url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://dsp.stackexchange.com/questions/86671/how-to-produce-self-adjoint-filters-using-gaussian-and-motion-blur-in-matlab | 1,716,388,892,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058557.6/warc/CC-MAIN-20240522132433-20240522162433-00271.warc.gz | 188,552,089 | 42,334 | # How to produce self-adjoint filters using Gaussian and motion blur in Matlab?
I am trying to write a Matlab code to produce motion and Gaussian blur. Here is my code:
f0=imread('cameraman.png');
[Nx,Ny,Nz]=size(f0);
if Nz>1;f0=double(rgb2gray(f0));
else
f0=double(f0);
end
blurfilter1 = fspecial('gaussian', [7 7], 1);
blurfilter2 = fspecial('motion',5,30);
Au = @(u) imfilter(u,blurfilter1,'symmetric');
Bu = @(u) imfilter(u,blurfilter2,'symmetric');
In above code, operators Au and Bu will produce Gaussian blur and motion blur respectively. However, I need these two operators to be self-adjoint. I have found the following code for producing self-adjoint opearots but I am not sure whether it works for both Gaussian and blur operators. Here is the code:
Au = @(u) imfilter(imfilter(u,blurfilter1,'symmetric'),blurfilter1','symmetric');
Bu = @(u) imfilter(imfilter(u,blurfilter2,'symmetric'),blurfilter2','symmetric');
I really appriciate it if anyone could please help me to build self-adjoint Au and Bu operators for Gaussian and motion blur.
# Gaussian
To start, fspecial('gaussian' is deprecated; the fspecial documentation explains how to do the same thing, but better.
Why do you use an n-dimensional convolution (imfilter) if you're actually after 2 sequences of 1-dimensional convolutions?
Something like (excuse me, this is untested, I uninstalled my Matlab recently)
x = [-3:3]
sigma = 1
gaussian = exp(- x.^ 2 / (2 * sigma ^ 2));
self_adjoint_filter = @(img, impresp1d) conv2(impresp1d, impresp1d, img)
Au = @(img) self_adjoint_filter(img, gaussian)
should do the trick.
# Motion blur
That's not a self-adjunct operation. It has a definite direction, so eigenvalues can't be the same magnitude.
• Thank you so much for your time and valuable answer. Yes, now I understand that motion blur is not a self-adjoint operator and I am working on the structure of its adjoint. Many thanks.
– eli
Feb 15, 2023 at 6:50
• @Marcus, Hi Marcus, maybe i'm wrong, but what if the motion blur kernel were symmetric around the origin ? Feb 15, 2023 at 8:32
• then it wouldn't be a motion blur. Feb 15, 2023 at 9:01
• as an example, what is wrong with a line centered around origin to be a motion blur kernel? Feb 15, 2023 at 9:50
• if you consider it in time domain. the adjoint operator is like the anti-causal version of causal filter. Feb 15, 2023 at 16:07
First there are functions convmtx and convmtx2 which returns the matrix operator form of any convolution kernel.
Second, if you consider the convolution operator, the complex sinusoidal are their eigenfunctions and the amplitude of transfer function is their eigenvalues. So, if the operator is selfadjoint, it must have real eigenvalues and in order to have a selfadjoint convolution operator the Fourier transform of kernel must have real values. If a function is real valued in Fourier space then it must have even symmetry in spatial domain.
• Thank you very much for your valuable answer and your time.
– eli
Feb 15, 2023 at 6:53 | 801 | 3,017 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-22 | latest | en | 0.825411 |
http://www.rhamill.com/RobertHamill/home-electric-use-matlab/ | 1,581,959,976,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875142603.80/warc/CC-MAIN-20200217145609-20200217175609-00399.warc.gz | 245,095,165 | 15,139 | # Home Electric Use and MATLAB
I started tracking home electric use in the 1990s. I wanted to see if the rhythm of temperatures matched the rhythm of usage. Since I got MATLAB, I want to see what I could do easily in it.
## Monthly Temperatures
Average Monthly temperatures
The plot of average monthly temperature for the past 25 years in mid-Maryland shows the yearly pulse. I early discovered that we used the most electricity for heating. The power company’s bill was twice as much in winter months.
## January Temperature Focus
With MATLAB, I was able to extract the January data and then plot it. That plot has a narrower temperature range, 24-44 degrees Fahrenheit. Interesting to note, its slight upward trend.
Average January Temperatures for 25 years
## Usage and January Temperature
Let’s see how our usage varied with January’s average temperature. First I tried a 3D plot of usage by temperature and date. The plot was created, but it was not easily understood. In MATLAB, the 3D chart can be viewed easily from changing positions, which I couldn’t save for display here, but this snapshot shows the difficulty of understanding the information in the plot below.
Jan electricity usage in past 25 years
## Improved Temp Usage Plot
When I looked at the 3D plot, I realized the year did not add useful information. So I dropped it. Now I plotted usage against the temperature (of January).
Usage temperature relationship
The MATLAB scatterplot clearly shows at low temperatures (towards the left), the most electricity is consumed (data points higher). The various near pairs (at 29, 31, 33, 35, and 42 degrees) are interesting. They indicate that our power consumption is a nearly repeatable amount. Other values at those temperatures indicate Januaries when we had guests or work done in the house or left the Christmas lights up until well after the Epiphany.
## Future MATLAB Investigations
I may investigate our baseline electricity consumption upon which heating and A/C contribute to the bill in winter and summer.
Probably the number one remaining task is to see how good a mathematical relationship I can fit to the data.
The upshot is that MATLAB has made analyzing the data easier. Voila!
## 1 comment for “Home Electric Use and MATLAB”
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 494 | 2,357 | {"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} | 3.515625 | 4 | CC-MAIN-2020-10 | latest | en | 0.947643 |
https://wikimili.com/en/Newton's_method | 1,632,394,561,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057421.82/warc/CC-MAIN-20210923104706-20210923134706-00350.warc.gz | 620,610,097 | 67,998 | # Newton's method
Last updated
In numerical analysis, Newton's method, also known as the Newton–Raphson method, named after Isaac Newton and Joseph Raphson, is a root-finding algorithm which produces successively better approximations to the roots (or zeroes) of a real-valued function. The most basic version starts with a single-variable function f defined for a real variable x, the function's derivative f, and an initial guess x0 for a root of f. If the function satisfies sufficient assumptions and the initial guess is close, then
## Contents
${\displaystyle x_{1}=x_{0}-{\frac {f(x_{0})}{f'(x_{0})}}}$
is a better approximation of the root than x0. Geometrically, (x1, 0) is the intersection of the x-axis and the tangent of the graph of f at (x0, f(x0)): that is, the improved guess is the unique root of the linear approximation at the initial point. The process is repeated as
${\displaystyle x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}}$
until a sufficiently precise value is reached. This algorithm is first in the class of Householder's methods, succeeded by Halley's method. The method can also be extended to complex functions and to systems of equations.
## Description
The idea is to start with an initial guess which is reasonably close to the true root, then to approximate the function by its tangent line using calculus, and finally to compute the x-intercept of this tangent line by elementary algebra. This x-intercept will typically be a better approximation to the original function's root than the first guess, and the method can be iterated.
More formally, suppose f : (a, b) → ℝ is a differentiable function defined on the interval (a, b) with values in the real numbers , and we have some current approximation xn. Then we can derive the formula for a better approximation, xn + 1 by referring to the diagram on the right. The equation of the tangent line to the curve y = f(x) at x = xn is
${\displaystyle y=f'(x_{n})\,(x-x_{n})+f(x_{n}),}$
where f′ denotes the derivative. The x-intercept of this line (the value of x which makes y = 0) is taken as the next approximation, xn + 1, to the root, so that the equation of the tangent line is satisfied when ${\displaystyle (x,y)=(x_{n+1},0)}$:
${\displaystyle 0=f'(x_{n})\,(x_{n+1}-x_{n})+f(x_{n}).}$
Solving for xn + 1 gives
${\displaystyle x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}.}$
We start the process with some arbitrary initial value x0. (The closer to the zero, the better. But, in the absence of any intuition about where the zero might lie, a "guess and check" method might narrow the possibilities to a reasonably small interval by appealing to the intermediate value theorem.) The method will usually converge, provided this initial guess is close enough to the unknown zero, and that f(x0) ≠ 0. Furthermore, for a zero of multiplicity 1, the convergence is at least quadratic (see rate of convergence) in a neighbourhood of the zero, which intuitively means that the number of correct digits roughly doubles in every step. More details can be found in the analysis section below.
Householder's methods are similar but have higher order for even faster convergence. However, the extra computations required for each step can slow down the overall performance relative to Newton's method, particularly if f or its derivatives are computationally expensive to evaluate.
## History
The name "Newton's method" is derived from Isaac Newton's description of a special case of the method in De analysi per aequationes numero terminorum infinitas (written in 1669, published in 1711 by William Jones) and in De metodis fluxionum et serierum infinitarum (written in 1671, translated and published as Method of Fluxions in 1736 by John Colson). However, his method differs substantially from the modern method given above. Newton applied the method only to polynomials, starting with an initial root estimate and extracting a sequence of error corrections. He used each correction to rewrite the polynomial in terms of the remaining error, and then solved for a new correction by neglecting higher-degree terms. He did not explicitly connect the method with derivatives or present a general formula. Newton applied this method to both numerical and algebraic problems, producing Taylor series in the latter case.
Newton may have derived his method from a similar but less precise method by Vieta. The essence of Vieta's method can be found in the work of the Persian mathematician Sharaf al-Din al-Tusi, while his successor Jamshīd al-Kāshī used a form of Newton's method to solve xPN = 0 to find roots of N (Ypma 1995). A special case of Newton's method for calculating square roots was known since ancient times and is often called the Babylonian method.
Newton's method was used by 17th-century Japanese mathematician Seki Kōwa to solve single-variable equations, though the connection with calculus was missing. [1]
Newton's method was first published in 1685 in A Treatise of Algebra both Historical and Practical by John Wallis. [2] In 1690, Joseph Raphson published a simplified description in Analysis aequationum universalis. [3] Raphson also applied the method only to polynomials, but he avoided Newton's tedious rewriting process by extracting each successive correction from the original polynomial. This allowed him to derive a reusable iterative expression for each problem. Finally, in 1740, Thomas Simpson described Newton's method as an iterative method for solving general nonlinear equations using calculus, essentially giving the description above. In the same publication, Simpson also gives the generalization to systems of two equations and notes that Newton's method can be used for solving optimization problems by setting the gradient to zero.
Arthur Cayley in 1879 in The Newton–Fourier imaginary problem was the first to notice the difficulties in generalizing Newton's method to complex roots of polynomials with degree greater than 2 and complex initial values. This opened the way to the study of the theory of iterations of rational functions.
## Practical considerations
Newton's method is a powerful technique—in general the convergence is quadratic: as the method converges on the root, the difference between the root and the approximation is squared (the number of accurate digits roughly doubles) at each step. However, there are some difficulties with the method.
### Difficulty in calculating derivative of a function
Newton's method requires that the derivative can be calculated directly. An analytical expression for the derivative may not be easily obtainable or could be expensive to evaluate. In these situations, it may be appropriate to approximate the derivative by using the slope of a line through two nearby points on the function. Using this approximation would result in something like the secant method whose convergence is slower than that of Newton's method.
### Failure of the method to converge to the root
It is important to review the proof of quadratic convergence of Newton's method before implementing it. Specifically, one should review the assumptions made in the proof. For situations where the method fails to converge, it is because the assumptions made in this proof are not met.
#### Overshoot
If the first derivative is not well behaved in the neighborhood of a particular root, the method may overshoot, and diverge from that root. An example of a function with one root, for which the derivative is not well behaved in the neighborhood of the root, is
${\displaystyle f(x)=|x|^{a},\quad 0
for which the root will be overshot and the sequence of x will diverge. For a = 1/2, the root will still be overshot, but the sequence will oscillate between two values. For 1/2 < a < 1, the root will still be overshot but the sequence will converge, and for a ≥ 1 the root will not be overshot at all.
In some cases, Newton's method can be stabilized by using successive over-relaxation, or the speed of convergence can be increased by using the same method.
#### Stationary point
If a stationary point of the function is encountered, the derivative is zero and the method will terminate due to division by zero.
#### Poor initial estimate
A large error in the initial estimate can contribute to non-convergence of the algorithm. To overcome this problem one can often linearize the function that is being optimized using calculus, logs, differentials, or even using evolutionary algorithms, such as the stochastic tunneling. Good initial estimates lie close to the final globally optimal parameter estimate. In nonlinear regression, the sum of squared errors (SSE) is only "close to" parabolic in the region of the final parameter estimates. Initial estimates found here will allow the Newton–Raphson method to quickly converge. It is only here that the Hessian matrix of the SSE is positive and the first derivative of the SSE is close to zero.
#### Mitigation of non-convergence
In a robust implementation of Newton's method, it is common to place limits on the number of iterations, bound the solution to an interval known to contain the root, and combine the method with a more robust root finding method.
### Slow convergence for roots of multiplicity greater than 1
If the root being sought has multiplicity greater than one, the convergence rate is merely linear (errors reduced by a constant factor at each step) unless special steps are taken. When there are two or more roots that are close together then it may take many iterations before the iterates get close enough to one of them for the quadratic convergence to be apparent. However, if the multiplicity ${\displaystyle m}$ of the root is known, the following modified algorithm preserves the quadratic convergence rate: [4]
${\displaystyle x_{n+1}=x_{n}-m{\frac {f(x_{n})}{f'(x_{n})}}.}$
This is equivalent to using successive over-relaxation. On the other hand, if the multiplicity m of the root is not known, it is possible to estimate ${\displaystyle m}$ after carrying out one or two iterations, and then use that value to increase the rate of convergence.
If the multiplicity m of the root is finite then g(x) = f(x) / f(x) will have a root at the same location with multiplicity 1. Applying Newton's method to find the root of g(x) recovers quadratic convergence in many cases although it generally involves the second derivative of f(x). In a particularly simple case, if f(x) = xm then g(x) = x / m and Newton's method finds the root in a single iteration with
${\displaystyle x_{n+1}=x_{n}-{\frac {g(x_{n})}{g'(x_{n})}}=x_{n}-{\frac {x_{n}/m}{1/m}}=0\,.}$
## Analysis
Suppose that the function f has a zero at α, i.e., f(α) = 0, and f is differentiable in a neighborhood of α.
If f is continuously differentiable and its derivative is nonzero at α, then there exists a neighborhood of α such that for all starting values x0 in that neighborhood, the sequence (xn) will converge to α. [5]
If the function is continuously differentiable and its derivative is not 0 at α and it has a second derivative at α then the convergence is quadratic or faster. If the second derivative is not 0 at α then the convergence is merely quadratic. If the third derivative exists and is bounded in a neighborhood of α, then:
${\displaystyle \Delta x_{i+1}={\frac {f''(\alpha )}{2f'(\alpha )}}(\Delta x_{i})^{2}+O(\Delta x_{i})^{3}\,,}$
where
${\displaystyle \Delta x_{i}\triangleq x_{i}-\alpha \,.}$
If the derivative is 0 at α, then the convergence is usually only linear. Specifically, if f is twice continuously differentiable, f(α) = 0 and f(α) ≠ 0, then there exists a neighborhood of α such that, for all starting values x0 in that neighborhood, the sequence of iterates converges linearly, with rate 1/2 [6] Alternatively, if f(α) = 0 and f(x) ≠ 0 for xα, x in a neighborhood U of α, α being a zero of multiplicity r, and if fCr(U), then there exists a neighborhood of α such that, for all starting values x0 in that neighborhood, the sequence of iterates converges linearly.
However, even linear convergence is not guaranteed in pathological situations.
In practice, these results are local, and the neighborhood of convergence is not known in advance. But there are also some results on global convergence: for instance, given a right neighborhood U+ of α, if f is twice differentiable in U+ and if f ≠ 0, f · f > 0 in U+, then, for each x0 in U+ the sequence xk is monotonically decreasing to α.
### Proof of quadratic convergence for Newton's iterative method
According to Taylor's theorem, any function f(x) which has a continuous second derivative can be represented by an expansion about a point that is close to a root of f(x). Suppose this root is α. Then the expansion of f(α) about xn is:
${\displaystyle f(\alpha )=f(x_{n})+f'(x_{n})(\alpha -x_{n})+R_{1}\,}$
(1)
${\displaystyle R_{1}={\frac {1}{2!}}f''(\xi _{n})(\alpha -x_{n})^{2}\,,}$
where ξn is in between xn and α.
Since α is the root, ( 1 ) becomes:
${\displaystyle 0=f(\alpha )=f(x_{n})+f'(x_{n})(\alpha -x_{n})+{\tfrac {1}{2}}f''(\xi _{n})(\alpha -x_{n})^{2}\,}$
(2)
Dividing equation ( 2 ) by f(xn) and rearranging gives
${\displaystyle {\frac {f(x_{n})}{f'(x_{n})}}+\left(\alpha -x_{n}\right)={\frac {-f''(\xi _{n})}{2f'(x_{n})}}\left(\alpha -x_{n}\right)^{2}}$
(3)
Remembering that xn + 1 is defined by
${\displaystyle x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}\,,}$
(4)
one finds that
${\displaystyle \underbrace {\alpha -x_{n+1}} _{\varepsilon _{n+1}}={\frac {-f''(\xi _{n})}{2f'(x_{n})}}(\,\underbrace {\alpha -x_{n}} _{\varepsilon _{n}}\,)^{2}\,.}$
That is,
${\displaystyle \varepsilon _{n+1}={\frac {-f''(\xi _{n})}{2f'(x_{n})}}\cdot {\varepsilon _{n}}^{2}\,.}$
(5)
Taking the absolute value of both sides gives
${\displaystyle \left|{\varepsilon _{n+1}}\right|={\frac {\left|f''(\xi _{n})\right|}{2\left|f'(x_{n})\right|}}\cdot {\varepsilon _{n}}^{2}\,.}$
(6)
Equation ( 6 ) shows that the rate of convergence is at least quadratic if the following conditions are satisfied:
1. f(x) ≠ 0; for all xI, where I is the interval [αr, α + r] for some r|αx0|;
2. f(x) is continuous, for all xI;
3. x0 is sufficiently close to the root α.
The term sufficiently close in this context means the following:
1. Taylor approximation is accurate enough such that we can ignore higher order terms;
2. ${\displaystyle {\frac {1}{2}}\left|{\frac {f''(x_{n})}{f'(x_{n})}}\right| for some C < ∞;
3. ${\displaystyle C\left|{\frac {f''(\alpha )}{f'(\alpha )}}\right|\varepsilon _{n}<1,}$ for n, n ≥ 0 and C satisfying condition b.
Finally, ( 6 ) can be expressed in the following way:
${\displaystyle \left|{\varepsilon _{n+1}}\right|\leq M{\varepsilon _{n}}^{2}\,}$
where M is the supremum of the variable coefficient of εn2 on the interval I defined in condition 1, that is:
${\displaystyle M=\sup _{x\in I}{\frac {1}{2}}\left|{\frac {f''(x)}{f'(x)}}\right|.\,}$
The initial point x0 has to be chosen such that conditions 1 to 3 are satisfied, where the third condition requires that M|ε0| < 1.
### Basins of attraction
The disjoint subsets of the basins of attraction—the regions of the real number line such that within each region iteration from any point leads to one particular root—can be infinite in number and arbitrarily small. For example, [7] for the function f(x) = x3 − 2x2 − 11x + 12 = (x − 4)(x − 1)(x + 3), the following initial conditions are in successive basins of attraction:
2.35288 converges to 4; 2.35284 converges to −3; 2.35284 converges to 4; 2.35284 converges to −3; 2.35284 converges to 1.
## Failure analysis
Newton's method is only guaranteed to converge if certain conditions are satisfied. If the assumptions made in the proof of quadratic convergence are met, the method will converge. For the following subsections, failure of the method to converge indicates that the assumptions made in the proof were not met.
In some cases the conditions on the function that are necessary for convergence are satisfied, but the point chosen as the initial point is not in the interval where the method converges. This can happen, for example, if the function whose root is sought approaches zero asymptotically as x goes to or −∞. In such cases a different method, such as bisection, should be used to obtain a better estimate for the zero to use as an initial point.
#### Iteration point is stationary
Consider the function:
${\displaystyle f(x)=1-x^{2}.}$
It has a maximum at x = 0 and solutions of f(x) = 0 at x = ±1. If we start iterating from the stationary point x0 = 0 (where the derivative is zero), x1 will be undefined, since the tangent at (0,1) is parallel to the x-axis:
${\displaystyle x_{1}=x_{0}-{\frac {f(x_{0})}{f'(x_{0})}}=0-{\frac {1}{0}}.}$
The same issue occurs if, instead of the starting point, any iteration point is stationary. Even if the derivative is small but not zero, the next iteration will be a far worse approximation.
#### Starting point enters a cycle
For some functions, some starting points may enter an infinite cycle, preventing convergence. Let
${\displaystyle f(x)=x^{3}-2x+2\!}$
and take 0 as the starting point. The first iteration produces 1 and the second iteration returns to 0 so the sequence will alternate between the two without converging to a root. In fact, this 2-cycle is stable: there are neighborhoods around 0 and around 1 from which all points iterate asymptotically to the 2-cycle (and hence not to the root of the function). In general, the behavior of the sequence can be very complex (see Newton fractal). The real solution of this equation is −1.76929235….
### Derivative issues
If the function is not continuously differentiable in a neighborhood of the root then it is possible that Newton's method will always diverge and fail, unless the solution is guessed on the first try.
#### Derivative does not exist at root
A simple example of a function where Newton's method diverges is trying to find the cube root of zero. The cube root is continuous and infinitely differentiable, except for x = 0, where its derivative is undefined:
${\displaystyle f(x)={\sqrt[{3}]{x}}.}$
For any iteration point xn, the next iteration point will be:
${\displaystyle x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}=x_{n}-{\frac {{x_{n}}^{\frac {1}{3}}}{{\frac {1}{3}}{x_{n}}^{{\frac {1}{3}}-1}}}=x_{n}-3x_{n}=-2x_{n}.}$
The algorithm overshoots the solution and lands on the other side of the y-axis, farther away than it initially was; applying Newton's method actually doubles the distances from the solution at each iteration.
In fact, the iterations diverge to infinity for every f(x) = |x|α, where 0 < α < 1/2. In the limiting case of α = 1/2 (square root), the iterations will alternate indefinitely between points x0 and x0, so they do not converge in this case either.
#### Discontinuous derivative
If the derivative is not continuous at the root, then convergence may fail to occur in any neighborhood of the root. Consider the function
${\displaystyle f(x)={\begin{cases}0&{\text{if }}x=0,\\x+x^{2}\sin {\frac {2}{x}}&{\text{if }}x\neq 0.\end{cases}}}$
Its derivative is:
${\displaystyle f'(x)={\begin{cases}1&{\text{if }}x=0,\\1+2x\sin {\frac {2}{x}}-2\cos {\frac {2}{x}}&{\text{if }}x\neq 0.\end{cases}}}$
Within any neighborhood of the root, this derivative keeps changing sign as x approaches 0 from the right (or from the left) while f(x) ≥ xx2 > 0 for 0 < x < 1.
So f(x)/f(x) is unbounded near the root, and Newton's method will diverge almost everywhere in any neighborhood of it, even though:
• the function is differentiable (and thus continuous) everywhere;
• the derivative at the root is nonzero;
• f is infinitely differentiable except at the root; and
• the derivative is bounded in a neighborhood of the root (unlike f(x)/f(x)).
In some cases the iterates converge but do not converge as quickly as promised. In these cases simpler methods converge just as quickly as Newton's method.
#### Zero derivative
If the first derivative is zero at the root, then convergence will not be quadratic. Let
${\displaystyle f(x)=x^{2}\!}$
then f(x) = 2x and consequently
${\displaystyle x-{\frac {f(x)}{f'(x)}}={\frac {x}{2}}.}$
So convergence is not quadratic, even though the function is infinitely differentiable everywhere.
Similar problems occur even when the root is only "nearly" double. For example, let
${\displaystyle f(x)=x^{2}(x-1000)+1.}$
Then the first few iterations starting at x0 = 1 are
x0 = 1
x1 = 0.500250376
x2 = 0.251062828
x3 = 0.127507934
x4 = 0.067671976
x5 = 0.041224176
x6 = 0.032741218
x7 = 0.031642362
it takes six iterations to reach a point where the convergence appears to be quadratic.
#### No second derivative
If there is no second derivative at the root, then convergence may fail to be quadratic. Let
${\displaystyle f(x)=x+x^{\frac {4}{3}}.}$
Then
${\displaystyle f'(x)=1+{\tfrac {4}{3}}x^{\frac {1}{3}}.}$
And
${\displaystyle f''(x)={\tfrac {4}{9}}x^{-{\frac {2}{3}}}}$
except when x = 0 where it is undefined. Given xn,
${\displaystyle x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}={\frac {{\frac {1}{3}}{x_{n}}^{\frac {4}{3}}}{1+{\tfrac {4}{3}}{x_{n}}^{\frac {1}{3}}}}}$
which has approximately 4/3 times as many bits of precision as xn has. This is less than the 2 times as many which would be required for quadratic convergence. So the convergence of Newton's method (in this case) is not quadratic, even though: the function is continuously differentiable everywhere; the derivative is not zero at the root; and f is infinitely differentiable except at the desired root.
## Generalizations
### Complex functions
When dealing with complex functions, Newton's method can be directly applied to find their zeroes. [8] Each zero has a basin of attraction in the complex plane, the set of all starting values that cause the method to converge to that particular zero. These sets can be mapped as in the image shown. For many complex functions, the boundaries of the basins of attraction are fractals.
In some cases there are regions in the complex plane which are not in any of these basins of attraction, meaning the iterates do not converge. For example, [9] if one uses a real initial condition to seek a root of x2 + 1, all subsequent iterates will be real numbers and so the iterations cannot converge to either root, since both roots are non-real. In this case almost all real initial conditions lead to chaotic behavior, while some initial conditions iterate either to infinity or to repeating cycles of any finite length.
Curt McMullen has shown that for any possible purely iterative algorithm similar to Newton's method, the algorithm will diverge on some open regions of the complex plane when applied to some polynomial of degree 4 or higher. However, McMullen gave a generally convergent algorithm for polynomials of degree 3. [10]
### Systems of equations
#### k variables, k functions
One may also use Newton's method to solve systems of k (nonlinear) equations, which amounts to finding the zeroes of continuously differentiable functions F : ℝk → ℝk. In the formulation given above, one then has to left multiply with the inverse of the k × k Jacobian matrix JF(xn) instead of dividing by f(xn):
${\displaystyle x_{n+1}=x_{n}-J_{F}(x_{n})^{-1}F(x_{n})}$
Rather than actually computing the inverse of the Jacobian matrix, one may save time and increase numerical stability by solving the system of linear equations
${\displaystyle J_{F}(x_{n})(x_{n+1}-x_{n})=-F(x_{n})}$
for the unknown xn + 1xn.
#### k variables, m equations, with m > k
The k-dimensional variant of Newton's method can be used to solve systems of greater than k (nonlinear) equations as well if the algorithm uses the generalized inverse of the non-square Jacobian matrix J+ = (JTJ)−1JT instead of the inverse of J. If the nonlinear system has no solution, the method attempts to find a solution in the non-linear least squares sense. See Gauss–Newton algorithm for more information.
### In a Banach space
Another generalization is Newton's method to find a root of a functional F defined in a Banach space. In this case the formulation is
${\displaystyle X_{n+1}=X_{n}-{\bigl (}F'(X_{n}){\bigr )}^{-1}F(X_{n}),\,}$
where F′(Xn) is the Fréchet derivative computed at Xn. One needs the Fréchet derivative to be boundedly invertible at each Xn in order for the method to be applicable. A condition for existence of and convergence to a root is given by the Newton–Kantorovich theorem. [11]
In p-adic analysis, the standard method to show a polynomial equation in one variable has a p-adic root is Hensel's lemma, which uses the recursion from Newton's method on the p-adic numbers. Because of the more stable behavior of addition and multiplication in the p-adic numbers compared to the real numbers (specifically, the unit ball in the p-adics is a ring), convergence in Hensel's lemma can be guaranteed under much simpler hypotheses than in the classical Newton's method on the real line.
### Newton–Fourier method
The Newton–Fourier method is Joseph Fourier's extension of Newton's method to provide bounds on the absolute error of the root approximation, while still providing quadratic convergence.
Assume that f(x) is twice continuously differentiable on [a, b] and that f contains a root in this interval. Assume that f(x), f(x) ≠ 0 on this interval (this is the case for instance if f(a) < 0, f(b) > 0, and f(x) > 0, and f(x) > 0 on this interval). This guarantees that there is a unique root on this interval, call it α. If it is concave down instead of concave up then replace f(x) by f(x) since they have the same roots.
Let x0 = b be the right endpoint of the interval and let z0 = a be the left endpoint of the interval. Given xn, define
${\displaystyle x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}},}$
which is just Newton's method as before. Then define
${\displaystyle z_{n+1}=z_{n}-{\frac {f(z_{n})}{f'(x_{n})}},}$
where the denominator is f(xn) and not f(zn). The iterations xn will be strictly decreasing to the root while the iterations zn will be strictly increasing to the root. Also,
${\displaystyle \lim _{n\to \infty }{\frac {x_{n+1}-z_{n+1}}{(x_{n}-z_{n})^{2}}}={\frac {f''(\alpha )}{2f'(\alpha )}}}$
so that distance between xn and zn decreases quadratically.
### Quasi-Newton methods
When the Jacobian is unavailable or too expensive to compute at every iteration, a quasi-Newton method can be used.
### q-analog
Newton's method can be generalized with the q-analog of the usual derivative. [12]
### Modified Newton methods
#### Maehly's procedure
A nonlinear equation has multiple solutions in general. But if the initial value is not appropriate, Newton's method may not converge to the desired solution or may converge to the same solution found earlier. When we have already found N solutions of ${\displaystyle f(x)=0}$, then the next root can be found by applying Newton's method to the next equation: [13] [14]
${\displaystyle F(x)={\frac {f(x)}{\prod _{i=1}^{N}(x-x_{i})}}=0.}$
This method is applied to obtain zeros of the Bessel function of the second kind. [15]
#### Hirano's modified Newton method
Hirano's modified Newton method is a modification conserving the convergence of Newton method and avoiding unstableness. [16] It is developed to solve complex polynomials.
#### Interval Newton's method
Combining Newton's method with interval arithmetic is very useful in some contexts. This provides a stopping criterion that is more reliable than the usual ones (which are a small value of the function or a small variation of the variable between consecutive iterations). Also, this may detect cases where Newton's method converges theoretically but diverges numerically because of an insufficient floating-point precision (this is typically the case for polynomials of large degree, where a very small change of the variable may change dramatically the value of the function; see Wilkinson's polynomial). [17] [18]
Consider ${\displaystyle f\in {\mathcal {C}}^{1}(X)}$, where ${\displaystyle X}$ is a real interval, and suppose that we have an interval extension ${\displaystyle F'}$ of ${\displaystyle f'}$, meaning that ${\displaystyle F'}$ takes as input an interval ${\displaystyle Y\subseteq X}$ and outputs an interval ${\displaystyle F'(Y)}$ such that:
{\displaystyle {\begin{aligned}F'([y,y])&=\{f'(y)\}\\[5pt]F'(Y)&\supseteq \{f'(y)\mid y\in Y\}.\end{aligned}}}
We also assume that ${\displaystyle 0\notin F'(X)}$, so in particular ${\displaystyle f}$ has at most one root in ${\displaystyle X}$. We then define the interval Newton operator by:
${\displaystyle N(Y)=m-{\frac {f(m)}{F'(Y)}}=\left\{\left.m-{\frac {f(m)}{z}}~\right|~z\in F'(Y)\right\}}$
where ${\displaystyle m\in Y}$. Note that the hypothesis on ${\displaystyle F'}$ implies that ${\displaystyle N(Y)}$ is well defined and is an interval (see interval arithmetic for further details on interval operations). This naturally leads to the following sequence:
{\displaystyle {\begin{aligned}X_{0}&=X\\X_{k+1}&=N(X_{k})\cap X_{k}.\end{aligned}}}
The mean value theorem ensures that if there is a root of ${\displaystyle f}$ in ${\displaystyle X_{k}}$, then it is also in ${\displaystyle X_{k+1}}$. Moreover, the hypothesis on ${\displaystyle F'}$ ensures that ${\displaystyle X_{k+1}}$ is at most half the size of ${\displaystyle X_{k}}$ when ${\displaystyle m}$ is the midpoint of ${\displaystyle Y}$, so this sequence converges towards ${\displaystyle [x^{*},x^{*}]}$, where ${\displaystyle x^{*}}$ is the root of ${\displaystyle f}$ in ${\displaystyle X}$.
If ${\displaystyle F'(X)}$ strictly contains ${\displaystyle 0}$, the use of extended interval division produces a union of two intervals for ${\displaystyle N(X)}$ ; multiple roots are therefore automatically separated and bounded.
## Applications
### Minimization and maximization problems
Newton's method can be used to find a minimum or maximum of a function ${\displaystyle f(x)}$. The derivative is zero at a minimum or maximum, so local minima and maxima can be found by applying Newton's method to the derivative. The iteration becomes:
${\displaystyle x_{n+1}=x_{n}-{\frac {f'(x_{n})}{f''(x_{n})}}.}$
### Multiplicative inverses of numbers and power series
An important application is Newton–Raphson division, which can be used to quickly find the reciprocal of a number a, using only multiplication and subtraction, that is to say the number x such that 1/x = a. We can rephrase that as finding the zero of f(x) = 1/xa. We have f′(x) = −1/x2.
Newton's iteration is
${\displaystyle x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}=x_{n}+{\frac {{\frac {1}{x_{n}}}-a}{\frac {1}{x_{n}^{2}}}}=x_{n}(2-ax_{n}).}$
Therefore, Newton's iteration needs only two multiplications and one subtraction.
This method is also very efficient to compute the multiplicative inverse of a power series.
### Solving transcendental equations
Many transcendental equations can be solved using Newton's method. Given the equation
${\displaystyle g(x)=h(x),}$
with g(x) and/or h(x) a transcendental function, one writes
${\displaystyle f(x)=g(x)-h(x).}$
The values of x that solve the original equation are then the roots of f(x), which may be found via Newton's method.
### Obtaining zeros of special functions
Newton's method is applied to the ratio of Bessel functions in order to obtain its root. [19]
### Numerical verification for solutions of nonlinear equations
A numerical verification for solutions of nonlinear equations has been established by using Newton's method multiple times and forming a set of solution candidates. [20] [21]
### CFD modeling
An iterative Newton-Raphson procedure was employed in order to impose a stable Dirichlet boundary condition in CFD, as a quite general strategy to model current and potential distribution for electrochemical cell stacks. [22]
## Examples
### Square root
Consider the problem of finding the square root of a number a, that is to say the positive number x such that x2 = a. Newton's method is one of many methods of computing square roots. We can rephrase that as finding the zero of f(x) = x2a. We have f′(x) = 2x.
For example, for finding the square root of 612 with an initial guess x0 = 10, the sequence given by Newton's method is:
${\displaystyle {\begin{matrix}x_{1}&=&x_{0}-{\dfrac {f(x_{0})}{f'(x_{0})}}&=&10-{\dfrac {10^{2}-612}{2\times 10}}&=&35.6\qquad \qquad \qquad \quad \;\,{}\\x_{2}&=&x_{1}-{\dfrac {f(x_{1})}{f'(x_{1})}}&=&35.6-{\dfrac {35.6^{2}-612}{2\times 35.6}}&=&{\underline {2}}6.395\,505\,617\,978\dots \\x_{3}&=&\vdots &=&\vdots &=&{\underline {24.7}}90\,635\,492\,455\dots \\x_{4}&=&\vdots &=&\vdots &=&{\underline {24.738\,6}}88\,294\,075\dots \\x_{5}&=&\vdots &=&\vdots &=&{\underline {24.738\,633\,753\,7}}67\dots \end{matrix}}}$
where the correct digits are underlined. With only a few iterations one can obtain a solution accurate to many decimal places.
Rearranging the formula as follows yields the Babylonian method of finding square roots:
${\displaystyle x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}=x_{n}-{\frac {x_{n}^{2}-a}{2x_{n}}}={\frac {1}{2}}{\biggl (}2x_{n}-{\Bigl (}x_{n}-{\frac {a}{x_{n}}}{\Bigr )}{\biggr )}={\frac {1}{2}}{\Bigl (}x_{n}+{\frac {a}{x_{n}}}{\Bigr )}}$
i.e. the arithmetic mean of the guess, xn and a/xn.
### Solution of cos(x) = x3
Consider the problem of finding the positive number x with cos(x) = x3. We can rephrase that as finding the zero of f(x) = cos(x) − x3. We have f′(x) = −sin(x) − 3x2. Since cos(x) ≤ 1 for all x and x3 > 1 for x > 1, we know that our solution lies between 0 and 1.
For example, with an initial guess x0 = 0.5, the sequence given by Newton's method is (note that a starting value of 0 will lead to an undefined result, showing the importance of using a starting point that is close to the solution):
${\displaystyle {\begin{matrix}x_{1}&=&x_{0}-{\dfrac {f(x_{0})}{f'(x_{0})}}&=&0.5-{\dfrac {\cos 0.5-0.5^{3}}{-\sin 0.5-3\times 0.5^{2}}}&=&1.112\,141\,637\,097\dots \\x_{2}&=&x_{1}-{\dfrac {f(x_{1})}{f'(x_{1})}}&=&\vdots &=&{\underline {0.}}909\,672\,693\,736\dots \\x_{3}&=&\vdots &=&\vdots &=&{\underline {0.86}}7\,263\,818\,209\dots \\x_{4}&=&\vdots &=&\vdots &=&{\underline {0.865\,47}}7\,135\,298\dots \\x_{5}&=&\vdots &=&\vdots &=&{\underline {0.865\,474\,033\,1}}11\dots \\x_{6}&=&\vdots &=&\vdots &=&{\underline {0.865\,474\,033\,102}}\dots \end{matrix}}}$
The correct digits are underlined in the above example. In particular, x6 is correct to 12 decimal places. We see that the number of correct digits after the decimal point increases from 2 (for x3) to 5 and 10, illustrating the quadratic convergence.
## Code
The following is an implementation example of the Newton's method in the Julia programming language for finding a root of a function f which has derivative fprime.
The initial guess will be x0 = 1 and the function will be f(x) = x2 − 2 so that f′(x) = 2x.
Each new iteration of Newton's method will be denoted by x1. We will check during the computation whether the denominator (yprime) becomes too small (smaller than epsilon), which would be the case if f′(xn) ≈ 0, since otherwise a large amount of error could be introduced.
x0=1# The initial guessf(x)=x^2-2# The function whose root we are trying to findfprime(x)=2x# The derivative of the functiontolerance=1e-7# 7 digit accuracy is desiredepsilon=1e-14# Do not divide by a number smaller than thismaxIterations=20# Do not allow the iterations to continue indefinitelysolutionFound=false# Have not converged to a solution yetfori=1:maxIterationsy=f(x0)yprime=fprime(x0)ifabs(yprime)<epsilon# Stop if the denominator is too smallbreakendglobalx1=x0-y/yprime# Do Newton's computationifabs(x1-x0)<=tolerance# Stop when the result is within the desired toleranceglobalsolutionFound=truebreakendglobalx0=x1# Update x0 to start the process againendifsolutionFoundprintln("Solution: ",x1)# x1 is a solution within tolerance and maximum number of iterationselseprintln("Did not converge")# Newton's method did not convergeend
## Notes
1. "Chapter 2. Seki Takakazu". Japanese Mathematics in the Edo Period. National Diet Library. Retrieved 24 February 2019.
2. Wallis, John (1685). A Treatise of Algebra, both Historical and Practical. Oxford: Richard Davis. doi:10.3931/e-rara-8842.
3. Raphson, Joseph (1697). Analysis Æequationum Universalis (in Latin) (2nd ed.). London: Thomas Bradyll. doi:10.3931/e-rara-13516.
4. "Accelerated and Modified Newton Methods". Archived from the original on 24 May 2019. Retrieved 4 March 2016.
5. Ryaben'kii, Victor S.; Tsynkov, Semyon V. (2006), A Theoretical Introduction to Numerical Analysis, CRC Press, p. 243, ISBN 9781584886075 .
6. Süli & Mayers 2003 , Exercise 1.6
7. Dence, Thomas (November 1997). "Cubics, chaos and Newton's method". Mathematical Gazette . 81 (492): 403–408. doi:10.2307/3619617. JSTOR 3619617.
8. Henrici, Peter (1974). "Applied and Computational Complex Analysis". 1.Cite journal requires |journal= (help)
9. Strang, Gilbert (January 1991). "A chaotic search for i". The College Mathematics Journal . 22: 3–12. doi:10.2307/2686733. JSTOR 2686733.
10. McMullen, Curt (1987). "Families of rational maps and iterative root-finding algorithms" (PDF). Annals of Mathematics. Second Series. 125 (3): 467–493. doi:10.2307/1971408. JSTOR 1971408.
11. Yamamoto, Tetsuro (2001). "Historical Developments in Convergence Analysis for Newton's and Newton-like Methods". In Brezinski, C.; Wuytack, L. (eds.). Numerical Analysis : Historical Developments in the 20th Century. North-Holland. pp. 241–263. ISBN 0-444-50617-9.
12. Rajkovic, Stankovic & Marinkovic 2002 [ incomplete short citation ]
13. Press et al. 1992 [ incomplete short citation ]
14. Stoer & Bulirsch 1980 [ incomplete short citation ]
15. Zhang & Jin 1996 [ incomplete short citation ]
16. Murota, Kazuo (1982). "Global Convergence of a Modified Newton Iteration for Algebraic Equations". SIAM J. Numer. Anal. 19 (4): 793–799. doi:10.1137/0719055.
17. Moore, R. E. (1979). Methods and applications of interval analysis (Vol. 2). Siam.
18. Hansen, E. (1978). Interval forms of Newtons method. Computing, 20(2), 153–163.
19. Krawczyk (1969) [ incomplete short citation ][ incomplete short citation ]
20. Colli, A. N.; Girault, H. H. (June 2017). "Compact and General Strategy for Solving Current and Potential Distribution in Electrochemical Cells Composed of Massive Monopolar and Bipolar Electrodes". Journal of the Electrochemical Society. 164 (11): E3465–E3472. doi:10.1149/2.0471711jes. hdl:.
## Related Research Articles
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Numerical certification is the process of verifying the correctness of a candidate solution to a system of equations. In (numerical) computational mathematics, such as numerical algebraic geometry, candidate solutions are computed algorithmically, but there is the possibility that errors have corrupted the candidates. For instance, in addition to the inexactness of input data and candidate solutions, numerical errors or errors in the discretization of the problem may result in corrupted candidate solutions. The goal of numerical certification is to provide a certificate which proves which of these candidates are, indeed, approximate solutions. | 11,811 | 45,517 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 86, "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} | 4.28125 | 4 | CC-MAIN-2021-39 | latest | en | 0.91772 |
https://education.launchcode.org/intro-to-professional-web-dev/chapters/booleans-and-conditionals/equality.html | 1,718,219,209,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861183.54/warc/CC-MAIN-20240612171727-20240612201727-00607.warc.gz | 207,744,833 | 4,869 | # 5.2. Equality¶
## 5.2.1. Loose Equality With `==`¶
In the section Booleans, we learned about the comparison operators `==` and `!=`, which test whether two values are equal or not equal, respectively. However, there are some quirks with using the `==` operator, which occur when we use `==` to compare different data types.
Example
```1 2 3``` ```console.log(7 == "7"); console.log(0 == false); console.log(0 == ''); ```
Console Output
```true
true
true
```
In order to properly make a comparison, the two operands must be the same type. If the two operands to `==` are of different data types, JavaScript will implicitly convert the operands so that the values are of the same data type before comparing the two. For this reason, the `==` operator is often said to measure loose equality.
Type conversions with `==` are carried out according to a complex set of rules, and while many of these conversions make some sense, others do not.
For example, `Number("7")` returns `7`, so it makes some sense that `7 == "7"` returns `true`. However, the following example leaves us scratching our heads.
Example
```1 2 3``` ```console.log('0' == 0); console.log(0 == ''); console.log('0' == ''); ```
Console Output
```true
true
false
```
The `==` operator is non-transitive. We think of equality as being transitive; for example, if A and B are equal and B and C are equal, then A and C are also equal. However, the example above demonstrates that that is not the case for the `==` operator.
Since `==` does not follow rules that we typically associate with equality, unexpected results may occur if `==` is used in a program. Thankfully, JavaScript provides another operator that returns more predictable results.
## 5.2.2. Strict Equality With `===`¶
The operator `===` compares two operands without converting their data types. In other words, if `a` and `b` are of different data types (say, `a` is a string and `b` is a number) then `a === b` will always be false.
Example
```1 2 3``` ```console.log(7 === "7"); console.log(0 === false); console.log(0 === ''); ```
Console Output
```false
false
false
```
For this reason, the `===` operator is often said to measure strict equality.
Just as equality operator `==` has the inequality operator `!=`, there is also a strict inequality operator, `!==`. The boolean expression `a !== b` returns `true` when the two operands are of different types, or if they are of the same type and have different values.
Tip
USE `===` AND `!==` WHENEVER POSSIBLE. In this book we will use these strict operators over the loose operators from now on.
## 5.2.3. Check Your Understanding¶
Question
What is the result of the following boolean expression?
```4 == "4"
```
1. `true`
2. `false`
3. `"true"`
4. `"false"`
Question
What is the difference between `==` and `===`?
1. There is no difference. They work exactly the same.
2. Only `===` throws an error if its arguments are of different types.
3. `==` converts values of different types to be the same type, while `===` does not.
4. `==` works with all data types, while `===` does not. | 788 | 3,108 | {"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-2024-26 | latest | en | 0.879099 |
https://leetcode.ca/2023-03-14-2661-First-Completely-Painted-Row-or-Column/ | 1,721,902,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763857355.79/warc/CC-MAIN-20240725084035-20240725114035-00352.warc.gz | 309,453,222 | 9,032 | # 2661. First Completely Painted Row or Column
## Description
You are given a 0-indexed integer array arr, and an m x n integer matrix mat. arr and mat both contain all the integers in the range [1, m * n].
Go through each index i in arr starting from index 0 and paint the cell in mat containing the integer arr[i].
Return the smallest index i at which either a row or a column will be completely painted in mat.
Example 1:
Input: arr = [1,3,4,2], mat = [[1,4],[2,3]]
Output: 2
Explanation: The moves are shown in order, and both the first row and second column of the matrix become fully painted at arr[2].
Example 2:
Input: arr = [2,8,7,4,1,3,5,6,9], mat = [[3,2,5],[1,4,6],[8,7,9]]
Output: 3
Explanation: The second column becomes fully painted at arr[3].
Constraints:
• m == mat.length
• n = mat[i].length
• arr.length == m * n
• 1 <= m, n <= 105
• 1 <= m * n <= 105
• 1 <= arr[i], mat[r][c] <= m * n
• All the integers of arr are unique.
• All the integers of mat are unique.
## Solutions
Solution 1: Hash Table + Array Counting
We use a hash table $idx$ to record the position of each element in the matrix $mat$, that is $idx[mat[i][j]] = (i, j)$, and define two arrays $row$ and $col$ to record the number of colored elements in each row and each column respectively.
Traverse the array $arr$. For each element $arr[k]$, we find its position $(i, j)$ in the matrix $mat$, and then add $row[i]$ and $col[j]$ by one. If $row[i] = n$ or $col[j] = m$, it means that the $i$-th row or the $j$-th column has been colored, so $arr[k]$ is the element we are looking for, and we return $k$.
The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here $m$ and $n$ are the number of rows and columns of the matrix $mat$ respectively.
• class Solution {
public int firstCompleteIndex(int[] arr, int[][] mat) {
int m = mat.length, n = mat[0].length;
Map<Integer, int[]> idx = new HashMap<>();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
idx.put(mat[i][j], new int[] {i, j});
}
}
int[] row = new int[m];
int[] col = new int[n];
for (int k = 0;; ++k) {
var x = idx.get(arr[k]);
int i = x[0], j = x[1];
++row[i];
++col[j];
if (row[i] == n || col[j] == m) {
return k;
}
}
}
}
• class Solution {
public:
int firstCompleteIndex(vector<int>& arr, vector<vector<int>>& mat) {
int m = mat.size(), n = mat[0].size();
unordered_map<int, pair<int, int>> idx;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
idx[mat[i][j]] = {i, j};
}
}
vector<int> row(m), col(n);
for (int k = 0;; ++k) {
auto [i, j] = idx[arr[k]];
++row[i];
++col[j];
if (row[i] == n || col[j] == m) {
return k;
}
}
}
};
• class Solution:
def firstCompleteIndex(self, arr: List[int], mat: List[List[int]]) -> int:
m, n = len(mat), len(mat[0])
idx = {}
for i in range(m):
for j in range(n):
idx[mat[i][j]] = (i, j)
row = [0] * m
col = [0] * n
for k in range(len(arr)):
i, j = idx[arr[k]]
row[i] += 1
col[j] += 1
if row[i] == n or col[j] == m:
return k
• func firstCompleteIndex(arr []int, mat [][]int) int {
m, n := len(mat), len(mat[0])
idx := map[int][2]int{}
for i := range mat {
for j := range mat[i] {
idx[mat[i][j]] = [2]int{i, j}
}
}
row := make([]int, m)
col := make([]int, n)
for k := 0; ; k++ {
x := idx[arr[k]]
i, j := x[0], x[1]
row[i]++
col[j]++
if row[i] == n || col[j] == m {
return k
}
}
}
• function firstCompleteIndex(arr: number[], mat: number[][]): number {
const m = mat.length;
const n = mat[0].length;
const idx: Map<number, number[]> = new Map();
for (let i = 0; i < m; ++i) {
for (let j = 0; j < n; ++j) {
idx.set(mat[i][j], [i, j]);
}
}
const row: number[] = Array(m).fill(0);
const col: number[] = Array(n).fill(0);
for (let k = 0; ; ++k) {
const [i, j] = idx.get(arr[k])!;
++row[i];
++col[j];
if (row[i] === n || col[j] === m) {
return k;
}
}
}
• use std::collections::HashMap;
impl Solution {
pub fn first_complete_index(arr: Vec<i32>, mat: Vec<Vec<i32>>) -> i32 {
let m = mat.len();
let n = mat[0].len();
let mut idx = HashMap::new();
for i in 0..m {
for j in 0..n {
idx.insert(mat[i][j], [i, j]);
}
}
let mut row = vec![0; m];
let mut col = vec![0; n];
for k in 0..arr.len() {
let x = idx.get(&arr[k]).unwrap();
let i = x[0];
let j = x[1];
row[i] += 1;
col[j] += 1;
if row[i] == n || col[j] == m {
return k as i32;
}
}
-1
}
} | 1,478 | 4,292 | {"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} | 3.90625 | 4 | CC-MAIN-2024-30 | latest | en | 0.751413 |
https://campusgrades.com/what-does-tolstoys-anna-karenina-reveal-about-women-in-ninteenth-century/ | 1,680,309,107,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949694.55/warc/CC-MAIN-20230401001704-20230401031704-00675.warc.gz | 183,211,410 | 13,979 | What does Tolstoy’s Anna Karenina reveal about women in ninteenth century?
# What does Tolstoy’s Anna Karenina reveal about women in ninteenth century?
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For what values of a does the quadratic equation ax^2 + 20x + 4 = 0 have equal roots?
The roots of a quadratic equation ax^2 + bx + c = 0 can be represented as [–b + sqrt (b^2 – 4ac)]/2a and [–b – sqrt (b^2 – 4ac)]/2a Now if the two are equal [–b + sqrt (b^2 – 4ac)]/2a =…
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Please provide a character analysis of Tybalt in Romeo and Juliet.
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• SILAS MARNER
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Did the Bank War divide the North and South economically?
I would say that the Bank War (I’m assuming you’re talking about Andrew Jackson and his “war” against the Bank of the United States) did not do anything to divide the North and the South. This was… | 3,808 | 14,077 | {"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-2023-14 | latest | en | 0.910536 |
http://stackoverflow.com/questions/2354736/when-is-the-appropriate-time-to-use-radix-sort/2354818 | 1,438,260,404,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042987174.71/warc/CC-MAIN-20150728002307-00034-ip-10-236-191-2.ec2.internal.warc.gz | 227,770,153 | 19,039 | # When is the appropriate time to use Radix Sort?
What are the constraints on your data for you to be able to use Radix sort?
If I'm sorting a large list of integers, would it be appropriate to use Radix sort? Why is Radix sort not used more?
-
do you have an example of a place you would expect it to be used, but it is not? – Peter Recore Mar 2 '10 at 4:03
A radix sort puts stronger requirements on the types being compared than the comparison sorts, and isn't always significantly faster. For integers, a radix is probably faster. – Mooing Duck May 7 '12 at 22:22
It's great when you have a large set of data with keys that are somehow constrained. For example, when you need to order a 1-million array of 64-bit numbers, it can be used to sort by 8 least significant bits, then by the next 8, and so on (applied 8 times). That way this array can be sorted in 8*1M operations, rather than 1M*log(1M).
-
But log(1M) is 6... – Yaniv Jan 11 at 15:08
He means log base 2 – N. McA. May 31 at 9:41
If you know the range of the integer values, and it's not too large,
maybe counting sort would be a better choice in your case.
-
One reason you might not see it as often as you'd think you would is that Radix sort is not as general purpose as comparison based sorts (quicksort/mergesort/heapsort). It requires that you can represent the items to be sorted as an integer, or something like an integer. When using a standard library, it is easy to define a comparison function that compares arbitrary objects. It might be harder to define an encoding that properly maps your arbitrary data type into an integer.
-
Bucket sorting is useful in situations where the number of discrete key values is small relative to the number of data items, and where the goal is to produce a re-sorted copy of a list without disturbing the original (so needing to maintain both the old and new versions of the list simultaneously is not a burden). If the number of possible keys is too large to handle in a single pass, one can extend bucket sort into radix sort by making multiple passes, but one loses much of the speed advantage that bucket sort could offer for small keys.
In some external-sorting scenarios, especially when the number of different key values is very small (e.g. two), a stable sort is required, and the I/O device can only operate efficiently with one sequential data stream, it may be useful to make K passes through the source data stream, where K is the number of key values. On the first pass, one copies all the items where the key is the minimum legitimate value and skips the rest, then copies all the items where the key is the next higher value, skipping the rest, etc. This approach will obviously be horribly efficient if there are very many different key values, but will be quite good if there are two.
- | 660 | 2,833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2015-32 | latest | en | 0.932808 |
https://www.assignmentexpert.com/homework-answers/programming-and-computer-science/cpp/question-15342 | 1,601,246,759,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401582033.88/warc/CC-MAIN-20200927215009-20200928005009-00428.warc.gz | 717,165,062 | 290,468 | 90 205
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# Answer to Question #15342 in C++ for reymark
Question #15342
write a program that will accept a number n and display the sum of all even numbers and the sum of all odd numbers from 1 to n.
1
2012-09-27T12:20:34-0400
#include<iostream.h>
bool even(int n){
if (n%2==0) return true; else return 0;
}
int sumOdd=0, sumEven=0;
void main(){
int m;
cout<<"Enter an integer: ";
cin>>m;
for (int i=0; i<=m; i++)
& if (even(i)) sumEven += i;
& else sumOdd += i;
cout<<"Sum of even numbers: "<<sumEven<<"\nSum of odd numbers: "<<sumOdd<<"\n";
}
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for any assignment or question with DETAILED EXPLANATIONS! | 232 | 754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-40 | latest | en | 0.691158 |
https://www.mathworks.com/matlabcentral/cody/problems/2393-measure-a-special-distance/solutions/514220 | 1,585,869,970,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370508367.57/warc/CC-MAIN-20200402204908-20200402234908-00003.warc.gz | 1,041,642,899 | 15,414 | Cody
# Problem 2393. Measure a Special Distance
Solution 514220
Submitted on 18 Oct 2014 by Pavan Toraty
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% cd=randi(100); gd=-randi(100); linef=(cd-gd)*rand(randi([50,100]),2)+gd; % n=size(linef,1); ldist=zeros(n); for a=1:n for b=1:n if linef(a,1)>0&&linef(b,2)>0 ldist(b,a)=(linef(a,1)+linef(b,2)-cd).^2; elseif linef(a,1)<0&&linef(b,2)<0 ldist(b,a)=(linef(a,1)+linef(b,2)-gd).^2; else ldist(b,a)=(abs(linef(a,1)-linef(b,2))-(cd-gd)).^2; end end end y_correct=ldist; f_result=measure_dist(linef,cd,gd); assert(all(abs(f_result(:)-y_correct(:))<0.1)) | 266 | 723 | {"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} | 3.109375 | 3 | CC-MAIN-2020-16 | latest | en | 0.444749 |
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From "Phil Steitz (JIRA)" <j...@apache.org>
Subject [jira] [Comment Edited] (MATH-1246) Kolmogorov-Smirnov 2-sample test does not correctly handle ties
Date Tue, 07 Jul 2015 03:59:04 GMT
```
[ https://issues.apache.org/jira/browse/MATH-1246?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel&focusedCommentId=14616134#comment-14616134
]
Phil Steitz edited comment on MATH-1246 at 7/7/15 3:59 AM:
-----------------------------------------------------------
I think the current implementation can be fixed as follows. If we move to a faster implementation,
the strategy below may not work.
What exactP does now is to exhaustively compute all possible D-statistics for all m-set /
n-set partitions of m+n and simply tally the number that exceed (strict) or are as large as
(not strict) the observed D. If there are ties in the data, it is not correct to look at
partitions of m+n, since not all partitions of an m+n set with duplicates are distinct and
the set of possible D values is different in the presence of ties. I think we can correctly
handle ties in the data if we compute and tally D statistics based on a combined multi-set
sample with duplicates in the positions corresponding to what is observed in the data. For
example, suppose that the two samples are x = [0, 3, 6, 9, 9, 10] and y = [1, 3, 4, 8, 11].
then the multi-set universe is U = [0, 1, 3, 3, 4, 6, 8, 9, 9, 10, 11]. As before, we generate
partitions of 11 into a 6-set and a 5-set, but instead of computing the D-statistics on the
subsets of 11, we use indexes into U instead. So if a generated split is mSet = [0, 2, 3,
7, 8, 9], nSet = [1, 4, 5, 6, 10], we compute D for [0, 3, 3, 9, 9, 10] and [1, 4, 6, 8, 11].
The rationale here is that the p-value is the probability that if U is split randomly into
a 5-set and a 6-set, the D-value exceeds the observed d.
was (Author: psteitz):
I think the current implementation can be fixed as follows. If we move to a faster implementation,
the strategy below may not work.
What exactP does now is to exhaustively compute all possible D-statistics for all m-set /
n-set partitions of m+n and simply tally the number that exceed (strict) or are as large as
(not strict) the observed D. If there are ties in the data, it is not correct to look at
partitions of m+n, since not all partitions of an m+n set with duplicates are distinct and
the set of possible D values is different in the presence of ties. I think we can correctly
handle ties in the data if we compute and tally D statistics based on a combined multi-set
sample with duplicates in the positions corresponding to what is observed in the data. For
example, suppose that the two samples are x = [0, 3, 6, 9, 9, 10] and y = [1, 3, 4, 8, 11].
then the multi-set universe is U = {0, 1, 3, 3, 4, 6, 8, 9, 9, 10, 11}. As before, we generate
partitions of 11 into a 6-set and a 5-set, but instead of computing the D-statistics on the
subsets of 11, we use indexes into U instead. So if a generated split is mSet = {0, 2, 3,
7, 8, 9}, nSet = {1, 4, 5, 6, 10}, we compute D for [0, 3, 3, 9, 9, 10] and [1, 4, 6, 8, 11].
The rationale here is that the p-value is the probability that if U is split randomly into
a 5-set and a 6-set, the D-value exceeds the observed d.
> Kolmogorov-Smirnov 2-sample test does not correctly handle ties
> ---------------------------------------------------------------
>
> Key: MATH-1246
> URL: https://issues.apache.org/jira/browse/MATH-1246
> Project: Commons Math
> Issue Type: Bug
> Reporter: Phil Steitz
>
> For small samples, KolmogorovSmirnovTest(double[], double[]) computes the distribution
of a D-statistic for m-n sets with no ties. No warning or special handling is delivered in
the presence of ties.
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View raw message | 1,176 | 3,980 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-22 | latest | en | 0.869315 |
http://www.uwplatt.edu/distance-education/management-decision-analysis | 1,441,315,663,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440645328641.76/warc/CC-MAIN-20150827031528-00294-ip-10-171-96-226.ec2.internal.warc.gz | 780,639,957 | 11,755 | # Management Decision Analysis
## Clear Float
This pane clears float!
Course Number: BUSADMIN 7110 Course Name: Management Decision Analysis Course Description: A presentation of theory and applications of quantitative decision methods used in the business setting. Topics include decision theory, linear programming, PERT/CPM, forecasting and inventory control. Prerequisites: A statistics course or consent of the instructor. Level: Graduate Credits: 3 Format: Online Program: MS in Project ManagementMS in Integrated Supply Chain Management
NOTE: The information below is representative of the course and is subject to change. The specific details of the course will be available in the Desire2Learn course instance for the course in which a student registers.
Learning Outcomes
This course is designed to
• Introduce you to the common quantitative decision-making tools used in business.
• Provide you with an opportunity to develop the skills necessary to use the quantitative models of management decision analysis.
• Expose you to current literature and research in the field of management decision analysis.
SPECIFIC COURSE OBJECTIVES: Upon completion of this course, you will be able to
• Identify and analyze the problem.
• Choose the appropriate model to analyze the problem.
• Apply the model to the problem.
• Interpret the results in a meaningful way, so as to aid in the decision making process.
• Write an article review using proper APA (or MLA) formatting.
Unit Descriptions
Unit 1: Introduction to Decision Models
After completing the readings and activities in this unit, you should be able to
• Construct a model that can be used in a business decision.
• Know and use the basic principles of probability theory.
• Examine how decisions are made under conditions of certainty, uncertainty, and risk.
Unit 2: Linear Programming and Its Applications
After completing the readings and activities in this unit, you should be able to
• Formulate, solve, and interpret linear programming problems to achieve an optimal solution to a business problem.
• Evaluate typical problems in your own business setting using linear programming techniques.
• Formulate and solve linear programming problems using spreadsheets (i.e. - MS Excel).
• Apply linear programming techniques to specialized applications.
Unit 3: Decision Applications
After completing the readings and activities in this unit, you should be able to
• Demonstrate how cost reduction and efficiencies can be gained using network flow diagrams.
• Determine critical activities and costs in a project using CPM /PERT techniques.
• Decide how many waiting lines are appropriate while considering the cost of each additional waiting line using queuing theory.
• Select the appropriate forecasting technique for various business situations.
• Select the appropriate inventory control models for dependent and independent demand.
Activities:
Class icebreaker: 10 points
Homework Problems (2 points each): 100 points
Article Reviews (3 @ 15 points each): 45 points
Collaborative Activities (3 @ 15 points each): 45 points
Exams (3 @ 100 points each): 300 points
Term Project: 100 points
TOTAL POINTS: 600
A: 90% or more of total points
B: 80% - 89% of total points
C: 70% - 79% of total points
D: 60% - 69% of total points
F: Less than 60% of total points
## Clear Float
This pane clears float! | 707 | 3,403 | {"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-2015-35 | longest | en | 0.876179 |
https://gitlab.mpi-sws.org/iris/iris/-/commit/a6921bfd2a47d795752e9c0f3fe01df2645617da | 1,653,690,682,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663006341.98/warc/CC-MAIN-20220527205437-20220527235437-00368.warc.gz | 331,293,969 | 26,766 | Commit a6921bfd by Jacques-Henri Jourdan
Lemmas for updates that take a step.
parent ed3b7e73
Pipeline #2919 passed with stage
in 9 minutes and 57 seconds
... ... @@ -34,19 +34,6 @@ Notation "P ={ E }=★ Q" := (P -★ |={E}=> Q)%I Notation "P ={ E }=★ Q" := (P ⊢ |={E}=> Q) (at level 99, E at level 50, Q at level 200, only parsing) : C_scope. Notation "|={ E1 , E2 }▷=> Q" := (|={E1,E2}=> (▷ |={E2,E1}=> Q))%I (at level 99, E1, E2 at level 50, Q at level 200, format "|={ E1 , E2 }▷=> Q") : uPred_scope. Notation "P ={ E1 , E2 }▷=★ Q" := (P -★ |={ E1 , E2 }▷=> Q)%I (at level 99, E1, E2 at level 50, Q at level 200, format "P ={ E1 , E2 }▷=★ Q") : uPred_scope. Notation "|={ E }▷=> Q" := (|={E,E}▷=> Q)%I (at level 99, E at level 50, Q at level 200, format "|={ E }▷=> Q") : uPred_scope. Notation "P ={ E }▷=★ Q" := (P ={E,E}▷=★ Q)%I (at level 99, E at level 50, Q at level 200, format "P ={ E }▷=★ Q") : uPred_scope. Section fupd. Context `{invG Σ}. Implicit Types P Q : iProp Σ. ... ... @@ -200,3 +187,41 @@ End proofmode_classes. Hint Extern 2 (coq_tactics.of_envs _ ⊢ _) => match goal with |- _ ⊢ |={_}=> _ => iModIntro end. (** Fancy updates that take a step. *) Notation "|={ E1 , E2 }▷=> Q" := (|={E1,E2}=> (▷ |={E2,E1}=> Q))%I (at level 99, E1, E2 at level 50, Q at level 200, format "|={ E1 , E2 }▷=> Q") : uPred_scope. Notation "P ={ E1 , E2 }▷=★ Q" := (P -★ |={ E1 , E2 }▷=> Q)%I (at level 99, E1, E2 at level 50, Q at level 200, format "P ={ E1 , E2 }▷=★ Q") : uPred_scope. Notation "|={ E }▷=> Q" := (|={E,E}▷=> Q)%I (at level 99, E at level 50, Q at level 200, format "|={ E }▷=> Q") : uPred_scope. Notation "P ={ E }▷=★ Q" := (P ={E,E}▷=★ Q)%I (at level 99, E at level 50, Q at level 200, format "P ={ E }▷=★ Q") : uPred_scope. Section step_fupd. Context `{invG Σ}. Lemma step_fupd_mask_frame_r E1 E2 Ef P : E1 ⊥ Ef → E2 ⊥ Ef → (|={E1,E2}▷=> P) ⊢ |={E1 ∪ Ef,E2 ∪ Ef}▷=> P. Proof. iIntros (??) "HP". iApply fupd_mask_frame_r. done. iMod "HP". iModIntro. iNext. by iApply fupd_mask_frame_r. Qed. Lemma step_fupd_mask_mono E1 E2 F1 F2 P : F1 ⊆ F2 → E1 ⊆ E2 → (|={E1,F2}▷=> P) ⊢ |={E2,F1}▷=> P. Proof. iIntros (??) "HP". iAssert (|={E2,E1}▷=> True)%I as "HE". { iApply fupd_mono. apply later_intro. by iApply fupd_intro_mask. } iAssert (|={F2,F1}▷=> True)%I as "HF". { iApply fupd_mono. apply later_intro. by iApply fupd_intro_mask. } iMod "HE". iMod "HP". iMod "HF". iModIntro. iNext. iMod "HF". iMod "HP". by iMod "HE". Qed. End step_fupd.
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Finish editing this message first! | 1,111 | 2,589 | {"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-2022-21 | latest | en | 0.52621 |
http://www.physicsforums.com/showthread.php?t=135 | 1,411,055,269,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657128304.55/warc/CC-MAIN-20140914011208-00144-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 740,157,454 | 7,285 | # 4 laws of thermodynamics
by rjsniz
Tags: laws, thermodynamics
P: n/a Help!! Wouls someone be so kind as to explain the 4 laws of thermodynamics in 'laymans terms', so that I could better understand them. A short but complete explanation would be appreciated! Thanks.
P: 683 There is a basic list here: http://www.health.uottawa.ca/biomech...ws/thermos.htm The 3 after zeroth are easy to remember. You can't win This is the law of energy conservation, or law number one. It states that in any closed system, you will always wind up with the same amount of energy you started with. For example, take X amount of matter, which has a certain rest energy amount. If you were to attempt to completely destroy the matter, it would merely be converted to another form of energy. (ie. heat, EM radiation) The total amount of energy from this heat would still be equal to the original amount the matter had. Basically, energy can neither be created nor destroyed. Hence the title, you can't win. You can't break even This is the second law dealing with entropy. The disorder of a system can never decrease. Take the example above of matter, for example. If you convert the matter into radiation, the total amount of energy in the system stays the same, but the potential energy has decreased. The radiation that carries the energy from the now absent matter cannot be used to do work. One example of entropy is the universe as a whole. As time goes by, most of the energy in the universe will be in the form of radiation, and cannot be used for work. Hence, the universe dies a "heat death" and humans are screwed. You can't leave the game The third law states that the temperature of absolute zero for a system is impossible. There was an article about this on Physics Post, at http://physicspost.com/articles.php?articleId=2 The explanations above are not the most detailed, but maybe someone here can provide more insight.
Emeritus Sci Advisor PF Gold P: 1,817 Eh's explanations are good. One comment: The second law states that for a closed system entropy can never decrease. It is certainly possible to decrease entropy by shuffling it to a different place. For example, a freezer freezing water. Ice has less disorder than water. To freeze it, you need to take the 'disorder' from the water and push it into a different place: The air outside the refrigerator. The zero'th law is really simple. If particle A is the same temperature as particle B, and particle B is the same temperature as particle C, then particle A is the same temperature as particle C. The point is that there is a rule which governs thermometers. It's a 'duh' type statement, but the three laws have nothing in them which say that a thermometer which reads 20 degrees for one object will actually read 20 degrees for another object.
Related Discussions Academic Guidance 10 Classical Physics 3 General Physics 1 Chemistry 2 Classical Physics 2 | 644 | 2,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2014-41 | latest | en | 0.95293 |
https://www.physicsforums.com/threads/finding-a-function-that-goes-through-given-points.837204/ | 1,508,540,094,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824357.3/warc/CC-MAIN-20171020211313-20171020231313-00172.warc.gz | 1,004,136,432 | 16,861 | # Finding a function that goes through given points
1. Oct 11, 2015
### iScience
1. The problem statement, all variables and given/known data
Find a function of a decaying logarithmic trend that passes through a set of 3 points (at x=1,2,3) whose sum is S, where S>0.
2. Relevant equations
3. The attempt at a solution
starting point: i assumed the equation had the form: $$y= -kln(x+1)+S$$ with the condition that must satisfy:
$$S=Σ(y(x))$$ from 1→3
but I don't know what to do next
2. Oct 11, 2015
### haruspex
In your proposed generic form, I see no reason to choose x+1 rather than x+an unknown constant. And I don't understand the final S; did you mean this to be the same as the given S or something different?
If we make it x+c as well as a final +d (say) then we have three unknowns (c, d, k) but effectively only one equation. So presumably the question expects a simpler generic form, but it is not clear what. If I had to guess, I would ditch d first.
Edit:After a little background reading, seems like you should keep d and ditch c. Logarithmic decay appears to be defined as the functional inverse of exponential decay, i.e. x=Ae-ky.
3. Oct 11, 2015
### iScience
Oh, no actually not necessary, I was just trying to find a convenient way to set up the graph on my calculator; my proposed form would have essentially looked like the same graph as that of x=Ae^(-ky)
Same S.
I tried your suggested formula. Since the A was left as a free variable to be determined, i had two equations two unknowns and was able to solve. much thanks! | 413 | 1,560 | {"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} | 3.203125 | 3 | CC-MAIN-2017-43 | longest | en | 0.963348 |
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A123871 Expansion of g.f.: (1+x+x^2)/(1-4*x-4*x^2). 2
1, 5, 25, 120, 580, 2800, 13520, 65280, 315200, 1521920, 7348480, 35481600, 171320320, 827207680, 3994112000, 19285278720, 93117562880, 449611366400, 2170915717120, 10482108334080, 50612096204800, 244376818155520, 1179955657441280, 5697329902387200 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..1000 A. Burstein and T. Mansour, Words restricted by 3-letter ..., Annals of Combinatorics 7 (2003), 1-14. arXiv:math.CO/0112281 Martin Burtscher, Igor Szczyrba, Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5. Index entries for linear recurrences with constant coefficients, signature (4,4). FORMULA a(n) = 4*a(n-1) + 4*a(n-2) for n>2. - Philippe Deléham, Sep 19 2009 MAPLE seq(coeff(series((1+x+x^2)/(1-4*x-4*x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Aug 08 2019 MATHEMATICA CoefficientList[Series[(1+x+x^2)/(1-4*x-4*x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Jun 27 2012 *) PROG (MAGMA) I:=[1, 5, 25]; [n le 3 select I[n] else 4*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jun 27 2012 (PARI) my(x='x+O('x^30)); Vec((1+x+x^2)/(1-4*x-4*x^2)) \\ G. C. Greubel, Aug 08 2019 (Sage) def A123871_list(prec): P. = PowerSeriesRing(ZZ, prec) return P( (1+x+x^2)/(1-4*x-4*x^2) ).list() A123871_list(30) # G. C. Greubel, Aug 08 2019 (GAP) a:=[1, 5, 25];; for n in [4..30] do a[n]:=4*a[n-1]+4*a[n-2]; od; a; # G. C. Greubel, Aug 08 2019 CROSSREFS Column 5 in A265584. Sequence in context: A089927 A269463 A068539 * A268453 A218989 A078717 Adjacent sequences: A123868 A123869 A123870 * A123872 A123873 A123874 KEYWORD nonn,easy AUTHOR N. J. A. Sloane, Nov 20 2006 STATUS approved
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Last modified August 8 19:29 EDT 2020. Contains 336298 sequences. (Running on oeis4.) | 915 | 2,357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2020-34 | latest | en | 0.441686 |
http://forums.cgsociety.org/archive/index.php?t-205905.html | 1,527,055,026,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865450.42/warc/CC-MAIN-20180523043959-20180523063959-00329.warc.gz | 114,331,726 | 6,503 | View Full Version : "If-test requires boolean class, got: 0"
Yvo01 January 2005, 11:40 AMHi everybody, I'm a beginner considering Maxscript (and new on this forum), so the following problem shoul be too easy: I'm making a special shape existing of ellipses: When i run this one in MaxScript Listener it gives the following error for the highlighted line: "If-test requires Boolean Class, got: 0" How can i resolve this problem? ------------------------------------------------ A = 6750.0 B = 9000.0 F = 3750.0 - 375.0 G = 6000.0 - 375.0 tga = 37.5 / 75. C = (A - F) * tga + 375 S = 100 ------------------------------------------------ dak = Sphere() dak.radius = C dak.segs= S converttomesh(dak) For n=1 to (getnumverts(dak)) do If (getvert dak n).x = 0 then H = A else H = 1/sqrt(1/A^2+ (getvert dak n).y^2/((getvert dak n).x*B)^2) If H >= A*sqrt(1-G^2/B^2) then J=tga*(H-F) else (If H <= F then J=tga*(B*sqrt(1-H^2/A^2)-G) else J = tga*sqrt((H-F)^2+(B*sqrt(1-H^2/A^2)-G)^2)) k = H/sqrt(1-J^2/C^2) l = k*(getvert dak n).y/(getvert dak n).x u = 1/sqrt(1/k^2 + (getvert dak n).z^2/((getvert dak n).x*C)^2) v = l*u/k w = C*sqrt(1-u^2/k^2) setvert dak n [u,v,w]
Yvo
01 January 2005, 09:07 AM | 439 | 1,190 | {"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} | 3.15625 | 3 | CC-MAIN-2018-22 | latest | en | 0.529813 |
http://www.ck12.org/arithmetic/Mixed-Numbers-as-Improper-Fractions/lesson/Mixed-Numbers-as-Improper-Fractions/ | 1,444,173,952,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736679281.15/warc/CC-MAIN-20151001215759-00246-ip-10-137-6-227.ec2.internal.warc.gz | 495,441,920 | 38,500 | <meta http-equiv="refresh" content="1; url=/nojavascript/">
# Mixed Numbers as Improper Fractions
## 2 and 3/4 = [(4 x 2)+3]/4=11/4
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Practice Mixed Numbers as Improper Fractions
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Mixed Numbers as Improper Fractions
### Let’s Think About It
Credit: Eli Christman
Source: https://www.flickr.com/photos/gammaman/6866396035/
License: CC BY-NC 3.0
Casey ordered eight pizzas for the drama club to enjoy. Each pizza had ten slices. At the end of the pizza party, there were two whole pizzas and two slices left. How many slices weren’t eaten? How can you express this as an improper fraction of pizza slices?
In this concept, you will learn how to rewrite mixed numbers as an improper fraction.
### Guidance
A mixed number is a number that has both wholes and parts in it. Here is a mixed number.
514
There are five whole items and one-fourth of a whole. The opposite of a mixed number is an improper fraction.
An improper fraction is a fraction that has a larger numerator than the denominator. Here is an improper fraction.
125
The denominator tells you how many parts the whole has been divided into. This whole has been divided into 5 parts. The numerator tells you the number of parts. In this case, there are twelve parts. There are more parts than there are in 1 whole.
To write a mixed number as an improper fraction, write the mixed number as a fraction in terms of parts instead of in terms of wholes and parts. Remember that a whole number can also be written as a fraction. The numerator is equal to the denominator.
1=55
Change 213\begin{align*}2\frac{1}{3}\end{align*} to an improper fraction.
First, multiply the whole number by the denominator to convert the whole to a fraction and add the numerator. This will give you the new numerator.
2×3+1=7
Then, put the sum over the denominator. The denominator is 3.
213=73
The mixed number 213\begin{align*}2 \frac{1}{3}\end{align*} is also written as 73\begin{align*}\frac{7}{3}\end{align*}.
Change the following mixed numbers to improper fractions.
### Guided Practice
Express 478\begin{align*}4 \frac{7}{8}\end{align*} as an improper fraction.
First, multiply the whole number by the denominator and add the numerator.
4×8+7=39
Then, put the sum over the denominator. The denominator is 8.
478=398
The mixed number 478\begin{align*}4 \frac{7}{8}\end{align*} is expressed as 398\begin{align*}\frac{39}{8}\end{align*}.
### Examples
#### Example 1
313\begin{align*}3\frac{1}{3}\end{align*}
First, multiply the whole number by the denominator and add the numerator.
3×3+1=10\begin{align*}3 \times 3 +1 = 10\end{align*}
Then, put the sum over the denominator. The denominator is 3.
313=103\begin{align*}3 \frac{1}{3} = \frac{10}{3}\end{align*}
The mixed number 313\begin{align*}3 \frac{1}{3}\end{align*} is expressed as 103\begin{align*}\frac{10}{3}\end{align*}.
#### Example 2
523\begin{align*}5\frac{2}{3}\end{align*}
First, multiply the whole number by the denominator and add the numerator.
3×5+2=17\begin{align*}3 \times 5 +2 = 17\end{align*}
Then, put the sum over the denominator. The denominator is 3.
313=103\begin{align*}3 \frac{1}{3} = \frac{10}{3}\end{align*}
The mixed number 523\begin{align*}5 \frac{2}{3}\end{align*} is expressed as 173\begin{align*}\frac{17}{3}\end{align*}.
#### Example 3
618\begin{align*}6 \frac{1}{8}\end{align*}
First, multiply the whole number by the denominator and add the numerator.
6×8+1=49\begin{align*}6 \times 8 +1 = 49\end{align*}
Then, put the sum over the denominator. The denominator is 3.
618=498\begin{align*}6 \frac{1}{8} = \frac{49}{8}\end{align*}
The mixed number 618\begin{align*}6 \frac{1}{8}\end{align*} is expressed as 498\begin{align*}\frac{49}{8}\end{align*}.
### Follow Up
Credit: rootseven
Source: https://www.flickr.com/photos/21742592@N03/7179163594/
License: CC BY-NC 3.0
Remember Casey and the pizzas? Convert 2 pizzas and 2 slices to an improper fraction to find the number of uneaten slices of pizza.
Two whole pizzas and two slices =2210\begin{align*}= 2 \frac{2}{10}\end{align*}
First, multiply the whole number by the denominator and add the numerator.
2×10+2=22\begin{align*}2 \times 10 +2 = 22\end{align*}
Then, put the sum over the denominator. The denominator is 10.
2210=2210\begin{align*}2 \frac{2}{10} = \frac{22}{10}\end{align*}
There were a total of 22 slices of pizza left uneaten.
### Video Review
http://www.youtube.com/watch?v=1xuf6ZKF1_I
http://www.youtube.com/watch?v=p_YRBcZ4u4g
### Explore More
Write each mixed number as an improper fraction.
1. 212\begin{align*}2\frac{1}{2}\end{align*}
2. 314\begin{align*}3\frac{1}{4}\end{align*}
3. 513\begin{align*}5\frac{1}{3}\end{align*}
4. 423\begin{align*}4\frac{2}{3}\end{align*}
5. 614\begin{align*}6\frac{1}{4}\end{align*}
6. 625\begin{align*}6\frac{2}{5}\end{align*}
7. 713\begin{align*}7\frac{1}{3}\end{align*}
8. 825\begin{align*}8\frac{2}{5}\end{align*}
9. 745\begin{align*}7\frac{4}{5}\end{align*}
10. 827\begin{align*}8\frac{2}{7}\end{align*}
11. 834\begin{align*}8\frac{3}{4}\end{align*}
12. 956\begin{align*}9\frac{5}{6}\end{align*}
13. 658\begin{align*}6\frac{5}{8}\end{align*}
14. 923\begin{align*}9\frac{2}{3}\end{align*}
15. 512\begin{align*}5\frac{1}{2}\end{align*}
16. 1614\begin{align*}16\frac{1}{4}\end{align*}
### Vocabulary Language: English
Equivalent
Equivalent
Equivalent means equal in value or meaning.
improper fraction
improper fraction
An improper fraction is a fraction in which the absolute value of the numerator is greater than the absolute value of the denominator.
Mixed Number
Mixed Number
A mixed number is a number made up of a whole number and a fraction, such as $4\frac{3}{5}$.
### Image Attributions
1. [1]^ Credit: Eli Christman; Source: https://www.flickr.com/photos/gammaman/6866396035/; License: CC BY-NC 3.0
2. [2]^ Credit: rootseven; Source: https://www.flickr.com/photos/21742592@N03/7179163594/; License: CC BY-NC 3.0
Please wait...
Please wait... | 1,929 | 5,980 | {"found_math": true, "script_math_tex": 40, "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": 1, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2015-40 | longest | en | 0.90003 |
https://www.excelforum.com/excel-formulas-and-functions/953408-copy-number-of-distinct-values-and-their-details-from-one-sheet-to-another.html | 1,656,408,765,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103360935.27/warc/CC-MAIN-20220628081102-20220628111102-00489.warc.gz | 816,470,726 | 17,055 | # Copy number of distinct values and their details from one sheet to another.
1. ## Copy number of distinct values and their details from one sheet to another.
Hi All,
Data:
There are two sheets : Sheet1 is Input and Sheet 2 is Output.
Sheet 1 has following columns.
Item Shop Buyer Quantity in Jan Quantity in Feb
In my sheet if I filter by a Shop for example "Shop 1" . I get the following data
Shop Vehicle Buyer Quantity in Jan Quantity in Feb
Shop1 Car Jim 1 5 5
Shop1 Car Nick 5 6 9
Shop1 Scooter Jim 7 8 0
Shop1 Cycle Mary 9 0 0
Now I need the following :
1.Number of distinct vehicles purchased from "Shop 1" and place it in the Sheet "Distinct".
2.Sum of "Quantity in Jan" in sheet "Input" for all vehicles purchased from Shop1 and place in sheet distinct.
Following is the example values in Sheet "Distinct"
Shop Number of Distinct Vehicles Total Quantity in Jan
Shop1 3 22
Shop2
Shop3
Also Attached is the sheet.
Thanks.
2. ## Re: Copy number of distinct values and their details from one sheet to another.
I re-aranged the data with an macro.
After that I used pivot table to get the result.
See the attached file.
3. ## Re: Copy number of distinct values and their details from one sheet to another.
your approach is very good, but here it is slightly deviating from my requirement. Can I please have data in Sheet "Distinct"> In the format specified below . Also Can I have a solution for this without the use of macros.? (If we have to use macro it will be fine but I need data in the format specified.)
Shop Number of Distinct Vehicles Total Quantity in Jan
Shop1 3 22
Shop2
Shop3
4. ## Re: Copy number of distinct values and their details from one sheet to another.
Maybe other members have a smart formula solution.
I would go for the given option.
5. ## Re: Copy number of distinct values and their details from one sheet to another.
Taj, try these.
Define two more names Shop & Vehicle with appropriate ranges. Then in;
Distinct A2:
=IFERROR(INDEX(Shop,MATCH(TRUE,INDEX(ISNA(MATCH(Shop,A\$1:A1,0)),),0)),"")
B2:
=IF(A2="","",SUMPRODUCT((Shop=A2)*(Vehicle<>"")*ISNUMBER(1/(MATCH(Vehicle&"_"&A2,Vehicle&"_"&Shop,0)=ROW(Shop)-MIN(ROW(Vehicle))+1))))
C2:
Copy A2:C2 then copy down.
IFERROR only works with Excel 2007 or above. A per you profile assuming you are using Excel 2010.
There are currently 1 users browsing this thread. (0 members and 1 guests) | 621 | 2,405 | {"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-2022-27 | latest | en | 0.893274 |
http://demonstrations.wolfram.com/PricingPutOptionsWithTheExplicitFiniteDifferenceMethod/ | 1,550,592,158,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247490225.49/warc/CC-MAIN-20190219142524-20190219164524-00623.warc.gz | 66,872,254 | 6,819 | # Pricing Put Options with the Explicit Finite-Difference Method
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This Demonstration shows a put option's value function versus the time to expiry and the asset's log price. The 3D plot shows the American (red dots) and the European (blue dots) put values at each grid point. The European put value is never greater than the American, because the American put can be exercised at any time up to maturity, while the European put can only be exercised at maturity. Use the controls to define the option parameters and the time discretization.
Contributed by: Michail Bozoudis (July 2014)
Suggested by: Michail Boutsikas
Open content licensed under CC BY-NC-SA
## Details
The explicit finite-difference method [1] is applied to solve the Black–Scholes–Merton partial differential equation using a uniform log price and time grid. The asset log price discretization is calculated automatically in order to ensure stability and convergence. Compared to the implicit finite-difference method, the explicit finite-difference method is considered simpler and faster but less stable and accurate. It is relatively accurate for a small number of steps. The explicit finite-difference method is similar to the trinomial method, in that both provide an explicit formula for determining future states of the option process in terms of the current state, whereas the implicit finite-difference method requires the iterative solution of linear equations linking consecutive time steps [2].
References
[1] M. Brennan and E. Schwartz, "Finite Difference Methods and Jump Processes Arising in the Pricing of Contingent Claims: A Synthesis," The Journal of Financial and Quantitative Analysis, 13(3), 1978 pp. 461–474. www.jstor.org/stable/2330152.
[2] P. Boyle, "Option Valuation Using a Three-Jump Process," International Options Journal, 3, 1986 pp. 7–12.
## Permanent Citation
Michail Bozoudis
Feedback (field required) Email (field required) Name Occupation Organization Note: Your message & contact information may be shared with the author of any specific Demonstration for which you give feedback. Send | 505 | 2,351 | {"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-2019-09 | latest | en | 0.838096 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-10-exponents-and-radicals-10-2-rational-numbers-as-exponents-10-2-exercise-set-page-641/34 | 1,537,696,926,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267159193.49/warc/CC-MAIN-20180923095108-20180923115508-00093.warc.gz | 743,097,474 | 15,321 | Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$27y^{9}$
Using $a^{m/n}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m$, the given expression, $(9y^6)^{3/2} ,$ is equivalent to \begin{array}{l}\require{cancel} \sqrt[]{(9y^6)^{3}} \\\\= \left( \sqrt[]{9y^6} \right)^3 \\\\= \left( \sqrt[]{(3y^3)^2} \right)^3 \\\\= \left( 3y^3 \right)^3 \\\\= 3^3y^{3(3)} \\\\= 27y^{9} .\end{array} | 171 | 392 | {"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} | 4.125 | 4 | CC-MAIN-2018-39 | longest | en | 0.468451 |
http://forums.wolfram.com/mathgroup/archive/2005/Nov/msg00828.html | 1,726,591,779,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651800.83/warc/CC-MAIN-20240917140525-20240917170525-00774.warc.gz | 10,430,857 | 7,793 | Re: 3D plot question
• To: mathgroup at smc.vnet.net
• Subject: [mg62631] Re: 3D plot question
• From: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig.de>
• Date: Wed, 30 Nov 2005 00:06:08 -0500 (EST)
• Organization: Uni Leipzig
• References: <dmh8kd\$8nm\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
```Hi,
a surface has the tickness zero -- otherwise it is
no surface.
Regards
Jens
"Tun Myint Aung" <tma at nus.edu.sg> schrieb im
Newsbeitrag news:dmh8kd\$8nm\$1 at smc.vnet.net...
|
| Dear MathGroup,
|
| I have found the solution. Use
PlotRange->All. Another question is
| that how I can plot to see the thickness of the
surface?
|
| With Regards,
|
| Tun Myint Aung
|
| _____
|
| From: Tun Myint Aung
To: mathgroup at smc.vnet.net
| Subject: [mg62631] 3D plot question
|
|
| Dear Mathgroup,
|
| When I plot a 3D plot in mathematica, I found
that some part of the
| surface has been cut off in the graph. For
example,
|
| z = x^5*y
|
| Plot3D[z, {x, -1, 1}, {y, -1, 1},
PlotPoints -> 40,
|
| Mesh -> False, BoxRatios -> {1, 1, 0.4}]
|
| Mathematica plot for z values only in the
range -0.2 to 0.2. Actually
| when x=1, y=1, z value will be 1. But in the
graph, the surface goes
| flat for z>0.2.Why? How can I get the graph for
full range of z?
|
|
|
| Best Regards,
|
| Tun Myint Aung
| National University of Singapore
| E1A #02-18
| Kent Ridge, 119260, Singapore
| E-mail g0202015 at nus.edu.sg
<mailto:g0202015 at nus.edu.sg>
|
|
```
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• Previous by thread: Re: 3D plot question
• Next by thread: Re: Re: FW: 3D plot question | 596 | 1,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-38 | latest | en | 0.804522 |
https://physics.aps.org/articles/v12/100 | 1,571,036,228,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986649232.14/warc/CC-MAIN-20191014052140-20191014075140-00355.warc.gz | 749,601,731 | 9,587 | # Focus: Teaching Energy Before Forces
Physics 12, 100
A calculus-based introductory physics course that postpones the challenges of vectors by starting with energy concepts improves students’ learning.
Forces, and Newton’s three laws that describe how they work, are usually regarded as central components of classical mechanics. But in a new proposal, a team of physicists suggests that students learning mechanics be introduced first to concepts involving energy and that forces be taught by deriving them from the energy equations. The researchers now present evidence that this “energy-first,” calculus-based introductory curriculum improves performance of students with weaker math skills and boosts grades in later coursework in physics and engineering.
Previous work has shown that one of the challenges in learning about mechanics is the need to master force vectors, with which many students struggle [1]. Christopher Fischer of the University of Kansas, Lawrence, and his co-workers reasoned that starting with a nonvector (scalar) quantity like energy might be easier than the traditional approach of learning forces first.
For example, the conservation of energy means that, under the right conditions, the sum of kinetic and potential energy remains a fixed quantity. A few basic applications of calculus (differentiation) lead from this energy equation to the equations of motion for the system, as encapsulated by Newton’s laws. This derivation of mechanics from energy concepts is not new; it’s equivalent to an alternative to the Newtonian formulation that was developed in the 19th century by Irish mathematician William Hamilton, known as Hamiltonian mechanics.
Fischer and colleagues have devised an introductory curriculum that takes this approach to explain many of the important topics in physics, from simple kinematics to oscillatory and rotational motion. In each case, differential calculus is involved in deriving equations of motion from energy equations. “We are still teaching them forces,” Fischer explains, “but only after students have time to understand classical mechanics in the framework of ‘easier’ mathematics,” meaning scalars rather than vectors. The curriculum also gives a calculus-based derivation of momentum and thermodynamics. Fischer has published two introductory textbooks that take the energy-first approach [2, 3].
To test the value of this approach, the researchers took advantage of the fact that the University of Kansas offers two introductory physics courses, PHSX210 and PHSX211. The courses have essentially the same content but slightly different structure and tutors, and the math ability of students in each course is about the same.
In 2015, the energy-first curriculum was introduced into PHSX211, but not PHSX210, and between 2016 and 2018, all students were given a standardized conceptual physics exam at the beginning and end of each semester. For students strongest in math, the participants in PHSX211 did only slightly better. But for students with the lowest math scores, the PHSX211 students performed significantly better than those who took PHSX210. What is more, those PHSX211 students who took subsequent physics and engineering courses got better grades on average in those later classes.
The researchers caution that more work will be needed on larger samples of students to be fully confident that the curriculum content, and not some other factor, was behind the differences in performance. In particular, the class size for PHSX210 was more than twice as large as that of PHSX211, and PHSX211 had an extra 50 minutes of class time per week; each of these differences could potentially influence student learning.
Assuming the new curriculum is responsible for the exam score differences, the team has not yet identified the reasons for the improvements, although they have some ideas. “Our current working hypothesis is that tasking the students with using calculus repeatedly throughout PHSX211 has helped the students to improve their understanding of and skills with both physics and calculus,” says Fischer.
There are other introductory physics curricula that place emphasis on teaching energy [4, 5], says Benjamin Dreyfus of George Mason University in Fairfax, Virginia, a specialist in physics education, “but this is the first one I have seen that is calculus focused,” using energy concepts to derive force laws. He sees this paper as “a proof of concept that it was successful in this one particular context” and agrees that further research is needed to determine how broadly the conclusions apply.
Physics educator David Meltzer of Arizona State University in Tempe is not convinced that the current data can establish that the improvements are due to the curriculum changes. He thinks that the differences between the two courses, especially in classroom time, make the comparison ambiguous and that the sample sizes are too small for strong conclusions to be drawn. However, he thinks there is a “strong case” for further analysis of whether energy-first truly makes a difference.
This research is published in Physical Review Physics Education Research.
–Philip Ball
Philip Ball is a freelance science writer in London. His latest book is How To Grow A Human (University of Chicago Press, 2019).
## References
1. N.-L. Nguyen and D. E. Meltzer, “Initial understanding of vector concepts among students in introductory physics courses,” Am. J. Phys. 71, 630 (2003).
2. C. J. Fischer, The Energy of Physics, Part I: Classical Mechanics and Thermodynamics, 2nd Edition (2019)[Amazon][WorldCat].
3. C. J. Fischer, The Energy of Physics, Part II: Electricity and Magnetism, 2nd Edition (2020)[Amazon][WorldCat].
4. E. Brewe, “Energy as a substancelike quantity that flows: Theoretical considerations and pedagogical consequences,” Phys. Rev. ST Phys. Educ. Res. 7, 020106 (2011).
5. J. Solbes, J. Guisasola, and F. Tarín, “Teaching energy conservation as a unifying principle in physics,” J. Sci. Ed. Technol. 18, 265 (2009).
## Subject Areas
Interdisciplinary Physics
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Calculations show that the number of white filaments springing from the top of a dandelion seed is optimized for steady flight, allowing the seeds to cruise long distances through the air. Read More » | 1,378 | 6,771 | {"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-2019-43 | longest | en | 0.9576 |
https://dsp.stackexchange.com/questions/63937/can-a-kalman-filter-be-applied-using-measurement-space-dependent-sensors | 1,701,342,693,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100184.3/warc/CC-MAIN-20231130094531-20231130124531-00360.warc.gz | 250,784,260 | 42,995 | # Can a Kalman Filter Be Applied Using Measurement-Space Dependent Sensors?
I'm currently attempting to apply a Kalman filter to track the angular position, velocity, and acceleration of a bike wheel, and I'm having a lot of trouble, so I want to check if I'm even applying the right tool for the job. As you will see, this is not a "classical" Kalman implementation, and I want to check if I can still use it.
I have a wheel with n reed switches evenly spaced around the circumference of the wheel. The way that I have the filter set up is that every time one of the reed switches is triggered by a magnet sitting on the bike's fork, a new measurement is fed to the filter that has the position and velocity (z = [x_prev + wheel_circumference/n ; wheel_circumference/(n*dt)] where x_prev is the previous position, wheel_circumference is the circumference of the bike wheel, and dt is the time since the last measurement). (I have tried this system using only position measurements, and I am still having trouble.)
Unlike the classic application of the Kalman filter, the measurements in this system occur periodically in position, not in time, i.e. the sensor(s) sends a measurement to the filter iff the wheel's position reaches a certain position. Compared to a traditional sensor, this implies a heavy correlation between position, velocity, and inter-measurement period. Does this correlation negatively affect the Kalman filter?
Yes, sorta kinda. Basically, if you can model your system accurately and the resulting model fits the Kalman filter paradigm, then yes.
However, I think it's a mistake to give the filter position and velocity -- were I doing this, I'd just give it the wheel position and let the Kalman filter determine the velocity -- which it will do well to the extent that your model matches the actual plant.
If it'll work at all, it'll be because you take the typical Kalman system model, $$\mathbf{x}_k = \mathbf{F}_k \mathbf{x}_{k-1} + \mathbf{B u}_k + \mathbf{w}_k$$, and you (probably, unless you have some prior idea of the acceleration being applied) let $$\mathbf{B} = \mathbf{0}$$, let $$\mathbf{w}_k$$ reflect your maximum expected acceleration, and let $$k$$ be the triggering instance.
The first fly in this ointment is that you have to calculate $$\mathbf{F}_k$$ and $$\mathbf{w}_k$$ according to the amount of time elapsed since the last trigger instant. So for $$\mathbf{x} = \left [ \begin{matrix} \theta & \omega \end{matrix} \right ]^T$$, where $$\theta$$ is angular position and $$\omega$$ is angular velocity, you'd get something like $$\mathbf{F}_k = \left [ \begin{matrix} 1 & t \\ 0 & 1 \end{matrix}\right ]$$ and $$\mathbf{w}_k = \left [ \begin{matrix} t^2 & t \\ t & 1 \end{matrix}\right ] \sigma_\alpha^2$$ where $$\sigma_\alpha$$ is your expected variation in acceleration.
(Note that I'm not giving the derivation of $$\mathbf{w}_k$$, for which I only sorta-kinda-apologize -- it's tedious for me to translate my hen-scratches to properly formatted math, but it'll be good for you to derive it yourself. Start with the definition of the Kalman-Bucy filter and grind through the math to find $$\mathbf{P}(t)$$ for a constant $$\mathbf{w}(t)$$ -- the above expression for the effective sampled-time $$\mathbf{w}_k$$ should drop out).
Note that the second fly in the ointment is that as the speed of the wheel drops the effect of the acceleration uncertainty ($$\sigma_\alpha$$, above) will get bigger and bigger compared to the position change from sample to sample. At the point where the likelihood of overrunning the sample gets significant, you can't really use a "pure" Kalman filter anymore -- logically, this is because if you haven't triggered a pulse, then you know darned well that your position and velocity excursion are limited, but there's no way to effectively communicate that to a pure Kalman application.
If you need to track a really slow wheel, then you need to come up with some other, nonlinear, state estimator.
PART II
On reflection, I thought of a possibly-better way. It's still not the best way, because you're jamming a nonlinear system with non-Gaussian uncertainty into a model of a linear system with Gaussian noise -- but it has the advantage both of being conceptually simpler, and of working down to zero speed.
The system model that you use to build a Kalman filter doesn't have to be time-invariant. So model the system as \begin{align} \mathbf{x}_k = \mathbf{F} \mathbf{x}_{k-1} + \mathbf{B u}_k + \mathbf{w} \\ \mathbf{y}_k = \mathbf{H} \mathbf{x}_k + \mathbf{v}_k \end{align}
Note that $$\mathbf{v}_k$$ is modeled as time-varying. In truth, the system is nonlinear (you get a pulse at specific values of $$\theta$$), but we'll just pretend.
Run the Kalman at a constant rate in time, with $$\mathbf{F}$$, $$\mathbf{B}$$, and $$\mathbf{w}$$ $$\mathbf{H} = \left [ \begin{matrix} 1 & 0 \end{matrix} \right ]$$ all constant.
But for each sample where a pulse is triggered, set $$\mathbf{y}$$ equal to the known position of the wheel, and $$\mathbf{v}_k$$ small, to reflect that you darned well know where the wheel is at that moment (size $$\mathbf{v}$$ to reflect any uncertainties in position due to mounting or irregular triggering or whatever).
For each sample where a pulse is not triggered, set $$\mathbf{v}_k$$ large, to reflect the fact that you don't know where between the triggers the wheel may be. I'd start with $$\mathbf{v}_k = \left( 3 \theta_0 \right)^2$$, with $$\theta_0$$ equal to the spacing between trigger points -- and note that the multiply by three is coming directly out of my ear, so feel free to experiment with it. Set $$\mathbf{y}$$ equal exactly halfway between the last known position of the wheel and the expected next position of the wheel (I'm assuming you know the wheel direction).
The net effect of this should be that if the wheel is turning fairly quickly the filter will have a darned good idea of the position and velocity, with "coasting" between readings. If the wheel is turning slowly, it should coast OK if it's going at a constant rate, it should update at least fairly well if it speeds up, and if it slows down or stops, the fact that you're telling the Kalman filter that it's between positions "somewhere" should cause the Kalman's velocity estimate to slowly decay to zero (which is your best guess, intuitively), while the position estimate slowly settles on halfway between triggers (which again, is intuitively your best guess). | 1,634 | 6,487 | {"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": 31, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-50 | longest | en | 0.931323 |
https://corfire.com/how-do-you-find-the-percentile-for-implied-volatility/ | 1,716,130,712,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057788.73/warc/CC-MAIN-20240519132049-20240519162049-00506.warc.gz | 155,693,648 | 8,735 | ## How do you find the percentile for implied volatility?
IV Percentile: To calculate percentile, you take the time-series and arrange it in order from highest to lowest, then you see what number of days the implied volatility was lower that it is today, and divide it by the total number of days.
What is considered a high implied volatility rank?
As a general rule of thumb, IV Ranks above 50 are considered expensive, and below 50 are considered cheap.
### What is IV and HV percentile?
IV is a forward-looking measure implied by the options market, and HV is backward looking. HV is a moving average of actual price variability in the stock over the previous 52 weeks.
How do you find the IV percentile of a stock?
Traders should compare high options volume to the stock’s average daily volume for clues to its origin. IV Rank is the at-the-money (ATM) average implied volatility relative to the highest and lowest values over the past 1-year. If IV Rank is 100%, this means the IV is at its highest level over the past 1-year.
## How do you read implied volatility on thinkorswim?
To find implied and historical volatility in the thinkorswim® platform from TD Ameritrade, pull up a chart and select Studies > Add Study > Volatility Studies.
How do you find high implied volatility of a stock?
Generally speaking, traders look to buy an option when the implied volatility is low, and look to sell an option (or consider a spread strategy) when implied volatility is high. Implied volatility is determined mathematically by using current option prices and the Binomial option pricing model.
### Is a higher delta better?
The rule of thumb here is the higher the delta is, the more likely it is the option ends up profitable. Out-of-the-money options have the lowest delta, while in-the-money options have the highest delta. So you’d want to avoid the out-of-the-money option that has the delta of 0.04 like the plague.
What is volatility percentile?
Implied Volatility percentile is a ranking method to compare implied volatility to its past values. The ranking is standardized from 0-100, where 0 is the lowest value in recent history, and 100 is the highest value. This value tells us how high or low the current value is compared with the past. | 483 | 2,266 | {"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-2024-22 | latest | en | 0.917511 |
https://number.academy/11930 | 1,653,326,997,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662560022.71/warc/CC-MAIN-20220523163515-20220523193515-00798.warc.gz | 493,138,327 | 12,195 | # Number 11930
Number 11,930 spell 🔊, write in words: eleven thousand, nine hundred and thirty . Ordinal number 11930th is said 🔊 and write: eleven thousand, nine hundred and thirtieth. The meaning of number 11930 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 11930. What is 11930 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 11930.
## What is 11,930 in other units
The decimal (Arabic) number 11930 converted to a Roman number is (X)MCMXXX. Roman and decimal number conversions.
#### Weight conversion
11930 kilograms (kg) = 26300.9 pounds (lbs)
11930 pounds (lbs) = 5411.4 kilograms (kg)
#### Length conversion
11930 kilometers (km) equals to 7413 miles (mi).
11930 miles (mi) equals to 19200 kilometers (km).
11930 meters (m) equals to 39140 feet (ft).
11930 feet (ft) equals 3637 meters (m).
11930 centimeters (cm) equals to 4696.9 inches (in).
11930 inches (in) equals to 30302.2 centimeters (cm).
#### Temperature conversion
11930° Fahrenheit (°F) equals to 6610° Celsius (°C)
11930° Celsius (°C) equals to 21506° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
11930 seconds equals to 3 hours, 18 minutes, 50 seconds
11930 minutes equals to 1 week, 1 day, 6 hours, 50 minutes
### Zip codes 11930
• Zip code 11930 Lomas de Reforma, Ciudad de México, Miguel Hidalgo, Mexico a map
### Codes and images of the number 11930
Number 11930 morse code: .---- .---- ----. ...-- -----
Sign language for number 11930:
Number 11930 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
## Mathematics of no. 11930
### Multiplications
#### Multiplication table of 11930
11930 multiplied by two equals 23860 (11930 x 2 = 23860).
11930 multiplied by three equals 35790 (11930 x 3 = 35790).
11930 multiplied by four equals 47720 (11930 x 4 = 47720).
11930 multiplied by five equals 59650 (11930 x 5 = 59650).
11930 multiplied by six equals 71580 (11930 x 6 = 71580).
11930 multiplied by seven equals 83510 (11930 x 7 = 83510).
11930 multiplied by eight equals 95440 (11930 x 8 = 95440).
11930 multiplied by nine equals 107370 (11930 x 9 = 107370).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 11930
Half of 11930 is 5965 (11930 / 2 = 5965).
One third of 11930 is 3976,6667 (11930 / 3 = 3976,6667 = 3976 2/3).
One quarter of 11930 is 2982,5 (11930 / 4 = 2982,5 = 2982 1/2).
One fifth of 11930 is 2386 (11930 / 5 = 2386).
One sixth of 11930 is 1988,3333 (11930 / 6 = 1988,3333 = 1988 1/3).
One seventh of 11930 is 1704,2857 (11930 / 7 = 1704,2857 = 1704 2/7).
One eighth of 11930 is 1491,25 (11930 / 8 = 1491,25 = 1491 1/4).
One ninth of 11930 is 1325,5556 (11930 / 9 = 1325,5556 = 1325 5/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
11930
#### Is Prime?
The number 11930 is not a prime number. The closest prime numbers are 11927, 11933.
#### Factorization and factors (dividers)
The prime factors of 11930 are 2 * 5 * 1193
The factors of 11930 are 1 , 2 , 5 , 10 , 1193 , 2386 , 5965 , 11930
Total factors 8.
Sum of factors 21492 (9562).
#### Powers
The second power of 119302 is 142.324.900.
The third power of 119303 is 1.697.936.057.000.
#### Roots
The square root √11930 is 109,224539.
The cube root of 311930 is 22,849681.
#### Logarithms
The natural logarithm of No. ln 11930 = loge 11930 = 9,386812.
The logarithm to base 10 of No. log10 11930 = 4,07664.
The Napierian logarithm of No. log1/e 11930 = -9,386812.
### Trigonometric functions
The cosine of 11930 is -0,196809.
The sine of 11930 is -0,980442.
The tangent of 11930 is 4,981697.
### Properties of the number 11930
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 11930 in Computer Science
Code typeCode value
11930 Number of bytes11.7KB
Unix timeUnix time 11930 is equal to Thursday Jan. 1, 1970, 3:18:50 a.m. GMT
IPv4, IPv6Number 11930 internet address in dotted format v4 0.0.46.154, v6 ::2e9a
11930 Decimal = 10111010011010 Binary
11930 Decimal = 121100212 Ternary
11930 Decimal = 27232 Octal
11930 Decimal = 2E9A Hexadecimal (0x2e9a hex)
11930 BASE64MTE5MzA=
11930 MD5bc3ff3ecffc6ba348bbc43cb98ee4dcf
11930 SHA1e02bbe5cd2bce29e136b63c6b83b94752cd11df1
11930 SHA224308fd26d4b76e2b1c5927aee01eccca244cd0c8d46f8ebbefcdc365e
11930 SHA38449eaccdfd5af3158966e3e430a44a27ec004026e20d8096330553456ae6eca0bd7c9904fdb3c46183774b2c50186e0c4
More SHA codes related to the number 11930 ...
If you know something interesting about the 11930 number that you did not find on this page, do not hesitate to write us here.
## Numerology 11930
### Character frequency in number 11930
Character (importance) frequency for numerology.
Character: Frequency: 1 2 9 1 3 1 0 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 11930, the numbers 1+1+9+3+0 = 1+4 = 5 are added and the meaning of the number 5 is sought.
## Interesting facts about the number 11930
### Asteroids
• (11930) Osamu is asteroid number 11930. It was discovered by T. Hioki; S. Hayakawa from Okutama on 2/15/1993.
### Distances between cities
• There is a 11,930 miles (19,199 km) direct distance between Córdoba (Argentina) and Xiamen (China).
• There is a 7,413 miles (11,930 km) direct distance between Fukuoka-shi (Japan) and Harare (Zimbabwe).
• There is a 7,413 miles (11,930 km) direct distance between Chengdu (China) and Phoenix (USA).
• There is a 11,930 miles (19,199 km) direct distance between Rosario (Argentina) and Shiyan (China).
## Number 11,930 in other languages
How to say or write the number eleven thousand, nine hundred and thirty in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 11.930) once mil novecientos treinta German: 🔊 (Anzahl 11.930) elftausendneunhundertdreißig French: 🔊 (nombre 11 930) onze mille neuf cent trente Portuguese: 🔊 (número 11 930) onze mil, novecentos e trinta Chinese: 🔊 (数 11 930) 一万一千九百三十 Arabian: 🔊 (عدد 11,930) أحد عشر ألفاً و تسعمائةثلاثون Czech: 🔊 (číslo 11 930) jedenáct tisíc devětset třicet Korean: 🔊 (번호 11,930) 만 천구백삼십 Danish: 🔊 (nummer 11 930) ellevetusinde og nihundrede og tredive Dutch: 🔊 (nummer 11 930) elfduizendnegenhonderddertig Japanese: 🔊 (数 11,930) 一万千九百三十 Indonesian: 🔊 (jumlah 11.930) sebelas ribu sembilan ratus tiga puluh Italian: 🔊 (numero 11 930) undicimilanovecentotrenta Norwegian: 🔊 (nummer 11 930) elleve tusen, ni hundre og tretti Polish: 🔊 (liczba 11 930) jedenaście tysięcy dziewięćset trzydzieści Russian: 🔊 (номер 11 930) одиннадцать тысяч девятьсот тридцать Turkish: 🔊 (numara 11,930) onbirbindokuzyüzotuz Thai: 🔊 (จำนวน 11 930) หนึ่งหมื่นหนึ่งพันเก้าร้อยสามสิบ Ukrainian: 🔊 (номер 11 930) одинадцять тисяч дев'ятсот тридцять Vietnamese: 🔊 (con số 11.930) mười một nghìn chín trăm ba mươi Other languages ...
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## Comment
If you know something interesting about the number 11930 or any natural number (positive integer) please write us here or on facebook. | 2,511 | 7,327 | {"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} | 2.984375 | 3 | CC-MAIN-2022-21 | latest | en | 0.602724 |
https://discusstest.codechef.com/t/prime-generator-using-sieve-algorithm/6022 | 1,627,481,716,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153729.44/warc/CC-MAIN-20210728123318-20210728153318-00114.warc.gz | 223,185,630 | 4,174 | # prime generator using sieve algorithm
Question->http://www.codechef.com/problems/PRIME1
My code is giving correct answer for all the test cases i have tried .Also for the extreme case i.e
when m=999900000 and n=1000000000 it running absolutely fine.I m not able to find the bug .Plz help
``````#include <iostream>
#include<vector>
#include<cmath>
using namespace std;
int main()
{
long long int i,n,k,t,m;
cin>>t;
if(t<=10){
while(t--)
{
cin>>m>>n;
vector<bool> v1;
for(i=0;i<=n-m;i++)
{
v1.push_back(0); //corresponding to each value assigning 0 value
}
for(i=2;i<=sqrt(n);i++)
{
if (m/i-i>=0){k=(m/i)-i;}//this if else condition help k to reach nearthe "m" value
else{k=m/i;} //so that we can skip the terms before m
while(i*i+k*i<m){k++;}
while(i*i+k*i<=n)
{
v1[i*i+k*i-m]=1; //making all composite no's 1
k++;
}
}
if(m==1){v1[0]=1;}// condition to eliminate 1 as a prime
for(i=0;i<=n-m;i++)
{
if(!v1[i]){cout<<i+m<<"\n";}// all the no's which are still set to 0 are prime
}
cout<<"\n";
}
}
return 0;
}``````
Hai, for this case:
1
2 10
2
3
4
5
7
‘4’ should not be there in the output as ‘4’ is not prime.
2 Likes
thanks @achait for finding the bug.
// | 396 | 1,186 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-31 | latest | en | 0.614548 |
https://www.tutorialcup.com/interview/linked-list/flattening-linked-list.htm | 1,585,550,916,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370496669.0/warc/CC-MAIN-20200330054217-20200330084217-00052.warc.gz | 1,216,238,519 | 281,750 | # Flattening a linked list
0
346
## In the given linked list, every node has two pointers :
1. Pointer to next node (Main linked list(right pointer))
2. Pointer to next node where this node is head (Down linked list(down pointer))
We need to flatten this linked list into one normal singly linked list. And the output linked list should be sorted. The down list and right lists should be sorted atready that is right and down are always greater than it`s value.
### Example
Time complexity : O(n)
## Algorithm
a. We use merge sort for merging linked lists.
b. We recursively flatten the lists by merge the current list with already flatten list.
c. We use down pointer to link nodes of the flattened list.
d. In merge function we compare data values of head nodes and put smaller one in flatten list.
## C++ Program
``````#include <bits/stdc++.h>
using namespace std;
struct LLNode
{
int data;
struct LLNode* right;
struct LLNode* down;
};
/* Function to insertAtBeginning ListA node */
void insertAtBeginning(struct LLNode** head, int dataToBeInserted)
{
struct LLNode* curr = new LLNode;
curr->data = dataToBeInserted;
curr->right = NULL;
//O(1) constant time
}
void display(struct LLNode**node)
{
struct LLNode *temp= *node;
while(temp!=NULL)
{
if(temp->down!=NULL)
cout<<temp->data<<"->";
else
cout<<temp->data;
temp=temp->down; //move to right node
}
//O(number of nodes)
cout<<endl;
}
//Merge sort of ListA and ListB
struct LLNode* MergeSort(struct LLNode* ListA, struct LLNode* ListB )
{
//Base cases
if (ListA == NULL)
{
return ListB;
}
if (ListB == NULL)
{
return ListA;
}
//compare data of heads and add smaller to root
struct LLNode* result;
if (ListA->data < ListB->data)
{
result = ListA;
result->down = MergeSort(ListA->down, ListB);
}
else
{
result = ListB;
result->down = MergeSort(ListA, ListB->down);
}
return result;
}
//Function that flattens the given list
struct LLNode* flatten(struct LLNode* root)
{
if (root == NULL || root->right == NULL)
{
return root;
}
//Merge right with already flattens
return MergeSort(root,flatten(root->right) );
}
//Main function
int main()
{
struct LLNode* root = NULL;
insertAtBeginning(&root, 14);
insertAtBeginning(&root, 8);
insertAtBeginning(&root, 7);
insertAtBeginning(&(root->right), 6);
insertAtBeginning(&(root->right), 4);
insertAtBeginning(&(root->right->right), 18);
insertAtBeginning(&(root->right->right), 13);
insertAtBeginning(&(root->right->right), 9);
insertAtBeginning(&(root->right->right->right), 18);
insertAtBeginning(&(root->right->right->right), 17);
insertAtBeginning(&(root->right->right->right), 15);
insertAtBeginning(&(root->right->right->right), 11);
insertAtBeginning(&(root->right->right->right->right), 16);
root = flatten(root);
cout<<"Output (sorted)flatten List is: ";
display(&root);
return 0;
}`````` | 754 | 2,824 | {"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} | 3.140625 | 3 | CC-MAIN-2020-16 | longest | en | 0.608303 |
http://jayisgames.com/archives/2013/05/banana_breakers.php | 1,411,178,110,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657132495.49/warc/CC-MAIN-20140914011212-00078-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 151,680,442 | 17,600 | Banana Breakers
• Currently 4.1/5
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| Comments (5) | Views (17)
What happens when you combine the logical deduction puzzles of Mastermind, the word-searching challenge of Boggle, and a handful of military monkeys ready to do battle? Everything goes bananas! Thank you, I'm here all night, tip your server, etc. Banana Breakers by Simian Logic is a quick twist on a logic/word hybrid puzzle where you've got to solve not just one word, but up to six words at once!
The grid of letters can be divided into a number of words (for example, the easy level has you finding four 4-letter words). Click or tap the letters to select them; they don't have to be touching, any letters in the grid are allowed. If you don't spell one of the target words picked by the computer, you're given clues to the position of the letters in one of the target words. A letter highlighted in green means it's a correct letter in the right position, and a letter in red means it's a correct letter but in a wrong position. Use these clues to help formulate your next guess.
However, unlike other word-Mastermind games like Wordspector, you're not just focusing on one word at a time, you're taking on all four (in easy mode) at once! The clues revealed to you always pertain to the hidden word that most closely matches your guess. You might use the clues from one word to formulate another guess, only to find it gives you new clues to a completely different word! Words that give you less new information than what is already shown won't give you any clues at all, but you'll still be credited for making a valid word with some coins.
For each hidden word you discover (or at least try), you'll earn some coins when you complete the board. Coins can be used to purchase clues such as a crossword-style definition of one of the hidden words, or outright spoil a word if you're stuck. Each board is a self-contained game, so you can quickly blaze through a few easy levels, or take your time with the medium (five 5-letter words) and hard (six 6-letter words) levels. Can you crack the code and deduce all the words before going bananas?
Play Banana Breakers
NOTE: This game was played and reviewed on the iPhone 4. Game was available in the North American market at the time of publication, but may not be available in other territories. Please see individual app market pages for purchasing info.
May 11, 2013 9:25 AM
You might mention that, at least on Android, it requires you to install Adobe Air. It's not a big file, but some of these cheapo Androids have seriously limited space. It triples the size of the download.
May 11, 2013 9:34 AM
That being said, this is fun!
May 11, 2013 10:26 PM
This is a fantastic melding of 2 of my favorite games! This one earns a place on my homepage.
Really great game, but my one complaint would be that it uses American spelling, which for the most part is no problem but had me completely stumped on
'DIALOG' as opposed to 'DIALOGUE'
June 21, 2014 7:53 PM
Sadly it checks for a specific version of flash player which is no longer current :( I lost Monster Mahjong the same way, back in the day. Standalone legacy players might do the trick though.
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Limit to the last 5 comments | 1,249 | 5,237 | {"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-2014-41 | longest | en | 0.939052 |
https://discuss.codecademy.com/t/credit-card-checker-challenge-project-javascript/462375?page=43 | 1,653,571,997,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662606992.69/warc/CC-MAIN-20220526131456-20220526161456-00657.warc.gz | 260,393,390 | 13,791 | # Credit Card Checker Challenge Project (JavaScript)
I’m burning my brain on this project, can someone take a look a my code please?
1 - in the Luhn algorithm talk about sum the last digit (wich first we remove) with the sum of all the element of the array, so how I make this happen? I can’t find this in some codes I looked.
2 - With my currently code here some of the valids arrays keep showing invalid when I execute the function, but with my noob eyes I’m can’t see the mistake, can someone point out?
``````// All valid credit card numbers
const valid1 = [4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8];
const valid2 = [5, 5, 3, 5, 7, 6, 6, 7, 6, 8, 7, 5, 1, 4, 3, 9];
const valid3 = [3, 7, 1, 6, 1, 2, 0, 1, 9, 9, 8, 5, 2, 3, 6];
const valid4 = [6, 0, 1, 1, 1, 4, 4, 3, 4, 0, 6, 8, 2, 9, 0, 5];
const valid5 = [4, 5, 3, 9, 4, 0, 4, 9, 6, 7, 8, 6, 9, 6, 6, 6];
// All invalid credit card numbers
const invalid1 = [4, 5, 3, 2, 7, 7, 8, 7, 7, 1, 0, 9, 1, 7, 9, 5];
const invalid2 = [5, 7, 9, 5, 5, 9, 3, 3, 9, 2, 1, 3, 4, 6, 4, 3];
const invalid3 = [3, 7, 5, 7, 9, 6, 0, 8, 4, 4, 5, 9, 9, 1, 4];
const invalid4 = [6, 0, 1, 1, 1, 2, 7, 9, 6, 1, 7, 7, 7, 9, 3, 5];
const invalid5 = [5, 3, 8, 2, 0, 1, 9, 7, 7, 2, 8, 8, 3, 8, 5, 4];
// Can be either valid or invalid
const mystery1 = [3, 4, 4, 8, 0, 1, 9, 6, 8, 3, 0, 5, 4, 1, 4];
const mystery2 = [5, 4, 6, 6, 1, 0, 0, 8, 6, 1, 6, 2, 0, 2, 3, 9];
const mystery3 = [6, 0, 1, 1, 3, 7, 7, 0, 2, 0, 9, 6, 2, 6, 5, 6, 2, 0, 3];
const mystery4 = [4, 9, 2, 9, 8, 7, 7, 1, 6, 9, 2, 1, 7, 0, 9, 3];
const mystery5 = [4, 9, 1, 3, 5, 4, 0, 4, 6, 3, 0, 7, 2, 5, 2, 3];
// An array of all the arrays above
const batch = [valid1, valid2, valid3, valid4, valid5, invalid1, invalid2, invalid3, invalid4, invalid5, mystery1, mystery2, mystery3, mystery4, mystery5];
// Add your functions below:
const validateCred = arr => {
let newArray = [];
let reversedArray = arr.slice(0, -1).reverse();
for(let i = 0; i < reversedArray.length; i++){
if(i % 2 !== 0) {
reversedArray[i] *= 2;
if (reversedArray[i] > 9){
reversedArray[i] -= 9;
}
}
newArray.push(reversedArray[i]);
let sum = newArray.reduce((accumulator, number) => {
accumulator + number, 0
});
if (sum % 10 === 0){
return true;
} else {
return false;
}
}
};
console.log(validateCred(valid1));
console.log(validateCred(valid2))
const findInvalidCards = arr => {
let invalidCards = [];
for (let i = 0; i < arr.length; i++) {
if (validateCred(arr[i]) === false) {
invalidCards.push(arr[i])
}
}
return invalidCards;
};
console.log(findInvalidCards(batch));
// output
true
false
[ [ 5, 5, 3, 5, 7, 6, 6, 7, 6, 8, 7, 5, 1, 4, 3, 9 ],
[ 3, 7, 1, 6, 1, 2, 0, 1, 9, 9, 8, 5, 2, 3, 6 ],
[ 4, 5, 3, 9, 4, 0, 4, 9, 6, 7, 8, 6, 9, 6, 6, 6 ],
[ 4, 5, 3, 2, 7, 7, 8, 7, 7, 1, 0, 9, 1, 7, 9, 5 ],
[ 5, 7, 9, 5, 5, 9, 3, 3, 9, 2, 1, 3, 4, 6, 4, 3 ],
[ 3, 7, 5, 7, 9, 6, 0, 8, 4, 4, 5, 9, 9, 1, 4 ],
[ 6, 0, 1, 1, 1, 2, 7, 9, 6, 1, 7, 7, 7, 9, 3, 5 ],
[ 5, 3, 8, 2, 0, 1, 9, 7, 7, 2, 8, 8, 3, 8, 5, 4 ],
[ 3, 4, 4, 8, 0, 1, 9, 6, 8, 3, 0, 5, 4, 1, 4 ],
[ 5, 4, 6, 6, 1, 0, 0, 8, 6, 1, 6, 2, 0, 2, 3, 9 ],
[ 4, 9, 2, 9, 8, 7, 7, 1, 6, 9, 2, 1, 7, 0, 9, 3 ],
[ 4, 9, 1, 3, 5, 4, 0, 4, 6, 3, 0, 7, 2, 5, 2, 3 ] ]
``````
Hi Julia,
You’re very welcome
Hahaha eventually you’ll want to know both front & back-end. I’m in the full-stack engineering course which teaches both front & back-end.
Oh nice, I’m working on that project now! Maybe we will see each other there .
Well once you get onto discord make sure you add me up I’m always looking to collaborate, for now we’ll keep in touch via this platform!
Happy coding!
Hi Reginag,
It seems on line 30 where you declare `reversedArray` and call `slice` with the parameters of (0, -1) you’re removing the last digit completely from the array that you’re creating.
Keep in mind for the Luhn Algorithm, we want to still add the last number from the original array to our sum function.
You can achieve this in multiple ways:
You can save the last digit into a variable, then add it on in your `sum` variable.
You can loop through the array backwards instead of using .slice(), .reverse() - for your loop initializer you can subtract the array.length by 1 which doesn’t mutate your original array, it just skips the last digit.
I’m sure there are many different solutions to the error you’re running into but those are the ones I can think of off the top of my head.
I would suggest you try to console.log() through your function so you can see exactly what your loops, variables, and return statements look like for the computer.
Happy coding!
I just finished it. You motivated me
No challenges though. I keep those for another day.
I posted under the mysterious organism project if you like to see it. Looking forward to see your solution!
1 Like
That’s awesome glad to hear I motivated you ! I just finished mine.
Tried the challenges, couldn’t figure out the last one.
I’ll take a look at yours shortly and post mine as-well!
Happy Coding!
Hey guys! I finished the credit card checker today and although at first everything seems to work flawlessly, it is unable to verify valid3- i know this happens because it is shorter than the other numbers and since I have not reversed the array but rather just iterated it backwards, there must be some kind of solution to how I duplicate and/or subtract every other number besides mine (on l. 30). My head is knotting itself in a twist though, can anyone give me some advice?
Hi gooddevenin,
I finally finish the project and your help was extremely important, I choose store the last digit in a variable and worked fine, but also haved a little mistake with my logic on the loop and well, took some time to figure out .lol
Here is my code and I accept any feedback.
happy coding!
``````// All valid credit card numbers
const valid1 = [4, 5, 3, 9, 6, 7, 7, 9, 0, 8, 0, 1, 6, 8, 0, 8];
const valid2 = [5, 5, 3, 5, 7, 6, 6, 7, 6, 8, 7, 5, 1, 4, 3, 9];
const valid3 = [3, 7, 1, 6, 1, 2, 0, 1, 9, 9, 8, 5, 2, 3, 6];
const valid4 = [6, 0, 1, 1, 1, 4, 4, 3, 4, 0, 6, 8, 2, 9, 0, 5];
const valid5 = [4, 5, 3, 9, 4, 0, 4, 9, 6, 7, 8, 6, 9, 6, 6, 6];
// All invalid credit card numbers
const invalid1 = [4, 5, 3, 2, 7, 7, 8, 7, 7, 1, 0, 9, 1, 7, 9, 5];
const invalid2 = [5, 7, 9, 5, 5, 9, 3, 3, 9, 2, 1, 3, 4, 6, 4, 3];
const invalid3 = [3, 7, 5, 7, 9, 6, 0, 8, 4, 4, 5, 9, 9, 1, 4];
const invalid4 = [6, 0, 1, 1, 1, 2, 7, 9, 6, 1, 7, 7, 7, 9, 3, 5];
const invalid5 = [5, 3, 8, 2, 0, 1, 9, 7, 7, 2, 8, 8, 3, 8, 5, 4];
// Can be either valid or invalid
const mystery1 = [3, 4, 4, 8, 0, 1, 9, 6, 8, 3, 0, 5, 4, 1, 4];
const mystery2 = [5, 4, 6, 6, 1, 0, 0, 8, 6, 1, 6, 2, 0, 2, 3, 9];
const mystery3 = [6, 0, 1, 1, 3, 7, 7, 0, 2, 0, 9, 6, 2, 6, 5, 6, 2, 0, 3];
const mystery4 = [4, 9, 2, 9, 8, 7, 7, 1, 6, 9, 2, 1, 7, 0, 9, 3];
const mystery5 = [4, 9, 1, 3, 5, 4, 0, 4, 6, 3, 0, 7, 2, 5, 2, 3];
// An array of all the arrays above
const batch = [valid1, valid2, valid3, valid4, valid5, invalid1, invalid2, invalid3, invalid4, invalid5, mystery1, mystery2, mystery3, mystery4, mystery5];
// Add your functions below:
const validateCred = arr => {
let newArray = [];
let reversedArray = arr.slice(0, -1).reverse();
let lastDigit = arr[arr.length - 1];
//console.log(reversedArray);
// console.log(lastDigit);
for(let i = 0; i < reversedArray.length; i++){
if(i % 2 == 0) {
reversedArray[i] *= 2;
if (reversedArray[i] > 9){
reversedArray[i] -= 9;
}
}
newArray.push(reversedArray[i])
//console.log(reversedArray[i]);
}
//console.log(newArray);
let sum = newArray.reduce((accumulator, number) => accumulator + number, 0) + lastDigit;
//console.log(sum);
return sum % 10 === 0 ? true : false;
};
//console.log(validateCred(valid1));
//console.log(validateCred(valid2));
//console.log(validateCred(valid3));
//console.log(validateCred(valid4));
//console.log(validateCred(valid5));
//console.log(validateCred(invalid2));
//console.log(validateCred(mystery1));
const findInvalidCards = arr => {
let invalidCards = [];
for (let i = 0; i < arr.length; i++) {
if (validateCred(arr[i]) === false) {
invalidCards.push(arr[i])
}
}
return invalidCards;
};
//console.log(findInvalidCards(batch));
const idInvalidCardCompanies = arr => {
let invalidArrays = findInvalidCards(arr);
let companies = [];
console.log(invalidArrays);
for (let i = 0; i < invalidArrays.length; i++){
switch (invalidArrays[i][0]){
case 3:
companies.push('Amex');
break;
case 4:
companies.push('Visa');
break;
case 5:
companies.push('Mastercard');
break;
case 6:
companies.push('Discover');
break;
default:
companies.push('Company not found');
break;
}
}
let invalidCardCompanies = [ ...new Set(companies) ];
return invalidCardCompanies;
};
console.log(idInvalidCardCompanies(batch));
//output
[ [ 4, 5, 3, 2, 7, 7, 8, 7, 7, 1, 0, 9, 1, 7, 9, 5 ],
[ 5, 7, 9, 5, 5, 9, 3, 3, 9, 2, 1, 3, 4, 6, 4, 3 ],
[ 3, 7, 5, 7, 9, 6, 0, 8, 4, 4, 5, 9, 9, 1, 4 ],
[ 6, 0, 1, 1, 1, 2, 7, 9, 6, 1, 7, 7, 7, 9, 3, 5 ],
[ 5, 3, 8, 2, 0, 1, 9, 7, 7, 2, 8, 8, 3, 8, 5, 4 ],
[ 3, 4, 4, 8, 0, 1, 9, 6, 8, 3, 0, 5, 4, 1, 4 ],
[ 6, 0, 1, 1, 3, 7, 7, 0, 2, 0, 9, 6, 2, 6, 5, 6, 2, 0, 3 ],
[ 4, 9, 2, 9, 8, 7, 7, 1, 6, 9, 2, 1, 7, 0, 9, 3 ] ]
[ 'Visa', 'Mastercard', 'Amex', 'Discover' ]
``````
1 Like
Woohoo, this was fun! I only had to reset all of my JS modules and re-do them all a second time before I had any thoughts about being able to complete what I thought was a very daunting project. Second time through and I didn’t even need to consult the help sections for ideas! My solution can be found here on github:
Hi Reginag,
That’s amazing! I’m glad I was able to help you out.
Happy Coding!
Phew, finished!!
Here’s what I came up with:
All the tests are coming back as they should!
Hey everybody sharing a project for the first time, completed this without peeping at the solution code, so an honest opinion on how to tackled this would be greatly appreciated, thanks in advance!
Here is my code for the challenge!
Hi, guys!
Here’s my solution code, so you guys can compare
Quick note: i tried to use ternary on the switch statement in the last function, but it kept having issues with the ‘continue’ keyword, so i had to implement long if…else statements, making the code a little too verbose. You’re free to improve it if you feel like it!
Kudos for completing your project as well, and if you’re stuck or interested, try taking a look on mine!
Here’s a repl.it with it: https://replit.com/@ProspektsMarch2/Validador-de-Cartao-Demoniaco
Here is my solution:
any feedback is appreciated
My initial attempt at the validateCred() function involved checking if the arrays were even or odd in length and using that info to modify every other digit. It worked when tested by itself but was somehow permanently modifying the arrays so that the following two functions would not work properly.
The idea to loop through the array backwards was what I went to as this was my original idea I thought of but I struggled for hours to figure out a method to modify every other digit correctly. I often find solutions to my issues on stack overflow or elsewhere but sometimes choose not to implement them as I can’t always understand the logic of why they work.
And when something like this is presented as, “something I found and would like to understand” we are only too happy to provide insight. Just so long as it is not something posted as a solution to an exercise.
That over with, definitely bring your questions to the forums. Others can learn from it, too. What we want to prevent is the ‘loss of our youth’ because we were too busy trying to be adults. There is so much to behold in the very rudimentary of code that we should never want to stray too far from it. It’s like riding a bike. That’s the perfect analog.
I think it works so I am chuffed
Hi,
Thanks for sharing your standard solution. I solved the last question by creating an object within a function. In general, is it a good practice generating an object within function? My code is below for your reference:
``````function idInvalidCardCompanies(nested) {
results = [];
const company_obj = {
3: 'Amex (American Express)',
4: 'Visa',
5: 'Mastercard',
6: 'Discover'
};
for (const arr of nested) {
//console.log(company_obj[arr[0]]);
//console.log(!(arr[0] in company_obj));
if (!(arr[0] in company_obj)) {
console.log('Company not found');
} else {
if (!(results.includes(company_obj[arr[0]]))) {
results.push(company_obj[arr[0]]);
}
}
}
return results;
}
``````
Hi, all. Just setup a new Github account and wanted to share my code with some debugging code still left inside. Long day but so just want to submit this but can’t wait to take a look at the official solution and others code. | 4,869 | 12,716 | {"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-2022-21 | latest | en | 0.611571 |
https://oeis.org/A320429 | 1,632,049,306,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056856.4/warc/CC-MAIN-20210919095911-20210919125911-00002.warc.gz | 477,441,120 | 4,096 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A320429 The length of the shortest prefix of the Thue-Morse word decomposable to not less than n palindromes. 0
1, 2, 6, 10, 26, 90, 154, 410, 1434, 2458, 6554, 22938, 39322, 104858, 367002, 629146, 1677722, 5872026, 10066330, 26843546, 93952410, 161061274, 429496730, 1503238554, 2576980378, 6871947674, 24051816858, 41231686042, 109951162778, 384829069722, 659706976666, 1759218604442, 6157265115546 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS A. E. Frid, Prefix palindromic length of the Thue-Morse word, arXiv:1906.09392 [cs.DM] (2019). FORMULA a(n+3) = 16*a(n) - 6. PROG (PARI) a(n)=if(n<=3, [1, 2, 6][n], 16*a(n-3)-6); \\ Joerg Arndt, Aug 27 2019 CROSSREFS Cf. Thue-Morse word A010060. Indices of new values of A307319. Sequence in context: A162581 A061547 A218791 * A319014 A190034 A119459 Adjacent sequences: A320426 A320427 A320428 * A320430 A320431 A320432 KEYWORD easy,nonn AUTHOR Anna Frid, Aug 27 2019 STATUS approved
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Last modified September 19 06:56 EDT 2021. Contains 347551 sequences. (Running on oeis4.) | 525 | 1,486 | {"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-2021-39 | latest | en | 0.575826 |
http://tutoringsanjose.net/?p=1352 | 1,656,377,868,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103344783.24/warc/CC-MAIN-20220627225823-20220628015823-00133.warc.gz | 65,495,924 | 7,326 | ## Using multiplicity of factors to characterize graphs of rational functions
Rational functions can be scary because there are so many details to manage. Check other posts on this website for information on how to graph rational functions. In this post, I look at one small clue that can help you figure out the behavior of a rational function as it approaches the vertical asymptotes. All you need to do is check the multiplicity of the factor in the denominator.
If the multiplicity of the factor is even, then the graph approaches +∞ from both sides of the asymptote, or it approaches -∞ from both sides of the asymptote.
If the multiplicity of the factor is odd, then the graph approaches +∞ on one side of the asymptote and approaches -∞ on the other side.
Here is an example that demonstrates this property:
$\text{Graph } \dfrac {(x-2)(x+1)}{(x-1)(x+2)^2}$
There are two vertical asymptotes for this function, at $x=-2$ and at $x=1.$ The $(x+2)$ factor is multiplicity 2 (even), so the graph approaches the same limit from both sides of the asymptote. The $(x-1)$ factor is multiplicity 1 (odd), so the graph approaches opposite limits on either side of the asymptote. Here is the graph of the function, demonstrating this property:
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Optimal Geometric Flows via Dual Programs
Courant Institute NYU.
Linköping University, Department of Science and Technology, Communications and Transport Systems. Linköping University, Faculty of Science & Engineering.
2014 (English)In: Annual Symposium on Computational Geometry, 2014, p. 100-109Conference paper, Published paper (Refereed)
Text
##### Abstract [en]
Considering potentials in the dual of a planar network has proved to be a powerful tool for computing planar maximum flows. In this paper we explore the use of potentials for giving algorithmic and combinatorial results on continuous flows in geometric domains -- a (far going) generalization of discrete flows in unit-capacity planar networks.
A continuous flow in a polygonal domain is a divergence-free vector field whose magnitude at every point is bounded by a given constant -- the domain's permeability. The flow enters the domain through a source edge on the outer boundary of the domain, and leaves through a sink edge on the boundary. The value of the flow is the total flow amount that comes in through the source (the same amount comes out of the sink, due to the vanishing divergence). The cost of the flow is the total length of its streamlines. The flow is called monotone if its streamlines are x-monotone curves.
Our main result is an algorithm to find (an arbitrarily close approximation to) the minimum-cost monotone flow in a polygonal domain, by formulating the problem as a convex program with an interesting choice of variables: one variable per hole in the domain. Our approach is based on the following flow of ideas: flow is the gradient of a potential function; a potential function can be extended from free space to holes so that it is constant over each hole; instead of the potential function (a value per point in the domain) we can thus speak of a potential vector (a value per hole); a potential uniquely (up to homotopic equivalence) defines "thick" paths in the domain; the paths define a flow. We show that the flow cost is convex in the potential; this allows us to employ the ellipsoid method to find the mincost flow, using holes potentials as variables. At each ellipsoid iteration the separation oracle is implemented by feasibility checks in the "critical graph" of the domain -- as we prove, the graph defines linear constraints on the potentials.
Using potentials and critical graphs we also prove MaxFlow/MinCut theorems for geometric flows -- both for monotone and for general ones. Formulating and proving the monotone version is a new result. The MaxFlow/MinCut theorem for non-monotone flows has been proved earlier by two different methods; the potentials technique provides a third proof.
2014. p. 100-109
##### National Category
Fluid Mechanics and Acoustics
##### Identifiers
ISBN: 978-1-4503-2594-3 (print)OAI: oai:DiVA.org:liu-128022DiVA: diva2:928739
##### Conference
SoCG 2014
Available from: 2016-05-16 Created: 2016-05-16 Last updated: 2016-05-26
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Polishchuk, Valentin
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• rtf | 989 | 3,950 | {"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-2018-09 | longest | en | 0.850082 |
https://betterlesson.com/lesson/435660/histogram-activity | 1,544,602,492,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823785.24/warc/CC-MAIN-20181212065445-20181212090945-00346.warc.gz | 561,415,962 | 27,229 | # Histogram activity
11 teachers like this lesson
Print Lesson
## Objective
The students will be able to understand that histogram is similar to a bar graph and be able to define mean and know how to apply it to a histogram.
#### Big Idea
Students will see how an actuary uses histograms to analyze the frequency of events to determine risk.
## Do Now
10 minutes
Watch a short video clip about Hurricane Katrina. I’m using this video clip to provide some background knowledge for students that may be unaware of what flooding does and why there is a need to keep tract of this data monetarily.
## Using Histograms in the real world
60 minutes
Read a short article about actuaries. After the students have completed the article, have them partner up to make a list of information learned about the actuary’s job. Once everyone has their list, you can do an “I have that” to go over list.
Resource: Actuary’s job description
What do they tell us?
How are they like a bar graph?
How are they different?
Which measure of central tendency can be found in a histogram?
Once you have reviewed this data display ask students to sketch the following example: A histogram showing the number of books read during several months. Allow students time to sketch their display. Once complete, do a Hands Up, Stand UP, Pair up to discuss sketches. Students should be instructed to look for differences in displays and decide if the differences change the way the data looks and/or if the data is still represented correctly (MP3)
Next, review the job of an actuary. Explain to students that actuaries can use histograms to compare ranges of data – ex., about populations – and graph the mean.
Review the definition of mean and explain that in this activity the mean represents the single amount of money each person filing a claim would have to pay so that the total amount would cover the costs of all claims, regardless of individual claim amounts. For example, if 5 different people have the following claim amounts: \$10, \$15, \$25, \$30, \$100 (have the students calculate the mean) Ask the students what this number tells us? (looking for students to say that \$36 would be the average amount people would have to pay so that all costs would be covered)
Distribute the Histograms Manage a Flood of Data worksheet. Allow students time to look over the sheet without working. Specifically, draw their attention to the table and ask them what information is being displayed in the table. Then, ask students what they are being asked to do? (create a histogram)
What do they notice about the histogram that they may not have seen before? (there is a break in data between 10,000 and 100,000)
Will this graph be misleading because of the break in data? (depends on how the break is shown on the graph)
Give students time to complete the activity worksheet. When they are done, have students partner up to share their responses. I always make sure students know that when sharing response and there isn’t a similar answer, they must go over through their thinking with their partner to decide if their response is accurate or not.
## Closure
15 minutes
Use the questions from the activity sheet to close up this lesson. The students have already partnered up to get feedback from their peers. So to use this as whole group discussion will be interesting to hear their responses. | 715 | 3,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-51 | longest | en | 0.919437 |
https://www.competoid.com/error/14-0-15474/Question_answers/ | 1,553,583,006,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204857.82/warc/CC-MAIN-20190326054828-20190326080828-00264.warc.gz | 715,566,788 | 9,829 | $$\Large 1\frac{3}{5} + 1\frac{8}{9} + 2\frac{4}{5}$$ = ?
A) $$\Large 6\frac{19}{45}$$ B) $$\Large 6\frac{16}{45}$$ C) $$\Large 6\frac{17}{45}$$ D) $$\Large 6\frac{13}{45}$$
D) $$\Large 6\frac{13}{45}$$
Description for Correct answer:
$$\Large 1\frac{3}{5}+1\frac{8}{9}+2\frac{4}{5} = \left(1+1+2\right) + \left(\frac{3}{5}+\frac{8}{9}+\frac{4}{5}\right)$$
$$\Large 4 + \frac{27+40+36}{45}$$
$$\Large 4 + \frac{103}{45} = 4 + 2\frac{13}{45} = 6\frac{13}{45}$$
Please provide the error details in above question
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3
Posted by8 years ago
Archived
## 2+2=4... prove me wrong.
57% Upvoted
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Comment deleted8 years ago(12 children)
[deleted]
2 points · 8 years ago
Could you explain? I've heard this quite a few times, but never really understood.
6 points · 8 years ago
2.3 is about 2, right? But 2.3 + 2.3 = 4.6, which is about 5.
[deleted]
1 point · 8 years ago
Danke.
I'm not a mathie :P
2 points · 8 years ago
It's not so much a math thing as an engineering thing; it's a good demonstration of why too much rounding is bad.
[deleted]
1 point · 8 years ago
It's also a good demonstration on how everything in the real world is rounded.
[deleted]
1 point · 8 years ago
it's a good demonstration of why too much rounding is bad.
ugh, should've told me that before I ate that Big Mac.
[deleted]
1 point · 8 years ago
[deleted]
2 points · 8 years ago
Gotcha. Thanks!
[deleted]
1 point · 8 years ago
Oh! So it's not a joke. I wasted a laugh point.
[deleted]
1 point · 8 years ago
apparantly it is a joke. I'm just not smart enough to get it
1 point · 8 years ago
for large enough values of 2
He didn't indicate that though. Wouldn't "2+2=4" be logically 2.0 + 2.0 = 4.0? By reading "2+2=4" I wouldn't think 2.41 + 2.34 or 2.13565 + 2.4912234.
1 point · 8 years ago
Depends on context - anyone that's involved in measurement, from a carpenter to an engineer to a physicist, should look at any quoted number and immediately assume that was a rounded value with implied error margins ... their next thought would be to question the source and the accuracy; a number like 2.13565 often has "drama digits" (more digits quoted than the measurement can support or justify).
[deleted]
10 points · 8 years ago
2+2 =11
in base 3
2 points · 8 years ago
I was about to post this, but then I ctrl-f'd and saw you had done the same.
An upvote for you, good sir
8 points · 8 years ago
Big brother says that you're wrong:
2 + 2 = 5.
That is all the proof you need.
8 points · 8 years ago
If two drops fall from a faucet, and then 2 more in the same spot, you have 1 larger drop. 2+2=1.
1 point · 8 years ago
If one googol drops fall from a faucet, then you'll end up with one motherfucking huge drop.
∴ 10100=1
3 points · 8 years ago
Over which field?
[deleted]
3 points · 8 years ago
2 goats + 2 wolves = 2 wolves.
6 points · 8 years ago
2 points · 8 years ago
This actually demonstrates when the rounding issue occurs IRL. Thank you sir, and fuck you Excel.
2 points · 8 years ago
If you want an answer to this kind of question you have to ask it in a certain way. What you have posed is meaningless dribble - as is this response unless specified correctly
Original Poster0 points · 8 years ago
OK, here's the context. I'm an idiot when it comes to philosophy, but I posed a question to the philosophy subreddit about truth, and how you can never say for certain that anything is true. well, predictably, one guy declared that 2+2=4 is true. My initial response was to tell him that if he thinks he's so sure, he could declare it true on Reddit, and see how wrong he was (Reddit will argue against anything you post. But we all know that.)
Instead, I decided to demonstrate it, then share the link with him.
3 points · 8 years ago
If you have 2 apples, and I give you two more, you'll have 4 apples, until you do something with them (like eat one, or give one/some away).
Always. Fact. Truth.
Owned.
[deleted]
1 point · 8 years ago
Math does contain many certain truths. The only arguments against 2+2=4 you are going to see will be a result of you not specifically defining the problem.
1 point · 8 years ago
Fair point and good luck! I would say that within the realm of my human experience i can develop a concept that to me is something i will call true and I make that thing 2+2=4
1 point · 8 years ago
Now that you have added some context to your original post, I would like to toss in my 2 cents.
I don't like it when someone makes a declarative statement such as "2+2=4 is true" because that invites other people to ask "how do you know it is true?" and the discussion begins to deteriorate.
I prefer saying "2+2=4 is an axiom". This helps avoid pointless circular arguments. Axioms are things which are taken as self-evidently true, and used as a basis upon which to build more complex theories.
So if someone asks me "how do you know 2+2=4 is true?", I reply "We do not need to debate the truth of this statement because it is an axiom."
1 point · 8 years ago
X,X + X,X = X,X,X,X Nope, everything checks out, you're good to go.
1 point · 8 years ago
Inflation.
1 point · 8 years ago
ellipses imply that there is an omission.
2+2=4 just 4, not 4-point-anything.
1 point · 8 years ago
Entropy
1 point · 8 years ago
2 + 2 = (1 + 1)(1 +1)
1 point · 8 years ago
2 + 2 = 4 can also be wrong if the values attributed to the "symbols" are changed. What if, to someone, "4" actually meant five? In Chinese languages, for example, they use lines to represent numerical values.
1 point · 8 years ago
2 cups of rice + 2 Cups of water = About 3 cups of rice when finished. The cooking part might not count so how about:
2 Chicken Nuggets + 2 Dipping Sauce Packets = 1 Happy Me
2 witnesses to a crime + 2 hitmen = 2 men just minding their own business, nothing to see here just move along.
2 bottles of Cheap Wine + 2 good reasons to be upset = 1 awful headache.
1 point · 8 years ago
``````declare "2"=100;
x="2"+"2";
print x;
end;
``````
200
1 point · 8 years ago
According to The Who, one plus doesn't equal two, one plus one is one.
There is no two.
1 point · 8 years ago
Numbers are just human constructions that allows us to quantify something. Say I was from another planet and in my math 2 + 2 = 5. This would be true if my number system was ...1, 2, 3, 5, 6, 7, 4, 9....
Thus your proof would be wrong.
1 point · 8 years ago
2+2=4... end of story. unless, you're mentally ill...
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Comments in [Serious] posts must be on topic | 1,861 | 6,578 | {"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-2018-30 | latest | en | 0.964198 |
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# Multiple Choice
This activity contains 9 questions.
## The value of the determinant is
db - ca. da - cb. dc + ba. db + ca. dc - ba.
## The correct determinant value for the determinant would be
(7)(5) - (8)(2). (7)(2) + (5)(8). (7)(8) - (5)(2). (7)(2)(5)(8). (7)(2) - (5)(8).
## The addition of the matrices gives a
3 x 3 matrix. 3 x 2 matrix. 2 x 3 matrix. 2 x 2 matrix. none of the above
## The multiplication of the matrices gives a
3 x1 matrix. 1 x 3 matrix. 1 x 1 matrix. 3 x 3 matrix. none of the above
## The multiplication of the matrices gives a
3 x 1 matrix. 1 x 3 matrix. 1 x 1 matrix. 3 x 3 matrix. none of the above
## Which of the following matrix multiplications would not be possible?
a 5×6 matrix with a 6×3 a 4×3 matrix with a 3×3 a 2×2 matrix with a 3×3 a 5×6 matrix with a 3×3 matrix none of the above
## When you multiply a matrix by the identity matrix, you obtain the
inverse matrix. the transpose matrix. adjoint matrix. cofactor matrix. original matrix.
## Multiply matrix
none of the above
## The inverse of the matrix is
undefined. none of the above | 373 | 1,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2019-22 | latest | en | 0.668244 |
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• Author: Solid
• Date: 28-01-2016, 21:56
28-01-2016, 21:56
### Excel 2016: Advanced Formulas and Functions
Category: Tutorials / Other
Excel 2016: Advanced Formulas and Functions
5h 59m | Video: h264, yuv420p, 1280x720 | Audio: aac, 44100 Hz, 2 ch | 993 MB
| Genre: eLearning | Language: English
Excel expert Dennis Taylor demystifies some of the most useful of the 450+ formulas and functions in Excel and shows how to put them to their best use. Dennis starts with a review of the more basic functions (SUM, AVERAGE, and MAX), and a few critical keyboard shortcuts that will let you locate and display formula cells and accelerate working with Excel formulas-even on multiple sheets. He then covers how to find and retrieve data with the VLOOKUP and INDEX functions, calculate totals with counting and statistical functions, extract data with text functions, and work with date, time, array, math and information functions. The course focuses on practical examples that will help viewers easily transition to using Excel's most powerful formulas and functions in real-world scenarios.
Topics include:
*Displaying and highlighting formulas
*Converting formulas to values
*Tabulating data from multiple sheets
*Understanding the hierarchy of operations in formulas
*Using absolute and relative references
*Creating and expanding nested IF statements
*Looking up information with VLOOKUP, MATCH, and INDEX
*Using the powerful COUNTIF family of functions
*Analyzing data with statistical functions
*Calculating dates and times
*Analyzing data with array formulas and functions
*Extracting data with text function
check my other posts
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Dear visitor, you went to the site as unregistered user. | 437 | 1,885 | {"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-2016-44 | longest | en | 0.809502 |
https://cstheory.stackexchange.com/questions/18437/intersection-between-sets/18450 | 1,560,760,348,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998462.80/warc/CC-MAIN-20190617083027-20190617105027-00169.warc.gz | 401,842,188 | 36,695 | # Intersection between sets
Assume that we have $p$ sets, with given sizes: $m_1,m_2,...,m_p$. The (distinct) elements in each set are taken from $N$ elements (where $m_1,m_2,...,m_p \le N$).
A combination is defined as an assignment of distinct elements (from $N$ possible ones) to each of the sets. For example, say that we have $p=2$ sets of sizes $m_1=2,m_2=3$ and that $N=4$. Then one possible combination is $\left\{ {1,2} \right\},\left\{ {1,2,3} \right\}$, another one is $\left\{ {2,3} \right\},\left\{ {2,3,4} \right\}$ and so on.
It is easy to note that there are $\prod\limits_{i = 1}^p {\left( {\begin{array}{*{20}{c}} N\\ {{m_i}} \end{array}} \right)}$ possible combinations.
My question is - given the sizes $m_1,m_2,...,m_p$, how many combinations have intersection of exact size $k=1,2,...,{\text {min}}(m_i)$ (denote this by $I(k)$)?
My thought was to find first how many combinations have an intersection of size at least $k=1,2,...,{\text {min}}(m_i)$ (denote this by $v(k)$), and then $I(k) = v(k) - v(k+1)$.
So far I know that for $k_{min}={\text {min}}(m_i)$:
$$v\left( {{k_{\min }}} \right) = \left( {\begin{array}{*{20}{c}} N \\ {{k_{\min }}} \end{array}} \right)\prod\limits_{i = 1}^p {\left( {\begin{array}{*{20}{c}} {N - {k_{\min }}} \\ {{m_i} - {k_{\min }}} \end{array}} \right)}$$ (and $I(k_{min})=v(k_{min})$).
However, the formula above for $k<k_{min}$ gives an over estimate.
Any help would be greatly appreciated.
• copy on Mathematics. Please link in both directions when cross-posting. – Kaveh Jul 24 '13 at 20:15
• Now deleted at math.SE. – András Salamon Sep 23 '13 at 14:19
I think you can finish off your computation using inclusion-exclusion.
If we fix a set $S$ of $k$ elements, let's count the number of ways to choose the $p$ sets so that they all contain $S$. Call this $n(k)$ [note that this number depends only on $k$ but not otherwise on $S$, justifying my notation]. You correctly showed how to compute $n(k)$, namely, $$n(k) = \prod_{i=1}^p {N-k \choose m_i-k}.$$
Now you should be able to use inclusion-exclusion to count the number of ways to choose the $p$ sets so that their intersection is of size exactly $k$.
For example, if $m=\min(m_1,\dots,m_p)$, then the number of ways to choose the $p$ sets so that their intersection is of size exactly $m$ is $$I(m) = {N \choose m} n(m)$$ [as you correctly noted in your question].
As another example, the number of ways to choose the $p$ sets so that their intersection is of size exactly $m-1$ is $$I(m-1) = {N \choose m-1} n(m-1) - m {N \choose m} n(m).$$ Here we've counted each set $S$ of size $m-1$ once; but then that overcounts things, because each set of size $m$ got counted $m$ times (once for each of its subsets of size $m-1$), so we account for the overcounting by subtracting off $m$ times the number of ways of getting an intersection of size $m$.
The number of ways to choose them so their intersection has size exactly $m-2$ is $$I(m-2) = {N \choose m-2} n(m-2) - (m-1) {N \choose m-1} n(m-1) + {m \choose 2} {N \choose m} n(m).$$ Here we've accounted for the overcounting by subtracting something related to sets of size $m-1$; but then that over-corrects and now undercounts some sets of size $m$, so we correct for that by adding an appropriate term. It's easier to re-discover the formula than to explain it, but hopefully you get the idea.
You should be able to take it from here.
• Many thanks for your very helpful answer. Can you explain the third term in the expression for $I(m-2)$? I don't see why the first factor should be $\binom{m-1}{2}$. – user17047 Jul 24 '13 at 10:14
• If I make it correct, this factor should be: $-\binom{m}{m-2}+(m-1)\binom{m}{m-1} = \binom{m}{m-2}$. – user17047 Jul 24 '13 at 10:47
• @Rami, oops, you are right, that's an error. I've edited the answer to fix that equation. I think it is correct now. (How I got it: consider a set $S$ of size $m$. How many times has it been counted by the first two terms? It is counted positively $m(m-1)/2$ times by the first term, since it has $m(m-1)/2$ subsets of size $m-2$. Then, since there are $m$ subsets of size $m-1$ and each one gets counted negatively $m-1$ times by the second term, the second term has counted it negatively $m(m-1)$ times. Now $m(m-1)/2 - m(m-1) = -m(m-1)/2$, so we need to correct by adding it $m(m-1)/2$ times.) – D.W. Jul 24 '13 at 16:03
• Can you give some more details on the way you choose the first terms for each $I(m-k)$ (this choice leads to the needed corrections)?. I tried to follow your logic for $m-3$ and afterwards, but I don't get the correct expressions. – user17047 Jul 24 '13 at 19:57
• @Rami, this site is for research-level questions in theoretical computer science; I'm not seeing how this is research-level. This seems like something that should fall out from standard undergraduate techniques for counting. Am I missing something deep? Perhaps you might want to migrate it to Computer Science? – D.W. Jul 24 '13 at 20:02
Hint: Another way of seeing the problem is in terms of integer binary boolean operations:
The intersection of a set B indexed by b (the binary representation of the set) and a set B' indexed by b' is given by:
bb'
where the operation is bit-by-bit multiplication.
The number of elements in the intersection is then:
\sum \sum bb' = k
For b and b' running from 1 to N. Your equation is then:
\sum \sum bb' <= k The solution is an iterated map... | 1,684 | 5,439 | {"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} | 4.21875 | 4 | CC-MAIN-2019-26 | longest | en | 0.824152 |
http://un-petrol.ru/search.php?q=cofunction%20sin%2845%29 | 1,521,400,497,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645943.23/warc/CC-MAIN-20180318184945-20180318204945-00769.warc.gz | 304,679,259 | 4,946 | # cofunctionsin(45)
Enter cofunction statement below:
Setup cofunction identity in degrees for sin(θ)
sin(θ) = cos(90 - θ)
Plug in θ = 45
sin(45) = cos(90 - 45)
sin(45) = cos(45) | 64 | 181 | {"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-2018-13 | latest | en | 0.479938 |
http://gmatclub.com/forum/there-are-5-pairs-of-socks-2-blue-2-white-2-black-2-yell-94882.html | 1,484,819,200,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280504.74/warc/CC-MAIN-20170116095120-00260-ip-10-171-10-70.ec2.internal.warc.gz | 117,821,964 | 60,908 | There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell : GMAT Problem Solving (PS)
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# There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell
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There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink]
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27 May 2010, 02:41
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There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yellow, 2 green. You select randomly four socks together. What is the probability that you'll get at least two of the same color?
I think it's
[Reveal] Spoiler:
13/21
The book says it's
[Reveal] Spoiler:
41/42
which is in my opinion a mistake.
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Re: Probability question: 10 socks [#permalink]
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27 May 2010, 03:59
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nonameee wrote:
There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yellow, 2 green. You select randomly four socks together. What is the probability that you'll get at least two of the same color?
I think it's
[Reveal] Spoiler:
13/21
The book says it's
[Reveal] Spoiler:
41/42
which is in my opinion a mistake.
I think you are right.
"Probability that you'll get at least two of the same color" - means at least one pair (one pair or two pairs out of 4 socks).
P(at least one pair)=1-P(no pair) --> no pair means all socks must be of different colors --> $$P=1-\frac{C^4_5*2^4}{C^4_{10}}=\frac{13}{21}$$.
$$C^4_5$$ - # of ways to choose 4 different colors out of 5, basically # of ways to choose which 4 color socks will give us one sock (in this case we obviously will have 4 different colors);
$$2^4$$ - each of 4 chosen colors can give left or right sock;
$$C^4_{10}$$ - total # of ways to choose 4 socks out of 10.
What solution was given in the book for 41/42?
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Re: Probability question: 10 socks [#permalink]
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27 May 2010, 04:02
Quote:
What solution was given in the book for 41/42?
I think they've forgotten to multiply by 2^4. But the solution was the same you gave.
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Re: Probability question: 10 socks [#permalink]
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28 May 2010, 20:26
Thanks for the explanation.
Bunuel wrote:
nonameee wrote:
There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yellow, 2 green. You select randomly four socks together. What is the probability that you'll get at least two of the same color?
I think it's
[Reveal] Spoiler:
13/21
The book says it's
[Reveal] Spoiler:
41/42
which is in my opinion a mistake.
I think you are right.
"Probability that you'll get at least two of the same color" - means at least one pair (one pair or two pairs out of 4 socks).
P(at least one pair)=1-P(no pair) --> no pair means all socks must be of different colors --> $$P=1-\frac{C^4_5*2^4}{C^4_10}=\frac{13}{21}$$.
$$C^4_5$$ - # of ways to choose 4 different colors out of 5, basically # of ways to choose which 4 color socks will give us one sock (in this case we obviously will have 4 different colors);
$$2^4$$ - each of 4 chosen colors can give left or right sock;
$$C^4_{10}$$ - total # of ways to choose 4 socks out of 10.
What solution was given in the book for 41/42?
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Re: Probability question: 10 socks [#permalink]
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28 May 2010, 21:07
Bunuel wrote:
nonameee wrote:
There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yellow, 2 green. You select randomly four socks together. What is the probability that you'll get at least two of the same color?
I think it's
[Reveal] Spoiler:
13/21
The book says it's
[Reveal] Spoiler:
41/42
which is in my opinion a mistake.
I think you are right.
"Probability that you'll get at least two of the same color" - means at least one pair (one pair or two pairs out of 4 socks).
P(at least one pair)=1-P(no pair) --> no pair means all socks must be of different colors --> $$P=1-\frac{C^4_5*2^4}{C^4_10}=\frac{13}{21}$$.
$$C^4_5$$ - # of ways to choose 4 different colors out of 5, basically # of ways to choose which 4 color socks will give us one sock (in this case we obviously will have 4 different colors);
$$2^4$$ - each of 4 chosen colors can give left or right sock;
$$C^4_{10}$$ - total # of ways to choose 4 socks out of 10.
What solution was given in the book for 41/42?
wait a sec i dont get it. Why isnt 5 choose 4 $$C^5_4$$ ?
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Re: Probability question: 10 socks [#permalink]
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29 May 2010, 02:31
Detran wrote:
wait a sec i dont get it. Why isnt 5 choose 4 $$C^5_4$$ ?
It's just another way of writing it $$C^5_4$$, $$C^4_5$$, $$5C4$$.
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Re: Probability question: 10 socks [#permalink]
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06 Mar 2014, 10:37
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Bunuel wrote:
nonameee wrote:
There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yellow, 2 green. You select randomly four socks together. What is the probability that you'll get at least two of the same color?
I think it's
[Reveal] Spoiler:
13/21
The book says it's
[Reveal] Spoiler:
41/42
which is in my opinion a mistake.
I think you are right.
"Probability that you'll get at least two of the same color" - means at least one pair (one pair or two pairs out of 4 socks).
P(at least one pair)=1-P(no pair) --> no pair means all socks must be of different colors --> $$P=1-\frac{C^4_5*2^4}{C^4_10}=\frac{13}{21}$$.
$$C^4_5$$ - # of ways to choose 4 different colors out of 5, basically # of ways to choose which 4 color socks will give us one sock (in this case we obviously will have 4 different colors);
$$2^4$$ - each of 4 chosen colors can give left or right sock;
$$C^4_{10}$$ - total # of ways to choose 4 socks out of 10.
What solution was given in the book for 41/42?
May i knw the logic behind this? m not getting the numerator part ..
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Re: Probability question: 10 socks [#permalink]
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07 Mar 2014, 00:21
sanjoo wrote:
Bunuel wrote:
nonameee wrote:
There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yellow, 2 green. You select randomly four socks together. What is the probability that you'll get at least two of the same color?
I think it's
[Reveal] Spoiler:
13/21
The book says it's
[Reveal] Spoiler:
41/42
which is in my opinion a mistake.
I think you are right.
"Probability that you'll get at least two of the same color" - means at least one pair (one pair or two pairs out of 4 socks).
P(at least one pair)=1-P(no pair) --> no pair means all socks must be of different colors --> $$P=1-\frac{C^4_5*2^4}{C^4_10}=\frac{13}{21}$$.
$$C^4_5$$ - # of ways to choose 4 different colors out of 5, basically # of ways to choose which 4 color socks will give us one sock (in this case we obviously will have 4 different colors);
$$2^4$$ - each of 4 chosen colors can give left or right sock;
$$C^4_{10}$$ - total # of ways to choose 4 socks out of 10.
What solution was given in the book for 41/42?
May i knw the logic behind this? m not getting the numerator part ..
If 4 out of 5 pairs of socks will give one sock, we won't have a matching pair, we'll have 4 different color socks and this is exactly what we are counting in the numerator.
Hope it's clear.
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Re: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink]
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08 Mar 2014, 04:02
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I solved this one be listing method
Prob of at least getting a pair = 1 -prob(all different at four attempts) = 1-( 1 x 8/9 x 6/8 x 4/7) = 1-8/21= 13/21
a) 1 st attempt = getting any color =1
b) 2nd attempt = not getting the color picked in a). = 8/9
c) 3rd attempt = not getting the two colors above = 6/8
d) not getting any of the four colors above = 4/7
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Re: Probability question: 10 socks [#permalink]
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12 Mar 2014, 10:52
Dear Bunuel
I did 10c1X1c1 X 8c1X6c1/ 10c4 + 10c1X1c1 X 8c1X1c1/10c4.
Thanks & regards
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Re: Probability question: 10 socks [#permalink]
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03 May 2014, 01:00
Why we took 2^4 ?
what if we want to solve it by straight forward method ?
that is considering P ( getting two socks atleast of same color ) ?
Bunuel wrote:
nonameee wrote:
There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yellow, 2 green. You select randomly four socks together. What is the probability that you'll get at least two of the same color?
I think it's
[Reveal] Spoiler:
13/21
The book says it's
[Reveal] Spoiler:
41/42
which is in my opinion a mistake.
I think you are right.
"Probability that you'll get at least two of the same color" - means at least one pair (one pair or two pairs out of 4 socks).
P(at least one pair)=1-P(no pair) --> no pair means all socks must be of different colors --> $$P=1-\frac{C^4_5*2^4}{C^4_{10}}=\frac{13}{21}$$.
$$C^4_5$$ - # of ways to choose 4 different colors out of 5, basically # of ways to choose which 4 color socks will give us one sock (in this case we obviously will have 4 different colors);
$$2^4$$ - each of 4 chosen colors can give left or right sock;
$$C^4_{10}$$ - total # of ways to choose 4 socks out of 10.
What solution was given in the book for 41/42?
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Posts: 36549
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Re: Probability question: 10 socks [#permalink]
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03 May 2014, 03:37
Expert's post
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sunny3011 wrote:
Why we took 2^4 ?
what if we want to solve it by straight forward method ?
that is considering P ( getting two socks atleast of same color ) ?
Bunuel wrote:
nonameee wrote:
There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yellow, 2 green. You select randomly four socks together. What is the probability that you'll get at least two of the same color?
I think it's
[Reveal] Spoiler:
13/21
The book says it's
[Reveal] Spoiler:
41/42
which is in my opinion a mistake.
I think you are right.
"Probability that you'll get at least two of the same color" - means at least one pair (one pair or two pairs out of 4 socks).
P(at least one pair)=1-P(no pair) --> no pair means all socks must be of different colors --> $$P=1-\frac{C^4_5*2^4}{C^4_{10}}=\frac{13}{21}$$.
$$C^4_5$$ - # of ways to choose 4 different colors out of 5, basically # of ways to choose which 4 color socks will give us one sock (in this case we obviously will have 4 different colors);
$$2^4$$ - each of 4 chosen colors can give left or right sock;
$$C^4_{10}$$ - total # of ways to choose 4 socks out of 10.
What solution was given in the book for 41/42?
Can you please tell me which part of below explanation is not clear?
$$C^4_5$$ - # of ways to choose 4 different colors out of 5, basically # of ways to choose which 4 color socks will give us one sock (in this case we obviously will have 4 different colors);
$$2^4$$ - each of 4 chosen colors can give left or right sock.
As for direct approach: it would be lengthier way to get the answer. You should count two cases:
A. one par of matching socks with two socks of different color;
B. two pairs of matching socks.
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Re: There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink]
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24 Jul 2014, 17:07
Total Number of Possible Combination= 10C4= 210
Total number of Combination where socks is not of same colour= 5C4= 5
Probability of Same= 1- 5/210= 41/42
PS: In socks there is nothing as left & right. So the factor of 2^4 is not required.
Hope it is clear & book is right!
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There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink]
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25 Jul 2014, 02:16
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Expert's post
Total Number of Possible Combination= 10C4= 210
Total number of Combination where socks is not of same colour= 5C4= 5
Probability of Same= 1- 5/210= 41/42
PS: In socks there is nothing as left & right. So the factor of 2^4 is not required.
Hope it is clear & book is right!
The correct answer is 13/21, not 41/42. The point is that each out of 4 color socks you are selecting can give either first or second sock. Please read the whole discussion above.
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There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell [#permalink] 25 Jul 2014, 02:16
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# There are 5 pairs of socks: 2 blue, 2 white, 2 black, 2 yell
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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®. | 4,726 | 15,093 | {"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": 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} | 4.1875 | 4 | CC-MAIN-2017-04 | latest | en | 0.883099 |
http://at.metamath.org/ileuni/elex2.html | 1,718,632,341,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861719.30/warc/CC-MAIN-20240617122241-20240617152241-00436.warc.gz | 2,418,841 | 3,644 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > elex2 GIF version
Theorem elex2 2570
Description: If a class contains another class, then it contains some set. (Contributed by Alan Sare, 25-Sep-2011.)
Assertion
Ref Expression
elex2 (𝐴𝐵 → ∃𝑥 𝑥𝐵)
Distinct variable groups: 𝑥,𝐴 𝑥,𝐵
Proof of Theorem elex2
StepHypRef Expression
1 eleq1a 2109 . . 3 (𝐴𝐵 → (𝑥 = 𝐴𝑥𝐵))
21alrimiv 1754 . 2 (𝐴𝐵 → ∀𝑥(𝑥 = 𝐴𝑥𝐵))
3 elisset 2568 . 2 (𝐴𝐵 → ∃𝑥 𝑥 = 𝐴)
4 exim 1490 . 2 (∀𝑥(𝑥 = 𝐴𝑥𝐵) → (∃𝑥 𝑥 = 𝐴 → ∃𝑥 𝑥𝐵))
52, 3, 4sylc 56 1 (𝐴𝐵 → ∃𝑥 𝑥𝐵)
Colors of variables: wff set class Syntax hints: → wi 4 ∀wal 1241 = wceq 1243 ∃wex 1381 ∈ wcel 1393 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 99 ax-ia2 100 ax-ia3 101 ax-5 1336 ax-gen 1338 ax-ie1 1382 ax-ie2 1383 ax-8 1395 ax-4 1400 ax-17 1419 ax-i9 1423 ax-ial 1427 ax-ext 2022 This theorem depends on definitions: df-bi 110 df-sb 1646 df-clab 2027 df-cleq 2033 df-clel 2036 df-v 2559 This theorem is referenced by: snmg 3486 oprcl 3573 exss 3963 onintrab2im 4244 regexmidlemm 4257 acexmidlem2 5509 enm 6294 ssfiexmid 6336 fin0 6342 fin0or 6343 diffitest 6344 diffisn 6350 caucvgsrlemasr 6874 gtso 7097 indstr 8536 negm 8550 fzm 8902 fzom 9020 r19.2uz 9591 resqrexlemgt0 9618 climuni 9814
Copyright terms: Public domain W3C validator | 785 | 1,420 | {"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-2024-26 | latest | en | 0.173776 |
http://planning.cs.uiuc.edu/node656.html | 1,513,449,381,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948588420.68/warc/CC-MAIN-20171216181940-20171216203940-00299.warc.gz | 222,142,538 | 5,730 | ### 13.1.1.3 Conversion from implicit to parametric form
There are trade-offs between the implicit and parametric ways to express differential constraints. The implicit representation is more general; however, the parametric form is more useful because it explicitly gives the possible actions. For this reason, it is often desirable to derive a parametric representation from an implicit one. Under very general conditions, it is theoretically possible. As will be explained shortly, this is a result of the implicit function theorem. Unfortunately, the theoretical existence of such a conversion does not help in actually performing the transformations. In many cases, it may not be practical to determine a parametric representation.
To model a mechanical system, it is simplest to express constraints in the implicit form and then derive the parametric representation . So far there has been no appearance of in the implicit representation. Since is interpreted as an action, it needs to be specified while deriving the parametric representation. To understand the issues, it is helpful to first assume that all constraints in implicit form are linear equations in of the form
(13.5)
which are called Pfaffian constraints. These constraints are linear only under the assumption that is known. It is helpful in the current discussion to imagine that is fixed at some known value, which means that each of the coefficients in (13.5) is a constant.
Suppose that Pfaffian constraints are given for and that they are linearly independent.13.1 Recall the standard techniques for solving linear equations. If , then a unique solution exists. If , then a continuum of solutions exists, which forms an -dimensional hyperplane. It is impossible to have because there can be no more than linearly independent equations.
If , only one velocity vector satisfies the constraints for each . A vector field can therefore be derived from the constraints, and the problem is not interesting from a planning perspective because there is no choice of velocities. If , then components of can be chosen independently, and then the remaining are computed to satisfy the Pfaffian constraints (this can be accomplished using linear algebra techniques such as singular value decomposition [399,961]). The components of that can be chosen independently can be considered as scalar actions. Together these form an -dimensional action vector, . Suppose without loss of generality that the first components of are specified by . The configuration transition equation can then be written as
(13.6)
in which each is a linear function of and is derived from the Pfaffian constraints after substituting for for each from to and then solving for the remaining components of . For some values of , the constraints may become linearly dependent. This only weakens the constraints, which means the dimension of can be increased at any for which independence is lost. Such points are usually isolated and will not be considered further.
Example 13..2 (Pfaffian Constraints) Suppose that , and there is one constraint of the form (13.5)
(13.7)
For this problem, and . There are two action variables because . The configuration transition equation is
(13.8)
in which the last component was obtained by substituting and , respectively, for and in (13.7) and then solving for .
The constraint given in (13.7) does not even depend on . The same ideas apply for more general Pfaffian constraints, such as
(13.9)
Following the same procedure, the configuration transition equation becomes
(13.10)
The ideas presented so far naturally extend to equality constraints that are not linear in . At each , an -dimensional set of actions, , is guaranteed to exist if the Jacobian (recall (6.28) or see [508]) of the constraint functions has rank at . This follows from the implicit function theorem [508].
Suppose that there are inequality constraints of the form , in addition to equality constraints. Using the previous concepts, the actions may once again be assigned directly as for all such that . Without inequality constraints, there are no constraints on , which means that . Since is interpreted as an input to some physical system, will often be constrained. In a physical system, for example, the amount of energy consumed may be proportional to . After performing the substitutions, the inequality constraints indicate limits on . These limits are expressed in terms of and the remaining components of , which are the variables , , . For many problems, the inequality constraints are simple enough that constraints directly on can be derived. For example, if represents scalar acceleration applied to a car, then it may have a simple bound such as .
One final complication that sometimes occurs is that the action variables may already be specified in the equality constraints: . In this case, imagine once again that is fixed. If there are independent constraints, then by the implicit function theorem, can be solved to yield (although theoretically possible, it may be difficult in practice). If the Jacobian has rank at , then actions can be applied to yield any velocity on a -dimensional hyperplane in . If , then there are enough independent action variables to overcome the constraints. Any velocity in can be achieved through a choice of . This is true only if there are no inequality constraints on .
Steven M LaValle 2012-04-20 | 1,066 | 5,430 | {"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-2017-51 | latest | en | 0.933073 |
http://www.chestnut.com/en/eg/5/mathematics/7/3/examples/11/ | 1,477,093,326,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988718311.12/warc/CC-MAIN-20161020183838-00000-ip-10-171-6-4.ec2.internal.warc.gz | 371,902,172 | 37,904 | # 5.7.3. Area of a Square in Terms of Its Diagonal Length
Given that is a rectangle, is a square, cm, and cm, calculate the area of the shaded part.
• A
• B
• C
• D
### Example
Given that is a rectangle, is a square, cm, and cm, calculate the area of the shaded part.
### Solution
Step 1: Find the area of the rectangle.
Step 2: Find the area of the square.
Use the formula for the area of a square using its diagonal. The length of the square's diagonal is the same length as ().
Step 3: To find the area of the shaded part, subtract the area of the square from the area of the rectangle.
0
correct
0
incorrect
0
skipped | 172 | 631 | {"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} | 3.609375 | 4 | CC-MAIN-2016-44 | latest | en | 0.874777 |
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USING THE SHARP EL 738 FINANCIAL CALCULATOR. Basic financial examples with financial calculator steps. Prepared by Colin C Smith 2010.
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THE SHARP EL-738 CALCULATOR is recommended for this module. THE advantage of this cal- culator is that it can do basic calculations, FINANCIAL calculations and
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Chapter 1 Notes on THE SHARP EL-738 CALCULATOR General THE SHARP EL-738 CALCULATOR is recommended for this module. THE advantage of this cal-culator is that …
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## Graphing Calculator El 9450 9400 Sharp Corporation
Key Operation Display Notes Example 6 When THE graph of y = x2 is changed, how does it affect THE equation? Change Graph change function helps students grasp THE ... | 784 | 3,229 | {"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-2019-43 | longest | en | 0.617089 |
https://invent.kde.org/multimedia/juk/-/raw/0660beadd2b530dca82ba5b586ebd6b690bc9ad8/stringhash.h | 1,590,526,922,000,000,000 | text/plain | crawl-data/CC-MAIN-2020-24/segments/1590347391309.4/warc/CC-MAIN-20200526191453-20200526221453-00039.warc.gz | 390,306,890 | 2,842 | /*************************************************************************** begin : Sun Feb 2 2003 copyright : (C) 2003 - 2004 by Scott Wheeler email : wheeler@kde.org ***************************************************************************/ /*************************************************************************** * * * This program is free software; you can redistribute it and/or modify * * it under the terms of the GNU General Public License as published by * * the Free Software Foundation; either version 2 of the License, or * * (at your option) any later version. * * * ***************************************************************************/ #ifndef STRINGHASH_H #define STRINGHASH_H #include #include "filehandle.h" /** * A simple hash representing an (un-mapped) set of data. */ template class Hash { friend class Iterator; public: Hash() : m_table(m_tableSize) {} ~Hash(); /** * To combine two operations into one (that takes the same amount as each * independantly) this inserts an item and returns true if the item was * already in the set or false if it did not. */ bool insert(T value); /** * Returns true if the set contains the item \a value. */ bool contains(T value) const; /** * Removes an item. Returns true if the item was present and false if not. */ bool remove(T value); QValueList values() const; int hash(T key) const; static inline int tableSize() { return m_tableSize; } protected: struct Node { Node(T value) : key(value), next(0) {} T key; Node *next; }; public: class Iterator { friend class Hash; public: Iterator(const Hash &hash) : m_hash(hash), m_index(0), m_node(hash.m_table[0]) {} const T &operator*() const { return m_node->key; } T &operator*() { return m_node->key; } bool operator==(const Iterator &it) const { return m_index == it.m_index && m_node == it.m_node; } bool operator!=(const Iterator &it) const { return !(it == *this); } T &operator++(); private: const Hash &m_hash; int m_index; Node *m_node; }; Iterator begin() const { Iterator it(*this); while(!it.m_node && it.m_index < m_tableSize - 1) { it.m_index++; it.m_node = m_table[it.m_index]; } return it; } Iterator end() const { Iterator it(*this); it.m_node = 0; it.m_index = m_tableSize - 1; return it; } protected: void deleteNode(Node *n); QPtrVector m_table; static const int m_tableSize = 5003; }; //////////////////////////////////////////////////////////////////////////////// // helper functions //////////////////////////////////////////////////////////////////////////////// inline char hashStringAccess(const QString &in, int index) { return in.unicode()[index].cell(); } inline char hashStringAccess(const QCString &in, int index) { return in[index]; } // Based on QGDict's hash functions, Copyright (C) 1992-2000 Trolltech AS template inline int hashString(const StringType &s) { uint h = 0; uint g; for(uint i = 0; i < s.length(); i++) { h = (h << 4) + hashStringAccess(s, i); if((g = h & 0xf0000000)) h ^= g >> 24; h &= ~g; } int index = h; if(index < 0) index = -index; return index; } //////////////////////////////////////////////////////////////////////////////// // specializations //////////////////////////////////////////////////////////////////////////////// // StringHash template<> inline int Hash::hash(QString key) const { return hashString(key) % tableSize(); } typedef Hash StringHash; // PtrHash template<> inline int Hash::hash(void *key) const { return long(key) % tableSize(); } typedef Hash PtrHash; // FileHandleHash template<> inline int Hash::hash(FileHandle key) const { return hashString(key.absFilePath()) % tableSize(); } class FileHandleHash : public Hash { public: FileHandleHash() : Hash() {} FileHandle value(const QString &key) const { int h = hashString(key) % tableSize(); Node *i = m_table[h]; while(i && i->key.absFilePath() != key) i = i->next; return i ? i->key : FileHandle::null(); } }; //////////////////////////////////////////////////////////////////////////////// // template method implementations //////////////////////////////////////////////////////////////////////////////// template Hash::~Hash() { for(int i = 0; i < m_tableSize; i++) deleteNode(m_table[i]); } template bool Hash::insert(T value) { int h = hash(value); Node *i = m_table[h]; Node *j = 0; while(i) { if(i->key == value) return true; else { j = i; i = i->next; } } if(j) j->next = new Node(value); else m_table.insert(h, new Node(value)); return false; } template bool Hash::contains(T value) const { int h = hash(value); Node *i = m_table[h]; while(i && i->key != value) i = i->next; return bool(i); } template bool Hash::remove(T value) { int h = hash(value); Node *previous = 0; Node *i = m_table[h]; while(i && i->key != value) { previous = i; i = i->next; } if(!i) return false; if(previous) previous->next = i->next; else { if(i->next) m_table.insert(h, i->next); else m_table.remove(h); } delete i; return true; } template QValueList Hash::values() const { QValueList l; Node *n; for(int i = 0; i < tableSize(); i++) { n = m_table[i]; while(n) { l.append(n->key); n = n->next; } } return l; } template void Hash::deleteNode(Node *n) { if(n) { deleteNode(n->next); delete n; } } template T &Hash::Iterator::operator++() { if(m_node) m_node = m_node->next; while(!m_node && m_index < m_tableSize - 1) { m_index++; m_node = m_hash.m_table[m_index]; } return m_node->key; } #endif | 1,310 | 5,336 | {"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-2020-24 | latest | en | 0.476041 |
http://www.castep.org/CASTEP/FAQSCF | 1,685,645,852,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648000.54/warc/CC-MAIN-20230601175345-20230601205345-00350.warc.gz | 63,542,270 | 4,876 | CASTEP
# SCF convergence questions
### What SCF solver methods are available in CASTEP?
The goal of the SCF solver is to find the ground-state wavefunctions (and density) which are the lowest-energy solutions to the Kohn-Sham equations (the fundamental equations of conventional density functional theory).
CASTEP has two main methods to solve the Kohn-Sham equations: density mixing (DM); and ensemble density functional theory (EDFT).
Density mixing is a fast, but potentially unstable method. Rather than attempting to compute the ground-state wavefunctions and density simultaneously, in the DM method the Kohn-Sham wavefunctions are first computed for a given input density, and then a separate "density mixing" algorithm is used to calculate an estimate of the ground-state density. This estimate is then used as the input density for a new iteration of the method. The algorithm continues until the estimated ground-state density is the same as the input density, which occurs when the ground state has been reached.
DM is CASTEP's default method unless a fixed-occupancy calculation is specified. You can force a calculation to use DM using
`elec_method : dm`
in the `param` file.
Ensemble density functional theory is a stable, but slow method of solving the Kohn-Sham equations. In EDFT the density is always computed directly from the estimated ground-state wavefunctions, and these wavefunctions are only ever updated in a way which brings them closer to the ground state.
You can select EDFT as the SCF method to use by setting
`elec_method : edft`
in the `param` file.
### My SCF failed to converge. What can I do?
Check your input makes physical/chemical sense
It is not unusual to find that underlying an SCF convergence failure is some mistake in specifying the system. Common causes include
• The atomic co-ordination does not make physical sense, for example atoms are too close.
• Specifiying `fix_occupancy=true` when the system is really metallic. This selects the allbands SCF solver by default, which will probably fail to converge.
• Specifying `spin=0` if the system is really magnetic. Even starting a spin-polarised calculation from a zero-spin initial state can cause SCF convergence failure, as the code does not know which way to break the spin symmetry. Atomic spin intialisation using `SPIN=s` in `%BLOCK POSITIONS_CART` will usually solve this.
Density Mixing
The density mixing algorithm used by Castep has several parameters that you can experiment with to improve its convergence. The main ones are:
• `mixing_scheme`
There are two main density mixing schemes, one due to Broyden and one to Pulay. The default is Broyden, but if your system is not converging it is always worth trying Pulay.
• `mix_cut_off_energy`
The density mixing is actually performed in reciprocal space, and only plane-waves whose energy is less than `mix_cut_off_energy` are mixed. If the high-energy components are causing problems, increasing `mix_cut_off_energy` will improve the convergence.
The default value is the plane-wave cut-off energy for the wavefunction and this is a sensible minimum value for the parameter; the maximum sensible value is given by cut_off_energy * (fine_grid_scale)2.
• `mix_charge_amp` (and `mix_spin_amp`)
The density mixing takes a fraction of the proposed new density and the old density, and this fraction is set by `mix_charge_amp`. The possible values of mix_charge_amp range from 0 to 1, where a smaller value can stabilise the algorithm but may lead to slow convergence, and a large value will generally either converge rapidly or fail to converge at all. For systems which are proving difficult to converge, sensible values are generally between 0.05 and 0.2.
If your system is spin-polarised then you may also want to investigate the corresponding parameter for the spin-density mixing, `mix_spin_amp`. This works in the same way as mix_charge_amp except that its values range from 0 to 2.
Ensemble DFT
EDFT is a very robust SCF solver, but much more expensive than density-mixing. But if DM still fails to converge after tweaking the parameters, then EDFT usually will. Select using `ELEC_METHOD=EDFT` | 900 | 4,177 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-23 | latest | en | 0.922659 |
https://inquiryintoinquiry.com/2014/03/25/peirces-1870-logic-of-relatives-%E2%80%A2-comment-10-3/ | 1,632,277,303,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057303.94/warc/CC-MAIN-20210922011746-20210922041746-00110.warc.gz | 353,749,295 | 42,022 | Peirce’s 1870 “Logic Of Relatives” • Comment 10.3
We’ve been using several different styles of picture to illustrate relative terms and the relations they denote. Let us now examine the relationships that exist among the variety of visual schemes. Two examples of relative multiplication that we considered before are diagrammed again in Figures 11 and 12.
(11) (12)
Figures 11 and 12 employ one style of syntax Peirce used for relative multiplication, to which I added lines of identity to connect the corresponding marks of reference. These forms are adapted to showing the anatomy of relative terms themselves, while the forms of analysis in Table 13 and Figure 14 are designed to highlight the structures of the objective relations they denote.
$\text{Table 13.} ~~ \text{Relational Composition}$
$\mathit{1}$ $\mathit{1}$ $\mathit{1}$
$L$ $X$ $Y$
$S$ $Y$ $Z$
$L \circ S$ $X$ $Z$
(14)
There are many ways that Peirce might have gotten from his 1870 Notation for the Logic of Relatives to his more evolved systems of Logical Graphs. It is interesting to speculate on how the metamorphosis might have been accomplished by way of transformations that act on these nascent forms of syntax and that take place not too far from the pale of its means, that is, as nearly as possible according to the rules and the permissions of the initial system itself.
In Existential Graphs, a relation is represented by a node whose degree is the adicity of that relation, and which is adjacent via lines of identity to the nodes that represent its correlative relations, including as a special case any of its terminal individual arguments.
In the 1870 Logic of Relatives, implicit lines of identity are invoked by the subjacent numbers and marks of reference only when a correlate of some relation is the relate of some relation. Thus, the principal relate, which is not a correlate of any explicit relation, is not singled out in this way.
Remarkably enough, the comma modifier itself provides us with a mechanism to abstract the logic of relations from the logic of relatives, and thus to forge a possible link between the syntax of relative terms and the more graphical depiction of the objective relations themselves.
Figure 15 demonstrates this possibility, posing a transitional case between the style of syntax in Figure 11 and the picture of composition in Figure 14.
(15)
In this composite sketch the diagonal extension $\mathit{1}$ of the universe $\mathbf{1}$ is invoked up front to anchor an explicit line of identity for the leading relate of the composition, while the terminal argument $\mathrm{w}$ has been generalized to the whole universe $\mathbf{1}.$ Doing this amounts to an act of abstraction from the particular application to $\mathrm{w}.$ This form of universal bracketing isolates the serial composition of the relations $L$ and $S$ to form the composite $L \circ S.$
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# Let A and B be events such that P(A ^ B) = 1/4, P(
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Let A and B be events such that P(A ^ B) = 1/4, P( [#permalink]
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04 Nov 2003, 13:40
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Let A and B be events such that P(A ^ B) = 1/4, P( ╦Ь A) = 1/3, and P(B) =
1/2. What is P(A U B)?
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04 Nov 2003, 19:52
does P(A ANd B) mean P(A OR B)
the formula that you are looking for i s
P(A OR B) = P(A) + P(B) - P (A AND B)
= 1 - P(~A) + P(B) - P(A AND B)
= 1 - 1/3 + 1/2 - 1/4
= 11/12
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05 Nov 2003, 00:38
agree
P(A OR B) = P(A) + P(B) - P (A AND B)
1/4=2/3+1/2-X
X=11/12
05 Nov 2003, 00:38
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# Problem 111. Find matching parenthesis
Solution 72454
Submitted on 8 Apr 2012 by Drew Weymouth
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
s = '()'; n = 1; y_correct = 2; assert(isequal(find_matching_paren(s, n),y_correct))
2 Pass
%% s = '()'; n = 2; y_correct = 1; assert(isequal(find_matching_paren(s, n),y_correct))
3 Pass
%% s = '()()'; n = 3; y_correct = 4; assert(isequal(find_matching_paren(s, n),y_correct))
4 Pass
%% s = '(()((a+b))())'; n = 13; y_correct = 1; assert(isequal(find_matching_paren(s, n),y_correct))
5 Pass
%% s = '(()((a+b))())'; n = 12; y_correct = 11; assert(isequal(find_matching_paren(s, n),y_correct))
6 Pass
%% s = '(()((a+b))())'; n = 10; y_correct = 4; assert(isequal(find_matching_paren(s, n),y_correct))
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# Solving Equations Using the Pythagorean Theorem
## Use a^2 + b^2 = c^2 to find missing side lengths of right triangles.
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Practice Solving Equations Using the Pythagorean Theorem
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Solving Equations Using the Pythagorean Theorem
What if you knew the lengths of two sides of a right triangle but not the third? How could you find the length of this missing side? After completing this Concept, you'll be able to use the Pythagorean Theorem to solve problems like this one where variables are involved.
### Guidance
In the previous concept, we learned about the Pythagorean theorem and how to use it to find the hypotenuse. In this concept, we will learn how to use the Pythagorean theorem to find any side of a right triangle.
#### Example A
Determine the value of the missing side. You may assume that each triangle is a right triangle.
Solution
Apply the Pythagorean Theorem.
#### Example B
Determine the values of the missing sides. You may assume that each triangle is a right triangle.
Solution
Apply the Pythagorean Theorem.
#### Example C
One leg of a right triangle is 5 units longer than the other leg. The hypotenuse is one unit longer than twice the size of the short leg. Find the dimensions of the triangle.
Solution
Let length of the short leg.
Then length of the long leg
And length of the hypotenuse.
The sides of the triangle must satisfy the Pythagorean Theorem.
The negative solution doesn’t make sense when we are looking for a physical distance, so we can discard it. Using the positive solution, we get: and .
Watch this video for help with the Examples above.
### Vocabulary
• The Pythagorean Theorem is a statement of how the lengths of the sides of a right triangle are related to each other. A right triangle is one that contains a 90 degree angle. The side of the triangle opposite the 90 degree angle is called the hypotenuse and the sides of the triangle adjacent to the 90 degree angle are called the legs.
• If we let and represent the legs of the right triangle and represent the hypotenuse then the Pythagorean Theorem can be stated as:
In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. That is: .
### Guided Practice
Determine the values of the missing sides. You may assume that each triangle is a right triangle.
Solution
Apply the Pythagorean Theorem.
### Practice
Find the missing length of each right triangle.
1. One leg of a right triangle is 4 feet less than the hypotenuse. The other leg is 12 feet. Find the lengths of the three sides of the triangle.
2. One leg of a right triangle is 3 more than twice the length of the other. The hypotenuse is 3 times the length of the short leg. Find the lengths of the three legs of the triangle.
3. Two sides of a right triangle are 5 units and 8 units respectively. Those sides could be the legs, or they could be one leg and the hypotenuse. What are the possible lengths of the third side?
### Vocabulary Language: English
Hypotenuse
Hypotenuse
The hypotenuse of a right triangle is the longest side of the right triangle. It is across from the right angle.
Legs of a Right Triangle
Legs of a Right Triangle
The legs of a right triangle are the two shorter sides of the right triangle. Legs are adjacent to the right angle.
Pythagorean Theorem
Pythagorean Theorem
The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by $a^2 + b^2 = c^2$, where $a$ and $b$ are legs of the triangle and $c$ is the hypotenuse of the triangle. | 916 | 3,859 | {"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": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 4, "texerror": 0} | 4.75 | 5 | CC-MAIN-2016-07 | latest | en | 0.799606 |
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Hey i medal and fan help plz Amy is playing a board game and rolls two number cubes. Let A = {the sum of the number cubes is odd}, and let B = {the sum of the number cubes is divisible by 3}. List the outcomes in A ∪ B. {3, 6, 9, 12} {1, 3, 5, 7, 9, 11} {3, 5, 6, 7, 9, 11, 12} {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
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scale invariance
a feature of objects or laws that do not change if length scales (or energy scales) are multiplied by a common factor.
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Matrix-method: compare use of different fixpoints Gottfried Ultimate Fellow Posts: 787 Threads: 121 Joined: Aug 2007 11/07/2007, 08:41 PM bo198214 Wrote:Gottfried Wrote:bo198214 Wrote:I want your acknowledgement on this. Well, you say it in the next sentence: let truncation aside. In all my considerations ...Is this a confirmation?I can only restate: in no situation I assumed the matrices as truncated by principle - all my considerations assume the unavoidable truncations in praxi as giving approximations in numerical evaluation as plcaeholder for the basically infinite matrices. I never discussed a finite (truncated) matrix being more than such an approximation for determination of any intermediate result. For instance: if the carleman-matrix is thought of finite size, then I don't see any relation between carleman matrices and any of my matrices. I don't know, whether this satisfies your request (if not, then please give even more explanation of what is the core of your question, I must then being unable to understand the relevant implications correctly) Quote:Gottfried though I am not 100% clear about your my-oh-my method, it seems as if it does nothing more than to find the regular iteration at a fixed point, we discussed something similar already here. That may all be - and whether it is "nothing more" or not: if it is the regular iteration, then fine; if not, then fine again. Quote:The regular iteration of a function $f$ with fixed point at 0 is given by $f^{\circ t} = \sigma^{-1}\circ \mu_{c^t} \circ \sigma$ with $c=f'(0)$, $\mu_d(x)=d\cdot x$ and $\sigma$ being the principal Schroeder function. If the fixed point is not at 0 but is at $a$ then $f=\tau_{-a}\circ \exp_b \circ \tau_a$ is a function with fixed point at 0 and $f'(0)=c=\ln(a)$, where $\tau_d(x)=d+x$. Putting this into the first equation we get: $\exp_b^{\circ t} = \tau_a\circ\sigma^{-1}\circ\mu_{{\ln(a)}^t}\circ\sigma\circ \tau_{-a}$. But if we translate this into matrix notation by simply replacing $\circ$ by the matrix multiplication and replacing each function by the corresponding Bell matrix (which is the transposed Carleman matrix) then we see a diagonalization of $B_b$ because the Bell matrix (and also the Carleman matrix) of $\mu_{\ln(a)}(x)=\ln(a)x$ is just your diagonal matrix ${_dV}(\ln(a))$! Yes, this seems so - but did we not already state the identity of the matrix B (or Bs) with the Bell/Carleman-transposes? I thought, that this had settled the question already? I was very happy, when you pointed out the relation in one of your previous posts - I couldn't have done it due to my lack of understanding of those concepts (described in elaborated articles, more than I could follow in detail). Quote:So it is nothing new, that we can diagonalize the untruncated $B_b$ for each fixed point $a$ with the diagonal matrix ${_dV}(\ln(a))$. As result we only get the plain old regular iteration at a fixed point. Hmm, for me this is an *achievement*, so my bottom up approach just out of the sandbox is then decoded into the terminology of "Schroeder-equation" and "regular iteration" - so: good! If from there some shortcomings of the method are already *known* then I would like to know them, too. Hmm, I've no more idea at this moment. I'll reread your post later this evening, perhaps I'm missing some point. Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »
Messages In This Thread Matrix-method: compare use of different fixpoints - by Gottfried - 11/04/2007, 12:38 PM RE: Matrix-method: compare use of different fixpoints - by bo198214 - 11/04/2007, 12:59 PM RE: Matrix-method: compare use of different fixpoints - by Gottfried - 11/04/2007, 01:28 PM RE: Matrix-method: compare use of different fixpoints - by bo198214 - 11/04/2007, 01:31 PM RE: Matrix-method: compare use of different fixpoints - by Gottfried - 11/04/2007, 01:40 PM RE: Matrix-method: compare use of different fixpoints - by bo198214 - 11/07/2007, 10:52 AM RE: Matrix-method: compare use of different fixpoints - by Gottfried - 11/07/2007, 01:33 PM RE: Matrix-method: compare use of different fixpoints - by bo198214 - 11/07/2007, 01:57 PM RE: Matrix-method: compare use of different fixpoints - by Gottfried - 11/07/2007, 02:10 PM RE: Matrix-method: compare use of different fixpoints - by bo198214 - 11/07/2007, 02:21 PM RE: Matrix-method: compare use of different fixpoints - by Gottfried - 11/07/2007, 02:59 PM RE: Matrix-method: compare use of different fixpoints - by bo198214 - 11/07/2007, 03:35 PM RE: Matrix-method: compare use of different fixpoints - by Gottfried - 11/07/2007, 04:31 PM RE: Matrix-method: compare use of different fixpoints - by bo198214 - 11/07/2007, 07:44 PM RE: Matrix-method: compare use of different fixpoints - by Gottfried - 11/07/2007, 08:41 PM RE: Matrix-method: compare use of different fixpoints - by bo198214 - 11/07/2007, 09:32 PM RE: Matrix-method: compare use of different fixpoints - by Gottfried - 11/11/2007, 06:05 PM RE: Matrix-method: compare use of different fixpoints - by bo198214 - 11/11/2007, 10:05 PM RE: Matrix-method: compare use of different fixpoints - by Gottfried - 11/12/2007, 01:53 AM RE: Matrix-method: compare use of different fixpoints - by Gottfried - 11/13/2007, 05:48 PM RE: Matrix-method: compare use of different fixpoints - by Gottfried - 11/12/2007, 07:48 AM RE: Matrix-method: compare use of different fixpoints - by bo198214 - 11/12/2007, 11:52 AM RE: Matrix-method: compare use of different fixpoints - by Gottfried - 11/12/2007, 03:13 PM RE: Matrix-method: compare use of different fixpoints - by andydude - 11/30/2007, 05:24 PM
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Frequent Visitor
## Cumulative sum of a measure
Hello,
I've been strugling with my problem for 2 weeks now. My background is excel, so i understand calculated columns, but not measures at all. What i'm trying to achieve here is a cumulative sum of a measure. First, let me expose the data :
- 1 table 'History_All' where 1 line = 1 part, and columns are ID / Plant / Engine / Status1 Entry / Status1 Exit / Status2 Entry / Status2 Exit (several status so several columns like the last ones)
- 1 calendar 'Jours' (= 'days' in french) table created from scratch with a column Date = CALENDAR (DATE(2017;1;1); DATE(2018;12;31))
I can make everything work using calculated columns, except that when i try to filter by plants or engine (or both) it does not filter, because i've found no way to use FILTERS involving these columns (hence my slicers are not 'connected' to these columns). When using measures, everything filters automatically when adding slicers
Goal :
- Show number of parts per day --> Achieved with the help of the measure [get date]
`get date = max(Jours[Date])Status1 = CALCULATE(counta(History_All[ID]);FILTER(History_All;History_All[Status1.Entry]<=[get date]&&History_All[Status.Exit]>=[get date]))`
- Show a prevision of parts in each status, which would be :
`for days > today, take the last known value, add the average of added parts (with Status1.Entry) per day in this status`
This is what i can't achieve because it would involve a recursive sum based on a measure (and not a calculated column).
I managed to get the last value :
`Status1 Last = CALCULATE([Status1];Jours[Date]=TODAY())`
I managed to get the average of last month :
```Added = CALCULATE(counta(History_All[ID]);FILTER(History_All;History_All[Status1.Entry]=[get date]))
Added Average = Calculate(AVERAGEX(Jours;[Added]);DATESINPERIOD(Jours[Date];lastdate(Jours[Date]);-30;DAY))Added Average last = CALCULATE([Added Average];Jours[Date]=TODAY())```
I was planning to make a cumulative value of [Added Average last] and add this to [Status1 last] and that would make my prevision line, but that only works in a 'calculated column' state of mind (SUMX for example, doesn't work with measure, and my problem is that i don't get why, because i can't picture what a measure is exactly.)
Any help would be appreciated, even a good link to understand what a measure is exactly.
Regards,
1 ACCEPTED SOLUTION
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Frequent Visitor
## Re: Cumulative sum of a measure
Hello,
Thanks for the useful link, however i could not apply it in my case.
But I have figured out how to make my trend line :
1. Get the last day with data :
`get date Last = CALCULATE([get date];Jours[Date]=TODAY())`
2. Extend the data past today : the formula below writes the [Status1] value if it has a value, otherwise it writes the last value (the one from today)
`Status1 Extended = if([get date] > [get date Last];CALCULATE([Status1];Jours[Date]=TODAY());[Status1]) `
3. Get the rolling average (30days) of added parts in this status
`Status1 added = CALCULATE(counta(History_All[ID]);FILTER(History_All;History_All[Status1.Entry]<=[get date] && History_All[Status1.Exit]>=[get date]))Status1 added avg = Calculate(AVERAGEX(Jours;[Status1 added]);DATESINPERIOD(Jours[Date];lastdate(Jours[Date]);-30;DAY))`
4. Get the last average and extend it beyond today :
`Status1 added avg extended = if([get date] > [get date Last];CALCULATE([Status1 added avg];Jours[Date]=TODAY());blank())`
5. Get the cumulative sum of this average (That was the missing measure i could not find)
EDIT : this formula is based on the method found at this link : https://blog.gbrueckl.at/2015/04/recursive-calculations-powerpivot-dax/
```Status1 Cumulated Addby =
Calculate(
sumx( values(Jours[Date]); [Status1 added avg extended]);
datesbetween(
Jours[Date];blank();max(Jours[Date])
)
)```
6. Addition between Status1 Extended and Status1 Cumulated Addby gives us the trend :
`Status1 Trend = if([get date] >= [get date Last];[Status1 Extended] + [Status1 Cumulated Addby];BLANK())`
This is a little bit dirty as i'm not DAX expert but it does the trick !
2 REPLIES 2
Super User IX
## Re: Cumulative sum of a measure
Check out this article, I am guessing that you might have a similar situation.
https://community.powerbi.com/t5/Community-Blog/Design-Pattern-Groups-and-Super-Groups/ba-p/138149
---------------------------------------
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## Re: Cumulative sum of a measure
Hello,
Thanks for the useful link, however i could not apply it in my case.
But I have figured out how to make my trend line :
1. Get the last day with data :
`get date Last = CALCULATE([get date];Jours[Date]=TODAY())`
2. Extend the data past today : the formula below writes the [Status1] value if it has a value, otherwise it writes the last value (the one from today)
`Status1 Extended = if([get date] > [get date Last];CALCULATE([Status1];Jours[Date]=TODAY());[Status1]) `
3. Get the rolling average (30days) of added parts in this status
`Status1 added = CALCULATE(counta(History_All[ID]);FILTER(History_All;History_All[Status1.Entry]<=[get date] && History_All[Status1.Exit]>=[get date]))Status1 added avg = Calculate(AVERAGEX(Jours;[Status1 added]);DATESINPERIOD(Jours[Date];lastdate(Jours[Date]);-30;DAY))`
4. Get the last average and extend it beyond today :
`Status1 added avg extended = if([get date] > [get date Last];CALCULATE([Status1 added avg];Jours[Date]=TODAY());blank())`
5. Get the cumulative sum of this average (That was the missing measure i could not find)
EDIT : this formula is based on the method found at this link : https://blog.gbrueckl.at/2015/04/recursive-calculations-powerpivot-dax/
```Status1 Cumulated Addby =
Calculate(
sumx( values(Jours[Date]); [Status1 added avg extended]);
datesbetween(
Jours[Date];blank();max(Jours[Date])
)
)```
6. Addition between Status1 Extended and Status1 Cumulated Addby gives us the trend :
`Status1 Trend = if([get date] >= [get date Last];[Status1 Extended] + [Status1 Cumulated Addby];BLANK())`
This is a little bit dirty as i'm not DAX expert but it does the trick !
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Top Solution Authors
Top Kudoed Authors | 1,821 | 6,984 | {"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-2020-40 | latest | en | 0.83492 |
https://www.physicsforums.com/threads/diccrete-math-problem-finding-a-proposition-given-a-specific-truth-table.639957/ | 1,519,510,631,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815951.96/warc/CC-MAIN-20180224211727-20180224231727-00661.warc.gz | 885,548,632 | 15,384 | # Diccrete Math problem: finding a proposition given a specific truth table
1. Sep 29, 2012
### nicnicman
1. The problem statement, all variables and given/known data
Find a proposition using only p, q, ¬ and the connective ∧ with the given truth table.
p q ?
T T F
T F F
F T T
F F T
2. Relevant equations
3. The attempt at a solution
There is a systematic way of solving these but this system requires the use of an 'OR'. Is there a systematic way of solving this with only 'ANDs' or do you just keep guessing till you get it right. There must be an easier way!
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution | 177 | 684 | {"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} | 2.796875 | 3 | CC-MAIN-2018-09 | longest | en | 0.804966 |
https://www.klayout.de/forum/discussion/1434/image-to-layout-alignment-using-3-landmarks-registrartion | 1,586,167,254,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371620338.63/warc/CC-MAIN-20200406070848-20200406101348-00479.warc.gz | 866,466,423 | 7,570 | # Image to layout alignment using 3 landmarks registrartion
Hello Matthias-
This question is more of a geometric question but I'm sure you will have a very short and simple code that can help a lot:
Say I want to align an image to a layout. Both the image and the layout have a set of 3 matching respective landmarks.
So the image landmarks set is: (Im1x,Im1y) (Im2x,Im2y) (Im3x,Im3y) and the layout landmark set is (La1x,La1y) (La2x,La2y) (La3x,La3y)
Can you suggest a code that will transpose the image to the layout using above landmarks coordinates?
Thanks for your time and effort! Itamar
• edited February 10
Hi Itamar,
The method which achieves this is "Matrix3d.adjust". Here is a sample:
``````# Landmarks image (in pixel coordinates)
landmarks_img = [
pya.DPoint(100, 100),
pya.DPoint(200, 200),
pya.DPoint(200, 101)
]
# Landmarks layout (in micrometers)
landmarks_layout = [
pya.DPoint(105.0, 25.0),
pya.DPoint(110.0, 30.0),
pya.DPoint(110.0, 25.0)
]
matrix = pya.Matrix3d()
print(str(matrix))
``````
The output is (beautified):
``````(0.05, 0, 100)
(-0.000505, 0.0505, 20)
(0, 0, 1)
``````
which is a displacement of 100, 20 plus a scaling by 0.05 plus a small shear I have induced because I used a y value of 101 for the third image point instead of 100.
This matrix you can then put into the "matrix" property of your pya.Image object to place it according to the landmark mapping.
Matthias
• Thank you so much for it Matthias | 452 | 1,480 | {"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-2020-16 | latest | en | 0.841574 |
http://www.toktalk.net/2007/12/24/what-are-thought-experiments/ | 1,498,646,340,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323604.1/warc/CC-MAIN-20170628101910-20170628121910-00146.warc.gz | 682,724,277 | 12,111 | ## What are Thought Experiments?
It is not always necessary to conduct real-life experiments to reach a valid scientific conclusion. Thought experiments may in some cases also suffice. In this edition I will illustrate you a thought experiment from physics: In a vacuum, all objects accelerate the same way and they both have the same velocity. Heavy objects will not fall faster. But how can we test this? We do not have a large vacuum chamber to test this. A thought experiment can be useful in this case.
In this edition of TOK-Talk I will explain you what a thought experiment is. Is it always necessary to conduct real-life experimets to reach a valid scientific conclusion? Listen to find out!
Do heavy or lighter objects fall to the ground faster? Which one will hit the ground first when dropped at the same time from the same height – a bowling ball or a ball of crumpled newspaper of the same size? The answer is: “it depends”! If you drop both objects from a tower on earth, then the bowling ball will hit the ground first. If you drop both objects on the moon, then both will hit the ground at the same time. Here on earth, the air resistance has a large influence on the falling velocity of an object. In a complete vacuum all objects hit the the ground at the same time. This may be difficult to imagine, but it is indeed the case.
The problem is that we can not go to the moon to test this physical law. We could construct a giant air-less vacuum chamber here on earth. But this is of course not practical – and not necessary either! I would like to introduce to you a thought experiment that demonstrates that all objects do indeed fall to the ground at the same rate. Now, it is very critical that we take the experiment step by step, otherwise it can become a bit confusing.
• Step 1: We start the thought experiment by making an assumption. We assume that heavy objects fall to the gound faster than light objects. I know that I just told you that in a vacuum all objects fall to the ground at the same rate, but for right now we need to take this assumption. Again: we assume that heavy objects fall to the ground faster. I take a square block of lead with a mass of 10kg into my left hand and a block of lead with a mass of only 1kg into my right hand. Actually you have to be pretty strong to carry the 10kg lead block in one hand, but as we do this only in our imagination, this is not a problem. I drop them both at the same time from the same height. The 10kg lead block will hit the ground before the 1kg block. It will accelerate faster. We hear two sounds as the two blocks hit the floor: boom – boom. The first boom is from the 10kg block and the second boom is from the 1kg block hitting the floor. This is what we assume. Close your eyes and try to imagine this situation. Just don’t drop them on your toes.
• Step 2: I now pick up both blocks of lead from the floor. Luckily it is only a thought experiment so no real damage has been done to the floor (or your toes). In your imagination put the larger 10kg block of lead on top of the 1kg block. The two blocks are now stacked on top of each other. You now have one chunk of lead, 11kg heavy. The large block is sitting on the small one. The next observation is crucial. Drop this chunk from the same height. What should happen? The smaller block will fall slower than the larger one. The larger block wants to fall faster, but is slowed down by the smaller block in front of it. The falling velocity of the 11kg combination block will therefore be between the velocity of the 10kg and the 1kg block when they are dropped separately. The combination block will fall slower than the 10kg block alone. This is because the small block breaks the falling of the larger block behind it. Actually the situation is similar to two people running, a fast person running behind a slow person. The fast runner bumps into the slower runner in front. The faster runner is slowed down.
• Step 3: Now it’s starting to become interesting. Remember our assumption from the beginning? We assumed that heavier objects fall to the ground faster than lighter objects. But is this not strange? If this assumption is correct then the 11kg combination block should actually fall faster than the 10kg block alone. But as we have just seen, we came to the conclusion that the combination block falls slower. Our conclusion contradicts the initial assumption. Something is wrong here. But what?
• The assumption is wrong. If we assume that all objects fall to the ground (in a vacuum of course) at the same rate, then we do not have this contradiction. It then does not matter if the block of lead has a mass of 1kg, 10kg or 11kg. All of them fall to the ground at the same rate. And indeed, this is what we observe in space. If we reach contradictory conclusions, then either we made a mistake in the deductive process or the assumptions are wrong. But we made no errors, so the assumption must be wrong.
This method of trying to contradicting the assumptions is also used in the field of Mathematics. Certain mathematical proofs are conducted in a similar way. Initially you make a false assumption and then try to disprove this assumption by generating a contradiction.
For those of you who still doubt that all objects fall to the ground the at same rate, I have another explanation. Objects with a larger mass also are pulled by gravity with a stronger force. A heavy shopping bag pulls on our arms with greater force than a light shopping bag. That’s why our arms start to hurt faster. So far so good. But does this mean that a heavy shopping bag also falls faster to the ground? Objects that have a larger mass are also more difficult to accelerate. In simple words, heavy objects don’t like to move from their place when you apply a force to them. This is why heavier cars need stronger engines to accelerate them. Otherwise you’ll get off the spot only very slowly. Now, this larger mass compensates the stronger pulling force. Example: a bowling ball is pulled to the ground more than a golf ball. But the bowling ball also wants to resist the movement more than the golf ball. The consequence is that both objects fall to the gound at the same rate again. Golf ball and bowling ball will contact the ground at the same time.
After so much imagination work a little final quote by Galileo Galilei: “All truths are easy to understand once they are discovered; the point is to discover them.”
Tags: , , , , , , , , | 1,401 | 6,487 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-26 | longest | en | 0.942292 |
https://compassionhamilton.org/and-pdf/349-public-finance-and-public-policy-gruber-solutions-pdf-595-152.php | 1,638,725,095,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363215.8/warc/CC-MAIN-20211205160950-20211205190950-00299.warc.gz | 261,905,314 | 9,663 | Thursday, May 6, 2021 3:06:57 AM
# Public Finance And Public Policy Gruber Solutions Pdf
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There are many regulations of academic honesty of your institution to be considered at your own discretion while using it. Public Finance and Public Policy, 5th Edition… tax from such falsification.
## Jonathan Gruber Public Finance And Public Policy Fifth Edition Pdf
Questions and Problems 1. What is the relative price of a gallon of gas, in terms of bus trips? A commuter could exchange 2. One way to sketch a linear demand function is to find the x Q and y P intercepts.
Price elasticity is the percent change in the quantity purchased divided by the percent change in price. To calculate these percentage changes, divide the change in each variable by its original value. Graph your budget constraint. The food intercept y in the accompanying figure is 25 units. Suppose that the government subsidizes clothing such that each unit of clothing is half-price, up to the first 5 units of clothing.
Graph your budget constraint in this circumstance. This budget constraint will have two different slopes. In other words, for less than 5 units of clothing, 1 unit of clothing costs the same as 1. The point where the line kinks, 5, The new x-intercept clothing intercept is The theory of diminishing marginal utility predicts that the more people eat the less utility they gain from each additional unit consumed. The marginal price of an additional unit of food at an all-you-can-eat buffet is zero; rational consumers will eat only until their marginal utility gain from an additional bite is exactly zero.
The marginal cost of remaining at the buffet is the value of the time spent on the best alternative activity. When the marginal benefit of that activity is greater than the marginal benefit of remaining at the buffet, diners will leave. Consumers optimize their choice when they are on the highest possible indifference curve given their budget constraint. Under these circumstances, the budget constraint must pass through the indifference curve where it intersects the chosen point.
There must then be at least a segment of the budget constraint that lies above up and to the right of the indifference curve associated with that choice. Any choice on that segment would yield higher utility. The single tangency point C in the figure is the only point at which this occurs.
Consider the utilitarian social welfare function and the Rawlsian social welfare function, the two social welfare functions described in this chapter. Which one is more consistent with a government that redistributes from rich to poor? Which is more consistent with a government that does not do any redistribution from rich to poor? The Rawlsian social welfare function is consistent with redistribution from the rich to the poor whenever utility is increasing in wealth or income.
The utilitarian social welfare function can also be consistent with a government that redistributes from the rich to the poor, for example, if utility depends only on wealth and exhibits diminishing marginal utility.
However, the Rawlsian social welfare function depends entirely on the least-well-off, so it will generally prescribe more redistribution than the utilitarian social welfare function. Think about your answer to part a.
Show that government redistribution from rich to poor can still be consistent with either of the two social welfare functions. If utility depends only on wealth and exhibits diminishing marginal utility, and if efficiency losses from redistribution are small, then both the utilitarian and Rawlsian social welfare functions can be consistent with government redistribution.
A simple example can illustrate this point. Because the free market competitive equilibrium maximizes social efficiency, why would the government ever intervene in an economy?
Efficiency is not the only goal of government policy. Equity concerns induce government to intervene to help people living in poverty, even when there are efficiency losses.
In economic terms, a society that willingly redistributes resources has determined that it is willing to pay for or give up some efficiency in exchange for the benefit of living in a society that cares for those who have fewer resources.
Social welfare functions that reflect this willingness to pay for equity or preference for equity may be maximized when the government intervenes to redistribute resources.
Draw the new budget constraint. The difference is that the all-leisure income is higher, but the slope of the line segment from hours of leisure to 2, hours of leisure is flatter. A higher income guarantee with a higher reduction rate is more likely to discourage work for two reasons.
First, not working at all yields a higher income. Draw a system of smooth indifference curves that bend the right way but would lead an agent to work more under the program you chose in part c than under the other program. Describe what seems extreme about these curves that leads to the unusual behavior. This is odd behavior.
Moreover, as the graph indicates, the part b system makes the worker richer leaving them with more money for any given amount of work less than 1, hours. One would typically expect that making someone richer would also discourage work by making earnings less pressing and encouraging extra leisure.
Economically speaking, this is happening because these indifference curves imply that leisure is an inferior good for this individual. Giving this individual more money is making them consume less leisure and therefore work more. And this inferior good effect is strong enough to outweigh the substitution effect coming from the lower effective wage in part c.
Such behavior is highly unlikely to happen in the real world. A good is called normal if a person consumes more of it when her income rises for example, she might see movies in theaters more often as her income rises.
It is called inferior if a person consumes less of it when her income rises for example, she might be less inclined to buy a used car as her income rises. Sally eats out at the local burger joint quite frequently. The burger joint suddenly lowers its prices. Suppose that, in response to the lower burger prices, Sally goes to the local pizza restaurant less often.
Can you tell from this whether or not pizza is an inferior good for Sally? You cannot. Since Sally eats at the burger joint quite a bit, falling burger prices imply that she is richer. If this was the only effect, you could indeed conclude that pizza is an inferior good—Sally gets richer and buys less pizza. But there is also a substitution effect here: the relative price of pizza has gone up.
This leads her to substitute away from pizza. If the substitution effect is bigger than the income effect for Sally, then she could respond in this way, even if pizza is a normal good.
Suppose instead that, in response to the lower burger prices, Sally goes to the burger joint less often. Since she actually consumes less of them, the income effect must be working in the opposite direction, leading her to consume fewer burgers and it must be stronger than the substitution effect.
Since the fall of burger prices made Sally richer, burgers must be an inferior good for Sally. Note: A good for which falling prices leads to reduced consumption is known as a Giffen good. Giffen goods are observed in reality very rarely, if at all. Advanced Questions The government is considering altering the program to improve work incentives. Its proposal has two pieces. Draw the budget constraint facing any worker under the original program.
The budget constraint for the original program is depicted in the graph that follows. The budget constraint for the proposed program is depicted in the following graph. The original budget constraint is also depicted by the dashed line. At this point, she will be back onto the budget line that she would have been on, absent the guarantee program, just like under the old program.
Which of the four workers do you expect to work more under the new program? Who do you expect to work less? Are there any workers for whom you cannot tell if they will work more or less?
Workers working fewer than hours see their hourly wage effectively doubled under the plan. The substitution effect therefore tends to make Alice, Bob, and Calvin all work more.
The income effect is thus different for these three workers. She consumes less of all normal goods, including leisure, so this also makes her work more.
We can unambiguously conclude that she will work more. We can conclude from the substitution effect alone that he too will work more. He will therefore tend to work less due to the income effect. We cannot tell if the substitution effect or the income effect is stronger, so we cannot tell if Calvin will work more or less.
Finally, Deborah was working hours before. There is no substitution effect for her. As the graph shows, however, she experiences an increase in income.
We conclude that she will work less. What is the value of consumer surplus? What is the value of producer surplus? The first step is to find the equilibrium price and quantity by setting quantity demanded equal to quantity supplied. Recall that the condition for equilibrium is that it is the price at which these quantities are equal. This solves to What is the consumer surplus now?
The producer surplus? Why is there a deadweight loss associated with the subsidy, and what is the size of this loss?
## MODERATORS
Read honest and unbiased product reviews from our users. Jonathan Gruber. Books and Manuscripts. Public Finance and Public Policy, 6 th Edition. New York: Worth Publishers, It was the first text written from the ground up to reflect current realities of public finance, enhancing its survey of traditional topics with an emphasis on empirical. This is why we allow the books compilations in this website.
You pay sales taxes on most of the goods you purchase. If you smoke or drink or drive a car, you pay taxes on cigarettes, alcohol, and gasoline. If you work, you may pay income and payroll taxes. What does the government do with the taxes it collects? If you go to a public school, you are a consumer of public sector services.
## Jonathan Gruber Public Finance And Public Policy Fifth Edition Pdf
We are currently engaged in the most fundamental debate about the role of government in decades, and who better than Jonathan Gruber to guide students through the particulars in the new edition of his best-selling text, Public Finance and Public Policy, 6e. The new edition detail The new edition details ongoing policy debates, with special focus on the largest tax reform in 30 years. New topics include universal basic income, the legalization of pot, and congestion pricing. And, of course, there is an extensive, in-depth discussion of the debate over health care.
New coverage of inefficiencies in the U. S health care system, has been. No need to wait for office hours or assignments to be graded to find out where you took a wrong turn.
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This fifth edition is being released at the right moment, as we are at the start of the new EU programming period No need to wait for office hours or assignments to be graded to find out where you took a wrong turn. He has published more than research articles and has edited 6 research volumes. During the academic year, Dr. Treasury Department.
Sapling can only be accessed if your instructor has set up a course at your University. Please only buy this code if your instructor has an active Sapling course. This product should only be purchased by International students at University of Illinois. Jonathan Gruber.
We are at a pivotal point in history with respect to the role of government in the U.
Некорректный ввод только ускорит процесс разрушения. Два некорректных ввода - и шифр навсегда захлопнется от нас на замок. Тогда всему придет конец.
Пустые, но мои, черт тебя дери. - Прошу прощения, - сказал Беккер, поворачиваясь, чтобы уйти. Парень загородил ему дорогу.
После чего вошел в номер и включил свет. Немец не ожидал такого оборота. - Wasmachst… - Помолчите! - Беккер перешел на английский.
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jonathan gruber public finance and public policy fifth edition pdf is important information accompanied by photo and HD pictures sourced from all websites in the. | 2,655 | 13,128 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-49 | latest | en | 0.939003 |
https://www.physicsforums.com/threads/explanation-of-rayleighs-work-on-vibrations-of-liquid-droplets.708382/ | 1,709,055,922,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474676.79/warc/CC-MAIN-20240227153053-20240227183053-00776.warc.gz | 944,479,582 | 15,719 | # Explanation of Rayleighs work on vibrations of liquid droplets
• M_1
In summary, the conversation discusses calculating the vibrational energy of a water droplet using Rayleigh's work from 1879, but with some simplifications and difficulties in understanding the calculations. The speaker also mentions looking for a more accessible presentation of Rayleigh's work in a thesis or master thesis. A link is provided for a thesis on the subject that may be helpful.
#### M_1
Hi,
I'm trying to calculate the vibrational energy of a water droplet for a given vibrational amplitude.
Rayleigh wrote about this in Proc. R. Soc. Lond., volume 29, 1879, "On the Capillary Phenomena of Jets", Appendix II, but he makes some simplifications (as I understand it) and I also have problems following parts of his calculus.
Is anyone aware of a thesis or master thesis or similar on this subject where the work of Rayleigh is presented in a more accesible way?
Thanks!
I don't know what level of detail you want to go into, but does the derivation involve the spherical wave equation? If you calculated the amplitude from that and then introduced the water tension to get the work wouldn't we get a relatively accurate measure of the vibrational energy?
1 person
## 1. What is the significance of Rayleigh's work on vibrations of liquid droplets?
Rayleigh's work on vibrations of liquid droplets is significant because it laid the foundation for understanding the behavior of fluids and their interactions with sound waves. This work led to the development of the Rayleigh-Plesset equation, which is used to model the oscillations of bubbles in a liquid, and has applications in various fields such as acoustics, meteorology, and oceanography.
## 2. What is the Rayleigh-Plesset equation and how is it used?
The Rayleigh-Plesset equation is a mathematical model that describes the dynamics of a gas bubble in a liquid. It takes into account factors such as surface tension, pressure, and viscosity to predict the behavior of the bubble under the influence of an external force, such as sound waves. This equation is used in various fields, including medical ultrasound imaging, underwater acoustics, and sonar technology.
## 3. How did Rayleigh's work contribute to our understanding of sound waves?
Rayleigh's work on vibrations of liquid droplets helped us understand the behavior of sound waves in fluids. He showed that when a droplet vibrates, it creates a sound wave that travels through the surrounding fluid. This discovery led to the understanding of how sound waves interact with different types of fluids, such as air and water, and how they can be used in various applications.
## 4. What are some practical applications of Rayleigh's work on vibrations of liquid droplets?
Rayleigh's work has practical applications in various fields, such as medical imaging, sonar technology, and weather forecasting. The understanding of sound waves in fluids has led to the development of medical ultrasound imaging, where sound waves are used to create images of internal body structures. Sonar technology, which uses sound waves to detect and locate objects underwater, also relies on the principles of Rayleigh's work. In addition, his work has contributed to our understanding of how sound waves affect weather patterns and ocean currents.
## 5. How did Rayleigh's work on vibrations of liquid droplets impact the field of acoustics?
Rayleigh's work had a significant impact on the field of acoustics. His study of the vibrations of liquid droplets led to the understanding of how sound waves travel through different mediums and how they can be manipulated. This work also paved the way for advancements in acoustics, such as the development of more accurate models for sound wave propagation and the design of more efficient acoustic devices. | 786 | 3,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-10 | latest | en | 0.938454 |
https://nursingwriters.org/financial-management-hw/ | 1,590,585,756,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347394074.44/warc/CC-MAIN-20200527110649-20200527140649-00357.warc.gz | 477,817,690 | 17,943 | July 11, 2019
###### DEFINE HOW ELECTRONIC MESSAGES SENT BETWEEN HEALTHCARE PROVIDERS ARE FORMATTED.
July 11, 2019
1. First City Bank pays 6 percent simple interest on its savings account balances, whereas Second City Bank pays 6 percent interest compounded annually.
If you made a \$60,000 deposit in each bank, how much more money would you earn from your Second City Bank account at the end of 10 years? (Do not round intermediate calculations and round your final answer to 2 decimal places. (e.g., 32.16))
Difference in accounts __\$?
2. For each of the following, compute the future value (Do not round intermediate calculations and round your final answers to 2 decimal places. (e.g., 32.16)):
Present Value Years Interest Rate Future Value
\$1,950 10 14 % \$___?___
\$8,152 8 8% \$___?___
\$70,355 15 13% \$___?___
\$177,796 6 5% \$___?___
3. For each of the following, compute the present value (Do not round intermediate calculations and round your final answers to 2 decimal places. (e.g., 32.16)):
Present Value Years Interest Rate Future Value
\$_____ 12 6 % \$14,751
\$_____ 3 12% \$44,557
\$_____ 28 13% \$879,073
\$_____ 30 10 % \$543,164
4. Solve for the unknown interest rate in each of the following (Do not round intermediate calculations and round your final answers to 2 decimal places. (e.g., 32.16)):
Present Value Years Interest Rate Future Value
\$170 3 __ % \$196
\$290 17 __% \$732
\$32,000 18 __% \$124,723
\$31,261 20 __ % \$226,140
5.Solve for the unknown number of years in each of the following (Do not round intermediate calculations and round your final answers to 2 decimal places. (e.g., 32.16)):
Present Value Years Interest Rate Future Value
\$510 __ 9 % \$1,212
\$760 __ 10% \$1,629
\$17,900 __ 17% \$260,563
\$21,000 __ 15 % \$391,887
6. Assume the total cost of a college education will be \$250,000 when your child enters college in 17 years. You presently have \$62,000 to invest.
What annual rate of interest must you earn on your investment to cover the cost of your child’s college education? (Do not round intermediate calculations and round your final answer to 2 decimal places. (e.g., 32.16))
Annual rate of interest ___% ?
7. At 5.70 percent interest, how long does it take to double your money? (Round your answer to 2 decimal places. (e.g., 32.16))
Length of time ____ years ?
At 5.70 percent interest, how long does it take to quadruple it? (Round your answer to 2 decimal places. (e.g., 32.16))
Length of time ____ years ?
8. Assume that in January 2010, the average house price in a particular area was \$277,400. In January 2001, the average price was \$194,300.
What was the annual increase in selling price? (Do not round intermediate calculations and round your final answer to 2 decimal places. (e.g., 32.16))
Annual increase in selling price ___ % ?
9. You’re trying to save to buy a new \$194,000 Ferrari. You have \$44,000 today that can be invested at your bank. The bank pays 5.2 percent annual interest on its accounts.
How long will it be before you have enough to buy the car? (Do not round intermediate calculations and round your final answer to 2 decimal places. (e.g., 32.16))
Number of years ____ ?
10. Imprudential, Inc. has an unfunded pension liability of \$572 million that must be paid in 25 years. To assess the value of the firm’s stock, financial analysts want to discount this liability back to the present.
If the relevant discount rate is 6.5 percent, what is the present value of this liability? (Enter your answer in dollars not in millions. Do not round intermediate calculations and round your final answer to 2 decimal places. (e.g., 32.16))
Present value \$ _____ ?
11. You have just received notification that you have won the \$1 million first prize in the Centennial Lottery. However, the prize will be awarded on your 100th birthday (assuming you’re around to collect), 74 years from now.
What is the present value of your windfall if the appropriate discount rate is 9 percent? (Do not round intermediate calculations and round your final answer to 2 decimal places. (e.g., 32.16))
Present value \$____ ?
12. Your coin collection contains 44 1952 silver dollars. If your grandparents purchased them for their face value when they were new, how much will your collection be worth when you retire in 2054, assuming they appreciate at a 5.7 percent annual rate? (Do not round intermediate calculations and round your final answer to 2 decimal places. (e.g., 32.16))
Future value \$___ ?
13. In 1895, the first Putting Green Championship was held. The winner’s prize money was \$180. In 2010, the winner’s check was \$1,380,000.
What was the percentage increase per year in the winner’s check over this period? (Do not round intermediate calculations and round your final answer to 2 decimal places. (e.g., 32.16))
Interest rate ____ % ?
If the winner’s prize increases at the same rate, what will it be in 2037? (Do not round intermediate calculations and round your final answer to 2 decimal places. (e.g., 32.16))
Future value ____ \$ ?
14. Assume that in 2010, a gold dollar minted in 1893 sold for \$127,000. For this to have been true, what rate of return did this coin return for the lucky numismatist? (Do not round intermediate calculations and round your final answer to 2 decimal places. (e.g., 32.16))
Rate of return _____ % ?
15. Although appealing to more refined tastes, art as a collectible has not always performed so profitably. During 2003, an auction house sold a sculpture at auction for a price of \$10,361,500. Unfortunately for the previous owner, he had purchased it in 2000 at a price of \$12,477,500.
What was his annual rate of return on this sculpture? (Negative amount should be indicated by a minus sign. Do not round intermediate calculations and round your final answer to 2 decimal places. (e.g., 32.16))
Annual rate of return _____ % ?
1b. Wainright Co. has identified an investment project with the following cash flows.
Year Cash Flow
1\$740
2\$970
3\$1,230
4\$1,325
If the discount rate is 9 percent, what is the present value of these cash flows? (Do not round intermediate calculations and round your final answer to 2 decimal places. (e.g., 32.16))
Present Value \$ ______ ?
What is the present value at 18 percent? (Do not round intermediate calculations and round your final answer to 2 decimal places. (e.g., 32.16))
Present value \$ ____ ?
What is the present value at 24 percent? (Do not round intermediate calculations and round your final answer to 2 decimal places. (e.g., 32.16))
Present value \$ ____ ?
2. Investment X offers to pay you \$4,200 per year for seven years, whereas Investment Y offers to pay you \$6,300 per year for four years.
Calculate the present value for Investment X and Y if the discount rate is 6 percent. (Do not round intermediate calculations and round your final answers to 2 decimal places. (e.g., 32.16))
Present value
Investment X\$ ______?
Investment Y\$ ______?
Calculate the present value for Investment X and Y if the discount rate is 16 percent. (Do not round intermediate calculations and round your final answers to 2 decimal places. (e.g., 32.16))
Present value
Investment X\$ ____?
Investment Y\$ ____?
3. Toadies, Inc., has identified an investment project with the following cash flows.
YearCash Flow
1\$1,625
2\$1,745
3\$1,830
4\$1,880
If the discount rate is 9 percent, what is the future value of the cash flows in year 4? (Do not round intermediate calculations and round your final answer to 2 decimal places. (e.g., 32.16))
Future value\$ _____ ?
If the discount rate is 10 percent, what is the future value of the cash flows in year 4? (Do not round intermediate calculations and round your final answer to 2 decimal places. (e.g., 32.16))
Future value\$ _____ ?
If the discount rate is 25 percent, what is the future value of the cash flows in year 4? (Do not round intermediate calculations and round your final answer to 2 decimal places. (e.g., 32.16))
Future value\$ _____ ?
4. First Simple Bank pays 8.2 percent simple interest on its investment accounts. If First Complex Bank pays interest on its accounts compounded annually, what rate should the bank set if it wants to match First Simple Bank over an investment horizon of 10 years? (Do not round intermediate calculations and round your final answer to 2 decimal places. (e.g., 32.16))
Interest rate _____ % ?
5. Marko, Inc. is considering the purchase of ABC Co. Marko believes that ABC Co. can generate cash flows of \$6,600, \$11,600, and \$17,800 over the next three years, respectively. After that time, they feel the business will be worthless. Marko has determined that a rate of return of 13 percent is applicable to this potential purchase. What is Marko willing to pay today to buy ABC Co.?
1. \$27,261.50
2. b. \$39,420.00
c. \$28,896.22
d. \$36,000.00
e. \$24,876.50
6. One year ago, the Jenkins Family Fun Center deposited \$5,200 in an investment account for the purpose of buying new equipment four years from today. Today, they are adding another \$7,000 to this account. They plan on making a final deposit of \$9,200 to the account next year. How much will be available when they are ready to buy the equipment, assuming they earn a 6 percent rate of return?
a. \$25,438.74
b. \$26,753.46
c. \$25,993.97
d. \$24,234.57
1. \$25,495.73
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financial management hw was first posted on July 11, 2019 at 4:46 pm. | 2,524 | 9,791 | {"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-2020-24 | latest | en | 0.575842 |
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04.12.2019
If you've ever used the Martingale strategy, you'll know it involves doubling down on your wagers, chasing the excitement of trying to secure a. These techniques are used to give you a bigger chance of winning or in some cases, just to earn a profit. One popular betting system is the Martingale. The Martingale Strategy is a strategy of investing or betting introduced by French mathematician Paul Pierre Levy. · It is based on the theory of. SLACK ETHEREUM
In most casinos that is European or single zero roulette. You make a one-unit wager on a bet that pays even money, usually red or black. In each series of bets you keep doubling your bet until you win, at which point you go back to the small starting bet. Most of the time you have a win within a few spins, but sometimes you have a string of losses.
If you do, one of two things happens. And those large bets arose from what seemed at first small betting. Every day thousands of unknowledgeable players around the world walk into casinos to give the Martingale Betting System a trial and most of them win a little bit. The shorter their time at the roulette table the more likely it is to have a small win.
That unbettable amount a wager of more than the table limit rarely shows up, but when it does, the player loses more than they have won, and usually considerably more. The longer the gambling session, the more likely it is that that limit is reached. That is, the more spins, the likelier that a losing streak of 10 bets in a row will occur. What Is a Martingale? In general, a martingale system assumes that the same bet is played repeatedly while raising its stake with a multiplier at every new game if the previous bet is lost.
If you win, stop and continue if you want with whatever bet. If you win, stop or continue as you want. The idea is to double your stake after each loss. In our example, the bet was on color, and the multiplier of the stake was 2; however, theoretically, you may choose any kind of bet and any multiplier to have a martingale.
Still, what does mathematics have to say about this? How is that possible? It is due to a simple algebraic certitude. This mathematical certainty also holds for other types of bets and other multipliers. With this mathematical proof, the system seems infallible: whatever succession of losses will end at some point, and the next win will make you a profit.
Furthermore, the probability of having a long succession of reds or blacks is very low if such information counts in any way. Constraints and Weaknesses In the mathematical relation above, we focused on the last term, reflecting the profit of the last bet, but the sum in the left-hand member reflects the cumulated loss. Hence, the net profit of such a progressive bet is S, whatever the number of consecutive failures, and this amount is actually the goal of the bettor.
Therefore, real-life circumstances seem to threaten the effective application of that mathematical certainty related to the martingale. First, it is about the size of your bankroll, which should be large enough to sustain a loss as a result of a hypothetical long succession of failures. Limiting the cumulated loss has imposed constraints on the parameters of the martingale.
ETHEREUM KERATITIS
So, what does a Martingale player do? The player doubles their previous wager after each loss, making it extremely easy to calculate progressions. If a win is generated, then the player bets his base stake instead. In the game of roulette, you are usually expected to lose three times out of four. So, using this strategy, you should be able to come out ahead and make profit. The Martingale Strategy may offer many advantages to the player, but it also has some risks.
It is easy to master, easily appliable to many games and a good way to generate profit. On the other hand, it is not advised to use the method for long term profit. Also, the player should have an unlimited bankroll and enough determination.
Summed up the pros and cons of this strategy are as follows: Img source: pexels. The player makes the most out of winning streaks. The player can set his base stake as low or high as he desires. Easy to generate money in the short term. Best with most popular games: Blackjack, Baccarat, Roulette Cons: It is not ideal for generating money in the long term.
Players may find themselves taking on a big risk for a small reward. Bet sizes increase exponentially. Requires an unlimited bankroll to guarantee success. Reaching the table limit is a risk. Challenging to recognize when to stop. Img source: pexels. Following are some of the most favoured ones by players.
It is for players who like the simplicity of the original but would like to win more than a single unit. The Grand Martingale system offers an even higher risk but a more reasonable win per ended streak. After each win, the player is although advised to regress and bet their base stake instead. A Martingale stacking strategy is an exceptionally straightforward approach to win cashback when you lose your underlying wager.
You double your stake when you lose to make some income. In the wake of losing the principal bet, you will stake twice as much for the second. On the off chance that you lose the subsequent bet, you would have the option to double your stake again on the third bet. If you win the bet, you would keep multiplying the stake. Despite its straightforwardness, the Martingale sports betting system is frequently interesting to novices.
There are two conditions under which the Martingale framework can be helpful. The first would be if you had limitless money if you lost numerous wagers and the second would be if internet wagering destinations permitted limitless deposits. As far as wagering techniques, the Martingale system in sports betting is viewed as high-hazard. Novices regularly place a lot of confidence in the Martingale sports betting system because of an inclination to misjudge the danger implied with a losing streak and an underestimation of the danger implied in upping the stakes with this kind of betting procedure.
As a rule, the Martingale system football betting is utilized in games with an equivalent shot at winning or losing. The market is anything but a lose-lose situation, which is critical to comprehend. Accordingly, before a methodology is applied to betting, it is generally adjusted. At the point when you lose a bet, you should double your stake.
That is the substance of the Martingale system in sports betting. You need to constantly double your bet with the desire for higher winnings. A unit is equivalent to the stake you start with. On the off chance that this appears to be excessively complex, you can utilize a calculator to make certain of what to stake. Chapter 3 What is Anti-Martingale?
An anti-Martingale procedure is a money management technique that builds volumes when profits are recorded while reducing volumes when misfortunes are recorded. The quantity of steps to follow differs for dealers utilizing a standard anti-Martingale methodology.
Money management MM is generally utilized by the two novices and experts. A merchant utilizing the anti-Martingale framework ought to rather double his position to gain by a series of wins before arriving at his betting limits. In the anti-Martingale system, expanded trade size is safer when the broker is beating the competition consistently than when they are losing because trading is more forceful during times of extensive development and when the dealer is on a losing streak.
Chapter 4 Advantages of Using the Martingale System Practically, all wagering frameworks enjoy their benefits and drawbacks. Martingale sports betting system enjoys the accompanying benefits: You make certain to build your net rewards by multiplying your bet each time you lose It is an incredible framework for present moment betting, particularly if you just mean to wager for a brief time frame Using the Martingale system football betting will assist you with recovering your misfortunes while you gain proficiency with a specific table game or online opening Using this procedure, a prepared online club player can slowly raise their wagers, recover their misfortunes, and create benefits Chapter 5 Risk of Using the Martingale System From an external perspective, stacking with this wagering methodology is a simple process.
Playing by this technique can convolute a few components altogether. The Martingale system football betting, for example, comes with risks that you ought to know about on the off chance that you choose to give it a shot for yourself. A decent spot to begin is by analyzing your bankroll and deciding your betting limits.
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# Lectures_Lect14 - Phys ic s 212 Le c ture 14 S lide Phys ic...
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Unformatted text preview: Phys ic s 212 Le c ture 14, S lide Phys ic s 212 Le c ture 14, S lide 1 Physics 212 Physics 212 Lecture 14 Lecture 14 Biot-Savart Law Biot-Savart Law 2 ˆ 4 I ds r dB r μ π × = :05 Phys ic s 212 Le c ture 14, S lide Phys ic s 212 Le c ture 14, S lide 2 Biot-Savart Law: Biot-Savart Law: :05 2 ˆ 4 I ds r dB r μ π × = We can use this law to calculate the magnetic field produced by We can use this law to calculate the magnetic field produced by ANY ANY current current distribution distribution BUT BUT Easy analytic calculations are possible only for a few distributions: Easy analytic calculations are possible only for a few distributions: Infinite Straight Wire Infinite Straight Wire Axis of Current Loop Axis of Current Loop Field at the center of a ring P ds r ds ˆ r d B P d B points out of page for each ds I 2 2 4 4 P P I I d s r d d B d B r r μ μ θ π π = = = 2 2 4 2 P I I B d B d r r π π μ μ θ π = = = ds r d θ = d θ 2 ˆ ˆ 4 I ds xr r d B r r r μ π = = Physics 212 Lecture 14, Slide Physics 212 Lecture 14, Slide 4 B fro m infinite line o f c urre nt Current I OUT • r = dis ta nc e fro m wire 7 4 10 / Tm A μ π- = 2 I B r μ π = r Ma g nitude : B :07 Dire c tio n: Thumb: on I Fingers: curl in direction of B Current in Wire Current in Wire Resulting B f ield Resulting B f ield I I Physics 212 Lecture 14, Slide Physics 212 Lecture 14, Slide 5 A long straight wire is carrying current from left to right. Two identical charges are moving with equal speed. Compare the magnitude of the force on charge a moving directly to the right, to the magnitude of the force on charge b moving up and to the right at the instant shown (i.e. same distance from the wire). a) |F a |> |F b | b) |F a |= |F b | c) |F a |< |F b | v I v (a) r r (b) 28 Currents + Charges F qv B = | | sin F qvB θ = Same q, |v|, B and θ (=90) :10 • B Forces are in different directions • F F Physics 212 Lecture 14, Slide Physics 212 Lecture 14, Slide 6 Two long wires carry equal and opposite currents What is the direction of the magnetic field above, and midway between the two wires carrying current – at the point marked “X”? x Magnetic Fields obey superposition A) Left B) Right C) Up D) Down E) Zero B x :13 Physics 212 Lecture 14, Slide Physics 212 Lecture 14, Slide 7 Force between current-carrying wires • I 1 F Conclusion: Currents in same direction attract....
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133,760 results
### Math
Logs are stacked up in apile as shown in the figure. The top row has 15 logs and the bottom row has 21 logs. How many logs are in the stack? a1= 1+14=15 an= 7+14=21 s1=(n/2)(a1+an) (n/2)(15+21) = 18 How do I figure out the n?
### Physics
Raft made of 2.2m long pine logs.Average diameter of logs 26cm.How many logs are needed to strap together to build raft to hold max 4 ?What percent of the logs is under water when you are on your raft in the water with your friend?
### Physics
Raft made of 2.2m long pine logs.Average diameter of logs 26cm.How many logs are needed to strap together to build raft to hold max 4 people? What percent of the logs is under water when you are on your raft in the water with your friend?
### Physics
PART 1 Raft made of 2.2m long pine logs with Average diameter of logs 26cm.How many logs are needed to strap together to build raft to hold max 4 people? Density of log = 600 kg/m^3 Each people weights 512N PART 2 What percent of the logs is under water when you are on your ...
### math - natural logs
Find the EXACT solution to e^(1-x) = 4^x You must use logs, not just a calculator.
### math logs
solve 3^(2x+1) = 9^(2x-1) using base 3 logs
### algebra
200 logs are stacked so that 20 logs at bottom row 19 in the next ,18 in the next and so on.In how many rows can 200 logs be arranged and how many are there on the top row ?
### Precalculus
.5=e^(-20t) I am assuming this involves logs but we did't really learn logs. Anyone care to walk me through with a calculator. Have the Ti-83
### Maths (Logs)
1.Write the following equation without logs log y = mx+ c This would be y = 10^(mx+c) right? 2. Explain why this fits the exponential model (in other words, it can be written in the form y = a x b^x where a and b are real numbers).
### Pre Cal (logs)
I need help with these logs 1: Log underscore4 16X 2: 2 Ln ( X(sqr. root)e 3:5^[2logUnderscore5(3x)] 4:log underscore2 ^[1280-logunderscore2 5) I don't understand how to do them thanks
### math
Mr Logan can saw 8 logs in 3 hours. How many logs can Mr Logan saw in one hour?
### chem
"Yule logs are simulated wood logs make of compressed paper that burn with a red and green flame.How would you produce a similar yule log?"
### Physics
A raft is made of 14 logs lashed together. Each is 29.5 cm in diameter and has a length of 6.50 m. How many people (whole number) can the raft hold before they start getting their feet wet, assuming the average person has a mass of 75.0 kg? Do not neglect the weight of the ...
### Quantative Method
Generate your Unit transportation costs (in excel) 1. Ship construction logs from 8 provinces in Canada to 50 states in the U.S. Minimize total shipping cost. 2. Set up the demand in each state of the U.S. 3. Set up availability of logs in the 8 provinces of Canada
### chemistry
help required in calculating the pH value of a .84 mols per litre solution of sulfuric acid. I understand that pH = -log 10 (2x 0.84), I am not very experienced with logs and need assistance as to which keys to press on the scientific calculator in relation to the logs. thank ...
### more pre cal logs
sorry I am so lost I missed the class that talked about logs since my sister was in the hospital giving birth to a still born baby.... now I am all lost... Expand the log Ln 3sqrootx^2y/x+3 (the 3 is part of the sqroot not infront like the 3rd power) I hope that makes since
### Physics
A raft is made of 15 logs lashed together. Each is 30.0 cm in diameter and has a length of 7.00 m. How many people (whole number) can the raft hold before they start getting their feet wet, assuming the average person has a mass of 66.0 kg? Do not neglect the weight of the ...
### Logs-Algebra
How do you graph the following logs? f(x)=log5 (x-2) f(x)=log5 x-2 f(x)= log5 x f(x)=log5 (x+2)
### math - logs
2 ln (x) = ln (20-x)
10^3x-8=2^5-x
### Statistic
At peak periods 15 of attempted logins to an email service fail. log-in attempts are independent and each has the name probability of failing. Ann logs in repeatedly until she succeeds A. Find the probability that Ann needs to log in atleast four times before she succeeds B. ...
### math
solve with logs 3^(1-2x)=4^(x)
### Math - logs
the log(10^ -1)of 3.66 = of 9.86 = of -2.8 = thank you
### math
how do i solve these logs? log1/2 (3x+1)^1/3= -2 3^(x^3)= 9^x
### Math 12 (Logs)
Solve: 3^(2a+3)=7^(a-4) Please Help and Thank you
### math- logarithms
how do i solve these logs? log1/2 (3x+1)^1/3= -2 3^(x^3)= 9^x
### math
How would I simplify this problem using logs? (25/16) -3/2
### math
how do you solve for x using logs: [(a + b)^x]/b^x I am not sure of the steps to get there.
### Math
Could you check these pleas? Firewood usually sold by a measure known as a cord. A full cord may be a stack 8x4x4ft or a stack 8x8x2ft 10. what is the volume of a full cord? 8*4*4=(128ft) 11. A short cord or a face cord of wood is 8x4xThe lenght of the logs. What is the volume...
### Science
Which of the following materials will burn the fastest in open air? A. a log, two feet in diameter B. two logs, each one foot in diameter C. a pile of small splinters made from a two-foot diameter log D. Both logs and the splinters will burn at the same rate.
### Science
Which of the following materials will burn the fastest in open air? A. a log, two feet in diameter B. two logs, each one foot in diameter C. a pile of small splinters made from a two-foot diameter log D. Both logs and the splinters will burn at the same rate.
### math - logs
if logx3 = a and logx25 = b, determine an expression for logx(9/5).
### LOGARITHMS
HI THE QUESTION STATES THAT ONE SHOULD CALCULATE THE LOGS WITHOUT THE USE OF A CALCULATOR...OKAY MY PROBLEM IS THIS... 2 3 2log 8 + 2log 8 what should i do? i had come to the piont of... 4 5 log 8 + log 8 But what's to do now i don't know..Think u can help?... cheers michelle ...
### math logs - please help
True or false: 2 ln x / ln 5 = ln x^2 - ln 5 for all positive values of x.
### math condense logs
write the expression as the logarithm of a single quantity 3[ln x-2 ln(x^2+1)]+2 ln 5
### math
5 divided by (3-e to the -x power)=4 5/(3- e^-x) = 4 5 = 12 - 4 e^-x 4 e^-x = 7 e^-x = 7/4 e^x = 4/7 Take natural logs of both sides, x = ln (4/7) = -0.5596
### math
also a problem that i cant figure out is how to do inverses of logs f(x)= 2log(base2)X
### Math (logs)
In x = -3 find x. I worked it out like this e^inx= e^-3 then I got x = 1/e^3. Is that correct?
### exponential and logarithmic graphs question
hi, i desperately need help solving this question: Find the values of A and k if the general equation of the graph shown is y=Ae^kx where the graph goes through the points (-1,-4) and (-2,-10) do i solve it using simultaneous equations? because i have attempted this yet failed...
### Math (check please)
Could you check these please? Firewood is usually sold by a measure known as a cord. A full cord may be a stack 8x4x4ft or a stack 8x8x2ft 10. what is the volume of a full cord? 8*4*4=(128ft) 11. A short cord or a face cord of wood is 8x4xThe lenght of the logs. What is the ...
### Logarithms
How do I solve this using logs? I am having a total math meltdown tonight! 10=2^10/n Thanks again!
### Math
solve for x without using a calculator: log5(3^x) = log25(9^(1-2x)) I don't know how to get rid of the logs for this one. Help? Thank you!
### math
How would you condense a logarithm expression into a single quantity if the logs have different bases?
### MATH 1111
College Algebra help with logarithms! Ln (x-4) - Ln (x+1) = 16 How do I take natural logs?? it's a study guide question for our final exams
### Math
1. Given the product law of logarithms, prove the product law of exponents. 2. Given the quotient law of logarithms, prove the quotient law of exponents. 3. Apply algebraic reasoning to show that a=b^(loga/logb) for any a,b>0 Please explain these to me. All I know is that ...
logs?
### Math logs
Solve the system of equations: y= (lnx)^2 + 2 (lnx^2) y = 3ln(1/x^2)+24 Im not sure what to do..
### calculus
Differentiate y=lnx You ought to have this memorized. y'=1/x for y= ln x dy/dx = 1/x This is the most basic and fundamental of log derivatives. If you are studying Calculus, and the topic is logs, I fail to understand why you asked that question. IS there a website that can ...
### Math Logs
logbase4(x+5)=3+logbase4(x-2) Tried so many times please help!
### Science
3. Which of the following conditions would likely cause the activation energy to be high? A.atoms are close together B.the temperature is hot C.atoms are not close together * D.a catalyst is present 4. Which of the following materials will burn the fastest in open air? A.a log...
### natural logs
Solve for x: ln(x+1)+ 2 = x^2 – 8x + 12
### natural logs
solve for x. xe^x = 9
### logs
How do you solve: log (1/3) 1 ?
Log3x=4
### Math
Which term is 13122? The geometric sequence is 2, -6, 18, -54 tn=ar^(n-1) 13122=2(-3)^(n-1) 6561=(-3)^(n-1) log6561=n-1log-3 You can't divide with negative logs, how do I isolate n?
### math. logs
anyone know how to solve this step by step? e^(3ln2)
### Math - logs
the log(10^ -1)of 3.66 = of 9.86 = of -2.8 = please help. (the 10^-1 is the subscript to log)
### maths
solve for x 8=6^x using logs
### precal
logs 9^(2x-5)=27(x+4) and 3log(underscore4)(2x-1)=15 Thank you
### precalculus logs
what is the domain of f(x)=-ln(-x) and the inverse
### algebra
solve e^x / 3 = 5^x using logs.
### LOGS (ALGEBRA)
FIND THE INVERSE: f(x) = 10^(5x+3) - 10 f^-1(x) =
### logs
e^1−10x=9 solve for x
### Exponents and Logs (Math)
Solve for x: 2^(-x+1) = 4(8^x) I try isolating the 4 but I can't seem to solve it correctly when using ln or log applications. The answer is supposed to be x=-1/4. Any help?
### Precalculus
This is for solving exponential/logarithmic functions: (This is a base e Logarithmic function I would assume): e^(4x)/10 =4^x-2 ? I understand the properties of logs for the most part, but I have a hard time figuring out the step-by-step process on how to solve exponential/log...
### Precalc
if you have how do u do 2^2x - 2^x - 6 =0 (calc)how do u do it? using logs??
### Logs
log6(x+7)−log6x=2 Solve for x
### logs 3
How do you create a table to graph log (3/2) x ???
### calculus
Solve the equation 14=65(19)x for x using logs.
Log3(x)=log9(6x)
### physics
You are planning to build a log cabin in northern Minnesota. You will pull the logs up a long, smooth hill to the building site by means of a rope attached to a winch. You need to buy a rope for this purpose, so you need to know how strong the rope must be. Stronger ropes cost...
### Logs
Solve for x. log5x=(1/2). Express the answer to the nearest hundredth.
### Algebra
express log (3x^5y^3/z^3) as the sum and difference of logs with no exponents.
### Math
I understand generally how to solve, by taking natural logs and the the derivative, but i am confused by the presence of the 3 in front... if f(x)= 3(sin(x))^x, find f'(1) should i move 3 to the y side or take the natural log of it? Thanks!
### Langauge
Underline the complete compound predicate. Circle the verbs.. 1/ Millworkers peel logs or cut boards. Complete pradicate: peel logs or cut boards. Verb: peel 2/ truckers load the boards and deliver them. C.p: load the board and deliver them. V.: load 3/ merchants sell products...
### natural logs
solve for x. xe^x = 9 please help i am stuck. i keep going in circles.
### logs
log (base 8) 1 = ??? I know it's 8 to some power gives me 1, but I can't figure it out.
### solving for x math
2e^(4x^2)=3 how do i solve for x. do i use the ln methode because i don't know how to do that. by calculator, i got x=.168 but i don't know how to show my work. please help me solve for x. Take natural logs of both sides. ln 2 + 4 x^2 = ln 3 4 x^2 = ln (3/2) = 0.40547 x^2 = 0....
### Math
How would you solve a problem like this with logarithims? I tried to start a couple of different ways but I kept getting stuck. Do you take the logs of both sides right away or do you do something else first? 17c^2/3 = 44
### Math-LOGS
Express x in terms of a,b and c. log x = 1/2 (log a + log b - log c) Please solve and explain how to do this type of problem, thank you!
### Math Repost (Logs)
.704=log(100/x) I tried this 10^.704=10^(100/x) But i not sure what to do now?
### Precalc (LOGS)
1. (e^x - 1 )/(x^2 - 1) < 0 2. The inverse of ln (3 - x) ? 3. Simplify x^3 - 3x^2 + 2 = 0 Any help would be GREATLY appreciated. Thank you so much for your time.
### math
If you can cut three logs in six minutes. How many minutes will it take to cut 8? 16 is not the answer.
### Math - logs
I need help solving this because it involves 2 bases: log (base 2) [log (base 5) 25]= ???
### social studies
mahogany logs from central America sometimes wash up in scotland. how do you explain this?
### Alg 2
How do you simplify this expression? It gets confusing b/c of logs and exponents. Can someone help me explain how to do this. 8^log(8)x
### PreCalculus
Assume that x y and a are positive numbers. Use the properties of logs to write the expression log a 9(squrt)xy.
### misplaced nodifier
we finally found the fire extinguisher we had been hunting for behind a pile of logs
### Precalc
I need help with logs. Thank you! Evaluate. 1. log base(3) 25 to the cubed root Condense. 2. 1/3logx + 2logy^3 -3log(z+1)
### Logs
Determine the domain of the following functions, write answer using interval notation. T(x)=log12(x^2+14x)
### math- exponents and logs
5,000=55.66 (1.114)^x (i.e. What is x?) ln 5,000= x ln 55.66 1.114 I keep getting 2 point something, but does this make sense? My calculator display looks like this: ln 5000/ln (55.66)(1.114)
### Math - Logarithmic
Is the following correct when restating logs with variables to exponential? Log: log_a(a^3) --- base = a Exp: a^3 = a I know when dealing with real numbers, this is what you do: Log: log(10^5) Exp: 10^5 = 100 000
### math
the population of Smallville in the year 1890 was 6250.assume the population increased at a rate of 2.75% per year. predict when the population reached 50,000 I'm not suppose to use logs to find the answer :(
### algebra 2
find the points of intersection of the following algebraically y=2^x + 4^x y=2^x+1 - 4^x+1 I set them equal to each other but not sure if you use logs or what to solve it
### math
I need an expression written for the following logs using "sum, difference, or multiple of logarithms." a) log[base(2)] ((x^2) - 9)/(x^3))^3 b) log[base(a)]5a^4 Hopefully they are clearly written. thanks. -roberto
### Natural Logs
When you are subtracting functions with ln, I know you divide the numbers after ln, but what do you do with constants in front of ln? For example, I'm currently dealing with: 4ln8 - 4ln3 Thanks!
### trigonometry
(3^m)(5^(m+1))=12^(2m-1) not sure how to solve this but I think you use natural logs? I'm just stuck because the first two are multiplying and not adding and we didn't go over this
### language arts
Three Indian elephants patiently towed the logs that had just been cut. is this an adverb clause
### math-logs
1. logaB = 2 and logbC = 4, evaluate logcA *the lowercase letters are bases ie. log(base a)B 2. solve for x: (log(base 2)x))^3 = log(base 2)x
### Math
What is the real value of x in the equation log(subscript 2)24-log(subscript 2)3=log(subscript 5)x? I really cannot remember at all how logs work. I know the answer is 125 but have no idea how to get there. Thanks in advance. | 4,702 | 15,627 | {"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} | 4.0625 | 4 | CC-MAIN-2017-34 | latest | en | 0.908955 |
https://www.formulas.today/ja/formulas/%E6%8C%87%E6%95%B0%E9%96%A2%E6%95%B0/ | 1,726,624,351,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00596.warc.gz | 710,729,139 | 5,259 | ## 指数関数の解明：公式、例、応用
ベース: 指数:
# Unraveling the Exponential Function: Formula, Examples, and Applications
Formula: `f(x) = a^x`
## Introduction to the Exponential Function
The exponential function is one of the most fascinating and widely used functions in mathematics. Represented as `f(x) = a^x`, where `a` is the base and `x` is the exponent, its application spans across various fields like finance, physics, and computer science. This article will delve deep into understanding what the exponential function is, how it works, and its real-life applications.
## Understanding the Exponential Function Formula
At its core, the exponential function can be defined as:
`f(x) = a^x`
Here:
• a: Base of the exponential function (must be a positive real number, typically not equal to 1).
• x: Exponent (can be any real number).
Essentially, the function takes a base number and raises it to the power of the exponent. The result is typically greater than the base for any positive exponent, between 0 and 1 for a negative exponent, and always equal to 1 when the exponent is 0.
## Real-Life Examples and Applications
Now that we have a basic understanding of the exponential function formula, let's explore some real-life examples and applications of this powerful mathematical tool.
### Finance
One of the most common applications of the exponential function is in finance, particularly in calculating compounded interest. The formula for compound interest is given by:
`A = P(1 + r/n)^(nt)`
Where:
• P: Principal amount (initial investment).
• r: Annual interest rate (as a decimal).
• n: Number of times interest is compounded per year.
• t: Time the money is invested for, in years.
Imagine you invested \$1,000 (P) at an annual interest rate of 5% (r = 0.05), compounded quarterly (n = 4), for 10 years (t). Using the exponential function, we can calculate:
`A = 1000(1 + 0.05/4)^(4*10)`
The result is approximately \$1,648.72, showing how investments grow exponentially over time.
### Physics
In the realm of physics, exponential functions often describe natural growth and decay processes. For instance, radioactive decay can be modeled with the formula:
`N(t) = N_0 e^(-λt)`
Where:
• N(t): Quantity of substance at time t.
• N_0: Initial quantity of substance.
• λ: Decay constant (determines the rate of decay).
• e: Euler's number, approximately equal to 2.71828.
This formula helps scientists predict how much of a substance will remain after a certain period, which is crucial for fields like nuclear physics and archeology.
### Biology
Exponential growth models in biology often describe how populations increase under ideal conditions. For example, the population of bacteria can grow exponentially under favorable conditions. The formula is similar to other exponential equations:
`N(t) = N_0 * 2^(t/T)`
Where:
• N(t): Population at time t.
• N_0: Initial population.
• T: Doubling time.
If a bacterial culture starts with a population of 500 (N_0) and doubles every 3 hours (T), the population after 9 hours can be calculated using this formula. Plugging in the values, we get:
`N(9) = 500 * 2^(9/3) = 500 * 2^3 = 500 * 8 = 4000`
Hence, the bacterial population grows to 4,000.
## Data Tables Illustrating Exponential Growth and Decay
### Example of Exponential Growth in Finance
Year Investment Value (USD)
0 1000
1 1050
2 1102.50
3 1157.63
### Example of Exponential Decay in Radioactive Material
Time Elapsed (Years) Remaining Substance (%)
0 100
1 81.87
2 67.03
3 54.88
• Q: What is an exponential function?
A: An exponential function is a mathematical expression of the form `f(x) = a^x`, where `a` is a positive constant called the base, and `x` is the exponent.
• Q: Where are exponential functions used in real life?
A: Exponential functions are used in various fields including finance (compound interest), physics (radioactive decay), biology (population growth), and more.
• Q: What is the significance of the base `e` in exponential functions?
A: The base `e` (approximately 2.71828) is a mathematical constant that appears naturally in many processes and is the base of natural logarithms. Functions with base `e` are called natural exponential functions.
• Q: How do we differentiate an exponential function?
A: If `f(x) = a^x`, then the derivative is `f'(x) = a^x * ln(a)`, where `ln(a)` is the natural logarithm of the base `a`.
## Conclusion
The exponential function is a powerful tool that models a variety of real-life phenomena. From calculating compound interest in finance to modeling population growth in biology, its applications are endless. By understanding the formula `f(x) = a^x`, we can unlock a wealth of knowledge that allows us to analyze and predict behavior in numerous scientific and financial contexts. The more we understand this function, the better we are equipped to harness its potential to solve real-world problems.
Tags: 数学, 指数関数, 実生活での応用 | 1,203 | 4,940 | {"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} | 4.78125 | 5 | CC-MAIN-2024-38 | latest | en | 0.857651 |
https://www.yaclass.in/p/mathematics-state-board/class-9/real-numbers-2951/real-numbers-and-its-visual-representation-2814/re-8f22fa4d-00db-461c-a55d-38a5d6b35d43 | 1,603,263,812,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107876136.24/warc/CC-MAIN-20201021064154-20201021094154-00693.warc.gz | 988,155,815 | 7,586 | ### Theory:
Real numbers are either rational or irrational
We learned that rational numbers satisfies the commutative, associative and distributive properties for addition and multiplication. Thus rational number is closed under addition and multiplication.
We also learned that irrational numbers also satisfy the commutative, associative and distributive properties for addition and multiplication.
But the sum, difference, product and division of two irrational numbers is not always irrational. The operations on irrational numbers results in rational or irrational.
Thus, it can be concluded that the irrational number is not closed under addition, subtraction, multiplication and division.
Let us recall some properties on irrationals from the previous topic and solve some problems on it.
1. Addition, subtraction, multiplication and division of two irrational number is may or may not be irrational.
2. Addition of rational and irrational number is always irrational.
3. Subtraction of rational and irrational number is always irrational.
4. Multiplication of rational and irrational is always irrational.
5. Division of rational and irrational is always irrational. | 214 | 1,179 | {"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} | 3.640625 | 4 | CC-MAIN-2020-45 | latest | en | 0.93313 |
http://math.tutorpace.com/statistics/median-in-math-online-tutoring | 1,490,561,136,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189252.15/warc/CC-MAIN-20170322212949-00001-ip-10-233-31-227.ec2.internal.warc.gz | 234,097,830 | 10,440 | # Median in Math
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
Median is the middle number of a set of observation.
Example: - Median of 6, 10, 20 is 10 because 10 is the middle value and 10 divides the set in two equal portion. So the number of values to the left of the median will be equal to the number of values to the right.
Following are the steps to find the median.
Step 1: Arrange the data.
Step 2: Find the position of the median by using the formula
(n + 1) / 2
If there are odd number of values in the data then you wil get the whole number by using
(n + 1) / 2. So there will be only one middle value which is the median.
If there are even numbers of values in the data then you will get a decimal number by applying (n + 1) / 2. So there will be two middle value and after averaging them you will get the median.
Example: - Find the median of
9 2 6 5 1
Solution: - Step 1: - Arrange data is
1 2 5 6 9
Step 2: - (n+1) /2= (5+1)/2= 3
Hence 3rd position is the median.
Median = 5
Example 2: Find the median of
9 2 6 5 1 3
Solution: - Arrange data is
1 2 3 5 6 9
(n+1)/2= (6+1)/2= 3.5
Therefore average of 3rd and 4th position value gives the median.
Median= (3+5)/2= 4 | 412 | 1,432 | {"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} | 4.53125 | 5 | CC-MAIN-2017-13 | longest | en | 0.739161 |
https://us.metamath.org/ileuni/blininf.html | 1,725,980,520,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00580.warc.gz | 555,704,371 | 7,557 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > blininf GIF version
Theorem blininf 12771
Description: The intersection of two balls with the same center is the smaller of them. (Contributed by NM, 1-Sep-2006.) (Revised by Mario Carneiro, 12-Nov-2013.)
Assertion
Ref Expression
blininf (((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋) ∧ (𝑅 ∈ ℝ*𝑆 ∈ ℝ*)) → ((𝑃(ball‘𝐷)𝑅) ∩ (𝑃(ball‘𝐷)𝑆)) = (𝑃(ball‘𝐷)inf({𝑅, 𝑆}, ℝ*, < )))
Proof of Theorem blininf
Dummy variable 𝑥 is distinct from all other variables.
StepHypRef Expression
1 xmetcl 12699 . . . . . . 7 ((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋𝑥𝑋) → (𝑃𝐷𝑥) ∈ ℝ*)
213expa 1182 . . . . . 6 (((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋) ∧ 𝑥𝑋) → (𝑃𝐷𝑥) ∈ ℝ*)
32adantlr 469 . . . . 5 ((((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋) ∧ (𝑅 ∈ ℝ*𝑆 ∈ ℝ*)) ∧ 𝑥𝑋) → (𝑃𝐷𝑥) ∈ ℝ*)
4 simplrl 525 . . . . 5 ((((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋) ∧ (𝑅 ∈ ℝ*𝑆 ∈ ℝ*)) ∧ 𝑥𝑋) → 𝑅 ∈ ℝ*)
5 simplrr 526 . . . . 5 ((((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋) ∧ (𝑅 ∈ ℝ*𝑆 ∈ ℝ*)) ∧ 𝑥𝑋) → 𝑆 ∈ ℝ*)
6 xrltmininf 11144 . . . . 5 (((𝑃𝐷𝑥) ∈ ℝ*𝑅 ∈ ℝ*𝑆 ∈ ℝ*) → ((𝑃𝐷𝑥) < inf({𝑅, 𝑆}, ℝ*, < ) ↔ ((𝑃𝐷𝑥) < 𝑅 ∧ (𝑃𝐷𝑥) < 𝑆)))
73, 4, 5, 6syl3anc 1217 . . . 4 ((((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋) ∧ (𝑅 ∈ ℝ*𝑆 ∈ ℝ*)) ∧ 𝑥𝑋) → ((𝑃𝐷𝑥) < inf({𝑅, 𝑆}, ℝ*, < ) ↔ ((𝑃𝐷𝑥) < 𝑅 ∧ (𝑃𝐷𝑥) < 𝑆)))
87pm5.32da 448 . . 3 (((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋) ∧ (𝑅 ∈ ℝ*𝑆 ∈ ℝ*)) → ((𝑥𝑋 ∧ (𝑃𝐷𝑥) < inf({𝑅, 𝑆}, ℝ*, < )) ↔ (𝑥𝑋 ∧ ((𝑃𝐷𝑥) < 𝑅 ∧ (𝑃𝐷𝑥) < 𝑆))))
9 xrmincl 11140 . . . 4 ((𝑅 ∈ ℝ*𝑆 ∈ ℝ*) → inf({𝑅, 𝑆}, ℝ*, < ) ∈ ℝ*)
10 elbl 12738 . . . . 5 ((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋 ∧ inf({𝑅, 𝑆}, ℝ*, < ) ∈ ℝ*) → (𝑥 ∈ (𝑃(ball‘𝐷)inf({𝑅, 𝑆}, ℝ*, < )) ↔ (𝑥𝑋 ∧ (𝑃𝐷𝑥) < inf({𝑅, 𝑆}, ℝ*, < ))))
11103expa 1182 . . . 4 (((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋) ∧ inf({𝑅, 𝑆}, ℝ*, < ) ∈ ℝ*) → (𝑥 ∈ (𝑃(ball‘𝐷)inf({𝑅, 𝑆}, ℝ*, < )) ↔ (𝑥𝑋 ∧ (𝑃𝐷𝑥) < inf({𝑅, 𝑆}, ℝ*, < ))))
129, 11sylan2 284 . . 3 (((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋) ∧ (𝑅 ∈ ℝ*𝑆 ∈ ℝ*)) → (𝑥 ∈ (𝑃(ball‘𝐷)inf({𝑅, 𝑆}, ℝ*, < )) ↔ (𝑥𝑋 ∧ (𝑃𝐷𝑥) < inf({𝑅, 𝑆}, ℝ*, < ))))
13 elbl 12738 . . . . . . 7 ((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋𝑅 ∈ ℝ*) → (𝑥 ∈ (𝑃(ball‘𝐷)𝑅) ↔ (𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑅)))
14133expa 1182 . . . . . 6 (((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋) ∧ 𝑅 ∈ ℝ*) → (𝑥 ∈ (𝑃(ball‘𝐷)𝑅) ↔ (𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑅)))
1514adantrr 471 . . . . 5 (((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋) ∧ (𝑅 ∈ ℝ*𝑆 ∈ ℝ*)) → (𝑥 ∈ (𝑃(ball‘𝐷)𝑅) ↔ (𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑅)))
16 elbl 12738 . . . . . . 7 ((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋𝑆 ∈ ℝ*) → (𝑥 ∈ (𝑃(ball‘𝐷)𝑆) ↔ (𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑆)))
17163expa 1182 . . . . . 6 (((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋) ∧ 𝑆 ∈ ℝ*) → (𝑥 ∈ (𝑃(ball‘𝐷)𝑆) ↔ (𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑆)))
1817adantrl 470 . . . . 5 (((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋) ∧ (𝑅 ∈ ℝ*𝑆 ∈ ℝ*)) → (𝑥 ∈ (𝑃(ball‘𝐷)𝑆) ↔ (𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑆)))
1915, 18anbi12d 465 . . . 4 (((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋) ∧ (𝑅 ∈ ℝ*𝑆 ∈ ℝ*)) → ((𝑥 ∈ (𝑃(ball‘𝐷)𝑅) ∧ 𝑥 ∈ (𝑃(ball‘𝐷)𝑆)) ↔ ((𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑅) ∧ (𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑆))))
20 elin 3286 . . . 4 (𝑥 ∈ ((𝑃(ball‘𝐷)𝑅) ∩ (𝑃(ball‘𝐷)𝑆)) ↔ (𝑥 ∈ (𝑃(ball‘𝐷)𝑅) ∧ 𝑥 ∈ (𝑃(ball‘𝐷)𝑆)))
21 anandi 580 . . . 4 ((𝑥𝑋 ∧ ((𝑃𝐷𝑥) < 𝑅 ∧ (𝑃𝐷𝑥) < 𝑆)) ↔ ((𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑅) ∧ (𝑥𝑋 ∧ (𝑃𝐷𝑥) < 𝑆)))
2219, 20, 213bitr4g 222 . . 3 (((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋) ∧ (𝑅 ∈ ℝ*𝑆 ∈ ℝ*)) → (𝑥 ∈ ((𝑃(ball‘𝐷)𝑅) ∩ (𝑃(ball‘𝐷)𝑆)) ↔ (𝑥𝑋 ∧ ((𝑃𝐷𝑥) < 𝑅 ∧ (𝑃𝐷𝑥) < 𝑆))))
238, 12, 223bitr4rd 220 . 2 (((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋) ∧ (𝑅 ∈ ℝ*𝑆 ∈ ℝ*)) → (𝑥 ∈ ((𝑃(ball‘𝐷)𝑅) ∩ (𝑃(ball‘𝐷)𝑆)) ↔ 𝑥 ∈ (𝑃(ball‘𝐷)inf({𝑅, 𝑆}, ℝ*, < ))))
2423eqrdv 2152 1 (((𝐷 ∈ (∞Met‘𝑋) ∧ 𝑃𝑋) ∧ (𝑅 ∈ ℝ*𝑆 ∈ ℝ*)) → ((𝑃(ball‘𝐷)𝑅) ∩ (𝑃(ball‘𝐷)𝑆)) = (𝑃(ball‘𝐷)inf({𝑅, 𝑆}, ℝ*, < )))
Colors of variables: wff set class Syntax hints: → wi 4 ∧ wa 103 ↔ wb 104 = wceq 1332 ∈ wcel 2125 ∩ cin 3097 {cpr 3557 class class class wbr 3961 ‘cfv 5163 (class class class)co 5814 infcinf 6915 ℝ*cxr 7890 < clt 7891 ∞Metcxmet 12327 ballcbl 12329 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-in1 604 ax-in2 605 ax-io 699 ax-5 1424 ax-7 1425 ax-gen 1426 ax-ie1 1470 ax-ie2 1471 ax-8 1481 ax-10 1482 ax-11 1483 ax-i12 1484 ax-bndl 1486 ax-4 1487 ax-17 1503 ax-i9 1507 ax-ial 1511 ax-i5r 1512 ax-13 2127 ax-14 2128 ax-ext 2136 ax-coll 4075 ax-sep 4078 ax-nul 4086 ax-pow 4130 ax-pr 4164 ax-un 4388 ax-setind 4490 ax-iinf 4541 ax-cnex 7802 ax-resscn 7803 ax-1cn 7804 ax-1re 7805 ax-icn 7806 ax-addcl 7807 ax-addrcl 7808 ax-mulcl 7809 ax-mulrcl 7810 ax-addcom 7811 ax-mulcom 7812 ax-addass 7813 ax-mulass 7814 ax-distr 7815 ax-i2m1 7816 ax-0lt1 7817 ax-1rid 7818 ax-0id 7819 ax-rnegex 7820 ax-precex 7821 ax-cnre 7822 ax-pre-ltirr 7823 ax-pre-ltwlin 7824 ax-pre-lttrn 7825 ax-pre-apti 7826 ax-pre-ltadd 7827 ax-pre-mulgt0 7828 ax-pre-mulext 7829 ax-arch 7830 ax-caucvg 7831 This theorem depends on definitions: df-bi 116 df-dc 821 df-3or 964 df-3an 965 df-tru 1335 df-fal 1338 df-nf 1438 df-sb 1740 df-eu 2006 df-mo 2007 df-clab 2141 df-cleq 2147 df-clel 2150 df-nfc 2285 df-ne 2325 df-nel 2420 df-ral 2437 df-rex 2438 df-reu 2439 df-rmo 2440 df-rab 2441 df-v 2711 df-sbc 2934 df-csb 3028 df-dif 3100 df-un 3102 df-in 3104 df-ss 3111 df-nul 3391 df-if 3502 df-pw 3541 df-sn 3562 df-pr 3563 df-op 3565 df-uni 3769 df-int 3804 df-iun 3847 df-br 3962 df-opab 4022 df-mpt 4023 df-tr 4059 df-id 4248 df-po 4251 df-iso 4252 df-iord 4321 df-on 4323 df-ilim 4324 df-suc 4326 df-iom 4544 df-xp 4585 df-rel 4586 df-cnv 4587 df-co 4588 df-dm 4589 df-rn 4590 df-res 4591 df-ima 4592 df-iota 5128 df-fun 5165 df-fn 5166 df-f 5167 df-f1 5168 df-fo 5169 df-f1o 5170 df-fv 5171 df-isom 5172 df-riota 5770 df-ov 5817 df-oprab 5818 df-mpo 5819 df-1st 6078 df-2nd 6079 df-recs 6242 df-frec 6328 df-map 6584 df-sup 6916 df-inf 6917 df-pnf 7893 df-mnf 7894 df-xr 7895 df-ltxr 7896 df-le 7897 df-sub 8027 df-neg 8028 df-reap 8429 df-ap 8436 df-div 8525 df-inn 8813 df-2 8871 df-3 8872 df-4 8873 df-n0 9070 df-z 9147 df-uz 9419 df-rp 9539 df-xneg 9657 df-seqfrec 10323 df-exp 10397 df-cj 10719 df-re 10720 df-im 10721 df-rsqrt 10875 df-abs 10876 df-psmet 12334 df-xmet 12335 df-bl 12337 This theorem is referenced by: blin2 12779
Copyright terms: Public domain W3C validator | 4,017 | 5,889 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-38 | latest | en | 0.204714 |
http://www.hpmuseum.org/software/41/41lagrn.htm | 1,534,555,346,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213247.0/warc/CC-MAIN-20180818001437-20180818021437-00243.warc.gz | 511,201,164 | 6,030 | The Museum of HP Calculators
# Lagrange's Interpolation Formula for the HP-41
This program is Copyright © 2003-2006 by Jean-Marc Baillard and is used here by permission.
This program is supplied without representation or warranty of any kind. Jean-Marc Baillard and The Museum of HP Calculators therefore assume no responsibility and shall have no liability, consequential or otherwise, of any kind arising from the use of this program material or any part thereof.
Overview:
1°) Univariate Interpolation
2°) Bivariate Interpolation ( new )
a) non-synthetic program ( new )
b) synthetic program ( new )
1°) Univariate Interpolation
-Let be a set of n data points: M1 ( x1 ; y1 ) , M2 ( x2 ; y2 ) , .......... , Mn ( xn ; yn )
There is a unique polynomial L of degree < n such that: L(xi) = yi i = 1 ; 2 ; ..... ; n
-In other words, L is the collocation polynomial for these arguments.
( xi values can be unequally-spaced.)
-The following program calculates L(x) = y1 L1(x) + y2 L2(x) + ..... + yn Ln(x)
where Li(x) = ( x - x1 ) ...... ( x - xi-1 ) ( x - xi+1 ) ...... ( x - xn ) / [ ( xi - x1 ) ...... ( xi - xi-1 ) ( xi - xi+1 ) ( xi - xn ) ]
are the Lagrange's multiplier functions.
-This formula is also applied to an extrapolation to the limit in example2.
Data Registers:
-You have to store the 2n numbers x1 , y1 , x2 , y2 , ........ , xn , yn into contiguous registers, starting at any register Rbb:
x1 x2 ........... xn y1 y2 ........... yn
into
Rbb Rbb+2 ............. Ree-1 Rbb+1 Rbb+3 ............. Ree
( ee = bb + 2n -1 )
Flags: /
Subroutines: /
-Thanks to synthetic programming, "LAGR" only uses the registers containing the xi and yi numbers,
but you can replace status registers M N O P Q by the standard R00 R01 R02 R03 R04 ( in this case, of course, bb > 04 ).
01 LBL "LAGR"
02 CLA
03 STO M
04 X<>Y
05 STO O
06 STO Q
07 LBL 01
08 RCL Q
09 STO N
10 SIGN
11 LBL 02
12 RCL M
13 RCL IND O
14 RCL IND N
15 ST- Z
16 -
17 X#0?
18 GTO 03
19 SIGN
20 STO Y
21 LBL 03
22 /
23 *
24 ISG N
25 ISG N
26 GTO 02
27 ISG O
28 RCL IND O
29 *
30 ST+ P
31 ISG O
32 GTO 01
33 RCL Q
34 RCL M
35 SIGN
36 X<> P
37 CLA
38 END
( 69 bytes / SIZE? )
STACK INPUTS OUTPUTS Y bbb.eee bbb.eee X x L(x) L / x
Example1:
Evaluate y(3) and y(5) for the collocation polynomial that takes the values prescribed below:
xi 0 1 2 4 7 yi 3 2 4 6 5
For instance, you store these 10 numbers ( 0 3 1 2 2 4 4 6 7 5 ) into R01 thru R10, then
1.010 ENTER^
3 XEQ "LAGR" produces: y(3) = 5.847619048 ( in 14 seconds )
RDN 5 R/S ---------- y(5) = 4.523809520
Example2:
-Lagrange's polynomial can also be used for an extrapolation to the limit:
Suppose we are using the trapezoidal rule to evaluate the integral: I = §02 ex dx
The interval [0;2] is divided into n subintervals and
n = 1 yields I1 = 8.389056101
n = 2 ----- I2 = 6.912809880
n = 4 ----- I4 = 6.521610110
n = 8 ----- I8 = 6.422297820
We would like to know the result with n = infinity! i-e 1/n = 0.
The error of the trapezoidal rule is nearly proportional to 1/n2 , therefore, we can use Lagrange interpolation formula with
x1 = 1 x2 = 1/4 x3 = 1/16 x4 = 1/64
y1 = I1 y2 = I2 y3 = I4 y4 = I8 and evaluate L (0)
If these 8 numbers are stored in R11 thru R18 for instance,
11.018 ENTER^
0 XEQ "LAGR" produces 6.389056387
Exact result is of course e2 - 1 = 6.389056099 to ten places
and the error of our Lagrange approximation is only 3. 10-7 although I8 was only exact to 1 decimal!
N.B: 1- If you are using Simpson's rule instead of the trapezoidal rule, store xn = 1/n4 : it's a 4th order method ( at least far from any singularity )
2- n is usually doubled at every step, but it's not necessary here.
3- The same idea can be applied to any other problem of this kind.
2°) Bivariate Interpolation
a) Non-synthetic program
-Now we have a grid of nxm values: f(xi,yj) with i = 1,2,......,n & j = 1,2,.....,m and we want to estimate f(x,y)
-"LAGR2" uses the Lagrange Polynomial in the directions of x-axis and y-axis.
-You have to store the data in registers Rbb to Ree as shown below ( bb > 10 ):
Data Registers: • R00 = bbb.eeenn ( Registers R00 and Rbb thru Ree are to be initialized before executing "LAGR2" )
R01 = x R03 = f(x,y)
R02 = y R04 thru R10: temp We must choose bb > 10
x \ y | • Rbb+n = y1 • Rbb+2n+1 = y2 .............................. • Ree-n = ym
---------------------------------------------------------------------------------------------------------------------
• Rbb = x1 | • Rbb+n+1 = f(x1,y1) • Rbb+2n+2 = f(x1,y2) .............................. • Ree-n+1 = f(x1,ym)
• Rbb+1 = x2 | • Rbb+n+2 = f(x2,y1) • Rbb+2n+3 = f(x2,y2) .............................. • Ree-n+2 = f(x2,ym)
.................... | .............................................................................................................................................
|
• Rbb+n-1 = xn | • Rbb+2n = f(xn,y1) • Rbb+3n+1 = f(xn,y2) ............................... • Ree = f(xn,ym)
Flags: /
Subroutines: /
01 LBL "LAGR2"
02 STO 01
03 X<>Y
04 STO 02
05 CLX
06 STO 03
07 RCL 00
08 ISG X
09 STO 07
10 INT
11 DSE X
12 .1
13 %
14 RCL 00
15 INT
16 +
17 STO 05
18 STO 06
19 E-5
20 ST+ 07
21 LBL 00
22 RCL 07
23 STO 08
24 STO 09
25 FRC
26 RCL 06
27 INT
28 +
29 ISG X
30 STO 10
31 CLX
32 STO 04
33 LBL 01
34 RCL IND 10
35 LBL 02
36 RCL 02
37 RCL IND 08
38 RCL IND 09
39 ST- Z
40 -
41 X#0?
42 GTO 02
43 SIGN
44 STO Y
45 LBL 02
46 /
47 *
48 ISG 09
49 GTO 02
50 ST+ 04
51 RCL 07
52 STO 09
53 ISG 10
54 CLX
55 ISG 08
56 GTO 01
57 RCL 05
58 STO 10
59 RCL 04
60 LBL 03
61 RCL 01
62 RCL IND 06
63 RCL IND 10
64 ST- Z
65 -
66 X#0?
67 GTO 03
68 SIGN
69 STO Y
70 LBL 03
71 /
72 *
73 ISG 10
74 GTO 03
75 ST+ 03
76 ISG 06
77 GTO 00
78 RCL 01 Lines 78-79-80 are only useful to save y in Y-register and x in L-register
79 SIGN Otherwise, these lines may be deleted.
80 RCL 02
81 RCL 03
82 END
( 121 bytes / SIZE? )
STACK INPUTS OUTPUTS Y y y X x f(x,y) L / x
Example: A function f(x,y) is only known by the following values ( meaning for example: f(1,4) = 3 f(4,6) = 9 ...etc... )
x \ y | 2 3 4 6 | R14 R18 R22 R26
-------------------------- if, for instance, you choose bb = 11 ---------------------------------
1 | 4 3 3 5 store these 19 numbers into R11 | R15 R19 R23 R27 respectively
2 | 3 1 2 6 R12 | R16 R20 R24 R28
4 | 1 0 4 9 R13 | R17 R21 R25 R29
-The control number of this tableau is 11.02903 so 11.02903 STO 00
( note that nn = 03 is the number of x-values which may be different from m = the number of y-values ( here m = 4 ) )
-Then, to compute, say f(3,5)
5 ENTER^
3 XEQ "LAGR2" >>>> f(3,5) ~ 5.8333 ( in 29 seconds )
-Likewise, to obtain f(1.6,2.7)
2.7 ENTER^
1.6 R/S >>>> f(1.6,2.7) ~ 1.9038
Note: Execution time is approximately proportional to nxm
b) Synthetic program
-Synthetic programming allows to store the data into registers R01 thru Rn.m+n+m
-However, synthetic register a is used, so this program cannot be called as more than a first level subroutine.
Data Registers: • R00 = n.mmm = n + m/103 ( Registers R00 thru Rnm+m+n are to be initialized before executing "BVI" )
x \ y | • Rnn+1 = y1 • R2n+2 = y2 .............................. • Rnm+m = ym
---------------------------------------------------------------------------------------------------------------------
• R01 = x1 | • Rnn+2 = f(x1,y1) • R2n+3 = f(x1,y2) .............................. • Rnm+m+1 = f(x1,ym)
• R02 = x2 | • Rnn+3 = f(x2,y1) • R2n+4 = f(x2,y2) .............................. • Rnm+m+2 = f(x2,ym)
........... | .............................................................................................................................................
|
• Rnn = xn | • R2n+1 = f(xn,y1) • R3n+2 = f(xn,y2) .............................. • Rnm+m+n = f(xn,ym)
Flags: /
Subroutines: /
01 LBL "BVI"
02 CLA
03 STO M
04 X<>Y
05 STO N
06 RCL 00
07 INT
08 STO P ( synthetic )
09 LBL 01
10 RCL 00
11 INT
12 LASTX
13 FRC
14 ENTER^
15 SIGN
16 10^X
17 CHS
18 10^X
19 SQRT
20 +
21 ST* Y
22 +
23 ISG X
24 STO Q
25 CLX
26 LBL 02
27 RCL Q
28 FRC
29 ISG X
30 STO a
31 SIGN
32 LBL 03
33 RCL N
34 RCL IND a
35 -
36 RCL IND Q
37 ST- L
38 X<> L
39 CHS
40 X#0?
41 GTO 03
42 SIGN
43 STO Y
44 LBL 03
45 /
46 *
47 ISG a
48 GTO 03
49 RCL P
50 RCL Q
51 +
52 RDN
53 RCL IND T
54 *
55 +
56 ISG Q
57 GTO 02
58 RCL 00
59 INT
60 STO Q
61 X<>Y
62 LBL 04
63 RCL M
64 RCL IND P
65 RCL IND Q
66 ST- Z
67 -
68 X#0?
69 GTO 04
70 SIGN
71 STO Y
72 LBL 04
73 /
74 *
75 DSE Q
76 GTO 04
77 ST+ O
78 DSE P
79 GTO 01
80 RCL M
81 SIGN
82 RCL N
83 RCL O
84 CLA
85 END
( 131 bytes / SIZE (n+1)(m+1) )
STACK INPUTS OUTPUTS Y y y X x f(x,y) L / x
Example: With the same data as above:
x \ y | 2 3 4 6 | R04 R08 R12 R16
-------------------------- --------------------------------
1 | 4 3 3 5 are to be stored into R01 | R05 R09 R13 R17 respectively
2 | 3 1 2 6 R02 | R06 R10 R14 R18
4 | 1 0 4 9 R03 | R07 R11 R15 R19
-Here, n = 3 and m = 4 so 3.004 STO 00
-Then, to compute f(3,5)
5 ENTER^
3 XEQ "BVI" >>>> f(3,5) ~ 5.8333 ( in 31 seconds )
-And to obtain f(1.6,2.7)
2.7 ENTER^
1.6 R/S >>>> f(1.6,2.7) ~ 1.9038 | 3,871 | 10,778 | {"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} | 3.109375 | 3 | CC-MAIN-2018-34 | longest | en | 0.517189 |
https://board.flashkit.com/board/showthread.php?670308-Particle-Question-how-to-attach&s=1f596643344ebb4c533bb1427666c32f&p=3519573&viewfull=1 | 1,624,559,689,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488556482.89/warc/CC-MAIN-20210624171713-20210624201713-00308.warc.gz | 137,243,317 | 16,127 | A Flash Developer Resource Site
# Thread: Particle Question: how to attach...
1. ## Particle Question: how to attach...
Hi, if somebody could solve my problem, I finally can start creating levels for my game. Almost everything is done, only the ragdollphysics. The ragdolls itself are done, but I need to attach limb pictures etc... to the ragdoll
I'm using the Flade physics engine (http://www.cove.org/flade)
There you can download flade and the most usefull thing for me is that you create two particles and attach a stupid drawing to it (for example a woodblock). So that the block follows the constraint between the particles, got it?
Here's an example of the ragdoll (simple) that I created with Flade : http://sniperware.50webs.com/ragdoll.swf
My question is pretty hard isn't it? I tried everything!
2. Lets say, for example, the upper arm of your ragdoll extends from 'pointA' (shoulder) to 'pointB' (elbow)
to attach a movie clip to that, try doing the following thing:
(i'm assuming that the axis of the movieclip is where the arm is supposed to turn(shoulder))
1. position the movieclip at pointA.x,pointA.y
2. ang = getAngle(pointA,pointB) //in degrees
3. movieclip._rotation = ang //add or subtract from this value to correct for angle offset
im guessing you know how to find the angle between two points
3. No, i dont
thanks for the help.
4. ## try this
Code:
```function getAngle(pA,pB){
var xDiff=pA.x-pB.x
var yDiff=pA.y-pB.y
var radians = Math.atan2(yDiff, xDiff);
if (yDiff<0) {
}
return degrees
}```
good luck
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OnlineConversion Forums cubic feet in a ton of coal
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#1
01-13-2008, 05:38 PM
Unregistered Guest Posts: n/a
cubic feet in a ton of coal
I have a coal with a density of 52 lbs/cu. ft. How many cubic feet in ]
a ton.
Pete
#2
01-13-2008, 06:04 PM
Sakura Guest Posts: n/a
Re: cubic feet in a ton of coal
Hi
1 Tonne (metric ton) = 36.8394 cubic ft.
#3
01-13-2008, 08:10 PM
Roy Nakatsuka Guest Posts: n/a
Re: cubic feet in a ton of coal
Quote:
Originally Posted by Unregistered I have a coal with a density of 52 lbs/cu. ft. How many cubic feet in ] a ton. Pete
Hello Pete,
There are three main types of ton. I don't know which type you're referring to, so here are the results for all of them:
short ton = 2000 lb
volume = 2000 ÷ 52 = 38.5 cu. ft. per short ton
long ton = 2240 lb
volume = 2240 ÷ 52 = 43.1 cu. ft. per long ton
metric ton = tonne = 1000 kg = 2204.6 lb
volume = 2204.6 ÷ 52 = 42.4 cu. ft. per metric ton
Poster #2 obtained a result different from any of mine, but without being able to see any steps in their calculation, I don't know how they arrived at the answer they did.
#4
01-14-2008, 10:58 AM
JohnS Moderator Join Date: Dec 2007 Location: SE Michigan, USA Posts: 9,550 Rep Power: 19
Re: cubic feet in a ton of coal
I also don't know where the figure in post 2 came from. It may be a "typical" figure from some reference. Working backwards, it requires coal of about 59 lb/cu ft density. While that certainly exists, it is not what the OP has.
Roy has it correct for the three possible "flavors" of ton.
#5
02-19-2008, 11:57 AM
Unregistered Guest Posts: n/a
Re: cubic feet in a ton of coal
Quote:
Originally Posted by JohnS I also don't know where the figure in post 2 came from. It may be a "typical" figure from some reference. Working backwards, it requires coal of about 59 lb/cu ft density. While that certainly exists, it is not what the OP has. Roy has it correct for the three possible "flavors" of ton.
bernie if you have 52 lb in a cubic foot. you bought rock. coal is lighter than rock. when coal is run through a braker, rock sinks coal floats.
#6
04-13-2008, 03:02 PM
fishnutz Guest Posts: n/a
Re: cubic feet in a ton of coal
which coal has the greatest volume per ton....stove coal or pea coal..
the stove coal is the biggest piece and the pea is the smallest....in a ton of coal there is more spaces between the bigger pieces ,than the smaller pea coal pieces..........but if you smash up a big piece of coal you can't put it back together in the same volume or density.......De Ken pigpenken@aol.com
#7
06-26-2008, 12:26 PM
Unregistered Guest Posts: n/a
Re: cubic feet in a ton of coal
So please: What is the density of nut coal?
I'm building a coal bin. I need to know (1) minimum delivery tons from the supplier and (2) then how many cubic feet I need for each ton. I thought I hit the jackpot until I see that there is uncertainty on the density of coal.
#8
06-26-2008, 05:57 PM
JohnS Moderator Join Date: Dec 2007 Location: SE Michigan, USA Posts: 9,550 Rep Power: 19
Re: cubic feet in a ton of coal
Quote:
Originally Posted by Unregistered So please: What is the density of nut coal? I'm building a coal bin. I need to know (1) minimum delivery tons from the supplier and (2) then how many cubic feet I need for each ton. I thought I hit the jackpot until I see that there is uncertainty on the density of coal.
This thread on a coal-oriented forum recommends allowing 40 ft³/ton for the bin design as a safe number (most coal will really be a little denser.) But it's just a Google result; I have zero experience with burning coal:
#9
10-03-2008, 01:05 PM
Re: cubic feet in a ton of coal
Rock (like granite) weighs about 185 lb per cubic foot.
Water weighs 62.4 lbs per cubic foot.
coal that weighs less than 62.4 lbs per cubic foot would float on water (and I've never seen coal that floats unless it is close to being lignite or peat). So unless there are a lot of openings and spaces around and in the coal (heavily broken up) it is not going to weigh 50-60 lbs per cubic foot. Even then the coal will still sink in water.
#10
10-05-2008, 05:17 AM
JohnS Moderator Join Date: Dec 2007 Location: SE Michigan, USA Posts: 9,550 Rep Power: 19
Re: cubic feet in a ton of coal
Quote:
Originally Posted by clinker Rock (like granite) weighs about 185 lb per cubic foot. Water weighs 62.4 lbs per cubic foot. coal that weighs less than 62.4 lbs per cubic foot would float on water (and I've never seen coal that floats unless it is close to being lignite or peat). So unless there are a lot of openings and spaces around and in the coal (heavily broken up) it is not going to weigh 50-60 lbs per cubic foot. Even then the coal will still sink in water.
For all materials broken into random shaped lumps or ground to powder, there is a significant difference (more or less 2:1, but considerable variation) between true particle density (air spaces excluded) and bulk density (the volume required to hold a given weight, including air space). For a given material, bulk density can vary depending on degree of vibration or compacting to "settle it".
For a coal bin, the bulk density is what is needed for design. This site may help
http://www.powderandbulk.com/resourc...ty_chart_c.htm
(scroll halfway down page to coal)
Note the considerable range for coal in different forms (mean particle size), but in ALL forms, the bulk density is less than water. For coal averaging 50 lb/cu ft, 40 cu ft/ton is sufficient for designing storage, but dust or extremely finely ground coal could be as low as 35 lb/cu ft, requiring around 58 cu ft/ton.
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http://www.studymode.com/essays/Physics-Lab-Answers-Hooke's-Law-933812.html | 1,490,843,158,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218191984.96/warc/CC-MAIN-20170322212951-00355-ip-10-233-31-227.ec2.internal.warc.gz | 713,520,270 | 18,398 | # Physics Lab Answers- Hooke's Law
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1.What are the spring constant (k) is for spring 1 and spring 2? What are the units of k? Which spring is stiffer? Fs= kx
k is the spring constant and it is measured in N/m or in kg•m/s2 Spring constant for Spring 1= .49=k(.01)= 49 k=49 N/m
Spring constant for spring 2= .49=k(.02)= 24.5 k=24.5 N/m Spring 1 is stiffer; you can see that by the smaller change in spring length, and the higher spring constant.
2.Does your data fall neatly on your best-fit line? What are the possible sources of error in your measurements? Our data falls fairly neatly on our best-fit line. The possible sources of error are incorrect positioning of the ruler, and possible incorrect reading of the measurements.
3.When we draw a line past the data points measured, we call that extrapolation. Why is this not advised in this case? Extrapolation is not advised in this case because extrapolating gets less precise measurements. Since the measurements here might not be 100% accurate in the first place, assumed measurements would not add to the data. There will be more uncertainty between the extrapolated points.
4.Why are the cables of a suspension bridge elastic? The cables are elastic so they act like springs and “bounce” back. This way there isn’t too much strain on the cables as heavy vehicles pass over it. If the cables are too stiff, then the brittleness will cause it to snap. The elastic allows for proper function.
5.Hooke’s Law is a direct relationship. What does this mean? This means that Hooke’s law is a direct relationship between an applied force and the change in the spring’s length due to that applied force. The more weight is placed on the spring, the greater the spring will stretch. | 427 | 1,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2017-13 | latest | en | 0.923761 |
www.technowize.com | 1,568,964,020,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573908.70/warc/CC-MAIN-20190920071824-20190920093824-00521.warc.gz | 1,042,328,778 | 165,779 | One of the most dreadful subjects in a student’s life is mathematics. Considered a challenging subject for majority of the students around the world, various companies launch ways to make maths easier. While learning algebra or any such advanced math, solving an equation can be confusing and complicated. Due to the advent of technology, various apps are available in the market which help in solving equations. But the app known as PhotoMath is way different than all the other conventional apps. This app helps one solve all complicated math problems just at the touch of a button in the literal sense. The PhotoMath app is available on the Google Play Store as well as the iOS store.
## Why is PhotoMath a boon for the students?
As the name suggests, PhotoMath solves all sorts of maths equations through the technique of a picture. All one has to do is click a picture of the problem which needs to be solved and the app produces its solution immediately. In order to use this app, one needs a smartphone with a camera and this app installed in it. The installation of this app is absolutely free. After this, scan the problem which needs solving in the camera of the app. This will be followed by the detailed solution of the given problem. PhotoMath also provides a proper explanation of the process of the equation.
## Why is the hype for PhotoMath 3.0 justified
Until now, the camera of PhotoMath could only scan printed equations and solve them. But this week, PhotoMath has released the 3.0 version of this app bringing a major update in its working. This new version can not only scan printed text but can also scan handwritten problems making things easier for the user. Now all one has to do is jot down the problem on a piece of paper. On scanning this problem, the solution is available making maths a piece of cake for the students.
This latest update also introduces a new math expert system known as PhotoMath+. This feature would solve problems close to the way humans solve them. It would also explain the users about how to solve it themselves. This is done by a step-by-step explanation and colourful illustrations. The idea behind this feature is that only providing the solution of the equation isn’t enough. It is more important for a person to understand the working of the problem. Earlier this app just helped students in completing their homework. But after this update, this app now also helps a student in understanding the subject better. Currently, this update is only available on iOS but it's introduction on the Android portal will take place by the end of this month.
## PhotoMath in the long run
PhotoMath founder Damir Sabol says that this app has been downloaded over 36 million times on Android and iOS. It is consistently one of the top 5 educational apps in the US. The PhotoMath+ service is available on an unpaid basis till November 30. After this, a premium charge will be applicable on its use. The finalization of the pricing and subscription model has not taken place yet. But the core services of this app like solving problems by scanning formulas through the camera will remain complimentary.
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http://blog.dnevnik.hr/12-bmx-bicycle/2011/11/1629652726/metric-bike-exhaust-bike-exhaust.html | 1,524,358,413,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945484.57/warc/CC-MAIN-20180422002521-20180422022521-00343.warc.gz | 40,290,791 | 8,497 | # 12 BMX BICYCLE
## 28.11.2011., ponedjeljak
### METRIC BIKE EXHAUST - BIKE EXHAUST
Metric bike exhaust - New balance 7.5 r recumbent bike - Rocket dirt bike.
## Metric Bike Exhaust
exhaust
• Expound on, write about, or explore (a subject or options) so fully that there is nothing further to be said or discovered
• gases ejected from an engine as waste products
• system consisting of the parts of an engine through which burned gases or steam are discharged
• wear out completely; "This kind of work exhausts me"; "I'm beat"; "He was all washed up after the exam"
• Drain (someone) of their physical or mental resources; tire out
• Use up (resources or reserves) completely
metric
• A system or standard of measurement
• measured: the rhythmic arrangement of syllables
• Metric units, or the metric system
• A binary function of a topological space that gives, for any two points of the space, a value equal to the distance between them, or to a value treated as analogous to distance for the purpose of analysis
• metric function: a function of a topological space that gives, for any two points in the space, a value equal to the distance between them
• based on the meter as a standard of measurement; "the metric system"; "metrical equivalents"
bike
• A bicycle or motorcycle
• motorcycle: a motor vehicle with two wheels and a strong frame
• bicycle: a wheeled vehicle that has two wheels and is moved by foot pedals
• bicycle: ride a bicycle
Vincent Tech Info 4
The second auto advance and retard died in Latvia but by then I knew just what to do (or so I thought!) and had the bike running again in no time at all. All went well for a while until after an hour running I noticed oil coming from the top rear section of the timing cover. I stopped the bike to see if it had made its way to my new rear tyre and was waiting for me to turn right. It had! I stopped and tightened the screws - which didn't work - then stopped again and took it all apart. There was a two inch piece of gasket missing (or so I thought) In reality it had dropped down the side of the gearbox cover and got stuck on some oil. I was convinced it was lost so I put it all back togther without any gasket using only Wellseal and as I started the engine the peice fell to the floor. Grrr! Later it started leaking once again but only a bit so I waited until we did a hundred miles more and did it properly.
I lost a front tank bolt on a Sunday moring in Estonia. I used one from the exhaust pipe where it meets the silencer, replacing that one with a smaller one whilst the exhaust was off in Latvia. (Graham was impressed as he thought we would not get back to England with only one bolt at the front. Grahams sestion was to wrap a small peice of the economist around a metric bolt and screw it in, which is what you can expect from a BMW owner.
Carbs went out of balance just as we entered Tallinn in Estonia (or so I thought). What had happened was that the cables had loosened enough for the outer of one of them to catch on the adjuster, therefore opening one carb before the other. Once realised, it was easily sorted. Now the bike once again will pull cleanly from 34 miles an hour in top gear. (It never did that back home).
The digital speedo has come back to life giving a fair approximation of my speed in kph.
All these niggling problems have been causing me lack of sleep which just adds to my well balanced veiw of the world when I'm at the side of the road. This is further enhaced by Graham's constant spouting on about the history of Lithuania, Latvia, Estonia, or Finland or wherever we happen to be at the time of the breakdown!
Exhausted
Reflected images of Tower Bridge on both of Santa's Scania exhaust pipes.
_DS30098aa
Please do not use this image without prior permission
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# Waves, Fields & Nuclear Energy - PowerPoint PPT Presentation
Waves, Fields & Nuclear Energy. Contents. Oscillations & Waves Capacitance Gravitational & Electric Fields Magnetic Effects of Currents Nuclear Applications. Circular Motion. Consider an object going round in a circle of radius r: - speed is constant - velocity changes s = r
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## PowerPoint Slideshow about ' Waves, Fields & Nuclear Energy' - guy-jenkins
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### Waves, Fields & Nuclear Energy
Contents
• Oscillations & Waves
• Capacitance
• Gravitational & Electric Fields
• Magnetic Effects of Currents
• Nuclear Applications
Circular Motion
• Consider an object going round in a circle of radius r:
- speed is constant
- velocity changes
s = r
- angular velocity
ω = 2f = r/v
- centripetal acceleration
a = v2/r = ω2r
- centripetal force
f = ma = mv2/r = mω2r
Oscillations
• Natural frequency: an object will swing freely at this frequency
• Free oscillation: an object oscillates independently
• Forced oscillation: a force causes an object to oscillate
• Resonant frequency: where maximum amplitude is attained
(car suspensions, bridges swaying, bells ringing)
• Damping: amplitude of oscillations exponentially decreases
- light damping reduces oscillations slowly
- heavy damping reduces oscillations quickly
- critical damping stops the oscillation within one cycle
SHM
• max. a and max. v: origin
• V = 0 at –A and +A
• max. PE at –A and +A
• max. KE at origin
• a = - (2f )2x a = - ω2x
• v = 2f (A2 – x2)
• s = A cos 2ft
• T = 2(l/g)
• Etot = PE + KE
SHM
• Mass on a spring:
• Fup = k(l + x) – mg
• a = -kx/m = - (2f )2x
• T = 2(m/k)
Progressive Waves
• Wave Equation:
v = fλ
v = velocity (m/s)
f = frequency (Hz) or (1/s)
λ = wavelength (m) λ
• Polarisation:
Superposition of Waves
• Superposition can only be applied to waves of the same kind
• The diagram shows a green wave added to a red wave. The result is the black wave, whose wavelength and amplitude reflects the sum of the two waves
Wave Behaviour
• Interference: When two waves collide, they superimpose
• Superposition affects the waveform and interference results
• Path difference: difference in distance between two sources. It is measured in half wavelengths
• Waves in phase interfere constructively (increased amplitude)
• Waves out of phase interfere destructively (cancellation)
• Constructive: even number of ½ λs
• Destructive: odd number of ½ λs
Wave Behaviour
• Diffraction Grating:
- Light is split by travelling through very thin slits called a diffraction grating
- Light is split because it is composed of different wavelengths
- Each of these wavelengths diffracts at a different angle
d sin = mλ
d = slit width
• = angle
m = spectrum order number (1st: m= 1, 2nd: m = 2 etc.)
λ = wavelength
NB: “m” is sometimes denoted as “n” instead
Wave Behaviour
• The more slits, the more defined the diffractions
• The more slits, the greater the intensity
• The more slits, the greater the angle (easier to measure!)
• There is a limited number of orders, as sin has a maximum value of 1
- therefore at maximum, d = mλ
Capacitors
• Capacitors: store charge for a short time
- consists of two metal plates separated by a layer of insulating material dielectric
• Electrons are pumped onto the –ve plate
• Electrons are repelled off the +ve plate
• A potential difference is formed thus a charge
• Capacitance: charge required to produce 1V of potential difference in a conductor
capacitance (F) = charge (C) /voltage (V)
C = Q / V
Capacitors
• Energy in a Capacitor: When a capacitor is charged up, a certain amount of charge moves through a certain voltage. Work is done on the charge to build up the electric field in the capacitor
energy = charge x voltage
capacitance = charge / voltage
Thus: E = ½CV2
• Discharge of a Capacitor: Charge decreases by the same fraction for each time interval, so that if it takes time, t, for the charge to decay to 50 % of its original level, the charge after 2t seconds is 25 % of the original
Capacitors
• Q = Q0e–t/RC
• V = V0e–t/RC
• I = I0e–t/RC
RC = time constant
• t½ = 0.693 RC
t½ = half life
Gravity Fields
• Newton’s Square Law of Gravitation:
- Every particle of matter in the Universe attracts every other particle with a gravitational force that is proportional to the products of the masses and inversely proportional to the square of the distance between them
Thus: F = -GMm/r2 G = 6.67x10-11Nm2kg-2
• a = F/m where a = gravity: g = F/m
Thus: g = -GM/r2 r = radius from centre of orbit!
Gravity Fields
• Heading towards the centre of the Earth…
• At centre: g = 0 as matter is pulled in all directions equally
Gravity Fields
• Gravitational Potential:
- Work done on a unit mass in moving it to that point from a point remote from all other masses
• Always negative, because this involves a closed system
- the zero point of gravitational potential is at infinity
Vg = -GM/r Vg = gravitational potential
• Vg is the area under the curve on the previous slide
• Potential Energy in space: Ep = -GMm/r
Electric Fields
• Electric field: region of force around a point charge
F = kQ1Q2/r2 k =
0 = 8.8510-12 C2N-1m-2 (F/m)
• Electric Field Strength: force per unit charge
E = F/Q
This is radial for point charges:
Electric Fields
• Electric Field Strength: is inversely proportional to the square of the radius
- uniform field: E = V/d
• Electric Potential: energy per unit charge
Magnetic Fields
• A current (I) has a magnetic field (B) around it
• A wire has a circular magnetic field around it
• If the current changes direction, so does the field
Magnetic Fields
• Magnets attract magnetic materials using a magnetic field
• The magnetic field surrounds the magnet, and gets weaker as the distance from the magnet increases
• Magnets should be called permanent magnets
the magnetism is always there
• Electricity makes a magnet much stronger
• This can be turned on and off
Magnetic Fields
Magnets pick up paper clips etc.
Electromagnets pick up cars etc.
strong
weak
Magnetic Fields
• The magnetic field around a coil electromagnet can be increased by:
- Increasing the current flowing through the wire
- Adding loops on the coil (loops are long lengths of wire)
- Placing an iron or steel core inside the coil
Basic electromagnet
Magnetic Fields
• The Motor Effect:
- When two magnets are placed close to each other, they the fields affect each other produce a force
• If a wire carrying a current is placed inside this magnetic field, a force is produced. This is called the motor effect
• The direction of the force will depend on the direction of the magnetic field and the direction of the current in the field
Magnetic Fields
• Fleming’s Left Hand Rule:
- When creating a force, use Fleming’s LH Rule to determine in which way the motor will spin
-
Magnetic Fields
• We can increase the force produced by:
- increasing the current
- increasing the number of coils
- increasing the magnetic field strength (stronger magnet)
Magnetic Fields
• When a magnet is moved into a coil, an electrical current is induced
• When the magnet stops,
• the induced current stops
• When the magnet reverses, the electrical current reverses
Magnetic Fields
• Increase the voltage? … 3 ways…
• Stronger magnet
• 2. Speed of magnet
• 3. Number of coils
Magnetic Fields
• To work out the force on a wire: use Fleming’s LH Rule
• Force is proportional to:
- current
- magnetic field strength
- length of wire inside magnetic field
F = BIl B = magnetic field strength or flux density
(Tesla)
When a wire is at an angle to the magnetic field… F = BIl sin
Magnetic Fields
• To work out the force on a charge: use Fleming’s LH Rule
• Force is proportional to:
- current (flow of charge)
- magnetic field strength
- velocity of charged particle
F = BqV B = magnetic field strength or flux density
(Tesla)
When a charge is at an angle to the magnetic field… F = BqV sin
F = mv2/r BqV = mv2/r V = Bqr/m
Magnetic Fields
• Magnetic Flux: Product between the magnetic flux density and the area when the field is at right angles to the area
• Ф = BA
• Flux Linkage: Ф multiplied by number of turns on a wire
• Ф = NBA
• It can be changed by:
- changing the strength of the magnetic field
- move the coil so it enters the field at an angle
• Lenz’s Law: direction of an induced current opposes the flux change that caused it
Mass & Energy
• 1 atomic mass unit (u) = 1.661 10-27 kg
• Atomic mass: mass of an atom
• Nuclear mass: mass of atom’s nucleus
E = mc2 c = 3x108m/s
(J) = (kgm2/s2)
• 1eV = 1.6x10-19J
• 1u = 931.3MeV
• Binding Energy per Nucleon: Energy required to remove a nucleon. Higher numbers more stable nuclei
Mass & Energy
• Fission: splitting up of a large nucleus which is rarely spontaneous
• The strong nuclear force acts between neighbouring nucleons
• The forces are now weak in this shape/formation
• Nucleus splits (rarely spontaneously)
• Induce fission: add thermal neutron whose kinetic energy:
1) isn’t too low (will bounce off nucleus)
2) isn’t too high (will go through nucleus)
3) is correct to be captured by the attractive force in between nucleons - this can result in a chain reaction
Mass & Energy
• Fusion: when light nuclei bind together which increases the binding energy per nucleon energy is released
• Each nucleus has to have sufficient energy to:
- overcome electrostatic repulsion from the protons
- overcome the repulsive strong force which is found outside the region of the strong force
• High temperatures are required (gas plasma)
• If it could be made to work, has advantages over fission:
- greater power per kilogram of fuel used
- raw materials are cheap and readily available
Nuclear Power
• Although the fission products are not easily predictable, three more neutrons are produced
• An uncontrolled chain reaction causes a violent explosion
• Minimum mass before chain reaction occurs: critical mass
• Nuclear power station:
• Reactor is housed in a concrete to prevent radiation from leaking
• Expensive to build
• Costly to run
• Very clean, no pollution
• Need very little fuel
• Produce dangerous waste
• Nuclear power France vs. England = 80% vs. 20%
Nuclear Power
• Safety:
- Strict regulations
- Serious accidents involving radiation leaks have occurred
- Disposal of radioactive waste must be carried out carefully
• Transmutation:
- Definition: changing the nuclei of elements by exposing them to particles
- Particles have to travel slow enough to be captured by the nucleus
- used in medicine
Summary
• Circular Motion
• Oscillations
• SHM
• Progressive Waves
• Superposition of Waves
• Wave Behaviour
• Capacitors
• Gravity Fields
• Electric Fields
• Magnetic Fields
• Mass & Energy
• Nuclear Power | 2,975 | 11,443 | {"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} | 2.875 | 3 | CC-MAIN-2017-43 | latest | en | 0.780335 |
https://www.physicsforums.com/threads/find-voltage-across-resistor-using-mesh-current-analysis.862539/ | 1,721,766,646,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518115.82/warc/CC-MAIN-20240723194208-20240723224208-00828.warc.gz | 806,311,869 | 17,005 | # Find voltage across resistor using mesh current analysis
• vizakenjack
In summary, The conversation discusses finding Vth using mesh current analysis. It is noted that Mesh A and B share a current source and are treated as one supermesh. The equation for this supermesh is given as +12V + ia(6Ω) + ib(6Ω + 4Ω) = 0. It is also mentioned that the loop with the open circuit can be ignored since there is no current flowing through it. The final values for ia and ib are calculated and the voltage across 4Ω (Vth) is found to be 40V instead of the expected 6V. In order to find the correct value, the voltages across all resistors are summed in a
vizakenjack
Thread moved from the technical forums, so no Homework Help Template is shown
I'd like to find Vth by mesh current analysis.
Mesh A and B share a current source, so they're treated as one supermesh.
What's the equation for this supermesh? Also, do I ignore the loop with the open circuit? Since there's no current flowing there...
2A = ib - ia
Anyway, is this the right equation:
+12V + ia(6Ω) + ib(6Ω + 4Ω) = 0?
if yes, then my ia ends up being -32V/(4Ω) = - 8A
ib = 2A + 8A = 10A
And the Voltage across 4Ω (Vth) = ib * 4Ω = 40 V!
But that's not correct, Vth should be 6V!
vizakenjack said:
+12V + ia(6Ω) + ib(6Ω + 4Ω) = 0?
With currents marked clockwise, the voltages across all resistors here sum clockwise as voltage falls, and you could pencil in a small + and - sign on each resistor as a reminder.
So summing in a clockwise path the voltage rises around the loop:
+12 - ia(6) - ib(6+4) = 0
vizakenjack
## 1. What is mesh current analysis?
Mesh current analysis is a method used to find the voltage across a specific resistor in a circuit. It involves assigning current values to different loops or "meshes" in the circuit and using Kirchhoff's Voltage Law (KVL) to solve for the voltage across the desired resistor.
## 2. How does mesh current analysis differ from other circuit analysis methods?
Unlike other analysis methods such as nodal analysis, mesh current analysis focuses on the current flow within specific loops in the circuit rather than the potential difference at specific nodes. This can be especially useful when dealing with circuits that have multiple current sources and multiple loops.
## 3. What are the steps involved in performing mesh current analysis?
The steps for mesh current analysis are as follows: 1. Identify the number of meshes in the circuit 2. Assign current variables (i1, i2, etc.) to each mesh 3. Apply Kirchhoff's Voltage Law (KVL) to each mesh, setting up equations 4. Solve the resulting system of equations for the desired voltage value.
## 4. Can mesh current analysis be used for all types of circuits?
Mesh current analysis can be used for most types of circuits, including those with multiple voltage sources, current sources, and resistors. However, it may become more complex and time-consuming for circuits with a large number of meshes and components.
## 5. What are some advantages of using mesh current analysis?
Some advantages of using mesh current analysis include the ability to handle circuits with multiple current sources and the fact that it can be applied to both planar and non-planar circuits. Additionally, the use of KVL makes it a straightforward and systematic method for solving for the voltage across a specific resistor in a circuit.
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Qualified Writers
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Question 1 (17 marks)
Table 1 Input and output forms of energy for some different types of
energy system
Type of system Input energy
form
Output energy
form
Hydroelectric dam generating
plant
Potential energy Electrical energy
(i) Gas boiler ? Heat
(ii) Torch battery Chemical energy ?
(iii) PV (solar) cell ? Electricity
(iv) Wind turbine Kinetic energy ?
(v) Electric table fan ? Kinetic energy
(vi) Radio transmitter Electrical energy ?
(6 marks)
TMA 01
1. Explain the difference between electrical energy and electrical power, and
state the units used for each.
(3 marks)
1. Table 1 below shows the input and output forms of energy for some different
types of energy system. Copy the table and complete it by filling in the six
blanks (i) to (vi) with the input energy form, or the output energy form, as
appropriate. The entry in the first row has been completed for you as an
example.
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Question 2 (12 marks)
Question 3 (16 marks)
1. A new pumped storage plant is proposed at Espejo de Tarapaca in Chile. It
will use sea water flowing to and from a new high level lake situated on top of
a cliff 600 m above sea level.
1. The high level lake is described as having an area of 375 hectares and
being capable of holding 14 billion (14 × 10 ) US gallons of water. What
will be the average depth of water in the lake in metres? (Conversion
factors between different units of area and volume are given in
Appendix A3 of Energy Book 1.) Give your answer to 3 significant
figures.
(3 marks)
9
1. Calculate the total stored potential energy in GWh assuming a sea
water density of 1.025 kg l , taking g as 9.81 m s and 1
(3 marks)
1
2
iii. Assuming a turbine efficiency of 85%, calculate the maximum number of
hours that the plant will be able to supply its rated output of 300 MW.
(2 marks)
1. Explain in your own words what is meant by the reserves/production (R/P)
ratio for a fossil fuel resource.
(2 marks)
1. Chapter 2 of Energy Book 1 describes the energy situation in six different
countries, including the USA and the UK.
List the key ways in which the USA and the UK differ in:
Make sure your list includes the following, along with any other points of
interest:
the role played by coal and oil in the two countries
the energy selfsufficiency
of the two countries in 2009, with comments
on their natural resources, imports and exports
differences in per capita energy consumption
the relative contributions of oil and natural gas to energy consumption.
(10 marks)
1. their energy histories between 1950 and 2009
and
1. their energy situations in 2009.
2. Write the chemical equation for the combustion of carbon.
(2 marks)
1. Explaining your reasoning, use this equation to show that the combustion of
5 kg of carbon should release just over 18 kg of carbon dioxide. (The relative
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Question 4 (12 marks)
atomic masses of carbon and oxygen are 12 and 16 respectively.)
(2 marks)
1. A high quality anthracite coal can be considered to be almost pure carbon.
The combustion of 1.0 kg of this coal produces 34 MJ of heat energy. Use
this fact, and the result of part (b) above, to find the mass of CO released
per GJ of heat produced.
(2 marks)
2
1. The combustion of 1.0 kg of wood, on the other hand, produces 15 MJ of
heat energy. If the density of the wood is 480 kg/m , calculate the heat
energy produced by burning 1.0 cubic metre of wood.
(2 marks)
3
1. A small allyear
holiday cabin in the United States has a rudimentary wood
stove for heating. In the USA, wood is commonly sold by the unit of the ‘cord’,
equivalent to 128 cubic feet of wood. Express 1.0 cord in cubic metres.
(Conversion factors for different units of volume are given in Appendix A3 of
Energy Book 1.)
(2 marks)
1. Typically the owners of the holiday cabin use 1.0 cord of wood every three
months during the colder months. How much energy in gigajoules will they
use over six winter months?
(2 marks)
1. In the USA, coal is usually sold by the ‘short ton’ of 2000 lb. Show that the
holiday cabin would require approximately 1.7 short tons of anthracite to
provide the same amount of heat as the wood in part (f). (Conversion factors
between different units of mass are given in Appendix A3 of Energy Book 1.)
(2 marks)
1. If the density of anthracite is 1100 kg/m , how much space would be
required to store this amount of coal?
(2 marks)
3
1. Figure 1 shows UK primary energy and delivered energy consumption for the
year 2009 (Figure 3.10 in Energy Book 1).
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Question 5 (15 marks)
Figure 1 2009 primary energy and delivered energy View larger image
Using your own words, define the terms primary energy and delivered
energy.
(2 marks)
1. In 2009 over 30% of UK primary energy consumption was ‘lost in conversion
and delivery’. Give two main areas within the UK energy system where these
losses occurred.
(4 marks)
1. The top bar of Figure 1 shows a large amount of primary energy is supplied
from coal. Explain why only a small amount of ‘solid fuel’ appears in the
second (delivered energy) bar.
(2 marks)
1. Assuming that transport uses only liquid fuel, roughly how much of the
energy delivered as liquid fuel was used for purposes other than transport?
You may also wish to look at Table 3.4 in Energy Book 1, page 93. Give your
answer to the nearest 100 PJ.
(2 marks)
1. Assuming all the delivered energy used for machinery is used by industry,
how much energy was used by industry for purposes other than machinery?
(2 marks)
1. Chapter 1 of Energy Book 1 describes three outlooks for future world energy
use: ‘growthist’, ‘peakist’ and ‘environmentalist’. Peakists and
environmentalists both believe that fossil fuel use will decline significantly by
2050 but for different reasons. Briefly explain in your own words the key
differences between these two outlooks.
(6 marks)
1. List three atmospheric pollutants in the flue gases of a coalfired
power
station, other than carbon dioxide.
(3 marks)
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Question 6 (13 marks)
Figure 1 below is a McKelvey diagram showing the relationship between reserves,
conditional resources and hypothetical resources.
Figure 2 McKelvey diagram (repeat of Energy Book 1
Figure 5.17 (a)).
Question 7 (15 marks)
1. For each of the three pollutants, briefly describe a cleanup
technology as
used in coalfired
power stations.
(6 marks)
1. Define each of the three terms reservesconditional resources and
hypothetical resources in your own words.
(3 marks)
1. A coalfield was suspected to exist on the edge of a currently exploited mine
but had not been confirmed. Exploration confirmed both that it existed and
that it would be economic and feasible to extract the coal. To which of the
three categories in the diagram would the coal have initially been assigned,
and to which would have it been moved after the exploration?
(4 marks)
1. In April 2016, the US coal mining company, Peabody Energy, filed for
bankruptcy saying that low world coal prices made much of its quoted coal
reserves uneconomic to mine. To which of the three categories above should
their coal reserves now be assigned?
(2 marks)
1. UK coal reserves are now only a small fraction of what they were in 1900.
Most of the remaining coal has not been extracted and there has been over
a century of development of new mining technology. Why have the UK coal
reserves fallen so dramatically, and what category in the diagram is the
remaining coal now in?
(4 marks)
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Learning outcomes
This TMA assesses the following learning outcomes.
Knowledge and understanding
The forms of energy and the various definitions of ‘sustainable’ energy, and
the relevance of sustainability to the design and implementation of energy
systems.
The principal sources of primary energy in use today, their development over
the past century, and the general patterns of world, regional and national
energy consumption.
The basic principles underlying the design and use of energy supply
systems, and the basic principles underlying the efficient use of energy.
The pollutants produced through the combustion of fossil fuels, their
environmental impacts and the amelioration of these (for coal).
Cognitive skills
Critically evaluate differing explanations and arguments for the relative
resource depletion impacts of different energy technologies (for coal).
Key skills
Use appropriate scientific and mathematical techniques applied to energy
related issues or contexts.
Analyse information about energy and its use, from a variety of sources and
using ICT tools.
1. A heat engine is working between an input temperature and an outlet
temperature T . State Carnot’s formula for its maximum possible efficiency.
What temperature scale must be used and how does this relate to the
Celsius scale?
(4 marks)
1
2
1. A new concentrating solar power (CSP) plant was completed in 2013 at
Ivanpah in Arizona. It has three power towers and is similar in design to the
Spanish plants shown in Figure 14.20 of Energy Book 1. It can produce
steam at a temperature of 550 °C. It uses air cooling and a final condenser
outlet temperature of 60 °C can be assumed. It has a design efficiency of
28.7%. Show that this is about a half of the theoretical Carnot efficiency for
these temperatures.
(4 marks)
1. The full Ivanpah site covers 16 square kilometres of land. On average, over a
year, the solar radiation falling on the site is 2000 kWh per square metre. In
2015 the solar power plant produced 650 GWh of electricity. Express this as
a percentage of the total incident solar radiation falling on the site.
(3 marks)
1. In 2009, what was the average per capita electricity consumption for the
USA? How does this compare to the equivalent figure for the UK? Calculate
the number of ‘average’ residents in each country whose electricity
consumption equals the output of the Ivanpah plant.
(4 marks)
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### Topic: defining wff
#### Patrick Thomas (Aug 03 2019 at 00:11):
I was wondering if the following definition can be formalized in Lean, and if so, how. I am thinking that a propositional variable might be defined as a natural number, but I am not sure how to introduce the connectives.
DEFINITION. The set of wffs is the intersection of all sets S of formulas such
that:
(i) p ∈ S for each propositional variable p.
(ii) For each formula A, if A ∈ S, then ~A ∈ S.
(iii) For all formulas A and B, if A ∈ S and B ∈ S, then [A ∨ B] ∈ S.
A wff is a member of the set of wffs.
I would like to use this definition in a formal proof of the following exercise:
Using the definition of the set of wffs, show that:
(a) Every formula consisting of a propositional variable standing alone is
a wff.
(b) If A is a wff, then ~A is a wff.
(c) If A and B are wffs, then [A ∨ B] is a wff.
#### Mario Carneiro (Aug 03 2019 at 00:27):
While it is possible to define wffs in this set-theoretic way, the much easier approach is to use an inductive type - this is what they are designed for. With that the theorem is trivial, as it is just the constructors of the inductive type
#### Floris van Doorn (Aug 03 2019 at 00:31):
Here is an example of using inductive types for this:
-- α is the type of propositional variables
inductive wff (α : Type) : Type
| var : α → wff
| neg : wff → wff
| or : wff → wff → wff
local prefix # := wff.var
local prefix ~ := wff.neg
local infix ∨ := wff.or
example {α} (p q : α) : wff α :=
#p ∨ ~ #q
#### Patrick Thomas (Aug 03 2019 at 00:40):
Thank you. I would still be interested in seeing how it might be done in the set-theoretic way, if you have a chance.
#### Floris van Doorn (Aug 03 2019 at 00:49):
In that case, I would define a character as a propositional variable or one of the 4 special characters (~ [ ] ∨) and then define S to be a set on the type of list of characters (S is the intersection of all sets satisfying ...)
#### Patrick Thomas (Aug 03 2019 at 00:58):
I'm sorry, I'm not sure I understand.
#### Andrew Ashworth (Aug 03 2019 at 01:16):
It's a much more raw interpretation. Your formulas are sequences of [ ] ~ \/ 1 2 3 4 5 6 7 8 9. Step 1 is to turn a sequence of digits into a var. Step 2 is to come up with a prop describing 1, 2, 3 in terms of this raw list of characters.
#### Floris van Doorn (Aug 03 2019 at 01:17):
import data.set.lattice
noncomputable theory
open sum
-- Var is the type of propositional variables
constant Var : Type
def character := Var ⊕ fin 4
def lbracket : character := inr 0
def rbracket : character := inr 1
def neg : character := inr 2
def disj : character := inr 3
local notation ⟦ := lbracket
local notation ⟧ := rbracket
local notation ~ := neg
local notation ∨ := disj
def wff : set (list character) :=
⋂₀ { S | (∀(p : Var), [inl p] ∈ S) ∧ (∀(A ∈ S), [ ~ ] ++ A ∈ S) ∧
∀(A B ∈ S), [ ⟦ ] ++ A ++ [ ∨ ] ++ B ++ [ ⟧ ] ∈ S }
#### Patrick Thomas (Aug 03 2019 at 01:24):
It might take me a bit to go through this, but thank you!
Last updated: May 14 2021 at 02:15 UTC | 962 | 3,097 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-21 | latest | en | 0.913624 |
https://www.sarthaks.com/1178450/bag-contains-black-white-balls-one-ball-drawn-random-what-probability-that-ball-drawn-black | 1,638,611,654,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362969.51/warc/CC-MAIN-20211204094103-20211204124103-00194.warc.gz | 1,094,391,856 | 14,875 | A bag contains 8 red, 2 black and 5 white balls. One ball is drawn at random. What is the probability that the ball drawn is not black?
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A bag contains 8 red, 2 black and 5 white balls. One ball is drawn at random. What is the probability that the ball drawn is not black?
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Total number of balls in the bag= 8+2+5 = 15.
Number of non-black balls = 8+5 = 13.
P(getting a non-black ball) = 13/15. | 133 | 442 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-49 | latest | en | 0.956391 |
https://aptitude.brainkart.com/aptitude/problem-solving-and-reasoning/mathematics-or-quantitative-aptitude/algebra/4/ | 1,624,006,330,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487635920.39/warc/CC-MAIN-20210618073932-20210618103932-00159.warc.gz | 111,697,362 | 11,097 | # Algebra (Test 4)
## Problem Solving And Reasoning : Mathematics Or Quantitative Aptitude
| Home | | Problem Solving And Reasoning | | Mathematics Or Quantitative Aptitude | | Algebra |
Algebra
| Algebra |
Q.1
If x2 + y2 = 29 and xy = 10, where x>0, y>0, x>y then the value of c is
A. -7/3
B. 7/3
C. 3/7
D. -3/7
Explaination / Solution:
Given, x2 + y2 = 29 and xy = 10, where x > 0, y > 0, x > y
Workspace
Report
Q.2
What will be the value of if a+b+c = 0 and a2+b2+c2 = 2abc ?
A. abc
B. a2b2c2
C. a3b3c3
D. 4abc
Explaination / Solution:
Given a+b+c = 0
⇒b+c = -a
⇒(b+c)2 = a2
⇒b2 + c2 + 2bc = a2
⇒a2 + b2 + c2 = 2(a2 - bc)
⇒(a2 - bc) = (1/2)(a2 + b2 + c2
Similarly,
(b2 - ac) = (1/2)(a2 + b2 + c2
And, (c2 - ab) = (1/2)(a2 + b2 + c2
Now,
Workspace
Report
Q.3
If p3 – q3 = (p – q) {(p + q )2 – x p q} then the value of x is
A. 1
B. -1
C. 2
D. -2
Explaination / Solution:
On comparing the coefficient, we get
Workspace
Report
Q.4
LCM of two polynomials is x3 + x2 - 4x - 4 is and their HCF is (x-2) and one of the polynomials is (x2-4) then which of the following is the 2nd polynomial?
A. x2-x-2
B. x2-3x+2
C. x2+x-6
D. x2-5x+6
Explaination / Solution:
LCM = x3 + x2 - 4x - 4 = x(x2-4)+1(x2-4) = (x+1)(x-2)(x+2)
HCF = (x-2)
Som 2nd polynomials = (x+1)(x-2)(x+2)(x-2)/(x2-4) = (x+1)(x-2) = x2-x-2
Workspace
Report
Q.5
If then the value of (a4 – a) is:
A. 0
B. 1
C. 2
D. -1
Explaination / Solution:
Workspace
Report
Q.6
What is the value of m in the quadratic equation x2 + mx +24=0 if one of its roots is
A. -45/2
B. 16
C. -21/2
D. -35/2
Explaination / Solution:
X = 3/2 in x2 + mx + 24 = 0
(3/2) + m*(3/2) + 24 = 0
(9/4) + (3m/2) + 24 = 0
Workspace
Report
Q.7
a = 1000, b = 1001, c = 1002. then find the value of .
A. 0
B. 1
C. 2
D. 3
Explaination / Solution:
Workspace
Report
Q.8
Sum of three consecutive even integers is 54. Find the least among them
A. 18
B. 15
C. 14
D. 16
Explaination / Solution:
Let three consecutive even integers by 2x, 2x + 2 and 2x + 4. respectively
2x + 2x + 2 + 2x + 4 = 54
6x + 6 = 54
Workspace
Report
Q.9
+ = ?
A. 1
B. 0
C. -1
D. 2
Explaination / Solution:
+
+ = 1
Workspace
Report
Q.10
If + = 2 and + = 1 than find (ZQ – PX)2 + (ZX + PQ)2
A. -1
B. 0
C. 4/9
D. 2 | 1,103 | 2,221 | {"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} | 4.625 | 5 | CC-MAIN-2021-25 | latest | en | 0.723202 |
https://oeis.org/A180311 | 1,709,634,035,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707948234904.99/warc/CC-MAIN-20240305092259-20240305122259-00146.warc.gz | 422,947,143 | 3,961 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A180311 Decimal expansion of -64/(9 Pi) + Pi. 1
8, 7, 8, 0, 5, 5, 6, 8, 5, 1, 7, 1, 7, 2, 6, 2, 4, 0, 8, 6, 0, 7, 4, 0, 9, 7, 0, 8, 7, 0, 4, 0, 9, 7, 3, 5, 2, 6, 2, 6, 3, 2, 2, 1, 5, 5, 1, 0, 8, 3, 6, 3, 2, 7, 6, 7, 4, 7, 8, 6, 8, 1, 0, 1, 3, 6, 8, 3, 9, 7, 2, 8, 8, 2, 1, 6, 5, 3, 8, 3, 2, 9, 0, 1, 9, 7, 1, 8, 5, 2, 6, 6, 6, 5, 1, 7, 6, 2, 3, 2, 7, 3, 5, 5, 6 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS Torsional rigidity constant for a full circular sector (sliced disk)-shaped shaft. LINKS Table of n, a(n) for n=0..104. Eric Weisstein's World of Mathematics, Torsional Rigidity. Index entries for transcendental numbers EXAMPLE 0.87805568517172624086... MATHEMATICA RealDigits[Pi - 64/(9 *Pi), 10, 100][[1]] (* Amiram Eldar, Jun 09 2021 *) PROG (PARI) Pi - 64/9/Pi \\ Charles R Greathouse IV, Oct 01 2022 CROSSREFS Cf. A180309, A180310, A180314, A180317. Sequence in context: A086911 A327854 A263179 * A289913 A103984 A203914 Adjacent sequences: A180308 A180309 A180310 * A180312 A180313 A180314 KEYWORD nonn,cons AUTHOR Eric W. Weisstein, Aug 26 2010 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified March 5 05:10 EST 2024. Contains 370537 sequences. (Running on oeis4.) | 702 | 1,599 | {"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} | 3.109375 | 3 | CC-MAIN-2024-10 | latest | en | 0.61136 |
https://www.crazy-numbers.com/en/80144 | 1,717,053,087,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059506.69/warc/CC-MAIN-20240530052602-20240530082602-00740.warc.gz | 606,334,027 | 3,502 | Warning: Undefined array key "numbers__url_substractions" in /home/clients/df8caba959271e8e753c9e287ae1296d/websites/crazy-numbers.com/includes/fcts.php on line 156
Number 80144: mathematical and symbolic properties | Crazy Numbers
Discover a lot of information on the number 80144: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 80144
Is 80144 a prime number? No
Is 80144 a perfect number? No
Number of divisors 10
List of dividers
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1, 2, 4, 8, 16, 5009, 10018, 20036, 40072, 80144
Sum of divisors 155310
Prime factorization 24 x 5009
Prime factors
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2, 5009
## How to write / spell 80144 in letters?
In letters, the number 80144 is written as: Eighty thousand hundred and forty-four. And in other languages? how does it spell?
80144 in other languages
Write 80144 in english Eighty thousand hundred and forty-four
Write 80144 in french Quatre-vingts mille cent quarante-quatre
Write 80144 in spanish Ochenta mil ciento cuarenta y cuatro
Write 80144 in portuguese Oitenta mil cento quarenta e quatro
## Decomposition of the number 80144
The number 80144 is composed of:
1 iteration of the number 8 : The number 8 (eight) represents power, ambition. It symbolizes balance, realization.... Find out more about the number 8
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1 iteration of the number 0 : ... Find out more about the number 0
1 iteration of the number 1 : The number 1 (one) represents the uniqueness, the unique, a starting point, a beginning.... Find out more about the number 1
2 iterations of the number 4 : The number 4 (four) is the symbol of the square. It represents structuring, organization, work and construction.... Find out more about the number 4 | 723 | 2,553 | {"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} | 2.84375 | 3 | CC-MAIN-2024-22 | latest | en | 0.655844 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=77&t=20735 | 1,597,239,090,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738892.21/warc/CC-MAIN-20200812112531-20200812142531-00292.warc.gz | 368,515,116 | 11,359 | ## 2014 Final 1C
Blake_Katsev_2E
Posts: 113
Joined: Wed Sep 21, 2016 2:57 pm
### 2014 Final 1C
Why would there be 3 sig figs in the answer if the question had 25 degrees C and 47 degrees C?
Stephanie_Thai_2C
Posts: 10
Joined: Fri Jul 22, 2016 3:00 am
### Re: 2014 Final 1C
I believe there are 3 sig figs in the answer because at one point in the equation, there is adding/subtracting, which has different sig fig rules from multiplying/dividing. When dealing with addition/subtracting, you take the smallest number of decimal places which will be 3 sig figs, so when you divide it later to find the enthalpy of vaporization, the answer will have 3 sig figs.
Sangita_Sub_3H
Posts: 23
Joined: Sat Jul 09, 2016 3:00 am
### Re: 2014 Final 1C
For this question, the solution in the course reader says q = m*C*delta T + 1/2*m*(enthalpy of vaporization). Why is the enthalpy of vaporization multiplied by 1/2?
Posts: 34
Joined: Wed Sep 21, 2016 3:00 pm
Been upvoted: 1 time
### Re: 2014 Final 1C
It is multiplied by 1/2 because the questions says that Ron "realizes that half of his 255 g forgetfulness potion has boiled away. Therefore, you would multiply the mass by 1/2 since only half of it was converted into vapor. | 387 | 1,225 | {"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-2020-34 | latest | en | 0.912369 |
https://calculatorschool.com/Numbers/MillionBillionConverter.aspx | 1,723,133,099,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640736186.44/warc/CC-MAIN-20240808155812-20240808185812-00228.warc.gz | 116,765,257 | 22,788 | # Million Billion Conversion
Convert a value of money such as from thousands, lakhs, crores to millions, billions, etc. This conversion is often confusing but this online million to billion converter will ease it for you.
• =
## Online Million Crore Billion Converter
The billion calculator offers several types of conversions including million to billion, crore to million, crore to billion, lakh to million, million to lakhs, trillion to crores, and the list goes on.
Using million to billion converter is easier than making a cup of tea. This tool can be very useful for students as well as individuals working in the finance industry. You can use it as a billion to million converter, number to million converter, millions to billions converter as per your conversion needs.
## How million crore billion calculator works?
Here’s how you can use this converter:
• Enter the number in the given input box.
• Select the notation from the first dropdown list.
• Select the notation from the second list to convert your number.
• Hit the Calculate button to get the conversion.
• Use the Reset button to enter new values.
If you want to convert a fraction to a million or billion, you can use our fraction to decimal converter to get a decimal value. Then, you can use that decimal value for conversion.
## Million to billion conversion
The terms million, billion, and trillion are used in the western numbering system. 1 billion is equal to 1000 million. Divide the number in million by 1000 to get the number in billion. Or simply, multiply it with 0.001.
1 billion = 1000 million
So, to convert million to billion,
Million to billion = Number in million × 0.001
Example:
Convert to billion: 100 million
100 × 0.001 = 0.1 Billion
Million to Billion Million Billion 1 Million 0.001 Billion 1.2 Million 0.0012 Billion 1.5 Million 0.0015 Billion 2 Million 0.002 Billion 3 Million 0.003 Billion 4 Million 0.004 Billion 5 Million 0.005 Billion 6 Million 0.006 Billion 7 Million 0.007 Billion 8 Million 0.008 Billion 9 Million 0.009 Billion 10 Million 0.01 Billion 100 Million 0.1 Billion 500 Million 0.5 Billion 1000 Million 1 Billion
## Billion to million conversion
1 million is equal to 0.001 billion. Multiply the number in billion with 1000 to get the number in millions.
1 million = 0.001 billion
Hence, to convert billion to million,
Billion to million = Number in billion × 1000
Example:
Convert to million: 0.8 billion
0.8 × 1000 = 800 million
Billion to Million Billion Million 1 Billion 1000 Million 1.2 Billion 1200 Million 1.3 Billion 1300 Million 1.5 Billion 1500 Million 2 Billion 2000 Million 2.4 Billion 2400 Million 3 Billion 3000 Million 4 Billion 4000 Million 5 Billion 5000 Million 6 Billion 6000 Million 7 Billion 7000 Million 7.7 Billion 7700 Million 8 Billion 8000 Million 9 Billion 9000 Million 10 Billion 10000 Million 0.001 Billion 1 Million 0.002 Billion 2 Million 0.003 Billion 3 Million 0.005 Billion 5 Million 0.01 Billion 10 Million 0.1 Billion 100 Million
## Lakh to crore conversion
1 crore is equal to 100 lakhs. Multiply the number in lakhs with 0.01 or divide the number by 100 to get the number in crore.
1 crore = 100 lakhs
Therefore, to convert lakh to crore,
Lakh to crore = Number in lakh × 0.01
Example:
Convert to crore: 75 lakhs
75 × 0.01 = 0.75 crores
## Million to lakh conversion
Want to know 1 million is equal to how many rupees? One million is equal to 10 lakhs.
So, a million means you are dealing with 10 lakhs. To convert million to lakh, multiply the number in million with 10 to get the number in lakh.
1 million = 10 lakhs
Thus, to convert million to lakh,
Million to lakh = Number in million × 10
Example:
Convert to lakh: 0.2 million
0.2 × 10 = 2 lakhs
Million to Lakh Million Lakhs 0.1 Million 1 Lakh 0.2 Million 2 Lakhs 0.5 Million 5 Lakhs 1 Million 10 Lakhs 1.2 Million 12 Lakhs 1.5 Million 15 Lakhs 2 Million 20 Lakhs 5 Million 50 Lakhs 5.6 Million 50 Lakhs 2 Million 20 Lakhs 2.5 Million 25 Lakhs 3 Million 30 Lakhs 10 Million 100 Lakhs 50 Million 500 Lakhs 70 Million 700 Lakhs 100 Million 1000 Lakhs 122 Million 1220 Lakhs 130 Million 1300 Lakhs 150 Million 1500 Lakhs
## Million to crore conversion
Want to convert 1 million in crores?
Try our million to crore conversion calculator online to save your valuable time. You can also convert 1 million in rupees using this tool.
1 crore is equal to 10 million. Multiply the number in million with 0.1 or divide the number by 10 to get the number in crores.
1 million = 0.1 crore
So, to convert million to crore,
Million to crore = Number in million × 0.1
Example:
Convert to crore: 50 million
50 × 0.1 = 5 crores
Million to Crore Million Crore 1 Million 0.1 Crore 2 Million 0.2 Crore 5 Million 0.5 Crore 10 Million 1 Crore 15 Million 1.5 Crores 20 Million 2 Crores 50 Million 5 Crores 100 Million 10 Crores 150 Million 15 Crores 200 Million 20 Crores 250 Million 25 Crores 300 Million 30 Crores 500 Million 50 Crores 780 Million 78 Crore 40 Million 4 Crore 70 Million 7 Crore 101 Million 10.1 Crore 1000 Million 100 Crore
## Billion to lakh conversion
1 billion is equal to 10,000 lakhs. Multiply the number in billion with 10,000 to get the number in lakh.
1 billion = 10,000 lakhs
Therefore, to convert billion to lakh,
Billion to lakh = Number in billion × 10,000
Example:
Convert to lakh: 0.5 billion
0.5 × 10,000 = 5000 lakh
You can also use the above converter to convert 1 billion in rupees at one click.
## Billion to crore conversion
Wondering 1 billion is equal to how many crores?
1 billion is equal to 100 crores. 1 billion in crores can be expressed as 100 crores. Multiply the number in billion with 100 to get the number in crores.
1 billion = 100 crores
So, to convert billion to crore,
Billion to crore = Number in billion × 100
Example:
Convert to crore: 10 billion
10 × 100 = 1000 crores
Convert any value from billion to crores instantly using online million billion converter.
Billion to Crore Billion Crore 1 Billion 100 Crores 1.2 Billion 120 Crores 1.3 Billion 130 Crores 1.324 Billion 132.4 Crores 1.339 Billion 133.9 Crores 1.5 Billion 150 Crores 1.8 Billion 180 Crores 1.9 Billion 190 Crores 2 Billion 200 Crores 3 Billion 300 Crores 5 Billion 500 Crores 7.6 Billion 760 Crores 7.7 Billion 770 Crores 30 Billion 3000 Crores 50 Billion 5000 Crores 100 Billion 10000 Crores
## Trillion to crore conversion
1 trillion is equal to 100,000 crores. Multiply the number in trillion with 100,000 to get the number in crores.
1 trillion = 100,000 crores
So, to convert trillion to crore,
Trillion to crore = Number in trillion × 100,000
Example:
Convert to crore: 0.2 trillion
0.2 × 100,000 = 20,000 crores
## FAQs
### How many thousands are in a million?
There are 1000 thousand in a million. A million is equal to one thousand thousand (1,000,000).
### How many millions are in a billion?
There are a total of 1000 million in a billion. A billion is equal to one thousand million.
### How much is 1 million?
1 million is equal to 1000 thousand or 10 lakhs. It can be expressed as 1,000,000 in the numeral system.
### What is the value of 1 million?
The value of 1 million is 1,000,000. It is a term in the numbering system that is used to express 1000 thousand in short form.
### References:
1. How Big Are Millions, Billions, and Trillions? by thoughtco.com | 2,066 | 7,388 | {"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-2024-33 | latest | en | 0.845431 |
https://vustudents.ning.com/group/cs602computergraphics/forum/topics/cs602-5th-assignment-reqired?commentId=3783342%3AComment%3A2191619&groupId=3783342%3AGroup%3A60025 | 1,642,416,797,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300533.72/warc/CC-MAIN-20220117091246-20220117121246-00469.warc.gz | 652,365,661 | 17,445 | www.vustudents.ning.com
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# CS602 Assignment No 05 Solution & Discussion Due Date:07-02-2013
Assignment Question No. 1 [10] In Bezier Curve if we want to add more control points between the starting and ending point of the Bezier Curve, then discuss with the help of the example, how shape of the Bezier Curve would be changed. Question No. 2 [5] How can we express the third degree Bezier Curve in matrix form, also write the parametric equation of Bezier Curve. You can send email at cs602@vu.edu.pk for any assignment related problem. Your work must be original. No marks will be given in case of cheating or copying from the internet, handouts or from any other student and strict action will be taken against that student.
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### Replies to This Discussion
its my idea if any one have more best solution then tell.
bht bura kr ry saqib bhai
what to do?
.doc file plx :(
:( plx upl0ad idea s0lution in .doc file
2nd q idea
Attachments:
mujy nai pata ye kitna correct ha bt agr yehi chalana ha to changing kar len plz mujy zero mily to i will cry.........
mjhy zero mily t0u i ll also cry :(
itx 1st question
yes 1st question
post 2 nd question
1
2
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# The value of $x$ such that ${3^{2x}} - 2\left( {{3^{x + 2}}} \right) + 81 = 0$ is$a.1 \\ b.2 \\ c.3 \\ d.4 \\ e.5 \\$
Last updated date: 28th May 2024
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Hint: Substitute ${3^x}$ to any other variable say (t) then factorize the equation to reach the answer.
The given equation is
${3^{2x}} - 2\left( {{3^{x + 2}}} \right) + 81 = 0$
Now simplify the above equation as ${3^{x + 2}}$ is written as ${3^x} \times {3^2}$ and ${3^{2x}}$ is written as ${\left( {{3^x}} \right)^2}$.
${3^{2x}} - 2\left( {{3^x} \times {3^2}} \right) + 81 = 0 \\ \Rightarrow {3^{2x}} - \left( {2 \times 9 \times {3^x}} \right) + 81 = 0 \\ \Rightarrow {\left( {{3^x}} \right)^2} - 18 \times {3^x} + 81 = 0 \\$
Now let ${3^x} = t...........\left( 1 \right)$, substitute this value in above equation
$\Rightarrow {t^2} - 18t + 81 = 0$
Now factorize this equation
$\Rightarrow {t^2} - 9t - 9t + 81 = 0 \\ \Rightarrow t\left( {t - 9} \right) - 9\left( {t - 9} \right) = 0 \\ \Rightarrow \left( {t - 9} \right)\left( {t - 9} \right) = 0 \\ \Rightarrow {\left( {t - 9} \right)^2} = 0 \\ \Rightarrow \left( {t - 9} \right) = 0 \\ \Rightarrow t = 9 \\$
Now from equation (1)
${3^x} = t = 9$
Now we know 9 is written as ${3^2}$
$\Rightarrow {3^x} = {3^2}$
So, on comparing
$x = 2$
Hence option (b) is correct.
Note: In such types of questions first simplify the given equation then substitute ${3^x} = t$ this makes the equation simple after this factorizes the equation and calculates the value of t, then re-substitute the value of t and simplifies, we will get the required answer. | 642 | 1,684 | {"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} | 4.65625 | 5 | CC-MAIN-2024-22 | latest | en | 0.707253 |
http://docs.pymc.io/notebooks/Diagnosing_biased_Inference_with_Divergences.html | 1,526,983,348,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864657.58/warc/CC-MAIN-20180522092655-20180522112655-00471.warc.gz | 83,747,855 | 17,857 | # Diagnosing Biased Inference with Divergences¶
In [1]:
%matplotlib inline
import pymc3 as pm
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
plt.style.use('seaborn-darkgrid')
from collections import defaultdict
%config InlineBackend.figure_format = 'retina'
SEED = [20100420, 20100234]
print('Runing on PyMC3 v{}'.format(pm.__version__))
Runing on PyMC3 v3.3
/home/osvaldo/anaconda3/lib/python3.6/site-packages/h5py/__init__.py:36: FutureWarning: Conversion of the second argument of issubdtype from float to np.floating is deprecated. In future, it will be treated as np.float64 == np.dtype(float).type.
from ._conv import register_converters as _register_converters
This notebook is a PyMC3 port of Michael Betancourt’s post on ms-stan. For detailed explanation of the underlying mechanism please check the original post and Betancourt’s excellent paper.
Bayesian statistics is all about building a model and estimating the parameters in that model. However, a naive or direct parameterization of our probability model can sometimes be ineffective, you can check out Thomas Wiecki’s blog post on the same issue in PyMC3. Suboptimal parameterization often leads to slow sampling, and more problematic, biased MCMC estimators.
More formally, as explained in the original post:
Markov chain Monte Carlo (MCMC) approximates expectations with respect to a given target distribution,
$\mathbb{E}{\pi} [ f ] = \int \mathrm{d}q \, \pi (q) \, f(q)$
using the states of a Markov chain, $${q{0}, \ldots, q_{N} }$$,
$\mathbb{E}{\pi} [ f ] \approx \hat{f}{N} = \frac{1}{N + 1} \sum_{n = 0}^{N} f(q_{n})$
These estimators, however, are guaranteed to be accurate only asymptotically as the chain grows to be infinitely long,
$\lim_{N \rightarrow \infty} \hat{f}{N} = \mathbb{E}{\pi} [ f ]$
To be useful in applied analyses, we need MCMC estimators to converge to the true expectation values sufficiently quickly that they are reasonably accurate before we exhaust our finite computational resources. This fast convergence requires strong ergodicity conditions to hold, in particular geometric ergodicity between a Markov transition and a target distribution. Geometric ergodicity is usually the necessary condition for MCMC estimators to follow a central limit theorem, which ensures not only that they are unbiased even after only a finite number of iterations but also that we can empirically quantify their precision using the MCMC standard error.
Unfortunately, proving geometric ergodicity is infeasible for any nontrivial problem. Instead we must rely on empirical diagnostics that identify obstructions to geometric ergodicity, and hence well-behaved MCMC estimators. For a general Markov transition and target distribution, the best known diagnostic is the split $$\hat{R}$$ statistic over an ensemble of Markov chains initialized from diffuse points in parameter space; to do any better we need to exploit the particular structure of a given transition or target distribution.
Hamiltonian Monte Carlo, for example, is especially powerful in this regard as its failures to be geometrically ergodic with respect to any target distribution manifest in distinct behaviors that have been developed into sensitive diagnostics. One of these behaviors is the appearance of divergences that indicate the Hamiltonian Markov chain has encountered regions of high curvature in the target distribution which it cannot adequately explore.
In this notebook we aim to identify divergences and the underlying pathologies in PyMC3.
## The Eight Schools Model¶
The hierarchical model of the Eight Schools dataset (Rubin 1981) as seen in Stan:
$\mu \sim \mathcal{N}(0, 5)$
$\tau \sim \text{Half-Cauchy}(0, 5)$
$\theta_{n} \sim \mathcal{N}(\mu, \tau)$
$y_{n} \sim \mathcal{N}(\theta_{n}, \sigma_{n}),$
where $$n \in \{1, \ldots, 8 \}$$ and the $$\{ y_{n}, \sigma_{n} \}$$ are given as data.
Inferring the hierarchical hyperparameters, $$\mu$$ and $$\sigma$$, together with the group-level parameters, $$\theta_{1}, \ldots, \theta_{8}$$, allows the model to pool data across the groups and reduce their posterior variance. Unfortunately, the direct centered parameterization also squeezes the posterior distribution into a particularly challenging geometry that obstructs geometric ergodicity and hence biases MCMC estimation.
In [2]:
# Data of the Eight Schools Model
J = 8
y = np.array([28., 8., -3., 7., -1., 1., 18., 12.])
sigma = np.array([15., 10., 16., 11., 9., 11., 10., 18.])
# tau = 25.
## A Centered Eight Schools Implementation¶
Stan model:
data {
int<lower=0> J;
real y[J];
real<lower=0> sigma[J];
}
parameters {
real mu;
real<lower=0> tau;
real theta[J];
}
model {
mu ~ normal(0, 5);
tau ~ cauchy(0, 5);
theta ~ normal(mu, tau);
y ~ normal(theta, sigma);
}
Similarly, we can easily implemented it in PyMC3
In [3]:
with pm.Model() as Centered_eight:
mu = pm.Normal('mu', mu=0, sd=5)
tau = pm.HalfCauchy('tau', beta=5)
theta = pm.Normal('theta', mu=mu, sd=tau, shape=J)
obs = pm.Normal('obs', mu=theta, sd=sigma, observed=y)
Unfortunately, this direct implementation of the model exhibits a pathological geometry that frustrates geometric ergodicity. Even more worrisome, the resulting bias is subtle and may not be obvious upon inspection of the Markov chain alone. To understand this bias, let’s consider first a short Markov chain, commonly used when computational expediency is a motivating factor, and only afterwards a longer Markov chain.
### A Dangerously-Short Markov Chain¶
In [4]:
with Centered_eight:
short_trace = pm.sample(600, chains=2, random_seed=SEED)
Auto-assigning NUTS sampler...
Multiprocess sampling (2 chains in 2 jobs)
NUTS: [theta, tau_log__, mu]
100%|██████████| 1100/1100 [00:02<00:00, 486.04it/s]
There were 10 divergences after tuning. Increase target_accept or reparameterize.
There were 9 divergences after tuning. Increase target_accept or reparameterize.
The estimated number of effective samples is smaller than 200 for some parameters.
In the original post a single chain of 1200 sample is applied. However, since split $$\hat{R}$$ is not implemented in PyMC3 we fit 2 chains with 600 sample each instead.
The Gelman-Rubin diagnostic $$\hat{R}$$ doesn’t indicate any problems (values are all close to 1). You could try re-running the model with a different seed and see if this still holds.
In [5]:
pm.summary(short_trace).round(2)
Out[5]:
mean sd mc_error hpd_2.5 hpd_97.5 n_eff Rhat
mu 4.18 3.31 0.18 -1.99 10.30 193.53 1.01
theta__0 6.37 5.94 0.26 -4.14 18.85 418.80 1.00
theta__1 4.99 4.65 0.21 -4.28 13.13 380.18 1.00
theta__2 3.41 6.07 0.25 -8.09 15.29 345.68 1.00
theta__3 4.50 4.89 0.19 -6.26 13.37 544.63 1.00
theta__4 3.20 4.89 0.20 -6.48 12.77 397.12 1.00
theta__5 3.91 4.86 0.21 -6.60 12.81 312.51 1.00
theta__6 6.24 5.17 0.23 -3.35 16.33 405.45 1.00
theta__7 4.81 5.85 0.24 -6.81 16.03 438.11 1.00
tau 4.28 3.03 0.20 0.89 10.31 123.81 1.02
Moreover, the trace plots all look fine. Let’s consider, for example, the hierarchical standard deviation $$\tau$$, or more specifically, its logarithm, $$log(\tau)$$. Because $$\tau$$ is constrained to be positive, its logarithm will allow us to better resolve behavior for small values. Indeed the chains seems to be exploring both small and large values reasonably well,
In [6]:
# plot the trace of log(tau)
pm.traceplot(short_trace, varnames=['tau'], transform=np.log);
Unfortunately, the resulting estimate for the mean of $$log(\tau)$$ is strongly biased away from the true value, here shown in grey.
In [7]:
# plot the estimate for the mean of log(τ) cumulating mean
logtau = np.log(short_trace['tau'])
mlogtau = [np.mean(logtau[:i]) for i in np.arange(1, len(logtau))]
plt.figure(figsize=(15, 4))
plt.axhline(0.7657852, lw=2.5, color='gray')
plt.plot(mlogtau, lw=2.5)
plt.ylim(0, 2)
plt.xlabel('Iteration')
plt.ylabel('MCMC mean of log(tau)')
plt.title('MCMC estimation of log(tau)');
Hamiltonian Monte Carlo, however, is not so oblivious to these issues as $$\approx$$ 3% of the iterations in our lone Markov chain ended with a divergence.
In [8]:
# display the total number and percentage of divergent
divergent = short_trace['diverging']
print('Number of Divergent %d' % divergent.nonzero()[0].size)
divperc = divergent.nonzero()[0].size / len(short_trace) * 100
print('Percentage of Divergent %.1f' % divperc)
Number of Divergent 19
Percentage of Divergent 3.2
Even with a single short chain these divergences are able to identity the bias and advise skepticism of any resulting MCMC estimators.
Additionally, because the divergent transitions, here shown in green, tend to be located near the pathologies we can use them to identify the location of the problematic neighborhoods in parameter space.
In [9]:
pm.pairplot(short_trace,
sub_varnames=['theta_0','tau_log__'],
divergences=True,
color='C3', figsize=(10, 5), kwargs_divergence={'color':'C2'})
plt.title('scatter plot between log(tau) and theta[0]');
It is important to point out that the pathological samples from the trace are not necessarily concentrated at the funnel: when a divergence is encountered, the subtree being constructed is rejected and the transition samples uniformly from the existing discrete trajectory. Consequently, divergent samples will not be located exactly in the region of high curvature.
In pymc3, we recently implemented a warning system that also saves the information of where the divergence occurs, and hence you can visualize them directly. To be more precise, what we include as the divergence point in the warning is the point where that problematic leapfrog step started. Some could also be because the divergence happens in one of the leapfrog step (which strictly speaking is not a point). But nonetheless, visualizing these should give a closer proximate where the funnel is.
Notices that only the first 100 divergences are stored, so that we don’t eat all memory.
In [10]:
divergent_point = defaultdict(list)
chain_warn = short_trace.report._chain_warnings
for i in range(len(chain_warn)):
for warning_ in chain_warn[i]:
if warning_.step is not None and warning_.extra is not None:
for RV in Centered_eight.free_RVs:
para_name = RV.name
divergent_point[para_name].append(warning_.extra[para_name])
for RV in Centered_eight.free_RVs:
para_name = RV.name
divergent_point[para_name] = np.asarray(divergent_point[para_name])
ii = 5
tau_log_d = divergent_point['tau_log__']
theta0_d = divergent_point['theta'][:, ii]
Ndiv_recorded = len(tau_log_d)
In [11]:
_, ax = plt.subplots(1, 2, figsize=(15, 6), sharex=True, sharey=True)
pm.pairplot(short_trace,
sub_varnames=['theta_0', 'tau_log__'],
divergences=True,
ax=ax[0],
color='C7', figsize=(12, 10), alpha=0.2,
kwargs_divergence={'color':'C2'})
plt.title('scatter plot between log(tau) and theta[5]')
pm.pairplot(short_trace,
sub_varnames=['theta_0', 'tau_log__'],
divergences=True,
ax=ax[1],
color='C7', alpha=0.2, figsize=(12, 10),
kwargs_divergence={'color':'C2', 'label':'Divergent samples'})
theta_trace = short_trace['theta']
theta0 = theta_trace[:, 0]
plt.plot([theta0[divergent == 1][:Ndiv_recorded], theta0_d],
[logtau[divergent == 1][:Ndiv_recorded], tau_log_d],
'k-', alpha=.25)
plt.scatter(theta0_d, tau_log_d,
color='C3', alpha=.5, label='Location of Energy error (start location of leapfrog)')
plt.title('scatter plot between log(tau) and theta[0]')
plt.legend();
There are many other ways to explore and visualize the pathological region in the parameter space. For example, we can reproduce Figure 5b in Visualization in Bayesian workflow
In [12]:
tracedf = pm.trace_to_dataframe(short_trace)
plotorder = ['mu', 'tau', 'theta__0', 'theta__1', 'theta__2', 'theta__3', 'theta__4',
'theta__5', 'theta__6', 'theta__7']
tracedf = tracedf[plotorder]
_, ax = plt.subplots(1, 2, figsize=(15, 4), sharex=True, sharey=True)
ax[0].plot(tracedf.values[divergent == 0].T, color='k', alpha=.025)
ax[0].plot(tracedf.values[divergent == 1].T, color='C2', lw=.5)
ax[1].plot(tracedf.values[divergent == 0].T, color='k', alpha=.025)
ax[1].plot(tracedf.values[divergent == 1].T, color='C2', lw=.5)
divsp = np.hstack([divergent_point['mu'][:,None],
np.exp(divergent_point['tau_log__'])[:,None],
divergent_point['theta']])
ax[1].plot(divsp.T, 'C3', lw=.5)
plt.ylim([-20,40])
plt.xticks(range(10), plotorder)
plt.tight_layout();
In [13]:
# A small wrapper function for displaying the MCMC sampler diagnostics as above
def report_trace(trace):
# plot the trace of log(tau)
pm.traceplot(trace, varnames=['tau'], transform=np.log)
# plot the estimate for the mean of log(τ) cumulating mean
logtau = np.log(trace['tau'])
mlogtau = [np.mean(logtau[:i]) for i in np.arange(1, len(logtau))]
plt.figure(figsize=(15, 4))
plt.axhline(0.7657852, lw=2.5, color='gray')
plt.plot(mlogtau, lw=2.5)
plt.ylim(0, 2)
plt.xlabel('Iteration')
plt.ylabel('MCMC mean of log(tau)')
plt.title('MCMC estimation of log(tau)')
plt.show()
# display the total number and percentage of divergent
divergent = trace['diverging']
print('Number of Divergent %d' % divergent.nonzero()[0].size)
divperc = divergent.nonzero()[0].size / len(trace) * 100
print('Percentage of Divergent %.1f' % divperc)
# scatter plot between log(tau) and theta[0]
# for the identifcation of the problematic neighborhoods in parameter space
pm.pairplot(trace,
sub_varnames=['theta_0', 'tau_log__'],
divergences=True,
color='C3', figsize=(10, 5), kwargs_divergence={'color':'C2'})
plt.title('scatter plot between log(tau) and theta[0]');
### A Safer, Longer Markov Chain¶
Given the potential insensitivity of split $$\hat{R}$$ on single short chains, Stan recommend always running multiple chains as long as possible to have the best chance to observe any obstructions to geometric ergodicity. Because it is not always possible to run long chains for complex models, however, divergences are an incredibly powerful diagnostic for biased MCMC estimation.
In [14]:
with Centered_eight:
longer_trace = pm.sample(4000, chains=2, tune=1000, random_seed=SEED)
Auto-assigning NUTS sampler...
Multiprocess sampling (2 chains in 2 jobs)
NUTS: [theta, tau_log__, mu]
100%|██████████| 5000/5000 [00:10<00:00, 455.96it/s]
There were 105 divergences after tuning. Increase target_accept or reparameterize.
There were 228 divergences after tuning. Increase target_accept or reparameterize.
The acceptance probability does not match the target. It is 0.5192263333577541, but should be close to 0.8. Try to increase the number of tuning steps.
The estimated number of effective samples is smaller than 200 for some parameters.
In [15]:
report_trace(longer_trace)
Number of Divergent 333
Percentage of Divergent 8.3
In [16]:
pm.summary(longer_trace).round(2)
Out[16]:
mean sd mc_error hpd_2.5 hpd_97.5 n_eff Rhat
mu 4.32 3.22 0.12 -1.98 10.62 458.33 1.02
theta__0 6.19 5.56 0.16 -4.27 18.22 432.74 1.00
theta__1 4.78 4.77 0.16 -4.43 14.66 113.48 1.02
theta__2 3.92 5.36 0.13 -7.19 14.98 1549.72 1.00
theta__3 4.69 4.70 0.12 -4.90 14.15 864.23 1.01
theta__4 3.46 4.64 0.13 -5.48 13.34 793.89 1.01
theta__5 4.06 4.70 0.12 -5.94 13.29 1261.23 1.01
theta__6 6.39 4.99 0.15 -2.45 17.32 707.78 1.00
theta__7 4.73 5.27 0.14 -5.62 15.67 850.21 1.01
tau 3.80 3.18 0.16 0.60 10.03 75.81 1.00
Similar to the result in Stan, $$\hat{R}$$ does not indicate any serious issues. However, the effective sample size per iteration has drastically fallen, indicating that we are exploring less efficiently the longer we run. This odd behavior is a clear sign that something problematic is afoot. As shown in the trace plot, the chain occasionally “sticks” as it approaches small values of $$\tau$$, exactly where we saw the divergences concentrating. This is a clear indication of the underlying pathologies. These sticky intervals induce severe oscillations in the MCMC estimators early on, until they seem to finally settle into biased values.
In fact the sticky intervals are the Markov chain trying to correct the biased exploration. If we ran the chain even longer then it would eventually get stuck again and drag the MCMC estimator down towards the true value. Given an infinite number of iterations this delicate balance asymptotes to the true expectation as we’d expect given the consistency guarantee of MCMC. Stopping after any finite number of iterations, however, destroys this balance and leaves us with a significant bias.
More details can be found in Betancourt’s recent paper.
Divergences in Hamiltonian Monte Carlo arise when the Hamiltonian transition encounters regions of extremely large curvature, such as the opening of the hierarchical funnel. Unable to accurate resolve these regions, the transition malfunctions and flies off towards infinity. With the transitions unable to completely explore these regions of extreme curvature, we lose geometric ergodicity and our MCMC estimators become biased.
Algorithm implemented in Stan uses a heuristic to quickly identify these misbehaving trajectories, and hence label divergences, without having to wait for them to run all the way to infinity. This heuristic can be a bit aggressive, however, and sometimes label transitions as divergent even when we have not lost geometric ergodicity.
To resolve this potential ambiguity we can adjust the step size, $$\epsilon$$, of the Hamiltonian transition. The smaller the step size the more accurate the trajectory and the less likely it will be mislabeled as a divergence. In other words, if we have geometric ergodicity between the Hamiltonian transition and the target distribution then decreasing the step size will reduce and then ultimately remove the divergences entirely. If we do not have geometric ergodicity, however, then decreasing the step size will not completely remove the divergences.
Like Stan, the step size in PyMC3 is tuned automatically during warm up, but we can coerce smaller step sizes by tweaking the configuration of PyMC3’s adaptation routine. In particular, we can increase the target_accept parameter from its default value of 0.8 closer to its maximum value of 1.
In [17]:
with Centered_eight:
fit_cp85 = pm.sample(5000, chains=2, tune=2000,
nuts_kwargs=dict(target_accept=.85))
Auto-assigning NUTS sampler...
Multiprocess sampling (2 chains in 2 jobs)
NUTS: [theta, tau_log__, mu]
100%|██████████| 7000/7000 [00:16<00:00, 431.92it/s]
There were 210 divergences after tuning. Increase target_accept or reparameterize.
There were 122 divergences after tuning. Increase target_accept or reparameterize.
The number of effective samples is smaller than 10% for some parameters.
In [18]:
with Centered_eight:
fit_cp90 = pm.sample(5000, chains=2, tune=2000,
nuts_kwargs=dict(target_accept=.90))
Auto-assigning NUTS sampler...
Multiprocess sampling (2 chains in 2 jobs)
NUTS: [theta, tau_log__, mu]
100%|██████████| 7000/7000 [00:15<00:00, 444.34it/s]
There were 108 divergences after tuning. Increase target_accept or reparameterize.
The acceptance probability does not match the target. It is 0.7681042731609842, but should be close to 0.9. Try to increase the number of tuning steps.
There were 186 divergences after tuning. Increase target_accept or reparameterize.
The number of effective samples is smaller than 10% for some parameters.
In [19]:
with Centered_eight:
fit_cp95 = pm.sample(5000, chains=2, tune=2000,
nuts_kwargs=dict(target_accept=.95))
Auto-assigning NUTS sampler...
Multiprocess sampling (2 chains in 2 jobs)
NUTS: [theta, tau_log__, mu]
100%|██████████| 7000/7000 [00:32<00:00, 214.59it/s]
There were 59 divergences after tuning. Increase target_accept or reparameterize.
There were 246 divergences after tuning. Increase target_accept or reparameterize.
The acceptance probability does not match the target. It is 0.8854456149540602, but should be close to 0.95. Try to increase the number of tuning steps.
The number of effective samples is smaller than 10% for some parameters.
In [20]:
with Centered_eight:
fit_cp99 = pm.sample(5000, chains=2, tune=2000,
nuts_kwargs=dict(target_accept=.99))
Auto-assigning NUTS sampler...
Multiprocess sampling (2 chains in 2 jobs)
NUTS: [theta, tau_log__, mu]
100%|██████████| 7000/7000 [00:58<00:00, 120.01it/s]
There were 20 divergences after tuning. Increase target_accept or reparameterize.
There were 51 divergences after tuning. Increase target_accept or reparameterize.
The acceptance probability does not match the target. It is 0.9655538350420538, but should be close to 0.99. Try to increase the number of tuning steps.
The number of effective samples is smaller than 10% for some parameters.
In [21]:
df = pd.DataFrame([longer_trace['step_size'].mean(),
fit_cp85['step_size'].mean(),
fit_cp90['step_size'].mean(),
fit_cp95['step_size'].mean(),
fit_cp99['step_size'].mean()], columns=['Step_size'])
df['Divergent'] = pd.Series([longer_trace['diverging'].sum(),
fit_cp85['diverging'].sum(),
fit_cp90['diverging'].sum(),
fit_cp95['diverging'].sum(),
fit_cp99['diverging'].sum()])
df['delta_target'] = pd.Series(['.80', '.85', '.90', '.95', '.99'])
df
Out[21]:
Step_size Divergent delta_target
0 0.375381 333 .80
1 0.304575 332 .85
2 0.252309 294 .90
3 0.073495 305 .95
4 0.032518 71 .99
Here, the number of divergent transitions dropped dramatically when delta was increased to 0.99.
This behavior also has a nice geometric intuition. The more we decrease the step size the more the Hamiltonian Markov chain can explore the neck of the funnel. Consequently, the marginal posterior distribution for $$log (\tau)$$ stretches further and further towards negative values with the decreasing step size.
Since in PyMC3 after tuning we have a smaller step size than Stan, the geometery is better explored.
However, the Hamiltonian transition is still not geometrically ergodic with respect to the centered implementation of the Eight Schools model. Indeed, this is expected given the observed bias.
In [22]:
_, ax = plt.subplots(1, 1, figsize=(10, 6))
pm.pairplot(fit_cp99, sub_varnames=['theta_1', 'tau_log__'], ax=ax,
color='C3', alpha=0.6, label='Centered, delta=0.85')
pm.pairplot(longer_trace, sub_varnames=['theta_1', 'tau_log__'], ax=ax,
color='C1', alpha=0.6, label='Centered, delta=0.99')
plt.title('scatter plot between log(tau) and theta[1]')
plt.legend();
In [23]:
logtau0 = longer_trace['tau_log__']
logtau2 = np.log(fit_cp90['tau'])
logtau1 = fit_cp99['tau_log__']
plt.figure(figsize=(15, 4))
plt.axhline(0.7657852, lw=2.5, color='gray')
mlogtau0 = [np.mean(logtau0[:i]) for i in np.arange(1, len(logtau0))]
plt.plot(mlogtau0, label='Centered, delta=0.85', lw=2.5)
mlogtau2 = [np.mean(logtau2[:i]) for i in np.arange(1, len(logtau2))]
plt.plot(mlogtau2, label='Centered, delta=0.90', lw=2.5)
mlogtau1 = [np.mean(logtau1[:i]) for i in np.arange(1, len(logtau1))]
plt.plot(mlogtau1, label='Centered, delta=0.99', lw=2.5)
plt.ylim(0, 2)
plt.xlabel('Iteration')
plt.ylabel('MCMC mean of log(tau)')
plt.title('MCMC estimation of log(tau)')
plt.legend();
## A Non-Centered Eight Schools Implementation¶
Although reducing the step size improves exploration, ultimately it only reveals the true extent the pathology in the centered implementation. Fortunately, there is another way to implement hierarchical models that does not suffer from the same pathologies.
In a non-centered parameterization we do not try to fit the group-level parameters directly, rather we fit a latent Gaussian variable from which we can recover the group-level parameters with a scaling and a translation.
$\mu \sim \mathcal{N}(0, 5)$
$\tau \sim \text{Half-Cauchy}(0, 5)$
$\tilde{\theta}_{n} \sim \mathcal{N}(0, 1)$
$\theta_{n} = \mu + \tau \cdot \tilde{\theta}_{n}.$
Stan model:
data {
int<lower=0> J;
real y[J];
real<lower=0> sigma[J];
}
parameters {
real mu;
real<lower=0> tau;
real theta_tilde[J];
}
transformed parameters {
real theta[J];
for (j in 1:J)
theta[j] = mu + tau * theta_tilde[j];
}
model {
mu ~ normal(0, 5);
tau ~ cauchy(0, 5);
theta_tilde ~ normal(0, 1);
y ~ normal(theta, sigma);
}
In [24]:
with pm.Model() as NonCentered_eight:
mu = pm.Normal('mu', mu=0, sd=5)
tau = pm.HalfCauchy('tau', beta=5)
theta_tilde = pm.Normal('theta_t', mu=0, sd=1, shape=J)
theta = pm.Deterministic('theta', mu + tau * theta_tilde)
obs = pm.Normal('obs', mu=theta, sd=sigma, observed=y)
In [25]:
with NonCentered_eight:
fit_ncp80 = pm.sample(5000, chains=2, tune=1000, random_seed=SEED,
nuts_kwargs=dict(target_accept=.80))
Auto-assigning NUTS sampler...
Multiprocess sampling (2 chains in 2 jobs)
NUTS: [theta_t, tau_log__, mu]
100%|██████████| 6000/6000 [00:06<00:00, 964.62it/s]
There were 5 divergences after tuning. Increase target_accept or reparameterize.
There were 3 divergences after tuning. Increase target_accept or reparameterize.
In [26]:
pm.summary(fit_ncp80).round(2)
Out[26]:
mean sd mc_error hpd_2.5 hpd_97.5 n_eff Rhat
mu 4.38 3.30 0.03 -2.49 10.57 11024.33 1.0
theta_t__0 0.33 0.98 0.01 -1.62 2.21 14375.77 1.0
theta_t__1 0.09 0.96 0.01 -1.82 1.91 13212.45 1.0
theta_t__2 -0.08 0.95 0.01 -1.95 1.76 14160.93 1.0
theta_t__3 0.06 0.96 0.01 -1.86 1.91 14159.81 1.0
theta_t__4 -0.17 0.93 0.01 -1.94 1.74 13618.34 1.0
theta_t__5 -0.08 0.94 0.01 -1.85 1.86 14357.32 1.0
theta_t__6 0.36 0.97 0.01 -1.55 2.18 11769.38 1.0
theta_t__7 0.08 0.98 0.01 -1.81 2.00 12260.39 1.0
tau 3.71 3.32 0.04 0.00 10.31 6344.00 1.0
theta__0 6.31 5.71 0.06 -4.55 17.98 9310.19 1.0
theta__1 4.94 4.69 0.04 -4.49 14.38 11068.91 1.0
theta__2 3.92 5.19 0.06 -7.36 13.73 10019.67 1.0
theta__3 4.76 4.84 0.05 -4.90 14.65 11157.51 1.0
theta__4 3.51 4.61 0.04 -6.24 12.18 10478.96 1.0
theta__5 3.98 4.82 0.05 -5.71 13.64 10618.70 1.0
theta__6 6.35 5.13 0.05 -3.60 16.64 9684.43 1.0
theta__7 4.86 5.27 0.05 -6.19 15.27 10132.72 1.0
As shown above, the effective sample size per iteration has drastically improved, and the trace plots no longer show any “stickyness”. However, we do still see the rare divergence. These infrequent divergences do not seem concentrate anywhere in parameter space, which is indicative of the divergences being false positives.
In [27]:
report_trace(fit_ncp80)
Number of Divergent 8
Percentage of Divergent 0.2
As expected of false positives, we can remove the divergences entirely by decreasing the step size,
In [28]:
with NonCentered_eight:
fit_ncp90 = pm.sample(5000, chains=2, tune=1000, random_seed=SEED,
nuts_kwargs=dict(target_accept=.90))
# display the total number and percentage of divergent
divergent = fit_ncp90['diverging']
print('Number of Divergent %d' % divergent.nonzero()[0].size)
Auto-assigning NUTS sampler...
Multiprocess sampling (2 chains in 2 jobs)
NUTS: [theta_t, tau_log__, mu]
100%|██████████| 6000/6000 [00:08<00:00, 746.35it/s]
Number of Divergent 0
The more agreeable geometry of the non-centered implementation allows the Markov chain to explore deep into the neck of the funnel, capturing even the smallest values of $$\tau$$ that are consistent with the measurements. Consequently, MCMC estimators from the non-centered chain rapidly converge towards their true expectation values.
In [29]:
_, ax = plt.subplots(1, 1, figsize=(10, 6))
pm.pairplot(fit_ncp80, sub_varnames=['theta_0', 'tau_log__'], ax=ax,
color='C0', alpha=0.6, label='Centered, delta=0.80')
pm.pairplot(fit_cp99, sub_varnames=['theta_0', 'tau_log__'], ax=ax,
color='C3', alpha=0.6, label='Centered, delta=0.99')
pm.pairplot(fit_cp90, sub_varnames=['theta_0', 'tau_log__'], ax=ax,
color='C1', alpha=0.6, label='Centered, delta=0.90')
plt.title('scatter plot between log(tau) and theta[1]')
plt.legend();
In [30]:
logtaun = fit_ncp80['tau_log__']
plt.figure(figsize=(15, 4))
plt.axhline(0.7657852, lw=2.5, color='gray')
mlogtaun = [np.mean(logtaun[:i]) for i in np.arange(1, len(logtaun))]
plt.plot(mlogtaun, color='C0', lw=2.5, label='Non-Centered, delta=0.80')
mlogtau1 = [np.mean(logtau1[:i]) for i in np.arange(1, len(logtau1))]
plt.plot(mlogtau1, color='C3', lw=2.5, label='Centered, delta=0.99')
mlogtau0 = [np.mean(logtau0[:i]) for i in np.arange(1, len(logtau0))]
plt.plot(mlogtau0, color='C1', lw=2.5, label='Centered, delta=0.90')
plt.ylim(0, 2)
plt.xlabel('Iteration')
plt.ylabel('MCMC mean of log(tau)')
plt.title('MCMC estimation of log(tau)')
plt.legend(); | 8,438 | 28,210 | {"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} | 2.9375 | 3 | CC-MAIN-2018-22 | longest | en | 0.758852 |
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# ESE319 Introduction to Microelectronics
1
2009 Kenneth R. Laker, updated 06Oct09 KRL
Common Base BJT Amplifier
Common Collector BJT Amplifier
Design Example
## Amplifier Input and Output Impedances
ESE319 Introduction to Microelectronics
2
2009 Kenneth R. Laker, updated 06Oct09 KRL
Basic Single BJT Amplifier Features
CE Amplifier CC Amplifier CB Amplifier
Voltage Gain (A
V
) moderate (-R
C
/R
E
) low (about 1) high
Current Gain (A
I
) moderate ( ) moderate ( ) low (about 1)
Input Resistance high high low
Output Resistance high low high
+1
CE BJT amplifier => CS MOS amplifier
CC BJT amplifier => CD MOS amplifier
CB BJT amplifier => CG MOS amplifier
VCVS CCCS
ESE319 Introduction to Microelectronics
3
2009 Kenneth R. Laker, updated 06Oct09 KRL
Common Collector ( Emitter Follower) Amplifier
In the emitter follower, the output voltage is taken between emitter
and ground. The voltage gain of this amplifier is nearly one the
output follows the input - hence the name: emitter follower.
v
s
v
o
R
E
R
E
R
1
R
2
C
B
V
CC
v
o
v
s
r
o
Current Bias Design
Voltage Bias Design
r
o
ESE319 Introduction to Microelectronics
4
2009 Kenneth R. Laker, updated 06Oct09 KRL
R
1
=R
2
For an assumed = 100:
R
B
=R
1
R
2
=
R
1
2
=(+1)
R
E
10
10 R
E
R
1
=R
2
=20 R
E
V
B
=
V
CC
2
R
E
=
V
E
I
E
=
V
CC
/ 20.7
I
E
Then, choose/specified I
E
, and
the rest of the design follows:
Vb
i
B
i
C
i
E
vout
## As with CE bias design, stable op.
pt. => R
B
(+1) R
E
, i.e.
Emitter Follower Biasing
Split bias voltage drops about
equally across the transistor
V
CE
(or V
CB
) and V
Re
(or V
B
).
For simplicity,choose:
V
B
R
E
V
CC
R
1
R
2
C
B
R
S
v
s
v
o
ESE319 Introduction to Microelectronics
5
2009 Kenneth R. Laker, updated 06Oct09 KRL
Typical Design
Choose: I
E
=1mA
V
CC
=12V
And the rest of the design
follows immediately:
R
E
=
V
E
I
E
=
12/ 20.7
10
3
=5.3k D
Use standard sizes:
R
1
=R
2
=100 k D
R
E
=5.1k D
5.1 k
100 k
100 k
12V
ESE319 Introduction to Microelectronics
6
2009 Kenneth R. Laker, updated 06Oct09 KRL
Equivalent Circuits
<=>
vout
vout
V
CC
/2
Rb
R
B
=R
1
R
2
R
E
R
B
C
B
V
B
v
s
v
o
R
S
V
CC
ESE319 Introduction to Microelectronics
7
2009 Kenneth R. Laker, updated 06Oct09 KRL
Multisim Bias Check
Identical results as expected!
<=>
Rb
+
-
VRb
V
Rb
=I
B
R
B
=
I
E
(+1)
R
B
=0.495V
i
B
Rb
vout
vout
ESE319 Introduction to Microelectronics
8
2009 Kenneth R. Laker, updated 06Oct09 KRL
Small signal mid-band circuit - where C
B
has negligible reactance
(above f
min
). Thevenin circuit consisting of R
S
and R
B
shows
effect of R
B
negligible, since it is much larger than R
S
.
Emitter Follower Small Signal Circuit
Mid-band equivalent circuit:
v
s
'
=
R
B
R
B
+R
S
v
s
=
50
50.05
v
s
v
s
R
TH
=R
S
R
B
=
50
50.05
R
S
R
S
Rb
vout
v
s
R
S
R
B
R
E
v
o
ESE319 Introduction to Microelectronics
9
2009 Kenneth R. Laker, updated 06Oct09 KRL
Follower Small Signal Analysis - Voltage Gain
Circuit analysis:
i
b
=
v
s
R
S
+r
n
+(+1) R
E
v
o
=
R
E
(+1) v
s
R
S
+r
n
+(+1) R
E
A
V
=
v
o
v
s
=
R
E
R
S
+r
n
(+1)
+R
E
1
v
s
=( R
S
+r
n
+(+1) R
E
) i
b
vout
i
b
i
e
Solving for i
b
v
s
R
S
R
E
v
o
v
o
=R
E
i
e
=R
E
(1+)i
b
for Current Bias Design
replace R
E
with r
o
||r
o
= r
o
/2 >> R
E
r
o
r
o
r
o
r
o
ESE319 Introduction to Microelectronics
10
2009 Kenneth R. Laker, updated 06Oct09 KRL
Small Signal Analysis Voltage Gain - cont.
v
o
v
s
=
R
E
R
S
+r
n
(+1)
+R
E
Since, typically:
R
S
+r
n
( +1)
R
E
A
V
=
v
o
v
s
R
E
R
E
=1
Note: A
V
is non-inverting
(or r
o
||r
o
= r
o
/2)
ESE319 Introduction to Microelectronics
11
2009 Kenneth R. Laker, updated 06Oct09 KRL
i
b
=
v
bg
r
n
+(+1) R
E
Use the base current expression:
To obtain the base to ground resistance of the transistor:
This transistor input resistance is in parallel with the 50 k
R
B
, forming the total amplifier input resistance:
R
in
=R
S
+R
B
r
bg
R
B
r
bg
=
515
(515+50)
50 k D=45.6 k DR
B
=50 k D
D
v
bg
=r
n
i
b
+R
E
i
E
=( r
n
+( +1)) i
b
vbg
+
-
R
in
r
bg
=
v
bg
i
b
=r
n
+(+1) R
E
(+1) R
E
=1015.1 k=515k D
Rb
R
B
=50k DR
S
R
S
=50D
Blocking Capacitor - C
B
- Selection
i
b
i
b
i
e
R
S
v
bg
R
E
R
B
R
in
v
s
C
B
v
o
ESE319 Introduction to Microelectronics
12
2009 Kenneth R. Laker, updated 06Oct09 KRL
C
B
Selection cont.
R
in
50 k D
Assume f
min
= 20 Hz
C
B
10
2n205010
3
1.59j F
Choose C
B
such that its reactance is 1/10 of R
in
at f
min
:
C
B
10
2n f
min
R
in
1
2n f C
B
=
R
in
10
Pick C
B
= 2 F (two 1 F caps in parallel), the nearest standard
value in the RCA Lab. We could be (unnecessarily) more precise
and include R
s
as part of the total resistance in the loop. It is very
small compared to R
in
.
j j
with
ESE319 Introduction to Microelectronics
13
2009 Kenneth R. Laker, updated 06Oct09 KRL
Final Design
2.0 uF
v
s
v
o
C
B
R
E
R
1
R
2
R
S
V
CC
ESE319 Introduction to Microelectronics
14
2009 Kenneth R. Laker, updated 06Oct09 KRL
Multisim Simulation Results
20 Hz Data
1 kHz Data
ESE319 Introduction to Microelectronics
15
2009 Kenneth R. Laker, updated 06Oct09 KRL
Of What value is a Unity Gain Amplifier?
To answer this question,
we must examine the small-signal
output impedance of the amplifier
and its power gain.
i
b
i
e
v
s
v
o
R
E
R
S
ESE319 Introduction to Microelectronics
16
2009 Kenneth R. Laker, updated 06Oct09 KRL
Emitter Follower Output Resistance
v
x
i
x
R
out
0
i
b
i
x
=i
b
i
b
=(1+) i
b
i
b
=
i
x
(1+)
v
x
=i
b
( R
S
+r
n
)=
R
S
+r
n
1+
i
x
R
out
=
v
x
i
x
=
R
S
+r
n
1+
r
n
1+
=r
e
Assume:
I
C
=1mAr
n
=
V
T
I
B
=
V
T
I
C
=2500D
=100 R
S
=50D
R
out
2550
100
=25.5D
R
B
=50k DR
S
R
out
is the Thevenin resistance looking
into the open-circuit output.
v
s
R
S
ESE319 Introduction to Microelectronics
17
2009 Kenneth R. Laker, updated 06Oct09 KRL
Multisim Verification of R
out
Multisim short circuit check
( = 100, v
o
= v
s
):
R
out
=
v
oc
i
sc
=
A
V
v
s(rms)
i
sc(rms)
=
1
0.0396
=25.25D
Thevenin equivalent for the
short-circuited emitter follower.
R
out
Av*vsig
A
V
= 1
<=>
i
sc
=i
x
i
sc
=i
x
R
in
i
x
=(1+)i
b
v
sig
=R
S
i
b
+r
n
i
b
R
out
=
A
V
v
sig
i
x
=
R
S
+r
n
1+
R
in
=R
S
+r
n
+(+1) R
E
(+1) R
E
+
-
v
oc
=A
V
v
s
v
s
R
S
=100
v
s
If = 200, as for most good
NPN transistors, R
out
would
be lower - close to 12 .
ESE319 Introduction to Microelectronics
18
2009 Kenneth R. Laker, updated 06Oct09 KRL
Equivalent Circuits with Load R
L
i
b
i
e
RL
+
-
+
-
R
out
=
v
s( rms)
i
sc (rms)
=
1
0.0396
=25.25D
<=>
Z
in
=
v
s
i
e
'
R
in
=R
S
+r
n
+(+1) R
E
R
L
(+1) R
L
R
out
R
L
R
E
v
s
v
o
R
S
R
E
R
L
v
s
A
v
v
s
v
o
R
in
ESE319 Introduction to Microelectronics
19
2009 Kenneth R. Laker, updated 06Oct09 KRL
Emitter Follower Power Gain
Consider the case where a R
L
= 50 load is connected through an infinite
capacitor to the emitter of the follower we designed. Using its Thevenin
equivalent:
v
o
=
R
L
A
V
v
s
R
L
+R
out
=
50
75
v
s
=
2
3
v
s
i
o
=
A
V
v
s
R
out
+R
L
=
v
s
75
p
o
=v
o
i
o
=
2
225
v
s
2
i
s
=i
b
=
v
s
R
in
v
s
(+1) R
E
R
L
v
s
10150
v
s
5000
p
s
=v
s
i
s
1
5000
v
s
2
D
+
-
-
vth=Gvsig
R
out
25D
R
L
50D
+
50 load is in parallel with 5.1k
R
E
and dominates:
D D
C=
A
V
1
A
pwr
=
p
o
p
s
=
2(5000)
225
=44.41
i
s
v
s
v
o
A
v
v
s
i
o
R
in
R
E
R
L
=5.1k D50D50D
ESE319 Introduction to Microelectronics
20
2009 Kenneth R. Laker, updated 06Oct09 KRL
The Common Base Amplifier
Voltage Bias Design Current Bias Design
ESE319 Introduction to Microelectronics
21
2009 Kenneth R. Laker, updated 06Oct09 KRL
Common Base Configuration
Both voltage and current biasing follow the same rules as
those applied to the common emitter amplifier.
As before, insert a blocking capacitor in the input signal path
to avoid disturbing the dc bias.
The common base amplifier uses a bypass capacitor or a
direct connection from base to ground to hold the base at
ground for the signal only!
The common emitter amplifier (except for intentional R
E
feedback) holds the emitter at signal ground, while the common
collector circuit does the same for the collector.
ESE319 Introduction to Microelectronics
22
2009 Kenneth R. Laker, updated 06Oct09 KRL
We keep the same bias that we established for the gain
of 10 common emitter amplifier.
All that we need to do is pick the capacitor values and
calculate the circuit gain.
Voltage Bias Common Base Design
ESE319 Introduction to Microelectronics
23
2009 Kenneth R. Laker, updated 06Oct09 KRL
Common Base Small Signal Analysis - C
IN
Determine C
IN
:
Find a equivalent impedance for
the input circuit, R
S
,
C
in
, and R
E2
:
4.7 k Ohm
470 Ohm
ideally
for
f f
min
1
2n f
min
C
in
R
S
+R
E2
r
e
1
2n f
min
C
in
=
R
S
+r
e
10
C
in
=
10
2n f
min
( R
S
+r
e
)
v
Re2
=
R
E2
r
e
R
E2
r
e
+R
S
+
1
j2n f C
in
v
s
v
Re2
=
R
E2
r
e
R
E2
r
e
+R
S
v
s
(let ) C
B
=
i
b
i
c
i
e
r
e
=
r
n
1+
r
e
v
s
NOTE:
R
B
is shorted
by C
B
=
v
Re2
ESE319 Introduction to Microelectronics
24
2009 Kenneth R. Laker, updated 06Oct09 KRL
Determine C
IN
cont.
2n f
min
C
in
( R
S
+r
e
)1C
in
10
2n f
min
( R
S
+r
e
)
=
10
2n2075
F
A suitable value for C
in
for a 20 Hz lower frequency:
C
in
=
10
125.675
1062j F !
Not too Practical!
Must choose smaller value of C
in
.
1. Choose:
2n f
min
C
in
( R
S
+r
e
)=1
or
2. Choose larger f
min
ESE319 Introduction to Microelectronics
25
2009 Kenneth R. Laker, updated 06Oct09 KRL
Small-signal Analysis - C
B
i'
b
i'
c
i'
e
Note the ac reference current
reversals (due to v
s
polarity)!
v
s
=R
S
i
e
'
+
(
r
n
+
1
j oC
B
)
i
b
'
v
s
=R
S
i
e
'
+
(
r
n
+
1
j oC
B
)
i
e
'
+1
i
e
'
=
+1
( +1) R
S
+r
n
+
1
j oC
B
v
s
Z
in
=
v
s
i
e
'
i
c
i
e
i
b
Determine
Z
in
Determine C
B
: (let ) C
in
=
R
E2
>> R
S
ib'
v
s
ignore R
B
ESE319 Introduction to Microelectronics
26
2009 Kenneth R. Laker, updated 06Oct09 KRL
Determine C
B
i'
b
i'
c
i'
e
i
e
'
=
+1
(+1) R
S
+r
n
+
1
j 2n f C
B
v
s
ideally
1
2n f C
B
(+1) R
S
+r
n
f f
min
R
E2
>> R
S
ib'
v
s
r
e
Z
in
(+1) R
S
+r
n
+1
=R
S
+
r
n
+1
or
f f
min
Choose (conservatively):
Z
in
=
v
s
i
e
'
C
B
10
2n f
min
(
(+1) R
S
+r
n
)
F
ignore R
B
ESE319 Introduction to Microelectronics
27
2009 Kenneth R. Laker, updated 06Oct09 KRL
Determine - C
B
cont.
i'
b
i'
c
i'
e
Choosing (conservatively):
vout
C
B
10
2n20( (100)50+2500)
=10.6j F
i.e.
R
E2
>> R
S
ib'
v
s
Choose (less conservatively):
C
B
10
2n f
min
(
(+1) R
S
+r
n
)
F
for f
min
= 20 Hz
C
B
1
2n20( (100)50+2500)
=1.06jF
ignore R
B
ESE319 Introduction to Microelectronics
28
2009 Kenneth R. Laker, updated 06Oct09 KRL
Small-signal Analysis Voltage Gain
i
e
'
1
R
S
+
r
n
+1
v
s
=
1
(
R
S
+r
e
)
v
s
v
out
=R
C
i
c
'
=oR
C
i
e
'
=
+1
R
C
R
S
+r
e
v
s
A
V
=
v
out
v
s
=
+1
R
C
R
S
+r
e
=
100
101
5100
50+25
67
Assume: C
B
=C
in
=
R
E2
>> R
S
ib'
v
s
ignore R
B
v
out
ESE319 Introduction to Microelectronics
29
2009 Kenneth R. Laker, updated 06Oct09 KRL
Multisim Simulation
1060 uF
v
s
10.6 uF
ESE319 Introduction to Microelectronics
30
2009 Kenneth R. Laker, updated 06Oct09 KRL
Multisim Frequency Response
20 Hz response
1 kHz Response | 4,925 | 10,836 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-21 | latest | en | 0.62305 |
https://governmentadda.com/profit-loss-tricks-1/ | 1,620,755,182,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991648.10/warc/CC-MAIN-20210511153555-20210511183555-00495.warc.gz | 310,287,912 | 66,534 | Profit Loss Tricks – 1
Profit/Loss is another easy topic of SSC CGL. Most of the questions can be solved in less than 30 seconds. First let me introduce a formula that will be used in solving 50% of the questions.
Where,
SP = Selling price
CP = Cost Price
f = Profit/loss factor
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What is this profit/loss factor ? It’s simple, ‘f’ depends on the profit/loss %
If profit% = 10, then f = 1.1
If profit% = 30, then f = 1.3
If profit% = 15, then f = 1.15
If loss% = 10, then f = 0.9
If loss% = 25, then f = 0.75
If loss% = 12.5, then f = 0.875
Please note that f depends on profit/loss percentage and not on the absolute value of Profit/Loss. So if in any question it is given that the profit is Rs. 30, then it doesn’t mean that f = 1.3
Let us see some SSC CGL questions that can be solved with this formula
1. 1
Let ‘s’ be the SP of 1 article and ‘c’ be the CP of 1 article.
Given, 6c = 4s
Therefore s/c = 1.5
Gain % = 50
Q .2.
Let ‘s’ be the SP of 1 metre of cloth and ‘c’ be the CP of 1 metre of cloth
Total SP = 20s, Profit = 4s
CP = SP – Profit = 16s
The ratio s/c = 20s/16s = 1.25
Gain % = 25
Q . 3.
SP of 1 article(s) = Rs. 10/8 = 5/4
CP of 1 article(c) =Rs. 8/10 = 4/5
s/c = 25/16 = 1.5625
Gain % = 56.25
1. Q) 4. Kunal sold a shirt at a loss of 10%. Had he sold it for Rs 60 more, he would have gained 5% on it. Find the CP of the shirt.
In this question we have to find the CP of the article
c = s/f
From basic mathematics or elementary science we know that putting delta (∆) sign in numerator and denominator doesn’t change the equation. ∆ stands for ‘change’
c = ∆s/∆f
where ∆s = change in SP
∆f = change in factor
Therefore ∆s = New SP – Old SP = Rs. 60
∆f = New factor – Old factor
New factor is the factor when profit is 5%. Old factor is the one with loss = 10%
So ∆f = 1.05 – 0.9 = 0.15
c = ∆s/∆f = 60/0.15 = Rs. 400
Answer : Rs 400 | 766 | 2,210 | {"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} | 4.46875 | 4 | CC-MAIN-2021-21 | longest | en | 0.904827 |
https://www.trainerroad.com/forum/t/heart-rate-curve-instead-of-power-curve/25865 | 1,652,984,307,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662529658.48/warc/CC-MAIN-20220519172853-20220519202853-00206.warc.gz | 1,230,320,471 | 8,646 | # Heart rate curve instead of power curve
So I was thinking about how best to pace efforts on my mountain bike, which doesn’t have a power meter. I wondered if there would be any point in using some kind of “heart rate curve”, similar to the power curve, except based on max HR over time rather than watts. In that way, if I know I have a 15 minute climb on a route, I can look on the curve to see what my historical highest HR was over 15 minutes and use that value to try to pace the effort. I looked on Google and couldn’t find anything… Does such a thing exist? Should it exist? Answers on a postcard…
Mark, this is not an unreasonable request. “HRcurve” is plotted on all common training software including strava. Also recall the powercal strap from powertap?..it estimated power from heart rate, once it had some baseline data. I recently did some work with David Tinker on the same thing, and he now has a model built into intervals.icu. However you want pacing info. If you were satisfied with training load from heart rate that is possible: look up the TRIMP here (https://fellrnr.com/wiki/TRIMP). However these models appear to work best with heart rate zones > HRR > HRmax. The difficulty is you want pacing feedback, which I assume you mean instant feedback on your pace? This could probably be achieved (approximately) by monitoring your HRzones during the first half of the effort. A more accurate model would be numerical comparison against your previous effort / personalized to you. The tricky bit s of course HR lag which is the minutes of effort that HR is “catching up” with actual power and or speed. Against intervals.icu has controlled for this mathematically, I suggest you try uploading some of your rides and see if their HR model works for you. bw alex
1 Like
Yes - you can use HR and RPE to pace climbs. Takes some experience to be able to “calibrate” appropriately - but very doable.
I mention RPE above - as I’ve found this is needed in addition to HR. Does HR look right; does RPE feel right; if so, you’ve got the right pacing.
2 Likes
Hi, yes but I would like to remove the need to have to calibrate or use RPE. The idea would be to consult the duration of the effort on the x axis and read off what my max heart rate is, in the same way that I can see on a power curve what my best power is over any given duration…
@ Alex_Fastfitnesstips I took a look at intervals ICU but couldn’t find anything that pulls historical max HR data into a graph…
MaxHR as a graph, no I don’t think they have that. If you think it is important then David Tinker might look at it. bw alex
I have added that to the todo list. Interested to see it for myself! I don’t have power in races (I have a powertap hub for training and deep carbon wheels for racing) and I pace climbs on HR. I know I need to stay under 181 unless there is only a minute or so left and if I go over 183 there had better be only seconds left.
2 Likes
Thanks David!! I’m super interested in seeing the results!!!
This has now been added. You see this info on each activity with HR. Intervals.icu is busy computing the curves for everyone which will take another day or two so yours might not be done yet.
4 Likes
This is really good!! It seems to do exactly what I had anticipated…Thank you!!
1 Like
Awesome stuff!
How is this calculated? Is this the highest average heart rate for a given duration?
Yes thats it exactly. | 785 | 3,426 | {"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-2022-21 | latest | en | 0.966148 |
https://philpeople.org/profiles/emmanuel-lopez/publications | 1,723,456,390,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641036895.73/warc/CC-MAIN-20240812092946-20240812122946-00485.warc.gz | 343,811,609 | 12,939 | # Emmanuel Lopez
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##### Sets, classes and the propositional calculus Manuscrito 28 (2): 417-448. 2005.
The propositional calculus AoC, “Algebra of Classes”,and the extended propositional calculus EAC, “Extended Algebra ofClasses” are introduced in this paper. They are extensions, by additionalpropositional functions which are not invariant under the biconditional,of the corresponding classical propositional systems. Theirorigin lies in an analysis, motivated by Cantor’s concept of the cardinalnumbers, of A. P. Morse’s impredicative, polysynthetic set theory
• 17
##### Chateaubriand on propositional logic Manuscrito 31 (1): 103-113. 2008.
In Logical Forms Part II, Chateaubriand begins the Chapter on “Propositional Logic” by considering the reading of the ‘conditional’ by ‘implies’; in fact he states that:There is a confusion, as a matter of fact, and it runs deep, but it is a confusion in propositional logic itself, and the mathematician’s reading is a rather sensible one.After a careful, erudite analysis of various philosophical viewpoints of logic, Chateaubriand comes to the conclusion that:Pure propositional logic, as just cha…Read more
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##### The Logic of Classes Logic Journal of the IGPL 15 (5-6): 689-706. 2007.
An extension of the Quantified Propositional Calculus1 obtained by the addition of two binary propositional functions is put forward as an inheritor of E. Schröder's “Algebra der Logik”. The formal system is itself not new, in fact it forms part of A. P. Morse's “A Theory of Sets”; although the latter is considered as a first-order system. Since the additional propositional functions are not invariant under the logical biconditional, this system–and many others naturally obtained from it–give us…Read more
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• 213 | 547 | 1,894 | {"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-2024-33 | latest | en | 0.80104 |
http://forum.allaboutcircuits.com/threads/need-info.6442/ | 1,485,096,276,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281426.63/warc/CC-MAIN-20170116095121-00266-ip-10-171-10-70.ec2.internal.warc.gz | 104,589,449 | 13,751 | # need info
Discussion in 'Math' started by shaider530, Jul 7, 2007.
1. ### shaider530 Thread Starter New Member
Jul 7, 2007
6
0
j exponential j, or j rise to j? what is the answer with solution.
2. ### hgmjr Moderator
Jan 28, 2005
9,030
214
Welcome to AAC, shaider530,
hgmjr
3. ### recca02 Senior Member
Apr 2, 2007
1,211
0
tried it on my calculator
got a math error which is the same error for an operation involving division by zero.
perhaps it is an invalid operation.u get the same error for anything(including real number) raised to i.
4. ### Papabravo Expert
Feb 24, 2006
10,340
1,850
Try
Code ( (Unknown Language)):
1.
2. exp( j*ln(j)) = 0.20788
3.
4. ln(j) = (j*pi)/2
5. j*ln(j) = (j²*pi)/2 = -pi/2
6. exp(-pi/2) = 0.20788
7.
This all comes from Euler's equation that
Code ( (Unknown Language)):
1.
2. exp(i*pi) = -1
3.
My TI-92 also gives that result without any sort of complaint.
Jul 7, 2007
6
0 | 322 | 919 | {"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} | 2.84375 | 3 | CC-MAIN-2017-04 | latest | en | 0.801618 |
https://solveforce.com/understanding-numerical-order-principles-applications-and-importance/ | 1,722,693,776,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640368581.4/warc/CC-MAIN-20240803121937-20240803151937-00440.warc.gz | 435,425,008 | 45,467 | Numerical order is a fundamental concept in mathematics and data organization, referring to the arrangement of numbers or numerical data according to their value, either in ascending (increasing) or descending (decreasing) sequence. This method of sorting is widely utilized in various fields, from mathematics to computer science, and is essential for efficient data processing, retrieval, and analysis.
### Principles of Numerical Order
1. Ascending Order: Numbers are arranged from the smallest to the largest. For example, sorting the set {7, 2, 9, 4} in ascending order results in {2, 4, 7, 9}.
2. Descending Order: Numbers are arranged from the largest to the smallest. The same set {7, 2, 9, 4}, when sorted in descending order, would be {9, 7, 4, 2}.
3. Significance of Order in Decimal Systems: In numerical systems, especially those involving decimals, the arrangement might also consider the placement of decimal points. For example, sorting {1.1, 1.01, 1.001, 2.1} in ascending order yields {1.001, 1.01, 1.1, 2.1}.
### Applications of Numerical Order
1. Mathematics: In mathematics education, learning to sort numbers numerically is foundational, helping students understand number value, perform calculations more effectively, and solve problems.
2. Data Analysis: Numerical ordering is crucial in statistics and data analysis, where it is used to organize data sets before performing statistical measures like median or quartile calculations.
3. Computer Science: Algorithms often require data to be sorted numerically to optimize search efficiency and data management. Sorting algorithms, such as quicksort and mergesort, are fundamental in computer programming.
4. Financial Services: In banking and finance, numerical order is used to organize transactions, account numbers, and financial statements, facilitating easier tracking and management of financial data.
5. Inventory Management: Businesses use numerical order to track and organize inventory, whether by serial number, part number, or other numerical identifiers. This helps in maintaining organized records and efficient retrieval of inventory items.
### Importance of Numerical Order
• Efficiency: Sorting numbers or data in numerical order can significantly enhance the efficiency of data retrieval and processing systems. It allows for quicker searches and easier access to records, which is critical in many technology-driven applications.
• Clarity and Usability: Numerical order simplifies the presentation of data, making it easier to understand and analyze. This is especially important in fields like education, where clear data presentation can enhance learning and comprehension.
• Error Reduction: In automated systems, using numerical order helps minimize errors in data handling. It ensures that algorithms function optimally by providing them with systematically organized data.
### Conclusion
Numerical order is a key principle in organizing data in virtually every field involving numbers. Its applications range from educational methodologies to sophisticated data algorithms in computer science. Understanding and implementing numerical order is crucial for maintaining system efficiency, improving data analysis, and enhancing operational accuracy across various disciplines. | 633 | 3,282 | {"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} | 4.40625 | 4 | CC-MAIN-2024-33 | latest | en | 0.866256 |
https://community.tableau.com/thread/200572 | 1,552,961,625,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912201882.11/warc/CC-MAIN-20190319012213-20190319034213-00148.warc.gz | 459,482,239 | 27,646 | 6 Replies Latest reply on Feb 11, 2016 4:06 AM by Shinichiro Murakami
# Replicate excel formula: week calendar - ron de bruin
Hello, I want to replicate a excel formula, and i can´t to replicate:
Excel:
=ENTERO((A2-FECHA(AÑO(A2-DIASEM(A2-1)+4);1;3)+DIASEM(FECHA(AÑO(A2-DIASEM(A2-1)+4);1;3))+5)/7)
My wrong tableau calculation replication (Ron de bruin):
([Fecha]-DATE(DATEPART('year',([Fecha]-DATEPART('weekday',[Fecha]-1)+4)))+
DATEPART('year',[Fecha]-(DATEPART('weekday',[Fecha]-1)+4))+5)/7
Ron de bruin info:
http://www.rondebruin.nl/win/s8/win001.htm
I attach excel calculation and my fail twbx.
Example calendars
http://www.calendario-365.es/calendario-2015.html
http://www.calendario-365.es/calendario-2016.html
thanks
• ###### 1. Re: Replicate excel formula: week calendar - ron de bruin
Hi Carlos,
Are you not just trying to get the current week which can be obtained using Datepart('week',[Your Date])
With this, you can modify as required to suit your calcs.
Steve
• ###### 2. Re: Replicate excel formula: week calendar - ron de bruin
Hi Steve,
Thanks for you response.
Datepart('Week',date) isn´t valid for us, because for example 01/01/2016 is week 53 for us.
Thanks!
• ###### 3. Re: Replicate excel formula: week calendar - ron de bruin
So this has got a bit messy for me as Tableau is refusing to accept Monday as the start of the week.
According to Wiki, the calc is Int((([Ordinal Position] - [day of week]) + 10) / 7)
Ordinal position is the day's position in the year so 10th Feb 10/02/2016 = 41
Day of week for 10th Feb assuming the first day of the week is Monday = 3
((41 - 3) + 10) / 7 = 6.857 for which the Int is 6 so the Week for 10th Feb is week 6
Now onto the calcs:
[Ordinal Date] Attr(datepart('dayofyear',[Order Date]))
[Week Day] (nb I had to correct this as Tableau is ignoring the start of the week for me)
If Attr(datepart('weekday',[Order Date])-1) = 0 Then 7
Else Attr(datepart('weekday',[Order Date])-1)
End
[ISO Week]
Int((([Ordinal Date]-[Weekday])+10)/7)
Steve
• ###### 4. Re: Replicate excel formula: week calendar - ron de bruin
Carlos,
I think I've done.
My company also uses workweeks and this is under my coverage
But please double check. The logic itself can be used, so you can modify if needed.
[Weeknumber]
int(
(datetrunc('week',[Fecha],"Monday")+3 // 12/28 = ww53, 12/29 = ww01
-
makedate(year(datetrunc('week',[Fecha],"Monday")+3),1,1) // same year January 1st
)
/7)
+1
Thanks,
Shin
.
1 of 1 people found this helpful
• ###### 5. Re: Replicate excel formula: week calendar - ron de bruin
Thanks SHINICHIRO MURAKAMI! It´s perfect for us. best regards, Carlos
• ###### 6. Re: Replicate excel formula: week calendar - ron de bruin
Sounds good
Shin | 854 | 2,767 | {"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-2019-13 | latest | en | 0.71715 |
http://drhuang.com/science/mathematics/software/complexplot.htm | 1,709,347,588,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475727.3/warc/CC-MAIN-20240302020802-20240302050802-00522.warc.gz | 9,012,399 | 3,847 | AI math handbook calculator - Fractional Calculus Computer Algebra System software
+ + + =
## Complex Function plot
With this tool you can visualize complex-valued functions
by assigning a color to each point in the complex plane
according to its argument/phase.
The identity function f(z)=z shows how colors are assigned.
Input box:
Enter any expression in z. Here are some example functions to try:
• z
• arctan(z)
• re(arctan(z)) real part
• im(arctan(z)) imag part
• arctan(z-random())
• z^4-1
• (z^2-i)/(i*z-1)^2
• sin(z)-e^(cos(z))
• log(z)-sech(z-i)
• (z-1)(conj(z)^2-conj(z)-1)
• 0.926(z-7.3857e-2*z^5-4.5458e-3*z^9)
• special function: gamma(z), zeta(z)
• Jacobi Elliptic: sn(z,0.3), cn(z,0.3), dn(z,0.3)
• Taylor Series: cos(z)=sum((-1)^n*z^(2n)/(2n)!,7) parameter must be n
• Atomic Singular Inner Function: prod(e^((z*(e^(2*pi*i/5))^n)/(z-(e^(2*pi*i/5))^n)),5) parameter must be n
• Iterated function: iter(z-z'^2,z,15)
Parameters: You can also use three parameters t,n,u in your function, e. g.
• t, where 0 ≤ t ≤ 1,
• u, where u = exp(i*s) and 0 ≤ s ≤ 2pi
• n, with 0 ≤ n ≤ 30
Zoom In/Out:
Press button (+) to zoom in or (-) to zoom out. Alternatively use the mouse wheel. You can also change the view by dragging the plot.
Its independent variable must be z. put mouse on the graph to show
Mouse z: (re,im) on upper left and values of f(z): (re,im) on upper right.
• random(),re(z),im(z),modulus(z),arg(z),recip(z),neg(z),conj(z),disk(z),floor(z),ceil(z),square(z),cube(z),sqrt(z),exp(z),log(z),
• trig function :
sin(z),cos(z),tan(z),cot(z),sec(z),csc(z),sinh(z),cosh(z),tanh(z),coth(z),sech(z),csch(z),asin(z),acos(z),atan(z),acot(z),asec(z),acsc(z),asinh(z),acosh(z),atanh(z),acoth(z),asech(z),acsch(z),arcsin(z),arccos(z),arctan(z),arccot(z),arcsec(z),arccsc(z),arcsinh(z),arccosh(z),arctanh(z),arccoth(z),arcsech(z),arccsch(z),
• inverse trig function :
asin(z)=arcsin(z),acos(z)=arccos(z),atan(z)=arctan(z),acot(z)=arccot(z),asec(z)=arcsec(z),acsc(z)=arccsc(z),asinh(z)=arcsinh(z),acosh(z)=arccosh(z),atanh(z)=arctanh(z),acoth(z)=arccoth(z),asech(z)=arcsech(z),acsch(z)=arccsch(z)
• special function :
gamma(z),pow(z,2),rationalBlaschke(z,2),mobius(z,2,3,4,5),psymbol(z,2),binet(z),joukowsky(z,2,3),zeta(z),dirichletEta(z),binomial(z,2),sn(z,0.2),cn(z,0.2),dn(z,0.2),sum(z,2),prod(z,2),blaschke(z,2),iter(z,z,3),
## Complex
1. complex - complex function - complex math
2. math handbook chapter 10 complex function
3. complex animate(z) for phase animation, the independent variable must be z.
4. complex plot(z) for phase and/or modulus, the independent variable must be z.
5. complex2D(x) for complex 2 curves of real and imag parts, the independent variable must be x.
6. complex3D(x) for 3 dimensional graph, the independent variable must be x.
7. color WebXR surface of complex function on complex plane
8. Riemann surface - Complex Branches - complex coloring
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https://www.homeworktiger.com/solution-info/8016/Chi--square-statistic | 1,542,264,489,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039742567.46/warc/CC-MAIN-20181115054518-20181115080518-00355.warc.gz | 895,230,340 | 8,606 | ### Question details
Chi -square statistic
\$ 40.00
• Take the data you have assembled from the second part of your Week Two assignment, namely, numbers of first-born boy and girl births in your state between 2007 and 2012, separately by racial group (i.e., American Indians, Asians, Blacks, and Whites). Form a two-by-four contingency table from these data: the two row categories are female (girl) and male (boy), and the four column categories are the four racial groups. Calculate the chi-square statistic from this contingency table, and interpret the result. See attached excel to use
My state is California
• Return to the CDC Wonder website, and obtain the numbers of births in your state between 2007 and 2012, by month. (Disregard gender, or race, or birth order—you want all births). Calculate a chi-square statistic to assess whether there is any seasonality to births. (Your null hypothesis is that births should be equally likely to occur in any of the 12 months. We are ignoring the varying lengths of the months to simplify calculations.) How would you interpret your findings? Explain in 500 words in APA format supported by scholarly sources.
• BONUS: Give a graphical representation of your findings for this portion highlighting what you consider significant.
This is the total I came up with from the website you can verify before you complete the assignment
Census Region
Results are sorted in by-variable order
Births Click to sort by Births ascending Click to sort by Births descending
Census Region 4: West (CENS-R4) 4,158,870
Total 4,158,870
### Jaimcw
Available solutions | 360 | 1,623 | {"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-2018-47 | latest | en | 0.904465 |
http://4chandata.org/sci-23-75-2 | 1,506,456,446,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818696681.94/warc/CC-MAIN-20170926193955-20170926213955-00497.warc.gz | 4,246,938 | 5,867 | [ 3 / a / adv / an / c / cgl / ck / cm / co / diy / fa / fit / g / i / ic / jp / k / lit / m / mlp / mu / n / o / p / po / q / sci / sp / tg / toy / trv / tv / v / vg / vp / w / wg / wsg / x ]
# /sci/ Science & Math
How do you live with knowing that not nearly enough is being done to stop our downward environmental spiral?
I have an interesting conundrum, /sci/. Today, at uni, we were supposed to prove this statement. I laid out my proof exactly like this, and just before the end of the class, the prof told me it was incorrect, so he had no time to elaborate on what was wrong. Can any /sci/entist figure this out?
What is the intuition behind \Prod_{n=1}^N\ (1+1/n) = 1+N ?
I just realised. HFA/Asperges snuggly fits as a combination of OCD, ADHD and pathological narcissism/aspd. Looks like another fake psychiatric criteria will go down the drain thanks to brain imaging.
/sci/ I'm sure you have seen problems that look like this in the past, as have I. However there is a difference in this one that I can't seem to explain. Right off the bat we have a stick with mass m and length l that is placed at an angle a at some starting point. There is no friction between the stick and the wall. At the starting moment the stick is stationary however due to no friction it will surely slide. Now the question is what the force of the stick on either wall is as a function of the angle a. Now, at first glance this seemed like something I should be able to solve however I am having some difficulty. One thing I can't seem to understand that I think is key is how this example would differ if there was no wall on the left, since there is no friction I would assume it wouldn't make any difference however I have a feeling that the center of mass would not move horizontally in the case of no wall. Does anyone have any ideas on how to approach this?
I know this is 5th grade math but I need help for my test/project. I need a 2nd grade graph function that intersect with x at -.99 and .99 and with y at -2.43.
### A career in scientific research
Is it worth it, or is it too risky?
Why has humanity never created cobalt bombs on any scale other than experimental / prototype stages? They are the ultimate weapons, period. They make thermonuclear bombs look like child's play in comparison. I'm not even joking when I say that cobalt bombs or "salted nukes" are doomsday devices, if you were to set off enough of them it would destroy all complex cellular life on the planet within the span of 24 hours. ALL complex cellular life. On the entire planet. 24 hours. The invention of the nuclear bomb and the cold war's stockpiling of enough to send humanity back to the stone age shows us that we're blatantly suicidal, so why has humanity never went the entire way? Why have we never taken the logical conclusion and just created the world's most efficient "destroy earth" weapon, when it's been within our means to do so since the 1960's? In for a penny in for a pound, I say. Why stop at thermonuclear bombs? You've gotten that far, why not take one last step further?
What are the most common research methods? The only methods i know of are empirical studies and grounded theory.
Hey /sci/, A few weeks ago one of my docents in solid-state physics mentioned a video showing the result of neutron spectroscopy used on a running engine. I think he said something about a nuclear reactor being used to produce the neutrons. Being as curious as unable to find said video: Does anyone know the video or similar videos? Our docent was unable to tell us the source.
Hey /sci/ i will soon be studying math D and i wanna know which place is the best to learn it on and could you provide a newfag some tips? also i will be studying math B since i failed it but managed to get C done.
What is the relationship between the molar mass of alkanols and the molar heat of combustion. How can this trend be explained by the structure and bonding of the alkanol, and how is this significant in regards to the use of the said alkanol as fuel?
[ 3 / a / adv / an / c / cgl / ck / cm / co / diy / fa / fit / g / i / ic / jp / k / lit / m / mlp / mu / n / o / p / po / q / sci / sp / tg / toy / trv / tv / v / vg / vp / w / wg / wsg / x ]
Contact me | All the content on this website come from 4chan.org. All trademarks and copyrights on this page are owned by their respective parties. Images uploaded are the responsibility of the Poster. Comments are owned by the Poster.
Dofus quêtes | 1,319 | 4,633 | {"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-2017-39 | longest | en | 0.95175 |
http://yogi--anand-consulting.blogspot.com/2014/08/yogistarting-with-given-reference-date.html | 1,566,409,792,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316150.53/warc/CC-MAIN-20190821174152-20190821200152-00050.warc.gz | 340,680,192 | 18,732 | ## Wednesday, August 13, 2014
### yogi_Starting With A Given Reference Date (or TODAY) Find Specified Number Of Weekdays (i.e. excluding Saturdays and Sundays)
Yogi Anand, D.Eng, P.E. ANAND Enterprises LLC -- Rochester Hills MI www.energyefficientbuild.com. Aug-13-2014
post by Mark6200:
If function with weekday/workday and skipping weekend
I have the following setup but something is wrong with the following dates. I am trying to achieve that I see the next five working days in the fields but somehow it breaks on certain days, this is due to the fact that today + # of days is not bulletproof, it needs to determine again if today + # of days is not weekend. I am aware of this but am stuck on how to fix it.
B2 =IF(WEEKDAY( TODAY(), 2)>5 , WORKDAY( TODAY(), -1) , TODAY())
C2=IF(WEEKDAY( TODAY()+1, 2)>5 , WORKDAY( TODAY()+1, +1) , TODAY()+1)
D2=IF(WEEKDAY( TODAY()+2, 2)>5 , WORKDAY( TODAY()+2, +1) , TODAY()+2)
E2=IF(WEEKDAY( TODAY()+3, 2)>5 , WORKDAY( TODAY()+3, +1) , TODAY()+3)
F2=IF(WEEKDAY( TODAY()+4, 2)>5 , WORKDAY( TODAY()+4, +1) , TODAY()+4)
I hope someone can help me out with this
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http://konesvrzno.cz/sling-blade-ouqq/voltage-controlled-voltage-source-example-bf0843 | 1,679,428,057,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943746.73/warc/CC-MAIN-20230321193811-20230321223811-00239.warc.gz | 28,177,486 | 6,026 | It is responding depending on the input voltage. The transresistance amplifier is often referred to as a transimpedance amplifier, especially by semiconductor manufacturers.. Based in Austin, Voltage Control designs and … Note that the dependent source is found in the The output of the op-amp turn on and control the MOSFET until the voltage drop across the shunt resistor became .4V. Here, we’ve specified a gain of 100, and a controlling current of R1.nA, meaning the current into terminal nA of resistor R1: Voltage-Controlled Current Source Circuit. A gyrator is shown in Fig. The first two examples specify a DC voltage source of 5 V connected between node 1 and ground. 14.8 b. 14.8 a. 14.12, the gyrator used in Sec. The direction of current is from node 1 to ground. So, it is a current source but instead of having a constant current value, it is current is always two times of the voltage drop across the at top left (). The third and fourth examples specify a 5 mA DC current source between node 1 and ground. In Sec. Voltage Control is a facilitation agency that helps teams work better together with custom-designed meetings and workshops, both in-person and virtual. Circuit with Dependent Source . Current Controlled Voltage Source CCVS H Analog Figure 1. A transresistance amplifier outputs a voltage proportional to its input current. Figure 1 above shows an example of a circuit with a current controlled voltage source. This type of current source can be very useful in industrial applications. Our master facilitators offer trusted guidance and custom coaching to companies who want to transform ineffective meetings, reignite stalled projects, and cut through assumptions. Current Controlled Voltage Source (CCVS) A CCVS is a voltage source where the voltage is controlled by a current elsewhere in a circuit. As a footnote, the only controlled source which has not been considered in the above worked examples is the current-controlled voltage source. AC Sources AC current and voltage sources are impulse functions used for an AC analysis. For example: The current source at right side bottom is a voltage controlled current source. It is scaled to provide a 20mA output current with a +1V input voltage. Voltage Controlled Dependent Voltage Source; Voltage Controlled Dependent Current Source; Current Controlled Dependent Current Source; Current Controlled Dependent Voltage Source; Using PSpice to find the Thévenin Equivalent Circuit; 3rd Tutorial: Subcircuits. 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Current with a +1V input voltage scaled to provide a 20mA output current with a current Controlled voltage CCVS. | 1,882 | 8,753 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-14 | latest | en | 0.906413 |
http://dev.physicslab.org/DocumentPrint.aspx?doctype=5&filename=Freefall_EnergyProjectiles2.xml | 1,527,293,107,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867254.84/warc/CC-MAIN-20180525235049-20180526015049-00631.warc.gz | 71,070,485 | 4,299 | Worksheet Energy Methods: More Practice with Projectiles
Refer to the following information for the next two questions.
A projectile is thrown straight downwards at 6 m/sec from the top of a 30-meter balcony.
Use energy methods to determine how fast it is moving just as it strikes the ground at the base of the balcony.
How much time does it spend in the air?
Refer to the following information for the next four questions.
A projectile is thrown straight upwards at 6 m/sec from the top of a 30-meter balcony.
Use energy methods to determine the maximum height, above the base of the balcony, reached by the projectile in its trajectory.
Use energy methods to determine how fast the projectile is traveling when it is halfway down; that is, when it is 15 meters above the ground at the base of the balcony?
Use energy methods to determine how fast it is moving just as it strikes the ground at the base of the balcony.
How much time does it spend in the air?
Refer to the following information for the next four questions.
A projectile is thrown horizontally at 6 m/sec off the top of a 30-meter balcony.
Use energy methods to determine how fast the projectile will be traveling just as it strikes the ground.
Determine the angle that the projectile strikes the ground at the base of the balcony.
How much time does the projectile take to reach the ground?
Will the projectile hit a bulls-eye placed 15 meters away from the base of the building?
Refer to the following information for the next four questions.
A projectile is released at 15 m/sec at an angle of 37º from the top of a 30 meter balcony.
How fast will it be moving as it passes through the apex of its trajectory?
How fast will it be traveling when it strikes the ground?
At what angle will it strike the ground?
A second projectile is thrown off the same building at the same speed but at an angle of 53º. Compare the following criteria: their impact speeds, their impact angles, their ranges, their total time spent in the air. | 428 | 2,023 | {"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} | 3.71875 | 4 | CC-MAIN-2018-22 | longest | en | 0.927192 |
https://mathoverflow.net/questions/120293/what-logic-is-modelled-by-generalized-boolean-algebra | 1,563,458,786,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525634.13/warc/CC-MAIN-20190718125048-20190718151048-00286.warc.gz | 452,254,437 | 35,827 | # What logic is modelled by generalized boolean algebra?
While it is well known that classical propositional logic is modelled by booleaan algebras, I have never heard of the logic modelled by generalized boolean algebras (GBA) defined by M.Stone (author of Stone's reprezentation theorem). Does anybody know anything about this logic or any attempts to find it? It would be nice to know at least what could be a good analog of implication in GBA. Since Stone axiomatized the GBAs in terms of equations, this is an equational logic, which I am not treating as proper logic - it is more algebra.But how this logic can be formulated in more regular manner by axioms in the form of implications rather than identities?
• I have never seen some particular name attached to this logic; the logic of generalized Boolean algebras is (at least to me) a good enough definition. It is clear (and well known) that this logic can be axiomatized using a presentation for Johansson's minimal logic (see for instance ncatlab.org/nlab/show/minimal+logic ) plus the involutive law (i.e., the axiom $p \Leftrightarrow \neg \neg p$). – boumol Jan 30 '13 at 12:38
• @boumol: this cannot be correct because the logic of GBA cannot have a negation. There is no complement in a GBA. – Andrej Bauer Jan 30 '13 at 14:51
• @Bauer: Sorry, I answered too quick and I wrote a wrong answer. The axiom one has to add to Johansson's minimal logic is the following one $(p \lor q) \Leftrightarrow ((p\Rightarrow q)\Rightarrow q)$. This axiom is the natural generalization of the involutive law (it is obtained when one replaces $q$ with $\bot$). One place where one can find more details is in the book "Residuated lattices" books.google.es/books?id=Gy46SP7px7gC&pg=PA158 – boumol Jan 30 '13 at 15:32
• @Andrej: Yes, there is an implication, namely $y-x$. You just have to realize that truth is $\bot$. And as I already mentioned, the book boumol uses as a reference defines GBAs upside down, so your $\bot$ is their $1$ (which is also $\top$ in this case, but residuated lattices are not in general assumed to have a top), and $y-x$ is their $x\to y=x\setminus y=y/x$. (The monoid operation in residuated lattices is not in general commutative, hence it has two residua, which are denoted $\setminus$ and $/$ in the book. They coincide in the commutative case, and then it is written with $\to$. ... – Emil Jeřábek Jan 31 '13 at 12:26
• ... GBA are commutative, but since Prop. 3.23 gives equivalent axiomatizations of GBA over the noncommutative base theory, it needs to emply $\setminus$ and $/$.) – Emil Jeřábek Jan 31 '13 at 12:29
Propositional logic captures the partial order of a Boolean algebra, i.e., logical entailment $p_1, \ldots, p_n \vdash q$ corresponds to $p_1 \land \cdots \land p_n \leq q$. To obtain a logic for a generalised Boolean algebra we should express its laws in terms of partial order. If the laws are written as adjunctions, the rules of inference will be clearly visible.
A GBA has finite binary meets $\land$ and joins $\lor$, a least element $\bot$, and relative complements $\setminus$. There will be no surprises in the rules for $\land$, $\lor$ and $\bot$.
Conjunction is the easiest. Meet is characterised as $$r \leq p \land q \quad\text{iff}\quad \text{r \leq p and r \leq q}.$$ Reading this from left to right gives us the elimination rules (where I write $\Gamma$ to indicate an arbitrary number of hypotheses $p_1, \ldots, p_n$) $$\frac{\Gamma \vdash p \land q}{\Gamma \vdash p} \quad\text{and}\quad \frac{\Gamma \vdash p \land q}{\Gamma \vdash q}$$ while the other direction gives the introduction rule $$\frac{\Gamma \vdash p \quad \Gamma \vdash q}{\Gamma \vdash p \land q}.$$ Notice how $\Gamma$ took the role of $r$.
A naive conversion of the characterisation of $\bot$, namely $$\bot \leq p,$$ would give us the rule $\bot \vdash p$. We actually want $\Gamma, \bot \vdash p$, but this is ok because $\bot$ is just as well characterised by $$r \land \bot \leq p.$$ This is a small point which becomes very important when we think about disjunctions. Again, a naive conversion of $$p \lor q \leq r \quad\text{iff}\quad \text{p \leq r and q \leq r}$$ would give us something without $\Gamma$. What we really need is the characterisation $$s \land (p \lor q) \leq r \quad\text{iff}\quad \text{s \land p \leq r and s \land q \leq r}.$$ But does this really characterize joins? Yes, thanks to the distributivity law! And so by writing $\Gamma$ instead of $s$ we get the laws $$\frac{\Gamma, p \lor q \vdash r}{\Gamma, p \vdash r} \quad\text{and}\quad \frac{\Gamma, p \lor q \vdash r}{\Gamma, q \vdash r}$$ and $$\frac{\Gamma, p \vdash r \quad \Gamma, q \vdash r}{\Gamma, p \lor q \vdash r}.$$ You may find these rules for $\lor$ a bit odd, but they are equivalent to whatever variant you are used to.
The laws for disjunction baked in just enough distributivity to make the distributivity law provable from the rules stated so far. So we need not worry about distributivity.
The interesting connective is the relative complement. If we secretly think of $p \setminus q$ as "$p$ and not $q$", then it would seem that the relative complement is to be characterised in terms of its lower bounds because it is like a conjunction. Indeed, we have $$r \leq p \setminus q \quad\text{iff}\quad \text{r \leq p and r \land q \leq \bot}$$ which suggests the rules $$\frac{\Gamma \vdash p \quad \Gamma, q \vdash \bot}{\Gamma \vdash p \setminus q}$$ and $$\frac{\Gamma \vdash p \setminus q}{\Gamma \vdash p} \quad\text{and}\quad \frac{\Gamma \vdash p \setminus q}{\Gamma, q \vdash \bot}$$ These seem perfectly reasonable to me.
The point here is not what precise rules I derived, but how I derived them in a principled way:
• logical entailment corresponds to the partial order
• logical operations correspond to the operations
• logical rules corespond to adjunctions that characterize the operations
By the way, there is of course no truth $\top$ in this calculus. If we add it, we get the usual classical propositional calculus, just like a GBA with a top element is a Boolean algebra.
• Does your calculus include any structural rules? In any case, I don’t understand the rationale why you chose the rules you did, and left out others, it all seems rather arbitrary. As far as I can see, if structural rules are included, your calculus describes the $\{\land,\lor,\bot,x\land\neg y\}$-fragment of intuitionistic logic. In particular, it does not derive e.g. $p\setminus(p\setminus q)\vdash q$, which is valid in the semantics given in the first sentence. – Emil Jeřábek Jan 30 '13 at 15:36
• Anyway, a sequent $\Gamma\vdash\phi$ is valid in your semantics iff $\Gamma\models\phi$ in the semantics #1 in my answer, so a complete calculus should be the respective fragment of classical logic. – Emil Jeřábek Jan 30 '13 at 15:56
• I assume that in a sequence $\Gamma \vdash p$ we have a set of propositions $\Gamma$. That's an alternative to fiddling with structural rules which I find more pleasant. My rules are direct transcripts of adjunctions, but it is quite possible I got the characterization of $p \setminus q$ wrong, and so the rules would be wrong as well. – Andrej Bauer Jan 30 '13 at 20:48
• Making antecedents of sequents sets takes care of contraction and exchange, but it will not give you $p\vdash p$, cut, or weakening. The equational theory of GBAs includes more than just adjunctions, that’s the point. – Emil Jeřábek Jan 31 '13 at 12:08
The question is not quite well posed. In algebraic logic, logics are not defined by algebras, but by logical matrices: these are pairs $\langle A,D\rangle$, where $D$ is a subset of $A$, termed the set of designated values. The logic of a class $K$ of matrices is then the consequence relation $\models_K$ such that for any formula $\phi$ and a set of formulas $\Gamma$, $\Gamma\models_K\phi$ holds iff for every $\langle A,D\rangle\in K$ and every homomorphism $v$ from the algebra of formulas to $A$: if $v(\Gamma)\subseteq D$, then $v(\phi)\in D$. In the most important situations, $D$ is equationally definable in $A$: $D=\{x\in A:A\models E(x)\}$ for a set of equations $E$, which reduces matrices back to pure algebraic language. However, there may be many different choices of $E$, hence it is not sufficient to specify just a class of algebras.
For example, the matrices for classical logic are $\langle A,\{1\}\rangle$, where $A$ is a Boolean algebra. Here, $\{1\}$ is definable by $E(x)=\{x\approx1\}$.
See R. Jansana’s SEP article Propositional consequence relations and algebraic logic for a comprehensive introduction to algebraic propositional logic.
Since you didn’t specify which sets of designated values to take in generalized Boolean algebras (and the most obvious choice doesn’t work as GBA do not need to have a top), the question does not necessarily admit a unique answer. Let me give some specific examples. For the following, I assume GBA formulated in the signature $\{\land,\lor,0,-\}$, where $x-y$ is the relative complement of $y$ in $[0,x]$:
1. Let $K$ be the class of matrices $\langle A,D\rangle$, where $A$ is a GBA, and $D$ is a nonempty filter in $A$. Then the logic of $K$ is the $\{\land,\lor,0,-\}$-fragment of classical logic.
2. Let $K$ be the class of matrices $\langle A,\{0\}\rangle$, where $A$ is a GBA. Then the logic of $K$ is a notational variant of the positive fragment (i.e., $\{\lor,\land,1,\to\}$) of classical logic, where the connectives have been renamed to their duals as indicated by the order in which I have written them.
3. Let $K$ be the class of matrices $\langle A,\varnothing\rangle$, where $A$ is a GBA. Then the logic of $K$ is the maximal logic with no theorems, i.e., $\Gamma\models_K\phi$ iff $\Gamma\ne\varnothing$.
4. Let $K$ be the class of matrices $\langle A,A\rangle$, where $A$ is a GBA. Then the logic of $K$ is the inconsistent logic, i.e., $\Gamma\models_K\phi$ for every $\Gamma,\phi$.
Choice #2 is better behaved than the other three as the matrices in question are (equationally definable and) reduced, and in particular, the logic obtained is (finitely, strongly, and regularly) algebraizable, with GBA being its equivalent semantics. (See Jansana’s article for the basic definitions.) While in principle the condition of algebraizability still does not lead to a unique logic from a given class of algebras, it means that the upside-down positive fragment of classical logic corresponds to GBA in as good a sense as the full classical logic corresponds to BA.
[EDIT 2: Let me qualify the previous sentence. It’s true that in general, more than one logic can be algebraizable wrt the same variety of algebras. However, in the case of GBA, there are not that many possible choices for $E(x)$, and it is in fact easy to check that positive classical logic is the unique logic algebraized by GBA (and the translations given below are also unique up to equivalence). Thus, the question does have a well-defined unique answer after all.]
EDIT: I will spell out explicitly what algebraizability of the positive fragment means, and how it provides a logic modelled by GBA. Consider the propositional logic $\vdash$ defined by the following Hilbert calculus: $$\begin{gather} (\phi-\psi)-\phi\\\\ ((\chi-\phi)-(\psi-\phi))-((\chi-\psi)-\phi)\\\\ \phi-(\phi-(\psi-\phi))\\\\ 0\\\\ \phi-(\phi\lor\psi)\\\\ \psi-(\phi\lor\psi)\\\\ ((\phi\lor\psi)-\phi)-\psi\\\\ (\phi\land\psi)-\phi\\\\ (\phi\land\psi)-\psi\\\\ ((\chi-(\phi\land\psi))-(\chi-\psi))-(\chi-\phi)\\\\ \phi,\psi-\phi\vdash\psi \end{gather}$$ Let $\let\sd\vartriangle\phi\sd\psi:=(\phi-\psi)\lor(\psi-\phi)$, and let $\models$ denote validity in GBA. Then we have:
• $\phi_1\approx\psi_1,\dots,\phi_n\approx\psi_n\models\phi\approx\psi$ iff $\phi_1\sd\psi_1,\dots,\phi_n\sd\psi_n\vdash\phi\sd\psi$
• $\phi_1,\dots,\phi_n\vdash\psi$ iff $\phi_1\approx0,\dots,\phi_n\approx0\models\psi\approx0$
• $\phi\approx\psi\models(\phi\sd\psi)\approx0$, $(\phi\sd\psi)\approx0\models\phi\approx\psi$
• $\phi\vdash\phi\sd0$, $\phi\sd0\vdash\phi$
Thus, the mappings $\phi\approx\psi\mapsto\phi\sd\psi$ and $\phi\mapsto\phi\approx0$ provide a bi-interpretation of the quasiequational theory of GBA with the logic given by $\vdash$. (This is what distinguishes this case from other choices of matrices based on GBA, such as #1,3,4 above. These choices give logics modeled in GBA’s, but they do not have matching translations of algebra into logic, hence passing from GBA to such logics is losing structure and information.)
• Actually, I did not find the question ill-posed. My view was that we are trying to describe a particular class of posets, namely GBAs. If at the outset we postulate that $\Gamma \vdash p$ is to be interpreted as "$p$ is an upper bound for the set $\Gamma$", then I think it is clear what the game should be: all inference rules must be valid for GBAs, and every GBA is a model for our inference rules. – Andrej Bauer Jan 30 '13 at 20:52
• @Andrej: That’s exactly the problem: the postulate is unwarranted. – Emil Jeřábek Jan 31 '13 at 12:00 | 3,643 | 13,068 | {"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} | 3.15625 | 3 | CC-MAIN-2019-30 | latest | en | 0.921824 |
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