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http://www.maths.usyd.edu.au/u/UG/JM/MATH1011/Quizzes/quiz3.html | 1,508,700,187,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825436.78/warc/CC-MAIN-20171022184824-20171022204824-00549.warc.gz | 507,045,802 | 8,275 | ## MATH1011 Quizzes
Quiz 3: Exponential and logarithmic functions
Question 1 Questions
Write $3sinx+4cosx$ in the form $Rsin\left(x+\alpha \right)$. Exactly one option must be correct)
a) $\sqrt{7}sin\left(x+0.64\right)$ b) $\sqrt{7}sin\left(x+0.93\right)$ c) $5sin\left(x+0.64\right)$ d) $5sin\left(x+0.93\right)$
Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
$Rcos\alpha =3$, $Rsin\alpha =4$, $R=\sqrt{{3}^{2}+{4}^{2}}=5$, $\alpha ={tan}^{-1}4∕3=0.93$
Write $\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx$ in the form $Rsin\left(x+\alpha \right)$. Exactly one option must be correct)
a) $sin\left(x+\frac{\pi }{6}\right)$ b) $sin\left(x-\frac{\pi }{6}\right)$ c) $sin\left(x+\frac{5\pi }{6}\right)$ d) $sin\left(x-\frac{5\pi }{6}\right)$
Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
$Rcos\alpha =-\frac{\sqrt{3}}{2}$, $Rsin\alpha =\frac{1}{2}$, $R=\sqrt{\frac{1}{3}+\frac{3}{4}}=1,\phantom{\rule{1em}{0ex}}tan\alpha =-\frac{1}{\sqrt{3}}$ and $\alpha$ is in the 2nd quadrant.
Choice (d) is incorrect
Which of the following graphs indicates that $y$ is proportional to ${x}^{2}$. Exactly one option must be correct)
a) b) c) d)
Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
The experimental data $\begin{array}{cccccc}\hfill t\hfill & \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill y\hfill & \hfill 0.9\hfill & \hfill 3.6\hfill & \hfill 8.1\hfill & \hfill 14.4\hfill & \hfill 22.5\hfill \end{array}$ appears to show that $y$ is proportional to ${t}^{2}$. Find the relationship between $y$ and $t$. Exactly one option must be correct)
a) $y=0.9{t}^{2}$ b) $y=2.7{t}^{2}$ c) $y=0.9t$ d) $y=2.7t$
Choice (a) is correct!
Scaling the data gives $\begin{array}{cccccc}\hfill {t}^{2}=T\hfill & \hfill 1\hfill & \hfill 4\hfill & \hfill 9\hfill & \hfill 16\hfill & \hfill 25\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill y\hfill & \hfill 0.9\hfill & \hfill 3.6\hfill & \hfill 8.1\hfill & \hfill 14.4\hfill & \hfill 22.5\hfill \end{array}$ A scatter plot of $y$ against $T$ is a straight line through (0,0) with slope $\frac{3.6-0.9}{4-1}=0.9$.
$\text{therefore,}y=0.9T=0.9{t}^{2}$.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Given the experimental data $\begin{array}{cccccc}\hfill t\hfill & \hfill 1.1\hfill & \hfill 1.6\hfill & \hfill 2.1\hfill & \hfill 2.6\hfill & \hfill 3.1\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill y\hfill & \hfill 2.9\hfill & \hfill 4.4\hfill & \hfill 5.9\hfill & \hfill 7.4\hfill & \hfill 8.9\hfill \end{array}$ which of the following statements is correct ? Exactly one option must be correct)
a) $y$ is proportional to $t$. b) $y$ is proportional to ${t}^{2}$. c) $y$ is a linear function of $t$. d) There is no recognisable relationship between $y$ and $t$.
Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The equation of the line satisfying the data is $y=3t-0.4$.
Choice (d) is incorrect
Solve $3{e}^{x}=5$. Exactly one option must be correct)
a) $x=\frac{ln5}{3}$ b) $x=\frac{ln5}{ln3}$ c) $x=ln5+ln\left(-3\right)$ d) $x=ln5-ln3$
Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
$\begin{array}{rcll}3{e}^{x}=5⇒{e}^{x}& =& \frac{5}{3}& \text{}\\ ln{e}^{x}& =& ln\left(\frac{5}{3}\right)& \text{}\\ x& =& ln5-ln3.& \text{}\end{array}$
Solve $\frac{1}{3}ln8{x}^{3}=2.5$ . Exactly one option must be correct)
a) $x=\frac{{e}^{2.5}}{2}$ b) $x=\sqrt[3]{7.5-ln8}$ c) $x=3{e}^{7.5-ln8}$ d) $x=\frac{1}{3}{e}^{7.5∕8}$
Choice (a) is correct!
$\begin{array}{rcll}\frac{1}{3}ln8{x}^{3}& =& 2.5& \text{}\\ ln{\left[{\left(2x\right)}^{3}\right]}^{1∕3}& =& 2.5& \text{}\\ ln2x=2.5& & & \text{}\\ 2x={e}^{2.5}& & & \text{}\\ x=\frac{{e}^{2.5}}{2}.& & & \text{}\end{array}$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Simplify ${e}^{2ln\sqrt{{x}^{3}}}$. Exactly one option must be correct)
a) $2{x}^{3}$ b) ${e}^{2}+{x}^{3∕2}$ c) ${x}^{3}$ d) $2\sqrt{{x}^{3}}$
Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
${e}^{2ln\sqrt{{x}^{3}}}={e}^{2ln{x}^{3∕2}}={e}^{ln{\left({x}^{3∕2}\right)}^{2}}={e}^{ln{x}^{3}}={x}^{3}$.
Choice (d) is incorrect
It is expected that the set of experimental data $\begin{array}{ccccccc}\hfill x\hfill & \hfill 0.8\hfill & \hfill 1.4\hfill & \hfill 2.0\hfill & \hfill 2.6\hfill & \hfill 3.2\hfill & \hfill 3.8\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill y\hfill & \hfill 1.0\hfill & \hfill 5.5\hfill & \hfill 16.0\hfill & \hfill 35.1\hfill & \hfill 65.5\hfill & \hfill 109.7\hfill \end{array}$ follows a power law. Find the best approximation for the given data. Exactly one option must be correct)
a) $y=1{0}^{-2}{e}^{7.5x}$ b) $y=2×1{0}^{-3}{e}^{7.5x}$ c) $y=1.1{x}^{1∕3}$ d) $y=2{x}^{3}$
Choice (a) is incorrect
This equation does not follow a power law.
Choice (b) is incorrect
This equation does not follow a power law.
Choice (c) is incorrect
Choice (d) is correct!
We do a log-log transformation and to 2 decimal places we find $\begin{array}{ccccccc}\hfill lnx=X\hfill & \hfill -0.22\hfill & \hfill 0.34\hfill & \hfill 0.69\hfill & \hfill 0.96\hfill & \hfill 1.16\hfill & \hfill 1.34\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill lny=Y\hfill & \hfill 0\hfill & \hfill 1.70\hfill & \hfill 2.77\hfill & \hfill 3.56\hfill & \hfill 4.18\hfill & \hfill 4.70\hfill \end{array}$ A scatterplot shows that the relationship between $X$ and $Y$ is linear.You should draw the scatterplot to verify this. So putting $Y=mX+b$ we find $m=3$ thus $Y=3X+b$. Now $0=-0.66+b$ therefore $y={e}^{0.66}{x}^{3}\approx 2{x}^{3}$.
It is expected that the set of experimental data $\begin{array}{cccccc}\hfill x\hfill & \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill y\hfill & \hfill 3.0\hfill & \hfill 8.5\hfill & \hfill 15.6\hfill & \hfill 24.0\hfill & \hfill 33.5\hfill \end{array}$ follows a power law. Find the best approximation for the given data. Exactly one option must be correct)
a) $y=1.1{e}^{x}$ b) $y=1.2{e}^{0.9x}$ c) $y=3{x}^{3∕2}$ d) $y=3{x}^{2∕3}$
Choice (a) is incorrect
This equation does not follow a power law.
Choice (b) is incorrect
This equation does not follow a power law.
Choice (c) is correct!
We do a log-log transformation and to 2 decimal places we find $\begin{array}{cccccc}\hfill lnx=X\hfill & \hfill 0\hfill & \hfill 0.69\hfill & \hfill 1.10\hfill & \hfill 1.39\hfill & \hfill 1.61\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill lny=Y\hfill & \hfill 1.10\hfill & \hfill 2.14\hfill & \hfill 2.75\hfill & \hfill 3.18\hfill & \hfill 3.51\hfill \end{array}$ A scatterplot shows that the relationship between $X$ and $Y$ is linear. You should draw the scatterplot to verify this. So letting $Y=mX+b$ we find $m=1.5$ and $b=1.1$. Thus $y={e}^{1.1}{x}^{1.5}\approx 3{x}^{1.5}=3{x}^{3∕2}$.
Choice (d) is incorrect | 3,043 | 6,927 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 89, "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-2017-43 | longest | en | 0.523268 |
http://www.leedsmathstuition.co.uk/2012/06/ | 1,558,936,632,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232261326.78/warc/CC-MAIN-20190527045622-20190527071622-00330.warc.gz | 285,973,212 | 7,153 | ## John solving the Rubik’s Cube
After re-discovering my Rubik’s cube a few days ago I decided to film myself solving it. I first learned to solve a Rubik’s cube about six years ago but have never actually seen myself solving one – until now.
I first decided to learn how to solve a Rubik’s cube after seeing someone playing with one in the maths department at Warwick University; up until then the Rubik’s cube that I owned had remained unused and unsolved after scrambling it when it first came out of the box and then naturally giving it up as a bad job a couple of days later. Well I started trying to solve the cube again and after a little bit of perseverance I managed to solve part of the cube but not all of it – I needed some help!!
Well now we are blessed with access to a whole range of resources and information on the internet and it only takes a few seconds to Google the phrase ‘Rubik’s cube solution’ and see hundreds if not thousands of web pages that tell you how to solve the cube – which I imagine makes things much easier than when the cube first became popular in the 1980s. It turns out that there’s dozens of different ways of solving the cube, some are faster but more difficult than other more straightforward but slower methods. The method that I ended up learning was a method developed by Lars Petrus, a champion Rubik’s cube solver (apparently); I didn’t make a conscious decision to learn this method over all of the others, I just wanted a way of solving the cube.
A week or so and a lot of practice later thanks to the Petrus method I managed to solve the Rubik’s cube for the very first time. I couldn’t believe it – it took me about 20 minutes from start to finish but I did it. I took the cube everywhere with me and spent hours solving it over and over – eventually getting my fastest solve down to about one minute. I haven’t really got any faster at solving it, in fact I’ve probably got slower at solving it but I love messing around with the Rubik’s cube every now and again.
Some of the fastest “speed-cubers” in the world can solve the cube in less than 10 seconds which is unbelievable – I’m not sure that I’d ever get that good and I’m not really sure if it would really be worth the time and effort but it’s always fun to watch the videos on youtube of the people that can do it; I think if I could get to 45 seconds I’d be quite happy with that. There’s even people that can solve the cube blindfolded, some of them solve the thing blindfolded faster than I can solve it with no blindfold – pretty impressive stuff. I’ll keep you all updated on my progress… | 576 | 2,610 | {"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} | 3.046875 | 3 | CC-MAIN-2019-22 | longest | en | 0.976921 |
https://physics.stackexchange.com/questions/tagged/constrained-dynamics?tab=newest&page=4 | 1,576,418,432,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541308149.76/warc/CC-MAIN-20191215122056-20191215150056-00099.warc.gz | 484,915,413 | 41,409 | # Questions tagged [constrained-dynamics]
A constraint is a condition on the variables of a dynamical problem that the variables (or the physical solution for them) must satisfy. Normally, it amounts to restrictions of such variables to a lower-dimensional hypersurface embedded in the higher-dimensional full space of (unconstrained) variables.
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I've seen a few other threads on here inquiring about what is the point of Lagrange Multipliers, or the like. My main question though is, how can I tell by looking at a system in a problem that ... | 384 | 1,785 | {"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} | 2.5625 | 3 | CC-MAIN-2019-51 | latest | en | 0.902016 |
https://school.careers360.com/ncert/ncert-exemplar-class-11-physics-solutions-chapter-3-motion-in-straight-line?Err | 1,719,296,574,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865560.33/warc/CC-MAIN-20240625041023-20240625071023-00240.warc.gz | 453,377,165 | 110,633 | NCERT Exemplar Class 11 Physics Solutions Chapter 3 Motion in a Straight Line
# NCERT Exemplar Class 11 Physics Solutions Chapter 3 Motion in a Straight Line
Edited By Safeer PP | Updated on Aug 09, 2022 03:00 PM IST
NCERT Exemplar Class 11 Physics solutions chapter 3 introduces Motion's basic concept and connects it to our daily activities. NCERT Exemplar Class 11 Physics chapter 3 solutions help understand how Motion is crucial to everything in the universe, be it the acceleration of a car or the earth's rotation. This chapter of NCERT Class 11 Physics Solutions revolves around Motion characteristics like position and displacement, introductory concepts of instantaneous velocity, and kinematic equations to analyse Motion from a fresh perspective of velocity.
## NCERT Exemplar Class 11 Physics Solutions Chapter 3 MCQI
Question:1
The correct answer is the option (b)
Explanation: In graph (b) displacements are in the opposite direction & when we add them, we get net displacement & average velocity as zero. It satisfies the condition of displacement for different timings.
Let us draw a parallel line from A to the time axis at . It intersects the graph at B & the change in displacement time is zero. So, the displacement from A to & hence the average velocity of the body also vanishes to 0.
Question:2
The correct answer is the option (a) x < 0, v < 0, a > 0
Explanation: The value of x becomes negative as the lift comes downward, i.e., from 8th to 4th floor, thus, x<0.
Velocity is downward, i.e., negative, thus, v<0.
The lift retard before reaching the 4th floor and hence the acceleration will be upwards, i.e., a>0.
Question:3
In one dimensional motion, instantaneous speed v satisfies
(a) The displacement in time T must always take non-negative values.
(b) The displacement x in time T satisfies –
(c) The acceleration is always a non-negative number.
(d) The motion has no turning points.
The correct answer is the option (b) The displacement x in time T satisfies -
Explanation: The magnitude & direction of max. & min. velocity can be used to determine the max. & min. displacement.
v0 is the maximum velocity in the positive direction as well as in the opposite direction (or we can say minimum velocity).
Thus, v0T is the maximum displacement in the positive direction & -v0T is the maximum displacement in the opposite direction.
Thus, -
Question:4
The correct answer is the option
Explanation: Let be the time taken in half distance,
Let be the time taken in half distance,
Therefore, the total time taken in distance will be equal to
$=\frac{L(v_{1}+v_{2})}{v_{1}v_{2}}$
& the total distance will be equal to
Thus, the average speed will be,
$\\=\frac{2L}{\frac{L(v_{1}+v_{2})}{v_{1}v_{2}}}\\\\=\frac{2V_1V_2}{V_1+V_2}$
Question:5
The answer is the option (b) 8m
Explanation: It is given that,
Now, we know that,
&
$a=\frac{d^{2}x}{dt^{2}}$
Now,
Now, distance is equal to the area between the time-axis graph and the (v-t) graph,
Hence, option (b).
Question:6
The answer is the option
Explanation: Let us consider L as the length of the escalator,
as the velocity of the girl w.r.t. the ground
& as the velocity of the escalator w.r.t. the ground
Now, w.r.t. the ground, the effective velocity of the girl will be,
$\\\frac{L}{t}=L[\frac{t_1+t_2}{t_1t_2}]\\t=\frac{t_1 t_2}{t_1+t_2}$
## NCERT Exemplar Class 11 Physics Solutions Chapter 3 MCQII
Question:7
The variation of quantity A with quantity B, plotted in Fig. 3.2 describes the motion of a particle in a straight line.
(a) Quantity B may represent time.
(b) Quantity A is velocity if motion is uniform.
(c) Quantity A is displacement if motion is uniform.
(d) Quantity A is velocity if motion is uniformly accelerated.
The correct answer is the option:
(a) Quantity B may represent time.
(c) Quantity A is displacement if the motion is uniform.
(d) Quantity A is velocity if the motion is uniformly accelerated.
Explanation:
Verification of opt (a) & (d)
If the quantity B would have represented velocity instead of time, then the graph would’ve become a straight line, viz., uniformly accelerated motion, hence the motion is not uniform.
Verification of opt (c)
If A represents displacement and B represents time, then the graph will be a straight line which would represent uniform motion.
Question:8
A graph of x versus t is shown in Fig. 3.3. Choose correct alternatives from below.
(a) The particle was released from rest at t = 0.
(b) At B, the acceleration a > 0.
(c) At C, the velocity and the acceleration vanish.
(d) Average velocity for the motion between A and D is positive.
(e) The speed at D exceeds that at E.
The correct answer is the option:
(a) The particle was released from rest at t = 0.
(c) At C, the velocity and acceleration vanish.
(e) The speed at D exceeds that at E.
Explanation: Now, we know that,
Slope of the x-t graph gives us
Verification of opt(a)-
is zero or particle is at rest at A since graph (x-t) is parallel to the time axis.
Slope increases after A and hence velocity also increases.
Verifying option (c) and rejecting opt (b)-
Now, or v = 0 since the tangent at B & C is graph (x-t), viz., parallel to the time axis. Hence, acceleration = 0.
Verifying opt (e)-
Speed at D is greater than speed at E since the slope at D is greater at D than at E.
Rejecting opt (d)-
Average velocity at A is zero as graph (x-t) is parallel to time axis, also displacement is negative at D, which makes it clear that the velocity at D is also negative.
Question:9
The correct answer is the option:
Explanation: Now,
We know that
Now,
vmax will be,
. vmin will be,
Thus, it is clear that v lies between 0 to 2 and option (d) is verified.
Now,
Thus, sin t lies between 1 and -1 for all t > 0.
Thus, x will always be positive and option (a) is verified.
Now,
&
Hereby opt (b) is discarded.
Now,
&
Thus, acceleration can be negative as well, and hence opt (c) is also discarded here.
Question:10
A spring with one end attached to a mass and the other to a rigid support is stretched and released.
a) magnitude of acceleration, when just released is maximum
b) magnitude of acceleration, when at equilibrium position is maximum
c) speed is maximum when mass is at equilibrium position
d) magnitude of displacement is always maximum whenever speed is minimum
The correct answer is the option:
(a) Magnitude of acceleration, when just released is maximum.
(c) Speed is maximum when mass is at the equilibrium position.
Explanation: Let us consider a spring lying on a frictionless table. Let k be the spring constant, viz., attached to a mass ‘m’ at one end and the other end is fixed at right support.
Now let us stretch the spring to a displacement x by force F,
Now, P.E. at
Since restoring force is proportional to x, Simple Harmonic Motion is executed here.
Therefore,
Or
&
Thus, when spring is released, the magnitude will be maximum. Hence, opt(a) is verified here.
At x = 0, the speed of mass is maximum.
Hence opt (c) is also verified.
At x = 0, magnitude of a = 0.
Hence, opt (b) is discarded.
The speed of mass may or may not be zero when it is at its maximum displacement.
Hence, opt (d) is also discarded here.
Question:11
A ball is bouncing elastically with a speed 1 m/s between walls of a railway compartment of size 10 m in a direction perpendicular to walls. The train is moving at a constant velocity of 10 m/s parallel to the direction of motion of the ball. As seen from the ground,
a) the direction of motion of the ball changes every 10 seconds
b) speed of ball changes every 10 seconds
c) average speed of ball over any 20 seconds intervals is fixed
d) the acceleration of ball is the same as from the train
The correct answer is the option:
(b) The speed of the ball changes every 10 seconds.
(c) average speed of the ball over any 20 seconds interval is fixed.
(d) the acceleration of the ball is the same as from the train.
Explanation: If we observe the motion from the ground, we will see that the ball strikes with the wall after every 10 seconds. The direction of the ball is the same since it is moving at a very small speed in the moving train, therefore, it will not change w.r.t. observer from the earth.
The speed of ball can change after a collision, hence, option (a) will be discarded and opt (b) is verified.
Average speed of the ball at any time remain same or is 1 m/s, i.e., it is uniform.
Hence opt (c) is also verified.
When the ball strikes to the wall, initial speed of the ball will be in the direction of the moving train w.r.t. the ground as well as its speed will also change (vTG)
Thus,
The speed of the ball after collision with a side of the train is in the opposite direction of the train
Thus, the magnitude of acceleration on both the walls of the compartment will be the same, but in opposite direction. Hence, opt (b), (c) & (d) are verified here.
NCERT Exemplar Class 11 Physics Solutions Chapter 3 very short answer
Question:12
Graph Characteristics a) i) has and throughout b) ii) has throughout and has a point with and a point with c) iii) has a point with zero displacement for d) iv) has and
(i) From the graph (d), it is indicated that the slope is always positive between to (tan ?).
Hence, (i) (d)
(ii) At point A, v = 0 & a = 0 as the slope is zero; thus, the graph always lies in +x direction.
Hence, (ii) (b).
(iii) There is zero displacement only in graph (a), where y = 0.
Hence, (iii) (a).
(iv) In the graph (c), since v < 0, the slope is negative here.
Hence, (iv) (c).
Question:13
A uniformly moving cricket ball is turned back by hitting it with a bat for a very short time interval. Show the variation of its acceleration with taking acceleration in the backward direction as positive.
When we hit a ball with a bat, the acceleration of the ball decreases, till its velocity becomes zero. Hence, the acceleration will be in the backward direction
After the velocity of the ball has been decreased to zero, it increases in the forward direction. Thus, the acceleration will be negative in the forward direction
Question:14
Give examples of a one-dimensional motion where
a) the particle moving along positive x-direction comes to rest periodically and moves forward
b) the particle moving along positive x-direction comes to rest periodically and moves backward
(a) Let us consider a motion where
Thus,
&
(ii) Let us consider a function of motion where,
Thus the displacement of the particle is in negative direction and it comes to rest periodically.
Thus,
is a periodic function.
Now,
After zero displacement, velocity changes periodically.
Thus, is the function required.
(i) Now, let us consider a function
Thus, the particle moves in a positive direction, periodically with zero displacements.
Hence is the required function.
Question:15
Give example of a motion where at a particular instant.
Let x(t) be the function of motion,
Here, & A are constant & B is the amplitude.
At time t the displacement is x(t),
Here
Thus,
Now,
Thus, we get,
x > 0, i.e., x is always positive, since A>B
v<0, i.e., v is always negative, since v<0
& a>0, i.e., a is always positive.
The value of varies from 0 to
Question:16
An object falling through a fluid is observed to have acceleration given by where g = gravitational acceleration and b is constant. After a long time of release, it is observed to fall with constant speed. What must be the value of constant speed?
The velocity becomes constant after a long time from when it is released, thus,
$\frac{dv(t)}{dt} = 0\; \; or\; \; a=0$
Thus,
Thus, is the constant speed after a long time of release.
## NCERT Exemplar Class 11 Physics Solutions Chapter 3 Short Answer
Question:17
A ball is dropped and its displacement vs time graph is as shown in the figure where displacement x is from the ground and all quantities are positive upwards.
a) Plot qualitatively velocity vs time graph
b) Plot qualitatively acceleration vs time graph
If we observe the graph we know that the displacement (x) is always positive. The velocity of the body keeps on increasing till the displacement becomes zero, after that the velocity decreases to zero in the opposite direction till the maximum value of x is reached, viz., smaller than earlier. When the body reaches towards x=0, the velocity increases and acceleration is in the downward direction. And when the body’s displacement is , i.e., the body moves upwards, the direction will be downwards and velocity will decrease, i.e., .
(a) Velocity time graph
(b) Acceleration time graph.
Question:18
A particle executes the motion described by where
a) Where does the particles start and with what velocity?
b) Find maximum and minimum values of . Show that and increase with time and decreases with time.
Here,
So,
$=+x_{0}\gamma e^{-\gamma t}\; \; \; \; \; \; ..........(i)$
&
+
Thus, is the starting point of the particle and its velocity is
(b) $x (t)\ is,$
Maximum at since $[x(t)]_{max} = \infty$
Minimum at since at ,
v(t) is,
maximum at since
minimum at since, ,
a(t) is,
maximum at since at ,
minimum at since at ,
Question:20
A man runs across the roof-top of a tall building and jumps horizontally with the hope of landing on the roof of the next building which is of a lower height than the first. If his speed is 9 m/s, the distance between the two buildings is 10 m and the height difference is 9 m, will he be able to land on the next building?
In vertical motion,
&
now,
thus,
Therefore,
Now, the horizontal distance covered by the person will be,
$u_{x}\times\; t=9\left ( \frac{3}{\sqrt{5}} \right )$
Therefore, the person will reach the building which s next farther the first edge by .
Question:21
A ball is dropped from a building of height 45 m. Simultaneously another ball is thrown up with a speed 40 m/s. Calculate the relative speed of the balls as a function of time.
For the first ball-
we know that,
(for second ball)
now,
Now relative velocity of the 1st ball w.r.t. 2nd
The speed of one ball increases and the speed of the other decreases with the same rate due to acceleration.
Hence, relative speed = 40 m/s.
Question:22
The velocity-displacement graph of a particle is shown in the figure.
a) Write the relation between v and x.
b) Obtain the relation between acceleration and displacement and plot it.
a) Consider the point P(x,v) at any time t on the graph such that angle ABO is such that
$\tan \theta = \frac{AQ}{QP} = \frac{(v_{0}-v)}{x} = \frac{v_{0}}{x_{0}}$
When the velocity decreases from to zero during the displacement, the acceleration becomes negative.
$v_{0}-v=\left ( \frac{v_{0}}{x_{0}} \right )x$
$v=v_{0}(1-\frac{x}{x_{0}})$
is the relation between v and x.
$a=\frac{-v_{0}}{x_{0}}v$
$a=\left ( \frac{v{_{0}}^{2}}{x{_{0}}^{2}} \right )x-\left ( \frac{v{_{0}}^{2}}{x_{0}}\right )$
x=x0
The points are
and
## NCERT Exemplar Class 11 Physics Solutions Chapter 3 Long Answer
Question:23
It is a common observation that rain clouds can be at about a kilometre altitude above the ground.
a) If a raindrop falls from such a height freely under gravity, what will be its speed? Also, calculate in km/h
b) A typical raindrop is about 4 mm diameter. Momentum is mass x speed in magnitude. Estimate its momentum when it hits ground.
c) Estimate the time required to flatten the drop.
d) Rate of change of momentum is force. Estimate how much force such a drop would exert on you.
e) Estimate the order of magnitude force on umbrella. Typical lateral separation between two raindrops is 5 cm.
Thus,
(a) Let's find out the velocity of raindrop on the ground’
$\\v^{2}=u^{2}-2as\\=u^{2}-2g(-h)\\=u^{2}+2gh\\=0^{2}+2(10)(1000)$
Thus,
$\\v=100\sqrt{2}\; m/s\\v=100\sqrt{2}\left ( \frac{18}{5} \right )km/hr$
(b) Momentum of the raindrop when it touches the ground
mass of drop(m) = Vol. × density
Now, density of water
Thus,
Now, we know that Momentum (p) = mv
(c) Time required for a drop to be flattened-
(d) Now, we know that,
(e) Here,
Thus, Area of umbrella
Now, the square area covered by one drop
$= (5 \times 10^{-2})^{2}$
Therefore, no. of drops falling on the umbrella
Therefore 314 drops fell on the umbrella.
Thus, the net force on the umbrella
Question:24
A motor car moving at a speed of 72 km/h cannot come to a stop in less than 3 s while for a truck this time interval is 5 s. On a highway the car is behind the truck both moving at 72 km/h. The truck gives a signal that it is going to stop at an emergency. At what distance the car should be from the truck so that it does not bump onto the truck. Human response time is 0.5 s.
Given : (for truck)
We know that,
i.e.,
thus,
Given : (for car)
Again,
$a_c = \frac{-20}{3} \; m/s^2$
A human takes atleast 0.5 seconds to respond, thus time taken by the car driver to respond is sec …. (car takes t time to stop)
$Vc = u + a{_{c}}{t}$
$0 = 20-\frac{20}{3}.(t - 0.5) \; \; \; \; \; \; .......... (i)$
There is no responding time for the truck driver so he applies breaks with passing signal to car back side, hence,
From (i) & (ii),
$20 - 4t = 20 - \frac{20}{3} ( t - 0.5)$
Thus,
Thus,
Now the distance covered by the car & the truck in sec will be,
$S = 20 \left ( \frac{5}{4} \right ) + 0.5(-4) \left ( \frac{5}{4} \right )\left ( \frac{5}{4} \right )$
…….
$=21.875 m$
First 0.5 seconds the car moves with uniform speed but after responding brakes are applied for 0.5 sec and the retarding motion of the car starts.
$=21.875 m$
Thus,
Therefore, the car must be 1.25 m behind the truck to avoid bumping into it.
Question:25
A monkey climbs up a slippery pole for 3 seconds and subsequently slips for 3 seconds. Its velocity at time t is given by and for on m/s. It repeats this cycle till it reaches the height of 20 m.
a) At what time is its velocity maximum?
b) At what time is its average velocity maximum?
c) At what times is its acceleration maximum in magnitude?
d) How many cycles are required to reach the top?
(a) for the velocity to be maximum,
Thus,
(b)Now, average velocity
Let us integrate both the sides from 0 to 3
Thus,
Now, average velocity
$t=2.36s$
Thus, the average velocity is maximum at 2.36 seconds.
(c) When the body returns at its mean position or changes direction in periodic motion, time for acceleration is maximum.
Thus, the acceleration is maximum at t = 3 sec.
(d)Now, for 3 to 6 sec
Integrate from 3 to 6 s
.............because distance is in downward direction
Thus, net distance
Thus, in three cycle
Remaining height will be
The monkey can climb up to 9m without slipping but in the 4th cycle it will slip and the height remaining to climb will be 6.5 m.
Net no. of cycle = 4.
Question:26
A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other after a time interval. The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.
Let be the speed of the 1st ball,
& be the speed of the second ball,
Let be the height of the two balls before coming to rest & be the height covered by 1 ball before coming to rest.
Now, we know that,
Thus,
and
Now,
Thus,
Calculating time for 1st ball,
Thus,
Now, calculating time for second ball,
This,
Thus, time intervals between these two balls will be,
To help comprehend the concepts well, students can take the help of NCERT Exemplar Class 11 Physics Solutions Chapter 3 PDF Download, which could be put to download by differrent web page download options
Well-versed academicians prepare these solutions to provide the students an insight into a creative learning process that would help them perform better in exams.
Also, check NCERT Solution subject wise -
Also, Read NCERT Notes subject wise -
## Class 11 Physics NCERT Exemplar Solutions Chapter 3 Includes The Following Topics:
3 Motion in a Straight Line
• 3.1 Introduction
• 3.2 Position, path length, and displacement
• 3.3 Average velocity and average speed
• 3.4 Instantaneous velocity and speed
• 3.5 Acceleration
• 3.6 Kinematic equations for uniformly accelerated Motion
• 3.7 Relative velocity
## What Will The Students Learn in NCERT Exemplar Class 11 Physics Solutions Chapter 3 Motion In A Straight Line?
• Students will be able to throw light on the significance of Motion in their typical activities.
• They would be able to utilise the concept of Motion to explain and observe the working of everything that surrounds them.
• NCERT Exemplar Class 11 Physics chapter 3 solutions enlighten the students by providing logical explanations behind situations.
• NCERT Exemplar Class 11 Physics solutions chapter 3 helps get a clear idea of average speed and velocity, which is a crucial part of our lives.
• Students are also introduced to numerical interpretations of these concepts with the help of kinematic expressions.
## NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
Chapter 2 Units and Measurement Chapter 4 Motion in a Plane Chapter 5 Laws of Motion Chapter 6 Work, Energy, and Power Chapter 7 Systems of Particles and Rotational Motion Chapter 8 Gravitation Chapter 9 Mechanical Properties of Solids Chapter 10 Mechanical Properties of Fluids Chapter 11 Thermal Properties of Matter Chapter 12 Thermodynamics Chapter 13 Kinetic Theory Chapter 14 Oscillations Chapter 15 Waves
## Important Topics To Cover For Exams in NCERT Exemplar Solutions For Class 11 Physics Chapter 3
· NCERT Exemplar Class 11 Physics solutions chapter 3 covers the properties and terms related to Motion, such as position, path length, the frame of reference, and a detailed description of displacement. This chapter provides an in-depth analysis of the study of the Motion of objects along a straight line.
· This chapter also helps students to differentiate between speed and velocity and know the terms like average speed, average velocity, instantaneous speed, and instantaneous velocity. It also offers a graphical representation using diverse examples of velocity and speed. It dives deep into the study of acceleration to understand how an object gains or loses speed.
· The mathematical and graphical analysis of Motion and the relative terms are introduced in this chapter through Kinematic equations for uniform accelerated Motion. NCERT Exemplar Class 11 Physics chapter 3 solutions would also help highlight the critical feature of Motion in terms of frames of reference, by the concept of relative velocity.
## NCERT Exemplar Class 11 Solutions
NCERT Exemplar Class 11 Mathematics Solutions NCERT Exemplar Class 11 Chemistry Solutions NCERT Exemplar Class 11 Physics Solutions NCERT Exemplar Class 11 Biology Solutions
### Check Class 11 Physics Chapter-wise Solutions
Chapter 1 Physical world Chapter 2 Units and Measurement Chapter 3 Motion in a straight line Chapter 4 Motion in a Plane Chapter 5 Laws of Motion Chapter 6 Work, Energy and Power Chapter 7 System of Particles and Rotational motion Chapter 8 Gravitation Chapter 9 Mechanical Properties of Solids Chapter 10 Mechanical Properties of Fluids Chapter 11 Thermal Properties of Matter Chapter 12 Thermodynamics Chapter 13 Kinetic Theory Chapter 14 Oscillations Chapter 15 Waves
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### Frequently Asked Question (FAQs)
1. How can these solutions help?
NCERT Exemplar Class 11 Physics Solutions Chapter 3 can help you to understand the applications of motions in real life and help you perform better in exams.
2. What are the essential topics of this chapter?
Essential topics of NCERT Exemplar Class 11 Physics Solutions Chapter 3 are the Motion in a Straight Line, position, path length, displacement, average velocity and average speed, Kinematic equations, relative velocity, instantaneous velocity and speed.
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Get answers from students and experts
A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
Option 1) Option 2) Option 3) Option 4)
A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms−2 :
Option 1) 2.45×10−3 kg Option 2) 6.45×10−3 kg Option 3) 9.89×10−3 kg Option 4) 12.89×10−3 kg
An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
Option 1) Option 2) Option 3) Option 4)
A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
Option 1) Option 2) Option 3) Option 4)
In the reaction,
Option 1) at STP is produced for every mole consumed Option 2) is consumed for ever produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .
How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?
Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2
If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will
Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.
With increase of temperature, which of these changes?
Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.
Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is
Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023
A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is
Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9 | 6,855 | 27,093 | {"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": 25, "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} | 4.53125 | 5 | CC-MAIN-2024-26 | latest | en | 0.83675 |
https://mathematica.stackexchange.com/questions/26773/how-to-improve-the-quality-of-the-listplot-output | 1,718,691,019,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861747.46/warc/CC-MAIN-20240618042809-20240618072809-00336.warc.gz | 340,104,353 | 43,831 | # How to improve the quality of the ListPlot output
I need to plot massive data stored in external ASCII files. Below is the Mathematica code I use for this job:
Clear["Global*"];
SetDirectory[" ... "];
dataRG = ReadList["dataRG.out", Number, RecordLists -> True];
dataCH = ReadList["dataCH.out", Number, RecordLists -> True];
Vd = -((Md*(1 + δ))/
Sqrt[(1 + δ)*(b^2 + y^2) +
x^2 + (a + Sqrt[h^2 + (1 + δ^2)*z^2])^2]);
Vn = -(Mn/Sqrt[x^2 + y^2 + z^2 + cn^2]);
Vh = -(Mh/Sqrt[x^2 + y^2 + z^2 + ch^2]);
V = Vd + Vn + Vh;
Md = 8200; b = 8; a = 3; h = 0.3; δ = 0.1;
Mn = 400; cn = 0.25;
Mh = 0; ch = 25;
E0 = -700;
z0 = 1;
f[x_, px_] := 1/2*px^2 + V /. {y -> 0, z -> z0};
xmax = 11;
pxmax = 41;
plrange = {{-xmax, xmax}, {-pxmax, pxmax}};
C0 = ContourPlot[Evaluate[f[x, px]], {x, -20, 20}, {px, -80, 80},
Contours -> {E0}, ContourStyle -> {Black, Thick}, AspectRatio -> 1,
ContourShading -> False, PlotPoints -> 200,
PerformanceGoal -> "Speed", PlotRange -> plrange];
S1 = ListPlot[Flatten[List /@ dataRG[[All, {1, 2}]], 1],
PlotStyle -> {GrayLevel[0.8], PointSize[0.003]}];
S2 = ListPlot[Flatten[List /@ dataCH[[All, {1, 2}]], 1],
PlotStyle -> {GrayLevel[0.05], PointSize[0.003]}];
P0 = Show[{S1, S2, C0}, Frame -> True, Axes -> False,
FrameLabel -> {"x", OverDot["x"]}, RotateLabel -> False,
FrameStyle -> Directive[FontSize -> 17, FontFamily -> "Helvetica"],
AspectRatio -> 1, PlotRange -> plrange, ImageSize -> 550]
which produces the following output
As you may see from the image two are the main issues:
(1). How can I get rid off the vertical blank gaps?
(2). Why there are some "bold horizontal lines" in the plot and again how can I get rid off them?
I observed, that both issues are influenced by the ImageSize option. In particular, if you play with this option between 400 and 600, it might eliminate the problems. However, I suspect that the cause must be something more profound. So, I would be very grateful, if you suggest me how can I solve these issues and also if you could provide me with a more correct way to plot my data files.
Both data files can be found here: dataRG and dataCH.
• Quite likely, those things you're seeing are moiré artifacts... Commented Jun 10, 2013 at 17:30
• @0x4A4D They could be artifacts, but when I export the plot in different formats (.eps, .pdf, .jpg) they are still present messing up with my plot. Commented Jun 10, 2013 at 17:59
• Though I didn't check your data, I agree with 0x4A4D it might be moiré pattern. One way is to blur them a little when the distance between your adjacent points comes near the DPI of your monitor. Commented Jun 10, 2013 at 18:24
Your data is on a regular grid, so it is also possible to put each data point into a matrix element, then use functions like ArrayPlot/MatrixPlot to efficiently plot it.
posCH = Reverse[{100, 73} + #/{1, 5}] & /@
Union[Round[10 dataCH[[All, 1 ;; 2]]]];
posRG = Reverse[{100, 73} + #/{1, 5}] & /@
Union[Round[10 dataRG[[All, 1 ;; 2]]]];
S12Array = SparseArray[{
posRG -> ConstantArray[.8, Length@posRG],
posCH -> ConstantArray[.05, Length@posCH]
}, {145, 200}, 1];
S12 = ArrayPlot[S12Array, ColorFunction -> GrayLevel, ColorFunctionScaling -> False]
In order to match C0 (of which I changed the color to Red for highlighting) on it, we need to transform C0:
C0trans = C0 /. {
GraphicsComplex[pts_, others__] :>
GraphicsComplex[{100, 73} + 10 #/{1, 5} & /@ pts, others],
(PlotRange -> _) :>
(PlotRange -> ({100, 73} + 10 #/{1, 5} & /@
(plrange\[Transpose])\[Transpose]))}
And we'll have to construct the FrameTicks manually:
xticks = Join[
{100 + 10 #, #, {.01, 0}} & /@ Range[-20, 20, 2],
{100 + 10 #, "", {.005, 0}} & /@ Range[-20, 20, 1/4]];
xticks2 = Join[
{100 + 10 #, "", {.01, 0}} & /@ Range[-20, 20, 2],
{100 + 10 #, "", {.005, 0}} & /@ Range[-20, 20, 1/4]];
yticks = Join[
{73 + (10 #)/5, #, {.01, 0}} & /@ Range[-80, 80, 5],
{73 + (10 #)/5, "", {.005, 0}} & /@ Range[-80, 80, 1]];
yticks2 = Join[
{73 + (10 #)/5, "", {.01, 0}} & /@ Range[-80, 80, 5],
{73 + (10 #)/5, "", {.005, 0}} & /@ Range[-80, 80, 1]];
Combine them:
Show[S12, C0trans,
FrameTicks -> {{yticks, yticks2}, {xticks, xticks2}},
FrameLabel -> {"x", OverDot["x"]}, RotateLabel -> False,
FrameStyle -> Directive[FontSize -> 17, FontFamily -> "Helvetica"],
AspectRatio -> 1, ImageSize -> 550]
• Also try the ArrayPlot with setting the PixelConstrained -> True option.
– shrx
Commented Jun 14, 2013 at 19:18
• @shrx Thanks. This option does make the plot prettier. I didn't adopt it because OP explicitly specified ImageSize -> 550, which will contradict the PixelConstrained -> True or PixelConstrained -> 1 option. Commented Jun 15, 2013 at 7:12
You are forcing a regular pattern (your data) on a regular grid with a different spacing. This is bound to lead to aliasing. In your data vertical steps are 5 times larger than the horizontal steps:
Differences@(dataRG[[All, 1]] // Union) // Union
{0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1}
Differences@(dataRG[[All, 2]] // Union) // Union
{0.5, 1., 2.5}
The AspectRatio->1 setting forces the data range ratio of ...
((Subtract @@ #2)/)Subtract @@ #1)) & @@ (P0 // PlotRange)
3.7272727273.72
to be 1, making the vertical step size a non-integer multiple of the horizontal one. That's trouble.
You will note that if you set AspectRatio->Automatic and the graphics size is sufficient the aliasing is gone.
• If I set AspectRatio->Automatic the plot is too elongated at the vertical axes and therefore distorted! I want to believe that there is a much simpler solution which we haven't thought yet. Commented Jun 10, 2013 at 21:01
• @Vaggelis_Z As Sylvia already mentioned you could blur your data. Commented Jun 10, 2013 at 21:17
Another possible way would be to use Interpolation (Caution: Very slow!):
posRG = Union[dataRG[[All, {1, 2}]]];
posCH = Union[dataCH[[All, {1, 2}]]];
posBoundary = Cases[C0, GraphicsComplex[pts_, __] :> pts, ∞][[1,
Most[Cases[C0, Line[pts__] :> pts, ∞][[1]]]]];
interpFunc = Interpolation[
Join[{#, .8} & /@ posRG,
{#, .05} & /@ posCH,
{#, .05} & /@ posBoundary]]
S12 = DensityPlot[interpFunc[x, y], {x, -10, 10}, {y, -40, 40},
PlotRange -> {0, 1}, PlotPoints -> 100,
ColorFunction -> GrayLevel, ColorFunctionScaling -> False,
RegionFunction -> Function[{x, y, z}, Evaluate[f[x, px] < E0 /. px -> y]]]
P0 = Show[{S12, C0}, Frame -> True, Axes -> False,
FrameLabel -> {"x", OverDot["x"]}, RotateLabel -> False,
FrameStyle -> Directive[FontSize -> 17, FontFamily -> "Helvetica"],
AspectRatio -> 1, PlotRange -> plrange, ImageSize -> 550]
You could render it as a bitmap at a higher resolution, then scale it back down to size. The image scaling algorithms will do a better job of representing the fine detail than ListPlot. This can be done easily with Rasterize. For example, you can do
Rasterize[
Show[{S1, S2, C0}, Frame -> True, Axes -> False,
FrameLabel -> {"x", OverDot["x"]}, RotateLabel -> False,
FrameStyle -> Directive[FontSize -> 17, FontFamily -> "Helvetica"],
AspectRatio -> 1, PlotRange -> plrange, ImageSize -> 550],
RasterSize -> 550*8]
`
Note that the resulting object is a bitmap, so won't look good if you try to resize it. | 2,348 | 7,153 | {"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} | 2.5625 | 3 | CC-MAIN-2024-26 | latest | en | 0.762721 |
http://thegatebook.in/qa/5188/dm-groups-and-lattice-q14?show=5232 | 1,580,251,932,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251783342.96/warc/CC-MAIN-20200128215526-20200129005526-00059.warc.gz | 159,746,796 | 5,623 | # DM-Groups and Lattice-Q14
+1 vote
The no of elements which do not have complements in the above lattices is ---?
asked Jul 8, 2019
reshown Jul 9, 2019
+1 vote
for complement LUB(x,y)=1 and GLB(x,y)=0 for above diagram
LUB(a,b)=1 GLB(a,b)=0 so a and b are complement to each other.
LUB(0,1)=1 GLB(0,1)=0 so 0,1 are complement to each other.
for rest complement is not possible.
c,d,f,h,g,e will have 0 complement. | 165 | 458 | {"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-2020-05 | longest | en | 0.841549 |
sunnemath.wordpress.com | 1,548,041,736,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583755653.69/warc/CC-MAIN-20190121025613-20190121051613-00561.warc.gz | 664,073,194 | 18,445 | Posted on
# Mathematical Mosaics
In your home, down a street, or even in ordinary objects, mosaics make their mark on our daily lives. A mathematical mosaic is a group of shapes (with sides of equal length and measurement) arranged together to form a repetitious pattern in which they share sides at each corner point. Each corner point contains the same neighbors throughout the pattern. (Mathematics: A Human Endeavor, 3rd Edition)
For example, everyday I walk over this.
At first glance it looks like ordinary tile. However, if you look closer it is mathematical mosaic composed of rectangles and squares. To make this arrangement you need four rectangles and one square. The four rectangles create the point all having interior right angles. The equation for the angles would be 90°+90°+90°+90° = 360°. The rectangles have rotational symmetry creating a square in the pattern.
Everyday I cuddle up to this seemingly innocent lap quilt.
But in reality it is a mathematical mosaic. The pattern is composed of eight isosceles triangles. The triangles have one right angle and two forty-five degree angles. The eight triangles around the point each have an interior angle of 45. Thus the equation would be 45°+45°+45°+45°+45°+45°+45°+45°= 360°
Incorporating activities like finding mathematical mosaics in everyday life into your lessons give the student not only an understanding of how mosaics influence our world but also a real world application of math. It helps the student understand that math can be applied to our everyday environment and lives not just in the classroom.
Here are some questions you might consider using in a lesson on mathematical mosaics.
How do you know if a mosaic is really a mathematical mosaic?
If you were given triangles and squares how many different math mosaics could you create in a 6×6 square?
If you have a mathematical mosaic made up of octagons and dodecagons, what other polygons could you use to complete the mosaic? Give reasons why you chose each polygon.
Do you think the number of sides of a polygon has any correlation between the corresponding interior angles of that polygon? Can you prove your answer?
## 3 responses to “Mathematical Mosaics”
1. Love your pictures and your questions!! The question about the sides and angles corresponding helps students to understand an important mathematical relationship!
Great Blog!
Kasey
2. Jennifer,
I spent 29 years looking at those same tiles in my school! I wonder if they were sold at a discount….. The real world connections that you have modelled for kids can be very powerful motivators in the classroom. They can take a confusing term and make sense of it for our students. Nicey done!
Pat
3. Jennifer,
That is a beautiful quilt! I asked a similar question to your second one, but I didn’t give a dimension for a guideline. I think that is much better. I will adjust my question! Thanks for the tip;)
Lauren | 651 | 2,941 | {"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} | 4.5625 | 5 | CC-MAIN-2019-04 | latest | en | 0.904524 |
http://forums.wesnoth.org/viewtopic.php?f=3&t=38319&p=552268 | 1,597,184,755,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738855.80/warc/CC-MAIN-20200811205740-20200811235740-00319.warc.gz | 46,256,282 | 9,266 | ## Wesnoth probability calculator
Share and discuss strategies for playing the game, and get help and tips from other players.
Moderators: Forum Moderators, Developers
Tondo
Posts: 15
Joined: February 28th, 2006, 3:33 pm
Contact:
### Wesnoth probability calculator
Hi,
This is a crude probability calculator for Wesnoth I doodled at work, I thought I'd post it here.
See if you like it or if it is any help at playing or at least settling disputes about good/bad luck.
Enjoy,
Tondo
Attachments
Wesnoth chance 100 v1.02 RTP.ods
Wesnoth probability calculator
tr0ll
Posts: 528
Joined: June 11th, 2006, 8:13 pm
### Re: Wesnoth probability calculator
that certainly shows the value of using terrain!
cautions:
- the table will be off if any of the attackers are marksmen or magic.
- assumes the target unit has infinite HP. if you kill the unit on the 1st hit, the rest of the strikes are wasted, so players shouldnt necessarily count on needing that many units to win a particular fight. this is where other factors like leadership, time of day, slow, resistance (weapon+unit+steadfast) come in.
Aelaris
Posts: 77
Joined: January 21st, 2010, 3:22 am
### Re: Wesnoth probability calculator
Actually, the table is off if the units have different attacks. Accuracy is one thing, but hit threshold doesn't take into account the difference between a Wose hitting a unit and a Elvish Scout hitting a unit. To kill, say, an Elvish Shaman, let's have a Wose and two scouts attack. It dies in two hits of a wose, one wose-hit and three scout hits, or six scout hits.
So what's the hit threshold?
I feel that comparing different units is sort of beyond the scope of his build here.
The question of how many units are needed is something you can break down by running multiple threshold calculations, and comparing the numbers. I'll do it:
Elvish Fighters (5-4 attack) VS Spearman (36HP) on flat terrain (40%)
Threshold is 8 hits (40 damage > 36 HP > 35 damage)
Two elvish fighters = 1.6% chance to kill
Three elvish fighters = 43% chance to kill, 1.6% chance you'll have one extra.
Four elvish fighters = 85% chance to kill, 41.4% chance you'll have one extra, 1.6% you'll have two extra.
And so forth.
All the bonuses, you'll just have to math into the hit threshold, but it's fine as long as they are all the same. | 613 | 2,324 | {"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-2020-34 | latest | en | 0.920318 |
https://www.excel-university.com/articles/on-balance/excel-and-access-tools-of-our-trade/ | 1,719,070,027,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862404.32/warc/CC-MAIN-20240622144011-20240622174011-00503.warc.gz | 664,963,044 | 15,722 | # Excel and Access: Tools of our Trade
## What would you bring if you were stranded on a desert island?
I would bring Microsoft Excel because it’s the only application I would need! It stores more than 17 billion cells in a single sheet, summarizes data with Pivot Tables, generates charts and has a built-in Web browser. Excel, in my opinion, is the greatest software application of all time. However, it has its limitations. In addition, it’s nice to know that when we run into an Excel limitation, we can turn to Microsoft Access.
## Excel Tips
When I was the accounting manager at a public company, my life was sometimes stressful because I had more work than time. I took a long time to complete my work because I didn’t use the right Excel functions or features. Once I figured out how to use Excel the right way, I was able to delegate many of my mechanical recurring tasks to Excel. The following are my top three favorite functions, features and shortcuts:
Functions
• SUBTOTAL: The SUBTOTAL function is similar to the SUM function, except the SUBTOTAL function excludes other SUBTOTAL functions in the range.
• SUMIFS: The SUMIFS function is similar to the SUM function, except it only includes the rows that meet a condition. For example, sum up the “amount” column, but only include those rows where the “account” column equals “cash.”
• VLOOKUP: The VLOOKUP function looks for matching rows in a related table, and returns a corresponding value. For example, look up account number “100” in the chart of accounts and return the matching account name.
## Features
These functions help improve my productivity, increase my accuracy, and make my workbooks more “bulletproof.”
• Tables: Tables solve one of the biggest pitfalls in Excel. Tables have special properties, including auto-expansion. Any new data typed or pasted directly under a table will expand the table, and will be included in any formulas that reference the table. (Insert > Table)
• PivotTables: PivotTables are reports. They aggregate and summarize transactional data, are interactive and allow you to analyze the underlying data effortlessly. This is a huge feature and the most powerful feature of Excel. (Insert > PivotTable)
## Shortcuts
Three time-saving shortcuts are:
• Fill-down: You can quickly fill a formula down a column when there is data in the adjacent column by double-clicking the lower right corner of the formula cell.
• Next worksheet: You can navigate to the next worksheet in the workbook by selecting Ctrl + PageDown. (The previous sheet is Ctrl + PageUp.)
• Arrow keys: You can use the arrow keys to navigate within a worksheet or to define function arguments. Holding down the Shift key with the arrow keys extends the current selection. Holding down the Ctrl key with the arrow keys jumps to the edge of the data region. Holding down both the Shift and Ctrl keys with the arrow keys extends the current selection to the edge of the data region.
## Introduction to Access
Excel can be a powerful tool, but there are times when certain job tasks are better suited for Access. Access is Microsoft’s smallest database application. A database is a collection of related data. The data is stored in “tables” within an Access file. A database table is similar to an Excel worksheet, as it has rows and columns. These are called records and fields in database lingo.
Although an Excel worksheet can technically store more than a million rows of data, it’s better to store large data sets inside a database, such as Access. Databases are designed to store and process many records, and have special settings designed to handle large data sets, such as indexing.
Excel doesn’t handle several people working on the same workbook at the same time. Databases like Access are designed for multiple concurrent users. Use Access when you have a project that requires several people to work with data simultaneously.
Excel has limited security settings, but Access has detailed user-level security settings. For example, we can allow the user to view certain tables, another user to edit data, and another user to run reports. In Access, user level security is well-defined.
We can centralize data in an Access database and pull it into Excel using the external data feature. Clicking the Refresh button pulls the changes into Excel. If there’s no software to help you automate a specific or unique business process, use Access to create your own application. Examples include a budget system or a commission reporting system. Access can help us overcome Excel’s limitations, improve Excel’s utility and help automate workflow and processes. Excel and Access are both tools of our trade.
Posted in
### Jeff Lenning
I love sharing the things I've learned about Excel, and I built Excel University to help me do that. My motto is: Learn Excel. Work Faster.
### Excel is not what it used to be.
You need the Excel Proficiency Roadmap now. Includes 6 steps for a successful journey, 3 things to avoid, and weekly Excel tips. | 1,027 | 5,031 | {"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-2024-26 | latest | en | 0.927262 |
http://mathhelpforum.com/calculus/67686-integration-question.html | 1,481,199,746,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542588.29/warc/CC-MAIN-20161202170902-00496-ip-10-31-129-80.ec2.internal.warc.gz | 173,790,965 | 9,806 | 1. Integration (?) question
a cuboid has the square base of side $xcm$ and height $hcm$. its volume is $120cm^3$.
i) find $h$ in terms of $x$. Hence show that the surface area, $Acm^2$, of the cuboid is given by $A = 2x^2 + \frac{480}{x}$ .
ii) find $\frac{dA}{dx}$ and $\frac{d^2A}{dx^2}$
iii) Hence find the value of $x$ which gives the minimum of surface area. Find also the value of the surface area in this case.
Thanks alot guys, as always your amazing :P
2. Originally Posted by coyoteflare
a cuboid has the square base of side $xcm$ and height $hcm$. its volume is $120cm^3$.
i) find $h$ in terms of $x$. Hence show that the surface area, $Acm^2$, of the cuboid is given by $A = 2x^2 + \frac{480}{x}$ .
ii) find $\frac{dA}{dx}$ and $\frac{d^2A}{dx^2}$
iii) Hence find the value of $x$ which gives the minimum of surface area. Find also the value of the surface area in this case.
$V = x^2h = 120$
solve for h ...
$h = \frac{120}{x^2}$
surface area is (top+bottom) + (4 sides) ...
$A = 2x^2 + 4xh$
sub in for h ...
$A = 2x^2 + 4x\left(\frac{120}{x^2}\right)$
$A = 2x^2 + \frac{480}{x}$
... try the rest of the problem yourself. | 407 | 1,151 | {"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": 25, "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-2016-50 | longest | en | 0.863843 |
https://fr.scribd.com/document/267078004/Frequentism-and-Bayesianism-Python | 1,579,437,330,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594603.8/warc/CC-MAIN-20200119122744-20200119150744-00190.warc.gz | 446,045,290 | 94,225 | Vous êtes sur la page 1sur 9
# PROC. OF THE 13th PYTHON IN SCIENCE CONF.
(SCIPY 2014)
Primer
Jake VanderPlas
## arXiv:1411.5018v1 [astro-ph.IM] 18 Nov 2014
AbstractThis paper presents a brief, semi-technical comparison of the essential features of the frequentist and Bayesian approaches to statistical inference, with several illustrative examples implemented in Python. The differences
between frequentism and Bayesianism fundamentally stem from differing definitions of probability, a philosophical divide which leads to distinct approaches
to the solution of statistical problems as well as contrasting ways of asking
discussion of these differences, we briefly compare several leading Python statistical packages which implement frequentist inference using classical methods
and Bayesian inference using Markov Chain Monte Carlo.1
Index Termsstatistics, frequentism, Bayesian inference
Introduction
## One of the first things a scientist in a data-intensive field
hears about statistics is that there are two different approaches:
frequentism and Bayesianism. Despite their importance, many
researchers never have opportunity to learn the distinctions
between them and the different practical approaches that result.
This paper seeks to synthesize the philosophical and pragmatic aspects of this debate, so that scientists who use these
approaches might be better prepared to understand the tools
available to them. Along the way we will explore the fundamental philosophical disagreement between frequentism and
Bayesianism, explore the practical aspects of how this disagreement affects data analysis, and discuss the ways that these
practices may affect the interpretation of scientific results.
This paper is written for scientists who have picked up some
statistical knowledge along the way, but who may not fully
appreciate the philosophical differences between frequentist
and Bayesian approaches and the effect these differences
have on both the computation and interpretation of statistical
results. Because this passing statistics knowledge generally
leans toward frequentist principles, this paper will go into
more depth on the details of Bayesian rather than frequentist
approaches. Still, it is not meant to be a full introduction
to either class of methods. In particular, concepts such as
the likelihood are assumed rather than derived, and many
* Corresponding author: jakevdp@cs.washington.edu
eScience Institute, University of Washington
c 2014 Jake VanderPlas. This is an open-access article distributed
unrestricted use, distribution, and reproduction in any medium, provided the
original author and source are credited.
1. This paper draws heavily from content originally published in a series
of posts on the authors blog, Pythonic Perambulations [VanderPlas2014].
## advanced Bayesian and frequentist diagnostic tests are left
out in favor of illustrating the most fundamental aspects of
the approaches. For a more complete treatment, see, e.g.
[Wasserman2004] or [Gelman2004].
The Disagreement: The Definition of Probability
## Fundamentally, the disagreement between frequentists and
Bayesians concerns the definition of probability.
For frequentists, probability only has meaning in terms of
a limiting case of repeated measurements. That is, if an
astronomer measures the photon flux F from a given nonvariable star, then measures it again, then again, and so on,
each time the result will be slightly different due to the
statistical error of the measuring device. In the limit of many
measurements, the frequency of any given value indicates
the probability of measuring that value. For frequentists,
probabilities are fundamentally related to frequencies of
events. This means, for example, that in a strict frequentist
view, it is meaningless to talk about the probability of the true
flux of the star: the true flux is, by definition, a single fixed
value, and to talk about an extended frequency distribution for
a fixed value is nonsense.
For Bayesians, the concept of probability is extended to
cover degrees of certainty about statements. A Bayesian
might claim to know the flux F of a star with some probability
P(F): that probability can certainly be estimated from frequencies in the limit of a large number of repeated experiments,
but this is not fundamental. The probability is a statement
of the researchers knowledge of what the true flux is. For
Bayesians, probabilities are fundamentally related to their
own knowledge about an event. This means, for example,
that in a Bayesian view, we can meaningfully talk about the
probability that the true flux of a star lies in a given range.
That probability codifies our knowledge of the value based on
prior information and available data.
The surprising thing is that this arguably subtle difference in
philosophy can lead, in practice, to vastly different approaches
to the statistical analysis of data. Below we will explore a
few examples chosen to illustrate the differences in approach,
along with associated Python code to demonstrate the practical
aspects of the frequentist and Bayesian approaches.
A Simple Example: Photon Flux Measurements
## First we will compare the frequentist and Bayesian approaches
to the solution of an extremely simple problem. Imagine that
## we point a telescope to the sky, and observe the light coming
from a single star. For simplicity, we will assume that the
stars true photon flux is constant with time, i.e. that is it
has a fixed value F; we will also ignore effects like sky
background systematic errors. We will assume that a series
of N measurements are performed, where the ith measurement
reports the observed flux Fi and error ei .2 The question is,
given this set of measurements D = {Fi , ei }, what is our best
estimate of the true flux F?
First we will use Python to generate some toy data to
demonstrate the two approaches to the problem. We will draw
50 samples Fi with a mean of 1000 (in arbitrary units) and a
(known) error ei :
>>> np.random.seed(2) # for reproducibility
>>> e = np.random.normal(30, 3, 50)
>>> F = np.random.normal(1000, e)
## In this toy example we already know the true flux F, but
the question is this: given our measurements and errors, what
is our best point estimate of the true flux? Lets look at a
frequentist and a Bayesian approach to solving this.
Frequentist Approach to Flux Measurement
approach. Given a single observation Di = (Fi , ei ), we can
compute the probability distribution of the measurement given
the true flux F given our assumption of Gaussian errors:
(Fi F)2
2 1/2
.
P(Di |F) = 2ei
exp
2e2i
This should be read the probability of Di given F equals ....
You should recognize this as a normal distribution with mean
F and standard deviation ei . We construct the likelihood by
computing the product of the probabilities for each data point:
N
i=1
## Here D = {Di } represents the entire set of measurements. For
reasons of both analytic simplicity and numerical accuracy, it
is often more convenient to instead consider the log-likelihood;
combining the previous two equations gives
(Fi F)2
1 N
.
log L (D|F) = log(2e2i ) +
2 i=1
e2i
We would like to determine the value of F which maximizes
the likelihood. For this simple problem, the maximization can
be computed analytically (e.g. by setting d log L /dF|F = 0),
which results in the following point estimate of F:
wi Fi
F =
; wi = 1/e2i
wi
The result is a simple weighted mean of the observed values.
Notice that in the case of equal errors ei , the weights cancel
and F is simply the mean of the observed data.
2. We will make the reasonable assumption of normally-distributed measurement errors. In a Frequentist perspective, ei is the standard deviation of the
results of the single measurement event in the limit of (imaginary) repetitions
of that event. In the Bayesian perspective, ei describes the probability
distribution which quantifies our knowledge of F given the measured value
Fi .
We can go further and ask what the uncertainty of our estimate is. One way this can be accomplished in the frequentist
approach is to construct a Gaussian approximation to the peak
likelihood; in this simple case the fit can be solved analytically
to give:
!1/2
N
F =
wi
i=1
## This result can be evaluated this in Python as follows:
>>> w = 1. / e ** 2
>>> F_hat = np.sum(w * F) / np.sum(w)
>>> sigma_F = w.sum() ** -0.5
## For our particular data, the result is F = 999 4.
Bayesian Approach to Flux Measurement
## The Bayesian approach, as you might expect, begins and ends
with probabilities. The fundamental result of interest is our
knowledge of the parameters in question, codified by the
probability P(F|D). To compute this result, we next apply
Bayes theorem, a fundamental law of probability:
P(F|D) =
P(D|F) P(F)
P(D)
## Though Bayes theorem is where Bayesians get their name,
it is important to note that it is not this theorem itself that is
controversial, but the Bayesian interpretation of probability
implied by the term P(F|D). While the above formulation
makes sense given the Bayesian view of probability, the
setup is fundamentally contrary to the frequentist philosophy,
which says that probabilities have no meaning for fixed model
parameters like F. In the Bayesian conception of probability,
however, this poses no problem.
Lets take a look at each of the terms in this expression:
## P(F|D): The posterior, which is the probability of the
model parameters given the data.
P(D|F): The likelihood, which is proportional to the
L (D|F) used in the frequentist approach.
P(F): The model prior, which encodes what we knew
about the model before considering the data D.
P(D): The model evidence, which in practice amounts
to simply a normalization term.
## If we set the prior P(F) 1 (a flat prior), we find
P(F|D) L (D|F).
That is, with a flat prior on F, the Bayesian posterior is
maximized at precisely the same value as the frequentist
result! So despite the philosophical differences, we see that
the Bayesian and frequentist point estimates are equivalent for
this simple problem.
You might notice that we glossed over one important piece
here: the prior, P(F), which we assumed to be flat.3 The prior
allows inclusion of other information into the computation,
which becomes very useful in cases where multiple measurement strategies are being combined to constrain a single model
(as is the case in, e.g. cosmological parameter estimation).
## The necessity to specify a prior, however, is one of the more
controversial pieces of Bayesian analysis.
A frequentist will point out that the prior is problematic
when no true prior information is available. Though it might
seem straightforward to use an uninformative prior like
the flat prior mentioned above, there are some surprising
subtleties involved.4 It turns out that in many situations, a
truly uninformative prior cannot exist! Frequentists point out
that the subjective choice of a prior which necessarily biases
the result should have no place in scientific data analysis.
A Bayesian would counter that frequentism doesnt solve
this problem, but simply skirts the question. Frequentism can
often be viewed as simply a special case of the Bayesian
approach for some (implicit) choice of the prior: a Bayesian
would say that its better to make this implicit choice explicit,
even if the choice might include some subjectivity. Furthermore, as we will see below, the question frequentism answers
is not always the question the researcher wants to ask.
Where The Results Diverge
## In the simple example above, the frequentist and Bayesian
approaches give basically the same result. In light of this,
arguments over the use of a prior and the philosophy of
probability may seem frivolous. However, while it is easy to
show that the two approaches are often equivalent for simple
problems, it is also true that they can diverge greatly in other
situations. In practice, this divergence most often makes itself
most clear in two different ways:
1. The handling of nuisance parameters: i.e. parameters which affect the final result, but are not otherwise of interest.
2. The different handling of uncertainty: for example,
the subtle (and often overlooked) difference between
frequentist confidence intervals and Bayesian credible regions.
We will discuss examples of these below.
Nuisance Parameters: Bayes Billiards Game
## We will start by discussing the first point: nuisance parameters.
A nuisance parameter is any quantity whose value is not
directly relevant to the goal of an analysis, but is nevertheless
required to determine the result which is of interest. For example, we might have a situation similar to the flux measurement
above, but in which the errors ei are unknown. One potential
approach is to treat these errors as nuisance parameters.
Here we consider an example of nuisance parameters borrowed from [Eddy2004] that, in one form or another, dates
all the way back to the posthumously-published 1763 paper
written by Thomas Bayes himself [Bayes1763]. The setting is
3. A flat prior is an example of an improper prior: that is, it cannot be
normalized. In practice, we can remedy this by imposing some bounds on
possible values: say, 0 < F < Ftot , the total flux of all sources in the sky. As
this normalization term also appears in the denominator of Bayes theorem,
it does not affect the posterior.
4. The flat prior in this case can be motivated by maximum entropy;
see, e.g. [Jeffreys1946]. Still, the use of uninformative priors like this often
raises eyebrows among frequentists: there are good arguments that even
uninformative priors can add information; see e.g. [Evans2002].
## a gambling game in which Alice and Bob bet on the outcome
of a process they cant directly observe.
Alice and Bob enter a room. Behind a curtain there is a
billiard table, which they cannot see. Their friend Carol rolls
a ball down the table, and marks where it lands. Once this mark
is in place, Carol begins rolling new balls down the table. If
the ball lands to the left of the mark, Alice gets a point; if
it lands to the right of the mark, Bob gets a point. We can
assume that Carols rolls are unbiased: that is, the balls have
an equal chance of ending up anywhere on the table. The first
person to reach six points wins the game.
Here the location of the mark (determined by the first roll)
can be considered a nuisance parameter: it is unknown and
not of immediate interest, but it clearly must be accounted for
when predicting the outcome of subsequent rolls. If this first
roll settles far to the right, then subsequent rolls will favor
Alice. If it settles far to the left, Bob will be favored instead.
Given this setup, we seek to answer this question: In a
particular game, after eight rolls, Alice has five points and
Bob has three points. What is the probability that Bob will get
six points and win the game?
Intuitively, we realize that because Alice received five of the
eight points, the marker placement likely favors her. Given that
she has three opportunities to get a sixth point before Bob can
win, she seems to have clinched it. But quantitatively speaking,
what is the probability that Bob will persist to win?
A Nave Frequentist Approach
## Someone following a classical frequentist approach might
reason as follows:
To determine the result, we need to estimate the location
of the marker. We will quantify this marker placement as a
probability p that any given roll lands in Alices favor. Because
five balls out of eight fell on Alices side of the marker, we
compute the maximum likelihood estimate of p, given by:
p = 5/8,
a result follows in a straightforward manner from the binomial
likelihood. Assuming this maximum likelihood probability, we
can compute the probability that Bob will win, which requires
him to get a point in each of the next three rolls. This is given
by:
P(B) = (1 p)
3
Thus, we find that the probability of Bob winning is 0.053, or
odds against Bob winning of 18 to 1.
A Bayesian Approach
## A Bayesian approach to this problem involves marginalizing
(i.e. integrating) over the unknown p so that, assuming the
prior is accurate, our result is agnostic to its actual value. In
this vein, we will consider the following quantities:
B = Bob Wins
D = observed data, i.e. D = (nA , nB ) = (5, 3)
p = unknown probability that a ball lands on Alices side
during the current game
We want to compute P(B|D); that is, the probability that
Bob wins given the observation that Alice currently has five
## points to Bobs three. A Bayesian would recognize that this
expression is a marginal probability which can be computed
by integrating over the joint distribution P(B, p|D):
P(B|D)
P(B, p|D)dp
This identity follows from the definition of conditional probability, and the law of total probability: that is, it is a fundamental consequence of probability axioms and will always
be true. Even a frequentist would recognize this; they would
simply disagree with the interpretation of P(p) as being a
measure of uncertainty of knowledge of the parameter p.
To compute this result, we will manipulate the above
expression for P(B|D) until we can express it in terms of other
quantities that we can compute.
We start by applying the definition of conditional probability
to expand the term P(B, p|D):
Z
P(B|D) =
P(B|p, D)P(p|D)d p
Z
P(B|D) =
P(B|p, D)
P(D|p)P(p)
dp
P(D)
## Finally, using the same probability identity we started with,
we can expand P(D) in the denominator to find:
R
P(B|D) =
P(B|p, D)P(D|p)P(p)d p
R
P(D|p)P(p)d p
## Now the desired probability is expressed in terms of three
quantities that we can compute:
## P(B|p, D): This term is proportional to the frequentist
likelihood we used above. In words: given a marker
placement p and Alices 5 wins to Bobs 3, what is the
probability that Bob will go on to six wins? Bob needs
three wins in a row, i.e. P(B|p, D) = (1 p)3 .
P(D|p): this is another easy-to-compute term. In words:
given a probability p, what is the likelihood of exactly 5
positive outcomes out of eight trials? The answer comes
from the Binomial distribution: P(D|p) p5 (1 p)3
P(p): this is our prior on the probability p. By the
problem definition, we can assume that p is evenly drawn
between 0 and 1. That is, P(p) 1 for 0 p 1.
R1
P(B|D) = R01
0
(1 p)6 p5 d p
(1 p)3 p5 d p
## These integrals are instances of the beta function, so we can
quickly evaluate the result using scipy:
Discussion
## The Bayesian approach gives odds of 10 to 1 against Bob,
while the nave frequentist approach gives odds of 18 to 1
against Bob. So which one is correct?
For a simple problem like this, we can answer this question
empirically by simulating a large number of games and count
the fraction of suitable games which Bob goes on to win. This
can be coded in a couple dozen lines of Python (see part II of
[VanderPlas2014]). The result of such a simulation confirms
the Bayesian result: 10 to 1 against Bob winning.
So what is the takeaway: is frequentism wrong? Not necessarily: in this case, the incorrect result is more a matter
of the approach being nave than it being frequentist.
The approach above does not consider how p may vary.
There exist frequentist methods that can address this by,
e.g. applying a transformation and conditioning of the data
to isolate dependence on p, or by performing a Bayesianlike integral over the sampling distribution of the frequentist
estimator p.
## Another potential frequentist response is that the question
itself is posed in a way that does not lend itself to the classical,
frequentist approach. A frequentist might instead hope to give
the answer in terms of null tests or confidence intervals: that
is, they might devise a procedure to construct limits which
would provably bound the correct answer in 100 (1 )
percent of similar trials, for some value of say, 0.05. We
will discuss the meaning of such confidence intervals below.
There is one clear common point of these two frequentist
responses: both require some degree of effort and/or special
expertise in classical methods; perhaps a suitable frequentist approach would be immediately obvious to an expert
statistician, but is not particularly obvious to a statistical layperson. In this sense, it could be argued that for a problem
such as this (i.e. with a well-motivated prior), Bayesianism
provides a more natural framework for handling nuisance
parameters: by simple algebraic manipulation of a few wellknown axioms of probability interpreted in a Bayesian sense,
we straightforwardly arrive at the correct answer without need
for other special statistical expertise.
Confidence vs. Credibility: Jaynes Truncated Exponential
## A second major consequence of the philosophical difference
between frequentism and Bayesianism is in the handling of
uncertainty, exemplified by the standard tools of each method:
frequentist confidence intervals (CIs) and Bayesian credible
regions (CRs). Despite their apparent similarity, the two approaches are fundamentally different. Both are statements of
probability, but the probability refers to different aspects of the
computed bounds. For example, when constructing a standard
95% bound about a parameter :
## >>> from scipy.special import beta
>>> P_B_D = beta(6+1, 5+1) / beta(3+1, 5+1)
winning.
## A Bayesian would say: Given our observed data, there
is a 95% probability that the true value of lies within
the credible region.
A frequentist would say: If this experiment is repeated
many times, in 95% of these cases the computed confidence interval will contain the true .5
## Notice the subtle difference: the Bayesian makes a statement
of probability about the parameter value given a fixed credible
region. The frequentist makes a statement of probability about
the confidence interval itself given a fixed parameter value.
This distinction follows straightforwardly from the definition
of probability discussed above: the Bayesian probability is
a statement of degree of knowledge about a parameter; the
frequentist probability is a statement of long-term limiting
frequency of quantities (such as the CI) derived from the data.
This difference must necessarily affect our interpretation
of results. For example, it is common in scientific literature
to see it claimed that it is 95% certain that an unknown
parameter lies within a given 95% CI, but this is not the
case! This is erroneously applying the Bayesian interpretation
to a frequentist construction. This frequentist oversight can
perhaps be forgiven, as under most circumstances (such as the
simple flux measurement example above), the Bayesian CR
and frequentist CI will more-or-less overlap. But, as we will
see below, this overlap cannot always be assumed, especially
in the case of non-Gaussian distributions constrained by few
data points. As a result, this common misinterpretation of the
frequentist CI can lead to dangerously erroneous conclusions.
To demonstrate a situation in which the frequentist confidence interval and the Bayesian credibility region do not
overlap, let us turn to an example given by E.T. Jaynes, a
20th century physicist who wrote extensively on statistical
inference. In his words, consider a device that
...will operate without failure for a time
because of a protective chemical inhibitor injected
into it; but at time the supply of the chemical is
exhausted, and failures then commence, following
the exponential failure law. It is not feasible to
observe the depletion of this inhibitor directly; one
can observe only the resulting failures. From data
on actual failure times, estimate the time of
guaranteed safe operation... [Jaynes1976]
Essentially, we have data D drawn from the model:
exp( x) , x >
P(x| ) =
0
, x<
where p(x| ) gives the probability of failure at time x, given
an inhibitor which lasts for a time . We observe some failure
times, say D = {10, 12, 15}, and ask for 95% uncertainty
bounds on the value of .
First, lets think about what common-sense would tell us.
Given the model, an event can only happen after a time .
Turning this around tells us that the upper-bound for must
be min(D). So, for our particular data, we would immediately
write 10. With this in mind, lets explore how a frequentist
and a Bayesian approach compare to this observation.
Truncated Exponential: A Frequentist Approach
## In the frequentist paradigm, wed like to compute a confidence
interval on the value of . We might start by observing that
5. [Wasserman2004], however, notes on p. 92 that we need not consider
repetitions of the same experiment; its sufficient to consider repetitions of
any correctly-performed frequentist procedure.
Z
E(x) =
xp(x)dx = + 1.
0
## So, using the sample mean as the point estimate of E(x), we
have an unbiased estimator for given by
1 N
= xi 1.
N i=1
In the large-N limit, the central limit theorem tells us that the
sampling distribution is normal with standard deviation given
by the standard error of the mean: 2 = 1/N, and we can write
the 95% (i.e. 2 ) confidence interval as
CIlarge N = 2N 1/2 , + 2N 1/2
For our particular observed data, this gives a confidence
interval around our unbiased estimator of CI( ) = (10.2, 12.5),
entirely above our common-sense bound of < 10! We
might hope that this discrepancy is due to our use of the
large-N approximation with a paltry N = 3 samples. A more
careful treatment of the problem (See [Jaynes1976] or part
III of [VanderPlas2014]) gives the exact confidence interval
(10.2, 12.2): the 95% confidence interval entirely excludes the
sensible bound < 10!
Truncated Exponential: A Bayesian Approach
P( |D) =
P(D| )P( )
.
P(D)
N
P(D| ) P(xi | )
i=1
## and, in the absence of other information, use an uninformative
flat prior on to find
N exp [N( min(D))] , < min(D)
P( |D)
0
, > min(D)
where min(D) is the smallest value in the data D, which
enters because of the truncation of P(xi | ). Because P( |D)
increases exponentially up to the cutoff, the shortest 95%
credibility interval (1 , 2 ) will be given by 2 = min(D), and
1 given by the solution to the equation
Z 2
P( |D)d = f
## which has the solution
1 = 2 +
i
1 h
ln 1 f (1 eN2 ) .
N
## For our particular data, the Bayesian credible region is
CR( ) = (9.0, 10.0)
which agrees with our common-sense bound.
Discussion
## Why do the frequentist CI and Bayesian CR give such different
results? The reason goes back to the definitions of the CI and
CR, and to the fact that the two approaches are answering
the value of itself (the probability that the parameter is in the
the procedure used to construct the CI (the probability that
any potential CI will contain the fixed parameter).
Using Monte Carlo simulations, it is possible to confirm
that both the above results correctly answer their respective
questions (see [VanderPlas2014], III). In particular, 95% of
frequentist CIs constructed using data drawn from this model
in fact contain the true . Our particular data are simply
among the unhappy 5% which the confidence interval misses.
But this makes clear the danger of misapplying the Bayesian
interpretation to a CI: this particular CI is not 95% likely to
contain the true value of ; it is in fact 0% likely!
This shows that when using frequentist methods on fixed
data, we must carefully keep in mind what question frequentism is answering. Frequentism does not seek a probabilistic
statement about a fixed interval as the Bayesian approach does;
constructed intervals, with the particular computed interval
just a single draw from among them. Despite this, it is common
to see a 95% confidence interval interpreted in the Bayesian
sense: as a fixed interval that the parameter is expected to
be found in 95% of the time. As seen clearly here, this
interpretation is flawed, and should be carefully avoided.
Though we used a correct unbiased frequentist estimator
above, it should be emphasized that the unbiased estimator is
not always optimal for any given problem: especially one with
small N and/or censored models; see, e.g. [Hardy2003]. Other
frequentist estimators are available: for example, if the (biased)
maximum likelihood estimator were used here instead, the
confidence interval would be very similar to the Bayesian
credible region derived above. Regardless of the choice of
frequentist estimator, however, the correct interpretation of the
CI is the same: it gives probabilities concerning the recipe
for constructing limits, not for the parameter values given
the observed data. For sensible parameter constraints from
a single dataset, Bayesianism may be preferred, especially if
the difficulties of uninformative priors can be avoided through
the use of true prior information.
Bayesianism in Practice: Markov Chain Monte Carlo
## Though Bayesianism has some nice features in theory, in
practice it can be extremely computationally intensive: while
simple problems like those examined above lend themselves to
relatively easy analytic integration, real-life Bayesian computations often require numerical integration of high-dimensional
parameter spaces.
A turning-point in practical Bayesian computation was the
development and application of sampling methods such as
Markov Chain Monte Carlo (MCMC). MCMC is a class
of algorithms which can efficiently characterize even highdimensional posterior distributions through drawing of randomized samples such that the points are distributed according
## to the posterior. A detailed discussion of MCMC is well
beyond the scope of this paper; an excellent introduction
can be found in [Gelman2004]. Below, we will propose a
straightforward model and compare a standard frequentist
approach with three MCMC implementations available in
Python.
Application: A Simple Linear Model
## As an example of a more realistic data-driven analysis, lets
consider a simple three-parameter linear model which fits a
straight-line to data with unknown errors. The parameters will
be the the y-intercept , the slope , and the (unknown)
For data D = {xi , yi }, the model is
y(x
i |, ) = + xi ,
and the likelihood is the product of the Gaussian distribution
for each point:
N
[yi y(x
i |, )]2
.
L (D|, , ) = (2 2 )N/2 exp
2 2
i=1
We will evaluate this model on the following data set:
import numpy as np
np.random.seed(42) # for repeatability
theta_true = (25, 0.5)
xdata = 100 * np.random.random(20)
ydata = theta_true[0] + theta_true[1] * xdata
ydata = np.random.normal(ydata, 10) # add error
## Below we will consider a frequentist solution to this problem
computed with the statsmodels package6 , as well as a Bayesian
solution computed with several MCMC implementations in
Python: emcee7 , PyMC8 , and PyStan9 . A full discussion of
the strengths and weaknesses of the various MCMC algorithms
used by the packages is out of scope for this paper, as is a
full discussion of performance benchmarks for the packages.
Rather, the purpose of this section is to show side-by-side
examples of the Python APIs of the packages. First, though,
we will consider a frequentist solution.
Frequentist Solution
A frequentist solution can be found by computing the maximum likelihood point estimate. For standard linear problems
such as this, the result can be computed using efficient linear
algebra. If we define the parameter vector, = [ ]T ; the
response vector, Y = [y1 y2 y3 yN ]T ; and the design matrix,
T
1 1 1 1
X=
,
x1 x2 x3 xN
it can be shown that the maximum likelihood solution is
= (X T X)1 (X T Y ).
6.
7.
8.
9.
## statsmodels: Statistics in Python http://statsmodels.sourceforge.net/
emcee: The MCMC Hammer http://dan.iel.fm/emcee/
PyMC: Bayesian Inference in Python http://pymc-devs.github.io/pymc/
PyStan: The Python Interface to Stan https://pystan.readthedocs.org/
## The confidence interval around this value is an ellipse in
parameter space defined by the following matrix:
2
= 2 (M T M)1 .
2
Here is our unknown error term; it can be estimated based
on the variance of the residuals about the fit. The off-diagonal
elements of are the correlated uncertainty between the
estimates. In code, the computation looks like this:
>>>
>>>
...
>>>
>>>
>>>
...
X = np.vstack([np.ones_like(xdata), xdata]).T
theta_hat = np.linalg.solve(np.dot(X.T, X),
np.dot(X.T, ydata))
y_hat = np.dot(X, theta_hat)
sigma_hat = np.std(ydata - y_hat)
Sigma = sigma_hat ** 2 *\
np.linalg.inv(np.dot(X.T, X))
## The 1 and 2 results are shown by the black ellipses in
Figure 1.
In practice, the frequentist approach often relies on many
more statistal diagnostics beyond the maximum likelihood and
confidence interval. These can be computed quickly using
convenience routines built-in to the statsmodels package
[Seabold2010]. For this problem, it can be used as follows:
>>>
>>>
>>>
>>>
>>>
>>>
## import statsmodels.api as sm # version 0.5
result = sm.OLS(ydata, X).fit()
sigma_hat = result.params
Sigma = result.cov_params()
print(result.summary2())
====================================================
Model:
OLS AIC:
147.773
Dependent Variable: y
BIC:
149.765
No. Observations:
20
Log-Likelihood:
-71.887
Df Model:
1
F-statistic:
41.97
Df Residuals:
18
Prob (F-statistic): 4.3e-06
R-squared:
0.70 Scale:
86.157
0.68
---------------------------------------------------Coef. Std.Err. t
P>|t| [0.025 0.975]
---------------------------------------------------const
24.6361 3.7871 6.5053 0.0000 16.6797 32.592
x1
0.4483 0.0692 6.4782 0.0000 0.3029 0.593
---------------------------------------------------Omnibus:
1.996
Durbin-Watson:
2.75
Prob(Omnibus):
0.369
Jarque-Bera (JB):
1.63
Skew:
0.651
Prob(JB):
0.44
Kurtosis:
2.486
Condition No.:
100
====================================================
## The summary output includes many advanced statistics which
we dont have space to fully discuss here. For a trained
practitioner these diagnostics are very useful for evaluating and
comparing fits, especially for more complicated models; see
[Wasserman2004] and the statsmodels project documentation
for more details.
Bayesian Solution: Overview
## The Bayesian result is encapsulated in the posterior, which
is proportional to the product of the likelihood and the
prior; in this case we must be aware that a flat prior is not
uninformative. Because of the nature of the slope, a flat prior
leads to a much higher probability for steeper slopes. One
## might imagine addressing this by transforming variables, e.g.
using a flat prior on the angle the line makes with the xaxis rather than the slope. It turns out that the appropriate
change of variables can be determined much more rigorously
by following arguments first developed by [Jeffreys1946].
Our model is given by y = + x with probability element
P(, )dd . By symmetry, we could just as well have
written x = 0 + 0 y with probability element Q( 0 , 0 )d 0 d 0 .
It then follows that ( 0 , 0 ) = ( 1 , 1 ). Computing the
determinant of the Jacobian of this transformation, we can
then show that Q( 0 , 0 ) = 3 P(, ). The symmetry of the
problem requires equivalence of P and Q, or 3 P(, ) =
P( 1 , 1 ), which is a functional equation satisfied by
P(, ) (1 + 2 )3/2 .
This turns out to be equivalent to choosing flat priors on the
alternate variables ( , ) = (tan1 , cos ).
Through similar arguments based on the invariance of
under a change of units, we can show that
P( ) 1/ ,
which is most commonly known a the Jeffreys Prior for
scale factors after [Jeffreys1946], and is equivalent to flat
prior on log . Putting these together, we find the following
uninformative prior for our linear regression problem:
1
(1 + 2 )3/2 .
## With this prior and the above likelihood, we are prepared to
numerically evaluate the posterior via MCMC.
P(, , )
## The emcee package [ForemanMackey2013] is a lightweight
pure-Python package which implements Affine Invariant Ensemble MCMC [Goodman2010], a sophisticated version of
MCMC sampling. To use emcee, all that is required is to
define a Python function representing the logarithm of the
posterior. For clarity, we will factor this definition into two
functions, the log-prior and the log-likelihood:
import emcee
# version 2.0
def log_prior(theta):
alpha, beta, sigma = theta
if sigma < 0:
return -np.inf # log(0)
else:
return (-1.5 * np.log(1 + beta**2)
- np.log(sigma))
def log_like(theta, x, y):
alpha, beta, sigma = theta
y_model = alpha + beta * x
return -0.5 * np.sum(np.log(2*np.pi*sigma**2) +
(y-y_model)**2 / sigma**2)
def log_posterior(theta, x, y):
return log_prior(theta) + log_like(theta,x,y)
## Next we set up the computation. emcee combines multiple
interacting walkers, each of which results in its own Markov
chain. We will also specify a burn-in period, to allow the
chains to stabilize prior to drawing our final traces:
## ndim = 3 # number of parameters in the model
nwalkers = 50 # number of MCMC walkers
nburn = 1000 # "burn-in" to stabilize chains
nsteps = 2000 # number of MCMC steps to take
starting_guesses = np.random.rand(nwalkers, ndim)
## Now we call the sampler and extract the trace:
sampler = emcee.EnsembleSampler(nwalkers, ndim,
log_posterior,
args=[xdata,ydata])
sampler.run_mcmc(starting_guesses, nsteps)
# chain is of shape (nwalkers, nsteps, ndim):
# discard burn-in points and reshape:
trace = sampler.chain[:, nburn:, :]
trace = trace.reshape(-1, ndim).T
## The result is shown by the blue curve in Figure 1.
Solution with PyMC
## The PyMC package [Patil2010] is an MCMC implementation
written in Python and Fortran. It makes use of the classic
Metropolis-Hastings MCMC sampler [Gelman2004], and includes many built-in features, such as support for efficient
sampling of common prior distributions. Because of this, it
requires more specialized boilerplate than does emcee, but the
result is a very powerful tool for flexible Bayesian inference.
The example below uses PyMC version 2.3; as of this
writing, there exists an early release of version 3.0, which is a
complete rewrite of the package with a more streamlined API
and more efficient computational backend. To use PyMC, we
first we define all the variables using its classes and decorators:
import pymc
# version 2.3
## alpha = pymc.Uniform(alpha, -100, 100)
@pymc.stochastic(observed=False)
def beta(value=0):
return -1.5 * np.log(1 + value**2)
@pymc.stochastic(observed=False)
def sigma(value=1):
return -np.log(abs(value))
# Define the form of the model and likelihood
@pymc.deterministic
def y_model(x=xdata, alpha=alpha, beta=beta):
return alpha + beta * x
y = pymc.Normal(y, mu=y_model, tau=1./sigma**2,
observed=True, value=ydata)
# package the full model in a dictionary
model = dict(alpha=alpha, beta=beta, sigma=sigma,
y_model=y_model, y=y)
## Next we run the chain and extract the trace:
S = pymc.MCMC(model)
S.sample(iter=100000, burn=50000)
trace = [S.trace(alpha)[:], S.trace(beta)[:],
S.trace(sigma)[:]]
## The result is shown by the red curve in Figure 1.
Solution with PyStan
PyStan is the official Python interface to Stan, a probabilistic programming language implemented in C++ and making
## use of a Hamiltonian MCMC using a No U-Turn Sampler
[Hoffman2014]. The Stan language is specifically designed
for the expression of probabilistic models; PyStan lets Stan
models specified in the form of Python strings be parsed,
compiled, and executed by the Stan library. Because of this,
PyStan is the least Pythonic of the three frameworks:
import pystan
# version 2.2
model_code = """
data {
int<lower=0> N; // number of points
real x[N]; // x values
real y[N]; // y values
}
parameters {
real alpha_perp;
real<lower=-pi()/2, upper=pi()/2> theta;
real log_sigma;
}
transformed parameters {
real alpha;
real beta;
real sigma;
real ymodel[N];
alpha <- alpha_perp / cos(theta);
beta <- sin(theta);
sigma <- exp(log_sigma);
for (j in 1:N)
ymodel[j] <- alpha + beta * x[j];
}
model {
y ~ normal(ymodel, sigma);
}
"""
# perform the fit & extract traces
data = {N: len(xdata), x: xdata, y: ydata}
fit = pystan.stan(model_code=model_code, data=data,
iter=25000, chains=4)
tr = fit.extract()
trace = [tr[alpha], tr[beta], tr[sigma]]
Comparison
## The 1 and 2 posterior credible regions computed with these
three packages are shown beside the corresponding frequentist
confidence intervals in Figure 1. The frequentist result gives
slightly tighter bounds; this is primarily due to the confidence
interval being computed assuming a single maximum likelihood estimate of the unknown scatter, (this is analogous to
the use of the single point estimate for the nuisance parameter
p in the billiard game, above). This interpretation can be
confirmed by plotting the Bayesian posterior conditioned on
the maximum likelihood estimate ; this gives a credible
region much closer to the frequentist confidence interval.
The similarity of the three MCMC results belie the differences in algorithms used to compute them: by default,
PyMC uses a Metropolis-Hastings sampler, PyStan uses a
No U-Turn Sampler (NUTS), while emcee uses an affineinvariant ensemble sampler. These approaches are known to
have differing performance characteristics depending on the
features of the posterior being explored. As expected for the
near-Gaussian posterior used here, the three approaches give
very similar results.
A main apparent difference between the packages is the
## FREQUENTISM AND BAYESIANISM: A PYTHON-DRIVEN PRIMER
Fig. 1: Comparison of model fits using frequentist maximum likelihood, and Bayesian MCMC using three Python packages: emcee,
PyMC, and PyStan.
Finally, we combined these ideas and showed several examples of the use of frequentism and Bayesianism on a more
realistic linear regression problem, using several mature packages available in the Python language ecosystem. Together,
these packages offer a set of tools for statistical analysis in
both the frequentist and Bayesian frameworks.
So which approach is best? That is somewhat a matter
of personal ideology, but also depends on the nature of the
problem at hand. Frequentist approaches are often easily
computed and are well-suited to truly repeatible processes
and measurements, but can hit snags with small sets of data
and models which depart strongly from Gaussian. Frequentist
tools for these situations do exist, but often require subtle
considerations and specialized expertise. Bayesian approaches
require specification of a potentially subjective prior, and
often involve intensive computation via MCMC. However,
they are often conceptually more straightforward, and pose
results in a way that is much closer to the questions a scientist
wishes to answer: i.e. how do these particular data constrain
the unknowns in a certain model? When used with correct
understanding of their application, both sets of statistical tools
can be used to effectively interpret of a wide variety of
scientific and technical results.
R EFERENCES
Python interface. Emcee is perhaps the simplest, while PyMC
requires more package-specific boilerplate code. PyStan is the
most complicated, as the model specification requires directly
writing a string of Stan code.
Conclusion
## This paper has offered a brief philosophical and practical
glimpse at the differences between frequentist and Bayesian
approaches to statistical analysis. These differences have their
root in differing conceptions of probability: frequentists define
probability as related to frequencies of repeated events, while
Bayesians define probability as a measure of uncertainty. In
practice, this means that frequentists generally quantify the
properties of data-derived quantities in light of fixed model
parameters, while Bayesians generally quantify the properties
of unknown models parameters in light of observed data. This
philosophical distinction often makes little difference in simple
problems, but becomes important within more sophisticated
analysis.
We first considered the case of nuisance parameters, and
showed that Bayesianism offers more natural machinery to
deal with nuisance parameters through marginalization. Of
course, this marginalization depends on having an accurate
prior probability for the parameter being marginalized.
Next we considered the difference in the handling of
uncertainty, comparing frequentist confidence intervals with
Bayesian credible regions. We showed that when attempting
to find a single, fixed interval bounding the true value of a
parameter, the Bayesian solution answers the question that
researchers most often ask. The frequentist solution can be
informative; we just must be careful to correctly interpret the
frequentist confidence interval.
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A. Gelman, J.B. Carlin, H.S. Stern, and D.B. Rubin.
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and Hall/CRC, Boca Raton, FL, 2004.
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M. Hardy. An illuminating counterexample. Am.
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M.C. Hoffman & A. Gelman. The No-U-Turn Sampler: Adaptively Setting Path Lengths in Hamiltonian
Monte Carlo. JMLR, submitted, 2014.
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E.T. Jaynes. Confidence Intervals vs Bayesian Intervals (1976) Papers on Probability, Statistics and
Statistical Physics Synthese Library 158:149, 1989
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H. Jeffreys An Invariant Form for the Prior Probability in Estimation Problems. Proc. of the Royal
Society of London. Series A 186(1007): 453, 1946
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A. Patil, D. Huard, C.J. Fonnesbeck. PyMC:
Bayesian Stochastic Modelling in Python Journal of
Statistical Software, 35(4):1-81, 2010.
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J.S. Seabold and J. Perktold. Statsmodels: Econometric and Statistical Modeling with Python Proceedings
of the 9th Python in Science Conference, 2010
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J. VanderPlas. Frequentism and Bayesianism. Fourpart series (I, II, III, IV) on Pythonic Perambulations
http://jakevdp.github.io/, 2014.
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L. Wasserman. All of statistics: a concise course in
statistical inference. Springer, 2004. | 11,182 | 46,501 | {"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-2020-05 | latest | en | 0.904902 |
https://community.wolfram.com/groups/-/m/t/227722 | 1,685,246,434,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643462.13/warc/CC-MAIN-20230528015553-20230528045553-00213.warc.gz | 214,718,783 | 23,083 | # How to remove curly brackets from NDSolve/Evaluate output
Posted 9 years ago
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s=NDSolve[{y'[x]==y[x]Cos[x+y]],y[0]0==1},y,{x,0,30}]{{y-> InterpolatingFunction[{{0.,30.}},<>]}}f[x_]:=Evaluate[y[x] /. s]f[1]{0.991387}Dear All,I am trying to get the output of a numerical solution to a differential euqation as shown in the example above. As usual I defined the function f using the Evaluate function to get the outpout from the resulting interpolating function. I notice that when substiuting a numerical value for the indepdenent variable, the output is surrounded by a curly bracket suggesting that the output is a list. Although the presence of the curly brackets doesn't seem to affect functions like Plot[f, {x,0,30}], I notice that it is causing me some trouble when I try to use output of the function f in other computations like in parametric plot or in peicewise functions. Please advise how to remove the curly brackets from this output?Thanks,Abdullah
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Posted 3 years ago
use f[x_]:=Evaluate[y[x] /. s[[1]]]instead
Posted 9 years ago
Dear Abdullah,I am not sure whether this answers your question, because your input does not seem to be valid Mathematica input. I changed it a bit to make it work on my comptuer.s = NDSolve[{y'[x] == y[x] Cos[x + y[x]], y[0] == 1}, y[x], {x, 0, 30}]If I do this and then useĀ f[x_] := Evaluate[y[x] /. s]f[1]I get your output.{0.991387}As you see this is a list of one element. If you only want that element you need to ask Mathematica for element one of that list.f[1][[1]]This gives 0.991387 without the curly brackets.M.
Posted 9 years ago
Thank you very much. That was helpful
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Attachments | 498 | 1,814 | {"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-2023-23 | longest | en | 0.818593 |
https://www.numbersaplenty.com/1524013554 | 1,709,412,741,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475897.53/warc/CC-MAIN-20240302184020-20240302214020-00432.warc.gz | 911,156,995 | 3,297 | Search a number
1524013554 = 23773497069
BaseRepresentation
bin101101011010110…
…1001100111110010
310221012200220120110
41122311221213302
511110121413204
6411120522150
752523621130
oct13265514762
93835626513
101524013554
117122a1a38
12366480356
131b397c48b
14106594750
158dbde889
1524013554 has 32 divisors (see below), whose sum is σ = 3531185280. Its totient is φ = 429466752.
The previous prime is 1524013553. The next prime is 1524013559. The reversal of 1524013554 is 4553104251.
It is not an unprimeable number, because it can be changed into a prime (1524013553) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 245469 + ... + 251600.
It is an arithmetic number, because the mean of its divisors is an integer number (110349540).
Almost surely, 21524013554 is an apocalyptic number.
1524013554 is a gapful number since it is divisible by the number (14) formed by its first and last digit.
1524013554 is an abundant number, since it is smaller than the sum of its proper divisors (2007171726).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
1524013554 is a wasteful number, since it uses less digits than its factorization.
1524013554 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 497154.
The product of its (nonzero) digits is 12000, while the sum is 30.
The square root of 1524013554 is about 39038.6161896141. The cubic root of 1524013554 is about 1150.7905223724.
The spelling of 1524013554 in words is "one billion, five hundred twenty-four million, thirteen thousand, five hundred fifty-four". | 497 | 1,690 | {"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-2024-10 | latest | en | 0.860946 |
https://www.cscodehelp.com/%E7%A7%91%E7%A0%94%E4%BB%A3%E7%A0%81%E4%BB%A3%E5%86%99/%E4%BB%A3%E5%86%99%E4%BB%A3%E8%80%83-eeee4122-electrical-electronics-%E7%94%B5%E5%AD%90%E7%94%B5%E6%B0%94/ | 1,722,717,622,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640377613.6/warc/CC-MAIN-20240803183820-20240803213820-00309.warc.gz | 576,218,595 | 16,296 | # 代写代考 EEEE4122 electrical & electronics 电子电气 – cscodehelp代写
The University of Nottingham
DEPARTMENT OF ELECTRICAL AND ELECTRONIC ENGINEERING
Sample Practice Paper 2
EEEE4122 ELECTRICAL & ELECTRONICS FUNDAMENTALS FOR MSC STUDENTS (20 credit) Time allowed 1.5 Hours
Candidates may complete the front cover of their answer book and sign their desk card but must NOT write anything else until the start of the examination period is announced.
Only silent, self-contained calculators with a Single-Line Display or Dual-Line Display are permitted in this examination.
Dictionaries are not allowed with one exception. Those whose first language is not English may use a standard translation dictionary to translate between that language and English provided that neither language is the subject of this examination. Subject specific translation dictionaries are not permitted.
No electronic devices capable of storing and retrieving text, including electronic dictionaries, may be used.
EEEE4122
DO NOT turn examination paper over until instructed to do so
Turn Over
SECTION A
Phasor Sketch:
(i) Calculate the rms value and phase angle of the supply voltage V. (ii) Sketch the phasor diagram for this system.
Answers: Vrms = 19V, phase angle = 21.8o.
2 EEEE4122
QUESTION 1
(a) Which of the following is untrue in describing the characteristics of transmission lines & cables? POSSIBLE ANSWERS:
(1) Series resistance & inductance; (2) Parallel capacitance; (3) Parallel conductance; (4) Parallel inductance. [1.5 Marks]
(b) Which of the following is untrue in describing a power-flow relationship in a transmission line? POSSIBLE ANSWERS:
(1) Transmission voltage drop depends primarily on the reactive power of load; (2) Transmission voltage drop depends primarily on the real power of load; (3) Phase difference between two ends of the line determines the real power flow; (4) Phase difference depends on the effective cross sectional area of the line. [1.5 Marks]
(c) For a given load, the T&D losses can be reduced by which of the following? POSSIBLE ANSWERS:
(1) Increasing the T&D voltage; (2) Increasing the T&D current; (3) Increasing the distance of transmission; (4) None of these. [1.5 Marks]
(d) Doubling the T&D voltage reduces the potential losses to which of the following? POSSIBLE ANSWERS:
(1) 0.10; (2) 0.25; (3) 0.50; (4) 0.75. [1.5 Marks]
QUESTION 2
(a) Two voltages are defined as VA = 115cos(2f.t + 25o) and VB = 95cos(2f.t + 45o) where the supply frequency is 60 Hz. Calculate the sum of these voltages using the phasor approach, and also convert your answer to the time domain.
Answers: Vrms = 146.2V, phase angle = 33.9o and V(t) = 206.8cos(377.t + 33.9o)
[4 Marks]
(b) A 50 Hz V volt sinusoidal voltage source is connected over a series combination of resistance and inductance as shown below. The voltage across the resistor and inductor are measured as VR = 25cos(ω.t) and VL = 10cos(ω.t + (/2)) respectively. Using Kirchoff’s Voltage Law and the phasor approach:
(c) A series connection of an inductor and resistor is subsequently connected in parallel with a capacitor as shown in diagram below. A sinusoidal voltage of frequency f (Hz) is applied across the combination. The current flowing in the inductor/resistor branch and the
[6 Marks]
Answers: Irms = 0.308A, phase angle = -6.6o.
QUESTION 3
[6 Marks]
3 EEEE4122
capacitor are measured as IRL = 0.5cos(2f.t – (30o)) and IC = 0.2cos(2f.t + 90o). Using Kirchoff’s Current Law and the phasor approach, calculate the rms value and phase angle of the supply current I, and sketch the phasor diagram for this system.
(a) A 2Ω resistor (Z1) is connected in series with a parallel combination of impedances Z2 and Z3. Z2 is a 0.01H inductor whilst Z3 consists of a 1Ω resistor in series with a 0.001F capacitor. This circuit is supplied from a voltage which is defined by VS = 162cos(377t). Calculate the supply current IS(t) and both the active and reactive power drawn from the said supply.
Answers: IS(t) = 18.21cos (377t + 21.5o), 1380W, -543Var.
Note: Numerical value of PF does not tell us whether current lags or leads the voltage. Hence we define Lagging PF (+Q) when I lags V and Leading PF (-Q) when I leads V.
– see slides 104 and 105 in “EEEE4122_CircuitANALYSIS_AlexCHONG” notes on Moodle.
[6 Marks]
(b) Two loads are connected in series and have impedances of Z1 = 4 – j3 and Z2 = 6 + j8. They are supplied with power from a 220V, 50Hz source. Calculate the following:
(i) The total apparent power required from the supply and the supply power factor
(ii) The active and reactive powers absorbed by each load.
Answers: 4334 VA, PF=0.89, P1=1552 W, Q1 = -1164 VAr, P2 = 2328 W, Q2 = +3104 VAr.
[6 Marks]
TOTAL FOR SECTION A [34 MARKS]
Phasor Sketch:
SECTION B
QUESTION 4
(a) What is the Laplace transform of 3t? POSSIBLE ANSWERS:
(1) (𝑠) = 3 ; (2) 𝐻(𝑠) = 3 ; (3) 𝐻(𝑠) = 3𝑠2; (4) 𝐻(𝑠) = 6 𝑠2 𝑠 𝑠2
[3 Marks]
.
(b) What is the closed loop transfer function 𝐶(𝑠) for the control loop in Fig. Q4b below?
𝐶(𝑠) = 𝑅(𝑠)
following statements about the Root Locus method are correct? (i) It can be easily used with measured data.
(ii) It explicitly shows all closed-loop poles.
(iii) It is a good indicator of the transient response.
(iv) Performance and stability can be inferred from the same plot.
(v) It is easy to handle time delays correctly. POSSIBLE ANSWERS:
(1) (i) and (ii); (2) (ii) and (iv); (3) (iv) and (v); (4) (ii) and (iii).
4
EEEE4122
POSSIBLE ANSWERS: (1) 𝐶(𝑠) = 𝐺(𝑠) ; (2)
𝑅(𝑠)
Fig. Q4b: Feedback control loop. 𝐺(𝑠) ; (3) 𝐶(𝑠) = 𝐺(𝑠) ; (4) 𝐶(𝑠) =
1+𝐺(𝑠)𝐻(𝑠) 𝑅(𝑠) 1+𝐺(𝑠) 𝑅(𝑠)
(c) The response of a system can be analysed using either Bode plots or the root locus method. Which of the
𝑅(𝑠)
𝐺(𝑠) . 1+𝐺(𝑠)
[3 Marks]
[3 Marks]
5 EEEE4122
QUESTION 5
(a) Which of the following statements is true?
(i) The energy stored in a capacitor is given by 𝐸𝑛𝑒𝑟𝑔𝑦 = 𝑣2.
2𝐶
(ii) The effects of source inductance can be mitigated with a decoupling capacitor.
(iii) Capacitors are often used as voltage integrators
(iv) If Δ𝑉 = 𝑉(𝑡2) − 𝑉(𝑡2) = 0 in a voltage waveform across a capacitor, then the area under the waveform of the current through the capacitor is zero.
(1) (i), (ii) and (iv); (2) (i) and (iv); (3) (ii) and (iv); (4) (ii) and (iii). [3 Marks]
(b) Consider a 3-phase transmission line whose phase voltages are given by: 𝑉 = 𝑉 ∠0∘, 𝑉 = 𝑉 ∠−120∘, and
𝑉 = 𝑉 ∠120∘. What is the phase of the line voltage VCA? 𝐶𝑁 𝑝
(1) 30; (2) 90; (3) -120; (4) 150.
𝐴𝑁 𝑝 𝐵𝑁 𝑝
(c) There are three classes of power switching device: uncontrolled devices; controlled devices; and latching devices? Which of the following is NOT a controlled device?
(1) Bipolar transistor; (2) MOSFET; (3) Thyristor; (4) IGBT.
[3 Marks]
[3 Marks]
QUESTION 6
(a)
Determine the Laplace transform 𝑣0(𝑠) (without initial conditions) of the transfer function of the circuit in 𝑖(𝑠)
(b)
Using the Final Value Theorem, calculate the steady state error ess of the closed loop system in Fig. Q5b for a unity step input keeping 2 significant digits.
Fig. Q5b: Unity feedback closed loop system.
ANSWER: ess = 0.80 (accept 0.79 to 0.81) [3 Marks]
Consider the circuit in Fig. Q5c. If I(0) = 0A and the inductor is ideal, answer the following questions:
Fig. Q5c: Circuit for question 5c.
i) What is the magnitude of the current (in amperes) at t = 30 s?
ANSWER: I= 6 (accept answers from 5.9 to 6.1) A. [1.5 Marks]
ii) How much energy is stored in the inductor at t = 90 s ?
ANSWER: Energy = 0 J. [1.5 Marks]
Fig. Q5a:
Fig. Q5a: Circuit for question 5a.
; (3) 𝑣0(𝑠) = 𝐶⁄𝑠 ; (4) 𝑣0(𝑠) = 𝐶 . [3 Marks] 𝑖(𝑠) 𝑠+𝐶⁄𝑅 𝑖(𝑠) 𝑅𝐶𝑠+1
6 EEEE4122
(1)
𝑣0(𝑠) = 𝐶 ; (2) 𝑖(𝑠) 𝑠+𝐶⁄𝑅
𝑣0(𝑠) = 𝐶 𝑖(𝑠) 𝑠+𝑅𝐶
(c)
QUESTION 7
7 EEEE4122
Consider the 415V 3-phase rectifier circuit and waveform diagram in Fig. Q7 below and answer the following questions:
(a) Which diodes conduct current in the time between 11.7ms and 15ms? [2 Marks]
ANSWER: Diode 3 and diode 4
(b) What is the average voltage applied across the load? [4 Marks]
TOTAL FOR SECTION B [33 MARKS]
SECTION C
QUESTION 8
(a) Which parameter of an ideal opamp is equal to infinity: POSSIBLE ANSWERS:
(1) input impedance; (2) output impedance; (3) NF; (4) input offset.
(b) The golden rules of opamp analysis are: POSSIBLE ANSWERS:
[1.5 Marks]
(c) A typical value of the open loop gain of a real opamp is: POSSIBLE ANSWERS:
(1) 1; (2) 10; (3) 100; (4) None of these.
(d) Single ended inverting amplifier gain is: POSSIBLE ANSWERS:
(1) equal to zero; (2) negative; (3) positive; (4) infinite.
(e) The gain of an amplifier is 40dB. If the voltage input to this amplifier is 5mV the output would be:
[3 Marks] [1.5 Marks]
[1.5 Marks]
[1.5 Marks]
(1) 20mV; (2) 200mV; (3) 0.5V; (4) 5V; QUESTION 9
8
EEEE4122
(1) The inputs are always at the same potential and the inputs draw no current; (2) The inputs are always at the same potential and the output draws no current; (3) The inputs are always at the same potential and the inputs draw infinite current; (4) The inputs are always at a negative potential and the inputs draw infinite current.
Figure Q9
(a) If Rx = 100k, Ry = 10k calculate the absolute value of the first stage gain for the instrumentation amplifier shown in Fig.Q9.
(b) If R3 = R4 = 10k, R1 = R2 = 5k calculate the absolute value of the second stage gain for the instrumentation amplifier shown in Fig.Q9.
2 (accept answers from 1.9 to 2.1)
(c) If the differential gain of the first stage of the instrumentation amplifier shown in the circuit above is 100 and the resistor values associated with the second stage are R1 = 10k, R2 = 10.5k, R3 = 100k and R4 = 101k, calculate the absolute value of the 2nd stage differential gain.
9 EEEE4122
21 (accept answers from 20.9 to 21.1)
[4 Marks]
[4 Marks]
10 (accept answers from 9.9 to 10.1)
(d) For the same condition as in Question (c) above, calculate the absolute value of the 2nd stage common mode gain.
0.035 (accept answers from 0.034 to 0.036) | 3,256 | 9,895 | {"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} | 2.671875 | 3 | CC-MAIN-2024-33 | latest | en | 0.862024 |
https://www.rasch.org/mra/mra-06-09.htm | 1,722,801,420,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640412404.14/warc/CC-MAIN-20240804195325-20240804225325-00768.warc.gz | 757,626,230 | 7,909 | MEASUREMENT RESEARCH ASSOCIATES TEST INSIGHTSJune 2009
Greetings With computer-based tests, test takers have the ability to go back and review items. This brief study investigates the relationship between the amount of time spent reviewing items and candidate test performance. Lidia Martinez Manager Test Development and Analysis
Time Usage and Candidate Performance
Computer based testing provides the opportunity to track the amount of time candidates spend responding to and reviewing each exam item. The time usage of a test was studied to understand the relationship between the amount of time a candidate takes to review items and their final score. For purposes of this study, scores are reported as percent correct without any consideration for calibrated item difficulty or test equating. The question is the impact of the amount of time used for review on candidate scores.
This candidate population took a multiple choice examination and was divided into three groups based on the mean amount of time they used to review items. Candidates in Group 1 used an average of 5 seconds or less per item to review. Candidates in Group 2 used an average of 5 - 20 seconds per item to review and candidates in Group 3 used an average of more than 20 seconds per item to review. These groups were compared by 1) mean time spent initially responding per item; 2) mean time spent reviewing per item; 3) total test percent correct. All time is given in seconds. An alpha level of .05 was used for all statistical tests.
An analysis of variance showed that there was a significant difference in the amount of time used to initially respond to items (p = .039). A post hoc analysis using Tukey's HSD test revealed that Group 3's average time spent initially responding to items was significantly less than Group 1's average time (p = .030).
Descriptive Statistics for Time Used to Initially Respond to Items
Group based on time used to review Mean Time per Item SD Min Max Group 1: Average Review Time ≤ 5 sec. 57.19 15.33 35.25 84.31 Group 2: 5 sec. < Avg. Rev. Time ≤ 20 sec. 54.39 13.04 31.25 75.33 Group 3: Average Review Time > 20 sec. 47.68 9.78 30.54 63.43 Total Population 54.22 13.87 30.54 84.31
An ANOVA showed that there was a significant difference in the amount of time used to review items (p < .001). A post hoc analysis revealed all groups were significantly different from one another (all p values < .001). Since the groups were divided based on amount of time taken to review, these results are not surprising.
Descriptive Statistics for Time Used to Review Items after Initial Response
Group based on time used to review Mean Time per Item SD Min Max Group 1: Average Review Time ≤ 5 sec. 1.25 1.37 .00 4.71 Group 2: 5 sec. < Avg. Rev. Time ≤ 20 sec. 12.05 4.35 5.31 19.76 Group 3: Average Review Time > 20 sec. 27.45 6.79 20.57 44.26 Total Population 10.55 10.71 .00 44.26
An ANOVA showed that there was no significant difference in percent correct scores based on the amount of time spent reviewing items (p = .335). Based on this study, the amount of time spent reviewing items does not seem to have an effect on candidate test performance.
Descriptive Statistics for Candidate Total Percent Correct Scores
Group based on time used to review Mean % Correct SD Min Max Group 1: Average Review Time ≤ 5 sec. 59% 8% 40% 75% Group 2: 5 sec. < Avg. Rev. Time ≤ 20 sec. 61% 7% 51% 74% Group 3: Average Review Time > 20 sec. 61% 8% 47% 74% Total Population 60% 8% 40% 75%
For this data sample, candidates who spent more time reviewing items, spent less time initially responding to items. This could be due to the fact that if more time is taken initially to view items, there will be less time remaining after the first view of the exam to review items. While candidates who spent more time reviewing items earned slightly higher percent correct scores, this is not a trend, since there was only a 2% difference in group performance. The mean percent correct for each group is statistically comparable.
Measurement Research Associates, Inc. 505 North Lake Shore Dr., Suite 1304 Chicago, IL 60611 Phone: (312) 822-9648 Fax: (312) 822-9650 www.MeasurementResearch.com
Rasch-Related Resources: Rasch Measurement YouTube Channel
Rasch Measurement Transactions & Rasch Measurement research papers - free An Introduction to the Rasch Model with Examples in R (eRm, etc.), Debelak, Strobl, Zeigenfuse Rasch Measurement Theory Analysis in R, Wind, Hua Applying the Rasch Model in Social Sciences Using R, Lamprianou El modelo métrico de Rasch: Fundamentación, implementación e interpretación de la medida en ciencias sociales (Spanish Edition), Manuel González-Montesinos M.
Rasch Models: Foundations, Recent Developments, and Applications, Fischer & Molenaar Probabilistic Models for Some Intelligence and Attainment Tests, Georg Rasch Rasch Models for Measurement, David Andrich Constructing Measures, Mark Wilson Best Test Design - free, Wright & Stone
Rating Scale Analysis - free, Wright & Masters
Virtual Standard Setting: Setting Cut Scores, Charalambos Kollias Diseño de Mejores Pruebas - free, Spanish Best Test Design A Course in Rasch Measurement Theory, Andrich, Marais Rasch Models in Health, Christensen, Kreiner, Mesba Multivariate and Mixture Distribution Rasch Models, von Davier, Carstensen
Rasch Books and Publications: Winsteps and Facets
Applying the Rasch Model (Winsteps, Facets) 4th Ed., Bond, Yan, Heene Advances in Rasch Analyses in the Human Sciences (Winsteps, Facets) 1st Ed., Boone, Staver Advances in Applications of Rasch Measurement in Science Education, X. Liu & W. J. Boone Rasch Analysis in the Human Sciences (Winsteps) Boone, Staver, Yale Appliquer le modèle de Rasch: Défis et pistes de solution (Winsteps) E. Dionne, S. Béland
Introduction to Many-Facet Rasch Measurement (Facets), Thomas Eckes Rasch Models for Solving Measurement Problems (Facets), George Engelhard, Jr. & Jue Wang Statistical Analyses for Language Testers (Facets), Rita Green Invariant Measurement with Raters and Rating Scales: Rasch Models for Rater-Mediated Assessments (Facets), George Engelhard, Jr. & Stefanie Wind Aplicação do Modelo de Rasch (Português), de Bond, Trevor G., Fox, Christine M
Exploring Rating Scale Functioning for Survey Research (R, Facets), Stefanie Wind Rasch Measurement: Applications, Khine Winsteps Tutorials - free
Facets Tutorials - free
Many-Facet Rasch Measurement (Facets) - free, J.M. Linacre Fairness, Justice and Language Assessment (Winsteps, Facets), McNamara, Knoch, Fan | 1,636 | 6,564 | {"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} | 2.953125 | 3 | CC-MAIN-2024-33 | latest | en | 0.925956 |
https://www.itdaan.com/tw/69be88c4de92e5e80c379372a896340d | 1,618,960,906,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039491784.79/warc/CC-MAIN-20210420214346-20210421004346-00597.warc.gz | 937,375,632 | 7,067 | # P1854 花店櫥窗布置
## 輸入輸出樣例
`3 57 23 -5 -24 165 21 -4 10 23-21 5 -4 -20 20`
`532 4 5重點：記錄轉移路徑用二維數組標記一下`
`#include<iostream>#include<cstdio>#include<cstring>#define N 109#define INF 200000000using namespace std;int F,V;int A[N][N];int go[N][N];int put[N];int f[N][N];int read(){int r=0,k=1;char c=getchar();for(;c<'0'||c>'9';c=getchar())if(c=='-')k=-1;for(;c>='0'&&c<='9';c=getchar())r=r*10+c-'0';return r*k;}int print(int x,int y){if(x==0)return 0;if(go[x][y]){put[x]=y;print(x-1,y-1);}else{print(x,y-1);}}int main(){F=read();V=read();for(int i=1;i<=F;++i){for(int j=1;j<=V;++j){A[i][j]=read();}}for(int i=1;i<=F;++i){for(int j=0;j<=V;++j){f[i][j]=-INF;}}for(int i=1;i<=F;++i){for(int j=i;j<=V;++j){if(f[i-1][j-1]+A[i][j]>f[i][j-1]){f[i][j]=f[i-1][j-1]+A[i][j];go[i][j]=1;}else{f[i][j]=f[i][j-1];go[i][j]=0;}}}printf("%d\n",f[F][V]);/*for(int i=1;i<=F;++i){for(int j=1;j<=V;++j){cout<<put[i][j]<<' ';}cout<<endl;}*/print(F,V);for(int i=1;i<=F;++i)printf("%d ",put[i]);return 0;}` | 479 | 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} | 2.6875 | 3 | CC-MAIN-2021-17 | latest | en | 0.216745 |
https://www.coursehero.com/tutors-problems/Chemistry/11856491-house-with-a-151-x-10-3-ft-2-floor-area-and-850-ft-high-ceilings/ | 1,547,826,083,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583660175.18/warc/CC-MAIN-20190118151716-20190118173716-00519.warc.gz | 734,567,333 | 21,045 | View the step-by-step solution to:
# house with a 15.1 x 10 3 ft 2 floor area and 8.50 ft high ceilings is to be heated with propane.
house with a 15.1 x 103 ft 2 floor area and 8.50 ft high ceilings is to be heated with propane. How many kilograms of propane would be needed to raise the temperature ofthe air inside the house from 62.0 o F to 74.0 o F ?
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Browse Documents | 182 | 739 | {"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-2019-04 | latest | en | 0.892279 |
https://3yellowsandpails.com/tag/estimation/ | 1,619,029,767,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039546945.85/warc/CC-MAIN-20210421161025-20210421191025-00503.warc.gz | 191,469,711 | 21,161 | # Posts tagged ‘estimation’
## Reflections on Summer Fun
The last few days spent in classes have been lovely. I knew they were going to be. Spending time in classes with kiddos and their teachers trying out some ideas that were new to my colleagues, and then chatting about what we saw and heard, and what we learned was just wonderful.
1. Kids (and their teachers) are super interested in containers filled with stuff. It is nearly impossible to resist joining in the conversation about “how much stuff” is in the container, especially when the stuff is the lovely orange cheese balls that you can buy in the mega-container at Target.
It offers kiddos who really, really, really struggle with math anxiety a way to join in the conversation. Everyone made guesses and built number lines and laughed and shared ideas.
These are the sights and sounds of what learning should be for our students.
2. Games provide an awesome arena for kids to explore numbers and their structure, especially the game we shared on Monday morning, What Fits Between? We taught them how to play the game. They had to figure out how the three numbers created are related (largest number, smallest number, and the number that fits between) and organize the results.
• We learned that if we leave them alone to figure things out, they do. The adage, be less helpful, comes to mind.
• Kids are able to explore ideas more deeply when they engage in conversation with classmates than when they engage, individually, with worksheets.
• It’s great fun to have a group of teachers in a classroom when you want to try out new things especially when you are able do some planning together before hand.
• It’s really nice to have a couple of days to work on new ideas and have time for people to talk together on each of the days.
• Kids are amazing. If we choose the right tasks, they aren’t even nervous and anxious about the math part of their day.
3. Using games as assessment is an informative practice. Notes and photos and conversation bring the next instructional steps to light.
• Of the 12 kiddos playing, nine were ready to move onto playing the next level of the game where each player must build a three-digit number and determine who had created the number this fits between the other two. Three students needed additional opportunities to play the original version of the game.
• The “top” math student got to play and interact with the ideas just like everyone else, as did the kiddos who are often not interested in math. All 12 students were invested in the game. The students who usually need to get a drink, sharpen a pencil, or use any number of distractions to help them get through math did not need to employ any of those strategies.
• During the reflection time at the end of the class period the kids decided that they wanted to change the criteria that determined who won the game. Now, in their class, the winner is determined after each player counts up the number of rounds s/he won. The player who has the number of wins that is between the other two, wins.
How do we use the learning we saw today to support tomorrow’s instruction?
The collaborative work started when we noticed that kids were having a difficult time estimating height in an estimation warm-up. We decided to address the manner in which they worked with the concept of estimation. We selected a graphic organizer for them to use, determined that the quantity of objects for them to estimate would be less than 100, provided some boundaries to the quantity, and put the cheese balls in clear, plastic containers so the students could see them. With these adjustments the students were more comfortable in making estimates and discussing their ideas. After the team talked about the work the students did that day, it was decided our next move would be to use a series of photographs from Estimation180.com. We selected the group that asks kids to estimate the number of cheese balls on plates and trays. The students will play What Fits Between? and do some other math work.
I am looking forward to hearing how the day went, and what the team is going to tackle next.
## Candy Cane Christmas Tree
How many candy canes did the design team of Curate Decor + Design use to make this fabulous tree in the lobby of The Modern Honolulu?
First, make a guess that is too high. Then, make a guess that is too low. Finally, make your guess. Be sure to write down each guess. (This nice system of having kids work with number and estimating is courtesy of Andrew Stadel’s work found at Estimation 180.)
If you need a little more information before you give a final estimate, here is the very top part of the tree.
Record any revisions. How confident are you that your estimate is really, really, really close?
I am sure that by now you JUST can’t stand it and desperately need to know how many candy canes in the Christmas tree.
That was where I was at as soon as I saw the tree–just had to know how many candy canes were used. The folks at the front desk at The Modern Honolulu were super helpful and they just told me. They didn’t even make me guess.
Are you ready? Are you sure you don’t want to revise your guess one more time? | 1,107 | 5,207 | {"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} | 2.859375 | 3 | CC-MAIN-2021-17 | latest | en | 0.969893 |
https://shakyradunn.com/slide/mechanics-of-materials-engr-350-lecture-0-wclhua | 1,603,508,269,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881640.29/warc/CC-MAIN-20201024022853-20201024052853-00223.warc.gz | 524,782,290 | 9,022 | # Mechanics of Materials Engr 350 - Lecture 0
Mechanics of Materials Engr 350 - Lecture 0 Some references taken from: Philpot, Timothy A. Mechanics of Materials: An Integrated Learning System, 4rd Edition. Wiley Approved calculators Casio FX-115 HP 33s and HP 35s TI -30X and TI-36X 2
Engineering Problem-Solving Documentation Given: (Background information about the problem being solved, figures, etc.)
Find: (The parameters you are being asked to solve for in the problem)What Expected Solution: (What do you know about the solution before calculating anything? This might include bounds of expected values, a direction, an expected failure mode, etc.) Plan and Assumptions: (What steps are you going to take to solve this problem? What equations or principles will be used? Do you have enough equations to solve for all the variables? Any engineer should be able to follow your plan and come up with the same results that you do.) Solution: (This is just crunching the numbers. Your solution should have units on every value, and conversions as necessary.) Check: (Come up with some way to validate that your solution makes sense. This may be an alternate calculation or method, a simplification that shows you are in a reasonable range, or comparison to known/accepted values) Reflection: (This is taking a broad look after you have thought through and
implemented the solution. What did you learn from this, and how might you use that in future problems? What additional things would you want to know about this problem if you were working on it as a paid engineering analysis? What insights did you gain about the system while working on this problem?) 3 FBDs 1. Establish a coordinate system, and moment sign convention 2. Decide which bodies are to be included and detach these from all others 3. Indicate all forces and moments on the FBD, including:
1. Those external and applied 2. Those resulting from the contact 3. Those resulting from removed bodies 4. Indicate magnitude and direction of any forces or moments if known 5. Include important dimensions 4 FBD Example Draw the approriate FBD of the forklift and crate.
5 FBD Draw the appropriate FBD of the bracket 6 FBD Draw the appropriate FBD of the crane arm and
cable 7 Review centroids The centroid is the geometric center of an area and can be determined with the following formula for composite areas use the formula
along with the centroids of common areas, Table A.1 8 Centroids example Determine the centroid location y_bar Where do you predict the centroid will be? Ai yi
yi*Ai 1 2 9 Moment of inertia Also known as the second moment of an area 10
Parallel axis theorem States that the moment of inertia for an area about an axis is equal to the areas moment of inertia about a parallel axis passing through the centroid plus the product of the area and the square of the distance between the two axes. For composite sections
11 Moment of inertia for composite sections example Determine the moment of inertia for the composite section Ici di di2Ai
Ib=Ici+di2Ai 1 2 12
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Lecture Notes for CS4400/5400
# Lambda Calculus
Lambda calculus is a theory of functions. What is a function? There are two basic views one can take when characterizing them:
1. Function as a graph
2. Function as a value
Considering a function $$f$$ as a graph is to consider it as a set of pairs – mappings between input and output values $$(x, f(x))$$. For example the square function on natural numbers $$^2 : \mathbb{N} \to \mathbb{N}$$ can be characterized as a set of pairs $$(n, n^2)$$:
$\{ (0, 0), (1, 1), (2, 4), (3, 9), (4, 16), (5, 25), ... \}$
Using a function as a graph is to find an output that corresponds to our input. The alternative view to take is to consider a function as rules – equations, which tell us how to compute the output of the function from its input. For example, the square function $$^2 : \mathbb{N} \to \mathbb{N}$$ is defined by the equation:
$n^2 = n \times n$
How do we use this function? We substitute an expression that looks like the left-hand side with the right-hand side, replacing the argument $$n$$ with the expression and then computing the resulting expression. For example, our calculation might proceed as follows:
\begin{aligned} {4^2}^2 + 3^2 &= (4 \times 4)^2 + 3^2\\ &= \left((4 \times 4) \times (4 \times 4)\right) + 3^2\\ &= \left((4 \times 4) \times (4 \times 4)\right) + (3 \times 3)\\ % &= \left(16 \times (4 \times 4)\right) + (3 \times 3)\\ % &= \left(16 \times (4 \times 4)\right) + 9\\ % &= \left(16 \times 16\right) + 9\\ % &= 256 + 9\\ &... \\ &= 265 \end{aligned}
Or, as follows: \begin{aligned} {4^2}^2 + 3^2 &= (4 \times 4)^2 + 3^2\\ &= 16^2 + 3^2\\ &= 16^2 + 9\\ &= 256 + 9\\ &... \\ &= 265 \end{aligned}
In any case, the important thing to note is that we replace any occurrence of $$n^2$$ for any $$n$$ using the defining equation. In general, if we define a function $$f$$ by the equation $$f(x) = E$$, where $$E$$ is some mathematical expression (potentially containing $$x$$), then we use (apply) this function by replacing any occurrence of $$f(D)$$ (where $$D$$ is a mathematical expression) by $$E[x := D]$$, that is the expression $$E$$ where all occurrences of $$x$$ are replaced by $$D$$. This is called substitution of a variable $$x$$ in an expression $$E$$ for another expression $$D$$. E.g., if
$f(x) = x + x$
then:
\begin{aligned} f(20) + f(2 \times 3) &= (x + x)[x := 20] + (x + x)[x := 2 \times 3] \\ &= (20 + 20) + ((2 \times 3) + (2 \times 3)) \\ ... \end{aligned} The next question is, how important is the name of the function? We use names as mnemonics, so that we can say we can
1. say “let $$f$$ be the function defined by the equation $$f(x) = E$$” (where $$E$$ is an arbitrary mathematical expression), and
2. replace any occurrence of $$f$$ applied to an argument with an instance of $$E$$ where $$x$$ is replaced with the argument expression.
We can do this without inventing names, by using functions as anonymous objects – just like we easily use numbers or strings or arrays. In mathematics an anonymous function will be written as $$x \mapsto E$$. For example, the square function is $$x \mapsto x \times x$$, the above function $$f$$ is $$x \mapsto x + x$$.
The above exposition applies to programming too. Basically, all sensible “higher-level” programming languages allow us to define functions to abstract a computation by replacing a concrete expression with a variable – a placeholder. In Python we might write:
def square(x):
return x * x
In C/C++:
int square(int x) {
return x * x;
}
square :: Integer -> Integer
square x = x * x
In Scheme:
(define (square x)
(* x x))
In any programming language we operate with the rough understanding that whenever square is invoked with an argument, that application might as well be replaced with the body of the function with the argument variable replaced with the actual argument (either before or after evaluating the argument itself). More and more programming languages, particularly those which allow passing functions as arguments, allow creating functions without naming them – so called anonymous functions. Python and Scheme have lambda:
lambda x : x * x
(lambda (x) (* x x))
OCaml has fun or function:
fun x => x * x
\x -> x * x
C++ has, well…
[](int x){ return x * x; }
As hinted by the Scheme and Python examples, Lambda calculus is the underlying theory behind these anonymous functions. In its pure form, it is exclusively concerned with what it means to apply an abstracted expression (as an anonymous function), to an argument. It studies this as a purely syntactic operation.
Where Python and Scheme have lambda, OCaml has fun and function, Lambda calculus has $$\lambda$$. That is an anonymous function with the formal parameter $$x$$ is constructed using $$\lambda x...$$ We can write the squaring function in lambda notation as
$\lambda x.\ x \times x$
We say that this is a lambda abstraction that binds the variable $$x$$ in $$x \times x$$. In other words, $$x$$ is bound in $$x \times x$$. An application is written (similarly to Scheme, OCaml, or Haskell) by writing the function and argument next to each other (juxtaposition). For example, where in Scheme we could write
((lambda (x) (* x x)) 10)
(\x -> x * x) 10
In lambda notation we write:
$(\lambda x.\ x \times x)\ 10$
As I mentioned before, Lambda calculus looks at the application of a function as a syntactic operation, in terms of substitution, as the process of replacing any occurrence of the abstracted variable with the actual argument. For the above, this is replacing any occurrence of $$x$$ in $$x \times x$$ with $$10$$:
\begin{aligned} (\lambda x.\ x \times x)\ 10 &= (x \times x)[x := 10]\\ &= 10 \times 10 \end{aligned}
Another way of thinking about the bound variable $$x$$ in the $$\lambda x.\ x \times x$$ as a placeholder or hole, where the argument “fits”.
$(\lambda \boxed{\phantom{x}}.\ \boxed{\phantom{x}} \times \boxed{\phantom{x}})\ 10 = \boxed{10} \times \boxed{10}$
## Pure Lambda Calculus
Here, we will look at the formal theory of pure Lambda Calculus. We will look at the syntax and a notion of computation.
### Syntax
The basic syntax of the calculus is really simple:
<Lambda> ::= <Variable>
| ( <Lambda> <Lambda> )
| ( λ <Variable> . <Lambda> )
That is all there really is:1
• variable reference, e.g. $$x$$, $$y$$, $$z$$, $$a$$, $$\mathit{square}$$
• application, e.g., $$(x\ y)$$, $$((\lambda x.\ x)\ (\lambda x.\ x))$$
• lambda abstraction, e.g.,
• $$(\lambda x.\ x)$$ – expressing the identity function
• $$(\lambda x.\ x\ x)$$ – a function that applies its argument to itself
You might ask: what can we do with such a minuscule language? Turns out a lot. As proven by A.M. Turing, this pure version of Lambda calculus is equivalent in computational power to Turing Machines. That means we are able to build up a programming language out of these three constructs. We will look at how to do that in the section on Programming in Pure Lambda Calculus below.
#### Syntax Conventions and Terminology
Terminology: Take a lambda abstraction:
$(\lambda x.\ N)$
1. $$\lambda x$$ is a binder binding $$x$$
2. $$N$$ is the body of the abstraction
To avoid writing too many parentheses, these conventions are usually taken for granted:
1. Outermost parentheses are usually dropped: x x, λx. x.
2. Application associates to the left. That is, (((a b) c) d) is the same as ((a b c) d) is the same as (a b c d), which is the same as a b c d (see previous rule).
3. Lambda abstraction bodies extend as far to the right as possible. That is, (λa. (λb. ((a b) c))) is the same as λa. λb a b c.
### Beta Reduction
Computation in pure lambda calculus is expressed in a single rule: the $$\beta$$-reduction rule:
$(\lambda x.\ M)\ N \longrightarrow_\beta M[x := N]$
The long arrow stands for “reduces to”. On the left-hand side, we have an application of a lambda abstraction to an arbitrary term. On the right-hand side, we substitute the abstraction’s bound variable with the argument. A term that matches the pattern on the left-hand side (that is, a lambda abstraction applied to something) is called a redex, short for reducible expression. For example:
• (λx. x) a -->β a
• (λx. x x) (λy. y) -->β (λy. y) (λy. y)
• the above reduces further: (λy. y) (λy. y) -->β (λy. y)
• not a redex: (x x)
• also not a redex: x (λy. y)
• also not a redex: (λy. (λx. x) y), although it does contain a redex ((λx. x) y)
### Variables: Bound, Free. Closed Expressions
We have already mentioned the notion of a bound variable. A variable is said to be bound in an expression, if it appears under a λ-abstraction binding that particular variable. Or, in other words, it is bound if it appears in the scope of a binder. For example:
• x is bound in (λx. x x) – it appears in the scope of the binder λx
• both x and y are bound in (λx. λy. x y)x appears in the scope of λx, y in the scope of λy
• x is not bound in (λy. x y), but y is – x does not appear in the scope of any binder here, while y appears in the scope of λy
A free variable is one which appears in a position where it is not bound. For example:
• x is free in x x, in λy. x y, or in (λy. y y) x
• x is not free in (λx. x x) (λx. x)
• x is both bound and free in (λx. x y) x, while y is only free
As you can see above, a variable might be both bound and free in an a expression.
An expression which contains no free variables is closed, for example:
• λx. x
• λx. λy. x y x
A closed lambda expression is also called a combinator.
A variable is called fresh for an expression, if it does not appear free in that expression. For example, x is fresh for y z or (λx. x x).
### Names of Bound Variables Don’t Matter
Intuitively, an identity function should be an identity function, no matter what we choose to name its bound variable. That is, (λx. x) should be considered the same as (λy. y) or (λz. z). This is captured in the notion of alpha equivalence: two expressions are α-equivalent, if they only differ in the names of their bound variables. This also means, that we are free to α-convert any lambda term by consistently renaming bound variables. However, the new names must differ from free variables under the particular binder. We are thus free to convert (λx. x) to, e.g., (λa. a); (λy. z y) to (λx. z x), but not to (λz. z z).
### Substitution
We were happy to use substitution in an informal manner up until now:
$$M[x := N]$$ means replacing occurrences of the variable $$x$$ in the expression $$M$$ with the expression $$N$$.
Here we want to pin it down. For that, we will need to consider the difference between bound and free variables. Let’s try to start with a naive definition of substitution.
#### Naive Substitution
There are three syntactic forms, we need to consider each form:
Variable: x[y := N] = ?
Application: (M1 M2)[y := N] = ?
Abstraction: (λx. M)[y := N] = ?
Variables are straightforward: we either find the variable to be substituted or we find a different one:
1. y[y := N] = N
2. x[y := N] = x if x $$\neq$$ y
Application is also relatively simple – we simply substitute in both left-hand and right-hand side:
1. (M1 M2)[y := N] = (M1[y := N] M2[y := N])
Now, for a lambda abstraction we need to consider the variables involved. We certainly don’t want to override the bound variable of a function:
1. (λy. M)[y := N] = (λy. M)
The remaining case seems simple enough too:
1. (λx. M)[y := N] = (λx. M[y := N]) if x $$\neq$$ y
If we test this substitution everything seems to be okay:
(x x)[x := (λy. y)] = (x[x := (λy. y)] x[x := (λy. y)])
= (λy. y) (λy. y)
((λx. x y) x)[x := (λy. y)] = (λx. x y)[x := (λy. y)] x[x := (λy. y)]
= (λx. x y) (λy. y)
However, what happens if the expression that we are substituting contains the bound variable?
(λy. x)[x := y] = λy. y
We see that in this case, we have just “captured” variable y and changed its status from free to bound. This changes the meaning of a variable – whereas the original meaning of y was given by the context of the left-hand side expression, now it is given by the binder λy. In particular, we changed a constant function—which, after the substitution should return a free y, no matter what argument it is applied to—to an identity function, that just returns whatever its argument is.
From this we see that we need substitution to behave differently when there the expression that we are trying to substitute, contains free variables that clash with variables bound by a lambda-abstraction.
#### Safe Substitution
To fix this we can restrict when the last case of our substitution applies:
1. (λx. M)[y := N] = (λx. M[y := N]) if x $$\neq$$ y and if x is not free in N
Now our substitution is “safe”. However, this turns it into a partial function – it is left undefined for cases where the bound variable x appears free in N. To go around this, we can make use of alpha-conversion: we consistently rename the bound variable x to one that doesn’t clash with y or the free variables in N or M. Only then do we perform the actual substitution of y.
1. (λx. M)[y := N] = (λx'. M[x := x'][y := N]) if x $$\neq$$ y and x' is fresh for y, N and M
Now substitution is a total function again. For an implementation, we just need to know how to pick a fresh variable. Notice how we replace the bound variable x with x' and also rename any ocurrence of x to x' in the body M. Since x' is chosen so that it does not appear free in M or N, we are avoiding any potential clashes.
### Reduction Strategies
Beta reduction tells us how to reduce a redex. The missing piece of the puzzle is how to decide where to look for a redex and apply the beta-reduction rule. This is given by reduction strategies.
(The following text is taken, with minor modifications, from Types and Programming Languages)
#### Full Beta-Reduction
Under this strategy, any redex may be reduced at any time. At each step we pick some redex, anywhere inside the term we are evaluating, and reduce it. For example, consider the term:
(λa. a) ((λb. b) (λz. (λc. c) z))
This term contains three redexes:
• (λa. a) ((λb. b) (λz. (λc. c) z))
• (λb. b) (λz. (λc. c) z)
• (λc. c) z
Under full beta-reduction, we might choose, for example, to begin with the innermost redex, then do the one in the middle, then the outermost:
U+10FC74
(λa. a) ((λb. b) (λz. (λc. c) z))
--> (λa. a) ((λb. b) (λz. z))
--> (λa. a) (λz. z)
--> λz. z
λz. z cannot be reduced any further and is a normal form.
Note, that under full beta-reduction, each reduction step can have more than possible one result, depending on which redex is chosen.
#### Normal Order
Under normal order, the leftmost, outermost redex is always reduced first. Our example would be reduced as follows:
(λa. a) ((λb. b) (λz. (λc. c) z))
--> (λb. b) (λz. (λc. c) z)
--> λz. (λc. c) z
--> λz. z
Again, λz. z is the normal form and cannot be reduced further.
Because each redex is chosen in a deterministic manner, each reduction step has one possible result – reduction thus becomes a (partial) function.
#### Call by Name
Call by name puts more restrictions on which redexes are fair game, and disallows reductions inside abstractions. For our example, we perform the same reduction steps as normal form, but stop short of “going under” the last abstraction.
(λa. a) ((λb. b) (λz. (λc. c) z))
--> (λb. b) (λz. (λc. c) z)
--> λz. (λc. c) z
Haskell uses an optimization of call by name, called call by need or lazy evaluation. Under call by name based strategies, arguments are only evaluated if they are needed.
#### Call by Value
Under call by value, only outermost redexes are reduced and each redex is only reduce after its right-hand side has been fully reduced to a normal form.
(λa. a) ((λb. b) (λz. (λc. c) z))
--> (λa. a) (λz. (λc. c) z)
--> λz. (λc. c) z
Evaluation strategies based on call by value are used by the majority of languages: an argument expression is evaluated to a value before it is passed into the function as an argument. Such a strategy is also called strict, because it strictly evaluates all arguments, regardless of whether they are used.
## Programming in Pure Lambda Calculus
### Multiple Arguments
So far, we have looked at lambda abstractions which only take a single argument. However, unary functions are only a small part of our experience with programming. We use functions with multiple arguments all the time. How do we pass more than one argument to a lambda?
One approach would be to extend the calculus with a notion of tuples. Perhaps throw in some pattern matching, for good measure:
$(\lambda (x, y).\ x\ y)\ (a, b)$
However, this means that we are abandoning the very minimal core lambda calculus with all its simplicity. And we don’t have to! As we know well by now, applying an abstraction simply replaces its bound variable with the argument that it’s applied to, as in this trivial example:
$(\lambda x. x\ y)\ b \longrightarrow (x\ y)[x := b] = (b\ y)$
What happens if the abstraction actually just returns another abstraction.
\begin{aligned} (\lambda x.\ (\lambda y.\ x\ y))\ b \longrightarrow (\lambda y.\ x\ y)[x := b] = (\lambda y.\ b\ y) \end{aligned}
Since neither of the bound variable of the inner abstraction ($$y$$) and the variable we are substituting for ($$x$$), nor the bound variable of the inner abstraction ($$y$$) and the term we are substituting ($$b$$) are in conflict, we simply substitute $$x$$ for $$b$$ inside the inner abstraction. This yields an abstraction which can be applied to another argument. That is applying $$(\lambda x.\ (\lambda y.\ x\ y))$$ to $$b$$ returned an abstraction which is “hungry” for another argument. We can now apply that abstraction to another argument:
$(\lambda y.\ b\ y)\ a \longrightarrow (b\ y)[y := a] = b\ a$
Let’s do the same in one expression:
\begin{aligned} (((\lambda x.\ (\lambda y.\ x\ y))\ b)\ a &\longrightarrow ((\lambda y.\ x\ y)[x := b])\ a \\ &= (\lambda y.\ b\ y)\ a \\ &\longrightarrow (b\ y)[y := a]\\ &= (b\ a) \end{aligned}
We just applied an abstraction to two arguments. To make this a little easier to see, we can use left-associativity of application and the fact that the scope of a binder goes as far right as possible to rewrite the original expression as
$(\lambda x.\ \lambda y.\ x\ y)\ b\ a$
This technique is called currying (after Haskell Curry, although he was not the first one to come up with it). It is so common that, usually a short-hand is introduced for abstractions with more than one argument:
\begin{aligned} (\lambda x\ y.\ ...) &\equiv (\lambda x.\ \lambda y.\ ...)\\ (\lambda x\ y\ z.\ ...) &\equiv (\lambda x.\ \lambda y.\ \lambda z.\ ...)\\ \text{etc.} \end{aligned}
If we allow arithmetic in our lambda expressions a nice example will be:
\begin{aligned} \left(\lambda x\ y.\ \frac{x + y}{y}\right) 4\ 2 &\longrightarrow \left(\lambda y.\ \frac{4 + y}{y}\right) 2 \\ &\longrightarrow \frac{4 + 2}{2} \end{aligned}
Currying is used as the default for functions of multiple arguments by Haskell and OCaml (determined mostly by their standard libraries). On the other hand, Standard ML’s library uses tuples as default.
### Data types
We see that we can represent functions with multiple arguments in PLC. Surely, for representing other kinds of data (such as booleans, numbers, data structures), we need to introduce extensions and add these as primitive operations? Not really…
#### Booleans
Many types of values can be represented using Church encodings. Booleans are probably the simplest and most straightforward:
\begin{aligned} \mathsf{true} &= \lambda t\ f.\ t &\qquad&(= \lambda t.\ \lambda f.\ t)\\ \mathsf{false} &= \lambda t\ f.\ f & &(= \lambda t.\ \lambda f.\ f)\\ \end{aligned}
What do these mean? The representation of $$\mathsf{true}$$ is a function that takes two arguments and returns the first one. On the other hand, $$\mathsf{false}$$ returns its second argument. To make sense of these, we need to put them to work and see how they work with boolean operations.
We start with the conditional: $$\textsf{if-else}$$. It should take three arguments and return its second one if the first one evaluates to $$\textsf{true}$$, and its third argument otherwise. That is we are looking for an expression:
$\textsf{if-then}\ \textsf{true}\ x\ y \longrightarrow ... \longrightarrow x$
and
$\textsf{if-then}\ \textsf{false}\ x\ y \longrightarrow ... \longrightarrow y$
Notice something?
\begin{aligned} \textsf{true}\ x\ y &\longrightarrow x\\ \textsf{false}\ x\ y &\longrightarrow y \end{aligned}
That means that all $$\textsf{if-then}$$ needs to do is to apply its first argument to its second and third argument, since the boolean representation takes care of the selection itself:
$\textsf{if-then} = \lambda b\ t\ f.\ b\ t\ f$
Let’s try to look at conjunction: and. We look for ??? to put in:
(λa b. ???) true true --> ... --> true
(λa b. ???) true false --> ... --> false
(λa b. ???) false true --> ... --> false
(λa b. ???) false false --> ... --> false
First note that true true x --> true for any $$x$$, so it seems that λa b. a b x could work if we find an appropriate $$x$$:
(λa b. a b x) true true --> (λb. true b x) true --> true true x --> ... --> true
Now note that in all but the first case and should reduce to false. In the second case,
(λa b. a b x) true false --> ... --> true false x --> ... --> false
for any $$x$$, so that still works. Now, how can we get false true x --> false? By taking $$x$$ to be false:
(λa b. a b false) false true --> ... --> false true false --> ... --> false
The final case also works:
(λa b. a b false) false false --> ... --> false false false --> ... --> false
Hence $\textsf{and} = \lambda a\ b.\ a\ b\ \textsf{false}$
Another way of thinking about the definition of and is to define it terms of if-then-else. E.g., in Haskell,
and :: Bool -> Bool -> Bool
and a b = if a then b else False
which just says that if the first argument is true then the result of and depends on the second one, and if its false the result will be false regardless of the second argument.
Based on this, we can express the and operation using if-else, which we defined above, and show that it is equivalent to the previous definition by simplifying it using normal order reduction:
and = λa b. if-else a b false
= λa b. (λb t f. b t f) a b false
--> λa b. (λt f. a t f) b false
--> λa b. (λf. a b f) false
--> λa b. a b false
Can you come up with a representation of $$\textsf{or}$$? $$\textsf{not}$$?
#### Pairs
Pairs can be encoded using an abstraction which “stores” its two arguments:
pair = λl r s. s l r
You can think of a s as a “chooser” function which either picks l or r. Selectors for the first and second element are then, respectively, defined as:
fst = λp. p (λl r. l)
snd = λp. p (λl r. r)
Take a look at the selector functions we pass to the pair representation. Are they familiar? (Hint: booleans)
#### Natural Numbers: Church Numerals
Natural numbers are Church-encoded as Church numerals:
zero = λs z. z
one = λs z. s z
two = λs z. s (s z)
three = λs z. s (s (s z))
...
A numeral for $$n$$ can be understood as a function that takes some representation of a successor function and some representation of zero and applies the successor to zero $$n$$ times.
How about operations on numerals? The successor of a numeral λs z... is computed by inserting one more application of s inside of the abstraction:
succ (λs z. z) --> ... --> λs z. s z
succ (λs z. s z) --> ... --> λs z. s (s z)
succ (λs z. s (s z)) --> ... --> λs z. s (s (s z))
...
We know that succ takes a numeral (which is an abstraction) and returns another numeral, which is again an abstraction:
succ = λn. (λs z. ...n...)
Taking λs z. z as an example input:
(λn. (λs z. ...n...)) (λs z. z)
--> (λs z. ...(λs z. z)...))
--> (λs z. s z)
We see that we need to apply an extra s under λs z.:
(λs z. s ...(λs z. z)...) --> ... --> (λs z. s z)
To do this we need to “open” the abstraction representing 0. This can be achieved by passing the outer s and z as arguments. We achieve what we wanted.
(λs z. s ...(λs z. z) s z...) --> (λs z. s ...z...) = (λs z. s z)
Working backwards, we arrive at our successor function:
(λs z. s z)
<-- (λs z. s ((λs z. z) s z))
<-- (λn. λs z. s (n s z)) (λs z. z)
= succ (λs z. z)
Successor can be thus defined as:
succ = λn. (λs z. s (n s z)) = λn s z. s (n s z)
Once we have a successor operation, defining addition is quite simple if we keep in mind that a Church numeral $$m$$ applies its first argument (s) to its second argument (z) $$m$$ times:
plus = λm n. m succ n
Multiplication follows the same principle:
$m * n = \underbrace{n + (... n}_{m \text{ times}} + 0)$
Hence:
times = λm n. m (plus n) zero
We can define subtraction via a predecessor function, which is surprisingly more tricky than the successor function. For a numeral λs z. s (s ... (s z)), the predecessor should return a numeral with one less s. One way of defining a predecessor is via a function that “counts” the number of s applications in a numeral, but also remembers the previous count, that is, one less than the total number of applications of s:
pred = λn. snd (n (λp. pair (succ (fst p)) (fst p)) (pair zero zero))
Here the numeral n (of which we want to compute the predecessor) is applied to two arguments:
1. The function (λp. pair (succ (fst p)) (fst p)). This function takes a pair (bound to p) containing two numerals. It returns a pair containing the successor of the first element of p, together with its original value. That means, everytime the function is applied to a pair containing numerals n and m, it returns a pair with numerals corresponding to n + 1 and n (m is discarded).
2. A pair containing two zeros: (pair zero zero).
Finally, the second element of the pair is returned – which contains the count of s applications, except for the last one.
Here is an example. We let f = (λp. pair (succ (fst p)) (fst p))
pred three
= (λn. snd (n f (pair zero zero))) (λs z. s (s (s z)))
--> snd ((λs z. s (s (s z))) f (pair zero zero))
--> snd ((λz. f (f (f z))) (pair zero zero))
--> snd (f (f (f (pair zero zero))))
= snd (f (f ((λp. pair (succ (fst p)) (fst p)) (pair zero zero))))
--> snd (f (f (pair (succ (fst (pair zero zero))) (fst (pair zero zero)))))
--> ...
--> snd (f (f (pair (succ (fst zero)) (fst zero zero))))
--> snd (f (f (pair (succ zero) zero)))
--> ...
--> snd (f (f (pair one zero)))
= snd (f ((λp. pair (succ (fst p)) (fst p)) (pair one zero)))
--> snd (f (pair (succ (fst (pair one zero))) (fst (pair one zero))))
--> snd (f (pair (succ one) one))
--> ...
--> snd (f (pair two one))
--> ...
--> snd (pair (succ two) two)
--> ...
--> snd (pair three two)
--> ...
--> two
To subtract n from m, we need to take 1 away from m n times.
minus = λm n. n pred m
For completeness, an alternative predecessor definition is as follows (TODO: explain):
pred' = λn f x. n (λg h. h (g f)) (λu. x) (λu. u)
We can check if a variable is zero:
is-zero = λn.n (λx. false) true
We can define $$\leq$$
leq = λm n. is-zero (minus m n)
And we can define equality:
equal = λm n. and (leq m n) (leq n m)
### Recursion
We have seen that we can define booleans, together with a conditional, and numbers, together with arithmetic operations in pure lambda calculus. However, to reach full Turing power, we lack one important ingredient: the ability to loop. To loop in a functional setting, we need the little brother of looping: self-reference.
To see that we can loop, let us look at a term, for which $$\beta$$-reduction never terminates in a normal form. This interesting term, called $$\Omega$$, is defined as follows:
Ω = (λx. x x) (λx. x x)
We see that we have an abstraction which applies its argument to itself and which is applied to itself. How does reduction proceed?
(λx. x x) (λx. x x) --> (x x)[x := (λx. x x)]
= (λx. x x) (λx. x x) --> (x x)[x := (λx. x x)]
= (λx. x x) (λx. x x) --> (x x)[x := (λx. x x)]
...
Immediately after the first reduction step, we are back where we started! Well, we see we can loop forever (diverge), but how is this useful?
In a programming language like OCaml, we are used to defining recursive functions which refer to themselves inside of their body:
let rec fact = fun n ->
if n = 0 then 1 else n * fact (n - 1)
How do we achieve this in lambda? While we have been freely using equations to define names for lambda expressions, these were just meta-definitions of names. That is, when we write
fact = λn. if-true (is-zero n) one (mult n (?fact? (pred n)))
we rely on our meta-language and our common understanding of it to replace any occurrence of ?fact? with the right-hand side of the above equation, as many times as needed. But this is not beta-reduction, that is we are not defining a recursive function as an object in lambda calculus. To get there, we can think of a recursive definition as follows: “Assuming we have a function to call in the recursive case, we can complete the definition”. In Haskell or OCaml, we can simply assume that we already have the function that we are defining. But what is really going on here, is that we can abstract the recursive call as an argument – which corresponds to saying “assuming we already have a function to call in the recursive case”:
fact = λf. λn. if-true (is-zero n) one (mult n (f (pred n)))
Now factorial does not refer to itself anymore, we just need to give it a function to call in the else branch. Easy:
fact = (λf. λn. if-true (is-zero n) one (mult n (f (pred n)))) (λn. if-true (is-zero n) one (mult n (f (pred n))))
Wait, but now what about f in the second case? Ah, no problem:
fact = (λf. λn. if-true (is-zero n) one (mult n (f (pred n))))
((λf. λn. if-true (is-zero n) one (mult n (f (pred n))))
(λn. if-true (is-zero n) one (mult n (f (pred n)))))
This apparently won’t work… unless we have a way of supplying an argument for f as many times as it’s needed. That is, a way to allow the function reference itself whenever it needs to. This is where fixpoint combinators come in.
In math, a fixed point of a function $$f$$ is an input for which the function returns the input itself:
$f(x) = x$
If the above holds, we say that $$x$$ is a fixed point of $$f$$. A fixpoint combinator (in general called $$\operatorname{fix}$$) is an operation that computes the fixed point of a function. That is, it is a function for which the following equation holds:
fix f = f (fix f)
This equation just spells out that when a function is applied to its fixpoint, the fixpoint shall be returned. Let’s use the above equation on itself, by replacing occurrences of fix f with the right-hand side:
fix f = f (fix f)
= f (f (fix f))
= f (f (f (fix f)))
= ...
Now glance above: “If only we had a way of supplying an argument for f as many times as it’s needed.” Seems we are onto something. Let’s replace f with our factorial:
fact = λf. λn. if-true (is-zero n) one (mult n (f (pred n)))
fix fact
= fact (fix fact)
= (λf. λn. if-true (is-zero n) one (mult n (f (pred n)))) (fix fact)
--> (λn. if-true (is-zero n) one (mult n ((fix fact) (pred n))))
This looks promising. The problem? We haven’t defined what fix is, we are just abusing our meta-notation again. In fact, there is more than one possible definition of fix. The simplest one is the Y combinator:
Y = λf. (λx. f (x x)) (λx. f (x x))
Notice how the structure is very similar to $$\Omega$$ above. We should check if it is a fixpoint combinator, that is, if it satisfies the fixpoint equation:
Y g = (λf. (λx. f (x x)) (λx. f (x x))) g
= (λx. g (x x)) (λx. g (x x)))
= g ((λx. g (x x)) (λx. g (x x)))
= g ((λf. ((λx. f (x x)) (λx. f (x x)))) g)
= g (Y g)
We have ourselves a fixpoint combinator. Let us try to use it to define our factorial function:
fact0 = (λf. λn. if-true (is-zero n) one (mult n (f (pred n))))
fact = Y fact0
What happens when we try to apply fact to a numeral?
fact three
= Y fact0 three
= (λf. (λx. f (x x)) (λx. f (x x))) fact0 three
--> (λx. fact0 (x x)) (λx. fact0 (x x)) three
--> fact0 ((λx. fact0 (x x)) (λx. fact0 (x x))) three
= fact0 (Y fact0) three
--> (λn. if-true (is-zero n) one (mult n ((Y fact0) (pred n)))) three
--> if-true (is-zero three) one (mult three ((Y fact0) (pred three)))
--> ...
--> mult three ((Y fact0) (pred three))
= mult three (fact0 (Y fact0) (pred three))
--> ...
--> mult three (fact0 (Y fact0) (if-true (is-zero (pred three)) one (mult (pred three) ((Y fact0) (pred (pred three)))))
...
-->
However, the $$Y$$ combinator is not universally applicable under any reduction strategy. Consider what happens with the $$Y$$ combinator, if we apply the CBV strategy.
Y g = (λf. (λx. f (x x)) (λx. f (x x))) g
--> (λx. g (x x)) (λx. g (x x))
--> g ((λx. g (x x)) (λx. g (x x)))
--> g (g (λx. g (x x)) (λx. g (x x)))
--> g (g (g (λx. g (x x)) (λx. g (x x))))
--> ...
For CBV, we need the Z combinator:
λf. (λx. f (λy. x x y)) (λx. f (λy. x x y))
#### Let bindings
The last useful notation to introduce are let-bindings. We have already implemented them as part of our arithmetic expressions language – both as a substitution-based and environment-based evaluator. Let bindings can be introduced to pure lambda-calculus as syntactic sugar – a construct that is defined by translation to a combination of other constructs in the language. Introducing a let-binging corresponds to creating a λ-abstraction and immediately applying it to the bound expression:
let x = e1 in e2 ≡ (λx. e2) e1
We have to define let as syntactic sugar – we cannot write it as a function, the way we did for if-then, add, etc. Why is that the case?
We can also define a recursive version of let – called let rec in OCaml, letrec in Scheme:
let rec f = e1 in e2 ≡ let f = fix (λf. e1) in e2
≡ (λf. e2) (fix (λf. e1))
Where fix is an appropriate fixpoint combinator (e.g., $$Y$$ under CBN, $$Z$$ under CBV and CBN).
Most languages also allow specifying function arguments to the left-hand side of the equal sign:
let f x y z = e1 in e2
let rec f x y z = e1 in e2
(define (f x y z) e1)
These can be translated as:
let f x y z ... = e1 in e2 ≡ let f = λx y z. e1 in e2
≡ (λf. e2) (λx y z. e1)
## Extensions
While it is useful to show how various important programming concepts and constructs can be expressed in pure lambda-calculus directly, in general, it is a rather inefficient approach.
The approach usually taken in designing lambda-calculus-based languages, is to take the calculus as a core language and add extensions to support various kinds of data.
#### Pairs
<Lambda> ::= <Variable>
| (<Lambda> <Lambda>)
| (λ <Variable> . <Lambda>)
| ( <Lambda> , <Lambda> )
| fst <Lambda>
| snd <Lambda>
Often just written informally as an extension of a previous BNF:
<Lambda> ::= ...
| ( <Lambda> , <Lambda> )
| fst <Lambda>
| snd <Lambda>
Together with the extension to syntax, we need to specify what the meaning of these new construct is. That is, we need to update the reduction strategy and provide reduction rules for the pairs. For primitive operations, these reduction rules are sometimes called $$\delta$$-rules or $$\delta$$-reduction rules:
fst (l, r) -->δ l
snd (l, r) -->δ r
## Reduction vs. Evaluation
### Reducing to a Normal Form
So far, in connection with Lambda calculus, we have only talked about reduction as a transformation of an expression containing redexes into another expression where the redex has been reduced. To actually evaluate a lambda-expression, we can simply iterate the reduction step (with a particular strategy), until we reach a normal form: an expression that cannot be reduced further. Note that some expressions (such as $$\Omega$$) do not have a normal form – under any reduction strategy. Formally, an iteration of reduction steps $$\longrightarrow$$ is written as $$\longrightarrow^{*}$$, which stands for “reduces in zero or more steps to”. Mathematically, it is a reflexive-transitive closure of $$\longrightarrow$$. It is defined by the following two rules:
1. M -->* M
2. M -->* M' there is a N, such that M --> N and N -->* M'
This can be lso expressed as:
• an expression reduces in zero or more steps to itself
• an expression $$M$$ reduces in zero or more steps to the expression $$M'$$, if there is an intermediate expression $$N$$ and $$M$$ reduces to $$N$$ and $$N$$ reduces in one or more steps to $$M'$$
In Haskell, this is expressed as iteration. Assuming that the one-step reduction function implementing a particular strategy has the type Lambda -> Maybe Lambda, we define an iteration function as:
iter :: (a -> Maybe a) -> a -> a
iter step e =
case step e of
Just e' -> iter step e'
Nothing -> e
That is, while we can perform reduction steps on an expression, we do so. If the input expression cannot be reduced (it contains no redex), just return it.
If we have an implementation of a reduction strategy, say, stepNormal :: Lambda -> Maybe Lambda, then we can build a “reducer” by passing it as an argument to iter:
iterNormal :: Lambda -> Lambda
iterNormal = iter stepNormal
We will call such a reducer an iterative evaluator, or normalizer.
### Evaluation
Instead of performing evaluation through gradual reduction (normalization) of lambda terms, we can write a recursive evaluator in the same style as we have done for our languages with arithmetic, booleans and let-bindings. This kind of evaluator takes an expression (term, expressing a computation) and either succeeds, returning a value, or fails. On the other hand, the iterative reduction function returned the same type of result as its input – an expression. The criterion for deciding when a lambda expression is reduced fully (evaluated) was that there was no further reduction possible – the expression has reached its normal form. That means, for the iterative evaluator, we didn’t have to worry about distinguishing values from expressions syntactically – once the evaluator finished reducing, we took the normal form and decided if it makes sense or not.
For a recursive evaluator of lambda terms, we need to decide what its return type should be. That is, we need to syntactically distinguish values from unevaluated expressions (computations). What are the values of pure lambda calculus?
A value is (a representation) of an entity, that has no computation left to perform and can only be inspected or used in another computation. It also does not depend on the context in which it appears – dependence on context would imply possible computation. For lambda calculus, we only have 3 syntactic forms, meaning we only have 3 candidates for what constitutes a value.
Application cannot be a value – it is a computation term, which can be, potentially, reduced. A single variable reference also cannot be a value – its meaning wholly depends on the context, in which it appears. The only option is lambda abstraction. In particular, an abstraction that is closed, i.e., does not contain any free variables. Having lambda-abstractions as values is consistent with Church encodings which we have explored above – each value is a closed abstraction containing only bound variables.
<LambdaValue> ::= λ <Variable> . <Lambda> -- constructor VLam
data LambdaValue = VLam Variable Lambda
As previously, we add values to the Lambda datatype:
data Lambda = Var Variable -- <Lambda> ::= <Variable>
| Lam Variable Lambda -- | ( λ <Variable> . <Lambda> )
| App Lambda Lambda -- | ( <Lambda> <Lambda> )
| Val LambdaValue -- | <LambdaValue>
Now we can implement a call-by-name evaluator for lambda terms, relying on substitution to deal with application.
evalCBN :: Lambda -> Maybe LambdaValue
evalCBN (Var x) = Nothing -- all variables should be substituted
evalCBN (Val v) = Just v
evalCBN (Lam x e) = Just (VLam x e)
evalCBN (App e1 e2) =
case evalCBN e1 of
Just (VLam x e) -> evalCBN (subst x e2 e) -- leave argument unevaluated
Nothing -> Nothing
For call-by-value, we use
evalCBV :: Lambda -> Maybe LambdaValue
evalCBV (Var x) = Nothing
evalCBV (Val v) = Just v
evalCBV (Lam x e) = Just (VLam x e)
evalCBV (App e1 e2) =
case evalCBV e1 of
Just (VLam x e) ->
case evalCBV e2 of -- evaluate the argument before substituting
Just v2 -> evalCBV (subst x (Val v2) e)
Nothing -> Nothing
Nothing -> Nothing
What is the benefit of a recursive evaluator vs. reduction-based iterative evaluator?
### Environment-based Evaluation
We can also use environments to implement a recursive evaluator for lambda calculus. Here is a basic implementation of a call-by-value one.
evalCBV :: Env LambdaValue -> Lambda -> Maybe LambdaValue
evalCBV env (Val v) = Just v
evalCBV env (Var x) = get x env
evalCBV env (Lam x e) = Just (VLam x e)
evalCBV env (App e1 e2) =
case evalCBV env e1 of
Just (VLam x e) ->
case evalCBV env e2 of
Just v2 -> evalCBV (add x v2 env) e -- bind the abstracted variable to the argument's value
Nothing -> Nothing
Nothing -> Nothing
### Scoping
Our evaluator might seem ok at first sight. However, there is a problem.
Using our intuition from Haskell (and Racket), what shall we expect to be the value resulting from the expression below?
let x = 2 in
let f y = y + x in
let x = 3 in
f 1
In Haskell, this expression evaluates to 3.
Using our desugaring rule introduced earlier, this expands to the pure lambda calculus expression
(λx. (λf. (λx. f 1) 3) (λy. (+ y x))) 2
In Haskell, this is represented as (for a lambda extended with numbers and addition as primitives):
e = App (Lam "x" -- let x = 2 in
(App (Lam "f" -- let f = ... in
(App (Lam "x" -- let x = 3 in
(App (Var "f") (Val (Num 1)))) -- f 1
(Val (Num 3))))
(Lam "y" ((Add (Var "y") (Var "x"))))))
(Val (Num 2))
What happens if we ask our evaluator to evaluate the above expression?
> evalCBV empty e
Just (Num 4)
What we implemented here is dynamic scoping. The value of a variable is determined by the most recently bound variable at runtime. In general, under dynamic scoping the value of a variable cannot be determined statically, that is, just by looking at the code. This is usually counterintuitive for us. We are used to lexical (or static) scoping, that is, the scope of a variable is given by its location in the source code of the program. Most programming language use lexical scoping, because this allows us to easily reason about the values of variables when we read source code. Examples of languages with dynamic scoping are: Lisp (Emacs), LaTeX, bash. Some languages allow choosing scope when defining a variables, such as Common Lisp or Perl. How do we ensure static scoping?
### Closures
• The problem arises because we are passing around a lambda value and the original association between its free variables and values is lost
• We need a way to keep this association: closures
Closures package up a lambda abstraction with the environment that was valid at the time of evaluating the abstraction. They can be implemented simply as a pairing of a lambda abstraction with an environment:
data LambdaValue = Clo Variable Lambda (Env LambdaValue)
data Lambda = Var Variable
| Lam Variable Lambda
| App Lambda Lambda
| Val LambdaValue
evalCBV :: Env LambdaValue -> Lambda -> Maybe LambdaValue
evalCBV env (Var x) = get x env
evalCBV env (Val v) = Just v
evalCBV env (Lam x e) = Just (Clo x e env) -- save the current environment
evalCBV env (App e1 e2) =
case evalCBV env e1 of
Just (Clo x e env') -> -- the closure remembers the corresponding environment
case evalCBV env e2 of
Just v2 -> evalCBV (add x v2 env') e -- evaluate in the closure's environment updated with the argument binding
Nothing -> Nothing
Nothing -> Nothing
Now all abstraction bodies are evaluated with the environment that belongs to them statically. And, indeed, the example expression evaluates as expected:
> evalCBV empty e
Just (Num 3)
## De Bruijn Indices
As we have talked about before, the name of a bound variable is insignificant – as long as there is a consistency between the binder and the variable reference. This is called alpha-equivalence. Using strings as variables seems natural: it corresponds to the mathematical with variable names. However, as a computer representation this is inefficient: there are infinitely many ways to represent any lambda term: λx. x, λy. y, λz. z, λzzzz. zzzz, λlongervariablename. longervariablename, …
Moreover representing variable names as strings forces us to complicate the definition of substitution and define functions for obtaining fresh variable names. In an implementation, we would ideally want to do away with these complications. Sure, we could simply use natural numbers to represent variables and this would simplify picking a fresh variable name – e.g., by taking the maximum of all free variables and adding 1. We sill complicate the substitution definition and we still have the problem of multiple representations of alpha-equivalent lambda terms. There is another alternative.
When we look at a lambda abstraction:
$\lambda x.\ (\lambda y.\ x\ y\ (\lambda x.\ \lambda z.\ x\ z\ y))$
we really use the occurrence of $$x$$ in the binder as a marker and a variable reference as a reference back to the marker, that is, each variable can be viewed as referring back to the binder that bound it:
+-------------------+
| |
+----+ +------+ |
| | | | |
λx. (λy. x y (λx. λz. x z y))
| | | |
+-------+ +----+
This means that we do not really need to use names to know where an argument value should be inserted:
+-------------------+
| |
+----+ +------+ |
| | | | |
λ_. (λ_. * * (λ_. λ_. * * *))
| | | |
+-------+ +----+
Now the question is, how do we represent these connections without using names. The Dutch mathematician Nicolaas Govert de Bruijn had the idea that each variable reference should represent the number of binders between it and the binder that bound it. If there is no other binder between the reference and its binder, the count 0 and we can refer to the binder by that number. If there is one binder between, we refer to the variable’s binder by 1, etc.
+-------------------+
| |
+----+ +------+ |
| | | | |
λ_. (λ_. 1 0 (λ_. λ_. 1 0 2))
| | | |
+-------+ +----+
This leads to a simplification of the syntax: since we use do not need to mark binders using variables, lambdas do not carry any variable names:
+-----------------+
| |
+----+ +-----+ |
| | | | |
λ. (λ. 1 0 (λ. λ. 1 0 2))
| | | |
+------+ +----+
Thus the syntax of Lambda expressions using de Bruijn indices is as follows:
<DLambda> ::= <Index> -- variable reference
| <DLambda> <DLambda> -- application is as before
| λ. <DLambda> -- lambda abstraction does refer to the bound variable explicitly
data DLambda = DVar Integer
| DApp DLambda DLambda
| DLam DLambda
Here are a few more examples:
• Any identity function (λx. x, λy. y, …) is λ. 0
• λx. x x is λ. 0 0
• Churchian for true and false is λ. λ. 1 and λ. λ. 0, respectively
• Church numerals are λ. λ. 0, λ. λ. 1 0, λ. λ. 1 (1 0), λ. λ. 1 (1 (1 0))
• The Y combinator, λf. (λx. f (x x)) (λx. f (x x)) is λ. (λ. 1 (0 0)) (λ. 1 (0 0))
What is the advantage of using de Bruijn indices? It certainly isn’t (human) readability. Maybe you have noticed, that for each alpha-equivalent term, there is only one representation. This is a major advantage when implementing lambda calculus, since we do not need to care about renaming of bound variables. Another advantage is that “environments” for evaluating lambda expressions are simplified – they are just stacks of values:
data DValue = DClo DLambda [DValue]
eval :: [DValue] -> DLambda -> Maybe DValue
eval env (DVar i) | i < length env = Just (env!!i) -- lookup the value
| otherwise = Nothing
eval env (DApp e1 e2) =
case eval env e1 of
Nothing -> Nothing
Just (DClo e env') -> case eval env e2 of
Nothing -> Nothing
Just v2 -> eval (v2 : env') e
eval env (DLam e) = Just (DClo e env)
# Imperative Features: Programming with State
The language features we have considered so far were pure in the sense that their evaluation didn’t yield any side-effect.[^Well, there was one possible side-effect: failure.]
Now we will look at features associated with imperative languages. In an imperative language computation proceeds by processing statements (or commands) which explicitly modify the machine state. These changes in state can me modifying memory, consuming input, producing output, etc.
## Imperative Variables I
The first side-effect that we introduce is modifying memory via imperative variables. How do these differ from applicative variables? Probably the simplest way to think about the difference is that applicative variables bind to (stand for) values, imperative variables stand for memory cells. How do we model them?
The simplest approach: modelling memory – the store – as a mapping between variables and values, like we did with environments. The difference is in how we use them. While environments were used as read-only entities – the evaluator (at each stage) would receive an environment and might perform recursive calls with a modified environment, but it could pass a modified environment back to its context. A store, on the other hand, is treated as a read and write entity. At each stage, the evaluator can not just read what is stored in a variable, but also modify and return an updated store.
We will demonstrate using a small imperative language with basic arithmetic, assignment and retrieval of variables. To compose statements, we introduce a sequencing command: Seq s1 s2 first executes statement s1 and then s2. Notice how the execution of s2 uses an updated store produced by the execution of s1. Note that our expressions are still pure: they can read from the store, but not update it. This is apparent from the type of the expression evaluator evalExpr :: Store -> Expr -> Maybe Value. Contrast this with the type of the evaluator for statements:
data Stmt = Assign String Expr
| Seq Stmt Stmt
deriving Show
data Expr = Val Value
| Var String
deriving Show
data Value = Num Integer
deriving Show
type Store = Env Value
evalExpr :: Store -> Expr -> Maybe Value
evalExpr sto (Val v) = Just v
evalExpr sto (Var x) = get x sto
evalExpr sto (Add e1 e2) =
case evalExpr sto e1 of
Just (Num n1) -> case evalExpr sto e2 of
Just (Num n2) -> Just (Num (n1 + n2))
_ -> Nothing
_ -> Nothing
execStmt :: (Store, Stmt) -> Maybe Store
execStmt (sto, Assign x e) =
case evalExpr sto e of
Just v -> Just (add x v sto)
Nothing -> Nothing
execStmt (sto, Seq c1 c2) =
case execStmt (sto, c1) of
Just sto' -> execStmt (sto', c2)
Nothing -> Nothing
Note, how the function execStmt does not return any value – the result of the computation is determined purely by the effect it had on the store.
## Printing and Output Streams
We model output streams using a list of values.
data Stmt = ...
| Print Expr
deriving Show
data Expr = ... -- same as before
data Value = ... -- as before
type Store = ... -- as before
type Out = [Value]
evalExpr :: Store -> Expr -> Maybe Value
evalExpr sto (Val v) = ... -- all cases as before
-- We need to handle the output stream. Notice the cases for Seq and Print
execStmt :: (Store, Stmt) -> Maybe (Store, Out)
execStmt (sto, Assign x e) =
case evalExpr sto e of
Just v -> Just (add x v sto, [])
Nothing -> Nothing
execStmt (sto, Seq c1 c2) =
case execStmt (sto, c1) of
Just (sto', out1) ->
case execStmt (sto', c2) of
Just (sto'', out2) -> Just (sto'', out1 ++ out2)
Nothing -> Nothing
Nothing -> Nothing
execStmt (sto, Print e) =
case evalExpr sto e of
Just v -> Just (sto, [v])
Nothing -> Nothing
## Loops
data Stmt = ...
| While Expr Stmt
deriving Show
data Expr = ...
| Le Expr Expr -- adding a boolean operation
deriving Show
data Value = ...
| Bool Bool -- booleans to be used in the condition expression
deriving Show
type Store = Map String Value
type Out = [Value]
evalExpr :: Store -> Expr -> Maybe Value
...
evalExpr sto (Le e1 e2) =
case evalExpr sto e1 of
Just (Num n1) -> case evalExpr sto e2 of
Just (Num n2) -> Just (Bool (n1 <= n2))
_ -> Nothing
_ -> Nothing
execStmt :: (Store, Stmt) -> Maybe (Store, Out)
...
execStmt (sto, While e c) =
case evalExpr sto e of
Just (Bool False) -> Just (sto, [])
Just (Bool True) ->
case execStmt (sto, c) of
Just (sto', out1) ->
case execStmt (sto', While e c) of
Just (sto'', out2) -> Just (sto'', out1 ++ out2)
_ -> Nothing
_ -> Nothing
_ -> Nothing
## Imperative Variables II: Block Scope
• Our first representation of memory was a direct mapping from names to stored values
• Consequence:
• A variable defined inside, e.g., a while loop or a branch of an if statement, is visible and accessible outside of that loop, or branch
• This is does not correspond to what happens in most common imperative languages
• Languages like Python, C, Java, … enforce block scope, that is a variable is only visible in the block in which it was defined (or declared)
int x = 20;
do {
int x = 10; // different variable - different storage cell
// inside this block, shadows the outer x
int y = 1;
} while (false);
// here, x is still 20
System.out.println(y); // error: y doesn't exist here
• Block scope is similar to let expressions
• Another issue: in languages like Python, or Java (for object types), the semantics of assignment is to set a reference – instead of copying the object stored in the memory location
• E.g., the Python program
x = [1, 2, 3]
y = x
print(y)
y.append(4)
print(x)
prints
[1, 2, 3]
[1, 2, 3, 4]
because x and y refer to the same object
How do we combine block scope with imperative variables?
• Block is captured well by (read-only) environments
• A store captures variables which can be modified
• Idea:
• Environment for mapping variables to “addresses”
• Store for mapping addresses to (maybe) values
type Address = Integer
type Store = Map Integer Value
type Env = Map String Integer
Implementation: we add a new construct Block which bundles variable declarations with statements that can use these variables. We keep our Assign construct, however, we’ll redefine its meaning (implementation). We will also change the semantics of variable references (Var)
data Stmt = Assign String Expr -- we're keeping our assignment statement
...
| Block Decl Stmt -- a block is a bunch of variable declarations followed by statements
deriving Show
data Decl = NewVar String Decl -- a declaration list can be
| Empty -- empty
deriving Show
data Expr = ... -- expressions remain the same...
| Var x -- except we will change the meaning of variable lookup
| ...
data Value = Num Integer -- value are the same
deriving Show
type Out = [Value]
-- There are multiple ways we can represent allocation
-- The questions to answer are:
-- a) how do we select the next address?
-- b) what do we store at the freshly allocated address?
-- Here we use the following:
-- a) We take the highest address plus one
-- b) We store the number 0. Note that this choice can have various consequences: Do we need to distinguish between initialized and unitialized storage cells?
alloc :: Store -> (Address, Store)
alloc [] = (first, add first (Num 0) empty)
where first = 0
alloc m = (new, add new (Num 0) m)
where new = (maximum (keys m)) + 1
keys [] = [] -- get all the keys from a map
keys ((k, _) : m) = k : keys m
-- process declarations
procDecl :: Store -> Env -> Decl -> (Store, Env)
procDecl sto env (NewVar x d) =
procDecl sto' env' d
where (addr, sto') = alloc sto -- allocate a new address in the store
procDecl sto env Empty = (sto, env)
evalExpr :: Env -> Store -> Expr -> Maybe Value
...
evalExpr env sto (Var x) = -- NEW
do addr <- get x env -- first we find the address of the variable
get addr sto -- then we find the value in the store
...
execStmt :: Env -> (Store, Stmt) -> Maybe (Store, Out)
execStmt env (sto, Assign x e) = -- NEW
case evalExpr env sto e of -- first evaluate the expression to assign
Just v ->
case get x env of -- find the address of the variable
_ -> Nothing
Nothing -> Nothing
execStmt env (sto, Seq c1 c2) = ... -- same as before
execStmt env (sto, Print e) = ... -- same as before
execStmt env (sto, Block d s) = -- NEW: the structure of a block is similar to let
let (sto', env') = procDecl sto env d -- process declarations (allocating variables in the store)
in execStmt env' (sto', s) -- execute statements in the store
# Types and Type Systems
This part is partially based on notes by Norman Ramsey and on the book Types and Programming Languages by Benjamin Pierce
What is a type?
• In a Haskell-like language (simplified by ignoring type classes)
1 + 5 :: Integer
"hello " ++ "world" :: String
n > 10 :: Bool -- if n :: Integer
if n > 10 then "Yes" else "No" :: String -- if n :: Integer
\x -> x + 10 :: Integer -> Integer
• Types classify program phrases according to the kinds of values they compute
• They are predictions about values
• Static approximation of runtime behavior – conservative
Why types?
• Static analysis: detect (potential) runtime errors before code is run:
1 + True
10 "hello" -- application of the number 10 to a string?
• E.g., Python is happy to accept the following function definition:
def f(x):
if x < 10:
x(x, 10)
else:
"Hello " + x
What happens at runtime, when the function is called as f(4)?
• Enforcing access policy (private/protected/public methods)
• Guiding implementation (type-based programming)
• Documentation: types tell us a lot about functions and provide a documentation that is repeatedly checked by the compiler (unlike comments)
• Help compilers choose (more/most) efficient runtime value representations and operations
• Maintenance: if we change a function’s type, the type checker will direct us to all use sites that need adjusting
What is a type system?
A tractable syntactic method for proving the absence of certain program behavior.
• They are studied on their own as a branch of mathematics/logic: type theory
• Original motivation: avoiding Russell’s paradox
• Type systems are generally conservative:
1 + (if True then 10 else "hello")
would behave OK at runtime, but is, nevertheless, typically rejected by a static type checker (e.g., Haskell’s)
What kind of errors are typically not detected by type systems?
• Division by zero
• Selecting the head of an empty list
• Out-of-bounds array access
• Non-termination
Consideration: a program which mostly runs numeric computations will benefit from a strong static type system less than a program which transforms various data structures
Terminology: type systems
Dynamic
types are checked at runtime, typically when operations are performed on values
Static
types are checked before program is (compiled and) run
What is a type safe language?
• Can a dynamically typed language be safe?
• Is a statically typed language automatically type safe?
Dynamic:
• Consider Python
>>> 1 + "hello"
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +: 'int' and 'str'
>>> "hello" + 1
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: can only concatenate str (not "int") to str
• Rejects applying + to incompatible arguments – protects from behavior that is incompatible with the abstractions of integers and strings
Static:
• Consider C:
int array[] = {1, 2, 3, 4};
char string[] = "Hello, world!";
printf("%d %d %d %d\n", array[0], array[1], array[2], array[3]);
array + 2;
string + 3;
Here the compiler will merely complain about unused values on lines 6 and 7. But what does the expression array + 2 mean? Of course, a C programmer knows, that arrays are just pointers, and so adding two to a pointer merely shifts where the pointer points to. But is it complatible with the abstraction of array?
So, again: what is a type safe-language?
A (type-)safe language is one that protects its own abstractions.
This means that a safe language won’t allow a programmer to apply operations which are not “sensible” for the given arguments, potentially leading to unexpected/undefined behavior at runtime.
## Specifying Type Systems
Abstract syntax + Types + Typing rules (+ auxiliary operations)
• A type system is typically specified as a set of rules which allow assigning types to abstract syntax trees – similarly to how evaluators assign values to abstract syntax trees
• The goal: determine what type an expression (program phrase) has – if it has one
• A type judgement – mathematically:
$\vdash e : t$
Read: “$$e$$ has type $$t$$
• A “type judgement” – in Haskell
typeOf e = t
• Typing rules tell us how to arrive at the above conclusion for a particular expression
• This is based on syntax – in other words type checking (and typing rules) are, typically, syntax-directed
• On paper, typing rules are usually expressed as inference rules:
$\frac{\text{1st premise} \quad \text{2nd premise} \quad \text{3rd premise ...}}{\text{conclusion}}$
• Such a rule can be read as “If 1st premise AND 2nd premise AND 3rd premise … are all true, then the conclusion is also true”
• If we can show that the premises hold, the rule allows us to conclude that is below the line
• If a rule has no premises, it is an axiom
• Here are some examples, written mathematically
$\frac{}{\vdash 3 : \text{Integer}}$
“The type of the value 3 is Integer”
$\frac{n \text{ is an integer value}}{\vdash n : \text{Integer}}$
“If $$n$$ is an integer value, then the type of $$n$$ is Integer”
$\frac{\vdash e_1 : \text{Integer} \quad \vdash e_2 : \text{Integer}}{\vdash e_1 + e_2 : \text{Integer}}$
“If the type of $$e_1$$ is Integer and the type of $$e_2$$ is Integer, then the type of expression $$e_1 + e_2$$ is also Integer”
• We can (and will) view these inference rules as a fancy way of writing Haskell functions
• Let us first define datatypes for expressions (for now only integer numbers and addition) and types (only integers)
data Expr = Num Integer
data Type = TyInt
• The above two rules as a Haskell type checker:
typeOf :: Expr -> Maybe Type -- an expression might not have a type
typeOf (Num n) = return TyInt
do TyInt <- typeOf e1
TyInt <- typeOf e2
return TyInt
• Note that the return type of typeOf is Maybe Type
• This is to allow for the possibility that an expression might not have a type (although in this trivial language, all expressions do)
• We use the do notation (together with return) to simplify the definition. The above is equivalent to:
typeOf :: Expr -> Maybe Type -- an expression might not have a type
typeOf (Num n) = Just TyInt
case typeOf e1 of
Just TyInt -> case typeOf e2 of
Just TyInt -> Just TyInt
_ -> Nothing
_ -> Nothing
• To make the connection a little more explicit, we will write inference rules as a mix of Haskell and math:
------------------
|- Num n : TyInt
|- e1 : TyInt |- e2 : TyInt
--------------------------------
|- Add e1 e2 : TyInt
• More typing rules (we add a few new expression shapes + a new type for booleans):
data Expr = ...
| Bool Bool
| And Expr Expr
| Not Expr
| Leq Expr
| If Expr Expr Expr
data Type = ...
| TyBool
--------------------
|- Bool b : TyBool
|- e1 : TyBool |- e2 : TyBool
---------------------------------
|- And e1 e2 : TyBool
|- e : TyBool
-------------------
|- Not e : TyBool
|- e1 : TyInt |- e2 : TyInt
-------------------------------
|- Leq e1 e2 : TyBool
|- e1 : TyBool |- e2 : t |- e3 : t
----------------------------------------
|- If e1 e2 e3 : t
How do we apply these rules?
• We build derivations!
• But what are derivations?
• A derivation is a (proof) tree built by consistently replacing variables in inference rules by concrete terms
• At the bottom of the tree is the typing judgment we are trying to show
Examples:
1. A numeric literal
------------------
|- Num 3 : TyInt
Nothing else needed here, since the rule is an axiom and doesn’t have any conditions (premises)
|- Num 3 : TyInt |- Num 3 : TyInt
-------------------------------------
|- Add (Num 3) (Num 4) : TyInt
3. Boolean expression:
|- Bool True : TyBool |- Bool False : TyBool |- Bool True : TyBool
----------------------------- -----------------------------------------------
|- Not (Bool True) : TyBool |- And (Bool False) (Bool True) : TyBool
-------------------------------------------------------------------------------
|- And (Not (Bool True)) (And (Bool False) (Bool True)) : TyBool
Prettier:
$\cfrac{\cfrac{\vdash \text{Bool True} : \text{TyBool}} {\vdash \text{Not (Bool True)} : \text{TyBool}} \quad \cfrac{\vdash \text{Bool False} : \text{TyBool} \quad \vdash \text{Bool True} : \text{TyBool}} {\vdash \text{And (Bool False) (Bool True)} : \text{TyBool}} } {\vdash \text{And (Not (Bool True )) (And (Bool False) (Bool True))} : \text{TyBool}}$
4. Conditional involving booleans and integers
|- Num 3 : TyInt Num 4 : TyInt
----------------------------------
|- Leq (Num 3) (Num 4) : TyBool
--------------------------------------
|- Not (Leq (Num 3) (Num4)) : TyBool |- Num 3 : TyInt |- Num 5 : TyInt
-----------------------------------------------------------------------------
|- If (Not (Leq (Num 3) (Num 4))) (Num 3) (Num 5)
5. A failing one:
|- Bool True : TyBool |- Num 3 : TyInt
-----------------------------------------
|- Add (Bool True) (Num 3) : ?
We have no rule to apply here. We would need Num 3 to have type TyBool and there is no rule that allows us to derive this. Hence, the above expression cannot be type-checked.
### Type-checking Involving Variables
Syntax extensions:
data Expr = ...
| Var Variable
| Let Variable Expr Expr
How do we deal with variables?
• We need to keep track of types assigned to variables
• Idea: Like for an (environment-based) evaluator for expressions, use an environment
• The environment maps variables to types
type TyEnv = Map Variable Type
Example rules:
t <- get x tenv
-------------------
tenv |- Var x : t
tenv |- e1 : t1 add x t1 env |- e2 : t2
--------------------------------------------
tenv |- Let x e1 e2 : t2
typeOf :: TyEnv -> Expr -> Maybe Type
...
typeOf tenv (Add e1 e2) = -- previous cases need to be refactored to use tenv
do TyInt <- typeOf tenv e1
TyInt <- typeOf tenv e2
return TyInt
...
typeOf tenv (Var x) = get tenv x -- NEW: variable lookup
typeOf tenv (Let x e1 e2) = -- NEW: let-binding
do t1 <- typeOf tenv e1 -- get type of e1
t2 <- typeOf (add x t1 tenv) e2 -- get the type of e2, assuming x : t1
return t2
### Simply-Typed Lambda Calculus (STLC)
Syntax extensions:
• Note that abstractions need to specify the type of the bound variable – there is no way for the type-checker to guess it (at this stage)
data Expr = ...
| Lam Variable Type Expr
| App Expr Expr
data Type = ...
| TyArrow Type Type
TyArrow:
• The new type constructor, TyArrow, represents a function type:
TyArrow TyInt TyBool is the type a function that takes an integer (TyInt) and returns a boolean (TyBool). In Haskell (also in some other languages and in type theory), this is written Integer -> Bool
TyArrow (TyArrow TyInt TyBool) (TyArrow TyInt TyBool) corresponds to (Integer -> Bool) -> (Integer -> Bool), that is, the type of a function that takes a function from integers to booleans and returns a function from integers to booleans.
Due to currying, we normally understand this as a function that takes a function from integers to booleans, then an integer and returns a boolean. Note that this also means that the arrow -> is right-associative and the above Haskell type can be equivalently written as (Integer -> Bool) -> Integer -> Bool. Also note, that this is opposite of how application associates, which is to the left.
Note on associativity:
Function type – RIGHT: t1 -> t2 -> t3 -> t4 is the same as t1 -> (t2 -> t3 -> t3) is the same as t1 -> (t2 -> (t3 -> t4))
Function application – LEFT: f a b c is the same as (f a) b c is the same as ((f a) b) c
Rules
add x t1 tenv |- e : t2
------------------------------------
tenv |- Lam x t1 e : TyArrow t1 t2
tenv |- e1 : TyArrow t2 t1 e2 : t2' t2 == t2'
----------------------------------------------------
tenv |- App e1 e2 : t1
The fixpoint operator:
• No fixpoint combinator (e.g., Y or Z) can be type-checked in STLC, so it has to be added as a primitive operation
data Expr = ...
| Fix Expr
tenv |- e : TyArrow t t' t == t'
------------------------------------
tenv |- Fix e : t
## Polymorphism
Limitations of monomorphic type systems:
• What we’ve seen so far are monomorphic types, meaning a type only represents one type
• What is the problem?
• identity
• operations on lists
• operations on pairs
• Consider length of a list – we need:
length_int :: [Integer] -> Integer
length_bool :: [Bool] -> Integer
Are we done?
What about length of lists of functions from integers to integers?
Length of lists of functions from integers to booleans?
Length of lists of functions from (function from integers to booleans) to integers?
Etc.
• Even simpler: identity
id_int :: Integer -> Integer
id_bool :: Bool -> Bool
id_intToBool :: (Integer -> Bool) -> (Integer -> Bool)
• Functions which perform operations such as counting the number of elements of a list, or swapping the elements of a pair, do not depend on the type of the elements in the list or pair
What do we need?
• Back to an untyped language? :-(
• We would really like to specify a type of functions that work for lists (or pairs) containing any type
• In other words, we need a way to say “for any type t, list of t”
• In yet other words,
length :: forall a. [a] -> Integer
id :: forall a. a -> a
• Ingredients:
1. Type variables
2. Abstracting type variables (quantifying)
• In Haskell (or ML, OCaml, …), polymorphic types are inferred for you and you (usually) do not need to say that you want a polymorphic function
• Another implementation of the same idea are Java generics
Basics:
type TyVariable = String
data Type = ...
| TyVar TyVariable
| TyForall TyVariable Type
• For TyForAll, we can use a more economical alternative:
data Type = ...
| TyVar TyVariable
| TyForall [TyVariable] Type
We are now able to abstract the type of a function. But how do we actually give the information to the type-abstracted function?
Idea: we pass what type we actually intend to use at runtime.
Consequence: We need type abstraction and type application on the expression level
New syntax:
data Expr = ...
| TAbs TyVariable Expr
| TApp Expr Type
We obtain: polymorphic lambda calculus (with extensions) aka System F.
How do we perform application? Substitute the types
Typing rules:
tenv |- e : t
---------------------------------
tenv |- TAbs a e : TyForall a t
tenv |- e : TyForall a t'
----------------------------------
tenv |- TApp e t : tsubst a t t'
Here we use type substitution to substitute types (careful about Forall!)
typeOf tenv (TAbs a e) =
do t <- typeOf tenv e
return (TyForall a t)
typeOf tenv (TApp e t) =
do TyForall a t' <- typeOf tenv e
tsubst a t t'
typeOf tenv (Add e1 e2) = -- previous cases need to be refactored to use tenv
do TyInt <- typeOf tenv e1
TyInt <- typeOf tenv e2
return TyInt
typeOf tenv (Num _) = return TyInt
tsubst :: TyVariable -> Type -> Type -> Maybe Type
tsubst a s (TyVar b) | a == b = return s
| otherwise = return (TyVar b)
tsubst a s (TyForall b t)
| a == b = return $TyForall b t | freeInType a s = Nothing | otherwise = TyForall b <$> tsubst a s t
tsubst a s (TyArrow t1 t2) =
do t1' <- tsubst a s t1
t2' <- tsubst a s t2
return (TyArrow t1' t2')
How do we evaluate type abstraction and type application? We can either use substitution or add another environment
## Other kinds of polymorphism
• What we talked about is parametric polymorphism – mention let polymorphism
• Other type of polymorphism: ad-hoc
• Allows a polymorphic value to exhibit different behaviors, depending on the actual type | 19,401 | 72,361 | {"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": 2, "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} | 4.5 | 4 | CC-MAIN-2019-51 | longest | en | 0.779126 |
http://saylordotorg.github.io/LegacyExams/MA/MA212/MA212-FinalExam-Answers.html | 1,542,766,525,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039746926.93/warc/CC-MAIN-20181121011923-20181121033923-00229.warc.gz | 298,359,233 | 25,945 | 1
Suppose that matrix is obtained by performing a sequence of row operations on . Can be obtained by performing row operations on ?
A. No B. In special cases C. Yes D. There is not enough information provided to determine this.
.
.
Question 2
If and , then which of the following cannot exist?
A. B. C. D.
.
.
Question 3
If and are symmetric matrices, then the product is also symmetric only when and are
A. matrices that commute. B. square matrices. C. Hermitian matrices. D. invertible matrices.
.
.
Question 4
Which of the following expresses the complex number in polar coordinates?
A. B. C. D.
.
.
Question 5
Let be a field. Then, is an ordered field if there exists an order that satisfies which of the following properties?
A. For any either or . B. If and either or , then . C. If , , then . D. All of these.
.
.
Question 6
has which of the following properties?
A. Boundedness B. Finiteness C. Archimedean D. All of these
.
.
Question 7
For , which of the following inequalities is true?
A. B. C. D. All of these
.
.
Question 8
Two matrices are row equivalent if and only if
A. one can be obtained from the other by a finite number of elementary row operations. B. one is the negative of the other. C. the product of the two matrices is zero. D. none of these.
.
.
Question 9
Which matrix below would you row reduce to solve the following system of equations?
A. B. C. D.
.
.
Question 10
The complex numbers are a field in part because
A. for every non-zero there is a such that . B. . C. is a real vector space. D. for all .
.
.
Question 11
Which of the following fields has the Archimedean property?
A. B. Every finite field C. The field of rational functions with real coefficients D. None of these
.
.
Question 12
Determine the angle between the vectors and in .
A. B. C. D.
.
.
Question 13
Which of the following computes the length of the vector ?
A. B. C. D.
.
.
Question 14
According to the fundamental theorem of algebra, how many complex solutions does the equation have, counted with multiplicity?
A. Eight B. At least one C. Exactly two D. Infinitely many
.
.
Question 15
Fill in the blanks. Consider the polynomial . Counted with multiplicity, this polynomial has ________ roots in and ________ roots in .
A. at least one; exactly five B. zero; exactly five C. exactly five; exactly five D. zero; exactly five
.
.
Question 16
Calculate the determinant of .
A. 24 B. 10 C. 12 D. 42
.
.
Question 17
Find the eigenvalues of .
A. , B. , C. , D. ,
.
.
Question 18
What are the eigenvalues of ?
A. Negative B. Real C. Complex D. All of these
.
.
Question 19
Calculate the eigenvalues of .
A. -1, degenerate B. 1, -1 C. 1 degenerate D. 0, -1
.
.
Question 20
17. If , then compute and determine whether or not is a normal matrix.
A. , is normal. B. , is not normal. C. , is normal. D. , is not normal.
.
.
Question 21
The eigenvalues of a real skew-symmetric matrix must be
A. zero. B. non-negative. C. purely real. D. purely imaginary.
.
.
Question 22
Determine the Jordan form for the matrix .
A. B. C. D. None of these
.
.
Question 23
Compute the Jordan canonical form for the matrix .
A. B. C. D.
.
.
Question 24
The determinant rank of the matrix is , where is the
A. smallest number such that some submatrix of has a non zero determinant. B. largest number such that some submatrix of has a non zero determinant. C. largest number such that some submatrix of has a nonzero determinant. D. none of these.
.
.
Question 25
Suppose is the zero map defined by for all . Then what is an eigenvector of ?
A. B. C. Every nonzero vector is an eigenvector of D. All of these.
.
.
Question 26
What does Schur’s Theorem say?
A. Every square matrix is unitarily similar to an upper triangular matrix. B. Every matrix is unitarily similar to a square matrix. C. Every matrix is unitarily similar to a triangular matrix. D. Every upper triangular matrix is similar to a square matrix.
.
.
Question 27
For the set , which of the following properties is not true?
A. Composition of permutations is associative. B. Composition of permutations is commutative. C. There is an identity element for composition. D. There is an inverse element for composition.
.
.
Question 28
Fill in the blanks. Let be an matrix. Then equals __________ and equals _________.
A. sum of the eigenvalues of ; negative of the sum of the eigenvalues of B. product of the eigenvalues of ; sum of the eigenvalues of C. sum of the eigenvalues of ; product of the eigenvalues of D. ;
.
.
Question 29
What does the second derivative test tell us?
A. Whether or not critical points of a function exist B. Where the derivative is zero for certain types of functions C. Whether a critical point of a function is a local minimum or a maximum D. All of these
.
.
Question 30
Suppose is a finite dimensional vector space and is a linear operator on . Then which of the following conditions must be true for a subspace to be an invariant subspace under ?
A. for all B. for all C. for all D. for all
.
.
Question 31
Suppose is a linear operator. Then is an eigenvalue of if and only if
A. is not injective. B. is not injective. C. is invertible. D. is surjective.
.
.
Question 32
Let be a finite dimensional vector space and let be invertible. Then is an eigenvalue for if and only if
A. is an eigenvalue for . B. is an eigenvalue for . C. is an eigenvalue for . D. all of these.
.
.
Question 33
If is a real matrix and one of its eigenvalues is with eigenvector , then what can be said about another one of its eigenvalues and eigenvectors?
A. and is the conjugate of B. and is the conjugate of C. and is the negative conjugate of D. None of these.
.
.
Question 34
Suppose one eigenvalue of a real matrix is and the corresponding eigenvector is . Then which of the following must also be an eigenpair of ?
A. and B. and C. and D. and
.
.
Question 35
Suppose that is an eigenpair of the invertible matrix . Then is an eigenpair of which of the following matrices?
A. B. C. D.
.
.
Question 36
A quadratic form in three dimensions can be written as , where satisfies which of the following properties?
A. is a symmetric matrix. B. is a orthogonal matrix. C. is a normal matrix. D. is a unitary matrix.
.
.
Question 37
Suppose at a critical point of a function , the Hessian matrix is given by . Then what does the second derivative test tell us about this critical point?
A. It is a local maximum. B. It is a local minimum. C. It is a saddle point. D. It is zero.
.
.
Question 38
Let where is an -dimensional -vector space. Then is guaranteed to have an eigenvalue when
A. the minimal polynomial for has a root in . B. the minimal polynomial for has no roots in . C. is an isomorphism. D. the only -invariant subspaces of are trivial.
.
.
Question 39
Let . Determine the Jordan normal form for .
A. B. C. D.
.
.
Question 40
What are the proper subspaces of ?
A. Trivial subspaces B. Lines through the origin C. Both A and B D. None of these
.
.
Question 41
Let be subspaces, then , if and only if two conditions hold. One is that . What is the other condition?
A. B. C. D.
.
.
Question 42
Which of the following subsets of is NOT a subspace of ?
A. Symmetric matrices B. Diagonal matrices C. Nonsingular matrices D. Upper triangular matrices
.
.
Question 43
Fill in the blank. The ___________ of orthogonal vectors is -dimensional.
A. collection B. span C. kernel D. transformation
.
.
Question 44
Which of the following is NOT a vector space associated with the matrix ?
A. B. C. D.
.
.
Question 45
Suppose is a nonsingular matrix. Then which of the following best describes its four fundamental subspaces?
A. and B. and C. and D.
.
.
Question 46
If and are subspaces of a vector space , then equals which of the following?
A. B. C. D.
.
.
Question 47
Consider the following linear transformation where rotates each vector 90 degrees about the -axis. Find the matrix representation of in the standard basis.
A. B. C. D.
.
.
Question 48
Which of the following is a basis of ?
A. B. C. D. All of these
.
.
Question 49
Let be a complex vector space. Let , and . Then, which of the following is true?
A. B. C. D. None of these
.
.
Question 50
Fill in the blank. A linear map is _____________, if and only if is injective and surjective.
A. singular B. invertible C. continuous D. one-to-one
.
.
Question 51
Let be such that for all written in the standard basis. Then, is a
A. zero map. B. identity map. C. vector space. D. surjective linear map.
.
.
Question 52
Consider the following linear transformation : → where rotates each vector 90 degrees counterclockwise about the -axis and then rotates 45 degrees counterclockwise about the -axis. Find the matrix representation of in the standard basis.
A. B. C. D.
.
.
Question 53
Consider the following linear transformation : → , where rotates each vector 90 degrees counterclockwise about the -axis and then rotates 45 degrees counterclockwise about the -axis. Find in the standard basis.
A. B. C. D.
.
.
Question 54
What is the characteristic polynomial of the matrix ?
A. B. C. D.
.
.
Question 55
Let and let be defined by . Then what is the matrix, , for using the standard basis?
A. B. C. D. None of these
.
.
Question 56
All have a singular value decomposition, meaning there exists an orthonormal basis and such that , where the 's are
A. any positive numbers. B. singular values of . C. complex values. D. obtained from the dot product of and .
.
.
Question 57
An matrix is called a Markov matrix if the following is satisfied:
A. for all and B. for all and C. for all and D. for all and
.
.
Question 58
What can be said about eigenvalues of a Markov matrix?
A. Every eigenvalue of a Markov matrix satisfies . B. Every eigenvalue of a Markov matrix satisfies . C. Every eigenvalue of a Markov matrix satisfies . D. All of these.
.
.
Question 59
What does it mean for a matrix to be stochastic?
A. has negative entries and . B. has non-negative entries and . C. has non-negative entries and . D.
.
.
Question 60
If is the union of the coordinate axes, then is NOT a real vector space because
A. it is not closed under addition. B. it is infinite. C. it is not closed under multiplication. D. it does not contain a zero vector.
.
.
Question 61
Suppose , , and are invertible matrices. Which of the following is false?
A. does not have full rank. B. C. is invertible. D.
.
.
Question 62
Fill in the blank. Columns of an matrix are an orthonormal basis for , if and only if is a _____________ matrix.
A. normal B. unitary C. symmetric D. square
.
.
Question 63
Using the standard inner product, which of the following pairs are orthogonal vectors in ?
A. , B. , C. , D. None of these
.
.
Question 64
The spectral theorem states which of the following?
A. Normal operators are diagonal with respect to an orthonormal basis. B. Normal operators are diagonal with respect to a singular set of vectors. C. Nilpotent operators are diagonal with respect to an orthonormal basis. D. Semi-symmetric operators are diagonal with respect to any basis.
.
.
Question 65
Given , the adjoint of is defined to be the operator , such that which is true?
A. for all . B. for all . C. for all . D. None of these
.
.
Question 66
Which of the following best describes every eigenvalue of a self-adjoint operator?
A. Real B. Complex C. Degenerate D. All of these
.
.
Question 67
According to the spectral theorem, if is a finite dimensional inner product space over and , then which of the following must be true?
A. is normal if and only if there exists an orthonormal basis for consisting of eigenvectors for . B. is self-adjoint if and only if the eigenvectors of are degenerate. C. is normal if and only if is diagonalizable. D. None of these.
.
.
Question 68
What are operators that preserve the inner product called?
A. Orthogonal B. Isometries C. Normal D. All of these
.
.
Question 69
Which of the following is an example of a normal operator?
A. Symmetric matrices B. Hermitian matrices C. Orthogonal matrices D. All of these
.
.
Question 70
In order to unitarily diagonalize an matrix , what do you need to do?
A. Find a unitary matrix and a diagonal matrix such that . B. Find a unitary matrix and a diagonal matrix such that . C. Find a unitary matrix and a unitary matrix such that . D. Find a diagonal matrix and a unitary matrix such that .
.
.
Question 71
It is possible for an operator to be normal but not satisfy which of the following conditions?
A. B. C. D. All of these
.
.
Question 72
For all , is
A. self-adjoint. B. nilpotent. C. orthogonal. D. diagonal.
.
.
Question 73
Let and let be a normal operator such that . Then, which of the following is true?
A. B. C. D. All of these
.
.
Question 74
If is a complex vector space and is a normal operator with only one distinct eigenvalue , then which of the following is true?
A. B. for every C. for every D. All of these
.
.
Question 75
Suppose and are two distinct eigenvalues of a real symmetric matrix . Then what are their corresponding eigenvectors?
A. Similar B. Equal C. Orthogonal D. None of these
.
.
Question 76
Which of the following properties should a norm on a normed linear space satisfy?
A. for all , and if and only if B. for all C. for all v,w \in V\$ D. All of these
.
.
Question 77
According to the Cauchy-Schwarz inequality, on any inner product space which of the following must be true?
A. B. C. D.
.
.
Question 78
Fill in the blanks. Suppose and are inner product spaces, , and . Then the tensor product is an element of ______ defined by _______.
A. ; B. ; C. ; D. ;
.
.
Question 79
What is the Riesz representation theorem?
A. Let where is an inner product space. Then there exists a unique such that for all , . B. Let where is an inner product space. Then there exists a unique such that for all , . C. Let where is an inner product space. Then there exist infinitely many such that for all , . D. None of these.
.
.
Question 80
A Hilbert space is
A. a complete inner product space. B. an inner product space. C. a normed space. D. none of these.
.
.
Question 81
In an abstract vector space, what does the norm measure?
A. The angle between two vectors B. The length of a vector C. The direction of a vector D. All of these
.
.
Question 82
Consider the vector space with basis vectors and . Applying the Gram-Schmidt process to produces which orthonormal basis for ?
A. B. C. D.
.
.
Question 83
If , , and are three linearly independent vectors in , then the volume of the parallelepiped determined by , , and is
A. . B. . C. . D. .
.
.
Question 84
Let be an matrix with characteristic polynomial . Which of the following is true?
A. B. is invertible. C. D.
.
.
Question 85
Let , , and . Which of the following is true?
A. is not orthogonal to either or . B. and are orthogonal. C. and are orthogonal. D. and are orthogonal.
.
.
Question 86
Let . Determine the range-nullspace decomposition of relative to .
A. B. C. D.
.
.
Question 87
Let . If is the singular-value decomposition of , then
A. B. C. D.
.
.
Question 88
Let be a linear transformation. Suppose the matrix for relative to a basis for is . Suppose is the transition matrix from another basis to . Then the matrix for with respect to is | 3,967 | 15,182 | {"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} | 3.140625 | 3 | CC-MAIN-2018-47 | latest | en | 0.922819 |
https://forums.mikeholt.com/threads/ambient-temperature.107659/ | 1,590,641,176,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347396495.25/warc/CC-MAIN-20200528030851-20200528060851-00021.warc.gz | 356,129,631 | 16,635 | # Ambient Temperature :)
#### larrybartowski
##### Member
From what I've read on other posts, I know that the definition of "ambient temperature" is not clear in the code book, but I'm going to throw this out there anyway.
I have a customer that tests generators. He has a open bottom ladder style cable tray feeder with (6) tri-tied sets of #500 MCM CT rated copper conductors protected by a 2000 amp breaker (100% rated) and feeding a load bank with a max load of 1972 amps (I've debated that the load will get that high. This is figured at 240 volts and I know they don't produce 240V 3-phase generators, but anyway). Another EC has told him that his cables are undersized.
The tray runs outside from the load bank and into a building, through an air conditioned space, and into a "test cell". It does get warm in the test cell, but it's only the last 10' of the cable tray run.
Per the last sentence of 392.80 (a)(2)(d) where it states "supported on a messenger" I'm using table 310.15(b)(20) that has #500 MCM rated for 496 amps at 75 degrees. When I de-rate for continuous load I'm using the 90 degree column that has 500's rated at 580....this gives me 580x6x.8 = 2,784 amps. Even if I use the 496 in the 75 degree column and derate that for continuous load I get 2380 amps.
Am I missing something? Do I need to derate for ambient temperature due to the last 10 or 20' of the run being in a warm test cell? If so can I use the 90 degree column?
#### GoldDigger
##### Moderator
Staff member
Do I need to derate for ambient temperature due to the last 10 or 20' of the run being in a warm test cell?
See 310.15(A)(2) Exception [2011]. If the total length of the run is 100' then you could get away with 10' in the test cell.
But even then you would need to derate based on ambient for any temperature limited terminations which are inside the test cell.
#### Smart \$
##### Esteemed Member
Do I need to derate for ambient temperature due to the last 10 or 20' of the run being in a warm test cell? If so can I use the 90 degree column?
Yes, you need to apply adjustments for ambient temperature... but see 310.15(A)(2) Exception. If the 10' through the test cell is equal or less than 10% of the circuit length, you get to ignore it. And provided the conductors are 90?C-rated, you can use that column's value as the basis for adjustment.
Last edited:
#### Smart \$
##### Esteemed Member
See 310.15(A)(2) Exception [2011]. If the total length of the run is 100' then you could get away with 10' in the test cell.
But even then you would need to derate based on ambient for any temperature limited terminations which are inside the test cell.
#### larrybartowski
##### Member
So 580 (90 deg) x 6 sets x .8 (for continuous load) x .87 (if ambient is determined at 105-113) = 2422 amps.....right?
Thanks for the help guys.
#### Smart \$
##### Esteemed Member
So 580 (90 deg) x 6 sets x .8 (for continuous load) x .87 (if ambient is determined at 105-113) = 2422 amps.....right?
Thanks for the help guys.
Perhaps I'm misunderstanding, but if you have a 100%-rated breaker, why are you figuring 125% for continuous load???
#### Smart \$
##### Esteemed Member
See 310.15(A)(2) Exception [2011]. If the total length of the run is 100' then you could get away with 10' in the test cell.
But even then you would need to derate based on ambient for any temperature limited terminations which are inside the test cell.
So 580 (90 deg) x 6 sets x .8 (for continuous load) x .87 (if ambient is determined at 105-113) = 2422 amps.....right?
Thanks for the help guys.
Also, I believe what GoldDigger is referring to is if your terminations are rated 75?C, your final ampacity after derating cannot exceed the 75?C column value of 310.15(B)(16).
#### larrybartowski
##### Member
That was a point of debate. You can add that to my list of questions:
1. Do the conductors need to be sized for 125% of the load? Someone smarter than I told me that the do.
2. Do I need to "double de-rate"? Meaning derate for continuous load and then for ambient? Or just worst case scenario?
3. When I'm done derating, if the value is higher than what I get in the 75 deg column, do I use the 75 deg rating? Edit: You just answered this, so disregard.
#### Smart \$
##### Esteemed Member
That was a point of debate. You can add that to my list of questions:
1. Do the conductors need to be sized for 125% of the load? Someone smarter than I told me that the do.
With a 100%-rated breaker, you figure continuous loads at 100%... not 125%
2. Do I need to "double de-rate"? Meaning derate for continuous load and then for ambient? Or just worst case scenario?
This question is rendered moot per the preceding answer... but we can discuss this later if you want.
3. When I'm done derating, if the value is higher than what I get in the 75 deg column, do I use the 75 deg rating?
If your terminations are 75?C-rated, you have no choice but to use the 75?C column value of Table 310.15(B)(16) as the maximum ampacity.
380A x 6 = 2280A
#### GoldDigger
##### Moderator
Staff member
If your terminations are 75?C-rated, you have no choice but to use the 75?C column value of Table 310.15(B)(16) as the maximum ampacity.
380A x 6 = 2280A
That is one basic hard and fast limitation regardless of ambient. But if the terminations are in an ambient of 60?C, then the ampacity correction because of ambient will have to be applied to that value, just as it would have to be factored in along with the ampacity adjustment in the other branch of the calculation where you are allowed to start at 90?C. The short distance exception may allow you to escape the ambient derating of the conductors, but will not, IMHO, allow you to ignore the ambient effect at the termination.
The latter point is arguable, since the code language does not explicitly say that, but if that is not the case, the whole issue of termination temperature could be bypassed just by sticking a 6" pigtail in line using a higher temperature rated butt connector.
PS: By that I mean a same-size pigtail. If you increase the wire size in the pigtail it should legitimately ease the termination temperature restriction.
Last edited:
#### larrybartowski
##### Member
It looks like no matter which way you slice it, the conductors are not undersized.
But......why would I go back to 310.15(b)(16) when 392.80(a)(2)(d) says that I can use 310.15(b)(20)?
#### petersonra
##### Senior Member
With a 100%-rated breaker, you figure continuous loads at 100%... not 125%
whether the breaker is 100% rated or not has no effect on the conductor ampacity required.
#### Smart \$
##### Esteemed Member
It looks like no matter which way you slice it, the conductors are not undersized.
But......why would I go back to 310.15(b)(16) when 392.80(a)(2)(d) says that I can use 310.15(b)(20)?
Because you can use 310.15(B)(20) for conductor ampacity. But 110.14(C) says you have to coordinate the conductor ampacity with the termination temperature limitations... which isn't really an ampacity, but rather a minimum permitted conductor size at the termination, and that is determined per 310.15(B)(16). The 75?C column value of that size conductor is the maximum permitted current allowed through the conductor, even if its ampacity is greater.
#### Smart \$
##### Esteemed Member
whether the breaker is 100% rated or not has no effect on the conductor ampacity required.
Well first off I have to say I oversimplified my statement. It has to be a 100%-rated assembly, not just a 100%-rated breaker.
Now that I got that out of the way, please refer to 210.19(A)(1) Exception and/or 215.2(A)(1) Exception No. 1.
Then come back and explain what you mean...
#### Smart \$
##### Esteemed Member
That is one basic hard and fast limitation regardless of ambient. But if the terminations are in an ambient of 60?C, then the ampacity correction because of ambient will have to be applied to that value, just as it would have to be factored in along with the ampacity adjustment in the other branch of the calculation where you are allowed to start at 90?C. The short distance exception may allow you to escape the ambient derating of the conductors, but will not, IMHO, allow you to ignore the ambient effect at the termination.
The latter point is arguable, since the code language does not explicitly say that, but if that is not the case, the whole issue of termination temperature could be bypassed just by sticking a 6" pigtail in line using a higher temperature rated butt connector.
PS: By that I mean a same-size pigtail. If you increase the wire size in the pigtail it should legitimately ease the termination temperature restriction.
Your assessment is logical and should be required, but as noted it is not. I had discussed collaborating on a proposal with Bob Alexander on the subject, but he never contacted me during the proposal submission period.
#### GoldDigger
##### Moderator
Staff member
Well first off I have to say I oversimplified my statement. It has to be a 100%-rated assembly, not just a 100%-rated breaker.
Now that I got that out of the way, please refer to 210.19(A)(1) Exception and/or 215.2(A)(1) Exception No. 1.
Then come back and explain what you mean...
Tell me what the heck "the assembly" refers to (maybe the OCPD and the enclosure it is part of?) and I will be happy to try to explain what those exceptions mean!
(Except that I then have a problem with just how the "before the application of any adjustment or correction factors" fits in.
The best I can get out of that part is that in addition to having to meet the ampacity requirements after derating, the conductors also have to meet the requirements before derating. That does not seem to be a particularly useful limitation. ):blink:
#### Smart \$
##### Esteemed Member
Tell me what the heck "the assembly" refers to (maybe the OCPD and the enclosure it is part of?) and I will be happy to try to explain what those exceptions mean!
(Except that I then have a problem with just how the "before the application of any adjustment or correction factors" fits in.
The best I can get out of that part is that in addition to having to meet the ampacity requirements after derating, the conductors also have to meet the requirements before derating. That does not seem to be a particularly useful limitation. ):blink:
Assembly means for example the enclosure (usually ventilated IIRC), bussing, and the breaker itself.
Regarding "before the application of any adjustment or correction factors", I wish those sections were revised as such...
Feeder/Branch-circuit conductors shall have an ampacity not
less than required to supply the load after the application of adjustment and
correction factors... The minimum feeder/branch-circuit
conductor size shall have an allowable ampacity not
less than the noncontinuous load plus 125 percent of the
#### GoldDigger
##### Moderator
Staff member
Assembly means for example the enclosure (usually ventilated IIRC), bussing, and the breaker itself.
Regarding "before the application of any adjustment or correction factors", I wish those sections were revised as such...
Thank you!
#### petersonra
##### Senior Member
Well first off I have to say I oversimplified my statement. It has to be a 100%-rated assembly, not just a 100%-rated breaker.
Now that I got that out of the way, please refer to 210.19(A)(1) Exception and/or 215.2(A)(1) Exception No. 1.
Then come back and explain what you mean...
I'll be darned.
#### larrybartowski
##### Member
Because you can use 310.15(B)(20) for conductor ampacity. But 110.14(C) says you have to coordinate the conductor ampacity with the termination temperature limitations... which isn't really an ampacity, but rather a minimum permitted conductor size at the termination, and that is determined per 310.15(B)(16). The 75?C column value of that size conductor is the maximum permitted current allowed through the conductor, even if its ampacity is greater.
Doesn't anyone see this as being odd? Triangular stacked cables with proper spacing in an open bottom/top ladder tray are limited to the same ampacity as if they are installed in a conduit? I would suggest that 392.80(a)(2)(d) allows us to use the 75 degree column "in accordance with 310.15(B)."
But what do I know... | 3,021 | 12,318 | {"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-2020-24 | latest | en | 0.940356 |
https://www.vanessabenedict.com/how-many-grams-of-silver-in-an-ounce/ | 1,718,395,265,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861568.20/warc/CC-MAIN-20240614173313-20240614203313-00681.warc.gz | 938,536,205 | 13,819 | # How many grams of pure silver is a ounce?
#### ByVanessa
Jun 9, 2022
Untitled Document
## How many grams of pure silver is a ounce
1 troy ounce contains exactly 31.1034768 grams of pure silver.
\$20
## What is the difference between an ounce and a troy ounce
What is the difference between an ounce and a troy ounce? A troy ounce is 2.75 grams larger than a regular coin. If you put it on a normal scale, it should be about 10% heavier than the standard unit of measure. To be precise, a regular ounce is considered to be 28.35 grams, while a troy ounce is usually 31.1 grams.
## Is a fluid ounce the same as an ounce
In his possible basic explanation, the fluid ounce (abbreviated fluid ounce) is used to measure liquids, and the bit (abbreviated ounce) is used for dry sizing. … A pint (English scale) is actually the same, so you can get 16 fluid ounces (usual US value).
## What’s the difference between an ounce and a troy ounce
Originally used in Troyes, France, one troy ounce is equal in the UK. Royal Mint thirty-one weighs 1034768 grams1. The standard ounce chosen for weighing other items such as sugar and grains weighs 28.35 grams, which is too much. … The troy ounce is often abbreviated as “t oz” or “t oz”.
## What is difference between troy ounce and ounce
Troy ounces versus ounces The ounce, commonly referred to by the abbreviation “ounce”, is also understood as the avoirdupois ounce. Its weight is approximately 28.35 grams or 1/16 of a pound. …one troy ounce. The von 31 weighs around 103 grams and can be much heavier than an avoirdupois ounce or just an ounce.
Untitled Document
## What is the difference between troy ounce and avoirdupois ounce
At 480 grains, most of the troy ounce is heavier than the regular avoirdupois ounce, which weighs 437.5 grains. The metric Troy the Bit weighs 31.1034768 grams, while the avoirdupois ounce is slightly smaller at 28.349523125 grams.
## What is the difference between 1 ounce and 1 troy ounce
Troy ounces versus ounces The ounce, commonly referred to by the abbreviation “ounce”, is still known as the avoirdupois ounce. It measures approximately 28.35 grams 1/16 lb. … A troy ounce weighs approximately 31.103 grams and is significantly heavier than a troy ounce, or just an ounce.
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# If 3^27^x = 27^3^x , then x is equal to
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If 3^27^x = 27^3^x , then x is equal to [#permalink]
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10 Mar 2017, 13:08
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If 3^27^x = 27^3^x , then x is equal to
A.−1
B. 1/2
C.1
D.2
E.-1/2
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If 3^27^x = 27^3^x , then x is equal to [#permalink]
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11 Mar 2017, 04:54
quantumliner wrote:
If 3^27^x = 27^3^x , then x is equal to
A.−1
B. 1/2
C.1
D.2
E.-1/2
I scratched my head a lot since I was rusty on these questions and had to view the answer first. I am partly convinced I have found the right path but would welcome any correction
RHS can be rewritten as
(3^3)^3^x
= 3 ^ (3.3^x)
equating powers of matching base from both LHS and RHS.
$$27^x = 3.3^x$$
27^x = 3^(x+1)
LHS can be rewritten.
(3^3)^x= 3^(x+1)
3^3x = 3^(x+1)
equating powers of matching base again
$$3x = x + 1$$
$$2x = 1$$
$$x = 1/2$$
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Re: If 3^27^x = 27^3^x , then x is equal to [#permalink]
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11 Mar 2017, 07:41
You lost me while solving RHS when you stated that 27^x = 3^(x+1).
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Re: If 3^27^x = 27^3^x , then x is equal to [#permalink]
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11 Mar 2017, 07:52
On RHS, we had $$3 x 3^x$$ which is same as $$3^1 x 3^x$$ and when base is same we just add the powers so it becomes 3^x+1
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Re: If 3^27^x = 27^3^x , then x is equal to [#permalink]
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11 Mar 2017, 20:23
Can VeritasPrepKarishma or Bunuel please explain this one to me step by step.
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Re: If 3^27^x = 27^3^x , then x is equal to [#permalink]
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11 Mar 2017, 21:15
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matthewsmith_89 wrote:
Can VeritasPrepKarishma or Bunuel please explain this one to me step by step.
Here is the step by step process:
Attachment:
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Note: When two numbers with the same base are multiplied, their exponents get added.
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Re: If 3^27^x = 27^3^x , then x is equal to &nbs [#permalink] 11 Mar 2017, 21:15
Display posts from previous: Sort by | 1,227 | 3,708 | {"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 | 4 | CC-MAIN-2018-43 | latest | en | 0.879564 |
http://drhuang.com/science/mathematics/math%20word/math/h/h143.htm | 1,721,701,964,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517931.85/warc/CC-MAIN-20240723011453-20240723041453-00705.warc.gz | 8,096,282 | 3,336 | ## Heegner Number
The values of for which Quadratic Fields are uniquely factorable into factors of the form . Here, and are half-integers, except for and 2, in which case they are Integers. The Heegner numbers therefore correspond to Discriminants which have Class Number equal to 1, except for Heegner numbers and , which correspond to and , respectively.
The determination of these numbers is called Gauss's Class Number Problem, and it is now known that there are only nine Heegner numbers: , , , , , , , , and (Sloane's A003173), corresponding to discriminants , , , , , , , , and , respectively.
Heilbronn and Linfoot (1934) showed that if a larger existed, it must be . Heegner (1952) published a proof that only nine such numbers exist, but his proof was not accepted as complete at the time. Subsequent examination of Heegner's proof show it to be essentially'' correct (Conway and Guy 1996).
The Heegner numbers have a number of fascinating connections with amazing results in Prime Number theory. In particular, the j-Function provides stunning connections between , , and the Algebraic Integers. They also explain why Euler's Prime-Generating Polynomial is so surprisingly good at producing Primes.
References
Conway, J. H. and Guy, R. K. The Nine Magic Discriminants.'' In The Book of Numbers. New York: Springer-Verlag, pp. 224-226, 1996.
Heegner, K. Diophantische Analysis und Modulfunktionen.'' Math. Z. 56, 227-253, 1952.
Heilbronn, H. A. and Linfoot, E. H. On the Imaginary Quadratic Corpora of Class-Number One.'' Quart. J. Math. (Oxford) 5, 293-301, 1934.
Sloane, N. J. A. Sequence A003173/M0827 in An On-Line Version of the Encyclopedia of Integer Sequences.'' http://www.research.att.com/~njas/sequences/eisonline.html and Sloane, N. J. A. and Plouffe, S. The Encyclopedia of Integer Sequences. San Diego: Academic Press, 1995. | 497 | 1,857 | {"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} | 2.78125 | 3 | CC-MAIN-2024-30 | latest | en | 0.819096 |
https://www.citehr.com/56793-tds-deduction-salary.html | 1,508,588,043,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824775.99/warc/CC-MAIN-20171021114851-20171021134851-00408.warc.gz | 916,741,507 | 11,340 | LOGIN NEW
Hi to all
I have one query on TDS.
Suppose a employee has a net salary (take home) of Rs. 25000/-. Is there any TDS Deduction applicable on this take home. If Yes !! Then how much should be deducted?
Thanks& Regards,
Vandana
6th November 2007 From India, New Delhi
Svandana9
Manager - Hr
SandhyaMukunda
Finance & Accounts
Ganesh330
Sales Executive
Praveenr22
Accountant
Prince0099
Service
Priya Abzer
Hr Manager
+4 Others
Hi All,
I am praveen from bangalore.My company has deputed some of our candidates in LGSI.We pay salary to them which is reimbursed by LGSI.
Should they deduct TDS on it or it is case where TDS is not applicable for reimbursement of expenses and whether service tax is applicable for reimbursement of food expenses.
7th November 2007 From India, Bangalore
Hi Vandana,
An employee drawing Rs. 25000 pm as salary, Employer has to deduct TDS assuming that he don't have any investments done.
Other thing, once the employee joins pls take from him Investment declaration form. Based on the Investment made, you will know what would be the amount of deduction.
Thanks
Sandhya
7th November 2007 From India, Bangalore
Hi Praveen,
Salary Payment made to employees, reimbursed by LGSI may be treated as professional receipts. TDS has to be deducted by LGSI for payment of professional charges.
Service tax is not applicable on reimbursement of food expenses.
Regards
Sandhya
7th November 2007 From India, Bangalore
Total Take home Amount 25000 X 12 =300000
Up to 160000 no tax
160001 to 300000 : 10%
300001 to 500000 : 20%
Above 5000000 : 30%
160000-300000= 140000
140000*10/100 = 14000
14000*2/100 = 280
14000*1/100 = 140
Total Taxble Amount = 14,420.00
26th April 2010 From India, Hyderabad
Hi!!
My employee is being paid take home salary as Rs. 70K per month. To my knowledge, he'll need to pay TDS at the rate 0f 30.3%.
Now, if he produces house rent receipts amounting to Rs. 16K per month as for HRA exemption, Rs. 1250 as medical, how much TDS he'll need to pay now??
Please provide me with calculations.
My email ID:
10th June 2010 From India, Mumbai
Dear All,
My monthly salary is 23500 from April'13 to Aug'13 and 26500 from Sep'13 to March'14. and also expecting annual incentive 1,97,000.
I have home rent slip of Rs. 96000 PA and LIC insurance 18612.
Therefore, kindly suggest me that how much TDS I have to pay.
Thanks,
Jai
6th October 2013 From India, Mumbai
HI, MY MONTHLY SALARY IS 23000/-. HOW MANY THE TDS WILL DEDUCT AND HOW CAN I GET IT BACK. PLEASE GIVE ME FULL PROCEDURE.
30th April 2015 From India, Pune
My salary pa is 300000 i want to know abt my deduction in tds. What amount will it be deducted and is it monthly or anually
1st November 2015 From India, undefined
My Employee Take home salary is 150000 per month in which he receives Convenience allowance + medical allowance + Ta (1600+1250+1000) and has produced doc against section 80c for rs 100000/- (per annum) and is paying rent fr rs 17000/ month. Please let me know his TDS/yr
5th January 2017 From India, Kochi
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Disclaimer: This network and the advice provided in good faith by our members only facilitates as a direction. The advice should be validated by proper consultation with a certified professional. The network or the members providing advice cannot be held liable for any consequences, under any circumstances. | 932 | 3,489 | {"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-2017-43 | latest | en | 0.947156 |
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If the tree grew by ten inches each year, how much higher would the nail be?
• Views : 70k+
• Sol Viewed : 20k+
# Math Riddles #7 - Number Eight Math Riddle
Difficulty Popularity
In a new Engineering Hostels containing 100 rooms. Ankit Garg was hired to paint the numbers 1 to 100 on the doors.
How many times will Ankit have to paint the number eight ?
• Views : 70k+
• Sol Viewed : 20k+
# Math Riddles #8 - Common Exam Problem
Difficulty Popularity
5+3+2 = 151022
9+2+4 = 183652
8+6+3 = 482466
5+4+5 = 202541
THEN ;
7+2+5 = ?
• Views : 70k+
• Sol Viewed : 20k+
# Math Riddles #9 - Mathematical Equation Riddle
Difficulty Popularity
Replace the '?' by any mathematical symbol to make the expression equal to 99.
18 ? 12 ? 2 ? 3 = 111
• Views : 30k+
• Sol Viewed : 10k+
# Math Riddles #10 - Car Meter Riddle
Difficulty Popularity
Today my car meter reads as 72927 kms. I notes that this is a palindrome. How many minimum kms i need to travel so my car meter find another palindrom.
• Views : 20k+
• Sol Viewed : 10k+
# Math Riddles #11 - Maths Number Riddle
Difficulty Popularity
Can you arrange four nines to make it equal to 100.
Hint: use two mathematical symbols.
• Views : 40k+
• Sol Viewed : 10k+
# Math Riddles #12 - Trick Maths Riddle
Difficulty Popularity
What is the value of 1/2 of 2/3 of 3/4 of 4/5 of 5/6 of 6/7 of 7/8 of 8/9 of 9/10 of 1000?
• Views : 50k+
• Sol Viewed : 20k+
# Math Riddles #13 - Divide Into Two parts Riddle
Difficulty Popularity
Divide 110 into two parts so that one will be 150 percent of the other. What are the 2 numbers?
• Views : 70k+
• Sol Viewed : 20k+
# Math Riddles #14 - Maths Logic Problem
Difficulty Popularity
Large number of people went to an party and they decided to make some fun at the bar.
The first person asks the barman for half a pint of beer.
The second person asks for a quarter pint of beer
The third person asks for one-eighth of beer and so on ...
How many pints of beer will the barman need to fulfill the people need of beer ?
• Views : 50k+
• Sol Viewed : 20k+
# Math Riddles #15 - Tricky Distance Riddle
Difficulty Popularity
Two villages owlna and kathur are exactly 2000 km apart.
Pramod leaves City "owlna" driving at 50 km/hr and Anmol leaves City "kathur" a half an hour later driving 90 km/hr. Who will be closer to City "owlna" when they meet?
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If 1 + 9 + 11 = 1, Then what is the valu... | 1,292 | 4,551 | {"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} | 2.78125 | 3 | CC-MAIN-2017-13 | latest | en | 0.922052 |
https://statsidea.com/r-how-to-merge-data-frames-by-column-names/ | 1,670,470,913,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711232.54/warc/CC-MAIN-20221208014204-20221208044204-00613.warc.gz | 577,615,279 | 13,968 | # R: How to Merge Data Frames by Column Names
You can use the following methods to merge data frames by column names in R:
Method 1: Merge Based on One Matching Column Name
```merge(df1, df2, by='var1')
```
Method 2: Merge Based on One Unmatched Column Name
`merge(df1, df2, by.x='var1', by.y='variable1')`
Method 3: Merge Based on Multiple Matching Column Names
`merge(df1, df2, by=c('var1', 'var2'))`
Method 4: Merge Based on Multiple Unmatched Column Names
`merge(df1, df2, by.x=c('var1', 'var2'), by.y=c('variable1', 'variable2'))`
The following examples show how to use each method in practice.
### Example 1: Merge Based on One Matching Column Name
The following code shows how to merge two data frames in R based on one matching column name:
```#define data frames
df1 <- data.frame(team=c('A', 'B', 'C', 'D'),
points=c(88, 98, 104, 100))
df2 <- data.frame(team=c('A', 'B', 'C', 'D'),
rebounds=c(22, 31, 29, 20))
#merge based on one column with matching name
merge(df1, df2, by='team')
team points rebounds
1 A 88 22
2 B 98 31
3 C 104 29
4 D 100 20
```
The result is one data frame that matched rows in each data frame using the team column.
### Example 2: Merge Based on One Unmatched Column Name
The following code shows how to merge two data frames in R based on one unmatched column name:
```#define data frames
df1 <- data.frame(team=c('A', 'B', 'C', 'D'),
points=c(88, 98, 104, 100))
df2 <- data.frame(team_name=c('A', 'B', 'C', 'D'),
rebounds=c(22, 31, 29, 20))
#merge based on one column with unmatched name
merge(df1, df2, by.x='team', by.y='team_name')
team points rebounds
1 A 88 22
2 B 98 31
3 C 104 29
4 D 100 20```
The result is one data frame that matched rows using the team column in the first data frame and the team_name column in the second data frame.
### Example 3: Merge Based on Multiple Matching Column Names
The following code shows how to merge two data frames in R based on multiple matching column names:
```#define data frames
df1 <- data.frame(team=c('A', 'A', 'B', 'B'),
position=c('G', 'F', 'G', 'F'),
points=c(88, 98, 104, 100))
df2 <- data.frame(team=c('A', 'A', 'B', 'B'),
position=c('G', 'F', 'G', 'F'),
rebounds=c(22, 31, 29, 20))
#merge based on multiple columns with matching names
merge(df1, df2, by=c('team', 'position'))
team position points rebounds
1 A F 98 31
2 A G 88 22
3 B F 100 20
4 B G 104 29
```
The result is one data frame that matched rows in each data frame using the team and position column in each data frame.
### Example 4: Merge Based on Multiple Unmatched Column Names
The following code shows how to merge two data frames in R based on multiple unmatched column names:
```#define data frames
df1 <- data.frame(team=c('A', 'A', 'B', 'B'),
position=c('G', 'F', 'G', 'F'),
points=c(88, 98, 104, 100))
df2 <- data.frame(team_name=c('A', 'A', 'B', 'B'),
position_name=c('G', 'F', 'G', 'F'),
rebounds=c(22, 31, 29, 20))
#merge based on multiple columns with matching names
merge(df1, df2, by.x=c('team', 'position'), by.y=c('team_name', 'position_name'))
team position points rebounds
1 A F 98 31
2 A G 88 22
3 B F 100 20
4 B G 104 29
```
The result is one data frame that matched rows using the team and position columns in the first data frame and the team_name and position_name columns in the second data frame. | 1,075 | 3,573 | {"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-2022-49 | latest | en | 0.726992 |
http://www.airspacemag.com/daily-planet/how-much-worlds-population-has-flown-airplane-180957719/ | 1,480,911,943,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541525.50/warc/CC-MAIN-20161202170901-00508-ip-10-31-129-80.ec2.internal.warc.gz | 326,550,794 | 17,807 | # How Much of the World’s Population Has Flown in an Airplane?
## Some numbers, and some guesses.
airspacemag.com
Thinking about your last flight, did it feel as if everyone in the world was on the airplane with you? The “load factor” for airlines worldwide—the amount of seats filled with fare-paying passengers—is currently running at about 80 percent, and the industry keeps breaking records for the number of people flying each year. The International Civil Aviation Organization reports that 3.5 billion passengers buckled up for takeoff in 2015, and the International Air Transport Association expects that number to jump to 3.8 billion next year.
Those of you inclined to do the math will realize that 3.5 billion is nearly half the world’s population. Does this mean that one out of two people on Earth flew sometime last year? Not at all.
“One person can be multiple passengers on a given day” explains John Heimlich, chief economist for Airlines for America, a trade association representing U.S. carriers. In other words, if your most recent journey involved a change of planes, you counted as two of the 3.5 billion passengers, or four if you flew round-trip.
No worldwide database keeps track of the number of discrete individual travelers taking to the air each year. Airlines may have that information for their own passengers, but they don’t share. So determining with any precision how many individual global citizens flew in a year, let alone what percentage of the world’s population has ever flown in an airplane, may be impossible.
Just because the number is unknowable does not mean it is un-guessable. One way to approach the question is to look at surveys, but the data are spotty at best, sometimes conflicting, and often out of date. In 2003, the U.S. Bureau of Transportation Statistics estimated, based on its Omnibus Household Survey, that one-third of U.S. adults had flown in the previous 12 months. And—the closest thing we’ve seen to the number we’re after—18 percent of Americans said they had never flown in their life, meaning that 82 percent had.
By 2009, the amount of people who flew a commercial airliner in the previous year had risen to 39.85 percent in the Omnibus Household Survey. Gallup polls give numbers that are a little higher. In 2012, 52 percent of respondents said they had flown at least once in the past year, the highest number in a decade.
Of course, these are only U.S. flyers, and for many parts of the world, information is even scarcer. A consumer survey conducted by Credit Suisse First Boston in 2004 found that 47 percent of respondents in eight large Chinese cities had ever flown in an airplane. Air travel is on the rise in China, as well as in other parts of Asia and Latin America. “As gross domestic product increases the number of trips per capita increases,” Heimlich says, citing numbers produced by airplane manufacturers. “What they show is, to a point, air travel increases exponentially. Places with large populations like China and India are of great interest.”
A country’s level of development may not always track with the number of air passengers. Some underdeveloped nations have a large number of flyers because of the inaccessibility of ground transport. Indonesia comes to mind.
According to Boeing’s market outlook for the next 20 years, “China and the Middle East once again led all regions [in 2014] with double-digit traffic growth. Europe traffic grew at five percent in 2014, far outpacing economic growth, while North America traffic grew more than two percent.”
What’s not clear from these statistics is whether more people were flying, or the same people were flying more often, or both. It seems likely that the percentage of global citizens who have ever flown is growing, and will continue to grow. We just don’t have the data to back it up.
Tom Farrier, an air safety specialist who contributes to the crowd-sourced information site, Quora, took a shot at answering the question a couple of years ago. He assumed most people fly round-trip, and that a number of flights are not point-to-point but involve a stop along the way. He then calculated that the majority of air tickets are purchased by people who travel for business. His final guess: Maybe six percent of the world’s population flew in a single year.
If you can be more precise than that, we’d love to hear from you. | 932 | 4,389 | {"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-2016-50 | longest | en | 0.938999 |
http://research.stlouisfed.org/fred2/series/KIPPPGGEA156NUPN?rid=258 | 1,432,358,357,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207927185.70/warc/CC-MAIN-20150521113207-00019-ip-10-180-206-219.ec2.internal.warc.gz | 201,555,405 | 19,559 | # Investment Share of Purchasing Power Parity Converted GDP Per Capita at constant prices for Georgia
2010: 13.04796 Percent (+ see more)
Annual, Not Seasonally Adjusted, KIPPPGGEA156NUPN, Updated: 2012-09-17 11:31 AM CDT
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For proper citation, see http://pwt.econ.upenn.edu/php_site/pwt_index.php
Source Indicator: ki
Source: University of Pennsylvania
Release: Penn World Table 7.1
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(a) Investment Share of Purchasing Power Parity Converted GDP Per Capita at constant prices for Georgia, Percent, Not Seasonally Adjusted (KIPPPGGEA156NUPN)
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Create your own data transformation: [+]
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Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
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``` University of Pennsylvania, Investment Share of Purchasing Power Parity Converted GDP Per Capita at constant prices for Georgia [KIPPPGGEA156NUPN], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/KIPPPGGEA156NUPN/, May 23, 2015. ```
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Name: Email: | 531 | 2,025 | {"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-2015-22 | latest | en | 0.778508 |
http://www.ck12.org/physical-science/Combining-Forces-in-Physical-Science/?difficulty=basic | 1,484,989,313,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281001.53/warc/CC-MAIN-20170116095121-00083-ip-10-171-10-70.ec2.internal.warc.gz | 397,523,422 | 16,891 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Combining Forces
## The net force acting on an object is the combination of all of the individual forces acting on it.
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
• Real World Application
## Ollie Up
by CK-12 //basic
Discover the physics behind the skateboarding trick called the ollie and how the skater combines forces to do the trick.
MEMORY METER
This indicates how strong in your memory this concept is
1
• Real World Application
## Tug of War
by CK-12 //basic
Learn what forces are involved in the game of tug of war and how the game illustrates Newton's first law of motion.
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5
• Real World Application
## Speeding Up by Falling Down
by CK-12 //basic
Discover how to find the terminal velocity of a penny dropped from a skyscraper and whether the penny will be traveling fast enough by the time it reaches the ground to damage materials or harm people.
MEMORY METER
This indicates how strong in your memory this concept is
6 | 363 | 1,647 | {"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-2017-04 | longest | en | 0.869961 |
https://studyadda.com/sample-papers/jee-main-sample-paper-29_q6/283/300607 | 1,723,433,740,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641028735.71/warc/CC-MAIN-20240812030550-20240812060550-00753.warc.gz | 426,628,380 | 21,119 | • # question_answer Let $\omega =\frac{-1}{2}+i\frac{\sqrt{3}}{2}$. Then the value of the determinant$\left| \begin{matrix} 1 & 1 & 1 \\ 1 & -1-{{\omega }^{2}} & {{\omega }^{2}} \\ 1 & {{\omega }^{2}} & {{\omega }^{4}} \\ \end{matrix} \right|$ is A) $3\omega$ B) $3\omega (\omega -1)$ C) $3{{\omega }^{2}}$ D) $3\omega (1-\omega )$
$D=\left| \begin{matrix} 1 & 1 & 1 \\ 1 & -1-{{\omega }^{2}} & {{\omega }^{2}} \\ 1 & {{\omega }^{3}} & {{\omega }^{4}} \\ \end{matrix} \right|=\left| \begin{matrix} 1 & 1 & 1 \\ 1 & +\omega & {{\omega }^{2}} \\ 1 & {{\omega }^{2}} & \omega \\ \end{matrix} \right|$ $=\left| \begin{matrix} 3 & 0 & 0 \\ 1 & \omega & {{\omega }^{3}} \\ 1 & {{\omega }^{2}} & \omega \\ \end{matrix} \right|$ $=3[{{\omega }^{2}}-{{\omega }^{4}}]=3({{\omega }^{2}}-\omega )=3\omega (\omega -1)$ | 370 | 879 | {"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} | 3.671875 | 4 | CC-MAIN-2024-33 | latest | en | 0.156618 |
https://tellybollyarab.com/2021/09/which-of-these-is-better/ | 1,653,680,137,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662675072.99/warc/CC-MAIN-20220527174336-20220527204336-00482.warc.gz | 622,647,115 | 13,507 | # Which of these is better?
This is a tricky one.
While there’s some clear winners, it’s hard to make a clear cut for each of these measuring instruments.
While you can use a simple linear or cubic measurement to get an idea of how many inches the earth is, the accuracy of these instruments is often affected by how accurate the measurement is.
In other words, the closer you get to the center of the earth, the more inaccurate the linear or cube measurement will be.
In order to figure out which of these measurement instruments is the best for you, it helps to know what your primary goals are and what measurement method you prefer.
Linear measuring instruments have a single, fixed axis.
For example, a simple measurement on a flat surface like a piece of paper is usually a linear measurement.
However, if you use an instrument with an adjustable axis (such as a table) and adjust the position of the axis, it will have to be adjusted to be a square.
For this reason, a linear or square measuring instrument with adjustable axis is often considered a good choice.
A cubic measuring instrument is a little different.
A linear measuring instrument has a variable axis that can vary by an amount proportional to the distance between the measuring point and the centerline.
For the most accurate measurements, you want the measuring points to be as close to the edge of the Earth as possible.
A square measuring device has a fixed axis that varies by an equal amount proportional.
For most measurements, a square measuring tool should be used.
A circular measuring device is the most common of these types of measuring instruments because it measures both the center and the circumference of the object at once.
This type of measuring instrument also has an adjustable point.
For measurements on a circular surface, the adjustable point will always be closer to the bottom of the circle, so you will see a square measurement when the point is at the top of the circular surface.
The best measuring instrument for you The most important thing to remember about measuring a piece is that it’s going to be measured by a single axis.
It doesn’t matter how many different axes you use, the axis will always have the same value.
This is because the measurement points are all in the same place on the measuring surface.
For a circle, you can measure the circumference by the distance from the center to the circumference.
For an ellipse, you measure the length by the diameter of the ellipsoid.
For any other type of measurement, you will need to adjust the axis to make sure the measurement comes out in the right location.
The most common type of linear measuring tools are a standard measuring device (called a table), and a circular measuring instrument.
The standard measuring instruments usually have a fixed, standard, or circular axis.
The circle or ellipth, for example, has a standard axis.
A standard measuring instrument uses an axis that is fixed.
The fixed axis is usually the same for all measuring axes.
For measuring distance, the standard is the length of the shortest distance between two measuring points.
For other measurements, the circle is measured by the width of the smallest distance between three measuring points (usually about one-half the width) or by the height of the tallest distance between five or more measuring points, usually about one meter.
For calculating a diameter, the diameter is the diameter between two measurements.
For finding the height, the height is measured from the measurement point to the measurement end.
For determining the radius, the radius is the distance in inches between two measurement points.
A circle can measure length (length divided by height).
A square can measure width (width divided by depth).
A circle has a center.
A elliphese has an area.
A cylinder has a volume.
The radius is a measure of the length between two of these points.
So a square can have a radius of 5.6 feet.
But the center can have only a radius equal to 1.8 feet.
So, for a circle with a center, a circle that is at a radius that is equal to one-fifth the diameter, a round square will measure width of 6.6 inches.
A table is a measuring instrument that is set up so that the axes are not in a perfect straight line.
The axes are spaced apart, and the distance they measure from the axis is measured in millimeters (m).
So a table with an axis of 10 inches will measure 6 inches (mm).
For calculating the height from the height measurement to the height measured, a table that is 20 inches (cm) in diameter will measure 2.1 feet (0.7 meters).
A cylinder is a type of circular measuring machine that has an axis in the middle of a cylinder.
The axis is in the center, and a point on the axis (called the axis mark) is at each end of the cylinder.
For some measurements, like distance, you need to use an axis with a radius less than one- | 996 | 4,926 | {"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} | 3.6875 | 4 | CC-MAIN-2022-21 | latest | en | 0.942922 |
http://images.planetmath.org/quadrant | 1,521,580,644,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647545.54/warc/CC-MAIN-20180320205242-20180320225242-00688.warc.gz | 160,911,426 | 3,289 | The $x$- and $y$-axes the $xy$-plane in four right-angle domains which are called the of the plane. They are numbered going round the origin in anticlockwise direction so that
$\{(x,\,y)\mid\;x>0,\;y>0\}$
$\{(x,\,y)\mid\;x<0,\;y>0\}$
the second quadrant, and so on.
Naturally, one can speak of the quadrants of the complex plane, too.
The lines $y=\pm x$ have as their slope angles $\pm 45^{\circ}$, thus halving the quadrant angles; they are called the quadrant bisectors. Cf. angle bisector as locus. | 161 | 511 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 7, "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} | 2.765625 | 3 | CC-MAIN-2018-13 | latest | en | 0.898869 |
http://www.chegg.com/homework-help/cylindrical-disk-volume-897-103-m3-mass-816-kg-disk-floating-chapter-9-problem-39p-solution-9780073512143-exc | 1,472,049,563,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982292330.57/warc/CC-MAIN-20160823195812-00163-ip-10-153-172-175.ec2.internal.warc.gz | 377,375,881 | 23,042 | View more editions
# College Physics With an Integrated Approach to Forces and Kinematics (4th Edition)Solutions for Chapter 9 Problem 39PProblem 39P: A cylindrical disk has volume 8.97 × 103 m3 and mass ...
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Chapter: Problem:
A cylindrical disk has volume 8.97 × 103 m3 and mass 8.16 kg. The disk is floating on the surface of some water with its flat surfaces horizontal. The area of each flat surface is 0.640 m2. (a) What is the specific gravity of the disk? (b) How far below the water level is its bottom surface? (c) How far above the water level is its top surface?
STEP-BY-STEP SOLUTION:
Chapter: Problem:
Corresponding Textbook
College Physics With an Integrated Approach to Forces and Kinematics | 4th Edition
9780073512143ISBN-13: 0073512141ISBN:
Alternate ISBN: 9780077437817, 9780077437886, 9780077437930, 9780077491109, 9780077598556 | 295 | 1,075 | {"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-2016-36 | longest | en | 0.822032 |
https://brilliant.org/discussions/thread/bashing-available-and-also-unavailableproving-a/ | 1,529,378,553,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267861752.19/warc/CC-MAIN-20180619021643-20180619041643-00052.warc.gz | 561,988,436 | 14,389 | # Bashing available and also unavailable:Proving a trigonometric result
I was studying trigonometry and I found a good prove problem which is
Prove that:$$\tan { A } +2\tan { 2A } +4\tan { 4A } +8\cot { 8A } =\cot { A }$$.
I have a non-bashing solution,want to see your approach.
Note by Shivamani Patil
2 years, 11 months ago
MarkdownAppears as
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## Comments
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Top Newest
Take tan on RHS and use the property
$$\displaystyle cotA - tanA = 2cot(2A)$$
Again take 2tan2A on RHS
$$2cot2A - 2tan2A = 4cot4A$$
Again bring 4tan4A on RHS
$$4cot4A - 4tan4A = 8cot8A$$ on RHS
I think there is a typo it should be 8cot8A instead of 8tan8A on LHS.
- 2 years, 11 months ago
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Ya I was talking of same non-bashing solution.Is there any shorter or equivalent method??
- 2 years, 11 months ago
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That's the shortest and simplest solution according to me, I cannot think of any 'shorter' solution right now.
- 2 years, 11 months ago
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According to me too .
- 2 years, 11 months ago
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×
Problem Loading...
Note Loading...
Set Loading... | 668 | 1,954 | {"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": 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} | 3.375 | 3 | CC-MAIN-2018-26 | latest | en | 0.82703 |
https://www.enotes.com/homework-help/what-integral-ln-e-3x-312994 | 1,516,379,113,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888041.33/warc/CC-MAIN-20180119144931-20180119164931-00119.warc.gz | 933,209,795 | 9,179 | # What is the integral of ln [e^(3x)]?
hala718 | Certified Educator
`int e^(3x) dx `
`==gt u= 3x ==gt du = 3 dx `
`==gt int e^(3x) dx = int e^u (du)/3 `
`==gt int e^(3x) dx = (1/3) int e^u du = (1/3) e^u + C `
`==gt int e^(3x) dx = (1/3) e^(3x) + C`
`` | 130 | 259 | {"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} | 3.234375 | 3 | CC-MAIN-2018-05 | latest | en | 0.565116 |
https://metanumbers.com/50031 | 1,638,473,167,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362287.26/warc/CC-MAIN-20211202175510-20211202205510-00556.warc.gz | 425,562,249 | 7,381 | # 50031 (number)
50,031 (fifty thousand thirty-one) is an odd five-digits composite number following 50030 and preceding 50032. In scientific notation, it is written as 5.0031 × 104. The sum of its digits is 9. It has a total of 5 prime factors and 16 positive divisors. There are 31,104 positive integers (up to 50031) that are relatively prime to 50031.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 9
• Digital Root 9
## Name
Short name 50 thousand 31 fifty thousand thirty-one
## Notation
Scientific notation 5.0031 × 104 50.031 × 103
## Prime Factorization of 50031
Prime Factorization 33 × 17 × 109
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 5559 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 50,031 is 33 × 17 × 109. Since it has a total of 5 prime factors, 50,031 is a composite number.
## Divisors of 50031
1, 3, 9, 17, 27, 51, 109, 153, 327, 459, 981, 1853, 2943, 5559, 16677, 50031
16 divisors
Even divisors 0 16 8 8
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 79200 Sum of all the positive divisors of n s(n) 29169 Sum of the proper positive divisors of n A(n) 4950 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 223.676 Returns the nth root of the product of n divisors H(n) 10.1073 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 50,031 can be divided by 16 positive divisors (out of which 0 are even, and 16 are odd). The sum of these divisors (counting 50,031) is 79,200, the average is 4,950.
## Other Arithmetic Functions (n = 50031)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 31104 Total number of positive integers not greater than n that are coprime to n λ(n) 432 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5133 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 31,104 positive integers (less than 50,031) that are coprime with 50,031. And there are approximately 5,133 prime numbers less than or equal to 50,031.
## Divisibility of 50031
m n mod m 2 3 4 5 6 7 8 9 1 0 3 1 3 2 7 0
The number 50,031 is divisible by 3 and 9.
• Arithmetic
• Deficient
• Polite
## Base conversion (50031)
Base System Value
2 Binary 1100001101101111
3 Ternary 2112122000
4 Quaternary 30031233
5 Quinary 3100111
6 Senary 1023343
8 Octal 141557
10 Decimal 50031
12 Duodecimal 24b53
20 Vigesimal 651b
36 Base36 12lr
## Basic calculations (n = 50031)
### Multiplication
n×y
n×2 100062 150093 200124 250155
### Division
n÷y
n÷2 25015.5 16677 12507.8 10006.2
### Exponentiation
ny
n2 2503100961 125232644179791 6265514420959123521 313469951995005908879151
### Nth Root
y√n
2√n 223.676 36.8479 14.9558 8.70658
## 50031 as geometric shapes
### Circle
Diameter 100062 314354 7.86372e+09
### Sphere
Volume 5.24573e+14 3.14549e+10 314354
### Square
Length = n
Perimeter 200124 2.5031e+09 70754.5
### Cube
Length = n
Surface area 1.50186e+10 1.25233e+14 86656.2
### Equilateral Triangle
Length = n
Perimeter 150093 1.08387e+09 43328.1
### Triangular Pyramid
Length = n
Surface area 4.3355e+09 1.47588e+13 40850.1
## Cryptographic Hash Functions
md5 d081111dbdee3c687d1439b444d64004 72c55f1fbcb29b10612a48faf24a16539ee9af69 b98ee8669a57be8fd683d10235a884f8c65b2e5bde5157c0b45d7ab825f11de0 010307b4cb80fd7fc909eff101bfc736670b78d1e32a5e83976315cdd0a1a123dc74a92d4aa5dae7a722edeb85a0f7f7278f5de49e0b787375520f3b01f4d177 48b2f308198fee550baa5ac68dd28a4713690492 | 1,466 | 4,091 | {"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} | 3.828125 | 4 | CC-MAIN-2021-49 | latest | en | 0.8046 |
http://www.jiskha.com/display.cgi?id=1270055971 | 1,496,044,284,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612036.99/warc/CC-MAIN-20170529072631-20170529092631-00373.warc.gz | 682,018,767 | 3,985 | # stats
posted by on .
A statistics practitioner wishes to test the following hypothesis: H0 : ì = 600 against H1 : ì < 600
A sample of 50 observations yielded the statistics: mean x = 585 and standard deviation sx = 45. The
test statistic of a test to determine whether there is enough evidence at the 10% significance level to reject
the null hypothesis is
(1) 2.3570
(2) 0.3333
(3) 16.6667
(4) 23.570
(5) −2.3570
• stats - ,
Judy, Sarah, Sam, Cindy, Joseph or whomever. First, we like for you to use the same screen name. It makes it easier for us to keep up with the questions.
Second, you are unlikely to get repsonses when you show no work.
• stats - ,
4. 23.570 | 203 | 674 | {"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} | 2.765625 | 3 | CC-MAIN-2017-22 | latest | en | 0.859982 |
https://www.r-bloggers.com/tag/recursion/ | 1,722,771,733,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640398413.11/warc/CC-MAIN-20240804102507-20240804132507-00671.warc.gz | 747,982,537 | 17,185 | # recursion
### the large half now
October 28, 2012 |
The little half puzzle proposed a “dumb’ solution in that players play a minimax strategy. There are 34 starting values less than 100 guaranteeing a sure win to dumb players. If instead the players maximise their choice at each step, the R code looks like this: and there are now 66 (=100-34, indeed!) ... [Read more...]
### Cross validated question
February 19, 2012 |
Another problem generated by X’validated (on which I spent much too much time!): given an unbiased coin that produced M heads in the first M tosses, what is the expected number of additional tosses needed to get N (N__M) consecutive heads? Consider the preliminary question of getting a ... [Read more...]
### ultimate R recursion
January 31, 2012 |
One of my students wrote the following code for his R exam, trying to do accept-reject simulation (of a Rayleigh distribution) and constant approximation at the same time: which I find remarkable if alas doomed to fail! I wonder if there exists a (real as opposed to fantasy) computer language ... [Read more...]
### More Recursion in R
May 26, 2009 |
I found another gem in R today. Earlier I commented about how R could do recursion, something that I love. I write some pretty complicated recursion functions in my research, but I also have a bad habit of compulsively reorganizing things. Now I've c... [Read more...] | 324 | 1,406 | {"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-2024-33 | latest | en | 0.964778 |
https://probstats.org/binomial.html | 1,718,783,757,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861806.64/warc/CC-MAIN-20240619060341-20240619090341-00211.warc.gz | 434,091,851 | 1,538 | # Binomial distribution
Tossing a coin $n$ times, how likely to get $k$ heads?
$n$
$p$
$$p(k) = \begin{cases} {n \choose k}p^k(1-p)^{n-k}, & \text{if k = 0, 1, \cdots, n,} \\ 0, & \text{otherwise.} \end{cases} \\[16pt] \text{where p is the probability the coin lands on heads.}$$ | 113 | 281 | {"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} | 3.21875 | 3 | CC-MAIN-2024-26 | latest | en | 0.651018 |
http://bach.istc.kobe-u.ac.jp/cgi-bin/seqprover/seqprover.cgi?output=pretty&sequent=%28p%28a%29%2F%5Cp%28b%29-%3Eq%29-%3EX%23%28p%28X%29-%3Eq%29 | 1,579,532,381,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250598800.30/warc/CC-MAIN-20200120135447-20200120164447-00315.warc.gz | 18,327,428 | 1,255 | Sequent Prover (seqprover)
Sequent:
Output style:
[Top page]
Trying to prove with threshold = 0 1
--------------------------- Ax --------------------------- Ax
p(a) --> p(a),q,X#(p(X)->q) p(b) --> p(b),q,X#(p(X)->q)
----------------------------- R-> ----------------------------- R-> ------------------------ Ax
--> p(a),p(a)->q,X#(p(X)->q) --> p(b),p(b)->q,X#(p(X)->q) p(Y),q --> q,X#(p(X)->q)
----------------------------- R# ----------------------------- R# ------------------------- R->
--> p(a),X#(p(X)->q) --> p(b),X#(p(X)->q) q --> p(Y)->q,X#(p(X)->q)
-------------------------------------------------------- R/\ ------------------------- R#
--> p(a)/\p(b),X#(p(X)->q) q --> X#(p(X)->q)
------------------------------------------------------------------------- L->
p(a)/\p(b)->q --> X#(p(X)->q)
--------------------------------- R->
--> (p(a)/\p(b)->q)->X#(p(X)->q)
# Proved in 0 msec.
Maintained by Naoyuki Tamura / Programmed by Naoyuki Tamura | 293 | 1,042 | {"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-2020-05 | latest | en | 0.259864 |
https://xlsxwriter.readthedocs.io/example_chart_clustered.html | 1,701,350,600,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100227.61/warc/CC-MAIN-20231130130218-20231130160218-00013.warc.gz | 1,197,988,477 | 7,120 | Example: Clustered Chart#
Example of creating a clustered Excel chart where there are two levels of category on the X axis.
The categories in clustered charts are 2D ranges, instead of the more normal 1D ranges. The series are shown as formula strings for clarity but you can also use the a list syntax.
```#######################################################################
#
# A demo of a clustered category chart in XlsxWriter.
#
# Copyright 2013-2023, John McNamara, jmcnamara@cpan.org
#
from xlsxwriter.workbook import Workbook
workbook = Workbook("chart_clustered.xlsx")
# Add the worksheet data that the charts will refer to.
headings = ["Types", "Sub Type", "Value 1", "Value 2", "Value 3"]
data = [
["Type 1", "Sub Type A", 5000, 8000, 6000],
["", "Sub Type B", 2000, 3000, 4000],
["", "Sub Type C", 250, 1000, 2000],
["Type 2", "Sub Type D", 6000, 6000, 6500],
["", "Sub Type E", 500, 300, 200],
]
for row_num, row_data in enumerate(data):
worksheet.write_row(row_num + 1, 0, row_data)
# Create a new chart object. In this case an embedded chart.
# Configure the series. Note, that the categories are 2D ranges (from column A
# to column B). This creates the clusters. The series are shown as formula
# strings for clarity but you can also use the list syntax. See the docs.
{
"categories": "=Sheet1!\$A\$2:\$B\$6",
"values": "=Sheet1!\$C\$2:\$C\$6",
}
)
{
"categories": "=Sheet1!\$A\$2:\$B\$6",
"values": "=Sheet1!\$D\$2:\$D\$6",
}
)
{
"categories": "=Sheet1!\$A\$2:\$B\$6",
"values": "=Sheet1!\$E\$2:\$E\$6",
}
)
# Set the Excel chart style.
chart.set_style(37)
# Turn off the legend.
chart.set_legend({"position": "none"})
# Insert the chart into the worksheet.
worksheet.insert_chart("G3", chart)
workbook.close()
``` | 519 | 1,749 | {"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} | 2.53125 | 3 | CC-MAIN-2023-50 | longest | en | 0.635226 |
https://www.r-bloggers.com/2010/09/effective-sample-size/ | 1,628,046,844,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154500.32/warc/CC-MAIN-20210804013942-20210804043942-00718.warc.gz | 956,540,827 | 22,568 | Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
In the previous days I have received several emails asking for clarification of the effective sample size derivation in “Introducing Monte Carlo Methods with R” (Section 4.4, pp. 98-100). Formula (4.3) gives the Monte Carlo estimate of the variance of a self-normalised importance sampling estimator (note the change from the original version in Introducing Monte Carlo Methods with R ! The weight W is unnormalised and hence the normalising constant $kappa$ appears in the denominator.)
$frac{1}{n},mathrm{var}_f (h(X)) left{1+dfrac{mathrm{var}_g(W)}{kappa^2}right}$
as
$dfrac{sum_{i=1}^n omega_i left{ h(x_i) - delta_h^n right}^2 }{nsum_{i=1}^n omega_i} , left{ 1 + n^2,widehat{mathrm{var}}(W)Bigg/ left(sum_{i=1}^n omega_i right)^2 right},.$
Now, the front term is somehow obvious so let us concentrate on the bracketed part. The empirical variance of the $omega_i$‘s is
$frac{1}{n},sum_{i=1}^nomega_i^2-frac{1}{n^2}left(sum_{i=1}^nomega_iright)^2 ,,$
the coefficient $1+widehat{mathrm{var}}_g(W)/kappa^2$ is thus estimated by
$n,sum_{i=1}^n omega_i^2 bigg/ left(sum_{i=1}^n omega_iright)^2,.$
which leads to the definition of the effective sample size
$text{ESS}_n=left(sum_{i=1}^nomega_iright)^2bigg/sum_{i=1}^nomega_i^2,.$
The confusing part in the current version is whether or not we use normalised W’s and $omega_i$‘s. I hope this clarifies the issue!
Filed under: Books, R, Statistics Tagged: effective sample size, importance sampling, Introducing Monte Carlo Methods with R | 493 | 1,592 | {"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": 18, "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} | 3.234375 | 3 | CC-MAIN-2021-31 | longest | en | 0.743255 |
https://mypages.unh.edu/jdmayer/graphics | 1,679,640,867,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945248.28/warc/CC-MAIN-20230324051147-20230324081147-00768.warc.gz | 480,498,824 | 8,372 | Graphics
This page contains some of the figures and other graphics that our lab uses to communicate our findings. Some of them are open-source, i.e., are protected under Creative Commons licenses.
Graphic 1. Some Basics Regarding Factor Models of Correlation Matrices
Mayer, J. D. and Bryan (2021, April 30). Some basics regarding factor models of correlation matrices. Personality Laboratory at the University of New Hampshire. https://mypages.unh.edu/jdmayer/graphics.
Notes: The table illustrates some basic principles regarding how correlations can be modeled in factor analysis. The spanner headings indicate that the first row depicts an example of a one-factor case; the second row, a two-factor case, and the third row depicts a hierachical relation with one factor at the top and subsidiary factors beneath it. Each of the three rows begins with a somewhat different matrix that represents the correlations among the four mental ability tasks, T1 through T4. In this hypothetical example, the four tasks were administered to 1000 participants. Any of the three correlation matrices of Column 1 might have arisen as a consequence.
The middle column of the table depicts the results of a factor analysis of the matrix. The roman numerals designate the first and second factors extracted for the one- and two-factor cases illustrated in the first two rows. The numbers beneath them represent the estimated correlations between the four individual tasks and the factor(s); those correlations are referred to as factor loadings. The final column represents a diagrammatic depiction of the results. More details: The three correlation matrices were factor analyzed using R. The exploratory solutions (top rows) were conducted with the psych package using a principal axis extraction followed, for the two-factor case, by a oblimin rotation. The bottom, hierarchical CFA was conducted in Lavaan using a Maximum Likelihood extraction. Fit statistics were 1.0 for both the CFI and TLI and 0 for the RMSEA.
The R code for the three examples can be found on this website here.
The following is an ancillary analysis of the open-source data from Bryan & Mayer, 2020 that illustrates how each of ten broad intelligences loads on a general intelligence factor (g).
Citation in APA Style:
Bryan, V. M., & Mayer, J. D. (2021, April 30). Ten broad Intelligences and their relations to general intelligence: An ancillary analysis of the open-source data from Bryan and Mayer, 2020. Personality Laboratory at the University of New Hampshire. https://mypages.unh.edu/jdmayer/graphics.
Graphic 3. The Four Areas of Personality Function from the Personality Systems Set
Here is an overview of the functional areas of the personality systems framework, as typically envisioned in our work.
Citation in APA Style: Mayer, 2020, "The Four Function Areas of the Personality Systems Set". Online lab graphic from https://mypages.unh.edu/jdmayer/graphics
Graphic 4. People-Centered Versus Thing-Centered Intelligences in the Three-Stratum Model of Intelligence
The following graphic elaborates on the model described in Mayer, J. D. (2018). Intelligences about things and intelligences about people. In R. J. Sternberg (Ed.). The nature of human intelligence (pp. 270-286). Cambridge, England: Cambridge University Press, especially on pp. 272 to 274, and 280.
Courtesy of Taylor & Francis Publishing...50 free PDF reprints
Limited Free Reprints of Recent Articles Available from Publishers
Posted 17 June 2020: From Taylor & Francis...50 free PDF reprints of just-published Mayer, J. D. (2019) An integrated approach to personality assessment based on the personality systems framework. Journal of Personality Assessment. Use this link for the PDF:
https://www.tandfonline.com/eprint/G6K5XYFIWCRTGKBYQMI2/full?target=10.1080/00223891.2018.1555539
Posted (approx.) June 2019...From Taylor & Francis Publishing...50 free PDF reprints of the just-published (2018) Employees High in Personal Intelligence Differ From Their Colleagues in Workplace Perceptions and Behavior (Journal of Personality Assessment). Use this link from the publisher: https://www.tandfonline.com/eprint/Y5VE8wQPj7aa8NYUA4we/full | 923 | 4,198 | {"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 | 0.930989 |
http://www.transum.org/Software/Mathswords/ | 1,550,477,587,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247484772.43/warc/CC-MAIN-20190218074121-20190218100121-00010.warc.gz | 450,214,618 | 7,823 | A pupil stands with their back to the board so they can't see the word(s) above. Their challenge is to guess the word from clues given by the class.
New word:
Comment recorded on the 10 September 'Starter of the Day' page by Carol, Sheffield PArk Academy:
"3 NQTs in the department, I'm new subject leader in this new academy - Starters R Great!! Lovely resource for stimulating learning and getting eveyone off to a good start. Thank you!!"
Comment recorded on the 12 July 'Starter of the Day' page by Miss J Key, Farlingaye High School, Suffolk:
"Thanks very much for this one. We developed it into a whole lesson and I borrowed some hats from the drama department to add to the fun!"
Comment recorded on the 9 May 'Starter of the Day' page by Liz, Kuwait:
"I would like to thank you for the excellent resources which I used every day. My students would often turn up early to tackle the starter of the day as there were stamps for the first 5 finishers. We also had a lot of fun with the fun maths. All in all your resources provoked discussion and the students had a lot of fun."
## Game 1
A pupil stands with his or her back to the screen and the teacher selects a word (Click the button above)
Members of the class give clues so that the person with their back to the screen can guess the word
## Game 2
If a large screen isn't available the teacher could select a word and a pupil could come up to the teacher's computer to read it.
This pupil then gives clues to the rest of the class to guess the word. Whoever guesses it first is the next to be the clue giver.
## Game 3
Pupils are divided into two teams.
One member of each team stands with their backs to the screen and the teacher selects a word (Click the button above).
The two teams take turns giving their team member a clue to the mystery word. The first team member to guess the word wins a point for their team.
## Game 4
Play as the party game "Charades". No talking allowed.
## Game 5
A pupil stands with his or her back to the screen and the teacher selects a word (Click the button above)
The pupil asks the rest of the class questions which they can only answer "Yes" or "No". The pupil attempts to guess the word with the minimum number of questions.
E.g. "Is it a shape?"
"Have we seen this word recently?"
"Is the word associated with probability?"
"Does it have less than five letters?"
## Game 6
Pupils could play the games suggested above in pairs if they have at least one computer, laptop, iPad or similar between them.
## Word Difficulty
The categories of Easy, Medium and Hard need explaining. The previous version of this application had the facility for teachers to vote on each word that was randomly selected from our database of 559 mathematical words and phrases. The votes indicated how suitable the teachers thought the words were for this game.
After collecting over 85000 of these votes we have arbitrarily divided up the database in to three sections accordingly. You will probably find some words are not in the category that you would put them in so we suggest you just skip those words and go on to select another.
#### Beat The Clock
It is a race against the clock to answer 30 mental arithmetic questions. There are nine levels to choose from to suit pupils of different abilities.
#### Hot Numbers
Move the numbered cards to form five 2-digit numbers which obey the given rules. A ten-stage numeracy challenge. So far this activity has been accessed 22683 times and 28 people have earned a Transum Trophy for completing it.
For Students:
For All: | 788 | 3,585 | {"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-2019-09 | latest | en | 0.97719 |
https://www.esaral.com/q/solve-each-of-the-following-systems-of-equations-by-the-method-of-cross-multiplication-92327 | 1,712,917,606,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296815919.75/warc/CC-MAIN-20240412101354-20240412131354-00815.warc.gz | 672,906,209 | 11,898 | # Solve each of the following systems of equations by the method of cross-multiplication :
Question:
Solve each of the following systems of equations by the method of cross-multiplication :
$\frac{x+y}{x y}=2, \frac{x-y}{x y}=6$
Solution:
GIVEN:
$\frac{x+y}{x y}=2$
$\frac{x-y}{x y}=6$
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
$\frac{x+y}{x y}=2$
$\frac{1}{x}+\frac{1}{y}=2$
$\frac{1}{x}+\frac{1}{3}-2=0$...(1)
$\frac{x-y}{x y}=6$
$\frac{1}{y}-\frac{1}{x}=6$
$\frac{1}{y}-\frac{1}{x}-6=0$ ...(2)
Let
$u=\frac{1}{x}$ and $\frac{1}{y}=v$
$u+v-2=0$...(3)
$-u+v-6=0$...$(4)$
By cross multiplication method we get
So $\frac{v}{8}=\frac{1}{2}$
$v=4$
We know that
$-2=\frac{1}{x}$ and $\frac{1}{y}=4$
$\Rightarrow x=-\frac{1}{2}$ and $y=\frac{1}{4}$
Hence we get the value of $x=-\frac{1}{2}$ and $y=\frac{1}{4}$ | 352 | 925 | {"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-2024-18 | longest | en | 0.676715 |
http://mathhelpforum.com/pre-calculus/40856-exponential-growth-decay-word-problem-print.html | 1,527,056,891,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865450.42/warc/CC-MAIN-20180523043959-20180523063959-00290.warc.gz | 185,712,674 | 2,944 | # exponential growth/decay word problem
• Jun 6th 2008, 08:38 PM
cityismine
exponential growth/decay word problem
When exposed to sunlight, the # of bacteria in a culture decreases exponentially at the rate of 10% per hour. What is the best approximation for the number of hours required for the initial # of bacteria to decrease by 50%.
• Jun 6th 2008, 09:54 PM
earboth
Quote:
Originally Posted by cityismine
When exposed to sunlight, the # of bacteria in a culture decreases exponentially at the rate of 10% per hour. What is the best approximation for the number of hours required for the initial # of bacteria to decrease by 50%.
Let A(0) denote the initial amount of bacteria and A(t) the amount of bacteria after t hours of exposion to sun light.
After 1 h there remains 90% of the bacteria. After t hours remains $\displaystyle 0.9^t$ of the bacteria.
You know that $\displaystyle \frac{A(t)}{A(0)}=\frac12$
Use the equation
$\displaystyle A(t) = A(0) \cdot (0.9)^t$
plug in the values you know and solve for t.
I've got t is roughly 6h 35 min
• Jun 6th 2008, 09:54 PM
TKHunny
Learning to translate is the point, here.
"# of bacteria in a culture"
Define this. B(t) = # of bacteria in a culture - at time t in hours.
"decreases exponentially"
A clue about $\displaystyle B(t) = ae^{bt}$
"decreases exponentially at the rate of 10% per hour."
A clue about the parameter: $\displaystyle B(1) = ae^{b} = 0.90*ae^{0} = B(0)$
"number of hours required for the initial # of bacteria to decrease by 50%."
B(x) = $\displaystyle 0.90^{x} = 0.50$ -- Solve for x. | 454 | 1,576 | {"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} | 4.1875 | 4 | CC-MAIN-2018-22 | latest | en | 0.867037 |
https://math.stackexchange.com/questions/3154011/subgroup-of-non-simple-group-necessarily-non-simple | 1,560,921,648,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998913.66/warc/CC-MAIN-20190619043625-20190619065625-00351.warc.gz | 510,076,070 | 34,219 | # Subgroup of non-simple group necessarily non-simple?
Let $$G$$ be an infinite group that is not simple. If $$B$$ is a subgroup of $$G$$, must $$B$$ necessarily be not simple as well?
I know that if $$N$$ is some proper normal subgroup of $$G$$, then $$B\cap N$$ is a normal subgroup of $$B$$. But we may also have that $$B$$ intersects trivially with $$N$$.
If no, is there a counter example? Would be grateful if anyone can enlighten me.
Hint: Try $$G = A_5 \times \mathbb Z$$. | 134 | 484 | {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2019-26 | latest | en | 0.852092 |
https://trafficsteed.com/qa/question-how-do-i-convert-kb-to-mb.html | 1,600,835,237,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400209665.4/warc/CC-MAIN-20200923015227-20200923045227-00115.warc.gz | 641,410,986 | 8,653 | # Question: How Do I Convert KB To MB?
## Is a MB 1000 or 1024 KB?
1MB is 1,024 kilobytes, or 1,048,576 (1024×1024) bytes, not one million bytes.
Similarly, one 1GB is 1,024MB, or 1,073,741,824 (1024x1024x1024) bytes.
A terabyte (TB) is 1,024GB; 1TB is about the same amount of information as all of the books in a large library, or roughly 1,610 CDs worth of data..
## How do you convert KB to MB?
1 Kilobyte is equal to 0.0009765625 megabytes (binary). 1 KB = 2-10 MB in base 2….Kilobytes vs Megabytes.Kilobytes (KB)Megabytes (MB)1,000 bytes1,000,000 bytes210 bytes (base 2)220 bytes (base 2)1,024 bytes1,048,576 bytes1,000 × 8 bits1,000,000 × 8 bits3 more rows
## How do you convert KB to GB manually?
If you want to convert mb to kb, just apply this formula: =A2*1024….Convert between kb and mb, gb, tb and vice versa with formulas.KB to GB:=A2/1024^2GB to KB:=A2*1024^2KB to TB:=A2/1024^3TB to KB:=A2*1024^3
## Is 1 kb a lot of data?
One kilobyte (KB) is a collection of about 1000 bytes. A page of ordinary Roman alphabetic text takes about 2 kilobytes to store (about one byte per letter). … In non-roman alphabets, such as Mandarin, the storage takes up 2 or 4 bytes per “letter” which is still pretty compact compared to audio and images.
## Why is it 1024 KB in a MB?
In most cases, this approximation is fine for determining how much space a file takes up or how much disk space you have. But there are really 1024 kilobytes in a megabyte. The reason for this is because computers are based on the binary system. That means hard drives and memory are measured in powers of 2.
## Is a KB smaller than MB?
The unit symbol of Megabyte is MB. 1 KB (Kilobyte) is equal to 0.001 MB in decimal and 0.0009765625 MB in binary. It also means that 1 megabyte is equal to 1000 kilobytes in decimal and 1024 kilobytes in binary. … So as you can see, a Megabyte is one thousand times bigger than a Kilobyte.
## How do you calculate KB MB GB TB?
FAQ >> Understanding file sizes (Bytes, KB, MB, GB, TB)1024 bytes. =1 KB.1024 KB. =1 MB.1024 MB. =1 GB.1024 GB. =1 TB.More items…
## Which is more MB or GB?
1 megabyte consists of one million bytes of information. … 1 Gigabyte is considered to be equal to 1000 megabytes in decimal and 1024 megabytes in binary system. As you can see, 1 Gigabyte is 1000 times bigger than a Megabyte. So, a GB is bigger than a MB.
## Is 1 MB 1000 KB or 1024 KB?
Therefore, 1 kB = 8000 bit. One thousand kilobytes (1000 kB) is equal to one megabyte (1 MB), where 1 MB is one million bytes.
## What is difference between KB and KB?
One kilobit (abbreviated “Kb”) contains one thousand bits and there are 1,000 kilobits in a megabit. Kilobits (Kb) are smaller than than kilobytes (KB), since there are 8 kilobits in a single kilobyte. Therefore, a kilobit is one-eighth the size of a kilobyte.
## Which is larger KB or GB?
A kilobyte (KB) is 1,000 bytes, and one megabyte (MB) is 1,000 kilobytes. One gigabyte (GB) is equal to 1,000 megabytes, while a terabyte (TB) is 1,000 gigabytes.
## How many KB is a GB of data?
For ease of calculation, we’ll say there are 1000 kilobytes (KB) in a megabyte (MB) and 1000 MB in a gigabyte (GB), often referred to as a gig of data. These figures are based on using mobile data on your smartphone and don’t take tethering into account.
## How many kb in a MB in a GB?
1,024Data Measurement ChartData MeasurementSizeKilobyte (KB)1,024 BytesMegabyte (MB)1,024 KilobytesGigabyte (GB)1,024 Megabytes5 more rows
## How do I reduce the MB and KB file size?
How to reduce the image size in KB/MB?To reduce the image size in KB or MB online, first upload it to ResizePixel’s website.Enter a desired file size and select the corresponding unit of measurement (KB or MB).Then proceed to Download page to get the image file.
## Is MB bigger than KB?
KB, MB, GB – A kilobyte (KB) is 1,024 bytes. A megabyte (MB) is 1,024 kilobytes. A gigabyte (GB) is 1,024 megabytes. A terabyte (TB) is 1,024 gigabytes. | 1,232 | 3,982 | {"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-2020-40 | latest | en | 0.821693 |
https://dev.to/bello/comments-and-code-refactor-276h | 1,621,261,266,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991772.66/warc/CC-MAIN-20210517115207-20210517145207-00419.warc.gz | 232,165,542 | 42,332 | ## DEV Community is a community of 620,905 amazing developers
We're a place where coders share, stay up-to-date and grow their careers.
Bello Updated on ・4 min read
There are two types of comments in JavaScript:
• Inline comments (`//...`) - inline to statements, expressions, etc
• Multi-line comments (`/*...*/`) - span multiple lines;
Comments are notes describing how and why the code works. It is crucial as a reminder to the creator of the code and when working with other developers.
If the code has anything subtle and counter-intuitive, it's definitely worth commenting.
There are times when novices use them wrongly.
See the example below:
``````function showPrimes(n) {
// nextPrime is just a label
nextPrime:
for (let i = 2; i < n; i++) {
/*
The iteration starts at 2 through to n. For example, if n is
10, iteration starts from iteration 2 to iteration 9. 9 is
less than 10
*/
for (let j = 2; j < i; j++) {
// iteration j should be less than i
if (i % j === 0 && n % 2 === 0) continue nextPrime;
/* if iteration of i and j modulus equals zero then it's a
prime number */
}
console.log(i); // 2, 3, ...7
}
}
showPrimes(10)
``````
Comments should be logical (when necessary) but short.
See the example below:
``````function showPrimes(n) {
nextPrime:
for (let i = 2; i < n; i++) {
// i = 2, 3, 4, 5, 6, 7, 8, 9; n = 10
// i = i2, ...i9
for (let j = 2; j < i; j++) {
// j = 2, (2, 3), (2, 3, 4), (2, 3, 4, 5)..., (2, 3, ...8)
// i2 = j1
// i3 = j1, j2
// i4 = j1, j2, j3
// i5 = j1, j2, j3, j4
// ...
// i9 = j1, ...j8
if (i % j === 0 /** && n % 2 === 0 **/) continue nextPrime;
// i % j => 3, 3, 3, 3, 5, 5, 7
}
console.log(i); // 2, 3, 5, 7
}
}
showPrimes(10);
``````
The comment above looks logical but overused.
See the example below of reduced comments:
``````function showPrimes(n) {
nextPrime:
for (let i = 2; i < n; i++) {
// i = i2, ...i(n-1)
for (let j = 2; j < i; j++) {
// in = j1, ...j(n-2)
if (i % j === 0 /** && n % 2 === 0 **/) continue nextPrime;
// R(i/j)
}
console.log(i);
}
}
showPrimes(10);
``````
Not only the comment above is short, but also precise.
• Describe the architecture
• Document function parameters and usage
``````/**
* Returns x raised to the n-th power.
*
* @param {number} x The number to raise.
* @param {number} n The power, must be a natural number.
* @return {number} x raised to the n-th power.
*/
function pow(x, n) {
// ...
}
``````
• such comments above make the function understandable without looking in its code
• some editors like WebStorm understand comments to provide autocomplete and some automatic code-checking to ease workflow
• tools like JSDoc 3 that can generate HTML-documentation from the comments. The official documentation is at usejsdoc.org/
Most of the time comments are used in development only but removed in production especially when minified.
## Code refactor
Code refactoring is the process of modifying the code structure without affecting the functionality for readability purposes.
See the example below:
``````function showPrimes(n) {
nextPrime:
for (let i = 2; i < n; i++) {
for (let j = 2; j < i; j++) {
if (i % j === 0) continue nextPrime;
}
console.log(i);
}
}
showPrimes(10);
``````
To refactor or make the code more readable above see the example below:
``````function showPrimes(n) {
for (let i = 2; i < n; i++) {
if (!isPrime(i)) continue;
console.log(i);
}
}
function isPrime(n) {
for (let i = 2; i < n; i++) {
if (n % i == 0) return false;
}
return true;
}
showPrimes(10);
``````
The code above looks easier to understand because it has lesser nesting and better factored out function `isPrime`.
The function itself becomes the comment. Such code is called self-descriptive.
Refactoring split code structure to make functions more understand.
Happy coding!
### TechStack | Domain
• Purchase a `.com` domain name as low as \$9.99.
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• Get cheaper domain names as low as \$3.
• Build a website with ease. | 1,207 | 4,002 | {"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-2021-21 | latest | en | 0.704741 |
http://www.math-only-math.com/sines-and-cosines-of-multiples-or-submultiples.html | 1,534,539,478,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221212910.25/warc/CC-MAIN-20180817202237-20180817222237-00412.warc.gz | 517,731,181 | 8,518 | # Sines and Cosines of Multiples or Submultiples
We will learn how to solve identities involving sines and cosines of multiples or submultiples of the angles involved.
We use the following ways to solve the identities involving sines and cosines.
(i) Take the first two terms of L.H.S. and express the sum of two sines (or cosines) as product.
(ii) In the third term of L.H.S. apply the formula of sin 2A (or cos 2A).
(iii) Then use the condition A + B + C = π and take one sine (or cosine) term common.
(iv) Finally, express the sum or difference of two sines (or cosines) in the brackets as product.
1. If A + B + C= π prove that,
sin A + sin B - sin C = 4 sin $$\frac{A}{2}$$ sin $$\frac{B}{2}$$ cos $$\frac{C}{2}$$
Solution:
We have,
A + B + C = π
⇒ C = π - (A + B)
⇒ $$\frac{C}{2}$$ = $$\frac{π }{2}$$ - ($$\frac{A + B}{2}$$)
Therefore, sin ($$\frac{A + B}{2}$$) = sin ($$\frac{π }{2}$$ - $$\frac{C}{2}$$) = cos $$\frac{C}{2}$$
Now, L.H.S. = sin A + sin B - sin C
= (sin A + sin B) - sin C
= 2 sin ($$\frac{A + B}{2}$$) cos ($$\frac{A - B}{2}$$) - sin C
= 2 sin ($$\frac{π - C}{2}$$) cos ($$\frac{A - B}{2}$$) - sin C
= 2 sin ($$\frac{π}{2}$$ - $$\frac{C}{2}$$) cos $$\frac{A - B}{2}$$ - sin C
= 2 cos $$\frac{C}{2}$$ cos $$\frac{A - B}{2}$$ - sin C
= 2 cos $$\frac{C}{2}$$ cos $$\frac{A - B}{2}$$ - 2 sin $$\frac{C}{2}$$ cos $$\frac{C}{2}$$
= 2 cos $$\frac{C}{2}$$[cos $$\frac{A - B}{2}$$ - sin $$\frac{C}{2}$$]
= 2 cos $$\frac{C}{2}$$[cos $$\frac{A - B}{2}$$ - sin ($$\frac{π}{2}$$ - $$\frac{A + B}{2}$$)]
= 2 cos $$\frac{C}{2}$$[cos ($$\frac{A - B}{2}$$) - cos ($$\frac{A + B}{2}$$)]
= 2 cos $$\frac{C}{2}$$[cos ($$\frac{A}{2}$$ - $$\frac{B}{2}$$) - cos ($$\frac{A}{2}$$ + $$\frac{B}{2}$$)]
= 2 cos $$\frac{C}{2}$$ [(cos $$\frac{A}{2}$$ cos $$\frac{B}{2}$$ + sin $$\frac{A}{2}$$ sin $$\frac{B}{2}$$) - (cos $$\frac{A}{2}$$ cos $$\frac{B}{2}$$ + sin $$\frac{A}{2}$$ sin $$\frac{B}{2}$$)]
= 2 cos $$\frac{C}{2}$$[2 sin $$\frac{A}{2}$$ sin $$\frac{B}{2}$$]
= 4 sin $$\frac{A}{2}$$ sin $$\frac{B}{2}$$ cos $$\frac{C}{2}$$ = R.H.S. Proved.
2. If A, B, C be the angles of a triangle, prove that,
cos A + cos B + cos C = 1 + 4 sin $$\frac{A}{2}$$ sin $$\frac{B}{2}$$ sin $$\frac{C}{2}$$
Solution:
Since A, B, C are the angles of a triangle,
Therefore, A + B + C = π
⇒ C = π - (A + B)
⇒ $$\frac{C}{2}$$ = $$\frac{π }{2}$$ - ($$\frac{A + B}{2}$$)
Thus, cos ($$\frac{A + B}{2}$$) = cos ($$\frac{π }{2}$$ - $$\frac{C}{2}$$) = sin $$\frac{C}{2}$$
Now, L. H. S. = cos A + cos B + cos C
= (cos A + cos B) + cos C
= 2 cos ($$\frac{A + B}{2}$$) cos ($$\frac{A - B}{2}$$) + cos C
= 2 cos ($$\frac{π}{2}$$ - $$\frac{C}{2}$$) cos ($$\frac{A - B}{2}$$) + cos C
= 2 sin $$\frac{C}{2}$$ cos ($$\frac{A - B}{2}$$) + 1 - 2 sin$$^{2}$$ $$\frac{C}{2}$$
= 2 sin $$\frac{C}{2}$$ cos ($$\frac{A - B}{2}$$) - 2 sin$$^{2}$$ $$\frac{C}{2}$$ + 1
= 2 sin $$\frac{C}{2}$$[cos ($$\frac{A - B}{2}$$) - sin $$\frac{C}{2}$$] + 1
= 2 sin $$\frac{C}{2}$$[cos ($$\frac{A - B}{2}$$) - sin ($$\frac{π}{2}$$ - $$\frac{A + B}{2}$$)] + 1
= 2 sin $$\frac{C}{2}$$[cos ($$\frac{A - B}{2}$$) - cos ($$\frac{A + B}{2}$$)] + 1
= 2 sin $$\frac{C}{2}$$ [2 sin $$\frac{A}{2}$$ sin $$\frac{B}{2}$$] + 1
= 4 sin $$\frac{C}{2}$$ sin $$\frac{A}{2}$$ sin $$\frac{B}{2}$$ + 1
= 1 + 4 sin $$\frac{A}{2}$$ sin $$\frac{B}{2}$$ sin $$\frac{C}{2}$$ Proved.
3. If A + B + C = π prove that,
sin $$\frac{A}{2}$$ +sin $$\frac{B}{2}$$ + sin $$\frac{C}{2}$$ = 1 + 4 sin $$\frac{π - A}{4}$$ sin $$\frac{π - B}{4}$$ sin $$\frac{π - C}{4}$$
Solution:
A + B + C = π
⇒ $$\frac{C}{2}$$ = $$\frac{π}{2}$$ - $$\frac{A + B}{2}$$
Therefore, sin $$\frac{C}{2}$$ = sin ($$\frac{π }{2}$$ - $$\frac{A + B}{2}$$) = cos $$\frac{A + B}{2}$$
Now, L. H. S. = sin $$\frac{A}{2}$$ +sin $$\frac{B}{2}$$ + sin $$\frac{C}{2}$$
= 2 sin $$\frac{A + B}{4}$$ cos $$\frac{A - B}{4}$$ + cos ($$\frac{π}{2}$$ - $$\frac{C}{2}$$)
= 2 sin $$\frac{π - C}{4}$$ cos $$\frac{A - B}{4}$$ + cos $$\frac{π - C}{2}$$
= 2 sin $$\frac{π - C}{4}$$ cos $$\frac{A - B}{4}$$ + 1 – 2 sin$$^{2}$$ $$\frac{π - C}{4}$$
= 2 sin $$\frac{π - C}{4}$$ cos $$\frac{A - B}{4}$$ - 2 sin$$^{2}$$ $$\frac{π - C}{4}$$ + 1
= 2 sin $$\frac{π - C}{4}$$ [cos $$\frac{A - B}{4}$$ - sin $$\frac{π - C}{4}$$] + 1
= 2 sin $$\frac{π - C}{4}$$ [cos $$\frac{A - B}{4}$$ - cos {$$\frac{π}{2}$$ - $$\frac{π - C}{4}$$}] + 1
= 2 sin $$\frac{π - C}{4}$$ [cos $$\frac{A - B}{4}$$ - cos ($$\frac{π}{4}$$ + $$\frac{C}{4}$$)] + 1
= 2 sin $$\frac{π - C}{4}$$ [cos $$\frac{A - B}{4}$$ - cos $$\frac{π + C}{4}$$] + 1
= 2 sin $$\frac{π - C}{4}$$ [2 sin $$\frac{A - B + π + C}{8}$$ sin $$\frac{π + C - A + B}{8}$$] + 1
= 2 sin $$\frac{π - C}{4}$$ [2 sin $$\frac{A + C + π - B}{8}$$ sin $$\frac{B + C + π - A}{8}$$] + 1
= 2 sin $$\frac{π - C}{4}$$ [2 sin $$\frac{π - B + π - B}{8}$$ sin $$\frac{π - A + π - A}{8}$$] + 1
= 2 sin $$\frac{π - C}{4}$$ [2 sin $$\frac{π - B}{4}$$ sin $$\frac{π - A}{4}$$] + 1
= 4 sin $$\frac{π - C}{4}$$ sin $$\frac{π - B}{4}$$ sin $$\frac{π - A}{4}$$ + 1
= 1 + 4 sin $$\frac{π - A}{4}$$ sin $$\frac{π - B}{4}$$ sin $$\frac{π - C}{4}$$ Proved.
4. If A + B + C = π show that,
cos $$\frac{A}{2}$$ + cos $$\frac{B}{2}$$ + cos $$\frac{C}{2}$$ = 4 cos $$\frac{A + B}{4}$$ cos $$\frac{B + C}{4}$$ cos $$\frac{C + A}{4}$$
Solution:
A + B + C = π
$$\frac{C}{2}$$ = $$\frac{π}{2}$$ - $$\frac{A + B}{2}$$
Therefore, cos $$\frac{C}{2}$$ = cos ($$\frac{π}{2}$$ - $$\frac{A + B}{2}$$) = sin $$\frac{A + B}{2}$$
Now, L. H. S. = cos $$\frac{A}{2}$$ + cos $$\frac{B}{2}$$ + cos $$\frac{C}{2}$$
= (cos $$\frac{A}{2}$$ + cos $$\frac{B}{2}$$) + cos $$\frac{C}{2}$$
= 2 cos $$\frac{A + B}{4}$$ cos $$\frac{A - B}{4}$$ + sin $$\frac{A + B}{2}$$ [Since, cos $$\frac{C}{2}$$ = sin $$\frac{A + B}{2}$$]
= 2 cos $$\frac{A + B}{4}$$ cos $$\frac{A - B}{4}$$ + 2 sin $$\frac{A + B}{4}$$ cos $$\frac{A + B}{4}$$
= 2 cos $$\frac{A + B}{4}$$[cos $$\frac{A - B}{4}$$ + sin $$\frac{A + B}{4}$$]
= 2 cos $$\frac{A + B}{4}$$ [cos $$\frac{A + B}{4}$$ + cos ($$\frac{π}{2}$$ - $$\frac{A + B}{4}$$)]
= 2 cos $$\frac{A + B}{4}$$ [2 cos $$\frac{\frac{A - B}{4} + \frac{π}{2} - \frac{A + B}{4}}{2}$$ cos $$\frac{\frac{π}{2} - \frac{A + B}{4} - \frac{A - B}{4}}{2}$$]
= 2 cos $$\frac{A + B}{4}$$ [2 cos $$\frac{π - B}{4}$$ cos $$\frac{π - A}{4}$$]
= 4 cos $$\frac{A + B}{4}$$ cos $$\frac{C + A}{4}$$ cos $$\frac{B + C}{4}$$, [Since, π - B = A + B + C - B = A + C; Similarly, π - A = B + C]
= 4 cos $$\frac{A + B}{4}$$ cos $$\frac{B + C}{4}$$ cos $$\frac{C + A}{4}$$. Proved. | 3,254 | 6,620 | {"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} | 4.875 | 5 | CC-MAIN-2018-34 | latest | en | 0.231433 |
https://academic-writing.net/2021/03/15/area-and-perimeter-problem-solving-worksheets_3v/ | 1,618,093,388,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038059348.9/warc/CC-MAIN-20210410210053-20210411000053-00320.warc.gz | 203,936,959 | 5,765 | Area and perimeter problem solving worksheets
By | March 15, 2021
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https://codehs.com/course/videogamedesign/outline | 1,716,114,469,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057786.92/warc/CC-MAIN-20240519101339-20240519131339-00138.warc.gz | 148,006,323 | 68,576 | 1. ## Programming with Karel
1. ### 1.1 Introduction to Programming With Karel
2. Video 1.1.1 Introduction to Programming With Karel
3. Check for Understanding 1.1.2 Karel Commands Quiz
4. Example 1.1.3 Our First Karel Program
5. Exercise 1.1.4 Your First Karel Program
6. Exercise 1.1.5 Short Stack
7. ### 1.2 More About Karel
8. Video 1.2.1 More About Karel
9. Check for Understanding 1.2.2 More Basic Karel Quiz
10. Example 1.2.3 Tennis Ball Square
11. Exercise 1.2.4 Make a Tower
12. Exercise 1.2.5 Pyramid of Karel
13. ### 1.3 Karel Can't Turn Right
14. Video 1.3.1 Karel Can't Turn Right
15. Check for Understanding 1.3.2 Karel Can't Turn Right Quiz
16. Example 1.3.3 Tower and Turn Right
17. Exercise 1.3.4 Slide Karel
18. Exercise 1.3.5 Fireman Karel
20. ### 1.4 Functions in Karel
21. Video 1.4.1 Functions in Karel
22. Check for Understanding 1.4.2 Functions in Karel Quiz
23. Example 1.4.3 Turn Around
24. Exercise 1.4.4 Pancakes
25. Exercise 1.4.5 Mario Karel
26. ### 1.5 The Main Function
27. Video 1.5.1 The Main Function
28. Check for Understanding 1.5.2 The Main Function Quiz
29. Example 1.5.3 Tower with Main Function
30. Exercise 1.5.4 Pancakes with Main
31. ### 1.6 Top Down Design and Decomposition in Karel
32. Video 1.6.1 Top Down Design and Decomposition
33. Check for Understanding 1.6.2 Top Down Design and Decomposition Quiz
34. Video 1.6.3 Top Down Design and Decomposition in Karel
35. Example 1.6.4 Hurdle Karel
36. Exercise 1.6.5 The Two Towers
37. ### 1.7 Commenting Your Code
38. Video 1.7.1 Commenting Your Code
39. Check for Understanding 1.7.2 Commenting Your Code Quiz
40. Example 1.7.3 Hurdle Karel
41. Exercise 1.7.4 The Two Towers + Comments
42. ### 1.8 Super Karel
43. Video 1.8.1 Super Karel
44. Check for Understanding 1.8.2 Super Karel Quiz
45. Example 1.8.3 Hurdle Karel (with Super Karel)
46. Exercise 1.8.4 The Two Towers + Super Karel
48. ### 1.9 For Loops
49. Video 1.9.1 For Loops
50. Check for Understanding 1.9.2 For Loops Quiz
51. Example 1.9.3 Repeated Move
52. Example 1.9.4 Put Down Tennis Balls
53. Exercise 1.9.5 Take 'em All
54. Exercise 1.9.6 Dizzy Karel
55. Exercise 1.9.7 Ball in Each Corner
56. Exercise 1.9.8 Lots of Hurdles
57. ### 1.10 If Statements and Conditionals
58. Video 1.10.1 If Statements and Conditionals
59. Check for Understanding 1.10.2 If Statements and Conditionals Quiz
60. Example 1.10.3 If Statements and Conditionals
61. Example 1.10.4 Safe Take Ball
62. Exercise 1.10.5 Is There a Ball?
63. Exercise 1.10.6 Don't Crash!
64. ### 1.11 If/Else Statements
65. Video 1.11.1 If/Else Statements
66. Check for Understanding 1.11.2 If/Else Statements Quiz
67. Example 1.11.3 If/Else Statements
68. Example 1.11.4 Opposite Day
69. Exercise 1.11.5 Right Side Up
70. Exercise 1.11.6 Right vs. Left Square
72. ### 1.12 While Loops
73. Video 1.12.1 While Loops
74. Check for Understanding 1.12.2 While Loops Quiz
75. Example 1.12.3 Move to Wall
77. Exercise 1.12.5 Lay Row of Tennis Balls
78. Exercise 1.12.6 Big Tower
79. ### 1.13 How to Indent Your Code
80. Video 1.13.1 How to Indent Your Code
81. Check for Understanding 1.13.2 How to Indent Your Code Quiz
82. Example 1.13.3 Dance and Clean Karel
83. Exercise 1.13.4 Diagonal
84. Exercise 1.13.5 Staircase
86. ### 1.14 Control Structures Example
87. Video 1.14.1 Control Structures Example
88. Check for Understanding 1.14.2 Control Structures Example Quiz
89. Example 1.14.3 Cleanup Karel
90. Exercise 1.14.4 Random Hurdles
91. ### 1.15 More Karel Examples and Testing
92. Video 1.15.1 More Karel Examples and Testing
93. Example 1.15.2 Move Tennis Ball Stack
94. Video 1.15.3 Live Coding: Climbing Karel
95. Example 1.15.4 Climbing Karel
96. Check for Understanding 1.15.5 Quiz: Which Control Structure?
97. Exercise 1.15.6 Opposite Corner
98. ### 1.16 Challenge Problems
99. Challenge 1.16.1 Fetch
100. Challenge 1.16.2 Racing Karel
101. Challenge 1.16.3 Tower Builder
102. Challenge 1.16.4 Super Cleanup Karel
103. Challenge 1.16.5 Double Tennis Balls
105. ### 1.17 Programming with Karel Quiz
106. Unit Quiz 1.17.1 Programming with Karel Quiz
2. ## JavaScript Basics
1. ### 2.1 Hello World
2. Video 2.1.1 Hello World
3. Check for Understanding 2.1.2 Hello World Quiz
4. Example 2.1.3 Hello World
5. Exercise 2.1.4 Your Name and Hobby
6. Exercise 2.1.5 ASCII Animals
7. ### 2.2 Variables
8. Video 2.2.1 Variables
9. Video 2.2.2 Live Coding: Variables
10. Check for Understanding 2.2.3 Variables Quiz
11. Example 2.2.4 Basic Variables
12. Exercise 2.2.5 Daily Activities
13. Debugging 2.2.6 Debugging Variables
14. ### 2.3 User Input
15. Video 2.3.1 User Input
16. Check for Understanding 2.3.2 User Input Quiz
17. Example 2.3.3 Basic User Input
18. Exercise 2.3.4 Dinner Plans
21. ### 2.4 Basic Math
22. Video 2.4.1 Basic Math
23. Check for Understanding 2.4.2 Basic Math Quiz
24. Example 2.4.3 Simple Calculator
25. Example 2.4.4 Dollars to Pounds
26. Example 2.4.5 Dividing Up Groups
27. Exercise 2.4.6 T-Shirt Shop
28. Exercise 2.4.7 Running Speed
29. ### 2.5 Collaborative Programming
30. Video 2.5.1 Pair-Programming
31. Check for Understanding 2.5.2 Pair-Programming
32. Connection 2.5.3 Why Practice Pair-Programming?
33. Free Response 2.5.4 Pair-Programming Reflection
34. ### 2.6 Random Numbers
35. Video 2.6.1 Random Numbers
36. Check for Understanding 2.6.2 Random Numbers Quiz
37. Example 2.6.3 Rolling a Die
38. Exercise 2.6.4 Treasure Chest Loot
39. Exercise 2.6.5 Multiplication Practice
41. ### 2.7 Basic Functions
42. Video 2.7.1 Basic Functions
43. Notes 2.7.2 Variables in Functions
44. Quiz 2.7.3 Basic Functions Quiz
45. Example 2.7.4 Function Flow
46. Exercise 2.7.5 Digital Business Card
47. Exercise 2.7.6 ASCII Karel
48. ### 2.8 JavaScript Basics Quiz
49. Unit Quiz 2.8.1 JavaScript Basics Quiz
3. ## The Canvas and Graphics
1. ### 3.1 Intro to the Canvas and Graphics
2. Video 3.1.1 Intro to Canvas and Graphics
3. Notes 3.1.2 Debug Mode for Positioning
4. Video 3.1.3 Live Coding: Circle and Rectangle
5. Connection 3.1.4 Canvas Coordinates
6. Quiz 3.1.5 Canvas and Graphics Quiz
7. Example 3.1.6 Creating a Circle
8. Example 3.1.7 A Circle and a Rectangle
9. Exercise 3.1.8 A Ball in a Box
10. Exercise 3.1.9 Raise the Flag
11. ### 3.2 More Graphics Objects
12. Video 3.2.1 More Graphics Objects
13. Video 3.2.2 Live Coding: More Graphics Objects
14. Quiz 3.2.3 Graphics Objects Quiz
15. Example 3.2.4 Cute Animals
16. Example 3.2.5 Greetings, Earth!
17. Exercise 3.2.6 Exploration: XY Plot
18. Exercise 3.2.7 Create Your Meme
19. Exercise 3.2.8 Saturday Mornings
20. ### 3.3 Positioning Graphics Objects
21. Video 3.3.1 Positioning Graphics Objects
22. Quiz 3.3.2 Positioning Quiz
23. Example 3.3.3 8 Ball
24. Exercise 3.3.4 Color the Rainbow
25. Challenge 3.3.5 Create Your Own Plant!
26. ### 3.4 JavaScript Graphics Quiz
27. Quiz 3.4.1 JavaScript Graphics Quiz
4. ## Graphics Challenges
1. ### 4.1 Graphics Challenges
2. Challenge 4.1.1 Ghost
3. Challenge 4.1.2 Fried Egg
4. Challenge 4.1.3 Draw Something
5. ## Control Structures
1. ### 5.1 Booleans
2. Video 5.1.1 Booleans
3. Check for Understanding 5.1.2 Booleans Quiz
4. Example 5.1.3 Boolean Exploration
5. Exercise 5.1.4 Do You Have a Dog?
6. Free Response 5.1.5 Booleans are Questions
7. Exercise 5.1.6 Best Day Ever
8. ### 5.2 If/Else Statements
9. Video 5.2.1 If Statements
10. Check for Understanding 5.2.2 If Statements Quiz
11. Example 5.2.3 Are You Logged In?
12. Exercise 5.2.4 Is It Raining?
13. Exercise 5.2.5 Mood Playlist
15. ### 5.3 Logical Operators
16. Video 5.3.1 Logical Operators
17. Check for Understanding 5.3.2 Logical Operators Quiz
18. Example 5.3.3 Light Switch
19. Example 5.3.4 Harry Potter
21. Example 5.3.6 Logical Operators Game
22. Exercise 5.3.7 Can You Graduate?
23. Exercise 5.3.8 Switching Players
24. Exercise 5.3.9 A Day of Decisions
25. ### 5.4 Comparison Operators
26. Video 5.4.1 Comparison Operators
27. Check for Understanding 5.4.2 Comparison Operators Quiz
28. Example 5.4.3 Great Names
30. Example 5.4.5 Even and Odd
31. Exercise 5.4.6 Rolling Dice
32. Exercise 5.4.7 Teenagers
33. Exercise 5.4.8 Rocket Launch Requirements
34. Exercise 5.4.9 Trivia Game
35. ### 5.5 Graphics and Conditionals
36. Notes 5.5.1 Graphics and Conditionals
37. Example 5.5.2 Circle or Rectangle?
38. Exercise 5.5.3 Correct or Incorrect?
39. Notes 5.5.4 Else If Statements
40. Example 5.5.5 Conditional Circle Color
41. Exercise 5.5.6 Odd or Even Shapes
42. Quiz 5.5.7 Graphics and Conditionals Quiz
43. Challenge 5.5.8 Interactive Modern Art
44. ### 5.6 While Loops
45. Video 5.6.1 While Loops
46. Check for Understanding 5.6.2 While Loops Quiz
47. Example 5.6.3 While Loop Countdown
48. Debugging 5.6.4 Debugging: Best Name Ever
49. Exercise 5.6.5 Level Up
50. Exercise 5.6.6 Inventory
51. ### 5.7 The Break Statement
52. Video 5.7.1 The Break Statement
53. Check for Understanding 5.7.2 The Break Statement Quiz
54. Example 5.7.3 Adding Up Numbers
55. Free Response 5.7.4 Break Statement Reflection
56. Exercise 5.7.5 Snake Eyes
57. Exercise 5.7.6 Better Password Prompt
58. Exercise 5.7.7 Riddle Machine
59. ### 5.8 While Loops and Graphics
60. Notes 5.8.1 While Loops and Graphics
61. Example 5.8.2 Lots of Circles
62. Example 5.8.3 Corners on Corners
63. Exercise 5.8.4 Concentric Circles
64. Debugging 5.8.5 Debugging: Circle Positions
65. Exercise 5.8.6 Growing Squares
66. ### 5.9 For Loops
67. Video 5.9.1 For Loops
68. Example 5.9.2 For Loop Exploration
69. Exercise 5.9.3 Chalkboard
70. Example 5.9.4 Count By Twos
71. Example 5.9.5 Eating Apples
72. Debugging 5.9.6 Countdown by Sevens
73. Check for Understanding 5.9.7 For Loops Quiz
74. Exercise 5.9.8 Lives Left
75. Example 5.9.9 For Loop Sum
76. Exercise 5.9.10 Jukebox
77. ### 5.10 For Loops and Graphics
78. Notes 5.10.1 For Loops and Graphics
79. Example 5.10.2 Lots of Circles Revisited
80. Exercise 5.10.3 Exploration: Confetti
81. Notes 5.10.4 Using i to Position Objects and Adjust Size
82. Example 5.10.5 Horizontal Stripes #1: Using i to Adjust Position
83. Example 5.10.6 Horizontal Stripes #2: Using i to Adjust Size
84. Debugging 5.10.7 Debugging: Colorful Bullseye
85. Exercise 5.10.8 Caterpillar
87. ### 5.11 Javascript Control Structures Quiz
88. Unit Quiz 5.11.1 JavaScript Control Structures Quiz
6. ## Control Structures Challenges
1. ### 6.1 Control Structures Challenges
2. Challenge 6.1.1 Guessing Game
3. Challenge 6.1.2 Landscape Generator
4. Challenge 6.1.3 Exploring RGB Color Codes
7. ## Functions
1. ### 7.1 Parameters
2. Video 7.1.1 Parameters
3. Video 7.1.2 Live Coding: Parameters
4. Quiz 7.1.3 Parameters Quiz
5. Example 7.1.4 Greetings
6. Example 7.1.5 Slope of a Line
7. Example 7.1.6 Draw Lots of Circles!
8. Exercise 7.1.7 Area of Triangle
9. Exercise 7.1.8 Rainbow Revisited
10. Exercise 7.1.9 Cityscape
11. ### 7.2 Return Values
12. Video 7.2.1 Return Values
13. Quiz 7.2.2 Return Values Quiz
14. Example 7.2.3 Mathematical Returns
15. Example 7.2.4 Offscreen Graphics
16. Exercise 7.2.5 Max
17. Exercise 7.2.6 Overlapping Graphics
18. Exercise 7.2.7 Is It Even?
19. ### 7.3 Default Parameter Values
20. Video 7.3.1 Default Parameter Values
21. Quiz 7.3.2 Default Parameter Values Quiz
22. Example 7.3.3 Default Printing
23. Debugging 7.3.4 Farming International
24. Exercise 7.3.5 Compound Interest
25. Exercise 7.3.6 Default Face
26. ### 7.4 Variable Scopes
27. Video 7.4.1 Variable Scope
28. Video 7.4.2 Live Coding: Variable Scope
29. Quiz 7.4.3 Variable Scope Quiz
30. Example 7.4.4 Scope of X
31. Exercise 7.4.5 Exploration: Scope of Ball
32. Free Response 7.4.6 Scope Reflection
33. Challenge 7.4.7 Choose Wisely Game
35. ### 7.5 Functions Quiz
36. Unit Quiz 7.5.1 Functions and Parameters Quiz
8. ## Functions Challenges
1. ### 8.1 Functions Challenges
2. Challenge 8.1.1 Global Travel Assistant
3. Challenge 8.1.2 Balloons
4. Challenge 8.1.3 Ghost Invasion!
9. ## Animation and Games
1. ### 9.1 Timers
2. Video 9.1.1 Timers
3. Check for Understanding 9.1.2 Timers Quiz
4. Example 9.1.3 Moving Ball
5. Example 9.1.4 Magic 8 Ball
6. Exercise 9.1.5 Crazy Ball
7. Exercise 9.1.6 Paint Splatter
8. Notes 9.1.7 Project: Evasion (Timers)
9. Free Response 9.1.8 Project Info and Links
10. ### 9.2 Stopping Timers
11. Video 9.2.1 Stopping Timers
12. Check for Understanding 9.2.2 Stop Timer Quiz
13. Example 9.2.3 Random Circles
14. Exercise 9.2.4 Growing Circle
15. Exercise 9.2.5 Brick Wall
16. Notes 9.2.6 Project: Evasion (Stop Timers)
18. ### 9.3 Collisions
19. Video 9.3.1 Collisions
20. Video 9.3.2 Live Coding: Collisions
21. Check for Understanding 9.3.3 Collisions Quiz
22. Example 9.3.4 Bouncing Ball
23. Exercise 9.3.5 Collision Simulation
24. Exercise 9.3.6 Carnival Game
25. Notes 9.3.7 Project: Evasion (Collisions)
26. ### 9.4 Mouse Click Events
27. Video 9.4.1 Mouse Click Events
28. Check for Understanding 9.4.2 Mouse Click Quiz
29. Example 9.4.3 Click For Circles
30. Exercise 9.4.4 Pausing the Carnival Game
31. Exercise 9.4.5 Dripping Paint
32. Notes 9.4.6 Project: Evasion (Mouse Click)
33. ### 9.5 More Mouse Events
34. Video 9.5.1 More Mouse Events
35. Check for Understanding 9.5.2 More Mouse Events Quiz
36. Example 9.5.3 Simple Painting
37. Example 9.5.4 Painting with Color
38. Exercise 9.5.5 Coordinates
39. Exercise 9.5.6 Target
40. Exercise 9.5.7 Drag and Drop
41. Notes 9.5.8 Project: Evasion (More Mouse)
43. ### 9.6 Key Events
44. Video 9.6.1 Key Events
45. Check for Understanding 9.6.2 Key Events Quiz
46. Example 9.6.3 Keyboard Character
47. Exercise 9.6.4 Basic Snake
48. Notes 9.6.5 Project: Evasion (Key Events)
49. Free Response 9.6.6 Project Reflection
50. ### 9.7 Animation and Games Quiz
51. Unit Quiz 9.7.1 Animation and Games Quiz
10. ## Animation Challenges
1. ### 10.1 Animation Challenges
3. Challenge 10.1.2 Increasing Number of Shapes
11. ## Project: Breakout
1. ### 11.1 Breakout
2. Notes 11.1.1 Breakout Introduction
3. Challenge 11.1.2 Bricks
4. Challenge 11.1.3 Ball and Paddle
5. Challenge 11.1.4 Collisions
12. ## Project: Snake
1. ### 12.1 Snake Game
2. Challenge 12.1.1 A Growing Snake
3. Challenge 12.1.2 Collisions
5. Challenge 12.1.4 Finishing Touches
13. ## Data Structures: Arrays
1. ### 13.1 Intro to Arrays
2. Video 13.1.1 Intro to Arrays
3. Notes 13.1.2 When to Use Arrays?
4. Check for Understanding 13.1.3 Intro to Arrays Quiz
5. Example 13.1.4 Array Basics
6. Exercise 13.1.5 Exploration: A Boxy Array
7. Exercise 13.1.6 List of Places to Travel
8. Exercise 13.1.7 Top Websites
9. ### 13.2 Adding & Removing from an Array
10. Video 13.2.1 Adding & Removing from an Array
11. Check for Understanding 13.2.2 Adding & Removing from an Array Quiz
12. Notes 13.2.3 A Note About Arrays as Parameters
13. Example 13.2.4 Temperature Array
14. Exercise 13.2.5 Exploration: Creating a To-Do List
15. Exercise 13.2.6 Stacking Barrels
16. Exercise 13.2.7 Key Logging
17. ### 13.3 Iterating Through an Array
18. Video 13.3.1 Iterate Through an Array
19. Check for Understanding 13.3.2 Iterate Through an Array Quiz
20. Example 13.3.3 Print Shopping Lists
21. Debugging 13.3.4 Test Average
22. Exercise 13.3.5 Reverse List
23. Exercise 13.3.6 Evens Only List
24. Challenge 13.3.7 Dice Roll Probabilities
25. Free Response 13.3.8 Dice Probability Reflection
26. ### 13.4 Array Iteration with Graphics
27. Notes 13.4.1 Array Iteration with Graphics
28. Quiz 13.4.2 Array Iteration with Graphics Quiz
29. Example 13.4.3 Snow Storm
30. Exercise 13.4.4 Exploration: Changing Properties
31. Exercise 13.4.5 Draw a Barcode
32. Exercise 13.4.6 Wind Turbines
33. Challenge 13.4.7 Parallax Challenge
34. ### 13.5 Array Methods
35. Notes 13.5.1 Array Methods
36. Example 13.5.2 Email List
37. Example 13.5.3 Weekly Temperatures
38. Example 13.5.4 Splitting Up Tasks
39. Exercise 13.5.5 Mutual Friends
40. Challenge 13.5.6 Scientific Data
14. ## Data Structures: Objects
1. ### 14.1 Intro to Objects
2. Video 14.1.1 Intro to Objects
3. Check for Understanding 14.1.2 Intro to Objects Quiz
4. Example 14.1.3 Phonebook
5. Exercise 14.1.4 Movie Database
6. Video 14.1.5 Object Literals & Properties
7. Example 14.1.6 Car Objects
8. Exercise 14.1.7 Two Player
9. Exercise 14.1.8 Shopping Cart
10. ### 14.2 Graphic Objects
11. Video 14.2.1 Graphic Objects
12. Notes 14.2.2 A Note About Objects as Parameters
13. Example 14.2.3 Super Bouncers
14. Example 14.2.4 Falling Blocks
15. Exercise 14.2.5 Exploration: Our Solar System
16. Exercise 14.2.6 Fireflies
17. Free Response 14.2.7 Firefly Reflection
18. Challenge 14.2.8 Level 1 Knight
19. ### 14.3 Object Methods
20. Video 14.3.1 Object Methods
21. Quiz 14.3.2 Object Methods Quiz
22. Example 14.3.3 Party Ball
23. Exercise 14.3.4 Exploration: Digi Pet
24. Exercise 14.3.5 Level 2 Knight
25. Exercise 14.3.6 Bank Account
26. ### 14.4 Iterating Through an Object
27. Video 14.4.1 Iterating Through an Object
28. Check for Understanding 14.4.2 Iterating Through an Object Quiz
29. Notes 14.4.3 Property or Method?
30. Example 14.4.4 Phonebook Extended
31. Example 14.4.5 Bouncing Emoji
32. Exercise 14.4.6 Starry Night
33. Exercise 14.4.7 Let's Go Birding
34. Exercise 14.4.8 Asteroids
35. ### 14.5 Object Constructors
36. Video 14.5.1 Object Constructors
37. Quiz 14.5.2 Object Constructors Quiz
38. Example 14.5.3 New Person
39. Example 14.5.4 CodeHS Graphics are Objects
40. Debugging 14.5.5 Musical Instruments
41. Exercise 14.5.6 Level 3 Knight
42. Exercise 14.5.7 Bank Account Constructor
43. Challenge 14.5.8 Hobby Constructors
44. Notes 14.5.9 Advanced Extension: Prototypes and Inheritance
15. ## Project: Tic Tac Toe
1. ### 15.1 Tic Tac Toe
2. Challenge 15.1.1 Tic Tac Toe: Part 1
3. Challenge 15.1.2 Tic Tac Toe: Part 2
4. Challenge 15.1.3 Tic Tac Toe: Full Game
16. ## Project: Helicopter Game
1. ### 16.1 Game Design: Helicopter
2. Video 16.1.1 Introduction to Helicopter
3. ### 16.2 Basics
4. Video 16.2.1 Moving the Helicopter
5. Exercise 16.2.2 Moving the Helicopter
8. Video 16.2.5 Smoother Movement
9. Exercise 16.2.6 Smoother Movement
11. ### 16.3 Improvements
12. Video 16.3.1 Colliding with Walls
13. Exercise 16.3.2 Wall Collisions
14. Video 16.3.3 Colliding with Obstacles
15. Exercise 16.3.4 Obstacle Collisions
18. Video 16.3.7 Moving the Terrain
19. Exercise 16.3.8 Moving the Terrain
20. ### 16.4 Polish
21. Video 16.4.1 Helicopter Image and Points!
22. Exercise 16.4.2 Image and Points
23. Video 16.4.3 Dust
24. Exercise 16.4.4 Dust
25. Video 16.4.5 More Obstacles
26. Exercise 16.4.6 More Obstacles
27. Challenge 16.4.7 Helicopter Extensions
17. ## Final Project: Your Own Game
1. ### 17.1 Project Prep and Development
2. Notes 17.1.1 Project Introduction
3. Free Response 17.1.2 Planning and Design
4. Pseudocode 17.1.3 Pseudocode
5. Challenge 17.1.4 Write the Code!
6. Presentation 17.1.5 Present your Project
18. ## Final Exam
1. ### 18.1 Final Exam
2. Final 18.1.1 JavaScript Final Exam Pt. 1: Multiple Choice
19. ## Midterm
1. ### 19.1 Midterm
2. Midterm 19.1.1 Midterm Pt 1: Multiple Choice
20. ## Project: Connect Four
1. ### 20.1 Connect Four
2. Exercise 20.1.1 Make The Board
3. Exercise 20.1.2 Take Turns
4. Exercise 20.1.3 Find the Winner
21. ## Project: Mastermind
1. ### 21.1 Mastermind
2. Connection 21.1.1 How to Play Mastermind
3. Demo 21.1.2 Mastermind: Demo
4. Exercise 21.1.3 Generate Number List
5. Exercise 21.1.4 Get User Guess
6. Exercise 21.1.5 Compare User to List!
7. Exercise 21.1.6 Finish the Game!
8. Exercise 21.1.7 Mastermind Board
22. ## Extension: Additional Data Structures
1. ### 22.1 Intro to Sets
2. Video 22.1.1 Intro to Sets
3. Check for Understanding 22.1.2 Intro to Sets Quiz
4. Example 22.1.3 Basic Sets
5. Exercise 22.1.4 Vowels
6. Exercise 22.1.5 Mutual Friends
7. Exercise 22.1.6 Total Network of Friends
8. ### 22.2 Intro to Grids
9. Video 22.2.1 Intro to Grids
10. Check for Understanding 22.2.2 Intro to Grids Quiz
11. Example 22.2.3 Grid Basics
12. Exercise 22.2.4 Building a Database
13. ### 22.3 Looping Over a Grid
14. Video 22.3.1 Looping Over a Grid
15. Check for Understanding 22.3.2 Looping Over a Grid Quiz
16. Example 22.3.3 Print Grid
17. Exercise 22.3.4 Summing Grid
18. ### 22.4 Grid Example: Get a Row
19. Video 22.4.1 Grid Example: Get a Row
20. Check for Understanding 22.4.2 Grid Example: Get a Row Quiz
21. Example 22.4.3 Get a Row
22. Exercise 22.4.4 Grid Diagonal
23. Challenge 22.4.5 Watercolor Grid
23. ## Practice: Karel
1. ### 23.1 Extra Karel Practice
2. Challenge 23.1.1 Functions Practice: K For Karel
3. Challenge 23.1.2 Functions Practice: Karel Plants A Tree
4. Challenge 23.1.3 Functions Practice: X Marks the Spot
5. Challenge 23.1.4 While Loop Practice: Blackout
6. Challenge 23.1.5 While Loop Practice: Move To Top
7. Challenge 23.1.6 While Loop Practice: Checkered Row
8. Challenge 23.1.7 For Loop Practice: Tall Hurdles
9. Challenge 23.1.8 Functions and While Loop Practice: Row and Back
10. Challenge 23.1.9 Functions and For Loop Practice: Opposite Squares
11. Challenge 23.1.10 Stairway To Heaven
13. Challenge 23.1.12 For Loop Practice: Square
14. ### 23.2 Extra Karel Puzzles
15. Challenge 23.2.1 Midpoint Karel
16. Challenge 23.2.2 Target Karel
17. Challenge 23.2.3 The Winding Yellow Road
18. Challenge 23.2.4 Super Random Hurdles
19. Challenge 23.2.5 Copy
20. Challenge 23.2.6 Multiply
21. Challenge 23.2.7 Fibonacci Karel
22. Challenge 23.2.8 Comparison Karel
23. Challenge 23.2.9 Swap
24. Challenge 23.2.10 Sorting Karel
24. ## Practice: Functions
1. ### 24.1 Functions and Parameters Practice
2. Challenge 24.1.1 Taking a Power
3. Challenge 24.1.2 Dot Rectangle
4. Challenge 24.1.3 Print the Date
6. Challenge 24.1.5 Concentric Circles
7. Challenge 24.1.6 Graphics Staircase
8. Exercise 24.1.7 The Weekend
25. ## Practice: Console Challenges
1. ### 25.1 Prime Numbers
3. Example 25.1.2 Voting Age
4. Example 25.1.3 Negative Numbers
5. Exercise 25.1.4 Fibonacci
6. Exercise 25.1.5 Better Sum
7. Exercise 25.1.6 Factorial
8. Exercise 25.1.7 All Dice Values
9. Exercise 25.1.8 Powers of Two
10. Challenge 25.1.9 Prime Numbers
11. Challenge 25.1.10 Find the Max
12. Challenge 25.1.11 Prime Factorization
13. Challenge 25.1.12 Fizz Buzz
14. Challenge 25.1.13 Grid Printer
15. Challenge 25.1.14 Number Sum
16. Challenge 25.1.15 Hailstone Sequence
17. Challenge 25.1.16 Pythagorean Triples
18. Challenge 25.1.17 Digit Array
26. ## Practice: Graphics and Animation
1. ### 26.1 Fun Graphics Challenges
2. Exercise 26.1.1 The Worm
3. Challenge 26.1.2 Happy Birthday!
4. Challenge 26.1.3 Balloons
5. Challenge 26.1.4 Broccoli
6. Challenge 26.1.5 Circles in Squares
8. Challenge 26.1.7 Circles in Circles
9. Challenge 26.1.8 Snowman Loop
11. ### 26.2 Animation Practice
12. Example 26.2.1 Random Ghosts
13. Example 26.2.2 Spinner
14. Example 26.2.3 Random Fireworks
15. Example 26.2.4 Drawing Lines
16. Example 26.2.5 Colorful Drag to Paint
17. Example 26.2.6 Keyboard Square
18. Example 26.2.7 Click For Ghosts
19. Exercise 26.2.8 Circle Wall
20. Exercise 26.2.9 Hotspot Ball
21. Exercise 26.2.10 Trail
22. Exercise 26.2.11 Teleporting Ball
23. Exercise 26.2.12 Leash
24. Exercise 26.2.13 Pause
25. ### 26.3 Crazy Ball Game
26. Video 26.3.1 Crazy Ball Game 1
27. Check for Understanding 26.3.2 Crazy Ball Game Quiz 1
28. Example 26.3.3 Crazy Ball Game 1
29. Video 26.3.4 Crazy Ball Game 2
30. Check for Understanding 26.3.5 Crazy Ball Game Quiz 2
31. Example 26.3.6 Crazy Ball Game 2
32. Exercise 26.3.7 Click for Collision
33. Exercise 26.3.8 Drag and Drop
27. ## Practice: Data Structures Challenges
1. ### 27.1 Conway's Game of Life
2. Exercise 27.1.1 Make The Grid
3. Exercise 27.1.2 Find Life
4. Exercise 27.1.3 Finish Up
28. ## Extra Quiz Questions
1. ### 28.1 Basic Javascript and Graphics
2. Check for Understanding 28.1.1 For Loop Examples Quiz
3. Quiz 28.1.2 Extra JavaScript Graphics Quesions
4. Quiz 28.1.3 Extra Logical Operators Questions
5. Quiz 28.1.4 Extra Functions and Return Values Questions
6. Quiz 28.1.5 Extra Local Variables and Scope Questions
7. Check for Understanding 28.1.6 General For Loop Quiz
8. ### 28.2 Animation and Games
9. Quiz 28.2.1 Extra Timers Questions
10. Quiz 28.2.2 Extra Bouncing Ball Questions
11. Quiz 28.2.3 Extra Mouse Events: Mouse Clicked Questions
12. Quiz 28.2.4 Extra Key Events Questions
13. Quiz 28.2.5 Extra Crazy Ball Game Questions
14. ### 28.3 Basic Data Structures
15. Quiz 28.3.1 Extra Intro to Lists/Arrays Questions
16. Quiz 28.3.2 Extra Indexing into an Array Questions
17. Quiz 28.3.3 Extra Adding/Removing from an Array Questions
18. Quiz 28.3.4 Extra Array Length and Looping Through Arrays Questions
19. Quiz 28.3.5 Extra Iterating Over an Array Questions
20. Quiz 28.3.6 Extra Finding an Element in a List Questions
21. Quiz 28.3.7 Extra Removing an Element from an Array Questions
22. Quiz 28.3.8 Extra Basics of Objects Questions
23. Quiz 28.3.9 Extra Intro to Sets Questions
24. Quiz 28.3.10 Extra Intro to Grids Questions
29. ## Extension: Visualizing Music
1. ### 29.1 Visualizing Music
2. Video 29.1.1 Visualizing Music
3. Example 29.1.2 Our First Visualization
4. Example 29.1.3 Vertical Bars
5. Example 29.1.4 Vertical Bars with Color
6. Example 29.1.5 Changing Circles
7. Exercise 29.1.6 Your First Visualization
8. Exercise 29.1.7 Changing Colors
9. Exercise 29.1.8 Custom Colors
10. Exercise 29.1.9 Create Your Own Music Visualization
30. ## Intro to CS: JavaScript Pretest
1. ### 30.1 Intro to CS: JavaScript Pretest
2. Notes 30.1.1 About the Pretest
3. Survey 30.1.2 Mindsets
4. Quiz 30.1.3 JavaScript Knowledge & Skills
31. ## JavaScript Level 1 Certification Practice
1. ### 31.1 JavaScript Syntax Update
2. Notes 31.1.1 JavaScript Syntax Update
3. Example 31.1.2 Syntax Comparison
4. Example 31.1.3 Re-declaring Variables: Let vs Var
5. Exercise 31.1.4 Name and Game
6. Quiz 31.1.5 JavaScript Syntax Update Quiz
7. ### 31.2 Practice #1: JavaScript Basics
8. Quiz 31.2.1 Quiz: JavaScript Basics
9. Notes 31.2.2 Practice #1 Reflection
10. ### 31.3 Practice #2: JavaScript Control Structures
11. Quiz 31.3.1 Quiz: JavaScript Control Structures
12. Notes 31.3.2 Practice #2 Reflection
13. ### 31.4 Practice #3: JavaScript Functions and Objects
14. Quiz 31.4.1 Quiz: JavaScript Functions and Objects
15. Notes 31.4.2 Practice #3 Reflection | 9,027 | 26,196 | {"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} | 3.34375 | 3 | CC-MAIN-2024-22 | latest | en | 0.386026 |
https://www.numwords.com/words-to-number/en/2479 | 1,563,382,688,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525355.54/warc/CC-MAIN-20190717161703-20190717183703-00480.warc.gz | 803,623,709 | 1,836 | NumWords.com
# How to write Two thousand four hundred seventy-nine in numbers in English?
We can write Two thousand four hundred seventy-nine equal to 2479 in numbers in English
< Two thousand four hundred seventy-eight :||: Two thousand four hundred eighty >
Four thousand nine hundred fifty-eight = 4958 = 2479 × 2
Seven thousand four hundred thirty-seven = 7437 = 2479 × 3
Nine thousand nine hundred sixteen = 9916 = 2479 × 4
Twelve thousand three hundred ninety-five = 12395 = 2479 × 5
Fourteen thousand eight hundred seventy-four = 14874 = 2479 × 6
Seventeen thousand three hundred fifty-three = 17353 = 2479 × 7
Nineteen thousand eight hundred thirty-two = 19832 = 2479 × 8
Twenty-two thousand three hundred eleven = 22311 = 2479 × 9
Twenty-four thousand seven hundred ninety = 24790 = 2479 × 10
Twenty-seven thousand two hundred sixty-nine = 27269 = 2479 × 11
Twenty-nine thousand seven hundred forty-eight = 29748 = 2479 × 12
Thirty-two thousand two hundred twenty-seven = 32227 = 2479 × 13
Thirty-four thousand seven hundred six = 34706 = 2479 × 14
Thirty-seven thousand one hundred eighty-five = 37185 = 2479 × 15
Thirty-nine thousand six hundred sixty-four = 39664 = 2479 × 16
Forty-two thousand one hundred forty-three = 42143 = 2479 × 17
Forty-four thousand six hundred twenty-two = 44622 = 2479 × 18
Forty-seven thousand one hundred one = 47101 = 2479 × 19
Forty-nine thousand five hundred eighty = 49580 = 2479 × 20
Sitemap | 416 | 1,443 | {"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} | 2.921875 | 3 | CC-MAIN-2019-30 | latest | en | 0.770727 |
https://congruentmath.com/product/perfect-squares-and-square-roots-guided-notes-doodles-8th-grade-sketch-notes-10061822/ | 1,721,660,381,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517878.80/warc/CC-MAIN-20240722125447-20240722155447-00111.warc.gz | 152,925,463 | 22,850 | # Perfect Squares and Square Roots Guided Notes & Doodles | 8th Grade Sketch Notes
## Overview
Introduce perfect squares and square roots with this no prep, guided notes with doodles resource. Contains 10 pages of printable worksheets - including sketch notes on perfect squares, square roots, and equations in the form x² = p, practice sheets with problem sets, color by number activity, and a real-life math application. It works well as graphic organizers, scaffolded notes, and interactive notebooks. And it's artsy—if your students love color by code or notes with doodles, they'll love this 8th Grade math lessons!
## Why you'll love this
Note: this lesson includes rational numbers only.
CCSS Standards: 8.EE.A.2
Planning a whole unit?
My 7th Grade Rational Numbers Pixel Art Bundle offers a full unit of digital practice for this and more—at 40% off.
What's included:
1. Guided notes. Teach the topic with these structured notes (perfect squares, square roots). Integrates checks for understanding to verify your students are on the right track. - 1 page
2. Practice worksheets. Worksheets, a maze activity, and color by code to practice perfect squares and square roots. - 3 pages
3. Real-life application. Read and write about the math concepts being applied to designers designing bathroom tiles. - 1 page
4. Answer key. Included for all of the worksheets. - 5 pages
Great for:
• Introductory Lessons
• Reteaching & Review
• Homework
• Sub Plans
• Quiz, Test & Exam Prep
What teachers say about my Guided Notes & Doodles lessons:
• ⭐️⭐️⭐️⭐️⭐️ “Great resource and a different way to take notes. Students were engaged and used their notes to help them with solving problems later.” - Heather P.
• ⭐️⭐️⭐️⭐️⭐️ “My students really enjoyed these notes. I needed an additional resource to reteach this material before our end of the year assessment, and this was perfect.” - Ashley H.
• ⭐️⭐️⭐️⭐️⭐️ “I used this resource with students who typically struggle to remain engaged in mathematics. They remained very engaged and didn’t hesitate to fix mistakes and complete their work. Great resource!” - Carissa S.
Want a free sample?
Something not right?
If you have a question, or something is off with the lesson, please email me or post a question on TPT. I’ll do everything I can to make it right. | 538 | 2,325 | {"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} | 3.046875 | 3 | CC-MAIN-2024-30 | latest | en | 0.915617 |
https://community.qlik.com/thread/288862 | 1,540,182,859,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583514497.14/warc/CC-MAIN-20181022025852-20181022051352-00173.warc.gz | 654,038,901 | 22,828 | 4 Replies Latest reply: Jan 26, 2018 8:43 AM by vikas mahajan
# How to use getfieldSelection in Aggr
Hi all,
With reference to discussion based on the link Market Share Calculation Month Wise Market Share Calculation in QS .
I have multiple dimension in my pivot chart where as I wanted to calculate market share based on many parameters
Like Company,Region1,RegionName,StateName,Territory etc
##### Sum ([CY Sale])/aggr(nodistinct Sum ({<[Company]=>}[CY Sale]),MonthName)
So i want to use getfieldselection dynamically when user will select Company then company wise market share will be calculated if user select Region1 then I should get Region1 wise MS.
Thanks
Vikas
• ###### Re: How to use getfieldSelection in Aggr
May be this
Sum([CY Sale])/Sum(TOTAL <\$(='[' & GetFieldSelections(FieldName, '], [') & ']')> {<Company>} [CY Sale])
or this
##### Sum([CY Sale])/Aggr(NoDistinct Sum({<[Company]=>}[CY Sale]), \$(='[' & GetFieldSelections(FieldName, '], [') & ']'))
• ###### Re: How to use getfieldSelection in Aggr
Hi Sunny
##### Sum ([CY Sale])/aggr(nodistinct Sum ({<[Company]=>}[CY Sale]), MonthName , GeographyDrill )
• ###### Re: How to use getfieldSelection in Aggr
This
##### Sum([CY Sale])/Aggr(NoDistinct Sum({<[Company]=>}[CY Sale]), MonthName, \$(='[' & GetFieldSelections(GeographyDrill, '], [') & ']'))
• ###### Re: How to use getfieldSelection in Aggr
Unfortunately this not working
my expression is
num(((Sum(Y2Sale)/aggr(nodistinct Sum ({<COMP=>}Y2Sale),Year,\$(='[' & GetFieldSelections(ReportingHierarchy, '], [') & ']')))),'#,##0.0%')
I am using ReportingHierarchy in drill of qlik sense.
Help appreciated.
Vikas | 468 | 1,665 | {"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-2018-43 | latest | en | 0.583179 |
https://www.calculushowto.com/frenet-frame/ | 1,643,051,969,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304600.9/warc/CC-MAIN-20220124185733-20220124215733-00265.warc.gz | 716,741,962 | 24,464 | # Frenet Frame: Simple Definition
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## What is a Frenet Frame?
The Frenet frame (also called the moving trihedron or Frenet trihedron) along a curve is a moving (right-handed) coordinate system determined by the tangent line and curvature. The frame, which locally describes one point on a curve, changes orientation along the length of the curve.
Every point on the curve can be described by three natural vectors.
## Formal Definition of Frenet Frame
More formally, the Frenet frame of a curve at a point is a triplet of three mutually orthogonal unit vectors {T, N, B}. In three-dimensions, the Frenet frame consists of [1]:
• The unit tangent vector T, which is the unit vector in the direction of what is being modeled (like velocity),
• The unit normal N: the direction where the curve is turning. We can get the normal by taking the derivative of the tangent then dividing by its length. You can think of the normal as being the place the curve sits in [2].
• The unit binormal B = T x N, which is the cross product of the unit tangent and unit normal.
A space curve; the vectors T, N and B; and the osculating plane spanned by T and N. Image: Popletibus | Wikimedia Commons.
The tangent and normal unit vectors span a plane called the osculating plane at F(s). In four-dimensions, the Frenet frame contains an additional vector, the trinormal unit vector [3]. While vectors have no origin in space, it’s traditional with Frenet frames to think of the vectors as radiating from the point of interest.
The Frenet frame {T, N, B} is called a discrete Frenet frame (DFF). As well as its importance in studying curves, the DFF has some specific uses, including studying shapes of long molecules like proteins [4].
## References
Image: Popletibus This W3C-unspecified vector image was created with Inkscape, CC BY-SA 4.0 <, via Wikimedia Commons [1] Green, P. & Rosenberg, J. (2000). Space Curves, FrenetFrames, and Torsion. Retrieved August 5, 2021 from: https://www.math.umd.edu/~jmr/241/curves2.htm [2] Kho, G. (2013). CS468 Notes.
[3] Frenet-Serret formulas. Retrieved August 5, 2021 from: https://ncatlab.org/nlab/show/Frenet-Serret+formulas
[4] u, Y. (2013). Discrete Frenet Frame with Application to Structural Biology and Kinematics. Retrieved August 5, 2021 from: https://diginole.lib.fsu.edu/islandora/object/fsu%3A183801
CITE THIS AS:
Stephanie Glen. "Frenet Frame: Simple Definition" From CalculusHowTo.com: Calculus for the rest of us! https://www.calculushowto.com/frenet-frame/
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willis January 26, 2012 08:31
Ground Effect
I am trying to model the flow around a cambered airfoil, taking ground effect into consideration.
Can anybody tell me how I would specify the right conditions for my lower boundary (the ground) in fluent, i.e a velocity and no slip condition?
Thanks very much for any help.
truffaldino January 26, 2012 09:29
You can put no-slip on the ground, but ground must move together with airflow, if your airfoil is stationary in fluent. Or simpler solution could be to put "reflection" of your airfoil in the ground and analyse double airfoil problem.
Truffaldino
willis January 26, 2012 10:26
Thanks Truffaldino,
I know that the ground needs to be given the same velocity as the flow, I just do not know how to do this in fluent. When defining my boundary conditions, if i select wall is there an option to specify a velocty as well?
truffaldino January 26, 2012 11:23
Hi Willis,
I have never used fluent with no-slip on moving surface, but I think fluent has something as "moving wall" bc. Just put the speed of wall to be equal to that of air at the inlet.
But I still think putting extra "reflection airfoil" is better solution of the problem.
Truffaldino
Martin Hegedus January 26, 2012 16:32
Out of curiosity, is the interest in race car aerodynamics, i.e. inverted cambered front wing of F1, or airplane ground effects?
lava12005 January 26, 2012 21:05
If we consider the case whereby the airfoil is moving, the speed at the ground (yes boundary layer is there) is equal to the speed of the farfield which is basically 0.
Am I right to say that the ground boundary condition for simulation (where the airfoil is fixed) is a velocity inlet where the velocity specified is equal in both magnitude and direction at the normal inlet?
Moreover, some of the experiment in quantifying ground effect is based on the double body test. But some other perform it using a moving belt as the ground, so is the double body experiment is 'correct'? Since the velocity at the symetry plane is not necessarily equal to the movement speed of the airfoil.
Martin Hegedus January 26, 2012 21:32
Quote:
Moreover, some of the experiment in quantifying ground effect is based on the double body test. But some other perform it using a moving belt as the ground, so is the double body experiment is 'correct'? Since the velocity at the symetry plane is not necessarily equal to the movement speed of the airfoil.
Depends. If the airfoil is a few chords from the surface and the upper surface normal is pointing away from the ground, then symmetry is probably OK.
On the other hand, if the airfoil is close to the ground and the upper surface normal vector is pointed towards the ground, i.e. the inverted race car airfoil and the flow over the "upper" surface is moving much faster than freestream, then it may be better to use a moving ground plane.
Edit: Oh, and it depends on Reynolds number. For the race car front wing, the lower the Reynolds number, the more likely the moving ground plane will be required.
All times are GMT -4. The time now is 04:50. | 799 | 3,327 | {"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} | 3.015625 | 3 | CC-MAIN-2017-26 | longest | en | 0.933064 |
https://socratic.org/questions/how-do-you-simplify-sqrt-125a-3 | 1,576,407,113,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541307813.73/warc/CC-MAIN-20191215094447-20191215122447-00268.warc.gz | 543,172,503 | 5,955 | # How do you simplify sqrt(125a^3)?
Jun 23, 2016
Factor it out.
#### Explanation:
Break down the expression into its factors:
$\sqrt{125 {a}^{3}} = \sqrt{5 \cdot 5 \cdot 5 \cdot a \cdot a \cdot a}$
Notice that we have a pair of $5$'s and $a$'s.
sqrt(5*5*5*a*a*a)=sqrt(5*(5*5)•a*(a*a)) =sqrt(5*(5)^2*a*(a^2))
We can take these (the pairs) out of the expression.
$\sqrt{5 \cdot {\left(5\right)}^{2} \cdot a \cdot \left({a}^{2}\right)} = 5 \cdot a \cdot \sqrt{5 \cdot a}$
Finally,
$5 \cdot a \cdot \sqrt{5 \cdot a} = 5 a \sqrt{5 a}$
$\sqrt{125 {a}^{3}} = 5 a \sqrt{5 a}$
Hope this helps! | 242 | 597 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "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} | 4.53125 | 5 | CC-MAIN-2019-51 | latest | en | 0.552279 |
http://mathoverflow.net/questions/67883/two-questions-from-hubbards-teichmuller-theory-book-vol-i-p-130-thm-4-4-1?sort=oldest | 1,469,744,525,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257829320.70/warc/CC-MAIN-20160723071029-00075-ip-10-185-27-174.ec2.internal.warc.gz | 163,179,403 | 15,276 | # Two questions from Hubbard's Teichmuller theory book Vol I, P. 130 , Thm 4.4.1, ( QC maps )
I was studying Theorem 4.4.1 from John H. Hubbard's Teichmuller Theory, vol I, Theorem 4.4.1 ( P. 129 ) which states :
Let $X,Y$ be two hyperbolic Riemann surfaces with hyperbolic metrics $d_X,d_Y$ respectively and let $K\geq 1$.Then there exists a function ( homeomorphism of positive real numbers) $\delta_K:(0,\infty)\to(0,\infty)$ such that $\lim_{\eta\to 0}\delta_K(\eta)=0$ such that for all $K$-q.c maps $f:X\to Y$, we have $dist_Y(f(x),f(y))\leq \delta_K(dist_X(x,y))$.
The way he proves it is the following : 1) It is enough to prove the statement for the universal cover ,i.e. the Poincare disk $D$, since a $K$-q.c. map lifts to a $K$-q.c map.
2) He defines $\delta_K(\eta)= M^{-1}(\frac{1}{K}M(\eta) )$, where $M$ is the modulus of the branched/ramified cover with ramification locus being the two-point set $P={z_1,z_2}$, which ( modulus ) he proves depends only on the hyperbolic distance $dist_D(z_1,z_2)$ . ( Lemma 4.4.2) and is a strictly decreasing homeomorphim of the positive real numbers.
My questions are :
1. Hubbard proves that the branched/ramified double cover of $D$ with ramification locus a two-point set is topologically a cylinder. But then how do we know that this cylinder has a finite modulus , i.e. the cylinder is not conformally equivalent to $C-{0},D-0$ ? Well, in proposition 4.4.6 ( P. 132 ),he proves it, but that is only after proving Thm 4.4.1.
2. I am unable to follow the lines 4.4.2 and 4.4.3, in the proof of lemma 4.4.2 ? What does he mean exactly by his notation $(D~_r)_ {0,z}$ ?
Is he scaling the standard hyperbolic metric on $D$ ? And why exactly the inclusion of the cylinders ( ramified covers ) in the line 4.4.3 true ? Please explain , thanks !
-
first the notation $d_{D_{r}}(0,z)$ means that you are considering the distance between $0$ and $z$ in the hyperbolic metric associated to the disk $D_{r}$ centered at the origin with radius $r$ (so if you consider two such disks, $D_{r_{1}}$, $D_{r_{2}}$ you will work with two different metrics). Now you can normalize the situation so that your two points $z_{1}, z_{2}$ are $\pm a$ and use the double cover of $\mathbb{C}$ (ramified above $\pm a$) given by $$\phi_{a}(z)=(a/2)(z+1/z).$$ The preimage by $\phi_{a}$ of a disk $D$ containing $\pm a$ will give you $\tilde{D}$ (because you obtain a double cover branched exactly where it should be). But since you have two such nested disks $D_{r_{2}} \subset D_{r_{1}}$ you will obtain an inclusion for their preimages as well $\tilde{D_{2}} \subset \tilde{D_{1}}$, as you wanted. | 830 | 2,639 | {"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} | 2.59375 | 3 | CC-MAIN-2016-30 | latest | en | 0.860102 |
https://mathforlove.com/2012/01/beautiful-mathematics/ | 1,709,382,757,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475825.14/warc/CC-MAIN-20240302120344-20240302150344-00228.warc.gz | 374,314,257 | 24,451 | # Beautiful Mathematics
January 20, 2012
“Beauty is the first test: there is no permanent place in the world for ugly mathematics.”
I’ve been reading Proofs from THE BOOK by Aigner and Ziegler, a paean to beautiful mathematics. The Book refers to a conceit of the mathematician Paul Erdős. From wikipedia:
[Erdős] spoke of “The Book”, an imaginary book in which God had written down the best and most elegant proofs for mathematical theorems. Lecturing in 1985 he said, “You don’t have to believe in God, but you should believe in The Book.”
Proofs from THE BOOK is a collection to these “book proofs,” some of the most elegant arguments in mathematics. One thing I like about this book is that it samples multiple, very diverse ideas to prove the same, often classic, result. For example, the book opens with six different proofs that there are infinitely many prime numbers. A number of these are classics, or variations on a classic: assume that there are only finitely many, and then find a contradiction by constructing a new prime, or showing one must exist. But some of the constructions are incredibly novel.
Now here’s the thing: this is a book that is decidedly not for the layperson. To read it, you need to be extremely comfortable with calculus and limits, infinite sums, and some serious subtleties, both conceptual and notational. But if you are, the surprises keep coming. The results on primes are nothing short of astonishing (especially to me, as a non-number theorist who’s always loved the subject).
For example, there are these incredible inequalities throughout the book. Bertrand’s Postulate states that there is always a prime number between any number n and its double 2n. I’ve always thought that this was a great result. But the proof is magical: a series of inequalities that leads to a contradiction that a quantity ends up growing larger than it should, unless there’s a prime between n and 2n. On the way, there’s this elegant proof of the remarkable fact that the product of all primes less than a number x is less than or equal to 4^{x-1}. The proof is too involved to get into here, but consider the depth of that statement: on one side, you’re going up number by number, multiplying primes when you come to them, ignoring nonprimes; on the other side, you’re putting in a factor of 4 for each number. No matter how large the primes get, the product of fours is always bigger.
These are the kind of details that at once make math rich and, sadly, inaccessible. For me, I read the book blown away, exclaiming aloud when a new idea is dropped into the mix, all the time with that sense of “how did anyone think of this?”
I suppose that’s what happens whenever you’re looking at great art. How did Stravinsky compose The Rite of Spring? How did Picasso make Guernica? It takes a huge amount of work to get a sense of how they thought of it (though there’s always a path). It’s inspiring, and also leads to the despair that comes from idealism. There’s a prime between n and 2n! Wow! And no one knows if there’s always a prime between n^2 and (n+1)^2? I’ll solve that! And then, of course, you realize just how hard these unsolved problems are. It’s easy to go from surveying the best work to deciding that you don’t have what it takes. Still, these works of art are worth seeing.
There’s another way to appreciate the wonder of it all, and that’s with the results that connect things that seem like they shouldn’t even be remotely connected. Consider Stirling’s Approximation. We could ask, how different are N! and N^N? If you need to adjust one of them, what is the adjustment? The answer: the adjustment factor is roughly \sqrt{2\pi N}/e^N. What in the world is \sqrt{2}, e, and \sqrt{\pi} doing there?! It seems like magic that these terms are all connected. If you look at the proof, it makes more sense. Nevertheless, there’s a way that I continue to carry a sense of wonder that all these pieces fit together, again and again, unexpectedly and inevitably.
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fshak92 March 12, 2012 11:07
Has the kind of mesh influence on the solving method used by software?
1) In a conjugate heat transfer case for finding the HTC and LocalHTC I've meshed my model(a body surrounded by air) in two ways(first polyhedral and second Trimmer).
There isn't any velocity and i've used the segregated solver and all the conditions are same for both cases.
In trimmer case Just the energy equation is solved but for polyhedral, the momentum and continuity equations are solved as well.
I wonder if someone tells me how the kind of mesh influences on solving equations in my case.
2)And also do you know how Heat transfer coefficient and LocalHTC are calculated by the software? the tutorial hasn't explained in a good way.
siara817 March 13, 2012 13:35
How was the quality of the mesh for the both cases?
If the quality of the mesh is low in any case, then the solution will not converge.
I hope you know how to check for example Cell quality in STAR CCM.
Good luck
MFitl March 14, 2012 03:07
Quote:
Originally Posted by omid88 (Post 348956) 1) 2)And also do you know how Heat transfer coefficient and LocalHTC are calculated by the software? the tutorial hasn't explained in a good way.
The definition of the HTC is:
=
Depending on the is different for the same .
For the HTC is the temperature you need to set in Tools|FieldFunctions|Heat Transfer Coefficient.
For the LocalHTC is the local temperature of the cell next to the wall.
abdul099 March 15, 2012 04:27
The solver should do the same for both meshes. Can you create a summary report on both cases and compare them for different settings (apart from mesh settings)?
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# chapter19[1].ppt
Measures of association
Measures of association
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### chapter19[1].ppt
1. 1. © 2006 The McGraw-Hill Companies, Inc., All Rights Reserved. McGraw-Hill/Irwin 19-1 Chapter 19 Measures of Association
2. 2. 19-2 Learning Objectives Understand . . . • how correlation analysis may be applied to study relationships between two or more variables • uses, requirements, and interpretation of the product moment correlation coefficient • how predictions are made with regression analysis using the method of least squares to minimize errors in drawing a line of best fit
3. 3. 19-3 Learning Objectives Understand . . . • how to test regression models for linearity and whether the equation is effective in fitting the data • nonparametric measures of association and the alternatives they offer when key assumptions and requirements for parametric techniques cannot be met
4. 4. 19-4 Exhibit 19-1 Measures of Association: Interval/Ratio Pearson correlation coefficient For continuous linearly related variables Correlation ratio (eta) For nonlinear data or relating a main effect to a continuous dependent variable Biserial One continuous and one dichotomous variable with an underlying normal distribution Partial correlation Three variables; relating two with the third’s effect taken out Multiple correlation Three variables; relating one variable with two others Bivariate linear regression Predicting one variable from another’s scores
5. 5. 19-5 Exhibit 19-1 Measures of Association: Ordinal Gamma Based on concordant-discordant pairs; proportional reduction in error (PRE) interpretation Kendall’s tau b P-Q based; adjustment for tied ranks Kendall’s tau c P-Q based; adjustment for table dimensions Somers’s d P-Q based; asymmetrical extension of gamma Spearman’s rho Product moment correlation for ranked data
6. 6. 19-6 Exhibit 19-1 Measures of Association: Nominal Phi Chi-square based for 2*2 tables Cramer’s V CS based; adjustment when one table dimension >2 Contingency coefficient C CS based; flexible data and distribution assumptions Lambda PRE based interpretation Goodman & Kruskal’s tau PRE based with table marginals emphasis Uncertainty coefficient Useful for multidimensional tables Kappa Agreement measure
7. 7. 19-7 Relationships “To truly understand consumers’ motives and actions, you must determine relationships between what they think and feel and what they actually do.” David Singleton, Vice President of Insights, Zyman Marketing Group
8. 8. 19-8 Pearson’s Product Moment Correlation r Is there a relationship between X and Y? What is the magnitude of the relationship? What is the direction of the relationship?
9. 9. 19-9 Exhibit 19-2 Scatterplots of Relationships
10. 10. 19-10 Exhibit 19-4 Scatterplots
11. 11. 19-11 Exhibit 19-7 Diagram of Common Variance
12. 12. 19-12 Interpretation of Correlations X causes Y Y causes X X and Y are activated by one or more other variables X and Y influence each other reciprocally
13. 13. 19-13 Exhibit 19-8 Artifact Correlations
14. 14. 19-14 Interpretation of Coefficients A coefficient is not remarkable simply because it is statistically significant! It must be practically meaningful.
15. 15. 19-15 Exhibit 19-9 Comparison of Bivariate Linear Correlation and Regression
16. 16. 19-16 Exhibit 19-10 Examples of Different Slopes
17. 17. 19-17 Concept Application X Average Temperature (Celsius) Y Price per Case (FF) 12 2,000 16 3,000 20 4,000 24 5,000 Mean =18 Mean = 3,500
18. 18. 19-18 Exhibit 19-11 Plot of Wine Price by Average Temperature
19. 19. 19-19 Exhibit 19-12 Distribution of Y for Observation of X
20. 20. 19-20 Exhibit 19-14 Wine Price Study Example
21. 21. 19-21 Exhibit 19-15 Least Squares Line: Wine Price Study
22. 22. 19-22 Exhibit 19-16 Plot of Standardized Residuals
23. 23. 19-23 Exhibit 19-17 Prediction and Confidence Bands
24. 24. 19-24 Testing Goodness of Fit Y is completely unrelated to X and no systematic pattern is evident There are constant values of Y for every value of X The data are related but represented by a nonlinear function
25. 25. 19-25 Exhibit 19-18 Components of Variation
26. 26. 19-26 Exhibit 19-19 F Ratio in Regression
27. 27. 19-27 Coefficient of Determination: r2 Total proportion of variance in Y explained by X Desired r2: 80% or more
28. 28. 19-28 Exhibit 19-20 Chi-Square Based Measures
29. 29. 19-29 Exhibit 19-21 Proportional Reduction of Error Measures
30. 30. 19-30 Exhibit 19-22 Statistical Alternatives for Ordinal Measures
31. 31. 19-31 Exhibit 19-23 Calculation of Concordant (P), Discordant (Q), Tied (Tx,Ty), and Total Paired Observations: KeyDesign Example
32. 32. 19-32 Exhibit 19-24 KDL Data for Spearman’s Rho _______ _____ Rank By_____ _____ _____ Applicant Panel x Psychologist y d d2 1 2 3 4 5 6 7 8 9 10 3.5 10.0 6.5 2.0 1.0 9.0 3.5 6.5 8.0 5.0 6.0 5.0 8.0 1.5 3.0 7.0 1.5 9.0 10.0 4.0 -2.5 5.0 -1.5 .05 -2 2.0 2.0 -2.5 -2 1.0 6.25 25.00 2.52 0.25 4.00 4.00 4.00 6.25 4.00 _1.00_ 57.00 .
33. 33. 19-33 Key Terms • Artifact correlations • Bivariate correlation analysis • Bivariate normal distribution • Chi-square-based measures • Contingency coefficient C • Cramer’s V • Phi • Coefficient of determination (r2) • Concordant • Correlation matrix • Discordant • Error term • Goodness of fit • lambda
34. 34. 19-34 Key Terms • Linearity • Method of least squares • Ordinal measures • Gamma • Somers’s d • Spearman’s rho • tau b • tau c • Pearson correlation coefficient • Prediction and confidence bands • Proportional reduction in error (PRE) • Regression analysis • Regression coefficients
35. 35. 19-35 Key Terms • Intercept • Slope • Residual • Scatterplot • Simple prediction • tau | 1,672 | 5,895 | {"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} | 2.59375 | 3 | CC-MAIN-2023-06 | latest | en | 0.738112 |
http://www.soft14.com/Graphic_Painting_and_Drawing/CAD_and_3D/MITCalc_-_Compression_Springs_6874_Review.html | 1,521,539,270,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647327.52/warc/CC-MAIN-20180320091830-20180320111830-00798.warc.gz | 476,674,975 | 8,193 | # MITCalc - Compression Springs Geometric and strength designs of helical compression cylindrical springs
O in short: Geometric and strength designs of helical compression cylindrical springs loaded with static or fatigue loading. Application supports Imperial and Metric units is based on ANSI ISO DIN standards and support many 2D and 3D CAD systems.
Easy software similar to MITCalc - Compression Springs, about:Spring, springs and cylindrical spring or free compression spring and cheap helical spring or the EN 13906 1, good DIN 2089 1 or also DIN 2095, static strength check and dynamic strength checkMITCalc - Compression SpringsClick here to get - Geometric and strength designs of helical compression cylindrical springs program
MITCalc - Compression Springs 1.17
Click to enlarge
Description:
The calculation is intended for the purposes of geometric and strength designs of helical compression cylindrical
developed in MS Excel, is multi-language, supports Imperial and Metric units and solves the following main tasks:
- Automatic design of a spring.
- Selection of an optimal alternative of spring design in view of strength, geometry and weight.
- Static and dynamic strength check.
- Support for Cold formed springs and Hot formed springs
- Calculation of working forces of a spring of known production and installation dimensions.
- Calculation of installation dimensions for known loading and production parameters of the spring.
- The application includes a table of commonly used spring materials according to ISO, ASTM/SAE, DIN, BS, JIS
and others.
calculation package (Autodesk Inventor, SolidWorks, SolidEdge, Pro/E).
The calculation is based on data, procedures, algorithms and data from specialized literature and standards EN
13906-1, DIN 2089-1, DIN 2095, DIN 2096.This module is a part of MITCalc - Mechanical and Technical
Calculation Package for gear, belt and chain drives, springs, beam, shaft, bolt connection, shaft connection,
tolerances and many others.
Price \$ 27 / 19
Purchase MITCalc - Compression Springs
Get it Now
Type: Shareware
File size: 1625 Kb
Date: 01/10/2010
Homepage
Install support: Install and Uninstall
OS: Win95, Win98, WinME, WinNT 3.x, WinNT 4.x, Windows2000, WinXP, Windows2003, Windows Tablet PC Edition 2005, WinME, Windows Vista Starter, Windows Vista Home Basic, Windows Vista Home Premium, Windows Vista Business, Windows Vista Enterprise, Windows Vista Ultimate, Windows Vista Home Basic x64, Windows Vista Home Premium x64, Windows Vista Business x64, Windows Vista Enterprise x64, Windows Vista Ultimate x64
System requirements: Microsoft Excel (2000, XP, 2007...)
Language: English, German, French, Italian, Spanish, Czech, Portuguese, Chinese Simplified, Chinese Traditional
Recent changes in this Minor Update:
Graphic Painting and Drawing: Graphic Painting and Drawing:CAD and 3D
MITCalc MITCalc is a multi-language calculation package includes solutions for gearing, belt, springs, beam, shaft, tolerances and many others. MITCalc support 2D and 3D CAD systems, Imperial and Metric units and many international standards ANSI, ISO, EN...
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DiagramStudio Diagram Studio is a tool for creating flowcharts, business and technical diagrams. Draw objects, various shapes and link them together. The program features user-controlled points of connection, color shadows, graphics import and export, curves.
DWG to IMAGE Converter MX DWG To Image Converter MX allows you convert DWG to Image, DXF to image and DWF to raster image directly without need of AutoCAD, it converts DWG, DXF and DWF files into raster image files, quick and easily.Support AutoCAD 2013 now.
Box Shot Maker Box Shot Maker is a 3D packaging design tool. You can easily add your own text and graphic to create the perfect 3D box for you product. With Box Shot Maker you can create: 2-side, 3-side Box Shot, Book, Cover, 3D screen-shot. www.533soft.com
See above information and user's reviews about MITCalc - Compression Springs Geometric and strength designs of helical compression cylindrical springs See also other good software tools: Graphic Painting and Drawing: Graphic Painting and Drawing:CAD and 3D
Search Soft14: Search: the Web within this Web site | 1,025 | 4,663 | {"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} | 2.53125 | 3 | CC-MAIN-2018-13 | longest | en | 0.846092 |
https://www.teacherspayteachers.com/Product/Math-Word-Wall-and-Games-1939635 | 1,487,703,642,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170823.55/warc/CC-MAIN-20170219104610-00650-ip-10-171-10-108.ec2.internal.warc.gz | 908,691,222 | 25,017 | Total:
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# Math Word Wall and Games
Subjects
Resource Types
Product Rating
Not yet rated
File Type
PDF (Acrobat) Document File
12 MB | 297 pages
### PRODUCT DESCRIPTION
Math Word Wall and Games:
This bundle is the perfect companion for your second grade math program. This 297 page packs includes math word walls for the year AND math worksheets that can be completed with each word wall theme. Also included are answer keys for all the worksheets.
The preview also includes some examples of the worksheets provided.
You can check out these two packs separately too.
Math Word Walls for the Year Pack
Looking for a quick and easy way to prep your math corner? Look no further! This 135 pack is filled with math word walls for your entire school year!
The pack includes 23 themed word walls.
The following word walls are included:
- Counting and Comparing Quantities
- Non-Numerical and Numerical Patterns
- Locating Objects in Space and in a Plane
- Representing Numbers (2)
- Skip Counting and Counting from a Given Number
- Describing Plane Figures and Describing Solids
- Comparing Numbers
- Estimating and Measuring Time
- Using a Table and Graph
- Finding a Missing Term in an Equation
- Temperature
- Fractions
- Representing and Decomposing Numbers
- Developing Processes for Written Computation
- Estimating and Measuring Dimensions
- Even and Odd Numbers
- Locating an Object in a Plane (2)
- Probability and Chance
- Rounding
- Developing Processes for Written Computation (2)
- Multiplication and Division
Math Word Wall Companion
Looking for a fun way to review math vocabulary and use a math word wall this year? This Math Word Wall Companion is sure to do the trick!
This 161 page packs includes two pages of practice and review for the following math concepts:
Counting and Comparing Quantities
Non-Numerical and Numerical Patterns
Locating Objects in Space and in a Plane
Representing Numbers (2)
Skip Counting and Counting from a Given Number
Describing Plane Figures and Describing Solids
Comparing Numbers
Estimating and Measuring Time
Using a Table and Graph
Finding a Missing Term in an Equation
Temperature
Fractions
Representing and Decomposing Numbers
Developing Processes for Written Computation
Estimating and Measuring Dimensions
Even and Odd Numbers
Locating an Object in a Plane (2)
Probability and Chance
Rounding
Developing Processes for Written Computation (2)
Multiplication and Division
Also included are answer keys for all the worksheets.
This is also the perfect companion for my Math Centers Bundle
Total Pages
297
N/A
Teaching Duration
N/A
### Average Ratings
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Total:
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\$6.50
User Rating: 4.0/4.0
(1,474 Followers)
\$6.50 | 655 | 2,800 | {"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-2017-09 | longest | en | 0.854779 |
http://www.kylesconverter.com/flow/acre--inches-per-second-to-gallons-(us-fluid)-per-hour | 1,643,284,812,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305260.61/warc/CC-MAIN-20220127103059-20220127133059-00016.warc.gz | 100,511,452 | 5,766 | # Convert Acre-inches Per Second to Gallons (us Fluid) Per Hour
### Kyle's Converter > Flow > Acre-inches Per Second > Acre-inches Per Second to Gallons (us Fluid) Per Hour
Acre-inches Per Second (ac in/s) Gallons (us Fluid) Per Hour (GPH) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Gallons (us Fluid) Per Hour to Acre-inches Per Second
(or just enter a value in the "to" field)
Please share if you found this tool useful:
Unit Descriptions
1 Acre-Inch per Second:
Volume flow rate of 1 acre-inch per second. Acre-inch being a volume of 660 feet by 66 feet by 1 inch; assuming an international foot of exactly 0.3048 meters. 1 ac in/s ≈ 102.79015312896 m3/s.
1 Gallon (US fluid) per hour:
1 Gallon (US fluid) per hour is approximately 1.051 503 273 333 x 10-6m3/s.
Link to Your Exact Conversion
Conversions Table
1 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 97755428.571570 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 6842880000.0022
2 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 195510857.142980 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 7820434285.7168
3 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 293266285.714490 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 8797988571.4314
4 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 391021714.2858100 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 9775542857.146
5 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 488777142.8573200 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 19551085714.292
6 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 586532571.4288300 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 29326628571.438
7 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 684288000.0002400 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 39102171428.584
8 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 782043428.5717500 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 48877714285.73
9 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 879798857.1431600 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 58653257142.876
10 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 977554285.7146800 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 78204342857.168
20 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 1955108571.4292900 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 87979885714.314
30 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 2932662857.14381,000 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 97755428571.46
40 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 3910217142.858410,000 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 977554285714.6
50 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 4887771428.573100,000 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 9775542857146
60 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 5865325714.28761,000,000 Acre-inches Per Second to Gallons (us Fluid) Per Hour = 97755428571460 | 1,007 | 3,070 | {"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-2022-05 | latest | en | 0.51488 |
http://www.nidokidos.org/threads/24571-Never-go-to-HR-for-help | 1,527,061,539,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865456.57/warc/CC-MAIN-20180523063435-20180523083435-00375.warc.gz | 425,085,493 | 10,628 | # Thread: Never go to HR for help
1. ## Never go to HR for help
After 2 years of selfless service, a man realized that he has not been promoted, no transfer, no salary increase no commendation and that the Company is not doing any thing about it. So he decided to walk up to His HR Manager one morning and after exchanging greetings, he told his HR Manager his observation. The boss looked at him, laughed and asked him to sit down saying. My friend, you have not worked here for even one day.
The man was surprised to hear this, but the manager went on to explain.
Manager:- How many days are there in a year?
Man:- 365 days and some times 366
Manager:- how many hours make up a day?
Man:- 24 hours
Manager:- How long do you work in a day?
Man:- 8am to 4pm. i.e. 8 hours a day.
Manager:- So, what fraction of the day do you work in hours?
Man:- (He did some arithmetic and said 8/24 hours i.e. 1/3(one third)
Manager:- That is nice of you! What is one-third of 366 days?
Man:- 122 (1/3x366 = 122 in days)
Manager:- Do you come to work on weekends?
Man:- No sir
Manager:- How many days are there in a year that are weekends?
Man:- 52 Saturdays and 52 Sundays equals to 104 days
Manager:- Thanks for that. If you remove 104 days from 122 days, how many days do you now have?
Man:- 18 days.
Manager:- OK! I do give you 2 weeks sick leave every year. Now remove that14 days from the 18 days left. How many days do you have remaining?
Man:- 4 days
Manager:- Do you work on New Year day?
Man:- No sir!
Manager:- Do you come to work on workers day?
Man:- No sir!
Manager:- So how many days are left?
Man:- 2 days sir!
Manager:- Do you come to work on the (National holiday )?
Man:- No sir!
Manager:- So how many days are left?
Man:- 1 day sir!
Manager:- Do you work on Christmas day?
Man:- No sir!
Manager:- So how many days are left?
Man:- None sir!
Manager:- So, what are you claiming?
Man:- I have understood, Sir. I did not realise that I was stealing Company money all these days.
Moral - NEVER GO TO HR FOR HELP!!!
Have a Nice Day.
HR = HIGH RISK
2. Lolz good one.
3. Hah hah haaaaaaa...........
Nice explanation.
4. hahahahahahhahaa......................
5. hmmmm.... poor employee...
6. Varun .. ur siggy pic is charming
7. lolzzzzzzzzzzz nice..wow
8. Thnks guys for rplying.....and thnks shainee for ur complimnt ......
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• | 694 | 2,575 | {"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} | 3.40625 | 3 | CC-MAIN-2018-22 | latest | en | 0.963908 |
https://drones.stackexchange.com/questions/2307/radius-of-drone-when-hovering-in-circles | 1,708,999,077,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474669.36/warc/CC-MAIN-20240226225941-20240227015941-00148.warc.gz | 226,142,565 | 28,251 | # Radius of drone when hovering in circles
According to Energy Minimization for Wireless Communication With Rotary-Wing UAV, it is better to fly in circles than to hover at a single place (if hovering is required). Flight requires less power. As the math model is simple, I can not find what is the "perfect" radius of the flight. Can rotary-wing UAV fly in lets say 2m radius with static speed $$s$$, or is there some law of physics (for example centrifugal force) that enforces the UAV to fly in wider circle? This is purely theoretical question, math is welcome.
• I would argue that increasing the centripetal force required to navigate would increase the effective load and take away from the efficiency gain of flying at Vme. Physics wise, you could mathematically determine the additional navigational thrust required to maintain a given radius knowing Vme, the radius of the circle, and the associated mass of the aircraft Mar 30, 2022 at 15:01 | 206 | 954 | {"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-10 | latest | en | 0.926147 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-concepts-through-functions-a-unit-circle-approach-to-trigonometry-3rd-edition/chapter-4-exponential-and-logarithmic-functions-section-4-5-properties-of-logarithms-4-5-assess-your-understanding-page-331/20 | 1,575,627,282,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540487789.39/warc/CC-MAIN-20191206095914-20191206123914-00170.warc.gz | 725,511,909 | 14,550 | ## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
$2$
$\because \log_a M+\log_a N = \log_a (MN)$ $\therefore \log_6 9 + \log_6 4 = \log_6 (9 \cdot 4) = \log_6 36$ With $36=6^2$, $\log_6 36 = \log_6 6^2$ Since $\log_a a^r =r$, then, $\log_6 6^2 = 2$ | 127 | 296 | {"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} | 3.921875 | 4 | CC-MAIN-2019-51 | latest | en | 0.530174 |
https://www.quantumstudy.com/the-frequency-of-a-particle-performing-shm-is-12-hz-its-amplitude-is-4cm-its-initial-displacement/ | 1,701,875,215,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100599.20/warc/CC-MAIN-20231206130723-20231206160723-00177.warc.gz | 1,053,240,998 | 48,260 | The frequency of a particle performing SHM is 12 Hz. Its amplitude is 4cm. Its initial displacement…
Q: The frequency of a particle performing SHM is 12 Hz. Its amplitude is 4cm. Its initial displacement is 2 cm toward positive extreme position. Its equation for displacement is
(a) x=0.04 cos(24πt+π/6)m
(b) x=0.04 sin(24πt)m
(c) x=0.04 sin(24πt+π/6)m
(d) x=0.04 cos(24πt)m
Ans: (c) | 134 | 389 | {"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-2023-50 | latest | en | 0.820037 |
https://poorgradpoorstudent.com/downloads/a-study-is-conducted-to-determine-if-the-traffic-light-patterns-at-a-busy-intersection-need-to-be-adjusted-to-help-the-flow-of-traffic-data-on-the-volume-of-traffic-and-the-traffic-patterns-during/ | 1,643,195,678,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304947.93/warc/CC-MAIN-20220126101419-20220126131419-00422.warc.gz | 519,689,299 | 18,480 | STATISTICS
1. A study is conducted to determine if the traffic light patterns at a busy intersection need to be
adjusted to help the flow of traffic. Data on the volume of traffic and the traffic patterns during
different times of day are recorded.
2
I. What is the population?
II. What is the sample?
III. Is the study observational or experimental? Justify your answer.
IV. If observational, identify ALL variables. If experimental, identify ALL independent and
dependent variables.
V. For each of the variables identified in part IV, list which of the four levels/scales of
measurement was used to obtain data on these variables?
VI. Classify each of the variables identified in part IV as either attribute or numerical.
Solution
I . Total Flow of the Traffic
II. The volume of Traffic & the patterns that are recorded.
III. It is an experimental study because the objective is to make changes and see how those
changes are effective and evaluate them. Since we are not merely drawing conclusions, it is not
observational.
IV. Variables are
1. Volume of traffic
2. Number of Vehicles
STATISTICS
3. Time
V. Volume of traffic – Ratio
Number of Vehicles – Ratio
Time – Nominal
3
2. Classify the following variable as nominal, ordinal, discrete, or continuous and state why
it is such a variable:
number of MP3 players sold each year
This is a discrete variable because there cannot be a value in decimals here.
3. Classify the following as an example of a nominal, ordinal, interval, or ratio level of
measurement, and state why it represents this level:
distance between two signposts
The distance between the signposts is a ratio level of measurement, because there is a set
definition of 0.0, as in no distance between these signposts. This has all the properties of an
interval variable and has a well defined value of 0.0, therefore such an example is a ratio level of
measurement.
4. Construct a frequency distribution table to organize the following set of data:
STATISTICS
4
101
102
103
105
106
107
108
109
110
2
2
2
1
3
1
2
6
1
5. Construct a grouped frequency distribution for the following 28 scores using a class width of
4:
74 80 79 74 77 64 77 81 63 68 83 71 75 71
70 69 83 71 82 86 77 86 79 75 81 86 78 77
Solution:
Range
63-67
68-72
73-77
78-82
83-87
Frequency
2
6
8
7
5
STATISTICS
6. Provide a different example for each of the following types of variables from your work or
home life:
5
A) A qualitative variable:
A qualitative variable could be the degree of honesty in each of my students during a
tutoring class.
B) A quantitative variable
A quantitative variable could be βheightβ for each of these students. | 666 | 2,634 | {"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} | 4.3125 | 4 | CC-MAIN-2022-05 | latest | en | 0.911991 |
https://tex.stackexchange.com/questions/223609/solids-of-revolution-using-tikz-or-pgf | 1,716,449,725,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058611.55/warc/CC-MAIN-20240523050122-20240523080122-00231.warc.gz | 488,195,304 | 35,465 | # Solids of revolution using tikz or pgf
I've been looking here on TEX Stackexchange for a solution to create Solids of Revolution but I could'n find anything. The analysis that I make on this problem is, first, draw a 2-D graph, second, draw the solid, and third, draw the slice (the disk).
Have you guys ever created something like that?
What I have so far is basically this:
\begin{figure}[h]
\centering
\begin{minipage}{.5\textwidth}
Analisando a região hachurada vemos que a integração deve ser feita com os limites $x = 1$ e $x = 2$. A equação da curva superior é $y = \left( 2 - \frac{x}{2} \right)$ e a curva inferior pode ser considerada o próprio eixo $x$ (ou seja, $y = 0$). O sólido obtido pela rotação da curva e sua unidade de volume estão representados abaixo.
\end{minipage}%
\begin{minipage}{.5\textwidth}
\centering
\begin{tikzpicture}
\begin{axis}[axis lines=middle,
xlabel=$x$,
ylabel=$y$,
enlargelimits,
ytick={2},
yticklabels={2},
xtick={1,2,4},
xticklabels={1,2,4}]
\addplot[name path=F,black,domain={-.2:4.3}] {2-x/2} node[pos=.8, above]{$f$};
\addplot[pattern=north east lines, pattern color=red]fill between[of=F and G, soft clip={domain=1:2}];
\end{axis}
\end{tikzpicture}
\end{minipage}
\end{figure}
\centering
\begin{minipage}{.5\textwidth}
Revolution Solid code
\end{minipage}
\begin{minipage}{.2\textwidth}
\centering
\begin{center}
\begin{tikzpicture}
\node [cylinder, cylinder uses custom fill, cylinder body fill=red, cylinder end fill=red!50, rotate=0, draw,
minimum height=0.01 cm, minimum width=5cm, color=black] (c) {};
% \draw[red, <->] (c.top) -- (c.bottom)
% node [at end, below, black] {height};
\draw[black, <->] ([xshift=20pt]c.north) -- ([xshift=20pt]c.center)
node [at start, above, black, xshift=35pt, yshift=-40pt] {raio: $\left( 2 - \frac{x}{2} \right)$};
\draw[black, dashed] ([xshift=10pt]c.north) -- ([xshift=20pt]c.north);
\draw[black, dashed] ([xshift=8pt]c.center) -- ([xshift=20pt]c.center);
\end{tikzpicture}
\end{center}
\end{minipage}
Thank you. | 703 | 2,019 | {"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} | 3.71875 | 4 | CC-MAIN-2024-22 | latest | en | 0.390416 |
https://www.splashlearn.com/math/measurement-games-for-5th-graders | 1,726,614,144,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.53/warc/CC-MAIN-20240917204739-20240917234739-00874.warc.gz | 914,678,288 | 32,668 | Measurement Games for 5th Grade Online
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## Conversion of Measurement Units Games
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• Conversion Of Measurement Units
##### Convert Metric Units of Length Game
Unearth the wisdom of mathematics by learning how to convert metric units of length.
5 5.MD.1
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• Conversion Of Measurement Units
##### Conversion Tables for Metric Units of Length Game
Help kids practice measurements with these conversion tables for metric units of length.
5 5.MD.1
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• Conversion Of Measurement Units
##### Conversion Tables for Metric Units of Weight Game
Take a look at conversion tables for metric units of weight with this measurement game.
5 5.MD.1
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• Conversion Of Measurement Units
##### Convert Metric Units of Weight Game
Begin the exciting journey of becoming a math wizard by learning to convert metric units of weight.
5 5.MD.1
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## Volume Games
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• Volume
##### Find the Volume using Unit Cubes Game
Kids must find the volume using unit cubes to practice geometry.
5 5.MD.3.b
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• Volume
##### Introduction to Volume Game
Introduce your child to the world of volume with this game.
5 5.MD.3.b
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• Volume
##### Use the 3D Shapes to Estimate the Volume Game
Begin the exciting journey of becoming a math wizard by using 3D shapes to estimate the volume.
4 5 5.MD.3.b
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• Volume
##### Estimate the Volume of a Given Shape Game
Kids estimate the volume of a given shape to practice their geometry skills.
4 5 5.MD.3.b
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## All Measurement Games
• Conversion Of Measurement Units
##### Conversion Tables for Metric Units of Capacity Game
Sharpen your measurement skills with conversion tables for metric units of capacity.
5 5.MD.1
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• Volume
##### Find Volume using the Formula Game
Shine bright in the math world by learning to use the formula to find the volume.
5 5.MD.5.b
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• Conversion Of Measurement Units
##### Convert Metric Units of Capacity Game
Add more arrows to your child’s math quiver by helping them convert metric units of capacity.
5 5.MD.1
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• Volume
##### Solve the Word Problems Related to Volume Game
Practice the superpower of math by learning how to solve the word problems related to volume.
5 5.MD.5.b
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• Conversion Of Measurement Units
##### Decimal Conversions for Metric Units of Length Game
Practice decimal conversions for metric units of length.
5 5.MD.1
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• Volume
##### Find the Volume of the 3D Shape by Iterating Game
Be on your way to become a mathematician by learning to find the volume of 3D shapes by iterating.
4 5 5.MD.5.c
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• Conversion Of Measurement Units
##### Decimal Conversions for Metric Units of Weight Game
Revise decimal conversions for metric units of weight with this game.
5 5.MD.1
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• Volume
##### Iterate and Find the Total Volume Game
Have your own math-themed party by learning how to iterate and find the total volume.
4 5 5.MD.5.c
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• Conversion Of Measurement Units
##### Decimal Conversions for Metric Units of Capacity Game
Recall decimal conversions for metric units of capacity with this game.
5 5.MD.1
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• Conversion Of Measurement Units
##### Conversion Tables for Customary Units of Length Game
Help your child master measurements with these conversion tables for customary units of length.
5 5.MD.1
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• Conversion Of Measurement Units
##### Convert Customary Units of Length Game
Enjoy the marvel of math-multiverse by exploring how to convert customary units of length.
5 5.MD.1
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• Conversion Of Measurement Units
##### Conversion Tables for Customary Units of Weight Game
Take a look at conversion tables for customary units of weight with this measurement game.
5 5.MD.1
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• Conversion Of Measurement Units
##### Convert Customary Units of Weight Game
Enjoy the marvel of mathematics by exploring how to convert customary units of weight.
5 5.MD.1
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• Conversion Of Measurement Units
##### Conversion Tables for Customary Units of Capacity Game
Sharpen your measurement skills with conversion tables for customary units of capacity.
5 5.MD.1
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• Conversion Of Measurement Units
##### Convert Customary Units of Capacity Game
Apply your knowledge of measurements to convert customary units of capacity.
5 5.MD.1
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• Measurement
##### Word Problems on Conversion of Metric Units Game
Apply your knowledge of measurements to solve word problems on conversion of metric units.
5 5.MD.1
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//
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4413+
4567+ | 1,080 | 4,761 | {"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} | 3.734375 | 4 | CC-MAIN-2024-38 | latest | en | 0.683078 |
https://mathoverflow.net/questions/264711/why-does-this-moir%C3%A9-pattern-look-this-way | 1,709,395,013,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475833.51/warc/CC-MAIN-20240302152131-20240302182131-00627.warc.gz | 401,795,678 | 28,588 | # Why does this Moiré pattern look this way?
I was making some gifs of Mobius transformations in Matlab, and some strange patterns began to appear. I'm not sure if a deeper knowledge of the filetype/algorithm is needed to understand this phenomenon, but I thought that there could perhaps be a purely mathematical explanation. The image is obtained by coloring the complex plane like a checkerboard, and then inverting it by taking the reciprocal of the complex conjugate. Here is the math psuedocode for the image with a given zoom $k$:
$\mbox{checkerboard}:\mathbb C \to\{\mbox{black},\mbox{white}\}$
$\mbox{checkerboard}(z):=\begin{cases} \mbox{black} & \mbox{if }\lfloor\Im(z)\rfloor+\lfloor\Re(z)\rfloor\equiv 0\mod 2\\ \mbox{white} & \mbox{if }\lfloor\Im(z)\rfloor+\lfloor\Re(z)\rfloor\equiv 1\mod 2 \end{cases}$
$\mbox{image} = \{z\in\mathbb C:|\Re(z)|,|\Im(z)|\leq 1\}$
$\mbox{color}:\mbox{image}\to\{\mbox{black},\mbox{white}\}$
$\mbox{color}(z):=\mbox{checkerboard}(k/\overline{z})$
And here are the pictures for $k=1$, $k=50$, and $k=200$. The resolution of each picture is 1000x1000.
EDIT: More specifically, why does the Moiré pattern 'sync up' with the resolution of the picture at certain points? Can the Moiré pattern be predicted?
EDIT (Partial answer): I posted this question on image processing stack exchange, and I got a decent answer for why the pattern syncs up at certain points. I would, however, love a more detailed mathematical explanation of why the pattern seems to behave differently at different such points. Here is a gif I made illustrating the interesting stuff going on when you zoom in: https://media.giphy.com/media/3og0IwUINwEQAYoUDK/source.gif
• I'm not sure what "pattern" you're referring to. The last image shows a significant amount of moiré patterns due to interference between the resolution's pixel grid and the pattern: at some places, the two are in a kind of resonance, so apparently undithered patches appear (the fact that both the pixel grid and the pattern are made essentially of squares makes sure that this resonance occurs in both dimensions simultaneously). Mar 15, 2017 at 18:12
• The pattern I'm referring to is the undithered patches! Why do the Moiré patterns "sync up" with the resolution at those places? Why is it sometimes solid black dot and sometimes another pattern? It seems as if these places are lattice points that are subject to the same inversion that defines this picture.
– B H
Mar 15, 2017 at 18:21
• en.wikipedia.org/wiki/Aliasing
– Dirk
Mar 15, 2017 at 18:44
• Similar but simpler effect: Type plot(sin(1:1000),'.') in Octave or MATLAB.
– Dirk
Mar 15, 2017 at 20:36
• I posted it! dsp.stackexchange.com/questions/38382/…
– B H
Mar 15, 2017 at 21:26
EXTENDED COMMENT
Can you specify what "sync up" means here? All of your drawings are chaotic near the origin since they are trying to pack an infinite amounts of information (the checkerboard) into a very small amount of Euclidean space (the pixels). The best you can do is approximate the Möbius transform (which has become rather chaotic) with the nearest integer value.
the Arnold Cat map is chaotic however if you have a finite amount of pixels it will become periodic with a period related to the Fibonacci numbers, as outlined in a paper of Curtis McMullen.
i think it's clear any good answer involves the word "aliasing" where if we undersample on wave can look like another. spectral theory in hyperbolic space is challenging but maybe there is a good picture here and a quantitative discussion
One might check for correlations with basis of Haar Wavelet, Discrete Fourier Transform or Walsh-Hadamard Transform. The basis is exactly the checkerboards of various sizes and scales. These inversion transforms behave like affine maps on a "curved" space. And we can try to work that out.
Some of the splotches or "aliasing" I am seeing look like hyperbolic hexagons or octagons.
So a reasonable goal would be to quantify where and how much the inverted checkerboard correlates with these patterns. Or how quickly they are converging to absolute chaos.
• i am on my phone and have no way to qualify any of this further Mar 15, 2017 at 22:47
• I am mostly interested in the points at which the pattern becomes clearer and not grey--in this picture some of them are white dots, some are black dots, some are in between.
– B H
Mar 15, 2017 at 23:00
(The following is not meant to be a full answer, but a few clues towards one, elaborating what I wrote in a comment. It's a bit too long for a comment.)
There are two things that overlap: one is the grid of square pixels, or more precisely, the "sample grid", namely the points where the "color" function has been evaluated (perhaps at the center of the square pixels, perhaps on one corner, it doesn't really matter, but certainly only at one point per pixel, that is, crucially, no anti-aliasing has been applied). The other is the pattern grid, defined by your function, and which is a conformal transformation (namely, a Möbius transformation) of a regular square grid; being a conformal transformation of a regular square grid, it looks locally like a checkerboard of squares, whose size and orientation are determined by the complex derivative of the conformal transformation applied (that is, of the Möbius tansformation).
Now to better understand what is going on, make two simplifying hypotheses: (A) that the pattern grid is, in fact, a simple square grid (colored in checkerboard pattern), not just locally like one, and (B) while we're at it, that we're only in $1$ dimension. In other words, you have an alternation of white and black intervals of equal length $\ell$ and you evaluate it at a given sample distance $d$. Clearly, if $d$ is equal to $2\ell$, the samples will systematically miss one color and "resonate" with the color, so you get a uniform color; if $d$ is only very close to $2\ell$, you get large intervals of one and the other color with length something like $d\cdot\ell/|d-2\ell|$ (an easy computation if you make a drawing — which I may or may not have gotten right, but it should look like this): it's perhaps clearer to say that the spatial frequency $1/\ell$ of the intervals is shifted by $2/d$, a phenomenon sometimes known as the Nyquist frequency.
Now if we let go of simplifying hypotheses (A) and (B), the places where you see a clear pattern patch emerge are those for which the complex derivative of the conformal transformation applied is such that the square grid which locally coincides with your pattern correctly matches up with the sample grid. One obvious possibility is that the sides of the pattern squares align with the axes of the sample grid, which happens for four different angles, and in each case for just the right size (the edge of the squares being one half the sample edge size), hence essentially in $8$ places arranged in a regular octagon since the derivative of a Möbius transformation is a degree $2$ map; another obvious one is that the diagonals of the pattern squares align with the axes of the sample grid, which also happens for four different angles (and this time the edge of the squares should be $1/\sqrt{2}$ times the sample edge size), so in another $8$ places in another octagon whose axes are shifted $\pi/8$ with respect to the other one and $\sqrt[4]{2}$ times larger. This "explains" the most prominent $16$ patches and why they form two regular octagons (and even the ratio of their sizes).
I don't think I can provide an explanation as to the color at the center of the patches, however, and I'm not sure there's much to be said on this subject.
• What can probably be explained along the same lines (and with a little more complex analysis than the first derivative) is the shape, rather than color, of the central colored region of each patch. (There is probably a complex analytic map with a singular point with a certain ramification that can be computed there.) Mar 15, 2017 at 23:10 | 1,954 | 7,982 | {"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} | 3.5625 | 4 | CC-MAIN-2024-10 | latest | en | 0.883484 |
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# M05- Question 28
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01 Mar 2010, 18:01
If $$x^2 = y + 5$$ , $$y = z - 2$$ and $$z = 2x$$ , is $$x^3 + y^2 + z$$ divisible by 7?
1. $$x \gt 0$$
2. $$y = 4$$
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01 Mar 2010, 20:16
Stmt 1) simplifying the given equation in terms of x,
x^3 + 4*x^2 - 14*x + 16.
for any value of x > 0, say x = 1, 2, it may or may not be divisible by 7. Insufficient.
Stmt 2) simplifying the above equation in terms of y,
(sqrt(y + 5))[y - 9] + [y + 21].
Substituting y = 4, (+/-3)(-5) + 25, no matter 3 is positive or negative, it is still not divisible by 7. Standard answer, confirmed not divisible - sufficient.
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02 Mar 2010, 17:36
OA is mentioned as D
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04 Mar 2010, 00:30
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Expert's post
ichha148 wrote:
If $$x^2 = y + 5$$ , $$y = z - 2$$ and $$z = 2x$$ , is $$x^3 + y^2 + z$$ divisible by 7?
1. $$x \gt 0$$
2. $$y = 4$$
We have system of equations with three equations and three unknowns, so we can solve it. As $$x$$ is squared we'll get two values for it and also two values for $$y$$ and $$z$$: two triplets. Hence we'll get two values for $$x^3 + y^2 + z$$, one divisible by 7 and another not divisible by 7. Each statement is giving us the info to decide which triplet is right, thus both statement are sufficient on its own.
Given:
$$x^2 = y + 5$$
$$y = z - 2$$ --> $$y=2x-2$$.
$$z = 2x$$
$$x^2 =2x-2+ 5$$ --> $$x^2-2x-3=0$$ --> $$x=3$$ or $$x=-1$$
$$x=3$$, $$y=4$$, $$z=6$$ - first triplet --> $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7;
$$x=-1$$, $$y=-4$$, $$z=-2$$ - second triplet --> $$x^3 + y^2 + z=-1+16-2=13$$, not divisible by 7.
(1) $$x \gt 0$$ --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.
(2) $$y = 4$$ --> --> we deal with first triplet. $$x^3 + y^2 + z=27+16+6=49$$, divisible by 7. sufficient.
Hope it helps.
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04 Mar 2010, 09:01
d
x=-1 or x=3
x>0
x=3
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06 Mar 2010, 13:44
Thanks a lot Bunuel , as always your explanation is great and lucid +1 from my side
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Re: M05- Question 28 [#permalink] 06 Mar 2010, 13:44
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# M05- Question 28
Moderator: Bunuel
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,460 | 4,180 | {"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} | 4.09375 | 4 | CC-MAIN-2017-04 | latest | en | 0.809051 |
https://www.physicsforums.com/threads/time-averaging-the-potential-energy.838079/ | 1,611,332,033,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703530835.37/warc/CC-MAIN-20210122144404-20210122174404-00226.warc.gz | 937,629,834 | 18,900 | Time-Averaging the Potential Energy
Homework Equations
Gravitational potential energy
The Attempt at a Solution
Isn't this the solution?
So Is <U> = U(r) where a = r? How do I incorporate the ##\frac {1}{\tau}## and the given "useful definite integral" from zero to 2π? Thanks
I tried taking the integral but I got undefined anwer
Gravitational potential energy = ##U = \frac {-GMm}{r^2}##
##-GMm \int_{0}^{a} \frac{1}{r^2}dr##
##-GMm (\frac {-1}{a} + \frac{-1}{0})##
I tried taking the integral but I got undefined anwer
Gravitational potential energy = ##U = \frac {-GMm}{r^2}##
##-GMm \int_{0}^{a} \frac{1}{r^2}dr##
##-GMm (\frac {-1}{a} + \frac{-1}{0})##
I thought the limits are final on top and initial on the bottom. We're going from the surface (0) to a distance a away from the planet. Why should the bottom limit be infinity?
I thought the limits are final on top and initial on the bottom. We're going from the surface (0) to a distance a away from the planet. Why should the bottom limit be infinity?
I think I understand now, in the second picture of the first post, they're bringing an object from infinitely far away to a radius of r. To clarify, I think ##F = \frac {GMm}{r^2}## is negative because it is an attractive force. Can this reasoning be justified? So for post #2, I'll just replace 0 with a , and the top with infinity. ##\int_{a}^{\infty}##. That would give me a positive value because I must put in work to move the object from a distance of a to infinity.
I think I understand now, in the second picture of the first post, they're bringing an object from infinitely far away to a radius of r. To clarify, I think ##F = \frac {GMm}{r^2}## is negative because it is an attractive force. Can this reasoning be justified? So for post #2, I'll just replace 0 with a , and the top with infinity. ##\int_{a}^{\infty}##. That would give me a positive value because I must put in work to move the object from a distance of a to infinity.
As a side note this seems to indicate that the formula for potential energy breaks down if I cross the Earth's core?
More þan one way to do it?
Now I have the solutions guide but it confuses me even more. Where do the marked definitions come from?
This is an online solutions guide, so I don't know what the teacher's lecture is.
Nathanael
Homework Helper
I think I understand now, in the second picture of the first post, they're bringing an object from infinitely far away to a radius of r.
Yes, this is because that is how gravitational potential energy is (typically) defined.
To clarify, I think ##F = \frac {GMm}{r^2}## is negative because it is an attractive force. Can this reasoning be justified?
Yes that is right. It's truly a vector equation. The force (from A on B) points along the radius vector (from A to B), but in the opposite direction (because it's attractive), and so there is the negative sign.
'll just replace 0 with a , and the top with infinity. ##\int_{a}^{\infty}##. That would give me a positive value because I must put in work to move the object from a distance of a to infinity.
You could do that. Now your distance of zero potential energy is "a." The reason it is typically defined the other way around (from infinity to a) is so that the distance of zero potential energy is infinity.
As a side note this seems to indicate that the formula for potential energy breaks down if I cross the Earth's core?
It breaks down before then, namely once you go below the surface of Earth. The reason is that the force equation assumes the Earth to be a point-object. Once you go below the surface, some of the mass will be above you, and so the force equation must be modified.
Isn't this the solution?
No it's not. That is the way to get the potential energy function from the law of gravity, but that is not what the problem asked for. Re-read the definition of the time average <f> of a function f. You want to apply this operation <U> to the potential energy U.
Calpalned
Here is the second page to the solution
It's frustrating when I don't understand the solutions guide
Nathanael
Homework Helper
I missed your post #7 at first.
The first equation you marked can be understood algebraically. ##\frac{1}{\frac{d\theta}{dt}}=\frac{dt}{d\theta}##.
Then they just used the dot notation (it's just shorthand convention for writing time derivatives) ##\dot \theta \equiv \frac{d\theta}{dt}##
The second equation you marked can be understood geometrically. h is (twice) the speed at which area is swept out by the position vector (sometimes called "areal velocity").
The understanding is geometrical, but I will try to give you the picture with words. In a short amount of time dt, the position vector sweeps out a thin triangle. The length of the long side will be R and the length of the short side will be Rdθ. The formula for the area of a triangle is half the product of these sides. So then the area per time dt is 0.5R2dθ/dt=0.5h
The fact that h is constant is one of Kepler's laws. It can also be understood in terms of angular momentum. (That law of Kepler's is equivalent to saying angular momentum is conserved.)
I hope you have been taught one of these two things (Kepler's law or that angular momentum is conserved in orbit). That is where that equation comes from.
It's frustrating when I don't understand the solutions guide
It's more fun to not check the solution and be confused than it is to check the solution and be frustrated
Calpalned
Just to clarify, it seems that time average <f(x)> is very similar to the average value of an integral
The second equation you marked can be understood geometrically. h
I can't find anything about h in my textbook
I hope you have been taught one of these two things (Kepler's law or that angular momentum is conserved in orbit). That is where that equation comes from.
Yes, I have been taught both, but I can't apply it to this context.
I know that the second law says "...a line between the sun and the planet sweeps equal areas in equal times."
Angular momentum is conserved if there is no net torque and it is ##L = I \times \omega ##
So then the area per time dt is 0.5R2dθ/dt=0.5h
So it looks like h is defined as #h = R^2 d\theta ## But if we let more time pass, then the angle theta will be greater, so shouldn't h be greater to? Thus h changes and isn't constant.
angular momentum is conserved
Kepler's second law is just another way to express angular momentum?
Nathanael
Homework Helper
Just to clarify, it seems that time average <f(x)> is very similar to the average value of an integral
Yes, they are exactly the same in logic. The only difference of course, is that the time average uses time as the independent variable.
I can't find anything about h in my textbook
So it looks like h is defined as ##h = R^2 d\theta ## But if we let more time pass, then the angle theta will be greater, so shouldn't h be greater to? Thus h changes and isn't constant.
You forgot the dt in there, ##h=R^2 \frac{d\theta}{dt}##
Kepler's second law is just another way to express angular momentum?
Yes, Kepler's law of areas implies conservation of angular momentum in orbit. The speed at which area is swept out is ##\frac{1}{2}R^2\frac{d\theta}{dt}=\text{constant by Kepler's law}## (I explain this formula in post #11). The equation for the angular momentum is ##mR^2\frac{d\theta}{dt}## which you can now see is 2m times the speed at which areas are swept out. Since this areal-speed is constant, and m is constant, that means that the angular momentum is constant.
Kepler's law is from observations, though. The theoretical reason angular momentum is conserved is because the force of gravity always acts along the radius vector, and so ##\text{torque}=\vec R \times \vec F= 0##
distance of zero potential energy is infinity
I thought the further away the object is from Earth, the more potential energy it has. For example, a rock held 10 feet above the ground has more potential energy that one held only one foot.
Is the solution from your textbook?
No it is not
Nathanael
Homework Helper
I thought the further away the object is from Earth, the more potential energy it has. For example, a rock held 10 feet above the ground has more potential energy that one held only one foot.
How much potential energy something has is really arbitrary and meaningless. How much potential energy something has is just an artifact of where you choose the zero to be. If you're dealing with earthly problems, you might choose the surface to be the location of zero potential energy. Typically with astronomical problems (like this one) you will choose the zero to be at infinity.
If it doesn't matter the magnitude of the potential energy, then what does matter? The important detail about potential energy functions is how they change. Changes in potential energy are independent of where you choose the zero to be.
Choosing a different zero just shifts the potential energy function, say from U to U+C where C is constant, but you see ##(U_f+C)-(U_i+C)=U_f-U_i## | 2,220 | 9,023 | {"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 | 4 | CC-MAIN-2021-04 | latest | en | 0.901545 |
https://byjus.com/question-answer/which-of-the-following-integers-are-less-than-9-8-7-6-5-4-11/ | 1,642,866,536,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303864.86/warc/CC-MAIN-20220122134127-20220122164127-00488.warc.gz | 211,839,549 | 20,460 | Question
# Which of the following integers are less than -9?
A
0, -1 ,-11 ,9 ,-9 ,99 ,-99 ,999 ,-999
B
-8, -7, -6, -5, -4, -11, -12, -13, -14
C
100, 1010, 11, 1, 12, 122, 1222, 121212, 3
D
-11, -12, -151, -121, -11111, -12121212, -1345, -1111123
Solution
## The correct option is C -11, -12, -151, -121, -11111, -12121212, -1345, -1111123 All integers to the left of -9 on the numberline are less than -9. Hence, -10,-11,-12,-13,-14,,-15,-16,-17,-18,-19,-20.....etc are less than -9. Mathematics
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# Calculation of wireless communication distance of Morse radio
Calculation method of wireless communication distance during free space propagation: The so-called free space propagation refers to the propagation of radio waves when the antenna is surrounded by an infinite vacuum. When a radio wave travels in free space, its energy is neither absorbed by the obstacle nor reflected or scattered.
Communication distance is related to transmit power, receiving sensitivity and operating frequency [Lfs] (dB) = 32.44 + 20lgd (km) + 20lgf (MHz)
Where Lfs is the transmission loss, d is the transmission distance, and the unit of frequency is calculated in MHz.
It can be seen from the above formula that the radio wave propagation loss (also known as attenuation) in free space is only related to the operating frequency f and the propagation distance d. When f or d doubles, [Lfs] will increase by 6dB, respectively.
The following formula illustrates the loss of radio wave propagation in free space
Los = 32.44 + 20lg d (Km) + 20lg f (MHz)
Los is the propagation loss in dB
d is the distance in Km
f is the operating frequency in MHz
The following illustrates the propagation distance of a system with a working frequency of 433.92MHz, a transmission power of + 10dBm (10mW), and a reception sensitivity of -105dBm in free space:
1. By transmitting power + 10dBm, receiving sensitivity is -105dBm
Los = 115dB
2. by Los, f
Calculated d = 9.7 km
This is the ideal transmission distance. In practical applications, it will be lower than this value. This is because wireless communication is affected by various external factors, such as the loss caused by the atmosphere, obstacles, and multipath. The reference value of is calculated into the above formula, and the approximate communication distance can be calculated. | 579 | 2,295 | {"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-2020-34 | latest | en | 0.866039 |
https://ferienwohnungen-rothfuss.de/boiler/rating/power/algorithm/ | 1,642,563,973,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301263.50/warc/CC-MAIN-20220119033421-20220119063421-00685.warc.gz | 277,626,436 | 5,156 | ## boiler rating power algorithm
### Understanding boiler specs - British Gas
The AFUE rating for an all-electric furnace or boiler is between 95% and . The lower values are for units installed outdoors because they have greater jacket heat loss. However, despite their high efficiency, the higher cost of electricity in most parts of the country makes all-electric furnaces or boilers an uneconomic choice.
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### Boiler Ratings by Brand - FurnaceCompare®
Aug 16, 2017 · This is estimated by dividing the Heating Capacity by the Fuel Input Rating (Btu/h). For example, if we have a natural gas boiler with an input rating of 100,000 Btu/h and a heating capacity of 85,000, we would have a steady state efficiency of 85 per cent.
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### Boiler Ratings | Spirax Sarco
This tutorial explains the three most commonly used boiler ratings: The 'From and at' rating for evaporation, the kW rating for heat output, and boiler horsepower. 'From and at' rating The 'from and at' rating is widely used as a datum by shell boiler manufacturers to give a boiler a rating which shows the amount of steam in kg/h which the boiler can create 'from and at 100°C',at atmospheric pressure.
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### NESHAP JJJJJJ - Area Source Boiler Rule
EPA - TTN EMC - Spectral Database - Reports - Coal Fired Boiler Emissi…
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### Making sense of boiler ratings - HPAC Magazine
Making sense of boiler ratings - HPAC Magazine
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### 1. ENERGY PERFORMANCE ASSESSMENT OF BOILERS
Jun 18, 2014 · For instance, if a boiler has an Input of 200,000 BTUH and a Gross Output of 160,000 BTUH, that boiler would be running at 80% combustion efficiency. It's not hard to figure this out. Just divide the big number into the little number and then multiply the result by 100 to get a percentage.
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### Boiler Capacity - Engineering ToolBox
Example: 8,368,000 BTU ÷ 33,472 = 250. Boiler load - The horsepower, lbs of steam per hour, or BTU is the rating indicating the maximum capacity of a boiler. When a boiler operates at its maximum rated capacity, it is referred to as maximum load. If the load varies from hour to hour, it …
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### Microwave Measurement Can Optimize Coal-Fired Boiler
Natural Gas Fired Boilers - innovativecombustion.com
Get A Quote | 552 | 2,296 | {"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} | 3.09375 | 3 | CC-MAIN-2022-05 | latest | en | 0.880266 |
http://math.stanford.edu/~church/notes.html | 1,534,699,186,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215261.83/warc/CC-MAIN-20180819165038-20180819185038-00107.warc.gz | 276,216,701 | 3,734 | ## Reading course on differential topology:
This page contains notes I wrote for a short reading course on de Rham cohomology and assorted topics in differential topology. I do not offer any guarantee that this will be useful, readable, or accessible to anyone; read at your own risk. Be warned that I intentionally avoid using standard terminology. I have not intentionally included any mistakes, but they may be present nonetheless.
Part 1: Introduction to de Rham cohomology (covector fields and d)
Part 2: Introduction to de Rham cohomology (2-covector fields)
Part 3: Introduction to de Rham cohomology (wedge of differential forms and Leibniz rule)
Part 4: Vector fields and derivations
Part 5: Vector fields, integral curves, and integral surfaces
Parts 1–3 are essentially a guided and expanded version of pages 1–4 of Bott–Tu, covering the basics of differential forms and de Rham cohomology (although the term "cohomology" does not appear in these notes!). Parts 4 and 5 ask the student to prove the fundamental theorems underlying vector fields as derivations, the existence and uniqueness of integral curves, the bracket of vector fields, and Frobenius' theorem on the integrability of distributions.
These notes will only be useful if you work out all the exercises as you go, setting the notes aside for as long as it takes you to solve them. Skimming through and promising yourself that you could do the exercises would be a complete waste of your time.
Be aware: These notes were written for a student about to start graduate math courses. Therefore although I try to maintain a down-to-earth perspective, these notes expect a high level of mathematical maturity. If you get lost, don't be discouraged! Come back in the summer before grad school, and you'll be surprised how much easier you find it.
Minimum background: Stanford students who have taken 62CM (previously called 52H) should be well-positioned to read these notes, possibly supplemented by my notes on wedge products (Sections 1 and 2). For other students, Parts 1–3 can in theory be done with a very solid computational understanding of multivariable calculus (say with A/A+ grades in 51, 52, and 113), though they may appear unmotivated without more context. Parts 4 and 5 would require experience with proofs at least at the level of 120, 171, or the 60XM series. You do not need to know what a manifold is.
For students elsewhere, to benefit from these notes you'd likely be at least a junior or senior honors math major. In any case, there is no reason to read this unless you have had a theoretical/honors multivariable calculus course, as well as a theoretical linear algebra course (including the wedge product of vector spaces, which is rarely covered in a first or second linear algebra course). I regret that I cannot respond to questions about these notes from anyone except current Stanford students. | 632 | 2,900 | {"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} | 3.625 | 4 | CC-MAIN-2018-34 | latest | en | 0.941229 |
https://mephistolessiveur.info/relationship/graphical-presentation-of-the-relationship-between-two-variables-is.php | 1,563,852,738,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528687.63/warc/CC-MAIN-20190723022935-20190723044935-00287.warc.gz | 475,228,198 | 9,746 | # Graphical presentation of the relationship between two variables is
A graphical presentation of the relationship between two quantitative variables is a. A pie chart b. a histogram c. a bar chart d. a scatter diagram ANS: D Answer to A _____ is a graphical presentation of the relationship between two quantitative variables. 1) histrogram 2) bar chart 3. We may wish to display two variables and their associations for a group of individuals. idea of the relationship between the variables across individuals is obtained. Even if the plot is not used in the final presentation, it may highlight outliers.
Dot plots can be used to look at whether values in one group are typically different from values in another group. In the example above, the plot shows it typically takes slightly longer for a scorpion to catch a prey with low activity than high activity.
Archives of Disease in Childhood,;71,F Horizontal bars denote medians for each group.
Linear Equations in 2 Variables – Graphs 01
The table below shows how social class varied between the two areas of the baby check scoring system. In both areas the mothers were mostly from social class III manual. This table shows how illness severity was related to baby-check score.
We can see the association of increasing severity with increasing score. The initial impression was not recorded for two babies. Neonatal morbidity and care-seeking behaviour in rural Bangladesh. Journal of Tropical Pediatrics,47, Amongst other things, the first table below shows how medically unqualified practitioners were used most often for all recorded forms of morbidity and for more than one in three skin rashes no care was sought.
In the second table we see that care from the district hospital appears to be the most expensive option, followed by private practitioners and village doctors. Three dimensional bar-charts can be used to show the numbers in each section of the table.
However, whilst these may look quite impressive, they do not generally make interpretation any simpler and may even 'lose the numbers'. Cardia et al, Outcome of craniocerebral trauma in infants and children, Childs Nerv.
The information shown above gender and age group could be given in a 2x3 table two rows: The three dimensional bar-chart replaces each of the six numbers with a bar of the appropriate height; however, because of the three dimensional aspect of the display it is not possible to read off the original numbers.
The display is used to impart only 6 figures, and it has lost those! It appears that for most of the years ozone was the major component of air quality standard. In sulphur dioxide was the main feature.
It is not possible to read off the actual figures. This data could have been shown as a 7x5 table.
These displays may look impressive, but they are not generally an effective way of imparting the information with minimal loss of relevant information.
Side-by-side or stacked bar charts may be an effective way of presenting data on two categorical variables. The following three examples of side by side bar charts show the data more effectively than the corresponding 5x4, 3x3 and 2x7 contingency tables would.
Farmer P et al. Community based treatment of advanced HIV disease: Bulletin of the World Health Organisation,79 12 Journal of Tropical Pediatrics,47, The following stacked bar chart is a very effective means of illustrating the fall in neural tube pregnancies over the years together with the increasing terminations, probably reflecting earlier detections in later years.
• The graphical relationship between a function & its derivative (part 1)
Hey et al, Use of local neural tube defect registers to interpret national trends, Archives of Disease in Childhood,;71,F Occasionally a picture may help the presentation.
Wound location by pathology group. Same data for the group with Charcot related ulceration. The slope seems to be positive, although it's not as positive as it was there.
### The graphical relationship between a function & its derivative (part 1) (video) | Khan Academy
So the slope looks like it is-- I'm just trying to eyeball it-- so the slope is a constant positive this entire time. We have a line with a constant positive slope.
So it might look something like this. And let me make it clear what interval I am talking about. I want these things to match up. So let me do my best. So this matches up to that. This matches up over here. And we just said we have a constant positive slope. So let's say it looks something like that over this interval. And then we look at this point right over here. So right at this point, our slope is going to be undefined. There's no way that you could find the slope over-- or this point of discontinuity.
## Graphical Displays: Two Variables
But then when we go over here, even though the value of our function has gone down, we still have a constant positive slope.
In fact, the slope of this line looks identical to the slope of this line. Let me do that in a different color. The slope of this line looks identical. So we're going to continue at that same slope. It was undefined at that point, but we're going to continue at that same slope.
And once again, it's undefined here at this point of discontinuity. So the slope will look something like that. And then we go up here. The value of the function goes up, but now the function is flat. So the slope over that interval is 0.
The slope over this interval, right over here, is 0. So we could say-- let me make it clear what interval I'm talking about-- the slope over this interval is 0. And then finally, in this last section-- let me do this in orange-- the slope becomes negative.
But it's a constant negative. And it seems actually a little bit more negative than these were positive. So I would draw it right over there. So it's a weird looking function. But the whole point of this video is to give you an intuition for thinking about what the slope of this function might look like at any point. | 1,249 | 6,020 | {"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-2019-30 | longest | en | 0.93747 |
https://brilliant.org/problems/matts-means/ | 1,660,184,187,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571232.43/warc/CC-MAIN-20220811012302-20220811042302-00568.warc.gz | 173,446,583 | 9,667 | # Matt's means
The arithmetic mean, geometric mean, and harmonic mean of $a,b,c$ are 8, 5, 3, respectively. What is the value of $a^2 + b^2 + c^2$?
This problem is posed by Matt.
Details and Assumptions:
The arithmetic mean, geometric mean, and harmonic mean of $n$ numbers $a_1, a_2, \ldots, a_n$ are (respectively)
$\frac {\sum_{i=1}^{n} a_n}{n},\quad \sqrt[n]{\prod_{i=1}^{n} a_n},\quad \left( \frac{n} { \sum_{i=1}^{n} \frac{1}{a_n} } \right) .$
× | 173 | 457 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "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} | 2.953125 | 3 | CC-MAIN-2022-33 | latest | en | 0.728671 |
http://www.chegg.com/homework-help/questions-and-answers/writing-passenger-side-mirror-car-says-warning-objects-closer-appear-warning-driver-s-mirr-q1994757 | 1,464,131,819,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049273667.68/warc/CC-MAIN-20160524002113-00024-ip-10-185-217-139.ec2.internal.warc.gz | 431,837,245 | 13,766 | The writing on the passenger-side mirror of your car says "Warning! Objects are closer than they appear." There is no such warning on the driver's mirror. Consider a typical convex passenger-side mirror with a focal length of -80 . A 1.5 -tall cyclist on a bicycle is 20 from the mirror. You are 1.0 from the mirror, and suppose, for simplicity, that the mirror, you, and the cyclist all lie along a line. How far are you from the image of the cyclist? What is the image height? | 115 | 478 | {"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-2016-22 | latest | en | 0.962294 |
http://www.dbai.tuwien.ac.at/marchives/fuzzy-mail/0645.html | 1,544,916,078,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827137.61/warc/CC-MAIN-20181215222234-20181216004234-00368.warc.gz | 363,906,900 | 2,851 | # Re:Fuzzy logic operator evaluation.
From: Walter Banks (walter@bytecraft.com)
Date: Tue Aug 21 2001 - 06:36:34 MET DST
• Next message: Stephan Lehmke: "Re: Thomas' Fuzziness and Probability"
WSiler@aol.com wrote:
> I<< > In article , Robert Dodier writes:
> > >
> >> Any such definition must ignore the relation between elements in a
> compound: if truth(B')=truth(B), then in any proposition containing A and B,
> I can swap in B' in place of B, and get exactly the same truth value for the
> compound; whether the elements are redundant, contradictory, or completely
> unrelated doesn't enter the calculation.
> >>
>
> This is easily achieved if we see that Short, Medium and Tall are all
> negatively associated. In my system, if A and B are not semantically
> inconsistent, we can use any multivalued logic we please including Zadeh's;
> but if A and B are semantically inconsistent, we MUST use A OR B = min(1, a +
> b), and A AND B = max(0, (a + b) - 1). Applying this to Earl's example, Short
> OR Medium OR Tall = 1, and Short AND Medium AND Tall = 0.
>
> This logic is not truth functional; we have to parse the (complex)
> proposition to see what logic we should use. But the results give us a
> multivalued logic which makes sense both mathematically and to the layman.
Most of us have at one time or another been shocked it the simplistic
nature of min and max to evaluate fuzzy 'AND and 'OR. We have been
equally surprised at the effectiveness of this method of combining
lingustic variables in practical applications.
The following table compares the results of the proposed operators
with the common min max inplimations. (The table was created with
an EXCEL spreadsheet I would be happy to email to anyone who
wishes to play with other ideas)
Fuzzy logic operator test
a b min(a,b) max(a,b) max(0,(a+b)-1) min(1,(a+b))
0 0 0 0 0 0
0 0.25 0 0.25 0 0.25
0 0.5 0 0.5 0 0.5
0 0.65 0 0.65 0 0.65
0 1 0 1 0 1
0.25 0.25 0.25 0.25 0 0.5
0.25 0.5 0.25 0.5 0 0.75
0.25 0.65 0.25 0.65 0 0.9
0.25 1 0.25 1 0.25 1
0.5 0.5 0.5 0.5 0 1
0.5 0.65 0.5 0.65 0.15 1
0.5 1 0.5 1 0.5 1
0.65 0.65 0.65 0.65 0.3 1
0.65 1 0.65 1 0 0.65 1
1 1 1 1 1 1
Walter Banks
############################################################################
This message was posted through the fuzzy mailing list.
(1) To subscribe to this mailing list, send a message body of
"SUB FUZZY-MAIL myFirstName mySurname" to listproc@dbai.tuwien.ac.at
(2) To unsubscribe from this mailing list, send a message body of
"UNSUB FUZZY-MAIL" or "UNSUB FUZZY-MAIL yoursubscription@email.address.com"
to listproc@dbai.tuwien.ac.at
(3) To reach the human who maintains the list, send mail to
fuzzy-owner@dbai.tuwien.ac.at
(4) WWW access and other information on Fuzzy Sets and Logic see
http://www.dbai.tuwien.ac.at/ftp/mlowner/fuzzy-mail.info
(5) WWW archive: http://www.dbai.tuwien.ac.at/marchives/fuzzy-mail/index.html
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March 9, 2011, 06:39 Boundary-layer thickness calculation. Scripting in Post. #1 New Member Andreas Join Date: Feb 2011 Location: Sweden Posts: 9 Rep Power: 8 Hey everyone, I'm currently working on on way to calculate the thickness of the momentum boundary-layer as well as the boundary layer of an airfoil. After reading around about the possibilites to do this in Post I've realised that there's no easy way to do this other than writing a script. What I want the script to do is sweep over the midspan of the airfoil creating lines normal to the surface and then saving necessary data to a file. I can then use this data in MATLAB to calculate the thickness and plot it against the relative position on the airfoil. So far I've used a script that creates a polyline along the midspan on the upper side of the airfoil and then creates a file containing x,y,z coordinates as well at the normal X and normal Y direction for 100 points along the line. (I should mention that the scripting is done in PERL) I then write a new script that looks something like this Code: ```! \$cnt=0 ! open(IN, "<./Axial-Profile_vane-0_sst_002.csv"); ! while (defined (\$line = )) {; ! @coord = split(',' \$line); ! \$coord[4] = \$coord[4] * -1; ! \$cnt = \$cnt +1; ! print "\$coord[0] \n"; PLANE: PlaneNorm # Here I create a plane with the coordinates and normal directions from axial-profile. .... CONTOUR: Contour Z #Here I create a contour plot on PlaneNorm with 3 contours in z-direction. .... POLYLINE: Polyline1 #Here I create a polyline in the normal direction at the midspan by creating it from Contour Z and contour level 2 ..... ! \$LineName = "Polylines/Line.".\$cnt.".csv"; EXPORT: #Here I export the x- and y-coordinates as well as the velocity of Polyline1 to the file \$LineName ..... ! } ! close(IN)``` The script above sweeps over the surface of the airfoil creating 100 polylines normal to the surface and then creates 100 files with the data I need to determine the B-L thickness in MATLAB. However... and this is where my problem is. The polylines that are created goes in both directions. That is, in the normal direction on both the upper AND lower side of the airfoil. What I wonder is if there's any easy way of manipulating the polylines so they're only created on the upper side of the airfoil and not across the entire domain. One possible solution I've thought of is to compare the actual y-coordinate in axial_profile each loop to the y-coordinates on each row in \$LineName. Then if the coordinates for the polyline are found to be below that from axial_profile, delete those rows. Thus giving me a file with only the points on the upper side of the airfoil. Is this something that's possible? and if so, how could I do it? My knowledge in programming is a bit limited and especially in PERL so any help will be greatly appreciated. Regards Andreas Last edited by a.ob; March 9, 2011 at 07:49.
March 11, 2011, 09:03 #2 New Member Andreas Join Date: Feb 2011 Location: Sweden Posts: 9 Rep Power: 8 I managed to come up with a solution for my problem by introducing a second while-loop within the while-loop. This loop went through all the lines of coordinates in the polyline file and compared the y-coordinate with the y-coordinate in the axial_profile file and then saved them to a new file. Thus giving me 100 polylines normal to the surface at the midspan on the suction side of the airfoil. All that's left now is to contruct a MATLAB program that calculates the B-L thickness from the data. If anyone is interested in knowing in more detail how I made the script, just send me a pm /Andreas vcvedant likes this.
March 30, 2011, 15:18 #3 New Member Irfan Join Date: Jan 2011 Location: Netherlands Posts: 16 Rep Power: 9 Hi, I am also working on the same problem. I have to calculate boundary layer characteristics of 2D airfoil (BL thickness, Displacement thickness and momentum thickness). So far I am not able to calculate them but I have an idea to calculate BL thickness which is simple but I am not sure whether is correct or not. Anyways I will share it. To calculate BL thickness: 1) Plot velocity contours in CFX-Post. For the range use option "Value List". For the values use two values of "0 and 0.99*U freestream (means the second value should be the 99% of the free stream velocity). 2) Create Polyline (say polyline 1). For method use option "From Contour". For the contour select the velocity contour and for the level select 2. This will create a polyline at 99% of the free stream velocity in the whoyle domain. 3) To extract the values of the BL, insert a Chart. For location' specify the polyline 1. For X variable use variable from the list of variables "X" and for y variable use "Y" variable from the list. This will give the coordinates of the BL. This method seems to work but I get some some strange values near the leading edge of the airfoil. Regards, Irfan
April 1, 2011, 03:40 #4 New Member Andreas Join Date: Feb 2011 Location: Sweden Posts: 9 Rep Power: 8 Hi Irfan, To calculate the displacement and momentum thickness I had to use Matlab where I imported the data from the lines normal to the surface that my script created. Each file contained X- and Y-coordinates along the line together with Total and Static Pressure. The stream situation in my simulations lacked real free stream conditions everwhere so I had to calculate my velocity from, v= sqrt(2*(P0-Pstat,wall)/rho). Because of this I didn't get very good results from your solution to finding 0,99*free stream velocity but I did get it to work very well for the flow over a flat plate. So thank you! I still haven't managed to figure out how to calculate the BL thickness in matlab but if you want some hints on how to calculate the displacement and momentum thickness, just let me know /Andreas
April 1, 2011, 11:47 #5 New Member Irfan Join Date: Jan 2011 Location: Netherlands Posts: 16 Rep Power: 9 Hi Andreas, It will be very helfpul, if you can share how to calculate momentum and displacement thickness. Actually I was trying to calculate it in some easy way, but guess there is no easy way to do. Can you give some info about it? Thanks in advance. Irfan
April 5, 2011, 04:50 #6 New Member Andreas Join Date: Feb 2011 Location: Sweden Posts: 9 Rep Power: 8 I sent you a PM /Andreas
October 8, 2013, 05:28 #7 Member Ye Zhang Join Date: Dec 2009 Location: Delft,Netherland Posts: 92 Rep Power: 10 Hello Andreas, Currently I am working on calculating the displacement thickness and momentum thickness on the airfoil. Could you please give me some hints to calculate it? and how to creat line normal to the airfoil surface? Thank you so much! Best regards, Ye
August 20, 2014, 11:18 Shape factor #8 Member venkatesh Join Date: May 2012 Posts: 93 Rep Power: 7 I am analyzing transition point over NACA 4412 airfoil at low Reynolds number using Transition Kw-SST model. I am using FLUENT for simulation. I dont know how to plot Shape factor from FLUENT result. I know that Shape factor is the displacement thickness divided by moment thickness. I read in a post that velocity profile can be obtained by creating a line perpendicular to the surface at the point of interest and with the help of that velocity variation can be plotted. I don't know how to plot shape factor in fluent. Can anybody please help me in this regard.
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Recently I watched this video about someone who made a "2D" Rubik's Cube and figured this would be easy enough to duplicate. I re-wrote the game in java because that is what I learned in school so I could do it during classes and am now converting it to C for the TI84+CE.
I haven't gotten around to all the graphics but here's a little video of the progress I have made (I've only spent one night on it so far and all I need to add is colors and a better layout)
Game has been completed!
Download Here!
matkeller19 wrote:
Recently I watched this video about someone who made a "2D" Rubik's Cube and figured this would be easy enough to duplicate. I re-wrote the game in java because that is what I learned in school so I could do it during classes and am now converting it to C for the TI84+CE.
I haven't gotten around to all the graphics but here's a little video of the progress I have made (I've only spent one night on it so far and all I need to add is colors and a better layout)
Oh, carykh! I watch that guys vids a TON because of his AI stuff. I didn't expect someone to port something like that to the CE, what about GOLAD? (Game of Life AND Death, it's also made by him)
I don't have much time to work on this but I finally got graphics somewhat working. The only problem is that some of the numbers in the array won't show:
The print function is
Code:
```void printBoard(int pos, bool selec){ int i, j; gfx_SetColor(0); gfx_FillRectangle(0,0,206,206); for(i = 0; i < BOARDY; i++){ for(j = 0; j < BOARDX; j++){ gfx_SetColor(colors[j][i]); gfx_FillRectangle(40*j+4,40*i+4,38,38); gfx_SetTextXY(i*40+14,14+j*40); gfx_PrintInt(board[j][i],2); gfx_SetColor(0xFF); if(selec){ gfx_SetColor(0xF8); } if(!home){ gfx_Rectangle(pos%10*40+3,pos/10*40+3,41,41); } } } } ```
I have figured out that the problematic line is gfx_FillRectangle(40*j+4,40*i+4,38,38 ); but I can't figure out a way to fix this. Help is appreciated
You do:
gfx_SetTextXY(i, j);
But set the rectangle with:
gfx_FillRectangle(j, i);
Quick Update. I spent the morning working on this a little bit more have finished almost everything I was planning for the game.
Things to add:
-Timer
-Move Counter
-High Score List for Time (Possibly)
-High Score List for Moves (Possibly)
-Speed everything up
The functioning part of the game is working now
P.S. Anyone have an idea for what color the selected box should be? Right now I have it outlined in pink but it doesn't stand out enough in my opinion.
matkeller19 wrote:
P.S. Anyone have an idea for what color the selected box should be? Right now I have it outlined in pink but it doesn't stand out enough in my opinion.
Maybe it should be a slightly darker shade of the color that's selected.
Also, I feel like there should be animations when you slide a tile.
Update:
I played around with the selected color and I'm not exactly set on how I want to color it (possibly just a bolder white outline OR add a symbol to the square selected such as "*"?)
Anyways, I took your advice and added sliding animations to the moves. Let me know how you think they look. I didn't want to make them too long as there will be a timer running with the game.
Anyways, let me know what you all think and what I should change/add
Suggestion: Can you make the width between the boxes be 3 not 2? Then the selection box would look centered.
Keep up the good work!
A few changes have been made. By executive decision, I chose to have the row and column highlighted gold when a box is selected. Also, a move counter has been implemented with a high score (or I guess low score) app var.
I still need to add in the timer and its appvar but this is more of a weekend project although I had little homework today and was able to complete a lot.
Looks great! My only observation is that the numbers in each square might need to come down 3 or 4 pixels to be centered vertically - if you know what I mean?
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Advertisement | 1,181 | 4,873 | {"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} | 2.984375 | 3 | CC-MAIN-2019-22 | latest | en | 0.915883 |
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how much backup can be obtained from 18650 (2600mah) battery and 5w led bulb as a load? Answered
hello;
i have some basic questions and want to clear my mind.
i want to create a portable power bank using 2*18650 (2600mah) rechargeable battery and 5w usb plugged led bulb (5v).
So keeping in mind the above values how can i determine the maximum backup this setup can provide?
i would highly appreciate if you provide some real life examples to calculate the above mentioned values.
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Use a proper battery pack or a suitable LED.
What you want is not really feasable and will not really be good for the lifespan of your battery.
If you need 5V you also need a converter that reduces the max power a little bit.
Makes me wonder why people always want to go overboard with LED's these days...
so you suggest to use 2 18650 batteries? so if i use 2 batteries how much backup will it provide for 5v led bulb?
LEDs have drop voltages around 3.2V or so, which is about how low you can safely discharge a lithium ion battery to! battery capacity is really a difficult thing to calculate accurately. The easiest way is to simply assume that you can derive the WH rating of the battery from the mAH or AH rating you see. Problem is though is that the voltage of the battery will change as it is discharged, and the discharge curve will depend on many, MANY factors, like the rate of discharge, the type of discharge (constant resistance, constant current, or constant power, or something more complicated?), the temperature, etc etc etc.
To calculate roughly how long your LED will be powered, you will need to first look into the LED driver you plan to use, if any. You will need to consider how much current the LED draws as a function of the voltage applied to it. Most good drivers will be roughly constant power, meaning that as the voltage applied to the driver sags, the current that it pulls will increase, such that the product of the voltage and current is constant. (hence the term, constant power load.) Once you know that, you need to find a discharge curve of the battery as close as possible to that 5W load, and then take the definite integral of that curve (figure out the area under the curve) to figure out the WH capacity. That can be done by drawing lots of thin tall rectangles inside the curve and adding up all the rectangle area's. Alternatively, you can import the data into a spreadsheet and calculate the WH that way. Then just extrapulate that WH figure to the load to see how long it will be powered. WH / W = H.
Even my phone cannot accurately judge how much charge is in the battery, when the percentage falls below 10%, the percentage starts dropping like a brick wall as I use my phone, it will within a minute or so get down to 1%, but if I stop using my phone, warm it up in a hot car, and/or just leave it alone, the apparent charge will climb back up to 15% or so! Similarly in cold weather the battery percentage is lower than it should be. That is because the voltage of the battery will recover slightly when not under load, (basicly it is like if you are working hard, outputting lots of work, and you get really tired and lose strength, until you are allowed to catch your breath. Same with batteries.)
thank you very much for detailed answer.
though its not that simple to know that..(scratcing my head)
Here is a great video on the topic, I think it will help you understand a little better!
1. Your battery is only 3.7 volts ( a single Lipo cell)
2. 3.7 x 2.6= 9.62 watts - So in principle 2 hours HOWEVER I have a 3 watt LED hand torch using this battery and only get 1/2 an hour of useful life from it (so much for theory)
how did you calculate 2 hours from 9.62 watts?
sorry for noob questions.
yea thanks for simplifying..stupid me...
Theory is right. Battery is rated at its 10 Hour discharge rate, capacity declines as a function of the rate of discharge.
Tanstafl and all that. | 931 | 4,057 | {"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} | 2.953125 | 3 | CC-MAIN-2020-50 | latest | en | 0.944517 |
https://math.stackexchange.com/questions/878793/a-subgroup-such-that-at-least-one-left-coset-is-a-right-coset | 1,563,515,297,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526064.11/warc/CC-MAIN-20190719053856-20190719075856-00114.warc.gz | 470,009,330 | 36,027 | A subgroup such that at least one left coset is a right coset
I saw an exercise that goes:
Let $G$ be a group of order 120, and $H$ be a subgroup of order 24. If at least one left coset of $H$ in $G$ is a right coset apart from $H$ itself, show that $H$ is normal.
Somehow I need to show that the remaining 3 left cosets are also right cosets. But I can't yet find a way to argue about that.
By assumption there exists an element $g \in G$ be such that $gH$ is a right coset of $H$, i.e. there is an element $g' \in G$ such that $gH = Hg'$. Then we also have $Hg=Hg'$ and thus $gH=Hg$, i.e. $g$ lies in the normalizer of $H$. Since 120/24=5 is prime, the normalizer of $H$ can either be $H$ or $G$ and since by assumption $g$ can be chosen outside of $H$, the normalizer of $H$ must be $G$ itself.
Hint: Let $g$ be an element of order 5 in your group. Show every left coset is of the form $g^kH$. So in particular if $hH$ ($h=g^k$) the non-trivial coset which is left and right, all left cosets are of the form $h^kH$.
An element $g$ of order 5 lies outside $H$ so the cosets are $Hg^i$, since one of these is also a left coset say $k=g^j$, $Hk=kH$, then the other cosets are $Hk^i=k^iH$, hence $H$ is normal. | 404 | 1,213 | {"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} | 3.828125 | 4 | CC-MAIN-2019-30 | latest | en | 0.914158 |
http://spotidoc.com/doc/177246/how-to-with-geometry-5 | 1,537,844,640,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160923.61/warc/CC-MAIN-20180925024239-20180925044639-00311.warc.gz | 226,553,358 | 15,418 | # How to with Geometry 5
```5
How to
Facts to Know
• • • • • • • • • • • • • • • Solve Word Problems
with Geometry
Basic Geometric Formulas
Perimeter
• Perimeter is the length around a closed shape.
It is computed by adding the length of all
the sides of the figure.
• The formula for finding the perimeter of
rectangles and other parallelograms is
P = (l + w) x 2 or P = 2 l + 2 w
Area
The area of a flat surface is a measure of how much space is covered by that surface. Area is measured
in square units.
• Area of a Rectangle
The area of a rectangle is computed by
multiplying the width of one side times the
A=lxw
• Area of a Triangle
A triangle is always one half of a rectangle
or a parallelogram. The area of a triangle is
computed by multiplying 1/2 of the base
times the height of a triangle.
A = 1– b x h
2
The area of a rectangle can also be
determined by multiplying the base
times the height.
A=bxh
• Area of a Parallelogram
The area of a parallelogram is computed by
multiplying the base times the height.
A=bxh
• Area of a Circle
To find the area of a circle, multiply π (3.14)
2
A=πr
Circumference
The circumference is the distance around a circle.
To find the circumference of a circle, multiply π
(which always equals 3.14) times the diameter or
multiply 2 times π (3.14) times the radius.
C = π d or C = 2 π r
Volume
• The formula for finding the volume of a rectangular prism, such as a box, is to multiply the length
times the width times the height. V = l x w x h
• The formula for finding the volume of a cylinder is to multiply π (3.14) times the radius squared
2
times the height. V = π x r x h
• Volume is always computed in cubic units. Use cubic inches or centimeters when determining
volume for small prisms and cylinders, and cubic feet or meters for larger ones.
21
1
Practice
• • • • • Recalling Information About Lines
and Geometry
Directions: Using the information on pages 5 and 6, choose the answer to each question.
1. Geometry comes from two Greek words meaning
a. “round” and “equal.”
b. “center” and “to measure.”
c. “earth” and “center.”
d. “earth” and “to measure.”
2. The difference between plane geometry and solid geometry is:
e. one is flat and the other is round.
f. plane is easier than solid.
g. plane has two dimensions and solid has three.
h. plane has three dimensions and solid has two.
3. A quadrilateral is a shape with
a. sides.
b. four straight sides.
c. three or more sides.
d. long sides.
4. A polygon is a shape with
e. curved sides.
f. irregular sides.
g. four straight sides.
h. three or more sides.
5. The beginning and end of a line segment are
a. capital letters.
b. points.
c. line segments that never end.
d. arrows.
6. What kind of shape is this?
e. polygon
g. regular
h. line segment
7. What kind of shape is this?
a. triangle
c. regular
d. line segment
7
▲
●
Pages 7 and 8
1. d
2. g
3. b
4. h
5. b
6. e
7. b
8. e
9. a
10. f
11. c
12. g
13. d
14. f
Pages 12 and 13
1. b
2. f
3. a
4. f
5. b
6. g
7. d
8. e
9. b
10. e
11. c
12. h
Page 17
1. 80°
2. 80°
3. 100°
4. 15°
5.
g
6.
f
7. 110°
8. 70°
9. 70°
10. 180°
11. 360°
12. 30°
13. 30°
14. 150°
15. 30°
■
• • • • • • • • • • • • • • • • • • • • • • Answer Key
6. parallelogram
7. 120 ft.
8. 36 ft.
9. 2.75 ft.
10. 7 ft.
11. 45 ft.
12. 14 ft.
Pages 36 and 37
1. 120 ft.2
2. 48 ft.2
3. 400 yds.2
4. 110.25 in.2, 176 in.2
5. 40 ft.2
6. 14.625 ft.2
7. 12 ft.2
8. 6 in.2
9. 6 ft.2
10. 21.85 ft.2
11. 37.1 ft.2
12. 117 in.2
Pages 40 and 41
1. 385 in.3
2. 125 in.3
3. 2,154 in.3
4. 565.2 in.3
5. 400 ft.3
6. 79.507 ft.3
7. 1,846.32 ft.3
8. 427⁄8 ft.3 or 42.875 ft.3
Pages 42 and 43
1. 21 m; 9.5 m2
2. 12 m; 9 m2
3. 36 m; 81 m2
4. 162 m
5. 13.72 m; 4.64 m2
6. 155 cm
7. 25.6 m; 40.87 m2
8. 195 m
9. 84 ft.2
10. 336 ft.2
11. 4 quarts
12. 13.58 m2
13. 43,560 ft.2
14. 4,840 yards2
15. 1.10 acres
Pages 20 and 21
2. diameter
3. chord
4. circumference
5. 4 ft.
6. 6 in.
7. 9 ft.
8. 8 1⁄2 in.
9. 13⁄4 in.
10. 110 ft.
11. 20.41 miles
12. 5 1⁄2 yds.
13. 452.16 ft.2
14. 615.44 in.2
15. 314 ft.2
Page 25
1. acute
2. equilateral
3. right
4. isosceles
5. obtuse
6. scalene
7. acute
8. isosceles
9. acute
10. scalene
11. acute
12. equilateral
Page 29
1. 60°
2. acute and scalene
3. 60°
4. acute and equilateral
5.
D = 55°
F = 55°
6. 50°
7. c = 2.5''
8. b = 12'
Pages 32 and 33
1. parallelogram
2. trapezoid
3. rhombus
4. rectangle
5. trapezoid
48
16. 3,780,000 pounds
17. A = 5,024 cm2
C = 251 cm
18. 16.75 minutes
19. r = 50 cm
A = 7,850 cm2
C = 314 cm
time = 20.93 min
20. r = 30 cm
A = 2,826 cm2
C = 188.4 cm
time = 12.56 min
Pages 44 and 45
1. 32 cm2 = 1,024 cm2
2. P = 2(4s) = 16 cm
3. P = 4(4s) = 32 cm
4. A = 4(1 x w) = 16 cm2
5. A = 16(1 x w) = 64 cm2
6. 50°
7. Let side of square A =
1 cm
Let the side of square
B = 4 cm
Area square A = 1 cm
Area square B = 16 cm
The area of square B
is 16 times greater
than the area of
square A.
8. Area of rectangle =
70 cm x 30 cm =
2,100 cm2
2,100 cm2 + 600 cm2
= 2,700 cm2
30 x 2,700 cm2 =
81,000 cm2 of wood
9. Yes, they have the
same area. Since you
multiply the base and
height, and these two
parallelograms use
the same numbers, so
it doesn’t matter
which is the base and
which is the height.
5
Practice
Geometry at Home
• • • • • • • • • • • • • Solving Word Problems
with Geometry
Geometry is a very important aspect of math around the home. Houses and property are measured in
geometric terms. Floor and wall coverings, heating systems, and the water supply all have a geometric
component.
For this practice page, you need to know the following:
• Wallpaper is sold in double rolls totaling 44 square feet.
• Carpeting is priced by the square yard.
• There are 9 square feet in 1 square yard.
• You cannot buy partial rolls of carpeting or wallpaper.
Directions: Use the formulas and information on page 21 and the information above to help you solve
these word problems.
1. Your mother said you can have new carpeting in your room if you compute the amount of
carpeting needed and the cost. The length of your room is 18 1– feet and the width is 17 feet.
2
The cost of one medium grade of carpeting is \$20.00 per square yard.
A. Compute the number of square feet in the room: __________________
B. Convert square feet to square yards (divide by 9): _________________
C. Compute the cost of carpeting needed (multiply by \$20.00): _________
2. You want to cover one wall of your room with neon-colored wallpaper that costs \$25.00 for a
double roll containing 44 square feet. The wall is 18 1– feet long and 10 feet high.
2
A. Compute the area of your wall in square feet. _________________
B. Determine how many rolls of wallpaper you need: _____________
C. Compute the cost of the wallpaper: _________________________
3. Your friend decided to paint the walls and the ceiling of her room with a lovely lavender paint.
One gallon of this paint will cover only 400 square feet and costs \$17.99 a gallon. These are the
dimensions of her room:
• Wall 1—21 1– feet long and 11 1– feet high
• Wall 3—21 1– feet long and 11 1– feet high
4
2
4
2
• Wall 2—20 feet long and 11 1– feet high
• Wall 4—20 feet long and 11 1– feet high
2
2
• Ceiling—21 1– feet long and 20 feet wide
4
A. Compute the area of each wall and ceiling in square feet.
Wall 1 ______ Wall 2 ______ Wall 3 ______ Wall 4 ______ Ceiling _________
B. Compute the total area in square feet: _________________________________
C. Determine how many gallons of paint are needed: _______________________
D. Compute the total cost of the paint: ___________________________________
22
5
Practice
Neighborhood Jobs
• • • • • • • • • Solving More Word Problems
with Geometry
You need money to supplement your allowance. You decide to pick up some jobs at home and in the
neighborhood so you can buy some necessities such as a scooter, a mountain bike, and a boom box.
Directions: Use the formulas and information on page 21 to help you solve these word problems.
front and back lawn. He will pay you \$0.01
a square foot. The front lawn is 62 feet long
and 38 feet wide.
5. A neighbor down the street offers to pay you
\$0.15 a square foot to paint his fence which
is 103 feet long and 6.25 feet high. He will
supply the paint.
A. What is the square footage? _________
B. How much will you be paid? ________
A. What is the square footage? __________
B. How much will you be paid? _________
trimming the edge of this lawn.
6. Your favorite uncle offers to pay you \$0.18 a
square foot to paint his board fence. It is
8 1– feet high and 26 feet long.
2
A. What is the square footage? __________
B. How much will you be paid? _________
A. What is the perimeter of the lawn? ____
B. How much will you be paid? _________
3. The back lawn is shaped like a
parallelogram. The base is 36 feet and the
height is 31 feet.
7. A neighboring mother wants you to paint a
dodge ball court with a 6-foot radius on her
driveway.
A. What is the square footage? __________
B. How much will you be paid? _________
A. What is the circumference of the court?
__________
B. What is the area in square feet of the
court? _________
4. Your next-door neighbor offers to pay you
the same price for edging and mowing his
circular lawn which has a radius of 5.5 feet.
A. What is the circumference of the lawn?
__________
B. How much will you be paid for edging?
__________
C. What is the area of the lawn in square
feet? _________
D. How much will you be paid for mowing
it? ___________
Extension
• Measure and compute the perimeter and area
• Measure and compute the perimeter and area
of a neighbor’s lawn.
23
5
Practice
• • • • • • • • • • • • • • • • • Solving Even More
Word Problems with Geometry
Directions: Use the formulas and information on page 21 to help you solve these word problems.
1. You decide to start your own sidewalk business after school selling candy bars. The candy bars
come packed in cartons which are 1 foot long, 1 foot wide, and 1 foot high (a cubic foot). How
many of these cartons could you pack into your closet which is 5 feet long, 4 feet wide, and
12 feet high? _______________
2. Your bedroom is 20 feet wide, 18 1– feet long, and 11 feet high. How many cubic feet of space are
2
3. The circular top of your water heater has a radius of 9 inches. The height of the cylinder is
8 feet 5 inches. How many cubic inches of water will the water heater hold? _________________
4. A can of cleanser has a radius of 4.5 cm and a height of 22.3 cm. How many cubic centimeters of
cleanser will the can hold? _______________
5. A closet in your parent’s bedroom is 9 1– feet long, 3 1– feet wide, and 12 feet high. How many
4
3
cubic feet of space does it have? _______________
6. This is a diagram of the living room in a house. Compute the number of cubic feet in the room.
(Hint: Do the problem in two sections.) _______________
51 feet
36 1
– feet
The height of the ceiling is 10 feet.
2
16 1
– feet
2
28 feet
7. A city water tower is 83 feet high with a radius of 25 feet. How many cubic feet of water can be
stored in the tower? ________________
8. A cubic foot of water weighs 62.38 pounds. What is the weight of the water that can be stored in
the water tower in problem #7? _______________
9. One cubic foot of water equals 7.48 gallons. How many gallons of water can be stored in the
water tower in problem #7? _______________
10. How many cubic inches of water will fit into a hose which is 50 feet long and has a radius of
1– inch? _________________
2
11. One silo or elevator for storing grain has a radius of 15 feet and is 120 feet high. How many
cubic feet of grain can be stored in it? _______________
24
• • • • • • • • • • • • • • • • • • • • • • Answer Key
Page 6
5. 405 in.2
1. 5 11⁄16"
6. 49.14 m2
2. 2 5⁄16"
7. 116.39 cm2
3. 6 3/4"
8. 86.45 m2
4. 6 7/16"
Page 16
1. 50.24 m2
Pages 7 and 8
2. 78.5 cm2
3. 314 cm2
4. 452.16 cm2
Page 10
5. 1,256 cm2
1. 18.2 cm
6. 615.44 ft.2
2. 26.2 cm
7. 706.5 in.2
3. 131⁄2 cm
8. 1,962.5 m2
4. 161⁄2 ft.
5. 151⁄4 in.
Page 18
6. 183⁄8 cm.
1. 105 m3
2. 720 ft.3
3. 343 cm3
Page 11
4. 165 in.3
1. 15.6 cm
5. 240 yd.3
2. 111⁄4 in.
6. 67.032 m3
3. 24.4 m
7. 92.736 m3
4. 183⁄4 ft.
8. 694.512 cm3
5. 74.4 m
9. 1,728 ft.3
6. 64 yd.
10. 86 6/8 ft.3
7. 137.4 cm
8. 105.3 m
Page 19
1. 351.68 m3
Page 12
2. 169.56 cm3
1. 19.1 m
3. 282.6 cm3
2. 22.6 m
4. 18.84 in.3
3. 26 in.
5. 50,240 cm3
4. 201⁄2 ft.
6. 1,538.6 ft.3
5. 25.12 m
6. 37.68 in.
Pages 20–23
7. 31.4 cm
8. 21.98 m
Page 24
Page 14
1. 6 lbs. 4 oz.
2
1. 41 m
2. 1 ton 300 lbs.
2. 126 yd.2
3. 4,000 cassettes
3. 67.5 cm2
4. 100 pills
4. 6.08 m2
5. 100,000 pills
2
5. 34 ft.
6. 2,000 dictionaries
6. 16 1/4 in.2
7. 12,000 staplers
7. 3,680 m2
8. 100 people
8. 7,500 mm2
9. 500 mg or 1/2 g
10. 220 kg
Page 15
11. 4,400 kg
1. 24 ft.2
12. 2,200 clips
2. 45 yd.2
13. 6,400 calculators
3. 11.66 cm2
14. 40 cameras
4. 27.72 cm2
Page 26
1. 8 fl. oz.
2. 16 fl. oz.
3. 32 fl. oz.
4. 48 fl. oz.
5. 64 fl. oz.
6. 72 fl. oz.
7. 32 fl. oz.
8. 64 fl. oz.
9. 160 fl. oz.
10. 96 fl. oz.
11. 4 qt.
12. 16 qt.
13. 128 fl. oz.
14. 60 qt.
15. 1,920 fl. oz.
16. 16 fl. oz.
17. 48 fl. oz.
18. 112 fl. oz.
19. 40 pints
20. 176 cups
21. 120 pints
22. 1,280 fl. oz.
23. 34 cups
24. 176 fl. oz.
25. 344 fl. oz.
Page 27
1. 30 mL
2. 240 mL
3. 1,000 mL
4. 960 mL
5. 40 mL
6. 480 mL
7. 3,840 mL
8. 3.84 L
9. 38.4 L
10. 69.1 L
11. 960 L
12. 96 L
13. 96 L
14. 1920
15. 360 L
Page 28
1. 2 qt.
2. 12 mL
3. 80 mL
4. 336 mL
5. 50 pennies
6. 432 mL
47
7.
8.
9.
10.
11.
12.
24 fl. oz.
384 mL
128 quarters
19.2 L
8 times
48 cups
Page 30
1. 40° acute
2. 120° obtuse
3. 180° straight
4. 90° right
5. 50° acute
6. 130° obtuse
7. 250° reflex
8. 215° reflex
9. 90° right
10. 80° acute
Page 31
1. <BAC = 100°
1. <CBA = 35°
1. <ACB = 45°
1. ▲ABC = 180°
2. <CDE = 50°
1. <ECD = 70°
1. <DEC = 60°
1. ▲DEC = 180°
3. <LMN = 90°
1. <MNL = 30°
1. <MLN = 60°
1. ▲LMN = 180°
4. <MNO = 25°
1. <OMN = 65°
1. <MON = 90°
1. ▲MNO = 180°
5. <XYZ = 60°
1. <ZXY = 60°
1. <YZX = 60°
1. ▲XYZ = 180°
6. <WPO = 154°
1. <POW = 11°
1. <PWO = 15°
1. ▲WPO = 180°
? ? ?
Page 6
1. change
subtraction
\$2.12
2. money spent
multiplication
\$36.64
3. split evenly
division
28 cards
4. amount needed
subtraction
\$10.33
5. total cost
\$129.17
6. how much saved
subtraction
\$2.21
7. total cost
multiplication
\$41.58
Page 7
1. change
subtraction
\$16.11
2. % discount
multiplication
\$59.80
3. total cost
\$50.73
4. times as much
multiplication
\$5,325
5. average
division
11.03 miles
6. total cost
\$1,342.97
7. times as much
multiplication
\$350.10
8. total
125.3 miles
Page 8
1. how much change
subtraction
\$8.05
• • • • • • • • • • • • • • • • • • • • • • Answer Key
2. how much saved
subtraction
\$6.95
3. product
multiplication
\$113.85
4. how much left
subtraction
\$25.41
5. split evenly
division
\$1.59
6. share evenly
division
27 CDs
7. discount
multiplication
\$3.19
8. difference
subtraction
\$3.11
Page 12
1. multiplication
\$22.68
\$8.97
3. multiplication
\$59.67
\$13.46
5. division
\$17.04
6. subtraction
\$2.70
Challenge:
\$70.20; 1 large
cola, 1 Double Bean
Taco; \$0.39
Page 14
1. 7/12 miles
2. 5/12 miles
3. 2 2/3 miles
4. 1/3 mile
5. 1 1/6 miles
6. 8 miles
7. 1 1/4 miles
8. 4 5/18 miles
9. 1/2 mile
10. 26 2/3 miles
Page 10
\$34.42
2. subtraction
\$2.55
3. subtraction
\$7.50
\$40.47
5. subtraction
\$3.50
\$78.41
vary.
Page 15
1. 3/4 pizza
2. 10 cups
3. 3 3/4 pizzas
4. 1 1/2 pizzas
5. 1/2 pizza
6. 1/10 cake
7. 15/16 cake
8. 14 cups
9. 5/8 pizza
10. 81 ounces
11. 338 ounces
12. 1 1/2 ounces
Page 11
1. multiplication
\$45.00
2. division
\$3.75
3. multiplication
\$126.50
4. multiplication
\$99.80
5. multiplication
\$119.25
6. division
\$1.79
Challenge: \$11.25; \$8.75
Extension: 4 2/3 pizzas
Page 16
1. 33 3/4 miles
2. 39/40 mile
46
3. 7/10 mile
4. 1/2 lb.
5. 14 2/3 miles
6. 9 lbs.
7. 4 5/3 miles
8. 1 13/40 sec.
9. 12 3/8 miles
10. 7 17/24 miles
vary.
Page 18
1. \$62.29; \$237.71
2. \$77.50; \$160.21
3. \$11.88; \$148.33
4. \$7.46; \$29.82;
\$118.51
5. \$57.94; \$60.57
6. \$10.00; \$60.00;
\$0.57
7. \$299.43
8. no
Page 19
1. 60%
2. 24 shots
3. 71% or 71.4%
4. 17 shots
5. 89% or 89.3%
6. 19 shots
7. 94% or 94.4%
8. 65% or 64.7%
9. 64% or 63.9%
10. 4 shots
vary.
Page 20
1. 0.625 gallons
2. 25.2 lbs.
3. 4.4 oz.
4. 43.2 lbs.
5. 2.4 qts.
6. 114.7 lbs.
7. 19.5 lbs.
8. 3.75 or 3 3/4 times
9. 56% or 55.6%
10. 41%
Page 22
1. A. 314.5 sq. ft.
B. 34.9 or
35 sq. yd.
• • • • • • • • • • • • • • • • • • • • • • Answer Key
? ? ?
C. \$698.00 or
\$700.00
2. A. 185 sq. ft.
B. 5 rolls
C. \$125
3. A. 244 3/8 sq. ft.
230 sq. ft.;
244 3/8 sq. ft.;
230 sq. ft.;
425 sq. ft.
B. 1,373 3/4 sq. ft.
or 1,374 sq. ft.
C. 4 gallons
D. \$71.96
Page 23
1. A.
B.
2. A.
B.
3. A.
B.
4. A.
B.
C.
D.
5. A.
B.
6. A.
B.
7. A.
B.
2,356 sq. ft.
\$23.56
200 ft.
\$6.00
1,116 sq. ft.
\$11.16
34.54 ft.
\$1.04
94.99 sq. ft.
\$0.95
643.75 sq. ft.
\$96.56
221 sq. ft.
\$39.78
37.68 ft.
113.04 sq. ft.
vary.
Page 24
1. 240 cartons
2. 4,070 cu. ft.
3. 25,688.34 cu. in.
4. 1,417.95 cu. cm
5. 370 cu. ft.
6. 14,820 cu. ft.
7. 162,887.5 cu. ft.
8. 10,160,922 lb.
9. 1,218,398.5 gallons
10. 471 cu. in.
11. 84,780 cu.ft.
Page 26
1. \$45.60
2. \$34.13
3. \$104.65
4.
5.
6.
7.
8.
9.
\$43.51
\$32.95
\$29.25
\$36.86
\$30,555.64
Monday and
Tuesday = Saturday
10. \$17,111.16
11. \$12,473.53
4.
5.
6.
7.
Page 27
1. \$101.47
2. \$12.27
DVD player;
\$179.67
\$5.96 change
4. \$786.15
machine/phone is
\$11.24 cheaper.
6. \$19.20
7. \$49.76
8. Boom Box City
\$25.46 less
9. \$16.30
10. 25%
8.
0 quarters,
2 half dollars
6, 9, 12, 15, 18
300, 350, 400, 450,
500
3 footballs, 6 tennis
balls, 3 baseballs, 2
Jack is 26 years
old
Marie is 22 years
old; Mother is 44
years old
Page 31
1. \$360.00
3. 240 total
16 skirts
32 jeans
64 shorts
128 blouses
4. \$372.00 total
Elaine \$12.00
Christina \$24.00
Alyse \$48.00
Doreen \$96.00
Melissa \$192.00
5. James 2 years old
Raymond 3 years
old
Brett 4 1/2 years
old
John 6 years old
Robert 11 years old
Page 28
1. 22.86 miles per day
2. 4 hr. 24 min.
3. 3 hr. 20 min.
4. 40 m.p.h.
5. 1 mile per minute
6. \$21.00
7. \$3.20
8. \$0.82
9. \$46.74
Page 32
1. 3 hr. 2 min.
2. 31 games
3. 81 times
4. 30 names
5. 20 points on 8th
game; 35 points on
14th game
6. 35 players are 13
years old
Page 30
1. 6 tops/4 skorts
2. 3 pennies, 3
nickels, 0 dimes, 3
quarters,
3. A. 1 penny, 0
nickels,
4 dimes,
4 quarters,
0 half dollars
B. 1 penny, 4
nickels,
2 dimes,
Page 34
1. n = 36–23
n = 13
13 years old
2. n = (4 x 15) + 2
47
n = 62
62 CDs
3. n = 216–122
n = 94
94 lb.
4. n = 25 x .60
n = 15
15 shots
5. n = 22 – 7
n = 15
15 minutes
6. n = 1,145 – 316
n = 829
829 words
7. n = 88 x 3/4
n = 66
66 minutes
vary.
Page 35
1. n + (n + 28) = 50
2n + 28 = 50
n = 11
Mother is 39 years
old.
Sarah is 11 years
old.
2. n + (n + 140)= 336
2n + 140= 336
n = 98
Joe weighs 98 lbs.
lbs.
3. n + 4n + 22 = 122
n = 25
Melissa has \$25.00.
Christina has \$97.00.
4. n + 2n = 669
3n = 669
n = 223
words.
words.
5. n + 4n = 15
5n = 15
n=3
Nicholas is 3 years
old.
Norman is 12 years
old.
10
Word
Problems
• • • • • • • • • • • • • • • • Real Life Geometry
The students at Wood Hill Elementary were surprised one day in gym class when the coach
handed out a math test.
“Good athletes have to be good students, too,” said the coach. “You don’t want to be
disqualified from a team because of poor grades. Answer these questions.” He gave them
each sheet of paper.
1. The volleyball net is 1 m wide and 9.50 m long. What are the perimeter and the area of the net?
perimeter = ____________ area = ____________
2. The service area in volleyball is 3 m long and 3 m wide.
What is the perimeter and the area of the service area? perimeter = ____________
area = ____________
3. The volleyball court is 18 m long and 9 m wide. It is divided into two halves. What are the
perimeter and the area of each half? perimeter = ____________ area = ____________
4. Tiffany runs 3 times around the volleyball court. How far does she run? ____________
5. The badminton net is 0.76 m wide and 6.10 m long. What are the perimeter and area of the net?
perimeter = ____________ area = ____________
6. The badminton net is 1.55 m high (1 m = 100 cm). What is its height in centimeters?
____________
7. The badminton court is 13.40 m long and 6.10 m wide. It is divided into two halves. What is the
perimeter of each half? ____________ How many square meters of material would it take to
cover one half? ____________
8. Dan runs 5 times around the badminton court. How far does he run? ____________
Ira has agreed to do a project for his father in exchange for a new snowboard this winter. Ira
needs to paint the garden shed in his backyard. His father needs to buy the paint for the
shed and has asked Ira to measure the size of each wall to determine the amount of paint
he should purchase. There are four walls to the shed.
9. After measuring the walls, Ira has determined that each wall is 7 feet high and 12 feet long. What
is the area of each wall? ____________
10. What is the total area of the walls around the shed? ____________
11. If a quart of paint covers 100 square feet, how many quarts of paint must Ira’s father purchase?
____________
12. Dan O’Leary has decided to plant a garden. He wants to make it 10.1 m long and 4.2 m wide.
However, in order to keep the rabbits out, Dan needs a fence surrounding the garden. He decides
to make the fence 11.2 m long and 5.0 m wide. What is the area between the fence and the
garden? ____________
(Hint: Find the area for the garden. Then, find the area of the space surrounded by the fence.)
42
10
Word
Problems
• • • • • • • • • • • • • • • • Real Life Geometry
The rod is an old unit of measurement of length. A rod is 16 1/2 feet long. A square rod is
a square plot of ground. Each side of the plot is 16 1/2 feet long. An acre is 160 square
rods.
13. How many square feet are in one acre? ____________
14. How many square yards are in one acre? ____________
15. A football field is 160 feet wide and 300 feet from goal line to goal line. What is the area of the
football field in acres? (round to the nearest hundredth) ____________
16. Mr. Anderson is a farmer. He has a 300-acre field. He expects to harvest about 225 bushels of
corn per acre. A bushel of corn weighs about 56 pounds. How many pounds of corn would Tom
get from his field? ____________
Mr. Peterson is a math teacher. Dinner at his house is
unusual. One night, after a pizza was delivered, he
posed the following questions to his hungry family.
“Before we eat this delicious pizza, let’s answer a few
interesting questions,” he said. Everyone groaned. “The
radius of a regular pizza is 40 cm. Now, listen closely to
my questions.”
17. Find the area and the circumference of the pizza.
area = _______ circumference = _______
18. If an ant walked 1 cm in 4 seconds, how long would it take for the ant to walk the
circumference? ____________
19. Repeat question #18 assuming the radius of the pizza is increased by 25%. Find the following
measurements:
area = _______
circumference = _______
ant’s time = _______
20. Repeat question #18 assuming the radius of the pizza is decreased by 25%. Find the following
measurements:
area = _______
circumference = _______
ant’s time = _______
43
11
• • • • • • • • • • • • Carpenters and Pyramids
Brain
Teasers
Directions: Answer these brain teaser questions.
1. Four strips of paneling 40 cm long and 4 cm wide are arranged to form a square, like a picture
frame. (Note: The ends of the strip of paneling will overlap.)
What is the area of the inner square in square cm? ____________
It’s “Challenge Day” in Mr. Peterson’s math class. “Take out a sheet of paper,” he says. “Now,
draw a 2 cm x 2 cm square. Listen closely.
2. What is the equation for finding the perimeter of a square that is twice the size of the original
square? _________________________
3. What is the equation for finding the perimeter of a square that is four times the size of the original
square? _________________________
4. What is the equation for finding the area of a square that is twice the size of the original
square? _________________________
5. What is the equation for finding the area of a square that is four times the size of the original
square? _________________________
6. In the following diagram of the front view of the Great Pyramid, the measure of PRQ is 120°
degrees, and the measure of PST is 110° degrees.
What is the measure of
RPS in degrees? _________________________
P
Q
R
(Hint: The sum of the angles in a triangle is
180 degrees. A straight line is 180 degrees.
Use the known angles to find the unknown
angles. See Unit 3 on supplementary angles
S
T
7. One side of square B is four times the length of one side of square A. How many times greater is
the area of square B than the area of square A? _________________________
B
A
44
11
Brain
Teasers
• • • • • • • • • • • • Carpenters and Pyramids
8. Two carpenters decided to design desks for students at James Hart Junior High. The dimensions
of the desks are as shown. How much wood in cm2 would they need for 30 desks? __________
70 cm
30 cm
20 cm
20 cm
(Hint: What is the area of one desk? Find the area of each part and add all the areas to find the
total area. If there are 30 desks, how much wood in square centimeters is needed?)
9. Do these parallelograms have the same area? How do you know? __________________________
_______________________________________________________________________________
_______________________________________________________________________________
2 cm
5 cm
2 cm
5 cm
(Hint: Review the formula for finding the area of a parallelogram.)
45
▲
●
Pages 7 and 8
1. d
2. g
3. b
4. h
5. b
6. e
7. b
8. e
9. a
10. f
11. c
12. g
13. d
14. f
Pages 12 and 13
1. b
2. f
3. a
4. f
5. b
6. g
7. d
8. e
9. b
10. e
11. c
12. h
Page 17
1. 80°
2. 80°
3. 100°
4. 15°
5.
g
6.
f
7. 110°
8. 70°
9. 70°
10. 180°
11. 360°
12. 30°
13. 30°
14. 150°
15. 30°
■
• • • • • • • • • • • • • • • • • • • • • • Answer Key
6. parallelogram
7. 120 ft.
8. 36 ft.
9. 2.75 ft.
10. 7 ft.
11. 45 ft.
12. 14 ft.
Pages 36 and 37
1. 120 ft.2
2. 48 ft.2
3. 400 yds.2
4. 110.25 in.2, 176 in.2
5. 40 ft.2
6. 14.625 ft.2
7. 12 ft.2
8. 6 in.2
9. 6 ft.2
10. 21.85 ft.2
11. 37.1 ft.2
12. 117 in.2
Pages 40 and 41
1. 385 in.3
2. 125 in.3
3. 2,154 in.3
4. 565.2 in.3
5. 400 ft.3
6. 79.507 ft.3
7. 1,846.32 ft.3
8. 427⁄8 ft.3 or 42.875 ft.3
Pages 42 and 43
1. 21 m; 9.5 m2
2. 12 m; 9 m2
3. 36 m; 81 m2
4. 162 m
5. 13.72 m; 4.64 m2
6. 155 cm
7. 25.6 m; 40.87 m2
8. 195 m
9. 84 ft.2
10. 336 ft.2
11. 4 quarts
12. 13.58 m2
13. 43,560 ft.2
14. 4,840 yards2
15. 1.10 acres
Pages 20 and 21
2. diameter
3. chord
4. circumference
5. 4 ft.
6. 6 in.
7. 9 ft.
8. 8 1⁄2 in.
9. 13⁄4 in.
10. 110 ft.
11. 20.41 miles
12. 5 1⁄2 yds.
13. 452.16 ft.2
14. 615.44 in.2
15. 314 ft.2
Page 25
1. acute
2. equilateral
3. right
4. isosceles
5. obtuse
6. scalene
7. acute
8. isosceles
9. acute
10. scalene
11. acute
12. equilateral
Page 29
1. 60°
2. acute and scalene
3. 60°
4. acute and equilateral
5.
D = 55°
F = 55°
6. 50°
7. c = 2.5''
8. b = 12'
Pages 32 and 33
1. parallelogram
2. trapezoid
3. rhombus
4. rectangle
5. trapezoid
48
16. 3,780,000 pounds
17. A = 5,024 cm2
C = 251 cm
18. 16.75 minutes
19. r = 50 cm
A = 7,850 cm2
C = 314 cm
time = 20.93 min
20. r = 30 cm
A = 2,826 cm2
C = 188.4 cm
time = 12.56 min
Pages 44 and 45
1. 32 cm2 = 1,024 cm2
2. P = 2(4s) = 16 cm
3. P = 4(4s) = 32 cm
4. A = 4(1 x w) = 16 cm2
5. A = 16(1 x w) = 64 cm2
6. 50°
7. Let side of square A =
1 cm
Let the side of square
B = 4 cm
Area square A = 1 cm
Area square B = 16 cm
The area of square B
is 16 times greater
than the area of
square A.
8. Area of rectangle =
70 cm x 30 cm =
2,100 cm2
2,100 cm2 + 600 cm2
= 2,700 cm2
30 x 2,700 cm2 =
81,000 cm2 of wood
9. Yes, they have the
same area. Since you
multiply the base and
height, and these two
parallelograms use
the same numbers, so
it doesn’t matter
which is the base and
which is the height.
``` | 10,639 | 27,717 | {"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} | 4.6875 | 5 | CC-MAIN-2018-39 | latest | en | 0.880319 |
http://uk.mathworks.com/help/images/ref/ordfilt2.html?nocookie=true | 1,433,347,317,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1433195036776.1/warc/CC-MAIN-20150601214356-00076-ip-10-180-206-219.ec2.internal.warc.gz | 171,817,337 | 11,111 | # ordfilt2
2-D order-statistic filtering
## Syntax
• `B = ordfilt2(A,order,domain)` example
• `B = ordfilt2(A,order,domain,S)`
• `B = ordfilt2(___,padopt)`
## Description
example
````B = ordfilt2(A,order,domain)` replaces each element in `A` by the `order`th element in the sorted set of neighbors specified by the nonzero elements in `domain`. This function supports code generation (see Tips).```
````B = ordfilt2(A,order,domain,S)` filters `A`, where `ordfilt2` uses the values of `S` corresponding to the nonzero values of `domain` as additive offsets. You can use this syntax to implement grayscale morphological operations, including grayscale dilation and erosion.```
````B = ordfilt2(___,padopt)` filters `A`, where `padopt` specifies how `ordfilt2` pads the matrix boundaries.```
## Examples
collapse all
### Filter Image with Maximum Filter
Read image into workspace and display it.
```A = imread('snowflakes.png'); figure imshow(A) ```
Filter the image and display the result.
```B = ordfilt2(A,25,true(5)); figure imshow(B) ```
## Input Arguments
collapse all
### `A` — Input matrix2-D, real, nonsparse, numeric or logical matrix
Input matrix, specified as a 2-D, real, nonsparse, numeric or logical array.
Example: `A = imread('snowflakes.png');`
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `uint8` | `uint16` | `uint32` | `logical`
### `order` — Element to replace the target pixelreal scalar integer
Element to replace the target pixel, specified as a real scalar integer of class `double`.
Example: `B = ordfilt2(A,25,true(5));`
Data Types: `double`
### `domain` — Neighborhoodnumeric or logical matrix
Neighborhood, specified as a numeric or logical matrix, containing `1`'s and `0`'s. `domain` is equivalent to the structuring element used for binary image operations. The 1-valued elements define the neighborhood for the filtering operation. The following table gives examples of some common filters.
Type of Filtering OperationMATLAB codeNeighborhood
Median filter`B = ordfilt2(A,5,ones(3,3))`
Minimum filter`B = ordfilt2(A,1,ones(3,3))`
Maximum filter`B = ordfilt2(A,9,ones(3,3))`
Minimum of north, east, south, and west neighbors`B = ordfilt2(A,1,[0 1 0; 1 0 1; 0 1 0])`
Example: `B = ordfilt2(A,25,true(5));`
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `logical`
### `S` — Additive offsetsmatrix
Additive offsets, specified as a matrix the same size as `domain`.
Data Types: `single` | `double` | `int8` | `int16` | `int32` | `int64` | `uint8` | `uint16` | `uint32` | `uint64` | `logical`
### `padopt` — Padding option`'zeros'` (default) | `'symmetric'`
Padding option, specified as either of the strings in this table. `'zeros'` or `'symmetric'`.
OptionDescription
`'zeros'`Pad array boundaries with `0`'s.
`'symmetric'`
Pad array with mirror reflections of itself.
Data Types: `char`
## Output Arguments
collapse all
### `B` — Output image2-D array
Output image, returned as a 2-D array of the same class as the input image `A`.
## More About
collapse all
### Tips
• This function supports the generation of C code using MATLAB® Coder™. Note that if you choose the generic `MATLAB Host Computer` target platform, the function generates code that uses a precompiled, platform-specific shared library. Use of a shared library preserves performance optimizations but limits the target platforms for which code can be generated. For more information, see Understanding Code Generation with Image Processing Toolbox.
When generating code, the `padopt` argument must be a compile-time constant.
• When working with large domain matrices that do not contain any zero-valued elements, `ordfilt2` can achieve higher performance if `A` is in an integer data format (`uint8`, `int8`, `uint16`, `int16`). The gain in speed is larger for `uint8` and `int8` than for the 16-bit data types. For 8-bit data formats, the domain matrix must contain seven or more rows. For 16-bit data formats, the domain matrix must contain three or more rows and 520 or more elements.
## References
[1] Haralick, Robert M., and Linda G. Shapiro, Computer and Robot Vision, Volume I, Addison-Wesley, 1992.
[2] Huang, T.S., G.J.Yang, and G.Y.Tang. "A fast two-dimensional median filtering algorithm.", IEEE transactions on Acoustics, Speech and Signal Processing, Vol ASSP 27, No. 1, February 1979
## See Also
Was this topic helpful?
Get trial now | 1,236 | 4,481 | {"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-2015-22 | latest | en | 0.593147 |
http://betterlesson.com/lesson/resource/3287094/totd-surfacearea-pptx | 1,487,597,939,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170562.59/warc/CC-MAIN-20170219104610-00126-ip-10-171-10-108.ec2.internal.warc.gz | 27,002,996 | 20,763 | ## TOTD.SurfaceArea.pptx - Section 5: Closing Summary
TOTD.SurfaceArea.pptx
TOTD.SurfaceArea.pptx
# What in the World is Surface Area?
Unit 5: Area & Volume
Lesson 9 of 14
## Big Idea: Pull up your net and see what surfaces: Using nets to find surface area
Print Lesson
2 teachers like this lesson
Standards:
65 minutes
### Michelle Braggs
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Environment: Urban | 237 | 972 | {"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} | 2.78125 | 3 | CC-MAIN-2017-09 | latest | en | 0.76618 |
https://learn.sparkfun.com/tutorials/series-and-parallel-circuits/calculating-equivalent-resistances-in-series-circuits | 1,529,306,983,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860089.13/warc/CC-MAIN-20180618070542-20180618090542-00083.warc.gz | 682,342,894 | 12,136 | # Series and Parallel Circuits
Pages
Contributors: Pete-O
## Calculating Equivalent Resistances in Series Circuits
Here’s some information that may be of some more practical use to you. When we put resistors together like this, in series and parallel, we change the way current flows through them. For example, if we have a 10V supply across a 10kΩ resistor, Ohm’s law says we’ve got 1mA of current flowing.
If we then put another 10kΩ resistor in series with the first and leave the supply unchanged, we’ve cut the current in half because the resistance is doubled.
In other words, there’s still only one path for current to take and we just made it even harder for current to flow. How much harder? 10kΩ + 10kΩ = 20kΩ. And, that’s how we calculate resistors in series – just add their values.
To put this equation more generally: the total resistance of N – some arbitrary number of – resistors is their total sum.
In 2003, CU student Nate Seidle blew a power supply in his dorm room and, in lieu of a way to order easy replacements, decided to start his own company. Since then, SparkFun has been committed to sustainably helping our world achieve electronics literacy from our headquarters in Boulder, Colorado.
No matter your vision, SparkFun's products and resources are designed to make the world of electronics more accessible. In addition to over 2,000 open source components and widgets, SparkFun offers curriculum, training and online tutorials designed to help demystify the wonderful world of embedded electronics. We're here to help you start something. | 349 | 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} | 3.1875 | 3 | CC-MAIN-2018-26 | longest | en | 0.931277 |
https://cboard.cprogramming.com/cplusplus-programming/155501-function-overloading-error.html | 1,505,951,812,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687582.7/warc/CC-MAIN-20170920232245-20170921012245-00234.warc.gz | 645,492,179 | 15,736 | I have the following code
Code:
```//illustrates overloading functions
#include <iostream>
using namespace std;
class rational{
private:
long a, q;
enum { BIG = 100};
public:
rational(int n = 0) : a(n), q(1) {}
rational(int i, int j) : a(i), q(j) {}
rational(double r) : a(static_cast<long> (r * BIG) ), q(BIG) {}
void print() const { cout << a << "/" << q; }
operator double() { return static_cast<double>(a) / q; }
};
//functions
inline int greater(int i, int j){
return ( i > j ? i : j);
}
inline double greater(double x, double y){
return (x > y ? x : y);
}
inline rational greater(rational w, rational z){
return (w > z ? w : z);
}
int main(void){
int i = 10, j = 5;
float x = 7.0;
double y = 14.5;
rational w(10), z(3.5), zmax;
cout <<"\ngreater(" << i << "," << j << ") = " << greater(i,j);
cout <<"\ngreater(" << x << "," << y << ") = " << greater(x,y);
cout <<"\ngreater(" << i << ",";
z.print();
cout <<") = " << greater(static_cast<rational>(i), z);
zmax = greater(w,z);
cout <<"\ngreater(";
w.print();
cout << ", ";
z.print();
cout << ") = ";
zmax.print();
cout << endl;
return 0;
}```
and getting the following errors.
Code:
```\$ g++ rational.cpp -o rational
rational.cpp: In function ‘int main()’:
rational.cpp:43:52: error: reference to ‘greater’ is ambiguous
rational.cpp:32:17: error: candidates are: rational greater(rational, rational)
rational.cpp:28:15: error: double greater(double, double)
rational.cpp:24:12: error: int greater(int, int)
In file included from /usr/include/c++/4.7/string:50:0,
from /usr/include/c++/4.7/bits/locale_classes.h:42,
from /usr/include/c++/4.7/bits/ios_base.h:43,
from /usr/include/c++/4.7/ios:43,
from /usr/include/c++/4.7/ostream:40,
from /usr/include/c++/4.7/iostream:40,
from rational.cpp:2:
/usr/include/c++/4.7/bits/stl_function.h:224:12: error: template<class _Tp> struct std::greater
rational.cpp:44:52: error: reference to ‘greater’ is ambiguous
rational.cpp:32:17: error: candidates are: rational greater(rational, rational)
rational.cpp:28:15: error: double greater(double, double)
rational.cpp:24:12: error: int greater(int, int)
In file included from /usr/include/c++/4.7/string:50:0,
from /usr/include/c++/4.7/bits/locale_classes.h:42,
from /usr/include/c++/4.7/bits/ios_base.h:43,
from /usr/include/c++/4.7/ios:43,
from /usr/include/c++/4.7/ostream:40,
from /usr/include/c++/4.7/iostream:40,
from rational.cpp:2:
/usr/include/c++/4.7/bits/stl_function.h:224:12: error: template<class _Tp> struct std::greater
rational.cpp:47:19: error: reference to ‘greater’ is ambiguous
rational.cpp:32:17: error: candidates are: rational greater(rational, rational)
rational.cpp:28:15: error: double greater(double, double)
rational.cpp:24:12: error: int greater(int, int)
In file included from /usr/include/c++/4.7/string:50:0,
from /usr/include/c++/4.7/bits/locale_classes.h:42,
from /usr/include/c++/4.7/bits/ios_base.h:43,
from /usr/include/c++/4.7/ios:43,
from /usr/include/c++/4.7/ostream:40,
from /usr/include/c++/4.7/iostream:40,
from rational.cpp:2:
/usr/include/c++/4.7/bits/stl_function.h:224:12: error: template<class _Tp> struct std::greater
rational.cpp:48:9: error: reference to ‘greater’ is ambiguous
rational.cpp:32:17: error: candidates are: rational greater(rational, rational)
rational.cpp:28:15: error: double greater(double, double)
rational.cpp:24:12: error: int greater(int, int)
In file included from /usr/include/c++/4.7/string:50:0,
from /usr/include/c++/4.7/bits/locale_classes.h:42,
from /usr/include/c++/4.7/bits/ios_base.h:43,
from /usr/include/c++/4.7/ios:43,
from /usr/include/c++/4.7/ostream:40,
from /usr/include/c++/4.7/iostream:40,
from rational.cpp:2:
/usr/include/c++/4.7/bits/stl_function.h:224:12: error: template<class _Tp> struct std::greater```
Seems to be saying I can't call the greater function? Just learing about function overloading so any help would be appreciated.
2. just replace
Code:
`using namespace std;`
with
Code:
```using std::cout;
using std::endl;```
and the compiler will know what to call.
It's never a good idea to give your own functions the same name like standard functions.
Kurt
3. This is why the using namespace std; clause is not recommended. You could also put your greater() functions into your own namespace then properly scope your namespace using the scope resolution operator::
Code:
```namespace hexDump
{
//functions
inline int greater(int i, int j) {
return ( i > j ? i : j);
}
inline double greater(double x, double y) {
return (x > y ? x : y);
}
inline rational greater(rational w, rational z) {
return (w > z ? w : z);
}
} // end of namespace```
Then when using your greater() functions:
Code:
`zmax = hexDump::greater(w,z);`
Of course you could always rename your functions to something else as well.
Jim
4. Excellent thanks guys. So what would u call this? A scoping or namespace error?
Also when was greater() added, id think my book would have known and anticipated this.
5. So what would u call this? A scoping or namespace error?
A name clash. These name clashes are the reasons namespaces were introduced.
Also when was greater() added, id think my book would have known and anticipated this.
Does your book use the "using namespace std;" clause or did you add that.
Jim
6. Originally Posted by jimblumberg
A name clash. These name clashes are the reasons namespaces were introduced.
Hmmm...you'd think I'd at least get a decent error message then. Something like "name clashes with std::greater" or something to that effect...
Does your book use the "using namespace std;" clause or did you add that.
Jim
It uses the "using namespace std;" convention. The book is C++ By Dissection: Ira Pohl: 9780201787337: Amazon.com: Books Not a big fan of it though, but I started it and I have mild OCD so I want to finish it and then move on to Prata's C++ Primer Plus and Accelerated C++.
I'm glad I got this problem though, this was good to know info.
7. > Hmmm...you'd think I'd at least get a decent error message then. Something like "name clashes with std::greater" or something to that effect...
You mean like this?
Code:
```/usr/include/c++/4.7/bits/stl_function.h:224:12: error: template<class _Tp> struct std::greater
rational.cpp:48:9: error: reference to ‘greater’ is ambiguous
rational.cpp:32:17: error: candidates are: rational greater(rational, rational)
rational.cpp:28:15: error: double greater(double, double)
rational.cpp:24:12: error: int greater(int, int)```
8. Originally Posted by Salem
> Hmmm...you'd think I'd at least get a decent error message then. Something like "name clashes with std::greater" or something to that effect...
You mean like this?
Code:
```/usr/include/c++/4.7/bits/stl_function.h:224:12: error: template<class _Tp> struct std::greater
rational.cpp:48:9: error: reference to ‘greater’ is ambiguous
rational.cpp:32:17: error: candidates are: rational greater(rational, rational)
rational.cpp:28:15: error: double greater(double, double)
rational.cpp:24:12: error: int greater(int, int)```
Hahahha yes, like that. Haven't gotten to templates and such yet so that was yiddish to me. Thanks though.
9. Originally Posted by hex_dump
Hahahha yes, like that. Haven't gotten to templates and such yet so that was yiddish to me. Thanks though.
Even if you don't seem to understand a compiler error, always read it.
Once I had a nasty template error and naturally, skipped it and just posted it here!
It turned out that the error message *literally* said what I had to do to get rid of it.
10. Originally Posted by hex_dump
Hmmm...you'd think I'd at least get a decent error message then. Something like "name clashes with std::greater" or something to that effect...
But it's not technically a clash. You brought in more symbols with the name greater from the std namespace, which peacefully coexists with what you wrote. But when you call it, the compiler doesn't know which one you meant, hence ambiguous.
11. If it's not a name clash then what is it?
12. I don't know the exact terminology, I was just responding to why the compiler shouldn't say "omg, your Y clashes with X."
13. Originally Posted by whiteflags
If it's not a name clash then what is it?
The correct terminology is "ambiguity" or (as used by the compiler in the original post) "ambiguous". The error message meant that, in the code which called the function, there is more than one possible candidate function to be called, and the compiler has no reason to prefer one over the other.
The compiler is correct in using that terminology, because that is the terminology used by the C++ standard. The standard is not using that terminology in some unusual manner either - in the English language, "ambiguity" is a noun meaning "doubtfulness or uncertainty of meaning or intention". The word "ambiguous" is an adjective meaning "open to or having several possible meanings or interpretations" or "of doubtful or uncertain nature".
It is often tiring in forums to see members debating the terminology for something that the standard defines. People seem to delight in introducing new terminology that - they claim - is clearer than the standard but, in reality, simply obfuscates what is going on.
Popular pages Recent additions | 2,543 | 9,495 | {"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} | 2.859375 | 3 | CC-MAIN-2017-39 | latest | en | 0.662219 |
https://number.academy/622592 | 1,680,130,088,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949035.66/warc/CC-MAIN-20230329213541-20230330003541-00240.warc.gz | 500,464,415 | 12,790 | # Number 622592
Number 622,592 spell 🔊, write in words: six hundred and twenty-two thousand, five hundred and ninety-two . Ordinal number 622592nd is said 🔊 and write: six hundred and twenty-two thousand, five hundred and ninety-second. Color #622592. The meaning of the number 622592 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 622592. What is 622592 in computer science, numerology, codes and images, writing and naming in other languages
## What is 622,592 in other units
The decimal (Arabic) number 622592 converted to a Roman number is (D)(C)(X)(X)MMDXCII. Roman and decimal number conversions.
#### Weight conversion
622592 kilograms (kg) = 1372566.3 pounds (lbs)
622592 pounds (lbs) = 282405.9 kilograms (kg)
#### Length conversion
622592 kilometers (km) equals to 386861 miles (mi).
622592 miles (mi) equals to 1001966 kilometers (km).
622592 meters (m) equals to 2042600 feet (ft).
622592 feet (ft) equals 189769 meters (m).
622592 centimeters (cm) equals to 245115.0 inches (in).
622592 inches (in) equals to 1581383.7 centimeters (cm).
#### Temperature conversion
622592° Fahrenheit (°F) equals to 345866.7° Celsius (°C)
622592° Celsius (°C) equals to 1120697.6° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
622592 seconds equals to 1 week, 4 hours, 56 minutes, 32 seconds
622592 minutes equals to 1 year, 3 months, 1 week, 5 days, 8 hours, 32 minutes
### Codes and images of the number 622592
Number 622592 morse code: -.... ..--- ..--- ..... ----. ..---
Sign language for number 622592:
Number 622592 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 622592
### Multiplications
#### Multiplication table of 622592
622592 multiplied by two equals 1245184 (622592 x 2 = 1245184).
622592 multiplied by three equals 1867776 (622592 x 3 = 1867776).
622592 multiplied by four equals 2490368 (622592 x 4 = 2490368).
622592 multiplied by five equals 3112960 (622592 x 5 = 3112960).
622592 multiplied by six equals 3735552 (622592 x 6 = 3735552).
622592 multiplied by seven equals 4358144 (622592 x 7 = 4358144).
622592 multiplied by eight equals 4980736 (622592 x 8 = 4980736).
622592 multiplied by nine equals 5603328 (622592 x 9 = 5603328).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 622592
Half of 622592 is 311296 (622592 / 2 = 311296).
One third of 622592 is 207530,6667 (622592 / 3 = 207530,6667 = 207530 2/3).
One quarter of 622592 is 155648 (622592 / 4 = 155648).
One fifth of 622592 is 124518,4 (622592 / 5 = 124518,4 = 124518 2/5).
One sixth of 622592 is 103765,3333 (622592 / 6 = 103765,3333 = 103765 1/3).
One seventh of 622592 is 88941,7143 (622592 / 7 = 88941,7143 = 88941 5/7).
One eighth of 622592 is 77824 (622592 / 8 = 77824).
One ninth of 622592 is 69176,8889 (622592 / 9 = 69176,8889 = 69176 8/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
622592
#### Is Prime?
The number 622592 is not a prime number. The closest prime numbers are 622577, 622603.
#### Factorization and factors (dividers)
The prime factors of 622592 are 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 19
The factors of 622592 are
1 , 2 , 4 , 8 , 16 , 19 , 32 , 38 , 64 , 76 , 128 , 152 , 256 , 304 , 512 , 608 , 1024 , 1216 , 2048 , 2432 , 4096 , 4864 , 8192 , 9728 , 622592 show more factors ...
Total factors 32.
Sum of factors 1310700 (688108).
#### Powers
The second power of 6225922 is 387.620.798.464.
The third power of 6225923 is 241.329.608.157.298.688.
#### Roots
The square root √622592 is 789,044992.
The cube root of 3622592 is 85,388853.
#### Logarithms
The natural logarithm of No. ln 622592 = loge 622592 = 13,341647.
The logarithm to base 10 of No. log10 622592 = 5,794204.
The Napierian logarithm of No. log1/e 622592 = -13,341647.
### Trigonometric functions
The cosine of 622592 is -0,829441.
The sine of 622592 is -0,558594.
The tangent of 622592 is 0,673458.
### Properties of the number 622592
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 622592 in Computer Science
Code typeCode value
PIN 622592 It's recommended that you use 622592 as your password or PIN.
622592 Number of bytes608.0KB
CSS Color
#622592 hexadecimal to red, green and blue (RGB) (98, 37, 146)
Unix timeUnix time 622592 is equal to Thursday Jan. 8, 1970, 4:56:32 a.m. GMT
IPv4, IPv6Number 622592 internet address in dotted format v4 0.9.128.0, v6 ::9:8000
622592 Decimal = 10011000000000000000 Binary
622592 Decimal = 1011122000222 Ternary
622592 Decimal = 2300000 Octal
622592 Decimal = 98000 Hexadecimal (0x98000 hex)
622592 BASE64NjIyNTky
622592 MD5d77195d0d9aa721d44e027ae598e84ef
622592 SHA19f582349281859707b39f0aaa0aeaba011c7ebd0
622592 SHA22435bb4a282dd8c41b51f3f824177d90cec68e689ce35223f91a819264
622592 SHA2560ee62fc45459415b1a109ddf1c8f1a1141f02559d22c282a5d81a0e02bf34590
622592 SHA384968f342c7630775862c08d125d41f12052490a76ebd1f6e4c84672b10a1a1cc4e51d27c8962ca90cf21648d5c15acbb2
More SHA codes related to the number 622592 ...
If you know something interesting about the 622592 number that you did not find on this page, do not hesitate to write us here.
## Numerology 622592
### Character frequency in the number 622592
Character (importance) frequency for numerology.
Character: Frequency: 6 1 2 3 5 1 9 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 622592, the numbers 6+2+2+5+9+2 = 2+6 = 8 are added and the meaning of the number 8 is sought.
## № 622,592 in other languages
How to say or write the number six hundred and twenty-two thousand, five hundred and ninety-two in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 622.592) seiscientos veintidós mil quinientos noventa y dos German: 🔊 (Nummer 622.592) sechshundertzweiundzwanzigtausendfünfhundertzweiundneunzig French: 🔊 (nombre 622 592) six cent vingt-deux mille cinq cent quatre-vingt-douze Portuguese: 🔊 (número 622 592) seiscentos e vinte e dois mil, quinhentos e noventa e dois Hindi: 🔊 (संख्या 622 592) छः लाख, बाईस हज़ार, पाँच सौ, बानवे Chinese: 🔊 (数 622 592) 六十二万二千五百九十二 Arabian: 🔊 (عدد 622,592) ستمائة و اثنان و عشرون ألفاً و خمسمائةاثنان و تسعون Czech: 🔊 (číslo 622 592) šestset dvacet dva tisíce pětset devadesát dva Korean: 🔊 (번호 622,592) 육십이만 이천오백구십이 Danish: 🔊 (nummer 622 592) sekshundrede og toogtyvetusindfemhundrede og tooghalvfems Dutch: 🔊 (nummer 622 592) zeshonderdtweeëntwintigduizendvijfhonderdtweeënnegentig Japanese: 🔊 (数 622,592) 六十二万二千五百九十二 Indonesian: 🔊 (jumlah 622.592) enam ratus dua puluh dua ribu lima ratus sembilan puluh dua Italian: 🔊 (numero 622 592) seicentoventiduemilacinquecentonovantadue Norwegian: 🔊 (nummer 622 592) seks hundre og tjue-to tusen, fem hundre og nitti-to Polish: 🔊 (liczba 622 592) sześćset dwadzieścia dwa tysiące pięćset dziewięćdzisiąt dwa Russian: 🔊 (номер 622 592) шестьсот двадцать две тысячи пятьсот девяносто два Turkish: 🔊 (numara 622,592) altıyüzyirmiikibinbeşyüzdoksaniki Thai: 🔊 (จำนวน 622 592) หกแสนสองหมื่นสองพันห้าร้อยเก้าสิบสอง Ukrainian: 🔊 (номер 622 592) шiстсот двадцять двi тисячi п'ятсот дев'яносто двi Vietnamese: 🔊 (con số 622.592) sáu trăm hai mươi hai nghìn năm trăm chín mươi hai Other languages ...
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If you know something interesting about the number 622592 or any other natural number (positive integer), please write to us here or on Facebook. | 2,728 | 7,786 | {"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} | 2.75 | 3 | CC-MAIN-2023-14 | latest | en | 0.678049 |
https://doczz.net/doc/9110552/experimenting-with-the-multiple-reflections-and-refractio... | 1,627,435,978,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153515.0/warc/CC-MAIN-20210727233849-20210728023849-00469.warc.gz | 229,549,109 | 7,613 | # Experimenting with the multiple reflections and refractions on a
## Transcription
Experimenting with the multiple reflections and refractions on a
```A STUDENT’S EXPERIMENT WITH MULTIPLE REFLECTIONS AND
REFRACTIONS ON A GLASS PLATE AND VALIDATION OF THE FRESNEL’S
EQUATIONS
N. Mahmudi1, S. Rendevski2, F. Ajredini1 and R. Popeski-Dimovski3
1Faculty of Natural Sciences and Mathematics, State University of Tetovo,
bul. Ilinden, b.b., 1200 Tetovo, Republic of Macedonia
2 Faculty of Electrical Engineering, University “Goce Delcev”,
ul. Krste Misirkov, b.b., 2000 Stip, Republic of Macedonia
3Faculty of Natural Sciences and Mathematics, University “Ss. Cyril and Methodius”,
ul. Gazi Baba, b.b., 1000 Skopje, Republic of Macedonia
1
Introduction
-Very important practical interest of experimenting with multiple reflections and
refractions on a layer in the last decade,
-the examination of multiple reflections and refractions in a single partially transmitting
layer with thickness D much larger than the wavelength of the incident light, .
-We choose the ray tracing method for calculating the transmission T, reflection R
and absorption coefficients A in such a layer .
-The other method that gives same equations for calculation of A, R and T
is the net radiation method described in the literature.
-mediums at the front and back side of the thick layer are decided to be air (index of
-refraction n1) and the thick layer to be a glass with index of refraction n2.
-We assumed that the thickness of the layer is thick enough that there is no interference
effect. Hence, the transmission and reflection of the interface 1 are the same as reflection
and transmission of the interface 2. The same is valid for interface 3 and interface 4.
Surface reflectance
of a layer if front and
back mediums are
different
Experimental ,Results and Discussion
-we define as reflection power, which is the ratio of energy reflected from a surface to
energy of the wave from the incident direction to that surface.
-If the front and the back environments of the layer are the same then the
surface reflectance and transmission on the top are the same as the bottom surface,
1 = 2 = .
-The reflection power for unpolarized incident light is defined according to the
Fresnel’s equation, where is derived from the Snell’s law sin = (1/n)·sin:
1 tan 2 sin 2
2
2
2 tan sin
On the upper surface (between interface 1 and interface 2)
an incident light beam with unity intensity is considered (Fig.1).
On the interface 1, an amount 1 is reflected and 1- 1 enter the
non-attenuating material.
Fig.1 Multiple reflections in a non-attenuating material
The fraction of incident energy reflected by the layer (material)
is the sum of the terms leaving the top surface:
R 1 1 12 2 1 22 12 23 ....
1 2 1 21
1 1 2
(1)
The fraction of incident energy transmitted by the layer (material)
is the sum of the terms leaving the bottom surface:
T 1 1 1 2 1 1 2 ....
2
1
2
2
1 1 1 2
1 1 2
Because there is no absorption the fraction of incident energy
absorbed by the layer is A = 0.
The equations (1) and (2) are known as Fresnel’s equations
for multiple internal reflections for non-attenuating material.
(2)
-The reflection power R and transmission power T depend only on the surface
reflectance. Also, the thickness of the non-attenuating material has no effect on
reflection and transmission power of the material.
-In a special case where 1 = 2 = , we have:
R 1 2 1 2 3 ....
T 1 1 2 1 4 ...
2
2
2
(3)
(4)
For attenuating materials with transmittance power ,
2
1
2
2
1
R 1 1 12 2 2 1 22 2 12 23 4 .... 1
1 1 2 2
1 1 1 2
2
2 2 4
T 1 1 1 2 1 1 2 1 2 ....
1 1 2 2
(5)
(6)
According to the energy conservation principle, the fraction of incident energy
absorbed in the layer A is:
A = 1 – R – T.
(7)
Experimental setup
glass with thickness of 5.0 mm and index of refraction n
of 1.50 for monochromatic laser light with wavelength 690 nm was used.
-Detection of reflected and transmitted laser beams from the glass
plate was made with PASCO ultrasensitive light sensor PS-2176.
Fig.2. Experimental setup
-We worked with seven different angles of incident light beam:
350, 400, 450, 500, 550, 600 and 650.
-Three light beams of multiple reflections going out of the top surface
of the glass plate had enough intensity to be measured.
-From the bottom surface of the glass plate, three light beams of multiple refractions
were detected in transmission from the bottom surface of the plate.
-In the experimental conditions limited to low power of the
laser pointer and light absorption properties of the glass plate,
it was hard to obtain fourth or higher order of reflected and transmitted light.
Fig.3. Dependences of R on the incident Fig.4. Dependences of R on the incident angle
angle for the first reflected light beam.
for the second reflected light beam.
Fig.5. Dependences of R on the incident angle
for the third reflected light beam.
Fig.6. Dependences of T on the incident
angle for the first transmitted light beam.
Fig.7. Dependences of T on the incident
angle for the second transmitted light beam.
Fig.8. Dependences of T on the incident angle
for the third transmitted light beam.
Conclusion
-Experimenting with the multiple reflections and refractions on a glass plate is suitable
for students learning physics at the university level.
-A practical validation of such experiment is in connection to ray tracing analysis in
single or multilayer materials for cooling and heating processes. | 2,003 | 5,817 | {"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} | 2.921875 | 3 | CC-MAIN-2021-31 | latest | en | 0.828257 |
https://www.cpalms.org/Public/PreviewStandard/Preview/5599 | 1,576,476,038,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575541317967.94/warc/CC-MAIN-20191216041840-20191216065840-00355.warc.gz | 671,086,196 | 20,076 | Help
# MAFS.912.G-CO.1.2
Represent transformations in the plane using, e.g., transparencies and geometry software; describe transformations as functions that take points in the plane as inputs and give other points as outputs. Compare transformations that preserve distance and angle to those that do not (e.g., translation versus horizontal stretch).
Subject Area: Mathematics
Domain-Subdomain: Geometry: Congruence
Cluster: Level 2: Basic Application of Skills & Concepts
Cluster: Experiment with transformations in the plane. (Geometry - Supporting Cluster) -
Clusters should not be sorted from Major to Supporting and then taught in that order. To do so would strip the coherence of the mathematical ideas and miss the opportunity to enhance the major work of the grade with the supporting clusters.
Date of Last Rating: 02/14
Status: State Board Approved
Assessed: Yes
### TEST ITEM SPECIFICATIONS
• Item Type(s): This benchmark may be assessed using: TI item(s)
• Also assesses:
MAFS.912.G-CO.1.4
• Assessment Limits :
Items may require the student to be familiar with using the algebraic
description for a translation, and
for a dilation when given the center of dilation.
Items may require the student to be familiar with the algebraic
description for a 90-degree rotation about the origin,
for a 180-degree rotation about the origin,
and for a 270-degree rotation about the origin,
Items that use more than one transformation may
ask the student to write a series of algebraic descriptions.
Items must not use matrices to describe transformations.
Items must not require the student to use the distance formula.
Items may require the student to find the distance between two
points or the slope of a line.
In items that require the student to represent transformations, at
least two transformations should be applied
• Calculator :
Neutral
• Clarification :
Students will represent transformations in the plane.
Students will describe transformations as functions that take points in
the plane as inputs and give other points as outputs.
Students will compare transformations that preserve distance and
angle to those that do not.
Students will use definitions of rotations, reflections, and translations
in terms of angles, circles, perpendicular lines, parallel lines, and line
segments.
• Stimulus Attributes :
Items may be set in real-world or mathematical context.
Items may ask the student to determine if a transformation is rigid.
Items may ask the student to determine if steps that are given can be
used to develop a definition of an angle, a circle, perpendicular lines,
parallel lines, or line segments by using rotations, reflections, and
translations.
• Response Attributes :
Items may require the student to give a coordinate of a transformed
figure.
Items may require the student to use a function, e.g.,
, to describe a transformation.
Items may require the student to determine if a verbal description of
a definition is valid.
Items may require the student to determine any flaws in a verbal
description of a definition.
Items may require the student to be familiar with slope-intercept
form of a line, standard form of a line, and point-slope form of a line.
Items may require the student to give a line of reflection and/or a
degree of rotation that carries a figure onto itself.
Items may require the student to draw a figure using a description of
a translation.
### SAMPLE TEST ITEMS (1)
• Test Item #: Sample Item 1
• Question:
Nicole, Jeremy, and Frances each perform a transformation on the triangle RST. Each recorded his or her transformation and the location of S' in the table. Point S of the triangle is located at *5,-7).
Complete the table to determine the values of a and b that make the algebraic descriptions of each person's transformation true.
• Difficulty: N/A
• Type: TI: Table Item | 797 | 3,869 | {"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-2019-51 | latest | en | 0.85728 |
http://stackoverflow.com/questions/4008287/basic-hexbin-with-r/4008351 | 1,467,160,023,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397428.37/warc/CC-MAIN-20160624154957-00144-ip-10-164-35-72.ec2.internal.warc.gz | 282,154,501 | 19,540 | Basic hexbin with R?
I have results from a survey. I am trying to create a graphic displaying the relationship of two variables: "Q1" and "Q9.1". "Q1" is the independent and "Q9.1" is the dependent. Both variables have responses from like scale questions: -2,-1,0,1,2. A typical plot places the answers on top of each other - not very interesting or informative. I was thinking that hexbin would be the way to go. The data is in lpp. I have not been able to use "Q1" and "Q9.1" for x and y. However:
``````> is.numeric("Q1")
[1] FALSE
q1.num <- as.numeric("Q1")
Warning message:
NAs introduced by coercion
``````
The values for Q1 are (hundreds of instances of): -2,-1,0,1,2
How can I make a hexbin graph with this data? Is there another graph I should consider?
Error messages so far:
``````Warning messages:
1: In xy.coords(x, y, xl, yl) : NAs introduced by coercion
2: In xy.coords(x, y, xl, yl) : NAs introduced by coercion
3: In min(x) : no non-missing arguments to min; returning Inf
4: In max(x) : no non-missing arguments to max; returning -Inf
5: In min(x) : no non-missing arguments to min; returning Inf
6: In max(x) : no non-missing arguments to max; returning -Inf
``````
-
It might help to provide some example data so we can see what structure it currently has. A helpful function for this is dput(...) which outputs a description that can be used to recreate an object. – PaulHurleyuk Oct 24 '10 at 20:41
@Donnied; You need to sort out your data first. Something isn't right here; you are introducing NAs when you coerce to numeric, during plotting, `xy.coords()` is creating NAs such that you have no non-NA data. Take a look at the output of `str(Q1)` etc for all your data - are they stored as numerics? Finally, your first two lines of R are wrong; you don't refer to an object by it's quoted name. If you wan to see if `Q1` is numeric you do `is.numeric(Q1)`. What you have done is ask if the string `"Q1"` is numeric, which inevitably is FALSE. You didn't do this in the `plot()` call did you? – Gavin Simpson Oct 25 '10 at 7:53
I apologize. I've just started using R. I have a csv file which I read in as data. "Q1" is one of the column headers / variables. – Donnied Oct 25 '10 at 12:24
Something such as this: d <- ggplot(lpp, aes(Q1, Q3a.8)) d + stat_binhex(bins = 11) works. Otherwise I'm having difficulty reading Q1 in as a numeric or converting to numeric. It seems that I've managed to simply store the characters Q1 in a numeric variable not the data. The data set 'lpp' – Donnied Oct 25 '10 at 12:28
is recognized as an object. However, I am not sure how to refer to the columns explicitly or use them as objects.I'm reading up on R but there are gaping deficits in what I know and need to know. – Donnied Oct 25 '10 at 12:31
How about taking a slightly different approach? How about thinking of your responses as factors rather than numbers? You could use something like this, then, to get a potentially useful representation of your data:
```# Simulate data for testing purposes
q1 = sample(c(-2,-1,0,1,2),100,replace=TRUE)
q9 = sample(c(-2,-1,0,1,2),100,replace=TRUE)
dat = data.frame(q1=factor(q1),q9=factor(q9))
library(ggplot2)
# generate stacked barchart
ggplot(dat,aes(q1,fill=q9)) + geom_bar()
```
You may want to switch q1 and q9 above, depending on the view of the data that you want.
-
This produced one solid bar. – Donnied Oct 30 '10 at 19:13
My apologies... Works great! Thank you. – Donnied Oct 30 '10 at 23:31
Great to hear that it worked for you. – seandavi Nov 1 '10 at 11:42
Perhaps ggplot2's stat_binhex could sort that one for you?
Also, I find scale_alpha useful for dealing with overplotting.
-
I really like the stat_binhex. I can't find how to add title though. labs, xl, and yl don't work. – Donnied Oct 24 '10 at 21:03
to label x axis you could try: qplot(x, y, data = data, xlab = "my label") or: ggplot(data, aes(x, y)) + geom_point() + scale_x_continuous("my label") – radek Oct 24 '10 at 21:37
The `ylab()` and `xlab()` functions also add axis labels, e.g. `ggplot(data, aes(x, y)) + geom_point() + ylab("my label")` – Gavin Simpson Oct 25 '10 at 12:09 | 1,222 | 4,130 | {"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-2016-26 | latest | en | 0.904787 |
http://markun.cs.shinshu-u.ac.jp/Mirror/mizar.org/JFM/Vol11/conlat_2.abs.html | 1,508,804,793,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187827662.87/warc/CC-MAIN-20171023235958-20171024015958-00633.warc.gz | 224,385,881 | 3,665 | Journal of Formalized Mathematics
Volume 11, 1999
University of Bialystok
Copyright (c) 1999 Association of Mizar Users
### A Characterization of Concept Lattices. Dual Concept Lattices
by
Christoph Schwarzweller
MML identifier: CONLAT_2
[ Mizar article, MML identifier index ]
```environ
vocabulary CONLAT_1, QC_LANG1, LATTICES, CAT_1, FUNCT_1, TARSKI, SETFAM_1,
BOOLE, LATTICE3, SUBSET_1, RELAT_1, FUNCT_3, LATTICE6, BHSP_3, GROUP_6,
WELLORD1, MOD_4, MCART_1, FILTER_1, ORDERS_1, CONLAT_2;
notation TARSKI, XBOOLE_0, ZFMISC_1, SUBSET_1, SETFAM_1, MCART_1, FUNCT_1,
DOMAIN_1, RELSET_1, PRE_TOPC, ORDERS_1, STRUCT_0, LATTICE2, LATTICE3,
LATTICE6, PARTFUN1, FUNCT_2, LATTICES, LATTICE4, CONLAT_1;
constructors DOMAIN_1, LATTICE2, LATTICE4, LATTICE6, CONLAT_1, MEMBERED,
PRE_TOPC;
clusters STRUCT_0, LATTICE3, PRE_TOPC, RELSET_1, LATTICE2, CONLAT_1, SUBSET_1,
SETFAM_1, LATTICES, FUNCT_2, PARTFUN1, XBOOLE_0;
requirements SUBSET, BOOLE;
begin
definition
let C be FormalContext;
let CP be strict FormalConcept of C;
func @CP -> Element of ConceptLattice(C) equals
:: CONLAT_2:def 1
CP;
end;
definition let C be FormalContext;
cluster ConceptLattice C -> bounded;
end;
theorem :: CONLAT_2:1
for C being FormalContext holds
Bottom (ConceptLattice(C)) = Concept-with-all-Attributes(C) &
Top (ConceptLattice(C)) = Concept-with-all-Objects(C);
theorem :: CONLAT_2:2
for C being FormalContext
for D being non empty Subset of bool(the Objects of C) holds
(ObjectDerivation(C)).(union D) =
meet({(ObjectDerivation(C)).O
where O is Subset of the Objects of C : O in D});
theorem :: CONLAT_2:3
for C being FormalContext
for D being non empty Subset of bool(the Attributes of C) holds
(AttributeDerivation(C)).(union D) =
meet({(AttributeDerivation(C)).A
where A is Subset of the Attributes of C : A in D});
theorem :: CONLAT_2:4
for C being FormalContext
for D being Subset of ConceptLattice(C) holds
"/\"(D,ConceptLattice(C)) is FormalConcept of C &
"\/"(D,ConceptLattice(C)) is FormalConcept of C;
definition
let C be FormalContext;
let D be Subset of ConceptLattice(C);
func "/\"(D,C) -> FormalConcept of C equals
:: CONLAT_2:def 2
"/\"(D,ConceptLattice(C));
func "\/"(D,C) -> FormalConcept of C equals
:: CONLAT_2:def 3
"\/"(D,ConceptLattice(C));
end;
theorem :: CONLAT_2:5
for C being FormalContext holds
"\/"({} ConceptLattice(C),C) = Concept-with-all-Attributes(C) &
"/\"({} ConceptLattice(C),C) = Concept-with-all-Objects(C);
theorem :: CONLAT_2:6
for C being FormalContext holds
"\/"([#] the carrier of ConceptLattice(C),C) = Concept-with-all-Objects(C) &
"/\"([#] the carrier of ConceptLattice(C),C) = Concept-with-all-Attributes(C);
theorem :: CONLAT_2:7
for C being FormalContext
for D being non empty Subset of ConceptLattice(C) holds
the Extent of "\/"(D,C) =
(AttributeDerivation(C)).((ObjectDerivation(C)).
union {the Extent of ConceptStr(#E,I#)
where E is Subset of the Objects of C,
I is Subset of the Attributes of C : ConceptStr(#E,I#) in D}) &
the Intent of "\/"(D,C) =
meet {the Intent of ConceptStr(#E,I#)
where E is Subset of the Objects of C,
I is Subset of the Attributes of C : ConceptStr(#E,I#) in D};
theorem :: CONLAT_2:8
for C being FormalContext
for D being non empty Subset of ConceptLattice(C) holds
the Extent of "/\"(D,C) =
meet {the Extent of ConceptStr(#E,I#)
where E is Subset of the Objects of C,
I is Subset of the Attributes of C : ConceptStr(#E,I#) in D} &
the Intent of "/\"(D,C) =
(ObjectDerivation(C)).((AttributeDerivation(C)).
union {the Intent of ConceptStr(#E,I#)
where E is Subset of the Objects of C,
I is Subset of the Attributes of C : ConceptStr(#E,I#) in D});
theorem :: CONLAT_2:9
for C being FormalContext
for CP being strict FormalConcept of C holds
"\/"({ConceptStr(#O,A#) where O is Subset of the Objects of C,
A is Subset of the Attributes of C :
ex o being Object of C st
o in the Extent of CP &
O = (AttributeDerivation(C)).((ObjectDerivation(C)).{o}) &
A = (ObjectDerivation(C)).{o}},
ConceptLattice(C)) = CP;
theorem :: CONLAT_2:10
for C being FormalContext
for CP being strict FormalConcept of C holds
"/\"({ConceptStr(#O,A#) where O is Subset of the Objects of C,
A is Subset of the Attributes of C :
ex a being Attribute of C st
a in the Intent of CP &
O = (AttributeDerivation(C)).{a} &
A = (ObjectDerivation(C)).((AttributeDerivation(C)).{a})},
ConceptLattice(C)) = CP;
definition
let C be FormalContext;
func gamma(C) -> Function of the Objects of C,
the carrier of ConceptLattice(C) means
:: CONLAT_2:def 4
for o being Element of the Objects of C holds
ex O being Subset of the Objects of C,
A being Subset of the Attributes of C st
it.o = ConceptStr(#O,A#) &
O = (AttributeDerivation(C)).((ObjectDerivation(C)).{o}) &
A = (ObjectDerivation(C)).{o};
end;
definition
let C be FormalContext;
func delta(C) -> Function of the Attributes of C,
the carrier of ConceptLattice(C) means
:: CONLAT_2:def 5
for a being Element of the Attributes of C holds
ex O being Subset of the Objects of C,
A being Subset of the Attributes of C st
it.a = ConceptStr(#O,A#) &
O = (AttributeDerivation(C)).{a} &
A = (ObjectDerivation(C)).((AttributeDerivation(C)).{a});
end;
theorem :: CONLAT_2:11
for C being FormalContext
for o being Object of C
for a being Attribute of C holds
(gamma(C)).o is FormalConcept of C & (delta(C)).a is FormalConcept of C;
theorem :: CONLAT_2:12
for C being FormalContext holds
rng(gamma(C)) is supremum-dense & rng(delta(C)) is infimum-dense;
theorem :: CONLAT_2:13
for C being FormalContext
for o being Object of C
for a being Attribute of C holds
o is-connected-with a iff (gamma(C)).o [= (delta(C)).a;
begin
theorem :: CONLAT_2:14
for L being complete Lattice
for C being FormalContext holds
ConceptLattice(C),L are_isomorphic
iff ex g being Function of the Objects of C, the carrier of L,
d being Function of the Attributes of C, the carrier of L
st rng(g) is supremum-dense &
rng(d) is infimum-dense &
for o being Object of C,
a being Attribute of C holds
o is-connected-with a iff g.o [= d.a;
definition
let L be Lattice;
func Context(L) -> strict non quasi-empty ContextStr equals
:: CONLAT_2:def 6
ContextStr(#the carrier of L, the carrier of L, LattRel L#);
end;
theorem :: CONLAT_2:15
for L being complete Lattice holds
ConceptLattice(Context(L)),L are_isomorphic;
theorem :: CONLAT_2:16
for L being Lattice holds
L is complete iff
ex C being FormalContext st ConceptLattice(C),L are_isomorphic;
begin :: Dual Concept Lattices
definition
let L be complete Lattice;
cluster L.: -> complete;
end;
definition
let C be FormalContext;
func C.: -> strict non quasi-empty ContextStr equals
:: CONLAT_2:def 7
ContextStr(#the Attributes of C, the Objects of C,
(the Information of C)~ #);
end;
theorem :: CONLAT_2:17
for C being strict FormalContext holds (C.:).: = C;
theorem :: CONLAT_2:18
for C being FormalContext
for O being Subset of the Objects of C holds
(ObjectDerivation(C)).O = (AttributeDerivation(C.:)).O;
theorem :: CONLAT_2:19
for C being FormalContext
for A being Subset of the Attributes of C holds
(AttributeDerivation(C)).A = (ObjectDerivation(C.:)).A;
definition
let C be FormalContext;
let CP be ConceptStr over C;
func CP.: -> strict ConceptStr over C.: means
:: CONLAT_2:def 8
the Extent of it = the Intent of CP &
the Intent of it = the Extent of CP;
end;
definition
let C be FormalContext;
let CP be FormalConcept of C;
redefine func CP.: -> strict FormalConcept of C.:;
end;
theorem :: CONLAT_2:20
for C being FormalContext
for CP being strict FormalConcept of C holds (CP.:).: = CP;
definition
let C be FormalContext;
func DualHomomorphism(C) ->
Homomorphism of (ConceptLattice(C)).:,ConceptLattice(C.:) means
:: CONLAT_2:def 9
for CP being strict FormalConcept of C holds it.CP = CP.:;
end;
theorem :: CONLAT_2:21
for C being FormalContext holds DualHomomorphism(C) is isomorphism;
theorem :: CONLAT_2:22
for C being FormalContext holds
ConceptLattice(C.:),(ConceptLattice(C)).: are_isomorphic;
``` | 2,390 | 7,937 | {"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-2017-43 | latest | en | 0.534665 |
https://gogeometry.blogspot.com/2013/03/problem-866-cyclic-quadrilateral-circle.html | 1,723,285,058,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640805409.58/warc/CC-MAIN-20240810093040-20240810123040-00197.warc.gz | 220,123,091 | 13,349 | ## Saturday, March 30, 2013
### Problem 866: Cyclic Quadrilateral, Circle, Rectangle, Center, Congruence, 90 Degrees
Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 866.
#### 1 comment:
1. http://img850.imageshack.us/img850/1512/98459089.png
construct a rectangle AMND (with center L) that AM=BC.
Angle ABC+ angle CDA=180. Because angle EBA+angle HBC=90+90=180 degrees ,is ,angle EBH+angle ABC=180.So, angle EBH=angleCDA . Additional, angle HBG=angle ADL and angle EBF=angle CDK.So, angle O1BO2=angle LDO3. The triangles O1BO2, LDO3 have additional,O1B=DO3 ,DL=BO2 ,therefore are equal ,so, O1O2=LO3 (1)
Angle BCD+ angle DAB=180. Because angle BCG+angle KCD=90+90=180 degrees ,is ,angleBCD +angle KCG=180 degrees . So, angle KCG=angleBAD . Additional, angle HCG=angle LAD and angle KCO3 =angleO1AB .So, angle O1AL=angle O2C O3. The triangles O1AL 2,
O2C O3 have additional,O1A=CO3 ,AL=CO2 ,therefore are equal ,so, O2O3=LO1 (2)
From (1),(2), O1O2O3L it is a parallelogram.
because triangle O1O2B = triangle LDO3 ,is , angle LO3D = angle O2O1B .But , angle EO1B= angle CO3D.Therefore ,angle EO1B-angle O2O1B= angle CO3D- angle LO3D ,so,
angle EO1O2=angle CO3L. Additional, angle CO3O2 =angle AO1L.
angle EO1O2+angle O2O1L+angle AO1L=180 ,therefore, 2angle O2O1L=180 degrees , angle O2O1L=90 degrees.So , O1O2O3L is, rectangle ,therefore angle O1O2O3L=90 degrees. Ο.ε.δ | 531 | 1,462 | {"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} | 4.34375 | 4 | CC-MAIN-2024-33 | latest | en | 0.542736 |
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