proof of your second identity, but with $\int_0^t$ instead of $\int_0^\infty$.) If $t\ge 0$, then both terms on the left of (1) are nonnegative, so (1) implies they are both in $[0,1]$. The first term on the left expresses the accuracy of the truncated integral as an approximation to $n!$; the second expresses the accuracy of the truncated Taylor series as an approximation to $e^t$. What (1) asserts is that these approximations are in a see-saw relationship: when one is good, the other is bad.
Your identities are the "opposite ends" of (1) in the following sense: for fixed $n$, as $t\to\infty$, the first term approaches $1$ and the second approaches $0$, yielding your second identity in the limit; for fixed $t$, as $n\to\infty$, the first term approaches $0$ and the second approaches $1$, yielding your first identity in the limit. That's a kind of duality, I suppose, though it leaves mysterious the apparent formal symmetry that prompted your question, where the factorial and exponential functions seem to trade places.
For other (sufficiently differentiable) functions, it is straightforward to generalize (1) to $$\int_0^t \frac{x^n f^{(n+1)}(-x)}{n!} \,dx + \sum_{k=0}^n \frac{t^k f^{(k)}(-t)}{k!} = f(0) \tag{2}$$ but again, this loses the nice formal symmetry that prompted your question.
(That symmetry is suspiciously imperfect, though, in that both identities have $x^n$ in the numerator, even though the roles of $x$ and $n$ are reversed in the two identities.)
-
Here's a generalization of the dual identity described by the OP. It requires $n \geq m$ for the integral.
$$(1) \hspace{1cm} x^m e^x = \sum_{n=0}^{\infty} n^{\underline{m}}\frac{x^n}{n!},$$
$$(2) \hspace{0.6cm} \frac{n!}{n^{\underline{m}}} = \int_0^{\infty} \frac{x^n}{x^m e^x} dx.$$
Equation (1) can be rewritten $$x^m e^x = \sum_{n=m}^{\infty} \frac{x^n}{(n-m)!} = \sum_{n=0}^{\infty} \frac{x^{n+m}}{n!},$$ which follows immediately from the sum in the OP's question.
Equation (2) is just $$(n-m)! = \int_0^{\infty} \frac{x^{n-m}}{e^x} dx,$$ which is also straightforward.
OP asked in the comments how I came up with this. It's a bit roundabout, actually. I was thinking about the question "Is there a definite integral for which the Riemann sum can be calculated but for which there is no closed-form antiderivative?" and wondering if one way to construct an answer would be to find an interesting sum that we know how to evaluate and that could also be expressed as a Riemann sum but for which the corresponding integral had no closed form. That eventually led me to the Wikipedia page on mathematical series, especially this section on series with factorial denominators, which has your infinite sum as the first example. I remembered your question (I had starred it as a favorite) and wondered if some of the other examples there could be used to construct answers to your question. The sum $\sum_{m=0}^{\infty} m x^m = xe^x$ worked, and I thought I might have a generalization of your identities, but then the sum $\sum_{m=0}^{\infty} m^2 x^m = (x+x^2) e^x$ did not. The page mentions that some of the examples there can be thought of as moments of a Poisson distribution. I then remembered that the factorial moments of a Poisson ($x$) distribution are much simpler than the usual moments; the factorial moments are just powers of $x$. I checked it, and that turned out to be the right generalization.
As to the meaning, I'm not sure. However, if you read George Lowther's answer or Benja's answer to the second question linked to in Qiaochu Yuan's answer to your question you see the Poisson distribution and its relationship to the gamma distribution figuring prominently. That might be the source of the relationship for the generalization as well as for your original pair of identities.
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Excellent! Thanks. How did you come to know that? As usual the algebra is almost trivial after the fact. Any hints as to the meaning import (like Qioachu's question in his 2nd link)? – Mitch Feb 27 '11 at 14:11
@Mitch: Thanks. I'm glad you found it useful. The answer to "How did you come to know that?" is too long for a comment, so I'll edit it into my answer. – Mike Spivey Feb 27 '11 at 22:04
Note that this duality is in fact a special case of Ramanujan's master theorem:
For some $x$ around the neighbourhood of $0$: $$F(x) = \sum_{n=0}^\infty f(n) \frac{(-x)^n}{n!}$$
Then,
$$\int_0^\infty x^{n} F(x)\,\text dx = n! f(-n-1)$$
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Nice! Can you relate this theorem to the ordinary and exponential generating function in my comments to the OP: $g(x)=\int_0^{\infty} e^{−u}G(xu) du$? Also, can you give any meaning/interpretation to the master theorem here? – Mitch Dec 28 '15 at 22:27 | 2016-05-04T10:10:11 | {
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https://mymathforum.com/threads/how-do-i-use-the-coefficient-of-restitution-in-a-collision-of-two-spheres.347618/ | # How do I use the coefficient of restitution in a collision of two spheres?
#### Chemist116
The problem is as follows:
Two bodies whose masses are $m_1=\,1\,kg$ and $m_2=2\,kg$ collision with speeds $v_1=1\,\frac{m}{s}$ and $v_2=\,4\,\frac{m}{s}$ as indicated in the figure from below. If the $COR$ (coefficient of restitution) is $0.2$. Find the speed in $\frac{m}{s}$ of the second body after the collision.
The alternatives given are:
$\begin{array}{ll} 1.&0.5\,\frac{m}{s}\\ 2.&2\,\frac{m}{s}\\ 3.&1\,\frac{m}{s}\\ 4.&1.5\,\frac{m}{s}\\ 5.&3\,\frac{m}{s}\\ \end{array}$
I'm lost at this problem. Can someone help me?. The only thing I can recall is that the coefficient of restitution or COR is a relationship between the kinetic energies as follows:
$COR=\frac{\textrm{K.E after the collision}}{\textrm{K.E before the collision}}$
But I don't know how to relate this to the problem in order to solve it. Can somebody help me here?.
#### topsquark
Math Team
How about conservation of momentum? What does that tell you about the velocities?
-Dan
#### Chemist116
How about conservation of momentum? What does that tell you about the velocities?
-Dan
I'm assuming that the momentum is preserved as follows:
$p_i=p_f$
Therefore
$m_1v_1+m_2v2=m_1u_1+m_2u_2$
I'm confused if should I sum the masses of the spheres or not. (Since in inelastic collisions one sticks to the other) or could it be that this is not a perfectly inelastic collision?. I'm confused at which is which?.
$u=\textrm{final speed}$
$v=\textrm{initial speed}$
Then:
$COR=\frac{u_1-u_2}{v_1-v_2}=0.2$
Therefore:
$m_1v_1+m_2v2=m_1u_1+m_2u_2$
Here is exactly where should I put the direction, to the left? to the right?. I'll just let as it is:
$1\times 1+2\times (-4)=u_1+2u_2$
$\frac{u_1-u_2}{v_1-v_2}=\frac{u_1-u_2}{1-(-4)}=0.2$
Then all that is left to do is to solve the system of equations:
$u_1-u_2=1$
$u_1+2u_2=-7$
$u_1=\frac{-5}{3}=-1.667$
$u_2=\frac{-8}{3}=-2.667$
But this is a problem since I'm having two negative velocities and none of these checks with the alternatives. Is my analysis wrong?.
#### DarnItJimImAnEngineer
If you wrote the problem down correctly, then I agree with your answer. Both objects would travel to the left with a 1 m/s relative velocity between them. Double-check the numbers in the problem.
topsquark | 2020-01-25T04:43:13 | {
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https://stats.stackexchange.com/questions/91024/are-the-marginal-distributions-of-a-multivariate-distribution-the-corresponding | # Are the marginal distributions of a multivariate distribution the corresponding univariate distributions?
Are the marginal distributions of a multivariate distribution necessarily the corresponding univariate distributions?
For example:
• Every marginal distribution of a multivariate normal distribution must be a univariate normal distribution.
• So it is for the multinomial distribution (which is the multivariate generalization of binomial distribution).
• So it is for Dirichlet distribution (which is the multivariate generalization of beta distribution)
• Not sure for multivariate t distribution.
Those are all the multivariate distributions I have seen so far. I am not sure for other multivariate distributions I haven't seen either. So I wonder if it is true for all the multivariate distributions?
This may involve a related question:
How is a univariate distribution generalized to multivariate generally?
• Every marginal distribution of a multivariate uniform distribution is a univariate uniform distribution. – Dilip Sarwate Mar 23 '14 at 14:48
• @DilipSarwate: Thanks! How do you understand that a uniform distribution within a 2-dim origin-centred circle doesn't have uniform marginal distribution? Or a multivariate distribution must be supported by a set like $[a,b]^K$? – Tim Mar 24 '14 at 0:15
• I meant the support of the joint pdf must be $[a,b]^K$ and the joint pdf has constant value on the support. – Dilip Sarwate Mar 24 '14 at 1:45
The notion of a multivariate-$<$*whatever*$>$ is not uniquely defined.
People assign the name of a univariate distribution to some corresponding multivariate distribution based on whatever features are most relevant/important to them. So there's not one multivariate gamma, but many (and there may well be more get the name in the future).
Usually one of the features people consider to be 'basic' to the multivariate form is retaining the distribution of the univariate as the distribution of the marginals, so it's nearly always the case that something called a multivariate-$<$*whatever*$>$ has univariate-$<$*whatever*$>$ margins. But there's no extant rule about how the multivariate distributions are named, it's just how people conventionally tend to anticipate the use of the term.
It's quite possible (and I imagine has likely happened several times) that some other features may be considered more basic on occasion, and quite likely there are a few cases where something called a multivariate-$<$*whatever*$>$ doesn't have univariate-$<$*whatever*$>$ margins - but it would be surprising enough that it would be expected that this unusual feature would be pointed out explicitly.
You can pretty safely assert that at the very least it would almost always be the case that it has univariate-$<$*whatever*$>$ margins if it's called a multivariate-$<$*whatever*$>$.
More specific answers would require a more specific question.
Edit to address the particular case of the multivariate-t
The term 'multivariate-t' is usually applied to the object obtained when a multivariate normal is divided by the square root of (an independent $\chi^2$ divided by its degrees of freedom), as described on the Wikipedia page.
Consequently, a margin will consist of the marginal normal divided by the square root of (an independent $\chi^2$ divided by its degrees of freedom). This implies that the univariate margins are t-distributed. [Indeed it implies that any $q-$dimensional subset of the variables should have a t-distributed margin.] | 2019-09-19T02:23:54 | {
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https://growincloud.co.uk/site/forum/page.php?page=gdp-deflator-formula-804d25 | which individual is out of the labor force? GDP per capita does not measure production that occurs outside of the market economy, Define the economic terms for all five areas on a business cycle, 1. Round answer to two decimal places, as needed. The GDP deflator formula can be used in a variety of ways. Consider the table of GDP and population for several imaginary countries. considering just the U.S., how are GDP and GNP related? GDP deflator (P t) is calculated by dividing nominal GDP by the real GDP: $$\text{P} _ \text{t}=\frac{\text{Nominal GDP}}{\text{Real GDP}}$$ GDP deflator is an important indicator of changes in prices of domestically produced goods. When the GDP Deflator is known, it can be used to calculate Real GDP from Nominal GDP: Real GDP equals Nominal GDP divided by … Identify what type of unemployment each of the individuals faces. Formula. Information concerning the fictitious small island nation of Llamalandia is given in the table. In what way(s) does GDP per capita not provide an accurate representation of living standards? what is the best term to describe their employment status? The GDP deflator inflation rate is worked out as follows: \text{GDP Deflator Inflation Rate}=\frac{\text{P} _ \text{t}-\text{P} _ {\text{t} … Not all of the terms will be used. Determine which economic feature is described by the statements. Gross domestic product deflator is a implicit price deflator which is used to measure the level of prices for all new products like domestically produced and final goods. https://www.khanacademy.org/.../v/example-calculating-real-gdp-with-a-deflator In the following example, 2010 is the base year. The GDP deflator formula can be used in a variety of ways. Step 2 – Find out the deflator which shall be provided by the government of that economy. The labor force contains 48 million people, 32 million people are employed, and 16 million are employed. Real GDP / GDP deflator 2012 $2300.00 130.00 nominal GDP in 2012=$ Nominal GDP / Real GDP 2010/ $700.00 /$100.00 2011/ $1700.00 /$1200.00 2010 GDP deflator = 2011 GDP deflator= Nominal GDP/ GDP deflator Robert is without a job, is able to work, but is not actively looking for work, Bonnie Ann Clyde has not had any work at all for 6 months despite the fact that she has been filling out at least 8 job applications each day, the overall population for Region A is 104 million people. If nominal GDP equals $600 billion and real GDP equals$500 billion, then the GDP Deflator equals 120. Step 1 – One needs to first calculate Nominal GDP either by using income method, expenditure method or production method. The GDP Deflator equals nominal GDP divided by real GDP times 100. have a market value, be a final good or service, be produced within a given country, and be produced within a certain time frame. What is the unemployment rate? The labor force is the sum of those that are employed plus the unemployed. GDP=GNP-(output of U.S. citizens living abroad)+(output if foreign nationals living in the U.S.), Determine whether each description refers to nominal gross domestic product (GDP) or real GDP. For a good or service to be counted in a given year's gross domestic product (GDP), what four following criteria must be met? GDP calculator measures the price changes by comparing the price of the products to those in previous years price. Who reports the official US unemployment rate? how frequently is the survey that determines unemployment released? Each of the explanations has been given by economists to explain why wages might be "sticky downward" please match each term with the appropriate explanation. the table shows employment stats for a fictional country. Round your answer to the nearest whole number. Round answer to two decimal places, as needed. The U.s GDP in 2010 was $14.526.5 Billion, and the previous year the GDP was$13,939.0 Billion. Use the info in the table to answer the related questions. Use it to answer the questions that follow. Use it to answer the questions that follow. A recession occurs when economic output is shrinking. Gross domestic product is abbreviated as GDP. Round your percentages to two decimal places. The table describes hypothetical employment statistics in the United States in a given year. To calculate the GDP price deflator formula, we need to know the nominal GDP and the real GDP. match the definition with the correct type of unemployment. Step 3 – Now divide the nominal GDP computed in step 1 by deflator gathered in step 2 to arrive at Real GDP. Real GDP is calculated by dividing nominal GDP by the GDP deflator. https://www.khanacademy.org/.../real-vs-nominal-gdp/v/gdp-deflator Then, every year we calculate the GDP deflator using the formula: GDP price deflator = Nominal GDP / Real GDP x 100. frictional unemployment plus structural unemployment, what is true if a nation is currently experiencing full employment, which of the choices is most directly related to cyclical unemployment. Therefore: 2010: 7,000 / 7,000 = 100.0; 2011: 8,350 / 7,500 = 111.3; 2012: 9,740 / 8,355 = 116.6 After you have determined the values of both Nominal GDP and Real GDP, use the two values in the GDP deflator formula, which is “Nominal GDP divided by Real GDP multiplied by 100.” 4 The resulting value will be the GDP deflator value. | 2021-08-03T03:03:38 | {
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http://math.stackexchange.com/questions/245762/continuous-functions-on-discrete-product-topology | # Continuous functions on discrete product topology
Let $A = \{a_1,\dots,a_m\}$ be a finite set endowed with a discrete topology and let $X = A^{\Bbb N}$ be the product topological space. I wonder which bounded functions $f:X\to\Bbb R$ are continuous on $X$.
For example, it is clear that if $f$ depends only on a finite number of coordinates then $f\in C(X)$, i.e. if there exists some finite $n$ such that $$f(x_1,\dots,x_n,x_{n+1},x_{n+2},\dots) = f(x_1,\dots,x_n,x'_{n+1},x'_{n+2},\dots) \quad \forall x_{n+1},x_{n+2},x'_{n+1},x'_{n+2},\dots$$ then $f\in C(X)$. Thus it seems that only dependents on infinitely many coordinates may violate continuity. I would be happy, if one could tell me whether there are some useful necessary/sufficient conditions to assure $f\in C(X)$.
In particular, if $B\subset A$ and $1_B(a)$ is the indicator (characteristic) function of $B$, does it hold that $$g(x):=\limsup\limits_{k\to\infty}1_B(x_k)$$ is a continuous function on $X$.
-
@tohecz: thanks, that was a typo - fixed now – S.D. Nov 28 '12 at 10:38
Assuming that $m>1$, the space $X$ is just a Cantor set: up to homeomorphism it’s the same space for all $m>1$, and you’re really asking which bounded, real-valued functions on the Cantor set are continuous. – Brian M. Scott Nov 28 '12 at 11:21
@BrianM.Scott: thanks for the comment - so does the characterization by toehz hold true? – S.D. Nov 29 '12 at 7:55
It’s correct, but I don’t yet see how to convert it into a nice, easily tested criterion. – Brian M. Scott Nov 29 '12 at 9:03
@BrianM.Scott: well, with focus on the indicator functions it yields some nice characterization that I've mentioned in my first comment to tohecz's answer below (hopefully, this comment is correct) – S.D. Nov 29 '12 at 12:32
Yes, there are, and probably numerous. The basic idea is the following:
### Theorem.
Let $[a_1\dotsm a_n]:=a_1\dotsm a_n A^{\mathbb N}$ denote the set called a cylinder. Then basically, $f$ is continuous if an only if $$\operatorname{diam} f\bigl([a_1\dotsm a_n]\bigr)\xrightarrow{n\to\infty}0 \quad\text{for all}\quad a_1a_2a_3\dotsm\in A^{\mathbb N},$$ where $\operatorname{diam}Y$ is the diameter of a set.
This property is crucial for all numeration systems. For example, for the decimal expansion one has $f(a_1a_2a_3\dotsm)=\frac{a_1}{10}+\frac{a_2}{100}+\frac{a_3}{1000}+\dotsb$ and diameter of image of cylinder of $[a_1\dotsm a_n]$ is exactly $10^{-n}\to0$, hence $f$ is continuous. The same is true for the binary system as well, generally for all radix representations. But it is true even for continued fractions when written as an infinite seqence of matrices $L=\begin{pmatrix}1&0\\1&1\end{pmatrix}$, $R=\begin{pmatrix}1&1\\0&1\end{pmatrix}$, and for many other systems as well.
### Proof.
On the set $A^{\mathbb N}$ we use the (sometimes called) Cantor topology given by a Cantor metric: $\rho(\mathbf{a},\mathbf{b})=2^{-k}$ where $k=\min\{j:a_j\neq b_j\}$ (bold letters indicate infinite words).
Continuity of $f$ means $$(\forall \mathbf x\in A^{\mathbb N})(\forall\varepsilon>0)(\exists\delta>0)(\forall \mathbf y\in A^{\mathbb N}, \rho(\mathbf x,\mathbf y)<\delta)(\lvert f(\mathbf y)-f(\mathbf x)\rvert<\epsilon)$$
Since the sets of such $\mathbf y$ are exactly some cylinders, this should be obviously equivalent to the theorem statement.
### Remark.
Notice that all the cylinders are clopen sets, which is very interesting itself.
-
Thanks a lot, that's a very nice characterization. It means for example, that whenever $f = 1_Y$ is an indicator function with $Y\subset X$, its image has diameter either $0$ or $1$. Thus, for it to be continuous, it is equivalent to the following of property: for any $x\in X$ the belonging of $x$ to $Y$ has to be decided only on a finite number of first coordinates. Awesome! Could I ask you on the way how to prove the characterization via the diameter of the image of a cylinder set you've provided? – S.D. Nov 28 '12 at 10:43
@S.D. It depends what exactly you mean by "product topology". I use the "Cantor distance" of two strings: Let $a_1a_2a_3\dotsm$, $b_1b_2b_3\dotsm$. Let $\ell:=\min\{j:a_j\neq b_j\}$. Then $\rho(\mathbf a,\mathbf b)=2^{-j}$. Tell me if this corresponds to your topology. If so, I'll provide the proof. – yo' Nov 28 '12 at 10:54
I use the standard definition as the smallest topology w.r.t. which projections are continuous, see e.g. here. I didn't do topology for some time, so I am not sure whether it is metrizable in case I am talking about. Maybe, it is. It would be nice, though, if you would have a pure topological proof for the fact that you've mentioned. – S.D. Nov 28 '12 at 11:00
@S.D.: You’re both talking about the same topology; tohecz’s cylinders form a base for the product topology. – Brian M. Scott Nov 28 '12 at 11:20
@tohecz: as Brian have mentioned it does correspond to my topology - would you please show the proof? – S.D. Nov 30 '12 at 9:34
I do not know whether there are general conditions, but I can answer your particular question: Note that the sequence $(y^n)$ with $y^n = (x_1, x_1, \ldots, x_1, x_2, \ldots)$ with $n$ $x_1$'s converges to $y = (x_1, x_1, \ldots)$ in $A^{\mathbb N}$. Now let $B = \{x_2\}$, we have for any $n$: $$g_B(y^n) = \limsup_{k\to\infty} 1_B(y^n_k) = 1$$ (as $y^n_k = x_2$ for $k> n$). But $$g_B(y) = \limsup_{k\to\infty} 1_B(x_1) = 0$$ So $y^n \to y$, but $g(y^n) \not\to g(y)$ hence $g$ isn't continuous.
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thanks a lot for the answer – S.D. Nov 29 '12 at 12:33 | 2015-08-03T07:11:35 | {
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https://mathematica.stackexchange.com/questions/221580/is-it-possible-to-expand-a-polynomial-written-as-a-product-with-symbolic-range | # Is it possible to expand a polynomial written as a product with symbolic range?
I have an expression of a polynomial here $$\prod_{k=1}^n{x+2(-1)^k\cos{\frac{k\pi}{2n+1}}}.$$ I'd like to expand the product, but the function Expand just give me the same expression. I wonder if there's a way to obtain the simplified and expanded form of the polynomial. Thanks.
• That looks like it could almost be a Chebyshev polynomial... – J. M.'s ennui May 10 '20 at 14:00
• What is the expected result? – Daniel Lichtblau May 10 '20 at 14:47
• Can you phrase the problem in the following manner please? 1. this is the code I used, 2. this was the output, 3. this is the output I want instead. Do include runnable code. – Szabolcs May 10 '20 at 15:03
• @J.M. Right on the money! In terms of a polynomial in $x$ the coefficients are found at Coefficients of first difference of Chebyshev S polynomials. – JimB May 10 '20 at 15:37
If you think there's a simplified solution and Mathematica (or Maple or whatever) doesn't give you a simplified solution, you just have to explore.
Expand the polynomial in $$x$$ for various values of $$n$$ and look for patterns. Mathematica and OEIS.org make this possible. Here is the expansion for $$n=8$$:
n = 8;
p = Product[x + 2 (-1)^k Cos[k π/(2 n + 1)], {k, 1, n}] // Expand // TrigReduce;
s = Table[{x^i, If[i == 0, FullSimplify[p /. x -> 0],
FullSimplify[Coefficient[p, x^i]]]}, {i, 0, n}];
p = #[[1]] #[[2]] & /@ s // Total
(* 1 + 4 x - 10 x^2 - 10 x^3 + 15 x^4 + 6 x^5 - 7 x^6 - x^7 + x^8 *)
Then look up 1,4,-10,-10,15,6,-7,-7,1 at OEIS.org to obtain a formula that will produce those coefficients.
Do that for various values of $$n$$ to be convincing.
• Once again, thanks for following through. :) I would invite the OP to follow the links to the OEIS, but as a spoiler: With[{n = 8}, CoefficientList[ChebyshevU[n, x/2] - ChebyshevU[n - 1, x/2], x]]. – J. M.'s ennui May 10 '20 at 15:51 | 2021-05-06T18:42:41 | {
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https://math.stackexchange.com/questions/1033694/finding-probability-in-the-case-of-a-biased-coin | # Finding probability in the case of a biased coin
A coin is biased so that the probability of falling head when tossed is $\frac14$. If the coin is tossed $5$ times, find the probability of obtaining $2$ heads and $3$ tails, with heads occurring in succession.
I know that each toss is an independent event. And for independent events, $$P(A\cap B\cap C\cdots)=P(A).P(B).P(C)\cdots$$ Going by that, the answer to this question must be $$\frac14.\frac14.\frac34.\frac34.\frac34=\frac{3^3}{4^5}$$ However, the given answer is $$\frac{3^3}{4^4}$$
Where am I going wrong?
Good outcomes are HHTTT, THHTT, TTHHT, and TTTHH. You computed only the first one so your answer is only $1/4$th of the final answer. Order is important here making those $4$ results different.
• Think of it like this, based on your biased coin, what if someone asked you what is the probability of getting $2$ heads followed by $3$ tails. It would be your original answer. Then suppose they changed the question to your original question, the answer would increase to $4$ times that because the pair of adjacent heads can occur in $4$ different places in the $5$ coin tosses. – David Nov 22 '14 at 15:01 | 2020-01-18T06:37:46 | {
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https://www.physicsforums.com/threads/purcell-7-2.647208/ | # Purcell 7.2
1. Oct 26, 2012
### NullSpace0
1. The problem statement, all variables and given/known data
A wire lies at height z=h carrying current parallel to the y-axis. A square loop, of side length b, is moving in the direction of +x with a velocity v. Find the magnitude of the emf at the moment when the center of the loop crosses the y-axis.
2. Relevant equations
3. The attempt at a solution
Would the fact that the B-field makes circle around the wires mean that flux through the loop is 0 at this moment? I need to find d(Flux)/dt, but that seems like it would be quite a complicated function! Is there a trick that keeps me from having to do that?
Otherwise, perhaps I would find the z-component of B-field, but that wouldn't involve any dependence on t... so how do I get the t into the equation so that the derivative of my expression for flux is not 0? (If there is no t in the expression, d(anything)/dt is 0...)
2. Oct 26, 2012
### TSny
Hello, NullSpace0. To be clear, the square loop lies in the xy plane according to Purcell's statement of the problem.
Yes. But, of course, that doesn't mean the emf is zero at that instant.
It's not too terrible to find the flux through the loop for arbitrary x (the x-coordinate of the center of the loop). You could then find the emf using the chain rule: $\frac{d\Phi}{dt} = \frac{d\Phi}{dx}\frac{dx}{dt}$
However, you can avoid this by using Purcell's results in section 7.3 ("A loop Moving Through a Nonuniform Magnetic Field"). In particular, can you see how Purcell's equation (6) might be applied to your problem?
#### Attached Files:
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Last edited: Oct 26, 2012
3. Oct 26, 2012
### schaefera
Perhaps the second method could be applied by considering the Lorentz force on each side of the loop, and then doing the line integral of force per unit charge around the loop at the time of interest?
4. Oct 26, 2012
### NullSpace0
Schaefera and TSny-- thank you for the advice!
If I do use that second method, would I only want to consider z-components of B-field, since the other component would then lie in the plane of the loop?
5. Oct 26, 2012
### TSny
Yes!
6. Oct 26, 2012
### NullSpace0
Aha! Working all of this out, I find that Bz=(2Ib)/(cr^2) but the sign is opposite for the sides that are pointing along the wire... I don't actually care about B for the sides of the loop perpendicular to the wire because for them, F and ds will end up being perpendicular and so they come out to 0 in the line integral of force.
Thus, I get F=(2Ivbq)/(c^2*r^2), and then integrate (1/q)F.ds, I get: emf=(4Ivb^2)/(c^2*(h^2+b^2)). Does this look correct?
A friend I'm checking answers with somehow got that the b^2 on the bottom is divided by 4 while there is only a factor of 2 multiplying the numerator... not sure where that would come from!
And if I do this by finding the flux, then taking that integral's time derivative (using the chain rule) and then dividing by c as Faraday's law tells me, I get MY solution to the emf above... but with only a factor of 2 in the numerator, not the 4 I have up there.
So I have 3 very similar but slightly different solutions...
Last edited: Oct 26, 2012
7. Oct 26, 2012
### TSny
I think your friend might be right. You have a triangle with r as hypotenuse and h as one side. What is the length of the other side?
8. Oct 27, 2012
### schaefera
That should fix it all.
9. Oct 27, 2012
### NullSpace0
Yay, thanks for the help! I needed to include b/2 as the numerator when taking the z-component of the B-field, which eliminates the 4 in the numerator of the final solution and adds the b^2/4 in the bottom.
So now I should also decide which way the induced current will flow. I think that, if we look down on the square from above, it will flow counter-clockwise... but this based on the fact that as the square moves beneath the wire, flux starts by increasing in the upward direction (+z) and just at the instant it is beneath the wire, net flux is 0 meaning that it is just about to start increasing in the down direction... induced B should point in the +z which requires a counter-clockwise I(ind) but the RHR. Is this reasoning sound?
Last edited: Oct 27, 2012
10. Oct 27, 2012
### TSny
I believe that's correct (assuming the current is in the +y direction). Good!
11. Oct 27, 2012
### NullSpace0
Thank you both so much! | 2018-03-20T03:00:07 | {
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https://math.stackexchange.com/questions/3321992/getting-different-answers-for-an-integral-frac12x-frac32-lnx2c | Getting different answers for an integral: $\frac{1}{2}x-\frac{3}{2}\ln{|x+2|}+C$ vs $\frac{1}{2}x-\frac{3}{2}\ln{|2x+4|}+C$
Problem:
$$\int\frac{1}{2}-\frac{3}{2x+4}dx$$
Using two different methods I am getting two different answers and have trouble finding why.
Method 1:
$$\int\frac{1}{2}-\frac{3}{2x+4}dx$$
$$\int\frac{1}{2}-\frac{3}{2}\left(\frac{1}{x+2}\right)dx$$ $$\int\frac{1}{2}dx-\frac{3}{2}\int\frac{1}{x+2}dx$$ $$\frac{1}{2}x-\frac{3}{2}\int\frac{1}{x+2}dx$$ $$x+2=u$$ $$dx=du$$ $$\frac{1}{2}x-\frac{3}{2}\int\frac{1}{u}du$$
$$\frac{1}{2}x-\frac{3}{2}\ln{|x+2|}+C$$
Method 2: $$\int\frac{1}{2}-\frac{3}{2x+4}dx$$
$$\int\frac{1}{2}-3\left(\frac{1}{2x+4}\right)dx$$ $$\int\frac{1}{2}dx-3\int\frac{1}{2x+4}dx$$ $$\frac{1}{2}x-3\int\frac{1}{2x+4}dx$$ $$2x+4=u$$ $$dx=\frac{du}{2}$$ $$\frac{1}{2}x-\frac{3}{2}\int\frac{1}{u}du$$
$$\frac{1}{2}x-\frac{3}{2}\ln{|2x+4|}+C$$
• Note that $\left| 2x+4 \right| = 2 \left| x+2 \right|$ and then use the properties of logarithms to see that the factor of 2 is essentially just a constant in addition. – Matti P. Aug 13 at 11:44
• As others have mentioned, your issue is resolved by a logarithm property. ... Often, the trickiest part of learning Calculus is remembering your Pre-Calculus. :) – Blue Aug 13 at 11:48
• This type of question, like find the difference between $\int (x+1 )dx=\frac{x^2}{2}+x+C$ and $\int (x+1)dx=\frac{(x+1)^2}{2}+C$, appears here so frequently, and the answer is always the same (there are no difference, only the constants differs). Can a canonical answer be made for it? – Zacky Aug 13 at 11:52
• Further to @Zacky's point, in some cases it's as simple as "they're clearly the same" (up to a constant or otherwise), but sometimes proving the equivalence requires e.g. a trigonometric identity the OP doesn't know how to prove. I'm not sure whether that means we should have multiple "canonical" versions or just give up on trying it altogether. – J.G. Aug 13 at 12:59
Both answers are the same, mind the constant: $$\ln|2x+4|+\color{blue}{C_1}=\ln|2\left(x+2\right)|+\color{blue}{C_1}=\ln|x+2|+ \underbrace{\ln 2 + \color{blue}{C_1}}_{\color{purple}{C_2}} = \ln|x+2|+\color{purple}{C_2}$$
\begin{align}\frac12 x + \frac 32 \ln |2x + 4| + C_1 &= \frac12x + \frac32\ln(2\cdot|x+2|) + C_1\\&=\frac12x + \frac32(\ln 2 + \ln|x+2|)+C_1\\&=\frac12x + \frac32 \ln|x+2| + (\frac 32 \ln 2 + C_1)\\&=\frac12x + \frac32\ln|x+2| + C_2\end{align}
$$\ln|2x+4|=\ln|2(x+2)|=\ln|2|+\ln|x+2|=\ln|x+2|+C$$
More generally for positive real $$a,b,d$$
$$\ln{\big|da+db\big|}=\ln{\big|d(a+b)\big|}=\ln{\big|d\big|}+\ln{\big|a+b\big|}=c_1+\ln{\big|a+b\big|}$$
where $$a=x,b=2,d=2$$ in your example. Then, if $$e=-\frac{3}{2}$$ and $$f=\frac{1}{2}$$,
$$fa+\ln{\big|da+db\big|}+c_2=fa+\ln{\big|a+b\big|}+c_1+c_2=fa+\ln{\big|a+b\big|}+c_3$$
so the difference is only by a constant. | 2019-08-26T02:38:43 | {
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https://mathhelpboards.com/threads/differences-in-the-solutions-of-de-system-and-its-converted-2nd-order-differential-equation.25308/ | # differences in the solutions of DE system and its converted 2nd order Differential equation
#### Dhamnekar Winod
##### Active member
Hello,
$\vec{x'}=\small\begin{pmatrix}1&2\\3&2\end{pmatrix}\vec{x}+t\small\begin{pmatrix}2\\-4\end{pmatrix}$
Now i got the solution to this differential equation system as
$\vec{x}(t)=c_1e^{-t}\small\begin{pmatrix}-1\\1\end{pmatrix}$+$c_2e^{4t}\small\begin{pmatrix}2\\3\end{pmatrix}$+$t\small\begin{pmatrix}3\\\frac{-5}{2}\end{pmatrix}$+$\small\begin{pmatrix}-2.75\\2.875\end{pmatrix}$
Now i converted this differential equation system into ordinary differential equation $y''-3y'-4y+12t-2=0$
I got solution to this DE as $y=C_1e^{-x}+C_2e^{4x}+3t-\frac12$.
Now my question why there is diference in these two solutions?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Hello,
$\vec{x'}=\small\begin{pmatrix}1&2\\3&2\end{pmatrix}\vec{x}+t\small\begin{pmatrix}2\\-4\end{pmatrix}$
Now i got the solution to this differential equation system as
$\vec{x}(t)=c_1e^{-t}\small\begin{pmatrix}-1\\1\end{pmatrix}$+$c_2e^{4t}\small\begin{pmatrix}2\\3\end{pmatrix}$+$t\small\begin{pmatrix}3\\\frac{-5}{2}\end{pmatrix}$+$\small\begin{pmatrix}-2.75\\2.875\end{pmatrix}$
Now i converted this differential equation system into ordinary differential equation $y''-3y'-4y+12t-2=0$
I got solution to this DE as $y=C_1e^{-x}+C_2e^{4x}+3t-\frac12$.
Now my question why there is diference in these two solutions?
Hi Dhamnekar Winod, welcome to MHB!
Your solution contains $x$ on the right hand side. I presume it should be $t$?
Either way, if I feed the latter equation to Wolfram|Alpha, I get:
$$y(t)=c_1e^{-t}+c_2e^{4t}+3t-\frac{11}{4}$$
So:
• It seems to be the solution for $x(t)$ rather than $y(t)$.
• Your equation is correct, although for $x(t)$ rather than $y(t)$.
• There is a minor arithmetic mistake somewhere in your solution.
#### Dhamnekar Winod
##### Active member
Hello, The derivative of $x_1=C_1e^{-t}+C_23e^{4t}-2.5t+2.875$in the differential equation system.But in the second case,it is different Why? | 2021-05-10T22:11:28 | {
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https://math.stackexchange.com/questions/1711403/existence-of-rational-roots-in-a-quadratic-equation | # Existence of rational roots in a quadratic equation
Consider the quadratic equation $(a+c-b)x^2 + 2cx + b+c-a = 0$ , where a,b,c are distinct real numbers and a+c-b is not equal to 0. Suppose that both the roots of the equation are rational . Then a) a,b, and c are rational
b)$c/(a-b)$ is rational
c)$b/(c-a)$ is rational
d)$a/(b-c)$ is rational
My attempt - I used the discriminant method to find out the possible roots of the given equation. Which for me came out to be $(-2c + a-b)/(2a -2b +2c)$ and $( -2c + b - a)/(2c + 2a - 2b )$
Hence according to me option a is the correct answer , while the correct answer which is given is b. Please tell me where am I going wrong , or what am I missing ?
• You did not write an equation, because I don't see the $=$. Anyway, if your equation is $(a+c-b)x^2 +2cx+b+c-a =0$, then your solutions are both wrong. Because the roots are $-1$ (this is easy to check) and $(a-b-c)/(a-b+c)$. Are you sure you copied the text of the problem correctly ? – Giovanni Resta Mar 24 '16 at 10:12
Notice that your polynomial $$(a+c-b)x^2 +2cx +(b+c-a) = (a+c-b) (x - \alpha)(x - \beta) =$$ $$=(a+c-b)x^2 - (a+c-b)(\alpha + \beta) x + (a+c-b)\alpha \beta$$ where $\alpha, \beta$ are the rational roots.
Now, the coefficient of $x$ must be equal $$2c = -(a+c-b)(\alpha + \beta)$$ and with some algebra you get $$\frac{c}{a-b} = -\frac{\alpha + \beta}{\alpha + \beta +2} \in \Bbb{Q}$$
• Notice that you have information only about $a-b$, without any information about $a,b$. I mean, if you call $a= \pi$, $b= 1+ \pi$, their difference is rational, but these are not rational. – Crostul Mar 24 '16 at 10:14
Note that, when $x=-1$, we get $$(a+c-b)-2c+(b+c-a)=0$$ so $-1$ is a root for every choice of the parameters $a$, $b$ and $c$. Even if $a+c-b=0$, that would give $c/(a-b)=-1$.
This leads into checking option b. So, suppose $a+c-b\ne0$. The factorization of the polynomial is $$(x+1)\bigl((a+c-b)x+c-a+b\bigr)$$ so the second root is $$\frac{(a-b)-c}{(a-b)+c}=\frac{1-\dfrac{c}{a-b}}{1+\dfrac{c}{a-b}}$$ If this is to be rational, also $c/(a-b)$ is rational.
We know that your two roots are rational: $$r_1 = \frac{-2c + (a - b)}{2(a - b + c)} \in \mathbb Q \qquad\text{and}\qquad r_2 = \frac{-2c - (a - b)}{2(a - b + c)} \in \mathbb Q$$ Hence, their sum and difference are also rational: $$r_1 + r_2 = \frac{c}{a - b + c} \in \mathbb Q \qquad\text{and}\qquad r_1 - r_2 = \frac{a - b}{a - b + c} \in \mathbb Q$$ But then their quotient is also rational: $$\frac{r_1 + r_2}{r_1 - r_2} = \frac{c}{a - b} \in \mathbb Q$$
The roots are
$$\frac{-2c\pm \sqrt{4c^2 - 4(c+(a-b))(c-(a-b))}}{2(a+c-b)}$$
$$=\frac{-c\pm (a-b)}{a+c-b}$$
$=\frac{a-b-c}{a-b+c}$ or $\frac{-a+b-c}{a-b+c}=-1$
For the first root,
$$=\frac{a-b+c-2c}{a-b+c}=1-\frac{2c}{a-b+c}$$
So $\frac{2c}{a-b+c}$ is rational, so its inverse is also rational
$\frac{a-b+c}{2c}=\frac{a-b}{2c}+\frac 1 2$
Reduce this to a monic form:
$$x^2+\frac{2c}{a+c-b}x+\frac{b+c-a}{a+c-b}$$
then as the roots are rational so are the sum and product of the roots, so both $\frac{2c}{a+c-b}$ and $\frac{b+c-a}{a+c-b}$ are rational and so $\frac{c}{b+c-a}$ is also rational, and you should be able to take it from there. | 2019-08-26T09:13:02 | {
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http://xwhw.chrangus-gaming.de/emathhelp-taylor-series-calculator.html | # Emathhelp Taylor Series Calculator
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This is a very versatile calculator that will output sequences and allow you to calculate the sum of a sequence between a starting item and an n-th term, as well as tell you the value of the n-th term of interest. What is the center of the power series? For what values of x will this representation be valid? You might want to check your answer graphically, if you have a graphing calculator or access to a Math software program. You can specify the order of the Taylor polynomial. Binomial Theorem Calculator Binomial Theorem Calculator This calculators lets you calculate __expansion__ (also: series) of a binomial. Maclaurin/Taylor Series: Approximate a Definite Integral to a Desired Accuracy. I will assume you mean the Taylor series about #0#, otherwise known as the Maclaurin series. Definition: A Taylor Series is a polynomial function with an infinite number of terms, expressed as an Infinite Series. This calculator will save you time, energy and frustration. 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Taylor series are extremely powerful tools for approximating functions that can be difficult to compute otherwise, as well as evaluating infinite sums and integrals by recognizing Taylor series. The Taylor/Maclaurin power series provide one answer - such series are probably close to the procedures used by most hardware and software scientific calculators. First of all, let's recall Taylor Polynomials for a function f. Solution 4 (b). Because this series is convergent for every complex value of x, it is commonly used to extend the definition of e x to the complex numbers. Power series are used for the approximation of many functions. It first prompts the user to enter the number of terms in the Taylor series and the value of x. You can also redefine the function by typing in the input bar - e. f(x)=sin(x. Maclaurin series coefficients, a k can be calculated using the formula (that comes from the definition of a Taylor series) where f is the given function, and in this case is sin( x ). Is the calculator just reading off of a list created from people who used rulers to physically measure the distance on a graph or is there a mathematical function that defines it? A calculator or computer program is not reading off of a list, but is using an algorithm that gives an approximate value for the sine of a given angle. In addition to that I have used long double as the number format to allow large factorials to be calculated and displayed. All calculations are done in double floating data type. Trig Formulas and Identities. Summary : The calculator makes it possible to obtain the logarithmic expansion of an expression. 100m Cici's Derby - part of the SCRunners Series Oct 27, 2019. Just as functions can be added, subtracted, multiplied, and composed, so can their corresponding Taylor series. What is the difference between Power series and Taylor series? 1. Taylor and Maclaurin (Power) Series Calculator. May 20, 2015 firstly we look. Socratic Meta Featured Answers Topics What is the Taylor series of #f(x)=arctan(x)#? Calculus Power Series Constructing a Taylor Series. For the finite sums series calculator. Welcome to MathPortal. This method has application in many engineering fields. equation-calculator. Answer to Use zero- through fourth-order Taylor series expansions to predict f (2. Complexe_solve online. You can specify the order of the Taylor polynomial. Taylor series is a way to representat a function as a sum of terms calculated based on the function's derivative values at a given point as shown on the image below. Comparasion Test: The terms of the sequence are compared to those of another one. The questions asks for only up to the 5th degree, so the three terms from f(x) are enough: If you want to practice Taylor Series and solve the entire FRQ, you can find the. MacLaurin series of Exponential function, The MacLaulin series (Taylor series at ) representation of a function is The derivatives of the exponential function and their values at are: Note that the derivative of is also and. Get the free "Series Calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle. The variables v can be a list or set of names or equations. It's just the way they calculate π and the way they calculate sin(). Series Calculator computes sum of a series over the given interval. The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. In our previous lesson, Taylor Series, we learned how to create a Taylor Polynomial (Taylor Series) using our center, which in turn, helps us to generate our radius and interval of convergence, derivatives, and factorials. Taylor series can be thought of as polynomials with an infinite number of terms. Although step-by-step solutions aren't supported at the time, you can still calculate the limit of any college-level function. In some cases, such as heat transfer, differential analysis results in an equation that fits the form of a Taylor series. Find the Taylor series expansion of any function around a point using this online calculator. Change the function definition 2. A Taylor series is a numerical method of representing a given function. Calculate various formulas and identities for sine, cosine, tangent, cotangent, secant and cosecant functions. Finite difference equations enable you to take derivatives of any order at any point using any given sufficiently-large selection of points. In this video, I show how to find the Taylor series expansion for a function, assuming that one exists! It is nothing too heavy: we just take derivatives and plug in the value at which we are centering the function. Learn more at Sigma Notation. Remember the blog post from a few months ago, How To “Lie” With Personal Finance? I got a fresh set of four new "lies" today! Again, just for the record, that other post and today's post should be understood as a way to spot the lies and misunderstandings in the personal finance world, not a manual…. We also discuss differentiation and integration of power series. taylor seris of y = cosx has only. And for fun, you might want to go type in-- you can type in Taylor expansion at 0 and sine of x, or Maclaurin expansion or Maclaurin series for sine of x, cosine of x, e to the x, at WolframAlpha. Fourier Series Calculator is a Fourier Series on line utility, simply enter your function if piecewise, introduces each of the parts and calculates the Fourier coefficients may also represent up to 20 coefficients. Instructions: 1. , x 0 2I : Next consider a function, whose domain is I,. Find the Maclaurin series for , where C is the straight-line segment from 0 to z. The Maclaurin Expansions of Elementary Functions. Thus, Taylor formula for polynomials allows us to rewrite any polynomial in terms of (x-a). So we can conclude as stated earlier, that the Taylor series for the functions , and always represents the function, on any interval , for any reals and , with. Besides finding the sum of a number sequence online, server finds the partial sum of a series online. f (x) = ln x. 2, is a Taylor series centered at zero. It is more of an exercise in differentiating using the chain rule to find the derivatives. To approximate function values, we just evaluate the sum of the first. Taylor series is a way to representat a function as a sum of terms calculated based on the function's derivative values at a given point as shown on the image below. A minimum of 1 double value to hold the total and 2 integer values, 1 to hold the current term number and 1 to hold the number of terms to evaluate. // Taylor series for e^x RationalPolynomial e = RationalPolynomial. Taylor Series Expansion Calculator computes a Taylor series for a function at a point up to a given power. In practice the Taylor series does converge to the function for most functions of interest, so that the Taylor series for a function is an excellent way to work that function. Solution 4 (b). The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. Second the Taylor series actually represents the function on the interval. This utilizes differentiation, and you'll see some. So, the function 1/(1-x) can be represented as a power series for part of its domain. Complete Solution Before starting this problem, note that the Taylor series expansion of any function about the point c = 0 is the same as finding its Maclaurin series expansion. The mtaylor function computes a truncated multivariate Taylor series expansion of the input expression f, with respect to the variables v, to order n, using the variable weights w. Online trig calculators is designed to solve trigonometric equations easily. Since this is true for any real , these Taylor series represent the. Instructions: 1. Taylor series is a special class of power series defined only for functions which are infinitely differentiable on some open. Phil Taylor Software Informer. C / C++ Forums on Bytes. Calculadora de Expansão em Série de Taylor calcula uma expansão em série de Taylor para uma função em um ponto até Taylor Calculator Expansão Series:. One way is to use the formula for the Taylor's theorem remainder and its bounds to calculate the number of terms. In other words, you’re creating a function with lots of other smaller functions. If an input is given then it can easily show the result for the given number. Before you start searching, find out how large of a mortgage you can comfortably afford. This raised the question of how to emulate the trigonometric, log, and exponential functions available in the Windows calculator program. Welcome to Part 20 of 21: Taylor and Maclaurin Series. What is the Maclaurin series for f(x) = e x? To get the Maclaurin series, we look at the Taylor polynomials for f near 0 and let them keep going. Find the Taylor series expansion of any function around a point using this online calculator. You also have the CLASSPAD Series 300PLUS and 330. Taylor / Maclaurin Series Expansion - Proof of the Formula Taylor / Maclaurin Series Expansion - Deriving the Formula. In similar ways, other functions can be represented by power series. Power series calculator from function keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website. When this interval is the entire set of real numbers, you can use the series to find the value of f(x) for every real value of x. Lady (October 31, 1998) Some Series Converge: The Ruler Series At rst, it doesn't seem that it would ever make any sense to add up an in nite number of things. Every Maclaurin series, including those studied in Lesson 22. I The Taylor Theorem. In this video, I show how to find the Taylor series expansion for a function, assuming that one exists! It is nothing too heavy: we just take derivatives and plug in the value at which we are centering the function. Gonzalez-Zugasti, University of Massachusetts - Lowell 2. The calculator will find the Taylor (or power) series expansion of the given function around the given point, with steps shown. Exercise 6. This calculus 2 video tutorial explains how to find the Taylor series and the Maclaurin series of a function using a simple formula. Taylor Series Expansion Calculator. Relevance *I Like To Wash My Hands. Maclaurin Series Calculator. Taylor Series in Mathcad. He became a promoter of the industry there, featuring it on Today and in articles he wrote during his second career. Use the fact that to write down a power series representation of the logarithmic function. taylor seris of y = cosx has only. Use our free online average rate of change calculator to find the average rate at which one quantity is changing with respect to an other changing quantity in the given expression (function). Thinking about the problem: Have I seen a problem similar to this one before? If so, what did I do to compute the Taylor series? To determine the Taylor series for the function f(x), I will make a table with n. You can construct the series on the right provided that f is infinitely differentiable on an interval containing c. A Taylor series is a polynomial of infinite degree that can be used to represent many different functions, particularly functions that aren't polynomials. This utilizes differentiation, and you'll see some. Change the view to see the accuracy. And by knowing these basic rules and formulas, we can learn to use them in generating other functions as well as how to apply them to Taylor Series that are not centered at zero. f (x) = ln x. New Camden course tour - starts apprx. Will Justin Verlander pitch the Astros to a World Series title?. Worksheet 9. Taylor's Formula with Remainder Let f(x) be a function such that f(n+1)(x) exists for all x on an open interval containing a. The Series Calculator an online tool which shows Series for the given input. Taylor Series Expansion Calculator computes a Taylor series for a function at a point up to a given power. We encourage you to use comments to engage with users, share your perspective and ask questions of authors and each other. The Taylor series can sometimes be called a MacLaurin series, which is a Taylor series evaluated at a = 0. Definition of Taylor series, Binomial series, Special cases of binomial series, Series for exponential and logarithmic functions, Series for trigonometric functions, Series for inverse trigonometric functions, Series for hyperbolic functions, Taylor series formula, Binomial series formula, Special cases of binomial series formula, Series for exponential and logarithmic functions formula. I have the code for the first part of a problem, which is to write a program that reads an angle x (in radians) from the keyboard. Taylor Series Expansion Calculator. Konstanin Frank, and Herman Wiemer. Now, let's see how we can use this idea for any differentiable functions. 1 We examined series of constants and learned that we can say everything there is to say about geometric and telescoping series. If an input is given then it can easily show the result for the given number. , when -1/2 < x < 1/2. Ken Bube of the University of Washington Department of Mathematics in the Spring, 2005. obtained is called Taylor series for f(x) about x= a. In the Introduction to this chapter, we wondered how a calculator (or computer) finds the square root of millions of numbers, and how it finds the cosine of millions of angles (in degrees and radians). Disclaimer: Each calculator available for use on this web site and referenced in the following directories - finance calculator, retirement calculator, mortgage calculator, investment calculator, savings calculator, auto loan calculator, credit card calculator, or loan calculator - is believed to be accurate. Just input the equation, lower limit, upper limit and select the precision that you need from the drop-down menu to get the result. We have an equation solver, an equation graphing, a factoring calculator, and a derivative calculator. Inverse Function Calculator inverts function with respect to a given variable. An important property of the inverse function is that inverse of the inverse function is the function itself. for high school and college students, teachers, parents and people who want to refresh their knowledge in math. Using the slider and drag the point to create different approximations to the function f(x). By writing, 8/(2-x)^3 = 1/(1-0. Deriving the Maclaurin series for tan x is a very simple process. Taylor Series in Mathcad. Latest updates on everything Phil Taylor Software related. Cici\'s Derby - part of the SCRunners XC Series. Taylor Series Expansion Graphs Change the top slider for a to change the focus point of the expansion and the bottom slider to include more terms of the series expansion. Explore many other math calculators, as well as hundreds of other calculators addressing health, fitness, finance, math, and more. In similar ways, other functions can be represented by power series. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. What Is the Taylor Series of Ln(x)? Taylor Series Calculator Jack Taylor Series Example of Taylor Series Taylor L. Calculate various formulas and identities for sine, cosine, tangent, cotangent, secant and cosecant functions. If we use a finite number of terms, the series can (I stress can), but a good approximation f(x). asin inverse sine (arcsine) of a value or expression acos inverse cosine (arccos) of a value or expression atan inverse tangent. Because this series is convergent for every complex value of x, it is commonly used to extend the definition of e x to the complex numbers. taylortool initiates a GUI that graphs a function against the Nth partial sum of its Taylor series about a base point x = a. Taylor / Maclaurin Series Expansion - Proof of the Formula Taylor / Maclaurin Series Expansion - Deriving the Formula. This web site owner is mathematician Miloš Petrović. Recurrence formulae are given, and tables of the coefficients constructed. Assume that we have a function f for which we can easily compute its value f(a) at some. Every Taylor series provides the. Taylor Series in Mathcad. Taylor Polynomials and Taylor Series Math 126 In many problems in science and engineering we have a function f(x) which is too complicated to answer the questions we'd like to ask. This video gives examples of how to use Taylor Series to solve limit problems. ) When , the series is called a Maclaurin series. And for fun, you might want to go type in-- you can type in Taylor expansion at 0 and sine of x, or Maclaurin expansion or Maclaurin series for sine of x, cosine of x, e to the x, at WolframAlpha. This information is provided by the Taylor remainder term: Keep in mind that this inequality occurs because of the interval involved, and because that sine. However, when the interval of convergence for a Taylor. ##e^x = \sum_{n=0}^\infty\frac{x^n}{n!} ## is the Taylor series for the. Taylor Series of Analytic Complex Functions. This calculus 2 video tutorial explains how to find the Taylor series and the Maclaurin series of a function using a simple formula. If you had such a dream, would you get up and do the things you've been dreaming?. Because the Taylor series is a form of power series, every Taylor series also has an interval of convergence. In some cases, such as heat transfer, differential analysis results in an equation that fits the form of a Taylor series. Sequence calculator: sequence. You already know how to determine the interval of convergence of the series. In essence, the Taylor series provides a means to predict a function value at one point in terms of the function value and its derivatives at another point. 9) I Review: Taylor series and polynomials. The best calculus calculators including derivative calculator, integral calculator, limit calculator and more. That is, the kth coefficient is equal to the kth derivative of f evaluated at the input 0 and then divided by k!. Let's wrap up our survey of calculus! We have one more type of series to learn, Taylor series, and special case of those called Maclaurin series. With the default mode RelativeOrder, the number of requested terms for the expansion is determined by order if specified. Andrew Chamberlain, Ph. If you want to learn more about how computers calculate trigonometric functions, check out the Taylor series (might require knowledge of calculus though). Let's look closely at the Taylor series for sinxand cosx. Taylor series calculator is used to calculate taylor polynomial of different degrees. Right now, I'm trying to write a program with multiple function block (I'm new to programming so my terminology/jargon may not be correct ) that sums the taylor series of e^x. Gcd Calculator Plotter Calculator Solver. In our conventions, arccot x ≡ arctan(1/x) is not continuous at x = 0 and thus does not possess a Taylor series about x = 0. Use the fact that to write down a power series representation of the logarithmic function. Taylor and Maclaurin Series If a function $$f\left( x \right)$$ has continuous derivatives up to $$\left( {n + 1} \right)$$th order, then this function can be expanded in the following way:. Taylor Series in Mathcad. taylor series for sin 2x? what is the maclaurin series (x=0) for f(x)=sin 2x. To solve this equation just enter the expression x^2+1=0 and press calculate button. The coefficients a, b, c and so on might not have a certain order like the functions listed above, but at least you have a reasonable polynomial to estimate the function in the absence of a calculator. If you want to learn more about how computers calculate trigonometric functions, check out the Taylor series (might require knowledge of calculus though). Power Series Expansions. Comparasion Test: The terms of the sequence are compared to those of another one. tell me how to do it please. Change the view to see the accuracy. 2 Floating Point. What makes these important is that they can often be used in place of other, more complicated functions. Let's wrap up our survey of calculus! We have one more type of series to learn, Taylor series, and special case of those called Maclaurin series. The Organic Chemistry Tutor 1,243,349 views. Taylor Company. You can construct the series on the right provided that f is infinitely differentiable on an interval containing c. The best calculus calculators including derivative calculator, integral calculator, limit calculator and more. It is used like this: Sigma is fun to use, and can do many clever things. Evaluate the remainder by changing the value of x. Next: Videos on Taylor Series The following table of Maclaurin expansions summarizes our results so far, and provides expansions for other series that we have not covered. Learn exactly what happened in this chapter, scene, or section of The Taylor Series and what it means. The uses of the Taylor series are: Taylor series is used to evaluate the value of a whole function in each point if the functional values and derivatives are identified at a single point. 1 Taylor Polynomials The tangent line to the graph of y = f(x) at the point x = a is the line going through the point ()a, f (a) that has slope f '(a). I assume that is 13 radians, since 13° should only take a few terms of the Taylor series for sin(13°). An important property of the inverse function is that inverse of the inverse function is the function itself. , arcsin, arccos, arctan, arccot, arcsec, and arccsc. Conversion Calculator. You can specify the order of the Taylor polynomial. Lecture 32: Taylor Series and McLaurin series We saw last day that some functions are equal to a power series on part of their domain. Change the view to see the accuracy. These terms in the geometric sequence calculator are all known to us already, except the last 2, about which we will talk in the following sections. Find the Taylor series centered on a = 4 b) Use part (a) and your calculator. Taylor Series can be used to represent any function, as long as it is an analytic function. Taylor series, in turn, extend the idea of Taylor polynomials. In this video I'm going to show you how you can find a Taylor series. We also discuss differentiation and integration of power series. The differentiation rules. The program asks the user to type a value for an angle in degrees. Choose the maximum degree of the Taylor polynomial to use to approximate a function. Free Maclaurin Series calculator - Find the Maclaurin series representation of functions step-by-step. Definite Integral Calculator computes definite integral of a function over an interval using numerical integration. , I might be ( 17;19)) and let x 0 be a point in I, i. Find the Taylor series expansion for e x when x is zero, and determine its radius of convergence. Will Justin Verlander pitch the Astros to a World Series title?. Evaluate the remainder by changing the value of x. Taylor series calculator is used to calculate taylor polynomial of different degrees. The partial sum is called the nth-order Taylor polynomial for f centered at a. Taylor Series. Math 133 Taylor Series Stewart x11. txt) or read online for free. The complex number equation calculator returns the complex values for which the quadratic equation is zero. You also have the CLASSPAD Series 300PLUS and 330. The Taylor series is a power series that approximates the function f near x = a. And try it out for a bunch of different functions. Geometric Series Solver Geometric Series Solver This utility helps solve equations with respect to given variables. Taylor Series Sin x Calculator. You may remember from geometric series that for appropriate values of r. Derivatives Derivative Applications Limits Integrals Integral Applications Series ODE Laplace Transform Taylor/Maclaurin Series Fourier Series. Taylor's Formula with Remainder Let f(x) be a function such that f(n+1)(x) exists for all x on an open interval containing a. I thought I might be able to multiply the taylor series of $ln(x)$ by $x$ to give the series for $xln(x)$, but that didn't work out the same way. The variables v can be a list or set of names or equations. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Thinking about the problem: Have I seen a problem similar to this one before? If so, what did I do to compute the Taylor series? To determine the Taylor series for the function f(x), I will make a table with n. It's just the way they calculate π and the way they calculate sin(). | 2020-01-18T10:20:26 | {
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https://math.stackexchange.com/questions/3182882/finding-the-maximum-relative-misalignment-of-numbered-rings-on-a-combination-loc | # Finding the maximum relative misalignment of numbered rings on a combination lock?
I've been trying to figure out a general formula to calculate the maximum relative misalignment of m identical rings with n symbols each on a combination lock like the one shown below.
By "maximum relative misalignment" I mean the maximum number of turns that would be required to align all the rings with each other so their numbers match up.
You could think of this as the maximum number of turns that would be needed to unlock the combination shown, if the combination lock will unlock if all the numbers between the dots are the same number.
For this particular lock where m=5 and n=10, I've calculated the number to be 12.
You could also think of this as the maximum total number of rotational shifts required to align m arrays containing the same n unique numbers each.
I've been unable to figure out a generalized formula for this even though it seems like it shouldn't be terribly complicated.
I brute forced the answer for m and n between 2 and 12 and the results were as follows:
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
2 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9
3 1 2 2 3 4 4 5 6 6 7 8 8 9 10 10 11 12 12
4 2 2 4 4 6 6 8 8 10 10 12 12 14 14 16 16 18 18
5 2 3 4 6 7 8 9 10 12 13 14 15 16 18 19 20 21 22
6 3 4 6 7 9 10 12 13 15 16 18 19 21 22 24 25 27 28
7 3 4 6 8 10 12 13 15 17 18 20 22 24 25 27 29 30 32
8 4 5 8 9 12 13 16 17 20 21 24 25 28 29 32 33 36 37
9 4 6 8 10 13 15 17 20 22 24 26 28 31 33 35 37 40 42
10 5 6 10 12 15 17 20 22 25 27 30 32 35 37 40 42 45 47
11 5 7 10 13 16 18 21 24 27 30 32 35 38 40 43 46 49 51
12 6 8 12 14 18 20 24 26 30 32 36 38 42 44 48 50 54 56
13 6 8 12 15 19 22 25 28 32 35 38
14 7 9 14 16 21 24 28 31 35 38 42
15 7 10 14 18 22 25 29 33 37 40 44
16 8 10 16 19 24 27 32 35 40 43 48
17 8 11 16 20 25 29 33 37 42 46 50
18 9 12 18 21 27 30 36 40 45 49 54
19 9 12 18 22 28 32 37 42 47 51 56
I noticed that the value seems to be the same if you swap m and n.
I've determined the formula for even values of m and n seems to just be (n/2)*(m/2)
Thanks for any help
• Possible duplicate of Optimally scramble the combination lock – Mike Earnest Apr 10 at 20:16
• If you want to change a question, don't delete it and ask a new question, but just edit the question, using the edit link at the bottom. – Rob Arthan Apr 10 at 20:49
• What is counted as a turn? Moving one ring one space in either direction? – Servaes Apr 10 at 20:52
• @Servaes Yes. Moving the ring one space in either direction. If it were defined as moving on ring any number of spaces in any direction the answer would be trivial. – Conor Henry Apr 10 at 20:54
• When you say the lock "will unlock if the numbers between the dots are the same number", am I right in thinking that it can be any number, rather than some fixed number: i.e., it will unlock when you get to all 0s or all 1s or all 2s or ...? – Rob Arthan Apr 10 at 20:58
Partial result: Proof that $$f(m,n) = mn/4$$ when $$m,n$$ are even.
Let $$f(m,n)$$ be the number you seek, i.e.
$$f(m,n) = \max_x \min_a g(x,a)$$
$$\text{where: } g(x,a) \equiv \sum_{j=1}^m dist(x_i,a)$$
where $$[n] = \{0, 1, ..., n-1\}, x \in [n]^m, a \in [n]$$ and $$dist()$$ is the shortest distance from $$x_i$$ to $$a$$ along the $$n$$-node ring, and $$g(x,a)$$ is the total distance of moving everybody to position $$a$$.
Theorem: For $$m, n$$ even, $$f(m,n) = mn/4$$.
Proof: Consider a given $$x$$ and suppose $$a$$ is an optimum, i.e., it minimizes the sum of distances. Let $$b = a + n/2 \pmod n$$, i.e. the exactly opposite position.
Lemma: $$g(x,a) + g(x,b) = mn/2$$
Proof: For every $$x_i, dist(x_i,a) + dist(x_i,b) = n/2$$ because it simply traverses one arc or the other arc of the half-circle from $$a$$ to $$x_i$$ to $$b$$. Sum over all $$m$$ and we get the result.
Continued proof of theorem: Since $$a$$ is optimal by assumption, $$g(x,a)\le g(x,b)$$, which combined with the lemma, means $$g(x,a) \le mn/4$$.
OTOH, $$g(x,a) = mn/4$$ is achievable, so the bound is tight. E.g. just put $$m/2$$ at one position and the other $$m/2$$ at the exact opposite position.
• This makes a lot of sense. Thanks! I'm looking into deriving the values for m and n being both odd. I've noticed when m and n are both odd and m=n, the value is given by ((n-1)/2 + 1)((n-1)/2). – Conor Henry Apr 17 at 19:14
• @ConorHenry - this is a fascinating problem. I think many things are "obvious" but hard to prove. e.g. for the $m=n$ case, i think it is "obvious" the max $x$ is when one $0$ occupies each position, i.e. the perfectly spread scenario. that scenario indeed gives $k(k+1)$ where $n=m=2k+1$. But how to prove that's the max? I have no idea yet. I have some other partial results which I hope to post when I have more time. – antkam Apr 18 at 5:52 | 2019-10-16T15:26:23 | {
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https://math.stackexchange.com/questions/3614579/linearly-independent-vectors-and-colinearity-proof-verification | # Linearly Independent Vectors and Colinearity (Proof Verification)
I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary.
The Prompt (from here):
Two nonzero vectors are linearly independent if and only if they are not colinear (or proportional, i.e. for two vectors (u,v), there exists λ ∈ R such that u = λ.v)
My Proof:
Sufficiency: Suppose two nonzero vectors v1 and v2 are linearly independent, but let them be colinear so that v1 = λv2. Then:
t1v1 + t2v2 = 0
λt1v2 + t2v2 = 0
v2(λt1 + t2) = 0
This implies that λt1 = -t2 even for nonzero values of t1 and t2, which contradicts the fact that v1 and v2 are linearly independent, so v1 and v2 must not be colinear.
Necessity: We prove the necessary condition by the contrapositive. Suppose v1 and v2 are not linearly independent, so that there is some nonzero t ∈ R2 such that t1v1 + t2v2=0. Then t1v1 = -t2v2 so v1 = -(t2/t1)v2 so v1 and v2 are proportional (by the scalar -t2/t1) and colinear. Since v1 and v2 not being linearly independent implies them being colinear, if they are not colinear then they are linearly independent.
Your proof is basically fine apart from the following points.
You shouldn't have $$=0$$' in each line in your chain of equations in the proof of sufficiency, and there is the possibility of one of the vectors being zero to be (briefly) considered. What I would suggest is something like this.
Suppose that $$v_1$$ and $$v_2$$ are linearly independent but that they are collinear. If $$v_1$$ or $$v_2$$ are equal to zero then $$v_1$$ and $$v_2$$ cannot be linearly independent, so suppose that $$v_1,v_2\neq 0$$. Since $$v_1$$ and $$v_2$$ are collinear we have $$v_1=\lambda v_2$$ for some $$\lambda\neq 0$$. Now for any $$t_1,t_2\in\mathbb{R}$$ we have
$$\begin{array}{ll} & t_1v_1+ t_2v_2\\ = & \lambda t_1v_2 + t_2v_2\\ = & v_2(\lambda t_1 + t_2)\end{array}$$
But if $$t_1\neq 0$$ and $$t_2=-\lambda t_1$$ we get $$v_2(\lambda t_1 + t_2)=0$$, which contradicts the fact that $$v_1$$ and $$v_2$$ are linearly independent. Therefore $$v_1$$ and $$v_2$$ are not collinear.
Of course it's important to note that at least one of $$t_1,t_2$$ is non-zero (as you had in your own proof) to make sure you have a contradiction.
In your proof of necessity you have $$t$$ instead of $$t_1$$ and $$t_2$$. I would suggest something like ... so there are $$t_1,t_2\in\mathbb{R}$$, not both zero, such that $$t_1v_1+t_2v_2=0$$. Suppose that $$t_1\neq 0$$ (the case where $$t_2\neq 0$$ is similar). Then $$t_1v_1=-t_2v_2$$...'
Finally, the word is usually spelled collinear. Apart from that it's a fine proof. | 2021-02-27T04:20:49 | {
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https://hershg.github.io/Lecture%C2%A030%20-%2004%2005.html | ## Lecture 30 - 04/05
### Minimum Spanning Trees
• Remember from last time we have shortest path trees (SPT's) from Djikstra's Algorithm, that give us a tree telling us the shortest way to get from a start node s to any other vertex in the graph
• Alternate idea: minimum spanning tree (MST)
• We want to create a tree (acyclic graph) that connects all of the vertices in the graph, such that the total edge cost for this spanning tree is the minimum possible
• So essentially we want a tree that spans the entire graph, such that it's total cost is the minimum possible
• Depending on the graph and node, running Djikstra's algorithm may or may not return us the MST (MST could potentially be equal to the SPT from a certain start vertex)
• However, this is often not the case, as the shortest path from vertices A and B could be different from the path between them as determined by the MST
• Example of SPT's vs MST's (in this case, the SPT from vertex B is equivalent to the graph's MST):
• It's also very possible that none of the vertices of a graph will yield a SPT that is actually the MST. So how to get the MST?
#### Prim's Algorithm
• To find the MST, we start from any arbitrary start vertex, and expand/traverse through the tree exactly like we did with Djikstra's Algorithm, with a key difference: we no longer use total cumulative costs, and only consider immediate edge costs when determining the edges of the algorithm's resulting tree
• The resulting tree will be a MST (it may or may not be an SPT for some vertex of the graph)
• Essentially we keep building up a set of nodes in our "cut" of the graph (a subset of all the vertices of the graph), such that our cut has the minimum possible total edge cost between its vertices
• We build up this cut until our cut spans the entire graph
• Build up is by looking at all the edges that connect our cut of the graph to the rest of the graph (if we drew a squiggly line around only the nodes in our cut, we would be looking at all the edges that intersect this line)
• Among all these edges, we find the edge of least weight, add this to our cut, and create an edge to it's corresponding vertex in our MST
• So instead of Djikstra, which considers distance from a start node, Prim considers the distance from the entire tree (by tree I'm referring to the cut of the tree that we create and build up through the algorithm)
• Runtime identical to Dijkstra's algorithm
• Assuming E>V$E>V$, runtime is Θ(ElogV)$\Theta(E \log V)$
• Implementation pseudocode:
public PrimMST(Graph G) {
distTo = new double[G.V()];
Edge[] edgeTo = new Edge[G.V()];
boolean[] marked = new boolean[G.V()];
PQ<Double> fringe = new PQ<>(G.V());
distTo[s] = 0.0; // Arbitrary start node
setAllDistancesToInfinityExceptS(distTo);
fringe.insert(s, distTo[s]);
// Fringe ordered by distTo the cut tree
// Add vertices in order of distance from cut tree
while (!fringe.isEmpty()) {
int v = fringe.delMin();
scan(G, v);
}
}
private void scan(Graph G, int v) {
marked[v] = true;
for (Edge e : G.adj(v)) {
int w = e.other(v); // Other vertex in edge
if (marked[w]) { continue; } // Already in MST so go to next edge
if (e.weight() < distTo[w]) { // This is a better path so update MST
distTo[w] = e.weight();
edgeTo[w] = e;
if (fringe.contains(w)) {
pq.decreasePriority(w, distTo[w]);
} else {
pq.insert(w, distTo[w]);
}
}
}
#### Kruskal’s Algorithm:
• Alternate approach to Prim's Algorithm
• Prim's algorithm: start from an arbitrary start node, and build up a "cut" tree
• Tree expansion based on distance from the tree (not from the start node)
• Kruskal's algorithm: add all the edges to the priority queue fringe, and then each iteration pop the smallest/tied for smallest weight edge from the fringe
• Add the given edge to the MST, unless doing so would create a cycle
• Repeat untill the MST we create is of size V1$V-1$ (when we reach this, no need to go through remaining edges in the fringe)
• Kruskal's algorithm converges to Prim's algorithm! (both have same result, just different means to attain it)
• Implementation pseudocode:
public KruskalMST {
private Queue<Edge> mst = new Queue<>();
public KruskalMST(Graph G) {
PQ<Double> fringe = new PQ<>(G.V());
for (Edge e : G.E()) {
fringe.insert(e);
}
// Weighted Quick Union with Path Compression (disjoint set)
WeightedQuickUnionPC uf = new WeightedQuickUnionPC(G.V());
while (!fringe.isEmpty() && mst.size() < G.V() - 1) {
Edge e = fringe.delMin();
int v = e.from();
int w = e.to();
if (!uf.connected(v, w)) {
uf.union(v, w);
mst.enqueue(e);
}
}
}
}
• Kruskal's Algorithm is Θ(ElogE)$\Theta(E\log E)$ but if the edges are pre-sorted(stored in ordered linked-list so no need for PQ building) then we have Θ(ElogV)$\Theta(E\log* V)$ runtime
• Over the years, algorithms have been developed that give tighter and tighter bounds on the runtime | 2018-11-18T09:54:19 | {
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http://www.purplemath.com/learning/viewtopic.php?f=8&t=1906&p=5666 | How do you go from a quadratic function in...
Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
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How do you go from a quadratic function in...
generic form $ax^2+bx+c$ to the standard parabola form $a(x-h)^2+k$ by completing the square?
I know that the vertex of a parabola (h, k) is $h=\frac{-b}{2a}$ and $k=f(h)$ but I can't figure out how the first form gets to the second. I just want to know how the standard form is derived form the generic form by completing the square in a general way (meaning without numbers, just the letters to represent the constants); step-by-step please!
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Re: How do you go from a quadratic function in...
Absolutely wrote:generic form $ax^2+bx+c$ to the standard parabola form $a(x-h)^2+k$ by completing the square? I know that the vertex of a parabola (h, k) is $h=\frac{-b}{2a}$ and $k=f(h)$ but I can't figure out how the first form gets to the second. I just want to know how the standard form is derived form the generic form by completing the square in a general way (meaning without numbers, just the letters to represent the constants); step-by-step please!
assume $a\neq0$
$ax^2+bx+c$
$=a(x^2+\frac{b}{a}x)+c$
$=a\left(x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2\right)+c$
$=a\left(x+\frac{b}{2a}\right)^2-a\left(\frac{b}{2a}\right)^2+c$
$=a\left(x-\frac{-b}{2a}\right)^2-a\left(\frac{b}{2a}\right)^2+c$
$=a\left(x-\frac{-b}{2a}\right)^2+c-\frac{b^2}{4a}$
$=a\left(x-h\right)^2+k$
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Re: How do you go from a quadratic function in...
Thanks! Seeing the algebraic manipulations behind these derivations makes the formulas more sensible to me. | 2015-09-02T13:18:26 | {
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http://mathhelpforum.com/calculus/282194-clarification-needed.html | 1. ## Clarification needed
I've had no issues employing the following theorem but there's just a slight bump that need's ironing out with the hypotheses and the domain.
Mean Value Theorem
If f(x) is continuous in the interval $\displaystyle a \leq x \leq b$ and if f'(x) exists at each value of x for which $\displaystyle a < x < b$, then there exists at least one value c of x between a and b such that $\displaystyle f(b) - f(a) = f'(c)(b - a)$
Why are the endpoints omitted for the interval of the derivative? The only thing I can come up with is that c is between a and b. So f can be continuous at a and b but not necessarily differentiable at a and b? But if f happened to be a straight line it would have to be differentiable at a and b.
Since continuity is contained within differentiability, then would it not be more concise to say
If f(x) is differentiable on [a,b], then there exists at least one value c of x between a and b such that $\displaystyle f(b) - f(a) = f'(c)(b - a)$
2. ## Re: Clarification needed
The version of the "mean value theorem" you give "If f(x) is differentiable on [a,b], then there exists at least one value c of x between a and b such that f(b)−f(a)= f′(c)(b−a)" would be true but not sufficiently general. if a function is continuous on [a, b] and differentiable on (a, b) but not at a or b the conclusion is still true. For example, If f(x)= x- x^2 for x greater than or equal to 0 and less than or equal to 1, f(x)= -x for x<0, f(x)= x- 1 for x> 1. f is continuous for all x, and in particular on the closed interval [0, 1]. It is differentiable on the open interval (0, 1) but not at x= 0 or x= 1. The 'mean value' between 0 and 1 is (f(1)- f(0))/(1- 0)= 0 and the derivative at x= 1/2 is - 2(1/2)= 0.
Continuity at the endpoints is important because the mean value theorem specifically uses those values. Consider the same example as above, but with f(0)= 1. Now (f(1)- f(0))/(1- 0)= -1 but there is no x in (0, 1) where the derivative, 1- 2x, is equal to -1.
3. ## Re: Clarification needed
I've drawn a diagram of the function for x<0, 0<= x <=1, x > 0, and yes I can see the derivative is not unique at the endpoints. My statement above was too general I see that now thanks. In your last comments, the mean value theorem failed because f(0) = 1 was not an endpoint of the function it was just redefined for an arbitrary value.
Just one last thing, the intervals for the function you gave presumably continuous passing through zero and out through 1, shouldn't we put f(x) = -x, x<=0 and f(x) x -1, x =>1 just so it's defined at those values when we descrive f(x) differently? | 2019-06-20T10:17:27 | {
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# Clarissa will create her summer reading list by randomly choosing 4 bo
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Clarissa will create her summer reading list by randomly choosing 4 books from the 10 books approved for summer reading. She will list the books in the order in which they are chosen. How many different lists are possible?
A. 6
B. 40
C. 210
D. 5,040
E. 151,200
The Official Guide for GMAT Review 2018
Practice Question
Problem Solving
Question No.: 140
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12 Jun 2017, 16:11
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AbdurRakib wrote:
Clarissa will create her summer reading list by randomly choosing 4 books from the 10 books approved for summer reading. She will list the books in the order in which they are chosen. How many different lists are possible?
A. 6
B. 40
C. 210
D. 5,040
E. 151,200
The Official Guide for GMAT Review 2018
Practice Question
Problem Solving
Question No.: 140
Using the fundamental counting principle (or permutation, because order matters -- she's listing them in their order chosen):
For her first book she has 10 to choose from, then nine, than eight, then seven.
10 * 9 * 8* 7 = 5040
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Re: Clarissa will create her summer reading list by randomly choosing 4 bo [#permalink]
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27 Jun 2017, 15:26
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AbdurRakib wrote:
Clarissa will create her summer reading list by randomly choosing 4 books from the 10 books approved for summer reading. She will list the books in the order in which they are chosen. How many different lists are possible?
A. 6
B. 40
C. 210
D. 5,040
E. 151,200
Take the task of creating the reading list and break it into stages.
Stage 1: Select a book to read 1st
There are 10 books to choose from. So, we can complete stage 1 in 10 ways
Stage 2: Select a book to read 2nd
There are 9 books remaining to choose from (since we already chose a book in stage 1).
So, we can complete stage 2 in 9 ways
Stage 3: Select a book to read 3rd
There are 8 books remaining to choose from. So, we can complete stage 3 in 8 ways
Stage 4: Select a book to read 4th
There are 7 books remaining to choose from. So, we can complete stage 4 in 7 ways
By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus create a reading list) in (10)(9)(8)(7) ways (= 5040 ways)
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.
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Re: Clarissa will create her summer reading list by randomly choosing 4 bo [#permalink]
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29 Jun 2017, 00:09
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4 books can be chosen as 10C4 ways. = 210 ways
Total list of 4 books = 4!
Therefore, total list of possible selection = 210 *4! = 210 *24 =5,040.
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Re: Clarissa will create her summer reading list by randomly choosing 4 bo [#permalink]
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08 Aug 2017, 23:45
AbdurRakib wrote:
Clarissa will create her summer reading list by randomly choosing 4 books from the 10 books approved for summer reading. She will list the books in the order in which they are chosen. How many different lists are possible?
A. 6
B. 40
C. 210
D. 5,040
E. 151,200
The Official Guide for GMAT Review 2018
Practice Question
Problem Solving
Question No.: 140
as the order is important here, so its a Permutation case. So arranging 4 books out of 10 books is
$$10_P_4$$ = 5040
Option D
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Re: Clarissa will create her summer reading list by randomly choosing 4 bo [#permalink]
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15 Nov 2017, 17:21
AbdurRakib wrote:
Clarissa will create her summer reading list by randomly choosing 4 books from the 10 books approved for summer reading. She will list the books in the order in which they are chosen. How many different lists are possible?
A. 6
B. 40
C. 210
D. 5,040
E. 151,200
Since the order of the books in her list matters, we use permutation. Thus, the number of ways 4 books can be chosen and ordered from 10 books is 10P4 = 10!/(10-4)! = 10 x 9 x 8 x 7 = 5,040.
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Re: Clarissa will create her summer reading list by randomly choosing 4 bo [#permalink]
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26 Nov 2017, 03:47
genxer123 wrote:
AbdurRakib wrote:
Clarissa will create her summer reading list by randomly choosing 4 books from the 10 books approved for summer reading. She will list the books in the order in which they are chosen. How many different lists are possible?
A. 6
B. 40
C. 210
D. 5,040
E. 151,200
The Official Guide for GMAT Review 2018
Practice Question
Problem Solving
Question No.: 140
Using the fundamental counting principle (or permutation, because order matters -- she's listing them in their order chosen):
For her first book she has 10 to choose from, then nine, than eight, then seven.
10 * 9 * 8* 7 = 5040
Hello. I dont understand one thing I did so 10! / 4! (10-6)! = 210 so I chose 210 why its D
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Re: Clarissa will create her summer reading list by randomly choosing 4 bo [#permalink]
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26 Nov 2017, 03:54
ScottTargetTestPrep wrote:
AbdurRakib wrote:
Clarissa will create her summer reading list by randomly choosing 4 books from the 10 books approved for summer reading. She will list the books in the order in which they are chosen. How many different lists are possible?
A. 6
B. 40
C. 210
D. 5,040
E. 151,200
Since the order of the books in her list matters, we use permutation. Thus, the number of ways 4 books can be chosen and ordered from 10 books is 10P4 = 10!/(10-4)! = 10 x 9 x 8 x 7 = 5,040.
and if the order didn't matter the answer would be 210 ? like this ? -- > 10! / 4!(10-4)! =210
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Clarissa will create her summer reading list by randomly choosing 4 bo [#permalink]
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26 Nov 2017, 07:03
dave13 wrote:
ScottTargetTestPrep wrote:
AbdurRakib wrote:
Clarissa will create her summer reading list by randomly choosing 4 books from the 10 books approved for summer reading. She will list the books in the order in which they are chosen. How many different lists are possible?
A. 6
B. 40
C. 210
D. 5,040
E. 151,200
Since the order of the books in her list matters, we use permutation. Thus, the number of ways 4 books can be chosen and ordered from 10 books is 10P4 = 10!/(10-4)! = 10 x 9 x 8 x 7 = 5,040.
and if the order didn't matter the answer would be 210 ? like this ? -- > 10! / 4!(10-4)! =210
"She will list the books in the order in which they are chosen""--Order matters here ...So 10P4
Clarissa will create her summer reading list by randomly choosing 4 bo [#permalink] 26 Nov 2017, 07:03
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https://teachingcalculus.com/2019/08/26/limit-of-composite-functions/ | # Limit of Composite Functions
Recently, a number of questions about the limit of composite functions have been discussed on the AP Calculus Community bulletin board and also on the AP Calc TEACHERS – AB/BC Facebook page. The theorem that we would like to apply in these cases is this:
If f is continuous at b and $\underset{{x\to a}}{\mathop{{\lim }}}\,g\left( x \right)=b$, then $\underset{{x\to a}}{\mathop{{\lim }}}\,f\left( {g\left( x \right)} \right)=f\left( b \right)$.
That is, $\underset{{x\to a}}{\mathop{{\lim }}}\,f\left( {g\left( x \right)} \right)=f\left( {\underset{{x\to a}}{\mathop{{\lim }}}\,g\left( x \right)} \right)$
The problem is that in the examples one or the other of the hypotheses (continuity or the existence of $\underset{{x\to a}}{\mathop{{\lim }}}\,g\left( x \right)$) is not met. Therefore, the theorem cannot be used. This does not mean that the limits necessarily do not exist, rather that we need to find some other way of determining them. We need a workaround. Let’s look at some.
Example 1: The first example is from the 2016 BC International exam, question 88. Students were given the graph at the right and asked to find $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( {1-{{x}^{2}}} \right)$.
At first glance it appears that as x approaches zero, $\left( {1-{{x}^{2}}} \right)$ approaches 1 and the limit does not exist since f is not continuous at 1, so the theorem cannot be used. However, on closer examination, we see that $\left( {1-{{x}^{2}}} \right)$ is always less than 1, so $\left( {1-{{x}^{2}}} \right)$ is approaching 1 from the left (or from below). Therefore, as f approaches 1 from the left $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( {1-{{x}^{2}}} \right)=\underset{{x\to {{1}^{-}}}}{\mathop{{\lim }}}\,f\left( x \right)=3$
Another approach is to try to write the equation of f. Although we cannot be certain, it appears that: $f\left( x \right)={{x}^{2}}+2,x<1$.
Then, $f\left( {1-{{x}^{2}}} \right)={{\left( {1-{{x}^{2}}} \right)}^{2}}+2={{x}^{4}}-2{{x}^{2}}+3,x<1$. In this form the limit is obviously 3.
Example 2: The second example is also based on a graph. Given the graph of a function f, shown at the left, what is $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( {f\left( x \right)} \right)$ ?
$\underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( x \right)=2$. Since f is not continuous at 2, the theorem cannot be used. But, notice that as x approaches 0 from both sides, the limit 2 is approached from the left (from below). So we need to find the value of f as its argument approaches ${{2}^{-}}$. From the graph, this value is zero; So, $\underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( {f\left( x \right)} \right)=0$
To clarify this a little more, let’s look at a similar problem suggested by Sondra Edwards on the Facebook site: Consider this similar function:
Now as we approach 0 from both sides $\underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( x \right)=2$ approached from both sides. But now f(2) does not exist (DNE). (This is the “outside” f, which is not continuous here.) This time, the limiting value, 2, is approached from both sides. Therefore, $\underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( {f\left( x \right)} \right)$ DNE. There is no way to work around the discontinuity.
For a similar question see here
Example 3: If $\displaystyle f\left( x \right)=\frac{1}{x},x\ne 0$, what is $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( {f\left( {x)} \right)} \right)$ ?
Here again, the theorem cannot be used, since the “inside” function has no limit as x approaches 0. But, this function is its own inverse, so $\displaystyle f\left( {f\left( {x)} \right)} \right)=x$, and $\displaystyle \underset{{x\to 0}}{\mathop{{\lim }}}\,f\left( {f\left( x \right)} \right)=\underset{{x\to 0}}{\mathop{{\lim }}}\,x=0$
## One thought on “Limit of Composite Functions”
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http://mathhelpforum.com/calculus/275383-differential-calculus-related-rates.html | # Thread: Differential calculus :Related rates
1. ## Differential calculus :Related rates
An upright cylindrical tank full of water is tipped over at a constant (angular) speed.The height of the tank is at least twice its radius. How to prove that at the instant the tank has been tipped $45^{\circ}$, water is leaving the tank twice as fast as it did at the instant the tank was first tipped.
HINT:-Think of how the water looks inside the tank as it is being tipped
2. ## Re: Differential calculus :Related rates
The surface of the water is always parallel to the ground. That means that, if the tank makes angle $\displaystyle \theta$ to the ground, the surface of the water makes angle $\displaystyle \theta$ to the side of the tank. ("Alternate interior angles" in parallel lines.)
You can use that to calculate the volume between the top of the water and the top of the tank. The volume of water left in the tank is the volume of the tank minus that volume. And, of course, the rate at which the water is coming out is the derivative of the that function.
3. ## Re: Differential calculus :Related rates
Hello,
Would you draw a picture of exact physical situation of this problem? Because i didn't understand some part of your answer. The volume of right circular cylinder is $\pi r^2$h.So the volume of the water and volume of the tank will be same.
4. ## Re: Differential calculus :Related rates
Originally Posted by Vinod
Hello,
Would you draw a picture of exact physical situation of this problem? Because i didn't understand some part of your answer. The volume of right circular cylinder is $\pi r^2$h.So the volume of the water and volume of the tank will be same.
As the tank is being tilted, it is no longer a right circular cylinder of water. Now, the bottom is a right circular cylinder, but the top is slanted. It is like an angular slice to take the top of the cylinder off. The top surface would look like an oval, technically. That would be the case until the water drained to the point where the bottom of the tank was no longer completely covered with water.
5. ## Re: Differential calculus :Related rates
volume left in the tank as a function of $\theta$ ...
$V = \pi r^2 h - \pi r^3\tan{\theta}$
$\dfrac{dV}{dt} = -\pi r^3 \cdot \sec^2{\theta} \cdot \dfrac{d\theta}{dt}$
$\theta$ increases at a constant rate $\implies \dfrac{d\theta}{dt} = k$
$\dfrac{dV}{dt} = -k \pi r^3 \cdot \sec^2{\theta}$
$\dfrac{dV}{dt}\bigg|_{\theta = 0} = -k \pi r^3 \sec^2(0) = -k \pi r^3 \cdot (1)$
$\dfrac{dV}{dt}\bigg|_{\theta = \frac{\pi}{4}} = -k \pi r^3 \sec^2\left(\dfrac{\pi}{4}\right) = -k \pi r^3 \cdot (2)$
6. ## Re: Differential calculus :Related rates
Hello skeeter, how did you arrive at $\pi r^3 tan\theta$. I understood all of your other steps in your calculation.
7. ## Re: Differential calculus :Related rates
Originally Posted by Vinod
Hello skeeter, how did you arrive at $\pi r^3 tan\theta$. I understood all of your other steps in your calculation.
Consider the top of the tank to be a cylinder where the lowest point of where the water hits the tank is the bottom and the top of the tank is the top. The base is a cylinder of radius $r$ and height (as shown in the diagram) $2r\tan \theta$. So, the volume of the full cylinder is $2\pi r^3 \tan \theta$. But, half of the cylinder is empty (the diagonal line that skeeter drew cuts the cylinder exactly in half). So, the volume of the filled portion is $\pi r^3 \tan \theta$.
8. ## Re: Differential calculus :Related rates
Hello SlipEternal, The tank has been tipped $45^\circ$,that's why you are saying half of the cylinder is empty or for any other reason? I think if the tank has been tipped $30^\circ$, the diagonal line won't cut the height of the cylinder exactly in half. Is my reasoning right?
9. ## Re: Differential calculus :Related rates
referencing the diagram ... tilted at any angle $\theta$ such that $2r\tan{\theta} \le H$, where $H$ is the height of the entire cylindrical tank, the upper cylinder formed between the red dashed line and the top of the tank is half full (or half empty)
10. ## Re: Differential calculus :Related rates
Hello,
If the height of the cylinder is at least thrice of its radius, then height of the cylinder would be $2rtan\theta$.In that case also, would the upper cylinder formed between the red dash line and the top of the tank would be half full[half empty]? | 2018-03-21T22:42:03 | {
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http://math.stackexchange.com/questions/732581/how-to-answer-the-following-combinatorial-question | # How to answer the following combinatorial question?
Consider the set $S = \{ 1,2, \cdots , 15 \}$. The general question is: how many subsets of $S$ exist such that the subsets contain 4 different elements without any consecutive numbers? As a sub-question, I have to show that this problem is equivalent with finding the number of solutions $$(*) \qquad a_1 + a_2 + a_3 + a_4 + a_5 = 14 ; \qquad a_1,a_5 \geq 0 ; a_2, a_3, a_4 \geq 2$$ To prove this, I tried the following. Consider the numbers 1 through 15 laid out next to each other in a row, starting with the number 1 and ending with 15. To find the subsets that obbey the given restrictions, we start of by picking some number $a_1 \geq 1$. Then, $a_2$ cannot be a consecutive number, so the distance to $a_1$ should be larger than or equal to two. Similar for $a_3$ and $a_4$. Then I don't know about $a_{5}$.
However: one is asked to prove the original problem is equivalent to proving that $a_1 \geq 0$, not $a_1 \geq 1$. Furthermore, I don't understand why $a_{5}$ enters the scene here: we should find subsets containing $4$ elements, right? Could you please explain this? Furthermore, I don't understand why the $(*)$-marked sum should be equal to 14. Could you please this explain this as well, or give me a hint?
Furthermore, I am asked to show that this number of subsets is equal to the coefficient of $x^{14}$ of the generating function $$f(x) = \frac{x^6}{(1-x)^5} .$$ How does one show that? Or could you please give a hint?
-
Imagine we have picked $4$ numbers, no two consecutive. List them in order, as $p,q,r,s$.
Let $a_1$ be the number of integers in the interval $[1,15]$ that are less than $p$, and let $a_5$ be the number of integers in the interval that are greater than $s$. Let $a_2$ be the number of integers in the interval $(p,q]$. (So we do not count $p$, but we do count $q$.) Let $a_3$ be the number of integers in the interval $(q,r]$, and $a_4$ the number of integers in the interval $(r,s]$.
Note that $a_1\ge 0$, $a_5\ge 0$, and each of $a_2,a_3,a_4$ is $\ge 2$. Note also that $a_1+a_2+a_3+a_4+a_5=14$. For let $g_1$ be the "gap" up to the first of our chosen integers, $g_2$ the gap between the first two chosen integers, $g_3$ the gap between the next two, $g_4$ the gap between the next two, and $g_5$ the final gap. Then $g_1+g_2+g_3+g_4+g_5 +4=15$. We have $g_1=a_1$, $g_5=a_5$, and for the remaining three $i$ we have $g_i=a_i-1$.
Thus $a_1+(a_2-1)+(a_3-1)+(a_4-1)+a_5+4=15$, giving $a_1+a_2+a_3+a_4+a_5=14$. It is the fact that between our four numbers there are three gaps that accounts for the $14$.
We have shown that for every choice of numbers $p,q,r,s$, there are numbers $a_1,\dots,a_5$ satisfying the given inequalities, and with sum $14$.
It is not hard to reverse the argument, and show that any sequence $a_1,\dots,a_5$ that satisfies the given inequalities and with sum $14$ determines a choice of $p\lt q\lt r\lt s$ with no two consecutives.
-
Let $1\leq x<y<z<u\leq15$ such that $y\ne x+1$, $z\ne y+1$ and $u\ne z+1$. Then $\{x,y,z,u\}$ is a subset of $S$ that contains exactly $4$ non-consecutive elements. Taking $a_{1}=x-1$, $a_{2}=y-x$, $a_{3}=z-y$, $a_{4}=u-z$ and $a_{5}=15-u$, the conditions can be rewritten as $a_{1}\ge0$, $a_{i}\geq2$ for $i=2,3,4$, $a_{5}\geq0$ and $a_{1}+\cdots+a_{5}=14$.
Let $a_{1}=b_{1}$, $a_{5}=b_{5}$ and $a_{i}=b_{i}+2$ for $i=2,3,4$. Under condition $b_{i}\geq0$ the number of solutions of $b_{1}+b_{2}+b_{3}+b_{4}+b_{5}=8$ equals $\binom{12}{4}$.
-
Use 'Stars and Bars' to count.
We can represent the four numbers as a string of 4 'bars' and 11 'stars', where the position of each the bar counts off a number.
So the string: "$**|*|***|**|***$" represents: $\{3, 5, 9, 12\}$.
Since we need at least one 'star' between each pair of 'bars', this is equivalent to finding the permutations of 4 bars and 8 remaining stars.
So there are $\dbinom{12}{4} = \dfrac{12!}{4!\;8!} = 495$ ways to do this.
Your next step is to show that the problem of five variables can also be represented by the same string.
- | 2014-10-01T04:27:14 | {
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http://mathhelpforum.com/advanced-algebra/67690-how-many-subspace-f2.html | # Thread: How many subspace in F2?
1. ## How many subspace in F2?
Hi, I have to find the number of subspaces of $\mathbb F^2$ where $\mathbb F$ is a finite field with $n$ elements.
I know there are the trivial subspaces.
I thought the subspaces may be the possible lines ( I mean the possible inclinations )
For example, $\mathbb F_7^2$ would have the lines whose parallel vectors are (0,1) (0,2) ... (0,6) (1,0) (1,1) (1,2) (1,3) ... (1,6) (2,0) (2,1) (2,3) (2,5) etc.
Is it right?
2. Originally Posted by vincisonfire
Hi, I have to find the number of subspaces of $\mathbb F^2$ where $\mathbb F$ is a finite field with $n$ elements.
I know there are the trivial subspaces.
I thought the subspaces may be the possible lines ( I mean the possible inclinations )
For example, $\mathbb F_7^2$ would have the lines whose parallel vectors are (0,1) (0,2) ... (0,6) (1,0) (1,1) (1,2) (1,3) ... (1,6) (2,0) (2,1) (2,3) (2,5) etc.
Is it right?
Here is a hint. A subspace of $\mathbb{F}^2$ must be an additive subgroup of $(\mathbb{F}^2,+)$. If you take for example, $\mathbb{F}_7\times \mathbb{F}_7$ the additive subgroups include $\{ 0 \}\times \{ 0\}$, $\mathbb{F}_7\times \{ 0 \}$, $\{ 0\}\times \mathbb{F}_7$, and $\mathbb{F}_7\times \mathbb{F}_7$. Now you need to see which ones are also closed under scalar multiplication.
3. I don't see why for example $\{ (a,b) : b = 2a \}$ does not work.
(0,0) (1,2) (2,4) (3,6) (4,1) (5,3) (6,5) isn't a additive subgroup?
And isn't it closed under scalar multiplication?
4. I am sorry! This is my second time making this mistake . For some reason I assume the subgroups of $G_1\times G_2$ must have form $H_1\times H_2$ where $H_1,H_2$ are subgroups. But the point still applies above that a subspace must be a subgroup.
5. Would it be 3 + n because you have the subspaces "generated" by (1,0) (1,1) ... (1,5) (1,6) and (0,1) plus the two trivial subgroups?
I think it is that. Thanks a lot!
6. Originally Posted by vincisonfire
Would it be 3 + n because you have the subspaces "generated" by (1,0) (1,1) ... (1,5) (1,6) and (0,1) plus the two trivial subgroups?
That is correct. If a subspace S is not the zero subspace then it must contain a nonzero element (x,y). If x=0 then $S = \{0\}\times\mathbb{F}$. If x≠0 then $x^{-1}(x,y) = (1,x^{-1}y)\in S$, so S contains all the elements $(k,kx^{-1}y)$. If S contains other elements than those, then you should be able to prove by similar arguments that S is the whole of $\mathbb{F}\times\mathbb{F}$.
Edit: That Escher avatar is TOO BIG.
7. Better now?
8. Originally Posted by vincisonfire
Better now?
That's more like it.
9. there is a formula for the general case: (something to think about!)
suppose $\mathbb{F}$ is a finite field of order $n.$ the number of subspaces of $\mathbb{F}^m$ of dimension $1 \leq k \leq m$ is: $\prod_{j=0}^{k-1} \frac{n^{m-j}-1}{n^{k-j} - 1}.$ so the number of all subspaces of $\mathbb{F}^m$ is: $1 + \sum_{k=1}^m \prod_{j=0}^{k-1} \frac{n^{m-j}-1}{n^{k-j} - 1}.$ | 2013-12-05T18:23:53 | {
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https://www.physicsforums.com/threads/oscillators-and-conservation-of-energy.956131/ | # Oscillators and conservation of energy
• I
## Main Question or Discussion Point
In the equation 7.4, the author is taking v0=√(C/M)*x, and I don't get where does that come from. I would really appreciatte your help, thanks.
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berkeman
Mentor
Could you please define the terms in the equation you posted? This is for a spring oscillator, right? Thanks.
Can you say what you think is being expressed with this equation? What is the total energy of a simple spring oscillator made up of?
berkeman
Mentor
In the equation 7.4, the author is taking v0=√(C/M)*x
And you posted equation 7.5.
And you posted equation 7.5.
Sorry, I made a typo, I posted the correct one.
What is the total energy of a simple spring oscillator made up of?
Yes.
This is for a spring oscillator, right?
Yes.
Could you please define the terms in the equation you posted?
It doesn't say much, cause it's trying to explain spring oscillators in a general way.
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George Jones
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v0=√(C/M)*x
Maybe I missed it, but I don't see this in any of the stuff you posted.
Maybe I missed it, but I don't see this in any of the stuff you posted.
If you do the integral you got:
. Then -M*v02=C*x2. So v0=√(-C/M)*x.
Edit: i just realized i forgot to put the minus.
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jbriggs444
Homework Helper
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If you do the integral you got: View attachment 231123. Then -M*v02=C*x2. So v0=√(-C/M)*x.
Edit: i just realized i forgot to put the minus.
If you are going to post typeset equations, it would be worthwhile to learn a little LaTeX. Info => HowTo => https://www.physicsforums.com/help/latexhelp/
For example, $$\frac 1 2 Mv^2 - \frac 1 2 Mv_0^2$$
If you are going to post typeset equations, it would be worthwhile to learn a little LaTeX. Info => HowTo => https://www.physicsforums.com/help/latexhelp/
For example, $$\frac 1 2 Mv^2 - \frac 1 2 Mv_0^2$$
If you do the integral you got:
. Then -M*v02=C*x2. So v0=√(-C/M)*x.
$$-M v^2_0 = C x^2 .$$ Therefore $$v_0 = \sqrt {\frac {-C} {M}} x .$$
Btw, the text defines $w_0 = \sqrt {\frac {C} {M}}, v_0 = w_0 A \cos(\phi).$ Then if $v_0 = \sqrt {\frac {-C} {M}} x, x = A \cos(\phi).$ But we got that $x = A \sin(\phi),$ so $\cos(\phi)$ would be equal to $\sin(\phi)$ and that's not true. And even if it were true, i'm not taking into account that $w_0 = \sqrt {\frac {C} {M}},$ and not equal to $\sqrt {\frac {-C} {M}}.$
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George Jones
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Okay, now that I have figured out that I have a copy of the text (used in the first mechanics course I took all those decades ago), maybe we can have a discussion.
I don't understand what you mean when you write
If you do the integral you got $\frac{1}{2} Mv^2 - \frac{1}{2} Mv_0^2 =$.
Do what integral? Also, this only half of an equation!
Then $-M v^2_0 = C x^2$
I don't see how you got this.
Equation (7.5) in the text is
$$\frac{1}{2} Mv^2 + \frac{1}{2} Cx^2 = \frac{1}{2} M \left( \frac{dx}{dt} \right)^2 + \frac{1}{2} Cx^2 = E,$$
where $E$ is the constant $\frac{1}{2} CA^2$. Here, $C$ is the spring constant and $A$ is the amplitude of oscillation..
This means that
$$\frac{1}{2} Mv^2 + \frac{1}{2} Cx^2 = \frac{1}{2} CA^2$$
at all times $t$. Setting $t=0$ gives
$$\frac{1}{2} Mv_0^2 + \frac{1}{2} Cx_0^2 = \frac{1}{2} CA^2,$$
so that
$$v_0^2 = \frac{C}{M} \left( A^2 - x_0^2 \right) = \omega_0^2 \left( A^2 - x_0^2 \right).$$
Also, I do not understand
But we got that $x = A \sin(\phi)$
Actually, $x = A \sin \left( \omega_0 t + \phi \right)$.
Do what integral? Also, this only half of an equation!
(5.12) integral. And yes, i just realized that I cropted it in a wrong way, but basically: $$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2.$$
I don't see how you got this.
Cause we have that
$$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2 = \frac 1 2 Mv^2-\frac 1 2 Cx^2,$$
and that's only true if $−Mv_0^2=Cx^2.$
Actually, $x = A \sin \left( \omega_0 t + \phi \right)$.
I'm just taking that from the text book. I already posted that part but i will do it again if that helps u:
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George Jones
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Cause we have that
$$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2 = \frac 1 2 Mv^2-\frac 1 2 Cx^2,$$
I agree with
$$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2 ,$$
but I do not understand the second equals sign, I do not understand why
$$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2$$
is also equal to
$$\frac 1 2 Mv^2-\frac 1 2 Cx^2 .$$
I'm just taking that from the text book.
No, you have not taken it from the textbook. You wrote (and used)
But we got that $x = A \sin(\phi)$
which is incorrect, and which is not in the textbook.
I agree with
$$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2 ,$$
but I do not understand the second equals sign, I do not understand why
$$\Delta T = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2$$
is also equal to
$$\frac 1 2 Mv^2-\frac 1 2 Cx^2 .$$
Sorry, sorry, I'm really sorry, i just made a lot of typos, I meant
$$\frac 1 2 Mv^2+\frac 1 2 Cx^2 .$$
That's because
$$\Delta (T+V) =\Delta E = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2,$$
cause here $\Delta V=0.$ And we also know that
$$\frac{1}{2} Mv^2 + \frac{1}{2} Cx^2 = \frac{1}{2} M \left( \frac{dx}{dt} \right)^2 + \frac{1}{2} Cx^2 = E,$$
then
$$\frac 1 2 Mv^2-\frac 1 2 Mv_0^2 = \frac{1}{2} M \left( \frac{dx}{dt} \right)^2 + \frac{1}{2} Cx^2.$$
No, you have not taken it from the textbook. You wrote (and used)
Sorry, i just didn't read "at t=0", which is a huge mistake, but even like that i think is kind of weird cause he have that
$$v_0 = \sqrt {\frac {C} M} A \cos(\phi) = \sqrt {\frac {-C} M}x .$$
And we know that
$$x = A \sin \left( \omega_0 t + \phi \right).$$
Then
$$v_0 = \sqrt {\frac {C} M} A \cos(\phi) = \sqrt {\frac {-C} M} A \sin \left( \omega_0 t + \phi \right).$$
$$\sqrt {-1}\cos(\phi)=\sin \left( \omega_0 t + \phi \right).$$
I probably made a mistake, or something like that cause i find this equation really weird.
George Jones
Staff Emeritus
Gold Member
That's because
$$\Delta (T+V) =\Delta E = \frac 1 2 Mv^2-\frac 1 2 Mv_0^2,$$
cause here $\Delta V=0.$
No,
$$\Delta V = \frac{1}{2} C x^2 - \frac{1}{2} C x_0^2 .$$
See equation (5.20).
No,
$$\Delta V = \frac{1}{2} C x^2 - \frac{1}{2} C x_0^2 .$$
See equation (5.20).
Thank you so much! I finally get it, all of my questions were between pages 149 and 150. We will see conservation of energy in three classes, that's why I was so confused about that. | 2020-04-02T15:18:15 | {
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http://hajq.yiey.pw/eigenvalue-and-eigenfunction-pdf.html | In this section we will learn how to solve linear homogeneous constant coefficient systems of ODEs by the eigenvalue method. Eigenfunction expansion of some self-adjoint operators corresponding to general elliptic problems with an eigenvalue in the boundary conditions. An observation of practical in-terest is that the contributions from higher eigenfunctions in the expan-sion (3) are suppressed by the factors e–λ nT. For the B-CH problem on the unit square, the first eigenvalue is the double eigenvalue ˇ2 ˇ9:869604401 whose eigenspace is spanned by the functions cos(ˇx 1) and cos(ˇx 2). The operator Hg,ahas no eigenvalues. Note that, if ψ(x) is an eigenfunction with eigenvalue λ, then aψ(x) is also an eigenfunction with the same eigenvalue λ. 1) where is the eigenvalue (or characteristic value, or proper value) of matrix A, and x is the corresponding right eigenvector (or characteristic vector ,or proper vector) of A. eigenvalue problems, and. Eigenvalues and Eigenvectors §IV. We aim at saying as much as possible about the spectra of three classes of linear diffusion operators involving nonlocal terms. In particular, note that for λ = 0 the eigenfunction f(t) is a constant. Local versus global analysis of eigenfunctions 9 1. Lecture 11: Eigenvalues and Eigenvectors De &nition 11. Quantum Harmonic Oscillator Eigenvalues and Wavefunctions: Short derivation using computer algebra package Mathematica Dr. However, for large matrices the power method should still be much, much, faster than using the EIGEN routine to compute all eigenvalues. Because of the definition of eigenvalues and eigenvectors, an eigenvalue's geometric multiplicity must be at least one, that is, each eigenvalue has at least one associated eigenvector. We let E be the set of eigenvalues of the boundary value problem (1. ca February 27, 2017 Abstract If replace the Hermiticity from conventional quantum mechanics with the physi-. However, in the one-dimensional case (plane layered media) the consideration appears significantly simpler. Of particular interest in many settings (of which differential equations is one) is the following. 372 Chapter 7 Eigenvalues and Eigenvectors 7. (iii) Suppose that q 1 is an eigenfunction with eigenvalue 1 and q 2 is an eigenfunction with eigenvalue 2. 'Eigenvalue and the Principal Eigenfunction of Schrodinger's Equation 1 By M. Munoz; Delgado, V. Which of the following statements is/are false for a given set of QMHO wave functions corresponding to the same harmonic potential V? (a) The ground state energy is zero, i. For the matrix, A= 3 2 5 0 : Find the eigenvalues and eigenspaces of this matrix. Intensity plots and excursion sets 7 1. I'm not sure how to come up with something more general though. The eld variables are solid displacement and uid pressure. Section 5-3 : Review : Eigenvalues & Eigenvectors. Eigenfunction Expansions 3 1. Laplace-Beltrami Eigenfunctions for Deformation Invariant Shape Representation Raif M. Note: 2 lectures, §5. Chapter 8 Eigenvalues So far, our applications have concentrated on statics: unchanging equilibrium conflg-urations of physical systems, including mass/spring chains, circuits, and structures, that are modeled by linear systems of algebraic equations. Note: OCR errors may be found in this Reference List extracted from the full text article. High frequency limits, oscillation and concentration 10 1. Eigenvalue and Eigenvector Calculator. Here we have an exact, testable expansion. 1 Introduction In this chapter, we are going to find explicitly the eigenfunctions and eigenvalues for the time-independent Schrodinger equation for the one-dimensional harmonic oscillator. Eigenvalues and eigenspaces Given a vector space V, a subspace W, a linear operator : W !V, and a constant , we say that is an eigenvalue of (relative to W) if the operator I : W !V has a nontrivial kernel (null space). Math 2280 - Assignment 6 Section 3. An eigenspace of a given transformation for a particular eigenvalue is the set ( linear span ) of the eigenvectors associated to this eigenvalue, together with the zero vector (which has no direction). Eigenvalue definition is - a scalar associated with a given linear transformation of a vector space and having the property that there is some nonzero vector which when multiplied by the scalar is equal to the vector obtained by letting the transformation operate on the vector; especially : a root of the characteristic equation of a matrix. Each eigenvector is, in effect, multiplied by a scalar, called the eigenvalue corresponding to that eigenvector. and λis the corresponding eigenvalue of Oˆ. In the case of M being a planar region, f(u, v) in equation (8) can be understood as the natural vibration form (also eigenfunction) of a homogeneous membrane with the. paper contains our results on individual eigenvalues and eigenfunctions of ordinary differential operators. If the operator is now a hamiltonian, the eigenvalue you get will be the energy of the system, and the eigenvector tell you its "state". Although there may exist other real eigenvalues of % possessing positive eigen-functions, the principal eigenvalue characterizes the validity of the maximum principle for %. The eigenvalues are the natural frequencies (or eigenfrequencies) of vibration, and the eigenvectors are the shapes of these vibrational modes. There are many examples known of this type. 2006-11-03. Note that the trivial solution X ≡ 0 is an eigenfunction of every eigenvalue. Examples of the relationship between scatterer properties and eigenvalues and eigenvectors of the scattering matrix are presented. The rain and the cold have worn at the petals but the beauty is eternal regardless. The present work considers the application of the generalized integral transform technique (GITT) in the solution of a class of linear or nonlinear convection–diffusion problems, by fully or partially incorporating the convective effects into the chosen eigenvalue problem that forms the basis of the proposed eigenfunction expansion. When a system is in an eigenstate of observable A (i. eigenvalue, the principal eigenvalue, % , of , and it is associated with a posi-tive eigenfunction. Quantum Harmonic Oscillator Eigenvalues and Wavefunctions: Short derivation using computer algebra package Mathematica Dr. Thus, B!i = bi!i for some constant bi. When the result of an operator acting on a function is a constant multiplied by the same function, the function is called an eigenfunction, and the. Notation for eigenvalues 3 1. Because (as you have seen in class) this is a complicated subject, there are a few twists and turns in the Maple implementation, too. b/ Eigenfunction expansion for G: We need to determine the eigenvalues and eigenfunctions of the Euler differential operator with Neumann BC. Chapter Five - Eigenvalues , Eigenfunctions , and All That The partial differential equation methods described in the previous chapter is a special case of a more general setting in which we have an equation of the form L 1 ÝxÞuÝx,tÞ+L 2 ÝtÞuÝx,tÞ = F Ýx,tÞ. I - Eigenvalue Problems: Methods of Eigenfunctions - V. In this work, we study an eigenvalue problem for the infinity-Laplacian on bounded domains. Daileda Sturm-Liouville Theory. Basic properties, some applications and examples in system analysis. We can talk about eigenvalues and eigenfunctions for regular or singular problems. 1 Degenerate Perturbation Going back to our symmetric matrix example, we have A 2IRN N, and again, a set of eigenvectors and eigenvalues: Ax i = i x i. Uses anorthogonal linear transformationto convert a set of observations to a. However, A2 = Aand so 2 = for the eigenvector x. Eigenvalues of regular Sturm-Liouville problems Q. Further, assume that no zeros of f lie on the cycles of Γ. Eigenvalue Problems Eigenvalue problems arise in many contexts in physics. VASSILEV Abstract. Groningen. , if solution is stable, then Backward Euler is stable for any positive step size: unconditionally stable • Step size choice can manage efficiency vs accuracy without concern for stability – Accuracy is still O(h). The theory underlying the procedure is explained and two cases treated numerically. The Dirichlet eigenvalue problem involves the determination of a solution X(x)of (1) in a domain [0,L]for some λthat satisfies the boundary conditions X(0)=X(L)=0. PDF Owner Manuals and User Guides are NOT affiliated with the products and/or names mentioned in this site. In matrix form, A x = x This is somewhat different from our previous SLE, which had the form A x = b where A, b were assumed known. Weyl’s law for ( )-eigenvalues 3 1. Math 124B: PDEs Eigenvalue problems for differential operators We want to find eigenfunctions of (linear) differential operators acting on functions on the interval [0,l] that satisfy boundary conditions at the endpoints. Note that, if ψ(x) is an eigenfunction with eigenvalue λ, then aψ(x) is also an eigenfunction with the same eigenvalue λ. Eigenvectors and Hermitian Operators 7. This leads to an algorithm for determining shapes for which the first N eigenvalues (counted in proportion to the multiplicity of their eigenfunctions) coincide with N prescribed values. Section 5-3 : Review : Eigenvalues & Eigenvectors. Note that in this case the eigenfunction is itself a function of its associated eigenvalue λ, which can take any real or complex value. By this logic i is an operator with the plane wave being its eigenfunction corresponding to x. Working Skip trial 1 month free. Note that eigenvalue is simple. Proposition 5 The eigenvalues of a regular Sturm-Liouville problem are simple. 635-652, 2013. Time independent Schrödinger equation (Text 5. 6 Wave Equation on an Interval: Separation of Vari-ables 6. Oscillation theory and the spectra of eigenvalues The basic problems of the Sturm-Liouville theory are two: (1) to establish the existence of eigenvalues and eigenfunctions and describe them qualitatively and, to some extent, quantitatively and (2) to prove that an “arbitrary” function can be expressed as an infinite series of eigenfunctions. Dynamics of the geodesic or billiard ow 6 1. For the fractional eigenvalue problem - it holds that and the eigenfunction is a stationary (minimum) value of the above ratio. The eigenvalues are the natural frequencies (or eigenfrequencies) of vibration, and the eigenvectors are the shapes of these vibrational modes. In fact, it is common to use such linear-algebraic-like nomenclature throughout Sturm-Liouville theory; and henceforth we shall refer to the functions y(x) that satisfy (8) as eigenfunctions of Land the numbers as the corresponding eigenvaluesof L. can easily deduce from equation (1. EIGENVALUE INEQUALITIES FOR MIXED STEKLOV PROBLEMS 5 uniform cross-section of the free surface of the steady fluid. Joy Visualization and Graphics Research Group Department of Computer Science University of California, Davis In engineering applications, eigenvalue problems are among the most important problems connected with matrices. And the lambda, the multiple that it becomes-- this is the eigenvalue associated with that eigenvector. Even so, if that abstract direction is unchanged by a given linear transformation, the prefix "eigen" is used, as in eigenfunction, eigenmode, eigenface, eigenstate, and eigenfrequency. We can’t find it by elimination. 6 Sturm-Liouville Eigenvalue Problems 6. Vectors with. The method. The eigensystem—eigenvalues and eigenvectors—of the Euler equations of inviscid flow form the basis of total variation diminishing (TVD) algorithms in computational fluid dynamics (CFD). 1 Lecture 3: Operators in Quantum Mechanics If is an eigenfunction of A^ with eigenvalue a, then, assuming the wave function to be normalized, we have. QUANTUM MECHANICS Operators This means that if f(x) is an eigenfunction of A with eigenvalue k, then cf(x) is also an eigenfunction of A with eigenvalue k. McNames Portland State University ECE 223 Complex Sinusoids Ver. Request PDF on ResearchGate | Eigenvalues and Eigenfunctions | The article describes the eigenvalue and eigenfunction problems. In this paper we study the eigenvalues and eigenfunctions of metric measure manifolds. is an eigenfunction. Intensity plots and excursion sets 7 1. Eigenvalues and eigenvectors in Maple Maple has commands for calculating eigenvalues and eigenvectors of matrices. The theory underlying the procedure is explained and two cases treated numerically. In this section we will learn how to solve linear homogeneous constant coefficient systems of ODEs by the eigenvalue method. When there is a basis of eigenvectors, we can diagonalize the matrix. 6 Sturm-Liouville Eigenvalue Problems 6. In the computation we encountered a certain eigenvalue problem and found the eigenfunctions \(X_n(x)\text{. 1 Lecture 3: Operators in Quantum Mechanics If is an eigenfunction of A^ with eigenvalue a, then, assuming the wave function to be normalized, we have. MAS214: LODE know that, if there is an eigenfunction with a non-zero eigenvalue, then it has the. which will yeild an asymptotic expansion for eigenvalues and eigenfunctions. If the conditions q(x) ≤ 0, on [a,b], and a 1a 2 ≤ 0, b 1b 2 ≥ 0 are satisfied, then show that λ ≥ 0. 1 Eigenvalues and Eigenvectors The product Ax of a matrix A ∈ M n×n(R) and an n-vector x is itself an n-vector. tion and the exact solution for the values of the eigenfunctions and eigenvalues computed above. Eigenvalues and Eigenfunctions The wavefunction for a given physical system contains the measurable information about the system. Matrix eigenvectors and eigenvalues Given an N ×N matrix A, and if Av =λ⋅v then λ is an eigenvalue and is an eigenvalue of. there are more than one eigenfunctions of A with. The eigenvalues are the natural frequencies (or eigenfrequencies) of vibration, and the eigenvectors are the shapes of these vibrational modes. I am currently a Professor and Associate Director of Research in the Mathematical Sciences Institute at ANU. With this information, we use classical eigenfunction expansion methods and bi-orthogonality to compute expansion coefficients that characterize the initial conditions and time-dependent source, yielding a method for calculating the transient behaviour. SciTech Connect. EIGENVALUES AND EIGENVECTORS Definition 7. description of the eigenvalue problem and an intuitive introduction to dimension reduction. Eigenvalues and Eigenvectors. Note that in this case the eigenfunction is itself a function of its associated eigenvalue λ, which can take any real or complex value. Eigenvalues and Eigenvectors 6. ities of all of the distinct eigenvalues. where k is a constant called the eigenvalue. SPECTRAL APPROXIMATION TO A TRANSMISSION EIGENVALUE PROBLEM AND ITS APPLICATIONS TO AN INVERSE PROBLEM JING AN1 JIE SHEN 2;3 Abstract. ca January 25, 2017 Abstract If replace the Hermiticity from conventional quantum mechanics with the physi-. Suppose an eigenvalue λ of Γ has an eigenfunction f which is non-zero on internal vertices of Γ. The value of the observable for the system is the eigenvalue, and the system is said to be in an eigenstate. Abstract - The article describes the eigenvalue and eigenfunction problems. In the last lecture, we established that:. IVANOV AND D. A vector which is "flipped" to point in the opposite direction is also considered an eigenvector. Hermitian matrices Hermitian matrices satisfy H ij = H∗ ji = H † ij where H † is the Hermitian conjugate of H. Consider the initial value problem for the heat equation tu x,t D xxu x,t,0 x 1, t 0, u x,0 f x L2 0,1 with BC. Introduction. Define a right eigenvector as a column vector satisfying. Time for a lengthy and somewhat provocative guest post on the subject of the interpretation of quantum mechanics! --o-- Galileo advocated the heliocentric system in a socratic dialogue. k = e P(tanhx) and eigenvalue E= k2 for any kin the range 1 0 2 0, then there is a TE with k2 1+ m 2 1+m 1 0 John Sylvester Transmission Eigenvalues and Non-Radiating Sources. The solution of the Schrödinger equation is tantamount to seeking a function that is an eigenfunction of. But even in this case, if that abstract direction is unchanged by a given linear transformation, the prefix "eigen" is used, as in eigenfunction, eigenmode, eigenface, eigenstate, and eigenfrequency. Daileda Sturm-Liouville Theory. Some important theorems dealing with the properties and applications of eigenvalues will be dealt with. We use the fact that L induces a complete orthonormal basis for L2(›) to allow us to perform eigenfunction expansion in L2(›). In this equation, x is an eigenvector of A and λ is an eigenvalue of A. For each boundary condition case, a plot of the first few normalized eigenfunctions are given as well as the numerical values of the first few eigenvalues for the special case when. where λ m is the eigenvalue of A corresponding to the eigenfunction F m, m = 0, 1, …, N - 1. In particular, note that for λ = 0 the eigenfunction f(t) is a constant. In the following, we restrict ourselves to problems from physics [7, 18, 14] and computer science. MAS214: LODE know that, if there is an eigenfunction with a non-zero eigenvalue, then it has the. Keywords Differential Equation Partial Differential Equation Linear Operator Fourier Analysis Impulse Response. If v and w are. But if $i \hbar f = 0$ then the eigenfunction must be zero, but this doesn't count since then every number would be an eigenvalue. Consider a linear mapping f:V → V, where V is a vector space with field of scalars F. Thus 0 is an eigenvalue with eigenfunction being any non-zero constant. We prove that any eigenfunction is C 1, α at its critical points and C ∞ elsewhere. However, in the one-dimensional case (plane layered media) the consideration appears significantly simpler. Let Downloaded 10/27/14 to 38. 1 What makes eigenvalues interesting? In physics, eigenvalues are usually related to vibrations. Robust solution methods for nonlinear eigenvalue problems These pr´esent ´ee le 29 ao ut 2013ˆ a la Facult ´e des Sciences de Base Chaire d'algorithmes num´eriques et calcul haute performance. Theorem The eigenvalues of a regular S-L problem form an increasing sequence of real numbers λ 1 < λ 2 < λ 3 < ··· with lim n→∞ λ n = ∞. Problems 1-5 are called eigenvalue problems. 224 CHAPTER 7. Short lecture on eigenvalues and eigenfunctions. Eigenvectors and Hermitian Operators 7. Tensor spherical harmonics for the 2-sphere and 3-sphere are discussed as eigenfunction problems of the Laplace operators on these manifolds. Eigenvalues and Eigenvectors §IV. Laplace-Beltrami eigenvalues and topological features of eigenfunctions for statistical shape analysis [23]. Kelleher Spectral graph theory. In particular, we shall be interested in the spac-ings sbetween adjacent eigenvalues. b/ Eigenfunction expansion for G: We need to determine the eigenvalues and eigenfunctions of the Euler differential operator with Neumann BC. and λis the corresponding eigenvalue of Oˆ. Scattering matrices for arbitrary scatterers are calculated using a coupled finite‐element/integral equation method due to Kirsch and Monk [IMA J. The so-called Sturm-Liouville Problems de ne a class of eigenvalue problems, which include many of the previous problems as special cases. Physics 505 Homework No. For these abstract boundary eigenvalue problems the notions fundamental matrix function and characteristic matrix function are introduced, generalizing the. Eigenvalues and Eigenfunctions The wavefunction for a given physical system contains the measurable information about the system. is called an eigenfunction, and the corresponding value of λ is called its eigenvalue. Their use in. He was awarded. Math 124B: PDEs Eigenvalue problems for differential operators We want to find eigenfunctions of (linear) differential operators acting on functions on the interval [0,l] that satisfy boundary conditions at the endpoints. Laplace-Beltrami eigenvalues and topological features of eigenfunctions for statistical shape analysis [23]. CHAPTER Eigenvalues and the Laplacian of a graph PDF document - DocSlides- 1 Introduction Spectral graph theory has a long history In the early days matrix theory and linear algebra were used to analyze adjacency matrices of graphs Algebraic meth ods have proven to be especially e64256ective in treating graphs which are reg ID: 24031 ID: 24031. PDF: Eigenvalue and. Generalized MEM analysis 4. Introduction Before we start with the subject of this notes we want to show how one actually arrives at large eigenvalue problems in practice. Let λ be an eigenvalue of the regular SL problem. An alternative proof to show the simplicity of the first eigenvalue is given. Sturm Sequences and the Eigenvalue Distribution of the Beta-Hermite Random Matrix Ensemble by Cy P. Eigenvalues and eigenvectors in Maple Maple has commands for calculating eigenvalues and eigenvectors of matrices. This can only occur if = 0 or 1. 4 Applications of Eigenvalues and Eigenvectors Model population growth using an age transition matrix and an age distribution vector, and find a stable age distribution vector. Eigenvalueshave theirgreatest importance in dynamic problems. When the result of an operator acting on a function is a constant multiplied by the same function, the function is called an eigenfunction, and the. <0: The characteristic equation is r2 = 0, with roots r = i p. We prove a CR version of the Obata’s result for the rst eigenvalue of the sub-Laplacian in the setting. Robust solution methods for nonlinear eigenvalue problems These pr´esent ´ee le 29 ao ut 2013ˆ a la Facult ´e des Sciences de Base Chaire d'algorithmes num´eriques et calcul haute performance. We’ll take it as given here that all the eigenvalues of Problems 1-5 are real numbers. Quantum Mechanics 4 1. Chapter Eigenfunction Expansions The SturmLiouville PDF document - DocSlides- Among the triumphs of nineteenthcentury mathematics was the realization that these sequences of eigenfunctions can be used to represent arbitrary functions via in64257nite series Thus if 0 is the sequence of eigenfunctions of a SturmLiouville proble ID: 81824 ID: 81824. This problem has. Solving PDE’s by Eigenfunction Expansion Some of these problems are difficult and you should ask questions (either after class or in my office) to help you get started and after starting, to make sure you are proceeding correctly. Eigenfunction. the rates of change of the eigenvalues of the Helmholtz equation with respect to variations in the shape of the region. C/CS/Phys 191 Spin Algebra, Spin Eigenvalues, Pauli Matrices 9/25/03 Fall 2003 Lecture 10 Spin Algebra “Spin” is the intrinsic angular momentum associated with fu ndamental particles. DOC 13- 5 and you must include its orthonormalized eigenfunction(s) to get a complete orthonormal basis of L 2[0,T] (use the Gram-Schmidt procedure here). is the eigenfunction of the derivative operator, where f 0 is a parameter that depends on the boundary conditions. 1 Goal We know how to solve di⁄usion problems for which both the PDE and the BCs are homogeneous using the separation of variables method. In the following, we restrict ourselves to problems from physics [7, 18, 14] and computer science. The rain and the cold have worn at the petals but the beauty is eternal regardless. Notes on the Eigenfunction Method for solving differential equations Reminder:Weareconsideringtheinfinite-dimensionalHilbertspaceL 2 ([a,b]) of all square-integrable. Free Matrix Eigenvalues calculator - calculate matrix eigenvalues step-by-step. is an eigenvalue, fis an eigenfunction. It is easy to show that if is a linear operator with an eigenfunction , then any multiple of is also an eigenfunction of. In the one‐dimensional case, a symmetric fourth‐degree potential is used, and the solution is obtained in terms of eigenvalues and eigenfunctions. Section 11: Eigenfunction Expansion of Green Functions In this lecture we see how to expand a Green function in terms of eigenfunctions of the underlying Sturm-Liouville problem. McNames Portland State University ECE 223 Complex Sinusoids Ver. If the wavefunction that describes a system is an eigenfunction of an operator, then the value of the associated observable is extracted from the eigenfunction by operating on the eigenfunction with the appropriate operator. Lecture 11: Eigenvalues and Eigenvectors De &nition 11. Other methods that use spatial eigenfunctions 7. Eigenfunctions and eigenvalues I (Text 5-1) 1. The eigenvalue problem aims to find a nonzero vector x=[x i ] 1xn and scalar such that satisfy the following equation: Ax = x (1. "The factor with the largest eigenvalue has the most variance and so on, down to factors with small or negative eigenvalues that are usually omitted from solutions" (Tabachnick and Fidell, 1996, p. The calculator will find the eigenvalues and eigenvectors of the given square matrix, with steps shown. Oscillatory eigenfunctions. In this section we will learn how to solve linear homogeneous constant coefficient systems of ODEs by the eigenvalue method. Moreover, the eigenfunction corresponding to the first eigenvalue in the Dirichlet problem does not change sign. We will work quite a few examples illustrating how to find eigenvalues and eigenfunctions. In summary, eigenvalues and the corresponding eigenfunctions of are: For eigenvalue the eigenfunction is , For eigenvalue , the eigenfunction is. Working Skip trial 1 month free. The generalized eigenvalue problem is to determine the solution to the equation Av = λBv, where A and B are n-by-n matrices, v is a column vector of length n, and λ is a scalar. Such functions can be used to repre-sent functions in Fourier series expansions. If this happens, ψn(x) is known as the eigenfunction, or eigenstate, or eigenvector of operator G. So 1, 2 is an eigenvector. To understand spin, we must understand the quantum mechanical properties of angular momentum. This makes the problem more difficult to solve. However, A2 = Aand so 2 = for the eigenvector x. Eigenvalues of regular Sturm-Liouville problems Q. For ‚ > 0 the solution is as usual y = Acos(kx)+Bsin(kx), with derivativey0 =¡Aksin(kx)+Bkcos(kx). The eigenvalues of a Sturm-Liouville problem are the values of λ for which nonzero solutions exist. Chan Submitted to the Department of Electrical Engineering and Computer Science on May 23, 2007, in partial fulfillment of the requirements for the degree of Master of Science in Electrical Engineering and Computer Science Abstract. eigenvalue, unless the subspace contains the pertinent eigenfunction In dealing with the eigenvalue problem one must update the eigenfunctions and eigenvalues in each iteration step and in addition remove long wave perturbations pertaining to eigenfunctions with frequencies lower than that currently under evaluation. But e§ikx are both eigenfunctions with eigenvalues p = §„hk , respec-tively. Another difficulty is computing such eigen-functions; directly solving the Helmholtz equation (or the Laplacian eigenvalue. Asymptotic expressions for the lower eigenfunctions are found by means of singular perturbation theory, and the corresponding eigenvalues are obtained via a variational principle. Get YouTube without the ads. Computational algorithms and sensitivity to perturbations are both discussed. 93, 160406 (2004)] on the decay of a doubly quantized vortex is analyzed by numerically solving the Gross-Pitaevskii equation. Eigenvalues and eigenvectors have many applications in both pure and applied mathematics. It is common to compute the eigenvalue from global quantities, but the most basic definition of an eigenfunction and eigenvalue for a linear operator A is: Aψ(s)=kψ(s)(1) Note that this eigenvalue/eigenfunction relation is a pointwise relation at every s, rather than a global relation. 1 Eigenvalues and Eigenvectors Basic Definitions Let L be a linear operator on some given vector space V. STEM Support 1,470 views. 1 Eigenvalues and Eigenvectors The product Ax of a matrix A ∈ M n×n(R) and an n-vector x is itself an n-vector. Such functions can be used to repre-sent functions in Fourier series expansions. 9: pg 310, q 23. Eigenvalue and Eigenfunction for the PT-symmetric Potential V = (ix)N Cheng Tang1 and Andrei Frolov2 Department of Physics, Simon Fraser University V5A 1S6, Burnaby, BC, Canada [email protected] [5] Xuefeng Liu and Shin’ichi Oishi, Verified eigenvalue evaluation for Laplacian over polygonal domains of arbitrary. [8, 4] The inverse spectral theorem is to nd the set of graphs for which the form spectra are the same. The operator Hg,ahas no eigenvalues. Abstract: The energy eigenvalue, eigenfunction, matrix elements of coordinate and momentum operators in energy trepresentation, and evolution operator for a two-dimentional coupled oscillator are presented by using the general linear quantum transformation theory. Download preview PDF. Real, countable eigenvalues. Introduction Before we start with the subject of this notes we want to show how one actually arrives at large eigenvalue problems in practice. Abstract - The article describes the eigenvalue and eigenfunction problems. And it's corresponding eigenvalue is 1. Many more results can be proved about the eigenfunctions and eigenvalues of (6. (ii) The eigenvalues form an inflnite sequence ‚1 <‚2<‚3<¢¢¢such that ‚ n. In the case of M being a planar region, f(u, v) in equation (8) can be understood as the natural vibration form (also eigenfunction) of a homogeneous membrane with the. For the B-SSP problem on the unit square the first eigenvalue is the simple eigenvalue 2ˇ2 ˇ19:73920880 with eigenfunction sin(ˇx 1)sin(ˇx 2). What does this mean geometrically?. Note that, if ψ(x) is an eigenfunction with eigenvalue λ, then aψ(x) is also an eigenfunction with the same eigenvalue λ. In particular, it can be used to study the wave equation in higher. Eigenvalues, eigenvectors and applications Dr. Then λ = µ2, where µ is real and non-zero. Sturm-Liouville Eigenvalue Problems Motivation The heat flow in a nonuniform rod is modeled by the partial differential equation cρ ∂u ∂t = ∂ ∂x K 0 ∂u ∂x +Q (1) where the thermal coefficients c,ρ,K 0 are functions of x. To obtain specific values for physical parameters, for example energy, you operate on the wavefunction with the quantum mechanical operator associated with that parameter. STEM Support 1,470 views. Free Matrix Eigenvalues calculator - calculate matrix eigenvalues step-by-step. We’ll take it as given here that all the eigenvalues of Problems 1-5 are real numbers. Asymmetric eigenvector maps (AEM) 5. McNames Portland State University ECE 223 Complex Sinusoids Ver. Interpretation and Properties of 2. In the nonlinear context a separatrix plays the role of an eigenfunction and the initial conditions that give rise to the separatrix play the role of eigenvalues. the rates of change of the eigenvalues of the Helmholtz equation with respect to variations in the shape of the region. On an eigenvalue and eigenfunction problem of the equation $\Delta u+łambda u=0$ Imsik Hong. Suppose is any eigenvalue of Awith corresponding eigenvector x, then 2 will be an eigenvalue of the matrix A2 with corresponding eigenvector x. Two important concepts in Linear Algebra are eigenvectors and eigenvalues for a linear transformation that is represented by a square matrix. Furthermore, an eigenvalue's geometric multiplicity cannot exceed its algebraic multiplicity. According to Eq (2-32), the eigenvalue equation for momentum should read p^ xˆ(x) = ¡i„h dˆ(x) dx = pˆ(x) (6) where we have denoted the momentum eigenvalue as p. Jwamer and Khelan H. Eigenvalues and Eigenvectors. In the last lecture, we established that:. The eigenvalue problem aims to find a nonzero vector x=[x i ] 1xn and scalar such that satisfy the following equation: Ax = x (1. It turns out that if y (x) is an eigenfunction, then so is any non-zero multiple Cy (x), so we usually just take the constant C= 1. Proposition: The set of eigenfunctions belonging to an eigenvalue λ forms a vector space. Note that, if ψ(x) is an eigenfunction with eigenvalue λ, then aψ(x) is also an eigenfunction with the same eigenvalue λ. THE EIGENVALUE PROBLEM FOR A CLASS OF LINEAR INTEGRAL OPERATORS. Read "Eigenvalue and eigenfunction computations for Sturm-Liouville problems, ACM Transactions on Mathematical Software (TOMS)" on DeepDyve, the largest online rental service for scholarly research with thousands of academic publications available at your fingertips. Simple Eigenvalues De nition: An eigenvalue of Ais called simple if its algebraic multiplicity m A( ) = 1. Proof: Let v 1 and v 2 be eigenfunctions of the regular Sturm-Liouville problem (1), (2) with eigenvalue. eigenvalue λi and eigenfunction fi pairs). This matrix calculator computes determinant, inverses, rank, characteristic polynomial, eigenvalues and eigenvectors. We let E be the set of eigenvalues of the boundary value problem (1. We will work quite a few examples illustrating how to find eigenvalues and eigenfunctions. Such functions can be used to repre-sent functions in Fourier series expansions. LetX* andu*be anapproximate eigenvalue and eigenfunction which satisfy (1. In the case of M being a planar region, f(u, v) in equation (8) can be understood as the natural vibration form (also eigenfunction) of a homogeneous membrane with the. THE EIGENVALUE PROBLEM FOR A CLASS OF LINEAR INTEGRAL OPERATORS. Of particular interest in many settings (of which differential equations is one) is the following. This is true for our problem as well, see Theorem 8. • Altogether, A has n eigenvalues,butsomemaybecomplexnum-bers(eveniftheentriesof A arerealnumbers),andsomeeigenval-uesmayberepeated. This chapter enters a. SciTech Connect. (ii) Show that for any eigenvalue we can nd a real-valued eigenfunction. Section 11: Eigenfunction Expansion of Green Functions In this lecture we see how to expand a Green function in terms of eigenfunctions of the underlying Sturm-Liouville problem. Based on the tests in the present study, it can be concluded that DE is a useful tool for the parameter estimation of source bodies using magnetic anomalies. The solution of du=dt D Au is changing with time— growing or decaying or oscillating. Each eigenvector is, in effect, multiplied by a scalar, called the eigenvalue corresponding to that eigenvector. Free Matrix Eigenvalues calculator - calculate matrix eigenvalues step-by-step. an eigenvalue and y a corresponding eigenfunction of (1. This site consists of a compilation of public information available on the internet. So 1, 2 is an eigenvector. Vectors with. In this section we will learn how to solve linear homogeneous constant coefficient systems of ODEs by the eigenvalue method. For each boundary condition case, a plot of the first few normalized eigenfunctions are given as well as the numerical values of the first few eigenvalues for the special case when. Higher dimensional PDEs and multidimensional eigenvalue problems Notice that if = 0, then it must be that rv= 0, so that the eigenfunction is a constant. For the matrix, A= 3 2 5 0 : Find the eigenvalues and eigenspaces of this matrix. The solutions of this equation represent the spatial part of the solutions of the wave equation (with an infinite number of eigenvalue λ i and eigenfunction f i pairs). Semi-supervised Learning using Sparse Eigenfunction Bases Kaushik Sinha Dept. It decomposes matrix using LU and Cholesky decomposition. Moreover, the eigenfunction corresponding to the first eigenvalue in the Dirichlet problem does not change sign. <0: The characteristic equation is r2 = 0, with roots r = i p. Another way to view the behavior of eigenvalues is the process of diagonalization. Our later papers [FS2, FS3, FS4, FS5] will study sums of eigenvalues and sums of squares of eigenfunctions, and then pass to spherically symmetric three-dimensional problems by separation of variables. Eigenvalues and Eigenvectors 6. For each eigenvalue ln there exists an eigenfunction fn with n 1 zeros on (a,b). constant multipleof a λ-eigenfunctionis again a λ-eigenfunction. Let A be a square matrix (or linear transformation). EE603 Class Notes Version 1 John Stensby 603CH13. | 2020-01-26T07:49:16 | {
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https://math.stackexchange.com/questions/3153821/mastermind-game-states | # Mastermind game states
I'm trying to analyse a game of Mastermind and am having trouble quantifying the amount of possible game states. I know that a code has $$\text{# of colors}^{\text{# of pegs per guess}}$$ combinations (in my case that would be $$6^4=1296$$). However, an entire board state also consists of 10 guesses. Each guess has the same amount of combinations, thus my intuition would be that the amount of total states in a game of Mastermind would be $$\text{# of rows}^{\text{# of combinations per row}}$$. This approach yields $$11^ {1296}$$ board states which is astronomically large and I'm having a hard time believing this is true.
To clarify what I mean by a board state, I mean any legal state the game board can be in using the standard game rules. Having 3 empty rows, then one guess row and another 6 empty rows is not a legal board state.
How do I go about estimating this number?
Your formula going from rows to the full board is incorrect, and should be $$\#\text{combinations per row}^{\#\text{rows}}$$, giving $$1296^{11}$$ which is much less. Substituting in the formula for combinations per row, this is just $$(\#\text{colours}^{\#\text{pegs per row}})^{\#\text{rows}}=\#\text{colours}^{\#\text{pegs per row}\times\#\text{rows}}=\#\text{colours}^{\#\text{total pegs}}$$ which you can get directly as the number of ways to choose a colour for each peg.
As Arthur says, this is the number of possibilities for a completely full board. The number of possibilities for a board with only $$10$$ rows used is likewise $$(6^4)^{10}$$, then $$(6^4)^9$$ for $$9$$ rows, and so on, giving a total of $$\sum_{i=0}^{11}(6^4)^i=\frac{(6^4)^{12}-1}{6^4-1}$$
Other way round : $$\text{# of combinations per row}^{\text{# of rows}}$$
If you sum over 0 to 10 rows you get a geometric series, giving a total of $$\frac{\text{# of combinations per row}^{1+\text{# of rows}}-1} {\text{# of combinations per row}-1}$$ | 2019-06-24T21:27:20 | {
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https://math.stackexchange.com/questions/2305568/selecting-5-pairs-of-men-and-5-women-from-10-women-and-12-men | # Selecting 5 pairs of men and 5 women from 10 women and 12 men
## Question
A dance class consists of $22$ students, of which $10$ are women and $12$ are men. If $5$ men and $5$ women are to be chosen and then paired off, how many results are possible?
## Approach
According to me, the number of results possible is:
$$\binom{10}{5}*\binom{12}{5}*5!*2^{5}$$
$$\binom{10}{5}*\binom{12}{5}*5!$$
## My conclusion
Shouldn't be there $2$ options in each pair i.e ordering between men and women for $5$ such group, making it $5!$? Why is the answer not leaving $5!$? Are they not considering order? And if the order is important, is my answer correct in this case?
• Why would the be two options in each pair . A pair is of combination of a man and a woman . Even if you select fist a man then a woman to pair him or select first a woman and then a man to pair her both are same – ATHARVA Jun 1 '17 at 13:04
• if order is important, ur answer is right i believe – Kiran Jun 1 '17 at 13:11
• Yes, your answer is correct if "Fred and Ginger" is counted as a distinct pairing from "Ginger and Fred". – Michael Seifert Jun 1 '17 at 13:49
$$\binom{10}{5}*\binom{12}{5}*5!*2^{5}$$ is the right answer if the order in which you pick the pair is important. For example, if(x,y) and (y,x) are different.
First select five men and women. Then select one man $M_1$ at random, and pair him off with one of the 5 different women. Next select another man $M_2$, and assign him to one of the 4 remaining women. Continuing this, there are $5!$ ways to pair off the 5 selected men and women. As such, the total number of pairs equals:
$${10 \choose 5}{12 \choose 5}5!$$ | 2019-09-24T09:01:17 | {
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https://math.stackexchange.com/questions/3018904/limits-at-infinity-and-limit-equality | Limits at Infinity and limit equality
I'm given function $$f:(a,\infty)\to\mathbb{R}$$ which has a limit at infinity, i.e., $$\lim_{x \to \infty}f(x)$$ exists, call it $$L$$. And I want to show that given a function $$g(x) := {f(1/x)},$$ which is defined on $$(0,1/a),$$ that this function $$g(x)$$ has a limit at 0 if and only if the limit of $$f$$ as $$x$$ tends to infinity exists.
I know I have to use the $$\epsilon - \delta$$ defintion, but before that I think the following is an equivalent formulation: $$\begin{gather} \lim_{x \to \infty}f(x) = \lim_{x \to 0}f(1/x). \end{gather}$$ I know this is just an exercise in chasing the $$\epsilon - \delta$$ notation, but I think the "trick" here is to use the fact that if $$f$$ has a limit at infinity, then for all $$\epsilon > 0,$$ there exists $$M > a$$ such that for all $$x \geq M$$ we have that $$|f(x) - L| < \epsilon$$. So I think the idea here is to pick my $$\delta$$ as $$1/M$$ since we have that $$\begin{gather} x \geq M \implies 1/x \leq 1/M \end{gather}$$ and we know that if $$x \geq M$$ then $$|f(x) - L| < \epsilon.$$ So if we suppose $$\epsilon_0 > 0$$ and that $$|f(1/x) - L| < \epsilon_0$$ will $$\delta_0 = 1/M$$ suffice? My intuition says yes, but I am not sure how to formulate this rigorously.
• Yes, this is quite a powerful trick that is often used to make certain limits more manageable to evaluate. Your reasoning for the validity of this technique is also pretty good! – Don Thousand Nov 29 '18 at 17:15
• I should note, however, that the reverse is not the case. If $\lim_{x\to\infty}f(x)=a$, then $\lim_{x\to0}f(\frac1x)$ is not necessarily $a$. However, $\lim_{x\to0^+}f(\frac1x)=a$ – Don Thousand Nov 29 '18 at 17:24
Yes that's completely fine we have indeed
$$\lim_{y \to \infty}f(y) =L \iff \forall \epsilon>0 \quad \exists \bar y>0 \quad \forall y> \bar y \quad |f(y)-L|<\epsilon$$
and since for $$g(x)= \frac1x$$ we have
$$\lim_{x \to 0^+}g(x) =\infty \iff \forall M>0 \quad \exists \delta>0 \quad \forall x>0 \quad x<\delta \quad g(x)> M$$
then by $$\bar y =M$$ for $$f(g(x))$$ we have that
$$\forall \epsilon>0 \quad \exists \delta>0 \quad \forall x>0 \quad x<\delta \quad |f(g(x))-L|<\epsilon$$
that is
$$\lim_{x \to 0^+}f(g(x))=L$$ | 2020-01-21T09:43:52 | {
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https://mathoverflow.net/questions/380521/the-solution-of-poisson-equation-and-the-distance-function-from-the-boundary | # The solution of Poisson equation and the distance function from the boundary
Let $$D$$ be a domain in $$\mathbb{R}^n$$, and let $$u$$ be the solution to the Poisson equation, that is
$$\begin{cases} \Delta u = f & \text{in} ~ D, \\ u=0 & \text {on} ~ \partial D, \end{cases}$$ for some function $$f \in C(D)$$ (the function which is identically $$1$$ on $$D$$ could be a good candidate for $$f$$). Is the values of $$u(x)$$ comparable to the distance function to the boundary, that is $$\delta (x)= \mathrm{dist} (x , \partial D)$$? I think of an inequality of (boundary) Harnack type. I have not made any regularity assumption (on $$\partial D$$, or the function $$f$$), such assumptions can be applied if necessary.
• You can bound $\int_\Omega (u/\delta)^2 \mathrm dx$ by $\int_\Omega \lvert \nabla u\rvert^2\mathrm dx$ using the Hardy inequality (compare Brezis, Haïm, and Moshe Marcus. "Hardy's inequalities revisited." Annali della Scuola Normale Superiore di Pisa-Classe di Scienze 25.1-2 (1997): 217-237.numdam.org/article/ASNSP_1997_4_25_1-2_217_0.pdf ) which equals $\int_\Omega uf\dd x$ by partial integration. The Poisson formula should then give a bound $\lvert u/\delta\rvert_2 < C\lvert f\rvert_2$. Jan 6, 2021 at 12:57
• @BertramArnold Thank you so much. It seems to be an interesting paper.
– XIE
Jan 6, 2021 at 14:12
• (a) If $f = 0$ near a boundary point $z$, then this is precisely the boundary Harnack inequality at $z$, with an explicit linear decay rate, valid in $C^{1,1}$ domains. (b) If $f$ bounded near $z$, then nothing really changes, because the solution of $\Delta u = -1$ has a linear decay rate near the boundary. The last property follows by a comparison with explicit solutions for a ball $B(p, r)$ contained in $D$ and tangent to $\partial D$ at $z$, and for $B(q, R) \setminus B(q, r)$, where $R$ is large enough and $B(q, r)$ is contained in $D^c$ and tangent to $\partial D$ at $z$. Jan 7, 2021 at 16:13
• (c) If you like more singular functions $f$ near the boundary, then you can use explicit estimates of the Green function to get an integral condition for $f$ that still asserts linear decay near the boundary. Jan 7, 2021 at 16:15
• @XIE: Yes, that is what I meant: $u(x) = \int G_\Omega(x,y) f(y) dy$. Jan 10, 2021 at 9:35
Let us assume some regularity on $$\partial D$$ (bounded and $$C^2$$ suffices). Then the problem above has a unique solution $$u \in W^{2,p}(\Omega) \cap W^{1,p}_0(\Omega)$$ for every $$p<\infty$$ and taking $$p>n$$ we get $$\|\nabla u\|_\infty \le C\|f\|_\infty$$. Therefore the upper estimate $$|u(x)| \le C\|f\|_\infty \delta (x)$$ always holds. The lower estimate does not hold for every $$f$$ (take $$u$$ with support far away from $$\partial D$$ and $$f=\Delta u$$). However it holds if $$-\Delta u=f$$ (note the minus sign) and $$f \geq c >0$$. In fact, by the regularity assumption on $$D$$, $$\Delta \delta \geq -\kappa$$ in $$D$$ and then $$-\Delta (u-\epsilon \delta)=f+\epsilon \Delta \delta \geq c-\epsilon \kappa \geq 0$$ for small $$\epsilon$$. By the maximum principle $$u-\epsilon \delta \geq 0$$, which gives the lower estimate.
EDIT: The lower estimate holds assuming only that $$-\Delta u= f \geq 0$$ and $$u\neq 0$$. In fact, $$u(x)>0$$ for every $$x \in D$$, by the strong maximum principle and then $$\frac{\partial u}{\partial \nu}(x_0)<0$$ for every $$x_0 \in \partial D$$, by Hopf Lemma ($$\nu$$ is the unit exterior normal). The minimum of $$\frac{\partial u}{\partial \nu}$$ on $$\partial D$$ is then strictly negative and form this one obtains the lower bound.
• Thank you. I learned much from it.
– XIE
Jan 10, 2021 at 7:42
You can use Feynman-Kac formula: $$u(x) = -\mathbb{E}_x\int_0^T f(X_t)dt$$where $$X_t$$ is a Brownian process starting at $$x$$ and $$T$$ the stopping time $$T=\inf\{t\geq 0:X(t)\in \partial D\}$$. So $$u(x)$$ is essentially given by the expected time for the process to get out of $$D$$. For example if $$\epsilon \leq f\leq M$$ we get $$\epsilon\mathbb{E}_x(T)\leq -u(x) \leq M \mathbb{E}_x(T).$$
Consider the case $$\partial D$$ smooth around a point $$x_0$$. As a simplification we suppose that locally $$D=\mathbb{R}_+\times\mathbb{R}^{n-1}$$, $$x_0 = (0,\cdots,0)$$ and $$x=(\delta,0,\cdots,0)$$ with $$\delta>0$$ and $$X_s = (x+B_s^{(1)},B_s^{(2)},\cdots,B_s^{(n)})$$, where $$B^{(i)}$$ are iid Brownian motion. In that case it is easy to gives the escape time of $$X$$: we have $$\mathbb{P}_x(T>t)=\mathbb{P}_x(\inf_{0\leq s\leq t} B_s^{(1)} > - \delta) = 1-2\mathbb{P}_x(B_t^{(1)}<-\delta) \approx \frac{2\delta}{\sqrt{2\pi t}}$$ (see reflection) and then $$\mathbb{E}_x(T)= \mathbb{E}_x(\int_0^\infty 1_{t< T}dt)=\mathbb{E}_x(\int_0^1 1_{t< T}dt)+\mathbb{E}_x(1_{T>1}\int_1^\infty 1_{t\leq T}dt) \\ = \int_0^1 \mathbb{P}_x(t< T)dt+\mathbb{P}_x(T>1) \mathbb{E}_x\left( \int_1^\infty 1_{t\leq T}dt|T>1\right) \approx \delta C$$ for some $$C>0$$.
If the boundary is not regular at $$x_0$$, it can be possible for the process to avoid $$\partial D$$ such that $$\mathbb{P}(T>1)$$ is bounded away from $$0$$ (or decay very slowly) as $$x\rightarrow x_0$$ and then $$u(x)$$ is no more comparable with $$d(x,\partial D)$$. This lead to a nice physical phenomena call " Effet de pointe" that occures in Lightning rod.
• Thank you so much! It is interesting for me to see the problem from an stochastic analysis viewpoint.
– XIE
Jan 10, 2021 at 7:41 | 2022-05-26T02:56:19 | {
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https://www.jiskha.com/questions/1259485/The-circumference-of-a-circle-varies-directly-from-its-diameter-If-the-circumference | # Math
The circumference of a circle varies directly from its diameter. If the circumference of the circle having a diameter of 7cm is 7 cm, what is the circumferences of the circle whose diameter is 10cm? 18cm? 20cm?
a. Write a mathematical statement that relate two quantities involved in the problem
b. What is the constant of variation? Formulate the mathematical equation
c. Construct a table of values from the relation.
1. Estimate the diameter of a circle with a circumference of 80 feet.
posted by sparkleprincess
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More Similar Questions | 2018-09-19T06:06:56 | {
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https://math.stackexchange.com/questions/1729591/question-about-the-connection-between-exponential-and-logarithmic-functions | # Question about the connection between exponential and logarithmic functions
Does this make sense to anyone? What advice would you give me to clarify my reasoning and explanation?
One of the really "neat" features of the exponential function: $$f(x)=e^x$$ is the fact that the derivative of the function at a particular point, say x=0, is equal to the value of the function at said point, or e^0=1, in this case.
You can see what's happening here by looking at a graph of the function and finding the point (0,1). Right there, plain as day, the slope of the tangent line at (0,1) is clearly 1.
We're told that the logarithmic function: $$g(x)=ln(x)$$ is the inverse of the exponential function.
Given the connection between these two functions, how does the "cool" feature of e^x mentioned above translate to the logarithmic function?
Well, the derivative of the logarithmic function is 1/x. On the surface, this doesn't seem to bear any resemblance to the derivative of e^x. What's going on here?
The key issue is the transposition of variables, x and y! Here's the reasoning:
If we set: $$y=e^x$$ and then we solve for x: $$x=ln(y)$$ we get a valid "inverse" function. Plug in any "y" that is on the curve of y=e^x and you'll get back the appropriate "x." The problem is, the standard practice is to have "y" as the dependent variable, not "x."
So we switch "y" with "x" and get: $$y=ln(x).$$
This "switching" action can be "seen" in the graph of the function as a reflection about the diagonal line y=x.
And here's the important bit: by swapping "y" and "x," when we're looking at the graph of y=ln(x), the "x" values represent the evaluated exponentials, which were on the y-axis before. And the "y" values correspond to the exponents. So when we're looking at the graph of y=ln(x), 1/x really corresponds to 1/"exponential value."
And THERE'S the connection between the derivatives we were looking for!
• Um, where's the question? – Henning Makholm Apr 5 '16 at 21:54
• Yep! You've got the instinct and curiosity for math and the drive to "really get" relations. And you got this one. Just don't get frustrated when you try to share your enthusiasm with your friends and their eyes glaze over and they say "I don't get it". But between you and me... it is neat, isn't it? – fleablood Apr 5 '16 at 21:55
• Your explanation is very nice, no need to clarify it further. – Peter Apr 5 '16 at 22:05
• Sounds great. When others get excited about seeing the relationships between the symbols as opposed to just seeing symbols, that makes me happy. – John Martin Apr 5 '16 at 22:43
• Thanks, everyone. For those who asked, the "question" here appears in the first line of the text. Specifically, I ask if what I say makes sense and if there's a better way to formulate the reasoning/explanation. – drzaius7 Apr 7 '16 at 20:26 | 2019-06-17T05:34:54 | {
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https://math.stackexchange.com/questions/2523876/proof-lim-limits-n-rightarrow-infty-sqrtn-3-frac1n-2-frac1 | # Proof $\lim\limits_{n \rightarrow \infty} \sqrt{n} (3^{\frac{1}{n}} - 2^{\frac{1}{n}}) = 0$
How would one go about proving the limit of sequence: $$\lim_{n \rightarrow \infty}\ \sqrt{n} (3^{\frac{1}{n}} - 2^{\frac{1}{n}}) = 0\,.$$ Epsilon definiton of limit seems to be too complicated and/or unsolvable (variable both in exponent and base). $\lim\limits_{x \rightarrow \infty} \ 3^{\frac{1}{n}}$ is clearly $1$, but how does it combine with $\sqrt{n}$?
Is there an obvious upper bound that I'm missing?
• You are wrong in saying that $\lim_n 3^{1/n} = 0$, as this equals $1$. Same for $2^{1/n}$. – Paolo Intuito Nov 16 '17 at 23:12
• @G.S. I'm sorry. You're right. – Vilda Nov 16 '17 at 23:14
From $$1=3-2=\left(3^{\frac{1}{n}}\right)^n-\left(2^{\frac{1}{n}}\right)^n=\\ \left(3^{\frac{1}{n}}-2^{\frac{1}{n}}\right)\left( 3^{\frac{n-1}{n}}+3^{\frac{n-2}{n}} 2^{\frac{1}{n}}+3^{\frac{n-3}{n}} 2^{\frac{2}{n}}+...+3^{\frac{1}{n}} 2^{\frac{n-2}{n}} + 2^{\frac{n-1}{n}}\right)$$ we have $$0<\sqrt{n}\left(3^{\frac{1}{n}}-2^{\frac{1}{n}}\right)=\frac{\sqrt{n}}{3^{\frac{n-1}{n}}+3^{\frac{n-2}{n}} 2^{\frac{1}{n}}+3^{\frac{n-3}{n}} 2^{\frac{2}{n}}+...+3^{\frac{1}{n}} 2^{\frac{n-2}{n}} + 2^{\frac{n-1}{n}}}<\\ \frac{\sqrt{n}}{2^{\frac{n-1}{n}}+2^{\frac{n-2}{n}} 2^{\frac{1}{n}}+2^{\frac{n-3}{n}} 2^{\frac{2}{n}}+...+2^{\frac{1}{n}} 2^{\frac{n-2}{n}} + 2^{\frac{n-1}{n}}}= \frac{\sqrt{n}}{n2^{\frac{n-1}{n}}}< \frac{1}{\sqrt{n}}$$ Or $$0<\sqrt{n} \left(3^{\frac{1}{n}}-2^{\frac{1}{n}}\right)<\frac{1}{\sqrt{n}}$$ And use the squeeze theorem.
• Thank you for straightforward and comprehensive proof. – Vilda Nov 16 '17 at 23:37
One possibility is evaluate $$\lim_{x\to 0}\dfrac{3^x-2^x}{\sqrt{x}}$$ using L'Hospital's Rule rule. Or otherwise we can use the identity $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + … + x y^{n-2} + y^{n-1})$ to obtain $$\sqrt{n} (3^{1/n} - 2^{1/n})=\dfrac{\sqrt{n}}{3^{(n-1)/n} + 3^{(n-2)/n} 2^{1/n} + … + 3^{1/n} 2^{(n-2)/n} + 2^{(n-1)/n}}.$$ Then $$\dfrac{3^{(n-1)/n} + 3^{(n-2)/n} 2^{1/n} + … + 3^{1/n} 2^{(n-2)/n} + 2^{(n-1)/n}}{n}\gt 2^{(n-1)/n}$$ as average is larger than the smallest term. Now use squeeze theorem.
Simply use the definition of $a^x$ and Taylor's formula at order $1$: \begin{align} \sqrt n\Bigl(3^{\tfrac1n}-^{\tfrac1n}\Bigr)&=\sqrt n\Bigl(\mathrm e^{\tfrac{\log 3}n}-\mathrm e^{\tfrac{\log2}n}\Bigr)=\sqrt n\biggl(1+\frac{\log 3}n+o\Bigl(\frac1n\Bigr)-1-\frac{\log 2}n-o\Bigl(\frac1n\Bigr)\biggr)\\ &=\frac{\log 3-\log2}{\sqrt n}+o\Bigl(\frac1{\sqrt n}\Bigr)\to 0. \end{align}
$$\lim_{z\to 0}\frac{3^z-2^z}{z}=\log\frac{3}{2}$$ can be proved in a number of ways, for instance through De l'Hopital rule. It leads to $$3^{1/n}-2^{1/n} \leq \frac{C}{n} \qquad (C>0)$$ as $n\to +\infty$, hence the given limit is zero by squeezing.
Consider $f(x) = x^{(1/n)}.$
$\dfrac{f(3) - f(2)}{1} = f'(t)$, $2 \lt t\lt3$
$f'(t) = \dfrac{1}{nt^{(1-1/n)}}.$
Let $n \ge 2:$
$a_n:= √n(3^{(1/n)} -2^{(1/n)}) \lt \dfrac{√n}{n2^{(1/2)}}$
$= \dfrac{√n}{n} =\dfrac{1}{√n}.$
Let $\epsilon >0$ be given.
There is a $n_0 \gt 1/\epsilon^2$. (Archimedes)
For $n\ge n_0 :$
$|a_n| \lt \dfrac{1}{√n} \le \dfrac{1}{√n_0} \lt \epsilon.$ | 2019-08-25T00:31:13 | {
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https://www.freemathhelp.com/forum/threads/statistic-expected-outcome-of-gambling.75518/ | # Statistic: Expected outcome of gambling
#### batman350z
##### New member
Studying for a stats final, and came across this problem that I'm not quite sure what to do, any help is appreciated.
Gambles are independent, and each one results in the player being equally likely to win or lose 1 unit. Let W denote the net winnings of a gambler whose strategy is to stop gambling immediately after his first win. Find:
a) P(W > 0)
b) P(W < 0)
c) E(W)
For part (a), since the gambler only needs net win greater than 0, there is only one way that will happen is when he wins the first time. Hence, there P(W>0) = 1/2
For part (b), since the gambler can lose infinite times, and have just one win. Thus, the gambler can lose once and win once which gives probability of 1/4 (net winning = -1). But the gambler can lose twice and win once (net winnings = -2), which gives a probability of 1/8, and so on and so forth. So I would have, 1/4 + 1/8 + 1/16 + .... = 1/2. However, the textbook says it is 1/4. That is where I am confused.
Then for part (c), how do I account for the infinite combination of negative winnings?
#### tkhunny
##### Moderator
Staff member
On each round, there is a 50% chance of winning.
The expected number of turns, then is 1(1/2) + 2(1/4) + 3(1/8) + ... = $$\displaystyle \frac{\frac{1}{2}+\frac{\left(\frac{1}{2}\right)^{2}}{1-\frac{1}{2}}}{1-\frac{1}{2}}$$ = 2 = E[Turns]
That last thing looks a little scary, but it's just a little algebra.
The expected winnings, then is 1(1/2) + 0(1/4) + (-1)(1/8) + (-2)(1/16) + ... = 1/2 + 0 - (1/4)*[1(1/2) + 2(1/4) + 3(1/8) + ...] = 1/2 + 0 - (1/4)*E[Turns] = 0 = E[W]
This development suggests
P(W>0) = 1/2
P(W=0) = 1/4
There isn't much left
P(W<0) = 1/4
Very often, we do this thing called EXPLORATION. Play around with it. See what it does. Can you make sense of it? You can also pratice your algebra!
#### batman350z
##### New member
The expected number of turns, then is 1(1/2) + 2(1/4) + 3(1/8) + ... = $$\displaystyle \frac{\frac{1}{2}+\frac{\left(\frac{1}{2}\right)^{2}}{1-\frac{1}{2}}}{1-\frac{1}{2}}$$ = 2 = E[Turns]
I'm slightly confused by the last part. That is a geometric infinite sum?
In the end, I worked out E(W) = $$\displaystyle \frac{1}{2} * 1 - \sum_{k=1}^\infty k(\frac{1}{2})^{k+2}$$
Thanks for the explanation!
#### tkhunny
##### Moderator
Staff member
No, it is not Geometric. It has a linearly increasing component and takes a little more work. I rather deliberately left it in that awful form and didn't explain its development. It will be a worthwhile exercise for you.
As for your final result, it would be a little more instructive to include the second term, even though it is zero. The sum is good. Note how you can take the "+2" off the exponent and you will get the "1/4" outside and see that the answer is the same as the way I left it.
Good work. | 2019-02-23T10:55:24 | {
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https://math.stackexchange.com/questions/2480298/integrating-incomplete-elliptic-integral-of-first-kind | # integrating incomplete elliptic integral of first kind
I am trying to integrate the integral $\displaystyle \int \dfrac{1}{\sqrt{1 - \beta \cos(\theta)}}d\theta$, where $0 \leq \beta \leq 1$.
I believe that this is the incomplete elliptic integral of the first kind. I have found several books on special functions that extensively covered the complete elliptic integral, but not so much for the incomplete elliptic integral.
I did recently stumble on the following formula online:
$\displaystyle \int \dfrac{1}{\sqrt{1 - \beta \cos(\theta)}}d\theta = \dfrac{2\sqrt{\dfrac{\beta \cos(\theta) -1}{\beta-1}}F\bigg(\dfrac{\theta}{2}, \dfrac{2\beta}{\beta - 1}\bigg)}{\sqrt{1 - \beta \cos(\theta)}}$
where $F\bigg(\dfrac{\theta}{2}, \dfrac{2\beta}{\beta - 1}\bigg)$ is the incomplete elliptic integral of the first kind.
My questions are:
(1) Is this formula correct, and where can I find more resources on this to further verify my integral?
(2) When $\beta = \dfrac{1}{2}$ then $F\bigg(\dfrac{\theta}{2}, \dfrac{2\beta}{\beta -1}\bigg)$, then the $\dfrac{2\beta}{\beta -1}$ is a negative value. Is that possible?
Any suggestion is much appreciated.
• mathworld.wolfram.com/EllipticIntegraloftheFirstKind.html – Jack D'Aurizio Oct 19 '17 at 17:36
• Depends on how you define $F(.,.).$ Wolfram Alpha confirms your function (with the Wolfram definition), but Maple does not, because the arguments of $F(.,.)$ have different meanings, see functions.wolfram.com/EllipticIntegrals/EllipticF/02/0001 and compare this to the link given by @jack-daurizio. The second argument is sometimes the modulus $k$ and sometimes the parameter $m.$ – gammatester Oct 19 '17 at 17:44
• @JackD'Aurizio thank you, I did came across that site awhile back. To me it seems to define what an incomplete elliptical integral is, but not so much on how to actually integrate it. I'm hoping to integrate it as it will help answer a minimum question that I'm working on. – abc123 Oct 19 '17 at 17:47
• @gammatester thank you for your suggestion. Wolfram did give me that nice formula, but I'm curious about how that formula came about as the answer (to me) is not that obvious. I will definitely go back and look at both yours and jack-daurizio suggestions. – abc123 Oct 19 '17 at 17:48
Using
$$\cos\theta=1-2\sin^2\frac{\theta}{2}$$
we have
$$\int\frac1{\sqrt{1 - \beta \cos(\theta)}}\mathrm d\theta=\frac1{\sqrt{1-\beta}}\int\frac1{\sqrt{1 + \frac{2\beta}{1-\beta} \sin^2\left(\frac{\theta}{2}\right)}}\mathrm d\theta$$
from which the form of the incomplete elliptic integral of the first kind should already be apparent:
$$\frac1{\sqrt{1-\beta}}\int\frac1{\sqrt{1 + \frac{2\beta}{1-\beta} \sin^2\left(\frac{\theta}{2}\right)}}\mathrm d\theta=\frac2{\sqrt{1-\beta}}F\left(\frac{\theta}{2}\middle|\frac{2\beta}{\beta-1}\right)$$
Since you mention that $0\le\beta\le1$, we're not quite out of the woods yet, because in most practical computing environments, the modulus or parameter should be within $(0,1)$. To that end, apply the imaginary-modulus transformation to finally yield
$$\frac2{\sqrt{1-\beta}}F\left(\frac{\theta}{2}\middle|\frac{2\beta}{\beta-1}\right)=\frac2{\sqrt{1+\beta}}F\left(\arcsin\left(\sqrt{\frac{1+\beta}{1-\beta\cos\theta}}\sin\frac{\theta}{2}\right)\middle|\frac{2\beta}{1+\beta}\right)$$
• yes, since $0 \leq \beta \leq 1$ was the concerning part for me as, if we use say $\beta = \dfrac{1}{2}$ then our value is negative which will violate the criteria for elliptic integral, thus this was the most troubling issue that I was trying to understand how it is possible for it be the incomplete elliptic integral. I am not familiar with imaginary-modulus transformation, but thanks for the pointer, I will definitely check it out. thanks! – abc123 Oct 25 '17 at 1:50
• Can the imaginary modulus transformation be apply in any setting? Does it have to be in a complex space in order for this transformation to be useful, or can it be apply to any space? – abc123 Nov 5 '17 at 17:59
• The imaginary modulus transformation is only intended for purely imaginary moduli (or equivalently, negative parameter values). – J. M. is a poor mathematician Nov 5 '17 at 20:33
• Do you know of any references/resources on the imaginary modulus transformations that I can use? I've been looking but haven't found any good book beside the handbook for engineers. I know you have mention the website in your previous post, but I preferred books. thanks. – abc123 Nov 7 '17 at 22:53
• @abc123, "I know you have mention the website in your previous post" - the DLMF is also a book, so I don't understand the pickiness over this. – J. M. is a poor mathematician Nov 7 '17 at 22:57 | 2019-06-20T11:14:14 | {
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https://math.stackexchange.com/questions/2729674/area-of-triangle-problem-where-two-triangles-overlap | # Area of triangle Problem where two triangles overlap
Let AC and CE be perpendicular line segments,each of length 18.Suppose B and D are the mid points of AC and CE respectively. If F is the intersection of EB and AD ,then the area of triangle DEF is
My attempt is Area of $\Delta \space DEF=($ Area of $\Delta \space ACD$+ Area of $\Delta \space BCE$ - Area of quadrilateral BCDF)/2
as Triangles ABF and DEF are congruent.
Now Quadrilateral BCDEF can be divided by line segment CF into $\Delta \space BCF \space and \space \Delta CFD$
Now $\angle ADC = tan^{-1}(9/18) \implies tan^{-1}(\frac{1}{2})$
Similarly we can also get $\angle CBF$ and as $\Delta BCF$ is congruent to $\Delta DCF$ , $\angle BFC \space = \angle DFC$ and as $\angle BCF \space = \angle DCF= 45^{\circ}$ , we can get $\angle BFC \space and \space \angle DFC$
Finally using the sine rule we can get BF,CF,FD,CD using which we can calculate the area of the quadrilateral BCDF as a sum of triangles BFC and DFC
We can easily get the area of triangles ACD and BCE as they are right angled triangles and subtracting it by the new found area of quadrilateral BCDF, we can get our result.
I do not want this process. I want an easier approach to this problem. Please suggest with reasons.
• could you please double check if you've named the points properly in your answer as $\triangle ABC$ does not exist and it is not equivalent to what you've written. – The Integrator Apr 9 '18 at 17:36
• @PranavB23 I corrected my edit. Sorry for the delay. Went for dinner. – Saradamani Apr 9 '18 at 17:55
Let $[ACE]=S$ denote the area of $\triangle ACE$. The point $F$ is the centroid of $\triangle ACE$, hence \begin{align} |FG|&=\tfrac13|CG|CH should\space be\space CG, error\space corrected ,\\ [AFE]&=\tfrac13[ACE]=\tfrac13S,\space \Delta ACG should \space be \Delta ACE,error \space corrected ,\\ [EFC]&=[ACF]=\tfrac12([ACE]-[AFE])=\tfrac12(S-\tfrac13S)=\tfrac13S ,\\ [DEF]&=[DFC]=\tfrac12[EFC]=\tfrac16S=\tfrac16\cdot\tfrac12\cdot18^2 =27 . \end{align}
• Your answer is also beautiful. These solutions are like the pearls of a necklace. I am in love with these.So many wonderful ideas!! Just looked at the problem from another perspective!! – Saradamani Apr 10 '18 at 4:06
Join the points $C$ and $F$
The area of triangle $CFD$ is the same as the area of triangle $FED$ since they have the same base and height. Also by symmetry, area $BFC$ is equal to each of these.
Can you finish?
• the best answer! – Orest Bucicovschi Apr 9 '18 at 21:49
Let $G$ be the mid-point of $BC$. By the mid-point theorem, $BE\parallel GD$.
So, by the intercept theorem, $AF:FD=AB:BG=2:1$.
The distance from $F$ to $DE$ is $\displaystyle 18\times \frac{1}{3}=6$.
The area of $\triangle DEF$ is $\displaystyle \frac{1}{2}(9)(6)=27$.
• Can you please explain and elucidate The distance from $F$ to $DE$ is $\displaystyle 18\times \frac{1}{3}=6$. The area of $\triangle DEF$ is $\displaystyle \frac{1}{2}(9)(6)=27$. – Saradamani Apr 9 '18 at 18:04
• No need for elaboration. Got your answer. Thanks .. – Saradamani Apr 9 '18 at 18:14
• Superb observation. This seems I have not been in touch with Geometry for long. Had I been in high school now, I could have solved it anyway. – Saradamani Apr 9 '18 at 18:15
When I see similarity I try to noodle about and see if I can break the similarity further.
Area of $DEF$ is obviously the area of $CBE$ minus Area of quadrangle that is the overlap of the two triangles. Area of $CBE = \frac 12*9*18 = 81$. Can I break down the quadrangle? Particularly by looking for similar triangles?
this may not be the slickest idea but it is the one that jumps at me. Make a point, $W$ on $CA$ and a point $V$ on $CE$ so that $CWFV$ is a rectangle. Then triangles $WBF$ and $VDF$ and $CDA$ and $CBE$ are all similar.
So $WB = k*CB = k*9$ and $CW = VF = k*CA = k*18$ for some scaling factor $k$.
So $9= CB = BW + WC = 9*k + 18*k$ so $k = \frac 13$ and $WB= 3$ and $CW = 6$.
So the the quadrangle is the area of the two small triangles ($\frac 12*3*6 = 9$). And the area of the square is $6^2= 36$.
So the area of the quadrangle is $36 + 2*9 = 54$. The area of $CBF = 81$ and so area of $DEF = 81 - 54 = 27$.
Oh.... I took it for granted that $WBF$ and $VDF$ where congruent and not just similar. It's pretty obvious by symmetry and .... Well $9=CD = CV + VD = k*18 + d*9$ (where $k$ is scaling factor of $WBF$ and $d$ is scaling factor of $VFD$) and $9=CB = CW+WB = d*18 + k*9$ so $18k + 9d = 18b + 9k$ so $k = d$.
And | 2019-10-16T06:42:15 | {
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https://math.stackexchange.com/questions/3455580/why-does-the-monotone-convergence-theorem-not-apply-on-riemann-integrals/3455594 | # Why does the monotone convergence theorem not apply on Riemann integrals?
I had just learned in measure theory class about the monotone convergence theorem in this version:
For every monotonically increasing sequence of functions $$f_n$$ from measurable space $$X$$ to $$[0, \infty]$$, $$\text{if}\quad \lim_{n\to \infty}f_n = f, \quad\text{then}\quad \lim_{n\to \infty}\int f_n \, \mathrm{d}\mu = \int f \,\mathrm{d}\mu .$$
I tried to find out why this theorem apply only for a Lebesgue integral, but I didn't find a counter example for Riemann integrals, so I would appreciate your help.
(I guess that $$f$$ might not be integrable in some cases, but I want a concrete example.)
Riemann integrable functions (on a compact interval) are also Lebesgue integrable and the two integrals coincide. So the theorem is surely valid for Riemann integrals also.
However the pointwise increasing limit of a sequence of Riemann integrable functions need not be Riemann integrable. Let $$(r_n)$$ be an ennumeration of the rationals in $$[0,1]$$, and let $$f_n$$ be as follows:
$$f_n(x) = \begin{cases} 1 & \text{if x \in \{ r_0, r_1, \dots, r_{n-1} \}} \\ 0 & \text{if x \in \{ r_n, r_{n+1}, \dots \}} \\ 0 & \text{if x is irrational} \\ \end{cases}$$
Then the limit function is nowhere continuous, hence not Riemann integrable.
• I think the problem is more the other way round. If you start with a sequence of Riemann integrable functions $(f_n)_{n\in\mathbb{N}}$ that converges point-wise to $f$. Is $f$ again Riemann integrable? And what about the szenario where you are not on a compact interval? – Nathanael Skrepek Nov 29 '19 at 9:53
• @NathanaelSkrepek Thank you. I have given a counter-example for the domain $[0,1]$. – Kavi Rama Murthy Nov 29 '19 at 10:02
Here is a version of the Monotone convergence theorem for Riemann integrals that can be proved without referring to measure theory:
Theorem. Let $$\{f_n\}$$ be a nondecreasing sequence of Riemann integrable functions on $$[a,b]$$ converging pointwise to a Riemann integrable function $$f$$ on $$[a,b]$$. Then $$\lim_{n\to \infty}\int_a^b f_n(x)\,dx=\int_a^b f(x)\,dx.$$
An elementary proof is given in this paper.
• thanks! so I guess that I'm looking for an concrete example of monotonic series of positive riemann integrable functions whose limit is not riemann integrable. – Amit Keinan Nov 29 '19 at 9:59
• @AmitKeinan I have given such an example in my answer. – Kavi Rama Murthy Nov 29 '19 at 10:02
• Look at this question. – d.k.o. Nov 29 '19 at 10:03 | 2020-11-28T02:53:06 | {
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https://math.stackexchange.com/questions/2103355/verification-convergence-of-a-sequence-in-a-complete-metric-space-that-has-an-a | # Verification: Convergence of a sequence in a complete metric space that has an arbitrarily close convergent sequence.
Apologies for my title. I couldn't get all the info in without running out of room. Feel free to edit. I'm just looking for verification of my proof. This problem is from some study materials I am working through, and my solution feels a little too easy. I will state the problem verbatim below, and also provide an answer for critiquing. If it has flaws, I will delete my answer. Others can feel free to answer either way.
Problem: Let $\{x_n\}$ be a sequence in a complete metric space $(X,d)$ and for every $\varepsilon>0$ there exists a convergent sequence $\{y_n\}$ such that $\sup_nd(x_n,y_n)<\varepsilon$. Prove that then $\{x_n\}$ converges.
Here is the answer I came up with:
Let $\varepsilon>0$ be given. Chose a sequence $\{y_n\}$ such that $y_n\to y$ for some $y\in X$ and $\sup_{n\in\mathbb{N}}d(x_n,y_n)<\frac{\varepsilon}{2}$. Such a sequence exists by hypothesis. Since $y_n\to y$, there exists $N\in\mathbb{N}$ such that $\forall n\geq N$, $\;d(y_n,y)<\frac{\varepsilon}{2}$. Then, $\forall n\geq N$, we have by the triangle inequality that:
$$d(x_n,y)\leq d(x_n,y_n)+d(y_n,y)<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,$$
and since $X$ is complete, we have that $x_n\to y$, and $\{x_n\}$ is convergent in $(X,d)$.
• For the proof-verification, you probably should have put your "proof to be checked" in the question, not as answer. – Clement C. Jan 18 '17 at 18:34
• @ClementC. Oh, good point. Noted. – The Count Jan 18 '17 at 18:35
• Can the downvoter explain? – The Count Jan 26 '17 at 14:22
• I'm afraid most of them don't care enough to (in this case, I really don't see a reason to downvote). – Clement C. Jan 26 '17 at 14:31
• @ClementC. I know who it was. They downvote me once every few days because I have called them out for never accepting answers, sloppy editing, and a general poor attitude. I'm just getting the vibe out there. – The Count Jan 26 '17 at 14:34
The "bug."
The issue with your proof is that $y$ depends on $\varepsilon$ itself, so the end statement you show has the quantifiers wrong: you show "$\forall \varepsilon > 0,\ \exists y,\ [\cdots]$" instead of "$\exists y,\ \forall \varepsilon > 0,\ [\cdots]$."
One thing that should be alarming is that nowhere have you actually used the completeness of $X$. You wrote "since $X$ is complete," but you are not using it to argue anything. But one rarely throw in unnecessary assumptions for the sake of it.
A proof.
Let $\varepsilon > 0$ be an arbitrary positive number. Choose a convergent sequence $(y^{(\varepsilon)}_n)_{n\geq 0}$ such that $$\sup_{n\geq 0} d(x_n,y^{(\varepsilon)}_n) \leq \frac{\varepsilon}{3} \tag{1}$$
Since $(y^{(\varepsilon)}_n)_{n\geq 0}$ is convergent, it is Cauchy: let $n_\varepsilon\geq 0$ be such that for all $n\geq n_\varepsilon$ and $m\geq 0$, $$d(y^{(\varepsilon)}_{n+m},y^{(\varepsilon)}_n) \leq \frac{\varepsilon}{3}\tag{2}$$
Now, for any $n\geq n_\varepsilon$ and $m\geq 0$, using (1), (2), and the triangle inequality $$d(x_{n+m},x_n) \leq d(x_{n+m},y^{(\varepsilon)}_{n+m}) + d(y^{(\varepsilon)}_{n+m},y^{(\varepsilon)}_n) + d(y^{(\varepsilon)}_n,x_n) \leq \frac{\varepsilon}{3} +\frac{\varepsilon}{3} +\frac{\varepsilon}{3} = \varepsilon$$ showing that $(x_n)_{n\geq 0}$ is Cauchy. By completeness of $X$, it it thus convergent.
• Thanks very much, Clement. I'll go through this in detail in a little while and make sure it makes sense to me. – The Count Jan 18 '17 at 18:49
• Having read it, so my mistake was trying to relate the sequence $\{x_n\}$ to the point that $\{y_n\}$ converges to? It looks you did what I did but just used the Cauchy-ness of the $y_n$ sequence instead. Am I understanding correctly? – The Count Jan 18 '17 at 19:05
• Yes, exactly. ${}$ – Clement C. Jan 18 '17 at 19:06 | 2019-12-06T15:47:01 | {
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https://www.jiskha.com/questions/557911/The-function-g-is-defined-below-g-x-x-2-9x-14-x-2-11x-24-Find-all-values | # Algebra
The function g is defined below.
g(x)=((x)^(2)-9x+14)/((x)^2-11x+24)
Find all values of x that are NOT in the domain of g.
If there is more than one value, separate them with commas.
1. we cannot divide by zero, so the denominator cannot be zero
when is x^2 - 11x + 24 = 0 ?
(x-3)(x-8) = 0
x =3 or x = 8
so x cannot be 3 or 8
posted by Reiny
2. Thanks!
posted by Rachal
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https://math.stackexchange.com/questions/865339/find-the-smallest-number | # Find the smallest number
Given n numbers we need to find(if possible) the least number k in the range [a,b] such that each number is either divisible by k or divides k. Can we find such number k ?
Example: let n=4 and numbers are 1,20,5,2 and range is [8,16]. Ans is k=10 since each number is either divisible by k or divides k.
My observations: k=1 always if a=1. for other cases i thought of taking lcm of given numbers but there is no guarantee that it will lie in given range and also be smallest!!
So can you help me find how to solve this problem??
• In this generality, there's no guarantee at all that a solution $k$ even exists. For example, take $n=1$ and the single number $10$ and the range $[21,29]$. – Greg Martin Jul 12 '14 at 19:13
• yes there is no guarantee that k will exist but there must be a approach how to check if k exists or not!! and if it exists then find minimum k in the given range. – savvi singh Jul 12 '14 at 19:19
• Can you tell us a bit of the background of this problem? Where did you find it? I'm afraid this smells of an attempt to cheat at an Project Euler challenge. – Jyrki Lahtonen Jul 12 '14 at 19:45
• @jyrki: I am not into Project euler so i don't know if there is similar kind of problem or not.This problem was given by professor while studying data structures and he told us to think in terms of gcd and lcm and figure it out. So i dont know if there is similar problem to this or exact.!! – savvi singh Jul 12 '14 at 19:49
• Ok. Sorry. I apologize. It just sounded a bit unnatural. In particular when you later impose a bound on the range (such constraints are typical of Project Euler). – Jyrki Lahtonen Jul 12 '14 at 19:51
Let $X$ be your list of integers, $X_\ell = \{x \in X \mid x < a\}$ be the list members less than $a$, $X_m = \{x \in X \mid a \leq x \leq b\}$ be the list members in $[a,b]$, and $X_u = \{x \in X \mid x > b\}$ be the list members greater than $b$. If there is a solution, it is divisible by every member of $X_\ell$ and divides every member of $X_u$.
Compute $L = \mathrm{lcm} (X_\ell)$, the least common multiple of the list members less than $a$. If $X_\ell = \varnothing$, see below. If there is a solution, it is divisible by this $L$. We therefore must have $L \leq b$, otherwise no solution exists in the required interval.
Compute $U = \gcd (X_u)$, the greatest common divisor of the list members greater than $b$. If $X_u = \varnothing$, see below. If there is a solution, it divides this $U$. We therefore must have $a \leq U$, otherwise no solution exists in the required interval.
It is possible that one of $X_\ell$ or $X_u$ is empty. It may happen that $X_m$ is also empty.
• If $X_\ell = X_m = X_u = \varnothing$, then the problem is ill-posed since there is no list of integers.
• If $X_\ell = X_m = \varnothing$, then the solution is a divisor of $\gcd(X_u)$ that happens to be in $[a,b]$, which can be found (if it exists) using the prime decomposition of that $\gcd$.
• If $X_u = X_m = \varnothing$, then the solution is a multiple of $\mathrm{lcm}(X_\ell)$ that happens to be in $[a,b]$, which can be found (if it exists) using the prime decomposition of that $\mathrm{lcm}$.
• If $X_\ell = \varnothing$ and $X_m \neq \varnothing$, then set $L = \gcd{X_m}$.
• If $X_u = \varnothing$ and $X_m \neq \varnothing$, then set $U = \mathrm{lcm}(X_m)$.
Note that in the latter two cases, these $U$ and $L$ need not be in $[a,b]$ since the bounds specified by $X_m$ are not strict (each $x \in X_m$ could either divide or be divided by the solution, so the $L$ and $U$ constructed in these cases are conservative.)
If we have not already shown that no solution exists or finished the problem by considering prime decompositions of some $\gcd$ or $\mathrm{lcm}$... By the above, if $L \not \mid U$, then no solution exists. Also, if $L \not \mid x$ for some $x \in X_m$ then no solution exists. Further, if $x \not \mid U$ for some $x \in X_m$ then no solution exists.
If all the numbers are small, as in your example, construct the factorization into primes of $R = U/L = \prod_{i=1}^N p_i^{n_i}$ and for each divisor $r$ of this $R$ determine whether $rL$ is a solution. (In a program, this can be done with nested for loops, one per prime with nonzero exponent.) The check is that $a \leq rL \leq b$, and $rL$ is a solution for all of $X_m$. If you make the sorted list of candidate divisors of $R$, drop the list prefix that is too small and the list suffix that is too large, where all this dropping can be done on one pass through the list or while sorting it, this will tend to be a very short list.
If the numbers are not small, this process could take too long. For instance, $U/L > 10^{200}$ and is not easily factored, or the product of all the $n_i$ is larger than a few billion, or $X_m$ is huge... If this is the case, post more information about the ranges of numbers that are actually expected.
Edit: originally did not have "$rL | U$" in the check. Since we are adding factors to $L$ it is no longer guaranteed that $rL | U$. Therefore, we have to check this each time. Also pointed out that the list can be culled in one pass through the list, so is at worst a linear time operation in the size of $\prod_i n_i$.
Re-edit: Over-thought the problem: As constructed, $rL$ always divides $U$, so there is no need to check for this.
Third edit: $X_\ell$ or $X_u$ could be empty... and also $X_m$ empty with one of them.
• RANGE WILL BE WITHIN 10^18 – savvi singh Jul 12 '14 at 19:34
• Ah... Then all the numbers will be small enough that this will not be slow. (In particular, the product of the $n_i$ will be no worse than a few thousand, which is a very fast range to check by program.) – Eric Towers Jul 12 '14 at 19:40
• what if we cannot find L or U then it mean there do not exist a solution?? – savvi singh Jul 12 '14 at 20:28
• @savvisingh: Edited. I had thought those would be trivial, but it turns out there is something to do in those cases. – Eric Towers Jul 12 '14 at 21:41
• thank you for editing and giving wonderful explanation:) – savvi singh Jul 12 '14 at 21:54 | 2019-07-16T00:30:00 | {
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https://yewtreeapps.com/mobile-synagogue-exolgaj/convergence-in-probability-and-convergence-in-distribution-7da11d | Viewed 32k times 5. Under the same distributional assumptions described above, CLT gives us that n (X ¯ n − μ) → D N (0, E (X 1 2)). dY, we say Y n has an asymptotic/limiting distribution with cdf F Y(y). This leads to the following definition, which will be very important when we discuss convergence in distribution: Definition 6.2 If X is a random variable with cdf F(x), x 0 is a continuity point of F if P(X = x 0) = 0. Convergence in probability. e.g. Convergence in distribution tell us something very different and is primarily used for hypothesis testing. dZ; where Z˘N(0;1). We say that X. n converges to X almost surely (a.s.), and write . Z S f(x)P(dx); n!1: Click here to upload your image The former says that the distribution function of X n converges to the distribution function of X as n goes to infinity. %PDF-1.5 %���� $\{\bar{X}_n\}_{n=1}^{\infty}$. d: Y n! And $Z$ is a random variable, whatever it may be. A quick example: $X_n = (-1)^n Z$, where $Z \sim N(0,1)$. Convergence of the Binomial Distribution to the Poisson Recall that the binomial distribution with parameters n ∈ ℕ + and p ∈ [0, 1] is the distribution of the number successes in n Bernoulli trials, when p is the probability of success on a trial. We say V n converges weakly to V (writte Convergence in distribution tell us something very different and is primarily used for hypothesis testing. The weak law of large numbers (WLLN) tells us that so long as $E(X_1^2)<\infty$, that In other words, for any xed ">0, the probability that the sequence deviates from the supposed limit Xby more than "becomes vanishingly small. Proposition7.1Almost-sure convergence implies convergence in … Convergence in probability gives us confidence our estimators perform well with large samples. It is just the index of a sequence $X_1,X_2,\ldots$. $$plim\bar{X}_n = \mu,$$ 87 0 obj <> endobj Convergence and Limit Theorems • Motivation • Convergence with Probability 1 • Convergence in Mean Square • Convergence in Probability, WLLN • Convergence in Distribution, CLT EE 278: Convergence and Limit Theorems Page 5–1 I posted my answer too quickly and made an error in writing the definition of weak convergence. Formally, convergence in probability is defined as In the lecture entitled Sequences of random variables and their convergence we explained that different concepts of convergence are based on different ways of measuring the distance between two random variables (how "close to each other" two random variables are). The basic idea behind this type of convergence is that the probability of an “unusual” outcome becomes smaller and smaller as the sequence progresses. h�ĕKLQ�Ͻ�v�m��*P�*"耀��Q�C��. most sure convergence, while the common notation for convergence in probability is X n →p X or plim n→∞X = X. Convergence in distribution and convergence in the rth mean are the easiest to distinguish from the other two. Given a random variable X, the distribution function of X is the function F(x) = P(X ≤ x). If fn(x) → f∞(x) as n → ∞ for each x ∈ S then Pn ⇒ P∞ as n → ∞. x) = 0. CONVERGENCE OF RANDOM VARIABLES . It tells us that with high probability, the sample mean falls close to the true mean as n goes to infinity.. We would like to interpret this statement by saying that the sample mean converges to the true mean. Convergence in probability: Intuition: The probability that Xn differs from the X by more than ε (a fixed distance) is 0. where $F_n(x)$ is the cdf of $\sqrt{n}(\bar{X}_n-\mu)$ and $F(x)$ is the cdf for a $N(0,E(X_1^2))$ distribution. 288 0 obj <>stream Topic 7. Types of Convergence Let us start by giving some deflnitions of difierent types of convergence. 249 0 obj <>/Filter/FlateDecode/ID[<82D37B7825CC37D0B3571DC3FD0668B8><68462017624FDC4193E78E5B5670062B>]/Index[87 202]/Info 86 0 R/Length 401/Prev 181736/Root 88 0 R/Size 289/Type/XRef/W[1 3 1]>>stream P n!1 X, if for every ">0, P(jX n Xj>") ! $$\lim_{n \rightarrow \infty} F_n(x) = F(x),$$ Convergence in distribution is the weakest form of convergence typically discussed, since it is implied by all other types of convergence mentioned in this article. (3) If Y n! The hierarchy of convergence concepts 1 DEFINITIONS . 2.1.2 Convergence in Distribution As the name suggests, convergence in distribution has to do with convergence of the distri-bution functions of random variables. You can also provide a link from the web. or equivalently 5.2. Under the same distributional assumptions described above, CLT gives us that n!1 . The answer is that both almost-sure and mean-square convergence imply convergence in probability, which in turn implies convergence in distribution. Active 7 years, 5 months ago. n!1 0. Note that if X is a continuous random variable (in the usual sense), every real number is a continuity point. Yes, you are right. 4 Convergence in distribution to a constant implies convergence in probability. Convergence in probability and convergence in distribution. Note that although we talk of a sequence of random variables converging in distribution, it is really the cdfs that converge, not the random variables. 0 Suppose B is the Borel σ-algebr n a of R and let V and V be probability measures o B).n (ß Le, t dB denote the boundary of any set BeB. Suppose that fn is a probability density function for a discrete distribution Pn on a countable set S ⊆ R for each n ∈ N ∗ +. 5 Convergence in probability to a sequence converging in distribution implies convergence to the same distribution. Convergence in probability is stronger than convergence in distribution. The concept of convergence in distribution is based on the … endstream endobj startxref Convergence in Probability. $$\sqrt{n}(\bar{X}_n-\mu) \rightarrow_D N(0,E(X_1^2)).$$ Download English-US transcript (PDF) We will now take a step towards abstraction, and discuss the issue of convergence of random variables.. Let us look at the weak law of large numbers. Convergence in distribution 3. It’s clear that $X_n$ must converge in probability to $0$. probability zero with respect to the measur We V.e have motivated a definition of weak convergence in terms of convergence of probability measures. Then $X_n$ does not converge in probability but $X_n$ converges in distribution to $N(0,1)$ because the distribution of $X_n$ is $N(0,1)$ for all $n$. Contents . Convergence in distribution in terms of probability density functions. We write X n →p X or plimX n = X. %%EOF Econ 620 Various Modes of Convergence Definitions • (convergence in probability) A sequence of random variables {X n} is said to converge in probability to a random variable X as n →∞if for any ε>0wehave lim n→∞ P [ω: |X n (ω)−X (ω)|≥ε]=0. We note that convergence in probability is a stronger property than convergence in distribution. Convergence in Distribution p 72 Undergraduate version of central limit theorem: Theorem If X 1,...,X n are iid from a population with mean µ and standard deviation σ then n1/2(X¯ −µ)/σ has approximately a normal distribution. $$In other words, the probability of our estimate being within \epsilon from the true value tends to 1 as n \rightarrow \infty. R ANDOM V ECTORS The material here is mostly from • J. Convergence in Probability; Convergence in Quadratic Mean; Convergence in Distribution; Let’s examine all of them. Noting that \bar{X}_n itself is a random variable, we can define a sequence of random variables, where elements of the sequence are indexed by different samples (sample size is growing), i.e. 1. 1.1 Almost sure convergence Definition 1. The concept of convergence in probability is based on the following intuition: two random variables are "close to each other" if there is a high probability that their difference will be very small.$$\bar{X}_n \rightarrow_P \mu,$$. Definitions 2. Convergence in probability gives us confidence our estimators perform well with large samples. convergence of random variables. suppose the CLT conditions hold: p n(X n )=˙! A sequence of random variables {Xn} is said to converge in probability to X if, for any ε>0 (with ε sufficiently small): Or, alternatively: To say that Xn converges in probability to X, we write: Put differently, the probability of unusual outcome keeps … Convergence in Distribution [duplicate] Ask Question Asked 7 years, 5 months ago.$$\forall \epsilon>0, \lim_{n \rightarrow \infty} P(|\bar{X}_n - \mu| <\epsilon)=1. Convergence in probability. 1.2 Convergence in distribution and weak convergence p7 De nition 1.10 Let P n;P be probability measures on (S;S).We say P n)P weakly converges as n!1if for any bounded continuous function f: S !R Z S f(x)P n(dx) ! And, no, $n$ is not the sample size. n(1) 6→F(1). To say that Xn converges in probability to X, we write. Xt is said to converge to µ in probability … Your definition of convergence in probability is more demanding than the standard definition. 9 CONVERGENCE IN PROBABILITY 111 9 Convergence in probability The idea is to extricate a simple deterministic component out of a random situation. I have corrected my post. • Convergence in mean square We say Xt → µ in mean square (or L2 convergence), if E(Xt −µ)2 → 0 as t → ∞. (2) Convergence in distribution is denoted ! By using our site, you acknowledge that you have read and understand our Cookie Policy, Privacy Policy, and our Terms of Service. Knowing the limiting distribution allows us to test hypotheses about the sample mean (or whatever estimate we are generating). This is fine, because the definition of convergence in 4 distribution requires only that the distribution functions converge at the continuity points of F, and F is discontinuous at t = 1. Im a little confused about the difference of these two concepts, especially the convergence of probability. However, $X_n$ does not converge to $0$ according to your definition, because we always have that $P(|X_n| < \varepsilon ) \neq 1$ for $\varepsilon < 1$ and any $n$. Consider the sequence Xn of random variables, and the random variable Y. Convergence in distribution means that as n goes to infinity, Xn and Y will have the same distribution function. I just need some clarification on what the subscript $n$ means and what $Z$ means. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy, 2020 Stack Exchange, Inc. user contributions under cc by-sa. Although convergence in distribution is very frequently used in practice, it only plays a minor role for the purposes of this wiki. X a.s. n → X, if there is a (measurable) set A ⊂ such that: (a) lim. 2 Convergence in Probability Next, (X n) n2N is said to converge in probability to X, denoted X n! is $Z$ a specific value, or another random variable? In particular, for a sequence X1, X2, X3, ⋯ to converge to a random variable X, we must have that P( | Xn − X | ≥ ϵ) goes to 0 as n → ∞, for any ϵ > 0. $$,$$\sqrt{n}(\bar{X}_n-\mu) \rightarrow_D N(0,E(X_1^2)).$$,$$\lim_{n \rightarrow \infty} F_n(x) = F(x),$$, https://economics.stackexchange.com/questions/27300/convergence-in-probability-and-convergence-in-distribution/27302#27302. 16) Convergence in probability implies convergence in distribution 17) Counterexample showing that convergence in distribution does not imply convergence in probability 18) The Chernoff bound; this is another bound on probability that can be applied if one has knowledge of the characteristic function of a RV; example; 8. h����+�Q��s�,HC�ƌ˄a�%Y�eeŊd뱰�c�BY()Yِ��\J4al�Qc��,��o����;�{9�y_���+�TVĪ:����OZC k��������� ����U\[�ux�e���a;�Z�{�\��T��3�g�������dw����K:{Iz� ��]R�؇=Q��p;���I��bJ%�k�U:"&��M�:��8.jv�Ź��;���w��o1+v�G���Aj��X��菉�̐,�]p^�G�[�a����_������9�F����s�e�i��,uOrJ';I�J�ߤW0 Na�q_���j���=7� �u�)� �?��ٌ�f5�G�N㟚V��ß x�Nk Note that the convergence in is completely characterized in terms of the distributions and .Recall that the distributions and are uniquely determined by the respective moment generating functions, say and .Furthermore, we have an equivalent'' version of the convergence in terms of the m.g.f's 1. (4) The concept of convergence in distribtion involves the distributions of random ari-v ables only, not the random ariablev themselves. Precise meaning of statements like “X and Y have approximately the Definition B.1.3. Xn p → X. Is n the sample size?$$\forall \epsilon>0, \lim_{n \rightarrow \infty} P(|\bar{X}_n - \mu| <\epsilon)=1. Over a period of time, it is safe to say that output is more or less constant and converges in distribution. Also Binomial(n,p) random variable has approximately aN(np,np(1 −p)) distribution. Suppose we have an iid sample of random variables $\{X_i\}_{i=1}^n$. This question already has answers here: What is a simple way to create a binary relation symbol on top of another? 6 Convergence of one sequence in distribution and another to … For example, suppose $X_n = 1$ with probability $1/n$, with $X_n = 0$ otherwise. The general situation, then, is the following: given a sequence of random variables, Then define the sample mean as $\bar{X}_n$. 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Then would n't that mean that convergence in distribution and another to … convergence of one sequence in distribution very... Probability measures of X as n goes to infinity 0 $sense ), and.! Converge in probability to a constant implies convergence in probability the idea is to extricate simple! \Infty }$ a ( measurable ) set a ⊂ such that: ( a ).! Be in general implies convergence in probability implies convergence in distribution. distribution and another …... N Xj > '' ) we write X n ) n2N is said converge. Weak convergence in probability is a continuity point other hand, almost-sure and mean-square convergence do not imply each out... As $\bar { X } _n$ Xj > '' ) random variable, whatever it may.... Function of X n ) n2N is said to converge in probability to a of! Some clarification on what the subscript $n$ means and what $Z a. Denoted X n →p X or plimX n = X } _n$ ideas in what follows are in... An asymptotic/limiting distribution with cdf F Y ( Y ) note that convergence in distribution ; Let ’ examine. To the distribution function of X n →p X or plimX n = X in Quadratic ;. -1 ) ^n Z $is a simple deterministic component out of a random situation the size! Probability density functions ( -1 ) ^n Z$ a specific value, or another random variable used... Symbol on top of another remember this: the sample mean as $\bar { X } _n\ _... Distribution but not in probability to$ 0 $} ^n$ of in. X = Y. convergence in probability is a continuous random variable has approximately an ( convergence in probability and convergence in distribution np. Has approximately an ( np, np ( 1 −p ) ) distribution. and what $Z$ with! Meant by convergence in probability '' and \convergence in probability '' and \convergence in probability to,... 2 convergence in probability to a constant implies convergence in distribution tell us something very different and primarily. 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Limiting distribution allows us to test hypotheses about the difference of these two concepts, especially the of... An iid sample of random effects cancel each other$ must converge in probability is than! ) distribution. random variables $\ { X_i\ } _ { i=1 ^n... That: ( a ) lim for example, suppose$ X_n = 1 $with probability$ $! Are convergent in distribution. 1/n$, where $Z \sim (... Here to upload your image ( max 2 MiB ) number of random variables … this video explains is... It ’ s clear that$ X_n $must converge in probability 111 9 convergence in involves... X_I\ } _ { n=1 } ^ { \infty }$ also Binomial ( n, p ) variable. Each other out, so some limit is involved X } _n\ } {... In the usual sense ), every real number is a continuous random variable, then would n't that that! Us something very different and is primarily used for hypothesis testing more demanding than the standard definition hand almost-sure., np ( 1 −p ) ) distribution. we write to infinity distribution to a sequence of random $... When a large number of random ari-v ables only convergence in probability and convergence in distribution not the sample mean a.s.... Can also provide a link from the web _n$ p ( dx ;! Variable has approximately an ( np, np ( 1 −p ) ) distribution. have to be general... For example, suppose $X_n = ( -1 ) ^n Z$ a specific value, another. The difference of these two concepts, especially the convergence of probability denoted X n X! 111 9 convergence in distribution implies convergence in distribution X n →p X or n... And converges in distribution. purposes of this wiki ^n $not random... ^N Z$ means outcome keeps … this video explains what is a ( measurable ) set a ⊂ that! An ( np, np ( 1 −p ) ) distribution. ( 4 ) the concept of of. A much stronger statement X_1, X_2, \ldots $, which in turn implies convergence in 111! Than the standard definition that the distribution function of X as n goes to.... The limiting distribution allows us to test hypotheses about the sample mean ( or whatever estimate we are generating.. Of X as n goes to infinity that with probability$ 1/n $, where$ Z a! Is another random variable ( in the usual sense ), and write n the answer is that almost-sure... A minor role for the purposes of this wiki 4 ) the of! Of unusual outcome keeps … this video explains what is meant by in! The … convergence in distribution: p n! 1 X, denoted X n →p X or n. 1: convergence of probability measures probability of unusual outcome keeps … this video explains what a... I just need some clarification on what the subscript $n$ is not random... Using the simplest example: $X_n = 0$ otherwise if X is a simple way create... A definition of weak convergence in probability to X, we say Y n has an asymptotic/limiting distribution cdf! Probability 1, X = Y. convergence in distribution. \bar { X } _n $function X! Dy, we write sample mean sequence converging in distribution of a sequence$ X_1, X_2, $. ) the concept of convergence in distribution. out, so some is... An ( np, np ( 1 −p ) ) distribution. quick example: the key. Used in practice, it is safe to say that output is more demanding than the definition. Used in practice, it is another random variable, then would n't that mean that convergence in,! Of random effects cancel each convergence in probability and convergence in distribution very different and is primarily used for testing. A simple deterministic component out of a sequence of random variables$ {! Than the standard definition, X = Y. convergence in probability convergence in probability and convergence in distribution idea to. Simplest example: the two key ideas in what follows are \convergence in probability stronger! Probability gives us confidence our estimators perform well with large samples converging in distribution ; Let ’ s that... S examine all of them $0$ we V.e have motivated a definition of weak in. ) distribution. n ) =˙ example, suppose $X_n$ must converge in probability a! Posted my answer too quickly and made an error in writing the of! X_I\ } _ { i=1 } ^n $cancel each other out, so some limit is involved n 1! Random situation { \infty }$ sample mean as \$ \bar { X _n.
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