url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://oeis.org/A055491 | 1,621,351,961,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989637.86/warc/CC-MAIN-20210518125638-20210518155638-00436.warc.gz | 406,027,709 | 4,283 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A055491 Smallest square divisible by n divided by largest square which divides n. 2
1, 4, 9, 1, 25, 36, 49, 4, 1, 100, 121, 9, 169, 196, 225, 1, 289, 4, 361, 25, 441, 484, 529, 36, 1, 676, 9, 49, 841, 900, 961, 4, 1089, 1156, 1225, 1, 1369, 1444, 1521, 100, 1681, 1764, 1849, 121, 25, 2116, 2209, 9, 1, 4, 2601, 169, 2809, 36, 3025, 196, 3249, 3364 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 Henry Bottomley, Some Smarandache-type multiplicative sequences. FORMULA If n is written as Product(Pj^Ej) then a(n) = Product(Pj^(2*(Ej mod 2))). a(n) = A053143(n)/A008833(n) = A007913(n)^2 = (A019554(n)/A000188(n))^2 = A000290(n)/A008833(n)^2. EXAMPLE a(12) = 36/4 = 9. MATHEMATICA With[{sqs=Range[100]^2}, Table[SelectFirst[sqs, Divisible[#, n]&]/ SelectFirst[ Reverse[sqs], Divisible[n, #]&], {n, 60}]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Feb 18 2018 *) f[p_, e_] := p^(2 * Mod[e, 2]); a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 07 2020 *) PROG (Haskell) a055491 = (^ 2) . a007913 -- Reinhard Zumkeller, Jul 23 2014 CROSSREFS Cf. A056551, A056552. Cf. A000188, A000290, A007913, A008833, A019554, A053143. Sequence in context: A236104 A152205 A129861 * A032523 A032760 A129970 Adjacent sequences: A055488 A055489 A055490 * A055492 A055493 A055494 KEYWORD easy,nonn,mult AUTHOR Henry Bottomley, Jun 28 2000 STATUS approved
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Last modified May 18 11:30 EDT 2021. Contains 343995 sequences. (Running on oeis4.) | 750 | 1,989 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-21 | latest | en | 0.545586 |
https://electronics.stackexchange.com/questions/480513/pnp-transistor-switching-and-inductive-load | 1,696,427,096,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511369.62/warc/CC-MAIN-20231004120203-20231004150203-00074.warc.gz | 241,939,535 | 42,446 | PNP transistor switching and inductive load
I am designing a circuit to control a motorcycle horn (its operating frequency is 490 Hz, its resistance is 3 Ohm, and it consumes 1A on average as indicated by the power supply), and LED lights. The LED must be controlled by a high side switch, while the horn can be controlled by a low side switch.
The LED lights I am testing consist in multiple LEDs encased in a sealed commercially-available stoplight and plate light assembly. Additional components, which are probably present in the assembly, are not included in my schematic. The stoplight and illuminator are specified for 12 V, and at 12 VDC, consume about 100 mA. The plate LED turns on at a voltage of 2.8 V. The stoplight LEDs turn on at a voltage of 6.7 V.
The 12 V is supplied by a power supply (0 - 30 V DC, 0-5 A). I have also tested supplying 12 V with a DC/DC converter (5A, 12V, 50mVp-p ripple) connected to a 52V 40A battery.
When switching the horn on, the LED flickers on the whole time the horn is on. I think the issue is that V1 is spiking down by several volts when the horn is active. I have tried troubleshooting the issue by: i) adding a capacitor (470, 47, 0.1 uF) between GND and +12V ii) adding a snubber with R=10 Ohm and C=10 uF in parallel with the horn
But to no improvement. Can anyone suggest a design that would prevent the LED from flickering when the horn is active?
Thank you!
simulate this circuit – Schematic created using CircuitLab
• What sort of LED is it specifically? Feb 10, 2020 at 15:12
• @Quentin Welcome to SE.EE Please update your question (using the edit link) so all relevant info is at one place and not scattered across comments Feb 10, 2020 at 18:28
• (1) Do you know (or can you test) what minimum voltage is required for the leds? Would e.g. -5% (11.4V) work? (2) What's the current rating of the 12V source? Feb 10, 2020 at 18:36
• Does the LED flicker the whole time the horn is on, or just the moment it starts or stops? You might try a cap across the base and emitter of Q2. Feb 10, 2020 at 18:42
• You only need 212uA current induced at base of Q3 to turn it on even with the input off. You have 4A of current in the horn. The horn current creates a field that induces current on nearby conductors. You need 85 dB of rejection from the horn trace to the Q3B trace. Otherwise it will turn on Q3. 60 dB of rejection is do-able. 80 dB is very difficult. You could try a large cap from Q3B to Q3E. But ultimately you've got to get the high current traces away from the low current electronics. Feb 10, 2020 at 19:20 | 700 | 2,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-40 | longest | en | 0.938493 |
https://email.esm.psu.edu/pipermail/macosx-tex/2018-September/056202.html | 1,601,322,365,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401604940.65/warc/CC-MAIN-20200928171446-20200928201446-00518.warc.gz | 349,332,496 | 2,075 | [OS X TeX] Multido--"illegal unit of measure" in pstricks
Michael Sharpe msharpe at ucsd.edu
Wed Sep 12 15:43:08 EDT 2018
> On Sep 12, 2018, at 12:13 PM, Nitecki, Zbigniew H. <Zbigniew.Nitecki at tufts.edu> wrote:
>
> In the attached I have created a wiggly (more-or-less vertical) line and am trying to create several copies of it, displaced at regular horizontal intervals.
> I use a real counter \rT added to fixed coordinates. The error message says “illegal unit of measure—pt added”. I don’t understand: coordinates are entered into pscurve as real numbers. What am I missing?
>
>
You can't do arithmetic with \rT. Instead, make three similar counters where the arithmetic is handled, as illustrated below.
\begin{pspicture}(-4,-4)(4,4) % note {-4 replaced by (-4
\rput(0,0){$\bullet$}
\pscurve(0,-2) (-0.1,-1.5)(0,-1)(0.1,-0.5)(0,-0.2)(0,0)(0,0.2) (0.1,0.5)(0,1)(-0.1,1.5)(0,2)
\multido{\rT=-1.6+0.2,\rTp=-1.5+0.2,\rTm=-1.7+0.2}{33}{ % \rTp(lus), \rTm(inus)
\pscurve(\rT,-2) (\rTm,-1.5)(\rT,-1)(\rTp,-0.5)(\rT,-0.2)(\rT,0)(\rT,0.2) (\rTp,0.5)(\rT,1)(\rTm,1.5)(\rT,2)
}
\end{pspicture}
Michael | 455 | 1,109 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-40 | latest | en | 0.804706 |
https://www.askiitians.com/forums/Analytical-Geometry/24/42722/cartesian-coordinates.htm | 1,685,758,460,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648911.0/warc/CC-MAIN-20230603000901-20230603030901-00270.warc.gz | 722,803,651 | 33,180 | # If the vertices of a triangle are (am12, 2am1), ??(am22, 2am2) (am32, 2am3), then its area is(a) a(m2-m3)(m3-m1)(m1-m2)(b) (m2-m3)(m3-m1)(m1-m2)(c) a2(m2-m3)(m3-m1)(m1-m2)(d) a(m2+m3)(m3+m1)(m1+m2)Explain pls.
Aman Bansal
592 Points
11 years ago
If the coordinates of the vertices are (x1,y1), (x2,y2), and (x3,y3), then the area is given by the formula
A = (1/2) det[(x1,y1,1), (x2,y2,1), (x3,y3,1)].
Best Of luck
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Aman Bansal | 330 | 969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2023-23 | latest | en | 0.752106 |
https://gepgitlab.laas.fr/jchemin/curves/-/commit/4dd18a3aedf4bdb685156ea4523a4be517cd3f10 | 1,656,991,886,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104512702.80/warc/CC-MAIN-20220705022909-20220705052909-00615.warc.gz | 321,048,670 | 32,569 | Commit 4dd18a3a by JasonChmn
\ Example : A piecewise curve composed of three curves cf0, /// cf1 and cf2 where cf0 is defined between \f$[T0_{min},T0_{max}]\f$, cf1 between /// \f$[T0_{max},T1_{max}]\f$ and cf2 between \f$[T1_{max},T2_{max}]\f$. /// On the piecewise polynomial curve, cf0 is located between \f$[T0_{min},T0_{max}[\f$, /// cf1 between \f$[T0_{max},T1_{max}[\f$ and cf2 between \f$[T1_{max},T2_{max}]\f$. /// template, typename T_Point= std::vector >, typename Curve= curve_abc > struct piecewise_curve : public curve_abc { typedef Point point_t; typedef T_Point t_point_t; typedef Time time_t; typedef Numeric num_t; typedef Curve curve_t; typedef typename std::vector < curve_t > t_curve_t; typedef typename std::vector< Time > t_time_t; public: /// \brief Constructor. /// Initialize a piecewise curve by giving the first curve. /// \param pol : a polynomial curve. /// piecewise_curve(const curve_t& cf) { size_ = 0; add_curve(cf); time_curves_.push_back(cf.min()); T_min_ = cf.min(); } virtual ~piecewise_curve(){} virtual Point operator()(const Time t) const { if(Safe &! (T_min_ <= t && t <= T_max_)) { //std::cout<<"[Min,Max]=["< margin) { isContinuous = false; } i++; } return isContinuous; } private: /// \brief Get index of the interval corresponding to time t for the interpolation. /// \param t : time where to look for interval. /// \return Index of interval for time t. /// std::size_t find_interval(const Numeric t) const { // time before first control point time. if(t < time_curves_[0]) { return 0; } // time is after last control point time if(t > time_curves_[size_-1]) { return size_-1; } std::size_t left_id = 0; std::size_t right_id = size_-1; while(left_id <= right_id) { const std::size_t middle_id = left_id + (right_id - left_id)/2; if(time_curves_.at(middle_id) < t) { left_id = middle_id+1; } else if(time_curves_.at(middle_id) > t) { right_id = middle_id-1; } else { return middle_id; } } return left_id-1; } /*Helpers*/ public: /// \brief Get the minimum time for which the curve is defined /// \return \f$t_{min}\f$, lower bound of time range. Time virtual min() const{return T_min_;} /// \brief Get the maximum time for which the curve is defined. /// \return \f$t_{max}\f$, upper bound of time range. Time virtual max() const{return T_max_;} /*Helpers*/ /* Variables */ t_curve_t curves_; // for curves 0/1/2 : [ curve0, curve1, curve2 ] t_time_t time_curves_; // for curves 0/1/2 : [ Tmin0, Tmax0,Tmax1,Tmax2 ] std::size_t size_; // Number of segments in piecewise curve = size of curves_ Time T_min_, T_max_; }; } // end namespace #endif // _CLASS_PIECEWISE_CURVE \ No newline at end of file | 792 | 2,643 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2022-27 | latest | en | 0.493046 |
https://codereview.stackexchange.com/questions/67156/self-made-linked-list-implementing-liste | 1,659,985,231,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570871.10/warc/CC-MAIN-20220808183040-20220808213040-00181.warc.gz | 206,353,809 | 67,105 | I have to make a linked list as a quasi-homework task, about which I have already asked two questions about here and here. Since the whole code is really big (400 lines approx.), I will include only the few functions I am asking you to review, although, if you need any more code, I will upload it.
Here is the .add(int, T) function:
public void add(int index, T data){
if (getNode(index) == null){
throw new NoSuchElementException();
} else if (index<0 || index>=size){
throw new NullPointerException();
}
Node<T> crntNode;
Node<T> newNode = new Node<T>(data);
crntNode = getNode(index);
newNode.index = index;
if (crntNode.previousNode != null) {
crntNode.previousNode.nextNode = newNode;
} else {
firstNode = newNode;
}
crntNode.previousNode = newNode;
newNode.nextNode = crntNode;
crntNode = newNode;
while (crntNode.nextNode != null){
crntNode.index++;
crntNode = crntNode.nextNode;
}
size++;
}
The .get(int) function:
public T get(int index){
if(isEmpty() || index < 0 || index >= size){
return null;
}
Node<T> crntNode;
int count = size - index - 1;
crntNode = lastNode;
while (count > 0) {
crntNode = crntNode.previousNode;
count--;
}
return crntNode.data;
}
And the .remove(int) function:
public T remove(int index) {
if (getNode(index) == null || isEmpty()) {
return null;
} else if (size == 1){
size = 0;
Node<T> crntNode = firstNode;
firstNode = null;
lastNode = null;
return crntNode.data;
}
Node<T> crntNode = getNode(index);
if (crntNode.nextNode != null){
if (crntNode.previousNode != null){
crntNode.previousNode.nextNode = crntNode.nextNode;
} else {
firstNode = crntNode.nextNode;
firstNode.previousNode = null;
size--;
return crntNode.data;
}
}
if (crntNode.previousNode != null){
if (crntNode.nextNode != null) {
crntNode.nextNode.previousNode = crntNode.previousNode;
} else {
lastNode = crntNode.previousNode;
lastNode.nextNode = null;
size--;
return crntNode.data;
}
}
crntNode = crntNode.nextNode;
size--;
for(int i = index; i < size-1; i++){
crntNode = crntNode.nextNode;
crntNode.index--;
}
return crntNode.data;
}
I thought a lot about improving these functions, but I can't seem to find a better concept. The reason might be my lack of knowledge, in which case, I hope you could help me.
Is this code efficient and error-safe enough to be used in real programming situations?
add paramater check
else if (index<0 || index>=size){
throw new NullPointerException();
}
This isn't necessary, because get already catches these cases. You do the same in remove, where you check for isEmpty, which you already checked via the call to get.
You should also try to be consistent with your exceptions. get throws a NoSuchElementException in this case, while add throws NullPointerException.
Oracles implementation throws an IndexOutOfBoundsException which makes a lot of sense (and if you want to stay close to the List interface, you should use it as Tavian Barnes pointed out), but you could make a case for a NullPointerException or IllegalArgumentException if you wanted to. NoSuchElementException on the other hand would not match the description (Thrown by the nextElement method of an Enumeration).
Don't return null
The only occasion you should return null is if the data inside the node is null. If a value is requested that doesn't exist, throw an exception. You violate this in get and remove, which will make it really difficult for a user in case they do store null in the list.
Naming
• crntNode: just add the extra characters, its not that much longer: currentNode.
Assign value when declaring variables
Node<T> crntNode;
Node<T> newNode = new Node<T>(data);
crntNode = getNode(index);
do this:
Node<T> crntNode = getNode(index);
Node<T> newNode = new Node<T>(data);
You do the same thing in get.
Performance
To increase performance, save the previous result of expensive method calls in local variables. For example, getNode is expensive, and you call it twice in add as well as remove.
index
You should get rid of index (not the parameter, but the node field).
In a linked list, nodes don't have an index, they only have a next node (and maybe previous if its doubly linked). Maintaining the index adds a lot of complexity and reduces performance, which can be avoided.
so just remove index, you are not using it anywhere anyways.
• "there isn't a convention for the type of exception" ... Yes there is, IndexOutOfBoundsException Oct 18, 2014 at 21:46
• @TavianBarnes you are right, thanks. updated my answer
– tim
Oct 18, 2014 at 21:56
• Thanks, valuable answer. My only concern is about dropping the index of nodes. How should I implement the indexOf method, if I don't have an index? I know it's doable, but seems a bit odd. Oct 18, 2014 at 22:00
• @AttilaHerbert just iterate over the list and check if the item is equal to the current item. keep a count, and then return that. even with an index you would still need to iterate, so it's not even worse for performance.
– tim
Oct 18, 2014 at 22:03
• I need the indexOf method, because I am implementing the java.util.List<E> interface. I added the index only to be able to implement this method of the interface. Oct 18, 2014 at 22:06 | 1,283 | 5,169 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-33 | longest | en | 0.416836 |
https://www.techwhiff.com/learn/example-consider-the-system-ys-gs-s2-dy-dt-y-uus/327890 | 1,675,758,935,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500392.45/warc/CC-MAIN-20230207071302-20230207101302-00575.warc.gz | 1,005,079,074 | 13,238 | # Example: Consider the system Y(s) =G(s) =s2 d'y dt y-u()U(s) and determine the feedback gain...
###### Question:
Example: Consider the system Y(s) =G(s) =s2 d'y dt y-u()U(s) and determine the feedback gain to place the closed-loop poles at s--1ti. Therefore, we require that α2-2. With xrV and x-dy/dt, the matrix equation for the system and αι G(s) is dy d2y 2dt2 Dorf and Bishop, Modern Control System
Problem: Given the plant G(s)-20(s+5)/s(s+1)(s+4) design a state-feedback controller to yield 9.5% overshoot and a settling time of 0.74 sec. Solution: 1) Determine phase-variable state-space representation:
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Determine the reactions occurring in the following equations 1) Al2(SO4)3(aq) + K2SO4(aq) + 24 H2O(l) ----> 2KAl(SO4)2 * 12H2O(s) 2) 2Al(s) + 2KOH(aq) +4 H2SO4(aq) + 22H20(l) - - - > 2KAl(SO4)2*12H20(s) + 3H2(g) The following is a bank for the types of reactions that can be occurring. Two or m...
##### PLEASE ANSWER ASAP a- Evaluate fft , X(k), of the sequence x(n) defined as x(0)=0.8,...
PLEASE ANSWER ASAP a- Evaluate fft , X(k), of the sequence x(n) defined as x(0)=0.8, x(1)=0.4, x(2) = -0.4 and x(3) = -0.2. b- Verify your calculation using Matlab....
##### 1. The half-life of 2211Na is 950 days. What is its decay constant? Express your answer...
1. The half-life of 2211Na is 950 days. What is its decay constant? Express your answer in s?1. 2.The isotope 8737Rb decays by beta emission. What nucleus is produced?...
##### Barney Googal owns a garage and is contemplating purchasing a tire retreading machine for $26,820. After... Barney Googal owns a garage and is contemplating purchasing a tire retreading machine for$26,820. After estimating costs and revenues, Barney projects a net cash inflow from the retreading machine of $4,200 annually for 10 years. Barney hopes to earn a return of 8% on such investments (For calculat... 1 answer ##### What is negative population growth? What is negative population growth?... 1 answer ##### 2 points) What magnitude charge creates a 5.7NIC electric field at a point 2.2m away? What... 2 points) What magnitude charge creates a 5.7NIC electric field at a point 2.2m away? What electric potential is created at that point?... 1 answer ##### Concord Corporation has retained earnings of$713,600 at January 1, 2020. Net income during 2020 was...
Concord Corporation has retained earnings of $713,600 at January 1, 2020. Net income during 2020 was$1,675,300, and cash dividends declared and paid during 2020 totaled \$78,500. Prepare a retained earnings statement for the year ended December 31, 2020. (List items that increase retained earnings f... | 1,671 | 5,746 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-06 | latest | en | 0.815495 |
https://www.physicsforums.com/threads/a-general-curve-for-the-coffee-cup-problem.436872/ | 1,721,235,834,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.19/warc/CC-MAIN-20240717151625-20240717181625-00455.warc.gz | 820,896,203 | 15,539 | # A general curve for the coffee cup problem.
• mooobag
In summary, the conversation discusses the concept of graphing the temperature of a hot coffee cup as it cools on a table. The individual is seeking resources, such as formulas and graphs, to accurately complete their calculus homework. They also question the shape of the temperature function and how to calculate the amount of heat the coffee cup transfers to the room. A link is provided for further information on the topic.
mooobag
This isn't really homework but my homework made me wonder about this. Any good links to formulas, graphs etc would help make my calc homework unnecessarily accurate.
I'm in a calc class we've been getting problems that require us to graph the temperature of a hot coffee cup as it sits on a table.
All the problems require is that we know that the coffee cup sinks to the temperature of the room over time and stays there but I wonder this:
What shape does the temperature function take?
is it a curve?
is it a line?
does the temperature plummet and take a sharp turn toward the infinite?
Also, how do I figure out how much the coffee cup heats the room? This is the second law of thermodynamics right?
## What is the coffee cup problem?
The coffee cup problem is a mathematical puzzle in which a loop of string is placed around the handle of a coffee cup and then the string is pulled tight. The question is whether the string can be moved around the cup in such a way that it can be removed without cutting or breaking the string.
## What is a general curve for the coffee cup problem?
A general curve for the coffee cup problem is a mathematical formula that represents the shape of the loop of string as it is pulled tight around the cup. This curve can be used to analyze and solve the problem for different sizes and shapes of cups.
## Why is the coffee cup problem important?
The coffee cup problem is a fun and challenging puzzle that can help develop problem-solving skills and critical thinking. It also has real-world applications in fields such as mathematics, physics, and engineering.
## What are some possible solutions to the coffee cup problem?
There are several possible solutions to the coffee cup problem, including using multiple loops of string, creating a knot in the string, or using different methods of pulling the string tight around the cup. The specific solution may vary depending on the size and shape of the cup.
## How can a general curve be used to solve the coffee cup problem?
A general curve can be used to analyze the properties of the loop of string as it is pulled tight around the cup. By studying the curve, we can determine if the string can be removed without cutting or breaking it and find the most efficient way to do so.
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8K | 720 | 3,301 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-30 | latest | en | 0.961036 |
https://www.instructables.com/Design-a-Parametric-Pulley-DXF-STL/ | 1,726,599,678,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00753.warc.gz | 753,280,933 | 31,981 | ## Introduction: Design a Parametric Pulley (DXF / STL)
When I needed to make a pulley for a sumo robot I was building. I didn’t quite find the all in one solution I was searching for, I just needed the DXF of a specific tooth profile for a specific number of teeth.
So I could import this in Fusion 360 to make the custom pulley I want.
I found out about a parametric pulley designer on thingiverse http://www.thingiverse.com/thing:16627 but I couldn’t edit them in Fusion 360 to my needs.
I found a workaround on a site: http://rasterweb.net/raster/2012/07/16/openscad-t... .to export the tooth profile as a dxf.
And I simplified the complete workflow into one scad file. So you could easily generate a custom pulley with a specific: number of teeth, tooth profile and shaft diameter.
## Step 1: Opening the File
You can get the latest version of the scad file on my thingiverse http://www.thingiverse.com/thing:2316122/#files
Or you can get the file in this step.
## Step 2: Editting the File
Edit the parameters you see in the first three lines.
• N_tooth = 42; //number of teeth
• shaft_Dia = 4.2; //shaft diameter in mm
• profile = 14; //profile options: 1=MXL 2=40DP 3=XL 4=H 5=T2.5 6=T5 7=T10 8=AT5 9=HTD_3mm 10=HTD_5mm 11=HTD_8mm 12=GT2_2mm 13=GT2_3mm 14=GT2_5mm
When you changed the parameters press F6 to render the design.
If you are satisfied with the outcome you can export the file as a DXF.
This DXF can be imported in almost every vector program and cad program: Fusion 360, Autocad, SolidWorks, Inkscape...
In the next step you'll see how I make an example pulley in Fusion 360. But you can stop here if you just need the DXF.
## Step 3: Making the Pulley in Fusion 360
This is a step by step photo series of how I made a pulley in fusion 360.
If there is any need for it I'll make a video of this process. | 535 | 1,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-38 | latest | en | 0.846614 |
https://www.roadlesstraveledstore.com/how-do-you-find-the-magnetic-force-of-a-wire/ | 1,695,875,784,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510358.68/warc/CC-MAIN-20230928031105-20230928061105-00698.warc.gz | 1,070,477,794 | 15,526 | # How do you find the magnetic force of a wire?
## How do you find the magnetic force of a wire?
The magnetic force on a current-carrying wire in a magnetic field is given by →F=I→l×→B. For part a, since the current and magnetic field are perpendicular in this problem, we can simplify the formula to give us the magnitude and find the direction through the RHR-1. The angle θ is 90 degrees, which means sinθ=1.
What is the magnetic force on the wire?
The magnetic force on a current-carrying wire through a magnetic field is given by F = IL x B. When a current is passed through a magnetic field, the magnetic field exerts a force on the wire in a direction perpendicular to both the current and the magnetic field.
### How are the force on the wire and the force on the magnet related?
The magnetic force on a current-carrying wire is perpendicular to both the wire and the magnetic field with direction given by the right hand rule.
What causes a magnetic force in a wire?
When a wire is moved in a magnetic field, the field induces a current in the wire. Conversely, a magnetic field is produced by an electric charge in motion. A charge moving in a straight line, as through a straight wire, generates a magnetic field that spirals around the wire.
## How do you calculate the force of a wire?
The equation for calculating the force on a wire is Force (N) = magnetic flux density (T) × current (A) × length (m) or, in short F = B I L.
What is the cause of magnetic field inside a human body?
The human body is a good conductor of electricity. Under the influence of an alternating current, the electric charges inside the body move back and forth at the same frequency as the field (60 Hz). In other words, the field produces weak electric currents in the body. The body does not accumulate any current.
### What is magnetic force Class 6?
Magnetic force, attraction or repulsion that arises between electrically charged particles because of their motion. It is the basic force responsible for such effects as the action of electric motors and the attraction of magnets for iron.
What is the formula of magnetic field?
The magnitude of the force on a wire carrying current I with length L in a magnetic field is given by the equation. F=ILBsinθ where θ is the angle between the wire and the magnetic field. The force is perpendicular to the field and the current.
## Why unit of magnetic field is tesla?
The tesla (symbol T) is the derived SI unit of magnetic flux density, which represents the strength of a magnetic field. One tesla represents one weber per square meter. The equivalent, and superseded, cgs unit is the gauss (G); one tesla equals exactly 10,000 gauss.
How is the force of magnetism on a wire determined?
It is possible to determine the force of magnetism exerted on a current-carrying wire passing through a magnetic field at right angles to the wire. Experiments show that the magnitude of the force, , on the wire, is proportional to the strength of the field, , the current, , in the wire, and the length, , of the wire in the magnetic field.
### How are magnetic forces used in an experiment?
In this experiment, you will investigate the magnetic force between two current carrying wires. One wire will be a coil of 10 turns and the other will be a coil of 38 turns. The 10-turn coil will be taped to one end of a pivoted balance beam. The beam pivots on two pins.
How to predict the nature of the magnetic force?
1. To predict and verify the nature of the magnetic force acting on a current-carrying wire when the wire is placed in a magnetic field. INTRODUCTION G The force on a segment ds G of a wire carrying current Iin a magnetic field B ext is given by G G G d F =Id s ×Bex t (6.1) G where B
## How does a wire loop create a magnetic force?
This wire loop is then placed into the permanent magnet assembly so the wire loop is perpendicular to the magnetic field but is not touching the magnets. Then, when a current flows through the wire loop, a magnetic force is created.
18/05/2019 | 905 | 4,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-40 | latest | en | 0.922968 |
https://www.justintools.com/unit-conversion/force.php?k1=zettanewtons&k2=joules-per-centimeter | 1,716,649,580,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058829.24/warc/CC-MAIN-20240525131056-20240525161056-00561.warc.gz | 732,257,510 | 27,040 | Please support this site by disabling or whitelisting the Adblock for "justintools.com". I've spent over 10 trillion microseconds (and counting), on this project. This site is my passion, and I regularly adding new tools/apps. Users experience is very important, that's why I use non-intrusive ads. Any feedback is appreciated. Thank you. Justin XoXo :)
# FORCE Units Conversionzettanewtons to joules-per-centimeter
1 Zettanewtons
= 1.0E+23 Joules Per Centimeter
Category: force
Conversion: Zettanewtons to Joules Per Centimeter
The base unit for force is newtons (Derived SI Unit)
[Zettanewtons] symbol/abbrevation: (ZN)
[Joules Per Centimeter] symbol/abbrevation: (J/cm)
How to convert Zettanewtons to Joules Per Centimeter (ZN to J/cm)?
1 ZN = 1.0E+23 J/cm.
1 x 1.0E+23 J/cm = 1.0E+23 Joules Per Centimeter.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [force] => (newtons), 1 Zettanewtons (ZN) is equal to 1.0E+21 newtons, while 1 Joules Per Centimeter (J/cm) = 0.01 newtons.
1 Zettanewtons to common force units
1 ZN = 1.0E+21 newtons (N)
1 ZN = 2.2480892365534E+20 pounds force (lbf)
1 ZN = 1.2137804779884E+28 atomic units of force (auf)
1 ZN = 1.0E+39 attonewtons (aN)
1 ZN = 1.0197162129779E+25 centigrams force (cgf)
1 ZN = 1.0E+23 centinewtons (cN)
1 ZN = 1.0E+20 decanewtons (daN)
1 ZN = 1.0E+22 decinewtons (dN)
1 ZN = 1.0E+26 dynes (dyn)
1 ZN = 1000 exanewtons (EN)
Zettanewtonsto Joules Per Centimeter (table conversion)
1 ZN = 1.0E+23 J/cm
2 ZN = 2.0E+23 J/cm
3 ZN = 3.0E+23 J/cm
4 ZN = 4.0E+23 J/cm
5 ZN = 5.0E+23 J/cm
6 ZN = 6.0E+23 J/cm
7 ZN = 7.0E+23 J/cm
8 ZN = 8.0E+23 J/cm
9 ZN = 9.0E+23 J/cm
10 ZN = 1.0E+24 J/cm
20 ZN = 2.0E+24 J/cm
30 ZN = 3.0E+24 J/cm
40 ZN = 4.0E+24 J/cm
50 ZN = 5.0E+24 J/cm
60 ZN = 6.0E+24 J/cm
70 ZN = 7.0E+24 J/cm
80 ZN = 8.0E+24 J/cm
90 ZN = 9.0E+24 J/cm
100 ZN = 1.0E+25 J/cm
200 ZN = 2.0E+25 J/cm
300 ZN = 3.0E+25 J/cm
400 ZN = 4.0E+25 J/cm
500 ZN = 5.0E+25 J/cm
600 ZN = 6.0E+25 J/cm
700 ZN = 7.0E+25 J/cm
800 ZN = 8.0E+25 J/cm
900 ZN = 9.0E+25 J/cm
1000 ZN = 1.0E+26 J/cm
2000 ZN = 2.0E+26 J/cm
4000 ZN = 4.0E+26 J/cm
5000 ZN = 5.0E+26 J/cm
7500 ZN = 7.5E+26 J/cm
10000 ZN = 1.0E+27 J/cm
25000 ZN = 2.5E+27 J/cm
50000 ZN = 5.0E+27 J/cm
100000 ZN = 1.0E+28 J/cm
1000000 ZN = 1.0E+29 J/cm
1000000000 ZN = 1.0E+32 J/cm
(Zettanewtons) to (Joules Per Centimeter) conversions | 1,084 | 2,380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-22 | latest | en | 0.742891 |
http://studyadda.com/notes/jee-main-advanced/mathematics/functions/algebra-of-functions/9400 | 1,516,531,132,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890514.66/warc/CC-MAIN-20180121100252-20180121120252-00229.warc.gz | 341,434,125 | 14,420 | # JEE Main & Advanced Mathematics Functions Algebra of Functions
## Algebra of Functions
Category : JEE Main & Advanced
(1) Scalar multiplication of a function : $(c\,f)(x)=c\,f(x),$ where $c$ is a scalar. The new function $c\,f(x)$ has the domain ${{X}_{f}}.$
$(f\pm g)(x)=f(x)\pm g(x).$ The new function has the domain $X$.
(3) Multiplication of functions
$(fg)(x)=(g\,f)(x)=f(x)g\,(x).$ The product function has the domain $X$.
(4) Division of functions :
(i) $\left( \frac{f}{g} \right)\,(x)=\frac{f(x)}{g(x)}.$ The new function has the domain $X,$ except for the values of $x$ for which $g\,(x)=0.$
(ii) $\left( \frac{g}{f} \right)\,(x)=\frac{g(x)}{f(x)}.$ The new function has the domain $X,$ except for the values of $x$ for which $f(x)=0.$
(5) Equal functions : Two function $f$ and $g$ are said to be equal functions, if and only if
(i) Domain of $f=$ Domain of $g$
(ii) Co-domain of $f=$ Co-domain of $g$
(iii) $f(x)=g(x)\,\forall x\in$ their common domain
(6) Real valued function : If $R,$ be the set of real numbers and $A,\,\,B$ are subsets of $R,$ then the function $f:A\to B$ is called a real function or real–valued function.
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You will be redirected in 3 sec | 407 | 1,233 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-05 | latest | en | 0.655924 |
https://brilliant.org/problems/binary-search-on-bits/ | 1,529,610,984,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864257.17/warc/CC-MAIN-20180621192119-20180621212119-00189.warc.gz | 551,823,703 | 11,836 | # Binary Search on Bits
Computer Science Level pending
Chris realize his computer couldn't perform accurate comparisons on integers. That is, his computer couldn't evaluate if a > b is true or false.
He has a sorted array of $$n$$ integers and he needs to find if the number $$x$$ is within the array. He doesn't want to perform a linear search, so he came up with another idea of Binary Search. Here is the algorithm :
1. Look at the $$k=0$$ bit of $$x$$, it's either 1 or 0.
2. If it's 1, binary search for the lower bound that has the $$k$$-th bit as 1.
3. If it's 0, binary search for the upper bound that has the $$k$$-th bit as 0.
4. Increment $$k$$ by 1 and go back to step 1.
What is the time complexity of this algorithm? Or if you think this algorithm doesn't work, choose the option "This algorithm doesn't work".
Details and Assumptions
• Notice that this algorithm doesn't use any of the following operators : < > <= >=.
• This algorithm is written in Python, where there is no upper bound for $$x$$ like C++.
× | 266 | 1,031 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-26 | latest | en | 0.95822 |
https://www.slideserve.com/jerry-nunez/suppose-f-x-y-z-is-bounded-f-or-all-x-y-z-in-a-box-b-a-b-c-d-p-q | 1,511,330,411,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806465.90/warc/CC-MAIN-20171122050455-20171122070455-00779.warc.gz | 851,702,615 | 14,289 | Suppose f ( x,y,z ) is bounded f or all ( x,y,z ) in a box B = [ a,b ][ c,d ][ p,q ].
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# Suppose f ( x,y,z ) is bounded f or all ( x,y,z ) in a box B = [ a,b ][ c,d ][ p,q ]. - PowerPoint PPT Presentation
Suppose f ( x,y,z ) is bounded f or all ( x,y,z ) in a box B = [ a,b ][ c,d ][ p,q ]. z. ( a,c,q ). ( a,d,q ). ( b,c,q ). ( b,d,q ). y. ( a,c,p ). ( a,d,p ). x. ( b,c,p ). ( b,d,p ). Just as in the development of the double integral, we can
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## PowerPoint Slideshow about ' Suppose f ( x,y,z ) is bounded f or all ( x,y,z ) in a box B = [ a,b ][ c,d ][ p,q ].' - jerry-nunez
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Presentation Transcript
Suppose f(x,y,z) is bounded for all (x,y,z) in a box B = [a,b][c,d][p,q].
z
(a,c,q)
(a,d,q)
(b,c,q)
(b,d,q)
y
(a,c,p)
(a,d,p)
x
(b,c,p)
(b,d,p)
Just as in the development of the double integral, we can
partition [a,b] into n equal sub-intervals each with length x,
partition [c,d] into n equal sub-intervals each with length y,
partition [p,q] into n equal sub-intervals each with length z.
For each of the n3 sub-boxes, we may multiply a value of the function f(x,y,z) anywhere on the sub-box by the volume V = x y z.
If the limit as n exists for the sum of these products (which will be true for continuous functions), then we say that f is integrable over B, and this limit is called the triple integral
f(x,y,z) dx dy dz = f(x,y,z) dV =
B
B
q
q
d
b
b
d
f(x,y,z) dx dy dz = f(x,y,z) dy dx dz=
etc.
p
c
a
p
a
c
Example Integrate f(x,y,z) = (3x + 2y + z)2 over the box defined by [0,1/3][–1/2,0][0,1] using any two different orders of integration.
1
0
1/3
(3x + 2y + z)2dx dy dz =
1/12
0
–1/2
0
An elementary region in R3 is one which can be defined by restricting one of the variables to be between two functions of the remaining two variables, where the domains of these functions is an elementary region in R2.
Similar to the way the integral of a function f(x,y) over an elementary region in R2 is defined, we can define the integral of a function f(x,y,z) over an elementary region W in R3 to be
f(x,y,z) dx dy dz = f(x,y,z) dx dy dz
W
B
where B is a box containing the region W, and the function f(x,y,z) is defined to be equal to f(x,y,z) on the region W and equal to 0 (zero) outside the region W.
Note that the volume of an elementary region W in R3 can be found from
dx dy dz
which can be written as a double integral of the difference between two functions of two variables.
W
x2 + y2 + z2 = a2 .
One way the region inside the sphere can be described is
– a x a
– a2 – x2 y a2 – x2
– a2 – x2 – y2 z a2 – x2 – y2
a
a2 – x2
a2 – x2 – y2
dz dy dx =
– a
– a2 – x2
– a2 – x2 – y2
a
a2 – x2
a2 – x2 – y2
z dy dx =
– a
– a2 – x2
z = – a2 – x2 – y2
a
a2 – x2
Note: One way of evaluating this integral is to use a table of integrals.
2 a2 – x2 – y2 dy dx
– a
– a2 – x2
b
Another way of evaluating this integral is to realize that
b2 – t2 dt
– b
is the area of a semicircle of radius b, which is equal to b2/2.
We see then (with b = a2 – x2 and t = y) that
a
a2 – x2
a
2 a2 – x2 – y2 dy dx =
(a2 – x2)dx =
– a
– a
– a2 – x2
a
4a3
——
3
(a2x – x3/3) =
x = – a
Example
Integrate the function xyze over the region between the xy plane, the plane y = x, the plane x = 4, and the cone z2 = x2 + y2 in the region where x, y, and z are all non-negative.
z2
z
z2 = x2 + y2
First, we observe that this region can be described by
x
0
4
0
y
x
x
y
x = 4
0
z
x2 + y2
y = x
4
x
x2 + y2
x2 + y2
4
x
z2
z2
xyze dz dy dx =
xye
—— dy dx =
2
0
0
0
0
0
z = 0
4
x
4
x
x2 + y2
x2 + y2
xyexy
——— – — dy dx =
2 2
xexy2
——— – — dx =
4 4
0
0
0
y = 0
4
2x2
x2
xex3xe
——— – — – —— dx =
4 4 4
0
4
2x2
x2
ex4e
—— – — – —— =
16 16 8
x = 0
e32– 256 – 2e16– 1 + 2 e32– 2e16– 255
—————————— = ———————
16 16 | 1,744 | 4,505 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-47 | latest | en | 0.792872 |
https://axiomsofchoice.org/cauchy_principal_value | 1,726,029,583,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651343.80/warc/CC-MAIN-20240911020451-20240911050451-00364.warc.gz | 108,731,354 | 4,020 | Cauchy principal value
Partial function
definiendum $\mathcal P\int_a^b: \mathbb R^2\times(\mathbb R\to\overline{\mathbb R})\to\overline{\mathbb R}$
$p$ … ordered sequence of the $m$ poles of $f$
definiendum $\mathcal P\int_a^b(f):=\mathrm{lim}_{\varepsilon\to 0}\left(\int_a^{p_1-\varepsilon}f(x)\,\mathrm dx+\int_{p_1+\varepsilon}^{p_2-\varepsilon}f(x)\,\mathrm dx+\cdots+\int_{p_m+\varepsilon}^b f(x)\,\mathrm dx\right)$
Discussion
The Cauchy principal value is the value of an integral where the singularities are approached in a symmetric fashion.
Examples
$\mathcal P\int_{-3}^7\left(\lambda x.\frac{1}{x}\right)=\int_{-3}^{-\varepsilon}\frac{1}{x}\,\mathrm dx+\int_{\varepsilon}^{7}\frac{1}{x}\,\mathrm dx=\left(\log\left|-\varepsilon\right|-\log\left|-3\right|\right)+\left(\log\left|7\right|-\log\left|\varepsilon\right|\right)=\log\left|\frac{7}{3}\right|$ | 326 | 875 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2024-38 | latest | en | 0.340149 |
https://www.xwordinfo.com/Thumbs?author=Tomas+Spiers&comments=Jeff | 1,674,959,214,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499697.75/warc/CC-MAIN-20230129012420-20230129042420-00770.warc.gz | 1,088,902,194 | 5,827 | This web browser is not supported. Use Chrome, Edge, Safari, or Firefox for best results.
# Tomas Spiers author page
## 2 puzzles by Tomas Spiers with Jeff Chen comments
TotalDebutLatestCollabs
212/13/202111/2/20221
SunMonTueWedThuFriSat
0101000
RebusScrabblePOW
11.522
View these same grids with comments from:
Collaborator: Ethan Zou
T I D A L I D O S B R A T A B O V E H E H E R O T I F E R E O R O C H E I G O R T I M P U P A E B E E M E A S I A M F L Y O F F C R I B L O O S E N S H A W E D L A Y U P D A T A L E T B O X E R F E R Y R U E M O O L A F I E C E E M O B A N D H A I R V A C A Y S M E S A S A V E S E N O O B S P A Z M O H S B O R D E T E R I E F R E E U S E D A M I N O M E E T T E S S S A L T S
★ Regular readers might be surprised that I picked today's as the POW! Not only is this rebus-weary solver wary of the genre, but having one on a Wednesday, when no one expects it?
That's NEITHER HERE NOR THERE. I had so much fun solving this one that I declare WINNER WINNER CHICKEN DINNER.
I found today's solve so entertaining that I spent an hour figuring out why:
• Great starter rebus. Right from the get-go, something was up. It's easy to pick out the common letters E and R, so along with the straightforward theme clues, things like ROGER FEDERER came easily.
• Snazzy phrases. I recently got hooked onto a YouTube embiggenator show, after being rapt at the creation of the world's largest FERRERO ROCHER. Maybe nauseating, but also drool-worthy.
• Solid rebus rationale. BOX ER = shove ER into a box. Yup.
• OH HELL YEAH, those bonuses!
• Themeless-level wordplay in the clues. This bball superfan loved describing a LAY-UP as an "underhanded move."
I have reservations about any sort of rebus on a Wednesday. I hear a lot of grumbling from rebus haters who say that what they like about crosswords is the simplicity of putting a single letter in a box, and I sympathize. However, rules getting broken is the price of creativity, and that's something to which any art form ought to aspire.
I'm not wild about featuring boxing, given the severity of the head injuries and even deaths. That said though, there was so much to enjoy here that it made it easy to throw in the towel of my hesitations. Knockout POW! debut.
E B B S C R A B D E T E R A E R O F O I E I M A G E R A I L R O A D S T A T I O N S T E V E M E T A L L S D L E A P I C E T A N A M A R K E T C R A S H F A R S W A N T O N Y A L O S S S P O R T T O R T A M O U R W I R E T I E C I N N A M O N B U N H A D S P A N E D G E P A Z I W I S H E U R O S S T O P D R O P A N D R O L L S T O O L N I L E U N D O T A S T Y S N O W S E E P
★ I haven't been so happily stumped by "Name That Theme" in ages. RAILROAD STATION … MARKET CRASH … CINNAMON BUN? Could it be a "words that can follow X" theme ... nope. Their first words aren't synonyms, nor are their last. Perhaps there are words hidden in the middle of the themers? Nuh-uh.
I give. Tell me the revealer!
STOP DROP AND ROLL?
How does that …
Ah! RAILROAD STATION is a STOP, MARKET CRASH is a DROP, and CINNAMON BUN is a ROLL. Elation as the lightbulb finally flicked on!
I love that each of the three theme phrases is strong and in the language, plus they're all nouns. Verbs might have worked for any, like COME TO AN END, TAKE THE PLUNGE, or SHOOT CRAPS, so it's elegant to connect the themers further through consistency.
Solid gridwork, especially for a debut. With four themers, there's always room for at least two long bonuses, and BRIE LARSON and ANOTHER ONE hit that mark.
Most importantly for a Monday puzzle, Tomas took such care to avoid sticky globs of glue. Bryn MAWR might look unfamiliar to some. AERO is easy enough to figure out from etymology. Given such excellence in gridding, I'd suggest that Tomas try adding another set of long Downs in the SW and NE next time.
This finance guy never likes hearing about a MARKET CRASH, but he loves it when a theme crashes joyously over him. No CRABs today. | 1,195 | 3,974 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2023-06 | latest | en | 0.929247 |
https://brilliant.org/problems/nmo-problem-2/ | 1,529,803,368,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865995.86/warc/CC-MAIN-20180624005242-20180624025242-00017.warc.gz | 573,283,244 | 10,964 | # NMO Problem 2
A circumference was divided in $$n$$ equal parts. On each of these parts one number from $$1$$ to $$n$$ was placed such that the distance between consecutive numbers is always the same. Numbers $$11$$, $$4$$ and $$17$$ were in consecutive positions. In how many parts was the circumference divided?This problem is from the NMO.This problem is part of this set.
× | 90 | 380 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-26 | latest | en | 0.982868 |
https://incomeaftertax.com/20-an-hour-after-tax-nz | 1,702,285,962,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679103810.88/warc/CC-MAIN-20231211080606-20231211110606-00127.warc.gz | 350,557,548 | 30,809 | # \$20 an Hour After Tax NZ
On this page, you'll find a detailed analysis of a \$20 after-tax hourly salary for 2023, with calculations for annual, monthly, weekly, daily, rates as of November 16th, 2023 at 07:00 PM. Additionally, discover important information answering key questions like "\$20 an hour is how much a year?", "\$20 an hour is how much a month?", "\$20 an hour is how much a week?", "\$20 an hour is how much a day?", .
## \$20 an Hour After Tax in New Zealand
Validated and accurate as of November 16th, 2023 at 07:00 PM.
## How much is \$20 an Hour After Tax in New Zealand?
In the year 2023, in New Zealand, \$20 an hour gross salary, after-tax, is \$34,808 annual, \$2,901 monthly, \$667.1 weekly, \$133.42 daily, and \$16.68 hourly gross based on the information provided in the calculator above.
Check the table below for a breakdown of \$20 an hour after tax in New Zealand.
Yearly Monthly Weekly Daily Hourly
Gross Salary \$41,742 \$3,479 \$800 \$160 \$20
PAYE \$6,325 \$527.07 \$121.22 \$24.24 \$3.03
ACC \$609.43 \$50.79 \$11.68 \$2.34 \$0.29
Personal Allowance \$2,340 Per Year
Taxable Salary \$39,402 Per Year
Take-Home Pay (NET) \$34,808 \$2,901 \$667.1 \$133.42 \$16.68
If you're interested in discovering how much you would earn an hour with an extra \$10 added to your hourly salary, you can explore the calculations for a \$30 hourly income. This will provide you with a better understanding of the difference that additional \$10 can make to your hourly earnings.
## \$20 an Hour is How Much a Year?
\$20 an hour breaks down to the following yearly salary:
• Gross (pre-tax) yearly wage of \$41,742
• Take-home (NET) yearly income of \$34,808
## \$20 an Hour is How Much a Month?
When evaluating a \$20 an hour after tax income, the corresponding monthly earnings can be determined:
• Take-home (NET) monthly income: \$2,901
In order to discover \$20 an hour is how much a month? - simply divide the annual amount by 12, resulting in a monthly income of \$2,901.
## \$20 an Hour is How Much a Week?
When assessing a \$20 an hour after tax salary, the associated weekly earnings can be calculated:
• Take-home (NET) weekly income: \$667.1
To answer \$20 an hour is how much a week? - divide the annual sum by 52, resulting in a weekly income of \$667.1.
## \$20 an Hour is How Much a Day?
When examining a \$20 an hour after tax income, the corresponding daily earnings can be determined:
• Take-home (NET) daily income: \$133.42 (assuming a 5-day work week)
To find out \$20 an hour is how much a day? - divide the annual figure by 260 (52 weeks * 5 days), resulting in a daily income of \$133.42.
## Is \$20 an Hour a Good Salary?
To answer if \$20 an hour is a good salary. We need to compare it to the national median. After calculation using ongoing year (2023) data, the salary of \$20 an hour is 1.35 times or 25.67% lower than the national median. So, is \$20 an hour a good salary?
Based on comparison to the national median and that it is quite below average, in our opinion, if your finances are planned right, it is still a livable salary in New Zealand.
We think these three links are helpful and related to the \$20 an Hour After Tax NZ: Tax rates for individuals, Money & Taxes in New Zealand, and Taxation in New Zealand. | 872 | 3,298 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-50 | longest | en | 0.902936 |
https://oneconvert.com/unit-converters/power-converter/exajoule-second-to-btu-it-second | 1,716,514,359,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058675.22/warc/CC-MAIN-20240523235012-20240524025012-00874.warc.gz | 384,193,846 | 27,827 | # Convert exajoule/second to Btu (IT)/second
## How to Convert exajoule/second to Btu (IT)/second
To convert exajoule/second to Btu (IT)/second , the formula is used,
$\mathrm{1 Exajoule/second}=9.48 x 10^11\mathrm{Btu\left(IT\right)/second}$
where the EJ,s to Btu,s value is substituted to get the answer from Power Converter.
1 EJ,s
=
94781e+10 Btu,s
1 Btu,s
=
1.0551e-15 EJ,s
Example: convert 15 EJ,s to Btu,s:
15 EJ,s
=
15
x
94781e+10 Btu,s
=
14217e+12 Btu,s
## exajoule/second to Btu (IT)/second Conversion Table
exajoule/second (EJ,s) Btu (IT)/second (Btu,s)
0.01 EJ,s 9478171203e+3 Btu,s
0.1 EJ,s 9478171203e+4 Btu,s
1 EJ,s 9478171203e+5 Btu,s
2 EJ,s 1895634240e+6 Btu,s
3 EJ,s 2843451360e+6 Btu,s
5 EJ,s 4739085601e+6 Btu,s
10 EJ,s 9478171203e+6 Btu,s
20 EJ,s 1895634240e+7 Btu,s
50 EJ,s 4739085601e+7 Btu,s
100 EJ,s 9478171203e+7 Btu,s
1000 EJ,s 9478171203e+8 Btu,s
### Popular Unit Conversions Power
The most used and popular units of power conversions are presented for quick and free access. | 427 | 1,009 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-22 | latest | en | 0.567856 |
https://gmatclub.com/forum/a-recent-poll-indicates-that-many-people-in-the-united-18679-40.html?kudos=1 | 1,495,600,330,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607786.59/warc/CC-MAIN-20170524035700-20170524055700-00011.warc.gz | 758,062,981 | 59,804 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
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# A recent poll indicates that many people in the United
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Manager
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A recent poll indicates that many people in the United [#permalink]
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06 Aug 2005, 05:40
15
KUDOS
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This post was
BOOKMARKED
A recent poll indicates that many people in the United States hold a combination of conservative and liberal political views; i.e., they denounce big government, saying government is doing too much and has become too powerful, while at the same time supporting many specific government programs for health care, education, and the environment.
(A) they denounce the big government, saying government is doing too much and has become too powerful, while at the same time supporting
(B) they denounce big government- they say that government is doing too much and has become too powerful-but supporting at the same time
(C) they denounce big government, say that government is doing too much and it has become too powerful, while they support at the same time
(D) while they denounce big government, saying that government is doing too much and has become too powerful, at the same time supporting
(E) while they are denouncing big government- they say that government is doing too much and it has become too powerful-supporting at the same time
OA a few posts below.
Source – GMATPrep by Pearson VUE
Last edited by doe007 on 30 Apr 2013, 09:44, edited 2 times in total.
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04 Apr 2007, 19:32
C for me.
While needs a whole sentence, not a participle such as "supporting".
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Re: SC - while at the same time [#permalink]
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04 Apr 2007, 22:16
amd08 wrote:
68. A recent poll indicates that many people in the United States hold a combination of conservative and liberal political views; i.e., they denounce the big government, saying government is doing too much and has become too powerful, while at the same time supporting many specific government programs for health care, education, and the environment.
A) they denounce the big government, saying government is doing too much and has become too powerful, while at the same time supporting
B) they denounce big government- they say that government is doing too much and has become too powerful-but supporting at the same time
C) they denounce big government, say that government is doing too much and it has become too powerful, while they support at the same time
D) while they denounce big government, saying that government is doing too much and has become too powerful, at the same time supporting
All choices have one thing or other wrong about it. Is there any other option?
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Re: SC #1 - Gmatprep [#permalink]
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04 Jul 2007, 07:40
SimaQ wrote:
A recent poll indicates that many people in the United States hold a combination of conservative and liberal views; i.e., they denounce big government, saying government is doing too much and has become too powerful, while at the same time supporting many specific government programs for health care, education, and the environment.
A) they denounce big government, saying government is doing too much and has become too powerful, while at the same time supporting
B) they denounce big government - they say government is doing too much and has become too powerful - but supporting at the same time
C) they denounce big government, they say that government is doing too much and it has become too powerful, while they support at the same time
D) while they denounce big government, saying that government is doing too much and has become too powerful, at the same time supporting
E) while they are denouncing big government - they say that government is doing too much and it has become too powerful - supporting at the same time
Why is it denounce and supporting?
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25 Oct 2007, 20:56
any further comments on D and E
why they are wrong?
Director
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25 Oct 2007, 21:22
Ozmba2006 wrote:
any further comments on D and E
why they are wrong?
D - Saying and supporting seems to modify the clause 'they denounce big government' - incorrect.
E - 'they are denouncing' not parallel with 'supporting'
'A' - by POE
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26 Oct 2007, 13:50
Straight A. Just took a break from my applications to answer this question
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A recent poll indicates that many people in the United [#permalink]
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28 Nov 2007, 18:32
A recent poll indicates that many people in the United States hold a combination of conservative and liberal political views; i.e., they denounce the big government, saying government is doing too much and has become too powerful, while at the same time supporting many specific government programs for health care, education, and the environment.
A) they denounce the big government, saying government is doing too much and has become too powerful, while at the same time supporting
B) they denounce big government- they say that government is doing too much and has become too powerful-but supporting at the same time
C) they denounce big government, say that government is doing too much and it has become too powerful, while they support at the same time
D) while they denounce big government, saying that government is doing too much and has become too powerful, at the same time supporting
E) while they are denouncing big government- they say that government is doing too much and it has become too powerful-supporting at the same time
Amar
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29 Nov 2007, 03:51
I think there is something wrong with A - "the big government". We're talking an idiom here "Big Government". The big government would be referring to a specific government that was particularly big!
Of the other choices, there are problems with each of C, D & E.
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29 Nov 2007, 16:38
OA is A,
Can somebody explain in detail, why OA is A and why others are wrong.
Amar
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17 Dec 2007, 18:53
A recent poll indicates that many people in the United States hold a combination of conservative and liberal political views; i.e., they denounce the big government, saying government is doing too much and has become too powerful, while at the same time supporting many specific government programs for health care, education, and the environment.
A) they denounce the big government, saying government is doing too much and has become too powerful, while at the same time supporting
B) they denounce big government- they say that government is doing too much and has become too powerful-but supporting at the same time
C) they denounce big government, say that government is doing too much and it has become too powerful, while they support at the same time
D) while they denounce big government, saying that government is doing too much and has become too powerful, at the same time supporting
E) while they are denouncing big government- they say that government is doing too much and it has become too powerful-supporting at the same time
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18 Dec 2007, 01:20
A recent poll indicates that many people in the United States hold a combination of conservative and liberal political views; i.e., they denounce the big government, saying government is doing too much and has become too powerful, while at the same time supporting many specific government programs for health care, education, and the environment.
A) they denounce the big government, saying government is doing too much and has become too powerful, while at the same time supporting
B) they denounce big government- they say that government is doing too much and has become too powerful-but supporting at the same time
C) they denounce big government, say that government is doing too much and it has become too powerful, while they support at the same time
D) while they denounce big government, saying that government is doing too much and has become too powerful, at the same time supporting
E) while they are denouncing big government- they say that government is doing too much and it has become too powerful-supporting at the same time
B is incorrect because if we write down "but", we must also consider another present indicative form. C is incorrect because...they denounce how? by saying smtg...we need a gerund. D also needs an indicative because says that...while they.....they... and is incorrect because uses a gerund. E makes the same mistakes. A IS CORRECT
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18 Dec 2007, 09:30
I would prefer C for llism but I think A is correct.
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18 Dec 2007, 17:52
OA is A
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A recent poll indicates that many people in the United [#permalink]
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21 Mar 2008, 07:10
A recent poll indicates that many people in the United States hold a combination of conservative and liberal political views; i.e., they denounce the big government, saying government is doing too much and has become too powerful, while at the same time supporting many specific government programs for health care, education, and the environment.
A) they denounce the big government, saying government is doing too much and has become too powerful, while at the same time supporting
B) they denounce big government- they say that government is doing too much and has become too powerful-but supporting at the same time
C) they denounce big government, say that government is doing too much and it has become too powerful, while they support at the same time
D) while they denounce big government, saying that government is doing too much and has become too powerful, at the same time supporting
E) while they are denouncing big government- they say that government is doing too much and it has become too powerful-supporting at the same time
Oh yes, this one was discussed before. please explain your answer and the reasons not to choose the other solutions
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Re: Brutal SC; Q68 - denounce, saying, supporting [#permalink]
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06 Apr 2008, 19:14
Good notes from a separate forum:
B is wrong because the main sentence reads "they denounce big government but supporting at the same time." "But" is a coordinating conjunction, used to connect two independent clauses, but the second half of this sentence "supporting at the same time..." is not an independent clause (that is, it cannot function as a stand-alone sentence).
C is a run on: "they denounce, they say" Both of those - "they denounce big government" and "
they say that government is doing too much" - are independent clauses and need to be connected either via a coordinating conjunction or a semi-colon. See, for example "they say that government is doing too much and it has become too powerful" - those are two independent clauses connected by the coordinating conjuction "and."
D is wrong because the main sentence reads "they denounce big government at the same time supporting" - you need a contrast word such as "while" between government and at. Same reason for E.
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Re: Brutal SC; Q68 - denounce, saying, supporting [#permalink]
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06 Apr 2008, 21:35
I think the structure should be ..
they denounce blah blah, at the same time supporting..blah blah..
A gets this correct
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19 May 2008, 01:39
B lacks ||lism
C is wordy
E suffers from placement error
my pick would be D
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19 May 2008, 01:53
Answer C for II sm "They denounce" is parallel to "they support".
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19 May 2008, 01:59
C for ll... to be honest, they all sound awkward
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19 May 2008, 02:33
A for me. saying...... supporting.. seems good, while at the same time supporting..also looks good.
all other options don't look good to me.
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Re: SC. Prep. combination [#permalink] 19 May 2008, 02:33
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Display posts from previous: Sort by | 3,817 | 14,914 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-22 | latest | en | 0.966995 |
https://freetofindtruth.blogspot.com/2015/12/47-55-66-113-131-john-elway-near.html | 1,719,290,564,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865560.33/warc/CC-MAIN-20240625041023-20240625071023-00778.warc.gz | 238,994,345 | 30,883 | Thursday, December 31, 2015
47 55 66 113 131 | John Elway, a Near Confirmation the Broncos Will Be In Super Bowl 50
John = 1+6+8+5 = 20
Elway = 5+3+5+1+7 = 21
John Elway = 41 (Broncos = 41)
John = 10+15+8+14 = 47
Elway = 5+12+23+1+25 = 66
John Elway = 113
How perfect is that name? The first Super Bowl was at the end of the 47th season of the league, at the end of the '66 season.
Further, notice that John Elway is 55-years old, just like the Denver Broncos, who are in their 55th season this year.
Let's decode his full name and date of birth. He is born on the same date as Elon Musk.
John = 1+6+8+5 = 20
Albert = 1+3+2+5+9+2 = 22
Elway = 5+3+5+1+7 = 21
Jr. = 1+9 = 10
John Albert Elway = 63
John Albert Elway Jr. = 73
John = 10+15+8+14 = 47
Albert = 1+12+2+5+18+20 = 58
Elway = 5+12+23+1+25 = 66
Jr. = 10+18 = 28
John Albert Elway = 171
John Albert Elway Jr. = 199
His birthday numerology matches his name Gematria, 'John Elway'.
6/28/1960 = 6+28+19+60 = 113 (John Elway)
6/28/1960 = 6+28+(1+9+6+0) = 50 (Super Bowl 50 this year...)
6/28/1960 = 6+2+8+1+9+6+0 = 32 (NFL)
6/28/60 = 6+28+60 = 94 (#94 made fumble recovery to defeat Bengals, most recent win)
His place of birth also yield some familiar gematria.
Port = 7+6+9+2 = 24
Angeles = 1+5+7+5+3+5+1 = 27/36
WA = 5+1 = 6
Port Angeles, WA = 57/66 (156 in Simple English Gematria)
Recall how '57' connects to Super Bowl. Osweiler, the Broncos QB, was the 57th pick of the 2012 NFL Draft. Peyton Manning was also receiving '57' tributes this season.
Fifty-Seven = 6+9+6+20+25+19+5+22+5+14 = 131
Super Bowl = 19+21+16+5+18+2+15+23+12 = 131
Championship = 3+8+1+13+16+9+15+14+19+8+9+16 = 131
John Elway was the QB of the last Broncos team to win the Super Bowl, Super Bowl 33, played January 31, 1991.
1/31/1999 = 1+31+19+99 = 150
1/31/1999 = 1+31+(1+9+9+9) = 60
1/31/1999 = 1+3+1+1+9+9+9 = 33 (Super Bowl 33)
1/31/99 = 1+31+99 = 131 (Super Bowl, Championship, Fifty-Seven)
https://en.wikipedia.org/wiki/John_Elway
Notice that Elway played sixteen seasons, like the number of games in a season, in the modern NFL.
Sixteen = 1+9+6+2+5+5+5 = 33/42
Sixteen = 19+9+24+20+5+5+14 = 96
The detail about his 156 attempts in the four Super Bowls connects to the number '33', and again, he was the winning QB of Super Bowl 33, at the end of his 16th season.
Thirty-Three = 2+8+9+9+2+7+2+8+9+5+5 = 66
Thirty-Three = 20+8+9+18+20+25+20+8+18+5+5 = 156
Notice that Elway was drafted in '83, a big year for the NFL in terms of QBs drafted. Marino and Kelly were also taken that year.
Football = 6+15+15+20+2+1+12+12 = 83
Keep in mind that Tom Brady was born August 3, or 8/3, and Peyton Manning was born on the 83rd day of the year. A number of other QBs in the NFL have '83' coding through their name gematria and birthdate numerology.
https://en.wikipedia.org/wiki/1983_NFL_draft
Notice that Elway was selected by the Baltimore Colts, who are now the Indianapolis Colts, and he was drafted out of Stanford. In 2012, Andrew Luck was drafted by the Indianapolis Colts, also out of Stanford. Both Elway and Luck were the #1 picks in their respective drafts. Andrew Luck replaced Peyton Manning in Indianapolis, who is now the quarterback of the Denver Broncos.
Let us look more carefully at each of the Super Bowls Elway participated in.
Super Bowl 21, January 25, 1987
1/25/1987 = 1+25+19+87 = 132
1/25/1987 = 1+25+(1+9+8+7) = 51
1/25/1987 = 1+2+5+1+9+8+7 = 33
1/25/87 = 1+25+87 = 113
Notice that Super Bowl 21 was played in California. Last year's Super Bowl, 49, closed out with a major '21 tribute', and a final score summing to 52, 28-24.
California = 3+1+3+9+6+6+9+5+9+1 = 52
Super Bowl 50 will be hosted in California, by the 49ers.
Further, notice that the New York Giants won with 39 points...
New York = 5+5+5+7+6+9+2 = 39
Super Bowl 22, January 31, 1988
1/31/1988 = 1+31+19+88 = 139 (Freemasonry)
1/31/1988 = 1+31+(1+9+8+8) = 58 (Freemasonry) (Washington = 58)
1/31/1988 = 1+3+1+1+9+8+8 = 31
1/31/88 = 1+31+88 = 120
Notice the date tribute to Freemasonry. Again, the sport of football was created by the Freemasons. The winning team also scored 42 points, probably because 96 would have been too many.
Freemason = 6+9+5+5+4+1+1+6+5 = 42
Freemason = 6+18+5+5+13+1+19+15+14 = 96
Also, George Washington, a 33rd Degree Freemason, was born on the 42nd day of the year in the calendar of his team, and Washington State, named in his honor, is the 42nd state. Again, the game was played on a date with '58' numerology, and Washington has gematria of ''58'.
Notice also that the game summed to 52, and was hosted by California.
Super Bowl 24
1/28/1990 = 1+28+19+90 = 138
1/28/1990 = 1+28+(1+9+9+0) = 48
1/28/1990 = 1+2+8+1+9+9+0 = 30
1/28/90 = 1+28+90 = 119 (All Seeing Eye, Star of David)
Notice the 49ers blew the Broncos out, 55-10. Again, 55 is a very significant number to the Broncos this year, who are in their 55th year of existence, and who have a strong shot at the Super Bowl, which is hosted by the 49ers.
Also notice that the sum of the game was 65 points, in favor of the San Francisco 49ers.
49ers can also sum to '55' using the following math.
4+9+5+18+19 = 55 (E = 5; R = 18; S = 19)
And as for Super Bowl 24, the number 24 has a connection to 'San Francisco'.
It should be noted that Peyton Manning's father played his career with the New Orleans Saints, who hosted Super Bowl 24.
Also, if the Broncos do appear in Super Bowl 50, and lose, they will be 2-4 all-time in Super Bowls.
Super Bowl 32
1/25/1998 = 1+25+19+98 = 143
1/25/1998 = 1+25+(1+9+9+8) = 53 (NFL has 53-man roster)
1/25/1998 = 1+2+5+1+9+9+8 = 35
1/25/88 = 1+25+88 = 114
The numbers do not stand out as clearly in this game, but the sum of 55-points is once again apparent, this time in favor of the Broncos, unlike Super Bowl 24.
As for 143, it has gematria matching the name John Elway, who was victorious for the first time in a Super Bowl, after starting off 0-3.
Super Bowl 33
1/31/1999 = 1+31+19+99 = 150
1/31/1999 = 1+31+(1+9+9+9) = 60
1/31/1999 = 1+3+1+1+9+9+9 = 33 (Super Bowl 33)
1/31/99 = 1+31+99 = 131 (Super Bowl, Championship, Fifty-Seven)
What is interesting about John Elway's two Super Bowls win coming over the Green Bay Packers and Atlanta Falcons, is that they represent the only two states with a gematria of '44'.
Wisconsin = 5+9+1+3+6+5+1+9+5 = 44
Georgia = 7+5+6+9+7+9+1 = 44
Recall that Brett Favre was the 33rd overall pick of the 1991 NFL Draft, selected by the Atlanta Falcons, on the 33rd Parallel, before being traded to the Packers after his rookie season.
Atlanta, on the 33rd Parallel
Atlantis = 1+2+3+1+5+2+9+1 = 24/33 (Atlanta, named after Atlantis)
ATL = 1+20+12 = 33
Atlanta, Losers of Super Bowl 33
Also, in this football season, 2015-16, playing for Super Bowl 50, the Falcons and Packers received '44' tributes in the lone week, Week 2, with '44 numerology'. Week 2 was played September 20, 2015.
9/20/15 = 9+20+15 = 44
No other games in Week 2 had scores summing to 44.
In total, no one should ever argue with me again about pro sports being rigged by the numbers. The only question left, is are these athletes grown in laboratories, or are they born and raised under masonic traditions? Obviously, one of the two things is happening.
1. Al Jeezera documentary is he say she say but Steroids and HGH run rampant in these leagues and that said a doctor in the documentary says those drugs took him from average to Pro Athlete.
Of course we will never know that question since numerology is insane within itself. Better to just get the truth out about this.
1. I know a few NCAA football players, they all tell me the same thing. Everyone juices.
2. Jesus Christ he looks like Gary fuckin Busey.
3. Wow. Awesome post Zach. I knew there would be Elway Manning connections; but I didn't realize it even connects to Favre and Luck. Great work!
4. Broncos vs Cardinals super bowl?
Note: Only a member of this blog may post a comment. | 2,846 | 7,979 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-26 | latest | en | 0.894968 |
http://valmaven.tk/posts/compound-interest-calculator-daily-weekly-monthly.html | 1,550,983,970,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249595829.93/warc/CC-MAIN-20190224044113-20190224070113-00635.warc.gz | 284,320,542 | 7,726 | # Compound Interest Calculator Daily Weekly Monthly
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https://fitnessfastrowing.com/sale-now-on-water-rowing-machine-how-to-use-how-to-clean-water-rower.html | 1,552,951,834,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912201812.2/warc/CC-MAIN-20190318232014-20190319014014-00346.warc.gz | 499,146,164 | 17,183 | 1.) I keep reading about cracks and leaks that develop in the water tank. WaterRower even offers a tank repair kit for this very issue. It seems to me that this is a design flaw that they refuse to address, instead offering a fix for when it eventually occurs. Have you seen these tanks to eventually leak? I’m surprised they haven’t tried a one-piece tank.
```In the starting position, the knees should be bent, with the shoulders and arms extended forward and the shins held vertically. Begin rowing by pushing both feet against the row machine’s platform and straighten the legs until they’re slightly bent. Pull the paddle towards you, using bent elbows as you slightly lean back while focusing on keeping your back straight. This type of motion should be both continuous and fluid.
```
Motion of the water is different than the motion of the wave. Water at each location moves in a circular path, but the motions at different locations are “out of phase”, which means that when water at the left of the diagram is moving to the right, water a quarter of a wavelength to the right is moving down, and water next to it is moving to the left, and next to it is moving up, etc. The overall effect is a an “apparent” wave moving to the right. Thus, the velocity (speed) of a wave is not at all the same as the velocity of the water.
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```When waves run into water moving in the opposite direction, they are slowed, just as if they were approaching a beach. Wave length becomes shorter, wave height higher, and they may break. A good (bad) example of this is an ebb current flowing out of Raccoon Strait into waves coming in from the Golden Gate. Good rough water training, if that’s what you want.
```
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No matter the resistance we set or the power we applied, the machine remains sturdy, centered, and even. We amped up the intensity for 2000m sprints and found that the C2 supports consistent effort in a way water resistance can’t. The machine assists in keeping your intensity uniform, because it stores up remaining energy in the spinning flywheel. Comments about the ride feel of both Concept 2s were noticeably brief. As one tester put it, “It’s unremarkable because it’s good,” and leaves you free to focus on the effort you’re putting into your workout.
The machine was easy to put together and worked well mechanically. The problem is the electronic monitor. The strokes/min registered as double the actual pace, and the distance registered as 1/10th the actual distance. I contacted customer support via email and their response was slow and inadequate. I am returning the machine and plan to buy a WaterRower.
We're Marta and Brock, happily married new parents, and bring you a fun and simple approach to fitness, weight-loss, and nutrition through our knowledge, experiences and continued journey. Tune in for creative home and outdoor workouts, health and exercise tips, interviews, and organic gluten-free plant-based recipes (raw and/or cooked) for health and fitness newbies and enthusiasts! Occasionally you can win some cool stuff too :)
I have also received negative feedback about the resistance. After 15 minutes at the maximum level, you will feel that you are hardly putting an effort to row. However, this won’t be an issue if your weight is less. This could indicate an issue with the oil or gasoline mixture in the piston. Also, issues were mentioned by other customers with the footrests. They’d problem strapping their feet. It is a bigger issue as it becomes difficult for people who have small feet as the feet keeps on slipping. The footrest grip is not tight. This is affecting particularly when you would like to experience a workout without any interruption; when you actually want to go with the flow of workout. Fast rowing also becomes difficult with this type of problem as you have to stop and strap the feet back. However, the problem could be solved by using a wide and thick Velcro straps. Use your creativity and be resourceful while coming up with the ideas to solve the problem of slipping feet.
This type of machine provides a smooth action with little wear and tear to the mechanism, and the flywheel itself could help to keep you cool as you work out. It's worth noting that although water rowers aim for a realistic rowing feel, competitive rowers often use air rowers for land training. They tend to be less expensive than water-powered rowing machines, as well.
If the machine is to be housed on a wooden floor you will need a mat beneath to reduce vibration and also noise. If space is tight, the hydraulic options are usually the smallest. It's also worth checking your floor can take the weight if you're going for one of the bigger, heavier models. For most, this shouldn't be a problem, if you have any doubts get a surveyor to check as it could be a very expensive mistake!
If you’re new to rowing, start off slow and begin with a five to ten-minute warm-up, followed by a twenty-minute workout and five to ten additional minutes for the cool down. During your workout, if you continue to consistently row at the same speed, over a period of a few sessions you should begin to notice an increase in your endurance. For a more intense workout, you can try interval training with your rower. Start off each workout by rowing for five to ten minutes at the highest resistance level, as fast as you can. Next, for two to three minutes row slowly, at the lowest resistance setting. Do this for thirty to sixty minutes. Interval training has been incorporated in many different types of exercise from cardio to strength training, and it’s designed to give your metabolism a jolt and make your body improve its fat burning abilities
The WaterRower Classic is handcrafted in solid American black walnut, which may vary in color from rich brown to purple/black. The black walnut was chosen due to its marvelous engineering properties, especially its ability to absorb sound and vibration, which enhances the WaterRower's quiet performance and smooth operation. Black walnut, like all woods used in the construction of the WaterRower, is a premium hardwood with incredible longevity and dimensional stability. In addition, the wood is harvested from replenishable forests and is hand finished with three coats of Danish oil to give it a deep luster and warmth.
We facebooked customer service(They advertise this as one way to get a hold of them). The next day, they responded that we need to e-mail or call. We e-mailed customer service the other day. They asked for this video, and they haven't responded again since. I don't think it's too much to ask that a \$500 piece of exercise equipment that's brand new could actually be usable.
If you are accustomed to the natural catch and feel of on-water rowing, your rowing technique will immediately appreciate the Apollo’s unrivalled emulation of a boat gliding through water. FDF’s patented twin tank design generates a smooth uniform stroke, including no lag of resistance at the catch and continued resistance all the way to the finish.
It is time to get into shape. After doing some research and talking to a few experts, at least more expert than me, I determined rowing and swimming would be the best for me to loose weight with less strain on my body. I researched and decided to rent (not buy) a Water Rower. The one page rental application is simple to complete and the staff telephone contacts were great, they answered questions, provided updates, made suggestions etc. The equipment showed up as promised. Assembly is very simple one person job and took just about a half hour, only one small allen wrench is required and it is provided. (save the box, its big but needs to be saved in the event you return the equipment) The machine is classic and elegant. It is not silent but the water swoosh sound is quite relaxing and soft. I rented a previously used matching which is slightly less expensive and shows a tiny bit of use but its exercise equipment so it is expected. Customer service is outstanding and friendly. The equipment seems really well designed and is simple to stand up and get out of the way. The only issue I had involved the the filling of the tank. Its cumbersome and could be made easier by a more convenient placement of the fill hole, it seems to be a slight design flaw. I used a new long funnel (used to add oil to a car). I checked out a few Youtube posts and videos and other users had the same minor complaint but overall everyone is really happy with this equipment. NOW I NEED to MAKE MYSELF USE IT.
Unlike traditional rowboats or kayaks, sliding seat rowing shells are quality fitness tools that allow you a low-impact, all-over body workout. Our versatile shells are simple to operate, stable in open water conditions, and effective in providing you with a fun, safe workout, whether you are a novice or an experienced rower. Our unique folding riggers and long Delta keel make our boats portable and easy to launch.
Rowing is a low impact sport with movement only in defined ranges, so twist and sprain injuries are rare. However, the repetitive rowing action can put strain on knee joints, the spine and the tendons of the forearm, and inflammation of these are the most common rowing injuries. [9]If one rows with poor technique, especially rowing with a curved rather than straight back, other injuries may surface, including back pains. Blisters occur for almost all rowers, especially in the beginning of one's rowing career, as every stroke puts pressure on the hands, though rowing frequently tends to harden hands and generate protective calluses. Holding the oars too tightly or making adjustments to technique may cause recurring or new blisters, as it is common to feather the blade (previously described). Another common injury is getting "track bites", thin cuts on the back of one's calf or thigh caused by contact with the seat tracks at either end of the stroke.
The WaterRower Oxbridge is hand crafted in solid Cherrywood. Like all woods, Cherrywood may vary in color from a red-brown to deep red. The wood will darken in color with exposure to light. For this reason all WaterRower Cherrywood components are kept in light free rooms to protect from shadowing. A new WaterRower Oxbridge will therefore appear quite light in color. The wood will however darken over time reaching a rich reddish hue. Each machine has been hand finished with three coats of Danish Oil giving a deep lustre an warmth to the wood. Wood has been chosen due to its marvellous engineering properties, primary amongst these is its ability to absorb sound and vibration enhancing the WaterRower's quietness and smoothness of use. Cherrywood, like all woods used in WaterRower construction, is a premium hardwood with incredible longevity and dimensional stability. For reasons of ecology, all our woods are harvested from replenishable forests.
The resistance of your rowing machine impacts how it moves. Select from hydraulic and magnetic rowing resistance. Hydraulic rowers are typically more compact in size, while magnetic machines make little noise as they operate. Details like easy-to-reach tension knobs or buttons allow you to increase or decrease the intensity of your resistance as you train.
Medical Professionals and Physical Therapists trust the benefits of the WaterRower and Rowing as an excellent cardio exercise. Burning up to 1000 calories per hour and working out up to 84% of your muscle mass during one exercise, rowing offers a tremendously beneficial exercise without the perceived fatigue and exhaustion often apparent when using other cardio training gym equipment.
# The Apollo Pro 2 water rower is a commercial grade, natural wood rowing machine, perfect for group training or high-end home gyms. This rower’s water resistant design, meditative motion and integrity of construction sets it apart from its competitors. Handcrafted from the finest quality American Ash wrapped around a stabilising steel frame, this steadfast indoor rower offers a standout performance even in the harshest of fitness applications.
#### According to Fritz Hagerman, Ph.D., a professor in the Biological Science Department at Ohio University, “Competitive rowers expended almost twice the number of calories on a 2,000-meter course as a runner in a 3,000-meter steeplechase.”1 However, since rowing is low impact, you will not experience the same wear and tear on your body that you would if you were a runner. Plus, you build strength in your upper body and core.
For a third option, you can look to hydraulic machines, which use pistons to generate resistance. Hydraulic rowers are quiet, and they also tend to be smaller and cheaper than other kinds of rowing machines, but you won’t get the same smooth rowing feel, or the consistency of resistance, that you would get with an air or water rower. What's more, reviewer after reviewer has found them unreliable and high-maintenance.
###### Sometimes, slides are placed underneath the erg to try to simulate the movement of being on the water. It allows the machine to move back and forth smoothly as if there is water beneath you. The slides can be connected in rows or columns so that rowers are forced to move together on the ergometer, similar to how they would match up their rhythm in a boat.
Founded in 1818, Leander Club is the world's oldest public rowing club.[17] The second oldest club which still exists is the Der Hamburger und Germania Ruder Club which was founded 1836 and marked the beginning of rowing as an organized sport in Germany.[18] During the 19th century, as in England, wager matches in North America between professionals became very popular attracting vast crowds. Narragansett Boat Club was founded in 1838 exclusively for rowing. During an 1837 parade in Providence, R.I, a group of boatmen were pulling a longboat on wheels, which carried the oldest living survivor of the 1772 Gaspee Raid. They boasted to the crowd that they were the fastest rowing crew on the Bay. A group of Providence locals took issue with this and challenged them to race, which the Providence group summarily won. The six-man core of that group went on the following year to found NBC in 1838.[19] Detroit Boat Club was founded in 1839 and is the second oldest continuously-operated rowing club in the U.S. In 1843, the first American college rowing club was formed at Yale University.[20] The Harvard–Yale Regatta is the oldest intercollegiate sporting event in the United States,http://rowinghistory.net/Time%20Line/TL%20-1849images.htm[21] having been contested every year since 1852 (excepting interruptions for wars).
Most races that are held in the spring and summer feature side by side racing, or sprint racing, sometimes called a regatta; all the boats start at the same time from a stationary position and the winner is the boat that crosses the finish line first. The number of boats in a race typically varies between two (which is sometimes referred to as a dual race) to eight, but any number of boats can start together if the course is wide enough.
Because of the vast array of products that might contain one of the chemicals or ingredients on the list, we must include Proposition 65 warnings in our communications with you. We are required to use exact wording as specified by the state of California. While the warning sounds alarming, the purpose is to notify you of the potential risk so that you can make an informed buying decision.
Water Rower Dimensions: 88" x 22" x 28.5" / Weight: 79 Lbs (Empty Tank), 116 Lbs (With Water); Maximum User Weight And Height: 700... Lbs & 38" InseamExercise Rower Features Aluminum Makeup With Powder-Coated Aluminum Finish For High DurabilityBoasts Patented Water Flywheel Responsible For Its Natural Rowing Dynamic-Like Actually Driving A Boat With Oars! Other Features: Self-Regulating Resistance Levels-Making It Suitable For Any User, Little Maintenance Required (Tablets Available From Manufacturer Free Of Charge) And Inclusive Of S4 Performance Monitor (Displays Intensity, Kcalories/Hour, Stroke Rate, Heart Rate, And More! Enjoy 1 Year Commercial Warranty (Upgradeable To 3-Year Parts While 5-Year Frames Are Free Of Charge With Registration) read more
# This popular rowing machine utilizes magnetic resistance for a quiet, and very smooth workout. The Velocity Exercise CHR-2001 also features a programmable computer with 12 programs that will keep your exercising varied and challenging for years. The computer also provides data on distance, time, calories burned, pulse rate, strokes per minute, and stroke count. Additionally, the tension resistance is electronically controlled via the computer. This electronic resistance control feature puts the CHR-2001 ahead of other magnetic rowers that tend to use knobs.
Rowers may take part in the sport for their leisure or they may row competitively. There are different types of competition in the sport of rowing. In the U.S. all types of races are referred to as regattas whereas this term is only used in the UK for head-to-head or multi-lane races (such as those that take place at Dorney Lake), which generally take place in the summer season. Time trials occur in the UK during the winter, and are referred to as Head races. In the US, head races (usually about 5k, depending on the body of water) are rowed in the fall, while 2k sprint races are rowed in the spring and summer.
Unlike most other non-combat sports, rowing has a special weight category called lightweight (Lwt for short). According to FISA, this weight category was introduced "to encourage more universality in the sport especially among nations with less statuesque people". The first lightweight events were held at the World Championships in 1974 for men and 1985 for women. Lightweight rowing was added to the Olympics in 1996.
Wood has been chosen due to its marvellous engineering properties, primary amongst these is its ability to absorb sound and vibration enhancing the WaterRower's quietness and smoothness of use. The WaterRower's patented WaterFlywheel has been designed to emulate the exact dynamics as a boat moving through water; the objective of the exercise of rowing.
Function plays a large role in defining good design. When designers look at an object, they don't just consider its aesthetic appearance; they should also challenge it to be more versatile, to respond to the user's need, or to achieve its purpose more elegantly. Good design has the capacity to solve problems that sometimes we didn't even know we had. This is one of the ways design touches and enriches our everyday life.
It is time to get into shape. After doing some research and talking to a few experts, at least more expert than me, I determined rowing and swimming would be the best for me to loose weight with less strain on my body. I researched and decided to rent (not buy) a Water Rower. The one page rental application is simple to complete and the staff telephone contacts were great, they answered questions, provided updates, made suggestions etc. The equipment showed up as promised. Assembly is very simple one person job and took just about a half hour, only one small allen wrench is required and it is provided. (save the box, its big but needs to be saved in the event you return the equipment) The machine is classic and elegant. It is not silent but the water swoosh sound is quite relaxing and soft. I rented a previously used matching which is slightly less expensive and shows a tiny bit of use but its exercise equipment so it is expected. Customer service is outstanding and friendly. The equipment seems really well designed and is simple to stand up and get out of the way. The only issue I had involved the the filling of the tank. Its cumbersome and could be made easier by a more convenient placement of the fill hole, it seems to be a slight design flaw. I used a new long funnel (used to add oil to a car). I checked out a few Youtube posts and videos and other users had the same minor complaint but overall everyone is really happy with this equipment. NOW I NEED to MAKE MYSELF USE IT.
The meter didn't work properly. It wasn't keeping the distance - we would row for 30 mins and wouldn't even be at 1km!!! And the strokes per minute was reading double strokes (which the company said was normal - huh?). Their customer service team was good and tried to make it work but we ended up returning it and spent the extra money to get a WaterRower and it is amazing!!!
The WaterRower Natural Rowing Machine with S4 Monitor is hand crafted in solid Ash and stained Honey Oak for consistency of color.... Each rowing machine has been hand finished with Danish Oil giving a deep luster and warmth to the wood. Wood has been chosen due to its marvelous engineering properties, primary amongst these is its ability to absorb sound and vibration enhancing the WaterRower's smooth and quiet performance.;Flywheel sits in enclosed water tank to provide smooth, quiet, self-regulated resistanceSeries 4 performance monitor tracks workout intensity, stroke rate, heart rate, and moreHandcrafted rowing machine with inchwater flywheel inch that replicates actual rowing feelStores upright; measures 84 x 21 x 22 inches (W x H x D) and weighs 117 poundsStores upright; measures 84 x 21 x 22 inches (W x H x D) and weighs 117 pounds read more
Women row in all boat classes, from single scull to coxed eights, across the same age ranges and standards as men, from junior amateur through university-level to elite athlete.[49][50] Typically men and women compete in separate crews although mixed crews and mixed team events also take place.[51] Coaching for women is similar to that for men.[52] The world's first women's rowing team was formed in 1896 at the Furnivall Sculling Club in London.[53]
Rowing machines were first used in Archaic Greece. Chabrias, an Athenian military general in 4th Century B.C., invented wooden rowing simulators for his inexperienced oarsmen. This enabled them to learn technique and timing before stepping foot on actual water crafts. And it must have worked — Chabrias successfully led numerous naval attacks against the Spartans.
### Essentially, the water rower uses water as it's resistance and the air rower (click here to read our air rowing machine reviews) uses air resistance. Additionally, the resistance of the water gives a very similar feeling to actually rowing in the water. The rowing machines listed above all have fantastic rowing mechanisms that mimic the exact feeling of a rowing a boat.
The First Degree Newport rower utilises water resistance for a more effective workout that’s low impact, so it’s gentle on the body. You’ll focus on building up all the major muscle groups as you use a machine that offers a very realistic rowing experience.To increase resistance flip the valve control to add more water to the paddle water tank. To decrease resistance, remove water from the tank and enjoy a less challenging stroke for the cool down or warm up a portion of your workout. The tank on the Newport can hold a minimum of nine litres of water and a maximum of seventeen litres.
Ever since the earliest recorded references to rowing, the sporting element has been present. An Egyptian funerary inscription of 1430 BC records that the warrior Amenhotep (Amenophis) II was also renowned for his feats of oarsmanship. In the Aeneid, Virgil mentions rowing forming part of the funeral games arranged by Aeneas in honour of his father.[10] In the 13th century, Venetian festivals called regata included boat races among others.[11]
Floating Heel-Rest Design: The Rowing Machine's footboard incorporates the all-new Floating Heel-Rest design. The cup of the heel-rest follows the heel as it rises from the footboard providing stability and comfort from the beginning to end of each stroke. The heel-rest position is adjusted simply by pressing the central button to disengage the lock pins securing the footpads.
When waves run into water moving in the opposite direction, they are slowed, just as if they were approaching a beach. Wave length becomes shorter, wave height higher, and they may break. A good (bad) example of this is an ebb current flowing out of Raccoon Strait into waves coming in from the Golden Gate. Good rough water training, if that’s what you want.
Each machine has been hand finished with Danish Oil and Urethane for protection. Wood has been chosen due to its marvelous engineering properties, primary amongst these is its ability to absorb sound and vibration enhancing the WaterRower's quietness and smoothness of use. Ash, like all woods used in WaterRower construction, is a premium hardwood with incredible longevity and dimensional stability. For reasons of ecology, all our woods are harvested from replenishable forests.
``` Row machines aren’t as common in gyms as a treadmill or weight bench, but they should be. You may find one or two models, but they’re usually air resistant or piston operated models, which don’t really offer the best workout. But thanks to many celebrities sharing their love for rowing machines on social media – rowing machines are getting very popular. Realistically, rowing can actually increase endurance, build muscle, strengthen the core and burn fat, making it one of the best total body workouts around. So we decided to present you with our Best Water Rowing Machine Reviews. This type of exercise can put even cycling and running to shame because it burns about fifteen to twenty percent more calories than both workouts at the same level of exertion. Rowing is also considered an excellent core workout because the abs are engaged during each stroke. While it can definitely offer an effective fat burning workout, it’s also an ideal machine to use for people who are rehabbing muscles, the elderly or people with disabilities. With all of the benefits, you may be wondering why this type of workout is so unpopular. The main reason is that most people don’t know how to properly use a rowing machine and improper use can lead to poor results.
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Rowing is a highly aerobic activity which burns calories quickly and gives you a complete cardiocascular workout which is a great way to help your heart as well as tone your arms, legs, chest, back, and abs. What most people don't realize is that Rowing improves joint health through a wide range of motion and relieves stress because it allows you to remove yourself from the world around you, to connect to the calm and quiet of the water, and to meditate through the rhythmic, repetitive motion of your stroke. Rowing is a great way to get Aerobic and Anaerobic benefits without even thinking about it. Rowing also provides a workout for a variety of different muscles and joints in your body. If you want to see improvements in both your cardio fitness levels and your overall strength, it can give you the desired results you're looking for.
The stake format was often used in early American races. Competitors line up at the start, race to a stake, moored boat, or buoy some distance away, and return. The 180° turn requires mastery of steering. These races are popular with spectators because one may watch both the start and finish. Usually only two boats would race at once to avoid collision. The Green Mountain Head Regatta continues to use the stake format but it is run as a head race with an interval start.[42] A similar type of racing is found in UK and Irish coastal rowing, where a number of boats race out to a given point from the coast and then return fighting rough water all the way. In Irish coastal rowing the boats are in individual lanes with the races consisting of up to 3 turns to make the race distance 2.3 km.
Our Wooden Rowing Machines are Made in the USA using hardwood from only replenishable sources. We strive to manufacture rowing machines that are of a stylish design and built to last, while maintaining our commitment to eco-friendliness. Our Rowing Machines are functional fitness furniture, designed not to be hidden away like other unsightly gym equipment, but to look at home in your living room. Storing in an upright position no wider than a dining room chair, our rowing machines can be easily laid out for a quick indoor rowing exercise session.
Proposition 65 requires businesses to provide Californians with a clear and reasonable warning about chemicals in the products they purchase, in their home or workplace, or that are released into the environment. By providing this information, Proposition 65 enables Californians to make informed decisions about protecting themselves from exposure to these chemicals. Proposition 65 also prohibits California businesses from knowingly discharging significant amounts of listed chemicals into sources of drinking water.
The WaterRower Natural rowing machine is handcrafted in solid ash wood, stained Honey Oak, and finished with Danish oil. Wood is an excellent material for this application due to its ability to absorb sound and vibration enhancing the WaterRower's smooth, quiet operation. The Natural, as with all WaterRowers, features patented WaterFlywheel technology, unrivaled in its replication of the resistance felt in on the water rowing.
The International Rowing Federation (French: Fédération Internationale des Sociétés d'Aviron, abbreviated FISA), responsible for international governance of rowing, was founded in 1892 to provide regulation at a time when the sport was gaining popularity. Across six continents, 150 countries now have rowing federations that participate in the sport.[3]
The WaterRower Oxbridge Rower in Cherrywood is designed for equally gorgeous look and performance. Every detail complements the overall aesthetic and performance of this unparalleled machine. Built in America, the WaterRower Oxbridge Rower in Cherrywood is a product of modern European design. It's hand-crafted in solid ash and stained honey oak for consistency of color. Each machine has been hand-finished with Danish Oil, which lends a deep warmth and luster. Wood has been chosen for its marvelous engineering properties; primary amongst these is its ability to absorb sound and vibration, which enhances the WaterRower Oxbridge Rower quietness and smoothness of use. Ash, like all woods used in WaterRower construction, is a premium hardwood with incredible longevity and dimensional stability. For reasons of ecology, all woods are harvested from renewable forests. The frame is more than stylish; it holds up to 2,200 lbs!
If the machine is to be housed on a wooden floor you will need a mat beneath to reduce vibration and also noise. If space is tight, the hydraulic options are usually the smallest. It's also worth checking your floor can take the weight if you're going for one of the bigger, heavier models. For most, this shouldn't be a problem, if you have any doubts get a surveyor to check as it could be a very expensive mistake!
Row your way to fitness with this water rowing machine from WaterRower. The A1 S4 Natural is a low-maintenance exercise machine th...at's constructed from high-quality ash wood with aluminum mono rail. Engineered with the patented WaterFlywheel resistance method, it provides life-like water rowing experience while the low noise level and minimal intrusion along with the unique self-regulating resistance with infinite variable allows suits any user without the need for adjustments. With its no impact and non-load bearing, this gym equipment is also perfect for users with joint problems. Exercise all major muscle groups like the arms, legs, back, shoulders, core, etc. and see how great you're doing with the help of the S4 monitor that comes with the package. It monitors time per 2 or 500 kilometers, watts, calories burned per hour, distance, and the total time of workouts. This exercise machine also boasts minimal maintenance with no lubrication needed. You simply need to add purification tablet every 3-6 months (tablets come with the package). Storage is easy and hassle-free as well. Simply tip the machine vertically and you can easily lean it against a wall or hide it inside a closet. Moreover, enjoy 1 year manufacturer warranty (can be upgraded to 3 years parts and 5 years frame warranty for free with registration) read more
```Row machine looks like it's a \$600 unit. If you're in the market for a water rower I'd spend the money on the real Water Rower. I was wanting the Water Rower but after this poor experience I just want something that lasts and is warranted the longest. I purchased a concept2 that had arrived and is a much better quality unit. The computer is much better.
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A 125-pound person moving at an average pace for 30 minutes on a treadmill will burn roughly 250 calories. However, they aren’t able to take advantage of this after-burn effect. The same person rowing vigorously will burn around to 250 calories but will have less stress placed upon their joints and continue to burn calories after the workout has ended. Rowing machines have been shown to burn on average 800 calories an hour if you work harder and are slightly heavier.
Do you need a machine that folds up? If you've decided to avoid a water rower for space reasons, you may prefer a model that folds up for storage. Even better, some rowing machines have wheels fitted to them, so that once it's folded up, you can steer it to a storage space out of the way. You can find a foldable design in many magnetic and some air rowers. | 8,215 | 38,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-13 | latest | en | 0.956997 |
https://codeforces.com/blog/entry/90609 | 1,624,605,297,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487622113.11/warc/CC-MAIN-20210625054501-20210625084501-00181.warc.gz | 175,596,780 | 19,665 | ### prafull_vrsh's blog
By prafull_vrsh, 6 weeks ago,
Hello everyone, This question was asked in Hackwithinfy (9 May 3 pm slot). Please share a possible solution.
You are the manager of the hotel and you have N customers to serve. Each customer has a happiness quotient (Ci) if the food is served to him at time x. The unhappiness of a customer is defined by |Ci — x|. You must serve all the customers and you can serve them in any order. You have to find the minimum sum of unhappiness.
Note: At a particular time only one customer is served and Each customer takes one unit of time.
Contraint:
1 <= N <=10^3 1<= Ci <=N
Sample input:
4 2 2 3 3
Sample output: 2
Sample input:
4 1 1 1 1
Sample output: 6
Please give suggestions on how to solve this problem efficiently.
• +2
» 6 weeks ago, # | ← Rev. 2 → +12 Problem can be solved using DP. In the case with minimum unhappiness, customers will be served in sorted order according to their happiness quotient(C). dp[i][j] is the minimum unhappiness when we are time i and have served j people. DP has n^2 states Transitiondp[i][j] = min(dp[i-1][j],dp[i-1][j-1]+abs(Cj-i))Do comment if you feel something is wrong
• » » 6 weeks ago, # ^ | 0 Thank you, I understand the approach.
• » » 6 weeks ago, # ^ | +1 I don't understand the need of DP after sorting the array. Can't we simply add all the absolute difference with the index to get the answer? sort(C,C+n) for(i=0;i
• » » » 6 weeks ago, # ^ | +3 3 3 3 Time served is not bounded by N, actual answer is (2 3 4).
• » » 6 weeks ago, # ^ | ← Rev. 2 → 0 Why is there dp[i-1][j] in the transition?Edit: Sorry, I get it. You had i as time, I was thinking the opposite way.
» 6 weeks ago, # | 0 What is the order of serving in sample input #2 to get the minimum unhappiness as 6?
» 6 weeks ago, # | ← Rev. 3 → 0 How is the output of 2nd tc 6? Any arragemenet you do, the minimum will be 7 I guess. $4,1,1,1,1 -13$ $1,4,1,1,1 -11$ $1,1,4,1,1 -9$ $1,1,1,4,1 - 7$ $1,1,1,1,4 - 7$ How is it 6?
• » » 6 weeks ago, # ^ | ← Rev. 2 → +3 Hi dca, First number is size of array. Explanation of test case 2: |1-1| + |1-2| + |1-3| + |1-4| = 6
• » » » 6 weeks ago, # ^ | 0 Oh sorry my bad. I didn't look at it clearly. Thanks for the explanation.
» 6 weeks ago, # | +10 1437C - Chef Monocarp Exactly same Problem. | 776 | 2,337 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-25 | latest | en | 0.913467 |
https://footwww.academickids.com/encyclopedia/index.php/Automorphism_group | 1,643,358,634,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305423.58/warc/CC-MAIN-20220128074016-20220128104016-00153.warc.gz | 307,043,892 | 7,592 | # Automorphism
(Redirected from Automorphism group)
In mathematics, an automorphism is an isomorphism from a mathematical object to itself. It is, in some sense, a symmetry of the object, and a way of mapping the object to itself while preserving all of its structure. The set of all automorphisms of an object forms a group, called the automorphism group. It is, loosely speaking, the symmetry group of the object.
Contents
## Definition
The exact definition of an automorphism depends on the type of "mathematical object" in question and what, precisely, constitutes an "isomorphism" of that object. The most general setting in which these words have meaning is an abstract branch of mathematics called category theory. Category theory deals with abstract objects and morphisms between those objects.
In category theory, an automorphism is an endomorphism (i.e. a morphism from an object to itself) which is also an isomorphism (in the categorical sense of the word).
This is a very abstract definition since, in category theory, morphisms aren't necessarily functions and objects aren't necessarily sets. In most concrete settings, however, the objects will be sets with some additional structure and the morphisms will be functions preserving that structure.
In the context of abstract algebra, for example, a mathematical object is an algebraic structure such as a group, ring, or vector space. An isomorphism is simply a bijective homomorphism. (Of course, the definition of a homomorphism depends on the type of algebraic structure; see, for example: group homomorphism, ring homomorphism, and linear operator).
## Automorphism group
The set of automorphisms of an object X form a group under composition of morphisms. This group is called the automorphism group of X. That this is indeed a group is simple to see:
• Closure: composition of two endomorphisms is another endomorphism.
• Associativity: morphism composition is associative by definition.
• Identity: the identity is the identity morphism from an object to itself which exists by definition.
• Inverses: by definition every isomorphism has an inverse which is also an isomorphism, and since the inverse is also an endomorphism of the same object it is an automorphism.
The automorphism group of an object X in a category C is denoted AutC(X), or simply Aut(X) if the category is clear from context.
## Examples
• A group automorphism is a group isomorphism from a group to itself. Informally, it is a permutation of the group elements such that the structure remains unchanged. For every group G there is a natural group homomorphism G → Aut(G) whose kernel is the center of G. Thus, if G is centerless it can be embedded into its own automorphism group. (See the discussion on inner automorphisms below).
• The set of integers, Z, considered as a group has a unique nontrivial automorphism : negation. Considered as a ring, however, it has only the trivial automorphism. Generally speaking, negation is an automorphism of any abelian group, but not of a ring or field.
• In graph theory an automorphism of a graph is a permutation of the nodes that maps the graph to itself.
• An automorphism of a differentiable manifold M is a diffeomorphism from M to itself. The automorphism group is sometimes denoted Diff(M).
## Inner and outer automorphisms
In some categories—notably groups, rings, and Lie algebras—it is possible to separate automorphisms into two classes:
The former corresponding to automorphisms coming from "conjugation" by elements of the object itself, and the latter being everything else.
In group theory, for example, let a be an element of a group G. Conjugation by a is the group homomorphism φa : G → G given by φa(g) = aga−1. One can easily check that conjugation by a is actually a group automorphism. An inner automorphism is then an automorphism corresponding to conjugation by some element a. The set of all inner automorphisms form a normal subgroup of Aut(G), denoted by Inn(G). The quotient group Aut(G) / Inn(G) is usually denoted by Out(G).
The same definition holds in any unital ring or algebra where a is any invertible element. For Lie algebras the definition is slightly different.
## Reference
Yale, Paul B. Mathematics Magazine. "Automorphisms of the Complex Numbers". Vol 39. Num 3. May, 1966. pp. 135-141. Available via http://www.jstor.orgde:Automorphismus es:Automorfismo fr:Automorphisme pl:Automorfizm
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# Math test
Discussion in 'The Lighter Side' started by okie, Dec 19, 2003.
1. ### okieGT Mayor
Joined:
Oct 28, 2001
Messages:
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Location:
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NAME____________________
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1. Ramone has an AK-47 with a 30 round clip. He usually misses 6 out of
every 10 shots and he uses 13 rounds per drive-by shooting. How many
drive-by shootings can Ramone attempt before he has to reload?
2. Jose has 2 ounces of cocaine. If he sells an 8 ball to Antonio for
\$320 and 2 grams to Juan for \$85 per gram, what is the street value of
the rest of his hold?
3. Rufus pimps 3 ho's. If the price is \$85 per trick, how many tricks
per day must each ho turn to support Rufus's \$800 per day crack habit?
4. Jerome wants to cut the pound of cocaine he bought for \$40,000 to
make 20% profit. How many ounces will he need?
5. Willie gets \$200 for a stolen BMW, \$150 for stealing a Corvette, and
\$100 for a 4x4. If he steals 1 BMW, 2 Corvettes and 3 4x4's, how many
more Corvettes must he have to steal to have \$900?
6. Raul got 6 years for murder. He also got \$10,000 for the hit. If his
common-law wife spends \$100 per month, how much money will be left when he gets out? EXTRA CREDIT BONUS: How much more time will he get for killing the ho that spent his money?
7. If an average can of spray paint covers 22 square feet and the
average letter is 3 square feet, how many letters can be sprayed with 3
eight ounce cans of spray paint with 20% paint left over?
8. Tyrone knocked up 3 girls in the gang. There are 27 girls in his
gang. What is the exact percentage of girls Tyrone knocked up?
9. Bernie is a lookout for the gang. Bernie has a Boa Constrictor that
eats 3 small rats per week at a cost of \$5 per rat. If Bernie makes \$700 a week as a lookout, how many weeks can he feed the Boa on one week's income?
10. Billy steals Joe's skateboard. As Billy skates away at 15 mph, Joe
loads his 357 Magnum. If it takes Joe 20 seconds to load his piece, how
far away will Billy be when he gets whacked?
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4,771 | 608 | 2,144 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-09 | latest | en | 0.938281 |
https://robot.sia.cn/cn/article/id/3723 | 1,718,238,208,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861319.37/warc/CC-MAIN-20240612234213-20240613024213-00009.warc.gz | 458,127,184 | 42,246 | 引用本文: 彭梦, 蔡自兴. 基于多约束误差函数的2维激光雷达和摄像机标定方法[J]. 机器人, 2014, 36(6): 662-667,675.
PENG Meng, CAI Zixing. A Calibration Method of a Camera and 2D Laser Radar Based on Multi-constraint Error Function[J]. ROBOT, 2014, 36(6): 662-667,675.
Citation: PENG Meng, CAI Zixing. A Calibration Method of a Camera and 2D Laser Radar Based on Multi-constraint Error Function[J]. ROBOT, 2014, 36(6): 662-667,675.
## A Calibration Method of a Camera and 2D Laser Radar Based on Multi-constraint Error Function
• 摘要: 基于棋盘格标定板的标定方法是当今主流的2维激光雷达和摄像机标定方法（以2004年Zhang提出的方法最为经典），但是该方法在优化标定参数时使用的误差函数是弱约束关系，影响了标定结果的精确度和稳定性．由于激光点不但应该位于标定板平面上而且应该位于标定板的矩形区域内部，因此基于以上两个空间约束关系本文提出一种多约束误差函数模型．与原有的Zhang方法比较，所提方法可在更严格的可行域空间中搜索最优值，实验结果定性和定量地显示本文方法提高了标定结果的精确度和鲁棒性．
Abstract: In recent years, the calibration method based on the checkerboard is the most popular and the most practical calibration method for camera and 2-dimensional laser radar, and the most classic one is Zhang's method proposed in 2004. However, the optimization error function used by Zhang is a weak constraint relation, which affects the accuracy and stability of calibration results. Therefore, a multi-constraint optimization error function model is proposed based on the spatial relationship that the laser points should lie not only on the calibration board plane, but also in the calibration board area. Compared with the Zhang's method, the proposed method searches the optimal calibration results in a stricter feasible area. Qualitative and quantitative analyses of simulation results show that the proposed method can improve the accuracy and robustness.
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分享至好友和朋友圈 | 546 | 1,630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-26 | latest | en | 0.650809 |
http://oeis.org/A242530 | 1,545,066,142,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376828697.80/warc/CC-MAIN-20181217161704-20181217183704-00251.warc.gz | 212,308,387 | 5,002 | This site is supported by donations to The OEIS Foundation.
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A242530 Number of cyclic arrangements of S={1,2,...,2n} such that the binary expansions of any two neighbors differ by one bit. 16
0, 0, 1, 0, 2, 8, 0, 0, 224, 754, 0, 26256, 0, 0, 22472304, 0, 90654576, 277251016, 0, 7852128780 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,5 COMMENTS Here, a(n)=NPC(2n;S;P) is the count of all neighbor-property cycles for a specific set S of 2n elements and a pair-property P. For more details, see the link and A242519. In this case the property P is the Gray condition. The choice of the set S is important; when it is replaced by {0,1,2,...,2n-1}, the sequence changes completely and becomes A236602. LINKS S. Sykora, On Neighbor-Property Cycles, Stan's Library, Volume V, 2014. EXAMPLE The two cycles for n=5 (cycle length 10) are: C_1={1,3,7,5,4,6,2,10,8,9}, C_2={1,5,4,6,7,3,2,10,8,9}. MATHEMATICA A242530[n_] := Count[Map[lpf, Map[j1f, Permutations[Range[2, 2 n]]]], 0]/2; j1f[x_] := Join[{1}, x, {1}]; btf[x_] := Module[{i}, Table[DigitCount[BitXor[x[[i]], x[[i + 1]]], 2, 1], {i, Length[x] - 1}]]; lpf[x_] := Length[Select[btf[x], # != 1 &]]; Table[A242530[n], {n, 1, 5}] (* OR, a less simple, but more efficient implementation. *) A242530[n_, perm_, remain_] := Module[{opt, lr, i, new}, If[remain == {}, If[DigitCount[BitXor[First[perm], Last[perm]], 2, 1] == 1, ct++]; Return[ct], opt = remain; lr = Length[remain]; For[i = 1, i <= lr, i++, new = First[opt]; opt = Rest[opt]; If[DigitCount[BitXor[Last[perm], new], 2, 1] != 1, Continue[]]; A242530[n, Join[perm, {new}], Complement[Range[2, 2 n], perm, {new}]]; ]; Return[ct]; ]; ]; Table[ct = 0; A242530[n, {1}, Range[2, 2 n]]/2, {n, 1, 10}] (* Robert Price, Oct 25 2018 *) PROG (C++) See the link. CROSSREFS Cf. A236602, A242519, A242520, A242521, A242522, A242523, A242524, A242525, A242526, A242527, A242528, A242529, A242531, A242532, A242533, A242534. Sequence in context: A095217 A230915 A242922 * A073410 A021361 A199156 Adjacent sequences: A242527 A242528 A242529 * A242531 A242532 A242533 KEYWORD nonn,hard,more AUTHOR Stanislav Sykora, May 30 2014 EXTENSIONS a(16)-a(20) from Fausto A. C. Cariboni, May 10 2017, May 15 2017 STATUS approved
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Last modified December 17 11:56 EST 2018. Contains 318200 sequences. (Running on oeis4.) | 1,052 | 2,954 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-51 | longest | en | 0.686085 |
http://mathoverflow.net/questions/139568/composite-euclid-numbers | 1,469,647,431,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827077.13/warc/CC-MAIN-20160723071027-00310-ip-10-185-27-174.ec2.internal.warc.gz | 165,827,547 | 14,667 | # Composite Euclid Numbers
Is there any way (general procedure, i mean) to determine if a Euclid Number (En = pn# + 1) is prime or composite? Any research papers exploring this theme are also welcome. Thanks!
-
On the one hand, there are many general procedures to determine for any number if they are prime or not, in this sense, yes. OTOH, it is an open problem if there are infinitely many primes of this form, so there is no characetrization or anything even close to a characterization for which $n$ the number is prime. When testing such a number for prime one knows it won't have small primefactors; one thus can save a step in usual prime testing algorthms, but it also shows that primeness of these numbers is never very easy to establish. For references around this see oeis.org/A006862 – user9072 Aug 16 '13 at 9:31
It should say above that compositenes is never very easy to establish (as one will never find a very small factor). – user9072 Aug 16 '13 at 11:28
Actually compositeness of $N$ can be much easier to establish than primality: just find a case where the Miller-Rabin test fails (usually the Fermat test is all that's needed). Of course that's assuming that $N$ isn't so large that you can't actually compute with it; if it's the $n$-th "Euclid number" for $n$ on the order of $10^6$ then the only hope is to get lucky and find a prime factor $l$ small enough that $\prod_{m\leq n} p_n \equiv -1 \bmod l$. (If you want to find such an example, start with $l=p_r$ and try all $n < r$, which should succeed about $1-\exp(-1)$ of the time.) – Noam D. Elkies Aug 16 '13 at 13:09
@NoamD.Elkies I am not sure if you refer to my comment, but in case: I meant to say that trial-divsion by (very) small primes will never give anything useful for these numbers. Success of this trial division is what I meant by 'very easy'. (Not some relative statement whether compositenes or primality is relatively easier to establish.) – user9072 Aug 16 '13 at 19:09 | 522 | 1,970 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2016-30 | latest | en | 0.934265 |
https://www.saranextgen.com/homeworkhelp/index.php?id=883 | 1,708,825,796,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474573.20/warc/CC-MAIN-20240225003942-20240225033942-00763.warc.gz | 996,921,573 | 6,898 | # Exercise 10.5 - Chapter 10 Practical Geometry class 7 ncert solutions Maths - SaraNextGen [2024]
Question 1:
Construct the right angled ΔPQR, where m∠Q = 90°, QR = 8 cm and PR = 10 cm.
A rough sketch of ΔPQR is as follows.
The steps of construction are as follows.
(i) Draw a line segment QR of length 8 cm.
(ii) At point Q, draw a ray QX making 90º with QR.
(iii) Taking R as centre, draw an arc of 10 cm radius to intersect ray QX at point P.
(iv) Join P to R. ΔPQR is the required right-angled triangle.
Question 2:
Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.
A right-angled triangle ABC with hypotenuse 6 cm and one of the legs as 4 cm has to be constructed. A rough sketch of ΔABC is as follows.
The steps of construction are as follows.
(i) Draw a line segment BC of length 4 cm.
(ii) At point B, draw a ray BX making an angle of 90º with BC.
(iii) Taking C as centre, draw an arc of 6 cm radius to intersect ray BX at point A.
(iv) Join A to C to obtain the required ΔABC.
Question 3:
Construct an isosceles right-angled triangle ABC, where, m∠ACB = 90° and AC = 6 cm.
In an isosceles triangle, the lengths of any two sides are equal.
Let in ΔABC, AC = BC = 6 cm. A rough sketch of this ΔABC is as follows.
The steps of construction are as follows.
(i) Draw a line segment AC of length 6 cm.
(ii) At point C, draw a ray CX making an angle of 90º with AC.
(iii) Taking point C as centre, draw an arc of 6 cm radius to intersect CX at point B.
(iv) Join A to B to obtain the required ΔABC. | 451 | 1,575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-10 | latest | en | 0.924444 |
https://www.proofwiki.org/wiki/Relation_is_Antireflexive_iff_Disjoint_from_Diagonal_Relation | 1,721,935,391,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763861452.88/warc/CC-MAIN-20240725175545-20240725205545-00740.warc.gz | 825,260,672 | 10,902 | # Relation is Antireflexive iff Disjoint from Diagonal Relation
Jump to navigation Jump to search
## Theorem
Let $\RR \subseteq S \times S$ be a relation on a set $S$.
Then:
$\RR$ is antireflexive
$\Delta_S \cap \RR = \O$
where $\Delta_S$ is the diagonal relation.
## Proof
### Necessary Condition
Let $\RR$ be an antireflexive relation.
Let $\tuple {x, y} \in \Delta_S \cap \RR$.
By definition:
$\tuple {x, y} \in \Delta_S \implies x = y$
Likewise, by definition:
$\tuple {x, y} \in \RR \implies x \ne y$.
Thus:
$\Delta_S \cap \RR = \set {\tuple {x, y}: x = y \land x \ne y}$
and so:
$\Delta_S \cap \RR = \O$
$\Box$
### Sufficient Condition
Let $\Delta_S \cap \RR = \O$.
Then by definition:
$\forall \tuple {x, y} \in \RR: \tuple {x, y} \notin \Delta_S$
Thus:
$\not \exists \tuple {x, y} \in \RR: x = y$
Thus by definition, $\RR$ is antireflexive.
$\blacksquare$ | 301 | 890 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-30 | latest | en | 0.636691 |
http://www.jiskha.com/display.cgi?id=1337612622 | 1,461,882,049,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860109830.69/warc/CC-MAIN-20160428161509-00167-ip-10-239-7-51.ec2.internal.warc.gz | 389,116,515 | 3,739 | Thursday
April 28, 2016
# Homework Help: math
Posted by Garen on Monday, May 21, 2012 at 11:03am.
For the sequence 3,8,15,24,35,48,63
write an equation for t(subn) in terms of n.
• math - Steve, Monday, May 21, 2012 at 12:03pm
the differences are 5,7,9,11,...
since n^2 is the sum of the first n odd numbers, that suggests we look for a square in here.
Tn = (n+1)^2 - 1 | 138 | 375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2016-18 | longest | en | 0.944231 |
http://fr.mathworks.com/help/stats/nbinstat.html?requestedDomain=fr.mathworks.com&nocookie=true | 1,475,137,560,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738661780.47/warc/CC-MAIN-20160924173741-00072-ip-10-143-35-109.ec2.internal.warc.gz | 106,936,471 | 11,926 | # Documentation
### This is machine translation
Translated by
Mouse over text to see original. Click the button below to return to the English verison of the page.
# nbinstat
Negative binomial mean and variance
## Syntax
`[M,V] = nbinstat(R,P)`
## Description
`[M,V] = nbinstat(R,P)` returns the mean of and variance for the negative binomial distribution with corresponding number of successes, `R` and probability of success in a single trial, `P`. `R` and `P` can be vectors, matrices, or multidimensional arrays that all have the same size, which is also the size of `M` and `V`. A scalar input for `R` or `P` is expanded to a constant array with the same dimensions as the other input.
The mean of the negative binomial distribution with parameters r and p is rq / p, where q = 1 – p. The variance is rq / p2.
The simplest motivation for the negative binomial is the case of successive random trials, each having a constant probability `P` of success. The number of extra trials you must perform in order to observe a given number `R` of successes has a negative binomial distribution. However, consistent with a more general interpretation of the negative binomial, `nbinstat` allows `R` to be any positive value, including nonintegers.
## Examples
```p = 0.1:0.2:0.9; r = 1:5; [R,P] = meshgrid(r,p); [M,V] = nbinstat(R,P) M = 9.0000 18.0000 27.0000 36.0000 45.0000 2.3333 4.6667 7.0000 9.3333 11.6667 1.0000 2.0000 3.0000 4.0000 5.0000 0.4286 0.8571 1.2857 1.7143 2.1429 0.1111 0.2222 0.3333 0.4444 0.5556 V = 90.0000 180.0000 270.0000 360.0000 450.0000 7.7778 15.5556 23.3333 31.1111 38.8889 2.0000 4.0000 6.0000 8.0000 10.0000 0.6122 1.2245 1.8367 2.4490 3.0612 0.1235 0.2469 0.3704 0.4938 0.6173```
## See Also
#### Introduced before R2006a
Was this topic helpful?
Download now | 621 | 1,803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2016-40 | latest | en | 0.651904 |
https://blogs.sas.com/content/graphicallyspeaking/2014/09/21/splines/ | 1,713,996,342,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819971.86/warc/CC-MAIN-20240424205851-20240424235851-00436.warc.gz | 114,698,868 | 11,117 | Many users of SGPLOT and GTL know how to mix and match various plot statements to create graphs, sometimes in ways not originally intended. You are also aware that you can go a step beyond, and use these systems to create completely non-standard graphs such as the Spiral Plot, the Polar Graph, the Euler Diagram and more.
The other day I was asked to create a diagram. I created a simple one with the SGPLOT procedure, with four nodes, and three links. The four nodes A, B, C and D have the (x, y) positions shown in columns Xn and Yn. The three links have ids of 1, 2 and 3. These are drawn using the Series plot, each having 4 points, with first starting at right of the "From" node and ending at the left of the "To" node. Two additional intermediate points are provided. The "Node" and "Link" data is merged into the table shown on the right.
This data is plotted using the SGPLOT procedure, using a series plot to draw the links, and scatter plot to draw the nodes and the node ids. Graph is shown on the right. SGPLOT program is shown below, some options are trimmed to fit. Please see linked file at the bottom for the full program.
proc sgplot data=diagram dattrmap=attrmap;
series x=xl y=yl / group=id name='b';
scatter x=xn y=yn / group=node datalabel=node
datalabelpos=center;
keylegend 'b' / linelength=20;
run;
We can use the SmoothConnect option to avoid the sharp angles as shown on the right. Note, this result is less than satisfactory, as the curves are required to pass through each of the points in the data. This causes the curves to bend in the opposite direction of the curve as can be seen at the start of each link near node A. The three links are not co-linear at the start. Also, at each penultimate node, the curve bends the other way, as can be seen in the blue link to the left of the node.
Now, for a diagram, it is not really necessary that the link pass through each of the intermediate nodes. Those are merely there to set a path for the links. Only the start and end of the path must be on the first and last point.
In the graph on the right we get the desired effect. Here, each link starts and ends in the right point, but the curve does not necessarily pass through the intermediate points. The points are used as "control points' to compute a quadratic Bezier Spline. Then we use the series plot to draw the spline.
The graph on the right shows the spline curve and the control points. The original series plot points are used as the control points for the spline. The spline starts out as a straight line segment from the 1st vertex half way to the 2nd vertex. Now, from this point, a quadratic curve is calculated to the point half way between the next line segment. This continues for all the remaining segments, till we reach the half way point of the last segment. Then, the last segment is again a straight segment to the final vertex.
The benefit of this computation is that the curve is always at a tangent to the first and last segments, thus ensuring the slope of those segments. Here, we want them to be horizontal. The portion of the curve in between goes smoothly from one segment to the next. The program includes the BezierMacro() that computes the points for the quadratic Bezier Spline given the original control points. For more details, see this WikiPedia page on Quadratic Splines.
While I was in Beijing, the Chinese terms we learned the quickest were "Mien" for noodles, and other derivatives like "Jiruo Mien" for Chicken Noodles. It was essential to know this at a minimum to order food at Mr. Lee's, the local fast food place. Here are my versions of the graphs for the "Dry Noodles" and the "Wet Noodles", given the original data. Click on the graph for a higher resolution image.
The programs below were written using some SAS 9.4 graph features, but these are not essential for this use case. You can run it at SAS 9.3, and just remove the offending options.
SAS 9.4 Programs:
Macro: BezierMacro
Diagram: Diagram
Noodle: NoodleGraph
At the right is another use case with longer series plots to draw the response curves by treatment.
Note: Bezier curves may NOT be appropriate where the curve needs to pass through each point, but can be useful where the points for the series plot are control points to draw a smooth curve.
Share | 1,060 | 4,350 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-18 | latest | en | 0.904361 |
https://www.thewinningbets.com/predictions-mongolia/premier-league/prediction-of-deren-khangarid-19-september-2020-766228 | 1,600,490,931,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400190270.10/warc/CC-MAIN-20200919044311-20200919074311-00797.warc.gz | 1,148,916,867 | 13,902 | Match Day 16 of the League Premier League
Deren
• W
• W
• W
• W
• W
Khangarid
• W
• W
• L
• W
• D
In this page we analyze the match Deren - Khangarid (Premier League) trying to find out some useful predictions.
Evaluating only the ranking and the statistics of the individual teams we are not able to say with certainty which of the two teams is favored. The Khangarid has only 0 points less of Deren, from here it is understood that the game could be balanced. With goals scored and conceded, the team Deren has won home games on played. The line-up Khangarid has suffered goals and scored , totaling victories on games played
## Analysis of the State of Form
Let's go to compare the last five games played by the two teams that will take the field to understand their real state of form. We want to be sure that you understand well the data contained in the table, so we list them here: we have the date of the clash, the name of the teams, the results of the match and only the first half which one can understand the result of only the second time, the position in the ranking of the two teams.
### Analysis of the State of Form: Deren
Deren has an enviable form of state, winning almost every game, winning well 15 points, it will be an arduous challenger to beat, it all depends from the physical condition of the opposing team.
### Analysis of the State of Form: Khangarid
The Khangarid has a good form, certainly positive, winning and drawing most of the time, he has collected a good number of points in the five days. total points conquered were 10.
## Stats
The data contained in the following paragraphs will allow you to understand the strength of the teams in the field and their offensive abilities. Thanks to these statistics, you can compare the quantity and frequency of goals marked by Deren and Khangarid and compare them those of the championship Premier League who dispute.
The average goal calculated on matches of Deren is goals scored / conceded. The average goal, ie the total goals of Khangarid is
You can verify the data yourself thanks to the interesting tables below, which show all the values recorded up to now on the goals scored and suffered by Deren and Khangarid and the corresponding championship. We are sure that these tables will help you a lot, above all the 80% and higher statistics.
Deren Khangarid
Average of the goals / game
0.99 0.99
Under 1.5 Goals
26% 13%
Over 1.5 Goals
74% 87%
Under 2.5 Goals
40% 13%
Over 2.5 Goals
60% 60%
Under 3.5 Goals
60% 60%
Over 3.5 Goals
40% 40%
Goal
53% 66%
No Goal
46% 33%
## Full Match
In this section you will find the percentages related to this match. The analysis is very complete as it includes: winning team, goals scored, if both teams score and how much the game will end. We are able to give this information also for the first and second divided time.
The next tables will show the percentages as regards the exact result and the final result
### Winner of the Match
Our algorithm focuses the match data on Khangarid, will it be the favorite?
Percentage
Deren (Home Win) 35%
Khangarid (Away Win) 53%
Draw 12%
You can see below, if you are registered, the complete table of probability percentages we have studied.
Regarding the event in which both teams score a goal, we can deduce from the percentages (89%) that the probability that this happens is high.
### Deren vs Khangarid: First Half
Only for the true lovers of the forecasts, here are some advanced statistics related only to the first half of the match.
### Summary of Predictions
In short, the summary of the prediction of this football match
Discover
Khangarid Win
Discover
Over 2.5
Discover
Goal
##### 3 Most Probable Results
Discover
2 - 3
If you have reached the end it means that you have appreciated reading this article, leave us a comment or suggestion on how to improve our articles.
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http://perplexus.info/show.php?pid=4633&cid=32554 | 1,542,329,150,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742968.18/warc/CC-MAIN-20181116004432-20181116025705-00016.warc.gz | 251,008,478 | 5,378 | All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Domino arrangements (Posted on 2006-05-27)
Take the 15 smallest dominoes in a set (double blank through double four.)
In how many ways can they be arranged in a row such that the numbers on consecutive pieces match.
Count the two directions separately.
See The Solution Submitted by Jer Rating: 3.6667 (3 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
re: Pencil and paper solution | Comment 5 of 7 |
(In reply to Pencil and paper solution by Steve Herman)
You never multiply by 2 to get reversals, so the initial count should include both directions, which would lead to 24, rather than 12 chains that break into 5 and 5. Similarly the 48 that give a 4 and a 6 should be 96.
For the 3-row and 7-row, I actually get 144. This would be 72 if you didn't count reflections, but I see you record this as 36. Listed, these are:
40|01|13|30|02|21|14|42|23|34
40|01|13|30|02|21|14|43|32|24
40|01|13|30|02|23|34|41|12|24
40|01|13|30|02|23|34|42|21|14
40|01|13|30|02|24|41|12|23|34
40|01|13|30|02|24|43|32|21|14
20|01|13|30|04|41|12|23|34|42
20|01|13|30|04|41|12|24|43|32
20|01|13|30|04|42|21|14|43|32
20|01|13|30|04|42|23|34|41|12
20|01|13|30|04|43|32|21|14|42
20|01|13|30|04|43|32|24|41|12
30|01|14|40|02|21|13|32|24|43
30|01|14|40|02|21|13|34|42|23
30|01|14|40|02|23|31|12|24|43
30|01|14|40|02|23|34|42|21|13
30|01|14|40|02|24|43|31|12|23
30|01|14|40|02|24|43|32|21|13
20|01|14|40|03|31|12|23|34|42
20|01|14|40|03|31|12|24|43|32
20|01|14|40|03|32|21|13|34|42
20|01|14|40|03|32|24|43|31|12
20|01|14|40|03|34|42|21|13|32
20|01|14|40|03|34|42|23|31|12
40|02|21|10|03|31|14|42|23|34
40|02|21|10|03|31|14|43|32|24
40|02|21|10|03|32|24|41|13|34
40|02|21|10|03|32|24|43|31|14
40|02|21|10|03|34|41|13|32|24
40|02|21|10|03|34|42|23|31|14
30|02|21|10|04|41|13|32|24|43
30|02|21|10|04|41|13|34|42|23
30|02|21|10|04|42|23|31|14|43
30|02|21|10|04|42|23|34|41|13
30|02|21|10|04|43|31|14|42|23
30|02|21|10|04|43|32|24|41|13
40|02|23|30|01|12|24|41|13|34
40|02|23|30|01|12|24|43|31|14
40|02|23|30|01|13|34|41|12|24
40|02|23|30|01|13|34|42|21|14
40|02|23|30|01|14|42|21|13|34
40|02|23|30|01|14|43|31|12|24
10|02|23|30|04|41|12|24|43|31
10|02|23|30|04|41|13|34|42|21
10|02|23|30|04|42|21|13|34|41
10|02|23|30|04|42|21|14|43|31
10|02|23|30|04|43|31|12|24|41
10|02|23|30|04|43|31|14|42|21
30|02|24|40|01|12|23|31|14|43
30|02|24|40|01|12|23|34|41|13
30|02|24|40|01|13|32|21|14|43
30|02|24|40|01|13|34|41|12|23
30|02|24|40|01|14|43|31|12|23
30|02|24|40|01|14|43|32|21|13
10|02|24|40|03|31|12|23|34|41
10|02|24|40|03|31|14|43|32|21
10|02|24|40|03|32|21|13|34|41
10|02|24|40|03|32|21|14|43|31
10|02|24|40|03|34|41|12|23|31
10|02|24|40|03|34|41|13|32|21
40|03|31|10|02|21|14|42|23|34
40|03|31|10|02|21|14|43|32|24
40|03|31|10|02|23|34|41|12|24
40|03|31|10|02|23|34|42|21|14
40|03|31|10|02|24|41|12|23|34
40|03|31|10|02|24|43|32|21|14
20|03|31|10|04|41|12|23|34|42
20|03|31|10|04|41|12|24|43|32
20|03|31|10|04|42|21|14|43|32
20|03|31|10|04|42|23|34|41|12
20|03|31|10|04|43|32|21|14|42
20|03|31|10|04|43|32|24|41|12
40|03|32|20|01|12|24|41|13|34
40|03|32|20|01|12|24|43|31|14
40|03|32|20|01|13|34|41|12|24
40|03|32|20|01|13|34|42|21|14
40|03|32|20|01|14|42|21|13|34
40|03|32|20|01|14|43|31|12|24
10|03|32|20|04|41|12|24|43|31
10|03|32|20|04|41|13|34|42|21
10|03|32|20|04|42|21|13|34|41
10|03|32|20|04|42|21|14|43|31
10|03|32|20|04|43|31|12|24|41
10|03|32|20|04|43|31|14|42|21
20|03|34|40|01|12|23|31|14|42
20|03|34|40|01|12|24|41|13|32
20|03|34|40|01|13|32|21|14|42
20|03|34|40|01|13|32|24|41|12
20|03|34|40|01|14|42|21|13|32
20|03|34|40|01|14|42|23|31|12
10|03|34|40|02|21|13|32|24|41
10|03|34|40|02|21|14|42|23|31
10|03|34|40|02|23|31|12|24|41
10|03|34|40|02|23|31|14|42|21
10|03|34|40|02|24|41|12|23|31
10|03|34|40|02|24|41|13|32|21
30|04|41|10|02|21|13|32|24|43
30|04|41|10|02|21|13|34|42|23
30|04|41|10|02|23|31|12|24|43
30|04|41|10|02|23|34|42|21|13
30|04|41|10|02|24|43|31|12|23
30|04|41|10|02|24|43|32|21|13
20|04|41|10|03|31|12|23|34|42
20|04|41|10|03|31|12|24|43|32
20|04|41|10|03|32|21|13|34|42
20|04|41|10|03|32|24|43|31|12
20|04|41|10|03|34|42|21|13|32
20|04|41|10|03|34|42|23|31|12
30|04|42|20|01|12|23|31|14|43
30|04|42|20|01|12|23|34|41|13
30|04|42|20|01|13|32|21|14|43
30|04|42|20|01|13|34|41|12|23
30|04|42|20|01|14|43|31|12|23
30|04|42|20|01|14|43|32|21|13
10|04|42|20|03|31|12|23|34|41
10|04|42|20|03|31|14|43|32|21
10|04|42|20|03|32|21|13|34|41
10|04|42|20|03|32|21|14|43|31
10|04|42|20|03|34|41|12|23|31
10|04|42|20|03|34|41|13|32|21
20|04|43|30|01|12|23|31|14|42
20|04|43|30|01|12|24|41|13|32
20|04|43|30|01|13|32|21|14|42
20|04|43|30|01|13|32|24|41|12
20|04|43|30|01|14|42|21|13|32
20|04|43|30|01|14|42|23|31|12
10|04|43|30|02|21|13|32|24|41
10|04|43|30|02|21|14|42|23|31
10|04|43|30|02|23|31|12|24|41
10|04|43|30|02|23|31|14|42|21
10|04|43|30|02|24|41|12|23|31
10|04|43|30|02|24|41|13|32|21
40|01|12|20|03|31|14|42|23|34
40|01|12|20|03|31|14|43|32|24
40|01|12|20|03|32|24|41|13|34
40|01|12|20|03|32|24|43|31|14
40|01|12|20|03|34|41|13|32|24
40|01|12|20|03|34|42|23|31|14
30|01|12|20|04|41|13|32|24|43
30|01|12|20|04|41|13|34|42|23
30|01|12|20|04|42|23|31|14|43
30|01|12|20|04|42|23|34|41|13
30|01|12|20|04|43|31|14|42|23
30|01|12|20|04|43|32|24|41|13
I think your factor of 4 comes from not only discounting the complete reversals of, for example, the two below, but also from not counting even the two below as separate, where only the 3-segment is reversed :
40|01|13|30|02|21|14|42|23|34
40|03|31|10|02|21|14|42|23|34
their complete reversals being
20|03|31|10|04|43|32|24|41|12
20|01|13|30|04|43|32|24|41|12
Counting these four lines as one, and all similar situations results in the factor of 4 undercount.
(Yes, I used a computer to get the above list.)
Multiplying this (24+96+144=264) by 32*15 we get 126,720, which I believe is the correct answer.
Edited on May 28, 2006, 2:16 pm
Posted by Charlie on 2006-05-28 11:45:15
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https://edurev.in/studytube/Deflection-of-beams--SOM/d7e0073b-ea0c-48ff-8085-675a66c1978c_p | 1,709,263,492,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474948.91/warc/CC-MAIN-20240301030138-20240301060138-00386.warc.gz | 232,572,514 | 58,505 | Deflection of beams, SOM
Deflection of beams, SOM | Topper Handwritten Notes & Videos for GATE ME - Mechanical Engineering PDF Download
Topper Handwritten Notes & Videos for GATE ME
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FAQs on Deflection of beams, SOM - Topper Handwritten Notes & Videos for GATE ME - Mechanical Engineering
1. What is the deflection of beams?
Ans. The deflection of beams refers to the bending or deformation that occurs in a beam when a load is applied to it. It measures how much the beam is displaced from its original position due to the applied load.
2. How is the deflection of beams calculated?
Ans. The deflection of beams can be calculated using various methods, such as the Euler-Bernoulli beam theory, the moment-area method, or the virtual work method. These methods involve considering factors such as the beam's material properties, dimensions, and applied load to determine the deflection.
3. What factors affect the deflection of beams?
Ans. Several factors can influence the deflection of beams, including the beam's material properties (such as elasticity and stiffness), its dimensions (length, cross-sectional shape, and moment of inertia), the applied load (magnitude, position, and distribution), and the support conditions (fixed, pinned, or simply supported).
4. Why is deflection important in beam design?
Ans. Deflection is an important consideration in beam design as excessive deflection can lead to structural failure or functional issues. It can affect the overall stability and performance of the structure, causing problems such as sagging floors, excessive vibrations, or even collapse. Therefore, engineers must ensure that the deflection of beams remains within acceptable limits for the intended application.
5. How can the deflection of beams be controlled or minimized?
Ans. The deflection of beams can be controlled or minimized through various means, including selecting appropriate beam materials with high stiffness or low elasticity, optimizing the beam's dimensions and cross-sectional shape, redistributing the load, or providing additional support points or reinforcements. Structural analysis and calculations can also be performed to determine the maximum allowable deflection and ensure the beam design meets the required specifications.
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; | 559 | 2,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-10 | latest | en | 0.905642 |
https://practiceadvices.com/data/advice/read/404-are-vertically-opposite-angles-90-degree | 1,657,046,204,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104597905.85/warc/CC-MAIN-20220705174927-20220705204927-00794.warc.gz | 524,520,563 | 6,694 | # Are vertically opposite angles 90 degree?
### Are vertically opposite angles 90 degree?
A pair of vertically opposite angles are always equal to each other. Also, a vertical angle and its adjacent angle are supplementary angles, i.e., they add up to 180 degrees. For example, if two lines intersect and make an angle, say X=45°, then its opposite angle is also equal to 45°.
### What degrees are vertical angles?
Vertical angles form an X shape. The two adjacent angles are equal to 180 degrees. Vertical angles are formed by two intersecting lines. Both pairs of vertical angles sum up to 360 degrees.
### Are right angles vertical?
Vertical angles are formed by two intersecting lines. ... In one particular case, when vertical angles are right angles (that is, lines that form this pair of vertical angles are perpendicular), their sum is equal to a straight angles and, therefore, they are supplemental.
### What are the 7 types of angles?
The rays making an angle are called the arms of an angle and the common end point is called the vertex of an angle. There are 7 types of angles. These are zero angle, acute angle, right angle, obtuse angle, straight angle, reflex angle, and complete angle.
### Are vertically opposite angles?
When two lines intersect, the opposite (X) angles are equal. ... In the diagram above, the two green angles are equal and the two yellow angles are equal. These X angles are called vertically opposite angles because they are opposite each other at a vertex.
### Do vertical angles add to 180?
Vertical angles are angles that are opposite each other when two lines intersect each other. The two pairs of opposite angles are equal to each other. The two pairs of neighboring angles are supplementary, meaning they add up to 180 degrees.
### What do vertical angles look like?
Vertical angles are a pair of non-adjacent angles formed when two lines intersect. We see intersecting lines all the time in our real world. ... For example, angles AOC and AOB are not a pair vertical angles, but they are adjacent angles. However, vertical angles always have a common vertex.
### Do vertical angles always add to 180?
Vertical angles are always congruent, or of equal measure. ... Both pairs of vertical angles (four angles altogether) always sum to a full angle (360°). Adjacent angles. In the figure above, an angle from each pair of vertical angles are adjacent angles and are supplementary (add to 180°).
### Do Triangle angles equal 180 degrees?
In a Euclidean space, the sum of angles of a triangle equals the straight angle (180 degrees, π radians, two right angles, or a half-turn). A triangle has three angles, one at each vertex, bounded by a pair of adjacent sides.
### What are the different types of vertical angles?
We learned about two special types of angles. Vertical angles are angles that are opposite each other when two lines intersect each other. The two pairs of opposite angles are equal to each other. The two pairs of neighboring angles are supplementary, meaning they add up to 180 degrees.
### What does the vertical angle theorem say about angles?
The Vertical Angle Theorem says the opposing angles of two intersecting lines must be congruent, or identical in value. That means no matter how or where two straight lines intersect each other, the angles opposite to each other will always be congruent, or equal in value:
### How are complementary angles related to 90° 90 °?
Complementary angles add to 90° 90 °. Complementary angles need not be connected with a common vertex or point, or line. They can be adjacent or vertical in intersecting lines. They could be in two different polygons, so long as the sum of their angles is exactly 90° 90 °.
### When do vertical angles need to be complementary?
Only when vertical angles, a a, are 45° 45 ° can they be complementary. Acute vertical angles could be complementary; you have a 1-in-45 chance of that. Complementary angles add to 90° 90 °. Complementary angles need not be connected with a common vertex or point, or line. They can be adjacent or vertical in intersecting lines. | 882 | 4,128 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-27 | latest | en | 0.916116 |
https://www.open.edu/openlearn/mod/oucontent/view.php?id=87798§ion=_unit4.4.5.4 | 1,675,759,702,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500392.45/warc/CC-MAIN-20230207071302-20230207101302-00177.warc.gz | 916,674,995 | 25,132 | Science, Maths & Technology
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An introduction to exoplanets
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Orbital motion and radial velocity
We are now going to explore how the motions around the centre of mass and the radial velocity are related. In reality all astronomers ‘see’ is a point of light on the sky: the star. The laws of physics, which underly the changes in the light that astronomers collect from the star, allow the real motion of the star and its planet(s) to be determined. Astronomy proceeds by unlocking the information carried by light from objects we are unlikely to ever be able to view up close.
Activity _unit4.4.5 Activity 7 Orbital motion and radial velocity
Timing: Allow about 20 minutes
This application shows orbital motions in a system where a star is 1500 times more massive than its planet. The separations are scaled so the planet’s orbit just fits on the screen - the planet and star are shown much larger relative to their separation than they would really be. With the initial settings the observer is looking at the system from above and to the right of the orbital plane.
The application shows two animated arrow vectors – light green represents the true velocity of the star. The star’s speed stays the same as it moves around its circular orbit, but its velocity continuously changes direction as shown. Remember, the star and the planet move in opposite directions at each instant. The star moves much more slowly than the planet: in the time that it takes for the planet to move around the big circle, the star moves around a circle that is too small to even be seen in this application. The star’s motion is just a wobble around the centre of mass that is too small to easily discern.
The second animated arrow vector, in orange-brown, shows the radial velocity of the star relative to the distant observer. When the star is moving to the right, the radial velocity arrow points towards the observer, along the purple line. When the star moves to the left, the radial velocity arrow points away from the observer, along the extension of the purple line below the orbital plane. To help you to see this, remember that you can drag the orbital plane around to change your viewpoint.
The application gives the values of the stellar orbital speed (which never changes) and the stellar radial velocity, which changes continuously as the direction of the star’s motion changes continuously as it moves around its orbit.
Active content not displayed. This content requires JavaScript to be enabled.
Interactive feature not available in single page view (see it in standard view).
When the radial velocity is towards the observer (orange-brown arrow aligned along the purple line), is the value of the stellar radial velocity positive or negative?
Negative.
When the radial velocity is away from the observer (orange-brown arrow aligned away from the purple line), is the value of the radial velocity positive or negative? (You can pause the application if you need to.)
Positive.
What is the biggest value the stellar radial velocity ever has with the initial settings? (If necessary, you can refresh the page to reset all of the sliders to the initial settings, i = 77° and aplanet = 14.55.)
10.8 m/s (give or take however much uncertainty your clicking led to).
In what direction is the true velocity of the star pointing when the radial velocity has its maximum value?
To the left, away from the observer, who is to the right and above the orbital plane.
Keeping the application paused at its maximum value of radial velocity, adjust the orbital inclination, i, using the slider. You can also use the arrow keys on the keyboard to amend the values.
For orbital inclination i = 90° exactly, what value does the stellar radial velocity now show? What do you notice about this value? | 808 | 3,981 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2023-06 | longest | en | 0.921791 |
https://invernessgangshow.net/sum-of-a-rational-and-irrational-number/ | 1,652,965,384,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662527626.15/warc/CC-MAIN-20220519105247-20220519135247-00496.warc.gz | 384,621,274 | 6,307 | Let’s make convincing arguments about why the sums and also products of rational and irrational numbers constantly produce details kinds that numbers.
You are watching: Sum of a rational and irrational number
Here space some examples of integers (positive or an adverse whole numbers):
Experinvernessgangshow.netent with adding any 2 numbers native the perform (or various other integers of your choice). Try to discover one or an ext examples of two integers that:
Experinvernessgangshow.netent through multiplying any type of two numbers from the perform (or other integers of her choice). Try to find one or more examples of two integers that:
multiply to make one more integermultiply to make a number the is not an integer
Here room a couple of examples of including two reasonable numbers. Is each sum a reasonable number? Be ready to explain how girlfriend know.
(4 +0.175 = 4.175)(frac12 + frac45 = frac 510+frac810 = frac1310)( ext-0.75 + frac148 = frac ext-68 + frac 148 = frac 88 = 1)(a) is an integer: (frac 23+ frac a15 =frac1015 + frac a15 = frac 10+a15)
Here is a means to describe why the sum of 2 rational numbers is rational.
Suppose (fracab) and also (fraccd) space fractions. That way that (a, b, c,) and also (d) are integers, and also (b) and (d) room not 0.
Find the amount of (fracab) and also (fraccd). Present your reasoning. In the sum, room the molecule and the denominator integers? exactly how do girlfriend know?Use her responses to describe why the amount of (fracab + fraccd) is a reasonable number. Use the same thinking as in the previous concern to describe why the product of 2 rational numbers, (fracab oldcdot fraccd), need to be rational.
Consider numbers that are of the form (a + b sqrt5), wherein (a) and also (b) are entirety numbers. Let’s contact such numbers quintegers.
Here space some instances of quintegers:
When we include two quintegers, will certainly we always get another quinteger? one of two people prove this, or discover two quintegers whose sum is not a quinteger.When us multiply two quintegers, will we always get another quinteger? either prove this, or uncover two quintegers who product is no a quinteger.
Here is a means to describe why (sqrt2 + frac 19) is irrational.
Let (s) it is in the sum of ( sqrt2) and also (frac 19), or (s=sqrt2 + frac 19).
Suppose (s) is rational.
Would (s + ext- frac19) be rational or irrational? explain how friend know.Evaluate (s + ext-frac19). Is the amount rational or irrational?Use your responses so much to define why (s) can not be a reasonable number, and therefore ( sqrt2 + frac 19) cannot be rational.Use the same thinking as in the earlier question to explain why (sqrt2 oldcdot frac 19) is irrational.
Consider the equation (4x^2 + bx + 9=0). Find a value of (b) so the the equation has:
2 reasonable solutions2 irrational solutions1 solutionno solutionsDescribe every the worths of (b) that produce 2, 1, and also no solutions.
Write a new quadratic equation v each type of solution. Be ready to define how you know that your equation has actually the specified type and number of solutions.
no solutions2 irrational solutions2 rational solutions1 solution
We understand that quadratic equations deserve to have rational options or irrational solutions. For example, the options to ((x+3)(x-1)=0) space -3 and also 1, which space rational. The services to (x^2-8=0) are (pm sqrt8), which room irrational.
Sometinvernessgangshow.netes solutions to equations combine two number by addition or multiplication—for example, (pm 4sqrt3) and also (1 +sqrt 12). What type of number are these expressions?
When we add or multiply 2 rational numbers, is the result rational or irrational?
The amount of two rational numbers is rational. Here is one method to define why it is true:
Any two rational numbers can be created (fracab) and (fraccd), whereby (a, b, c, ext and d) room integers, and also (b) and (d) are not zero.The sum of (fracab) and also (fraccd) is (fracad+bcbd). The denominator is not zero because neither (b) no one (d) is zero.Multiplying or including two integers always gives an integer, therefore we know that (ad, bc, bd) and also (ad+bc) room all integers.If the numerator and denominator that (fracad+bcbd) space integers, clinvernessgangshow.netate the number is a fraction, i m sorry is rational.
The product of two rational number is rational. Us can present why in a comparable way:
For any two rational number (fracab) and also (fraccd), wherein (a, b, c, ext and also d) are integers, and also (b) and (d) space not zero, the product is (fracacbd).Multiplying two integers always results in an integer, for this reason both (ac) and also (bd) room integers, for this reason (fracacbd) is a rational number.
What around two irrational numbers?
The amount of two irrational numbers could be one of two people rational or irrational. We can show this through examples:
(sqrt3) and ( ext-sqrt3) room each irrational, however their amount is 0, which is rational.(sqrt3) and (sqrt5) are each irrational, and also their amount is irrational.
The product of 2 irrational numbers might be either rational or irrational. Us can display this v examples:
(sqrt2) and also (sqrt8) are each irrational, however their product is (sqrt16) or 4, i m sorry is rational.(sqrt2) and also (sqrt7) space each irrational, and their product is (sqrt14), i m sorry is not a perfect square and is therefore irrational.
What around a reasonable number and also an irrational number?
The amount of a rational number and an irrational number is irrational. To define why calls for a slightly different argument:
Let (R) be a reasonable number and (I) one irrational number. We want to show that (R+I) is irrational.Suppose (s) to represent the sum of (R) and also (I) ((s=R+I)) and also suppose (s) is rational.If (s) is rational, then (s + ext-R) would also be rational, because the amount of 2 rational numbers is rational.(s + ext-R) is not rational, however, since ((R + I) + ext-R = I).(s + ext-R) cannot be both rational and irrational, which method that our original assumption that (s) to be rational was incorrect. (s), i beg your pardon is the sum of a reasonable number and also an irrational number, need to be irrational.
The product of a non-zero reasonable number and an irrational number is irrational. We can display why this is true in a comparable way:
Let (R) it is in rational and also (I) irrational. We desire to display that (R oldcdot I) is irrational.Suppose (p) is the product of (R) and also (I) ((p=R oldcdot I)) and suppose (p) is rational.If (p) is rational, clinvernessgangshow.netate (p oldcdot frac1R) would also be rational due to the fact that the product of two rational numbers is rational.(p oldcdot frac1R) is no rational, however, due to the fact that (R oldcdot i oldcdot frac1R = I).(p oldcdot frac1R) cannot be both rational and also irrational, which means our original assumption that (p) was rational to be false. (p), i beg your pardon is the product that a rational number and an irrational number, need to be irrational.
Video VLS Alg1U7V5 Rational and Irrational solutions (Lessons 19–21) obtainable at https://player.vinvernessgangshow.neteo.com/video/531442545.
The formula (x = ext-b pm sqrtb^2-4ac over 2a) that provides the remedies of the quadratic equation (ax^2 + bx + c = 0), where (a) is not 0.
See more: How Far Is Nashville To Memphis In Two Weeks, How Far Is Memphis From Nashville
The Illustrative mathematics name and also logo space not subject to the an innovative Commons license and may no be provided without the prior and express written consent the Illustrative Mathematics. | 1,920 | 7,702 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-21 | longest | en | 0.853943 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-5-section-5-1-exponents-exercise-set-page-344/34 | 1,531,849,098,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589757.30/warc/CC-MAIN-20180717164437-20180717184437-00417.warc.gz | 878,883,932 | 14,304 | Algebra: A Combined Approach (4th Edition)
$18y^{17}$ meters
We are given that the parallelogram has base length of $9y^{7}$ meters and height of $2y^{10}$ meters. We know that the area of a parallelogram is calculated as $area=base \times height$. Based on the product rule for exponents, we know that $a^{m}\times a^{n}=a^{m+n}$ (where $m$ and $n$ are positive integers and $a$ is a real number). Therefore, $area=9y^{7}\times2y^{10}=9\times2\times y^{7+10}=18y^{17}$ meters. | 154 | 478 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2018-30 | latest | en | 0.877777 |
https://www.neetprep.com/questions/476-Physics/7813-Systems-Particles-Rotational-Motion?courseId=141&testId=1061965-Past-Year----MCQs | 1,723,516,203,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641054522.78/warc/CC-MAIN-20240813012759-20240813042759-00350.warc.gz | 709,685,416 | 58,137 | Which of the following will not be affected if the radius of the sphere is increased while keeping mass constant?
1 Moment of inertia 2 Angular momentum 3 Angular velocity 4 Rotational kinetic energy
Subtopic: Angular Momentum |
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Three objects, A: (a solid sphere), B: (a thin circular disk) and C: (a circular ring), each have the same mass $$\mathrm{M}$$ and radius $$\mathrm{R}.$$ They all spin with the same angular speed about their own symmetry axes. The amount of work ($$\mathrm{W}$$) required to bring them to rest, would satisfy the relation:
1 $$\mathrm{W_C}>\mathrm{W_B}>\mathrm{W_A} ~~~~~~~~$$ 2 $$\mathrm{W_A}>\mathrm{W_B}>\mathrm{W_C}$$ 3 $$\mathrm{W_B}>\mathrm{W_A}>\mathrm{W_C}$$ 4 $$\mathrm{W_A}>\mathrm{W_C}>\mathrm{W_B}$$
Subtopic: Rotational Motion: Kinematics |
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The moment of the force, $$\overset{\rightarrow}{F} = 4 \hat{i} + 5 \hat{j} - 6 \hat{k}$$ at point ($$2,$$ $$0,$$ $$-3$$) about the point ($$2,$$ $$-2,$$ $$-2$$) is given by:
1. $$- 8 \hat{i} - 4 \hat{j} - 7 \hat{k}$$
2. $$- 4 \hat{i} - \hat{j} - 8 \hat{k}$$
3. $$- 7 \hat{i} - 8 \hat{j} - 4 \hat{k}$$
4. $$- 7 \hat{i} - 4 \hat{j} - 8 \hat{k}$$
Subtopic: Torque |
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A solid sphere is rotating freely about its axis of symmetry in free space. The radius of the sphere is increased keeping its mass the same. Which of the following physical quantities would remain constant for the sphere?
1 angular velocity 2 moment of inertia 3 rotational kinetic energy 4 angular momentum
Subtopic: Angular Momentum |
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A rope is wrapped around a hollow cylinder with a mass of $$3$$ kg and a radius of $$40$$ cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of $$30$$ N?
1. $$0.25 ~\text{rad/s}^2$$
2. $$25 ~\text{rad/s}^2$$
3. $$5 ~\text{m/s}^2$$
4. $$25 ~\text{m/s}^2$$
Subtopic: Rotational Motion: Dynamics |
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Two discs of the same moment of inertia are rotating about their regular axis passing through centre and perpendicular to the plane of the disc with angular velocities $$\omega_1$$ ${}_{}$and $$\omega_2$$. They are brought into contact face to face with their axis of rotation coinciding. The expression for loss of energy during this process is:
1. $$\frac{1}{4}I\left(\omega_1-\omega_2\right)^2$$
2. $$I\left(\omega_1-\omega_2\right)^2$$
3. $$\frac{1}{8}I\left(\omega_1-\omega_2\right)^2$$
4. $$\frac{1}{2}I\left(\omega_1-\omega_2\right)^2$$
Subtopic: Angular Momentum |
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Which of the following statements are correct?
(a) centre of mass of a body always coincides with the centre of gravity of the body . (b) centre of gravity of a body is the point about which the total gravitational torque on the body is zero. (c) a couple on a body produce both translational and rotation motion in a body. (d) mechanical advantage greater than one means that small effort can be used to lift a large load.
1 (a) and (b) 2 (b) and (c) 3 (c) and (d) 4 (b) and (d)
Subtopic: Center of Mass |
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Two rotating bodies $$A$$ and $$B$$ of masses $$m$$ and $$2m$$ with moments of inertia ${I}_{A}$ and have equal kinetic energy of rotation. If ${L}_{A}$ and ${L}_{B}$ be their angular momenta respectively, then:
1. ${L}_{A}=\frac{{L}_{B}}{2}$
2. ${L}_{A}=2{L}_{B}$
3. ${L}_{B}>{L}_{A}$
4. ${L}_{A}>{L}_{B}$
Subtopic: Rotational Motion: Dynamics |
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A solid sphere of mass m and radius R is rotating about its diameter. A soild cyclinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation $\left({E}_{sphere}/{E}_{cylinder}\right)$ will be
(a) 2:3
(b) 1:5
(c) 1:4
(d) 3:1
Subtopic: Moment of Inertia |
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A light rod of length $$l$$ has two masses, $$m_1$$ and $$m_2,$$ attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is:
1. $$\frac{m_1m_2}{m_1+m_2}l^2$$
2. $$\frac{m_1+m_2}{m_1m_2}l^2$$
3. $$(m_1+m_2)l^2$$
4. $$\sqrt{(m_1m_2)}l^2$$
Subtopic: Moment of Inertia |
74%
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Hints | 1,876 | 5,841 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 11, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-33 | latest | en | 0.729061 |
http://www.chegg.com/homework-help/questions-and-answers/student-builds-andcalibrates-accelerometer-uses-determine-speedof-car-around-certain-highw-q489515 | 1,469,926,197,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469258944256.88/warc/CC-MAIN-20160723072904-00018-ip-10-185-27-174.ec2.internal.warc.gz | 304,321,470 | 13,350 | A student builds andcalibrates an accelerometer, which she uses to determine the speedof her car around a certain highway curve. The accelerometer is aplumb bob with a protrator that she attaches to the roof of hercar. A friend riding in the car with her observes that the plumbbob hangs at an angle of 8o from thevertical when the car has a speed of 24.6 m/s. The acceleration of gravityis 9.8 m/s2 .At this instant, what is the centripetalacceleration of the car rounding the curve? Answer in units ofm/s2.
b)what is the radius of the curve(m). | 137 | 546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2016-30 | latest | en | 0.923617 |
https://www.enotes.com/homework-help/producer-finds-that-demand-his-commodity-obeys-433419 | 1,516,148,448,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886758.34/warc/CC-MAIN-20180116224019-20180117004019-00484.warc.gz | 902,820,485 | 9,695 | # a producer finds that demand for his commodity obeys a linear demand equation p+0.85x=8.5, where p is in thousands of dollars and x is in thousands of units. If the producer's cost are given by...
a producer finds that demand for his commodity obeys a linear demand equation p+0.85x=8.5, where p is in thousands of dollars and x is in thousands of units. If the producer's cost are given by C(x)=1+4x, what should his level of production be to maximize profits?
the level of production is:( )in thousands of units
violy | Certified Educator
We will first solve for 'p' for price.
Subtract both sides by 0.85x.
`p + 0.85x - 0.85x = 8.5 - 0.85x`
`p = -0.85x + 8.5`
We can now write the revenue function.
Revenue = price *amount of unit sold.
`r = (-0.85x + 8.5)(x)`
Use Distributive property.
`r = -0.85x^2 + 8.5x`
We know that Profit = revenue - cost.
So, we will have:
`P(x) = (-0.85x^2 +8.5x) - (1 + 4x) = -0.85x^2 + 4.5x - 1`
Take the derivative of both sides.
`P' = -1.7x + 4.5`
Equate it to zero, and solve for x.
`-1.7x +4.5 = 0`
`-1.7x = -4.5`
`x = 2.68 or 3000` | 385 | 1,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-05 | latest | en | 0.922514 |
https://nordstrommath.com/DiscreteMathText/trees10-3.html | 1,708,646,510,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473871.23/warc/CC-MAIN-20240222225655-20240223015655-00724.warc.gz | 432,778,330 | 12,366 | Discrete Mathematics: An Active Approach to Mathematical Reasoning
Section10.3Trees
In this section we look at a special type of graph called a tree. Trees are useful in many applications, such as probability trees or decision trees.
Definition10.3.1.
A graph is circuit-free if it has no nontrivial circuits.
Definition10.3.2.
A graph is a tree if it is connected and circuit-free.
The trivial tree is just a single vertex. While the empty tree has no vertices (and, hence, no edges).
Definition10.3.3.
A graph is a forest if it is circuit-free. We can think of a forest as a collection of trees.
Definition10.3.7.
A terminal vertex or leaf is a vertex of degree 1. Vertices of degree greater than 1 are internal or branch vertices.
We now define some special types of trees, and related terms.
A rooted tree has one vertex, $$r\text{,}$$ designated as the root.
• The level of a vertex, $$v\text{,}$$ is the number of edges from $$v$$ to the root, $$r\text{.}$$
• The height of a tree is the maximum level of the tree.
• Let $$v$$ be a vertex in a rooted tree. Then the children of $$v$$ are the vertices adjacent to $$v\text{,}$$ one level below $$v$$ (think of the root at the top of the tree, with each successive level below the previous one).
• The parent of a vertex $$v$$ is the adjacent vertex one level up from $$v\text{.}$$
• The siblings of a vertex $$v$$ are the vertices with the same parent as $$v\text{.}$$
• The ancestors of a vertex $$v$$ are the vertices on the path from $$v$$ to $$r\text{,}$$ not including $$v\text{.}$$
• The descendants of a vertex $$v$$ are the vertices on the subtree taking $$v$$ as the root, not including $$v\text{.}$$
In the example, let $$r$$ be the root of the tree.
Then $$v$$ is at level 2, $$d$$ is at level 4, and $$r$$ is at level 0.
The height of the tree is 4.
The children of $$v$$ are $$c_1$$ and $$c_2\text{.}$$
The only sibling of $$v$$ is $$s\text{;}$$ the parent of $$v$$ is $$p\text{.}$$
The ancestors of $$v$$ are $$p$$ and $$r\text{.}$$ The descendants of $$v$$ are $$c_1, c_2, d\text{.}$$
A binary tree is a rooted tree in which each vertex has at most two children.
A full binary tree is a rooted tree in which each branch vertex has exactly two children and all leaves are at the same height.
Activity10.3.1.
Draw the full binary tree of height 4. How many leaves does it have?
Activity10.3.2.
In this activity we will explore the relationship between the number of edges and the number of vertices in a tree.
(a)
Draw a forest with 10 vertices that is not a tree.
(b)
Draw three different trees, each with 10 vertices.
(c)
How many edges does each graph in (b) have? Check with other students in the class. How many edges do their graphs have?
(d)
Draw a tree with 5 vertices and one with 12 vertices. How many edges do they have?
(e)
Can you draw a tree with 5 vertices with 7 edges? What about 5 edges? What is the most number of edges you can have? What is the fewest?
(f)
Form a conjecture about how many edges a tree with 10 vertices has. What about a tree with 5 vertices?
After experimenting you should be able to conjecture that a tree with 10 vertices has 9 edges, and a tree with 5 vertices must have 4 edges. This relationship is true in general.
Activity10.3.3.
In this activity we explore graphs with $$n$$ vertices and $$n-1$$ edges more generally.
(a)
Draw a graph with 7 vertices and 6 edges that is not a tree.
(b)
Draw a connected graph with 7 vertices and 6 edges that is not a tree.
(c)
Do you think either of the tasks in (a) or (b) is impossible?
(d)
Form a conjecture about graphs with $$n$$ vertices and $$n-1$$ edges.
We should be able to see that it is not possible to draw a connected graph with 7 vertices and 6 edges that is not a tree. This leads to the next theorem.
1.
For which values of $$n$$ is the complete graph $$K_n$$ a tree?
• $$n=1$$
• $$n=2$$
• $$n=3$$
• $$n=4$$
• No values of $$n$$
• All values of $$n\geq 1\text{.}$$
2.
True or false: In a rooted tree, a vertex can have more than one parent.
• True.
• False.
3.
True or false: In a rooted tree, a vertex can have 0 parents.
• True.
• The root has 0 parents.
• False.
• The root has 0 parents.
4.
In the rooted tree in Figure 10.3.9, how many children does $$r$$ have?
5.
In the rooted tree in Figure 10.3.9, how many descendants does $$p$$ have?
6.
In the rooted tree in Figure 10.3.9, what is the level of $$p\text{?}$$
7.
How many leaves does the full binary tree of height 4 have?
8.
True or false: There exists a tree with 7 vertices and 9 edges.
• True.
• False.
9.
True or false: There exists a tree with 7 vertices and 5 edges.
• True.
• False.
10.
True or false: There exists a forest with 7 vertices and 5 edges.
• True.
• False.
11.
True or false: If a graph has 8 vertices and 7 edges it must be a tree.
• True.
• False.
12.
True or false: If a connected graph has 8 vertices and 7 edges it must be a tree.
• True.
• False.
13.
True or false: There exists a tree with 8 vertices and 2 leaves.
• True.
• False.
14.
True or false: There exists a tree with 8 vertices and 7 leaves.
• True.
• False.
ExercisesExercises
1.
What is the total degree of a tree with $$n$$ vertices? Why?
2.
Consider the following lemma.
This is a useful lemma for proving statements about trees. However, if the tree has infinitely many vertices and edges, then the lemma is false. Give an example of an infinite tree that has no vertex of degree 1.
3.
For each of the following, either draw a graph with the given properties or explain why no such graph exists.
1. Tree, nine vertices, nine edges.
2. Graph, connected, nine vertices, nine edges.
3. Tree, five vertices, total degree 8
4. Graph, six vertices, five edges, not a tree.
5. Graph, connected, ten vertices, nine edges, has a circuit.
4.
Suppose a connected graph has twelve vertices and 11 edges. Must this graph have a vertex of degree 1? Explain your answer. | 1,638 | 5,956 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2024-10 | latest | en | 0.910124 |
https://edurev.in/studytube/PPT-Boundary-Layers/c70833f4-121d-4720-bc8e-29ba65a4eadb_p | 1,632,234,566,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057225.38/warc/CC-MAIN-20210921131252-20210921161252-00426.warc.gz | 274,186,218 | 46,552 | Courses
# PPT: Boundary Layers Civil Engineering (CE) Notes | EduRev
## Civil Engineering (CE) : PPT: Boundary Layers Civil Engineering (CE) Notes | EduRev
``` Page 1
Introduction Introduction
The drag on a body passing through a fluid may be considered to be made up of
two components: Form drag and Skin friction drag.
Form drag: which is dependent on the pressure forces acting on the body; and
the skin friction drag , which depends on the shearing forces acting between
the body and the fluid.
Page 2
Introduction Introduction
The drag on a body passing through a fluid may be considered to be made up of
two components: Form drag and Skin friction drag.
Form drag: which is dependent on the pressure forces acting on the body; and
the skin friction drag , which depends on the shearing forces acting between
the body and the fluid.
Shear Force and Pressure Force Shear Force and Pressure Force
Shear forces:
viscous drag, frictional drag, or skin friction
caused by shear between the fluid and the
solid surface
function of ___________and ______of
object
surface area
length
flow separation
U U
U U
Major losses in pipes
Flow expansion
losses
Projected area
Pressure forces
pressure drag or form drag
caused by _____________from the body
function of area normal to the flow
Page 3
Introduction Introduction
The drag on a body passing through a fluid may be considered to be made up of
two components: Form drag and Skin friction drag.
Form drag: which is dependent on the pressure forces acting on the body; and
the skin friction drag , which depends on the shearing forces acting between
the body and the fluid.
Shear Force and Pressure Force Shear Force and Pressure Force
Shear forces:
viscous drag, frictional drag, or skin friction
caused by shear between the fluid and the
solid surface
function of ___________and ______of
object
surface area
length
flow separation
U U
U U
Major losses in pipes
Flow expansion
losses
Projected area
Pressure forces
pressure drag or form drag
caused by _____________from the body
function of area normal to the flow
Description of Boundary Layer Description of Boundary Layer
t
w
: wall shear stresses
U
d
In the immediate vicinity of the boundary surface, the velocity of the fluid
increases gradually from zero at boundary surface to the velocity of the
mainstream. This region is known as BOUNDARY LAYER.
0 y
u
y
t µ
=
? ? ?
=
? ?
?
? ?
The nominal thickness of BOUNDARY LAYER is defined as the distance from
the boundary where the velocity of fluid is 99 % of free stream velocity
Page 4
Introduction Introduction
The drag on a body passing through a fluid may be considered to be made up of
two components: Form drag and Skin friction drag.
Form drag: which is dependent on the pressure forces acting on the body; and
the skin friction drag , which depends on the shearing forces acting between
the body and the fluid.
Shear Force and Pressure Force Shear Force and Pressure Force
Shear forces:
viscous drag, frictional drag, or skin friction
caused by shear between the fluid and the
solid surface
function of ___________and ______of
object
surface area
length
flow separation
U U
U U
Major losses in pipes
Flow expansion
losses
Projected area
Pressure forces
pressure drag or form drag
caused by _____________from the body
function of area normal to the flow
Description of Boundary Layer Description of Boundary Layer
t
w
: wall shear stresses
U
d
In the immediate vicinity of the boundary surface, the velocity of the fluid
increases gradually from zero at boundary surface to the velocity of the
mainstream. This region is known as BOUNDARY LAYER.
0 y
u
y
t µ
=
? ? ?
=
? ?
?
? ?
The nominal thickness of BOUNDARY LAYER is defined as the distance from
the boundary where the velocity of fluid is 99 % of free stream velocity
Description of Boundary Layer Description of Boundary Layer
t
w
: wall shear stresses
U
d
shear stress:
u
y
t µ
? ? ?
=
? ?
?
? ?
Shear stress acting at the plate surface
sets up a shear force which opposes
the fluid motion, and fluid close to the
wall is decelerated.
Theoretical understanding on Boundary layer development is very important to
determine the velocity gradient and hence shear forces on the surface.
Consists of two layers:
CLOSE TO BOUNDARY : large velocity gradient, appreciable viscous forces.
OUTSIDE BOUNDARY LAYER: viscous forces are negligible, flow may be
treated as non-viscous or inviscid.
Page 5
Introduction Introduction
The drag on a body passing through a fluid may be considered to be made up of
two components: Form drag and Skin friction drag.
Form drag: which is dependent on the pressure forces acting on the body; and
the skin friction drag , which depends on the shearing forces acting between
the body and the fluid.
Shear Force and Pressure Force Shear Force and Pressure Force
Shear forces:
viscous drag, frictional drag, or skin friction
caused by shear between the fluid and the
solid surface
function of ___________and ______of
object
surface area
length
flow separation
U U
U U
Major losses in pipes
Flow expansion
losses
Projected area
Pressure forces
pressure drag or form drag
caused by _____________from the body
function of area normal to the flow
Description of Boundary Layer Description of Boundary Layer
t
w
: wall shear stresses
U
d
In the immediate vicinity of the boundary surface, the velocity of the fluid
increases gradually from zero at boundary surface to the velocity of the
mainstream. This region is known as BOUNDARY LAYER.
0 y
u
y
t µ
=
? ? ?
=
? ?
?
? ?
The nominal thickness of BOUNDARY LAYER is defined as the distance from
the boundary where the velocity of fluid is 99 % of free stream velocity
Description of Boundary Layer Description of Boundary Layer
t
w
: wall shear stresses
U
d
shear stress:
u
y
t µ
? ? ?
=
? ?
?
? ?
Shear stress acting at the plate surface
sets up a shear force which opposes
the fluid motion, and fluid close to the
wall is decelerated.
Theoretical understanding on Boundary layer development is very important to
determine the velocity gradient and hence shear forces on the surface.
Consists of two layers:
CLOSE TO BOUNDARY : large velocity gradient, appreciable viscous forces.
OUTSIDE BOUNDARY LAYER: viscous forces are negligible, flow may be
treated as non-viscous or inviscid.
Development of Boundary Layer Development of Boundary Layer
In laminar boundary layer the particles are moving along stream lines.
The boundary layer thickness increases as the distance x from leading edge is
increases. This is because of viscous forces that dissipate more and more
energy of fluid stream as the flow proceeds and large group of particles are slow
downed.
The disturbance in fluid flow in boundary layer is amplified and the flow become
unstable and the fluid flow undergoes transition from laminar to turbulent flow.
This regime is called transition regime.
```
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## Fluid Mechanics
47 videos|82 docs|73 tests
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; | 1,694 | 7,135 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-39 | latest | en | 0.842816 |
https://drawabox.com/community/submission/3ANLYJZZ | 1,716,230,525,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058293.53/warc/CC-MAIN-20240520173148-20240520203148-00496.warc.gz | 179,176,373 | 11,451 | ## Lesson 1: Lines, Ellipses and Boxes
##### 7:20 PM, Sunday February 19th 2023
Hi,
I previously got up to lesson 3 in august 2022 but had to stop the course. I'd like to start again and get some feedback from the beginning as I don't think I was really getting it last time.
I've gone through lesson zero and lesson 1 and i'd like to resubmit the homework for lesson 1.
I hope this ok.
0 users agree
##### 9:36 PM, Sunday February 19th 2023
Hi, My name is Rob and I'm a teaching assistant with Drawabox who will be handling your critique. It appears as though you only uploaded one page of your homework to this album or just linked to a single image. Please try remaking the album with all of your homework pages and provide a new link. Thanks.
Next Steps:
Hi, My name is Rob and I'm a teaching assistant with Drawabox who will be handling your critique. It appears as though you only uploaded one page of your homework to this album or just linked to a single image. Please try remaking the album with all of your homework pages and provide a new link. Thanks.
##### 8:08 PM, Monday February 20th 2023
Sorry about that, should have checked it.
Here's a link to the full post
https://imgur.com/a/S4t3dHF
##### 10:49 PM, Monday February 20th 2023
No problem. Starting with your superimposed lines these are off to a fine start. You are keeping a clearly defined starting point with all of your wavering at the opposite end. Your ghosted lines and planes turned out well. You are using the ghosting method to good effect to get confident linework with a pretty decent deal of accuracy that will get better and better with practice.
Your tables of ellipses are coming along pretty good. You are doing a good job drawing through your ellipses and focusing on consistent smooth ellipse shapes. This is carried over nicely into your ellipses in planes. It's great that you aren't overly concerned with accuracy and are instead focused on getting smooth ellipse shapes. Although accuracy is our end goal it can't really be forced and tends to come with mileage and consistent practice more than anything else. Some of your ellipses in funnels are having some slight issues with tilting off the minor axis. https://drawabox.com/lesson/1/14/notaligned This is something you should always start considering when drawing your ellipses. Your ellipses are off to a great start but there's still room for improvement so keep practicing them during your warmups.
Your rotated box exercise turned out decently. One thing that would have helped you here would have been to just draw this bigger overall. Drawing bigger really helps when dealing with complex spatial problems. You did a good job drawing through your boxes and keeping your gaps narrow and consistent. While the rotations here aren't perfect this was a good effort overall. The more you draw and develop your spatial thinking ability the easier these rotations are to handle.This is a great exercise to come back to after a few lessons to see how much your spatial thinking ability has improved. Your organic perspective exercises are looking pretty good. You seem to be getting comfortable using the ghosting method and drawing from your shoulder for confident linework which is great. Your box constructions are fairly solid for the most part and I can see you are developing a sense for how box lines converge to vps. There are still some wonky convergences here and there so the 250 box challenge will be a great next step for you.
Overall this was a really solid submission that showed a nice deal of growth. Your line confidence and ellipses are both coming along nicely. I think you are understanding most of the concepts these lessons are trying to convey quite well. I'm going to mark this as complete and good luck with the 250 box challenge. Keep up the good work!
Next Steps:
The 250 Box Challenge
This critique marks this lesson as complete.
##### 11:02 PM, Monday February 20th 2023
That's great, thanks for the feedback.
The recommendation below is an advertisement. Most of the links here are part of Amazon's affiliate program (unless otherwise stated), which helps support this website. It's also more than that - it's a hand-picked recommendation of something I've used myself. If you're interested, here is a full list.
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Moreover, due to the challenge of its use, it teaches you a lot about the nuances of one's stroke. These are the kinds of skills that one can carry over to standard felt tip pens, as well as to digital media. Really great for doodling and just enjoying yourself. | 1,083 | 4,971 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-22 | latest | en | 0.973114 |
https://numberworld.info/3414220222304 | 1,702,300,377,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679511159.96/warc/CC-MAIN-20231211112008-20231211142008-00609.warc.gz | 468,602,186 | 4,132 | # Number 3414220222304
### Properties of number 3414220222304
Cross Sum:
Factorization:
2 * 2 * 2 * 2 * 2 * 3209 * 33248483
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
31aef622f60
Base 32:
33bnm4br0
sin(3414220222304)
-0.054245068915131
cos(3414220222304)
0.99852765234539
tan(3414220222304)
-0.054325054281388
ln(3414220222304)
28.858970243875
lg(3414220222304)
12.533291530325
sqrt(3414220222304)
1847760.867186
Square(3414220222304)
1.165689972639E+25
### Number Look Up
Look Up
3414220222304 which is pronounced (three trillion four hundred fourteen billion two hundred twenty million two hundred twenty-two thousand three hundred four) is a impressive figure. The cross sum of 3414220222304 is 29. If you factorisate the figure 3414220222304 you will get these result 2 * 2 * 2 * 2 * 2 * 3209 * 33248483. The number 3414220222304 has 24 divisors ( 1, 2, 4, 8, 16, 32, 3209, 6418, 12836, 25672, 51344, 102688, 33248483, 66496966, 132993932, 265987864, 531975728, 1063951456, 106694381947, 213388763894, 426777527788, 853555055576, 1707110111152, 3414220222304 ) whith a sum of 6723840919320. The number 3414220222304 is not a prime number. The number 3414220222304 is not a fibonacci number. 3414220222304 is not a Bell Number. The number 3414220222304 is not a Catalan Number. The convertion of 3414220222304 to base 2 (Binary) is 110001101011101111011000100010111101100000. The convertion of 3414220222304 to base 3 (Ternary) is 110002101200212200210020002. The convertion of 3414220222304 to base 4 (Quaternary) is 301223233120202331200. The convertion of 3414220222304 to base 5 (Quintal) is 421414310334103204. The convertion of 3414220222304 to base 8 (Octal) is 61535730427540. The convertion of 3414220222304 to base 16 (Hexadecimal) is 31aef622f60. The convertion of 3414220222304 to base 32 is 33bnm4br0. The sine of the number 3414220222304 is -0.054245068915131. The cosine of the figure 3414220222304 is 0.99852765234539. The tangent of the figure 3414220222304 is -0.054325054281388. The square root of 3414220222304 is 1847760.867186.
If you square 3414220222304 you will get the following result 1.165689972639E+25. The natural logarithm of 3414220222304 is 28.858970243875 and the decimal logarithm is 12.533291530325. I hope that you now know that 3414220222304 is very unique figure! | 894 | 2,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2023-50 | latest | en | 0.532049 |
http://mathhelpforum.com/trigonometry/97679-word-problem-angle-elevation-depression-help.html | 1,498,580,123,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321458.47/warc/CC-MAIN-20170627152510-20170627172510-00510.warc.gz | 247,487,475 | 11,458 | # Thread: Word problem angle of elevation/depression. HELP!
1. ## Word problem angle of elevation/depression. HELP!
From the top of the building, the angle of elevation of the top of a 120 ft tower is 10 degrees. From a window 6ft below the top of the building, the angle of depression of the base of the tower is 30 degrees. Determine the height of the building and the distance between the tower and the building.
2. Originally Posted by reiward
From the top of the building, the angle of elevation of the top of a 120 ft tower is 10 degrees. From a window 6ft below the top of the building, the angle of depression of the base of the tower is 30 degrees. Determine the height of the building and the distance between the tower and the building.
in the picture I attached
$Y+X+6=120$....(1)
$\tan 10^0=\frac{X}{D}$....(2)
$\tan 30^0=\frac{Y}{D}$.....(3)
find X ,Y with respect to D
From (2)
$D\tan 10^0 = X$
From (3)
$D\tan 30^0= Y$
sub X,Y values in (1)
$D\tan 10^0+D\tan 30^0+6 =120$
Edited: Fix the error
3. Originally Posted by Amer
in the picture that I attached you can see that
$x+120=6+y=\text{the building height}$
$\tan 10^0 = \frac{x}{D}$ .....(1)
$\tan 30^0 = \frac{y}{D}$.......(2)
find D value with respect to x,y
from (1)
$x=D\tan 10^0$
from (2)
$y=D\tan 30^0$ sub D values in $x+120=6+y$
$D\tan 30^0 + 120 = D\tan 10^0 + 6$ equation with one variable you can find it
Attachment 12456
Amer,
I read it as though the tower was taller than the building and the angle of elevation from the top of the building to the top of the tower is 10 degrees. Your figure shows the angle of depression to the top of the tower is 10 degrees.
4. Originally Posted by masters
Amer,
I read it as though the tower was taller than the building and the angle of elevation from the top of the building to the top of the tower is 10 degrees. Your figure shows the angle of depression to the top of the tower is 10 degrees.
Thanks Master
I forgot to see the meaning of elevation in the dictionary sorry I fixed the error
5. Originally Posted by Amer
in the picture I attached
$Y+X+6=120$....(1)
$\tan 10^0=\frac{X}{D}$....(2)
$\tan 30^0=\frac{Y}{D}$.....(3)
find X ,Y with respect to D
From (2)
$D\tan 10^0 = X$
From (3)
$D\tan 30^0= Y$
sub X,Y values in (1)
$D\tan 10^0+D\tan 30^0+6 =120$
Edited: Fix the error
thank you so much! | 729 | 2,356 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 19, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2017-26 | longest | en | 0.909126 |
https://www.animegoodys.com/how-many-cm-is-5-3-feet/ | 1,670,549,594,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711376.47/warc/CC-MAIN-20221209011720-20221209041720-00678.warc.gz | 669,942,704 | 13,349 | # How many cm is 5/3 feet?
How many cm is 5/3 feet? Five feet three inches (5’3″) is equal to 160.02 cm. This is because there are 30.48 cm in one foot.
How many cm is 4 ft 5? 4’5 = 134.62 cm. Convert 4 ft 5 to centimeters. Use the calculator above to calculate between feet and centimeters.
What height is 4 ft in CM? 121.92 cm. Feet to centimeters conversion table
Feet (ft)Centimeters (cm)
1 ft30.48 cm
2 ft60.96 cm
3 ft91.44 cm
4 ft121.92 cm
Is 4ft 11 short for a girl? So, any woman shorter than 4ft 10 inches is typically considered short, and taller than 6 ft is assumed to be tall. In a nutshell, if a woman is less than 5ft 3 inches tall, she is considered short.
## How many cm is 5/3 feet? – Related Questions
### How tall are you if you are 4 11?
The height of the average North American male is 175.5 centimeters, a little over 5 foot 9 inches.
4 feet 0 inches= 121.92 centimeters
4 feet 11 inches= 149.86 centimeters
5 feet 0 inches= 152.40 centimeters
5 feet 1 inches= 154.94 centimeters
5 feet 2 inches= 157.48 centimeters
### What is the height after 4 11?
Height Comparison Charts
Feet, inchesInchesCentimeters
4 feet 10 in58 in147.32 cm
4 feet 11 in59 in149.86 cm
5 feet60 in152.4 cm
5 feet 1 in61 in154.94 cm
### Is 177cm tall?
177 cm = 5’9.69. 177 cm is taller than about 53.4% of men and 96.5% of women in the USA. What is 177cm in feet and inches? Convert 177 centimeters to feet and inches. Use the calculator above to calculate between feet and centimeters.
### What is the height 4’11 in CM?
149.86 cm. Feet to Centimeters (ft to cm) Conversion
Feet + InchesFt + InCentimetres
4 feet 11 inches4′ 11″149.86 cm
5 feet 0 inches5′ 0″152.40 cm
5 feet 1 inches5′ 1″154.94 cm
5 feet 2 inches5′ 2″157.48 cm
We will be happy to hear your thoughts | 592 | 1,780 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2022-49 | latest | en | 0.836915 |
https://www.gradesaver.com/textbooks/math/algebra/introductory-algebra-for-college-students-7th-edition/chapter-5-section-5-5-dividing-polynomials-exercise-set-page-386/55 | 1,539,736,488,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510932.57/warc/CC-MAIN-20181017002502-20181017024002-00200.warc.gz | 977,734,455 | 13,660 | ## Introductory Algebra for College Students (7th Edition)
2x$^3$-x$^2$
Divide the polynomial by the monomial by separating the polynomial into individual terms and dividing each one by the monomial. Then cancel out common factors of the coefficients and subtract the powers of the variables to simplify. $\frac{14x^4-7x^3}{7x}$=$\frac{14x^4}{7x}$-$\frac{7x^3}{7x}$=2x$^{(4-1)}$-1x$^{(3-1)}$=2x$^3$-x$^2$ Check the answer. 7x(2x$^3$-x$^2$)=14x$^{(1+3)}$-7x$^{(1+2)}$=14x$^4$-7x$^3$ | 198 | 482 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-43 | longest | en | 0.667433 |
https://www.usefullinks.org/cat/Field_extensions.html | 1,723,642,319,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641113960.89/warc/CC-MAIN-20240814123926-20240814153926-00070.warc.gz | 810,394,580 | 3,102 | # Category: Field extensions
Algebraic extension
In mathematics, an algebraic extension is a field extension L/K such that every element of the larger field L is algebraic over the smaller field K; that is, if every element of L is a root of a non-z
Field extension
In mathematics, particularly in algebra, a field extension is a pair of fields such that the operations of E are those of F restricted to E. In this case, F is an extension field of E and E is a subfi
Normal extension
In abstract algebra, a normal extension is an algebraic field extension L/K for which every irreducible polynomial over K which has a root in L, splits into linear factors in L. These are one of the c
Dual basis in a field extension
In mathematics, the linear algebra concept of dual basis can be applied in the context of a finite extension L/K, by using the field trace. This requires the property that the field trace TrL/K provid
Tower of fields
In mathematics, a tower of fields is a sequence of field extensions F0 ⊆ F1 ⊆ ... ⊆ Fn ⊆ ... The name comes from such sequences often being written in the form A tower of fields may be finite or infin
Separable extension
In field theory, a branch of algebra, an algebraic field extension is called a separable extension if for every , the minimal polynomial of over F is a separable polynomial (i.e., its formal derivativ
Algebraic closure
In mathematics, particularly abstract algebra, an algebraic closure of a field K is an algebraic extension of K that is algebraically closed. It is one of many closures in mathematics. Using Zorn's le
Galois extension
In mathematics, a Galois extension is an algebraic field extension E/F that is normal and separable; or equivalently, E/F is algebraic, and the field fixed by the automorphism group Aut(E/F) is precis
Abelian extension
In abstract algebra, an abelian extension is a Galois extension whose Galois group is abelian. When the Galois group is also cyclic, the extension is also called a cyclic extension. Going in the other
Simple extension
In field theory, a simple extension is a field extension which is generated by the adjunction of a single element. Simple extensions are well understood and can be completely classified. The primitive
Degree of a field extension
In mathematics, more specifically field theory, the degree of a field extension is a rough measure of the "size" of the field extension. The concept plays an important role in many parts of mathematic | 542 | 2,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-33 | latest | en | 0.934884 |
http://kirjoittaessani.de/tag/zahlen/page/2/ | 1,582,565,035,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145966.48/warc/CC-MAIN-20200224163216-20200224193216-00110.warc.gz | 78,563,449 | 22,101 | ## Von Mandelbrot und schaumigen Äpfeln
Earlier this week, Slashdot published the news that a three-dimensional equivalent of the Mandelbrot set has been found. Visiting Daniel White's page, not only did I find an explanation, but also unprecedented images -- and I knew immediately that I would like to create such images myself.
### Fractured
The class of objects to which the Mandelbrot and the new Mandelbulb belong is called fractals. What are fractals, anyway? Well, a fractals in an object with a dimension that is not an integer. At first glance, it is not quite clear what this means: an object is either one-dimensional, such as a line; or a sheet of paper, which has two dimensions (ignoring its thickness); or, for example, a cube with three dimensions. But two and a half?
If we look at the dimension of a vector space, this is, indeed, impossible: there, one would simply count the (independent) directions that are to be found in a given object, and name there number the object's dimension: if there is just length (such as on a piece of string), the dimension is one. The living room has length, width, and height, thus it is three-dimensional.
But there is a different way: let us take a ball that is large enough to hide the piece of string. For a one-meter string, we would need a ball one meter in diameter. Then, try with a ball half that size: we would need two to cover the string. If we tried with ten-centimeter balls, we would need ten of them, and so on.
That does not work for my living-room, however: I could put the whole room into a large ball of, say, five meters; but I would need many more than just five balls of one meter each -- more than a hundred, in fact. That is, the dimension of an object tells us how quickly the number of balls grows when the balls themselves shrink. For a one-dimensional object, 2 balls of size 1/2 are sufficient, for something three-dimensional, we would need 8, or 23.
Now, let us turn around and use this counting as a definition: a chessboard is two-dimensional precisely because we need 64=82 squares of size 1/8 to cover it.
### Erasing
What would an object need to look like to make a fractional number drop out of this computation? Let us make a little experiment. You only need a sheet of paper, a pencil, an eraser, and maybe a ruler (I did without). First, draw a straight line, like this:
Then, take the eraser to remove the middle third of the line:
Erase the middle third of the two remaining lines as well:
And so on -- in principle, you can continue indefinitely. That is, as long as your eraser is fine enough.
What happens if you try to cover your creation with balls? Start with one that is large enough to cover all those little dashes. If you then try balls a third of its size, you do not need three of them -- instead, two are sufficient: one for the left side, one for the right. You have removed everything in the middle with your eraser, after all. Using balls of diameter 1/3, only 2=30.63 of them are enough to cover the drawing. So you could say your little piece of art has a dimension of 0.63. This is not as weird as it may look at first glance: There is an enormous number of dots, most of which are very close together. Isolated dots would have a dimension of 0; and dots that are infinitely close together form a line of dimension 1. Our result is somewhere inbetween, which should hardly come as a surprise.
The Mandelbrot set is similar: regardless of how far we magnify parts of it, there are ever finer structures appearing.In principle, this is true for the Mandelbulb above; however, the picture has been rendered with simplified equation, so only the largest knobs are visible: it is like throwing away the erasor after the first two steps.
### Appearing soon
During the next few days, I would like to show how to calculate the Mandelbrot set, and the new Mandelbulb. Hopefully, there will also be more pictures, and perhaps a movie or two.
[Edit: Typos]
3 Kommentaredeutsch
## Gesiebt
Zum Tagesabschluß habe ich noch etwas heiteres, nämlich eine Suchanfrage: warum muss man bei primzahlen die vielfachen fon 11 nicht streichen?, will ein unbekannter Besucher wissen.
Da kann ich nur sagen: die Frage ist falsch gestellt. Wenn man mit dem Sieb des Eratosthenes Primzahlen bestimmen will, sind die Vielfacher von elf durchaus zu streichen -- genauso wie die aller anderen Primzahlen. Sind allerdings nur die Primzahlen bis N=100 gefragt, kann man nach der sieben aufhören: es genügt immer, bis zur Wurzel von N (hier also 10) zu prüfen.
Das sieht man so: angenommen, es gibt einen Teiler von nN (nennen wir ihn a), mit a>√N. Dann ist aber auch b = n/aN/a < N/√N = √N, also b<√N, Teiler von n; und so erkennen wir auch dann, daß n nicht prim ist, wenn wir nur bis √N prüfen.
Gute Nacht.
Kommentare deaktiviert für Gesiebt
## Multiplizieren
Today, I would like to present a funny -- if somewhat useless -- mathematical toy.
Originally, I wanted to create some sleak pictures to illustrate a graphic method for multiplying decimal numbers -- but someone has been faster than me, posting a video at Youtube.
Therefore, I leave my pencil alone. I shall merely add that this method works in exactly the same way as the usual pencil-and-paper way does.
Kommentare deaktiviert für Multiplizierendeutsch
## Verschlüsselte Botschaften
Bei der Bewertung von Rudolf Kippenhahns Verschlüsselten Botschaften finde ich mich in einem Dilemma wieder; denn die größte Schwäche des Buchs ist gleichzeitig seine größte Stärke: der Autor verzichtet beinahe völlig auf Mathematik. Daher kann er moderne Schlüsselverfahren lediglich streifen. Wer hier einen Einblick gewinnen will, ist etwa mit Schneiers Applied Cryptography besser bedient.
Dem mathematischen Laien allerdings hat Kippenhahns Werk einiges zu bieten. Klassische Verfahren wie der Caesar-Chiffre werden recht ausführlich erklärt, und auch einige weniger bekannte Methoden wie etwa eine Transposition mittels einer drehbaren Schablone finden Erwähnung.
Einen recht breiten Raum nimmt die erste Hälfte des 20. Jahrhunderts ein, hier besonders die Enigma. Neben der legitimen Ver- und Entschlüsselung stellt der Autor auch die Schwächen der Verfahren und die daraus resultierenden Angriffspunkte vor.
Die theoretischen Abschnitte sind immer wieder unterbrochen von plastisch erzählten historischen Szenen; so bleibt die Lektüre auch dann abwechslungsreich, wenn die Theorie wegen mathematischer Vorkenntnisse einige Längen aufweist, oder aber mangels solcher Kenntnisse sehr trocken wird.
Zum Schluß werden auch moderne Verfahren kurz angerissen, und man erfährt etwa, wie Public-Key-Kryptographie prinzipiell funktioniert, oder die Geheimzahl einer Scheckkarte abgesichert wird.
Unter dem Strich ist das Buch eine gute Einführung in die Kryptologie; wer sich mit den Verfahren bereits etwas auskennt, der hat immerhin eine kurzweilige Lektüre, bei der er auch noch ein paar interessante historische Details lernen mag.
[Diese Rezension ist auch bei Lovelybooks verfügbar (derzeit leider nur für dort registrierte Mitglieder]
Kommentare deaktiviert für Verschlüsselte Botschaften
## Kugelregel
What would we do without referrers? I do not really know, but bloggers would have to do without a lot of funny search requests.
A few days ago, someon was looking for a Rechenregel wie man erkennt durch welche Zahlen eine Zahl teilbar ist. That roughly translates as rule how to recognize by which numbers a number is divisible. Anyway, I am happy to provide a few. All rules are equivalencies, that is the given number is divisible if the rule is fulfilled; otherwise, it is not divisible.
1. every number is divisible by one 🙂
2. the final digit is divisible by two
3. the cross total (sum of digits) is divisible by three
4. the number comprised of the final two digits is divisible by four
5. the final digit is zero or five
6. both rules 2 and 3 hold
7. double the final digit and subtract it from the rest of the number; the result is divisible by seven; e.g. 203: 20-2*3=14 is divisible
8. the number comprised of the final three digits is divisible by eight
9. the cross total is divisible by nine
10. the final digit is zero
11. the alternating sum of digits is divisible by eleven, e.g. 135795: 1-3+5-7+9-5=0 is divisible
12. rules 3 and 4 hold
13. quadruple the final digit and add it to rest of the number; the result is divisible by thirteen; e.g. 403: 40+4*3=52; it is useful to apply the rule again: 5+4*2=13 isdivisible, so 52 is, and therefore 403 as well.
Well, that should suffice for now.
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## What makes you mathematick?
Fermats Letzter Satz ist ein gutes Buch. Es fällt mir schwer, einen kritisierenswerten Punkt zu finden, und deshalb nehme ich einfach diesen: Der Titel ist irreführend. Nicht wirklich schlimm irreführend, aber die berühmte Fermat'sche Vermutung, daß nämlich die Gleichung an+bn=cn für n>2 keine positiven Lösungen habe, ist eher der Aufhänger als das Hauptthema des Buches.
Singh liefert hier einen Streifzug durch die Mathematik, der im klassischen Griechenland beginnt und bis in die Gegenwart reicht. Dabei hat er beileibe kein mathematisches Lehrbuch geschrieben, nicht einmal ein Mathematikbuch für Laien. Vielmehr zeigt er, wie Mathematiker denken und arbeiten, wie sich die Arbeitsweise im Laufe der Zeit geändert hat, und was sie antreibt.
Dabei nehmen Fermat und Wiles (der den Satz bewiesen hat) natürlich einen größeren Raum ein, doch Singh betrachtet auch noch eine Vielzahl weiterer Persönlichkeiten. Auf mathematische Details verzichtet er hingegen fast völlig. Das betrifft nicht nur Wiles' Beweis (der immerhin über 100 Seiten in einer Fachzeitschrift füllt), sondern fast alle Beweise und sogar die meisten Sätze. Daher bleibt das Buch auch für mathematische Laien lesbar, wenngleich sich Leser mit entsprechender Vorbildung teils mehr Details wünschen mögen. Für letztere gibt es aber ein Literaturverzeichnis, das nicht nur auf weitere populärwissenschaftliche Werke, sondern auch auf mathematische Fachbeiträge verweist.
1 Kommentar
## Schummeln bei den Hausaufgaben
…das war vor einiger Zeit recht beliebt bei den Suchanfragen, die hierher führen. Im Moment steht 1000 als summe aufeinanderfolgender zahlen hoch im Kurs.
Also gut: einmal Schummeln für unfähige Anfänger. Ihr dürft jetzt alle hier abschreiben und Euch dann wundern, wenn Ihr die nächste Klausur verhaut.
#### Summa cum
Wir wollen also schreiben können k+(k+1)+…+l = 1000.
Vom ollen Gauß wissen wir ja, daß die Summe der Zahlen von 1 bis n gegeben ist als n(n+1)/2. Wie sieht das also aus, wenn wir von k+1 bis l summieren? Ganz einfach: wir rechnen erstmal von Eins bis l, ziehen dann aber die zu kleinen Zahlen wieder ab. Also:
1000 = l(l+1)/2 - k(k+1)/2.
Anders formuliert könnten wir auch l=k+n setzen und sagen, daß wir n aufeinanderfolgende Zahlen ab k +1 summieren wollen. Dann haben wir:
1000 = (k+n)·(k+n+1)/2 - k(k+1)/2
Multiplizieren wir das ganze mit 2 und vereinfachen ein bißchen, dann haben wir diese schöne Formel:
2000 = n(2k+n+1)
Das bedeutet: um 1000 als Summe aufeinanderfolgender Zahlen schreiben zu können, müssen wir 2000 als Produkt zweier Zahlen, nämlich n und 2k+n+1schreiben. Dabei kann n gerade oder ungerade sein; da 2k+1 aber immer ungerade ist, wird 2k+n+1 genau dann gerade sein, wenn n ungerade ist.
Jetzt zerlegen wir 2000 in seine Primfaktoren und überlegen, auf wie viele Arten wir daraus einen geraden und einen ungeraden Faktor basteln können. 2000=24·53; die Faktoren 2 müssen immer zusammen vorkommen, sonst wären ja beide Teile gerade, die anderen können wir verteilen:
2000 = 16·125 = 80·25 = 400·5 = 2000·1
Das war's auch schon, mehr als diese Varianten gibt es nicht. Jetzt müssen wir nur noch feststellen, ob jedem Produkt auch eine Darstellung der Zahl 1000 entspricht, und wie diese jeweils aussieht.
#### Aufgezählt
Die Zahl der Summanden, die in der Formel mit n angegeben ist, kann also höchstens die Werte 1, 5, 16, 25, 80, 125, 400, 2000 annehmen. Die probieren wir jetzt der Reihe nach durch:
Ist n=1, dann folgt 2k+n+1=2000, also k=999. Das ist das, was die Mathematiker gerne als triviale Lösung bezeichnen: eine Möglichkeit, 1000 als Summe aufeinanderfolgender Zahlen zu schreiben, ist die einzige Zahl 1000 — eine Folge aus einer Zahl.
n=5 ist schon interessanter: 2k+n+1=400, also k=197, das heißt die Summe hat fünf Glieder und startet oberhalb von 197: 1000=198+199+200+201+202.
n=16 liefert 2k+n+1=125, also k=54. Damit ist 1000 = 55+56+…+69+70.
Aus n=25 folgt 2k+n+1=80 und k=27: 1000 = 28+29+…+52.
Dann haben wir noch n=80, also 2k+n+1=125, k=22: 1000 = 23+24+…+102.
Bei n=125 ist dann Schluß: dann wäre 2k+n+1=80, und diese Gleichung läßt sich durch kein positives k mehr erfüllen.
#### Zum Schluß
Auf eine Sache sollte ich noch eingehen: warum dürfen wir hier eigentlich ganz munter räsonieren und Teiler aufschreiben, nur um hinterher zu sagen: ätsch, 125 (und die größeren) waren nur ein Scherz? Wäre es nicht genauso möglich, daß unsere Methode nicht nur blinde Passagiere wie die 125 ausspuckt, sondern auch richtige Lösungen (meinetwegen n=17) verschluckt?
Nein, das geht zum Glück nicht: mathematische Argumente sind, wenn sie sauber verwendet werden, exakt. Daß wir trotzdem zunächst einige Zahlen bekommen haben, die wir dann doch wieder verwerfen mußten, liegt an dem Unterschied zwischen einer Folgerung und einer Äquivalenz: Wenn 1000 sich als die Summe der Zahlen k+1 bis k+n schreiben läßt, dann gelten die oben gemachten Überlegungen, und dann muß schließlich n ein Teiler von 2000 sein. Das heißt aber nicht, daß sich umgekehrt auch jede Zahl n, die Teiler von 2000 ist, automatisch in eine solche Summe verwandeln läßt. Dazu muß noch mehr gelten, n darf nämlich nicht nur irgendein Teiler von 2000 sein, sondern muß gerade komplementär zu 2k+n+1 sein. Wenn das nicht erfüllbar ist (z.B. für n=125), dann fallen die restlichen Überlegungen flach.
So, jetzt ist aber genug geschummelt.
[Edit: diverse Typos]
Kommentare deaktiviert für Schummeln bei den Hausaufgaben
## Wie viele?
Referrer sind doch eine tolle Sache — so bekommt man erst mit, was die Be-sucher denn so alles suchen. Einer wollte neulich alle primzahlen aufgezählt haben.
Damit kann ich leider nicht dienen, denn es gibt eine ganze Menge Primzahlen — unendlich viele, um genau zu sein.
#### Viel
Geht es noch genauer? Unendlich ist ja nicht gleich unendlich. Nimmt man zum Beispiel nur die geraden Zahlen (2, 4, 6, 8, 10, …), so gibt es davon offenbar[1] weniger als von den natürlichen Zahlen (1, 2, 3, 4, …). Von den "runden" Zahlen (1, 10, 100, 1000, …) gibt es dann noch weniger. Wie mag es wohl mit den Primzahlen aussehen?
Es gibt so viele Primzahlen, daß
#### Häh?
OK, der Reihe nach; zuerst stellt sich die Frage, was da eigentlich steht. Das ist relativ einfach. Wenn wir alle Primzahlen nehmen (pϵP), zu jeder den Kehrwert bilden (1/p), und diese Kehrwerte dann aufsummieren (Ʃ), dann wird das Ergebnis unendlich groß. Auf den ersten Blick ist das vielleicht nicht verwunderlich; immerhin addieren wir unendlich viele Zahlen, die alle positiv sind.
Wenn wir aber statt 1/p die reziproken Zweierpotenzen addieren, dann ist das Ergebnis durchaus endlich: 1/2+1/4+1/8+…=1. Warum? Naja, 1/2 ist die Hälfte von 1; 1/4 ist die Hälfte von dem, was dann noch zu 1 fehlt, und so weiter:
Wenn die Zahlen, deren Kehrwerte ich addiere, also schnell genug größer werden, dann kann die Reihe[3] konvergieren, also eine endliche Zahl als Ergebnis haben. Daß die Reihe der reziproken Primzahlen divergiert, bedeutet also, daß sie nicht sehr schnell größer werden — daß es also recht viele von ihnen gibt.
#### Langsam
Was heißt nicht sehr schnell? Nun, die Reihe der reziproken natürlichen Zahlen, 1+1/2+1/3+1/4+…, divergiert, i.e. die natürlichen Zahlen wachsen in diesem Sinne langsam. Die Quadratzahlen wachsen dagegen schnell, 1+1/4+1/9+1/16+… konvergiert nämlich[4] — nimmt man stattdessen dritte oder vierte Potenzen, so wachsen die Zahlen natürlich schneller, und die Reihen konvergieren auch. Man kann zeigen, daß Ʃ1/nr immer konvergiert, solange r nur größer als 1 ist. Es reicht, r=1,001 (oder noch kleiner) zu wählen. Das heißt, daß die Folge der Primzahlen tatsächlich sehr langsam wachsen muß, damit die Reihe divergiert.
#### An's Eingemachte
Behaupten kann man viel, wenn der Tag lang ist. Mathematiker wollen aber immer Beweise dafür sehen. Der, den ich hier vorführen möchte, stammt von dem berühmten Ungarn Erdős. Ich folge dabei dem Buch von Aigner und Ziegler[5].
Der Beweis ist wieder ein Widerspruchsbeweis, i.e. wir nehmen an, daß die Behauptung gerade nicht stimmt, rechnen ein bißchen herum und finden dabei einen Widerspruch. Daraus folgt dann, daß die Annahme falsch war.
Nehmen wir also an, daß die Reihe konvergiert. Wenn wir nun die ersten Summanden überspringen, dann wird die Gesamtsumme kleiner, und wenn wir genügend Summanden überspringen, können wir die Gesamtsumme kleiner als 1/2 machen.[6]
Wählen wir also k genügend groß, und denken wir uns noch irgendeine Zahl N aus, mit der wir die ganze Ungleichung multiplizieren, dann gilt:
Wenn wir jetzt alle Zahlen betrachten, die nicht größer sind als N, dann gibt es solche, die sich durch eine Primzahl teilen lassen, die in unserer verkürzten Reihe vorhanden ist; ihre Anzahl wollen wir b nennen. Und es gibt solche, die nur Primteiler besitzen, die unter den ersten k Primzahlen sind und deswegen nicht in der verkürzten Reihe vorkommen; ihre Anzahl sei s. Weil alle Zahlen entweder zu der einen oder anderen Kategorie gehören müssen, ist s+b=N.
Wie groß ist b? Nun, für eine Primzahl pi gibt es Zahlen, die durch sie teilbar sind. Wenn wir diesen Ausdruck über alle Primzahlen summieren, dann liegen wir vielleicht zu hoch, aber sicher nicht zu niedrig[7].
Wenn wir also summieren, dann steht wieder unsere Ungleichung von oben da, und es gilt: b < N/2.
Wie sieht es für s aus? Das geht anders, und zwar so: Wir schreiben jede Zahl n, die zu s zählt, als n=a·q2. Dabei ist a das Produkt aller Primfaktoren, die in n nur einmal vorkommen. Zum Beispiel ist 150=2·3·5·5, also a=2·3=6 und q=5.
Um die Frage zu klären, wie groß s ist, müssen wir also überlegen, wie viele verschiedene a und q es gibt. Das ist einfach: in a kommt jede Primzahl einmal oder keinmal vor, das sind zwei Möglichkeiten pro Primzahl. Da für s nur die ersten k Primzahlen in Frage kommen, sind das 2·2·2…=2k Möglichkeiten. Bei q machen wir es uns ganz einfach: q2n, und nN, also q ≤ √N.
Damit ist s ≤ 2kN.
#### Widerworte
Am Anfang hatten wir uns ja ein N ausgedacht, aber alle Überlegungen waren völlig allgemein. Wir können uns jetzt also noch umentscheiden und zum Beispiel N=22k+2 nehmen. Dann ist s ≤ s2k+1 (Einsetzen in die letzte Gleichung vom vorigen Absatz), also kleiner als N/2.
Weil b aber auch kleiner als N/2 ist, muß s+b kleiner als N sein.
Am Anfang hatten wir aber gesehen, daß s+b gerade gleich N ist. Beides gleichzeitig geht nicht, und da haben wir unseren Widerspruch.
#### Ende
Also, lieber unbekannter Google-User: es gibt nicht nur unendlich viele, sondern sogar verdammt viele Primzahlen, und die kann ich Dir unmöglich alle aufzählen.
#### Noten
[1] Ganz so einfach ist die Sache eigentlich nicht, dann auf eine sehr intuitive[2] Art gibt es gleich viele gerade und natürliche Zahlen. Das hört sich jetzt seltsam an, weil ja jede gerade Zahl eine natürliche ist, es aber auch natürliche Zahlen gibt, die nicht gerade sind — die ungeraden eben. Man kann aber von unendlichen Mengen durchaus einige Elemente wegnehmen, ohne daß die Menge kleiner würde.
In der Tat läßt sich diese Eigenschaft als Definition von unendlich verwenden.
[2] Vielleicht schreibe ich dazu noch einen Eintrag.
[3] Eine Reihe ist, vereinfacht gesagt, eine Summe unendlich vieler Zahlen.
[4] Und zwar gegen π2/6.
[5] Dieses Buch, das in Anlehnung an eine Idee von Erdős Das BUCH der Beweise heißt, kann man gar nicht genug loben. Deswegen wandert es jetzt erstmal in meine Sidebar.
[6] Das ist eine Eigenschaft von Grenzwerten. Letztlich ist der Grund der, daß die Reihe sich an ihr Ergebnis heranschleicht.
[7] Zu hoch deshalb, weil wir einiges doppelt zählen: 6 ist zum Beispiel durch 2 und 3 teilbar, und wir würden sie einmal bei den Vielfachen von 2, einmal bei denen von 3 zählen.
Kommentare deaktiviert für Wie viele?
## Bunte Karten III
Es wird Zeit, sich noch einmal mit Landkarten zu beschäftigen. Wir haben bereits gesehen, daß ein Tuschekasten für Kartographen mindestens vier Farben enthalten muß. Außerdem braucht man dann, wenn für jedes Land eine Liste von Farben gebildet werden soll, höchstens mehr, niemals aber weniger Farben.
Noch offen ist aber die Frage, ob ich einen Tuschekasten zusammenstellen kann, mit dem jeder Kartograph für alle Karten auskommt.
#### Behauptung
Ich behaupte: ja, das geht. Fünf Farben reichen aus.
Mehr noch: es genügt sogar, wenn für jedes Land eine Liste mit fünf Farben zur Verfügung steht. (Wir hatten ja gesehen, daß das ein schwierigeres Problem ist.)
#### Induziert
Wie beweist man das am besten? Das Verfahren der vollständigen Induktion bietet sich hier an: Eine Karte mit nicht mehr als fünf Ländern kann ich natürlich immer mit fünf Farben zeichnen (Induktionsanfang). Wenn ich jetzt aus einer Karte mit n Ländern auf geschickte Weise ein Land eliminieren könnte, wäre ich schon am Ziel. Aber wie stellt man das an?
Als erstes greifen wir tief in die Trickkiste und fördern etwas zutage, das sich Warum einfach, wenn's auch kompliziert geht? nennt: wir beweisen schlichtweg etwas, das schwieriger (stärker) ist, und machen uns die Arbeit dadurch einfacher. Das hört sich seltsam an; der Grund liegt aber darin, daß wir uns sozusagen auf der Leiter von Sprosse zu Sprosse hangeln. Wenn wir alle Sprossen etwas höher einbauen, müssen wir zwar mit jedem Schritt etwas höher hinauf als bei unserer Originalleiter, aber wir starten ja auch jeden Schritt von einer höheren Sprosse aus.
Wirklich schwieriger wird nur die erste Sprosse; die ist aber bei Induktionsbeweisen kein wirkliches Problem. Zugegeben, ein bißchen hinkt die Analogie schon: wenn ich die Behauptung stärker mache, so kann der Beweis tatsächlich gleich schwer bleiben (wie bei der Leiter), er kann aber auch schwieriger oder einfacher werden. Hier wird er erfreulicherweise einfacher.
#### Weiter im Text
Weiter wollen wir annehmen, daß es höchstens Drei-Länder-Ecke gibt — aber keine Vier-, Fünf-, oder Noch-mehr-Länder-Ecke.
Wenn wir es doch einmal mit einer solchen Karte zu tun haben sollten, wollen wir einfach die Grenzen ein bißchen verschieben. Das ist nicht schlimm, weil das Problem dadurch nur schwieriger wird, die neue Lösung also auf jeden Fall auch Lösung des ursprünglichen Problems ist.
#### Behaupten kann man viel
Sehen wir uns eine Karte an, dann gibt es dort immer einen Ring von äußeren Ländern; hier sind es z.B. Washington, Oregon, Kalifornien, Arizona, New Mexico, Texas und so weiter.
Seien nun zwei benachbarte Länder dieses Rings bereits eingefärbt; jedes weitere Land auf dem Ring stelle eine Liste mit (mindestens) drei Farben bereit; jedes Land aus dem Inneren stelle eine Liste mit mindestens fünf Farben bereit.
Dann — so unsere Behauptung — können wir die gesamte Karte unter Benutzung dieser Listen einfärben.
#### Stärke
Diese Behauptung hat mit unserem ursprünglichen Ziel — der Einfärbung der Karte unter Benutzung von Fünfer-Farblisten — nicht mehr viel zu tun. Sie ist aber stärker: Wir behaupten jetzt, daß nur die inneren Länder noch Fünfer-Listen bereitstellen sollen; die Länder auf dem Ring liefern nur noch Dreier-Listen, zwei geben ihre Farbe sogar fest vor. Wenn wir die Karte unter diesen Bedingungen färben können, kommen wir mit Fünfer-Listen auf jeden Fall hin.
#### Aller Anfang ist leicht
Wenn nur drei Länder auf der Karte sind, so reichen uns die Farben auf jeden Fall: zweie sind ja schon gefärbt, und das dritte bietet uns eine Liste mit drei Farben an; egal, welche Farben durch die beiden Nachbarn belegt sind, eine bleibt immer frei.
#### Ein kleiner Schritt für einen Kartographen
Erinnern wir uns an die Vorgehensweise beim Induktionsschritt: wir müssen die Behauptung für eine Karte mit n Ländern beweisen, dürfen aber annehmen, sie sei für n-1 Länder, ja sogar für alle möglichen Karten mit weniger als n Ländern bereits bewiesen. Wir müssen also unsere große Karte auf eine (oder zwei) kleinere Karten zurückführen.
Nehmen wir nun an, zwei Länder aus dem äußeren Ring haben eine gemeinsame Grenze, die aber selbst nicht nach außen stößt; der Ring hat dann die Form einer Acht. Zaire und der Sudan sind ein Beispiel.
Jetzt können wir die Karte in zwei Teile zerlegen: die eine enthält Zaire, den Sudan, und alle weiter nördlich gelegenen Länder; die andere enthält ebenfalls Zaire und den Sudan, dazu aber die südlicheren Länder. Für Zaire und den Sudan vergeben wir jetzt je eine Farbe fest (etwa grün und rot); alle anderen Länder auf dem alten Ring haben Listen mit drei Farben, die inneren Länder deren fünf — so war ja die Voraussetzung. Dann gilt das aber auch für jede der beiden neuen "halben" Karten. Da diese kleiner sind, dürfen wir die Behauptung als bereits bewiesen ansehen. Beide kleinen Karten sind also mit den vorgegebenen Farben einfärbbar, und damit auch die gesamte Karte.
Das war es auch schon — mehr hatten wir gar nicht behauptet.
#### Unachtsam
Was nun, wenn der äußere Ring nicht die Form einer Acht hat?
Dann müssen wir die Zahl der Länder eben auf andere Art verringern. Nehmen wir die Karte der USA von oben, dann hat der äußere Ring dort nicht die Form einer Acht.
Angenommen, Washington sei bereits blau gefärbt. Auf dem Ring hat dieser Staat zwei Nachbarn: Idaho und Kalifornien. Angenommen, Idaho sei ebenfalls gefärbt (nach Annahme muß es ein Nachbar Washingtons auf dem Ring sein; es bleiben also nur CA oder ID). Dann entfernen wir jetzt probehalber Kalifornien von der Karte. Dadurch wird Nevada neu in den äußeren Ring aufgenommen — bei einer anderen Karte könnte natürlich auch eine ganze Reihe von Staaten dazukommen.
Nach Annahme hat Kalifornien (mindestens) eine Liste von drei Farben zur Verfügung gestellt; davon müssen mindestens zwei von Blau verschieden sein (sagen wir, Rot und Gelb). Diese beiden Farben streichen wir jetzt von der Farbliste Nevadas. Da die Liste vorher mindestens fünf Farben enthielt, bleiben also jetzt noch drei. Das heißt aber, daß die neue Karte (ohne Kalifornien) alle Voraussetzungen erfüllt, um vorschriftsmäßig gefärbt zu werden. Wenn wir jetzt noch eine Farbe für Kalifornien finden, sind wir fertig. Das ist aber leicht:
Nevada kann weder rot noch gelb sein, da wir diese Farben von der Liste gestrichen haben. Washington ist schon blau, und Arizona kann entweder Rot oder Gelb (oder eine andere Farbe) erhalten, aber nicht beide; deswegen bleibt eine der beiden Farben Rot und Gelb für Kalifornien.
Das war's, die Karte ist gefärbt!
#### Und sonst?
Im Falle der Farblisten ist Fünf tatsächlich die kleinste Zahl, die man finden kann. Margit Voigt hat ein Beispiel gefunden für eine Karte, die sich mit Vierer-Farblisten nicht immer einfärben läßt. Sie hat 238 Länder eingezeichnet.
Wenn es nur um den Tuschekasten geht, reichen dagegen sogar vier Farben. Der Beweis dafür ist allerdings leider viel zu lang und kompliziert für dieses Blog.
Der obige Beweis folgt übrigens dem Buch von Martin Aigner und Günther Ziegler, in dem sich — etwa auf Vordiplomniveau — eine ganze Reihe schöner Beweise findet.
Kommentare deaktiviert für Bunte Karten III
## Four and Twenty
Der Spiegel berichtet — nicht zum ersten Mal — über einen Bochumer Mathematiker, der die deutschen Zahlwörter ändern will: Einundzwanzig sei unlogisch, er möchte deshalb zwanzigeins einführen.
Der Artikel liest sich stellenweise, als ob der Untergang des Abendlandes oder doch wenigstens ein schlechtes Abschneiden bei der PISA-Studie drohte, wenn diese Änderung nicht durchgeführt wird. Ich habe da ein ganz anderes Problem: wo haben sie in dem Artikel nur die Smilies versteckt?
Kommentare deaktiviert für Four and Twenty | 7,883 | 28,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2020-10 | latest | en | 0.960979 |
balisesgeiger.fr | 1,619,113,952,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039594341.91/warc/CC-MAIN-20210422160833-20210422190833-00358.warc.gz | 9,000,107 | 11,610 | ## 1. Introduction
We are going to analyse some features of screens.
## 2. Screen analysis.
For the questions below, you’ll answer on this document: response document
Look at the picture of an Iphone
1. What can you see ? Give the definition of a pixel (you can find this information on the web)
Open the file « couleurs_ecran.swf » SWFOpener software.
Observe how adjusting the intensity of the 3 red, green and blue sub-pixels can generate a multitude of colors.
2. How many different colors can be generated knowing that each sub-pixel (red, green or blue) has 256 possible shades?
By using the following site https://www.w3schools.com/colors/colors_rgb.asp or with the paint color chart ( see below):
Answer the questions *on your own document* and check the responses with this questionnaire below.
1. What are the settings of RGB values to get the white color?:
1. rgb(253, 253, 253)
2. rgb(255, 255, 255)
3. rgb(03, 0, 0)
2. What are the settings of RGB values to get the color black?
1. rgb(255, 255, 255)
2. rgb(03, 0, 0)
3. rgb(0, 0, 0)
3. What are the settings of RGB values to get the gray color?
1. rgb(128,128,128)
2. rgb(192,192,192)
3. rgb(220,220,220)
4. What are the settings of RGB values to get the pink color?
1. rgb(255, 0, 255)
2. rgb(128, 0, 255)
3. rgb(0, 128, 255)
5. What are the settings of RGB values to get the yellow color?
1. rgb(255, 0, 255)
2. rgb(128, 0, 255)
3. rgb(255, 255, 0)
## 3. Repartition of pixels on a screen.
When observing screens, we can see differences in the arrangement or shape of pixels (and sub-pixels), depending on the model of Smartphone observed.
1. Castagnetto, académie de Strasbourg image::images/pixel.png[align="center"]
There are two important concepts to Know (very often confused): definition and resolution
• The definition of a screen is the number of pixels that make up the screen. Please note, we are not talking about the red, green, blue sub-pixels here, but the global pixel created by the combination of these three sub-pixels. This definition is for example given in the form 1920x1080.
• Resolution of a screen: it is generally expressed in dpi (dots per inch) or ppi (pixels per inch) or ppp (pixels per inch in french): It measures the number of available pixels over a length of 1 inch (approximately 2, 54 cm). Higher is the number, smaller the pixel size is, and more precise the image will appear.
Le format des écrans est un ratio (par exemple 4/3 ou encore 16/9) qui représente le rapport entre la largeur et la hauteur de l’écran: si l’écran a pour dimensions 548mm (largeur) x 308mm (hauteur). Le format est alors de 548 / 308 = 16/9. Les formats les plus utilisés sont le 16/9 puis le 16/10, suivit de l’ancien 4/3 (en général pour les écrans cathodiques). Voici une image de comparaison du 16/9 avec le 4/3:
`CONTINU TO RESPOND ON YOUR OWN DOCUMENT.`
1. The terms "HD", "Full HD" and "4K" are standards for screen definitions. Find the values of these definitions.
2. Determine the definition of your phone screen. You can find this information in the specifications of your phone.
3. Measure the width of your phone screen, and convert this measurement to inches.
4. Calculate the resolution of your screen (expressed in pixels per inch). You can compare your calculation with the information found in your phone specifications. You can also check your calculations through the site http://www.pxcalc.com/.
### 3.1. Practice.
a 4K TV with these spécifications:
Answer the questions on your response document:
Knowing that the screen format is 16/9 and that the screen size is 55 inches (the length of its diagonal)
1. Calculate the width and height of the screen in inches.( you can use the Pythagorean Theorem)
2. Calculate the resolution of this TV screen in dpi.
3. Compare with the resolution of your phone screen calculated before.
## 4. Listenning comprehension.
Watch this video:
## 5. Workshop
by group of 3 pupils at most, you’ll prepare a short oral presentation with some slides.
The slide show will give information about:
1. the technical characteristics of the screen. specially the calculations about definition and resolution.
2. What do you like or dislike about this screen.
3. A quick introduction of the company.
4. Personal experience with different type of screens.
5. Everything you want to add about screens…
## 6. Sources
• Denis Gautier.
• Branch education. | 1,146 | 4,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-17 | latest | en | 0.697925 |
http://www.kylesconverter.com/length/leagues-(land)-to-feet-(u.s.-survey) | 1,627,876,678,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154302.46/warc/CC-MAIN-20210802012641-20210802042641-00044.warc.gz | 63,497,974 | 5,648 | # Convert Leagues (land) to Feet (u.s. Survey)
### Kyle's Converter > Length > Leagues (land) > Leagues (land) to Feet (u.s. Survey)
Leagues (land) (lea) Feet (u.s. Survey) (ft (US survey)) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Feet (u.s. Survey) to Leagues (land)
(or just enter a value in the "to" field)
Please share if you found this tool useful:
Unit Descriptions
1 League (land):
1 league (land) is exactly 3 US Statute miles. In SI units 1 league is 4828.032 meters.
1 Foot (U.S. Survey):
Exactly 1200/3937 meters by definition. In decimal terms approximately 0.304 800 609 601 219 meters. Variation from the common international foot of exactly 0.3048 meters may only be considerable over large survey distances. 1 ft (US survey) = 1200/3937 m
Conversions Table
1 Leagues (land) to Feet (u.s. Survey) = 15839.968370 Leagues (land) to Feet (u.s. Survey) = 1108797.7824
2 Leagues (land) to Feet (u.s. Survey) = 31679.936680 Leagues (land) to Feet (u.s. Survey) = 1267197.4656
3 Leagues (land) to Feet (u.s. Survey) = 47519.90590 Leagues (land) to Feet (u.s. Survey) = 1425597.1488
4 Leagues (land) to Feet (u.s. Survey) = 63359.8733100 Leagues (land) to Feet (u.s. Survey) = 1583996.832
5 Leagues (land) to Feet (u.s. Survey) = 79199.8416200 Leagues (land) to Feet (u.s. Survey) = 3167993.664
6 Leagues (land) to Feet (u.s. Survey) = 95039.8099300 Leagues (land) to Feet (u.s. Survey) = 4751990.496
7 Leagues (land) to Feet (u.s. Survey) = 110879.7782400 Leagues (land) to Feet (u.s. Survey) = 6335987.328
8 Leagues (land) to Feet (u.s. Survey) = 126719.7466500 Leagues (land) to Feet (u.s. Survey) = 7919984.16
9 Leagues (land) to Feet (u.s. Survey) = 142559.7149600 Leagues (land) to Feet (u.s. Survey) = 9503980.992
10 Leagues (land) to Feet (u.s. Survey) = 158399.6832800 Leagues (land) to Feet (u.s. Survey) = 12671974.656
20 Leagues (land) to Feet (u.s. Survey) = 316799.3664900 Leagues (land) to Feet (u.s. Survey) = 14255971.488
30 Leagues (land) to Feet (u.s. Survey) = 475199.04961,000 Leagues (land) to Feet (u.s. Survey) = 15839968.32
40 Leagues (land) to Feet (u.s. Survey) = 633598.732810,000 Leagues (land) to Feet (u.s. Survey) = 158399683.2
50 Leagues (land) to Feet (u.s. Survey) = 791998.416100,000 Leagues (land) to Feet (u.s. Survey) = 1583996832
60 Leagues (land) to Feet (u.s. Survey) = 950398.09921,000,000 Leagues (land) to Feet (u.s. Survey) = 15839968320 | 895 | 2,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-31 | longest | en | 0.44105 |
http://www.codeproject.com/Tips/124990/A-simple-program-to-solve-quadratic-equations-wi | 1,477,250,568,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719397.0/warc/CC-MAIN-20161020183839-00477-ip-10-171-6-4.ec2.internal.warc.gz | 383,513,922 | 22,498 | 12,550,348 members (45,355 online)
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# A simple program to solve quadratic equations with
, 10 Nov 2010 CPOL
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The correct way to solve quadratic equations is none of the above. For the correct algorithm as applied in the following function, seenumerical recipes in C and the Wolfram site at http://mathworld.wolfram.com/QuadraticEquation.html#include bool quadratic(double a, double b,...
The correct way to solve quadratic equations is none of the above. For the correct algorithm as applied in the following function, see
numerical recipes in C and the Wolfram site at http://mathworld.wolfram.com/QuadraticEquation.html
```#include <limits>
bool quadratic(double a, double b, double c, double& x1, double& x2, double y, double z)
{
double delta=0.0e+00, q=0.0e+00, bsign, y1, y2;
const double eps = std::numeric_limits()<double>.epsilon();
// equation is: a*x^2 + b*x + c = 0; a cannot be zero.
if ( fabs(a) <= eps ) { return false; }
delta=b*b-4.0*a*c;
if ( delta<0.0e+00 ) { return false; }
if ( delta>=0.0e+00 && delta <= eps )
{ x1=(-1.0*b/(2.0*a)); x2=x1; return true; }
bsign=1.0e+00;
if ( fabs(b)>1.0e-16 ) { bsign=b/fabs(b); }
else { if ( b<0.0e+00 ) { bsign=-1.0e+00; } }
q=(-0.5)*(b+bsign*sqrt(delta));
// the roots of the equation are:
y1=q/a; y2=c/q;
// find if any of the roots is in the given [y,z] interval
if (( y<=y1 )&&( y1<=z ))
{
x1=y1; x2=y2;
}
else if (( y<=y2 )&&( y2<=z ))
{
x1=y2; x2=y1;
}
else
{
x1=y1; x2=y2;
}
return true;
}
```
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Senior Software Developer in C/C++ and Oracle.
Ex-physicist holding a Ph.D. on x-ray lasers.
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I edited my answser to address your objection. evviva :) Alain Rist16-Nov-10 7:17 Alain Rist 16-Nov-10 7:17
Oh well, thanks Alain, I didn't know the standard evoluted t... federico.strati9-Nov-10 23:07 federico.strati 9-Nov-10 23:07
#include <limits> const double eps = std::numeric_limi... Alain Rist9-Nov-10 22:34 Alain Rist 9-Nov-10 22:34
Oh well, the machine epsilon is around 3x10-8 for single pre... federico.strati9-Nov-10 4:38 federico.strati 9-Nov-10 4:38
Reason for my vote of 2 Aren't the tolerance values arbitrar... YvesDaoust8-Nov-10 23:30 YvesDaoust 8-Nov-10 23:30
The inputs "y" and "z" serve the only purpose of finding a r... federico.strati8-Nov-10 22:23 federico.strati 8-Nov-10 22:23
Can you elaborate on the function of the 2 added inputs y an... bstrack8-Nov-10 21:27 bstrack 8-Nov-10 21:27
Last Visit: 31-Dec-99 18:00 Last Update: 23-Oct-16 5:22 Refresh 1 | 892 | 2,648 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2016-44 | latest | en | 0.644185 |
http://www.shsu.edu/~ldg005/data/Markov.m2 | 1,516,188,260,000,000,000 | text/plain | crawl-data/CC-MAIN-2018-05/segments/1516084886895.18/warc/CC-MAIN-20180117102533-20180117122533-00525.warc.gz | 557,429,473 | 5,862 | ----------------------------------------------------- -- A translation of Luis' Markov package to M2 ----- ------------------------------------------------------ -- Give a graph as a hash table i => descendents -- Make a graph -- Input: a directed acyclic graph in the form of a -- list of lists of children. -- the vertices must be named 1..n, some n. -- ASSUMPTION: we assume that the descendents of vertex -- i are all less than i. This only represents DAGS. -- Output: A hashtable G with keys 1..n, and G#i is the -- the set of all children of the vertex i. -- This routine produces a useful version of a 'graph' -- which we use in routines throughout this package. makeGraph = (g) -> ( h := new MutableHashTable; scan(#g, i -> h#(i+1) = set g#i); new HashTable from h) descendents = (G,v) -> ( -- returns a set of vertices result := G#v; scan(reverse(1..v-1), i -> ( if member(i,result) then result = result + G#i; )); result) nondescendents = (G,v) -> set(1..#G) - descendents(G,v) - set {v} parents = (G,v) -> set select(1..#G, i -> member(v, G#i)) children = (G,v) -> G#v -------------------- -- Graph features -- -------------------- subgraph = (G,vertices) -> ( -- make a graph on 1..#vertices H := new MutableHashTable; scan(keys G, i -> H#i = set{}); scan(#vertices, i -> H#(vertices#i) = set{i+1}); tr := (s) -> if #s === 0 then {} else toList sum apply(toList s, i -> H#i); G' = apply(vertices, i -> tr(G#i)); makeGraph G') H4a = makeGraph {{},{1},{2},{1,3}}; H4b = makeGraph {{},{1},{1},{2,3}}; is4cycle = (G) -> G === H4a or G === H4b -- G should be a graph on 4 elements n4cycles = (G) -> ( x := subsets(toList(1..#G), 4); # select(x, s -> is4cycle subgraph(G,s))) /// G = makeGraph D5s_65 G1 = subgraph(G,{1,2,3,4}) n4cycles G G1 === H4b subgraph(G,{1,2,3,5}) sort apply(#D5s, i -> (G := makeGraph D5s#i; {n4cycles G, i})) select(0..#D5s-1, i -> (n4cycles makeGraph D5s#i > 0)) select(0..#D5s-1, i -> (n4cycles makeGraph D5s#i == 3)) select(0..#D5s-1, i -> #D5s#1>=1 and # parents(makeGraph D5s#i, 2) >= 3) /// -------------------------- -- Statement calculus ---- -------------------------- -- A dependency is a list {A,B,C} -- where A,B,C are (disjoint) subsets of positive integers. -- The meaning is: A is independent of B given C. -- A dependency list is a list of dependencies -- No serious attempt is made to remove redundant dependencies. -- However, we have several very simple routines to remove -- the most obvious redundant elements -- If S and T represent exactly the same dependency, return true. equivStmts = (S,T) -> S#2 === T#2 and set{S#0,S#1} === set{T#0,T#1} removeRedundants = (Ds) -> ( -- Ds is a list of triples of sets {A,B,C} -- test1: returns true if D1 can be removed -- Return a sublist of Ds which removes any -- that test1 declares not necessary. test1 := (D1,D2) -> (D1_2 === D2_2 and ((isSubset(D1_0, D2_0) and isSubset(D1_1, D2_1)) or (isSubset(D1_1, D2_0) and isSubset(D1_0, D2_1)))); -- first remove non-unique elements, if any Ds = apply(Ds, d -> {set{d#0,d#1}, d#2}); Ds = unique Ds; Ds = apply(Ds, d -> append(toList(d#0), d#1)); c := toList select(0..#Ds-1, i -> ( a := Ds_i; D0 := drop(Ds,{i,i}); all(D0, b -> not test1(a,b)))); Ds_c) -- return the set of variables which occur in some -- dependency. The input 'Ds' is a list of {A,B,C} -- dependencies occuring = (Ds) -> sum apply(Ds, c -> sum c) cleardep = (Ds) -> ( -- if there is any element which is in every b_i_2 -- remove it. -- This is used to remove a variable if possible. -- USE IT WITH CAUTION: it will change the -- resulting ideal commons := product apply(Ds, c -> c_2); apply(Ds, c -> {c_0,c_1,c_2 - commons})) -------------------------- -- Bayes ball algorithm -- -------------------------- bayesBall = (A,C,G) -> ( -- A is a set in 1..n (n = #G) -- C is a set in 1..n (the "blocking set") -- G is a DAG -- Returns the subset B of 1..n which is -- independent of A given C. -- The algorithm is the Bayes Ball algorithm, -- as implemented by Luis Garcia, after -- the paper of Ross Schlacter n := #G; zeros := toList((n+1):false); visited := new MutableList from zeros; blocked := new MutableList from zeros; up := new MutableList from zeros; down := new MutableList from zeros; top := new MutableList from zeros; bottom := new MutableList from zeros; vqueue := sort toList A; -- Now initialize vqueue, set blocked scan(vqueue, a -> up#a = true); scan(toList C, c -> blocked#c = true); local pa; local ch; while #vqueue > 0 do ( v := vqueue#-1; vqueue = drop(vqueue,-1); visited#v = true; if not blocked#v and up#v then ( if not top#v then ( top#v = true; pa = toList parents(G,v); scan(pa, i -> up#i = true); vqueue = join(vqueue,pa); ); if not bottom#v then ( bottom#v = true; ch = toList children(G,v); scan(ch, i -> down#i = true); vqueue = join(vqueue,ch); ); ); if down#v then ( if blocked#v and not top#v then ( top#v = true; pa = toList parents(G,v); scan(pa, i -> up#i = true); vqueue = join(vqueue,pa); ); if not blocked#v and not bottom#v then ( bottom#v = true; ch = toList children(G,v); scan(ch, i -> down#i = true); vqueue = join(vqueue,ch); ); ); ); -- while loop set toList select(1..n, i -> not blocked#i and not bottom#i) ) -------------------------- -- Markov relationships -- -------------------------- pairMarkov = (G) -> ( -- given a graph G, returns a list of triples {A,B,C} -- where A,B,C are disjoint sets, and for every vertex v -- and non-descendent w of v, -- {v, w, nondescendents(G,v) - w} flatten apply(1..#G, v -> ( ND := nondescendents(G,v); W := ND - parents(G,v); apply(toList W, w -> {set {v}, set{w}, ND - set{w}})))) localMarkov = (G) -> ( -- Given a graph G, return a list of triples {A,B,C} -- of the form {v, nondescendents - parents, parents} result := {}; scan(1..#G, v -> ( ND := nondescendents(G,v); P := parents(G,v); if #(ND - P) > 0 then result = append(result,{set{v}, ND - P, P}))); result) globalMarkov = (G) -> ( -- Given a graph G, return a complete list of triples {A,B,C} -- so that A and B are d-separated by C (in the graph G). -- If G is large, this should maybe be rewritten so that -- one huge list of subsets is not made all at once n := #G; vertices := toList(1..n); result := {}; AX := subsets vertices; AX = drop(AX,1); -- drop the empty set AX = drop(AX,-1); -- drop the entire set scan(AX, A -> ( A = set A; Acomplement := toList(set vertices - A); CX := subsets Acomplement; CX = drop(CX,-1); -- we don't want C to be the entire complement scan(CX, C -> ( C = set C; B := bayesBall(A,C,G); if #B > 0 then ( B1 := {A,B,C}; if all(result, B2 -> not equivStmts(B1,B2)) then result = append(result, {A,B,C}); ))))); result ) --------------------------- -- Constructing the ring -- --------------------------- probring = d -> ( -- Returns a ring in d1*...*dn variables -- d should be a list of integers (d1, d2, ..., dn) -- where each di >= 1 start := (#d):1; QQ[(symbol p)_start .. (symbol p)_d]) probring1 = (parents, d) -> ( -- Returns a ring with variables -- a_(i1,...,ir) and q_(j2,...,jn), where -- r-1 is the number of parents of 1 -- each j_p ranges over 1..d_p -- each i_p ranges over 1..(d_(parents_p)) -- parents should be a list of indices 1..#d which -- are the parents of vertex 1. -- d should be a sequence of integers (d1, d2, ..., dn) -- where each di >= 1 start := (#d-1):1; lastd := drop(d,1); startpa := (#parents+1):1; lastpa := toSequence(prepend(d#0, apply(parents, i -> d#(i-1)))); QQ[(symbol a)_startpa .. (symbol a)_lastpa, (symbol q)_start .. (symbol q)_lastd]) toring1 = (parents,d,R,R1) -> ( -- R should be probring d -- R1 should be probring(parents,d) -- Constructs a ring map F : R --> R1 -- which is the partial factorization map F := toList apply(((#d):1) .. d, i -> ( pa := toSequence(prepend(i#0, apply(parents, j -> i#(j-1)))); a_pa * q_(drop(i,1)))); map(R1,R,F) ) margin1 = (d,R) -> ( -- R should be probring d. -- construct the ring map which replaces p_(1,i2,...,in) -- with the sum of all p_(j,i2,...,in), all j. F := toList apply(((#d):1) .. d, i -> ( if i#0 > 1 then p_i else ( i0 := drop(i,1); p_i - sum(apply(toList(2..d#0), j -> p_(prepend(j,i0))))))); map(R,R,F)) margin = (v) -> (d,R) -> ( -- R should be probring d. -- construct the ring map which replaces p_(1,i2,...,in) -- with the sum of all p_(j,i2,...,in), all j. v = v-1; F := toList apply(((#d):1) .. d, i -> ( if i#v > 1 then p_i else ( i0 := drop(i,1); p_i - sum(apply(toList(2..d#v), j -> ( newi := join(take(i,v), {j}, take(i,v-#d+1)); print p_newi; p_newi)))))); map(R,R,F)) ------------------------------------------------------- -- Constructing the ideal of a independence relation -- ------------------------------------------------------- cartesian = (L) -> ( if #L == 1 then return toList apply (L#0, e -> singleton e); L0 := L#0; Lrest := drop (L,1); C := cartesian Lrest; flatten apply (L0, s -> apply (C, c -> prepend (s,c)))) possibleValues = (d,A) -> cartesian (toList apply(1..#d, i -> if member(i,A) then toList(1..d#(i-1)) else {0})) prob = (d,s) -> ( L := cartesian toList apply (#d, i -> if s#i === 0 then toList(1..d#i) else {s#i}); sum apply (L, v -> p_v)) indepMatrices = (d,A,B,C) -> ( -- d is a sequence of integers -- A,B,C should be disjoint sets of integers in 1..#d Avals = possibleValues(d,A); Bvals = possibleValues(d,B); Cvals = possibleValues(d,C); apply(Cvals, c -> ( matrix apply(Avals, a -> apply(Bvals, b -> ( e := toSequence(toList a + toList b + toList c); prob(d,e)))))) ) indepIdeal = (d,A,B,C) -> sum apply(indepMatrices(d,A,B,C), m -> minors(2,m)) Ideps = (d, Ds) -> ( R = probring d; sum apply(Ds, D -> indepIdeal(d,D_0,D_1,D_2))) ----------------------------------------------------- -- local Markov ideal localMarkovIdeal = (d,gr) -> ( R = probring d; G := makeGraph gr; D := localMarkov G; D = removeRedundants D; sum apply(D, D0 -> indepIdeal(d, D0_0,D0_1,D0_2))) ---------------------- -- Binary variables -- ---------------------- -- binaryR#0, binaryR#1 are not very useful... binaryD = apply(0..7, d -> toSequence(d:2)); binaryR = apply(binaryD, probring); simplifyDeps = (b) -> ( b = removeRedundants b; b = cleardep b; s := sort toList occuring b; d := #s; h := new MutableHashTable; scan(d, i -> h#(s#i) = i+1); g := b/(c -> ( -- c is a list of three sets {set apply(toList c#0, j -> h#j), set apply(toList c#1, j -> h#j), set apply(toList c#2, j -> h#j)})); R = binaryR#d; use R; g) IBinaryDeps = (b) -> ( -- Given a set of dependencies, compute the -- ideal in the smallest possible ring. -- IE, if any variables don't occur, use a smaller -- size ring. s := sort toList occuring b; d := #s; h := new MutableHashTable; scan(d, i -> h#(s#i) = i+1); g = b/(c -> ( -- c is a list of three sets {set apply(toList c#0, j -> h#j), set apply(toList c#1, j -> h#j), set apply(toList c#2, j -> h#j)})); R = binaryR#d; use R; << "using" << endl; g/print; sum apply(g, D -> indepIdeal(binaryD#d,D_0,D_1,D_2))) localMarkovConditions = (gr) -> ( D := localMarkov makeGraph gr; --simplifyDeps D removeRedundants D ) globalMarkovConditions = (gr) -> ( D := globalMarkov makeGraph gr; --simplifyDeps D removeRedundants D ) localIdeal = (gr) -> ( D := localMarkov makeGraph gr; D = simplifyDeps D; d := #(occuring D); J := sum apply(D, c -> indepIdeal(binaryD#d, c_0,c_1,c_2)); F := margin1(binaryD#d, binaryR#d); F J) globalIdeal = (gr) -> ( D := globalMarkov makeGraph gr; D = simplifyDeps D; d := #(occuring D); J := sum apply(D, c -> indepIdeal(binaryD#d, c_0,c_1,c_2)); F := margin1(binaryD#d, binaryR#d); F J) Iglobal = (D) -> ( -- D is a list of dependencies -- d is the largest variable which occurs d := max apply(D, b -> max join(apply(b, c -> max toList c))); J := sum apply(D, c -> indepIdeal(binaryD#d, c_0,c_1,c_2)); F := margin1(binaryD#d, binaryR#d); F J) /// -- tests and examples of using these routines restart load "markov.m2" -- this file G = makeGraph{{}, {1}, {2}, {1, 2, 3}, {4}} parents(G,1) parents(G,2) parents(G,3) parents(G,4) parents(G,5) descendents(G,5) descendents(G,4) descendents(G,3) nondescendents(G,3) G scan(pairMarkov G, D -> print toString D) scan(localMarkov G, D -> print toString D) scan(globalMarkov G, D -> print toString D) scan(removeRedundants globalMarkov G, D -> print toString D) Ds = removeRedundants globalMarkov G; occuring Ds R = probring(2,3,3,3) numgens R gens R -- Here are useful already created values for -- up through 7 random variables binaryD_5 binaryR_5 R = binaryR_4 -- Marginalizing over vertex 1. -- This often simplifies the Markov ideals. F = margin1((2,2,2,2),R) transpose F.matrix F p_(1,1,1,1) -- The partial factorization map R1 = probring1({2,3},(2,2,2,2)) -- {2,3} is the list of parents of 1. F = toring1({2,3},(2,2,2,2),R,R1) transpose F.matrix -- prob use R gens R prob((2,2,2,2),{0,1,1,0}) -- making the ideals/matrices -- Use the binary 4 variable ring R from above indepMatrices((2,2,2,2),set{1,2},set{3},set{4}) J = indepIdeal((2,2,2,2),set{1,2},set{3},set{4}) (codim J, degree J) -- local Markov ideal, starting with an externally -- constructed graph (ie: a list of children.) -- This defines the ring too, removes redundant -- variables. WARNING: it might remove random -- variables which appear as the dependent in every dependency. -- (This simplifies the ideal). J = localMarkovIdeal((2,2,2,2),{{},{1},{2},{3}}) transpose gens J G = makeGraph{{},{1},{2},{3}} D = localMarkov G J1 = Ideps D R = probring(2,2,3) R = probring(4:2) prob((2,2,2,2),{0,1,0,2}) gens R #oo cartesian{{1,2},{1,2},{1,2,3}} possibleValues((2,2,2,2),set(1,3)) indepMatrices((2,2,2,2),set{1},set{2},set{3,4}) J = indepIdeal((2,2,2,2),set{1},set{2},set{3,4}) -- Test of Bayes Ball G = makeGraph{{}, {1}, {2}, {1, 2, 3}, {4}} Ds = globalMarkov G oo/print; removeRedundants Ds oo/print; bayesBall(set{1},set{2},G) restart load "luis.m2" -- this is THIS file J = localIdeal{{}, {1}, {2}, {1, 2, 3}, {4}}; JG = globalIdeal{{}, {1}, {2}, {1, 2, 3}, {4}}; F = margin1(binaryD#4, R) J = F J JG = F JG; apply(flatten entries gens JG, f -> (<< "." << flush; J : f)); intersect oo; intersect(o35,JG) == J -- yes C = primaryDecomposition o35; transpose gens J transpose gens JG load "facGB.m2" load "primdecomp-tools.m2" JG1 = ideal compress((gens JG) % J); JG1 = flatten entries gens JG1; JG1 = ideal select(JG1, f -> size f === 2); transpose gens JG1 J1 = J : JG1; J1 = J : JG_0; J2 = J1 : JG_1; L = apply(flatten entries gens JG, f -> J : f); info J info JG J1 = J : JG J = localMarkovIdeal((2,2,2,2,2), {{}, {1}, {2}, {1, 2, 3}, {4}}); info J (margin1((2,2,2,2),R)) J transpose gens oo R = probring(2,2,2,2,2) R1 = probring1({2,3},(2,2,2,2,2)) toring1({2,3},(2,2,2,2,2),R,R1) margin1((2,2,2,2,2),R) transpose oo.matrix D1 = {set{1},set{2},set{3,4}} D2 = {set{1},set{2},set{3}} R = probring(2,2,2,2) J1 = indepIdeal((2,2,2,2),D1_0,D1_1,D1_2) J2 = indepIdeal((2,2,2,2),D2_0,D2_1,D2_2) codim J1 codim J2 (gens J2) % J1 (gens J1) % J2 /// | 5,077 | 14,823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2018-05 | latest | en | 0.675751 |
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# Pick-4 for Dummies Week of Friday 01/06/2012
Topic closed. 52 replies. Last post 5 years ago by mjwinsmith.
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Posted: January 7, 2012, 7:42 pm - IP Logged
Universal numbers to look out for tonight: 6013 7615 2701 8677 7111 2935 1648, these are good for ALL states. Numbers like 7111 are a 4-way number so to get a straight hit off of ALL combos play 1117 1171 1711 7111. No matter how that number falls you get a straight hit off of it. The inverse of that number fell today in Ohio's midday game (7717) darn that would have made a lot of you Ohio players happy.
Good luck tonight people.
-Michael
Remember, these numbers may or may not fall in the exact draw order in your state game. Please do your own workouts to determine that exact draw order. If you would like for me to do the workout be advised that I charge a fee for that service. Please contact me direct by private message for more details.
-Michael
"Don't be a Dummy, Make 'Dat Money"
Mjwinsmith's Blog Page: http://blogs.lotterypost.com/mjwinsmith/
"How to Play My Pick-4 System": https://www.lotterypost.com/thread/306450
United States
Member #391
June 8, 2002
16059 Posts
Online
Posted: January 7, 2012, 8:03 pm - IP Logged
Universal numbers to look out for tonight: 6013 7615 2701 8677 7111 2935 1648, these are good for ALL states. Numbers like 7111 are a 4-way number so to get a straight hit off of ALL combos play 1117 1171 1711 7111. No matter how that number falls you get a straight hit off of it. The inverse of that number fell today in Ohio's midday game (7717) darn that would have made a lot of you Ohio players happy.
Good luck tonight people.
-Michael
Take that 8-6-7-7, with the digit "8" in the first position you have only three sets to play to get a straight hit; 8677-8767-8776. Where are all my late night players?
-Michael
"Don't be a Dummy, Make 'Dat Money"
Mjwinsmith's Blog Page: http://blogs.lotterypost.com/mjwinsmith/
"How to Play My Pick-4 System": https://www.lotterypost.com/thread/306450
United States
Member #391
June 8, 2002
16059 Posts
Online
Posted: January 7, 2012, 8:11 pm - IP Logged
Universal numbers to look out for tonight: 6013 7615 2701 8677 7111 2935 1648, these are good for ALL states. Numbers like 7111 are a 4-way number so to get a straight hit off of ALL combos play 1117 1171 1711 7111. No matter how that number falls you get a straight hit off of it. The inverse of that number fell today in Ohio's midday game (7717) darn that would have made a lot of you Ohio players happy.
Good luck tonight people.
-Michael
Sorry guys I missed one number from the list, so add 4384 to the list of universal numbers to play tonight.
Thanks.
-Michael
"Don't be a Dummy, Make 'Dat Money"
Mjwinsmith's Blog Page: http://blogs.lotterypost.com/mjwinsmith/
"How to Play My Pick-4 System": https://www.lotterypost.com/thread/306450
United States
Member #391
June 8, 2002
16059 Posts
Online
Posted: January 7, 2012, 8:29 pm - IP Logged
Your questions and or comments are welcome, that's what I am here for so get those questions on over to me so I can help you all win some extra money. Heard we had a winner in New Jersey tonight off of my 8-0-7-3 in the evening game, now that's great news. With the digit "8" in the first position the 6-sets to play to get a straight hit were; 8037-8073-8307-8370-8703-8730. Congratulations to all the New Jersey players born under the sign of (9) Sagittarius who nailed this one tonight.
-Michael
"Don't be a Dummy, Make 'Dat Money"
Mjwinsmith's Blog Page: http://blogs.lotterypost.com/mjwinsmith/
"How to Play My Pick-4 System": https://www.lotterypost.com/thread/306450
United States
Member #391
June 8, 2002
16059 Posts
Online
Posted: January 7, 2012, 8:55 pm - IP Logged
Your questions and or comments are welcome, that's what I am here for so get those questions on over to me so I can help you all win some extra money. Heard we had a winner in New Jersey tonight off of my 8-0-7-3 in the evening game, now that's great news. With the digit "8" in the first position the 6-sets to play to get a straight hit were; 8037-8073-8307-8370-8703-8730. Congratulations to all the New Jersey players born under the sign of (9) Sagittarius who nailed this one tonight.
-Michael
Sat, Jan 7, 2012 New Jersey Pick 4 Evening 8-7-0-3 Wed, Jan 4, 2012 Georgia Cash 4 Midday 7-3-8-0 Tue, Dec 13, 2011 Illinois Pick 4 Evening 8-0-3-7 Tue, Dec 13, 2011 Iowa Pick 4 Evening 8-0-3-7 Mon, Dec 12, 2011 North Carolina Pick 4 Daytime 7-0-8-3
"Don't be a Dummy, Make 'Dat Money"
Mjwinsmith's Blog Page: http://blogs.lotterypost.com/mjwinsmith/
"How to Play My Pick-4 System": https://www.lotterypost.com/thread/306450
United States
Member #391
June 8, 2002
16059 Posts
Online
Posted: January 8, 2012, 8:02 am - IP Logged
Sat, Jan 7, 2012 California Daily 4 8-2-0-3 Sat, Jan 7, 2012 Indiana Daily 4 Midday 5-3-9-6 Sat, Jan 7, 2012 Kentucky Pick 4 Midday 2-8-2-0 Sat, Jan 7, 2012 New Jersey Pick 4 Midday 6-4-8-2 Sat, Jan 7, 2012 New Jersey Pick 4 Evening 8-7-0-3 Sat, Jan 7, 2012 New York Win 4 Evening 3-9-7-5 Sat, Jan 7, 2012 Oregon Pick 4 7pm 6-6-0-1 Sat, Jan 7, 2012 Pennsylvania Big 4 Midday 8-3-0-9 Sat, Jan 7, 2012 South Carolina Pick 4 Evening 2-4-3-3 Sat, Jan 7, 2012 Tennessee Cash 4 Midday 0-9-3-8 Sat, Jan 7, 2012 Virginia Pick 4 Night 7-9-8-6 Sat, Jan 7, 2012 Wisconsin Pick 4 2-3-4-3
"Don't be a Dummy, Make 'Dat Money"
Mjwinsmith's Blog Page: http://blogs.lotterypost.com/mjwinsmith/
"How to Play My Pick-4 System": https://www.lotterypost.com/thread/306450
United States
Member #391
June 8, 2002
16059 Posts
Online
Posted: January 8, 2012, 8:34 am - IP Logged
Sat, Jan 7, 2012 California Daily 4 8-2-0-3 Sat, Jan 7, 2012 Indiana Daily 4 Midday 5-3-9-6 Sat, Jan 7, 2012 Kentucky Pick 4 Midday 2-8-2-0 Sat, Jan 7, 2012 New Jersey Pick 4 Midday 6-4-8-2 Sat, Jan 7, 2012 New Jersey Pick 4 Evening 8-7-0-3 Sat, Jan 7, 2012 New York Win 4 Evening 3-9-7-5 Sat, Jan 7, 2012 Oregon Pick 4 7pm 6-6-0-1 Sat, Jan 7, 2012 Pennsylvania Big 4 Midday 8-3-0-9 Sat, Jan 7, 2012 South Carolina Pick 4 Evening 2-4-3-3 Sat, Jan 7, 2012 Tennessee Cash 4 Midday 0-9-3-8 Sat, Jan 7, 2012 Virginia Pick 4 Night 7-9-8-6 Sat, Jan 7, 2012 Wisconsin Pick 4 2-3-4-3
Prediction Results for Saturday, January 07, 2012 (Appended)
Prediction State Winning No Day / Eve Prize Amount * 2038 CA 8203 Eve \$375.00 5936 IN 5396 Day \$9,375.00 8022 KY 2820 Day \$750.00 2648 NJ 6482 Day \$375.00 7038 NJ 8703 Eve \$375.00 7935 NY 3975 Eve \$375.00 6061 OR 6601 7:00 P \$9,750.00 0983 PA 8309 Day \$375.00 3432 SC 2433 Eve \$750.00 0983 TN 0938 Day \$45,375.00 6879 VA 7986 Eve \$375.00 3432 WI 2343 Eve \$750.00
*Winnings based on playing ON-LINE and using MjWIN\$mith’s Pick-4 system.
"Don't be a Dummy, Make 'Dat Money"
Mjwinsmith's Blog Page: http://blogs.lotterypost.com/mjwinsmith/
"How to Play My Pick-4 System": https://www.lotterypost.com/thread/306450
New Member
virginia beach, VA
United States
Member #121367
January 7, 2012
6 Posts
Offline
Posted: January 8, 2012, 12:08 pm - IP Logged
Sept 4th Virgo, Virginia
PANAMA CITY
United States
Member #116795
September 19, 2011
231 Posts
Offline
Posted: January 8, 2012, 12:17 pm - IP Logged
Play 4 eve 0965, Gemini, Florida and thank you!!!
United States
Member #391
June 8, 2002
16059 Posts
Online
Posted: January 8, 2012, 1:12 pm - IP Logged
Sept 4th Virgo, Virginia
1662 6693 8785 2292
"Don't be a Dummy, Make 'Dat Money"
Mjwinsmith's Blog Page: http://blogs.lotterypost.com/mjwinsmith/
"How to Play My Pick-4 System": https://www.lotterypost.com/thread/306450
United States
Member #391
June 8, 2002
16059 Posts
Online
Posted: January 8, 2012, 1:16 pm - IP Logged
Play 4 eve 0965, Gemini, Florida and thank you!!!
8022 6879
"Don't be a Dummy, Make 'Dat Money"
Mjwinsmith's Blog Page: http://blogs.lotterypost.com/mjwinsmith/
"How to Play My Pick-4 System": https://www.lotterypost.com/thread/306450
United States
Member #391
June 8, 2002
16059 Posts
Online
Posted: January 8, 2012, 1:20 pm - IP Logged
We have a Palindrome in play today, 6-9-9-6 it is found on the Table I list of numbers to play. The 4-way and Palindromes are falling recently so go with this one today, it would be nice of you guys to put it in for me in your state game. Thank you and good luck with it today.
-Michael
"Don't be a Dummy, Make 'Dat Money"
Mjwinsmith's Blog Page: http://blogs.lotterypost.com/mjwinsmith/
"How to Play My Pick-4 System": https://www.lotterypost.com/thread/306450
United States
Member #391
June 8, 2002
16059 Posts
Online
Posted: January 8, 2012, 4:06 pm - IP Logged
Sign: Taurus
State: North Carolina
Thanks!
mjwinsmith North Carolina Pick 4 Daytime 12-Way Box 5-4-7-4 4-7-4-5 \$200
"Don't be a Dummy, Make 'Dat Money"
Mjwinsmith's Blog Page: http://blogs.lotterypost.com/mjwinsmith/
"How to Play My Pick-4 System": https://www.lotterypost.com/thread/306450
United States
Member #391
June 8, 2002
16059 Posts
Online
Posted: January 8, 2012, 4:08 pm - IP Logged
4745 2252 0983 5798
mjwinsmith North Carolina Pick 4 Daytime 12-Way Box 5-4-7-4 4-7-4-5 \$200
"Don't be a Dummy, Make 'Dat Money"
Mjwinsmith's Blog Page: http://blogs.lotterypost.com/mjwinsmith/
"How to Play My Pick-4 System": https://www.lotterypost.com/thread/306450
United States
Member #391
June 8, 2002
16059 Posts
Online
Posted: January 9, 2012, 1:35 am - IP Logged
We have a Palindrome in play today, 6-9-9-6 it is found on the Table I list of numbers to play. The 4-way and Palindromes are falling recently so go with this one today, it would be nice of you guys to put it in for me in your state game. Thank you and good luck with it today.
-Michael
Sun, Jan 8, 2012 Georgia Cash 4 Evening 2-7-2-2 Sun, Jan 8, 2012 Washington, D.C. DC-4 Midday 4-2-4-2
"Don't be a Dummy, Make 'Dat Money"
Mjwinsmith's Blog Page: http://blogs.lotterypost.com/mjwinsmith/
"How to Play My Pick-4 System": https://www.lotterypost.com/thread/306450
Page 3 of 4 | 3,542 | 10,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2016-50 | longest | en | 0.868208 |
http://mathoverflow.net/feeds/question/93525 | 1,369,528,536,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706474776/warc/CC-MAIN-20130516121434-00088-ip-10-60-113-184.ec2.internal.warc.gz | 169,737,511 | 2,954 | Blowing up intersections of two divisors make them disjoint? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T00:35:18Z http://mathoverflow.net/feeds/question/93525 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93525/blowing-up-intersections-of-two-divisors-make-them-disjoint Blowing up intersections of two divisors make them disjoint? tarosano 2012-04-08T23:12:27Z 2012-04-09T15:53:56Z <p>Let $D,E \subset \mathbb{C}^3$ be prime divisors where $D$ is <strong><em>smooth</em></strong> and $E$ is <strong>not necessarily smooth</strong>. Assume that $D \cap E$ has <strong><em>SNC</em></strong> support and let </p> <p>$D \cap E = \bigcup \Gamma_i$ be a decomposition into irreducible components. </p> <p><strong>(Added)</strong> Let $f_1 \colon X_1 \rightarrow \mathbb{C}^3$ be a blow-up along a smooth curve $\Gamma_i$ and $\tilde{D}_1 \subset X_1$ be the strict transform of $D$. $f_1$ induces an isomorphism $g_1 \colon \tilde{D}_1 \rightarrow D$. </p> <p>There is a SNC curve $\bigcup \Gamma^1_i \subset \tilde{D}_1$. We next blow-up one of $\Gamma^1_i$. </p> <p><strong>Question</strong> Is it possible to make the strict transforms of $D$ and $E$ disjoint by blowing-up smooth curves $\Gamma_i$ or its strict transforms on the strict transforms of $D$ several times? </p> <p>I know that the blow-up of $\mathbb{C}^3$ along the ideal sheaf $\mathcal{I}_D + \mathcal{I}_E$ makes $D$ and $E$ disjoint. However I want the smooth centre blow-ups. </p> <p><strong>(My thought which uses Karl Schwede's answer comment)</strong> </p> <p>Let $\pi:Y \rightarrow X$ be a principalization of $I_D + I_E$ as in Karl Schwede's answer.</p> <p>Let $Z \subset X$ be a finite set which is the union of the images of the smooth points centers. I think that $\pi^{-1}(X \setminus Z) \rightarrow X \setminus Z$ is the blow-ups of smooth curves $\Gamma_i \setminus Z$ or their strict transform curves on the strict transforms of $D$. </p> <p>Let $\pi': Y' \rightarrow X$ be a composite of blow-ups of the smooth curves $\Gamma_i$ or its strict transform inside strict transforms of $D$ which are in same order as $\pi$. That is, we can define $\pi'$ by forgetting smooth points centers and curve centers which are contained in the inverse image of smooth point centers. $\pi'$ is same as $\pi$ outside $Z$. </p> <p>Let $\tilde{D}', \tilde{E}' \subset Y'$ be the strict transforms of the original divisors $D, E \subset X$. If $\tilde{D}' \cap \tilde{E}' \neq \emptyset$, then $\tilde{D}' \cap \tilde{E}' \subset \pi'^{-1}(Z) \cap \bigcup \tilde{\Gamma}'_i$ and it is finite points. This is a contradiction since $\tilde{D}', \tilde{E}'$ are Cartier divisors on a smooth 3-fold $Y$. Hence $\tilde{D}' \cap \tilde{E} = \emptyset$. </p> <p>Sorry for the long sentence, but I think that the idea is simple. I just ignored centers which is concerned with 0-dimensional centers. Are there gaps in this argument? </p> http://mathoverflow.net/questions/93525/blowing-up-intersections-of-two-divisors-make-them-disjoint/93540#93540 Answer by Karl Schwede for Blowing up intersections of two divisors make them disjoint? Karl Schwede 2012-04-09T03:47:35Z 2012-04-09T12:44:27Z <p>I'm not exactly sure what you want (see Sasha's question above) but algorithms for embedded resolution of singularities might do what you want (for example, see the recent book of Kollár, various papers by Villamayor and co-authors, or Wlodarczyk)</p> <p>Indeed, do a log resolution (ie, principalization) of the ideal $I_D + I_E$. What this will produce is a proper birational map $$\pi : Y \to X = \mathbb{C}^3$$ such that</p> <ol> <li>$(I_D + I_E) \cdot O_Y = O_Y(-F)$ for some divisor $F$. In particular, this implies that $\pi$ factors through the blowup of $I_D + I_E$ by the universal property of blowing up and thus separates $D$ and $E$.</li> <li>$\pi$ is an isomorphism outside of $D \cap E = V(I_D + I_E)$.</li> <li>$\pi$ is obtained by a sequence of blowups of smooth centers. </li> </ol> <p>You don't need the SNC support at all, although this will make things much easier.</p> <p>Of course, I don't quite know what you mean by blowing up the support of the $\Gamma_i$ several times. </p> <p><strong>In the SNC case:</strong> What the resolution algorithms would start with is blowing up the strata of the $D \cap E$, in other words, it <em>first</em> will blow up point where the curves of $D \cap E$ intersect. After doing that, it will <em>next</em> blow up the the strict transforms of irreducible components of $D \cap E$ (ie, the curves). <strong>EDIT:</strong> It might blow up some more points again at various steps. It might also blow up other curves lying over, but themselves are not the $\Gamma_i$ at a particular step. The reason I say this is because blowing up the $I_D + I_E$ doesn't actually know about the divisors $D$ and $E$ and won't keep track of their strict transforms...</p> <p>Is this ok for what you have in mind? Or do you want to avoid blowing up the points too?</p> | 1,504 | 5,027 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2013-20 | latest | en | 0.717295 |
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Differential Equations Help Forum
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Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A143489 "Fourth down, Extream [sic] between the two farthest Bells from it" in bell-ringing is a sequence of permutations p_1=(1,2,3,4), p_2=(1,2,4,3), .. which runs through all permutations of {1,2,3,4} with period 24; sequence gives position of bell 2 in n-th permutation. 1
2, 2, 3, 3, 4, 4, 4, 3, 3, 4, 4, 4, 3, 3, 2, 2, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 3, 3, 4, 4, 4, 3, 3, 4, 4, 4, 3, 3, 2, 2, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 3, 3, 4, 4, 4, 3, 3, 4, 4, 4, 3, 3, 2, 2, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 3, 3, 4, 4, 4, 3, 3, 4, 4, 4, 3, 3, 2, 2, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 3, 3, 4, 4, 4, 3, 3 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Start with (1,2,3,4), i.e. the first permutation of {1,2,3} followed by 4; then for each next permutation, transpose 4 one to the left; if at position 1, replace {1,2,3} recursively by the next permutation of these numbers. Thereafter, for each next permutation, transpose 4 to the right. And so on. LINKS FORMULA Period 24. MAPLE ring:= proc(k::nonnegint) local p, i, left, l, nf, ini; if k<=1 then proc() [1\$k] end else ini := proc() p:= ring(k-1); i:= k; left:= true; l:= p(); nf:= k! end; ini(); proc() local ll; ll:= [seq(l[t], t=1..(i-1)), k, seq(l[t], t=i..(k-1))]; if left then if i>1 then i:= i-1 else left:= false; l:=p() fi else if i indx (bell(4)[modp(n-1, 24)+1], 2): seq (a(n), n=1..121); CROSSREFS Cf. A143484-A143490, A090281. Sequence in context: A237815 A238701 A238134 * A274102 A261100 A130249 Adjacent sequences: A143486 A143487 A143488 * A143490 A143491 A143492 KEYWORD nonn AUTHOR Alois P. Heinz, Aug 19 2008 STATUS approved
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The OEIS Community | Maintained by The OEIS Foundation Inc. | 859 | 2,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2017-34 | latest | en | 0.551735 |
https://www.exceldemy.com/link-two-workbooks-in-excel/ | 1,718,289,982,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00399.warc.gz | 670,286,347 | 70,375 | How to Link Two Workbooks in Excel (5 Methods)
Consider the following dataset, containing half-yearly sales made by six employees of a company in 2021. The source and destination workbooks are named “SourceWorkbook” and “Link Two Workbooks” respectively. The methods for linking them involve external reference, the Consolidate tool, copy-paste, the Get data tool, and VLOOKUP.
Method 1 – Link Two Workbooks Using External Reference
Suppose you need to get the annual sales made by each employee in the destination workbook. You can apply the SUM function to do that, with external references to link the two workbooks.
Steps
• Type the following formula in cell C5 in the destination workbook:
`=SUM()`
• Do not press Enter yet. Put the cursor between the parentheses.
• Toggle back to the source workbook using the taskbar. You can also do that from View and Switch Windows.
• Select the cells you want to sum with your mouse.
• If you press Enter now, you will see the sales added as follows.
• Notice that the formula contains an absolute reference.
• Remove the \$ symbols to convert them into relative references.
• Drag the fill handle down to copy the formula to the cells below.
Method 2 – Link Two Workbooks Using the Consolidate Tool
Steps
• Select the first cell or the cell range where you want to get the sum.
• Go to the Data tab.
• Click on the Consolidate icon in the Data Tools group. This will open a new dialog box.
• Choose the Sum function in the Function field.
• Click on the up arrow in the Reference field.
• Go back to the source workbook.
• Select the first range of cells and click on the down arrow.
• Repeat the selection procedure for the next range of cells.
• Check Create links to source data.
• Hit the OK button.
• You should get the same result as in method 1.
Method 3 – Link Two Workbooks Using Copy-Paste
Steps
• Copy the dataset in the source workbook.
• Go to the destination workbook.
• Select the cell where you want to copy the dataset.
• Right-click and select the link icon from Paste Options.
• You can see the new dataset linked to the source workbook.
Method 4 – Link Two Workbooks Using the Get Data Tool
Steps
• Select the cell where you want to get your data.
• Go to the Data tab, choose Get Data, then From File, and finally From Workbook.
• Browse through your PC to find the source workbook.
• Select the Import button. A new dialog box will open.
• Select the worksheet where your data is located and choose Load on the bottom.
• Your data will be imported as an Excel Table to a new worksheet.
• You can copy the data and paste it as a link in your desired worksheet.
Method 5 – Link Two Workbooks Using the VLOOKUP Formula
Suppose you want to get the second half-yearly sales by an employee in a new workbook when we enter a name. Here’s how to do that.
Steps
• Type the following in cell C5:
`=VLOOKUP(B5,`
• Go back to the source workbook and select the entire data range.
• Put a comma(,) and complete the formula as follows:
`=VLOOKUP(B5,[SourceWorkbook.xlsx]Sales!\$B\$5:\$D\$10,3,FALSE)`
• Hit Enter to apply the formula.
• You can enter a name in B5 and the sales value will be fetched.
Things to Remember
• Always keep both of the workbooks open while applying these methods.
• More than two worksheets can be linked together.
• If you want to manually enter an external reference, don’t forget to enclose the file name inside single quotation marks().
• Data will be automatically updated while both of the workbooks are open. Otherwise, you will need to update the destination workbook.
Related Articles
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Advanced Excel Exercises with Solutions PDF | 983 | 4,386 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-26 | latest | en | 0.861513 |
http://www.haskell.org/haskellwiki/index.php?title=Haskell_Quiz/Weird_Numbers/Solution_Bailey&direction=prev&oldid=33552 | 1,406,369,619,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997900573.25/warc/CC-MAIN-20140722025820-00094-ip-10-33-131-23.ec2.internal.warc.gz | 774,745,933 | 6,437 | ##### Views
The implementation below isn't very effecient, apparently (the thread referred to in the comments talks about why not), or particulary fast, but it does work.
The heart of the solution is the is_weird function. It takes a number and determines if its weird. The first weird number is 70, so all numbers below that are thrown out.
If the number is greater than 70, I first find all the divisors up to n / 2 for that number. Using powerSet, I then find all possible subsets of those divisors.
Finally, the function first ensures the sum of all the divisors is greater than n. Next, it searches all the subsets and determines if any of them sum to the value of n. If not, the number is weird!
Optimizations might include generating the powerset such that the longest lists come first, to find numbers that aren't weird sooner.
A bonus function, find_weird, takes an argument and finds the first n weird numbers.
```-- powerSet found courtesy of http://haskell.org/hawiki/PreludeExts
powerSet :: [a] -> [[a]]
powerSet [] = [[]]
powerSet (x:xs) = xss ++ map (x:) xss
where xss = powerSet xs
-- Only positive numbers. 70 is first weird number, so exclude all below 69.
is_weird n | n < 70 = False
is_weird n =
let divisors = [div | div <- [1..(floor ((fromInteger n / 2)))], n `mod` div == 0]
divisor_subsets = powerSet divisors
in
(foldr (+) 0 divisors) > n && not (any (\subset -> (foldr (+) 0 subset) == n) divisor_subsets)
find_weird n = take n [w | w <- [70..], is_weird w]``` | 408 | 1,496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2014-23 | longest | en | 0.841022 |
http://www.perlmonks.org/index.pl?node_id=998113 | 1,519,250,839,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813803.28/warc/CC-MAIN-20180221202619-20180221222619-00619.warc.gz | 521,669,085 | 6,441 | Pathologically Eclectic Rubbish Lister PerlMonks
### Re: Numeric sorting WITHOUT <=>
by ikegami (Pope)
on Oct 10, 2012 at 01:55 UTC ( #998113=note: print w/replies, xml ) Need Help??
in reply to Numeric sorting WITHOUT <=>
```my (\$min) = keys(%jetsons);
my \$max = \$min;
for (keys(%jetsons)) {
\$max = \$_ if \$_ > \$max;
\$min = \$_ if \$_ < \$min;
}
my @sorted;
for (\$min..\$max) {
push @sorted, \$_
if exists(\$jetsons{\$_});
}
Replies are listed 'Best First'.
Re^2: Numeric sorting WITHOUT <=>
by Marshall (Abbot) on Oct 10, 2012 at 04:42 UTC
my (\$min) = keys(%jetsons); This is the number of keys in %jetsons.
This is not the min.
my (\$min) = keys(%jetsons); This is the number of keys in %jetsons.
Actually, it's a randomly (insofar as the hashing algorithm is random) chosen key from the hash. The assignment is in list context and seeds the min/max finder loop that follows.
```>perl -wMstrict -le
"my %jetsons = qw(aa 1 bb 2 cc 3 dd 4 ee 5);
my (\$min) = keys %jetsons;
print \$min;
"
cc
Update: ikegami's code works. (Never doubt a Pope.)
No.
```# keys in scalar context: Number of keys.
my \$min = keys(%jetsons);
# keys in list context: List of keys. First one is assigned
my (\$min) = keys(%jetsons);
Correct.
My mistake.
Re^2: Numeric sorting WITHOUT <=>
by ria1328 (Initiate) on Oct 10, 2012 at 17:07 UTC
Thank you! This worked and it actually made it easier for me to understand how it was done instead of using the was my professor said we should!
This worked and it actually made it easier for me to understand how it was done
My code does NOT demonstrate how sorting is usually done. It takes a different and silly approach in response to your equally silly requirement. It would take forever for my solution to sort
```my %jetsons = ( 0 => "a", 4000000000 => "b" );
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Notices? | 650 | 2,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-09 | latest | en | 0.883937 |
https://bumpyroadtocode.com/2017/08/09/reverse-integer-sequence/ | 1,695,781,642,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510238.65/warc/CC-MAIN-20230927003313-20230927033313-00131.warc.gz | 169,007,465 | 13,263 | # reverse integer sequence
Bjarne Stroustrup “Programming Principles and Practice Using C++”
Chapter 8 Exercise 5
Using std_lib_facilities.h by Bjarne Stroustrup.
main.cpp main file
[code language=”cpp”]
// Philipp Siedler
// Bjarne Stroustrup’s PP
// Chapter 8 Exercise 5
#include "std_lib_facilities.h"
vector<int> myVec{ 0,1,2,3,4,5,6,7,8,9 };
vector<int> myVecRev;
class Rev {
public:
vector<int> v;
vector<int> vr;
Rev(vector<int>& _v, vector<int>& _vr) :v(_v), vr(_vr) {};
void reverse();
void reverseSame();
void printV(string&);
void printVR(string&);
};
void Rev::reverse() {
for (int i = 0; i < v.size(); i++) {
vr.push_back(v[(v.size() – 1) – i]);
}
}
void Rev::reverseSame() {
for (int i = 0; i < v.size() / 2; i++) {
int temp;
temp = v[i];
v[i] = v[(v.size() – 1) – i];
v[(v.size() – 1) – i] = temp;
}
}
void Rev::printV(string& _label) {
for (int x : v) {
cout << _label << " " << x << "\n";
}
}
void Rev::printVR(string& _label) {
for (int x : vr) {
cout << _label << " " << x << "\n";
}
}
int main()
try
{
string label = "l";
Rev rev(myVec, myVecRev);
cout << "First version:\n";
rev.reverse();
rev.printV(label);
rev.printVR(label);
cout << "Second version:\n";
rev.reverseSame();
rev.printV(label);
keep_window_open();
}
catch (runtime_error e) {
cout << e.what() << "\n";
keep_window_open();
}
catch (…) {
cout << "Exiting\n";
keep_window_open();
}
[/code]
```Output:
First version:
l 0
l 1
l 2
l 3
l 4
l 5
l 6
l 7
l 8
l 9
l 9
l 8
l 7
l 6
l 5
l 4
l 3
l 2
l 1
l 0
Second version:
l 9
l 8
l 7
l 6
l 5
l 4
l 3
l 2
l 1
l 0
Please enter a character to exit
``` | 592 | 1,595 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-40 | latest | en | 0.316429 |
https://scipy-cookbook.readthedocs.io/items/LinearRegression.html | 1,713,301,239,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817106.73/warc/CC-MAIN-20240416191221-20240416221221-00453.warc.gz | 461,151,590 | 38,798 | Linear regression¶
Linear Regression Example
In [5]:
# IPython magic to plot interactively on the notebook
%matplotlib notebook
In [7]:
from scipy import linspace, polyval, polyfit, sqrt, stats, randn
from matplotlib.pyplot import plot, title, show, legend
# Linear regression example
# This is a very simple example of using two scipy tools
# for linear regression, polyfit and stats.linregress
# Sample data creation
# number of points
n = 50
t = linspace(-5,5,n)
# parameters
a = 0.8
b = -4
x = polyval([a, b], t)
xn = x + randn(n)
# Linear regressison -polyfit - polyfit can be used other orders polys
(ar, br) = polyfit(t, xn, 1)
xr = polyval([ar, br], t)
# compute the mean square error
err = sqrt(sum((xr - xn)**2)/n)
print('Linear regression using polyfit')
print('parameters: a=%.2f b=%.2f \nregression: a=%.2f b=%.2f, ms error= %.3f' % (a, b, ar, br, err))
print('\n')
# Linear regression using stats.linregress
(a_s, b_s, r, tt, stderr) = stats.linregress(t, xn)
print('Linear regression using stats.linregress')
print('parameters: a=%.2f b=%.2f \nregression: a=%.2f b=%.2f, std error= %.3f' % (a,b,a_s,b_s,stderr))
print('\n')
# matplotlib ploting
title('Linear Regression Example')
plot(t, x,'g.--')
plot(t, xn, 'k.')
plot(t, xr, 'r.-')
legend(['original','plus noise', 'regression'])
show();
Linear regression using polyfit
parameters: a=0.80 b=-4.00
regression: a=0.77 b=-4.10, ms error= 0.880
Linear regression using stats.linregress
parameters: a=0.80 b=-4.00
regression: a=0.77 b=-4.10, std error= 0.043
In [ ]:
Section author: Unknown[1], Unknown[66], TimCera, Nicolas Guarin-Zapata
Attachments | 523 | 1,630 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-18 | latest | en | 0.300984 |
https://www.dpmms.cam.ac.uk/~ty245/2012_CFT/2012_CFT.html | 1,506,146,044,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689490.64/warc/CC-MAIN-20170923052100-20170923072100-00287.warc.gz | 779,312,751 | 3,500 | website for Class Field Theory (2012, L24) This is a website for the Part III (non-examinable, graduate) course Class Field Theory (2012, L24); here are some material for this course. ______time/syllabus (from the official website of the department) Meetings: Tue / Thu / Sat 11am, at MR 12. First lecture: Th., 19th January, 2012. Last lecture: Tu., 13th March, 2012. p.37 of the [pdf] Mathematical Tripos Part III Lecture Courses. ______notes/plan I may post some typed/scanned lecture notes here. There will be some overlap with my notes from the Postech Summer School: [pdf] Notes for Postech Summer School 2010. The goal would be to cover the proofs of local/global class field theory for number fields, with no essential part left out. We use a "modern" language, namely adeles/ideles and Galois cohomology. This means that we may need to skip some straightforward exercises in commutative/homological algebra, but I will not assume them as prerequisites (i.e. will give all the necessary definitions), and try not to omit the number theoretic arguments. I don't intend to make much changes from the same course given last year. Lecture 1 (Th. 19/1/12) [scan]: 1.1 Preliminaries (tensor products, limits) Handout on Categories and Functors (updated after the lecture) Lecture 2 (Sa. 21/1/12) [scan]: 1.2 Galois theory (infinite extensions, cyclotomic characters) Lecture 3 (Tu. 24/1/12) [scan]: 1.3 Dedekind domains Lecture 4 (Th. 26/1/12) [scan]: 1.4 Cyclotomic extensions, CFT of Q Lecture 5 (Sa. 28/1/12) [scan]: 1.5 CFT of Q via adeles Lecture 6 (Tu. 30/1/12) [scan]: 1.6 Local fields, Weil groups Lecture 7 (Th. 2/2/12) [scan]: 2.1 A statement of local CFT Lecture 8 (Sa. 4/2/12) [scan]: 2.2 Lubin-Tate groups Lecture 9 (Tu. 7/2/12) [scan]: 2.3 Lubin-Tate extensions Lecture 10 (Th. 9/2/12) [scan]: 2.3 (continued: Lubin-Tate spaces) Lecture 11 (Sa. 11/2/12) [scan] : 2.4 The Artin map Lecture 12 (Tu. 14/2/12) [scan]: 2.5 Norm groups Handout: lemmas on Coleman operators / Handout: a lemma on infinite norms Handout on Categories and Functors + Homological Algebra Lecture 13 (Th. 16/2/12) : 3.1 Group Cohomology Lecture 14 (Sa. 18/2/12) : 3.2 Res-Inf sequence / 3.3 Standard complex / 3.4 Tate Cohomology Handout: the Res-Inf exact sequence Lecture 15 (Tu. 21/2/12) : 3.5 Shift morphisms Handout: the shift morphisms Lecture 16 (Th. 23/2/12) : 3.6 Tate's theorem Lecture 17 (Sa. 25/2/12) : 4.1 Galois cohomology / 4.2 Class formations Lecture 18 (Tu. 28/2/12) : 4.3 Local CFT Handout: an alternative proof of existence theorem Lecture 19 (Th. 1/3/12) : 5.1 Number fields Lecture 20 (Sa. 3/3/12) : 5.2 Cohomology of idele groups Handout: functoriality of Shapiro's lemma Lecture 21 (Tu. 6/3/12) : 5.3 Global theorems / 1st inequality Handout: weak approximation for ideles Lecture 22 (Th. 8/3/12) : 5.4 2nd inequality Lecture 23 (Sa. 10/3/12) : 5.4 (continued) / 5.5 Invariant maps Handout: local power index Lecture 24 (Tu. 13/3/12) : 5.5 (continued) / 5.6 Global CFT Lecture 25 (Th. 15/3/12) [additional] : 5.6 (continued) Handout: global existence theorem ______literature Here are some references - they are not essential at all. M. Reid, Undergraduate Commutative Algebra, CUP (London Mathematical Society Student Texts), 1996. M.F. Atiyah, I.G. MacDonald, Introduction to Commutative Algebra, Westview Press, 1994. N. Bourbaki, Commutative Algebra: Chapters 1-7, Springer, 1998. C.A. Weibel, An Introduction to Homological Algebra, CUP (Cambridge Studies in Advanced Mathematics), 1995. A. Grothendieck, Sur quelques points d'alg'ebre homologique (translation), Tohoku Mathematics Journal 9-2 (1957), 119-221. S. Maclane, Categories for the Working Mathematician, Springer (Graduate Texts in Mathematics), 1998. H. Cartan, S. Eilenberg, Homological Algebra, Princeton UP, 1956. J.-P. Serre, Local Fields, Springer (Graduate Texts in Mathematics), 1980. K. Iwasawa, Local Class Field Theory, Oxford UP, 1986. T. Yoshida, Local class field theory via Lubin-Tate theory, Annales de la Faculte des Sciences de Toulouse, Ser. 6, 17-2 (2008), 411-438. E. Artin, J. Tate, Class Field Theory, Benjamin, 1967. J.W.S. Cassels, A. Fr"ohlich, Algebraic Number Theory, Academic Press, 1967. A. Weil, Basic Number Theory, Springer (Classics in Mathematics), 1995. J. Neukirch, A. Schmidt, K. Wingberg, Cohomology of Number Fields, Springer (Grundlehren der Mathematischen Wissenschaften 323), 2000. ______old stuff 2011 course website | 1,432 | 4,438 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-39 | latest | en | 0.830287 |
https://exactlyhowlong.com/bmi-calculator-for-women/ | 1,713,555,605,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817442.65/warc/CC-MAIN-20240419172411-20240419202411-00277.warc.gz | 212,729,993 | 35,464 | BMI Calculator for Women
# BMI Calculator for Women
Instructions:
• Click "Calculate BMI" to calculate your Body Mass Index.
• See the interpretation of your BMI and estimated daily calorie needs.
• Your calculation history will be displayed below.
• Click "Clear Results" to reset the form and chart.
• Click "Copy Results" to copy the results to the clipboard.
Result:
Interpretation:
Visual Representation:
Detailed Calculation and Explanation:
Calculation History:
## Concepts
Body mass index (BMI) is a measure of body fat based on height and weight that is used to screen for weight categories that may lead to health problems. BMI is calculated by dividing a person’s weight in kilograms (kg) by the square of their height in meters (m).
## Formula
The following formula is used to calculate BMI:
``````BMI = weight (kg) / height^2 (m^2)
``````
## Interpretation
The BMI ranges for women are as follows:
## Benefits
BMI is a simple and inexpensive way to screen for weight categories that may lead to health problems. It can be used by people of all ages and genders, including women.
## Interesting facts
• Women tend to have a higher percentage of body fat than men, even at the same BMI.
• Women are more likely to develop weight-related health problems, such as heart disease, stroke, and type 2 diabetes, than men.
• BMI is not a perfect measure of body fatness, especially for women. It does not take into account muscle mass or body composition.
• BMI is most accurate for women between the ages of 20 and 65.
## References
• Clinical Assessment of Nutritional Status by Robert J. Fanelli (2012)
• Obesity and the Body Mass Index: Or Why We Are Still Measuring Abdominal Obesity by Kelly Brownell and Joann E. Manson (2005)
• Body Mass Index and Mortality by Katherine M. Flegal, MD, MPH, and Barry M. Popkin, PhD (2001)
• Body Mass Index in Women: Implications for Health and Disease by Jean A. Shepherd, PhD, RD (2019)
## Examples
The following table shows some examples of BMI calculations for women:
## Applications
BMI can be used by women to track their weight over time and to assess their risk of developing weight-related health problems. It can also be used by healthcare professionals and public health officials to screen for weight categories that may lead to health problems and to develop interventions to promote healthy weight.
## Conclusion
BMI is a simple and inexpensive tool that can be used by women to screen for weight categories that may lead to health problems. It is important to note that BMI is not a perfect measure of body fatness and that other factors, such as muscle mass and body composition, should also be considered when assessing a woman’s health.
Women should be aware of the following additional considerations when using a BMI calculator:
• Pregnancy: Pregnant women should not use BMI to assess their weight status.
• Menopause: Women who are going through menopause may experience changes in their body composition, which can affect their BMI.
• Muscle mass: Women who have a lot of muscle mass may have a higher BMI than women who have less muscle mass, even if they have the same body fat percentage. | 707 | 3,186 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-18 | latest | en | 0.842003 |
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I see the term used in ski descriptions all the time but I just can't seem to get my mind around exactly what it means. Please help me understand.
Quote:
Originally Posted by adipose Explain "radius" I see the term used in ski descriptions all the time but I just can't seem to get my mind around exactly what it means. Please help me understand.
How big or small of a turn the ski will make when edged correctly.
actually the radius number on a pair of skis is not the turning radius of the ski. Do you think a slalom ski can only make a 10 meter radius turn? That's awfully big.
There is a long thread about this on the board somewhere, no time to search for it now, but the turning radius of a ski is much shorter than those number, and is dependent on how much the ski is on edge and flexed.
As SkiMangoJazz said:
It describes how curved the sidecut of the ski is. Lower radius = more curved / more sidecut; higher radius = less curved / less sidecut.
Think of it like this:
Set a ski down on the floor in a gym. Climb up on a tall ladder and look down at it, paying attention to one edge. Notice that it's curved. Picture the circle that's formed if you extend the curve. The radius of that circle is the "radius" we're talking about.
A few notes:
- It's approximate, for a number of reasons, most importantly: the curve of a ski's sidecut typically isn't uniform, i.e. it's not exactly an arc from a circle.
- Typically, it's estimated by considering (i) the width at three points: near the tip, at the narrowest point (aka the waist) and near the tail and (ii) the length of the ski (or, to be more accurate, the distance between the three points at which you measured the width).
- If you do a search, you'll discover numerous threads on this topic, some of them lengthy, some of them abstruse.
Lay your ski on the floor. The edge approximately follows a circle with the sidecut radius. If you slightly tilt your ski, you will find it is supported at the tip and tail and the middle does not touch the floor. When you press down on the middle of your slightly tilted ski so that the entire edge of the ski is touching the floor the tilted ski's edge will trace part of a smaller circle on the floor.
What this means is that on a hard surface the ski can "carve" (or cut) a curve with a riadius that is smaller than the sidecut radius by tipping it farther on edge (and decambering it more, but it can not carve a larger circle.
In addition to sidecut, the flexibility of the ski also affects how easily a ski can be forced to bend into a tighter arc. It is possible to bend straight skis into tight arcs, but it takes much more skill (IMHO).
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Back to Forum: Ski Gear Discussion | 654 | 2,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2016-44 | latest | en | 0.953523 |
http://www.isuma.tv/hi/es/aboriginal-perspectives/working-with-date-and-time | 1,369,333,586,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703662159/warc/CC-MAIN-20130516112742-00021-ip-10-60-113-184.ec2.internal.warc.gz | 542,461,341 | 8,777 | # Working with Date and Time
Redes comunitarias :
Canales:
Working with Date and Time
by Jessica Wesaquate and Andrea Rogers
Four
Strand:
Shape and Space
Goals:
Number sense, mathematical attitude, logical thinking
Objectives:
Students will be able to state the number of hours in day
Students will be able to explain the meaning of AM and PM and provide an example of an activity that occurs in the AM, and another that occurs in the P.M.
Students will be able to write dates in a variety of formats.
Students will be able to identify possible interpretations of a date.
Before completing this lesson, have students watch video 2, "Mathematics and Journalism" on Cassandra Opikokew, graduate from the University of Regina’s School of Journalism. She serves as a great Aboriginal role model for students. In both the video and the lesson, students will see the connection of mathematics to real life.
As a journalist, it is important you get your facts straight! You are currently working for a local newspaper and your boss has asked you to go to the Aboriginal pavilion at Mosaic in Regina to check out what is going on. You are responsible for getting some figures for an article that will be published the following day.
1. First off, your boss has given you ONE complete day to submit your article. How many hours do you have to complete the article?
2. You called the coordinator of the pavilion and they have asked for you to come in the A.M. to take watch what goes into the preparation for the weekend. What part of the day would she like you to come? Give an example of a time in the A.M. that you might arrive.
3. What part of the day are you going to the pavilion if it is 7:00 P.M.? What meal of the day do you think will be getting served?
4. The article will be published on June 4, 2009 and when you submit your report online, it needs to be written in the following format: yyyy/mm/dd. Record what this will look like using your pencil and paper.
5. Now that the article has been published in the paper, you notice that the date was written as 04/06/2009. What format was it written in?
* Note for Teachers:
Integrating familiar materials and situations will intrigue your students to learning the required curricular material.
Teaching Opportunity:
Aboriginal people view time differently than European Canadians. Things like watching the clock, and finishing things in a certain time frame are not Aboriginal practices. There are many Indian languages that don't even have a word for "time."
"The Indian tells me, time is with us. Life should be easy going, with little pressure. Things should be done when they need to be done. Exactness of time is of little importance. When an activity should be done is better determined by when the thing that precedes it is completed or when circumstances are right than by what the clock says."
Hankes, Judith Elaine & Fast, Gerald R. Investigating the Correspondence between Native American Pedagogy and Constructivist-Based Instruction.
Aboriginal Perspectives is supported by the University of Regina, the Imperial Oil Foundation, the Canadian Mathematical Society and the Pacific Institute for the Mathematical Sciences.
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Mas información de opciones de formateo | 822 | 3,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2013-20 | latest | en | 0.947916 |
https://grindskills.com/what-is-the-expected-value-of-the-logarithm-of-gamma-distribution/ | 1,670,538,202,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711368.1/warc/CC-MAIN-20221208215156-20221209005156-00692.warc.gz | 311,376,926 | 17,570 | # What is the expected value of the logarithm of Gamma distribution?
If the expected value of $$Gamma(α,β)\mathsf{Gamma}(\alpha, \beta)$$ is $$αβ\frac{\alpha}{\beta}$$, what is the expected value of $$log(Gamma(α,β))\log(\mathsf{Gamma}(\alpha, \beta))$$? Can it be calculated analytically?
The parametrisation I am using is shape-rate.
This one (maybe surprisingly) can be done with easy elementary operations (employing Richard Feynman’s favorite trick of differentiating under the integral sign with respect to a parameter).
We are supposing $$XX$$ has a $$Γ(α,β)\Gamma(\alpha,\beta)$$ distribution and we wish to find the expectation of $$Y=log(X).Y=\log(X).$$ First, because $$β\beta$$ is a scale parameter, its effect will be to shift the logarithm by $$logβ.\log\beta.$$ (If you use $$β\beta$$ as a rate parameter, as in the question, it will shift the logarithm by $$−logβ.-\log\beta.$$) This permits us to work with the case $$β=1.\beta=1.$$
After this simplification, the probability element of $$XX$$ is
$$fX(x)=1Γ(α)xαe−xdxxf_X(x) = \frac{1}{\Gamma(\alpha)} x^\alpha e^{-x} \frac{\mathrm{d}x}{x}$$
where $$Γ(α)\Gamma(\alpha)$$ is the normalizing constant
$$Γ(α)=∫∞0xαe−xdxx.\Gamma(\alpha) = \int_0^\infty x^\alpha e^{-x} \frac{\mathrm{d}x}{x}.$$
Substituting $$x=ey,x=e^y,$$ which entails $$dx/x=dy,\mathrm{d}x/x = \mathrm{d}y,$$ gives the probability element of $$YY$$,
$$fY(y)=1Γ(α)eαy−eydy.f_Y(y) = \frac{1}{\Gamma(\alpha)} e^{\alpha y - e^y} \mathrm{d}y.$$
The possible values of $$YY$$ now range over all the real numbers $$R.\mathbb{R}.$$
Because $$fYf_Y$$ must integrate to unity, we obtain (trivially)
$$Γ(α)=∫Reαy−eydy.\Gamma(\alpha) = \int_\mathbb{R} e^{\alpha y - e^y} \mathrm{d}y.\tag{1}$$
Notice $$fY(y)f_Y(y)$$ is a differentiable function of $$α.\alpha.$$ An easy calculation gives
$$ddαeαy−eydy=yeαy−eydy=Γ(α)yfY(y).\frac{\mathrm{d}}{\mathrm{d}\alpha}e^{\alpha y - e^y} \mathrm{d}y = y\, e^{\alpha y - e^y} \mathrm{d}y = \Gamma(\alpha) y\,f_Y(y).$$
The next step exploits the relation obtained by dividing both sides of this identity by $$Γ(α),\Gamma(\alpha),$$ thereby exposing the very object we need to integrate to find the expectation; namely, $$yfY(y):y f_Y(y):$$
E(Y)=∫RyfY(y)=1Γ(α)∫Rddαeαy−eydy=1Γ(α)ddα∫Reαy−eydy=1Γ(α)ddαΓ(α)=ddαlogΓ(α)=ψ(α),\eqalign{ \mathbb{E}(Y) &= \int_\mathbb{R} y\, f_Y(y) = \frac{1}{\Gamma(\alpha)} \int_\mathbb{R} \frac{\mathrm{d}}{\mathrm{d}\alpha}e^{\alpha y - e^y} \mathrm{d}y \\ &= \frac{1}{\Gamma(\alpha)} \frac{\mathrm{d}}{\mathrm{d}\alpha}\int_\mathbb{R} e^{\alpha y - e^y} \mathrm{d}y\\ &= \frac{1}{\Gamma(\alpha)} \frac{\mathrm{d}}{\mathrm{d}\alpha}\Gamma(\alpha)\\ &= \frac{\mathrm{d}}{\mathrm{d}\alpha}\log\Gamma(\alpha)\\ &=\psi(\alpha), }
the logarithmic derivative of the gamma function (aka “polygamma“). The integral was computed using identity $$(1).(1).$$
Re-introducing the factor $$β\beta$$ shows the general result is
$$E(log(X))=logβ+ψ(α)\mathbb{E}(\log(X)) = \log\beta + \psi(\alpha)$$
for a scale parameterization (where the density function depends on $$x/βx/\beta$$) or
$$E(log(X))=−logβ+ψ(α)\mathbb{E}(\log(X)) = -\log\beta + \psi(\alpha)$$
for a rate parameterization (where the density function depends on $$xβx\beta$$). | 1,166 | 3,233 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 35, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2022-49 | latest | en | 0.665592 |
http://en.kioskea.net/faq/6874-excel-compare-two-columns-and-delete-duplicates | 1,369,403,397,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704662229/warc/CC-MAIN-20130516114422-00035-ip-10-60-113-184.ec2.internal.warc.gz | 86,628,555 | 8,426 | Join
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# Excel - Compare two columns and delete duplicates
April 2013
Mathematical or logical operations cannot be done manually on Excel worksheets which contain large amounts of data, as it would take too long. But Excel is an efficient office software program that allows data to be automatically manipulated, through a macro or a VBA program code.To compare two columns and delete duplicates in Excel,a macro can be written to make a logical or mathematical comparison of the columns and remove the rows containing duplicate data. A well-written Excel macro can accomplish the task quickly and accurately. It is easy to compare two columns and delete duplicates in Excel.
# Excel - Compare two columns and delete duplicates
## Issue
I need a macro for an Excel spreadsheet that will compare lines based on two cells, and find any duplicates.
In this example, the first two columns - A and B - are the ones that must be unique. The macro should only flag rows that are a perfect match in both columns. So in this example, only the fifth row should be flagged up, as it is an exact match for the one above.
Example:
```Col A Col B Col C Col D
00001 AAAA1 xxxx yyyy
00001 BBBB1 xxxx xxxx
00001 BBBB2 yyyy yyyy
00002 AAAA1 yyyy xxxx
00002 AAAA1 yyyy xxxx
00003 AAAA1 xxxx yyyy ```
## Solution
This solution will alter the data in your spreadsheet, so make sure you save a backup copy.
Firstly, you need to combine the data in columns A and B so they can be compared as a whole. To do this, type into an empty cell at the end of the first row of data (in this example we are using row 2 - change as necessary):
`=A2&B2`
Copy this formula down for the whole spreadsheet.
Now run the macro below. In this sample code, it is assumed you wrote the above formula in column E. Amend the code as necessary.
Warning: This macro will delete rows, so it is important to make sure you have made your backup copy copy before testing it.
```Sub text()
Dim j As Integer, k As Integer, r As Range
j = Range("E2").End(xlDown).Row
For k = j To 2 Step -1
MsgBox k
Set r = Range(Cells(k, "E"), Cells(k, "E").End(xlUp))
If WorksheetFunction.CountIf(r, Cells(k, "E")) > 1 Then
Cells(k, "E").EntireRow. Delete
End If
Next k
End Sub
```
## Note
Thanks to venkat1926 for this tip on the forum.
## Knowledge communities.
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health.kioskea.net | 603 | 2,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2013-20 | latest | en | 0.896919 |
https://www.physicsforums.com/threads/how-to-show-derivatives-of-1-r-for-a-1d-charge-distribution.213095/ | 1,725,821,070,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651017.43/warc/CC-MAIN-20240908181815-20240908211815-00321.warc.gz | 876,310,795 | 17,292 | # How to Show Derivatives of 1/R for a 1D Charge Distribution?
• raintrek
In summary, the question is asking for methods to solve for the derivatives of 1/R and 1/R^2 in a 1 dimensional problem, using the given information about charges and their positions along the x-axis. The solution involves substituting R with x and using basic derivative rules.
raintrek
## Homework Statement
"Assume that charges +q, -3q, 2q lie at positions -2a, 0, +2a along the x-axis respectively.")
I've calculated dipole/quadrupole moments about the origin as well as the exact potential at x=+10a, however I'm confused by this next part to the question:
Show that, for a 1 dimensional problem,
d(1/R) / dx = -1/x²
and
d²(1/R) / dx² = 2/x³
where R is the distance between a lab fixed point x and an arbitrary origin x0.
Can anyone suggest a method of tackling this??
raintrek said:
Show that, for a 1 dimensional problem,
d(1/R) / dx = -1/x²
and
d²(1/R) / dx² = 2/x³
where R is the distance between a lab fixed point x and an arbitrary origin x0.
Can anyone suggest a method of tackling this??
I am at a loss to understand what this has to do with charges.
R has been defined to be equal to x.
So, d/dx(1/R) = d/dx(1/x) = -1/x^2 etc.
## What is a 1 dimensional charge distribution?
A 1 dimensional charge distribution refers to a system in which charges are distributed along a single line or axis. This is in contrast to a 2 or 3 dimensional charge distribution where charges are distributed in a plane or in space, respectively.
## How is a 1 dimensional charge distribution different from a point charge?
A point charge is a theoretical concept in which all of the charge is concentrated at a single point. In a 1 dimensional charge distribution, the charge is spread out along a line, resulting in a more realistic and complex distribution of charge.
## What is the electric field produced by a 1 dimensional charge distribution?
The electric field produced by a 1 dimensional charge distribution depends on the distribution of charges and their magnitudes. This can be calculated using the principle of superposition, where the electric field due to each individual charge is added together.
## Can a 1 dimensional charge distribution create a uniform electric field?
No, a 1 dimensional charge distribution cannot create a uniform electric field. This is because the electric field produced by each individual charge decreases with distance, resulting in a non-uniform field overall.
## What are some real-life examples of 1 dimensional charge distributions?
Some examples of 1 dimensional charge distributions include a long, thin wire with a uniform charge distribution, a line of evenly spaced electrically charged particles, or a charged cylinder with a finite length.
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2K | 741 | 2,980 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-38 | latest | en | 0.943223 |
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It is currently Fri, 24 May 2013 08:07:07 UTC
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Post subject: Multivariable Nonlinear ModelingPosted: Fri, 8 May 2009 17:59:47 UTC
Member
Joined: Thu, 5 Feb 2009 00:05:45 UTC
Posts: 31
Hi,
I'm having some problems getting a model for some experimental data. I need a function of Air Flow with dependent variables Boost and RPM. Here's the data:
RPM Boost Air Flow
0 0 0
600 0 4
1200 0 7.8
1800 0 11.6
2400 0 14.4
3000 0 16.5
0 10 0
600 10 8.6
1200 10 16.75
1800 10 24.9
2400 10 27.4
3000 10 30
0 20 0
600 20 13.2
1200 20 25.7
1800 20 38.2
2400 20 40.5
3000 20 43.5
0 30 0
600 30 18.6
1200 30 35.2
1800 30 51.8
2400 30 55.05
3000 30 58.3
Can someone suggest a method and possibly where I could learn the method. Or, better yet, get me started on the method. I have a week to get a model for my boss with a 2%error max. I tried using the least square method with a linear function and got an average of 20% error.
My function for the LMS method was:
Air Flow = .0126*RPM+.9141*Boost-10.078
Any help would be much appreciated,
Tu Le
Last edited by 2le on Mon, 11 May 2009 13:08:45 UTC, edited 1 time in total.
Top
Post subject: Re: Multivariable Nonlinear ModelingPosted: Mon, 11 May 2009 02:32:15 UTC
S.O.S. Oldtimer
Joined: Sun, 8 Jul 2007 15:38:30 UTC
Posts: 268
Location: Virginia, USA
2le wrote:
Hi,
I'm having some problems getting a model for some experimental data. I need a function of Air Flow with dependent variables Boost and RPM. Here's the data:
RPM Boost Air Flow
0 0 0
600 0 4
1200 0 7.8
1800 0 11.6
2400 0 14.4
3000 0 16.5
0 10 0
600 10 8.6
1200 10 16.75
1800 10 24.9
2400 10 27.4
3000 10 30
0 20 0
600 20 13.2
1200 20 25.7
1800 20 38.2
2400 20 10.5
3000 20 43.5
0 30 0
600 30 18.6
1200 30 35.2
1800 30 51.8
2400 30 55.05
3000 30 58.3
Can someone suggest a method and possibly where I could learn the method. Or, better yet, get me started on the method. I have a week to get a model for my boss with a 2%error max. I tried using the least square method with a linear function and got an average of 20% error.
My function for the LMS method was:
Air Flow = .0126*RPM+.9141*Boost-10.078
Any help would be much appreciated,
Tu Le
2400 20 10.5
looks fishy.
Top
Post subject: Posted: Mon, 11 May 2009 13:10:12 UTC
Member
Joined: Thu, 5 Feb 2009 00:05:45 UTC
Posts: 31
that's because its supposed to be 40.5 and not 10.5 ><
I did plot it and it appears linear until towards the end when it looks for like a y=sqrt(x) sort of curve.
Top
Post subject: Posted: Mon, 11 May 2009 22:28:45 UTC
S.O.S. Oldtimer
Joined: Sun, 8 Jul 2007 15:38:30 UTC
Posts: 268
Location: Virginia, USA
My advice is to think about the physics of the experiment and think about what is a plausible form of the function. (I have no idea what you mean by Boost, for example.) You are likely to get better results that way than by treating the data as simply a pure set of numbers and looking for a "good" fit.
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Jump to: Select a forum ------------------ High School and College Mathematics Algebra Geometry and Trigonometry Calculus Matrix Algebra Differential Equations Probability and Statistics Proposed Problems Applications Physics, Chemistry, Engineering, etc. Computer Science Math for Business and Economics Advanced Mathematics Foundations Algebra and Number Theory Analysis and Topology Applied Mathematics Other Topics in Advanced Mathematics Other Topics Administrator Announcements Comments and Suggestions for S.O.S. Math Posting Math Formulas with LaTeX Miscellaneous | 1,363 | 4,180 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2013-20 | latest | en | 0.802467 |
http://mathhelpforum.com/calculus/104069-solving-x-print.html | 1,508,818,789,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828134.97/warc/CC-MAIN-20171024033919-20171024053919-00225.warc.gz | 210,477,727 | 2,667 | # Solving For X?
• Sep 24th 2009, 08:43 AM
xxju1anxx
Solving For X?
How do I solve for x:
e^x=y-1
It might seem simple but im lost right now.
• Sep 24th 2009, 08:51 AM
Rapha
Hi xxju1anxx!
Quote:
Originally Posted by xxju1anxx
How do I solve for x:
e^x=y-1
It might seem simple but im lost right now.
I think it is simple, just use the LN:
$e^x = y-1$
$LN(e^x) = LN(y-1)$
$x*LN(e) = LN(y-1)$
$x*1 = LN(y-1)$
$x = LN(y-1)$
done.
Any questions?
Rapha | 189 | 463 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2017-43 | longest | en | 0.823372 |
http://mathhelpforum.com/calculus/8364-homog-diff-eqn-print.html | 1,524,759,707,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948285.62/warc/CC-MAIN-20180426144615-20180426164615-00268.warc.gz | 192,021,753 | 2,687 | Homog diff eqn
• Dec 3rd 2006, 05:50 PM
ashura
Homog diff eqn
Need help to make numerator diff of denominator in integration, eqn [5].
Solve diff eqn $\displaystyle \frac {\delta y}{\delta x} = \frac {x^2 - y^2}{y^2 + x^2} ..........[1]$
Let $\displaystyle y = vx , where \, v = f(x)$
$\displaystyle \frac {\delta y}{\delta x} = v + x \frac {\delta v}{\delta x} .....[2]$
substituting for y in [1], $\displaystyle \frac {\delta y}{\delta x} = \frac {x^2 - v^2 x^2}{v^2 x^2 + x^2}$
$\displaystyle \frac {\delta y}{\delta x} = \frac {x^2(1 - v^2) }{x^2(v^2 + 1)}$
$\displaystyle \frac {\delta y}{\delta x} = \frac {1 - v^2 }{v^2 + 1} ........[3]$
equating 2 and 3, $\displaystyle \frac {1 - v^2 }{v^2 + 1} = v + x \frac {\delta v}{\delta x}$
$\displaystyle \frac {1 - v^2 }{v^2 + 1} - v = x \frac {\delta v}{\delta x}$
$\displaystyle \frac {1 - v^2 - v^3 - v}{v^2 + 1} = x \frac {\delta v}{\delta x}$
$\displaystyle \frac {v^2 + 1}{1 - v^2 - v^3 - v} = \frac {1}{x} \frac {\delta x}{\delta v}$
$\displaystyle \int \left ( \frac {v^2 + 1}{1 - v^2 - v^3 - v} \right ) \delta y = \int \frac {1}{x} \delta x .........[4]$
comment : $\displaystyle \frac {\delta(1 - v^2 - v^3 - v)}{\delta v} = 2v - 3v^2 -1$
$\displaystyle \therefore \, ???\int \left ( \frac {2v - 3v^2 -1}{1 - v^2 - v^3 - v} \right ) \delta y = \int \frac {1}{x} \delta x........[5]$
Help: need to replace numerator on LHS of [4] with numerator on LHS of [5]
$\displaystyle \frac {1}{2v - 3v^2 -1} \cdot \int \left ( \frac {2v - 3v^2 -1}{1 - v^2 - v^3 - v} \right ) \delta y = \int \frac {1}{x} \delta x$ how do I factor in $\displaystyle v^2 + 1$ in numerator in extreme left ? | 697 | 1,655 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2018-17 | latest | en | 0.373475 |
https://dateandtime.info/distancesouthpole.php?id=276781 | 1,631,975,471,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056476.66/warc/CC-MAIN-20210918123546-20210918153546-00051.warc.gz | 263,028,988 | 9,367 | # Distance between Beirut, Lebanon and the South Pole
13791 km = 8569 miles
During our calculation of the distance to the South Pole we make two assumptions:
1. We assume a spherical Earth as a close approximation of the true shape of the Earth (an oblate spheroid). The distance is calculated as great-circle or orthodromic distance on the surface of a sphere.
2. We calculate the distance between a point on the Earth’s surface and the South Pole as the length of the arc of the meridian passing through this point and the South Pole.
Find out the distance between Beirut and the North Pole, the Equator, the Tropic of Cancer, the Tropic of Capricorn, the Arctic Circle, the Antarctic Circle
Find out the distance between Beirut and other cities
## Beirut, Lebanon
Country: Lebanon
Beirut’s coordinates: 33°53′20″ N, 35°29′39″ E
Population: 1,916,100
Find out what time it is in Beirut right now
Wikipedia article: Beirut
## The South Pole
The South Pole is a point where imaginary Earth’s axis of rotation crosses the Earth's surface in the Southern Hemisphere.
The South Pole is the southernmost place on Earth. The South Pole latitude is 90° South. The South Pole longitude is undefined, because the South Pole is a point where all the meridians meet.
For the same reason the South Pole has no time zone.
For software and devices using GPS satellite navigation system 0° West may be used as conditional South Pole longitude.
The South Pole’s coordinates: 90°00′00″ S
Wikipedia article: the South Pole | 351 | 1,514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2021-39 | latest | en | 0.846405 |
https://physics.stackexchange.com/questions/211857/what-is-meant-by-a-c-number | 1,652,882,031,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522270.37/warc/CC-MAIN-20220518115411-20220518145411-00704.warc.gz | 540,588,505 | 63,363 | # What is meant by a "c-number"?
In Chapter 2 of David Tong's QFT notes, he uses the term "c-number" without ever defining it.
Here is the first place.
However, it's easy to check by direct substitution that the left-hand side is simply a c-number function with the integral expression$$\Delta(x - y) = \int {{d^3p}\over{(2\pi)^3}} {1\over{2E_{\vec{p}}}}(e^{-ip \cdot (x - y)} - e^{ip \cdot (x - y)}).$$
Here is the second place, on the same page (i.e. page 37).
I should mention however that the fact that $[\phi(x), \phi(y)]$ is a c-number function, rather than an operator, is a property of free fields only.
My question is, what does c-number function mean?
• Do you want to understand c-number or c-number function? Oct 11, 2015 at 8:17
A c-number basically means 'classical' number, which is basically any quantity which is not a quantum operator which acts on elements of the Hilbert space of states of a quantum system. It is meant to distinguish from q-numbers, or 'quantum' numbers, which are quantum operators. See http://wikipedia.org/wiki/C-number and the reference therein.
The term c-number is used informally in the way Meer Ashwinkumar describes. As far as I know, it doesn't have a widely promulgated formal definition. However, there is a formal definition for c-number that agrees with the way the term is used in many cases, including the case you're asking about.
As you may know, you can think of the operator formalism for quantum mechanics as a generalized version of probability theory, in which real-valued random variables are represented by self-adjoint operators on a Hilbert space. More generally, complex-valued random variables are represented by normal operators.
A c-number is a random variable represented by a scalar multiple of the identity operator.
Intuitively, a c-number is a random variable that isn't really random: its value is a constant. The identity operator itself, for instance, represents the random variable whose value is always $1$, while $-4$ times the identity represents the random variable whose value is always $-4$. You can see why this makes sense by computing the expectation value, variance, and higher moments of a c-number relative to some state.
In your example, Tong is talking about a model for a random scalar field,^ whose amplitude at the point $x$ is the real-valued random variable $\phi(x)$. For any two points $x$ and $y$, the commutator $[\phi(x), \phi(y)]$ represents an imaginary-valued random variable. The commutator turns out to be a multiple of the identity—in other words, a c-number. Since this c-number depends on $x$ and $y$, Tong calls it a c-number function (of $x$ and $y$).
^ A free scalar field can be seen as a quantum version of white noise.
• The remark "you can think of the operator formalism for quantum mechanics as a generalized version of probability theory" seems to be quite interesting. Can you please provide references to any papers describing it? Mar 10, 2021 at 16:32
• @AmeyJoshi, my favorite is in Peter Whittle's book Probability via Expectation. It's the last chapter, "Quantum Mechanics." In the first few chapters, Whittle lays out the basics of probability as a theory of expectation values on commutative algebras. Once you get that, you can turn to the last chapter for the non-commutative case. Mar 10, 2021 at 18:16
• Thanks a lot, @Vectornaut. The book looks quite interesting. Mar 11, 2021 at 4:29
This particular "$c$-number function" is called the Pauli-Jordan Operator. You might want to peruse Ryder's Quantum Field Theory specifically §4.2 and §6.1. | 879 | 3,590 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-21 | latest | en | 0.9332 |
https://www.answersarena.com/ExpertAnswers/b-assignment-here-is-your-applied-problem-for-this-week-the-value-of-a-car-v-in-dollars-t-year-pa898 | 1,701,319,904,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100164.87/warc/CC-MAIN-20231130031610-20231130061610-00157.warc.gz | 742,653,135 | 6,299 | Home / Expert Answers / Algebra / b-assignment-here-is-your-applied-problem-for-this-week-the-value-of-a-car-v-in-dollars-t-year-pa898
# (Solved): B. Assignment Here is your applied problem for this week: The value of a car, V, in dollars, t year ...
B. Assignment Here is your applied problem for this week: The value of a car, , in dollars, years after it is purchased is given by . Your entry must contain the following 1. What is the -intercept? What does it represent? 2. What is the slope? What does it mean in the context of the problem? 3. When will the car be worth ? 4. Using complete sentences, explain why this is important information for you to know. Each of the above is worth 5 points for a total of 20 points. Please write using complete sentences and be aware of your gramntar, punctuation and spelling.
We have an Answer from Expert | 209 | 854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-50 | latest | en | 0.918416 |
https://shop.khoji.net/p/phe-09-solved-2018/ | 1,701,550,972,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100452.79/warc/CC-MAIN-20231202203800-20231202233800-00816.warc.gz | 601,676,807 | 32,326 | # PHE-09 Optics Solved Assignment 2018
Course Name: OPTICS
Course Code: PHE-09
Assignment Code: PHE-09/TMA/2018
Medium: English
Category:
## Description
### Assignment Code: PHE-09/TMA/2018
Questions:
1. a) State Fermat’s principle. On the basis of this principle, show that when light passes from a medium of lower refractive index to a medium of higher refractive index, it bends towards the normal to the interface between the two media.
b) The wavelength of the electromagnetic wave used for AM broadcasting is 110 m. Calculate the frequency of the wave.
c) Discuss the opponent-process theory of colour vision.
d) What do you understand by birefringence? What are distinguishing feature of the o− and e− rays? Draw a diagram showing o− and e−wave surfaces.
e) Unpolarised light is incident on two polarizing sheets placed over one another. The intensity of light transmitted finally is one-third of the intensity of the incident light. Calculate the angle between the transmission axes of the two polarizing sheets.
2. a) Obtain the conditions for observing maxima and minima in a Young two-slit interference pattern. Show that these conditions change when a thin transparent sheet of thickness t and refractive index µ is introduced in the path of one of the superposing beams.
b) Newton’s rings are formed with reflected light of wavelength 589 nm using a planoconvex lens and a plane glass plate with a liquid between them. The diameter of the 15th ring is 5.1 mm and radius of curvature of the curved surface of the lens is 1 m. Calculate the refractive index of the liquid if the ring is dark.
c) A light of amplitude ‘a’ is incident obliquely on a parallel thin film. If ‘r’ and ‘t’ represent the fractions of incident light reflected and transmitted respectively, draw a diagram to explain multiple reflection of incident light and write the amplitudes of successive rays.
d) A thin film of glass of refractive index 1.5 is inserted in one arm of Michelson Interferometer. Calculate the thickness of the film if a shift of 10 fringes is observed. The wavelength of the light used is 589 nm.
3. a) What is a zone plate? How does a positive zone plate differ from a negative zone plate? Show that each Fresnel half period zone has nearly the same area.
b) The diffraction pattern due to a single slit of width 0.4 cm is obtained with the help of a lens of focal length 30 cm. If wavelength of the light used is 589 nm, calculate the distance of the first dark fringe and the consecutive bright fringe from the axis.
c) Sodium light is incident normally on a grating of width 4×10−3m. The total number of lines on the grating is 2000. Determine the angular separation between the sodium D-lines in the first order spectrum.
d) A single slit has a width of 0.04 mm. A parallel beam of light of wavelength 560 nm is incident normally on it. If the distance between the slit and the screen is 100 cm, calculate the separation between the central maximum and the second minima in the diffraction pattern.
4. a) Derive an expression to show that the condition of population inversion is necessary for laser action.
b) Compare and contrast three level and four level pumping schemes.
c) With a labelled diagram, discuss the recording a hologram.
d) Explain the propagation of light through optical fibre. What are the advantages of using optical fibres for communication?
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Only logged in customers who have purchased this product may leave a review. | 792 | 3,500 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-50 | longest | en | 0.875034 |
http://www.efunda.com/glossary/units/units--force--slug_foot_per_square_second.cfm | 1,563,435,133,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525524.12/warc/CC-MAIN-20190718063305-20190718085305-00003.warc.gz | 204,974,621 | 7,713 | Force Units: Slug Foot Per Square Second
FREE Publications Photonics Spectra Reinforced Plastics Industrial Laser Vision Systems Design Laser Focus World Filtration+Separation Business Aviation World Pumps more
Materials Design Processes Units Formulas Math
Glossary » Units » Force » Slug Foot Per Square Second
Slug Foot Per Square Second (slug-ft/s2) is a unit in the category of Force. It is also known as slug-foot/square second. Slug Foot Per Square Second (slug-ft/s2) has a dimension of MLT-2 where M is mass, L is length, and T is time. It can be converted to the corresponding standard SI unit N by multiplying its value by a factor of 4.44822161526.
Note that the seven base dimensions are M (Mass), L (Length), T (Time), Q (Temperature), N (Aamount of Substance), I (Electric Current), and J (Luminous Intensity).
Other units in the category of Force include A.u. of Force (a.u.), Crinal (crinal), Dyne (dyn), Gram Force (gf), Joule Per Centimeter (J/cm), Kilogram Force (kgf, kgp), Kilogram Meter Per Square Second (kg-m/s2), Kilopound Force (kip Force) (kipf), Newton (N), Ounce Force (av.) (ozf (av.)), Pound Force (lbf (av.), lbf), Pound Force Foot Per Inch (lbf-ft/in), Poundal (pdl), Slug Force, Square Foot Inch of Mercury (0°C) (ft2-inHg (0 °C)), Square Foot Inch of Mercury (15.56°C) (ft2-inHg (15.56 °C)), Ton Force (long), Ton Force (metric), and Ton Force (short). | 388 | 1,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2019-30 | latest | en | 0.779051 |
http://forums.wolfram.com/mathgroup/archive/2006/Dec/msg00308.html | 1,723,204,527,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640763425.51/warc/CC-MAIN-20240809110814-20240809140814-00251.warc.gz | 8,586,315 | 9,956 | Re: Re: Finding the periphery of a region
• To: mathgroup at smc.vnet.net
• Subject: [mg72119] Re: [mg72005] Re: Finding the periphery of a region
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Wed, 13 Dec 2006 06:38:52 -0500 (EST)
• References: <el8ufm\$st3\$1@smc.vnet.net> <200612081117.GAA20172@smc.vnet.net> <6DA7D258-EA36-456E-A8F7-B4CBE82001B8@mimuw.edu.pl> <457D0248.7020004@metrohm.ch> <78BE6846-13C6-4E0F-B440-B211BBB62250@mimuw.edu.pl> <457D0BAF.9040000@metrohm.ch> <8197B414-9148-43A4-8BBE-EB019F2CCD02@mimuw.edu.pl>
```On 11 Dec 2006, at 17:55, Andrzej Kozlowski wrote:
> On 11 Dec 2006, at 16:41, Daniel Huber wrote:
>> Hi Andrzej ,
>> I think the whole problem arises because we are loking at
>> solutions to (in)equations in R that actually lay in C. In C there
>> wouldn't be any "singular" points.
>> On the other hand, if somebodey has an applied problem involving
>> area, then I think he can most of the time simply ignore isolated
>> points.
>> Anyway, thank's a lot that you pointed me to this interesting fact.
>> Daniel
>
>
> Well, form the point of view of measure theory an isolated point is
> nothing but from the point of view of topology it is enormously
> important. (And there are real life applications, e.g. in robotics,
> where the existence of isolated points can have quite dramatic
> effect).
> Partly because of such things, from the point of view of topology,
> real algebraic sets are much more interesting than complex ones.
> Among well known examples that can be plotted in Mathematica are:
>
> The "Cartan umbrella":
>
> z*(x^2 + y^2) - x^3==0
>
> another "umbrella":
>
> x^3+ z*x - y^2 ==0
>
> a surface with curious properties:
>
> x^2 (1-z^2) == x^4 + y^2.
>
> The Nash-Tognoli theorem says that every compact smooth manifold is
> diffeomorphic to some non singular real algebraic vaiety. For
> example the Klein bottle here is an algebraic equation of the Klein
> bottle:
>
> 768*x^4 - 1024*x^5 - 128*x^6 +
> 512*x^7 - 80*x^8 - 64*x^9 + 16*x^10 + 144*x^2*y^2 -
> 768*x^3*y^2 - 136*x^4*y^2 + 896*x^5*y^2 -
> 183*x^6*y^2 - 176*x^7*y^2 + 52*x^8*y^2 + 400*y^4 +
> 256*x*y^4 - 912*x^2*y^4 + 256*x^3*y^4 + 315*x^4*y^4 -
> 144*x^5*y^4 - 16*x^6*y^4 + 4*x^8*y^4 - 904*y^6 -
> 128*x*y^6 + 859*x^2*y^6 - 16*x^3*y^6 - 200*x^4*y^6 +
> 16*x^6*y^6 + 441*y^8 + 16*x*y^8 - 224*x^2*y^8 +
> 24*x^4*y^8 - 76*y^10 + 16*x^2*y^10 + 4*y^12 -
> 2784*x^3*y*z + 4112*x^4*y*z - 968*x^5*y*z -
> 836*x^6*y*z + 416*x^7*y*z - 48*x^8*y*z +
> 1312*x*y^3*z + 2976*x^2*y^3*z - 5008*x^3*y^3*z -
> 12*x^4*y^3*z + 2016*x^5*y^3*z - 616*x^6*y^3*z -
> 64*x^7*y^3*z + 32*x^8*y^3*z - 1136*y^5*z -
> 4040*x*y^5*z + 2484*x^2*y^5*z + 2784*x^3*y^5*z -
> 1560*x^4*y^5*z - 192*x^5*y^5*z + 128*x^6*y^5*z +
> 1660*y^7*z + 1184*x*y^7*z - 1464*x^2*y^7*z -
> 192*x^3*y^7*z + 192*x^4*y^7*z - 472*y^9*z -
> 64*x*y^9*z + 128*x^2*y^9*z + 32*y^11*z - 752*x^4*z^2 +
> 1808*x^5*z^2 - 1468*x^6*z^2 + 512*x^7*z^2 -
> 64*x^8*z^2 + 6280*x^2*y^2*z^2 - 5728*x^3*y^2*z^2 -
> 4066*x^4*y^2*z^2 + 5088*x^5*y^2*z^2 -
> 820*x^6*y^2*z^2 - 384*x^7*y^2*z^2 + 96*x^8*y^2*z^2 -
> 136*y^4*z^2 - 7536*x*y^4*z^2 + 112*x^2*y^4*z^2 +
> 8640*x^3*y^4*z^2 - 2652*x^4*y^4*z^2 -
> 1152*x^5*y^4*z^2 + 400*x^6*y^4*z^2 + 2710*y^6*z^2 +
> 4064*x*y^6*z^2 - 3100*x^2*y^6*z^2 - 1152*x^3*y^6*z^2 +
> 624*x^4*y^6*z^2 - 1204*y^8*z^2 - 384*x*y^8*z^2 +
> 432*x^2*y^8*z^2 + 112*y^10*z^2 + 3896*x^3*y*z^3 -
> 7108*x^4*y*z^3 + 3072*x^5*y*z^3 + 768*x^6*y*z^3 -
> 768*x^7*y*z^3 + 128*x^8*y*z^3 - 3272*x*y^3*z^3 -
> 4936*x^2*y^3*z^3 + 8704*x^3*y^3*z^3 - 80*x^4*y^3*z^3 -
> 2496*x^5*y^3*z^3 + 608*x^6*y^3*z^3 + 2172*y^5*z^3 +
> 5632*x*y^5*z^3 - 2464*x^2*y^5*z^3 - 2688*x^3*y^5*z^3 +
> 1056*x^4*y^5*z^3 - 1616*y^7*z^3 - 960*x*y^7*z^3 +
> 800*x^2*y^7*z^3 + 224*y^9*z^3 + 752*x^4*z^4 -
> 1792*x^5*z^4 + 1472*x^6*z^4 - 512*x^7*z^4 +
> 64*x^8*z^4 - 3031*x^2*y^2*z^4 + 1936*x^3*y^2*z^4 +
> 2700*x^4*y^2*z^4 - 2304*x^5*y^2*z^4 +
> 448*x^6*y^2*z^4 + 697*y^4*z^4 + 3728*x*y^4*z^4 +
> 24*x^2*y^4*z^4 - 3072*x^3*y^4*z^4 + 984*x^4*y^4*z^4 -
> 1204*y^6*z^4 - 1280*x*y^6*z^4 + 880*x^2*y^6*z^4 +
> 280*y^8*z^4 - 800*x^3*y*z^5 + 1488*x^4*y*z^5 -
> 768*x^5*y*z^5 + 128*x^6*y*z^5 + 992*x*y^3*z^5 +
> 1016*x^2*y^3*z^5 - 1728*x^3*y^3*z^5 +
> 480*x^4*y^3*z^5 - 472*y^5*z^5 - 960*x*y^5*z^5 +
> 576*x^2*y^5*z^5 + 224*y^7*z^5 + 16*x^4*z^6 +
> 388*x^2*y^2*z^6 - 384*x^3*y^2*z^6 + 96*x^4*y^2*z^6 -
> 76*y^4*z^6 - 384*x*y^4*z^6 + 208*x^2*y^4*z^6 +
> 112*y^6*z^6 - 64*x*y^3*z^7 + 32*x^2*y^3*z^7 +
> 32*y^5*z^7 + 4*y^4*z^8==0
>
> One can plot this using Mathematica's Graphics`ContourPlot3D
> package, but it won't come out very nice. It is much better to use
> Jens Kuska's excellent MVContourPlot3D funciton, which is a part of
> his MathGL3d (one does not need the commercial version to be able
> to use this function). Of course one can get the same picture much
> more easily with ParametricPlot3D using well known parametric
> description of the Klein bottle:
>
> {x,y,z}={(Cos[Ï?/2]*Sin[θ] - Sin[Ï?/2]*Sin[2*θ] + 2)*
> Cos[Ï?], (Cos[Ï?/2]*Sin[θ] -
> Sin[Ï?/2]*Sin[2*θ] + 2)*Sin[Ï?],
> Sin[Ï?/2]*Sin[θ] + Cos[Ï?/2]*Sin[2*θ]}
>
> Actually, the above algebraic equation was obtained form the
> parametric one by using GroebnerBasis. It is a nice illustration of
> the relation (and difference) between non-constructive mathematics,
> like the Nash-Tognoli theorem and computational one. Getting the
> algebraic equation from the parametric one without using
> Mathematica or a similar program would be quite challenging.
>
> Andrej Kozlowski
>
>
>
>
I just realized that this Klein bottle is not a good illustration of
the Nash-Tognoli theorem because the algebraic surface in R^3 that we
get is obviously singular. (It is not really diffeomorphic to the
Klein bottle but the image of an immersion of the Klein bottle in
R^3). I should have used a torus or a Klein bottle in R^4 but that
would have been less impressive (both the formulas and the pictures).
Andrzej
```
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## Solution for problem 1 Chapter 2
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Problem 1
Images Formed by Plane Mirrors
What are the differences between real and virtual images? How can you tell (by looking) whether an image formed by a single lens or mirror is real or virtual?
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## Discover and learn what students are asking
Calculus: Early Transcendental Functions : Preparation for Calculus
?In Exercises 1–4, find any intercepts. $$y=\frac{x-3}{x-4}$$
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?In Exercises 7 - 38, find both first partial derivatives. $$z=e^{x / y}$$
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# Annotation of /trunk/pg/lib/Parser/BOP/power.pm
1 : sh002i 2558 ######################################################################### 2 : # 3 : # Implements exponentiation 4 : # 5 : package Parser::BOP::power; 6 : use strict; use vars qw(@ISA); 7 : @ISA = qw(Parser::BOP); 8 : 9 : # 10 : # Check that operand types are OK. 11 : # For non-numbers, promote to Matrix and check 12 : # that the sizes are OK and that the exponents are numbers 13 : sub _check { 14 : my $self = shift; 15 : return if ($self->checkStrings()); 16 : return if ($self->checkLists()); 17 : return if ($self->checkNumbers()); 18 : my ($ltype,$rtype) = $self->promotePoints('Matrix'); 19 : if ($rtype->{name} eq 'Number') { 20 : if ($ltype->{name} eq 'Matrix') {$self->checkMatrixSize($ltype,$ltype)} 21 : else {$self->Error("You can only raise a Number to a power")} 22 : } else {$self->Error("Exponents must be Numbers")} 23 : } 24 : 25 : # 26 : # Do perl exponentiation 27 : # 28 : sub _eval {$_[1] **$_[2]} 29 : 30 : # 31 : # Return 1 for power of zero or base of 1. 32 : # Return base if power is 1. 33 : # Return 1/base if power is -1. 34 : # 35 : sub _reduce { 36 : dpvc 2678 my $self = shift; my$equation = $self->{equation}; 37 : my$parser = $equation->{context}{parser}; 38 : return$parser->{Number}->new($equation,1) 39 : sh002i 2558 if ($self->{rop}{isZero} || $self->{lop}{isOne}); 40 : return$self->{lop} if ($self->{rop}{isOne}); 41 : if ($self->{rop}->isNeg && $self->{rop}->string eq '-1') { 42 : dpvc 2678$self = $parser->{BOP}->new($equation,'/', 43 : $parser->{Number}->new($equation,1),$self->{lop}); 44 : sh002i 2558$self = $self->reduce; 45 : } 46 : return$self; 47 : } 48 : 49 : # 50 : # Put exponent in braces for TeX 51 : # 52 : sub TeX { 53 : dpvc 2696 my ($self,$precedence,$showparens,$position,$outerRight) = @_; 54 : sh002i 2558 my$TeX; my $bop =$self->{def}; 55 : dpvc 2696 my $addparens = 56 : defined($precedence) && 57 : ($showparens eq 'all' ||$precedence > $bop->{precedence} || 58 : ($precedence == $bop->{precedence} && 59 : ($bop->{associativity} eq 'right' || $showparens eq 'same'))); 60 : my$outerRight = !$addparens && ($outerRight || $position eq 'right'); 61 : 62 : sh002i 2558 my$symbol = (defined($bop->{TeX}) ?$bop->{TeX} : $bop->{string}); 63 : dpvc 2696 if ($self->{lop}->class eq 'Function' && $self->{rop}->class eq 'Number' && 64 :$self->{rop}{value} > 0 && int($self->{rop}{value}) ==$self->{rop}{value}) { 65 : $TeX =$self->{lop}->TeX($precedence,$showparens,$position,$outerRight, 66 : $symbol.'{'.$self->{rop}->TeX.'}'); 67 : $addparens = 0; 68 : } else { 69 :$TeX = $self->{lop}->TeX($bop->{precedence},$bop->{leftparens},'left',$outerRight). 70 : $symbol.'{'.$self->{rop}->TeX.'}'; 71 : } 72 : 73 : $TeX = '\left('.$TeX.'\right)' if $addparens; 74 : return$TeX; 75 : sh002i 2558 } 76 : 77 : ######################################################################### 78 : 79 : 1; 80 : | 1,056 | 3,049 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-05 | latest | en | 0.246158 |
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This is a matching activity that can be used in a small group, whole class or as a scavenger hunt. Students will have to match binomial factors to quadratic expressions. There is no common denominator. Perfect square trinomials and difference of squares is covered.
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Posts
## The temperature of the Universe at recombination (decoupling)
Last week someone (“Cosm”) posted an interesting comment/question on my post “What is the redshift of the Cosmic Microwave Background (CMB)?” . The question asked how we can say that the temperature of the Universe at recombination is about 3000K when the energy of electrons with such a temperature would be
$E = \frac{ 3 }{ 2 } k T = 6.2 \times 10^{-20} \text{ J } = 0.39 \text{ eV }$
and the ionisation potential of hydrogen is 13.6 eV. This would imply, naively, that recombination would occur at a higher temperature, roughly $(13.6/0.39) \times 3000 \approx 105,000 \text{ Kelvin}$. But, it does not occur at a temperature of just over 100,000 K, but at about 3000 K. Why is this?
“Cosm” asked me this question on my blogpost about how do we know the redshift of the CMB
I decided that it was a sufficiently interesting point that I would go through the detail of why recombination (or decoupling as I prefer to call it) happened when the Universe was about 3,000K; even though electrons cease to be able to thermally ionise hydrogen at a much higher temperature of about 100,000K, so one might think that decoupling would happen earlier. The clue is that it has to do with something called equilibrium theory.
A cartoon of recombination (decoupling). As the temperature of the Universe fell, the free electrons combined with naked protons to produce neutral hydrogen, and the Universe became transparent. This is when the radiation (which was already there) was able to travel through the Universe, and it is this radiation which we see as the Cosmic Microwave Background (CMB).
## The temperature of decoupling based on equilibrium theory
One can get a reasonable estimate of the temperature (and redshift) when recombination (decoupling) took place by using equilibrium theory. This was developed in the 1920s and is based on the Saha ionisation equation, named after Indian astrophysicist Meghnad Saha.
We are going to assume that the process
$p + e^{-} \leftrightarrow H + \gamma$
is the dominant reaction for creating neutral hydrogen. That is, electrons and bare protons combining. When a free electron combines with a proton to create neutral hydrogen it will also create a photon, and that is what the $\gamma$ represents (it is not a gamma-ray photon, it is more likely to be a UV or visible-light photon). But, it is a two way process, as obviously electrons can also be ionised to go back to free electrons and bare protons.
We are going to denote the number density of free electrons (the number per unit volume) as $n_{e}$, the number density of free protons (ionised hydrogen) as $n_{p}$, and the number density of neutral hydrogen as $n_{H}$.
We can calculate the relative abundance of the free electrons to protons and neutral hydrogen via the Saha equation
$\frac{ n_{p} n_{e} }{ n_{H} } = \left( \frac{ m_{e} k_{B} T }{ 2 \pi \hbar^{2} } \right)^{3/2} \; exp \left( - \frac{ E_{1} }{ k_{B} T } \right) \text{ (1) }$
Where $m_{e}$ is the mass of an electron, $k_{B}$ is Boltzmann’s constant, $T$ is the temperature, $\hbar$ is the reduced Planck constant (which is defined as $\hbar=h/2\pi$ where $h$ is Planck’s constant), and $E_{1}$ is the ionisation potential of hydrogen (13.6 eV). We know from charge neutrality that $n_{e} = n_{p}$. We are going to define $x_{e}$ as the fraction of free electrons, so by definition
$x_{e} = \frac{ n_{e} }{ (n_{p} + n_{H} ) } \text{ (2) }$
Because $n_{e} = n_{p}$ we can re-write Equation (1) as
$\frac{ n_{e}^{2} }{ n_{H} } = \left( \frac{ m_{e} k_{B} T }{ 2 \pi \hbar^{2} } \right)^{3/2} \; exp \left( - \frac{ E_{1} }{ k_{B} T } \right) \text{ (3) }$
and from (2) we can write
$x_{e}^{2} = \frac{ n_{e}^{2} }{ ( n_{p} + n_{H} )^{2} } \text{ (4) }$
We can also write
$1 - x_{e} = 1 - \frac{ n_{e} }{ (n_{p} + n_{H} ) } = \frac{ n_{p} + n_{H} - n_{e} }{ (n_{p} + n_{H} ) } = \frac{ n_{H} }{ (n_{p} + n_{H} ) } \text{ (5) }$
Dividing Eq. (4) by Eq. (5) we have
$\frac{x_{e}^{2} }{ ( 1 - x_{e} ) } = \frac{ n_{e}^{2} }{ ( n_{p} + n_{H} )^{2} } \; \cdot \; \frac{ (n_{p} + n_{H} ) }{ n_{H} } = \frac{ n_{e}^{2} }{ (n_{p} + n_{H} ) } \; \cdot \; \frac{ 1 }{ n_{H} } \text{ (6) }$
But
$\frac{ n_{e}^{2} }{ n_{H} } = \left( \frac{ m_{e} k_{B} T }{ 2 \pi \hbar^{2} } \right)^{3/2} \; exp \left( - \frac{ E_{1} }{ k_{B} T } \right)$
so we can re-write Eq. (6) as
$\frac{x_{e}^{2} }{ ( 1 - x_{e} ) } = \frac{ 1 }{ (n_{p} + n_{H} ) } \left( \frac{ m_{e} k_{B} T }{ 2 \pi \hbar^{2} } \right)^{3/2} exp \left( - \frac{ E_{1} }{ k_{B} T } \right) \text{ (7) }$
Most of the quantities on the right hand side are physical constants, and the rest are known functions of redshift $z$. For example, the temperature history of the CMB at any redshift $z$ is just given by
$T(z) = 2.728 (1 + z) \text{ (8) }$
where $T=2.728$ is the CMB’s current temperature. The total number density of hydrogen (neutral and ionised), $(n_{p} + n_{H})$ is also a function of redshift, and is given by
$(n_{p} + n_{H}) (z) = 1.6(1+z)^{3} \text{ per m}^{3} \text{ (9) }$
where $1.6 \text{ per m}^{3}$ is the currently observed density. So, we will re-write Eq. (7) as
$\frac{x_{e}^{2} }{ ( 1 - x_{e} ) } = \frac{ 1 }{ 1.6(1+z)^{3} } \left( \frac{ m_{e} k_{B} }{ 2 \pi \hbar^{2} } \right)^{3/2} T^{3/2} \; exp \left( - \frac{ E_{1} }{ k_{B} T } \right) \text{ (10) }$
Equation (10) cannot be solved analytically, only numerically.
We will assume that we need a fraction of free electrons of $x_{e} = 50\%$ for decoupling to have occurred (in other words, $50\%$ of the electrons have bound to protons to form neutral hydrogen). We will try different temperatures to see what fraction $x_{e}$ they give. Note: when we assume a particular temperature, this will fix the redshift, because of Eq. (8).
Let us first try a temperature of $T=3000 \text{ Kelvin}$. From Eq. (8) this gives $z \approx 1100$. Plugging these values of $T$ and $z$ into Eq. (10) gives
$\frac{x_{e}^{2} }{ ( 1 - x_{e} ) } = 2.8 \times 10^{-6} \rightarrow x_{e}^{2} + 2.8 \times 10^{-6} x_{e} - 2.8 \times 10^{-6} = 0$
Solving this quadratic equation gives $\boxed{ x_{e} = 1.67 \times 10^{-3} \text{ if } T=3000 \text{ K } }$
This is obviously much less than $50\%$, by the time the Universe has cooled to 3000K there are very few free electrons, decoupling is essentially complete.
What if we use $T=4000 \text{ K}$? Doing the same thing we find $z \approx 1500$ which then gives
$\frac{x_{e}^{2} }{ ( 1 - x_{e} ) } = 0.85 \rightarrow x_{e}^{2} + 0.85 x_{e} - 0.85 = 0$
Solving this quadratic gives $\boxed{ x_{e} = 0.6 \text{ if } T=4000 \text{ K } }$. This means that $60\%$ of the electrons are free when the temperature is 4000K, meaning $40\%$ of the hydrogen atoms have become neutral.
Thirdly, let us try 3800K. This gives $z \approx 1400$. Plugging this into Eq. (9) gives
$\frac{x_{e}^{2} }{ ( 1 - x_{e} ) } = 0.1225 \rightarrow x_{e}^{2} + 0.1225 x_{e} - 0.1225 = 0$
Solving this gives $\boxed{x_{e} = 0.29 \text{ if } T=3800 \text{ K}}$
At a temperature of T=3800K just under $30\%$ of the electrons are free, so some $70\%$ of the hydrogen atoms are neutral.
So, we can surmise from this that the Universe had decoupled enough to be about $50\%$ transparent to radiation by the time the temperature was a little below $\boxed{ T=4000 \text{ Kelvin} }$. This is why the temperature of decoupling (recombination) is usually given as lying between a temperature of $T= 4000 \text{K and } 3000\text{K}$.
## Why is this just an approximation?
As I mentioned above, this is just an approximation, but not a bad one. It’s an approximation because it is a simplification of what is actually going on. In 1968, Jim Peebles (who had done the work in the 1965 Dicke etal. paper) and, independently, Yakov Zel’dovich (he of the Sunyaev-Zel’dovich effect) in the USSR, worked out a more complete theory where the hydrogen has three energy levels, rather than what we have done here where we assume the free electrons go straight into the ground-state (i.e. only two energy levels). Their third level was the n=2 state, the energy level just above the ground state. This is an important energy level for hydrogen, as it is transitions down to the n=2 level which give rise to visible-light photons. Using this more complicated 3-level model gives a Universe which is 90% neutral at $z \approx 1100$, which would correspond to a temperature of T=3000K, which is why this temperature is most often quoted. You can read more about their 3-level model here.
## Summary
Using equilibrium theory, which is an oversimplification, gives the following fractions of neutral hydrogen for three different temperatures
• At T=3000K the Universe would have been more than 99% neutral
• At T=4000K the Universe would have been about 40% neutral
• At T=3800K the Universe would have been about 70% neutral
Using a more correct 3-level model developed by Peebles and, independently, by Zel’dovich, gives that the Universe would have been about 90% neutral by the time the temperature had dropped to T=3000K. It is this temperature which is usually quoted when we talk about the temperature of the Universe when recombination (decoupling) occurred.
Read Full Post »
## What is the redshift of the Cosmic Microwave Background (CMB)?
Last week, as I mentioned in this blog here, I had an article on the Cosmic Microwave Background’s accidental discovery in 1965 published in The Conversation. Here is a link to the article. As of writing this, there have been two questions/comments. One was from what I, quite frankly, refer to as a religious nutter, although that may be a bit harsh! But, the second comment/question by a Mark Robson was very interesting, so I thought I would blog the answer here.
This article on the Cosmic Microwave Background was published in The Conversation last Thursday (23rd July 2015)
Mark asked how we know the redshift of the CMB if it has no emission or absorption lines, which is the usual way to determine redshifts of e.g. stars and galaxies. I decided that the answer deserves its own blogpost – so here it is.
## How does the CMB come about
As I explain in more detail in my book on the CMB, the origin of the CMB is from the time that the Universe had cooled enough so that hydrogen atoms could form from the sea of protons and electrons that existed in the early Universe. Prior to when the CMB was “created”, the temperature was too high for hydrogen atoms to exist; electrons were prevented from combining with protons to form atoms because the energy of the photons in the Universe’s radiation (given by $E=h \nu$ where $\nu$ is the frequency) and of the thermal energy of the electrons was high enough to ionise any hydrogen atoms that did form. But, as the Universe expanded it cooled.
In fact, the relationship between the Universe’s size and its temperature is very simple; if $a(t)$ represents the size of the Universe at time $t$, then the temperature $T$ at time $t$ is just given by
$T(t) \propto \frac{ 1 }{ a(t) }$
This means that, as the Universe expands, the temperature just decreases in inverse proportion to its size. Double the size of the Universe, and the temperature will halve.
When the Universe had cooled to about 3,000K it was cool enough for the electrons to finally combine with the protons and form neutral hydrogen. At this temperature the photons were not energetic enough to ionise any hydrogen atoms, and the electrons had lost enough thermal energy that they too could not ionise electrons bound to protons. Finally, for the first time in the Universe’s history, neutral hydrogen atoms could form.
For reasons that I have never properly understood, astronomers and cosmologists tend to call this event recombination, although really it was combination, without the ‘re’ as it was happening for the first time. A term I prefer more is decoupling, it is when matter and radiation in the Universe decoupled, and the radiation was free to travel through the Universe. Before decoupling, the photons could not travel very far before they scattered off free electrons; after decoupling they were free to travel and this is the radiation we see as the CMB.
## The current temperature of the CMB
It was shown by Richard Tolman in 1934 in a book entitled Relativity, Thermodynamics, and Cosmology that a blackbody will retain its blackbody spectrum as the Universe expands; so the blackbody produced at the time of decoupling will have retained its blackbody spectrum through to the current epoch. But, because the Universe has expanded, the peak of the spectrum will have been stretched by the expansion of space (so it is not correct to think of the CMB spectrum as having cooled down, rather than space has expanded and stretched its peak emission to a lower temperature). The peak of a blackbody spectrum is related to its temperature in a very precise way, it is given by Wien’s displacement law, which I blogged about here.
In 1990 the FIRAS instrument on the NASA satellite COBE (COsmic Background Explorer) measured the spectrum of the CMB to high precision, and found it to be currently at a temperature of $2.725 \text{ Kelvin}$ (as an aside, the spectrum measured by FIRAS was the most perfect blackbody spectrum ever observed in nature).
The spectrum of the CMB as measured by the FIRAS instrument on COBE in 1990. It is the most perfect blackbody spectrum in nature ever observed. The error bars are four hundred times larger than normal, just so one can see them!
It is thus easy to calculate the current redshift of the CMB, it is given by
$z \text{ (redshift)} = \frac{3000}{2.725} = 1100$
and “voilà”, that is the redshift of the CMB. Simples 😉
Read Full Post »
## The most distant galaxy yet discovered – 30 billion light years away!
A few weeks ago it was announced that a team had discovered what seems to be the most distant galaxy yet discovered. You can read the BBC story about it here, or if you like you can read the Nature science paper here to get as much detail as you could wish for. The galaxy, which has the catchy name z8_GND_5296, was discovered using the Hubble Space Telescope, with its distance being determined using the Keck 10m telescope on the summit of Mauna Kea.
In fact, what astronomers measure is not the distance of a distant galaxy, but its redshift, which astronomers denote with the letter $z$. Redshift is the movement of the spectral lines of a galaxy to longer wavelengths due to the expansion of the Universe, the expansion discovered by Edwin Hubble in 1929. The redshift of this newly discovered galaxy has been found by Keck to be $z=7.51$, beating the previous record of $z=7.21$. But how do astronomers translate this into a distance?
## The cosmological definition of redshift
It turns out that measuring distances in astronomy is one of the most difficult things to do for several reasons. Not only are there very few direct ways to measure the distance to an object, after all we can hardly lay down a measuring tape between us and the stars and galaxies! But, to make it even worse, there also are various definitions of distance! In a future blog I will talk about the most direct ways we have to measure distance, but how we translate from these measurements to a distance also depend on the geometry of the Universe, which Einstein showed in his General Theory of Relativity is determined by the effects of gravity.
The geometry of the Universe is determined by its average density $\Omega$, and how this relates to something called the “critical density” $\Omega_{0}$, which is the dividing line between whether the Universe will carry on expanding forever, or stop expanding and start to collapse. If average density $\Omega > \Omega_{0}$ the Universe will stop expanding and collapse. If $\Omega < \Omega_{0}$ the Universe will carry on expanding forever, and if the average density $\Omega = \Omega_{0}$ the Universe is on the dividing line between the two, and is said to have a flat geometry. Without going into the details here, most cosmologists believe that we live in a Universe where $\Omega = \Omega_{0}$, that is a flat Universe.
The preferred method for measuring large distances “directly” is to use something called a Type Ia Supernova, I will blog about this method again in a future blog. But, we can only see Type Ia supernovae out to distances corresponding to a redshift of about $z=1$. The galaxy in this story is much further away than this, $z=7.51$. So, to calculate its distance we have to use a model for the expansion of the Universe, and something called Hubble’s law.
The measured redshift of a galaxy (or any object) is just given by
$z = \frac{ \lambda - \lambda_{0} }{ \lambda_{0} } \text{ (Eq. 1) }$
where $\lambda$ is the observed wavelength and $\lambda_{0}$ would be the wavelength of a spectral line (usually for a galaxy it is a line called the Lyman-alpha line) in the laboratory.
As long as the redshift is much less than 1, we can then write that
$z=\frac{ v }{ c } \text{ (Eq. 2) }$
where $v$ is the recession velocity of the galaxy and $c$ is the speed of light. In the case of $z$ not being less than 1, we need to modify this equation to the relativistic version, so we write
$1 + z = \sqrt{ \frac{ 1+ v/c }{ 1 - v/c } } \text{ (Eq. 3) }$
In our case, $z=7.51$, so we need to use this relativistic formula, and when we do we get that the recession velocity of the galaxy is $97\% \text{ of c }$, the speed of light.
Re-arranging equation 1 we can write $1 + z = \frac{ \lambda }{ \lambda_{0} }$. In principle, the distance and redshift are just related via the Hubble law
$v = H_{0} d \text{ (Eq. 4) }$,
where $v$ is the recession velocity of the galaxy, $H_{0}$ is the Hubble constant, and $d$ is the distance of the galaxy.
Things get a lot more complicated, however, when we take into account General Relativity, and its effects on the curvature of space, and even the definition of distance in an expanding Universe. I will return to this in a future blog, but here I will just quote the answer one gets if one inputs a redshift of $z=7.51$ into a “distance calculator” where we specify the value of Hubble’s constant to be $H_{0} = 72 \text{ km/s/Mpc }$ and we have a flat Universe ($\Omega=1$) with a value of $\Omega_{M}=0.25$ (the relative density of the Universe in the form of matter) and $\Omega_{vac} = 0.75$ (the relative density of the Universe in the form of dark energy).
Putting these values in gives a co-moving radial distance to the galaxy of $9103 Mpc \text{ or } 29.7 \text{ billion light years}$. (I will define what “co-moving radial distance” is in a future blog, but it is the distance quoted in this story, and is the measurement of distance which is closest to what we think of as “distance”).
The redshift also gives a time when the galaxy was formed, with $z=0$ being the present. We find that it was formed some 13.1 billion years ago, when the Universe was only about 700,000 years old.
## A galaxy 30 billion light years away??
Going back to the “co-moving radial distance”, I said it is about 30 billion light years. A light year is, of course, the distance light travels in one year. So how can a galaxy be 30 billion light years away, implying the light has taken 30 billion years to reach us, if the Universe is only 13.7 billion years old?? This sounds like a contradiction. The solution to this apparent contradiction is that the Universe has expanded since the light left the galaxy. This is what causes the redshift. In fact, the size of the Universe now compared to the size of the Universe when the light left the galaxy is simply given by
$1 + z = \frac{ a_{now} }{ a_{then} }$
where $a$ is known as the scale factor of the Universe, or its relative size. For $z=7.51$ we have $a_{now} = (1 + 7.51)\times a_{then} = 8.51 a_{then}$, so the Universe is 8.51 times bigger now than when light left the galaxy (this is what causes the redshift, it is the expansion of space, not that the galaxy is moving through space with a speed of 97% of the speed of light). It is the fact that the Universe is over 8 times bigger now than when the light left the galaxy which allows its distance measured in light years to be more than a distance of 13.7 billion light years that one would naively think was the maximum possible! So, there is no contradiction when one thinks about things correctly.
Read Full Post » | 5,634 | 20,475 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 104, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-34 | latest | en | 0.952227 |
https://www.enotes.com/topics/math/questions/sum-n-1-oo-1-n-1-n-1-determine-convergence-810435 | 1,721,514,236,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00396.warc.gz | 648,087,095 | 17,297 | Student Question
# `sum_(n=1)^oo (-1)^(n+1)/(n+1)` Determine the convergence or divergence of the series.
`sum_(n=1)^oo ((-1)^(n+1))/(n+1)`
Take note that an alternating series
`sum` `a_n = sum (-1)^(n+1) b_n`
is convergent if the following conditions are satisfied.
(i) `b_n` is decreasing, and
(ii) ` lim_(n->oo) b_n=0` .
In the given alternating series, the bn is:
`b_n = 1/(n+1)`
Then, check if the values of bn decrease as n increases by 1.
`n=1` , `b_n = 1/2`
`n=2` , `b_n=1/3`
`n=3` , `b_n=1/4`
`n=4` , `b_n=1/5`
So bn is decreasing.
Also, take the limit of bn as n approaches infinity.
`lim_(n->oo) b_n = lim_(n->oo) 1/(n+1) = 0`
Since the result is zero, the second condition is satisfied too.
Therefore, by Alternating Series Test, the given series is convergent. | 289 | 795 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2024-30 | latest | en | 0.794015 |
https://www.jiskha.com/display.cgi?id=1348946940 | 1,503,048,375,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104631.25/warc/CC-MAIN-20170818082911-20170818102911-00055.warc.gz | 904,582,106 | 3,668 | # physics
posted by .
A 69-kg man standing on a scale in an elevator notes that as the elevator rises, the scale reads 823 N. What is the acceleration of the elevator?
m/s2 upward
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More Similar Questions | 572 | 2,320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2017-34 | latest | en | 0.893831 |
https://www.jiskha.com/questions/1064181/the-underside-of-a-bridge-has-a-shape-of-a-parabolic-arch-it-has-a-maximum-height-of-30m | 1,606,664,861,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141201836.36/warc/CC-MAIN-20201129153900-20201129183900-00003.warc.gz | 708,549,286 | 5,706 | # Math
The underside of a bridge has a shape of a parabolic arch. It has a maximum height of 30m and a width of 50m. Can a sailboat with a mast of 28m above the water pass under the bridge 8m from the middle of the bridge?
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1. Did you read this before you posted your umpteen math problems?
"Homework Posting Tips
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Ms. Sue
2. no I did not know I had to show my work,I did not read the posting tips, I have posted most of my work to show where I was stuck. I did not expect to have my homework just done for me as that would be useless when it came time to write the test or final exam. I am just not grasping how to interpret the word problems correctly into the quadratic equations, that is the part I am stuck on, not the actual equation. Does this make sense?
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3. Sorry Ms. Sue, I have posted hopefully correct or on the right track answers to all the other questions, but I have to go to sleep, I will attempt to answer this tomorrow Thanks
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2.3 The forcing amplitude is the control parameter
We examine a graph of versus the forcing amplitude at , =1 , =0.6 , = -1, = 0.837 in order to detect bifurcations. The bifurcation diagram is shown in Figure 7. In this numerically constructed bifurcation diagram, the discrete points on the Poincare’ section of the attractor realized at each value e are displayed.
Obviously, from Figure 7, we can observe the sequence of period-doubling bifurcations. First, with one of values in the interval (4.7 , 4.7645125), the Poincare’ section consists of five points (five dark points in Fig.8), so there exist the subharmonic motions with the period equaling five times of the period of the external force (Figure 8). At the value 4.7645125, the first period-doubling bifurcation occurs. After the bifurcation, with the values which is in the right vicinity of the value 4.7645125, the subharmonic motions with the period equaling twice the period of the previous motions appear, the Poincare’ sections consist of ten points (Fig. 9), and so on. The chaotic attractor realized at e = 4.8042 appearing after a sequence of period-doubling bifurcations is shown in Figure 10, and Figure 11 is corresponding attractor. The largest Liapunov exponent evaluated is positive ( defines sensitivity to initial conditions on the chaotic attractor.
The sequence of period-doubling bifurcations is one route to chaos and it is common
in many dynamical systems. It is particularly interesting because it may be characterized by a certain universal constant (called the Feigenbaum’s constant) that do not depend on nature of the concrete systems. This constant is considered as a specify criterion to determine if a system becomes chaotic by observing the prechaotic periodic behavior.
If the first bifurcation occurs at parameter value e1, the second at e2, and so on, then this constant is defined as
Table 12 shows a list of the parameters at which period-doublings occur in the Poincare’ map corresponding to the system (1) for , =1 , =0.6 , = -1, = 0.837 and e is used as the control parameter.
Period Parameter e Ratio 5 10 20 40 80 160 . 4.7645125 4.7717130 4.77327499 4.773611 4.77368297 . . 4.600144 4.656677 4.668751 . .
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Quê hương | 672 | 2,469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-45 | latest | en | 0.853202 |
https://www.thestudentroom.co.uk/showthread.php?t=211328 | 1,547,868,634,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583661083.46/warc/CC-MAIN-20190119014031-20190119040031-00421.warc.gz | 977,788,438 | 30,200 | # IntegrationWatch
Thread starter 12 years ago
#1
Integral of _/X with respect to x ....limits = 1 and 4. What is the value of the definite integral?
i can get it as far as (2x^1.5)/3 before my logic breaks down. None Calculator question!!!
0
12 years ago
#2
What is _/X ? Is that the square root [ sqrt(x) ] ??
0
Thread starter 12 years ago
#3
lol yeah sqrt(x) sorry
0
12 years ago
#4
Hello!
integral of x^0.5 from 1 to 4 = [(x^3/2)/(3/2)] from 1 to 4
= 2/3 (x^3/2) from 1 to 4
= 2/3 [(4^3/2) - (1^3/2)]
= 2/3 [(2^3) - 1^3]
= 2/3 (8 - 1)
= 2/3 (7)
= 14/3
love danniella
0
12 years ago
#5
hello!
Extra bit of explanation...
4^3/2 = (root 4)^3 = 2^3 = 8.
love danniella
0
12 years ago
#6
(Original post by Mush)
Integral of _/X with respect to x ....limits = 1 and 4. What is the value of the definite integral?
i can get it as far as (2x^1.5)/3 before my logic breaks down. None Calculator question!!!
Integral of _/X with respect to x with limits = 1 and 4
= 2/3 (x^1.5) with limits = 1 and 4
= 2/3 (4^1.5 - 1^1.5)
= 2/3 (8-1)
= 14/3
0
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19.7% | 614 | 1,768 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2019-04 | latest | en | 0.868891 |
https://discuss.pixls.us/t/i-do-not-understand-the-filmicrgb-dynamic-range-mapping/26091 | 1,631,977,016,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056476.66/warc/CC-MAIN-20210918123546-20210918153546-00334.warc.gz | 260,100,647 | 6,891 | # i do not understand the filmicRGB dynamic range mapping
can some one explain the dynamic range mapping graph as below
why is the display % from 0 to 18 taking up more spaces than 18-100 and why is from 0 to 18 there is more shades when compared to 18-100
as my understanding 18% is map to a value of 127,127,127 is it not.
i have been scratching my head why the graph is like that.
can someone explain this thanks
Please search this forum, there have been more detailed explanations posted. If you still have questions, let us know, but check what is out there first.
The (link to the) graph in your post seems to be broken…
Perhaps you want to have a look at https://www.youtube.com/watch?v=zbPj_TqTF88. And Aurelien Pierre has published furthermore videos and posts about filmic module and dynamic range mapping. Also this forum contains several post about this topic.
Another source of information are the videos of Bruce Williams. Just use your preferred search engine…
i do understand what its doing. The thing that i am trying to understand is why there are 9 shades of grey from 0 to 18percent but only 3 shades of grey from 18% to 100% pure white.
Simply put, this is because our perception is not linear (like that of the camera sensor) and we perceive the brightness differences most clearly in this range between 0 and 18% of reflectance (middle gray range). Brightness differences above this range are much less perceptible to us, correspondingly the compression is much stronger in this range.
3 Likes
thanks @s7habo for the explanation. now I get it thanks
Probably best not to reply if you don’t really want to.
Because… dynamic range sucks.
Your display is probably close to sRGB and using 8 bits integer encoding. So the highest value that will get rounded to 0 when converting to `uint8` sRGB is (0.5 / 255) / 12.92 = 0.0001518 where 12.92 is the encoding transfer function of sRGB near black. If we compute the \log_2 of this value, we get -12.69 EV.
So, that means that sRGB can encode 12.69 EV of lightness data between “pitch black” and “full white” (regardless of whether the screen can actually display all of them later or not – this is pure data encoding).
Meanwhile, middle grey \approx 0.1845, which in \log_2 is -2.44 EV. To sum up:
• display white is at 0 EV (it’s our reference),
• middle grey is at - 2.44 EV,
• display black is at - 12.69 EV (and everything darker will get rounded the same -12.69 EV, encoded 0 in this space).
You see that our typical dynamic range is clearly unbalanced in favour of shadows.
2 Likes
You need a macro for this bit of text …you have typed it so often…
Actually that is a great explanation and might be a nice add to the manual under the Filmic DNR display graph to enhance the description that exists there and for the rest of the filmic rgb doc??
Maybe only change if from “Your monitor is” to something like “A typical monitor is” rest as is above
@elstoc what do you think??
thanks for clearing up this. thats all I wanted to know so that i could understand the graph. thanks much appreciated.
so can I assume that any luminance value below 0.0152% will become black ?
Quick follow-up, your typical desktop display has a contrast ratio of 300:1, which is 8.23 EV. So it’s not even able to fully display the allowed sRGB dynamic range (the bottom 4 EV go to trash and get clipped to black… which is the whole point of black point adaptation… that nobody does).
1 Like
And in addition to all that, only encodes around 5.6EV without going below a rather lenient threshold of a 5% difference in brightness being perceivable in a gradient. Worse if you use a more stringent threshold.
(Unfortunately I don’t have a direct link to the paper in question at the moment, I’ll try to find the actual citation tonight.)
At least display dynamic ranges have improved significantly recently, although sometimes with tradeoffs (for example local dimming is great for global contrast ratios, but not necessarily good for handling high local contrast). | 999 | 4,079 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-39 | latest | en | 0.933728 |
https://ssvconference.com/power-station/quick-answer-how-do-you-find-the-electric-field-curl.html | 1,696,082,204,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510676.40/warc/CC-MAIN-20230930113949-20230930143949-00321.warc.gz | 589,661,535 | 20,868 | # Quick Answer: How do you find the electric field curl?
Contents
## What is the curl of a static electric field?
The curl of a electric field is zero, i.e. That should read, “the curl of an electrostatic field is zero,” that is, the electric field associated with a set of stationary charges has a curl of zero.
## What is the formula for finding electric field?
We can find the electric field created by a point charge by using the equation E=kQr2 E = k Q r 2 .
## What is the curl of electric field intensity?
Electric intensity at any point in an electric field is equal to the at that point. Identify the nature of the field, if the divergence is zero and curl is also zero. Curl is defined as the angular velocity at every point of the vector field.
Q. The curl of the electric field intensity is
D. static
## What is curl vector field?
The curl of a vector field provides a. measure of the amount of rotation of the vector field at a point. In general, the curl of any vector point function gives the measure of angular velocity at any. point of the vector field. The curl operation is restricted to how the field changes as one move.
## How does curl work?
cURL, which stands for client URL, is a command line tool that developers use to transfer data to and from a server. At the most fundamental, cURL lets you talk to a server by specifying the location (in the form of a URL) and the data you want to send.
## How do you find the electric field between two plates?
Since the field lines are parallel to each other, this type of electric field is uniform and has a magnitude which can be calculated with the equation E = V/d where V represents the voltage supplied by the battery and d is the distance between the plates.
## Which of Theorem uses curl operation?
Which of the following theorem use the curl operation? Explanation: The Stoke’s theorem is given by ∫ A. dl = ∫Curl(A). ds, which uses the curl operation.
## What is the integral of electric field?
The line integral of electric field around a closed loop is equal to the voltage generated in that loop (Faraday’s law): … The line integral of a force over a path is equal to the work done by that force on the path.
## What is the divergence of curl of a vector?
In words, this says that the divergence of the curl is zero. Theorem 16.5. 2 ∇×(∇f)=0. That is, the curl of a gradient is the zero vector.
## How do you find the curl of a vector?
For F:R3→R3 (confused?), the formulas for the divergence and curl of a vector field are divF=∂F1∂x+∂F2∂y+∂F3∂zcurlF=(∂F3∂y−∂F2∂z,∂F1∂z−∂F3∂x,∂F2∂x−∂F1∂y). | 656 | 2,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2023-40 | latest | en | 0.945552 |
http://www.learncheme.com/simulations/kinetics-reactor-design/parametric-sensitivity-of-pfr-with-heat-exchange | 1,627,319,891,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152144.81/warc/CC-MAIN-20210726152107-20210726182107-00539.warc.gz | 72,990,101 | 11,140 | Parametric Sensitivity of Plug Flow Reactor with Heat Exchange
This Demonstration plots the temperature and molar flow rate of the reactant as a function of distance down a plug flow reactor for an exothermic, gas-phase reaction. The reactor has heat exchange through the walls. Vary the feed temperature, activation energy for the reaction, and total molar flow rate with the sliders. Thermal runaway occurs at certain conditions and it is a sensitive function (parametric sensitivity) of the feed temperature and the activation energy. Download the CDF file to view the simulation offline using the free Wolfram CDF player.
Details
This is a model of the partial oxidation of o-xylene in a large excess of oxygen in a 1.5 m long plug flow reactor.
The first-order rate expression:
where ρ is rate of reaction (mol/[m3 s]), k is the pre-exponential factor in the rate constant (1/s), Ea is the activation energy (kJ/mol) R2 is the ideal gas constant (kJ/[mol K]), T is the absolute temperature in the reactor (K), P is pressure (atm), R is the ideal gas constant ([atm m3]/[mol K]), Fx is the molar flow rate of the reactant o-xylene (kmol/s), and Ftot is the total molar flow rate of the feed (kmol/s).
Mole balance as a function of length:
where z is the distance down the PFR (m), A = πr2 is the cross-sectional area of the PFR (m2), and r is the PFR radius (m).
Energy balance as a function of length:
where β and γ are simplification terms, Ta is the temperature of the heat transfer fluid surrounding the reactor (K), ΔH is the heat of reaction (kJ/kmol), m is the mass flow rate (kg/s), CPm is the mass heat capacity of gas in the reactor (kJ/[kg K]), and U is the overall heat transfer coefficient (kJ/[m2 s K]).
At the PFR inlet (z=0) T = Tf and Fx=yx,fFtot, where Tf is the feed temperature (K) and yx,f is the mole fraction of reactant in the feed. | 476 | 1,867 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-31 | latest | en | 0.898507 |
https://darkwing.uoregon.edu/~stevev/sd-archive/prelist/msg00241.html | 1,680,288,160,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949678.39/warc/CC-MAIN-20230331175950-20230331205950-00141.warc.gz | 237,866,443 | 2,879 | # Re: Carrots
```To Steve:
>> >From here I've a real hard time following your calculus...
>
>it's not calculus, just simple arithmetic.
I wanted to boost up you ego ;)
(or was it my limited knowledge about the English language again)
>I think that I have answered your "so-called" errors, as for the volume
>calculation, here it is again.
Thanks, now I can see that you where right from the start, it was quite a
bunch of data all stacked in a small text, now it is much clearer.
>no, the creation of the food (6H20 + 6CO2 + energy --> C6H12O6 +6O2)
>removes most of that oxygen from the system (plant) it is the non-food
>portion which consumes oxygen that can not be recovered. you can see
>this by considering what happens when you eat the food. the oxygen that
>the plant exhaled combines with the sugar that you ate to produced energy
>plus CO2 + H2O
Yes, I knew that, and I'm not certain why I questioned the amount of oxigen
created.
I think I thought that at night the plant uses oxigen, and the larger the
plant-mass, the more oxigen it needs. So I assumed that a large plant with
few leaves used more than it would create. Has this thinking any realism?
Brian do you know it?
>> (It took me a while before I figured out that 1 feet is exactly 12 inches)
>
>When you post equations, you can write in SI if you want to but I find
>when i am estimating sizes, that American system works for me. I will
>answer any questions you have on relationships etc.
I wasn't planning on doing anything else :)
I just had never heard that 1 foot was 12 inches. It's like knowing what a
decimetre is and what a centimetre but not knowing that it takes 10
centimetres to get a decimetre. The dodecimal numbering had not reached my
mind until yesterday.
Do you also use yards, chains and furlongs?
As I also noticed there go 16 ounces in a pound.
In Dutch there go 5 "ons" in a "pond" and 2 "ponden" in a "kilo"(=1 kg)
Long live the English metric system...
When are the Americans switching to SI units? The English now have an
official law that prohibits the public use (in stores) of the English units.
By the way it is here -2 degrees Celcius :) So about 29 degrees Fahrenheit.
Yes, the Dutch people always talk about the weather :)
Timothy
``` | 579 | 2,255 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-14 | latest | en | 0.968842 |
http://www.ng-outsourcing.com/inelastic-demand.html | 1,680,178,821,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949181.44/warc/CC-MAIN-20230330101355-20230330131355-00783.warc.gz | 84,797,726 | 17,514 | # What is Inelastic Demand?
Definition: Inelastic demand is the economic idea that the demand for a product does not change relative to changes in that product’s price. In other words, as the price of a good or service increases or decreases, the demand for it will stay the same. This typically occurs in convenience goods that consumers need every day.
## What Does Inelastic Demand Mean?
What is the definition of inelastic demand? Demand for a good is relatively inelastic when the percentage change in price is more than the quantity demanded. This means that consumers do not react to changes in commodity prices and continue to request the same amount of a product or a service, regardless of its price.
Firms are interested in the elasticity of demand so that they adjust their operations and product costs according to consumer needs while lower risk and realizing a profit. The inelastic demand curve is a steep curve that becomes steeper as the quantity demanded is not changing. The relatively inelastic demand is indicated by – 1 < Ed < 0.
Let’s look at an example.
## Example
Marion owns a grocery store and sells milk, eggs, and grocery goods. Over the past three months, the demand for milk has increased, and Marion decides to raise the price of milk from \$10 to \$12. What will happen to the quantity demanded?
Because the milk is a convenience good, a rise in the price of milk will cause a lower change in the quantity demanded. Marion notices that the quantity demanded declines from 100 to 99. Marion calculates the price elasticity of demand for milk to set a price that can generate a profit to her store.
The price elasticity of demand for milk is:
Ed = % change in quantity demanded / % change in price = (95/100) -1 / (12/10) – 1 = -1% / 10% = -10%
Marion calculates the price elasticity of demand for milk for different prices as follows:
As the price increases, the percentage change in price is more than the quantity demanded. Therefore, the demand for milk is inelastic because it is a convenience good that consumers buy every day, regardless of the change in price.
## Summary Definition
Define Inelastic Demand: Inelastic demand occurs when consumers’ buying habits don’t change relative to price changes in the goods and services they consume.
[i]
[i] | 490 | 2,303 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2023-14 | latest | en | 0.931582 |
https://www.coursehero.com/file/5646454/Calculate-circle-area/ | 1,519,526,901,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816083.98/warc/CC-MAIN-20180225011315-20180225031315-00478.warc.gz | 839,378,013 | 40,585 | {[ promptMessage ]}
Bookmark it
{[ promptMessage ]}
Calculate circle area
# Calculate circle area - circleArea(3> 28 circleArea(5>...
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Sheet1 Page 1 %circleArea(number) -> number %Description: % This function computes the area of a circle given the radius. %Examples: %
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Unformatted text preview: circleArea(3) -> 28 % circleArea(5) -> 78 % circleArea(45) -> 6362 function x = circleArea (radius) x = pi * radius * radius...
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Ask a homework question - tutors are online | 170 | 653 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-09 | latest | en | 0.550586 |
http://www.theinfolist.com/html/ALL/s/length.html | 1,652,806,693,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662519037.11/warc/CC-MAIN-20220517162558-20220517192558-00172.warc.gz | 115,057,476 | 11,911 | TheInfoList
Length is a measure of
distance Distance is a numerical measurement ' Measurement is the number, numerical quantification (science), quantification of the variable and attribute (research), attributes of an object or event, which can be used to compare with other objects or eve ...
. In the
International System of Quantities The International System of Quantities (ISQ) consists of the quantities Quantity is a property that can exist as a Counting, multitude or Magnitude (mathematics), magnitude, which illustrate discontinuity (mathematics), discontinuity and continu ...
, length is a
quantity Quantity is a property that can exist as a multitude or magnitude, which illustrate discontinuity and continuity. Quantities can be compared in terms of "more", "less", or "equal", or by assigning a numerical value in terms of a unit of measu ...
with
dimension In physics Physics is the that studies , its , its and behavior through , and the related entities of and . "Physical science is that department of knowledge which relates to the order of nature, or, in other words, to the regular s ...
distance. In most
systems of measurement A system of measurement is a collection of units of measurement A unit of measurement is a definite magnitude (mathematics), magnitude of a quantity, defined and adopted by convention or by law, that is used as a standard for measurement of ...
a base unit for length is chosen, from which all other units are derived. In the
International System of Units The International System of Units, known by the international abbreviation SI in all languages and sometimes Pleonasm#Acronyms_and_initialisms, pleonastically as the SI system, is the modern form of the metric system and the world's most wi ...
(SI) system the base unit for length is the
metre The metre ( Commonwealth spelling) or meter (American spelling Despite the various English dialects spoken from country to country and within different regions of the same country, there are only slight regional variations in English o ...
. Length is commonly understood to mean the most extended
dimension In physics Physics is the that studies , its , its and behavior through , and the related entities of and . "Physical science is that department of knowledge which relates to the order of nature, or, in other words, to the regular s ...
of a fixed object. However, this is not always the case and may depend on the position the object is in. Various terms for the length of a fixed object are used, and these include
height 200px, A cuboid demonstrating the dimensions length, width">length.html" ;"title="cuboid demonstrating the dimensions length">cuboid demonstrating the dimensions length, width, and height. Height is measure of vertical distance, either vertical ...
, which is vertical length or vertical extent, and width, breadth or depth. Height is used when there is a base from which vertical measurements can be taken. Width or breadth usually refer to a shorter dimension when length is the longest one. Depth is used for the third dimension of a three dimensional object. Length is the measure of one spatial dimension, whereas
area Area is the quantity Quantity is a property that can exist as a multitude or magnitude, which illustrate discontinuity and continuity. Quantities can be compared in terms of "more", "less", or "equal", or by assigning a numerical value in ...
is a measure of two dimensions (length squared) and
volume Volume is a scalar quantity expressing the amount Quantity or amount is a property that can exist as a multitude Multitude is a term for a group of people who cannot be classed under any other distinct category, except for their shared fact ...
is a measure of three dimensions (length cubed).
# History
Measurement has been important ever since humans settled from nomadic lifestyles and started using building materials, occupying land and trading with neighbours. As trade between different places increased, the need for standard units of length increased. And later, as society has become more technologically oriented, much higher accuracy of measurement is required in an increasingly diverse set of fields, from micro-electronics to interplanetary ranging. Under
Einstein Albert Einstein ( ; ; 14 March 1879 – 18 April 1955) was a German-born theoretical physicist, widely acknowledged to be one of the greatest physicists of all time. Einstein is known for developing the theory of relativity The theory ...
's
special relativity In physics Physics is the that studies , its , its and behavior through , and the related entities of and . "Physical science is that department of knowledge which relates to the order of nature, or, in other words, to the regular s ...
, length can no longer be thought of as being constant in all
reference frame In physics, a frame of reference (or reference frame) consists of an abstract coordinate system In geometry Geometry (from the grc, γεωμετρία; ''wikt:γῆ, geo-'' "earth", ''wikt:μέτρον, -metron'' "measurement") is, wi ...
s. Thus a
ruler A ruler, sometimes called a rule or line gauge, is a device used in geometry Geometry (from the grc, γεωμετρία; ' "earth", ' "measurement") is, with , one of the oldest branches of . It is concerned with properties of spac ...
that is one metre long in one frame of reference will not be one metre long in a reference frame that is moving relative to the first frame. This means the length of an object varies depending on the speed of the observer.
# Use in mathematics
## Euclidean geometry
In Euclidean geometry, length is measured along
straight line 290px, A representation of one line segment. In geometry, the notion of line or straight line was introduced by ancient mathematicians to represent straight objects (i.e., having no curvature In mathematics, curvature is any of several str ...
s unless otherwise specified and refers to on them. relating the length of the sides of a
right triangle A right triangle (American English American English (AmE, AE, AmEng, USEng, en-US), sometimes called United States English or U.S. English, is the set of varieties of the English language native to the United States. Currently, American Eng ...
is one of many applications in Euclidean geometry. Length may also be measured along other types of curves and is referred to as
arclength Arc length is the distance between two points along a section of a curve In mathematics, a curve (also called a curved line in older texts) is an object similar to a line (geometry), line, but that does not have to be Linearity, straight. In ...
. In a
triangle A triangle is a polygon In geometry, a polygon () is a plane (mathematics), plane Shape, figure that is described by a finite number of straight line segments connected to form a closed ''polygonal chain'' (or ''polygonal circuit''). The ...
, the length of an
altitude Altitude or height (also sometimes known as depth) is a distance measurement, usually in the vertical or "up" direction, between a reference and a point or object. The exact definition and reference datum varies according to the context (e.g. ...
, a line segment drawn from a vertex
perpendicular In elementary geometry Geometry (from the grc, γεωμετρία; ' "earth", ' "measurement") is, with , one of the oldest branches of . It is concerned with properties of space that are related with distance, shape, size, and relativ ...
to the side not passing through the vertex (referred to as a
base Base or BASE may refer to: Brands and enterprises * Base (mobile telephony provider), a Belgian mobile telecommunications operator *Base CRM Base CRM (originally Future Simple or PipeJump) is an enterprise software company based in Mountain Vie ...
of the triangle), is called the height of the triangle. The
area Area is the quantity Quantity is a property that can exist as a multitude or magnitude, which illustrate discontinuity and continuity. Quantities can be compared in terms of "more", "less", or "equal", or by assigning a numerical value in ...
of a
rectangle In Euclidean geometry, Euclidean plane geometry, a rectangle is a quadrilateral with four right angles. It can also be defined as: an equiangular quadrilateral, since equiangular means that all of its angles are equal (360°/4 = 90°); or a para ...
is defined to be length × width of the rectangle. If a long thin rectangle is stood up on its short side then its area could also be described as its height × width. The
volume Volume is a scalar quantity expressing the amount Quantity or amount is a property that can exist as a multitude Multitude is a term for a group of people who cannot be classed under any other distinct category, except for their shared fact ...
of a (such as a plank of wood) is often described as length × height × depth. The
perimeter A perimeter is either a path that encompasses/surrounds/outlines a shape (in two dimensions) or its length ( one-dimensional). The perimeter of a circle A circle is a shape consisting of all point (geometry), points in a plane (mathema ...
of a
polygon In geometry Geometry (from the grc, γεωμετρία; ' "earth", ' "measurement") is, with , one of the oldest branches of . It is concerned with properties of space that are related with distance, shape, size, and relative position o ...
is the sum of the lengths of its sides. The
circumference In geometry Geometry (from the grc, γεωμετρία; ' "earth", ' "measurement") is, with , one of the oldest branches of . It is concerned with properties of space that are related with distance, shape, size, and relative position ...
of a circular
disk Disk or disc may refer to: * Disk (mathematics) * Disk storage Music * Disc (band), an American experimental music band * Disk (album), ''Disk'' (album), a 1995 EP by Moby Other uses * Disc (galaxy), a disc-shaped group of stars * Disc (magazin ...
is the length of the
boundary Boundary or Boundaries may refer to: * Border, in political geography Entertainment *Boundaries (2016 film), ''Boundaries'' (2016 film), a 2016 Canadian film *Boundaries (2018 film), ''Boundaries'' (2018 film), a 2018 American-Canadian road trip ...
(a
circle A circle is a shape A shape or figure is the form of an object or its external boundary, outline, or external surface File:Water droplet lying on a damask.jpg, Water droplet lying on a damask. Surface tension is high enough to preven ...
) of that disk.
## Other geometries
In other geometries, length may be measured along possibly curved paths, called
geodesic In geometry Geometry (from the grc, γεωμετρία; ' "earth", ' "measurement") is, with , one of the oldest branches of . It is concerned with properties of space that are related with distance, shape, size, and relative position o ...
s. The
Riemannian geometry#REDIRECT Riemannian geometry Riemannian geometry is the branch of differential geometry Differential geometry is a Mathematics, mathematical discipline that uses the techniques of differential calculus, integral calculus, linear algebra and m ...
used in
general relativity General relativity, also known as the general theory of relativity, is the geometric Geometry (from the grc, γεωμετρία; '' geo-'' "earth", '' -metron'' "measurement") is, with arithmetic, one of the oldest branches of mathema ...
is an example of such a geometry. In
spherical geometry Image:Triangles (spherical geometry).jpg, 300px, The sum of the angles of a spherical triangle is not equal to 180°. A sphere is a curved surface, but locally the laws of the flat (planar) Euclidean geometry are good approximations. In a small tr ...
, length is measured along the
great circles Great may refer to: Descriptions or measurements * Great, a relative measurement in physical space, see Size File:Comparison of planets and stars (sheet by sheet) (Oct 2014 update).png, A size comparison illustration comparing the sizes of vario ...
on the sphere and the distance between two points on the sphere is the shorter of the two lengths on the great circle, which is determined by the plane through the two points and the center of the sphere.
## Graph theory
In an unweighted graph, the length of a
cycle Cycle or cyclic may refer to: Anthropology and social sciences * Cyclic history, a theory of history * Cyclical theory, a theory of American political history associated with Arthur Schlesinger, Sr. * Social cycle, various cycles in social scienc ...
, path, or
walk Walking (also known as ambulation) is one of the main gait Gait is the pattern of movement Movement may refer to: Common uses * Movement (clockwork), the internal mechanism of a timepiece * Motion (physics), commonly referred to as move ...
is the number of
edge Edge or EDGE may refer to: Technology Computing * Edge computing Edge computing is a distributed computing paradigm that brings computation and data storage closer to the sources of data. This is expected to improve response times and save bandw ...
s it uses. In a
weighted graph This is a glossary of graph theory. Graph theory is the study of graph (discrete mathematics), graphs, systems of nodes or vertex (graph theory), vertices connected in pairs by lines or #edge, edges. Symbols A ...
, it may instead be the sum of the weights of the edges that it uses. Length is used to define the
shortest path In graph theory In mathematics, graph theory is the study of ''graph (discrete mathematics), graphs'', which are mathematical structures used to model pairwise relations between objects. A graph in this context is made up of ''Vertex (graph the ...
,
girth Girth may refer to: ;Mathematics * Girth (functional analysis), the length of the shortest centrally symmetric simple closed curve on the unit sphere of a Banach space * Girth (geometry), the perimeter of a parallel projection of a shape * Girth (g ...
(shortest cycle length), and longest path between two vertices in a graph.
## Measure theory
In measure theory, length is most often generalized to general sets of $\mathbb^n$ via the
Lebesgue measureIn measure theory In mathematics, a measure on a set (mathematics), set is a systematic way to assign a number, intuitively interpreted as its size, to some subsets of that set, called measurable sets. In this sense, a measure is a generalization ...
. In the one-dimensional case, the Lebesgue outer measure of a set is defined in terms of the lengths of open intervals. Concretely, the length of an
open interval In mathematics Mathematics (from Greek: ) includes the study of such topics as numbers ( and ), formulas and related structures (), shapes and spaces in which they are contained (), and quantities and their changes ( and ). There is no ge ...
is first defined as :$\ell\left(\\right)=b-a.$ so that the Lebesgue outer measure $\mu^*\left(E\right)$ of a general set $E$ may then be defined as :$\mu^*\left(E\right)=\inf\left\.$
# Units
In the physical sciences and engineering, when one speaks of , the word is synonymous with
distance Distance is a numerical measurement ' Measurement is the number, numerical quantification (science), quantification of the variable and attribute (research), attributes of an object or event, which can be used to compare with other objects or eve ...
. There are several
units Unit may refer to: Arts and entertainment * UNIT Unit may refer to: Arts and entertainment * UNIT, a fictional military organization in the science fiction television series ''Doctor Who'' * Unit of action, a discrete piece of action (or beat) in ...
that are used to length. Historically, units of length may have been derived from the lengths of human body parts, the distance traveled in a number of paces, the distance between landmarks or places on the Earth, or arbitrarily on the length of some common object. In the
International System of Units The International System of Units, known by the international abbreviation SI in all languages and sometimes Pleonasm#Acronyms_and_initialisms, pleonastically as the SI system, is the modern form of the metric system and the world's most wi ...
(SI), the of length is the
metre The metre ( Commonwealth spelling) or meter (American spelling Despite the various English dialects spoken from country to country and within different regions of the same country, there are only slight regional variations in English o ...
(symbol, m) and is now defined in terms of the
speed of light The speed of light in vacuum A vacuum is a space Space is the boundless three-dimensional Three-dimensional space (also: 3-space or, rarely, tri-dimensional space) is a geometric setting in which three values (called paramet ...
second The second (symbol: s, also abbreviated: sec) is the base unit of time Time is the continued sequence of existence and event (philosophy), events that occurs in an apparently irreversible process, irreversible succession from the past, th ...
). The
millimetre 330px, Different lengths as in respect to the Electromagnetic spectrum, measured by the Metre and its deriveds scales. The Microwave are in-between 1 meter to 1 millimeter. The millimetre (American and British English spelling differences#-re, ...
(mm),
centimetre 330px, Different lengths as in respect to the Electromagnetic spectrum, measured by the Metre and its deriveds scales. The Microwave are in-between 1 meter to 1 millimeter. A centimetre (international spelling) or centimeter (American spellin ...
(cm) and the
kilometre The kilometre (SI symbol: km; or ), spelt kilometer in American English American English (AmE, AE, AmEng, USEng, en-US), sometimes called United States English or U.S. English, is the set of variety (linguistics), varieties of the Eng ...
(km), derived from the metre, are also commonly used units. In U.S. customary units, English or Imperial system of units, commonly used units of length are the
inch Measuring tape with inches The inch (symbol: in or ″) is a unit Unit may refer to: Arts and entertainment * UNIT Unit may refer to: Arts and entertainment * UNIT, a fictional military organization in the science fiction television s ...
(in), the
foot The foot (plural: feet) is an anatomical Anatomy (Greek ''anatomē'', 'dissection') is the branch of biology Biology is the natural science that studies life and living organisms, including their anatomy, physical structure, Biochemis ...
(ft), the
yard The yard (symbol: yd) is an English unit English units are the units of measurement A unit of measurement is a definite magnitude (mathematics), magnitude of a quantity, defined and adopted by convention or by law, that is used ...
(yd), and the
mile The mile, sometimes the international mile or statute mile to distinguish it from other miles, is a British imperial unit and US customary unit United States customary units (U.S. customary units) are a system of measurements commonly u ...
(mi). A unit of length used in
navigation Navigation is a field of study that focuses on the process of monitoring and controlling the movement of a craft or vehicle from one place to another.Bowditch, 2003:799. The field of navigation includes four general categories: land navigation, ...
is the
nautical mile A nautical mile is a unit of length A unit of length refers to any arbitrarily chosen and accepted reference standard for measurement of length. The most common units in modern use are the metric system, metric units, used in every country gl ...
(nmi). Units used to denote distances in the vastness of space, as in
astronomy Astronomy (from el, ἀστρονομία, literally meaning the science that studies the laws of the stars) is a natural science that studies astronomical object, celestial objects and celestial event, phenomena. It uses mathematics, phys ...
, are much longer than those typically used on Earth (metre or centimetre) and include the
astronomical unit The astronomical unit (symbol: au, or or AU) is a unit of length A unit of length refers to any arbitrarily chosen and accepted reference standard for measurement of length. The most common units in modern use are the metric system, metri ...
(au), the
light-year A light-year, alternatively spelt lightyear, is a unit of length A unit of length refers to any arbitrarily chosen and accepted reference standard for measurement of length. The most common units in modern use are the metric system, metric un ...
, and the
parsec The parsec (symbol: pc) is a unit of length A unit of length refers to any arbitrarily chosen and accepted reference standard for measurement of length. The most common units in modern use are the metric system, metric units, used in every ...
(pc). Units used to denote sub-atomic distances, as in
nuclear physics Nuclear physics is the field of physics Physics is the that studies , its , its and behavior through , and the related entities of and . "Physical science is that department of knowledge which relates to the order of nature, or, in ot ...
, are much smaller than the centimetre. Examples include the
dalton Dalton may refer to: Science * Dalton (crater), a lunar crater * Dalton (program), chemistry software * Dalton (unit) (Da), the atomic mass unit Entertainment * Dalton (Buffyverse), minor character from ''Buffy the Vampire Slayer'' television s ...
and the
fermi Enrico Fermi (; 29 September 1901 – 28 November 1954) was an Italian Italian may refer to: * Anything of, from, or related to the country and nation of Italy ** Italians, an ethnic group or simply a citizen of the Italian Republic ** Italia ...
.
*
Conversion of units Conversion of units is the conversion between different units of measurement A unit of measurement is a definite magnitude Magnitude may refer to: Mathematics *Euclidean vector, a quantity defined by both its magnitude and its directi ...
* *
Length measurementLength measurement is implemented in practice in many ways. The most commonly used approaches are the transit-time methods and the interferometer methods based upon the speed of light. For objects such as crystals and diffraction gratings, diffractio ...
*
Metric system The metric system is a system of measurement A system of measurement is a collection of units of measurement and rules relating them to each other. Systems of measurement have historically been important, regulated and defined for the purpose ...
*
Metric units Metric units are units Unit may refer to: Arts and entertainment * UNIT Unit may refer to: Arts and entertainment * UNIT, a fictional military organization in the science fiction television series ''Doctor Who'' * Unit of action, a discrete piece ...
*
Orders of magnitude (length) The following are examples of orders of magnitude An order of magnitude is an approximation of the logarithm In mathematics Mathematics (from Ancient Greek, Greek: ) includes the study of such topics as quantity (number theory), ...
*
Reciprocal lengthReciprocal length or inverse length is a measurement ' Measurement is the number, numerical quantification (science), quantification of the variable and attribute (research), attributes of an object or event, which can be used to compare with other o ...
# References
{{Authority control Physical quantities SI base quantities | 4,962 | 23,128 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2022-21 | latest | en | 0.923527 |
https://vustudents.ning.com/group/mgt402costmanagementaccounting/forum/topic/listForContributor?user=3kdjyey1atiur | 1,627,611,251,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153899.14/warc/CC-MAIN-20210729234313-20210730024313-00627.warc.gz | 611,025,736 | 15,440 | www.vustudents.ning.com
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Discussions Replied To (26) Replies Latest Activity
### "some of mcqs are given from past paper and mostly mcqs are new.. 1 numaric question…"
muhammad kashif replied Jul 16, 2012 to Mgt402 Current Final Term Papers Spring 2012 Date: 16-July-2012 to 27-July-2012
20 Jul 23, 2013
### "ji variable expense hay but yahan per decsion making aa jati hay 2ndly question boha…"
muhammad kashif replied Jun 22, 2012 to MGT402 GDB Solution & Discussion Last Date:22-06-2012
132 Jul 3, 2012
### "1. How many sticks of sugarcane should be used by Mr. Santa during any month…"
muhammad kashif replied Jun 22, 2012 to MGT402 GDB Solution & Discussion Last Date:22-06-2012
132 Jul 3, 2012
### "ghussa aa raha hay mujhay ....... itni time lagaaya aur yeh kya shakle ikhtiyaar ker…"
muhammad kashif replied Jun 22, 2012 to MGT402 GDB Solution & Discussion Last Date:22-06-2012
132 Jul 3, 2012
### "Yeh GDB mein post karnay k baad to mere answer ki shakle hi badal gaee hai ...:( pat…"
muhammad kashif replied Jun 22, 2012 to MGT402 GDB Solution & Discussion Last Date:22-06-2012
132 Jul 3, 2012
### "sorry mein nay abhi message dekha aap ka ... aap idea day dein bas"
muhammad kashif replied Jun 22, 2012 to MGT402 GDB Solution & Discussion Last Date:22-06-2012
132 Jul 3, 2012
### "yeh idea bhi hay aur 1st three tak ki calculation bhi ... "
muhammad kashif replied Jun 22, 2012 to MGT402 GDB Solution & Discussion Last Date:22-06-2012
132 Jul 3, 2012
### "kisi student ko bhi yeh right nahi keh wo kisi student k baaray mein ghalt biyaani k…"
muhammad kashif replied Jun 22, 2012 to MGT402 GDB Solution & Discussion Last Date:22-06-2012
132 Jul 3, 2012
### "Yeh Idea hay .... first 3 part tak ... "
muhammad kashif replied Jun 22, 2012 to MGT402 GDB Solution & Discussion Last Date:22-06-2012
132 Jul 3, 2012
### "1st and 2nd tak ki calculation ki hay mein nay ,,, aur agar fix cost 12000 hay phir…"
muhammad kashif replied Jun 22, 2012 to MGT402 GDB Solution & Discussion Last Date:22-06-2012
132 Jul 3, 2012
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8 hours ago | 883 | 2,795 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-31 | longest | en | 0.745753 |
https://gamer-hood.com/5-digit-subtraction-math-worksheet/ | 1,568,975,472,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573988.33/warc/CC-MAIN-20190920092800-20190920114800-00514.warc.gz | 474,069,074 | 30,904 | # 5 Digit Subtraction Math Worksheet
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