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A160826 Improvement of A125852 over A053416, A053479 and A053417 1
0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 5, 4, 3, 0, 0, 0, 1, 0, 0, 0, 0, 2, 4, 5, 1, 3, 1, 0, 3, 2, 3, 4, 3, 4, 5, 6, 9, 4, 3, 0, 1, 0, 0, 0, 2, 4, 3, 4, 5, 10, 14, 3, 6, 0, 7, 0, 4, 5, 1, 8, 6, 0, 4, 7, 8, 6, 5, 11, 5, 9, 12, 12, 4, 0, 11, 7, 12, 0, 3, 1, 0, 1, 5, 0, 6, 2, 10, 11, 25, 17, 3, 2, 0, 9, 0, 12, 5, 0, 4, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,10 COMMENTS How many more lattice points of a hexagonal lattice can be covered by placing a disk of diameter n at an optimal center instead of one of the three obvious centers (a lattice point, midpoint between two lattice points, barycenter of a fundamental triangle)? The first difference occurs at n=9, when a diameter 9 disc around e.g. (1/2, 4*sqrt(5)) covers more lattice points than one around (0,0) or (1/2,0) or (1/2,sqrt(3)/6). Clearly a(n) = O(n) as all "extra" points have norm approximately n^2/4 if the optimal center is chosen near (0,0). Does a(n)/n converge? Are there only finitely many n with a(n)=0? LINKS H. v. Eitzen, Table of n, a(n) for n=1..1000 FORMULA a(n) = A125852(n) - max(A053416(n),A053479(n),A053417(n)) EXAMPLE For diameters n=2,4,6,8 a disc around (0,0) and for n=1,3,5,7 a disc around(1/2,0) happens to be optimal (covers as many points as possible); therefore a(1)=a(2)=...=a(8)=0. a(9) = A125852(9) - max(A053416(9),A053479(9),A053417(9)) = 77 - max(73,69,76) = 1. CROSSREFS Sequence in context: A049060 A092462 A256357 * A057149 A087499 A100046 Adjacent sequences: A160823 A160824 A160825 * A160827 A160828 A160829 KEYWORD nonn AUTHOR Hagen von Eitzen, May 27 2009 STATUS approved
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Last modified December 16 09:34 EST 2018. Contains 318160 sequences. (Running on oeis4.) | 927 | 2,375 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-51 | longest | en | 0.783885 |
https://nl.mathworks.com/matlabcentral/cody/solutions/2030010 | 1,590,992,993,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347414057.54/warc/CC-MAIN-20200601040052-20200601070052-00434.warc.gz | 458,721,826 | 15,852 | Cody
# Problem 4. Make a checkerboard matrix
Solution 2030010
Submitted on 21 Nov 2019 by Lauren Busby
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
n = 5; a = [1 0 1 0 1; 0 1 0 1 0; 1 0 1 0 1; 0 1 0 1 0; 1 0 1 0 1]; assert(isequal(a,checkerboard(n)))
2 Pass
n = 4; a = [1 0 1 0; 0 1 0 1; 1 0 1 0; 0 1 0 1]; assert(isequal(a,checkerboard(n)))
3 Pass
n = 7; a = [1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1]; assert(isequal(a,checkerboard(n))) | 321 | 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} | 3.015625 | 3 | CC-MAIN-2020-24 | latest | en | 0.635604 |
https://study.com/academy/topic/8th-grade-math-operations-with-decimals.html | 1,561,477,008,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999853.94/warc/CC-MAIN-20190625152739-20190625174739-00217.warc.gz | 600,650,679 | 25,003 | # Ch 15: 8th Grade Math: Operations with Decimals
Use the succinct and engaging videos found in this chapter to teach your eighth grader how to divide, multiply, add and subtract decimals. Interactive worksheets and end-of-lesson quizzes ensure that your child is getting everything they need out of the lessons.
## 8th Grade Math: Operations with Decimals - Chapter Summary
This chapter covers the basics of using operations with decimals, including tips on how to estimate solutions and solve two-step problems. Middle school students will also get a chance to hone their skills in mapping decimal differences and simplifying expressions. At the end of the chapter, students have a chance to take a comprehensive test that indicates how well they retained the presented information.
## Chapter Lessons and Objectives
Lesson Objective
Inequalities with Decimals Students are presented with decimal inequalities and instructed on how to solve them.
Converting Decimals to Mixed Numbers This lesson shows how to change a decimal into a mixed number.
Adding and Subtracting Decimals: Examples & Word Problems Using word problems, this lesson demonstrates how to add and subtract decimals.
Multiplying and Dividing Decimals: Examples & Word Problems Students study how to multiply and divide decimals.
How to Estimate with Decimals to Solve Math Problems Instructors provide examples of how to estimate decimals to solve problems.
Practice Estimating Sums, Differences, and Products of Decimals At the end of this lesson, students should have a firm grasp of estimating decimals.
Solving Problems using Decimal Numbers Students solve problems involving decimals and fractions.
Solving One and Two Operation Problems Involving Positive Decimals With Estimation Instructors show students how to use estimation to solve operation problems with decimals.
Solving One and Two Operation Problems Involving Positive Decimals With Guess and Check This lesson guides students through solving problems with lists and tables.
Solving One and Two Operation Problems Involving Positive Decimals by Looking for a Pattern With this lesson, students learn how to look for patterns to solve decimal-related operations problems.
How to Solve Multi-step Inequalities with Decimals This lesson covers how to figure out multi-step inequalities.
Maps with Decimal Distances Students view problems involving decimal maps and are instructed on how to solve them.
Simplify Expressions Involving Decimals Complicated expressions can be simplified, and this lesson shows your student how.
4 Lessons in Chapter 15: 8th Grade Math: Operations with Decimals
Test your knowledge with a 30-question chapter practice test
Chapter Practice Exam
Test your knowledge of this chapter with a 30 question practice chapter exam.
Not Taken
Practice Final Exam
Test your knowledge of the entire course with a 50 question practice final exam.
Not Taken
### Earning College Credit
Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 622 | 3,242 | {"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-2019-26 | longest | en | 0.881575 |
http://devguis.com/13-filters-and-attenuators-network-analysis-and-synthesis.html | 1,632,449,152,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057496.18/warc/CC-MAIN-20210924020020-20210924050020-00271.warc.gz | 19,911,416 | 35,848 | 13. Filters and Attenuators – Network Analysis and Synthesis
13. Filters and Attenuators
CHAPTER OBJECTIVES
After carefully studying this chapter, you should be able to do the following:
Explain the basic function of a filter circuit.
Distinguish between a passive filter and an active filter.
Classify passive filters and explain function of each type of filter.
Explain the parameters of a filter.
Draw and explain basic filter networks in T and π sections.
Carryout analysis of filter networks in both T-section and π-section.
Make analysis of constant K-type or proto-type filters.
Solve problems on constant K-type filters.
Make comparison of parameters of constant K-type low-pass and high-pass filters.
State the limitations of constant K-type filters.
Modify constant K-type filters to obtain m-derived filters.
Analyse all types of m-derived filters.
Develop composite filters using constant K-type and m-derived filters.
13.1 INTRODUCTION
A filter blocks unwanted signals or noise signals and passes wanted or desired signals. A filter is basically frequency selective network that allows signals of a particular band of frequencies and rejects or attenuates signals of other frequencies. For example, a low-pass filter (LPF) passes low frequency signals and attenuates all frequencies above a selected cut-off frequency.
A high-pass filter (HPF) network will pass only those input signals whose frequencies are above the selected cut-off frequency.
A band-pass filter (BPF) passes signals of a selected frequency band, while a band-stop filter (BSF) blocks frequencies within its band. The band of frequencies passed by a filter is called pass band and the band of frequency that separates pass band and stop band of a filter is called cut-off frequency. Figure 13.1 shows various types of filters, their input and output wave forms and also their frequency response.
An LPF passes low frequency signals and blocks or attenuates signals that have frequencies above a given cut-off frequency (fc). The input wave form for an LPF, as shown in Figure 13.1(a), is composed of a low frequency signal and high frequency unwanted signal. The filter will allow the low frequency signal. This low frequency signal will appear at the output of the filter. However, the high frequency unwanted signal (noise signal) will be stopped or drastically reduced at the output.
Figure 13.1(a) Low-Pass Filter
The gain versus frequency (f) response graph for the filter and the attenuation versus frequency response have been shown in the Figure.
An HPF allows high frequency input signals and stops or attenuates the low frequency signals as shown in Figure 13.1(b).
Figure 13.1(b) High-Pass Filter
Wave forms of BPF and BSF have been shown in Figure 13.1(c) and (d), respectively.
Figure 13.1(c) Band-Pass Filter
Figure 13.1(d) Band-Stop Filter
13.1.1 Measurement in Decibels
The change in the output from a filter is measured in decibels (dB). The decibel is one-tenth of a bel. Since bel is large, the unit of decibel is used. Let the output voltage of a filter unit changes from v1 to v2 as the frequency changes. The change in power is expressed as the log of the ratio of power that changes from p1 to p2. The change in power (ΔP) is expressed as follows:
If power is assumed to be dissipated in the load resistance RL, then ΔP is given as in the following:
Assuming power as i2 RL, ΔP can be expressed as follows:
The change in output power, voltage and current can be measured using the above three expressions respectively.
13.2 TYPES OF FILTERS
Basically filters are of two types: Active filters and Passive filters.
Active filters are the filters having active elements like OP-AMP and transistor, in addition to resistor and capacitor. An active filter not only passes or stops a particular band of frequency but also amplifies the signal that passes through it.
Passive filters are made up of only passive components like inductor and capacitor. Such filters cannot amplify the signal that passes through them.
In this chapter, we will describe passive filters only.
13.3 CLASSIFICATION OF PASSIVE FILTERS
On the basis of functions they perform, passive filters are classified as follows:
1.Low-Pass Filters (LPF)
2.High-Pass Filters (HPF)
3.Band-Pass Filters (BPF)
4.Band-Stop Filters (BSF)
13.3.1 Low-Pass Filters
These are the filters that pass all the frequencies lower than the selected cut-off frequency fc and attenuate/stop/suppress the signals whose frequency is greater than fc. Attenuation characteristic for an ideal LPF is shown in Figure 13.2. From the attenuation characteristic of the LPF, it is clear that attenuation is zero in pass band and attenuation of signal is maximum in stop band.
It is to be noted that the characteristic shown is for an ideal LPF. For the low-pass filter the pass-band is from 0 to fc and stop-band is from fc to ∞.
Figure 13.2 Attenuation Characteristic of an LPF
13.3.2 High-Pass Filters
It is a filter that passes all the signals whose frequency is higher than the cut-off frequency and stops the signals whose frequency is less than fc.
The attenuation characteristics of HPF is shown in Figure 13.3.
Therefore, for a high-pass filter (HPF) the pass band is from fc to ∞ and the stop-band is from 0 to fc.
Figure 13.3 Attenuation Characteristic of an HPF
13.3.3 Band-Pass Filters
It is a filter that passes a particular band of frequencies and stops all other frequencies.
It has two cut-off frequencies: lower cut-off frequency (f1) and higher cut-off frequency (f2). This filter passes all those signals whose frequency lies inside the band f1 to f2 and stops all other frequency signals. The attenuation characteristics for BPF is shown in Figure 13.4. Therefore, for a BPF, the following are given:
Pass band: f1 to f2
Stop band: 0 to f1 and f2 to ∞.
Figure 13.4 Attenuation Characteristic of a BPF
13.3.4 Band-Stop or Band-Elimination Filter
It is a filter that stops particular band of frequencies and passes all other frequencies. It is just opposite to that of a band-pass filter (BPF). The attenuation characteristics for a band-stop filter is shown in Figure 13.5.
Figure 13.5 Attenuation Characteristic of a Band-Stop Filter
Therefore, for a BPF, the following are given:
Pass band: 0 to f1 and f2 to ∞
Stop band: f1 to f2.
13.4 PARAMETERS OF A FILTER
There are four important parameters that are necessary to analyse the performance of a filter network. They are as follows:
1.Propagation constant γ
2.Attenuation constant α
3.Phase shift constant β
4.Characteristic impedance Z0
13.4.1 Propagation Constant (γ)
For any two-port network terminated by characteristic impedance Z0, as shown in Figure 13.6, we can write the following:
Figure 13.6 A Two-port Network Terminated by Characteristics Impedance Z0
where γ is known as propagation constant. Propagation constant determines the propagation performance of any two-port network.
Moreover,
where α is real part of γ and is known as attenuation constant of the filter, and β is imaginary part of γ and is known as phase constant.
13.4.2 Attenuation Constant
Whenever a signal passes through a passive network/filter, it gets attenuated, because passive components like capacitors and inductors consume some of the signal energy. The attenuation constant determines the attenuation of the signal when it passes through the filter.
Units of Attenuation
Attenuation can be expressed in decibels or nepers.
Neper: It is defined as the natural log of the ratio of input current or voltage or power to the output current or voltage or power.
Decibel: It is defined as the ten times the common log of the ratio of input current/voltage/power and output current/voltage/power.
Decibel (D) can be written as follows:
Relation Between Nepers and Decibel
13.4.3 Phase Shift Constant (β)
When the signal passes through the filter, it gets some shift in phase. Phase shift constant signifies the phase shift in the signal when it passes through the filter.
Units of Phase Shift
The unit of phase shift is radians or degrees. The relation between radians and degrees can be written as follows:
13.4.4 Characteristic Impedance (Z0)
Characteristic impedance is the image impedance of a two-port network. For symmetric networks, the image impedance at port 1-1′ is equal to the image impedance at port 2-2′. They are equal to the characteristic impedance Z0.
13.5 FILTER NETWORKS
A filter is constructed from reactive elements such as inductors and capacitors. Filter produces no attenuation to pass band or transmission band and provides complete attenuation to all other frequencies called attenuation band or stop band. Filters are used in the communication systems.
Filters are made of symmetrical T- or π-Sections. T- and π-Sections are the combination of asymmetrical L networks, as shown in Figures 13.7 and 13.8, respectively.
13.5.1 Formation of Symmetrical T-Network
Figure 13.7 Symmetrical and T-section Network is Made from Asymmetrical L-networks
13.5.2 Formation of Symmetrical π-Network
Figure 13.8 Symmetrical π-Network Made from Asymmetrical L-networks
A ladder network is a cascade or series connection of many T- and π-Sections. There are two types of ladder networks.
These are shown in Figures 13.9 and 13.10, respectively.
It is formed by connecting many T-sections in series, as shown in Figure 13.9.
It is formed by connecting several sections of π-networks in cascade, as shown in Figure 13.10.
13.6 ANALYSIS OF FILTER NETWORKS
In this section, we will explain how the basic parameters of T- and π-networks are determined.
13.6.1 Symmetrical T-network
We know that the basic parameters of a filter network are its characteristic impedance, propagation constant, attenuation constant and phase shift constant. These are calculated as follows.
Characteristic Impedance (Z0)
If a two-port network is symmetrical, the image impedance Zi1 at port 1-1′ is equal to the image impedance Zi2 at port 2-2′ and that image impedance is called the characteristic impedance Z0.
Figure 13.11 A T-network
When the network is terminated with Z0 (characteristic impedance), then input impedance
Consider the T-network shown in Figure 13.11.
The input impedance of the T-network shown in figure is written as follows:
Now, substitute Zin = Z0 (using equation (13.7))
The expression is for characteristic impedance of a T-network.
Z0 (characteristic impedance) can also be expressed in terms of ZOC (open-circuit impedance) and ZSC (short-circuit impedance).
Let us find open-circuit impedance of T-network. From Figure 13.12, it is observed that the open-circuit impedance is written as follows:
Figure 13.12 Open-circuit Impedance of a T-network
From Figure 13.13, the short-circuit impedance of the T-network is calculated as follows:
Figure 13.13 Short-circuit Impedance of a T-network
Multiplying equations (13.9) and (13.10), we get the following:
Therefore, the characteristic impedance for a T-network, ZOT is given as follows:
Figure 13.14(a) A Symmetrical T-network Connected to a Load Impedance Z0
Propagation Constant (γ)
As shown in Figure 13.14, I1, I2, V1 and V2 are input current, output current, input voltage and output voltage, respectively.
Applying KVL in mesh II, as shown in Figure 13.14(a), we get the following:
Now, by definition, we get the following form:
Substituting this value in equation (13.12), the following form can be obtained:
Substituting this value in the equation, we get the following form:
Squaring both sides, the equation can be written as follows:
Dividing both sides by 2, we get the following:
This is the required expression for propagation constant of symmetrical T-network.
Attenuation Constant (α) and Phase Shift Constant (β)
From equation (13.14), propagation constant is a complex function represented as γ = α + . The real part α is a measure of change in the magnitude of the current or the voltage in the network and the imaginary part β is a measure of difference in the phase between the input and output currents or voltages.
Substituting γ = α+, in the equation, we get the following:
Case I: When α = 0, equation (13.15) can be written as follows:
Considering the magnitude,
Case II: To get expression for α, substitute β = π in equation (i)
Cut-off Frequency (fc)
We have
For pass band, α = 0
Substituting γ=jβ in the equation, we get the following form:
Now, the limits of cos θ are ±1
Therefore, in pass band, the equation can be written as follows:
Cut-off frequency can be obtained by substituting the following:
Figure 13.14(b)
The graphical representation of the equations is shown in the following Figure 13.14(b).
Summary of Equations of T-network
1. : characteristic impedance
2. : propagation constant
3. : attenuation constant
4. : phase shift
5.Z1 + 4Z2 = 0 : equation to obtain cut-off frequency.
13.6.2 Analysis of π-Network
Let us find the characteristic impedance for π-network shown in Figure 13.15.
Figure 13.15 A π-network
From the figure, the input impedance of the network is given as follows:
By definition, Zin = Z0, substituting this value in the equation, we get the following:
Multiplying and dividing by Z1, we get the following:
Now, we will prove the following:
where ZOC and ZSC are, respectively, the open-circuit and short-circuit impedance of the π-Network.
Open-circuit Impedance (Zoc) of π-Network
Figure 13.16 Determination of Open-circuit Impedance of a π-Network
Let us find short-circuit impedance of π-network:
From Figure 13.17 and its equivalent circuit shown in Figure 13.18, we get,
Figure 13.17 Determination of Short-circuit Impedance of a π-Network
Multiplying equation (13.17) and (13.18), we get the following:
Figure 13.18 Equivalent Circuit of Figure 13.17
Multiplying and dividing by Z1 we write the equation as follows:
The relation between the characteristic impedances of T-network ZOT and π-network Z are given as in the following:
We have
Substituting the value of from equation (13.19) in equation (13.20), we get the following form:
Hence, the product of characteristic impedance of T-network and π-network is equal to the product of series and shunt impedances.
Here, Z1 is equal to the total series impedance of the filter network and Z2 is equal to the total shunt impedance of the filter network.
13.6.3 Summary of Parameters of Filter Network
Parameters of filter network are given in Table 13.1.
Table 13.1 Parameters of T-network and π-Network
13.7 CLASSIFICATION OF FILTERS
So far, we have seen four types of filters, that is, LPF, HPF, BPF and BSF. When a number of signals are transmitted along a line, filters need to be used to separate them. For example, a speech channel using a carrier frequency of 50 kHz requires a bandwidth of say, 46 to 54 kHz. A band-pass filter (BPF) should pass freely signal of any frequency within this band of frequencies, that is, 46 kHz to 54 kHz and reject all other signals outside this range.
Thus, the output of a filter varies with the frequency. This is called the frequency response of a filter. If the range of frequency or the range of variation of the signal amplitude is large, logarithmic scale is used to plot the frequency response.
We have also known that filters are of two basic types: passive filters and active filters. In both cases, filters are designed to select or reject a band of frequencies. This is achieved by using series or parallel combination of R, L and C. In active filters, operational amplifiers or transistors and R, L and C are used.
Filters may be classified to be of constant K-type or m-Derived type.
Depending upon the relationship between Z1 and Z2, filters are classified as follows:
1.Constant K-type or prototype filters.
2.m-Derived filters
These are described in this section.
13.8 CONSTANT K-TYPE OR PROTOTYPE FILTERS
A constant K-type filter is a filter that satisfies the following relationship:
where Z1 is the series arm impedance.
Z2 is the shunt arm impedance.
K is the design impedance or nominal impedance or zero characteristic impedance.
Constant K-type filters can be of T-type or π-type. These filters are also called prototype filters because other complex type of filters can be derived from constant K-type filters. Constant K-type filters may be of low-pass type, high-pass type, band-pass type or band-stop type. These are discussed in the following sections.
13.8.1 Constant K-type Low-Pass Filters (LPF)
In LPF, the series element is inductor and shunt element is capacitor. T- and π-Section for constant K-type LPF are shown in Figure 13.19.
Figure 13.19 Constant K-type LPF
Expressions for Different Parameters of Constant K-type LPF
Design Impedance (K). In constant K-type LPF,
Total series impedance, Z1 = jωL and
Total shunt impedence,
Further, for constant K-type filters, we have, Z1Z2 = K2
Substituting the values of Z1 and Z2 in the equation, we get the following:
Cut-off Frequency (fc). We have limits of pass band as given in the following:
Now, we have fc = 0 and the condition .
Substituting j2 = −1, we have
Attenuation Constant (α). To find attenuation constant, we have the following relation:
Substituting Z1 = jωL and in the equation, we get the following:
Substitute ω = 2πf
Now, from equation (13.23), we have the following form:
Substituting this value in equation (13.24), we get the following forms:
and attenuation in pass band α = 0.
Phase Constant β. In stop band, β = π and in pass band,
The performance characteristic of a constant K-type LPF has been shown in Figure 13.20.
Figure 13.20 Performance Characteristic of a Constant K-type LPF
Characteristic Impedance (Z0). We have
Substituting Z1 = jωL and , we get the following:
Further, for LPF
and
Substituting the value of Z1 = jωL
Using equation (13.26), we get the following form:
Design Parameters. We have
Equating the two equations, that is, equation (iii) and (iv), we get the following form:
Substituting this value in equation (iii), the equation can be written as follows:
Summary of Constant K-type LPF
1.Design impedance:
2.Design parameters:
3.Cut-off frequency:
4.Attenuation:
5.Phase constant:
6.Characteristic impedance:
Solved Numerical on Constant K-type LPF
Example 13.1 Design a low-pass T-section filter having a cut-off frequency of 1.5 kHz to operate with a design impedance of 600 Ω.
Figure 13.21(a)
Figure 13.21(b)
Figure 13.21(c)
Solution: Given
Now, for constant K-type LPF we write the equation as follows:
The required design is shown in Figure 13.21a.
Example 13.2 Figure 13.21(b) shows a passive filter section. Find its cut-off frequency and characteristic impedance at f = 0.
Solution: Given filter is T-section LPF of Constant K-type. By comparing Figure 13.21(b) with Figure 13.21(c), we get,
Now, for LPF, fc can be written as follows:
Example 13.3 Design a constant K-type LPF having a cut-off frequency of 2000 Hz and a zero-frequency characteristic impedance of 200 Ω. Draw T- and π-Section of the filter.
Solution: Given
Now, for an LPF, L and C can be calculated as follows:
Required design is given in Figure 13.22.
Figure 13.22
Example 13.4 A constant K-type LPF composed of T-section has 63.6 mH inductance in each series arm and 0.088 μF in the shunt arm. Find (1) cut-off frequency and (2) attenuation in β at 5000 Hz.
Solution: The circuit is given in Figure 13.23. For comparison, the general design has also been shown.
Figure 13.23
1.Cut-off frequency
2.Attenuation at 5000 Hz, that is, f = 5000 Hz and we have
Example 13.5 Each arm of a symmetrical T-section LPF consists of 6 mH inductor, while the shunt arm is a 0.03 μF capacitor. Find the design impedance and cut-off frequency.
Solution: Given circuit is shown in Figure 13.24
Now for LPF, the design impedance
and
Figure 13.24
Example 13.6 Design the T- and π-Section of a constant K-type LPF having a cut-off frequency of 10 kHz and design impedance of 450 Ω. Further, find its characteristic impedance and phase constant at 5 kHz as well as determine the attenuation at 12 kHz.
Solution: Given
Now for LPF, L and C can be calculated as follows:
The constant X-type LPF for both T- and π-Sections are shown in Figure 13.25.
Figure 13.25
Characteristic impedance at f = 5 kHz = 1500 Hz
We now calculate Phase shift at f = 5 kHz = 5000 Hz as
Attenuation α at f = 12 kHz, is calculated as below
For constant K-type LPF, we have the value of α as,
13.8.2 Constant K-type High-Pass-Filters (HPF)
In an HPF, the series element is a capacitor and the shunt arm element is an inductor, that is, in case of HPF.
The circuit configuration of constant K-type HPF, both T-type and π-type, have been shown in Figure 13.26.
Figure 13.26 Circuit Configurations of HPF
Expression of parameters for HPF are given as follows.
Design Impedance (K)
We know that for constant K-type HPF, the following can be written as follows:
Substituting the values of Z1 and Z2, we get the following:
Cut-off Frequency (fc)
In pass band, we get the following:
That is,
Attenuation Constant (α)
In pass band, α = 0 and in stop band
We get
Now for HPF, we have the following:
Substituting this value in equation (13.31), we get:
Phase Constant (β)
In pass band, β = π
In stop band,
Performance of HPF
The variation of α with frequency and variation of β with frequency have been shown in Figure 13.27 and Figure 13.28, respectively.
Figure 13.27 Variation of Attenuation with Frequency
Characteristic Impedance (Z0)
We have the following:
Figure 13.28 Variation Phase Shift with Frequency
Substituting
Design Parameters
We have derived the following for HPF:
and
From equations (13.33) and (13.34), it is clear that C can be written as follows:
Substituting this value in equation (13.33), we get the value of L as follows:
Summary of Constant K-type HPF
Attenuation:
Phase constant:
Characteristic impedance:
13.8.3 Comparison of Constant K-type LPF and HPF
We have, so far, discussed the constant K-type low-pass and high-pass filters in both π-section and T-section. The design parameters of these filters are shown in a consolidated manner in Table 13.2.
Table 13.2 Comparative Table of Design Parameters of Constant K-type LPF and HPF
Solved Numericals on Constant K-type HPF
Example 13.7 Design a constant K-type HPF having a cut-off frequency of 5500 Hz and a design impedance of 750Ω. Draw T-section filter and π-Section filter.
Solution: Given
Now for an HPF, L and C can be calculated as follows:
Therefore, the T-section and π-section filters are as shown in Figure 13.29.
Figure 13.29
Example 13.8 A T-section HPF has a cut-off frequency of 3000 Hz and infinite frequency characteristic impedance of 500 Ω. Find characteristic impedance at 5000 Hz.
Solution: Given
Now for HPF, we get the following:
Example 13.9 Figure 13.30 shows an HPF section. Find the cutoff frequency and characteristic impedance at f = ∞.
Figure 13.30
Solution: Given circuit is a T-type HPF and we know the general circuit configuration for T-type HPF is shown in Figure 13.31.
Figure 13.31
Therefore, given
Characteristic impedance at f = ∞
Example 13.10 Figure 13.32 shows a high-pass filter section. Find cut-off frequency and characteristic impedance at f = ∞.
Figure 13.32
Solution: The general circuit configuration of π-Section high-pass filter is as shown in Figure 13.33.
By comparing the general configuration with the given π-Section we get
and
Characteristic impedance at f equal to ∞ is calculated as
Figure 13.33
Example 13.11 Design the T- and π-Section of a constant K-type high-pass filter having cutoff frequency of 20 kHz and design impedance of 450 Ω. Also, find its characteristic impedance and phase constant at 25 kHz as well as determine the attenuation at 4 kHz.
Solution: Given
Now for an HPF, we get the following:
Therefore, the designs are shown in Figure 13.34.
Figure 13.34
The characteristic impedance at f = 25,000 Hz
The phase constant at 25 kHz, that is, at f = 25,000 Hz.
We have
The attenuation at f = 4 kHz, that is, at 400 Hz
For constant K-type HPF, we have the following:
13.8.4 Constant K-type Band-Pass Filter
A band-pass filter can be obtained by connecting a LPF and a HPF in cascade as shown in Figure 13.35.
The circuit configuration of constant K-type BPF has been shown in Figure 13.36.
Figure 13.35 Block Diagram of a Constant K-type Band-pass Filter
T-section
After connecting the T-sections in cascade (series arm is made up of series resonant circuit and short arm is made up of parallel resonant circuit), we get a band-pass filter as shown in Figure 13.37.
Figure 13.36 A Band-pass Filter Obtained by Connecting in Series an LPF and an HPF
Figure 13.37 A Band-pass Filter in T-section Obtained by Connecting in Series an LPF and an HPF
π-Section
After connecting π-sections in cascade (series), we get band-pass filter in π-section as shown in Figure 13.38.
Figure 13.38 A Band-pass Filter in π-Section Obtained by Connecting in Series a π-Section of Constant K-type LPF and a Constant K-type HPF
Analysis of Constant K-type BPF
For series arm, resonant frequency = and for shunt arm, resonant frequency =
In band-pass filter, the resonant frequency of series arm is equal to the resonant frequency of shunt arm.
Design Impedance (K)
and Z2 (total shunt arm impedance)
Now, for constant K-type filters, Z1 Z2 = K2
Substituting the value of Z1 and Z2 in the equation, we get the following:
Using equation (13.35), that is, L1C1 = L2C2, we get the following:
Using equation (13.36), that is, using ,
Cut-off Frequencies
In BPF, there are two cut-off frequencies: low cut-off frequency (f1) and higher cut-off frequency (f2). Let us derive the expressions for f1 and f2 for pass band,
Cut-off frequency can be obtained by the following equations:
Multiplying both sides by Z1, we get
Case 1: Take Z1 = j2K.
Therefore,
This is the one cut-off frequency of BPF.
Case II: When Z1 = − j2K
Then
Taking the negative value, we get the following:
This is the second cut-off frequency of BPF.
Attenuation (α)
Substituting the values of Z1 and Z2 from equations (13.37) and (13.38)
Phase Constant (β)
Resonant Frequency (fo)
We know cut-off frequencies occur when Z1 + 4Z2 = 0 or, Z1 = −4Z2
Multiplying both sides by Z1, we get the following:
That is, Z1 at f1 = Z1 at f2 [∴ Magnitude of Z1 is same at f1 and f1, the difference is only of equation]
Now, when ω0 is resonant free
Characteristic Impedance (Z0)
For a band-pass filter, the following equations can be written as,
Design Parameters
Now, at f1, that is, higher cut-off frequency
Therefore, we get
where f1 is upper cut-off frequency and f2 is lower cut-off frequency.
Further, we have to calculate the value of K.
Substituting the value of C1 in the equation, we get the following form:
Further, we know
Substituting the value of C1 in the equations, we get the value of L1.
and we have
Summary of Band-Pass-Filter:
Circuit Configuration.
Figure 13.39 Band-pass Filter in T-section and in π-section
Design Impedance.
Cut-off Frequencies.
Resonant Frequency.
Phase Constant (β).
Characteristic Impedances.
Design Parameters.
Performance of Constant K-type BPF.
Figure 13.40 α Versus f Characteristic
Figure 13.41 β Versus f Characteristic
Numericals on BPF
Example 13.12 Design the T-section and π-section of a constant K-type BPF that has a pass band from 1500 to 5500 Hz and characteristic resistance of 200 Ω. Further, find resonant frequency of series and shunt arms.
Solution: Given f1 (lower cut-off frequency) = 1500 Hz; f2 (upper cut-off frequency) = 5500 Hz
and K = 200 Ω
Now, for a constant K-type BPF,
Required design is given as shown in Figure 13.42.
Figure 13.42
Now, resonant frequency of series arm
13.8.5 Constant K-type Band-Stop/Band-Elimination Filter
By interchanging the series and shunt arms of the band-pass-filter, we can obtain the band-stop-filter as has been shown in Figure 13.43(a) and (b).
Circuit Configuration
Figure 13.43 Band-stop Filter Developed from Band-pass Filter (a) T-section; (b) π-section
That is, series elements have been connected in parallel and shunt elements have been placed in series.
Further, in this filter, the components, that is, L1, C1, L2 and C2 are so selected that their resonant frequencies are same. This frequency is known as the resonant frequency or centre frequency or rms frequency of the filter. So, we will have
For a band-stop filter, Z1 and Z2 can be calculated as:
and
The various parameters of a band-stop filter are calculated as,
Design impedance (K): Know that for constant K-type filters, Z1 Z2 = K2
Substituting the value of Z1 and Z2 from equations (13.40) and (13.41), we get the following form:
Substituting, L1C1 = L2C2 from equation (13.39), in the above, we get
From equation (13.39), L1C1 = L2C2 or
Substituting this value in the equation, we get the value of K.
Cut-off frequency: Cut-off frequencies can be obtained from the equation
Multiplying both sides by Z1, we get
Case I: When Z1 = j2K (let us assume that it is at f1)
Then, using equation (13.40), we have the following form:
or Expression for lower cut-off frequency of BPF,
Case II: When Z1 = − j2K (let us assume that it is at f2), then the following equation can be obtained
Considering the positive value only, we get f2 as follows:
Expression for higher cut-off frequency,
Attenuation (α): We have
Using equations (13.40) and (13.41), we get
Putting, from equation (13.39), we get
Phase shift (β):
Resonant frequency (f0):
Characteristic impedance (Z0): We have
Using equations (13.40) and (13.41), we get the following
Design parameters:
Putting we get
Further, we have or
Summary of Band-Stop Filter
Circuit configuration:
Figure 13.44 Band-stop Filter in T and p-sections
Design impedance:
Cut-off frequencies:
Attenuation (α):
Phase shift (β):
Resonant frequency(f0):
Characteristic impedance (Z0):
Design parameters:
Example 13.13 Design a passive constant K-type BSF having a design impedance of 200 Ω and cut-off frequency 2000 Hz and 6000 Hz.
Solution: Given K= 200 Ω; f1 = 2000 Hz; f2 = 6000 Hz
Now, for a constant K-type BSF, we have the following form:
Required design is shown as in Figures 13.45 and 13.46.
Figure 13.45
Figure 13.46
13.8.6 Comparison of Constant K-type Filters
After discussing all the constant K-type filters, we now present their circuit parameters in a consolidated way in Table 13.3.
Table 13.3 Parameters of Constant K-type Filters
13.8.7 Limitations of Constant K-type Filters
The constant K-type filters have mainly the following two limitations:
1.The characteristic impedance of the filter circuit does not remain constant over the pass band and it is a function of frequency, that is, it varies with frequency.
We know filter gives ideal performance only if it is terminated by a resistance equal to the characteristic impedance.
Since in constant K-type filters, characteristic impedance is different at different frequencies, a mismatch occurs.
2.Its attenuation does not rise abruptly beyond the cut-off frequency as shown in Figure 13.47 and Figure 13.48 for a low-pass filter.
Figure 13.47 α Versus f Characteristic (Actual Case)
Figure 13.48 α Versus f Characteristic (Ideal Case)
To overcome these limitations of constant K-type filters, m-derived filters are used.
13.9 m-DERIVED FILTERS
In m-derived filters, prototype or constant K-type filter is used.
Following modifications are done to obtain m-derived filters from constant K-type filters.
1.The series impedance is multiplied by a factor m
2.The shunt impedance is divided by a factor m
3.An additional impedance of the opposite sign is added either in series or in parallel
The resonant frequency of the shunt arm is so chosen such that it is slightly higher than that of the constant K-type filter.
At the frequency f, the impedance of shunt branch is zero, that is, the attenuation is infinite. This can be seen from the expression for attenuation as
At f = f, Z2 = 0 and hence α = ∞. This produces very sharp cut-off or attenuation.
Here, m is a constant and 0 < m < 1.
13.9.1 m-Derived T-section
As mentioned earlier, m-derived T-section as shown in Figure 13.50 can be obtained from constant K-type filter shown in Figure 13.49 by making the following modifications:
Figure 13.49 Constant K-type
Figure 13.50 m-Derived T-Section
1.Multiply series impedance, that is, by m.
2.Impedance Z2 is replaced by , such that the value of Z0 (characteristic impedance) should be same for both the cases.
Now, for constant K-type, T-section, we have the following:
For m-derived T-section, we have the following:
Now, equating the characteristic impedances,
The m-derived T-section is as shown in Figure 13.51. Also transformation from constant K-type to m-derived type has been illustrated in Figure 13.52.
Figure 13.51 Final m-Derived T-Section
Figure 13.52 Transformation from Constant K-type to m-Derived Type Filter
13.9.2 m-Derived π-Section
The constant K-type filter and the corresponding m-derived filter in π-sections are shown in Figure 13.53.
Figure 13.53 Constant K-type Filter and its m-Derived in π-section
Therefore,
So, is the parallel combination of mZ1 and .
Therefore, final m-derived π-section is redrawn as shown in Figure 13.54.
Figure 13.54 m-Derived π-section Filter
13.9.3 m-Derived Low-Pass Filter
In a Low-pass filter, series element is inductor, that is, Z1 = jωL and shunt element is the capacitor, that is,
Representing Z1 as L and Z2 as the circuit configurations are drawn as shown in Figures 13.55 and 13.56.
Circuit Configurations
Figure 13.55 m-Derived Low-pass Filter in T-section
Figure 13.56 π-Section for m-Derived Low-Pass Filter
Analysis of m-derived LPF is given in the following:
Frequency of infinite attenuation (f):
We know (cut-off frequency for LPF)
Equating both sides, equation (13.46) we get
Attenuation (α):
For m-derived LPF,
Now, for m-derived LPF,
Substituting the value of (1-m2),
Alteration Constant, where fc > f >
Phase shift (β) is calculated as,
Now, substituting the value of from equation (13.48) in the equation, we get
Variation of attenuation (α) and phase shift (β ) with frequency are shown in Figure 13.57.
Figure 13.57 Shows the Variation of Attenuation and Phase Shift of m-Derived Filter
Therefore, m-derived LPF produces very sharp attenuation but the attenuation falls off beyond f. This is the disadvantage of m-derived filters.
Characteristic impedance: For T-section, the following equation can be obtained:
Using equation (13.47), we have the following form:
For π-Section, we get the following form:
13.9.4 Summary of m-Derived Low-Pass Filter
Circuit configurations:
Figure 13.58 m-Derived Low-pass Filters in T and π-Sections
Expression for m:
Cut-off frequency:
Design parameters:
Attenuation (α):
Phase shift ( β):
Characteristic impedance:
13.9.5 m-Derived High-Pass Filter
Circuit configurations:
Figure 13.59 m-Derived High-pass Filters in T and π-sections
Expression for f: At f, the transmission through filter is zero and attenuation is infinite.
Now,
Now, for constant K-type HPF, we get the following form:
That is, f < fC Expression for m: Now, we have
Attenuation (α):
Firstly, let us find from T-section:
Therefore,
Therefore, attenuation
Phase shift (β): We have
Using equation (13.50), we get the value of β.
The variations of α and β with frequency have been shown in Figure 13.60.
Figure 13.60 Variation of Attenuation and Phase-shift with Frequency
Characteristic impedance (Z0): We have the following form:
Using equation (13.49), we have
Now, Z1Z2 = K2
13.9.6 Summary of m-Derived HPF
Circuit configurations:
Figure 13.61 m-Derived High-pass Filters in T and π-sections
Expression for m:
Cut-off frequencies:
Component values:
Attenuation (α):
Phase shift (β):
Characteristic impedance:
13.9.7 Comparison of m-Derived LPF and HPF
After discussing m-Derived LPF and HPF, we are now in a position to present these filters with their design parameters in a consolidated way as in Table 13.4.
Table 13.4 Design Parameters of m-derived Low-pass and High-pass Filters
13.9.8 m-Derived Band-Pass Filter
Circuit configuration:
Figure 13.62 m-Derived Band-pass Filters Obtained by Cascading m-Derived LPF and HPF in T-sections
After connecting in cascade T-section of m-derived low-pass filter and T-section of m-derived high-pass filter, we will obtain the m-derived band-pass filter as has been shown in Figure 13.63.
Figure 13.63 T-Section of m-Derived BPF
After cascading T-section of m-derived low-pass filter with π-section of m-derived low-pass filter as shown in Figure 13.64(a), we get a π-section band-pass filter as has been shown in Figure 13.64(b).
Figure 13.64 A Band-pass Filter in π-section as in Figure (b) is Formed by Cascading T-section and π-section of m-Derived Low-pass Filters as Shown in Figure (a).
Various useful results for m-derived BPF are as follows:
Expression for m:
where f2 (upper cut-off frequency) and f1 (lower cut-off frequency)
are the frequency of infinite attenuation.
Component values:
13.9.9 m-Derived Band-Stop Filter
m-derived band-stop filters can be made from m-derived band-pass filters both in T-section and π-section after changing their series and shunt arms as shown in Figure 13.65 and Figure 13.66, respectively.
Figure 13.65 m-Derived Band-stop Filter in T-section Obtained from T-section Band-pass Filter
Figure 13.66 m-Derived Band-stop Filter in π-section Obtained from π-section of m-Derived Band-pass Filter
Some useful results for m-derived BSF are as follows:
Component values:
13.10 COMPOSITE FILTERS
We have just studied constant K-type and m-derived filters. It has been observed that a constant K-type filter does not give sharp cut-off/attenuation. We have also seen that an m-derived filter has a sharp-cut-off or attenuation, but its attenuation decreases for frequencies beyond f. The decrease in attenuation beyond f is a limitation of m-derived filters that can be overcome by cascading the m-derived filter with that of constant K-type filter having a rising attenuation beyond cut-off.
Attenuation characteristics of constant K-type filter cascaded with m-derived filter are shown in Figure 13.67.
Figure 13.67 Attenuation Characteristics (α versus f) of Constant K-type Filters
Therefore, the composite filter is a filter that consists of a number of m-derived and prototype sections with two terminating half sections. Terminating half sections are used for proper impedance matching. Terminating half sections are used at both source end and load end with m = 0.6.
A composite filter consists of the following:
1.One or more constant K-type sections to produce cut-off between the pass band and the stop band at a specified frequency fc
2.One or more m-derived sections to given infinite attenuation at a frequency f near to fc
3.Two terminating m-derived half section with m = 0.6 to provide matching with the source and the load.
Block diagram representation of a composite filter is shown in Figure 13.68.
Figure 13.68 A Composite Filter Shown in Block Diagram Form
13.10.1 Composite Low-Pass Filter
Figure 13.69 shows a composite low-pass filter T-section which consists of two terminating half sections at both source and load, one constant K-type section, and one m-derived section. Figure 13.70 shows a similar composition of a composite low-pass filter in π-section.
Figure 13.69 Composite LPF in T-section
Figure 13.70 Composite LPF in π-section
Procedure to Design Composite LPF
Step 1: Design constant K-type section.
For this, find and
and design T- or π-section using following configurations as shown in Figure 13.71.
Figure 13.71 Constant K-section of the Composite Filter in T and π-sections
Step 2: Design m-derived sections.
For this, firstly find , and then, find the component values and hence design T- or π-Section as shown in Figure 13.72.
Figure 13.72 m-Derived Sections of the Composite Low-pass Filter in T and π-sections
Step 3: Design terminating half section using m = 0.6 as shown in Figure 13.73.
Figure 13.73 Terminating Half sections of Composite Low-pass Filter in T and π-sections
Step 4: Assemble all the sections of the designed composite filter.
13.10.2 Composite High-Pass Filter
On the same line of designing composite LPF, a high-pass composite filter can be developed in both T and π-sections as shown in Figures 13.74 and 13.75, respectively.
Figure 13.74 A Composite High-pass Filter in π-section
Figure 13.75 A Composite High-pass Filter in π-section
13.11 ADDITIONAL SOLVED NUMERICALS ON FILTERS
13.11.1 Problems on m-Derived Low-pass Filters
Example 13.14 An m-derived LPF has a cut-off frequency of 2000 Hz. If m = 0.3, find the frequency of infinite attenuation.
Solution: Given
Now, for m-derived LPF, we get the following equation:
Substituting values,
Example 13.15 An m-derived LPF has fc = 1500 Hz and f = 2000 Hz. Find the parameter.
Solution: Given
Now, for m-derived LPF, we have the following equation:
Example 13.16 Design the T-section and π-section of m-derived LPF having a design impedance of 500 Ω, cut-off frequency 2200 Hz and frequency of infinite attenuation of 2500 Hz.
Solution: Given
Now, for m-derived LPF, we can write the equation as follows:
Now, for LPF, we can calculate L and C as follows:
Now, the design of T-type m-derived LPF and component values are as follows:
The required T-section is shown in Figure 13.76
Figure 13.76
The design of π-type m-derived LPF and the component values are given as follows:
The required π-Section is shown in Figure 13.77.
Figure 13.77
Example 13.17 Design an m-derived LPF with T-section having cut-off frequency of 7.2 kHz and infinite attenuation at 7.5 kHz and design impedance of 500 Ω.
Solution: Given
Now, for LPF, we calculate the value of L and C as follows:
and
Now, let us find the component values of T-section m-derived LPF as in the following:
The required design of m-derived low-pass filter in T-section is shown in Figure 13.78.
Figure 13.78
13.11.2 Problems on m-Derived High-pass Filters
Examples 13.18 An m-derived HPF has fc = 1400 Hz and f = 1300 Hz. Find the value of m.
Solution: Given
Now, for m-derived HPF, m can be calculated as follows:
Example 13.19 An m-derived HPF has fc = 1700 Hz and m = 0.6. Find the frequency of infinite attenuation.
Solution: Given
Now, for HPF, f can be calculated as follows:
Example 13.20 Design m-derived HPF having a design impedance of 300 W, cut-off frequency of 2000 Hz and frequency of infinite attenuation of 1700 Hz.
Solution: Given
Now, for m-derived HPF, the following can be obtained:
T-section:
Component values for T-section are as follows:
Therefore, the required design is shown in Figure 13.79. π-Section:
Figure 13.79
Firstly, let us find component values as in the following:
The required design is shown in Figure 13.80.
Figure 13.80
13.11.3 Problems on Composite Filters
Example 13.21 Design a composite LPF to work into 150Ω with cut-off frequency of 1.5 kHz and very high attenuation at 2 kHz.
Solution: Given
Step 1: Let us design constant K-type filter
Now, for constant K-type LPF, component values are as follows:
Therefore, designs are as shown in Figure 13.81:
Figure 13.81
Step 2: Let us design m-derived LPF and for m-derived LPF.
T-section: Component values are as follows.
The circuit designed is shown in Figure 13.82. π-Section:
Figure 13.82
Component values of the π-section as in Figure 13.83 are as follows:
Figure 13.83
Step 3: Let us design terminating half sections.
Terminating half section for T-type as shown in Figure 13.84: For terminating half sections, m = 0.6
Figure 13.84
Terminating half section for π-type as shown in Figure 13.85: Now,
Figure 13.85
Step 4: Now assemble the composite low-pass filter in T-section as shown in Figure 13.86 and in π-section as shown in Figure 13.87.
Figure 13.86
Figure 13.87
Example 13.22 Design and draw the T-section of a composite LPF having the following specifications:
1.Required cut-off frequencies: 2 kHz; 2. Design impedances: 500 Ω
Assume suitable value of m for m-derived full section and half section.
Solution: Given
Step 1 : Let us design constant K-type LPF.
For constant K-type LPF, the value of L and C can be calculated as follows:
Therefore, required T-section of constant K-type LPF is shown in Figure 13.88
Figure 13.88
Step 2: Design m-derived LPF (T-section)
For m-derived full section, if value of m’ is not given to us, then it is assumed as 0.4.
Therefore, let m’ = 0.4
Now,
Therefore, required T-section of m-derived LPF filter is shown in Figure 13.89.
Figure 13.89
Step 3: Design the terminating half section of T-section m-derived LPF with m = 0.6 as shown in Figure 13.90. Component values are as follows:
Figure 13.90
Step 4: Assemble the designed composite filter as shown in Figure 13.91.
Figure 13.91
Example 13.23 Design a composite HPF to work into 500 Ω resistances with cut-off frequency of 1500 Hz and with high attenuation of 1400 Hz.
Solution: Given
Now, for m-derived FPF, the following can be calculated:
Therefore, clearly, for full m-derived section, we will use m = 0.359 and for terminating half section of m-derived filter, we will use m = 0.6
Step 1 : Design constant K-type HPF.
For HPF, L and C can be calculated as follows:
Therefore, the required design of constant K-type HPF in T-section and in π-section are shown, respectively, in Figures 13.92 and 13.93.
Figure 13.92
Step 2: Design m-derived HPF.
For this, we will use m = 0.359.
T-section: Component values are as follows.
The design is shown in Figure 13.94.
π-Section: Component values are as follows.
Figure 13.93
Figure 13.94
Figure 13.95
The design is shown in Figure 13.95.
Step 3: Design of terminating half section with m = 0.6.
For T-section: Component values can be calculated as follows.
The design is shown in Figure 13.96.
Figure 13.96
For π-section: The calculation of component values are given as follows.
The design is shown in Figure 13.97.
Figure 13.97
Design of Composite HPF
Now we are able to make the composite high-pass filter with two terminating half sections, the constant-K section and a m-derived section in both T-section and π-section with the calculated values as has been shown in Figure 13.98 and Figure 13.99, respectively.
Figure 13.98 Composite High-pass Filter Network in T-section
Figure 13.99 Composite High-pass Filter Network in π-section
13.12 ATTENUATORS
Attenuators are used to reduce the signal level required in various applications in the field of electronics. For example, attenuators may be used as volume controllers in a radio station or to obtain a small voltage required in a testing of laboratory, etc.
13.12.1 Introduction
Attenuator is a network, normally a two-port resistive network that reduces the signal level to a desired value. It is inserted between a source and a load to reduce current, voltage and power.
Attenuators may be symmetrical or asymmetrical. In this section, we will discuss only symmetrical resistance attenuators. A resistance attenuator is a network consisting of resistors and is designed to reduce the voltage, current or power by known amount. The block diagram of an attenuator has been shown in Figure 13.100.
Figure 13.100
Units of Attenuation Produced by Attenuator
Attenuation produced by an attenuator may be expressed in decibels (dB) or in nepers (N)
Relation between decibels and nepers can be given as follows:
Basically, there are four types of attenuators. They are as follows:
1.T-type Attenuator;
2.π-Type Attenuator
3.Lattice Attenuator;
4.Bridged T-type Attenuator
These four types of attenuators are described in the following sections.
13.12.2 T-type Attenuator
Symmetrical T-type attenuator is shown in Figure 13.101.
Here, R0 is the design impedance.
Let us find some parameters of T-type attenuator.
Figure 13.101 T-type Attenuator
Attenuation in Nepers (N)
By definition,
Now, applying KVL in mesh II, we get the following:
Characteristic Impedance (R0)
(input resistance of T-type network from Figure)
By definition,
Design Parameters
From equation (13.53), we have the following:
Now, to find the value of R2, let us consider equation (13.52),
Now, substituting the value of R1 from equation (13.53) in the equation, we get the following:
Summary of T-type Attenuator
1.
2.
3.Design parameters are as follows:
Example 13.24 Find the characteristic impedance for the T-network shown in Figure 13.102.
Figure 13.102
Solution: Given a T-attenuator, let us compare it with the general network of T-type attenuator as shown in Figure 13.103.
Figure 13.103
We get
Now, for T-type attenuator, we have the following.
By definition,
Example 13.25 Design a T-type actuator to given attenuation of 20 dB and to work in a line of 200 Ω impedance.
Solution: Given D = 20 dB; R0 = 200 Ω
Now, we have
Further, for T-type attenuator, we get the following forms:
and
Therefore, the required attenuator is shown in Figure 13.104.
Figure 13.104
13.12.3 π-Type Attenuator
Circuit Configuration
Now, we should know that for an attenuator shown in Figure 13.105, the following can be written
Figure 13.105
To find the values of R1 and R2, firstly, let us bisect the π-network as shown in Figure 13.106.
Figure 13.106
Now,
Open-circuit impedance of half section
Now, we have general equations (13.55) and (13.56)
Substituting the value of R2 from equation (13.58) in equation (13.59), we get the following:
From equation (13.58), we have the following:
Summary
Therefore, the design equations for π-type attenuators are as follows:
Example 13.26 Design a π-type attenuator to give an attenuation of 3 N and to work in a line impedance of 375 Ω.
Solution: Given
Now, for π-type attenuator, we get
Substituting the values of N and R0, we get
Substituting N = 3 and R0 = 375 Ω, the following are calculated.
Therefore, the required π-type attenuator is shown in Figure 13.107.
Figure 13.107
13.12.4 Lattice Attenuator
Circuit configuration for symmetric lattice attenuator is shown in Figure 13.108.
Figure 13.108
Now, the network shown in Figure 13.108 can also be drawn as shown in Figure 13.109.
Figure 13.109
Figure 13.110
We calculate can the open-circuit impedance ROC from Figure 13.110.
Figure 13.111
To find short-circuit impedance, we refer to Figure 13.111
Further, from Figure 13.109,
Now, we have
Substituting the values of ROC and RSC from equations (13.62) and (13.63) in the equation, we get the following:
Now, from equation (13.64), we get the following:
Applying components and dividend, we get
Now, from equation (13.65), R2 can be calculated as follows:
Here, substituting the values of R1, we get the following equation:
Summary
Therefore, for lattice attenuator, design equations are as follows:
Example 13.27 Design a symmetrical lattice attenuation to give an attenuation of 20 dB and to work in a line of 200 Ω impedance.
Solution: Given
Now, for lattice attenuation, we get the following equation:
Substituting the value of N and R0 in the equation, we have the following equation:
Therefore, required lattice attenuator is shown in Figure 13.112.
Figure 13.112
13.12.5 Bridged T-type Attenuator
Circuit configuration: The circuit configuration of a bridged T-type attenuator has been shown in Figure 13.113.
Figure 13.113 Bridged T-network
Bisected half section of the bridged T-network is shown in Figure 13.114.
Figure 13.114 Bisected Half Section of Bridged T-network
Now, from Figure 13.114,
Circuit to find RSC is shown in Figure 13.115.
Now, from equation (13.67) and the circuit shown in Figure 13.115, we can write
Figure 13.115
Figure 13.116
Further, from equation (13.66), we get the following equation:
Therefore, we get the following set of equations:
Therefore, for bridged T-type attenuator, the equation is written as follows:
Example 13.28 Design a bridged T-type attenuator for attenuation of 10 dB and load resistance of 200 Ω.
Solution: Given
Now, a bridged T-type attenuator we can write as
Therefore, the required bridged T-type attenuator is shown in Figure 13.117.
Figure 13.117
13.13 MORE SOLVED PROBLEMS ON FILTERS AND ATTENUATORS
Example 13.28 Given two capacitors of 1 μF each and coil L of 10 mH, compute the following:
1.Cut-off frequency and characteristic impedance at infinite frequency for an HPF.
2.Cut-off frequency and characteristic impedance at zero frequency for an LPF.
Draw the constructed sections of filters using these elements.
Solution: Given HPF is shown in Figure 13.118.
Figure 13.118
ZOT at infinite frequency = R0
Given
LPF from the given capacitors and inductor can be drawn as shown in Figure 13.119.
Figure 13.119
Example 13.29 For the network shown in Figure 13.120, find the characteristic impedance
Figure 13.120
Solution: To find Z0:
Now, we know the following:
Example 13.30 A symmetrical T-section has the following data:
Determine the T-section parameters and represent the two-port network.
Solution: We have to find Z1 and Z2 of the network shown Figure 13.121
Figure 13.121
From the expression of ZSC and ZOC as above, we get
Substituting the given values, we can calculate the value of Z2. as
From equation (13.67), we obtain the following form:
Substituting the values of Z0C and Z2 in the equation, we get the following set of equations:
Therefore, required T-section parameters are as follows:
Example 13.31 Design a T-type symmetrical attenuation network that offers 40 dB attenuation with a load of 400 Ω.
Solution: Given attenuation is 40 dB
Series arm resistance can be given as follows:
Shunt arm resistance can be written as follows:
Figure 13.122
The designed network in shown in Figure 13.122.
Example 13.32 Design a constant K-type BPF section having a cut-off frequency of 2 kHz and 5 kHz and a nominal impedance of 600 Ω. Draw the configuration of the filter.
Solution: Now, for constant K-type BPF, we get the value of L1, L2, C1 and C2 as follows:
Therefore, the required configuration is shown in Figure 13.123.
Figure 13.123
Example 13.33 Design the symmetrical bridge T-type attenuator with an attenuation of 40 dB and an impedance of 600 Ω.
Solution: Given
Therefore,
We have the following form;
Therefore,
Therefore, the required symmetrical bridge T-type attenuator is shown in Figure 13.124.
Figure 13.124
Example 13.34 Design an m-derived T-section (high-pass) filter with a cut-off frequency fc = 20 kHz, f = 16 kHz and a design impedance R0 = 600 Ω.
Solution: We know for m-derived filter, the following equation can be obtained:
For HPF, L and C can calculated as follows:
The components of T-section (m-derived HPF) are given as in the following:
Therefore, required filter is shown in Figure 13.125.
Figure 13.125
Example 13.35 Design a symmetrical T-section having parameters of ZOC = 1000 Ω & ZSC = 600Ω.
Solution: Given
Now, for a symmetrical T-section, we have the following:
Therefore, required symmetrical T-section is shown in Figure 13.126.
Figure 13.126
Example 13.36 Design an m-derived LPF (T- and π-Section) having a design impedance of 500Ω and cut-off frequency 1500 Hz and an infinite attenuation frequency of 2000 Hz.
Solution: Given
We know, for an LPF, m can be calculated as follows:
Now, components of T-section (m-derived LPF) can be given as follows:
The values of elements of π-section (m-derived LPF) can be written as follows:
Therefore, configurations in T-section and in π-section are shown in Figure 13.127.
Figure 13.127
Example 13.37 Design a symmetrical bridged T-type attenuator to provide attenuation of 60 dB and to work into a line of characteristic impedance of 600 Ω.
Solution: Given
Therefore, required attenuator is shown in Figure 13.128.
Example 13.38 In a symmetrical T-network, if the ratio of input and output power is 6.76. Calculate the attenuation in neper and dB. Further, design this attenuator operating between source and load resistance of 100 Ω.
Figure 13.128
Solution: Given
Series arm resistance is given as follows:
Shunt arm resistance is calculated as in the following:
Therefore, the required symmetrical T-type network is shown in Figure 13.129.
Figure 13.129
1.Design an m-derived LPF (T- and π-Section) having a design resistance of R0 = 500 Ω and the cut-off frequency (fc) of 1500 Hz and an infinite attenuation frequency of 2000 Hz.
[Ans. As shown in Figure 13.130]
Figure 13.130
2.Design a symmetrical bridged T-type attenuator to provide attenuation of 60 dB and to work into a line of characteristic impedance 600 Ω.
[Ans. As shown in Figure 13.131]
Figure 13.131
3.Design a symmetrical bridge T-type attenuator with an attenuation of 40 dB and an impedance of 600 Ω
[Ans. As shown in Figure 13.132]
Figure 13.132
4.Design a m-derived T-section (HPF) with a cut-off frequency fc = 20 kHz, f = 16 kHz and a design impedance 600 Ω
[Ans. As shown in Figure 13.133]
Figure 13.133
5.Design a symmetrical T-section having parameters of ZOC = 1000 Ω and ZSC = 600 Ω
[Ans. As shown in Figure 13.134]
Figure 13.134
6.Design a T-type symmetrical attenuator that offers 40 dB attenuation with a load of 400 Ω
[Ans. As shown in Figure 13.135]
Figure 13.135
7.Design a constant K-type BPF section having cut-off frequencies of 2 kHz and 5 kHz and a nominal impedance of 600 Ω. Draw the configuration of the filter.
[Ans. As shown in Figure 13.136]
Figure 13.136
8.Given the network for HPF
Figure 13.137
Find the cut-off frequency and characteristic impedance at infinity frequency for an HPF.
[Ans. 1.12 kHz, 141.4 Ω]
9.Given π-Section LPF
Figure 13.138
Compute the cut-off frequency and characteristic impedance at zero frequency for an LPF.
[Ans. 4.5 kHz, 707 Ω]
10.Design a constant K-type BPF T-section having cut-off frequencies 2 kHz and 5 kHz and a nominal impedance of 600 Ω. Draw the configuration of the filter.
[Ans. As shown in Figure 13.139]
Figure 13.139
11.Design a prototype LPF, assuming cut-off frequency fc
13.What is an attenuator? Classify and state its applications.
14.Derive the relationships between neper and decibel units
15.Derive the expression for characteristic impedance of a symmetrical bridged T-type network
16.What are the disadvantages of the prototype filters? How are they overcome in composite filters?
17.Discuss the characteristics of a filter.
18.Differentiate between the attenuator and the amplifier. List the practical applications of attenuators.
19.With the help of frequency response curves, give the classification of filters.
20.Derive an expression for design impedance of a symmetrical T-type attenuator.
21.Explain how the reactance and impedance of an HPF varies with frequency.
22.What is an attenuator? Define the terms decibel and neper. Derive the relation between the two.
23.Write short notes on the following
1.LPF and its design
2.T- and π-Section attenuators.
24.What are the requirements of an ideal filter? Distinguish between an LPF and an HPF.
25.What is the propagation constant of a symmetrical T-section and symmetrical π-Section? What is its significance?
26.How do you choose the resonant frequency of an m-derived LPF and m-derived HPF.
27.What is a band-elimination filter?
MULTIPLE CHOICE QUESTIONS
1.A 26 dB output in watts is equal to
(a)2.4 w
(b)0.26 w
(c)0.156 w
(d)0.4 w
2.The characteristic impedance of an LPF in attenuation band is
(a)Purely imaginary
(b)Zero
(c)Complex frequency
(d)Real value
3.The real part of the propagation constant shows:
(a) Variation of voltage and current on basic unit
(b) Variation of phase shift/position of voltage
(c) Reduction in voltage, current value of signal amplitude.
(d) Reduction of only voltage amplitude.
4.The purpose of an attenuator is to
(a)Increase signal strength
(b)Decrease reflections
(c)Decrease value of signal strength
(d)Provide impedance matching
5 In a transmission line terminated by characteristic impedance, Z0
(a) There is no reflection of the incident wave
(b) The reflection is maximum due to termination.
(c) There are a large number of maximum and minimum on the line.
(d) The incident current is zero for any applied signal.
6.An attenuator is a
(a)Resistive network
(b)R-L network
(c)R-C network
(d)L-C network.
7.If ‘α’ attenuation in nepers, then
(a)Attenuation in dB =
(b)Attenuation in dB = 8.686α
(c)Attenuation in dB = 0.1α
(d)Attenuation is dB = 0.01α
8.For a constant K-type high-pass π-Section filter, characteristic impedance Z0 for f < fc is
(a)Resistance
(b)Inductive
(c)Capacitive
(d)Inductive or capacitive
9.An ideal filter should have
(a)Zero attenuation in the pass band
(b)Zero attenuation in the attenuation band
(c)Infinite attenuation in the pass band
(d)Infinite attenuation in the attenuation band
10.For an m-derived HPF, the cut-off frequency is 4 kHz and the filter has an infinite attenuation at 3.6 kHz, the value of m is
(a)0.436
(b)4.36
(c)0.34
(d)0.6
11.If ZOC = 120 Ω and ZSC = 30 Ω, the characteristic impedance is
(a)30 Ω
(b)60 Ω
(c)120 Ω
(d)150 Ω
12.If ZOC = 100 Ω and ZSC = 64 Ω, the characteristic impedance is
(a)400 Ω
(b)60 Ω
(c)80 Ω
(d)170 Ω
13.The frequency of infinite attenuation (f) of a low-pass m-derived section is
(a)Equal to fc
(b)f = ∞
(c)Close to but greater than fc
(d)Close to but less than fc of the filter
14.A constant K-type BPF has pass band from 1000 Hz to 4000 Hz. The resonance frequency of shunt and series arm is a
(a)2500 Hz
(b)500 Hz
(c)2000 Hz
(d)3000 Hz
15.A constant K-type low-pass T-section filter has Z0 = 600 Ω at zero frequency. At f=fc, the characteristic impedance is
(a)600 Ω
(b)0 Ω
(c)∞ Ω
(d)more than 600 Ω
16.In m-derived terminating half section, m is equal to
(a)0.1
(b)0.3
(c)0.6
(d)0.9
17.In a symmetrical T-type attenuator with attenuation N and characteristic impedance R0, the resistance of each series arm is equal to
(a)R0
(b)(N − 1) R0
(c)
(d)
18.For a transmission line, open-circuit and short-circuit impedance are 20 Ω and 5 Ω. The characteristic impedance of the line is
(a)10 Ω
(b)20 Ω
(c)25 Ω
(d)100 Ω
19.For a prototype LPF, the phase constant β in the attenuation band is
(a)α
(b)0
(c)
(d)
20.In the m-derived HPF, the resonant frequency is to be chosen so that it is
(a)Above the cut-off frequency
(b)Below the cut-off frequency
(c)Equal to the cut-off frequency
(d)Infinitely high
21.In a symmetrical π-section attenuator with attenuation N and characteristic impedance R0, the resistance of each shunt arm is equal to
(a)R0
(b)(N − 1) R0
(c)
(d)
22.Bridged T-type network can be used as
(a)Attenuator
(b)LPF
(c)HPF
(d)BPF
23.One neper is equal to
(a)0.8686 dB
(b)8.686 dB
(c)0.115 dB
(d)86.86 dB
24.Terminating half sections used in composite filters are built with the following value of m
(a)0.3
(b)0.8
(c)0.1
(d)0.6
25.A transmission line works as
(a)Attenuator
(b)LPF
(c)HPF
(d)None of (a), (b) and (c).
1.d
2.a
3.c
4.c
5.a
6.a
7.b
8.d
9.a
10.a
Because for m-derived HPF
11.b
Because
12.c
13.c
14.c
15.b
16.c
17.c
18.a
19.a
20.b
21.d
22.a
23.c
24.d
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Question 1
Time: 00:00:00
A watch was sold at a loss of 10%. If it was sold for Rs.140 more, there would have been a gain of 4%. What is the cost price?
Rs.1000
Rs.1000
Rs.1140
Rs.1140
Rs.860
Rs.860
Rs.760
Rs.760
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Question 2
Time: 00:00:00
A man sells a horse for Rs.800 and loses something, if he had sold it for Rs.980, his gain would have been double the former loss. Find the cost price of the horse?
Rs.900
Rs.900
Rs.875
Rs.875
Rs.850
Rs.850
Rs.860
Rs.860
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Question 3
Time: 00:00:00
By selling a house for Rs.45000, it was found that 1/8 of the outlay was gained, what ought the selling to price to have been in order to have lost 5 p.c?
Rs.38750
Rs.38750
Rs.38000
Rs.38000
Rs.40000
Rs.40000
Rs.42000
Rs.42000
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Question 4
Time: 00:00:00
The C.P of 10 pens is equal to the S.P of 12 pens. Find his gain % or loss%?
16 2/3% profit
16 2/3% profit
50/6% profit
50/6% profit
16 2/3% loss
16 2/3% loss
100/3% loss
100/3% loss
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Question 5
Time: 00:00:00
By selling 12 pencils for a rupee a man loses 20%. How many for a rupee should he sell in order to gain 20%?
8
8
9
9
12
12
16
16
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Question 6
Time: 00:00:00
By selling 50 meters of cloth. I gain the selling price of 15 meters. Find the gain percent?
0.35
0.35
0.3
0.3
0.4
0.4
42 6/7%
42 6/7%
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Question 7
Time: 00:00:00
A reduction of 25% in the price of oil enables a house wife to obtain 5kgs more for Rs.800, what is the reduced price for kg?
Rs.20
Rs.20
Rs.30
Rs.30
Rs.40
Rs.40
Rs.25
Rs.25
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Question 8
Time: 00:00:00
A man buys two articles for Rs.1980 each and he gains 10% on the first and loses 10% on the next. Find his total gain or loss percent?
1% gain
1% gain
1% loss
1% loss
10% loss
10% loss
no gain or no loss
no gain or no loss
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Question 9
Time: 00:00:00
At what price must an article costing Rs.47.50 be marked in order that after deducting 5% from the list price. It may be sold at a profit of 25% on the cost price?
Rs.62.50
Rs.62.50
Rs.72.50
Rs.72.50
Rs.75.00
Rs.75.00
Rs.80.00
Rs.80.00
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Question 10
Time: 00:00:00
A shopkeeper buys mangoes at the rate of 4 a rupee and sells them at 3 a rupee. Find his net profit or loss percent?
0.2
0.2
0.28
0.28
33 1/3 %
33 1/3 %
35 2/7 %
35 2/7 %
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What is NPS?
NPS, which stands for Net Promoter Score, provides insights into customer loyalty and satisfaction. It measures the difference between the number of promoters (satisfied customers) and detractors (unsatisfied customers) relative to the total number of respondents. You can easily calculate the NPS score using Microsoft Excel. In this article, we’ll demonstrate two methods to calculate NPS in Excel. If you’re curious, follow along with our practice workbook.
Overview of NPS Score
1. Promoters: These are customers who are extremely satisfied with your goods or services. They actively promote your brand and encourage others to do the same.
2. Passives: Passives fall between promoters and detractors. They are neutral or passive individuals who don’t exhibit strong emotions.
3. Detractors: These are customers who give poor ratings. They are dissatisfied and unlikely to recommend your brand.
The mathematical expression of the NPS score is:
Moreover, you can also re-write the expression as shown below:
Dataset Overview
To illustrate these methods, we’ll examine the math test scores of 50 students listed in column B. Additionally, we’ll outline our NPS score criteria in column C. We define the Promoters’ range as 80-100, the Passive or Neutral range as 50-70, and the Detractors’ range as 0-40.
Method 1 – Using the Conventional Formula
• Calculate the number of students per criteria:
• Select cell D5.
• Use the COUNTIF function to count students with a score of 0:
`=COUNTIF(B:B,"=0")`
• Press Enter.
• Similarly, calculate counts for other score ranges.
• In cell D16, use the SUM function to get the total number of students:
`=SUM(D5:D15)`
• Press Enter.
• Estimate the percentage of students per criteria:
• In cell E5, calculate the percentage for the 0-score range:
`=D5/\$D\$16`
• Press Enter.
• Drag the formula down to fill cells E6 to E15.
• In cell E16, use SUM to get the total percentage:
`=SUM(E5:E15)`
• Press the Enter.
• Evaluate the three NPS categories:
• Promoters (scores >= 80):
`=COUNTIF(B:B,">=80")`
• Press Enter.
• Detractors (scores <= 40):.
`=COUNTIF(B:B,"<=40")`
• Press Enter.
• Neutral (deduct from total):
`=D16-F5-H5`
• Press Enter.
• Calculate the percentage of each category:
• Promoters:
`=F5/\$D\$16`
• Press Enter.
• Drag the Fill Handle icon to your right to copy the formula up to cell H6.
• Finally, compute the NPS Score:
`=((F5-H5)/D16)*100`
• Press Enter.
You will get the value of the NPS score.
Things You Should Know
To calculate the Net Promoter Score (NPS), focus on the percentage values of Promoters and Detractors, found in cells F6 and H6. The difference between these two percentages directly corresponds to the NPS score.
Method 2 – Combining COUNTIF and COUNT Functions
In this process, we are going to use the COUNTIF and COUNT functions to calculate the NPS score.
Follow these steps to implement this method:
• Select cell D5.
• Enter the following formula into the cell:
`=((COUNTIF(B:B,">=80")-COUNTIF(B:B,"<=40"))/COUNT(B:B))*100`
• Press Enter to execute the formula.
• The NPS score will be displayed.
To conclude, we confirm the successful functioning of our formula, enabling us to calculate the NPS score in Excel.
Breakdown of the Formula
We are breaking down the formula for cell D5.
COUNTIF(B:B,”>=80″): This function tallies the occurrences greater than or equal to 80 in column B. In this case, the count is 22.
COUNTIF(B:B,”<=40″): Similarly, this function counts the instances less than or equal to 40 in column B, yielding a count of 4 for this cell.
COUNT(B:B): The COUNT function enumerates the total entities in column B, which amounts to 50 in this context.
((COUNTIF(B:B,”>=80″)-COUNTIF(B:B,”<=40″))/COUNT(B:B))*100: Ultimately, this formula subtracts the count of values less than or equal to 40 from those greater than or equal to 80, then divides the result by the total count and multiplies by 100 to derive the NPS score. In this instance, the score is 36.
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Soumik Dutta
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Advanced Excel Exercises with Solutions PDF | 1,160 | 4,853 | {"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} | 3.53125 | 4 | CC-MAIN-2024-30 | latest | en | 0.884888 |
https://walkwithbud.com/solution/in-which-of-the-following-sentences-is-work-used-in-the-scientific-sense-537 | 1,701,470,877,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100308.37/warc/CC-MAIN-20231201215122-20231202005122-00858.warc.gz | 679,962,821 | 9,468 | # In which of the following sentences is work used in the scientific sense of the word? A scientist works on an experiment in the laboratory.* B Sam and Rachel pushed, but they could do no work on the car. ????
2 months ago
## Solution 1
Guest #6877099
2 months ago
B) obviously. Work in a scientific sense is the ability to make a displacement on an object in this case.
In A) the scientist can work on an experiment as your uncle can work on the field planting tomatos.
## đ Related Questions
Question
According the kinetic theory of matter, which of the following statements is true about matter
Solution 1
Hi
a. the atoms and molecules of a substance are always in motion
b. the higher the temperature of a substance, the faster its particles move
c. more massive particles move more slowly than less massive particles.
these are the main points of kinetic theory of matter.
Kinetic Theory of Matter
The kinetic theory of matter states that all matter is made of small particles that are in random motion and that have space between them. This means that no matter what phase matter is in, it is made of separate, moving particles.
This theory sounds pretty simple, but it actually explains a lot about the physical properties of matter and how it behaves. You might be surprised to learn that the particles of a solid are actually moving, just not enough for you to see. This type of vibrational movement is why a solid won't change shape no matter what kind of container you put it in.
Remember how liquid particles have more energy than solids? The extra energy in this state allows the particles to move around more freely, and they spread out more than those of a solid, putting more space between those particles. This is why a liquid will take the shape of its container up to its surface.
And since gases have even more energy than liquids, their particles are moving around a lot more, too. This is why a gas will expand to fill its entire container, not just to its surface like a liquid. Not only do the particles of a solid not move very much, but they're also held very close to each other by strong attractive forces. These forces are what hold the particles in place and are what give a solid its fixed size and shape.
On the other hand, the particles of a gas are so far apart that the attractive forces between them are assumed to be negligible. The particles of a gas are viewed as independent from each other, meaning that the gas is the opposite of a solid and has neither a fixed size nor shape.
Since the movement of liquid particles is in between a solid and a gas, the attractive forces between its particles are also in a middle range of the other two phases. Liquid particles have more freedom than solid particles, which is why a liquid can flow freely. This means that like a gas, a liquid has no fixed shape. But because the particles aren't quite as free as those of a gas, a liquid does have a fixed volume.
Question
Ball A has mass 5.0 kg and is moving at -3.2 m/s when it strikes stationary ball B, which has mass 3.9 kg, in a head-on collision. If the collision is completely inelastic, what is the common velocity of balls A and B?
Solution 1
Momentum conservation in a completely inelastic collision works like this:
mA*vA+mB*vB=v(mA+mB)
Since B was initially stationary pB=0.
Plugging in numbers gives:
5*(-3.2)=v(5+3.9)
v= -1.79 m/s to the left.
Solution 2
A.) Â -0.40 m/s
B.) -3.60 m/s
C.) -1.80 m/s
Explanation:
By the law of momentum conservation:-
=>m1u1 + m2u2 = m1v1 + m2v2 Â
=>5 x (-3.2) + 0 = 5v1 + 3.9v2
=>1.28v1 + v2 = -4.10 ---------------(i) Â
for elastic collision:-
=>v1 - v2 = u2 - u1 Â
=>v1 - v2 = 0 - (-3.2)
=>v1 - v2 = 3.2 ------------------(ii) Â
by (i) + (ii) :-
=>2.28v1 = -0.90
=>v1 = -0.40 m/s
=>v2 = -3.60 m/s
(c) By the law of momentum conservaation:-
=>m1u1+m2u2 = (m1+m2)v
=>5 x (-3.2) + 0 = (5 + 3.9) x v
=>v = -1.80 m/s
Question
The root word terra means âlandâ or âterritory.â Terrestrial planets are the planets closest to the Sun. Why are they are called terrestrial planets?
Solution 1
A Terrestrial plant, Tell-uric planet, or rock planet is a planet that is composed primarily of silicate rocks or metals.
Solution 2
They are made of rock and metal.
Explanation:
The planets closest to the Sun are called terrestrial planets because they are made of rock and metal. Plato
Question
How did scientists discover the Earth had a liquid outer core and solid inner core?
Solution 1
Dr. Inge discovered the make up of the earths inner core by studying how an earthquakes waves bounced off the core. And Inge Lehmann was studying the waves of a 1929 earthquake when she found them acting inconsistently with solid mantle crustÂ
hope it helps you
Question
Which of these statements best describes the relationship between matter, pure substances, and mixtures?
Solution 1
Question
A car initially at rest undergoes uniform acceleration for 6.32 seconds and covers a distance of 120 meters. What is the approximate acceleration of the car?
Solution 1
Answer : The approximate acceleration of the car is,
Solution : Given,
Initial velocity of the car = 0 m/s
Distance covered = 120 meter
Time taken = 6.32 second
Using second law of motion,
where,
s = distance covered
u = initial velocity
t = time taken
a = acceleration
Now put all the given values in the above equation, we get
Therefore, the approximate acceleration of the car is,
Solution 2
Using kinematic equation s=ut + 1/2 at^2(u = initial velocity=0, s=120m, t= 6.32s), 120 = 0(t) + 1/2 a(6.32)^2. a = 120x2/(6.32)^2 = 6m/s^2. Â
Question
When Nina jumps into the air, the Earth pulls her back down through the force of gravity. What else occurs? Ninaâs gravitational force pushes away from the Earth. Nina exerts a gravitational pull on the Earth, but the Earth does not accelerate at all. Nina pulls the Earth toward her, but the Earthâs acceleration is too small to notice. The force of gravity pulling Nina back down to Earth is the only force occurring.
Solution 1
It is A. hope this helps
Solution 2
1) WRONG. Since gravitational force is an attractive force.
2) WRONG. The Earth does accelerate but the acceleration is so small that it's negligible. However it DOES ACCELERATE.
3) CORRECT.
4) WRONG. Mass means gravitational field, so Nina might be small compared to the Earth but it still creates a gravitational field around her. This answer is also wrong because it's breaking Newton's third law of action-reaction. If Earth pulls Nina, then Nina pulls the Earth back with the same force. Yes, with the SAME FORCE, however F=m*a, so the great difference in mass makes Nina's acceleration much bigger than the Earth's one.
Question
The speed of an object determines the amount of inertia? True or False
Solution 1
The answer to this question is false.
Solution 2
Question
A plane is traveling with an air velocity of 720 kilometers/hour due east. It experiences a headwind of 16 kilometers/hour. Find the resultant velocity and direction of the plane with respect to the ground.
Solution 1
The resultant velocity and direction of the plane with respect to the ground is 704 km/h due east.
Explanation:
Given that,
Air velocity vâ= 720 km/h
The air velocity due to east is greater then the head wind.
So, the direction of the velocity respect to the ground will be due to east.
The velocity of plane respect to ground
The resultant velocity will be
Hence, The resultant velocity and direction of the plane with respect to the ground is 704 km/h due east.
Solution 2
If it is a headwind it means it's travelling against the motion of the plane. This means it's velocity is simply v=720-16=704 km/h due east.
Question
Select all the correct answers. In which situations is work being done? A. Hannah leans against her locker, but it doesnât move. B. Andyâs car rolls down a hill with the engine off. C. A wind turbine stands still on a sunny, windless day. D. A balloon expands as it is heated by the Sun. E. James pedals a stationary exercise cycle that doesnât move.
Solution 1
E. James pedals a stationary exercise cycle that doesnât move.
B. Andyâs car rolls down a hill with the engine off.
These Answer's Are Correct For It Show's Work Being Done A Car Moving Is Working, Pedaling A Exercise Machine Is Working Even If It Doesn't Move It Is Meant To Help You Work Out Pedaling Help's There Is No Need For It To Move.
Solution 2
Explanation:
Work is done when the force is applied to an object then there will be displacement by the object.
In the option A, Â Hannah leans against her locker, but it doesnât move. There is no displacement in this case. The work done is said to be zero in this case.
In the option B, Â Andyâs car rolls down a hill with the engine off. There is displacement in this case. The work is said to done in this case. At the hill, the car has a maximum potential energy.
When it rolls down then the kinetic energy of the car will increase. From the work energy theorem, the change in the kinetic energy is equal to the work done.
In the option C, A wind turbine stands still on a sunny, windless day. Â The work done is said to zero in this case.
In the option, Â A balloon expands as it is heated by the Sun. There is expansion of the balloon. The work is done in this case as the volume and the pressure will also increase on heating the balloon.
In the option E, James pedals a stationary exercise cycle that doesnât move. No work is done in this case.
Therefore, the statements B and D are correct answers.
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17108
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10987
3389 | 2,423 | 9,713 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-50 | latest | en | 0.952986 |
https://answers.yahoo.com/question/index?qid=20090919144803AAESu2E | 1,610,980,111,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514796.13/warc/CC-MAIN-20210118123320-20210118153320-00022.warc.gz | 237,060,421 | 28,656 | Nick N. asked in Science & MathematicsEngineering · 1 decade ago
# Series-parallel circuit confusion?
I have a 24 volt circuit that the first leg of the circuit is in series with a 10ohm resistor then the circuit goes into parallel with a 6ohm on left and 12ohm on right then the circuit goes back into a series with one 8 ohm resistor. Here is were I am lost I know on a series the volts will drop and in parallel the amps will drop. In circuit 24 V, 22 Ohm, 1.0909 Amp. But when the parallel circuit is calculated the amps are not equaling to the total circuit amps? I am getting a volt drop through the first 10ohm of 10.0909 Vd so as the remaining 13.0909 volts run through the parallel the amps are not dropping. If someone can give me info on were my calculations might be wrong or a different way to calculate.
Update:
Ok here is a drawing of the circut.
http://www.filedropper.com/circuit
### 4 Answers
Relevance
• 1 decade ago
Favorite Answer
You did not show your calculations so I am not sure what you are asking or where you may have made a mistake. I will try to show what I think you might be asking.You have calculated the correct total resistance and the correct total current. Current through a branch of a parallel circuit is inversely proportional to the resistance in that branch.
Thus current through the 6 Ohm resistor equals 1.0909A x ( 12 Ohm)/ (6 Ohm + 12 Ohm) = .7273 Amps.
The current through the 12 Ohm resistor equals 1.0909A x (6 Ohm) / (6 Ohm + 12 Ohm) = .3636 Amps.
Of course the current through the 6 Ohm resistor plus the current. through the12 Ohm resistor must equal the total current. So .7272A + .3636A = 1.0908A ( I lost .0001A somewhere in rounding off)
You were not correct in calculating the Voltage drop across the first 10 Ohm resistor as 10.0909 Volts. This Voltage should be 1.0909A x 10 Ohms = 10.909 Volts. Now you have 24V - 10.909V = 13.091V remaining to be dropped across the remainder of the circuit. Six Ohms in parallel with 12 Ohms = 4 Ohms. So now the resistance of the remainder of the circuit equals 4 Ohms + 8 Ohms = 12 Ohms. The remaining Voltage divided by the remaining resistance should equal the total current. 13.091V / 12 Ohms = 1.0909 Amps
The Voltage drops around the circuit are as follows;
V of 10 Ohm resistor = [(10 Ohm) / (10 + 4 +8) Ohms] x 24V = 10.909V.
V of 6 Ohms in parallel with 12 Ohms = V of 4 Ohms = [(4 Ohms) / (10 + 4 + 8) Ohms] x 24V = 4.364V.
V of 8 Ohm resistor = [(8 Ohms) / ( 10 + 4 + 8)] Ohms x 24V = 8.727V
Total Voltage equals the sum of the drops = 10.909V + 4.364V + 8.727V = 24 Volts.
• Anonymous
5 years ago
I can completely understand your frustration on this circuit. Even though this is technically a parallel circuit, it is also close to a bridge circuit. If you add up the resistances that are in each branch, then yes the voltage would be equal. However, it's asking for the difference between P and Q which are before those secondary resistors are "considered". If you simplify the circuit, it will make a lot more sense. It's not really 10 ohm and 2 ohm in parallel it's 15 and 20. You are right the current doesn't change, but the voltage going through the resistors do. Think about the circuit like a ladder circuit instead. I swear it will make sense after that.
• 1 decade ago
Not sure about the circuit Is the last 8 ohm resistor in parallel with the other two or is the parallel combination of the 6 & 12 ohm in series with the 10 ohm & 8 ohm.
• 1 decade ago
6 and 12 in parallel are 6*12/18 = 4 ohms. total resistance is 4+10+8 = 22 ohms.
Current is 24/22 = 1.09 amps
V10 = 10.9 volts (voltage across the 10 ohm resistor)
V8 = 8.73 volts (voltage across the 8 ohm resistor)
V4 = 4.36 volts (voltage across the parallel two)
and they add up to 24 volts.
Your comments I have trouble understanding, so I'll leave it at that.
.
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https://solvedlib.com/figure-4-a-firm-p-marginal-cost-21-18-average,438116 | 1,679,593,661,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945182.12/warc/CC-MAIN-20230323163125-20230323193125-00410.warc.gz | 578,727,454 | 17,395 | # Figure 4: A Firm P Marginal Cost 21 18 Average Total Cost 15 14 12 Marginal...
###### Question:
Figure 4: A Firm P Marginal Cost 21 18 Average Total Cost 15 14 12 Marginal Revanue Demand 0 30 40 50 60
b) (3 points) Assuming that this firm can not price discriminate, what is the profit maximizing quantity for the firm? c) (3 points) Assuming that this firm can not price discriminate, what is the profit maximizing price for this firm?
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##### Calculate A) heat energy q (kJ)&B) Enthalpy change ΔHrxn (kJ/mol)For the following reaction given the information:Hydrochloric acid and ammonium hydroxide HCl (aq) + NH4OH (qa) -> NH4Cl (aq) + H2O (l)50 mL 2 M HCl mixed with 50 mL 2 M NH4OHdensity - 1.03 g/mL specific heat - 4.184 J/(g·°C)calorimeter constant - 25.5 J/°CTmax (°C) = 29.5°CTinitial (°C) =20.1°C
Calculate A) heat energy q (kJ) & B) Enthalpy change ΔHrxn (kJ/mol) For the following reaction given the information: Hydrochloric acid and ammonium hydroxide HCl (aq) + NH4OH (qa) -> NH4Cl (aq) + H2O (l) 50 mL 2 M HCl mixed with 50 mL 2 M NH4OH density - 1.03 g/mL specific heat - 4....
##### Itroduatlon tqullblumt ccnalante duumhnentldloni cheicoi jcjolonCcann 0ntaFo" (aqh rongSCNoql FhloceztauFuscte iah lnlecvanald ronl[EsscN Fo * FlscNeslablsher rotwoon tteze (vo Icns and the Whon Fo" and SCN- are comnirad equituni iinocriyan know !he concentmiont ordorto Orual Ttoi In reicicn FeSCN = nqmldnum IFeSCN? [SCN [Fe" trr Ihrfe lons wil 3ustins ceainina dfereni cojconttillons cquiionmt icaa Will axpermentaly Uccituno_ equilibnum conceniralions Inc Lnroc pn-nared indeed con
Itroduatlon tqullblumt ccnalante duumhnent ldloni cheicoi jcjolon Ccann 0nta Fo" (aqh rong SCNoql Fhloceztau Fuscte iah lnlecvanald ronl [EsscN Fo * FlscN eslablsher rotwoon tteze (vo Icns and the Whon Fo" and SCN- are comnirad equituni iinocriyan know !he concentmiont ordorto Orual Ttoi I...
##### D (72x) = 2x: 72x-1 dxTrue False
d (72x) = 2x: 72x-1 dx True False...
##### One significant aspect of providing a high quality service to the individuals is the process of...
One significant aspect of providing a high quality service to the individuals is the process of reviewing and modifying our own work performance. 2.1) Briefly describe a method you would use to identify your personal performance in the workplace in terms of self-assessment (in 30-50 words each). 2.2...
##### Use the moment-generating function for a binomial R.V. X:M(t)[(1 − p) + pet ] n to find the mean µ and variance σ 2 of abinomial R.V.
Use the moment-generating function for a binomial R.V. X: M(t)[(1 − p) + pet ] n to find the mean µ and variance σ 2 of a binomial R.V....
##### H )chloride; (FW 53.491 dissuciation W SOIom on7hatt 1 9eution _
H )chloride; (FW 53.491 dissuciation W SOIom on7hatt 1 9e ution _...
##### Find f(x) using f'(x)=sin x (cos x + 1) using f(0)=10
find f(x) using f'(x)=sin x (cos x + 1) using f(0)=10...
##### Jenna titrates a 15.00 mL sample of amystery monoprotic weak acid solution with 0.35mol/L aqueous NaOH. She initially records the buret at0.23 mL. She knotes that the phenolphthalein indicator turns thesolution pink when the buret is at 29.58 mL.What volume of 0.35 mol/L NaOH was used?How many moles of NaOH were delivered?How many moles of weak acid were present in thissolution?What was the molarity of the unknown weak acidsolution?
Jenna titrates a 15.00 mL sample of a mystery monoprotic weak acid solution with 0.35 mol/L aqueous NaOH. She initially records the buret at 0.23 mL. She knotes that the phenolphthalein indicator turns the solution pink when the buret is at 29.58 mL. What volume of 0.35 mol/L NaOH was used? How man...
##### ") Find all the second order partial derivatives f(1y)-=xy+y31+4 b) Evaluate d: by partial fractions decomposition method X-1-21
") Find all the second order partial derivatives f(1y)-=xy+y 31+4 b) Evaluate d: by partial fractions decomposition method X-1-21...
##### I have found answers to part a and b and just really need help with part...
I have found answers to part a and b and just really need help with part c! and the extra if you have time. A= for part a then for part b, I have 5. Wave mechanics: (10 points) Suppose to have the following wave function (-oo 〈 x 〈 +00) r2 a for constants A and a a) Determine A, by ...
##### Magic Realm, Inc., has developed a new fantasy board game. The company sold 27,900 games last year at a selling price of...
Magic Realm, Inc., has developed a new fantasy board game. The company sold 27,900 games last year at a selling price of $65 per game. Fixed expenses associated with the game total$465,000 per year, and variable expenses are \$45 per game. Production of the game is entrusted to a printing contractor...
##### Betty DeRose, Inc. operates two departments, the handling department and the packaging department. During April, the...
Betty DeRose, Inc. operates two departments, the handling department and the packaging department. During April, the handling department reported the following information: % complete DM 38% % complete conversion 71% work in process, April 1 units started during April work in process, April 30 units...
##### Z dz, where L is the line segment from 2 = ~[ toz = ~1 + 2iSelect one: a.2 _ 2ib -2 + 2iNone of thesed.4 -2ie.1 + i
z dz, where L is the line segment from 2 = ~[ toz = ~1 + 2i Select one: a.2 _ 2i b -2 + 2i None of these d.4 -2i e.1 + i... | 4,481 | 15,763 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-14 | longest | en | 0.858251 |
https://math.stackexchange.com/questions/1502952/finding-a-basis-for-the-orthogonal-complement-of-a-vector-space | 1,719,298,007,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865560.33/warc/CC-MAIN-20240625041023-20240625071023-00364.warc.gz | 351,341,200 | 34,779 | Finding a basis for the orthogonal complement of a vector space.
Let $W$ be a subspace of the complex vector space $\mathbb{C}^4$. Given that the inner product on $\mathbb{C}^4$ is defined as $$\langle a,b\rangle=a_1\bar{b_1}+a_2\bar{b_2}+a_3\bar{b_3}+a_4\bar{b_4},$$ and the basis for $W$ is $$\mathcal{B}=\{(1,0,-1,0),(0,1,0,i)\},$$ find the basis for the orthogonal complement $W_{\perp}$.
I was wondering if there is a quick way to solve a problem of this kind, or do I have to define arbitrary vectors $a,b\in W_{\perp}$ such that each vector is orthogonal to each vector in $\mathcal{B}$ and are themselves linearly independent?
• This particular problem seems to want you to just choose two other vectors in a clever way. In particular, a vector which is zero in its first and third components will immediately be orthogonal to your first vector, so choose its second and fourth components to make the dot product with the second vector vanish. A similar trick (making the second and fourth components zero) will give you a second vector. Commented Oct 29, 2015 at 5:16 | 305 | 1,079 | {"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} | 3.359375 | 3 | CC-MAIN-2024-26 | latest | en | 0.913702 |
http://forums.wolfram.com/mathgroup/archive/2005/Feb/msg00182.html | 1,716,242,262,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00483.warc.gz | 13,855,496 | 8,386 | Re: NDSolve and differential equation system
• To: mathgroup at smc.vnet.net
• Subject: [mg54027] Re: NDSolve and differential equation system
• From: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig.de>
• Date: Tue, 8 Feb 2005 05:31:02 -0500 (EST)
• Organization: Uni Leipzig
• References: <cu1vjh\$ltj\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
```Hi,
NDSolve[] has a step size control, any numerical integration
assume that the right hand side has smooth derivatives and
use this information to estimate the step size.
Your UnitStep[] functions introduce nonsmooth derivatives
at the right hand side and mix up the step size control.
You can a) use the wonderful advanced interface of NDSolve[]
to use a constant step size or b) use smooth functions like
Erf[] or ArcTan[] instead of UnitStep[]
Regards
Jens
"Christian Moosmann" <some@Adress> schrieb im Newsbeitrag
news:cu1vjh\$ltj\$1 at smc.vnet.net...
> Hi MathGroup
>
> I have a question concerning NDSolve and a system of differential
> equations.
>
> consider an equation system A x'[t] + (K1 + v[t] * K2).x[t]=B, where
> A,K1,K2 are Matrices, B and x are vectors and v[t] is a defined skalar
> function. I want to compute this system with NDSolve, however, it takes
> a very long time.
>
> To try it you should use Theodor Gray's ShowStatus function in the
> frontend.
>
> ShowStatus[status_] :=
> SetNotebookStatusLine[FrontEnd`EvaluationNotebook[],
> ToString[ status ] ] ];
>
> Dim = 40;
>
> I use random Matrices here. They show the same basic behaviour as the
> matrices I usually use (those are also dense), so this problem should
> not be related directly to the matrices.
>
> K1 = Table[Random[], {i, Dim}, {j, Dim}];
> K2 = Table[Random[], {i, Dim}, {j, Dim}];
> Ainv = Inverse[Table[Random[], {i, Dim}, {j, Dim}]];
> B = Table[Random[], {i, Dim}];
>
> tStart = 0.;
> tEnd = 1;
>
> At first we solve without the function v[t]:
>
> Timing[NDSolve[{D[ x[t], t] + Ainv.( K1 + K2).x[t] == B ,
> x[tStart] == Table[0. , {Dim}]} , x , { t, tStart, tEnd },
> SolveDelayed -> True, StepMonitor :> ShowStatus[t] ] ]
>
> {0.080987 Second, {{x -> InterpolatingFunction[{{0., 1.}}, <>]}}}
>
> is quite fast, now we would like to include the function:
>
> v[t_] := UnitStep[t - 0.2]*4*(t - 0.2) - UnitStep[t - 0.45]*4*(t - 0.45);
>
> Timing[NDSolve[{D[ x[t], t] + Ainv.( K1 + v[t]*K2).x[t] == B ,
> x[tStart] == Table[0. , {Dim}]} , x , { t, tStart, tEnd },
> SolveDelayed -> True, StepMonitor :> ShowStatus[t] ] ]
>
> {137.341 Second, {{x -> InterpolatingFunction[{{0., 1.}}, <>]}}}
>
> Well, this is a factor of > 1000, and what I would like to compute is at
> least Dimension 100. If you use the stepmonitor you also may notice,
> that the time counts up, then stops for some time, counts up again...
>
>
> Does anyone have any suggestions how to deal with that?
> | 916 | 2,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-22 | latest | en | 0.712719 |
http://oeis.org/A160334 | 1,660,906,960,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573667.83/warc/CC-MAIN-20220819100644-20220819130644-00322.warc.gz | 32,813,587 | 4,077 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A160334 Numerator of Hermite(n, 23/31). 1
1, 46, 194, -167900, -8842004, 884083016, 125639477176, -4415829390416, -1893481677885040, -19202364475675424, 31870137298174352416, 1835095760938501860416, -589384037754831073199936, -69436314367007836275831680, 11532279106459848726285343616 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,2 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..200 FORMULA From G. C. Greubel, Jul 12 2018: (Start) a(n) = 31^n * Hermite(n, 23/31). E.g.f.: exp(46*x - 961*x^2). a(n) = numerator(Sum_{k=0..floor(n/2)} (-1)^k*n!*(46/31)^(n-2*k)/(k!*(n-2*k)!)). (End) EXAMPLE Numerators of 1, 46/31, 194/961, -167900/29791, -8842004/923521, ... MATHEMATICA Numerator[HermiteH[Range[0, 20], 23/31]] (* Harvey P. Dale, Aug 10 2014 *) Table[31^n*HermiteH[n, 23/31], {n, 0, 30}] (* G. C. Greubel, Jul 12 2018 *) PROG (PARI) a(n)=numerator(polhermite(n, 23/31)) \\ Charles R Greathouse IV, Jan 29 2016 (MAGMA) [Numerator((&+[(-1)^k*Factorial(n)*(46/31)^(n-2*k)/( Factorial(k) *Factorial(n-2*k)): k in [0..Floor(n/2)]])): n in [0..30]]; // G. C. Greubel, Jul 12 2018 CROSSREFS Cf. A009975 (denominators). Sequence in context: A235767 A192850 A211334 * A235550 A083358 A254229 Adjacent sequences: A160331 A160332 A160333 * A160335 A160336 A160337 KEYWORD sign,frac AUTHOR N. J. A. Sloane, Nov 12 2009 STATUS approved
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Last modified August 19 06:48 EDT 2022. Contains 356216 sequences. (Running on oeis4.) | 689 | 1,835 | {"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-2022-33 | latest | en | 0.454406 |
https://e-eduanswers.com/mathematics/question3856411 | 1,627,102,515,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150129.50/warc/CC-MAIN-20210724032221-20210724062221-00666.warc.gz | 236,345,095 | 16,456 | What is the equation of a circle with center (-2, 3) and radius 4? a) (x+2)^2 + (y-3)^2 = 16 b) (x-2)^2 + (y+3)^2 = 16 c) (x+2)^2 + (y-3)^2
, 18.10.2019 06:30, SunsetPrincess
# What is the equation of a circle with center (-2, 3) and radius 4? a) (x+2)^2 + (y-3)^2 = 16 b) (x-2)^2 + (y+3)^2 = 16 c) (x+2)^2 + (y-3)^2 = 4 d) (x+2)^2 - (y-3)^2 = 16
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If the sub shop sells at least 35 subs in a day they make a profit write an inequality for the number of the subs to make a profit
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Identify one charcteristic of exponential growth | 340 | 942 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-31 | latest | en | 0.771576 |
https://plainmath.org/differential-equations/1415-solve-and-classify-equation-k-plus-1-dy-plus-kydx-equal-0 | 1,723,025,782,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640690787.34/warc/CC-MAIN-20240807080717-20240807110717-00805.warc.gz | 344,563,929 | 24,037 | 2020-12-01
Solve and classify equation $\left(k+1\right)dy+kydx=0$
estenutC
$\left(k+1\right)dy+kydx=0$
Let y be the dependent variable
$\left(k+1\right)dy/dx+ky=0$
Substituing with y'
$\left(k+1\right){y}^{\prime }+ky=0$
$\left(k+1\right)/y{y}^{\prime }=-k$
The solution is
$y={e}^{\left(-kx+{c}_{1}\right)/\left(k+1\right)}$
This is exact form
Do you have a similar question? | 159 | 381 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 30, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-33 | latest | en | 0.649759 |
https://www.coursehero.com/file/6656279/practice-problem-set3/ | 1,513,640,558,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948627628.96/warc/CC-MAIN-20171218215655-20171219001655-00397.warc.gz | 739,792,767 | 108,824 | practice_problem_set3
# practice_problem_set3 - EEL3105 Fall2011...
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EEL 3105 Fall 2011 Practice Problem Set 3 1. Find Laplace transform of ( ) 7 ( 4) 4 ( 2) 3sin(6 / 4) ft Ut t t Here U(t) is the unit step function, (t) is the unit impulse function. 2. Find inverse Laplace transform for the following functions: a. F(s) = 2 2 (2 9 ) s ss s b. F(s) = 2 2 2 9 ) s s e s 3. Consider the differential equation 2 2 51 0 38 d y dy du yu dt dt dt Find the transfer function H(s)= () . Ys Us 4. Consider the differential equation 2 2 0 d y dy du dt dt dt Find A, B, C for a state space representation of the form . dx Ax t Bu t dt yt Cxt 5. Consider the differential equation 32 90 . dy d y dt dt dt We are given that y(t)=4e 3t is a solution to this differential equation. Find the value of .
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6. Consider the differential equation 2 2 41 6( ) . dy d y y ut dt dt Suppose u=unit impulse function and y(0)=0, dy/dt(0)=0. Find y(t). You can use any method to solve this problem. 7. Consider the differential equation 3() () . dy y tu t dt Suppose u(t) is the unit step function. Calculate
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Ask a homework question - tutors are online | 492 | 1,568 | {"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} | 3.765625 | 4 | CC-MAIN-2017-51 | latest | en | 0.67512 |
https://paizo.com/threads/rzs2nxcj&page=22?Did-you-know | 1,580,236,223,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251779833.86/warc/CC-MAIN-20200128153713-20200128183713-00353.warc.gz | 584,816,436 | 24,996 | # Did you know...?
### Off-Topic Discussions
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IANABioengineer, so I'll link the relevant article >HERE<. It does have a lot of implications RE future data storage and computing.
Patrick Curtin wrote:
1,000 Gigabytes. Or 1,000,000 Megabytes . Or 1,000,000,000 Kilobytes.
To get an idea of the volume of data, that's the same as having 14,000 50 Gigabyte Blu-Ray movie discs worth of data in an area smaller than your pinky tip.
That boggles the mind! I still can barely wrap my head around the idea of having 1TB in my harddrive, let alone 700 times that in the tippiest tip of my teeniest finger!
Al Copeland (founder of Popeyes Fried Chicken) once drove a white Lamborghini with a red interior.
The dot on top of the letter "i" is called a tittle.
1 person marked this as a favorite.
Patrick Curtin wrote:
1,000 Gigabytes. Or 1,000,000 Megabytes . Or 1,000,000,000 Kilobytes.
To get an idea of the volume of data, that's the same as having 14,000 50 Gigabyte Blu-Ray movie discs worth of data in an area smaller than your pinky tip.
Now it's my turn to nitpic! (But maybe somebody will find this interesting... It's sort of it's own "did you know" fact.)
Bytes use the standard metric prefixes (kilo, mega, giga) but are not actually metric. Each of level represents 1024 of the unit below it. This is because of the use of binary code; 1000 is not a exponent of 2, but 1024 is (2^10).
Therefore, there are 1024 bytes in a kilobyte, 1024 kilobytes in a megabyte, 1024 megabytes in a gigabyte, and 1024 gigabytes in a terabyte.
So in your example, 1 terabyte is actually 1,073,741,824 kilobytes.
Celestial Healer wrote:
Patrick Curtin wrote:
1,000 Gigabytes. Or 1,000,000 Megabytes . Or 1,000,000,000 Kilobytes.
To get an idea of the volume of data, that's the same as having 14,000 50 Gigabyte Blu-Ray movie discs worth of data in an area smaller than your pinky tip.
Now it's my turn to nitpic! (But maybe somebody will find this interesting... It's sort of it's own "did you know" fact.)
Bytes use the standard metric prefixes (kilo, mega, giga) but are not actually metric. Each of level represents 1024 of the unit below it. This is because of the use of binary code; 1000 is not a exponent of 2, but 1024 is (2^10).
Therefore, there are 1024 bytes in a kilobyte, 1024 kilobytes in a megabyte, 1024 megabytes in a gigabyte, and 1024 gigabytes in a terabyte.
So in your example, 1 terabyte is actually 1,073,741,824 kilobytes.
I did not know that. It makes perfect sense. Thank you for teaching me something new =)
Celestial Healer wrote:
Bytes use the standard metric prefixes (kilo, mega, giga) but are not actually metric. Each of level represents 1024 of the unit below it. This is because of the use of binary code; 1000 is not a exponent of 2, but 1024 is (2^10).
Therefore, there are 1024 bytes in a kilobyte, 1024 kilobytes in a megabyte, 1024 megabytes in a gigabyte, and 1024 gigabytes in a terabyte.
So in your example, 1 terabyte is actually 1,073,741,824 kilobytes.
I'd understood that manufacturers will often use them in the standard SI sense (1 TB = 1000000000 KB), to make it sound like the product has more memory, but operating systems report in binary (1 TB = 1073741824 KB) as you described. Therefore, if you have a 1 TB operating system, your 1 TB processor is 73,741,824 KB short!
RPG Superstar 2010 Top 16
1 person marked this as a favorite.
The number of water molecules in a thimble (5 ml) approximately equals the number of thimblefuls of water in all of earth's oceans.
Kirth Gersen wrote:
Celestial Healer wrote:
Bytes use the standard metric prefixes (kilo, mega, giga) but are not actually metric. Each of level represents 1024 of the unit below it. This is because of the use of binary code; 1000 is not a exponent of 2, but 1024 is (2^10).
Therefore, there are 1024 bytes in a kilobyte, 1024 kilobytes in a megabyte, 1024 megabytes in a gigabyte, and 1024 gigabytes in a terabyte.
So in your example, 1 terabyte is actually 1,073,741,824 kilobytes.
I'd understood that manufacturers will often use them in the standard SI sense (1 TB = 1000000000 KB), to make it sound like the product has more memory, but operating systems report in binary (1 TB = 1073741824 KB) as you described. Therefore, if you have a 1 TB operating system, your 1 TB processor is 73,741,824 KB short!
If you look at the fine print, they admit to doing this. But yeah, it's a scam to make the stuff look like it has more capacity than it does.
Kajehase wrote:
That's because we steal the best stuff you make and then add the best from our own traditions (Lex Beatles).
Funny, I was going to say the same for US.
The electric chair was invented by a dentist.
3 people marked this as a favorite.
Aberzombie wrote:
The electric chair was invented by a dentist.
Yep, I can believe it.
The guillotine was invented by a surgeon.
Vive le Galt!!
5 people marked this as a favorite.
Most government forms and paperwork were invented by slaadi.
Only the ones that have to be filled out in triplicate, right?
The name for the Australian marsupial Kangaroo came about when some of the first white settlers saw this strange animal hopping along and they asked the Aborigines what it was called. They replied with 'Kanguru', which in the native language meant 'I don't know' .
1 person marked this as a favorite.
Kip84 wrote:
The name for the Australian marsupial Kangaroo came about when some of the first white settlers saw this strange animal hopping along and they asked the Aborigines what it was called. They replied with 'Kanguru', which in the native language meant 'I don't know' .
I doubt the veracity of this one, but it's still entertaining.
In the same vein:
When the explorer Eric Shipton made his famous journey through the Himalayas in the 1930s, he reportedly glimpsed a strange creature on a far-off mountainside. The apocryphal encounter apparently went something like this:
Shipton: "You see that? What's that?"
Sherpa guide: "Yeh-teh? [That thing, over there?]"
And so the Yeti was born.
Wow a five minute check shows you to be correct Mr. Shiny. That's a little disappointing as its a story I've grown up with that was always presented as fact...
Starfinder Superscriber
Bob's Burgers was originally going to be about a family of cannibals that ran a fast food resturant.
DJEternalDarkness wrote:
Bob's Burgers was originally going to be about a family of cannibals that ran a fast food resturant.
I think I'd rather watch that.
1 person marked this as a favorite.
DJEternalDarkness wrote:
Bob's Burgers was originally going to be about a family of cannibals that ran a fast food resturant.
{strums guitar, sings:} "You can get anyone you want, at Bobby's Restaurant"
1 person marked this as a favorite.
I don't want a pickle, I just wanna ride my motorsickle.
Doodlebug Anklebiter wrote:
I don't want a pickle, I just wanna ride my motorsickle.
That phrase is one of those thing my dad has said since I was an infant. It's a "dadism" in my house.
How old are you? My dad never said it, but they were my mother's Arlo records.
2 people marked this as a favorite.
Klaus van der Kroft wrote:
Juan Valdez is my builder.
Doodlebug Anklebiter wrote:
How old are you? My dad never said it, but they were my mother's Arlo records.
I'm 32. Dad is in his 70's.
Yeah, I'm 35. Dirty, stinking, long-haired hippie parents.
Doodlebug Anklebiter wrote:
Yeah, I'm 35. Dirty, stinking, long-haired hippie parents.
Dad actually joined the Air Force so he wouldn't have to go to the Vietnam war. Odd as that sounds.
He was in college for much of it so that kept him out. So when he graduated he signed up to be a pilot. The pilot training was long enough that between that, Officer candidate school, and college the war was over. Plus now he could fly heavy aircraft.
Meh. 45 with stinking New Deal Boston Irish Democrat teacher parents.
Anyway, did you know?
There were only six civilian casualties from Japanese bombs on continental US soil in WWII. They came about from a Japanese terror bomb campaign that sent 9,000 'fire baloons' over the Pacific armed with incendiaries to try and ignite the West Coast and cause terror in the populace. Unfortunately for the plan, the season was very wet, and few fires were reported. Only a fraction made it to American soil, and only one caused any deaths.
My father joined the Air force and was shipped to Berlin during the Vietnam "conflict"
My dad was an Air Force Colonel. He also has a killer goatee and teaches law. Ever heard of the Purple Elephant?
Purple Elephant:
When driving home today after work Do not think about Purple Elephants. You cannot think about Purple Elephants, don't even try to think about Purple Elephants. What ever you do, don't think of Purple Elephants.
Why:
In the court of law, whenever the judge says don't think of something, its going to be one of the first things the jury thinks about when they have to discuss. Its a pretty big tactic if it works.
Crimson Jester wrote:
My father joined the Air force and was shipped to Berlin during the Vietnam "conflict"
That makes sense.
Germany is the jumping off point for europe and the middle east. Lots of supplies, etc go through there. Also several of the bases are joint command, meaning several branches work together for operations.
I was a kid in germany when the Berlin wall came down. I used to have a piece of it, but since I was a kid I lost it somewhere. I suspect it's in the attic of my old house in Ohio.
Tirq wrote:
My dad was an Air Force Colonel. He also has a killer goatee and teaches law. Ever heard of the Purple Elephant?
** spoiler omitted **
** spoiler omitted **
Heard of and used the purple elephant before. Not in a courtroom, but mostly just to annoy friends.
"Don't think of (insert something graphic)." Usually while eating.
I would actually call b&@%~++# on this one, but it's too funny to pass up....
You're more likely to be killed by a champagne cork than by a poisonous spider.
A crocodile can't stick out its tongue.
The Carpenter's hit We've Only Just Begun started out as just two verses and a bridge to be played in a TV-ad for a bank who needed to attract a younger clientèle.
Aberzombie wrote:
I would actually call b\$+\$#*@* on this one, but it's too funny to pass up....
You're more likely to be killed by a champagne cork than by a poisonous spider.
How many busy people today have room for both deadly champagne corks and poisonous spiders?! What a hassle!
But wait...
What if I told you that you could have both a champagne cork and spider in one!!! (NSFthe spider squeamish) How much would you pay? [/smarmy infomercial voice]
I think I just found my next PC killer scenario...
Tirq wrote:
Ever heard of the Purple Elephant?
** spoiler omitted **
** spoiler omitted **
I have heard of something similar, something from Tolstoy about sitting in the corner and trying NOT to think of a white elephant. Or maybe it was a white bear, I don't remember. Anyway, apparently it can't be done.
Patrick Curtin wrote:
Meh. 45 with stinking New Deal Boston Irish Democrat teacher parents.
Anyway, did you know?
There were only six civilian casualties from Japanese bombs on continental US soil in WWII. They came about from a Japanese terror bomb campaign that sent 9,000 'fire baloons' over the Pacific armed with incendiaries to try and ignite the West Coast and cause terror in the populace. Unfortunately for the plan, the season was very wet, and few fires were reported. Only a fraction made it to American soil, and only one caused any deaths.
There's a neat story here, which I'm typing from memory so errors are likely:
Back in the Twenties, a Japanese scientist discovered the jet stream. He thought this was hot stuff and published in Japanese, but he knew that would not get the eyes he wanted on his work. So like a good scientist he publishes in an accessible international language: Esperanto. You may be astonished to learn that nobody read that.
The Japanese remembered the meteorologist's work, so they decided they'd set the great American forests and cities afire by attaching the aforementioned incendiary bombs to some balloons like the meteorologist used to discover the jet stream. Late in the war, they launched thousands of the things from the east coast of Honshu. One made it as far as Michigan. The one that caused the deaths burned a church group in Oregon.
3 people marked this as a favorite.
Kryzbyn wrote:
The BBC Europe correspondent described the painting's current state as resembling "a crayon sketch of a very hairy monkey in an ill-fitting tunic."
LOL
The folks on the SA forums have started posting edits of the messed up Jesus painting.
This one is my favorite so far.
Aberzombie wrote:
I would actually call b\$+\$#*@* on this one, but it's too funny to pass up....
You're more likely to be killed by a champagne cork than by a poisonous spider.
How many busy people today have room for both deadly champagne corks and poisonous spiders?! What a hassle!
But wait...
What if I told you that you could have both a champagne cork and spider in one!!! (NSFthe spider squeamish) How much would you pay? [/smarmy infomercial voice]
WTF!!!!
Samnell wrote:
Patrick Curtin wrote:
Meh. 45 with stinking New Deal Boston Irish Democrat teacher parents.
Anyway, did you know?
There were only six civilian casualties from Japanese bombs on continental US soil in WWII. They came about from a Japanese terror bomb campaign that sent 9,000 'fire baloons' over the Pacific armed with incendiaries to try and ignite the West Coast and cause terror in the populace. Unfortunately for the plan, the season was very wet, and few fires were reported. Only a fraction made it to American soil, and only one caused any deaths.
There's a neat story here, which I'm typing from memory so errors are likely:
Back in the Twenties, a Japanese scientist discovered the jet stream. He thought this was hot stuff and published in Japanese, but he knew that would not get the eyes he wanted on his work. So like a good scientist he publishes in an accessible international language: Esperanto. You may be astonished to learn that nobody read that.
The Japanese remembered the meteorologist's work, so they decided they'd set the great American forests and cities afire by attaching the aforementioned incendiary bombs to some balloons like the meteorologist used to discover the jet stream. Late in the war, they launched thousands of the things from the east coast of Honshu. One made it as far as Michigan. The one that caused the deaths burned a church group in Oregon.
ohh so close, a minister's wife and his family on a fishing trip.
The balloons did kill six Americans. A minister and his wife had taken some children on a fishing trip in southern Oregon, east of the Cascades, when they discovered a balloon bomb that exploded while they were gathered around it, killing the woman and five children. Japanese propaganda broadcasts announced great fires and an American public in panic, declaring casualties as high as 10,000, but the six people killed in Oregon were the only casualties inflicted by the enemy on the American mainland in World War II.
And honestly after our attempts at BAT BOMBS I can see them trying many things.
Also William Shatner made a film in Esperanto. Why I don't know. Maybe because he was out of work.
Crimson Jester wrote:
Also William Shatner made a film in Esperanto. Why I don't know. Maybe because he was out of work.
Ĉar li estas iom freneza?
Crimson Jester wrote:
Also William Shatner made a film in Esperanto. Why I don't know. Maybe because he was out of work.
In the sci-fi world of Red Dwarf everything is Esperanto bilingual.
Rubber bands last longer when kept refrigerated.
Women blink twice as much as men.
1,051 to 1,100 of 6,213 << first < prev | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | next > last >> | 3,963 | 16,105 | {"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-2020-05 | latest | en | 0.882583 |
https://greatgreenwedding.com/what-is-the-relative-age-of-fossils/ | 1,723,209,441,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640763425.51/warc/CC-MAIN-20240809110814-20240809140814-00895.warc.gz | 220,886,842 | 10,127 | ## What is the relative age of fossils?
Relative dating is used to determine a fossils approximate age by comparing it to similar rocks and fossils of known ages. Absolute dating is used to determine a precise age of a fossil by using radiometric dating to measure the decay of isotopes, either within the fossil or more often the rocks associated with it.
A fossil is evidence of plants and animals that are no longer here. Any type of remains or evidence of past life is a fossil. Some fossils are parts of organisms. Shells and bones are examples of fossil remains.
### What does the relative age of a fossil tell you?
Relative age dating tells us which fossils are older and which fossils are younger. The fossil species below the ash must be slightly older than 507 million years, and the species above the ash must be slightly younger. If rocks in different places contain the same fossil species, they must be similar in age.
What is relative dating fossils?
Scientists use two approaches to date rocks and fossils. Relative age dating is used to determine whether one rock layer (or the fossils in it) are older or younger than another base on their relative position: younger rocks are positioned on top of older rocks.
#### What is relative age?
1. n. [Geology] The approximate age determination of rocks, fossils or minerals made by comparing whether the material is younger or older than other surrounding material.
How do you find the relative age of a rock?
Superposition of rock units is a very simple and straightforward method of relative age determination. The principle states that in a sequence of undeformed sedimentary rocks the oldest beds are at the bottom and the youngest ones are at the top.
## What is a relative age?
Which fossil is the youngest?
A dinosaur fossil believed to be the youngest ever found was discovered by Yale scientists in Montana’s Hell Creek formation, a study published in Biology Letters revealed. The 45-centimetre horn is understood to be from a triceratops.
### How is fossil age identified?
To establish the age of a rock or a fossil, researchers use some type of clock to determine the date it was formed. Geologists commonly use radiometric dating methods, based on the natural radioactive decay of certain elements such as potassium and carbon, as reliable clocks to date ancient events.
What is relative age dating?
Relative dating puts geologic events in chronological order without requiring that a specific numerical age be assigned to each event. Second, it is possible to determine the numerical age for fossils or earth materials.
#### What is relative age example?
For example, the grains within a sedimentary rock are older than the rock; a fragment of sandstone incorporated within a mudstone is older than the mudstone; a fossil bone found in a limestone is older than the limestone.
How do you determine relative age?
In the process of relative dating, scientists do not determine the exact age of a fossil or rock but look at a sequence of rocks to try to decipher the times that an event occurred relative to the other events represented in that sequence. The relative age of a rock then is its age in comparison with other rocks.
## What kind of rocks are fossils found in?
Fossils are found in sandstone, so Jane was excited to search through the rocks! Just as she began to dig, she found fossilized bones. The bones had turned to stone through a
What do you call fossils that have been turned to stone?
How do you know? Fascinating fossils Vocabulary Words Fossils – Remains of ancient life that have been turned to stone Body Fossil – Actual parts of plants and animals that have been turned to stone (i.e. bone, shells, leaves) Coprolite- Fossilized dung (scat)
### Which is the best description of a trace fossil?
Trace fossil- A fossilized sign that a plant or animal once lived in an area (i.e. footprints, coprolite) Porous – Full of tiny holes that water, air, and light can pass through Mineralization- To convert into a mineral substance; to fill with a mineral substance
What did Jane Austen do with the fossils?
Jane carefully removed the fossil from the rocks and donated them to a museum. For the questions below, circle the correct answer. Each question is for the station of the same letter. This fossil is. | 878 | 4,327 | {"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} | 2.90625 | 3 | CC-MAIN-2024-33 | latest | en | 0.944481 |
https://ai.stackexchange.com/questions/9990/confusing-on-gan-loss-function/10192 | 1,586,305,524,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371806302.78/warc/CC-MAIN-20200407214925-20200408005425-00451.warc.gz | 338,426,719 | 35,452 | # Confusing on GAN loss function
I was trying to understand the loss function of GANs, while I found a little mis-match between different papers.
This is the screen-shot from the original paper of Goodfellow at https://arxiv.org/pdf/1406.2661.pdf: ,
And equation (1) in this version of pix2pix paper at https://arxiv.org/pdf/1611.07004.pdf
Putting aside the fact that pix2pix is using conditional GAN, which introduces a second term $$y$$, the 2 formulas are quite resemble, except that in the pix2pix paper, they try to get minimax of $${\cal{L}}_{cGAN}(G, D)$$, which is defined to be $$E_{x,y}[...] + E_{x,z}[...]$$, whereas in the original paper, they define $$\min\max V(G, D) = E[...] + E[...]$$.
I am not coming from a good math background, so I am quite confused. I'm not sure where the mistake is, but assuming that $$E$$ is expectation (correct me if I'm wrong), the version in pix2pix makes more sense to me, although I think it's quite less likely that Goodfellow could make this mistake in his amazing paper. Maybe there's no mistake at all and it's me who do not understand them correctly.
• Suggest using blockquotes of text and LaTeX instead of not very well prepped images. Otherwise a good Q. – han_nah_han_ Jan 21 '19 at 4:01
The question is about a mismatch between the loss function in two papers on GANs. The first paper is Generative Adversarial Nets Ian J. Goodfellow et. al., 2014, and the excerpt image in the question is this.
The adversarial modeling framework is most straightforward to apply when the models are both multilayer perceptrons. To learn the generator’s distribution $$p_g$$ over data $$x$$, we define a prior on input noise variables $$p_z (z)$$, then represent a mapping to data space as $$G (z; \theta_g)$$, where $$G$$ is a differentiable function represented by a multilayer perceptron with parameters $$\theta_g$$. We also define a second multilayer perceptron $$D (x; \theta_d)$$ that outputs a single scalar. $$D (x)$$ represents the probability that $$x$$ came from the data rather than pg. We train $$D$$ to maximize the probability of assigning the correct label to both training examples and samples from $$G$$. We simultaneously train $$G$$ to minimize $$\log (1 − D(G(z)))$$:
In other words, $$D$$ and $$G$$ play the following two-player minimax game with value function $$V (G, D)$$:
$$\min_G \, \max_D V (D, G) = \mathbb{E}_{x∼p_{data}(x)} \, [\log \, D(x)] \\ \quad\quad\quad\quad\quad\quad\quad + \, \mathbb{E}_{z∼p_z(z)} \, [\log \, (1 − D(G(z)))] \, \text{.} \quad \text{(1)}$$
The second paper is Image-to-Image Translation with Conditional Adversarial Networks, Phillip Isola Jun-Yan Zhu Tinghui Zhou Alexei A. Efros, 2018, and the excerpt image in the question is this.
The objective of a conditional GAN can be expressed as
$$\mathcal{L}_{cGAN} (G, D) = \mathbb{E}_{x, y} \, [\log D(x, y)] \\ \quad\quad\quad\quad\quad\quad\quad + \mathbb{E}_{x, z} \, [\log \, (1 − D(x, G(x, z))], \quad \text{(1)}$$
where $$G$$ tries to minimize this objective against an adversarial $$D$$ that tries to maximize it, i.e.
$$G^{∗} = \arg \, \min_G \, \max_D \mathcal{L}_{cGAN} (G, D) \, \text{.}$$
To test the importance of conditioning the discriminator, we also compare to an unconditional variant in which the discriminator does not observe $$x$$:
$$\mathcal{L}_{GAN} (G, D) = \mathbb{E}_y \, [\log \, D(y)] \\ \quad\quad\quad\quad\quad\quad\quad + \mathbb{E}_{x, z} \, [\log \, (1 − D(G(x, z))] \, \text{.} \quad \text{(2)}$$
In the above $$G$$ refers to the generative network, $$D$$ refers to the discriminative network, and $$G^{*}$$ refers to the minimum with respect to $$G$$ of the maximum with respect to $$D$$. As the question author tentatively put forward, $$\mathbb{E}$$ is the expectation with respect to its subscripts.
The question of discrepancy is that the right hand sides do not match between the first paper's equation (1) and the second paper's equation (2) which is absent of the condition involving $$y$$.
First paper:
$$\mathbb{E}_{x∼p_{data}(x)} \, [\log \, D(x)] \\ \quad\quad\quad\quad\quad\quad\quad + \, \mathbb{E}_{z∼p_z(z)} \, [\log \, (1 − D(G(z)))] \, \text{.} \quad \text{(1)}$$
Second paper:
$$\mathbb{E}_y \, [\log \, D(y)] \\ \quad\quad\quad\quad\quad\quad\quad + \mathbb{E}_{x, z} \, [\log \, (1 − D(G(x, z))] \, \text{.} \quad \text{(2)}$$
The second later paper further states this.
GANs are generative models that learn a mapping from random noise vector $$z$$ to output image $$y, G : z \rightarrow y$$. In contrast, conditional GANs learn a mapping from observed image $$x$$ and random noise vector $$z$$, to $$y, G : {x, z} \rightarrow y$$.
Notice that there is no $$y$$ in the first paper and the removal of the condition in the second paper corresponds to the removal of $$x$$ as the first parameter of $$D$$. This is one of the causes of confusion when comparing the right hand sides. The others are use of variables and degree of explicitness in notation.
The tilda $$~$$ means drawn according to. The right hand side in the first paper indicates that the expectation involving $$x$$ is based on a drawing according to the probability distribution of the data with respect to $$x$$ and the expectation involving $$z$$ is based on a drawing according to the probability distribution of $$z$$ with respect to $$z$$.
The removal of the observation of $$x$$ from the second right hand term of the second paper's equation (2), which is the first parameter of $$G$$, the replacement of that equation's $$y$$ variable with the now freed up $$x$$ variable, and the acceptance of the abbreviation of the tilda notation used in the first paper then brings both papers into exact agreement.
• Thank you for your reply and sorry for the late response. Regarding this paragraph: "The question of discrepancy is that the right hand sides do not match between the first paper's equation (1) and the second paper's equation (2) which is absent of the condition involving 𝑦." No, it is not what confuses me. The question of discrepancy is the mismatch of the left hand sides of the 2 papers: the first paper defines the Expectations to be the results of the min-max operation, where as the second paper suggests to take the min-max of the Expections. – AugLe Feb 25 '19 at 10:39
What is meant by both papers is that we have two agents (generator and discriminator) playing a game with the value function V defined as a sum of the expectations (i.e. an expectation of the outcome value defined as a sum of two terms, or actually a logarithm of a product...). The generator uses a strategy G encoded in the parameters of its neural network (θg), the discriminator uses a strategy D encoded in the parameters of its neural network (θd). Our goal is to (hopefully) find such a pair of strategies (a pair of parameter sets θgmin and θdmax) that produce the minimax value.
While trying to find the (θgmin, θdmax) pair using gradient descent, we actually have two loss functions: one is the loss function for G, parameterized by θg, another is the loss function for D, parameterized by θd, and we train them alternatively on minibatches together.
If you look at the Algorithm 1 in the original paper, the loss function for the discriminator is -log(D(x; θd)) - log(1 - D(G(z); θd), and the loss function for the generator is log(1 - D(G(z; θg)) (in both cases, in the original paper, x is sampled from the reference data distribution and z is sampled from noise):
The ideal value for the loss function of the discriminator is 0, otherwise it's greater than 0. The "loss" function of the generator is actually negative, but, for better gradient descent behavior, can be replaced with -log(D(G(z; θg)), which also has the ideal value for the generator at 0. It is impossible to reach zero loss for both generator and discriminator in the same GAN at the same time. However, the idea of the GAN is not to reach zero loss for any of the game agents (this is actually counterproductive), but to use that "double gradient descent" to "converge" the distribution of G(z) to the distribution of x.
• Thanks for your reply. Yet, I am more concerned of the whole equations themselves. The pix2pix paper wants to minimize the expectation while the GAN paper shows that the expectation is the result of the minimization. – AugLe Jan 15 '19 at 10:05
• Yes, and they are different that one is defined to be the min of max of expectation (pix2pix) while the other is defined as the min of max of value function (GAN paper). Does that make sense? – AugLe Jan 15 '19 at 10:37
• So let me rephrase the question like this: what is the target function to be optimize for a general GAN? In the GAN paper, Goodfellow uses $V(D, G)$, but what is $V(D, G)$ exactly? Is it $V(D,G) = E(\log(D(y)) + E(1-\log(D(G(x, z))$ – AugLe Jan 17 '19 at 11:04
• @AugLe Edited again. Hopefully now it's clearer what is actually optimized in GANs. – Kit. Jan 17 '19 at 20:39
I'm not sure I understand your question. However responding to your question in the comments. The difference between the two objectives is that:
In an ordinary GAN, we want to push $$p(G)$$ to be as close as possible to $$p(data)$$
In a conditional GAN, we have a context $$c$$. If we imagine for ease of understanding that $$c=[1,2,3]$$ is a discrete variable where all the data can be categorised under one of these c values , then we want to: push $$p(G|c=1)$$ as close as possible to $$p(data|c=1)$$ push $$p(G|c=2)$$ as close as possible to $$p(data|c=2)$$ push $$p(G|c=3)$$ as close as possible to $$p(data|c=3)$$ | 2,620 | 9,568 | {"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": 65, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2020-16 | latest | en | 0.906452 |
http://www.talkstats.com/showthread.php/29307-mixture-of-normals-inequality?s=eb1a8a75faf2059cf01d21eab9e43ace&p=98484 | 1,368,947,599,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696384213/warc/CC-MAIN-20130516092624-00009-ip-10-60-113-184.ec2.internal.warc.gz | 727,248,056 | 10,249 | # Thread: mixture of normals inequality
1. ## mixture of normals inequality
Let C be a convex and compact set in R^n, and let T = {theta_1,...,theta_k} be a set of points not in C such that T and C are hyperplane separated. Let {c_1,...,c_k} be, respectively, the closest points in C to {theta_1,...,theta_k}. WLOG, suppose ||c_1 - theta_1|| >= ||c_i - theta_i|| for all i <= k.
Suppose we are interested in testing
H_1: X comes from an equally-weighted mixture of normals with covariance = I_n and means = T
against
H_0: X comes from an equally-weighted mixture of normals with covariance = I_n and means = {c_1,...,c_k}
Let the power of this level-alpha likelihood ratio test be P
Let the power of the level-alpha likelihood ratio test of H_1: X ~ N(theta_1,I_n) against H_0: X ~ N(c_1,I_n) be Q.
Is it possible to show that P <= Q?
2. ## Re: mixture of normals inequality
Originally Posted by student?
Let C be a convex ...
...Is it possible to show that P <= Q?
Yes, I think so. In fact, I suspect you can also show it if C is quasi-convex...because your're using a weak inequality.
3. ## Re: mixture of normals inequality
Originally Posted by Dragan
Yes, I think so. In fact, I suspect you can also show it if C is quasi-convex...because your're using a weak inequality.
ah..yes, actually just notice that the test has less power than the one in which the mean is known to be one of {c_1,...,c_k}
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• You may not edit your posts | 425 | 1,561 | {"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-2013-20 | latest | en | 0.913438 |
https://www.sscadda.com/previous-year-di-questions-for-ssc-cg | 1,627,226,878,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151699.95/warc/CC-MAIN-20210725143345-20210725173345-00272.warc.gz | 1,067,880,666 | 26,681 | # Previous Year D.I. Questions For SSC CGL 2017
Dear Readers, Here We are providing a quant Quiz of 15 D.I. questions in accordance with the syllabus of SSC CGL. Most of these Questions are Previous Year asked in SSC CGL. It will give you implicit idea of Math Question Paper of SSC CGL 2017
Directions (1-5): Percentage of the candidates en-rolled in seven different schools for the C.B.S.E 10th examination and candidates who passed the examination is given by following pie-graph.
Q1. What is the ratio of candidates passed to the candidates enrolled from the school A?
(a) 6 : 11
(b) 11 : 6
(c) 6 : 7
(d) 7 : 6
Q2. What percentage of candidates passed to the candidates enrolled for the schools B and C together?
(a) 60
(b) 72
(c) 76
(d) 80
Q3. Which school has the highest percentage of the candidates passed to the candidates enrolled?
(a) B
(b) C
(c) E
(d) F
Q4. The number of candidates passed from the schools D and A together exceeds the number of candidates enrolled from the schools E and C together?
(a) 299
(b) 379
(c) 399
(d) 439
Q5. What percentage of candidates passed in the examination from school E out of total number of candidates enrolled from the same school?
(a) 60
(b) 75
(c) 65
(d) 80
Directions (6-10): Monthly budget plan of a man, having monthly income Rs. 18000 is given below in the pie chart and answer the questions.
Q6. Monthly expenditure on food, in rupees, is
(a) 3500
(b) 5250
(c) 4000
(d) 3250
Q7. Monthly savings is more than the monthly expenditure on education by
(a) 750
(b) 500
(c) 700
(d) 450
Q8. Monthly expenditure on education is less than that on rent by
(a) 750
(b) 250
(c) 500
(d) 200
Q9. Expenditure on rent is more than that on miscellaneous purpose by
(a) 25(1/3)%
(b) 42(6/7)%
(c) 62.5%
(d) 75%
Q10. Monthly expenditure on miscellaneous, in rupees, is
(a) 2000
(b) 3500
(c) 2500
(d) 3250
Directions (11-15): The pie-chart given below shows the spending of a family on various heads during a month, study the graph & answer the following questions.
Q11. Which three expenditures together have a central angle of 108°?
(a) fuel, clothing & others
(b) fuel, education & others
(c) clothing, rent & others
(d) Education, rent & others
Q12. If the total income of the family is 25000 then the amount spent on rent and food together is.
(a) 17250
(b) 14750
(c) 11250
(d) 8500
Q13. What is the ratio of the expenses on education to the expenses on food?
(a) 1 : 3
(b) 3 : 1
(c) 3 : 5
(d) 5 : 3
Q14. Expenditure on rent is what percent of expenditure on fuel?
(a) 135%
(b) 156%
(c) 167%
(d) 172%
Q15. What is the ratio of the expenses on education to the expenses on rent.
(a) 15 : 14
(b) 14 : 15
(c) 3 : 4
(d) 2 : 5
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Are you sure you want to skip this step? | 1,301 | 4,675 | {"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} | 3.59375 | 4 | CC-MAIN-2021-31 | latest | en | 0.860352 |
https://nl.mathworks.com/matlabcentral/cody/problems/166-kaprekar-numbers/solutions/1250742 | 1,571,291,045,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672723.50/warc/CC-MAIN-20191017045957-20191017073457-00003.warc.gz | 602,163,725 | 16,357 | Cody
# Problem 166. Kaprekar numbers
Solution 1250742
Submitted on 13 Aug 2017 by Ilias Patelakis
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
x = 16; tf_correct = false; assert(isequal(kap(x),tf_correct))
2 Pass
x = 704; tf_correct = false; assert(isequal(kap(x),tf_correct))
3 Pass
x = 9 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 9 tf_correct = logical 1
4 Pass
x = 45 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 45 tf_correct = logical 1
5 Pass
x = 55 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 55 tf_correct = logical 1
6 Pass
x = 99 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 99 tf_correct = logical 1
7 Pass
x = 297 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 297 tf_correct = logical 1
8 Pass
x = 703 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 703 tf_correct = logical 1
9 Pass
x = 999 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 999 tf_correct = logical 1
10 Pass
x = 2223 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 2223 tf_correct = logical 1
11 Pass
x = 2728 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 2728 tf_correct = logical 1
12 Pass
x = 4950 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 4950 tf_correct = logical 1
13 Pass
x = 5050 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 5050 tf_correct = logical 1
14 Pass
x = 7272 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 7272 tf_correct = logical 1
15 Pass
x = 7777 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 7777 tf_correct = logical 1
16 Pass
x = 9999 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 9999 tf_correct = logical 1
17 Pass
x = 17344 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 17344 tf_correct = logical 1
18 Pass
x = 22222 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 22222 tf_correct = logical 1
19 Pass
x = 77778 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 77778 tf_correct = logical 1
20 Pass
x = 82656 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 82656 tf_correct = logical 1
21 Pass
x = 95121 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 95121 tf_correct = logical 1
22 Pass
x = 99999 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 99999 tf_correct = logical 1
23 Pass
x = 142857 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 142857 tf_correct = logical 1
24 Pass
x = 148149 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 148149 tf_correct = logical 1
25 Pass
x = 181819 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 181819 tf_correct = logical 1
26 Pass
x = 187110 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 187110 tf_correct = logical 1
27 Pass
x = 208495 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 208495 tf_correct = logical 1
28 Pass
x = 318682 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 318682 tf_correct = logical 1
29 Pass
x = 329967 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 329967 tf_correct = logical 1
30 Pass
x = 351352 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 351352 tf_correct = logical 1
31 Pass
x = 356643 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 356643 tf_correct = logical 1
32 Pass
x = 390313 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 390313 tf_correct = logical 1
33 Pass
x = 461539 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 461539 tf_correct = logical 1
34 Pass
x = 466830 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 466830 tf_correct = logical 1
35 Pass
x = 499500 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 499500 tf_correct = logical 1
36 Pass
x = 500500 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 500500 tf_correct = logical 1
37 Pass
x = 533170 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 533170 tf_correct = logical 1
38 Pass
x = 538461 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 538461 tf_correct = logical 1
39 Pass
x = 609687 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 609687 tf_correct = logical 1
40 Pass
x = 643357 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 643357 tf_correct = logical 1
41 Pass
x = 648648 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 648648 tf_correct = logical 1
42 Pass
x = 670033 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 670033 tf_correct = logical 1
43 Pass
x = 681318 tf_correct = true assert(isequal(kap(x),tf_correct))
x = 681318 tf_correct = logical 1
44 Pass
x = 681319 tf_correct = false assert(isequal(kap(x),tf_correct))
x = 681319 tf_correct = logical 0
45 Pass
x = 681320 tf_correct = false assert(isequal(kap(x),tf_correct))
x = 681320 tf_correct = logical 0 | 1,651 | 5,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-43 | latest | en | 0.321722 |
https://www.numbersaplenty.com/1050903881 | 1,716,104,339,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057774.18/warc/CC-MAIN-20240519070539-20240519100539-00099.warc.gz | 824,531,279 | 3,196 | Search a number
1050903881 is a prime number
BaseRepresentation
bin111110101000111…
…000010101001001
32201020110110002222
4332220320111021
54123012411011
6252140303425
735020351466
oct7650702511
92636413088
101050903881
1149a231705
12253b41b75
13139950b21
149d7dc66d
15623d8ddb
hex3ea38549
1050903881 has 2 divisors, whose sum is σ = 1050903882. Its totient is φ = 1050903880.
The previous prime is 1050903853. The next prime is 1050903883. The reversal of 1050903881 is 1883090501.
1050903881 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 834343225 + 216560656 = 28885^2 + 14716^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-1050903881 is a prime.
It is a super-2 number, since 2×10509038812 = 2208797934201724322, which contains 22 as substring.
Together with 1050903883, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (1050903883) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 525451940 + 525451941.
It is an arithmetic number, because the mean of its divisors is an integer number (525451941).
Almost surely, 21050903881 is an apocalyptic number.
It is an amenable number.
1050903881 is a deficient number, since it is larger than the sum of its proper divisors (1).
1050903881 is an equidigital number, since it uses as much as digits as its factorization.
1050903881 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 8640, while the sum is 35.
The square root of 1050903881 is about 32417.6476783865. The cubic root of 1050903881 is about 1016.6879243905.
The spelling of 1050903881 in words is "one billion, fifty million, nine hundred three thousand, eight hundred eighty-one". | 604 | 2,013 | {"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-2024-22 | latest | en | 0.868129 |
https://math-physics-problems.wikia.org/wiki/Projectile_Motion | 1,627,179,995,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151563.91/warc/CC-MAIN-20210725014052-20210725044052-00340.warc.gz | 393,773,031 | 28,287 | 303 Pages
## Problem
Consider a projectile fired at an initial velocity at firing angle of above the ground. Ignore air resistance or the effects of a rotational effects of the Earth.
Part 1: Calculate the time it takes for the projectile to land at the same altitude.
Part 2: Calculate the range of the projectile in terms of , , and .
Part 3: Calculate the maximum altitude that the projectile is able to attain.
## Solution
This problem must be analyzed in two directions.
Treat the initial point . There is no acceleration in the x-direction, and the acceleration in the x-direction is .
Part 1
Since the projectile lands at the same altitude, the displacement in the y-direction is 0.
This means the two times in which are and . Of course the second time is the landing time.
Part 2
.
Since , the range of the projectile is
.
Part 3
The top of the trajectory occurs at half the time of landing, so
.
Substitute this time to the displacement equation in the y-direction.
.
Community content is available under CC-BY-SA unless otherwise noted. | 230 | 1,066 | {"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-2021-31 | latest | en | 0.883816 |
https://www.doorsteptutor.com/Exams/IMO/Class-1/Questions/Topic-Time-4/Part-7.html | 1,529,530,240,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863886.72/warc/CC-MAIN-20180620202232-20180620222232-00282.warc.gz | 818,877,399 | 14,019 | # Time (IMO- Mathematics Olympiad (SOF) Class 1): Questions 49 - 55 of 57
Get 1 year subscription: Access detailed explanations (illustrated with images and videos) to 533 questions. Access all new questions we will add tracking exam-pattern and syllabus changes. View Sample Explanation or View Features.
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## Question number: 49
» Time
MCQ▾
### Question
Which date is the second Wednesday of March 2011?
### Choices
Choice (4) Response
a.
March
b.
March
c.
March
d.
March
## Question number: 50
» Time
MCQ▾
### Question
If Riya walk 2 kilometers in a day, then she walks ________kilometers in two weeks.
### Choices
Choice (4) Response
a.
Kilometers
b.
Kilometers
c.
Kilometers
d.
Kilometers
## Question number: 51
» Time
MCQ▾
### Question
Which of the following month comes just after seventh month of a year?
### Choices
Choice (4) Response
a.
August
b.
July
c.
September
d.
October
## Question number: 52
» Time
MCQ▾
### Question
If yesterday was Friday, then today is ________.
### Choices
Choice (4) Response
a.
Thursday
b.
Tuesday
c.
Saturday
d.
Friday
## Question number: 53
» Time
MCQ▾
### Question
What is Month of a year?
### Choices
Choice (4) Response
a.
November
b.
August
c.
May
d.
October
## Question number: 54
» Time
MCQ▾
### Question
If Sikha’s birthday is third Wednesday of November 2012, then when will Sikha celebrate her birthday?
### Choices
Choice (4) Response
a.
November
b.
November
c.
November
d.
November
## Question number: 55
» Time
MCQ▾
### Question
Which date is the third Thursday of May 2012?
### Choices
Choice (4) Response
a.
may
b.
may
c.
may
d.
may
f Page | 491 | 1,716 | {"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} | 2.875 | 3 | CC-MAIN-2018-26 | latest | en | 0.798825 |
https://www.enotes.com/homework-help/prove-statement-using-epsilon-delta-definition-382663 | 1,490,434,995,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188914.50/warc/CC-MAIN-20170322212948-00265-ip-10-233-31-227.ec2.internal.warc.gz | 883,228,423 | 12,678 | # prove the statement using the epsilon delta definition of a limit. lim (as x--> -2) of (3x+5)= -1
Asked on by ambybear
degeneratecircle | High School Teacher | (Level 2) Associate Educator
Posted on
We need to show that given any `epsilon>0,` we can find some `delta>0` such that `|(3x+5)-(-1)|=|3x+6|<epsilon` whenever `0<|x-(-2)|=|x+2|<delta.`
As usual in this type of proof, we essentially work backwards to get a value of `delta` and then reverse our steps to come up with an acceptable proof.
`|3x+6|=|3(x+2)|=3|x+2|<epsilon` implies `|x+2|<epsilon/3,` which means that given some value `epsilon,` we should try choosing `delta=epsilon/3.` Then we reverse the above steps to come up with the following proof:
Let `epsilon>0` be given. Set `delta=epsilon/3.` Then if `0<|x+2|<epsilon/3,` we have
`epsilon>3|x+2|=|3x+6|=|(3x+5)-(-1)|,` which is what we needed to show.
Sources:
We’ve answered 319,633 questions. We can answer yours, too. | 313 | 952 | {"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} | 3.796875 | 4 | CC-MAIN-2017-13 | longest | en | 0.854397 |
https://www.kopykitab.com/blog/rd-sharma-solutions-class-12-maths-chapter-1-exercise-1-1/ | 1,643,194,488,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304947.93/warc/CC-MAIN-20220126101419-20220126131419-00073.warc.gz | 887,632,455 | 30,456 | # RD Sharma Solutions for Class 12 Maths Exercise 1.1 Chapter 1 Relation (Updated for 2021-22)
RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1: The RD Sharma Class 12 Solution PDF of Chapter 1 Exercise 1.1 Relations can be downloaded from the link given. Solutions for correct answer practice are prepared by experts in the best possible way and are easily understood by the students. In this practice, students will gain knowledge about relations and types. RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1 are detailed and detailed to make learning easier for students. In this exercise, there are two levels according to the increasing order of difficulty.
## Download RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1 PDF
RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1
### Access answers to RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1
1. Let A be the set of all human beings in a town at a particular time. Determine whether the following relation is reflexive, symmetric, and transitive:
(i) R = {(x, y): x and y work at the same place}
(ii) R = {(x, y): x and y live in the same locality}
(iii) R = {(x, y): x is wife of y}
(iv) R = {(x, y): x is father of y}
Solution:
(i) Given R = {(x, y): x and y work at the same place}
Now we have to check whether the relation is reflexive:
Let x be an arbitrary element of R.
Then, x ∈R
⇒ x and x work at the same place is true since they are the same.
⇒(x, x) ∈R [condition for reflexive relation]
So, R is a reflexive relation.
Now let us check the Symmetric relation:
Let (x, y) ∈R
⇒x and y work at the same place [given]
⇒y and x work at the same place
⇒(y, x) ∈R
So, R is a symmetric relation.
Transitive relation:
Let (x, y) ∈R and (y, z) ∈R.
Then, x and y work at the same place. [Given]
y and z also work at the same place. [(y, z) ∈R]
⇒ x, y, and z all work at the same place.
⇒x and z work at the same place.
⇒ (x, z) ∈R
So, R is a transitive relation.
Hence R is reflexive, symmetric, and transitive.
(ii) Given R = {(x, y): x and y live in the same locality}
Now we have to check whether the relation R is reflexive, symmetric, and transitive.
Let x be an arbitrary element of R.
Then, x ∈R
It is given that x and x live in the same locality is true since they are the same.
So, R is a reflexive relation.
Symmetry:
Let (x, y) ∈ R
⇒ x and y live in the same locality [given]
⇒ y and x live in the same locality
⇒ (y, x) ∈ R
So, R is a symmetric relation.
Transitivity:
Let (x, y) ∈R and (y, z) ∈R.
Then,
x and y live in the same locality and y and z live in the same locality
⇒ x, y, and z all live in the same locality
⇒ x and z live in the same locality
⇒ (x, z) ∈ R
So, R is a transitive relation.
Hence R is reflexive, symmetric, and transitive.
(iii) Given R = {(x, y): x is wife of y}
Now we have to check whether the relation R is reflexive, symmetric, and transitive.
First, let us check whether the relation is reflexive:
Let x be an element of R.
Then, x is the wife of x cannot be true.
⇒ (x, x) ∉R
So, R is not a reflexive relation.
Symmetric relation:
Let (x, y) ∈R
⇒ x is the wife of y
⇒ x is female and y is male
⇒ y cannot be the wife of x as y is the husband of x
⇒ (y, x) ∉R
So, R is not a symmetric relation.
Transitive relation:
Let (x, y) ∈R, but (y, z) ∉R
Since x is the wife of y, but y cannot be the wife of z, y is the husband of x.
⇒ x is not the wife of z
⇒(x, z) ∈R
So, R is a transitive relation.
(iv) Given R = {(x, y): x is father of y}
Now we have to check whether the relation R is reflexive, symmetric, and transitive.
Reflexivity:
Let x be an arbitrary element of R.
Then, x is the father of x cannot be true since no one can be the father of himself.
So, R is not a reflexive relation.
Symmetry:
Let (x, y) ∈R
⇒ x is a father of y
⇒ y is the son/daughter of x
⇒ (y, x) ∉R
So, R is not a symmetric relation.
Transitivity:
Let (x, y) ∈R and (y, z) ∈R.
Then, x is a father of y and y is a father of z
⇒ x is the grandfather of z
⇒ (x, z) ∉R
So, R is not a transitive relation.
2. Three relations R1, R2, and R3 are defined on a set A = {abc} as follows:
R1 = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}
R2 = {(a, a)}
R3 = {(b, c)}
R4 = {(a, b), (b, c), (c, a)}.
Find whether or not each of the relations R1, R2, R3, R4 on is (i) reflexive (ii) symmetric, and (iii) transitive.
Solution:
(i) Consider R1
Given R1 = {(a, a), (a, b), (a, c), (b, b), (b, c), (c, a), (c, b), (c, c)}
Now we have check R1 is reflexive, symmetric and transitive
Reflexive:
Given (a, a), (b, b) and (c, c) ∈ R1
So, R1 is reflexive.
Symmetric:
We see that the ordered pairs obtained by interchanging the components of R1 are also in R1.
So, R1 is symmetric.
Transitive:
Here, (a, b) ∈R1, (b, c) ∈R1, and also (a, c) ∈R1
So, R1 is transitive.
(ii) Consider R2
Given R2 = {(a, a)}
Reflexive:
Clearly (a, a) ∈R2.
So, R2 is reflexive.
Symmetric:
Clearly (a, a) ∈R
⇒ (a, a) ∈R.
So, R2 is symmetric.
Transitive:
R2 is clearly a transitive relation since there is only one element in it.
(iii) Consider R3
Given R3 = {(b, c)}
Reflexive:
Here,(b, b)∉ R3 neither (c, c) ∉ R3
So, R3 is not reflexive.
Symmetric:
Here, (b, c) ∈R3, but (c, b) ∉R3
So, Ris not symmetric.
Transitive:
Here, R3 has only two elements.
Hence, R3 is transitive.
(iv) Consider R4
Given R4 = {(a, b), (b, c), (c, a)}.
Reflexive:
Here, (a, a) ∉ R4, (b, b) ∉ R4 (c, c) ∉ R4
So, R4 is not reflexive.
Symmetric:
Here, (a, b) ∈ R4, but (b, a) ∉ R4.
So, R4 is not symmetric
Transitive:
Here, (a, b) ∈R4, (b, c) ∈R4, but (a, c) ∉R4
So, R4 is not transitive.
3. Test whether the following relation R1, R2, and Rare (i) reflexive (ii) symmetric, and (iii) transitive:
(i) R1 on Q0 defined by (a, b) ∈ R1 ⇔ a = 1/b.
(ii) R2 on Z defined by (a, b) ∈ R2 ⇔ |a – b| ≤ 5
(iii) Ron R defined by (a, b) ∈ R3 ⇔ a2 – 4ab + 3b2 = 0.
Solution:
(i) Given R1 on Q0 defined by (a, b) ∈ R1 ⇔ a = 1/b.
Reflexivity:
Let a be an arbitrary element of R1.
Then, a ∈ R1
⇒ a ≠1/a for all a ∈ Q0
So, R1 is not reflexive.
Symmetry:
Let (a, b) ∈ R1
Then, (a, b) ∈ R1
Therefore we can write ‘a’ as a =1/b
⇒ b = 1/a
⇒ (b, a) ∈ R1
So, R1 is symmetric.
Transitivity:
Here, (a, b) ∈R1 and (b, c) ∈R2
⇒ a = 1/b and b = 1/c
⇒ a = 1/ (1/c) = c
⇒ a ≠ 1/c
⇒ (a, c) ∉ R1
So, R1 is not transitive.
(ii) Given R2 on Z defined by (a, b) ∈ R2 ⇔ |a – b| ≤ 5
Now we have to check whether R2 is reflexive, symmetric, and transitive.
Reflexivity:
Let a be an arbitrary element of R2.
Then, a ∈ R2
On applying the given condition we get,
⇒ | a−a | = 0 ≤ 5
So, R1 is reflexive.
Symmetry:
Let (a, b) ∈ R2
⇒ |a−b| ≤ 5 [Since, |a−b| = |b−a|]
⇒ |b−a| ≤ 5
⇒ (b, a) ∈ R2
So, R2 is symmetric.
Transitivity:
Let (1, 3) ∈ R2 and (3, 7) ∈R2
⇒|1−3|≤5 and |3−7|≤5
But |1−7| ≰5
⇒ (1, 7) ∉ R2
So, R2 is not transitive.
(iii) Given Ron R defined by (a, b) ∈ R3 ⇔ a2 – 4ab + 3b2 = 0.
Now we have check whether R2 is reflexive, symmetric and transitive.
Reflexivity:
Let a be an arbitrary element of R3.
Then, a ∈ R3
⇒ a− 4a × a+ 3a2= 0
So, R3 is reflexive
Symmetry:
Let (a, b) ∈ R3
⇒ a2−4ab+3b2=0
But b2−4ba+3a2≠0 for all a, b ∈ R
So, R3 is not symmetric.
Transitivity:
Let (1, 2) ∈ R3 and (2, 3) ∈ R3
⇒ 1 − 8 + 6 = 0 and 4 – 24 + 27 = 0
But 1 – 12 + 9 ≠ 0
So, R3 is not transitive.
4. Let A = {1, 2, 3}, and let R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}, R2 = {(2, 2), (3, 1), (1, 3)}, R3 = {(1, 3), (3, 3)}. Find whether or not each of the relations R1, R2, R3 on A is (i) reflexive (ii) symmetric (iii) transitive.
Solution:
Consider R1
Given R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}
Reflexivity:
Here, (1, 1), (2, 2), (3, 3) ∈R
So, R1 is reflexive.
Symmetry:
Here, (2, 1) ∈ R1,
But (1, 2) ∉ R1
So, R1 is not symmetric.
Transitivity:
Here, (2, 1) ∈R1 and (1, 3) ∈R1
But (2, 3) ∉R1
So, R1 is not transitive.
Now consider R2
Given R2 = {(2, 2), (3, 1), (1, 3)}
Reflexivity:
Clearly, (1, 1) and (3, 3) ∉R2
So, R2 is not reflexive.
Symmetry:
Here, (1, 3) ∈ R2 and (3, 1) ∈ R2
So, R2 is symmetric.
Transitivity:
Here, (1, 3) ∈ R2 and (3, 1) ∈ R
But (3, 3) ∉R2
So, R2 is not transitive.
Consider R3
Given R3 = {(1, 3), (3, 3)}
Reflexivity:
Clearly, (1, 1) ∉ R3
So, R3 is not reflexive.
Symmetry:
Here, (1, 3) ∈ R3, but (3, 1) ∉ R3
So, R3 is not symmetric.
Transitivity:
Here, (1, 3) ∈ R3 and (3, 3) ∈ R3
Also, (1, 3) ∈ R3
So, R3 is transitive.
5. The following relation is defined on the set of real numbers.
(i) aRb if a – b > 0
(ii) aRb iff 1 + a b > 0
(iii) aRb if |a| ≤ b.
Find whether a relation is reflexive, symmetric, or transitive.
Solution:
(i) Consider aRb if a – b > 0
Now for this relation, we have to check whether it is reflexive, transitive, and symmetric.
Reflexivity:
Let a be an arbitrary element of R.
Then, a ∈ R
But a − a = 0 ≯ 0
So, this relation is not reflexive.
Symmetry:
Let (a, b) ∈ R
⇒ a − b > 0
⇒ − (b − a) >0
⇒ b − a < 0
So, the given relation is not symmetric.
Transitivity:
Let (a, b) ∈R and (b, c) ∈R.
Then, a − b > 0 and b − c > 0
a – b + b − c > 0
⇒ a – c > 0
⇒ (a, c) ∈ R.
So, the given relation is transitive.
(ii) Consider aRb iff 1 + a b > 0
Now for this relation, we have to check whether it is reflexive, transitive, and symmetric.
Reflexivity:
Let a be an arbitrary element of R.
Then, a ∈ R
⇒ 1 + a × a > 0
i.e. 1 + a2 > 0 [Since, square of any number is positive]
So, the given relation is reflexive.
Symmetry:
Let (a, b) ∈ R
⇒ 1 + a b > 0
⇒ 1 + b a > 0
⇒ (b, a) ∈ R
So, the given relation is symmetric.
Transitivity:
Let (a, b) ∈R and (b, c) ∈R
⇒1 + a b > 0 and 1 + b c >0
But 1+ ac ≯ 0
⇒ (a, c) ∉ R
So, the given relation is not transitive.
(iii) Consider aRb if |a| ≤ b.
Now for this relation, we have to check whether it is reflexive, transitive, and symmetric.
Reflexivity:
Let a be an arbitrary element of R.
Then, a ∈ R [Since, |a|=a]
⇒ |a|≮ a
So, R is not reflexive.
Symmetry:
Let (a, b) ∈ R
⇒ |a| ≤ b
⇒ |b| ≰ a for all a, b ∈ R
⇒ (b, a) ∉ R
So, R is not symmetric.
Transitivity:
Let (a, b) ∈ R and (b, c) ∈ R
⇒ |a| ≤ b and |b| ≤ c
Multiplying the corresponding sides, we get
|a| × |b| ≤ b c
⇒ |a| ≤ c
⇒ (a, c) ∈ R
Thus, R is transitive.
6. Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
Solution:
Given R = {(a, b): b = a + 1}
Now for this relation, we have to check whether it is reflexive, transitive, and symmetric Reflexivity:
Let a be an arbitrary element of R.
Then, a = a + 1 cannot be true for all a ∈ A.
⇒ (a, a) ∉ R
So, R is not reflexive on A.
Symmetry:
Let (a, b) ∈ R
⇒ b = a + 1
⇒ −a = −b + 1
⇒ a = b − 1
Thus, (b, a) ∉ R
So, R is not symmetric on A.
Transitivity:
Let (1, 2) and (2, 3) ∈ R
⇒ 2 = 1 + 1 and 3
2 + 1 is true.
But 3 ≠ 1+1
⇒ (1, 3) ∉ R
So, R is not transitive on A.
7. Check whether the relation R on R defined as R = {(ab): a ≤ b3} is reflexive, symmetric, or transitive.
Solution:
Given R = {(ab): ≤ b3}
It is observed that (1/2, 1/2) in R as 1/2 > (1/2)3 = 1/8
∴ R is not reflexive.
Now,
(1, 2) ∈ R (as 1 < 23 = 8)
But,
(2, 1) ∉ R (as 2 > 13 = 1)
∴ R is not symmetric.
We have (3, 3/2), (3/2, 6/5) in “R as” 3 < (3/2)3 and 3/2 < (6/5)3
But (3, 6/5) ∉ R as 3 > (6/5)3
∴ R is not transitive.
Hence, R is neither reflexive, nor symmetric, nor transitive.
8. Prove that every identity relation on a set is reflexive, but the converse is not necessarily true.
Solution:
Let A be a set.
Then, Identity relation IA=IA is reflexive, since (a, a) ∈ A ∀a
The converse of it need not be necessarily true.
Consider the set A = {1, 2, 3}
Here,
Relation R = {(1, 1), (2, 2) , (3, 3), (2, 1), (1, 3)} is reflexive on A.
However, R is not an identity relation.
9. If A = {1, 2, 3, 4} define relations on A which have properties of being
(i) Reflexive, transitive but not symmetric
(ii) Symmetric but neither reflexive nor transitive.
(iii) Reflexive, symmetric and transitive.
Solution:
(i) The relation on A having properties of being reflexive, transitive, but not symmetric is
R = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)}
Relation R satisfies reflexivity and transitivity.
⇒ (1, 1), (2, 2), (3, 3) ∈ R
And (1, 1), (2, 1) ∈ R ⇒ (1, 1) ∈ R
However, (2, 1) ∈ R, but (1, 2) ∉ R
(ii) The relation on A having properties of being reflexive, transitive, but not symmetric is
R = {(1, 1), (2, 2), (3, 3), (4, 4), (2, 1)}
Relation R satisfies reflexivity and transitivity.
⇒ (1, 1), (2, 2), (3, 3) ∈ R
And (1, 1), (2, 1) ∈ R ⇒ (1, 1) ∈ R
However, (2, 1) ∈ R, but (1, 2) ∉ R
(iii) The relation on A having properties of being symmetric, reflexive, and transitive is
R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1)}
The relation R is an equivalence relation on A.
## RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1: Important Topics
Let us have a look at important topics covered in this exercise.
• Types of relations
• Void relation
• Universal relation
• Identity relation
• Reflexive relation
• Symmetric relation
• Transitive relation
• Antisymmetric relation
We have provided complete details of RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1. If you have any queries related to CBSE Class 12 Exam, feel free to ask us in the comment section below.
## FAQs on RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1
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You can download RD Sharma Solutions Class 12 Maths Chapter 1 Exercise 1.1 free PDF from the above article. | 6,062 | 14,129 | {"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} | 4.875 | 5 | CC-MAIN-2022-05 | latest | en | 0.613753 |
https://questions.examside.com/past-years/jee/question/int-x2-1-over-x3sqrt-2x4-2x2-1-dx-jee-advanced-2006-marks-3-lorv1tncb9prmgrp.htm | 1,721,209,970,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514759.39/warc/CC-MAIN-20240717090242-20240717120242-00125.warc.gz | 435,829,276 | 40,039 | 1
IIT-JEE 2006
MCQ (Single Correct Answer)
+3
-0.75
$$\int {{{{x^2} - 1} \over {{x^3}\sqrt {2{x^4} - 2{x^2} + 1} }}dx = }$$
A
$${{\sqrt {2{x^4} - 2{x^2} + 1} } \over {{x^2}}} + c$$
B
$${{\sqrt {2{x^4} - 2{x^2} + 1} } \over {{x^3}}} + c$$
C
$${{\sqrt {2{x^4} - 2{x^2} + 1} } \over {{x}}} + c$$
D
$${{\sqrt {2{x^4} - 2{x^2} + 1} } \over {{2x^2}}} + c$$
2
IIT-JEE 2005 Screening
MCQ (Single Correct Answer)
+3
-0.75
If $$\int\limits_{\sin x}^1 {{t^2}f\left( t \right)dt = 1 - \sin x,}$$ then f$$\left( {{1 \over {\sqrt 3 }}} \right)$$ is
A
$${1 \over 3}$$
B
$${{1 \over {\sqrt 3 }}}$$
C
$$3$$
D
$${\sqrt 3 }$$
3
IIT-JEE 1995 Screening
MCQ (Single Correct Answer)
+3
-0.75
The value of the integral $$\int {{{{{\cos }^3}x + {{\cos }^5}x} \over {{{\sin }^2}x + {{\sin }^4}x}}} \,dx\,$$ is
A
$$\sin x - 6{\tan ^{ - 1}}\left( {\sin x} \right) + c$$
B
$$\sin x - 2{\left( {\sin x} \right)^{ - 1}} + c$$
C
$$\sin x - 2{\left( {\sin x} \right)^{ - 1}} - 6{\tan ^{ - 1}}\left( {\sin x} \right) + c$$
D
$$\,\sin x - 2{\left( {\sin x} \right)^{ - 1}} + 5{\tan ^{ - 1}}\left( {\sin x} \right) + c$$
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Class 12 | 632 | 1,256 | {"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} | 3.296875 | 3 | CC-MAIN-2024-30 | latest | en | 0.284386 |
https://www.mrexcel.com/archive/formulas/combining-if-and-and-functions/ | 1,534,817,545,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221217909.77/warc/CC-MAIN-20180821014427-20180821034427-00299.warc.gz | 960,444,571 | 9,252 | # Combining IF and AND functions
Posted by Holle on December 20, 2001 11:05 AM
How do I combine IF and AND functions in a formula? For example: =IF(A5=2,I5). I also want to add the condition =AND(A6=I6,10 to 15). How would I write this formula?
Posted by Joe Was on December 20, 2001 11:19 AM
You can only nest seven levels of IF statments, but you can reference a cell that includes additional IF statments to extend this limit.
The IF statement has the syntax: Result=IF(test,Then,Else)
So, with this in mind you can replace the "Else" part with a new "IF" up to seven times. You can also mix functions within your IF statements, like:
=IF(AND(I5>0,I5<5),1,IF(AND(I5>4,I5<9),2,IF(AND(I5>8,I5<13),3,IF(AND(I5>12,I5<17),4,IF(AND(I5>16,I5<21),5,IF(AND(I5>20,I5<25),6,0))))))
Note: AND(I5>0,I5<5) is: test1.
,1, is: Then 1.
IF(And(I5>4,I5<9) is: test2.
,2, is: Then 2.
...
If(AND(I5>20,I5<25) is test6.
,6, is: Then 6. and
,0)))))) is: the last Else.
This groups up to 24 collections into groups of no more than 4. With nested IF statements the trick is the "()" parentheses order. JSW
Posted by Jacob on December 20, 2001 11:19 AM
Hi
Try this
=if(and(this=that,something else=something else),do this if true,do this if false)
Jacob
Posted by Aladin Akyurek on December 20, 2001 11:19 AM
Holle --
It's unclear (to me) what you want. But, here an example for bulding up a condition argument with AND in an IF formula:
=IF(AND(B1>C1,B1 < D1),B1*5%,0)
which says: if the value in B1 is between the values in C1 and D1 exclusive, multiply the value in B1 with 5%, otherwise return 0. | 566 | 1,625 | {"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} | 3.25 | 3 | CC-MAIN-2018-34 | longest | en | 0.793799 |
https://oswalpublishers.com/ncert-solutions/ncert-solutions-class-10-mathematics/chapter-2-polynomials/ | 1,719,295,873,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865560.33/warc/CC-MAIN-20240625041023-20240625071023-00324.warc.gz | 377,936,110 | 51,194 | # NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials
## NCERT Solutions for Class 10 Mathematics Chapter 2 Free PDF Download
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Exercise 2.1
1. The graphs of y = p(x) are given in figures below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.
Sol. (i) The graph do not intersects the x-axis.
Then Number of zeroes = No zeroes.
(ii) The graph intersects the x-axis at one point only.
Then Number of zeroes = 1
(iii) The graph intersects the x-axis at three points.
Then Number of zeroes = 3
(iv) The graph intersects the x-axis at two points.
Then Number of zeroes = 2
(v) The graph intersects the x-axis at four points.
Then Number of zeroes = 4
(vi) The graph intersects the x-axis at three points.
Then Number of zeroes = 3
Exercise 2.2
1. Find the zeroes of the following quadratic polynoials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(vi) 4u2 + 8u
(v) t2 – 15
(vi) 3x2 – x – 4
Sol. (i) Let be
F(x) = x2 – 2x – 8
F(x) = x2 – 4x + 2x – 8
F(x) = x(x – 4) + 2(x – 4)
F(x) = (x – 4)(x + 2)
So, the value of x2 – 2x – 8 is zero.
F(x) = 0
(x – 4)(x + 2) = 0
When
x – 4 = 0, x = 4
x + 2 = 0, x = – 2
Therefore, the zeroes of x2 – 2x – 8 are 4
and – 2
Now,
Sum of zeroes = 4 + (– 2)
= 2
$$\text{Sum of zeroes}=\frac{−(coefficient of x)}{−(coefficient of x^2)}\\=\frac{-(-2)}{1}\\=2\\\text{Product of zeroes} = 4 × – 2 = – 8\\\text{Product of zeroes}=\frac{\text{constant term}}{\text{termcoefficient of}\space x^2}\\=\frac{-8}{1}\\=-8$$
(ii) Let be F(s) = 4s2 – 4s + 1
F(s) = 4s2 – 2s – 2s + 1
= 2s(2s – 1) – 1(2s – 1)
= (2s – 1)(2s – 1) The value of F(s) = 0
(2s – 1)(2s – 1) = 0
$$(2s – 1) = 0, s =\frac{1}{2}\\(2s – 1) = 0, s =\frac{1}{2}\\\text{Therefore, the zeroes of}\space4s^2– 4s + 1\space are\space and\space\frac{1}{2}\\\text{Sum of zeroes} =\frac{1}{2}+\frac{1}{2}=1\\\text{Sum of zeroes}=\frac{−(coefficient of S)}{(coefficient of S^2)}\\=\frac{-(-4)}{4}=1\\\text{Product of zeroes}=\frac{1}{2}×\frac{1}{2}=\frac{1}{4}\\\text{Product of zeroes}=\frac{\text{constant term}}{\text{coefficient of} \space S^2}\\=\frac{1}{4}\\\text{Relationship between the zeroes and the coefficients is true.}$$
(iii) Let be
F(x) = 6x2 – 3 – 7x
⇒ F(x) = 6x2 – 7x – 3
⇒ F(x) = 6x2 – 9x + 2x – 3
⇒ F(x) = 3x(2x – 3) + 1(2x – 3)
⇒ F(x) = (2x – 3)(3x + 1)
Value of F(x) = 0
⇒ (2x – 3)(3x + 1) = 0
⇒ 2x – 3 = 0, 3x + 1 = 0
⇒ 2x = 3, 3x = – 1
$$x=\frac{3}{2},\space\space\space\space\space\space\space\space\space\space x=\frac{-1}{3},\\\text{Therefore, the zeroes of}\space 6x^2 – 3 – 7x\space\text{are}\frac{3}{2}\space\text{and}\space\frac{-1}{3}\\\text{Sum of zeroes}=\frac{3}{2}+(\frac{-1}{3})\\=\frac{9-2}{6}=\frac{7}{6}\\\text{Sum of zeroes}=\frac{\text{−(coefficient of x)}}{\text{(coefficient of}\space x^2)}\\=\frac{-(-7)}{6}\\=\frac{7}{6}\\\text{Product of zeroes}=\frac{3}{2}×\frac{-1}{3}\\=-\frac{1}{2}\\\text{Relationship between the zeroes and the coefficients is true.}$$
(iv) Let be
F(u) = 4u2 + 8u
⇒ F(u) = 4u (u + 2)
The value of F(u) = 0
⇒ 4u(u + 2) = 0
⇒ 4u (u + 2) = 0
⇒ 4u = 0, u + 2 = 0
⇒ u = 0, u = – 2
Therefore, the zeroes of 4u2 + 8u are 0 and – 2.
Now
sum of zeores = 0 + (– 2) = – 2
$$\text{Sum of zeroes}=\frac{\text{−(coefficient of u)}}{\text{(coefficient of}\space u^2)}\\=\frac{8}{4}=-2\\\text{Sum of zeroes}= 0 × – 2 = 0\\\text{Sum of zeroes}=\frac{\text{constant term}}{\text{term coefficient of}\space u^2}\\=\frac{0}{4}=0\\\text{Relationship between the zeroes and the coefficients is true.}$$
(v) Let be
F(t) = t2 – 15
The value of F(t) = 0
t2 – 15 = 0
(t2- 152)=0
(t- 15)(t+ 15)=0
Then,
t− 15 = 0
t+ 15 = 0
t = 15
t = -15
Therefore, the zeroes of t2 – 15 are 15 and −15.
Sum of zeroes = 15−15= 0
$$\text{Sum of zeroes}=\frac{\text{−(coefficient of t)}}{\text{(coefficient of}\space t^2)}\\=\frac{-0}{1}=0\\\text{Product of zeroes}= \sqrt[]{15}×-\sqrt[]{15}=-15\\\text{Product of zeroes}=\frac{\text{constant term}}{\text{coefficient of}\space t^2}\\=\frac{-15}{1}=-15\\\text{Relationship between the zeroes and the coefficients is true.}$$
(vi) Let be
F(x) = 3x2 – x – 4
F(x) = 3x2 – 4x + 3x – 4
F(x) = x(3x – 4) + 1(3x – 4)
F(x) = (3x – 4)(x + 1)
So, the value of F(x) is zero.
F(x) = 0
(3x – 4)(x + 1) = 0
3x – 4 = 0, x + 1 = 0
3x = 4, x = – 1
x=4/3
$$\text{Therefore, the zeroes of the}\space 3x^2 – x – 4\text{are}\frac{4}{3}\text{and}-1\\\text{Sum of zeroes}=\frac{4}{3}+(-1)\\\frac{4-3}{3}\\=\frac{1}{3}\\\text{Sum of zeroes}=\frac{\text{−(coefficient of x)}}{\text{(coefficient of}\space x^2)}\\=\frac{-(-1)}{3}=\frac{1}{3}\\\text{Product of zeroes}=\frac{4}{3}×-1\\=\frac{-4}{3}\\\text{Product of zeroes}=\frac{\text{(coefficient of x)}}{\text{(coefficient of}\space x^2)}\\\text{Product of zeroes}=\frac{-4}{3}\\\text{Relationship between zeroes and coefficients are true.}$$
2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
$$(i)\frac{1}{4},-1\\(ii)\sqrt{2},\frac{1}{3}\\(iii)0,\sqrt{5}\\(iv)1,1\\(v)\frac{-1}{4},\frac{1}{4}\\(vi)4,1$$
Sol. (i) Given that,
Sum of zoeres = 1/4
Product of zeroes = – 1
x2 – (sum of zeroes)x + product of zeroes = 0
$$x^2-\frac{1}{4}×x+(-1)=0\\\frac{4x^2-x-4}{4}=0\\4x^2-x-4=0\\\text{Hence, the Quadratic polynomial is}\\F(x) =4x^2-x-4.$$
$$(ii)\text{Given that,}\\\text{Sum of zeroes}=\sqrt{2}\space\text{and}\\\text{Product of zeroes}=\frac{1}{3}\\\text{Quadratic polynomial is :}\\x^2 –\text{(sum of zeroes)x + product of zeroes = 0}\\x^2-\sqrt{2x}+\frac{1}{3}=0\\{3x^2-3\sqrt{2x}+1=0}\\\text{Hence, the polynomial is :}\\F(x) ={3x^2-3\sqrt{2x}+1=0}$$
(iii) Given that,
Sum of zeroes = 0
Product of zeroes = 5
x2 – (sum of zeroes)x + product of zeroes = 0
x2-0 × X+ 5 0
x2+ 5=0
Hence, the polynomial is
F(x) =x2+ 5
(iv) Given that,
Sum of zeroes = 1
Product of zeroes = 1
x2 – (sum of zeroes)x + product of zeroes = 0
x2 – x + 1 = 0
F(x) = x2 – x +1
$$(v) \text{Given that,}\\\text{Sum of zeroes}=\frac{-1}{4}\\\text{Product of zeroes}=\frac{1}{4}\\\text{Quadratic polynomial is}\\\text{x2 – (sum of zeroes)x + product of zeroes = 0}\\x^2-(\frac{-1}{4})x+\frac{1}{4}=0\\x^2+\frac{x}{4}+\frac{1}{4}=0\\\frac{4x^2+x+1}{4}=0\\4x^2 + x + 1 = 0\\4x^2 + x + 1 = 0\\\text{Hence, the polynomial is}\\ F(x) = 4x^2 + x + 1$$
(vi) Given that,
Sum of zeroes = 4
Product of zeroes = 1
x2 – (sum of zeroes)x + product of zeroes = 0
x2 – 4x + 1 = 0
Hence, the polynomial is
F(x) = x2 – 4x + 1.
Exercise 2.3
1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following :
(i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
(ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2
Sol. (i) P(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
x2 – 2 ) x3 – 3x2 + 5x – 3 (x – 3
x3 – 2x
$$x^3-2x\\\frac{-\space\space\space\space\space\space\space +}{– 3x^2 + 7x – 3}\text{By subtract}\\– 3x^2 + 6\\\frac{+\space\space\space\space\space\space\space -}{7x – 9}\text{By subtract}$$
Quotient q(x) = x – 3
Remainder R(x) = 7x – 9
(ii) p(x) = x4 – 3x2 + 4x + 5
g(x) = x2 + 1 – x
Rearranging g(x)
g(x) = x2 – x + 1
x2 – x + 1 ) x4 – 3x2 + 4x + 5 (x2 + x – 3
x4 – x3 + x2
$$\frac{-\space\space\space+\space\space\space\space -}{x^3 – 4x^2 + 4x + 5}\text{By subtract}\\x^3 – x^2 + x\\\frac{-\space\space\space\space+\space\space\space\space -}{– 3x^2 + 3x + 5}\\– 3x^2 + 3x + 5\\\frac{+\space\space\space\space-\space\space\space\space +}{8}\text{By subtract}\\\text{Quotient q(x)} = x^2 + x – 3\\\text{Remainder r(x)} = 8$$
(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2
– x2 + 2) x4 – 5x + 6 (– x2 – 2
x4 – 2x
$$\frac{- \space\space\space\space+}{2x^2–5x+6}\\2x^2–4\\\frac{-\space\space\space\space +}{–5x+10}\\\text{Quotient q(x)} = – x^2 – 2\\\text{Remainder R(x)} = – 5x + 10$$
2. Check whether the first polynomial is a factor of the second polynoial by dividing the second polynomial by the first polynomial
(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12
(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
Sol. (i) Let be
I Polynomial p(t) = t2 – 3
II Polynomial q(t) = 2t4 + 3t3 – 2t2 – 9t – 12
t2 – 3) 2t4 + 3t3 – 2t2 – 9t – 12 (2t2 + 3t + 4
2t4– 6t
$$\frac{- \space\space\space\space+}{3t^3 + 4t^2 – 9t – 12}\\3t^3 – 9t\\\frac{-\space\space\space\space +}{4t^2 – 12}\\4t^2 – 12\\\frac{-\space\space\space\space +}{0}\\\text{Remainder }= 0\\\text{Then First polynomial is a factor of the second polynomial.}$$
(ii) Let be
I polynomial p(x) = x2 + 3x + 1
II polynomial q(x) = 3x4 + 5x3 – 7x2 + 2x + 2
x2 + 3x + 1) 3x4 + 5x3 – 7x2 + 2x + 2 (3x2 – 4x + 2
3x4 + 9x3 + 3x
$$\frac{- \space\space-\space\space-}{-4x^3 + 10x^2 +2x +2}\\-4x^3 – 12x^2-4x\\\frac{-\space\space-\space\space -}{2x^2 +6x+2}\\2x^2 +6x+2\\\frac{-\space\space-\space\space -}{0}\\\text{Remainder }= 0\\\text{Therefore First polynomial is a factor of the second polynomial.}$$
(iii) Let be
I Polynomial p(x) = x3 – 3x + 1
II Polynomial q(x) = x5 – 4x3 + x2 + 3x + 1
x3 – 3x + 1) x5 – 4x3 + x2 + 3x + 1 (x2 – 1
x5 – 3x3 + x
$$\frac{- \space\space+\space\space-}{-x^3 +3x +1}\\-x^3 +3x+1\\\frac{+\space\space-\space\space +}{2}\\\text{Remainder }= 2\\\text{Therefore,}\\\text{ First polynomial is not a factor of the second polynomial.}$$
3. Obtain all other zoroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are √5/3 and -√ 5/3
$$\text{Sol. Since two zeroes are}\space\sqrt{\frac{5}{3}}\space\text{and}-\space\sqrt{\frac{5}{3}}\\\Biggl(x-\sqrt{\frac{5}{3}}\Biggr)\Biggl(x+\sqrt{\frac{5}{3}}\Biggr)\\=x^2-\frac{5}{3}\\\text{is a factor of the given polynomial. Now, we divide the given polynomial by}\space x^2-\frac{5}{3}\\ x^2-\frac{5}{3}\Biggr)3x^4+6x^3-2x^2-10x-5\Biggl(3x^2+6x+3\\3x^2-5x^2\\\frac{- \space\space\space\space+}{6x^3 +3x^2 -10x-5}\\6x^3 -10x\\\frac{-\space\space\space\space +}{3x^2-5}\\3x^2-5\\\frac{- +}{0}\\\text{so,}3x^4 + 6x^3 – 2x^2 – 10x – 5\\=\Biggl(x^2-\frac{5}{3}\Biggr)(3x^2+6x+3)\\\text{We factorise}(3x^2+6x+3)\\3x^2 + 6x + 3 = 3x^2 + 3x + 3x + 3\\= 3x (x + 1) + 3(x + 1)\\= (x + 1)(3x + 3)\\\text{So, its zeroes are given by x = – 1 and x = – 1. Therefore, the zeroes of the given} \\\text{polynomial are}\space\sqrt{\frac{5}{3}},-\sqrt{\frac{5}{3}}\space– 1 and – 1 .$$
4. On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and – 2x + 4, respectively. Find g(x).
Sol. We know that
Dividend = Divisor × Quotient + Remainder
Given
Dividend = x3 – 3x2 + x + 2 Divisor = g(x) = ?
Quotient = x – 2
Remainder = – 2x + 4
Then
x3 – 3x2 + x + 2 = g(x) × (x – 2) + (– 2x + 4)
x3 – 3x2 + x + 2 = g(x) × (x – 2) – 2x + 4
g(x)(x – 2) = x3 – 3x2 + x + 2 + 2x – 4
g(x) = x3 – 3x2 + 3x – 2
$$g(x)=\frac{x^3-3x^2+3x-2}{x-2}\\x – 2) x^3 – 3x^2 + 3x – 2 (x^2 – x + 1\\x^3-2x^2\\\frac{-\space\space\space\space+}{-x^2 +3x-2}\\-x^2 +2x\\\frac{+\space\space\space\space -}{3x^2-5}\\3x^2-5\\\frac{+ -}{x-2}\\x-2\\\frac{- +}{0}\\g(x) = x^2 – x + 1$$
5. Given examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Sol. (i) p(x) = 3x3+ 6x2 + 9x + 12
g(x) = 3
3) 3x3 + 6x2 + 9x + 12 (x3 + 2x2 + 3x + 4
3x2
$$\frac{-}{6x^2 +3x+12}\\6x^2\\\frac{-}{9x+12}\\9x\\\frac{-}{12}\\12\\\frac{-}{0}\\q(x) = x^3 + 2x^2 + 3x + 4, r(x) = 0\\\text{deg p(x) = deg q(x)}$$
(ii) p(x) = x4 + x3 + x + 4, g(x) = x3
x3 ) x4 + x3 + x + 4 ( x + 1
x
$$\frac{-}{x^3 +x+4}\\x^3\\\frac{-}{x+4}\\q(x) = x + 1, r(x) = x + 4\\\text{deg q(x) = deg r(x)}$$
(iii) p(x) = x4 + 2, g(x) = x + 1
x + 1 ) x4 + 2 ( x3 – x2 + x – 1
x4 + x
$$\frac{-\space-}{-x^3 +2}\\-x^3-x^2\\\frac{+\space-}{x^2+2}\\x^2+x\\\frac{-\space-}{-x+2}\\-x-1\\\frac{+ +}{3}\\q(x) = x3 – x2 + x – 1\\r(x) = 3\\\text{deg r(x) = 0}$$
Play Video about Chapter 2 Polynomials
Exercise 2.4
1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case :
(i) 2x3 + x2 – 5x + 2; 1212,,−
(ii) x3 – 4x2 + 5x – 2; 2, 1, 1
Sol. (i) Let
P(x) = 2x3 + x2 – 5x + 2
$$p\biggl(\frac{1}{2}\biggr)=2×\biggl(\frac{1}{2}\biggr)^3+\biggl(\frac{1}{2}\biggr)^2-5×\frac{1}{2}+2\\=2×\frac{1}{8}+\frac{1}{4}-\frac{5}{2}+2\\=\frac{1}{2}-\frac{5}{2}\\=0\\P(1) = 2 × 1^3 + 1^2 – 5 × 1 + 2\\= 2 × 1 + 1 – 5 + 2\\= 5 – 5\\=0\\P(– 2) = 2 × (– 2)^3 + (– 2)^2 – 5 × (– 2) + 2\\= – 16 + 4 + 10 + 2\\= – 16 + 16\\= 0\\Hence,\frac{1}{2},1\text{and – 2 are zeroes of polynomial P(x). According to question,}\\a = 2, b = 1, c = – 5, d = 2\\\text{Zeroes}, α =\frac{1}{2},β= 1, γ = – 2.\\\text{Therefore,}\\α + β + γ =\frac{-b}{a}\\\frac{1}{2}+1-2=\frac{-1}{2}\\\frac{-1}{2}=\frac{-1}{2}\\αβ + βγ + γα =\frac{c}{a}\\\frac{1}{2}×1+1×(-2)+(-2)×\frac{1}{2}=\frac{-5}{2}\\\frac{1}{2}-2-1=\frac{-5}{2}\\\frac{1}{2}-3=\frac{-5}{2}\\\frac{-5}{2}=\frac{-5}{2}\\\text{And,}\\αβγ =\frac{-d}{a}\\\frac{1}{2}×1×(-2)=\frac{-2}{2}\\– 1 = – 1\\\text{Relationship between the zeroes and the coefficients in each case is true.}$$
(ii) Let P(x) = x3 – 4x2 + 5x – 2
P(2) = 23 – 4 × 22 + 5 × 2 – 2
P(2) = 8 – 16 + 10 – 2
P(2) = 0
P(1) = 13 – 4 × 12 + 5 × 1 – 2
P(1) = 1 – 4 + 5 – 2
= 0
Hence, 2, 1 and 1 are zeroes of polynomial P(x).
According to question,
a = 1, b = – 4, c = 5, d = – 2
α= 2, β = 1, γ = 1
Now
$$α + β + γ =\frac{-b}{a}\\2 + 1 + 1 =\frac{-(-4)}{1}\\4 = 4\\\text{Again}\\αβ + βg + γα=\frac{c}{a}\\2 × 1 + 1 × 1 + 1 × 2=\frac{5}{1}\\2 + 1 + 2\\5 = 5\\\text{And,}\\αβγ =\frac{-d}{a}\\2 × 1 ×1 =\frac{-(-2)}{1}\\2 = 2\\\text{Relationship betw een the zeroes and the coefficients in each case is true.}$$
2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time and the product of its zeroes as 2, – 7, – 14 respectively.
Sol. Let be cubic polynomial
P(x) = ax3 + bx2 + cx + d and its zeroes are α,βandγ .
Then according to question
α + β +γ = 2
$$\frac{-b}{a}=2\\a = 1, b = – 2\\αβ + βγ + γα =-7\\\frac{c}{a}=-7\\c = – 7a\\a = 1, c = – 7\\αβγ = – 14\\\frac{-d}{a}=-14\\d = 14\\\text{Put the value of a, b, c and d in polynomial P(x).}\\P(x) = x^3 – 2x^2 – 7x + 14$$
3. If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b find a and b.
Sol. Given Polynomial
P(x) = x3 – 3x2 + x + 1
Compare
with Ax3 + Bx2 + Cx + D
Then
A = 1, B = – 3, C = 1, D = 1
Given,
α = a – b, β= a, γ = a + b
We know that
$$α + β + γ =\frac{-B}{A}\\a – b + a + a + b =\frac{+3}{1}\\3a = 3\\a = 1\\α β γ =\frac{-D}{A}\\(a – b) × a × (a + b) =\frac{-1}{1}\\(a^2 – b^2) × a = – 1\text{Put the value of a}\\(1^2 – b^2) × 1 = – 1\\1 – b^2 = – 1\\– b^2 = – 2\\b^2 = 2\\b = ±\sqrt{2}$$
4. If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2± 3, find other zeroes.
Sol. Since, two zeroes are 23+ and 2+ 3 and 2- 3,(x-2- 3 )(x-2+ 3 )=x2 – 4x + 1 is a factor of the given polynomial. Now, we divide the given polynomial x2 – 4x + 1.
x2 – 4x + 1 ) x4 – 6x3 – 26x2 + 138x – 35 (x2 – 2x – 35
x4 – 4x3 + x
$$\frac{-+-}{– 2x^3 – 27x^2 + 138x – 35}Subtract\\– 2x^3 + 8x^2 – 2x\\\frac{+-+}{– 35x^2 + 140x – 35}Subtract\\– 35x^2 + 140x – 35\\\frac{+-+}{0}Subtract\\So,\\ x^4 – 6x^3 – 26x^2 + 138x – 35 = (x^2 – 4x + 1)(x^2 – 2x – 35)\\\text{We factorise}x^2 – 2x – 35 as (x – 7)(x + 5).\\\text{So, its zeroes are given by x = 7 and x = – 5.}\\\text{Therefore,the zeroes of the given polynomial are}\\2+\sqrt{3},2- \sqrt{3}\space\text{and} – 5.$$
5. If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k,the remainder comes out to be x + a, find k and a.
Sol. x2 – 2x + k ) x4 – 6x3 + 16x2 – 25x + 10 ( x2 – 4x + (8 – k)
x4 – 2x3 + kx2
$$\frac{-+-}{– 4x^3 + (16 – k)x^2 – 25x + 10}\\– 4x^3 + 8x^2 – 4kx\\\frac{+-+}{(16 – k – 8)x^2 + (4k – 25)x + 10}\\(8 – k)x^2 + (4k – 25)x + 10\\(8 – k)x^2 – 2(8 – k)x + k(8 – k)\\\frac{-+-}{(4k – 25 + 16 – 2k)x + 10 – k(8 – k)}\\$$
(2k – 9)x + 10 – k(8 – k)
Quotient q(x) = x2 – 4x + (8 – k)
Remainder r(x) = (2k – 9)x + 10 – k(8 – k)
Given remainder r(x) = x + a
Compare both remainder
2k – 9 = 1
2k = 10
k = 5
but
a = 10 – k(8 – k)
Put k = 5
⇒ a = 10 – 5(8 – 5)
⇒ a = 10 – 5 × 3
⇒ a = 10 – 15
⇒ a = – 5.
Hence, k = 5 and a = – 5. | 7,823 | 16,189 | {"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} | 4.84375 | 5 | CC-MAIN-2024-26 | latest | en | 0.733453 |
https://www.physicsforums.com/threads/maxwell-equations-in-lorenz-gauge.596330/ | 1,532,031,499,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591216.51/warc/CC-MAIN-20180719183926-20180719203926-00008.warc.gz | 988,648,179 | 17,865 | # Maxwell equations in Lorenz gauge
1. Apr 13, 2012
### jjustinn
In the Lorenz gauge, the Maxwell equations reduce to four inhomogenous wave equations, with the charge density acting as the source for V, and the current density for A.
For now, just take a static charge distribution -- say, a point charge at the origin.
It is well known that a static charge distribution leads to the electrostatic field; for our point charge at the origin, V(r, t=infinity) = 1/r.
However, if the charge is static, V would be constantly increasing...wouldn't it? So at t=infinity, V would be infinity everywhere? Now, running a simulation (because I suck at PDEs), it appears that while this appears to be true, E = gradV does appear to stay constant...so perhaps this is a red herring, but I would feel much more confident if I could find an analytical solution (or a logical explanation).
Any takers?
2. Apr 14, 2012
### fzero
If the charge distribution is static, then $\dot{\rho}(\mathbf{x})=0$. The scalar potential is
$$V = \int \frac{\rho(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|} d\mathbf{x}',$$
so $\dot{V}=0$ as well. So $V$ is static. Note that when the potential is static, the inhomogenous wave equation reduces to the Poisson equation of electrostatics. For the point charge at the origin, $V\sim 1/r$ at all times. The only divergence is at the origin.
3. Apr 14, 2012
### jjustinn
Right - but that assumes that it was always static, and therefore you're assuming what I want to derive: that's the green function for the Laplacian, not the Wave equation.
Put another way, you're assuming it was static for all time; But if the point charge was turned on any time after t=-infinity, the static approximation won't hold (except possibly at t=infinity, but it seems that at any large but finite time, V will be much larger than 1/r...
So to restate the problem (all scalar factors set to 1)
V,tt = V,xx + V,yy + V,zz + p
p(t<0) = 0
p(t>=0) = point charge at origin
I'm wondering if one trick might be putting a boundary condition so V(r=infinity) = 0; since it only falls off as 1/r that would (I think?) be an extra assumption with no apparent justification (other than the fact it might give the right answer ;) ).
This sort of reminds me of how I understand the process of renormalization in QFT, which of course is exactly that sort of unjustifiable hack, but again, I could be way off base here.
Thanks,
Justin
4. Apr 14, 2012
### vanhees71
Take the retarded Green's function of the wave equation and use it to calculate the electromagnetic field for a static charge distribution. On the one hand that's a weird way to calculate the Coulomb potential, on the other it furthers the understanding of the matter quite well :-).
5. Apr 14, 2012
### fzero
I assumed that it was static because that's what you said it was. If we introduce a time-dependent charge distribution, we have to use the retarded Green function to solve the wave equation. The solution is now
$$V(\mathbf{x},t) = \int \frac{\delta(t'-t + |\mathbf{x}-\mathbf{x}'|/c)}{|\mathbf{x}-\mathbf{x}'|} \rho(\mathbf{x}',t') d\mathbf{x}' dt' .$$
This is derived in, for example, Ch 6 of Jackson.
There's no reason to introduce a boundary condition, since the initial condition is enough. We can do the integral above to find
$$V(\mathbf{x},t) = \begin{cases} 0, & t<|\mathbf{x}|/c, \\ \frac{\rho_0}{|\mathbf{x}|}, & t\geq |\mathbf{x}|/c. \end{cases}$$
At a given distance from the origin, the potential is zero until late enough times that a signal traveling at light speed can reach the point. After that, we have the ordinary static potential. This is as expected. By causality, we can't measure the potential until the signal has reached us. Once it does, the charge is effectively static, so we should measure the static field.
Renormalization is actually conceptually deeper than just removing infinities. It is the realization that physics can depend crucially on the energy scale that we use to probe a system. You can find some context here: http://en.wikipedia.org/wiki/Renormalization_group This Wilsonian viewpoint generally isn't covered until grad-level statistical mechanics or an advanced QFT course.
6. Apr 14, 2012
### vanhees71
This cannot be since then you have an electric field popping out of nothing instantaneously and instantly at any point in space. Of course this comes from the totally unphysical assumption of "jjustinn" concerning his charge distribution: You cannot pop up a point charge out of nothing at $t=0$. This violates not only causality but also the necessary condition of charge conservation and thus gauge invariance. What you can have, is a charge sitting at the origin forever (which is of course also an unrealistic assumption, but it's not in contradiction with any physical laws). Then you get immediately the Coulomb field as it shoud be (here written in its form using non-rationalized Gaussian units).
7. Apr 14, 2012
### jjustinn
First, apologies for my poor phrasing: I didn't mean to suggest you should have read my mind.
I'm familiar with the derivation of the Wave equation green function, and that's another thing that's puzzled me -- the Green function seems to say, as you point out, that the potential at point r goes from 0 to 1/r, then stays at 1/r for all time -- which not only sounds reasonable (since the charge appearing at t=0 is jut as discontinuous), but agrees with observations.
So perhaps my problems are stemming from my lattice-based simulation, where the charge is acting as a constant source, adding to the V around it, and since there is no dissipation / damping, it never goes away.
Or, it might also have something to do with the fact I'm only updating it in the plane, and the 2D Green function isn't a spreading delta function, but IIRC a log function of some sort.
To give some background, I originally thought my simulation was broken, since V was monotonically increasing everywhere (while E stabilized to a constant value) -- but I started reading this (http://psych.colorado.edu/~oreilly/papers/OReilly05_md.pdf), and he mentions adding a empirically-determined damping factor at the (artificial, Sommerfield-condition-ed) boundary "to achieve the 1/r dependence"; before I'd seen this, I'd found that adding a damping term at/near the boundary had a similar effect, but rejected it outright as non-physical.
Anyway, thanks for taking the time to respond. I'll play with this some more and will undoubtedly be back with more dumb questions with crucial unstated assumptions ;)
8. Apr 14, 2012
### jjustinn
Ah hah! I think this explains my problem: (from http://ieeexplore.ieee.org/xpl/logi...l5/97/5487427/05492190.pdf?arnumber=5492190):
Of course, you need a subscription to see the paper >:/
So...this actually brings up anther question, for a new thread: shouldn't our point charge at origin generate the same waves in the XY plane (or any plane through the origin) as a 2D charge would, via radial symmetry?
9. Apr 14, 2012
### jjustinn
Hey vanhees -- I missed yr replies earlier, sandwiched between the other ones. I've also been concerned with the non-physicality of the situation, but it seemed like mathematically it should have been fine.
However, I did find something interesting when trying to make it more physical: by adding an equal/opposite negative charge, its outward monotonically-decreasing negative wave cancels the monotonically-increasing wave from the positive source, making the average value stable, rather than increasing to infinity...but it's not too useful to have a simulation where the net charge has to be zero. This is what initially made me think of "renormalization" (used in the lay 'removing infinities' sense) -- it wasn't until later I found that gradV actually stayed constant.
In any case, would you care to expand on that last comment? If you are uncomfortable with a single charge coming out of nowhere, how about a positron/electron pair pulled out of the vacuum? Or since this is explicitly classical, how about if I separate the charges on a glass rod and a bit of fur at t=0, then fire the glass rod from a cannon into the distance at t= 1? | 2,000 | 8,186 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2018-30 | latest | en | 0.944319 |
https://metanumbers.com/1521247000000 | 1,600,463,789,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400188841.7/warc/CC-MAIN-20200918190514-20200918220514-00523.warc.gz | 479,512,237 | 8,243 | ## 1521247000000
1,521,247,000,000 (one trillion five hundred twenty-one billion two hundred forty-seven million) is an even thirteen-digits composite number following 1521246999999 and preceding 1521247000001. In scientific notation, it is written as 1.521247 × 1012. The sum of its digits is 22. It has a total of 16 prime factors and 784 positive divisors. There are 472,780,800,000 positive integers (up to 1521247000000) that are relatively prime to 1521247000000.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 13
• Sum of Digits 22
• Digital Root 4
## Name
Short name 1 trillion 521 billion 247 million one trillion five hundred twenty-one billion two hundred forty-seven million
## Notation
Scientific notation 1.521247 × 1012 1.521247 × 1012
## Prime Factorization of 1521247000000
Prime Factorization 26 × 56 × 7 × 13 × 73 × 229
Composite number
Distinct Factors Total Factors Radical ω(n) 6 Total number of distinct prime factors Ω(n) 16 Total number of prime factors rad(n) 15212470 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,521,247,000,000 is 26 × 56 × 7 × 13 × 73 × 229. Since it has a total of 16 prime factors, 1,521,247,000,000 is a composite number.
## Divisors of 1521247000000
784 divisors
Even divisors 672 112 56 56
Total Divisors Sum of Divisors Aliquot Sum τ(n) 784 Total number of the positive divisors of n σ(n) 4.72831e+12 Sum of all the positive divisors of n s(n) 3.20706e+12 Sum of the proper positive divisors of n A(n) 6.03101e+09 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.23339e+06 Returns the nth root of the product of n divisors H(n) 252.238 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,521,247,000,000 can be divided by 784 positive divisors (out of which 672 are even, and 112 are odd). The sum of these divisors (counting 1,521,247,000,000) is 4,728,308,226,880, the average is 60,310,053,91.,428.
## Other Arithmetic Functions (n = 1521247000000)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 472780800000 Total number of positive integers not greater than n that are coprime to n λ(n) 17100000 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 56297096841 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 472,780,800,000 positive integers (less than 1,521,247,000,000) that are coprime with 1,521,247,000,000. And there are approximately 56,297,096,841 prime numbers less than or equal to 1,521,247,000,000.
## Divisibility of 1521247000000
m n mod m 2 3 4 5 6 7 8 9 0 1 0 0 4 0 0 4
The number 1,521,247,000,000 is divisible by 2, 4, 5, 7 and 8.
• Abundant
• Polite
• Practical
• Frugal
## Base conversion (1521247000000)
Base System Value
2 Binary 10110001000110001011000110001010111000000
3 Ternary 12101102121022121120101111
4 Quaternary 112020301120301113000
5 Quinary 144411003213000000
6 Senary 3122503510413104
8 Octal 26106130612700
10 Decimal 1521247000000
12 Duodecimal 2069b2661194
20 Vigesimal 2j899df000
36 Base36 jeun6pj4
## Basic calculations (n = 1521247000000)
### Multiplication
n×i
n×2 3042494000000 4563741000000 6084988000000 7606235000000
### Division
ni
n⁄2 7.60624e+11 5.07082e+11 3.80312e+11 3.04249e+11
### Exponentiation
ni
n2 2314192435009000000000000 3520458299180136223000000000000000000 5355486626252884688830081000000000000000000000000 8147017963727322074228694231007000000000000000000000000000000
### Nth Root
i√n
2√n 1.23339e+06 11500.9 1110.58 273.174
## 1521247000000 as geometric shapes
### Circle
Diameter 3.04249e+12 9.55828e+12 7.27025e+24
### Sphere
Volume 1.47465e+37 2.9081e+25 9.55828e+12
### Square
Length = n
Perimeter 6.08499e+12 2.31419e+24 2.15137e+12
### Cube
Length = n
Surface area 1.38852e+25 3.52046e+36 2.63488e+12
### Equilateral Triangle
Length = n
Perimeter 4.56374e+12 1.00207e+24 1.31744e+12
### Triangular Pyramid
Length = n
Surface area 4.0083e+24 4.1489e+35 1.24209e+12
## Cryptographic Hash Functions
md5 c05c0f42487b12879cb8f20c407b2b9b 30ed2369782d3f5d5431fcc1a8b440cc936a49ed 723fb8a76ccaa58fcd6b51797aa9f47b7def007ac062b9d65734b34f8b859fb4 1fb4e35904e06a8e87b90c0ac09c836fedc53df4017fbe6122310e3600991157f8f4638299222d04ab38d5f7014ab969a813785b68287e300e511a3affea3563 6464ebbeef337e6c653fb884972141118f2071ea | 1,741 | 4,796 | {"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} | 3.453125 | 3 | CC-MAIN-2020-40 | latest | en | 0.769976 |
https://brainmass.com/statistics/quantative-analysis-of-data/391110 | 1,516,439,806,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889542.47/warc/CC-MAIN-20180120083038-20180120103038-00260.warc.gz | 648,960,997 | 18,951 | Share
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# Mean, median or mode to represent average
You are responsible for planning the parking needed for a new 256-unit apartment complex, and you're told to base the needs on the statistic 'average number of vehicles per household is 1.9'. Which average (mean, median, mode, midrange) will be helpful to you? Explain.
#### Solution Preview
Hi there,
I agree with you, I think that the mean is the best method of central tendency to use in this case. However, I feel that you should multiply the number of units by the average number of cars to come up with your ...
#### Solution Summary
A discussion of which representation of central tendency is the most appropriate: the mean, median or mode.
\$2.19 | 160 | 730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2018-05 | longest | en | 0.953308 |
https://boxes.khanacademy.org/math/cc-eighth-grade-math/cc-8th-linear-equations-functions/cc-8th-function-intro/v/testing-if-a-relationship-is-a-function | 1,542,453,120,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743353.54/warc/CC-MAIN-20181117102757-20181117124757-00379.warc.gz | 572,295,169 | 44,910 | # □□□□□□□ □□ □ □□□□□□□□□□□□ □□ □ □□□□□□□□
## □□□□□ □□□□□□□□□□
We're asked: Do the points on the graph below represent a function? So in order for the points to represent a function, for every input into our function, we can only get one value. So if we look here, they've graphed the point--it looks like negative 1, 3, so that's the point negative 1, 3. So if we assume that this is our x-axis and that is our f of x axis, and I'm just assuming it's a function, I don't know whether it really is just now, this point is telling us that if you put negative 1 into our function, or that thing that might be a function, or maybe our relation, you'll get a 3 So it's telling us that f of negative 1 is equal to 3. So far it could be a reasonable function. You give me negative 1 and I will map it to 3. Then they have if x is 2, then our value is negative 2. This is the point 2, negative 2, so that still seems consistent with being a function. If you pass me 2, I will map you or I will point you to negative 2. Seems fair enough. Let's see this next value here. This is the point 3 comma 2 right there. So once again, that says that, look, if you give me 3 into my function, into my black box, I will output a 2. Pretty reasonable. No reason why these points can't represent a function so far. Now, what about when we input 4 into the function? Let me do this in magenta. So what happens if I input 4 into my function? So this is 4 right here. Well, according to these points, there's two points that relate to 4 that 4 can be mapped to. I could map it to the point 4 comma 5. So that says if you give me a 4, I'll give you a 5. But it also says if you give me a 4, I could also give you a negative 1 because that's the point 4, negative 1 So this is not a function. It cannot be a function if for some input into the function you could give me two different values. And you can see that right here. And an easy test is to just see, look, for one value I have two points for this relationship. So this cannot be a function. So this is not a function! I'll put an exclamation mark. | 575 | 2,082 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-47 | longest | en | 0.946698 |
http://cosmoquest.org/forum/showthread.php?13591-quot-Relative-motion-quot-falls-apart-when-applied-to-plan/page8&p=288642 | 1,386,521,543,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163066444/warc/CC-MAIN-20131204131746-00008-ip-10-33-133-15.ec2.internal.warc.gz | 40,164,978 | 28,843 | # Thread: "Relative motion" falls apart when applied to plan
1. ## Re: Fortis, Grand Vizier, Maksuta
Originally Posted by A Thousand Pardons
Originally Posted by Fortis
... As F=ma, then the only way that the earth can remain static (i.e. with the acceleration, a=0) for any finite force, F, is if the inertial mass, m, is infinite.
I disagree. Wouldn't that be the only way if it were only just the Sun and Earth in the Universe?
I think the point was that unless Earth's mass is infinite, the Sun and Earth would revolve around a common centre of gravity, in which case the Earth could not be considered "static".
But if that were the point, then that would be begging the question--in other words, asserting what you are trying to show.
2. Order of Kilopi
Join Date
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4,139
## Re: Fortis, Grand Vizier, Maksuta
Originally Posted by A Thousand Pardons
Originally Posted by A Thousand Pardons
Originally Posted by Fortis
... As F=ma, then the only way that the earth can remain static (i.e. with the acceleration, a=0) for any finite force, F, is if the inertial mass, m, is infinite.
I disagree. Wouldn't that be the only way if it were only just the Sun and Earth in the Universe?
I think the point was that unless Earth's mass is infinite, the Sun and Earth would revolve around a common centre of gravity, in which case the Earth could not be considered "static".
But if that were the point, then that would be begging the question--in other words, asserting what you are trying to show.
Unless the forces on the earth (from everything else in the universe) exactly cancel out (which would be a fairly miraculous outcome) then there would be a resultant force acting on the earth. If the inertial mass of the earth is finite (which seems to me to be the sensible assumption to make ), then the earth will experience an acceleration. If something is accelerating, then you'd be hard pushed to argue that it also remains static.
I don't think that this is begging the question.
3. Originally Posted by Grand Vizier
If things're getting away from fluid dynamics, has anyone gone into retrograde planetary motion and the notion of epicycles yet? It's an important historical point - the problems with these were part and parcel of the downfall of Ptolemaic geocentricity even before Kepler and Newton arrived on the scene.
What explanation does the good Dr Jones give for the planets now and then moving backwards in the sky?
I started getting into the issue of the geocentric versus heliocentric views of the solar system and the nature of the planets, the sun, etc. in the geocentric view - but Gary said that all takes care of itself, so I didn't go much further with it. I was curious to see if he would bring up epicycles himself - but no.
Gary has provided no explanation for what anything in space is supposed to be or any method by which you can calculate the motion of any object in space, but he doesn't seem to see why that is a fundamental problem with his argument.
I liked the planetary motion issue because it can be personally tested.
We didn't even begin to get into the nature of stars.
4. Originally Posted by Van Rijn
Originally Posted by Grand Vizier
If things're getting away from fluid dynamics, has anyone gone into retrograde planetary motion and the notion of epicycles yet? It's an important historical point - the problems with these were part and parcel of the downfall of Ptolemaic geocentricity even before Kepler and Newton arrived on the scene.
What explanation does the good Dr Jones give for the planets now and then moving backwards in the sky?
I started getting into the issue of the geocentric versus heliocentric views of the solar system and the nature of the planets, the sun, etc. in the geocentric view - but Gary said that all takes care of itself, so I didn't go much further with it. I was curious to see if he would bring up epicycles himself - but no.
Gary has provided no explanation for what anything in space is supposed to be or any method by which you can calculate the motion of any object in space, but he doesn't seem to see why that is a fundamental problem with his argument.
I liked the planetary motion issue because it can be personally tested.
We didn't even begin to get into the nature of stars.
That's extraordinary. I should have thought that, if these guys really wanted to make a case, finding a refutation of the original arguments against the geocentric view would be the logical place to start.
(For example, with modern computers, the multiplication of epicycles within epicycles might yield to computational solutions. It seems to go against Occam's Razor, true, but the Razor is just a guideline, not a disproof. A proponent of geocentrism might retort that, since we cannot solve the n-body problem, we have to apply endless corrections to our orbit solutions anyway if we are to maintain their predictive accuracy way into the future. But maybe I shouldn't give them ideas [-X )
However, I did enjoy reading Dr Jones' re-evaluation of distance scales based on the number of photons hitting a retinal cell. It was fun looking for the flaw, which is oddly hard to find, but seems obvious once it clicks. But I think that has already been dealt with fully on this thread.
5. ## Re: Fortis, Grand Vizier, Maksuta
Originally Posted by Fortis
Unless the forces on the earth (from everything else in the universe) exactly cancel out (which would be a fairly miraculous outcome) then there would be a resultant force acting on the earth. If the inertial mass of the earth is finite (which seems to me to be the sensible assumption to make ), then the earth will experience an acceleration. If something is accelerating, then you'd be hard pushed to argue that it also remains static.
I don't think that this is begging the question.
Of course it is. Not only are you assuming Newtonian mechanics hold exactly--which the Geocentrists deny--we know that they don't.
6. [quote="Grand Vizier"][quote="Van Rijn"]
Originally Posted by Grand Vizier
If things're getting away from fluid dynamics, has anyone gone into retrograde planetary motion and the notion of epicycles yet? It's an important historical point - the problems with these were part and parcel of the downfall of Ptolemaic geocentricity even before Kepler and Newton arrived on the scene.
Unfortunately, the majority of Geocentrist debates boil down to, "I know the Truth, why don't you just believe me?" :-?
I've only read a few who actually bother to get into the physics of things. Most either point to their religious texts, or take the stance that "it doesn't make sense for the Earth to move."
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## ToSeek,AGN Fuel, Grand Vizier, TinFoilHat
To Seek, AGN Fuel,
Thanks for the clarity of the explanation of the solar eclipse. I believe I can follow you both. It does seem to allow the cause to be something I had previously ruled out. I'll have to continue thinking about it, though. (I will post them to Dr. Jones' site for everyone's benefit there.)
Grand Vizier,
What about the question on my last post?
TinFoilHat wrote:
I'm wondering how Gary (or anyone else using his arguments) can explain the sucessful landing of the multiple probes we've put on the surface of Mars. We can see that Mars rotates. From the rate of rotation and radius of Mars, a quick calculation shows that the surface of mars at the equator is moving at a speed of over 500 miles per hour, relative to the planet as a whole, due to its rotation. We know that Mars has an atmosphere. If that atmosphere did not rotate along with Mars, any point on or near the equator of mars would feel a constant 500MPH wind blowing it sideways. None of the multiple probes we have landed on mars has detected this wind. None of them would be even capable of surviving a landing with a 500MPH crosswind in the first place! What they have shown us is that the atmosphere of Mars, though having intermittent local winds and turbulence, is as a whole rotating at the same speed as the planet Mars, and most likely has been for quite some time.
We can see from Mars that the fixed-earther's understanding of the mechanics of a rotating atmosphere is insufficient to describe the real world.
TinFoilHat, thanks for a good point. But it brings up another question. If, supposing everyone of you is correct on this board that the earth is indeed rotating as you say, as is Mars, does that mean the atmosphere is turning at the same angular velocity as the planet it is on? Like, when you say that the probes to Mars did not encounter 500 mph winds. I believe you. But in accord with what you say, I assume the atmosphere is indeed rotating RTTS (relative to the stars) and is generally still but with localized winds RTTP (relative to the planet), even on Mars, right?
So if that is all true, what is the effect of ENTERING the spinning atmosphere from outer space on these probes? Would they not necessarily have to go from the near perfect stillness of space to the spinning violence of the atmosphere suddenly upon entering it? Would they not encounter the body of the atmosphere, as a whole, that was turning at horrible rotation rates? Wouldn't this have to happen at some time for the probe to land on Mars?
Sincerely,
Gary Shelton
8. ## Re: ToSeek,AGN Fuel, Grand Vizier, TinFoilHat
Originally Posted by Gary Shelton
TinFoilHat, thanks for a good point. But it brings up another question. If, supposing everyone of you is correct on this board that the earth is indeed rotating as you say, as is Mars, does that mean the atmosphere is turning at the same angular velocity as the planet it is on? Like, when you say that the probes to Mars did not encounter 500 mph winds. I believe you. But in accord with what you say, I assume the atmosphere is indeed rotating RTTS (relative to the stars) and is generally still but with localized winds RTTP (relative to the planet), even on Mars, right?
Yes.
Originally Posted by Gary Shelton
So if that is all true, what is the effect of ENTERING the spinning atmosphere from outer space on these probes? Would they not necessarily have to go from the near perfect stillness of space to the spinning violence of the atmosphere suddenly upon entering it? Would they not encounter the body of the atmosphere, as a whole, that was turning at horrible rotation rates? Wouldn't this have to happen at some time for the probe to land on Mars?
They enter the atmosphere at whatever relative speed they have to it. If the probe is descending with a 5,000 mph (just an example, not necessarily a realistic speed) horizontal speed in the same direction as the rotation, the rotation is 500 mph, and the wind is calm, the probe would encounter a 4,500 mph headwind. Probes are designed to withstand that. Remember, too, that the atmosphere is MUCH thinner where the probe first encounters it. A 5,000 mph wind at the edge of the atmosphere doesn't provide much resistance. [Edited to add:] What was meant in the earlier post was that the probe did not find 500 mph winds relative to the planet.
9. ## Re: ToSeek,AGN Fuel, Grand Vizier, TinFoilHat
Originally Posted by Gary Shelton
So if that is all true, what is the effect of ENTERING the spinning atmosphere from outer space on these probes? Would they not necessarily have to go from the near perfect stillness of space to the spinning violence of the atmosphere suddenly upon entering it? Would they not encounter the body of the atmosphere, as a whole, that was turning at horrible rotation rates? Wouldn't this have to happen at some time for the probe to land on Mars?
Sincerely,
Gary Shelton
Yes, of course. However, the rotation speed is a modest issue as far as velocity is concerned. The relative speed difference between the spacecraft and planet is much more substantial. (And why is this rotation rate "horrible"?). Anyway, from:
The lander should come streaking in through the martian atmosphere, going about 12,000 miles per hour. Given atmospheric friction, the outside surface of the heat shield will be as hot as the surface of the sun (1,447 degrees Celsius, or 2,637 degrees Fahrenheit), but the rover will be protected by the heat shield and will stay at about room temperature inside the lander. The heat shield also aerodynamically acts as the first "brake" for the spacecraft, slowing Spirit down by thousands of miles per hour.
The spacecraft comes in at a shallow angle to the planet's surface to increase the time in the upper atmosphere.
Do a google search on "reentry heating" "reentry friction" "heat shield" and "aerobraking" - this is a standard issue for landing spacecraft.
10. Originally Posted by Grand Vizier
(For example, with modern computers, the multiplication of epicycles within epicycles might yield to computational solutions. It seems to go against Occam's Razor, true, but the Razor is just a guideline, not a disproof. A proponent of geocentrism might retort that, since we cannot solve the n-body problem, we have to apply endless corrections to our orbit solutions anyway if we are to maintain their predictive accuracy way into the future. But maybe I shouldn't give them ideas
Now you've done it! Coming to a website near you: The Neo-Ptolemaic model of the solar system. Actually, it wouldn't surprise me if somebody could come up with quite a good software orrery using many, many "wheels within wheels," basing the approximations on direct observation. But it wouldn't be able to predict the effects of future perterbations or allow you to map the orbit of a newly found asteroid after a few observations, because it wouldn't be based on physical principles.
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## Re: Fortis, Grand Vizier, Maksuta
Originally Posted by A Thousand Pardons
Originally Posted by Fortis
Unless the forces on the earth (from everything else in the universe) exactly cancel out (which would be a fairly miraculous outcome) then there would be a resultant force acting on the earth. If the inertial mass of the earth is finite (which seems to me to be the sensible assumption to make ), then the earth will experience an acceleration. If something is accelerating, then you'd be hard pushed to argue that it also remains static.
I don't think that this is begging the question.
Of course it is. Not only are you assuming Newtonian mechanics hold exactly--which the Geocentrists deny--we know that they don't.
So, a non-relativistic body (and I think the earth can be regarded as non-relativistic ), of finite inertial mass, will not accelerate when it experiences a net force? Even in QFT, where forces are mediated by the exchange of particles such as photons, etc., momentum is conserved at the vertices.
I'm pretty sure that the argument doesn't require Newtonian mechanics to be 100% correct. Even if the acceleration experienced by the earth is ten orders of magnitude less than that predicted with classical mechanics, then it still moves.
Finally, if geocentrists just deny anything that doesn't fit in with their world-view, then there is absolutely no point in arguing with them. (And perhaps that is the case.)
12. ## Re: Fortis, Grand Vizier, Maksuta
Originally Posted by Fortis
I'm pretty sure that the argument doesn't require Newtonian mechanics to be 100% correct. Even if the acceleration experienced by the earth is ten orders of magnitude less than that predicted with classical mechanics, then it still moves.
I agree. ATP/milli, we know that Newtonian mechanics is not a complete theory of the cosmos. We know that General Relativity is the more accurate explanation. But the plain fact is, it's harder to understand GR, the maths is a lot more complex, and for non-relativistic conditions, Newton is a very, very good first approximation when we're trying to demonstrate basic physics to geocentrists who haven't even got that skill yet.
Some parts of this thread reminded me of the famous exchange in scene 7 of Monty Python's The Life of Brian:
FRANCIS: Yeah. I think Judith's point of view is very valid, Reg, provided the Movement never forgets that it is the inalienable right of every man--
STAN: Or woman.
FRANCIS: Or woman... to rid himself--
STAN: Or herself.
FRANCIS: Or herself.
REG: Agreed.
FRANCIS: Thank you, brother.
STAN: Or sister.
FRANCIS: Or sister. Where was I?
REG: I think you'd finished.
FRANCIS: Oh. Right.
REG: Furthermore, it is the birthright of every man--
STAN: Or woman.
REG: Why don't you shut up about women, Stan. You're putting us off.
Calm. :-#
13. ## Re: Fortis, Grand Vizier, Maksuta
Originally Posted by Yorkshireman
Originally Posted by Fortis
I'm pretty sure that the argument doesn't require Newtonian mechanics to be 100% correct. Even if the acceleration experienced by the earth is ten orders of magnitude less than that predicted with classical mechanics, then it still moves.
I agree. ATP/milli, we know that Newtonian mechanics is not a complete theory of the cosmos. We know that General Relativity is the more accurate explanation. But the plain fact is, it's harder to understand GR, the maths is a lot more complex, and for non-relativistic conditions, Newton is a very, very good first approximation when we're trying to demonstrate basic physics to geocentrists who haven't even got that skill yet.
Indeed! And virtually all geocentrtics refute relativity anyway and assume Newtonian mechanics is 100% accurate for the sake of their deomonstrations that the earth can not be rotating. If you're going to show that a proof is inconsistent you must stick to the premises of the proof.
If they want to demonstrate that the eath can not be moving according to GR that's a whole different kettle of fish.
14. ## Re: Fortis, Grand Vizier, Maksuta
Originally Posted by Fortis
Originally Posted by A Thousand Pardons
Originally Posted by Fortis
Unless the forces on the earth (from everything else in the universe) exactly cancel out (which would be a fairly miraculous outcome) then there would be a resultant force acting on the earth. If the inertial mass of the earth is finite (which seems to me to be the sensible assumption to make ), then the earth will experience an acceleration. If something is accelerating, then you'd be hard pushed to argue that it also remains static.
I don't think that this is begging the question.
Of course it is. Not only are you assuming Newtonian mechanics hold exactly--which the Geocentrists deny--we know that they don't.
So, a non-relativistic body (and I think the earth can be regarded as non-relativistic ), of finite inertial mass, will not accelerate when it experiences a net force? Even in QFT, where forces are mediated by the exchange of particles such as photons, etc., momentum is conserved at the vertices.
You are deriving a general principle where none has been asserted.
I'm pretty sure that the argument doesn't require Newtonian mechanics to be 100% correct. Even if the acceleration experienced by the earth is ten orders of magnitude less than that predicted with classical mechanics, then it still moves.
If general relativity allows any reference frame, then it allows even the Earth-fixed, Earth-centered one, and it doesn't move, in that reference frame. If one ignores that possibility, then one is ignoring basic notions of commonly accepted physics. If one assumes something that we know isn't true, then of course you can prove anything you want--but that doesn't make it true, or the proof valid.
Finally, if geocentrists just deny anything that doesn't fit in with their world-view, then there is absolutely no point in arguing with them. (And perhaps that is the case.)
That is my point. One shouldn't deny the use of general relativity just because it annoyingly doesn't fit one's own argument. Nevermind that one or another geocentrist also rejects general relativity, that doesn't mean that we can.
Originally Posted by Yorkshireman
Originally Posted by Fortis
I'm pretty sure that the argument doesn't require Newtonian mechanics to be 100% correct. Even if the acceleration experienced by the earth is ten orders of magnitude less than that predicted with classical mechanics, then it still moves.
I agree.
I disagree, see my answer to Fortis. Unless you have a good proof that General Relativity is not the best explanation of physics, to date, you're out of luck on that.
ATP/milli, we know that Newtonian mechanics is not a complete theory of the cosmos. We know that General Relativity is the more accurate explanation. But the plain fact is, it's harder to understand GR, the maths is a lot more complex, and for non-relativistic conditions, Newton is a very, very good first approximation when we're trying to demonstrate basic physics to geocentrists who haven't even got that skill yet.
I think that is the wrong approach--denying the results of generally accepted physics.
Some parts of this thread reminded me of the famous exchange in scene 7 of Monty Python's The Life of Brian:
::snip::
REG: Why don't you shut up about women, Stan. You're putting us off.
Calm. :-#
I assumed that that last part was the point you were making. Sorry to be so inconvenient.
Originally Posted by worzel
Indeed! And virtually all geocentrtics refute relativity anyway and assume Newtonian mechanics is 100% accurate for the sake of their deomonstrations that the earth can not be rotating. If you're going to show that a proof is inconsistent you must stick to the premises of the proof.
If they want to demonstrate that the eath can not be moving according to GR that's a whole different kettle of fish.
We have the same responsibilities.
15. ATP, shall we take this argument to another topic? We're going off thread here now.
Cheers, Rob.
16. Originally Posted by Yorkshireman
ATP, shall we take this argument to another topic? We're going off thread here now.
It's been discussed before many times, and I've been involved in a lot of them, so I'd probably join another.
But I've only responded to comments here, when others have felt they were germane to the argument of the thread, so I guess I'm not the one to ask.
17. Originally Posted by Van Rijn
Originally Posted by Grand Vizier
(For example, with modern computers, the multiplication of epicycles within epicycles might yield to computational solutions. It seems to go against Occam's Razor, true, but the Razor is just a guideline, not a disproof. A proponent of geocentrism might retort that, since we cannot solve the n-body problem, we have to apply endless corrections to our orbit solutions anyway if we are to maintain their predictive accuracy way into the future. But maybe I shouldn't give them ideas
Now you've done it! Coming to a website near you: The Neo-Ptolemaic model of the solar system. Actually, it wouldn't surprise me if somebody could come up with quite a good software orrery using many, many "wheels within wheels," basing the approximations on direct observation. But it wouldn't be able to predict the effects of future perterbations or allow you to map the orbit of a newly found asteroid after a few observations, because it wouldn't be based on physical principles.
Agreed on the physical principles. But we have to take ourselves back to a time before Kepler and Newton. Those physical principles have not been elucidated, all we've got is some gravitational acceleration stuff from Galileo. Thought experiment - suppose we worked up such a program and sent it back in time on a laptop - no need to go any further, geocentricity proven...
Incidentally, I was thinking that such a program might relate to Fourier analysis. I'm totally handwaving - but sine waves, circular epicycles - simple harmonic motion seems to creep in here. I can sort of envisage the simple elliptical orbits we use as being somehow the sum of a infinitely converging series of epicycles... (In other words, maybe they are actually the same thing - in a mathematical, not a physical sense, of course.) But I don't know if this would fly.
I don't know what to do about perturbations. Now we're in physical theory, means linkages, means gravity (or springs or something ) and the whole house of cards falls over.
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Meteora wrote:
They enter the atmosphere at whatever relative speed they have to it. If the probe is descending with a 5,000 mph (just an example, not necessarily a realistic speed) horizontal speed in the same direction as the rotation, the rotation is 500 mph, and the wind is calm, the probe would encounter a 4,500 mph headwind. Probes are designed to withstand that. Remember, too, that the atmosphere is MUCH thinner where the probe first encounters it. A 5,000 mph wind at the edge of the atmosphere doesn't provide much resistance.
Van Rijn wrote (Quoting on the Spirit Landing on Mars):
Quote:
The lander should come streaking in through the martian atmosphere, going about 12,000 miles per hour. Given atmospheric friction, the outside surface of the heat shield will be as hot as the surface of the sun (1,447 degrees Celsius, or 2,637 degrees Fahrenheit), but the rover will be protected by the heat shield and will stay at about room temperature inside the lander. The heat shield also aerodynamically acts as the first "brake" for the spacecraft, slowing Spirit down by thousands of miles per hour.
If space probes can slow down to the 500 mph of Mars, that's one thing. But have we ever successfully sent probes in the Jovian atmosphere? Someone in this thread reported wind speeds of 28,000 mph there.
I suppose we have spacecraft able to meet that speed (how fast have we gone, BTW?), but it seems to me that the differences between localized currents there would be so great as to cause more problems than slowing down to the speed of the Martian atmosphere. Yes? No?
Sincerely,
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Originally Posted by Gary Shelton
If space probes can slow down to the 500 mph of Mars, that's one thing. But have we ever successfully sent probes in the Jovian atmosphere? Someone in this thread reported wind speeds of 28,000 mph there.
Actually, yes. Galileo.
Originally Posted by Gary Shelton
I suppose we have spacecraft able to meet that speed (how fast have we gone, BTW?), but it seems to me that the differences between localized currents there would be so great as to cause more problems than slowing down to the speed of the Martian atmosphere. Yes? No?
As far as I know, there's no evidence of this for Mars.
20. Originally Posted by Gary Shelton
I suppose we have spacecraft able to meet that speed (how fast have we gone, BTW?)
Without doing the math, I can't tell you an exact upper limit for our current tech, but assuming the fuel, we could get pretty fast (though it would take time to accelerate) because there is no (okay extremely little to the point of negigable) air resistance in space. This means that thrust forward is totally turned into acceleration (compare with a plane were at a certain speed the drag cannot be overcome by acceleration). This only stops when the craft's speed is at relativistic speeds and the increased accelertion is countered by the increase of mass due to the increased speed (obviously minus the fuel as well.) At the moment we can't send huge amounts of fuel into space because it plainly isn't worth it, so instead we slingshot probes about planets, transferring a tiny amount of the planet's momentum to the probe. This can get a probe up to very good speeds Cassini's flyby Jupiter accelerated it by 4,500 mph. The Huygen's probe entered Titin's atmosphere at 13,275 mph so these speeds are pretty good, and we could get them faster in needed, the big problem with doing it is that we need to slow down to go into orbit, and that requires a heap of fuel (or air braking which require heat shields etc.) For instance Cassini needed to fire its engines for 94 minutes to reduce its velocity by the 1,400 mph it needed to achieve orbit.
Originally Posted by Gary Shelton
but it seems to me that the differences between localized currents there would be so great as to cause more problems than slowing down to the speed of the Martian atmosphere. Yes? No?
Nope, there is no difference. What you do is enter the atmosphere at about 35,000 mph and air brake until you reach the local air speed (obviously using parachuttes for the last part of that.) What it would affect is the landing area, but since there is not really a solid surface in Jupiter, any probe we were dropping into it wouldn't be heading for a landing zone anyways so that wouldn't matter. Once the probe was down to the local speed it would be stationary compared to the air about it so it wouldn't matter if the air was travelling at 26,000mph or 19,000mph.
Originally Posted by Taibak
Actually, yes. Galileo.
I guess that depends on if you mean "successful" as in survived entry, or "successful" as in it did what it was expected to do.
21. Actually you don't need to go very far to see some serious speed. At the end of its missions the Space Shuttle reenters the earths atmosphere at around 17,000 mph.
22. I'd also add that at the end of the Apollo missions, the Command Module entered the Earth's atmosphere at 24,500 mph, but had shed almost all that speed by the time the drogue chutes deployed at 5 miles altitude. But since Gary thinks Apollo is faked I might be wasting my time here...!
23. Originally Posted by Gary Shelton
If space probes can slow down to the 500 mph of Mars, that's one thing. But have we ever successfully sent probes in the Jovian atmosphere? Someone in this thread reported wind speeds of 28,000 mph there.
No, Jovian wind speeds can be pretty high (a few hundred mph), but nothing like that. Now, the velocity of the Galileo atmospheric probe was a bit under 30,000 mph on entry. Obviously, a spacecraft can only take these conditions for a limited period of time, but it only needs to do it long enough to drop the velocity to something where a parachute, rocket, etc. comes into play.
24. Originally Posted by Van Rijn
Originally Posted by Gary Shelton
If space probes can slow down to the 500 mph of Mars, that's one thing. But have we ever successfully sent probes in the Jovian atmosphere? Someone in this thread reported wind speeds of 28,000 mph there.
No, Jovian wind speeds can be pretty high (a few hundred mph), but nothing like that. Now, the velocity of the Galileo atmospheric probe was a bit under 30,000 mph on entry.
[...]
I've seen Gary make this mix-up somewhere else. He is confusing local wind speeds (generally low) with the velocity of spacecraft relative to planetary bodies (generally high). Perhaps the distinction should be made more explicit.
25. Originally Posted by Yorkshireman
I'd also add that at the end of the Apollo missions, the Command Module entered the Earth's atmosphere at 24,500 mph, but had shed almost all that speed by the time the drogue chutes deployed at 5 miles altitude. But since Gary thinks Apollo is faked I might be wasting my time here...!
The crew of Apollo 10 reached 24,791 miles per hour at re-entry on 26th May 1969. This remains the fastest that man has ever travelled (relative to the Earth, anyhow!).
Must be true - it's in the Guiness Book of Records!
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Grand Vizier wrote:
I've seen Gary make this mix-up somewhere else. He is confusing local wind speeds (generally low) with the velocity of spacecraft relative to planetary bodies (generally high). Perhaps the distinction should be made more explicit.
I understand your point. I should have stated that the atmosphere of Jupiter was turning with the planet at those speeds. Relative to the "ground" of Jupiter, it might only be like what Van Rijn says, yes.
I have one more question for you all. If we have ascertained that a space craft can enter even the Jovian atmosphere without problem, because of its ability to have its flight trajectory and speed altered to "merge" with the planet's airspeed, what about meteors and other things striking the atmosphere of earth?
These things would not have the ability to adjust to the earth's turning atmosphere. They just come at us, though I'm sure the angles vary. Would we not see them deflected substantially upon a straight in entry?
Sincerely,
27. Originally Posted by Gary Shelton
These things would not have the ability to adjust to the earth's turning atmosphere. They just come at us, though I'm sure the angles vary. Would we not see them deflected substantially upon a straight in entry?
The velocities due to rotation reach a maximum near the equator, around 1000 mph, whereas the meteoroids come in much, much faster than that. Still, some break apart, and produce elements that zigzag a bit sometimes.
28. Originally Posted by Gary Shelton
I have one more question for you all. If we have ascertained that a space craft can enter even the Jovian atmosphere without problem, because of its ability to have its flight trajectory and speed altered to "merge" with the planet's airspeed, what about meteors and other things striking the atmosphere of earth?
These things would not have the ability to adjust to the earth's turning atmosphere. (snip)
In case we gave the impression that the Shuttle, Apollo, Mars landers etc 'steer' and adjust their speed into the planets' atmosphere, this is not really accurate. Essentially all of the braking is caused by the atmosphere itself and the spacecraft is quite a passive recipient of this resisting force. It's true that the Shuttle can steer to an extent (it does some big S-turns coming in through the atmosphere, to shed speed faster), but it's more a case of trying to keep a level of control over what the atmosphere is doing to the spacecraft anyway.
Spacecraft like Apollo has much less ability to 'steer' and in effect re-entered the atmosphere on a ballistic trajectory, like a big meteor. They could effect a small amount of control by angling their heatshield and firing attitude jets, but that was it. They had nothing like enough fuel to substantially alter their speed, unaided. There is not much difference between that and a 'natural' meteor, (apart from the heatshield of course.)
Having said that, meteors tend to enter the atmosphere even faster than the examples we've given above. I don't have a figure to hand, but some come in at 40-50 miles per second or even faster.
Some do skip out of the atmosphere entirely, like a rock skipping off a pond (grazers), others break up as ATP says above.
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Rob wrote:
Having said that, meteors tend to enter the atmosphere even faster than the examples we've given above. I don't have a figure to hand, but some come in at 40-50 miles per second or even faster.
A T P wrote:
The velocities due to rotation reach a maximum near the equator, around 1000 mph, whereas the meteoroids come in much, much faster than that. Still, some break apart, and produce elements that zigzag a bit sometimes.
I just figured what 40 miles per second means. It is 144,000 miles per hour. Wow, pretty fast. So to clarify your point. You two are saying that due to the tremendous velocities of these things that they overcome the relatively minor speed of the air into which they enter, right?
Assuming your response in the positive, I guess the question that is begged is have there ever been any sufficiently slower moving objects strike the atmosphere such that they were visibly "grabbed" by the atmosphere and deflected in their path in the direction of the atmosphere's rotation? Is there filmed evidence of such an event?
Further, have any space missions reported seeing any slow moving space objects "grabbed" and hurtled along with the air in its rotational flow?
Sincerely,
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Originally Posted by Gary Shelton
Assuming your response in the positive, I guess the question that is begged is have there ever been any sufficiently slower moving objects strike the atmosphere such that they were visibly "grabbed" by the atmosphere and deflected in their path in the direction of the atmosphere's rotation? Is there filmed evidence of such an event?
I'd say any landing of a spacecraft would qualify. They come in at high speed relative to the atmosphere and lose speed until they descend vertically hanging from parachutes. Or landing like a plane, in the case of the Space Shuttle.
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I. Basic math.
II. Pricing and Hedging.
III. Explicit techniques.
IV. Data Analysis.
V. Implementation tools.
1 Finite differences.
2 Gauss-Hermite Integration.
3 Asymptotic expansions.
4 Monte-Carlo.
5 Convex Analysis.
A. Basic concepts of convex analysis.
B. Caratheodory's theorem.
C. Relative interior.
D. Recession cone.
E. Intersection of nested convex sets.
F. Preservation of closeness under linear transformation.
G. Weierstrass Theorem.
H. Local minima of convex function.
I. Projection on convex set.
J. Existence of solution of convex optimization problem.
K. Partial minimization of convex functions.
L. Hyperplanes and separation.
M. Nonvertical separation.
N. Minimal common and maximal crossing points.
O. Minimax theory.
P. Saddle point theory.
Q. Polar cones.
R. Polyhedral cones.
S. Extreme points.
T. Directional derivative and subdifferential.
U. Feasible direction cone, tangent cone and normal cone.
V. Optimality conditions.
W. Lagrange multipliers for equality constraints.
X. Fritz John optimality conditions.
Y. Pseudonormality.
Z. Lagrangian duality.
[. Conjugate duality.
VI. Basic Math II.
VII. Implementation tools II.
VIII. Bibliography
Notation. Index. Contents.
Lagrange multipliers for equality constraints.
e are considering the following problem.
(Minimization with equality constraints)
where the and are smooth functions .
Proposition
(Existence of Lagrange multipliers for equality constraints). Let be a local minimum of the problem ( Minimization with equality constraints ) and then there are scalars such that
Proof
The condition implies where the matrix consists of the columns Hence, Equivalently, Therefore, We next show that Indeed, we already established that is a subspace, hence, . Therefore, if then . But and we proved already that
Therefore, According to the proposition ( Minimum of a smooth function ), hence, and the conclusion of the proposition follows.
The condition states that the consists of directions tangent to the level surfaces of crossing the . For example,
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## Two "count(if" functions in one formula
Hey everyone,
i need a function which calculates two "if"s in one forumla:
Count(if(medium='cpc',order_id)) -->> i need another if which is (order_status_id=2,5,order_id)
thanks alot!
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1 Solution
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## Re: Two "count(if" functions in one formula
Try this.
Count({<Medium = {"CPC"},Order_Status_ID = {"2","5"}>} Order_ID)
Regards,
Kaushik Solanki
5 Replies
Not applicable
## Re: Two "count(if" functions in one formula
Hi,
Try with this:
Count(if(medium='cpc', order_id, if(order_status_id=2, 5, order_id)))
or
Count(if(medium='cpc', if(order_status_id=2, 5, order_id)))
That depends on what you need.
Regards,
Ricardo
Not applicable
## Re: Two "count(if" functions in one formula
True, it depends very much on what's needed - which is not quite clear. Another version of Ricardo's logic:
Count(if(medium='cpc' AND order_status_id=2, 5, order_id))
Regards,
Michael
Not applicable
## Re: Two "count(if" functions in one formula
Hey,
i want to count the orders ( order_id) that have in their table the value for "medium=cpc" and for "order_status_id=2,5"
Count(if(medium='cpc' AND order_status_id=2, 5, order_id)) -> that gives me all the orders.
Not applicable
## Re: Two "count(if" functions in one formula
Try this.
Count({<Medium = {"CPC"},Order_Status_ID = {"2","5"}>} Order_ID)
Regards,
Kaushik Solanki
Not applicable
## Re: Two "count(if" functions in one formula
Thanks everyone for helping out!!!!
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posted by .
A car travels along a straight stretch of road. It proceeds for 14 mi at 50 mi/h, then 28.3 mi at 48 mi/h, and finally 37.4 mi at 38.3 mi/h.
What is the car’s average velocity during
the entire trip?
Can u show the steps and answer to it, thanks.
• physics -
To get the average speed during the entire trip, divide the total distance by the total time.
part 1: 14mi/50mph = 0.28 hr
part 2: 28.3mi/48mph = 0.59 hr
part 3: 37.4mi/38.3mph = 0.98 hr
total: 79.7mi/1.85hr = 43.1mi/hr
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More Similar Questions | 859 | 2,726 | {"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-2018-05 | latest | en | 0.907974 |
https://physics.stackexchange.com/questions/335721/difference-between-greens-function-and-density-matrix/564508 | 1,632,005,322,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056578.5/warc/CC-MAIN-20210918214805-20210919004805-00644.warc.gz | 518,558,421 | 38,006 | # Difference between green's function and density matrix
What is the basic difference between green's function or propagator of given system and density matrix (in the position basis) of the same system ? Can some one explain the difference between with some example, say one dimensional harmonic oscillator.
The one-particle Green function in the time domain (omitting time ordering) of a non-interacting system can be written in terms of the eigenfunctions of the Hamiltonian $$\psi_n$$
$$G(\mathbf{r},t;\mathbf{r}',t') = \sum_n \psi_n(\mathbf{r}) \, \psi_n^*(\mathbf{r}') \; e^{-i/\hbar E_n (t-t')}$$,
where the exponential at the end is the time evolution operator $$U(t,t')$$ that dictates the evolution of the wavefunction $$\Psi(\mathbf{r}',t')$$ to $$\Psi(\mathbf{r},t)$$ in infinitesimal time intervals.
Compare the previous equation with the first order density matrix of a monodeterminantal wavefunction
$$\gamma_1(\mathbf{r};\mathbf{r}') = \sum_n \psi_n(\mathbf{r}) \, \psi_n^*(\mathbf{r}')$$.
The first order density matrix is obtained from the Green function by contour integration
$$\gamma_1(\mathbf{r};\mathbf{r}') = - \frac{1}{2\pi i} \int_C G(\mathbf{r},t;\mathbf{r}',t) \, \mathrm{d}t$$.
A way to think about the Green function is as a correlation function. While the Green function measures spatio-temporal correlations, the density matrix measures spatial correlations.
• of course you are assuming pures states here... Aug 11 '20 at 13:23
Density matrix can be represented as a Green's function at equal times. E.g., in one-particle case: $$\rho(\mathbf{r},\mathbf{r}',t)=G(\mathbf{r},t; \mathbf{r}',t'=t).$$
Those familiar with diagrammatic expansion might enjoy the review by Rammer (not the better known by Rammer&Smith) where the diagrammatic expansion is obtained for the density matrix. | 501 | 1,826 | {"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": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-39 | latest | en | 0.804911 |
https://priorprobability.com/2017/02/23/what-is-a-light-year/ | 1,558,702,725,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257624.9/warc/CC-MAIN-20190524124534-20190524150534-00042.warc.gz | 592,166,353 | 23,145 | What is a light-year?
The journal Nature recently published this letter announcing the exciting discovery of seven exoplanets 40 light-years away from Earth: only 235 trillion miles away! Does this mean it would take us 40 Earth years to reach those planets if we could travel as fast as the speed of light? The infographic below says light-years are a measure of distance, not a measure of time. Bonus question: Why can’t we travel faster than the speed of light, or can we?
Credit: niks1rocks, via SlideShare
When I’m not blogging, I am a business law professor at the University of Central Florida.
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4 Responses to What is a light-year?
1. CHC says:
From Wikipedia: “The fastest outward-bound spacecraft yet sent, Voyager 1, has covered 1/600th of a light-year in 30 years and is currently moving at 1/18,000th the speed of light. At this rate, a journey to Proxima Centauri would take 80,000 years.” Proxima C. is our closest star system.
• Love Wikipedia … But what does it mean to say a light-year is a unit of distance and not a unit of time, since it takes X amount of time to travel Y distance?
• Craig C says:
Well, one can argue with the folks who named it a light-year I guess. But it is a distance. IF one could travel at the speed of light, THEN one could traverse this distance in a year. Otherwise, it would take longer, depending on one’s speed.
I propose a new measure that combines distance and speed, called the light-bagel. It is defined by the seven feet that you are away from the front of the counter at Brugger’s Bagels, and all you want is one sesame and one everything, in a bag, to go, but you have to wait in line for all the people who want theirs toasted, with a smear of this or that, and so you are waiting in line, only seven feet away from talking to the uniformed minimum-wage bagel-prepper, for what seems like seven light-bagels, to place your simple order.
• So close, yet so far … | 473 | 1,994 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2019-22 | latest | en | 0.95506 |
http://www.jiskha.com/display.cgi?id=1342809409 | 1,495,552,783,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607647.16/warc/CC-MAIN-20170523143045-20170523163045-00200.warc.gz | 536,048,675 | 3,703 | # Trig
posted by on .
There is an airplane at an altitude of 12000 ft. The angle of depression is 1 degree. How far on the ground is the plane.
• Trig - ,
Did you make a sketch?
if so , you will see that
tan 1° = 12000/x , where x is the distance along the ground
x = 12000/tan 1° = .....
"How far on the ground is the plane ? " sounds like a very strange question. | 108 | 371 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-22 | latest | en | 0.941401 |
https://www.lmfdb.org/GaloisGroup/30T42 | 1,723,205,634,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640763425.51/warc/CC-MAIN-20240809110814-20240809140814-00225.warc.gz | 678,883,312 | 11,162 | # Properties
Label 30T42 Degree $30$ Order $180$ Cyclic no Abelian no Solvable yes Primitive no $p$-group no Group: $S_3\times D_{15}$
Show commands: Magma
magma: G := TransitiveGroup(30, 42);
## Group action invariants
Degree $n$: $30$ magma: t, n := TransitiveGroupIdentification(G); n; Transitive number $t$: $42$ magma: t, n := TransitiveGroupIdentification(G); t; Group: $S_3\times D_{15}$ Parity: $-1$ magma: IsEven(G); Primitive: no magma: IsPrimitive(G); magma: NilpotencyClass(G); $\card{\Aut(F/K)}$: $1$ magma: Order(Centralizer(SymmetricGroup(n), G)); Generators: (16,18,17)(19,21,20)(22,24,23)(25,27,26)(28,30,29), (2,3)(4,15)(5,14)(6,13)(7,12)(8,11)(9,10)(16,28)(17,30)(18,29)(19,27)(20,26)(21,25)(22,23), (1,17,6,19,7,23,12,25,13,28)(2,18,4,20,8,24,10,26,14,29)(3,16,5,21,9,22,11,27,15,30) magma: Generators(G);
## Low degree resolvents
|G/N|Galois groups for stem field(s)
$2$: $C_2$ x 3
$4$: $C_2^2$
$6$: $S_3$ x 2
$10$: $D_{5}$
$12$: $D_{6}$ x 2
$20$: $D_{10}$
$30$: $D_{15}$
$36$: $S_3^2$
$60$: $D_5\times S_3$, $D_{30}$
Resolvents shown for degrees $\leq 47$
## Subfields
Degree 2: $C_2$
Degree 3: None
Degree 5: $D_{5}$
Degree 6: $S_3^2$
Degree 10: $D_{10}$
Degree 15: None
## Low degree siblings
45T13
Siblings are shown with degree $\leq 47$
A number field with this Galois group has no arithmetically equivalent fields.
## Conjugacy classes
Label Cycle Type Size Order Index Representative 1A $1^{30}$ $1$ $1$ $0$ $()$ 2A $2^{15}$ $3$ $2$ $15$ $( 1,24)( 2,22)( 3,23)( 4,27)( 5,25)( 6,26)( 7,29)( 8,30)( 9,28)(10,16)(11,17)(12,18)(13,20)(14,21)(15,19)$ 2B $2^{15}$ $15$ $2$ $15$ $( 1,24)( 2,23)( 3,22)( 4,19)( 5,21)( 6,20)( 7,18)( 8,17)( 9,16)(10,28)(11,30)(12,29)(13,26)(14,25)(15,27)$ 2C $2^{14},1^{2}$ $45$ $2$ $14$ $( 1, 3)( 4,14)( 5,13)( 6,15)( 7,11)( 8,10)( 9,12)(16,29)(17,28)(18,30)(19,25)(20,27)(21,26)(22,24)$ 3A $3^{10}$ $2$ $3$ $20$ $( 1, 3, 2)( 4, 6, 5)( 7, 9, 8)(10,12,11)(13,15,14)(16,17,18)(19,20,21)(22,23,24)(25,26,27)(28,29,30)$ 3B $3^{10}$ $2$ $3$ $20$ $( 1, 2, 3)( 4, 5, 6)( 7, 8, 9)(10,11,12)(13,14,15)(16,17,18)(19,20,21)(22,23,24)(25,26,27)(28,29,30)$ 3C $3^{5},1^{15}$ $4$ $3$ $10$ $(16,18,17)(19,21,20)(22,24,23)(25,27,26)(28,30,29)$ 5A1 $5^{6}$ $2$ $5$ $24$ $( 1,12, 6,13, 7)( 2,10, 4,14, 8)( 3,11, 5,15, 9)(16,27,21,30,22)(17,25,19,28,23)(18,26,20,29,24)$ 5A2 $5^{6}$ $2$ $5$ $24$ $( 1, 6, 7,12,13)( 2, 4, 8,10,14)( 3, 5, 9,11,15)(16,21,22,27,30)(17,19,23,25,28)(18,20,24,26,29)$ 6A $6^{5}$ $6$ $6$ $25$ $( 1,22, 3,24, 2,23)( 4,25, 6,27, 5,26)( 7,30, 9,29, 8,28)(10,17,12,16,11,18)(13,21,15,20,14,19)$ 6B $6^{5}$ $30$ $6$ $25$ $( 1,23, 3,24, 2,22)( 4,21, 6,19, 5,20)( 7,17, 9,18, 8,16)(10,30,12,28,11,29)(13,25,15,26,14,27)$ 10A1 $10^{3}$ $6$ $10$ $27$ $( 1,18, 6,20, 7,24,12,26,13,29)( 2,16, 4,21, 8,22,10,27,14,30)( 3,17, 5,19, 9,23,11,25,15,28)$ 10A3 $10^{3}$ $6$ $10$ $27$ $( 1,26, 7,18,13,24, 6,29,12,20)( 2,27, 8,16,14,22, 4,30,10,21)( 3,25, 9,17,15,23, 5,28,11,19)$ 15A1 $15^{2}$ $2$ $15$ $28$ $( 1, 8,15, 6,10, 3, 7,14, 5,12, 2, 9,13, 4,11)(16,23,29,21,25,18,22,28,20,27,17,24,30,19,26)$ 15A2 $15^{2}$ $2$ $15$ $28$ $( 1,10, 5,13, 8, 3,12, 4,15, 7, 2,11, 6,14, 9)(16,25,20,30,23,18,27,19,29,22,17,26,21,28,24)$ 15A4 $15^{2}$ $2$ $15$ $28$ $( 1,14,11, 7, 4, 3,13,10, 9, 6, 2,15,12, 8, 5)(16,28,26,22,19,18,30,25,24,21,17,29,27,23,20)$ 15A7 $15^{2}$ $2$ $15$ $28$ $( 1, 4, 9,12,14, 3, 6, 8,11,13, 2, 5, 7,10,15)(16,19,24,27,28,18,21,23,26,30,17,20,22,25,29)$ 15B1 $15,5^{3}$ $4$ $15$ $26$ $( 1, 7,13, 6,12)( 2, 8,14, 4,10)( 3, 9,15, 5,11)(16,24,28,21,26,17,22,29,19,27,18,23,30,20,25)$ 15B2 $15,5^{3}$ $4$ $15$ $26$ $( 1,12, 6,13, 7)( 2,10, 4,14, 8)( 3,11, 5,15, 9)(16,26,19,30,24,17,27,20,28,22,18,25,21,29,23)$ 15C1 $15^{2}$ $4$ $15$ $28$ $( 1, 5, 8,12,15, 2, 6, 9,10,13, 3, 4, 7,11,14)(16,19,24,27,28,18,21,23,26,30,17,20,22,25,29)$ 15C2 $15,5^{3}$ $4$ $15$ $26$ $( 1, 6, 7,12,13)( 2, 4, 8,10,14)( 3, 5, 9,11,15)(16,20,23,27,29,17,21,24,25,30,18,19,22,26,28)$ 15C4 $15^{2}$ $4$ $15$ $28$ $( 1,11, 4,13, 9, 2,12, 5,14, 7, 3,10, 6,15, 8)(16,25,20,30,23,18,27,19,29,22,17,26,21,28,24)$ 15C7 $15,5^{3}$ $4$ $15$ $26$ $( 1,13,12, 7, 6)( 2,14,10, 8, 4)( 3,15,11, 9, 5)(16,29,25,22,20,17,30,26,23,21,18,28,27,24,19)$ 30A1 $30$ $6$ $30$ $29$ $( 1,27, 9,18,14,23, 6,30,11,20, 2,25, 7,16,15,24, 4,28,12,21, 3,26, 8,17,13,22, 5,29,10,19)$ 30A7 $30$ $6$ $30$ $29$ $( 1,30,15,26,10,23, 7,21, 5,18, 2,28,13,27,11,24, 8,19, 6,16, 3,29,14,25,12,22, 9,20, 4,17)$ 30A11 $30$ $6$ $30$ $29$ $( 1,21,11,29, 4,23,13,16, 9,26, 2,19,12,30, 5,24,14,17, 7,27, 3,20,10,28, 6,22,15,18, 8,25)$ 30A13 $30$ $6$ $30$ $29$ $( 1,16, 5,20, 8,23,12,27,15,29, 2,17, 6,21, 9,24,10,25,13,30, 3,18, 4,19, 7,22,11,26,14,28)$
magma: ConjugacyClasses(G);
Malle's constant $a(G)$: $1/10$
## Group invariants
Order: $180=2^{2} \cdot 3^{2} \cdot 5$ magma: Order(G); Cyclic: no magma: IsCyclic(G); Abelian: no magma: IsAbelian(G); Solvable: yes magma: IsSolvable(G); Nilpotency class: not nilpotent Label: 180.29 magma: IdentifyGroup(G); Character table:
1A 2A 2B 2C 3A 3B 3C 5A1 5A2 6A 6B 10A1 10A3 15A1 15A2 15A4 15A7 15B1 15B2 15C1 15C2 15C4 15C7 30A1 30A7 30A11 30A13 Size 1 3 15 45 2 2 4 2 2 6 30 6 6 2 2 2 2 4 4 4 4 4 4 6 6 6 6 2 P 1A 1A 1A 1A 3A 3B 3C 5A2 5A1 3B 3A 5A2 5A1 15A7 15A2 15A4 15A1 15C1 15C4 15B1 15C2 15B2 15C7 15A1 15A7 15A4 15A2 3 P 1A 2A 2B 2C 1A 1A 1A 5A2 5A1 2A 2B 10A3 10A1 5A2 5A2 5A1 5A1 5A2 5A2 5A1 5A1 5A2 5A1 10A1 10A3 10A1 10A3 5 P 1A 2A 2B 2C 3A 3B 3C 1A 1A 6A 6B 2A 2A 3B 3B 3B 3B 3C 3C 3A 3C 3A 3C 6A 6A 6A 6A Type 180.29.1a R $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ 180.29.1b R $1$ $−1$ $−1$ $1$ $1$ $1$ $1$ $1$ $1$ $−1$ $−1$ $−1$ $−1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $−1$ $−1$ $−1$ $−1$ 180.29.1c R $1$ $−1$ $1$ $−1$ $1$ $1$ $1$ $1$ $1$ $−1$ $1$ $−1$ $−1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $−1$ $−1$ $−1$ $−1$ 180.29.1d R $1$ $1$ $−1$ $−1$ $1$ $1$ $1$ $1$ $1$ $1$ $−1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ $1$ 180.29.2a R $2$ $0$ $2$ $0$ $−1$ $2$ $−1$ $2$ $2$ $0$ $−1$ $0$ $0$ $2$ $2$ $2$ $2$ $−1$ $−1$ $−1$ $−1$ $−1$ $−1$ $0$ $0$ $0$ $0$ 180.29.2b R $2$ $2$ $0$ $0$ $2$ $−1$ $−1$ $2$ $2$ $−1$ $0$ $2$ $2$ $−1$ $−1$ $−1$ $−1$ $2$ $2$ $−1$ $−1$ $−1$ $−1$ $−1$ $−1$ $−1$ $−1$ 180.29.2c R $2$ $−2$ $0$ $0$ $2$ $−1$ $−1$ $2$ $2$ $1$ $0$ $−2$ $−2$ $−1$ $−1$ $−1$ $−1$ $2$ $2$ $−1$ $−1$ $−1$ $−1$ $1$ $1$ $1$ $1$ 180.29.2d R $2$ $0$ $−2$ $0$ $−1$ $2$ $−1$ $2$ $2$ $0$ $1$ $0$ $0$ $2$ $2$ $2$ $2$ $−1$ $−1$ $−1$ $−1$ $−1$ $−1$ $0$ $0$ $0$ $0$ 180.29.2e1 R $2$ $2$ $0$ $0$ $2$ $2$ $2$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $2$ $0$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ 180.29.2e2 R $2$ $2$ $0$ $0$ $2$ $2$ $2$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $2$ $0$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ 180.29.2f1 R $2$ $−2$ $0$ $0$ $2$ $2$ $2$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $−2$ $0$ $−ζ5−1−ζ5$ $−ζ5−2−ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $−ζ5−2−ζ52$ $−ζ5−1−ζ5$ $−ζ5−2−ζ52$ $−ζ5−1−ζ5$ 180.29.2f2 R $2$ $−2$ $0$ $0$ $2$ $2$ $2$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $−2$ $0$ $−ζ5−2−ζ52$ $−ζ5−1−ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $ζ5−2+ζ52$ $ζ5−1+ζ5$ $−ζ5−1−ζ5$ $−ζ5−2−ζ52$ $−ζ5−1−ζ5$ $−ζ5−2−ζ52$ 180.29.2g1 R $2$ $2$ $0$ $0$ $2$ $−1$ $−1$ $ζ15−6+ζ156$ $ζ15−3+ζ153$ $−1$ $0$ $ζ15−3+ζ153$ $ζ15−6+ζ156$ $ζ15−7+ζ157$ $ζ15−1+ζ15$ $ζ15−2+ζ152$ $ζ15−4+ζ154$ $ζ15−3+ζ153$ $ζ15−6+ζ156$ $ζ15−2+ζ152$ $ζ15−4+ζ154$ $ζ15−7+ζ157$ $ζ15−1+ζ15$ $ζ15−4+ζ154$ $ζ15−2+ζ152$ $ζ15−1+ζ15$ $ζ15−7+ζ157$ 180.29.2g2 R $2$ $2$ $0$ $0$ $2$ $−1$ $−1$ $ζ15−6+ζ156$ $ζ15−3+ζ153$ $−1$ $0$ $ζ15−3+ζ153$ $ζ15−6+ζ156$ $ζ15−2+ζ152$ $ζ15−4+ζ154$ $ζ15−7+ζ157$ $ζ15−1+ζ15$ $ζ15−3+ζ153$ $ζ15−6+ζ156$ $ζ15−7+ζ157$ $ζ15−1+ζ15$ $ζ15−2+ζ152$ $ζ15−4+ζ154$ $ζ15−1+ζ15$ $ζ15−7+ζ157$ $ζ15−4+ζ154$ $ζ15−2+ζ152$ 180.29.2g3 R $2$ $2$ $0$ $0$ $2$ $−1$ $−1$ $ζ15−3+ζ153$ $ζ15−6+ζ156$ $−1$ $0$ $ζ15−6+ζ156$ $ζ15−3+ζ153$ $ζ15−4+ζ154$ $ζ15−7+ζ157$ $ζ15−1+ζ15$ $ζ15−2+ζ152$ $ζ15−6+ζ156$ $ζ15−3+ζ153$ $ζ15−1+ζ15$ $ζ15−2+ζ152$ $ζ15−4+ζ154$ $ζ15−7+ζ157$ $ζ15−2+ζ152$ $ζ15−1+ζ15$ $ζ15−7+ζ157$ $ζ15−4+ζ154$ 180.29.2g4 R $2$ $2$ $0$ $0$ $2$ $−1$ $−1$ $ζ15−3+ζ153$ $ζ15−6+ζ156$ $−1$ $0$ $ζ15−6+ζ156$ $ζ15−3+ζ153$ $ζ15−1+ζ15$ $ζ15−2+ζ152$ $ζ15−4+ζ154$ $ζ15−7+ζ157$ $ζ15−6+ζ156$ $ζ15−3+ζ153$ $ζ15−4+ζ154$ $ζ15−7+ζ157$ $ζ15−1+ζ15$ $ζ15−2+ζ152$ $ζ15−7+ζ157$ $ζ15−4+ζ154$ $ζ15−2+ζ152$ $ζ15−1+ζ15$ 180.29.2h1 R $2$ $−2$ $0$ $0$ $2$ $−1$ $−1$ $ζ15−6+ζ156$ $ζ15−3+ζ153$ $1$ $0$ $−ζ15−3−ζ153$ $−ζ15−6−ζ156$ $ζ15−7+ζ157$ $ζ15−1+ζ15$ $ζ15−2+ζ152$ $ζ15−4+ζ154$ $ζ15−3+ζ153$ $ζ15−6+ζ156$ $ζ15−2+ζ152$ $ζ15−4+ζ154$ $ζ15−7+ζ157$ $ζ15−1+ζ15$ $−ζ15−4−ζ154$ $−ζ15−2−ζ152$ $−ζ15−1−ζ15$ $−ζ15−7−ζ157$ 180.29.2h2 R $2$ $−2$ $0$ $0$ $2$ $−1$ $−1$ $ζ15−6+ζ156$ $ζ15−3+ζ153$ $1$ $0$ $−ζ15−3−ζ153$ $−ζ15−6−ζ156$ $ζ15−2+ζ152$ $ζ15−4+ζ154$ $ζ15−7+ζ157$ $ζ15−1+ζ15$ $ζ15−3+ζ153$ $ζ15−6+ζ156$ $ζ15−7+ζ157$ $ζ15−1+ζ15$ $ζ15−2+ζ152$ $ζ15−4+ζ154$ $−ζ15−1−ζ15$ $−ζ15−7−ζ157$ $−ζ15−4−ζ154$ $−ζ15−2−ζ152$ 180.29.2h3 R $2$ $−2$ $0$ $0$ $2$ $−1$ $−1$ $ζ15−3+ζ153$ $ζ15−6+ζ156$ $1$ $0$ $−ζ15−6−ζ156$ $−ζ15−3−ζ153$ $ζ15−4+ζ154$ $ζ15−7+ζ157$ $ζ15−1+ζ15$ $ζ15−2+ζ152$ $ζ15−6+ζ156$ $ζ15−3+ζ153$ $ζ15−1+ζ15$ $ζ15−2+ζ152$ $ζ15−4+ζ154$ $ζ15−7+ζ157$ $−ζ15−2−ζ152$ $−ζ15−1−ζ15$ $−ζ15−7−ζ157$ $−ζ15−4−ζ154$ 180.29.2h4 R $2$ $−2$ $0$ $0$ $2$ $−1$ $−1$ $ζ15−3+ζ153$ $ζ15−6+ζ156$ $1$ $0$ $−ζ15−6−ζ156$ $−ζ15−3−ζ153$ $ζ15−1+ζ15$ $ζ15−2+ζ152$ $ζ15−4+ζ154$ $ζ15−7+ζ157$ $ζ15−6+ζ156$ $ζ15−3+ζ153$ $ζ15−4+ζ154$ $ζ15−7+ζ157$ $ζ15−1+ζ15$ $ζ15−2+ζ152$ $−ζ15−7−ζ157$ $−ζ15−4−ζ154$ $−ζ15−2−ζ152$ $−ζ15−1−ζ15$ 180.29.4a R $4$ $0$ $0$ $0$ $−2$ $−2$ $1$ $4$ $4$ $0$ $0$ $0$ $0$ $−2$ $−2$ $−2$ $−2$ $−2$ $−2$ $1$ $1$ $1$ $1$ $0$ $0$ $0$ $0$ 180.29.4b1 R $4$ $0$ $0$ $0$ $−2$ $4$ $−2$ $2ζ5−2+2ζ52$ $2ζ5−1+2ζ5$ $0$ $0$ $0$ $0$ $2ζ5−1+2ζ5$ $2ζ5−2+2ζ52$ $2ζ5−1+2ζ5$ $2ζ5−2+2ζ52$ $−ζ5−1−ζ5$ $−ζ5−2−ζ52$ $−ζ5−1−ζ5$ $−ζ5−2−ζ52$ $−ζ5−1−ζ5$ $−ζ5−2−ζ52$ $0$ $0$ $0$ $0$ 180.29.4b2 R $4$ $0$ $0$ $0$ $−2$ $4$ $−2$ $2ζ5−1+2ζ5$ $2ζ5−2+2ζ52$ $0$ $0$ $0$ $0$ $2ζ5−2+2ζ52$ $2ζ5−1+2ζ5$ $2ζ5−2+2ζ52$ $2ζ5−1+2ζ5$ $−ζ5−2−ζ52$ $−ζ5−1−ζ5$ $−ζ5−2−ζ52$ $−ζ5−1−ζ5$ $−ζ5−2−ζ52$ $−ζ5−1−ζ5$ $0$ $0$ $0$ $0$ 180.29.4c1 R $4$ $0$ $0$ $0$ $−2$ $−2$ $1$ $2ζ15−6+2ζ156$ $2ζ15−3+2ζ153$ $0$ $0$ $0$ $0$ $2ζ15−7+2ζ157$ $2ζ15−1+2ζ15$ $2ζ15−2+2ζ152$ $2ζ15−4+2ζ154$ $−ζ15−3−ζ153$ $−ζ15−6−ζ156$ $−ζ15−2−ζ152$ $−ζ15−4−ζ154$ $−ζ15−7−ζ157$ $−ζ15−1−ζ15$ $0$ $0$ $0$ $0$ 180.29.4c2 R $4$ $0$ $0$ $0$ $−2$ $−2$ $1$ $2ζ15−6+2ζ156$ $2ζ15−3+2ζ153$ $0$ $0$ $0$ $0$ $2ζ15−2+2ζ152$ $2ζ15−4+2ζ154$ $2ζ15−7+2ζ157$ $2ζ15−1+2ζ15$ $−ζ15−3−ζ153$ $−ζ15−6−ζ156$ $−ζ15−7−ζ157$ $−ζ15−1−ζ15$ $−ζ15−2−ζ152$ $−ζ15−4−ζ154$ $0$ $0$ $0$ $0$ 180.29.4c3 R $4$ $0$ $0$ $0$ $−2$ $−2$ $1$ $2ζ15−3+2ζ153$ $2ζ15−6+2ζ156$ $0$ $0$ $0$ $0$ $2ζ15−4+2ζ154$ $2ζ15−7+2ζ157$ $2ζ15−1+2ζ15$ $2ζ15−2+2ζ152$ $−ζ15−6−ζ156$ $−ζ15−3−ζ153$ $−ζ15−1−ζ15$ $−ζ15−2−ζ152$ $−ζ15−4−ζ154$ $−ζ15−7−ζ157$ $0$ $0$ $0$ $0$ 180.29.4c4 R $4$ $0$ $0$ $0$ $−2$ $−2$ $1$ $2ζ15−3+2ζ153$ $2ζ15−6+2ζ156$ $0$ $0$ $0$ $0$ $2ζ15−1+2ζ15$ $2ζ15−2+2ζ152$ $2ζ15−4+2ζ154$ $2ζ15−7+2ζ157$ $−ζ15−6−ζ156$ $−ζ15−3−ζ153$ $−ζ15−4−ζ154$ $−ζ15−7−ζ157$ $−ζ15−1−ζ15$ $−ζ15−2−ζ152$ $0$ $0$ $0$ $0$
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http://kwiznet.com/p/takeQuiz.php?ChapterID=293&CurriculumID=3&NQ=8&Num=3.5 | 1,718,537,427,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861659.47/warc/CC-MAIN-20240616105959-20240616135959-00561.warc.gz | 19,873,412 | 3,816 | Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
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Quiz Timer | 385 | 1,435 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-26 | latest | en | 0.82425 |
https://blogs.scientificamerican.com/roots-of-unity/jordan-ellenbergs-favorite-theorem/ | 1,695,288,339,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233505362.29/warc/CC-MAIN-20230921073711-20230921103711-00687.warc.gz | 169,721,454 | 18,126 | Today’s episode of My Favorite Theorem features University of Wisconsin mathematician Jordan Ellenberg. You can listen here or at kpknudson.com. We had a few audio issues in this episode, so if you missed anything, there’s a transcript here.
Ellenberg’s favorite theorem is Fermat’s Little Theorem, a not-so-little result in number theory. It states that if p is a prime number (a positive whole number that is divisible by exactly two distinct factors: itself and 1), and a is any whole number, then ap has a remainder of a when divided by p. As shorthand, mathematicians would say ap is congruent to a mod p. (Mod, short for modulo, just means the remainder when divided by a number. So the number 19 is congruent to 4 mod 5 or 1 mod 3 because it’s 4 more than 15, which is a multiple of 5, and 1 more than 18, which is a multiple of 3.)
Ellenberg stated the theorem in the special case that a=2. Then the theorem becomes the statement that when p is prime, 2p is congruent to 2 mod p, or 2has a remainder of 2 when divided by p. You can verify this easily for yourself for a few small prime numbers p: 23=8, which has a remainder of 2 when divided by 3. 25=32, which has a remainder of 2 when divided by 5, and so on.
As you’ll hear in the episode, Ellenberg relates Fermat's little theorem to the coefficients of Pascal’s triangle. This triangle (poorly named because it was studied by Persian, Indian, and Chinese mathematicians centuries before Pascal got his hands on it) shows up in a variety of different guises in mathematics. To get the entries in Pascal’s triangle, you start with a 1 by itself in row 0. (Labeling this row 0 instead of 1 makes bookkeeping a little more convenient.) Then each subsequent entry is the sum of the numbers diagonally above it.
Remembering that we start with row 0, the numbers in row n sum up to 2n.
Ellenberg describes how you can look at the entries in prime-numbered rows and practically observe the theorem, at least for the a=2 case, falling out in front of you.
As a resident of Wisconsin, Ellenberg felt a duty to pair his theorem with cheese. He went with gjetost, a caramel-colored Norwegian cheese. You’ll have to listen to the episode to find out why this is the perfect match for Fermat’s Little Theorem.
You can find Ellenberg online at his website JordanEllenberg.com and his Twitter account @JSEllenberg. He is the author of the pop math book How Not to Be Wrong.
You can find more information about the mathematicians and theorems featured in this podcast, along with other delightful mathematical treats, at kpknudson.com and here at Roots of Unity. A transcript is available here. You can subscribe and review the podcast on iTunes and other podcast delivery systems. We love to hear from our listeners, so please drop us a line at myfavoritetheorem@gmail.com. Kevin Knudson’s handle on Twitter is @niveknosdunk, and mine is @evelynjlamb. The show itself also has a Twitter feed: @myfavethm and a Facebook page. Join us next time to learn another fascinating piece of mathematics.
Previously on My Favorite Theorem: | 753 | 3,085 | {"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} | 4.4375 | 4 | CC-MAIN-2023-40 | latest | en | 0.945882 |
http://whjm.math.uzh.ch/Vol39-3.html | 1,539,987,591,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512460.41/warc/CC-MAIN-20181019212313-20181019233813-00268.warc.gz | 402,194,550 | 8,694 | ## HOUSTON JOURNAL OF MATHEMATICS
### Electronic Edition Vol. 39, No. 3 , 2013
Editors: D. Bao (San Francisco, SFSU), D. Blecher (Houston), Bernhard G. Bodmann (Houston), H. Brezis (Paris and Rutgers), B. Dacorogna (Lausanne), K. Davidson (Waterloo), M. Dugas (Baylor), M. Gehrke (LIAFA, Paris7), C. Hagopian (Sacramento), R. M. Hardt (Rice), Y. Hattori (Matsue, Shimane), W. B. Johnson (College Station), M. Rojas (College Station), Min Ru (Houston), S.W. Semmes (Rice).
Managing Editor: K. Kaiser (Houston)
Contents
Lee, Tsiu-Kwen, National Taiwan University, Taipei 106, Taiwan (tklee@math.ntu.edu.tw) and Liu, Kun-Shan, National Taiwan university, Taipei 106, Taiwan (kunshanliu@ntu.edu.tw).
Generalized skew derivations with algebraic values of bounded degree, pp. 733-740.
ABSTRACT.Let R be a prime algebra over a field F and let g be a nonzero generalized skew derivation of R. We show that if g(x) is algebraic over F of bounded degree for all x in R, then R is a primitive ring with a minimal idempotent e such that eRe is a finite-dimensional central division algebra.
Chun, Sangmin, Seoul National University, Seoul 151-747, Republic of Korea (schun@snu.ac.kr), and Anderson, D. D., The University of Iowa, Iowa City, IA 52242 (dan-anderson@uiowa.edu).
Irreducible elements in commutative rings with zero-divisors, II, pp. 741-752.
ABSTRACT. Let R be a commutative ring with identity. For elements a and b of R, a and b are associates (resp., strong associates, very strong associates) if (a) = (b) (resp., a = ub for some unit u, a and b are associated and either a = b = 0 or a is nonzero and a = rb implies r is a unit). A nonunit a of R is irreducible (resp., strongly irreducible, very strongly irreducible) if a = bc implies a is associate to b or c (resp., a is strongly associate to b or c, a is very strongly associate to a or c) and a is m-irreducible if (a) is a maximal element of the set of proper principal ideals of R. The ring R is said to be atomic (resp., strongly atomic, very strongly atomic, m-atomic) if each nonzero nonunit of R is a finite product of irreducible (resp., strongly irreducible, very strongly irreducible, m-irreducible) elements. In this paper we collect the known various characterizations of the different types of irreducible elements and give a number of new ones. We also continue the investigation of the various forms of atomicity.
Di Bartolo, Alfonso, Dipartimento di Matematica e Informatica, Via Archirafi 34, I-90123 Palermo, Italy (alfonso@math.unipa.it), Falcone, Giovanni, Dipartimento di Matematica e Informatica, Via Archirafi 34, I-90123 Palermo, Italy (giovanni.falcone@unipa.it), and Strambach, Karl, Department Mathematik, Cauerstrasse 11, D-91058 Erlangen, Germany (strambach@mi.uni-erlangen.de).
Near-rings and groups of affine mappings, pp. 753-780.
ABSTRACT. We classify semi-topological locally compact and semi-algebraic near-rings R where the set of non-invertible elements of R forms an ideal I of R such that the multiplicative group of R/I acts sharply transitively on the set of non-zero elements of I. To achieve our results we use as a main tool the classi cation of locally compact and algebraic (2, 2)-transformation groups given in two previous papers.
Escassut, Alain, Laboratoire de Mathematiques, UMR 6620, Université Blaise Pascal, 63171 Aubiere, France (alain.escassut@math.univ-bpclermont.fr) and Ojeda, Jacqueline, Departamento de Matemática, Facultad de Ciencias Físicas y Matemáticas, Universidad de Concepcion, Concepcion, Chile ( jacqojeda@udec.cl).
Branched values and quasi-exceptional values for p-adic meromorphic functions, pp. 781-795.
ABSTRACT. Let K be an algebraically closed field of characteristic 0, complete with respect to an ultrametric absolute value. We show that a transcendental meromorphic function in K or an "unbounded" meromorphic function inside an open disk cannot admit more than 4 perfectly branched values and a transcendental meromorphic function in K cannot admit more that 3 values aj such that all zeroes of f-aj are multiple. An unbounded analytic function inside an open disk cannot admit more than 2 perfectly branched values. And an entire function cannot admit more than 1 perfectly branched value. Completing a previous result by K. Boussaf and J. Ojeda, we prove that given a transcendental meromorphic function f in K, if f admits 0 and ∞ as perfectly branched values, then the function assumes all non-zero values infinitely often. Similarly, if f is an "unbounded" meromorphic function in an "open" disk, if the residue characteristic p is different from 2 and if all zeroes and poles are of even order, but finitely many, then the function assumes all non-zero values infinitely often.
Teresa Monteiro Fernandes, CMAF, Universidade de Lisboa Bloco C6, P 2, Campo Grande, 1749-16, Lisboa, Portugal (tmf@ptmat.fc.ul.pt).
Microsupport of tempered solutions of shd-modules associated to smooth morphisms, pp. 797-821.
ABSTRACT. Let f from X to Y be a smooth morphism of complex manifolds. Let G be any constructible object on X locally isomorphic to the inverse image by f of a constructible object on Y. We prove that the microsupport of the solution complex of a coherent D-module in Kashiwara-Schapira's tempered holomorphic functions (resp.tempered holomorphic microfunctions) associated to G is contained in the 1-characteristic variety (resp. in the 1-microcharacteristic variety) associated to f.
Czerwińska, Małgorzata M., The University of Memphis, Memphis, TN 38152 (mmczrwns@memphis.edu), Kamińska, Anna, The University of Memphis, Memphis, TN 38152 (kaminska@memphis.edu) and Kubiak, Damian, The University of Memphis, Memphis, TN 38152 (dmkubiak@memphis.edu).
Exposed and strongly exposed points in symmetric spaces of measurable operators, pp. 823-852.
ABSTRACT. We investigate the relationships between exposed or strongly exposed points of the unit ball of an order continuous symmetric function space E, and of the unit ball of the space of τ-measurable operators E(M,τ) associated to a semifinite von Neumann algebra M with trace τ. We prove that an operator x is an exposed or strongly exposed point of the unit ball of the symmetric space of measurable operators E(M,τ) if and only if its singular value function μ(x) is an exposed or strongly exposed point of the unit ball in E, respectively.
Dosi Anar, Middle East Technical University Northern Cyprus Campus, Guzelyurt KKTC, Mersin 10, Turkey (dosiev@metu.edu.tr), (dosiev@yahoo.com)
Quantum cones and their duality, pp. 853-887.
ABSTRACT. In this paper we investigate the duality properties of quantum cones. We prove a bipolar theorem for quantum cones and classify all inflated quantum orders over Paulsen's construction as quantum topologies compatible with the given duality.
Mikhail I. Ostrovskii, Department of Mathematics and Computer Science, St. John’s University, 8000 Utopia Parkway, Queens, NY 11439, USA (ostrovsm@stjohns.edu).
Different forms of metric characterizations of classes of Banach spaces, pp. 889-906.
ABSTRACT. For each sequence X of finite-dimensional Banach spaces there exists a sequence H of finite connected unweighted graphs with maximum degree 3 such that the following conditions on a Banach space Y are equivalent: (1) Y admits uniformly isomorphic embeddings of elements of the sequence X. (2) Y admits uniformly bilipschitz embeddings of elements of the sequence H.
Reinhold, Karin, University at Albany, 1400 Washington Avenue, Albany, NY, 12222 (reinhold@albany.edu) and Savvopoulou, Anna K., Indiana University South Bend, 1700 Mishawaka Avenue, South Bend, In, USA, 46634 (annsavvo@iusb.edu).
Variation and Oscillation inequalities for convolution products, pp. 907-918.
ABSTRACT. We establish variation and oscillation inequalities for convolution products of probability measures defined on the integers.
Yasheng Ye, Dept. of Math, University of Shanghai for Science and Technology, Shanghai 200093, P.R.China (yashengye@yahoo.com.cn), Hiaojun Liu, Dept. of Math, University of Shanghai for Science and Technology, Shanghai 200093, P.R.China (xiaojunliu2007@hotmail.com) , and Xuecheng Pang, Dept. of Math，East China Normal University, Shanghai 200241, P.R.China (xcpang@math.ecnu.edu.cn).
A normal criterion about two families of meromorphic functions concerning shared values, pp. 919-928.
ABSTRACT. In this paper, we discuss the normality of two families of functions concerning shared values. In particular, we prove the following result: Let F be a family of meromorphic functions on a domain D, all of whose zeros have multiplicity at least k, and for every f(z) in F, f=0 iff the k-th derivative of f(z)=1, where k is a positive integer. Let F1 and F2 be two subsequences of F, and F2 be normal on D. If for every f(z) in F1, there exists g(z) in F2, such that f=0 iff g=0, then F1 is normal.
Barroso, Ana Cristina, Faculdade de Ciências da Universidade de Lisboa, Departamento de Matemática and CMAF, Av. Prof. Gama Pinto, 2, 1649-003 Lisboa, Portugal (abarroso@ptmat.fc.ul.pt), Croce, Gisella, LMAH, Le Havre University, 25 rue Philippe Lebon, 76063 Le Havre, France (gisella.croce@univ-lehavre.fr), and Ribeiro, Ana Margarida, Departamento de Matemática and CMA, Faculdade de Ciências e Tecnologia da Universidade Nova de Lisboa, Quinta da Torre, 2829-516 Caparica, Portugal (amfr@fct.unl.pt).
Sufficient conditions for existence of solutions to vectorial differential inclusions and applications, pp. 929-967.
ABSTRACT.In this paper we discuss the existence of solutions to vectorial differential inclusions, refining a result proved in Dacorogna and Marcellini [1996, Théorèmes d'existence dans les cas scalaire et vectoriel pour les équations de Hamilton-Jacobi, C. R. Acad. Sci. Paris Sér. I Math., 322(3):237--240]. We investigate sufficient conditions for existence, more flexible than those available in the literature, so that important applications can be fitted in the theory. We also study some of these applications.
Jianjun, Zhang, Mathematics and Information Technology School, Jiangsu Institute of Education, Nanjing , 210013 P. R.China (zhangjianjun1982@163.com) and Liangwen, Liao, Department of Mathematics, Nanjing University, Nanjing, 210093 P. R. China (maliao@nju.edu.cn).
On Malmquist type theorem of complex difference equations, pp. 969-981.
ABSTRACT. In this paper, we will give a result on complex difference equations which is reminiscent of the classical Malmquist theorem in complex differential equations.
Ralowski, Robert and Szymon Zeberski, Wroclaw University of Technology, 50-370 Wroclaw, Poland, (robert.ralowski@pwr.wroc.pl), (szymon.zeberski@pwr.wroc.pl).
Generalized Luzin sets, pp. 983-993.
ABSTRACT. In this paper we investigate the notion of a generalized (I,J) - Luzin set. This notion generalizes the standard notion of a Luzin set and a Sierpinski set. We find set theoretical conditions which imply the existence of a generalized (I,J) - Luzin set.
We show how to construct a large family of pairwise non-equivalent (I,J) - Luzin sets. We find a class of forcings which preserves the property of being a (I,J) - Luzin set.
Escobedo, Raúl, López, María de J.,Facultad de Ciencias Físico Matemáticas, Benemérita Universidad Autónoma de Puebla, Ave. San Claudio y Río Verde, Ciudad Universitaria, San Manuel, Puebla, Pue., 72570, México, (escobedo@fcfm.buap.mx), (mtoriz@fcfm.buap.mx) and Tenorio, Jesús F., Universidad Tecnológica de la Mixteca, carretera a Acatlima km. 2.5, Huajuapan de León, Oaxaca, 69000, México, (jtenorio@mixteco.utm.mx).
Universality of maps on suspensions over products of span zero continua, pp. 995-1004.
ABSTRACT. We prove that the induced map to the topological suspension of a product of maps from continua onto span zero continua is universal. It follows that topological suspension and cone over a product of span zero (chainable) continua have the fixed point property.
Pol, Elzbieta, Institute of Mathematics, University of Warsaw, Banacha 2, 02-097 Warszawa, Poland (E.Pol@mimuw.edu.pl) and Pol, Roman, Institute of Mathematics, University of Warsaw, Banacha 2, 02-097 Warszawa, Poland (R.Pol@mimuw.edu.pl).
A connected, locally connected infinite metric space without separable sets of positive dimension, pp. 1005-1012.
ABSTRACT. We give an example of a non-separable connected and locally connected metrizable space whose all nonempty connected separable subspaces are singletons. This answers a question of T.Banakh, M.Vovk and M.R.Wojcik.
Bonanzinga, Maddalena, Dipartimento di Matematica e Informatica, Universita' di Messina, Via F. Stagno d'Alcontres N.31, 98166 Messina, Italy (mbonanzinga@unime.it).
On the Hausdorff number of a topological space, pp. 1013-1030.
ABSTRACT.A new cardinal function, called Hausdorff number, is defined. Some known cardinal inequalities for Hausdorff spaces are generalized in terms of the Hausdorff number.
Niemiec, Piotr, Jagiellonian University, Institute of Mathematics, ul. Lojasiewicza 6, 30-348 Kraków, Poland (piotr.niemiec@uj.edu.pl)
Ultrametrics, extending of Lipschitz maps and nonexpansive selections, pp. 1031-1050.
ABSTRACT.Ultrametric spaces are characterized (among all metric spaces) in aspects of extending of Lipschitz maps and existence of certain nonexpansive selections. Metric spaces every closed nonempty subset of which is a nonexpansive retract are fully characterized. Total orders on ultrametric spaces which induce metrics and partial orders in the class of compact metric spaces (up to isometry) are investigated. Certain results on isometric embeddings of the hyperspace of an ultrametric space into the space of nonexpansive maps of the space into itself and on upper bounded families of compact spaces are proved.
R. A. McCoy, Department of Mathematics, Virginia Tech, Blacksburg VA 24061-0123, U.S.A. (mccoy@math.vt.edu), Varun Jindal, Department of Mathematics, Indian Institute of Technology Delhi, New Delhi 110016, India (varunjindal.iitd@gmail.com), and S. Kundu, Department of Mathematics, Indian Institute of Technology Delhi, New Delhi 110016, India (skundu@maths.iitd.ernet.in).
Homeomorphism spaces under uniform and fine topologies, pp. 1051-1066.
ABSTRACT. A study is made of the countability and connectedness properties of the space H(X) of self-homeomorphisms from a metric space X onto itself, where H(X) has either the uniform topology or the fine topology. Also for the case that X is Euclidean n-space, three different natural compatible metrics are used to generate three different uniform topologies on H(X). These three uniform homeomorphism spaces are shown to be not homeomorphic to each other when n is greater than 1; and, for such X, these spaces are also compared to H(X) with the fine, point-open and compact-compact-open topologies.
Chatyrko, Vitalij , Dept. of Mathematics, Linkoping University, 581 83 Linkoping, Sweden (vitja@mai.liu.se) and Karassev, Alexandre, Dept. of Computer Science and Mathematics, Nipissing University, 100 College Drive, P.O. Box 5002, North Bay, ON, P1B 8L7, Canada (alexandk@nipissingu.ca).
On mtrizable remainders of locally compact separable metrizable spaces, pp. 1067-1081.
ABSTRACT. In this paper we describe those locally compact noncompact separable metrizable spaces X for which the class R(X) of all metrizable remainders of X consists of all metrizable non-empty compacta. Then we show that for any pair X and X'of locally compact noncompact connected separable metrizable spaces, either R(X) ⊂ R(X') or R(X') ⊂ R(X). | 4,104 | 15,539 | {"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-2018-43 | latest | en | 0.786138 |
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### Topic: IXZ-500 Gyro Issues (Read 1 time)previous topic - next topic
#### hunter7890
##### Feb 15, 2012, 12:04 am
Hello Everyone,
I'm trying to calculate angles using the IXZ-500 gyro but I seem to be having issues properly calculating the angles. For example if I rotate the gyro 90* my calculated angle ends up being around 30. It also seems to be calculating different rates in the negative direction opposed to positive, ie: returning to "zero" positions doesn't give a calculated angle of zero. My Code is below, I'm using trapezoid rule integration with a sampling time of 100 ms. Where the angle should be:
angle=angle + (Z_old+Z_new)*.1
-------------------
2
Where Z_old and Z_new are angles using the conversion: z*(5000mv) (*/s)
---------- * --------
1024bit 9.1mv
I'm using the amplified Z-Output connected to analog pin 1 with VREF connected to analog pin 2.
Does anyone have experience doing something similar with gyros? I'm not sure if I'm converting my values properly or the gyro isn't accurate enough.
Thanks!
Code: [Select]
`#include <LiquidCrystal.h>LiquidCrystal lcd(12, 11, 5, 4, 3, 2);int z1;int z2;int zo;int zn;float angle;int zero_o;int zero_n;float anglerate1;float anglerate2;void setup(){lcd.begin(16, 2);delay(1000);zero_o=analogRead(2);zo=zero_o;delay(1000);delay(1000);}void loop(){ zn = analogRead(1); zero_n=analogRead(2); z1=zero_o-zo; z2=zero_n-zn; anglerate1=z1*(5000/1024)/9.1; anglerate2=z2*(5000/1024)/9.1; angle=angle+(anglerate1+anglerate2)*.05; lcd.clear(); lcd.setCursor(0, 0); lcd.print("angle: "); lcd.print(angle); lcd.setCursor(0, 1); lcd.print("Angle Rate: "); lcd.print(anglerate1); delay(100); zo=zn; zero_o=zero_n;} `
Gyro Datasheet: http://invensense.com/mems/gyro/documents/PS-IXZ-0500B-00-03.pdf
#### johnwasser
#1
##### Feb 15, 2012, 01:51 amLast Edit: Feb 15, 2012, 01:55 am by johnwasser Reason: 1
Note that "(5000/1024)" is an integer divide that will produce the answer "4", not 4.8828125. Try "(5000.0/1024.0)".
Also note that the LCD output will take a significant amount of time so assuming that the delay(100) is the ONLY delay will put your time way off. you might want to note the time (millis()) when you take a sample and subtract the previous time to find out how long it has actually been since the previous sample.
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#### MarkT
#2
##### Feb 15, 2012, 12:35 pm
The code doesn't make sense - why are you using an old sample the next time round? You just need to sum the zero-corrected output each time round. Do any scaling afterwards, you don't want conversion quantization noise mixed in with the input to the integration.
You can do better with the zero-cancellation too (and it definitely affects drift performance). Take say 16 readings to get a zero-point reading in setup(), then you can get a less biased estimate of each sample by multiplying it by 16 and subtracting the zero-point reading. This should get your drift down (taking a single reading for zero-point means you are at the mercy of the noise in that single sample). In fact you can do far better at estimating the zero point if you're prepared to spend a few seconds running the integration loop.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]
#### johnwasser
#3
##### Feb 15, 2012, 04:02 pm
The code doesn't make sense - why are you using an old sample the next time round?
I think he is using the average of the initial and final values to get a better estimate of the average rotation rate over the time period.
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#### MarkT
#4
##### Feb 15, 2012, 04:47 pm
No its using a sample in one loop and then adding it again in the next, so it's just being counted 1/2 a sample time late on average - and more to the point introducing quantization error from the scaling before the accumulate step, which can't improve stability to my mind.
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#### hunter7890
#5
##### Feb 15, 2012, 08:34 pm
I'm using a numerical integration method rather than a simple angle = angle + rotational rate * time. The method I'm using is the trapezoid rule ( http://en.wikipedia.org/wiki/Trapezoidal_rule ) which I'm hoping will improve my accuracy. Although I will try the simpler method to get a comparison to see which is more consistent.
Just to note I don't necessarily need accurate angle calculations, more so consistent ones. If I can get the code to consistently zero out to a set a position -/+5* that should be fine as well. Right now that's my biggest issue.
Thanks for your reply's so far hopefully I can get this working soon!
#### johnwasser
#6
##### Feb 15, 2012, 10:16 pm
This is how I would do such a thing:
Code: [Select]
`const int ZeroPointPin = A0;const int XAxisRatePin = A1;const int YAxisRatePin = A2;const int ZAxisRatePin = A3;const float MillivoltsPerDegreePerSecond = 9.1;const int ZeroSmoothing = 16; // Use a power of 2 between 4 and 64int ZeroPoint;class Axis {public: Axis(const int pin) { dataPin = pin; previousValue = 0; previousUpdateTime = 0; currentAngle = 0; }; void update(); int angle() { return currentAngle; }; void resetAngle() { currentAngle = 0; };private: int dataPin; int previousValue; unsigned long previousUpdateTime; float currentAngle;};void Axis::update(void){ unsigned long updateTime = millis(); // Get a value multiplied by half the ZeroSmoothing factor. int newValue = analogRead(dataPin) * (ZeroSmoothing/2); // Add together the last two readings to get the average * the ZeroSmoothing factor. // Then subtract the current ZeroPoint which is alraedy * the ZeroSmoothing factor. int rateCounts = (previousValue + newValue) - ZeroPoint; if (rateCounts != 0) { // Convert to millivolts. Multiply by ADC full scale voltage (5000 mV) and divide by // ADC full scale count * ZeroSmoothing float rateMillivolts = (rateCounts * 5000L) / (ZeroSmoothing * 1024.0); // Convert to degrees per second by dividing by the scale factor. float rateDegreesPerSecond = rateMillivolts / MillivoltsPerDegreePerSecond; // Convert to degrees by multiplying by the number of seconds float degrees = rateDegreesPerSecond * ((updateTime - previousUpdateTime)/1000.0); currentAngle += degrees; } previousUpdateTime = updateTime; previousValue = newValue;}Axis XAxis(XAxisRatePin);Axis YAxis(YAxisRatePin);Axis ZAxis(ZAxisRatePin);void setup(){ Serial.begin(9600); ZeroPoint = 0; for (int i = 0; i<ZeroSmoothing; i++) ZeroPoint += analogRead(ZeroPointPin);}void loop(){ // Update the zero point to protect against zero point drift ZeroPoint = ZeroPoint * (ZeroSmoothing-1) + analogRead(ZeroPointPin); XAxis.update(); YAxis.update(); ZAxis.update(); Serial.print("Angles: X="); Serial.print(XAxis.angle()); Serial.print(" Y="); Serial.print(YAxis.angle()); Serial.print(" Z="); Serial.println(ZAxis.angle()); delay(100);} `
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#### hunter7890
#7
##### Feb 17, 2012, 05:42 pm
Thanks for your help. Unfortunately I was getting better results with the code I was using before. Although I'm not a great programmer and I'm not familiar with classes so I haven't attempted to debug the code.
I appreciate the effort, I'll work on understanding what you've got done there and see if I can get it work!
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Problem 1479 --Split Sticks
## 1479: Split Sticks
Time Limit: 2 Sec Memory Limit: 512 MB
Submit: 70 Solved: 20
[Submit][Status][Web Board]
## Description
Alice has a bundle of sticks. She wants to select some sticks and split them into $$k$$ rows.
For beauty, the difference between the lengths of all sticks in the same row shall not exceed 1 and each row has the same number of sticks. The length of each stick is a positive integer from 1 to $$n$$.
For every possible length, you know the amount of stick with that length.
Please calculate the maximum number of sticks Alice can select.
## Input
The 1st line is a positive integer $$T(1⩽ T ⩽ 10000)$$which is the number of test cases.
For each test case:
The first line contains two integers $$n(1⩽ n ⩽ 30000)$$ and $$k(1⩽ k ⩽ 10^{12})$$, representing the number of different length of sticks and the number of rows the stick needs to be divided into, respectively
The second line contains n integers $$C_1, C_2 ... C_n$$ $$(0⩽ C_i ⩽ 10^{12})$$), representing the number of sticks in each length
Ensure that the sum of n of all cases does not exceed 30000
## Output
For each case, output the maximum number of sticks Alice can select.
## Sample Input
5
3 4
7 1 13
1 1
100
1 3
100
2 1
1000000000000 1000000000000
4 1
10 2 11 1
## Sample Output
16
100
99
2000000000000
13
## HINT
1. The arrangement is [3,3,3,3], [1,2,1,1], [1,1,1,1], [3,3,3,3](each list represents a row)
2. All the sticks can be arranged in the same row
3. 33 sticks with length 1 in each 3 rows
4. All the sticks can be arranged in the same row
5. All sticks with lengths of 2 and 3 are arranged in the same row
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2016-09-20, 15:34 #837
Puzzle-Peter
Jun 2009
2·5·67 Posts
Quote:
Originally Posted by Puzzle-Peter I do not understand the following LLR output: My input looks like this: Code: 23566399112 59508 -1 23566399112 59543 -1 23566399112 59568 -1 and lresults.txt looks like this: Code: 23566399112*3^59508-1 is not prime. RES64: DBDD7F48F65B3879. OLD64: A7B882071AD77B5A Time : 9.327 sec. Iter: 94407/94407, ERROR: ROUND OFF (0.5) > 0.4 Continuing from last save file. Unrecoverable error, Restarting with next larger FFT length... U((N+1)/3) is coprime to N! Time : 58.448 sec. 23566399112*3^59568-1 is not prime. RES64: B3A7CCFC753017E5. OLD64: 1AF766F55F9047AC Time : 9.324 sec. So what does this tell me about n=59543? Prime or no prime? Thanks Peter
Updating to the most recent version of LLR fixed the issue
2016-09-25, 08:41 #838
Siemelink
Jan 2006
Hungary
4148 Posts
Quote:
Originally Posted by Siemelink As a tribute to this massive effort, I'll take a reservation too. Reserving k=22G to 23G, n=25K to 100K. Willem.
My pc died in July, so the progress was halted for a while. Continuing.
Willem.
2016-10-17, 11:03 #839 MyDogBuster May 2008 Wilmington, DE B2416 Posts R3 Reserving >63G n=25K-100K
2016-11-10, 13:29 #840 MyDogBuster May 2008 Wilmington, DE 22×23×31 Posts R3 R3 >63G tested to n=100K 282 primes found Results emailed - range released
2016-11-16, 16:03 #841 rebirther Sep 2011 Germany 23×32×5×7 Posts R3 tested to n=500k (300-500k) (4-6G) 467 primes found, 876 remain in this range Results emailed, Base released
2016-11-16, 16:14 #842 rebirther Sep 2011 Germany 23·32·5·7 Posts Reserving R3 to n=250k (200-250k) (2.147-4G) for BOINC
2016-12-01, 21:20 #843 rebirther Sep 2011 Germany 23×32×5×7 Posts R3 tested to n=250k (200-250k) (2.147-4G) 267 primes found, 1554 remain in this range Results emailed, Base released
2016-12-10, 08:26 #844 rebirther Sep 2011 Germany 1001110110002 Posts Reserving R3 to n=125k (100-125k) (6-10G) for BOINC
2016-12-31, 13:58 #845 KEP Quasi Admin Thing May 2005 3·313 Posts R3 13G-21G is complete to n=100K... results e-mailed
2016-12-31, 17:17 #846
VBCurtis
"Curtis"
Feb 2005
Riverside, CA
112E16 Posts
Quote:
Originally Posted by KEP R3 13G-21G is complete to n=100K... results e-mailed
Excellent way to end the year, Kenneth! Rather a monumental task.
I'll begin a sieve for 10-15G once the 6-10G work reaches 200 or 250k, midyear I suppose.
2017-01-01, 16:21 #847
KEP
Quasi Admin Thing
May 2005
3×313 Posts
Quote:
Originally Posted by VBCurtis Excellent way to end the year, Kenneth! Rather a monumental task. I'll begin a sieve for 10-15G once the 6-10G work reaches 200 or 250k, midyear I suppose.
Thanks!
It was about 40M tasks and it resulted in ~34800 primes (number may be wrong, is taken on top of my head)
So yes, it was a monumental task, however now you have plenty to sieve for the next many years. Really wish that Riesel base 3 could become a focus project for more users and maybe for BOINC
Unfortunantly I have no ressources available to help you sieve.
Happy new year everyone. Take care!
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A copy of the license is included in the FAQ. | 1,424 | 4,154 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-45 | latest | en | 0.726224 |
https://merlindictionary.com/what-does-absolute-value-mean/ | 1,696,321,531,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511055.59/warc/CC-MAIN-20231003060619-20231003090619-00062.warc.gz | 424,079,168 | 76,277 | absolute value means: A number that is numerically equal to a real number, regardless of its sign. The absolute value for -4, written as -4, is 4 (in Merlin Dictionary)
### What else does absolute value mean?
• Modulus is the square root of all the squares of real and imagined components of complex numbers. (in Merlin Dictionary) | 75 | 333 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2023-40 | latest | en | 0.904525 |
http://mathhelpforum.com/calculus/45310-solved-points-discontinuity-print.html | 1,495,689,327,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607998.27/warc/CC-MAIN-20170525044605-20170525064605-00135.warc.gz | 237,131,272 | 2,799 | [SOLVED] Points of discontinuity
• Aug 4th 2008, 08:28 PM
Snowboarder
[SOLVED] Points of discontinuity
Hi everyone.
I've been having problem with the one function
h(x)=(x^3-1)/(x^3-x)
I need to find all points of discontinuity.
Thx for any help
• Aug 4th 2008, 08:56 PM
Chris L T521
Quote:
Originally Posted by Snowboarder
Hi everyone.
I've been having problem with the one function
h(x)=(x^3-1)/(x^3-x)
I need to find all points of discontinuity.
Thx for any help
Did you try to factor it?
$h(x)=\frac{x^3-1}{x^3-x}=\frac{x^3-1}{x(x^2-1)}=\frac{(x-1)(x^2+x+1)}{x(x+1)(x-1)}$
Can you take it from here? It should be no problem now.
--Chris
• Aug 4th 2008, 08:58 PM
nikhil
Quote:
Originally Posted by Snowboarder
Hi everyone.
I've been having problem with the one function
h(x)=(x^3-1)/(x^3-x)
I need to find all points of discontinuity.
Thx for any help
Points of discontinuity are
at x=0,1,-1 as function's value is undefined at these points | 326 | 953 | {"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": 1, "wp_latex": 0, "mimetex.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} | 3.171875 | 3 | CC-MAIN-2017-22 | longest | en | 0.926596 |
https://blog.adafruit.com/2021/08/03/how-many-numbers-exist/ | 1,713,048,770,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816853.44/warc/CC-MAIN-20240413211215-20240414001215-00035.warc.gz | 124,815,420 | 16,938 | ## How Many Numbers Exist?
The number of numbers might be a number.
Infinite sets of numbers mess with our intuition about size, so as a warmup, compare the natural numbers {1, 2, 3, …} with the odd numbers {1, 3, 5, …}. You might think the first set is bigger, since only half its elements appear in the second set. Cantor realized, though, that the elements of the two sets can be put in a one-to-one correspondence. You can pair off the first elements of each set (1 and 1), then pair off their second elements (2 and 3), then their third (3 and 5), and so on forever, covering all elements of both sets. In this sense, the two infinite sets have the same size, or what Cantor called “cardinality.” He designated their size with the cardinal number ℵ0 (“aleph-zero”).
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https://www.rieselprime.de/ziki/Multiplication | 1,643,068,465,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304686.15/warc/CC-MAIN-20220124220008-20220125010008-00572.warc.gz | 1,015,745,714 | 7,379 | Note: Due to changes in the Riesel prime template, most of those pages (and related) are not shown properly.This will take some time!Wanna help? Please move any Riesel prime page first, then edit/add the base parameter.
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# Multiplication
Multiplication is the process of calculating the result when a number a is added to itself b times. The result of a multiplication is called the product of a and b, and each of the numbers is called a factor of the product ab.
Multiplication is denoted a×b,(a)(b), or simply ab. The symbol '×' is known as the multiplication sign.
The result of multiplying no numbers (empty product) is always 1 (the multiplicative identity, see below). The most common occurences are in exponentiation ($\displaystyle{ a^0=1 }$) and factorial numbers (0!=1).
## Multiplication properties
### Associative property
$\displaystyle{ (x*y)z=x(y*z) }$
### Commutative property
$\displaystyle{ x*y=y*x }$
### Distributive property
$\displaystyle{ x(y+z)=xy+xz }$
### Identity element
$\displaystyle{ 1x=x }$
### Multiplication by zero
$\displaystyle{ 0x=0 }$
## Multiplication algorithms
When the product fits in a variable supported by the programming language, or in a register when programming in assembler, the multiplication is trivial. However if the largest available register is n bits wide the factors can only be n/2 bits wide each.
Otherwise there are several algorithms used to calculate products, depending on the size of the factors: | 375 | 1,568 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2022-05 | longest | en | 0.796746 |
https://beyondnews.net/1147-celsius-to-fahrenheit/ | 1,721,087,402,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514724.0/warc/CC-MAIN-20240715224905-20240716014905-00478.warc.gz | 114,401,059 | 47,343 | What is 1147 degrees Celsius in Fahrenheit?
1147 c equals 2096.6 f 2096.6 Fahrenheit
The C to F formula is (C * 9/5) + 32 = F
1147 x 9 = 10323
1147 / 5 = 2064.6
2064.6 + 32 = 2096.6
1147 degrees Celsius equals 2096.6 degrees Fahrenheit. 1147ctof 1147c to f
Celsius Fahrenheit
1147.0 2096.6
1147.1 2096.78
1147.2 2096.96
1147.3 2097.14
1147.4 2097.32
1147.5 2097.5
1147.6 2097.68
1147.7 2097.86
1147.8 2098.04
1147.9 2098.22
1147.10 2096.78
1147.11 2096.798
1147.12 2096.816
1147.13 2096.834
1147.14 2096.852
1147.15 2096.87
1147.16 2096.888
1147.17 2096.906
1147.18 2096.924
1147.19 2096.942
1147.20 2096.96
1147.21 2096.978
1147.22 2096.996
1147.23 2097.014
1147.24 2097.032
1147.25 2097.05
1147.26 2097.068
1147.27 2097.086
1147.28 2097.104
1147.29 2097.122
1147.30 2097.14
1147.31 2097.158
1147.32 2097.176
1147.33 2097.194
1147.34 2097.212
1147.35 2097.23
1147.36 2097.248
1147.37 2097.266
1147.38 2097.284
1147.39 2097.302
1147.40 2097.32
1147.41 2097.338
1147.42 2097.356
1147.43 2097.374
1147.44 2097.392
1147.45 2097.41
1147.46 2097.428
1147.47 2097.446
1147.48 2097.464
1147.49 2097.482
1147.50 2097.5
1147.51 2097.518
1147.52 2097.536
1147.53 2097.554
1147.54 2097.572
1147.55 2097.59
1147.56 2097.608
1147.57 2097.626
1147.58 2097.644
1147.59 2097.662
1147.60 2097.68
1147.61 2097.698
1147.62 2097.716
1147.63 2097.734
1147.64 2097.752
1147.65 2097.77
1147.66 2097.788
1147.67 2097.806
1147.68 2097.824
1147.69 2097.842
1147.70 2097.86
1147.71 2097.878
1147.72 2097.896
1147.73 2097.914
1147.74 2097.932
1147.75 2097.95
1147.76 2097.968
1147.77 2097.986
1147.78 2098.004
1147.79 2098.022
1147.80 2098.04
1147.81 2098.058
1147.82 2098.076
1147.83 2098.094
1147.84 2098.112
1147.85 2098.13
1147.86 2098.148
1147.87 2098.166
1147.88 2098.184
1147.89 2098.202
1147.90 2098.22
1147.91 2098.238
1147.92 2098.256
1147.93 2098.274
1147.94 2098.292
1147.95 2098.31
1147.96 2098.328
1147.97 2098.346
1147.98 2098.364
1147.99 2098.382
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and yours is the greatest I’ve found out till now. | 1,771 | 4,791 | {"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} | 2.75 | 3 | CC-MAIN-2024-30 | latest | en | 0.348271 |
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# Carpenters, dentists, sewing machine operators,
Author Message
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Director
Joined: 05 Jan 2008
Posts: 701
Followers: 5
Kudos [?]: 414 [0], given: 0
Carpenters, dentists, sewing machine operators, [#permalink]
### Show Tags
25 Apr 2008, 23:37
00:00
Difficulty:
(N/A)
Question Stats:
77% (01:41) correct 23% (00:22) wrong based on 72 sessions
### HideShow timer Statistics
Carpenters, dentists, sewing machine operators, needlepointers, piano players, and indeed anyone who works with their hands for long hours can get carpal tunnel syndrome.
(A) anyone who works
(B) anyone working
(C) workers
(D) those for whom work is
(E) any people who work
_________________
Persistence+Patience+Persistence+Patience=G...O...A...L
If you have any questions
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25 Apr 2008, 23:43
Its a straight B here
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27 Apr 2008, 16:20
E for correct SV and keeping the meaning consistent (note the their)
prasannar wrote:
Carpenters, dentists, sewing machine operators, needlepointers, piano players, and indeed anyone who works with their hands for long hours can get carpal tunnel syndrome.
(A) anyone who works
(B) anyone working
(C) workers
(D) those for whom work is
(E) any people who work
Re: SC 1000: Carpenters [#permalink] 27 Apr 2008, 16:20
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JANUARY 2006
The Dean’s Column:
A Juicy Math Problem
By Alfred Posamentier, Ph.D.
When students are challenged by a problem, they often set it aside if it involves too much reading, for fear that the concentration required would be too exhausting to make the problem pleasurable. Although this problem does require a fair bit of reading it is rather easy to explain to a class, and could even be dramatized. Once past the statement of the problem, it is very easy to understand, but quite difficult to solve by conventional means.
This is where the beauty of the problem comes it. The solution—as unexpected as it is—almost makes the problem trivial. That is, the problem and its conventional solution will not get much of an enthusiastic reaction from students, but after having struggled with a solution attempt, the novel approach we will present here will gain you much favor with the class.
So let’s state the problem:
We have two one-gallon bottles. One contains a quart of grape juice and the other, a quart of apple juice. We take a tablespoonful of grape juice and pour it into the apple juice. Then we take a tablespoon of this new mixture (apple juice and grape juice) and pour it into the bottle of grape juice. Is there more grape juice in the apple juice bottle, or more apple juice in the grape juice bottle?
To solve the problem, we can figure this out in any of the usual ways—often referred to as “mixture problems”—or we can use some clever logical reasoning and look at the problem’s solution as follows:
With the first “transport” of juice there is only grape juice on the tablespoon. On the second “transport” of juice, there is as much apple juice on the spoon as there is grape juice in the “apple juice bottle.” This may require students to think a bit, but most should “get it” soon.
The simplest solution to understand and the one that demonstrates a very powerful strategy is that of using extremes. We use this kind of reasoning in everyday life when we resort to the option: “ such and such would occur in a worst case scenario….”
Let us now employ this strategy for the above problem. To do this, we will consider the tablespoonful quantity to be a bit larger. Clearly the outcome of this problem is independent of the quantity transported. So we will use an extremely large quantity. We’ll let this quantity actually be the entire one quart. That is, following the instructions given the problem statement, we will take the entire amount (one quart of grape juice), and pour it into the apple juice bottle. This mixture is now 50% apple juice and 50% grape juice. We then pour one quart of this mixture back into the grape juice bottle…The mixture is now the same in both bottles. Therefore, there is as much apple juice in the grape juice bottle as there is grape juice in the apple juice bottle!
We can consider another form of an extreme case, where the spoon doing the juice transporting has a zero quantity. In this case the conclusion follows immediately: There is as much grape juice in the apple juice bottle as there is apple juice in the grape juice bottle, that is, zero!
Carefully presented, this solution can be very significant in the way students approach future mathematics problems and even how they may analyze everyday decision-making.#
You may find other such examples in Math Wonders: To Inspire Teachers and Students, by Alfred S. Posamentier (ASCD, 2003) see: www.ascd.org. or Math Charmers: Tantalizing Tidbits for the Mind. By Alfred S. Posamentier (Prometheus Books, 2003) see: www.prometheusbooks.com. If you wish to learn more about p, see: p: A Biography of the world’s Most Mysterious Number, By Alfred S. Posamentier (Prometheus Books, 2004) see: www.prometheusbooks.com.
Dr. Alfred S. Posamentier is Dean of the School of Education at City College of NY, author of over 35 books on math, and member of the NYS Standards Committee on Math. | 983 | 4,390 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2018-51 | latest | en | 0.918872 |
https://icme.hpc.msstate.edu/mediawiki/index.php/Surface_Generation_Code | 1,591,147,206,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347426956.82/warc/CC-MAIN-20200602224517-20200603014517-00209.warc.gz | 386,005,660 | 7,637 | # Purpose
This code is written in the compute language Python. Based on user input, the script will build an aluminum super-cell and create a corresponding POSCSR file.
# Implementation
This script accepts either 2 or 7 input arguments.The first 2 inputs are required while the additional 5 are optional. The arguments must occur in the following order:
```lattice parameter, surface under consideration, vacuum length, x-atoms, y-atoms, z-atoms, adatom
```
• Lattice parameter (angstrom) - length of the equilibrium lattice constant found in previous section.
• Surface under consideration - for this assignment only the (111) plane is being considered, so for "surface under consideration" 111 should be used.
• Vacuum length (angstrom)- length of vacuum.
• X-atoms - number of atoms along the x-axis (must be an integer).
• Y-atoms - number of atoms along the y-axis (must be an integer).
• Z-atoms - number of atoms along the z-axis (must be an integer).
The following is what a sample input looks like:
```./<name_of_script> 3.57 111 4 1 1 1 0
```
# Output
This script will generate a POSCAR file to be used in VASP DFT calculations.
# Source Code
```#!/usr/bin/env python
# Purpose: Calculate FCC (111) surface energies
# Author: Sungho Kim and Laalitha Liyanage
# Edited: W.B. Lawrimore II
import os
import sys
import math
usage="""
Usage: ./gen_fcc_surface.py a surf vacuum nx ny nz adatom
Mandatory arguments
-------------------
a - equilibrium lattice constant
surf - Type of surface 100, 110 or 111
Optional arguments
-------------------
vacuum - length of vacuum; DEFAULT = 8.0 angstroms
nx,ny,nz - periodicity of supercell; DEFAULT (1,1,1)
adatom - 1/0 (True/False); DEFAULT = 0 (False)
"""
#Default setting
#--------------------------------------------------------------------
vacuum = 8.0
#--------------------------Surface (111)-----------------------------
def gen_data_for_111_fcc(a,nx=2,ny=2,nz=4):
""" Generate datafile of FCC surface: 110:x, 112:y, 111:z """
xa=[]; ya=[]; za=[]
ax = a*math.sqrt(2)/2
ay = a*math.sqrt(6)/2
az = a*math.sqrt(3)
x0 = 0.0
x2 = math.sqrt(2)/4 * a
y2 = math.sqrt(6)/4 * a
y3 = math.sqrt(6)/6 * a
y4 = math.sqrt(6)*5/12 * a
y5 = math.sqrt(6)*2/6 * a
y6 = math.sqrt(6)/12 * a
bx,by,bz = ax*nx, ay*ny, az*nz+vacuum
for i in range(nx):
for j in range(ny):
layer = 0
for k in range(nz):
xa.append(x0+i*ax); ya.append(x0+j*ay); za.append(layer/3.0*az)
xa.append(x2+i*ax); ya.append(y2+j*ay); za.append(layer/3.0*az); layer += 1
xa.append(x0+i*ax); ya.append(y3+j*ay); za.append(layer/3.0*az)
xa.append(x2+i*ax); ya.append(y4+j*ay); za.append(layer/3.0*az); layer += 1
xa.append(x0+i*ax); ya.append(y5+j*ay); za.append(layer/3.0*az)
xa.append(x2+i*ax); ya.append(y6+j*ay); za.append(layer/3.0*az); layer += 1
xa.append(bx/2.); ya.append(by/2.); za.append(x0+nz*az)
return xa,ya,za,bx,by,bz
#----------------------------POSCAR generation------------------------------------------------
def gen_poscar(xa,ya,za,bx,by,bz):
fout = open("POSCAR","w")
fout.write("Fe\n")
fout.write("1.0\n")
fout.write(" %22.16f %22.16f %22.16f\n"%(bx,0,0))
fout.write(" %22.16f %22.16f %22.16f\n"%(0,by,0))
fout.write(" %22.16f %22.16f %22.16f\n"%(0,0,bz))
fout.write("%d\n"%len(xa))
# fout.write("Selective Dynamics\n")
fout.write("Cart\n")
for i in range(len(xa)):
fout.write("%22.16f %22.16f %22.16f\n"%(xa[i],ya[i],za[i]))
# fout.write("%22.16f %22.16f %22.16f F F T\n"%(xa[i],ya[i],za[i]))
fout.close()
return len(xa)
#-------------------------------Main program---------------------------------------------------
if len(sys.argv) > 2:
if len(sys.argv) == 3:
a_latt = float(sys.argv[1])
surf = sys.argv[2]
if surf == '111':
xa,ya,za,bx,by,bz = gen_data_for_111_fcc(a_latt)
gen_poscar(xa,ya,za,bx,by,bz)
else:
print "Only calculate surface energies for the (111) surface"
elif len(sys.argv) == 8:
a_latt = float(sys.argv[1])
surf = sys.argv[2]
vacuum = float(sys.argv[3])
nx = int(sys.argv[4])
ny = int(sys.argv[5])
nz = int(sys.argv[6])
if surf == '111':
xa,ya,za,bx,by,bz = gen_data_for_111_fcc(a_latt,nx,ny,nz)
gen_poscar(xa,ya,za,bx,by,bz)
else:
print "Only calculate surface energies for the (111) surface"
else:
print "Error: wrong number of arguments!!!"
print usage
``` | 1,370 | 4,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2020-24 | latest | en | 0.61634 |
https://assignwrite.com/2021/07/09/determine-the-total-heat-transfer-and-the-outlet-temperatures-of-water-and-oil-cp-of-water-4186-j-kgk/ | 1,653,008,331,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662530553.34/warc/CC-MAIN-20220519235259-20220520025259-00612.warc.gz | 151,804,504 | 14,470 | # Determine the total heat transfer and the outlet temperatures of water and oil. (cp of water = 4186 J/kgK).
A. Water enters a counterflow double-pipe heat exchanger at 15ºC flowing at the rate of 1700 kg/h. It is heated by oil (cp = 2000 J/kgK) flowing at the rate of 550 kg/h from an inlet temperature of 94°C. For an area of 1 m2 and an overall heat transfer coefficient of 1075 W/m2K, determine the total heat transfer and the outlet temperatures of water and oil. (cp of water = 4186 J/kgK).
B. A simple counterflow heat exchanger operates under the following conditions: Fluid A: inlet and outlet temperatures 80°C and 40°C, Fluid B: inlet and outlet temperatures 20°C and 40°C. The exchanger is cleaned, causing an increase in the overall heat transfer coefficient by 10% and inlet temperature of fluid B is changed to 30°C. What will be the new outlet temperatures of fluid A and of fluid B. Assume that the capacity rates remain unaltered.
The post Determine the total heat transfer and the outlet temperatures of water and oil. (cp of water = 4186 J/kgK). appeared first on Best Custom Essay Writing Services | EssayBureau.com.
Posted in Uncategorized | 288 | 1,172 | {"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-2022-21 | latest | en | 0.874434 |
https://www.physicsforums.com/threads/calculating-deflection.736364/ | 1,527,372,789,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867904.94/warc/CC-MAIN-20180526210057-20180526230057-00445.warc.gz | 819,854,968 | 20,607 | # Homework Help: Calculating deflection
1. Feb 3, 2014
### Samppa
Hi,
I'd very much like to know how to calculate deflection of a beam by integrating when flexural rigidity EI is not constant. I tried finding an example on the internet but couldn't find any. Can anyone provide me with one? The problem is driving me nuts.
2. Feb 3, 2014
### Staff: Mentor
Hi Samppa. Welcome to Physics Forums!!
I assume you know EI as a function of distance x along the beam, correct? And, you know the bending moment as a function of x. You divide M by EI, and you have a new function of x. That's the function you integrate.
3. Feb 4, 2014
### Samppa
Yes, but the problem is how do I make EI a fucntion of x when it changes half-way in the beam. I know the equation of moment obviously. I mean I don't know what to do with the changing EI.
4. Feb 4, 2014
### Staff: Mentor
It doesn't have to be continuous. Just integrate it across the discontinuous change.
Chet
5. Feb 4, 2014
### SteamKing
Staff Emeritus
With stepped beams or other non-prismatic members, you may have to write more than one integral.
After all, $\int^{b}_{a}$ = $\int^{c}_{a}$ + $\int^{b}_{c}$
6. Feb 5, 2014
### Samppa
I sort of did something like that but I don't get it right. That's why I was hoping someone could provide and example so I could see where I'm going wrong.
7. Feb 5, 2014
### SteamKing
Staff Emeritus
It would be easier if you show what you have been doing. That would give you a quicker diagnosis to your actual problem.
8. Feb 5, 2014
### Samppa
Fine.
I attached a picture of the structural model.
F = 23 kN
L1 = 2.1 m
E = 210 000 MPa
IA-B = 7*10^-4 m^4
IB-C = 3.5*10^-4 m^4
EIA-B = 147000 kNm^2
EIB-C = 73500 kNm^2
Equation of moment: M''(X) = -23X + 96.6
M(X) = (-23/6)X^3 + 48.3X^2
Then I tried: M(2.52) = {[(-23/6)/147000]*2.52^3 + (48.3/147000)*2,52^2} + {[(-23/6)/73500]*2.52^3 + (48.3/73500)*2,52^2} = ....
This is just one of the ways I tried to calculate.
File size:
5.9 KB
Views:
115
9. Feb 5, 2014
### Staff: Mentor
Your equation for the moment us incorrect. What you call M''(x) is actually the bending moment M(x). What boundary conditions are you using at x = 0 for the displacement?
10. Feb 5, 2014
### Samppa
It is not incorrect it is exactly what it is supposed to be and as you say it is a bending moment.
Displacement at X = 0 is 0. This is why there are no C1X and such in case that is what you are wondering. The support is fixed.
11. Feb 5, 2014
### Samppa
I marked it M''(x) just to make it simple so I don't have to put those integration marks.
12. Feb 5, 2014
### SteamKing
Staff Emeritus
You have to include the constants of integration at each step. Since this is a cantilever, you will have two constants of integration in your expression for the deflection. To determine the value of these constants, you will have to apply the boundary conditions for this problem: one BC for the slope, and one BC for the deflection.
13. Feb 5, 2014
### Samppa
I don't quite know what these boundary conditions would be.
14. Feb 5, 2014
### Staff: Mentor
SteamKing: Do you agree with his determination of the bending moment? I get 23(4.2-x) for the bending moment. In this case, I would have EIy" =23(4.2-x).
Samppa: Again, what are the two boundary conditions on the displacement y for this cantilever beam at x = 0?
Chet
15. Feb 5, 2014
### Samppa
Dude come on...
23(4.2-x) = -23X + 96.6
I said: Displacement at X = 0 is 0. Since the support is fixed this means M'(0) = 0.
The only problem is what to do once the integration has been done. There is no reason to waist time with "wrong" markings, this isn't preliminary school. At least I haven't been there for quite some time.
Last edited by a moderator: Feb 6, 2014
16. Feb 5, 2014
### Staff: Mentor
Actually, Dude, you said M''(x) is equal to -23x +96.6, not M(x). I assumed you meant that M''(x) is the second derivative of M(x), where M is the moment.
Regarding the boundary conditions at x = 0, I felt that you failed to recognize a very important one that I was attempting to get you to identify. That boundary condition is that dy/dx = 0 at a "built-in" cantilever boundary. Is that what you meant by M'(0)? I assumed you were saying that the derivative of the moment was equal to zero at x = 0. Again, I though you were using the symbol M for moment.
If you want me to do the whole problem for you, I can do that. But, one of the objectives here is to help the person figure out how to do it themselves. Here is my next hint:
$y'(x)=\int_0^x{\frac{M(x)}{147000}dx}$ for x≤2.1
$y'(x)=\int_0^{2.1}{\frac{M(x)}{147000}dx}+\int_{2.1}^x{\frac{M(x)}{73500}dx}$ for 2.1<x≤4.2
where M(x) is the moment.
Hope this helps.
Last edited by a moderator: Feb 6, 2014
17. Feb 6, 2014
### Samppa
I can't believe this, again with the markings.
I told you that I marked it M''(X) so I don't have to use integration marks, meaning integration S if this is more understandable. In my calculations I also state that M''(X) is the moment. Please read the posts before you answer. This shouldn't even need to be said.
Regarding the boundary conditions at X = 0, those are obvious and that is why I did not tell them the first time you asked. If you looked at my calculations you would see that they have been taken into account. I also said: "Displacement at X = 0 is 0. This is why there are no C1X and such in case that is what you are wondering. The support is fixed".
Chet please, tell me where do I exactly ask you to solve "the whole problem"? You can see my calculations, you know where I've gotten and all along I've been asking what to do at the end. You've finally understandably answered, thank you for that.
I tried to be civil, I really tried but you've been harassing me Chet.
Last edited by a moderator: Feb 6, 2014
18. Feb 6, 2014
### Staff: Mentor
Dear Samppa,
Please excuse me if you felt that I was harassing you. That was not my intention.
The source of my confusion had to do with your use of the symbol M. In a beam bending problem, when someone uses the symbol M, it is natural to assume that they are talking about the bending moment. Yes, to be fair, you did identify M'' as the bending moment, but I just naturally thought that this was an oversight. (I still haven't figured out what the symbol M represents physically in your terminology, other than the bending moment integrated twice.)
The people who help on PF are expected to be patient with the people they are trying to help, even if the people they are helping use confusing terminology. By the same token, the people who are being helped need to be patient with the helpers until everyone is on the same page. I'll try to be more patient in the future. I hope you will agree to do the same.
So anyhow, was my hint helpful to you? Were you able to get to a correct final answer?
Chet
19. Feb 6, 2014
### Samppa
As you say M(x) is usually the bending moment but I didn't want to bother with the integration marks so I decided to name it M''(x). That's all there is to it. Why do you need M to mean something so badly. It shouldn't make a difference if I had named the bending moment SANTA or REINDEER.
I had given you the explanation what M''(X) was and you can see the equation and it clearly is the bending moment.
Well my calculations get stuck at the point when I've calculated y'(x) (as chest named it). Just integrating y'(x) didn't give the right answer.
Last edited by a moderator: Feb 6, 2014
20. Feb 6, 2014
### Staff: Mentor
I should have mentioned that y' is the derivative of the displacement y. After you got y', did you try to integrate again to get y? If not, please try and see how this works.
Chet
21. Feb 6, 2014
### Samppa
I thought I just said that I integrated it...
How should I have said it to be understandable, please tell me?
I suppose since you say I should integrate, I should be getting the right anwers.
I'll tell what I get:
At 2.1< X ≤ 4.2
y' = (1/73500) * (-11.5X^2 + 96.6X - 152,145) + (1/147000) * (152,145)
And yes I skipped some phases, I'd rather not show all the phases unless you absolutely insist.
22. Feb 6, 2014
### Samppa
Before I tell you what I got from the integration I wanted to make sure that y' is correct.
23. Feb 6, 2014 | 2,381 | 8,284 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-22 | latest | en | 0.923098 |
https://topclasswriter.com/3398-2/ | 1,726,539,189,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651722.42/warc/CC-MAIN-20240917004428-20240917034428-00757.warc.gz | 529,583,355 | 16,343 | # 3398
breakfast cereals…
DIET PROBLEM. Grant Winfield is a 71-year-old grandfather who likes to mix breakfast cereals together for taste and as a means of getting at least 50% of the recommended daily allowances (RDA) of five different vitamins and minerals. Concerned about his sugar intake, he wishes his mixture to yield the lowest possible amount of sugar. For taste, each of the cereals listed in the following table must make up at least 1 0 % of the total mixture. The table shows the amounts of the vitamins, minerals, and sugar contained in one ounce together with 1 / 2 cup of skim milk.
a. Formulate and solve for the number of ounces in each cereal that should be mixed together in order to minimize total sugar intake while providing at least 50% of the RDA for each of Vitamins A, C, D, B6 , and iron. How much sugar would be consumed in the process?
b. How much total cereal does Grant need to eat to achieve the minimum 50% RDA in all five categories? How much milk does he consume in doing this?
c. Determine the shadow prices for this problem. Interpret the shadow prices and the corresponding ranges of feasibility.
d. If Grant eliminates the restriction that each cereal must account for at least 1 0 % of the mixture, then, by inspection, why wouldn’t any Frosted Bran be included in the mix? Verify this conclusion by deleting these constraints from the original formulation and re-solving. | 317 | 1,419 | {"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-2024-38 | latest | en | 0.941904 |
http://applet-magic.com/BEbetalimits.htm | 1,620,719,639,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991904.6/warc/CC-MAIN-20210511060441-20210511090441-00239.warc.gz | 3,224,692 | 4,958 | The Limits to the Number of Neutrons in the Isotopes of Each Element with Proton Numbers Less Than or Equal to 82
San José State University
applet-magic.com
Thayer Watkins
Silicon Valley,
& the Gateway
to the Rockies
USA
The Limits to the Number of Neutrons
in the Isotopes of Each Element with
Proton Numbers Less Than or Equal to 82
Nuclei are composed of nucleons (protons and neutrons) in variable proportions. There are almost three thousand nuclides that are stable enough to have their mass measured and their binding energies computed. But most are unstable. The following beautiful display from Wikipedia shows the nature of their instabilities.
As can be seen from the display, up to proton number 82 overwhelming the mode of decay is either the ejection of an electron (β) or a positron (β+). Only the ones shown in black in the middle of the distribution are stable.
The ejection of an electron occurs because a neutron converts into a proton and an electron. This conversion releases energy. The positron ejection accompanies the conversion of a proton into a neutron, but this conversion requires an input of energy.
What are sought here are the changes in binding energies involved in these proton and neutron conversions. More basically the purpose of this material is to show the relationship between binding energy and the mode of radioactive decay or stability.
## Binding Energy and Radioactive Decay Products
For the decay of a neutron into a proton and an electron the relevant binding energies are:
#### BE(p, n) => BE(p+1, n−1)
For positron ejection the transformation is in the opposite direction.
#### BE(p, n) => BE(p−1, n+1)
What is plotted in the following graphs are the binding energies in millions of electron volts (MeV) for an arbitrarily selected complete sequence of decay products as a function of the number of neutrons in the decay product.
The binding energy reaches a maximum at p=56 and n=81. This is Ba137 which is a stable isotope of Barium. To the left of this maximum binding energy positron emission occurs. To the right electron emission occurs which decreases the neutron number.
Here is a table of the data for this sequence of beta transitions.
The Data for a Complete Beta Transition Sequence
Nuclide Number
of Protons
Number
of Neutrons
Binding
Energy
Incremental
Binding
Energy
Sn137 50 87 1117.2
Sb137 51 86 1126.1 8.90
Te137 52 85 1134.65 8.55
I137 53 84 1140.81 6.16
Xe137 54 83 1145.903 5.093
Cs137 55 82 1149.293 3.39
Ba137 56 81 1149.686 0.393
La137 57 80 1148.3 -1.386
Ce137 58 79 1146.3 -2.00
Pr137 59 78 1142.82 -3.48
Nd137 60 77 1138.34 -4.48
Pm137 61 76 1131.9 -6.44
Sm137 62 75 1125.22 -6.68
Eu137 63 74 1116.8 -8.42
Gd137 64 73 1107.3 -9.50
Here is the graph of the increments in binding energy that would result from an increase in the number of neutrons due to the conversion of a proton into a neutron and the emission of a positron.
The relationship shown in the above graph is very regular. It is linear with a shift at n=82.
When the increment is positive a positron emission occurs. When it is negative the opposite beta transition occurs; the conversion of a neutron into a proton and the emission of an electron.
The regression equation for the data on incremental binding energies is
#### IBE = 100.8839889 −1.32744n − 1.74397u(n-82) [21.3] [-20.5] [-3.0]
The u(z) function in the above equation is the unit step function; i.e., u(z)=0 if z<0 and u(z)=1 if z≥0.
The numbers in the square brackets below the regression coefficients are their t-ratios; i.e., the ratio of the coefficient to its standard deviation. The coefficient of determination (R²) for the regression is 0.992.
Here is the graph of the regresssion equation estimates
The regression equation could be used to predict what the incremental binding energy would be for a nuclide beyond the range of the existing isotopes. What appears to be the case is that if the incremental binding energy exceeds a critical level the nuclide does not exist long enough for its mass to be measured. The critical level of binding energy for positron emission appears to be about 10 MeV for p=50. The critical level for electrom emission appears to be about 10 MeV for p=64. The critical level varies with the proton number.
An approximation of the critical level for the maximum of neutrons can be obtained by looking at the change in binding energy that occurs when the nuclide with the maximum number of neutrons is created by a beta transition. Similarly an approximation of the critical level for the minimum of neutrons can be obtained by looking at the change in binding energy that occurs when the nuclide with the minimum number of neutrons is created by a beta transition. These incremental binding energy (IBE) values are tabulated in Increments for the maximums and Increments for the minimums.
What was found in those tables is that the values are significantly higher when the proton number is odd than when it is even. This undoubtedly occurs when the beta transition creates a proton-proton spin pair or breaks up a neutron-neutron spin pair. There is a similar effect when the neutron number and the proton number are nearly equal. In that case the beta transition may promote the formation of a neutron-proton spin pair and in other cases break up such a spin pair.
Here are the graphs of the incremental binding energies (IBE) only for the even proton numbers.
What this data shows is a constant level of IBE from the low proton numbers up to p=20 then a transition up to a range from p=26 to p=42 then another transition up to p=72 and a low constant level to p=82. The average value of the IBE from p=2 to p=40 is 16.66 MeV. From p=26 to p=42 the average is 9.28 MeV. For the final range for the maximums the average is a small fraction of 1 MeV if the anomalous value at p=80 is left out.
As shown below there are notable similarities and notable differences for the results from the nuclides with a minimum number of neutrons and those from the nuclides with a maximum number of neutrons.
## Conclusions
The true values of the critical levels of the incremental binding energy that determines the maximum and minimum numbers of neutrons in the isotopes of an element varies with the proton number. It is about 17 MeV for small proton numbers and for the maximums goes down to a bit above 9 MeV for proton numbers in a midrange and then for the maximums down to essentially zero for proton numbers approaching 82. | 1,592 | 6,494 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-21 | latest | en | 0.881226 |
https://alison.com/it/argomento/impara/102799/funzione-di-vendita-e-marketing-mix-elements | 1,695,478,289,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506481.17/warc/CC-MAIN-20230923130827-20230923160827-00445.warc.gz | 113,027,461 | 38,570 | La nuova app di Alison è ora disponibile su iOS e Android!
Study Reminders
Support
Welcome to this class, we will look at in this class the Sales Function with respect to two
Marketing Mix elements, in this particular slide. What we are looking at is the expenses
one company is going to incur on advertising and also on sales promotion. Two types of
graphs are presented here.
(Refer Slide Time: 01:09)
The graph here is an isometric view. If, you see there is A constant mixed line, there is A
constant budget line what does it indicate? Suppose, the budget you are asked to keep it
constant that is you cannot vary the budget, then what can you do as a marketing
manager? You can start in this constant budget to adjust the amounts spent on
advertising and the amount spent on sales promotion.
This graph which is in the isometric view that is the three dimensional graph, it presents
to you the constant budget line, it also presents to you the constant mixed line.
In this constant budget line if you see, you have an A 1 and then S 1, that is for the level
of A 1 that is a advertisement expenditure, you have these corresponding sales promotion
expenditure at S 1.
Similarly, with respect to A 2 and S 2, similarly with respect to A 2 and S 3. Depending
on what you are going to do as the adjustments for A and S you will have different
qualities of sales that is quantity of sales, you have it at Q 1 for A 1 S 1, Q 2 for A 2 S 2,
Q 3 for A 3 S 3.
You have 3 quantities of sales coming for different advertising and sales promotion
you can get slightly different quantities of sales. If, you see your Q 2 is higher compared
to Q 1 and Q 3, you are getting a slightly higher quantity of sales presented through Q 2.
A marketing mix manger is likely to adopt or go to this combination of Q 2 and A 1 A 2
and S 2 to get that Q 2. So, he may like to have his advertising expenditure at A 2, his
sales promotion expenditure at S 2 so, that you can realize this Q 2 sales. Now, all these
could also be represented by A 2 dimensional graph that is you have got advertising on
the Y axis and sales promotion on the X axis.
Now, you draw the 3 sales curves that is Q equal to 75, Q equal to 100 and Q equal to
150. Now, in order to get at the corresponding combination of AS and SS for getting this
sales quantity, what you are going to do is draw a tangent to each of these curves, that is
each of these ISO sales curves you draw a tangent. And, when you draw a tangent the
point at which this tangent cuts this ISO sales curves, it gives the constant budget line
where you can get the combination of A and S; the combination of A and S.
This is very useful to the marketing manager, because he has to make decisions, what is
he trying to get by doing this he is getting optimal marketing mix for a given marketing
budget. In this first diagram that is the three dimensional diagram, what we have looked
at is the relation of sales to different marketing mixes of advertising and promotion. what
we are looking at in the next diagram is finding the optimal marketing mix for a given
marketing budget.
Here you are representing it as A 3 dimensional diagram, here you are representing it as
A 2 dimensional graph and whichever is more understood by you can have, you can take
that particular method of representation if you think that, it is better for you to have A 2
dimensional graph, where you can see with reference to different sales curves, that is the
ISO sales curves. You can get a combination of the advertising and the sales promotion
expenditures.
In other words what we are trying to do in this is for the marketing manager creating
different scenarios ok. You take two elements advertising and sales promotion. And, now
you look at the constant your budget is constant you have the mix constant, mix line, you
find out how you can do different for different mixes of advertising and sales promotion,
what could be the sales?
Now, when your budget is constant here in the second graph, you can look at drawing a
tangent to each of the sales curves that is the ISO sales curves to get the optimal
marketing mix for a given marketing budget.
(Refer Slide Time: 08:13)
From this we move up further; why we are doing all these? This is the next question we
have to answer. We are doing all these to device different business unit strategies. And,
from the different business unit’s strategies, you go to the corporate strategy, but kindly
note that if there are 5 business units, if you summit that 5 business unit strategies, you
will not get the corporate strategy.
In other words, corporate strategy is not the summation of the individual business
strategies, that formula does not work there. When, we are looking at strategic
marketing, we are looking at the business unit strategy. When we are looking at the
corporate strategy, we are looking at the strategy of the company as a whole.
The business unit strategy looks at the individual business units, the corporate strategy
looks at the company as a whole. There are few things which are very critical to any
company; one is what is called the strategic decision. What is the strategic decision? This
decision is likely to have a long term effect, it is not that you took this decision today and
tomorrow you can reverse this decision it is not possible.
Compare it with the tactical decision; tactical decision is in order to achieve the strategic
marketing objectives, you will make some adjustments in the operational strategy. Those
decisions can be dynamic in nature you can take it on a data to day basis and when you
take it on a day to day basis, along with those decision you will have different budgets
scenarios coming in, you will have different mixes scenarios coming, but all with the
ultimate objective of achieving this strategic objective with respect to the business unit.
When you are looking at a business unit you are looking at strategic marketing with
respect to the business unit, when you are looking at the company as a whole you are
looking at corporate strategy. A strategic decision cannot be reversed a tactical decision
is dynamic in nature, and it depends on the situation that is prevalent in the market.
Depending on what type of situation comes in the market you keep sometimes
proactively acting to that or reactively responding to that.
It is a dynamic decision which is taken to achieve the strategic objectives which you had
earlier envisaged with respect to the business unit. So, in this slide here what we have
tried to present is; one is the business unit strategy, second is the what strategic
marketing is concerned with, third is the difference between the strategic decision and
the tactical decision.
(Refer Slide Time: 12:35)
Let us look at the silent features of strategic marketing; what is that we are trying to do,
when we are looking at strategy in when we are looking at strategic marketing. The
important aspect that we are looking at is emphasis of long-term implications. As a
marketing manager you may have to take decisions, which are spur of the movement,
that spur of the movement decisions are not coming into strategic decisions kindly note.
You might have one customer might bring to your knowledge that a particular product is
available at a much lower price then is being sold in your store. Then, you may have to
take a decision to decrease the price of the product in your store, that is a reaction to
what is happening in the market place. Kindly note that this is not a strategic decision,
this is a tactical decision where you are saying as a marketing manager I have taken the
decision to decrease the price of this product.
You can see in a place like big bazaar, take these vegetables. The price of vegetables are
constantly changing, today you may have 1 price for tomatoes tomorrow it may be
increasing or it may be decreasing depending on the supply that is coming in. A very
good example could be your onion a few days back the price of onion had shot up to
more than 60 rupees per kg. Now, with the supplies increasing due to several
government measures like putting a ban on export of onions, and also taxing the import
of onions what has happened is you have a higher supply of onions to the market.
If you looked at the price of onion, today you are getting it around 30 to 35 rupees per
kg. What was some 60 plus rupees per kg, you are getting it at 30 to 35 the indications
are that it is slightly to come down further. As a marketing manager if you are in charge
of this unit that is the vegetables unit, you have to be always on the go; on the go
meaning that you have to keep a watch on what your competitors are selling, at what
price they are selling?
When you are doing this, when you are taking this type of decision there are some
implications. You reacted to a situation by saying since the other competitor is offering
at a lower price, I am trying to match his price at best or even offer lower, but you have
to look at what is going to be the long term implications of such decision making.
If you take an outlet like Big Bazaar, on Wednesdays he offers many of these vegetables
at a much lower price compare to his competitors. He puts the condition that one can
pick up only 2 kgs or 3 kgs of onions.
What does the usual cart vendors or others who do business of these onions or different
vegetables do? They come to this big bazaar in large numbers that is their own group
comes, and takes these 2 kgs, 3 kgs at a much lower price, and he will sell that good at a
much higher price in his own cart or anywhere in this open market.
This type of implication also the marketing manager has got to study. Suppose the price
of a particular good is at x, for a particular day if you say I am going to reduce it by 0.5
x, it becomes instead of x 0.5 x, what are the long term implications, what is going to
happen? So, many of these vegetables may be perishable in nature and these vegetables
have to be sold during a particular time frame.
You may be at an advantage or disadvantage with respect to the pricing decisions, which
the marketing manager has to watch very closely. In other words, the job of the strategic
marketing manager is a tough job, it is not an easy job. One is, he has to be constantly
looking at, what is happening in the market place. Second is whatever decisions he is
going to take, it will have implications. Implications with respect to the company as a
whole and he has the obligation to explain, why he took this decision if asked by the
management.
He has to convince the management as to why these decisions are taken from his side. In
other words, this strategic marketing manager always has to be very objective in his
decision making. Clarity in decision making is called for.
In order to enable him to do this he may get inputs from the corporate also, that is the
company might provide him lot of inputs saying that take example of Big Bazaar. They
may provide him reliance fresh is pricing this product at this rate, Wal-Mart is pricing at
this rate all those types of things might be, these inputs might be provided to him from
the corporate body of this big bazaar itself.
Depending on the type of inputs that he is getting, he may have to change the roles for
different products and markets. For different products and markets, you may have to
change adopt different roles for a perishable market a different role, for a durable market
a different role. Consumer durables may call for a different pricing strategy, consumer
perishables may call for a different pricing strategy. In other words, this marketing
mangers job that is the strategic marketing mangers job is extremely challenging and he
has to be watch full always.
The business landscape in which he is operating is poses for him so many challenges on
an hour to hour basis, when this challenges are coming to him on an hour to hour basis,
he has to many times act pro actively or reactively. When you go to many of these stores
you may find so, many sales people also there, you may also find so, many middle level
managers also being there assisting the sales people. Why they are assisting the sales
people? They are helping them with many decision making inputs.
Whenever they are suppose some supply of some material is not there the job of this
middle level manager is to find out how can this supply get augmented or immediately
replenished. All these types of decisions are taken by middle level marketing managers.
This is in other words, strategic marketing and corporate strategy, they are intertwined;
you cannot separate, but as I mentioned earlier when you are looking at a strategic
marketing you are looking at a business unit, by just summating this strategies of the
different business units you are not going to get the corporate strategy, that is just the
some are not going to result in the corporate strategy.
When you are looking at the corporate strategy you look at a more holistic picture,
compared to what you are looking at with respect to the business unit. So, this is how the
strategic marketing and the corporate strategy are linked and also slightly different.
(Refer Slide Time: 23:17)
What are the strategic implications of all these? When you are looking at strategic
implications, you are looking at from marketing aggregation to targeting. You may like
to aggregate different market segments, saying for this particular product we can look at
so, many segments 1 2 3 4 5.
Take for example, the washing machine. You may have different models of washing
machines; one could be semiautomatic, second could be fully automatic, one could be
semiautomatic with not full drying semi drying, a fully automatic not with full drying,
but 80 percent drying, then there could be a separate drying unit itself. So, you may take
the cloths which are dried to get into 100 percent drying by putting into this drying unit.
There may be suppose it is a cold region you may like to opt or the consumer might like
to opt for this drying unit also, because there is he may not have place, or he may not get
that sun shine, to get that remaining 20 percent drying. He may say I am using this
washing machine here and I am putting it also to the drying, you find this very
commonly in place in most of the western countries you have a drier. And, that drier
makes it 100 percent dry your cloths are 100 percent dried in this.
When you are looking at segments, you are looking at market aggregation unto targeting.
That drying unit may be very useful in the western context, it may not be so, useful in the
Indian context. You may you may find that the Indian consumer is quite happy or
contented with the washing machine. He may not require the drying unit separately. He
may be happy with his 70 percent or 80 percent drying, which the washing machine
gives, he puts it for drying outside and in a few hours this gets dried fully.
That what is the this is the first implication he has to get from market aggregation to
targeting markets, that is segmenting the markets according to particular targets. All
these terminologies are becoming or is getting used differently in the market now, you
call it a niche market segment. When you say it is a niche market segment, you have to
find out what is this niche which this particular product is targeting on.
Take the example of this Nirma washing powder. So, it is it was concentrating on a niche
marketing, that is it was concentrating on a market which was highly susceptible to price.
That is people whole could not afford your surf, but wanted to use a washing powder at a
much lower price, that is the lower income strata of the Indian society, the Nirma was
targeting. He was targeting such that this section could afford this particular washing
powder.
But, since this particular segment the sales in this particular segment of this Nirma
washing powder increased to such a level, it became a threat to surf only. The result was
Hindustan lever had to come out with promotional campaigns as to differentiate the
cloths, which were washed by Nirma and that by surf. This advertising or promotion
helped to find surf it is due recognition in the market place.
The second one is what are we trying to do - what is the strategic marketing manager
trying to do? He is with respect to the organization or the brand or the product he is
putting a positioning map in the intended customers mind. He is saying- take the same
example which I gave you Nirma, a middle class user or a higher income segment user,
would never look at Nirma. He would look at surf only or products in that range, why he
would look at it because the attributes of surf (Refer Time: 29:06) Nirma, they are
different according to him in his intended mind set.
Similarly, if you look at different brands with respect to TVs you may say LG is at this
level compared to that your Panasonic is at a different level, compared to that the Sony is
at a different level. Which one you can afford LG may be available to you at a lower rate
Panasonic, a little higher Sony much higher which one do you want to take. Depends on
the customer’s budget and where he wants to get associated the brand; the brand
positioning.
In this whole scenario what it implies is the brand equity of a product? So, a brand equity
of a product comes into play and that brand equity gets depending on the value of the
brand a higher or a lower value to your product. In order to get to this positioning in the
intended customers mind, you many times marketing managers make use of symbolism,
that is strategic use of symbolism.
If you see the mobile phones now, the latest version of the android phone, you get a
symbol suppose your phone is not answered you get a symbol where a figure comes
saying that your contact may be out of reach. So, he shows by a symbol. You do not have
to really look at what is the voice which is coming in the symbolism is enough to tell
you, you are not able to reach your the person to whom you phoned.
The implications of all this is terrific with respect to product innovations and value
decisions. Keep on innovating your products. In other words, what is happening is in a
competitive market scenario, there is a lot of value which gets which gets attached to the
feedback of the customer. And, depending on the type of feedback which comes from the
customer side the product innovations start coming into play.
You start innovating on your product continuously this is not a onetime process. It is a
continuous process and where this value decisions come into play suppose, I innovate
like this what is going to be the effect on price that is the impingement on the price of the
product. Now, what is happening in the present market scenario? The markets have
become digital online.
Two things are happening here; one is your e shopping is becoming more pronounced,
but many times in e shopping you may not get the feel of the product. The consumer
might like to have a feel of the product, then sometimes he may say I would like to go to
store physically and feel the product before making a decision to buy the product. Then
e-shopping fails to address that particular criterion.
Now, what are you seeing with respect to all these digital marketing and other types of
online marketing? You are seeing a terrific spurt with respect to business to business
markets and also consumer markets. You may your seeing so, many online marketing
companies, catering to different consumer needs, even your whether it is your
vegetables, or your groceries, or many other items like your cloths. The online market is
coming in straightly to your house and telling this is available to you, why do not you
order.
e-shopping, B to B markets and consumer markets have changed the scenario, in which
the present marketing is operating. You have a lot of changes which has come into the
market place. The result is you have tremendous opportunities, if you are able to adapt to
this situation. Tremendous opportunities are getting open for you if you are able to adapt
to this change as a player, we stop here we will continue in the next class. | 4,428 | 20,153 | {"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} | 3.3125 | 3 | CC-MAIN-2023-40 | latest | en | 0.969692 |
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Sp11_Quiz7_soln
# Sp11_Quiz7_soln - EGN 3420 Quiz 7 solution Sp 2011 For the...
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EGN 3420 Quiz 7 solution Sp 2011 For the table of data points given below x i y i log ii ux log zy 2 i u uz 1 1 0 0 0 0 2 6 0.3010 0.7782 0.0906 0.2342 3 13 0.4771 1.1139 0.2276 0.5315 4 24 0.6021 1.3802 0.3625 0.8310 5 40 0.6990 1.6021 0.4886 1.1198 2.0792 i u 4.8744 i z 2 1.1693 i u 2.7165 i) A power equation model b y ax is used for interpolation. Fill in the table and solve for the coefficients a and b . (8 pts) a) 0.8615, 2.5405 ab b) 1.2582, 2.3559 c) 1.0810, 2.2631 d) 1.5000, 2.0000 e) 0.7505, 2.7519 d) none of the above 01 2 i i i i z a a u na u a z u a u a u z 5 2.0792 4.8744 2.0709 1.1693 2.7165 aa 0 0.0338 1 0.0338, 2.2631 10 10 1.0810 2.2631 a a ba ii) Use the power equation to estimate y when
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Ask a homework question - tutors are online | 590 | 1,366 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2018-09 | latest | en | 0.524128 |
https://algebra-answer.com/tutorials-2/converting-fractions/beginning-algebra-2.html | 1,618,397,881,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077810.20/warc/CC-MAIN-20210414095300-20210414125300-00378.warc.gz | 203,227,421 | 9,022 | # BEGINNING ALGEBRA
Course Description:
Provide the student with an understanding of elementary algebra. The course will cover topics
such as operations with signed numbers, factoring, and solving linear and quadratic equations
and graphing linear equations. Successful completion will prepare the student for MAT 020
Intermediate Algebra.
Developmental Education – MAT 015 will not apply toward graduation requirements.
Course Prerequisite:
To enroll in MAT 015, Beginning Algebra, a student must meet one of the following:
1. MAT 014, Beginning Mathematics, with a grade of “C” or above.
2. Qualifying scores on the ACT, COMPASS or ASSET tests.
Required Text:
Beginning Algebra, Lial, Hornsby, McGinnis, 9th edition, 2004.
Course Objectives/Outcomes:
The primary objective of this course is to prepare the student to be successful in MAT 020,
Intermediate Algebra. On completion, the student should have achieved these outcomes:
I. Arithmetic and Algebra Manipulation
A. Evaluate arithmetic expressions (including absolute values) using the order of operations
and properties of real numbers.
B. Evaluate algebraic expressions.
C. Apply the laws of exponents to simplify expressions containing integer exponents.
D. Express numbers in scientific notation.
E. Perform addition, subtraction, multiplication and division on polynomial expression .
F. Factor expressions with common factors, expression that require grouping , trinomial
expressions, and difference of square expressions.
G. Perform addition, subtraction, multiplication and division on rational expressions.
H. Evaluate radicals, approximating those that are irrational.
I. Simply numeric radicals using the product and quotient rules.
II. Equations and Inequalities
A. Solve linear equations in one variable.
B. Solve proportion equations .
C. Solve linear inequalities in one variable showing solutions on a number line.
D. Solve literal equations that do not require factoring.
E. Solve quadratic equations by factoring.
F. Develop and solve mathematical models including number, geometry, and percent
applications.
III. Graphs on a Coordinate Plane
A. Plot points on a coordinate plane.
B. Graph linear equations, by plotting points.
C. Graph linear equations using intercepts.
D. Graph linear equations using the y- intercept and slope .
IV. Analysis of Equations and Graphs
A. Identify the x-intercept, y-intercept, and slope of the line given its graph.
B. Identify the x-intercept, y-intercept, and slope of the line given its equation.
C. Determine the equation of a line given its graph, its slope and y-intercept, and its slope
and a point.
D. Determine equations of both horizontal and vertical lines.
E. Determine whether or not an equation is linear.
F. Calculate the slope of a line passing through two given points.
Method of Instruction :
A study assignment sheet will be provided at the beginning of each chapter or unit. Everyone
must keep up with the homework assignments. Homework must be completed prior to the next
class meeting. This is essential to understanding the problems and test preparation. A
homework quiz - selected homework problems, will need to be recopied in class and turned in
for homework credit during a select timeframe, usually 5 minutes for each homework section.
These select problems will be determined by the instructor and handed in at the end of the
announced timeframe. You will not be able to use your book or your neighbor, only your
homework papers or notebook. You must be in class for these homework credit opportunities.
Those who miss the quiz cannot make it up, unless it is due to a school-sponsored activity and
then it must be accomplished during office hours within two class days of the missed quiz.
Don’t miss those homework points, they are the only scores for your homework grade.
Homework problems will not be accepted if only answers appear on the homework quiz -
work must be shown. The minimum acceptable work for credit on problems is as follows:
1. State the problem.
2. Show any necessary work steps.
Class time is used to answer student questions from the previous assignment and to
introduce/provide explanation of the current assignment. Reviews are accomplished prior to the
chapter or unit test. No homework quizzes will be given on a test day. Pop quizzes may be
given besides homework quizzes, be prepared! Calculators are permitted.
Students with particular questions on material covered in class or assigned problems may get
help from the instructor. Office hours will be posted. Additionally, math tutors, software
tutorials on computers, and videos are available in the Instructional Resource Center (IRC).
These should be used as additional resource material. They can be a great aid to your success.
You are highly encouraged to take advantage of these aids. In fact, the instructor may inform
you that tutoring should be started. If this happens – start tutoring, you need help at that point
rather than later, if you realize it or not. Check in the IRC for times and availability . It’s free!
Method of Evaluation:
Allen County Community College has implemented a student academic gains assessment
program. The purpose of this program is to ensure instructional excellence and to maintain
regional accreditation. As a part of this program all students are expected to take an Entrance
exam upon enrolling and an Exit exam upon graduating. Students will also be asked by their
instructors to participate in some form of pre/post evaluation in each class.
Class grades are determined by: Homework quizzes- 20%; Chapter or unit - 80%; with the
A 90% or above
B 80 to 89%
C 70 to 79%
D 60 to 69%
F below 60%.
The class is not graded on the curve, nor or any tests graded on the curve.
Testing policy: Chapter or unit tests will be announced in advance. Therefore, the student is
expected to be in class to take the test on that day. Only through prior arrangements will you be
excused from taking the test at the announced time. Otherwise, you are not excused, and your
grade for the test is a zero . Do not expect a make-up test! Since this is a developmental course
with the primary objective of preparing the student for Intermediate Algebra, students may retake
an exam, and receive the better of the two scores. A student with a test score less than 80% is
expected to retake. Those with higher grades may request to take the retake. Retakes must be
taken during the announced retake window, generally 4-5 school days after the initial test is
returned to the students in class. If you miss this window of opportunity, you miss the chance for
the retake. Retakes are a special opportunity, not a guarantee. Plan your life accordingly.
Once a test is started, there is no stopping. It must be completed prior to leaving the room. This
includes testing in the IRC. The final exam will be taken in the announced timeframe and cannot
be retaken. Pencils will be used in this class for exams and homework quizzes. If you don’t
have a pencil, get one. Partial credit cannot be given if work is not shown on exams.
Dishonesty will result in failure for the course and or academic dismissal from class. Do your
own work when taking quizzes and tests. Teamwork is encouraged during other times, both
inside and outside of class.
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https://flex.phys.tohoku.ac.jp/english/pukiwiki-e/index.php?cmd=backup&action=source&page=Moya-NanoJapan-2013%28Open%29&age=27 | 1,669,524,408,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710192.90/warc/CC-MAIN-20221127041342-20221127071342-00701.warc.gz | 296,483,316 | 11,866 | Started by Hasdeo.2013.05.31
This is Moya's logbook for Nano Japan project from 2012.6.2-2012.7.31.
#contents
* Daily schedule [#daily]
- 09:00-10:00 Discussion (1)
- 10:00-12:00 Solve the problem and write the progress on pukiwiki
- 12:00-13:30 Lunch
- 13:30-14:30 Discussion (2)
- 14:30-16:30 Solve the problem and write the progress on pukiwiki
- 16:30-17:30 e-mail report and problems
* Goal of the project. [#goal]
- To describe the thermal conductivity of Nanotubes and graphene. The output will be an animation of atomic vibration and its propagation in Mathematica
- Keywords: Thermal conductivity, carbon nanotubes, Differential equation, Fourier Transform, Solid-state physics, Mathematica
* Schedule for discussion [#discussion]
|Time and Day|Mon|Tue|Wed|Thu|Fri|
|09:00-10:00|Hasdeo|Tatsumi|Mizuno|Thomas|Nugraha|
|13:30-14:30|Thomas|Mizuno|Tatsumi|[group meeting]|Hasdeo|
* Division of tutors [#v0d44b28]
|Name|Lesson to teach|Misc|
|Hasdeo|Phonon in Nanotubes| How to make a presentation, and Japanese class|
|Tatsumi|Nanotubes| Printer and Copy machine|
|Thomas|Physical Mathematics| Library|
|Mizuno|Physical Computation| Computer in Lab|
|Nugraha|Phonon in Nanotubes| |
- Tutorial takes place in Coffee room, since visitor room will be occupied with many people
* Questions and Answers [#QA]
- This section is for posting questions from Moya-san and answers from other group members.
- Please list here with some simple reasons or details.
- For every problem, give a tag double asterisks (**) in the code so that it will appear in the table of contents.
- For the answer, give a tag triple asterisks (***) in the code below the problem in order to make a proper alignment.
- List from new to old.
* Report: [#d6196c2a]
** June 3: [#k93c5c39]
Today, Hasdeo-san picked me up from Urban Castle Kawauchi (UCK) with Thomas-san. I learned how to get to the lab by campus bus and have been introduced to the lab, Saito-sensei and the secretaries. Hadeo-san taught me about coupled oscillators, where I learned how to solve for the motion of two masses connected by springs. The goal is to model heat in CNT's as atoms vibrating with springs. Thomas-san also lectured me and taught me some math. He taught me how to solve the 2-dimensional Laplacian using separation of variables. He also provided some insight on the applications solving the laplacian. So far, everyone seems very nice and I am very excited to start working on thermal conductivity, which was a surprise.
** June 4: [#p63feebc]
Today, I walked from Urban Castle Kawauchi to the lab and managed to find my way, but I might have strayed a bit! I came early to skype my family back home, but began my lessons with Tatsumi-san at 9:00. He taught me about the graphene and carbon nanotube lattice and how vectors compose the unit lattice to form the Chiral Vector and Translational Vector. He also showed me formulas which describe the nanotube lattice. For homework, I was assigned to build a carbon nanotube out of paper and I have already drawn the Chiral Vector, but have not finished the job as I need to borrow scissors and tape. I also spoke with Saito-sensei and discussed the oscillator between two walls. I have a much better understanding of this situation and Saito-sensei assigned me to work on a similar model including a drag force. I believe I understand this problem, but am not to sure about the graphs which show resonance.
After lunch, Mizuno-san introduced me to Mathematica. I do not understand it that well, but for homework I am to research it online and learn different functions. I also spent some time playing with the functions and getting the hang of working with differential equations and plotting. I am also supposed to work on an animation, but this seems rather difficult at the time. Saito-sense also showed me how to calculate different aspects of the CNT today. I learned a lot from our meeting and am very excited to work on the problems he sets up for me. I am to work on a time dependent oscillating function for homework tonight and hopefully I can get this one right and continue to expand.
** June 5: [#gf3952b1]
I walked to the lab again today and met with Mizuno-san again to discuss Mathematica. He taught me about Rules and Patterns in programming and showed me several functions in mathematica. For homework I am to study graphics and attempt to make a figure of a particle between springs. I am also to study animation and manipulation, so I can modify this graphic. These are not very easy tasks as I do not know much about programming and I plan on looking for material to help me on mathematica. I have not begun the homework yet as I met with Saito-sensei to discuss more on a particle between two springs with an applied force. I believe I am starting to get the hang of solving these differential equations as I am beginning to recall what I learned last semester in my Differential Equations class. Homework from Saito-sensei involved me recalculating the solutions to the equations and plotting them on mathematica. I think I have successfully plottted them, but plan to ask Saito-sensei if it is a correct graph. I will try to solve the differential equation of the particle with friction and later try to derive the calculations for the properties of the CNT that Saito-sensei showed me yesterday.
After lunch, I worked on the plot of the solution on mathematica and went online to learn different functions of mathematica. I learned some shortcuts in typing and how to edit graphics, but I still have no idea how to make good graphics or animations.I met with Tatsumi-san and he explained some basic CNT properties like chiralities and metallic/semiconductor conditions. After this we discussed relationships between energy and temperature and energy and wavelength. We then went on to attempt to solve the Heat Equation. I understood the steps to solve the heat equation up until we had to solve for the constants. Solving for the constants required some ugly integrals with summations and trigonometry tricks that I do not know. I hope to look into this more when I go home and get a better understanding of this. Afterwards I continued to study the motion of the particle and work on its plot in Mathematica. I showed this plot to Saito-sensei and he explained the physical interpretation of the graph which dealt with beats and periods. he also went on about the imaginary part of damped oscillations. I have some homework for tonight on the imaginary solution as the imaginary part is key to understanding the transformation of heat in these situations.
** June 6: [#e24a9280]
Today I arrived at the lab early and worked on solving the damped oscillator equation and plotting the solution in mathematica. I successfully plotted the solution and was able manipulate the variables, so I could study the physical event. However, Saito-sensei would later explain this graph in more detail giving me a much better understanding. I meant with Thomas-san and he had me solve the 3-D Laplacian with boundary equations. This problem worked into a real ugly system of equations, but I understood the concept behind it. I continued to reteach myself some differential equations with Thomas-san and he later explained his presentation to me. He taught me about the solving the Laplacian in spherical coordinates, but I couldn't quite understand it and I plan on looking into it again. I then attended to 'group meeting' and I got to learn from Thomas-san's and Nugraha-san's presentations.I learned about the tough standards Saito-sensei has for presentations and a little about CNT's that Nugraha-san talked about. Later on, Saito-sensei explained the damped oscillator and has assigned me to study a coupled oscillator. I think I successfully modeled the non-damped homogeneous solution, but I have had some trouble solving the driven equation and plan to ask Hasdeo-san or Saito-sensei for help later. The goal is still to understand the transfer of energy through the springs.
** June 7: [#wae9b44d]
Today I met with Nugraha-san and he helped me solve the driven coupled oscillator that Saito-sensei assigned me. He taught me a better trial solution to use which let me understand the phase difference. However, I think we set up our differential equations wrong and I plan on solving them again. Nugraha-san also introduced infinite coupled oscillators. We have not yet introduced a force to them, but have worked on solving the differential equations. He explained the wavevector and lent me a book to that will help explain what I learned. He assigned me to study an infinite series of different coupled masses. I plan to work on this over the weekend and show it to him when I have time. I met with Saito-sensei as well and he showed me the error in our differential equations for the coupled oscillator and introduced some concepts behind the wavevector and optical/acoustic phonons.
** June 10: [#a77f1b0a]
Today,Hasdeo-san helped me finish the homework that Nugraha-san had assigned. This was to solve for the displacement of particles (coupled by springs) in an infinite lattice where the lattice cell is constructed of two particles of different mass as opposed to one. We found the solutions to the system of differential equations and he introduced the concept of zone boundaries and which phonons are acoustic or optical. I understand the math behind these conclusions, but am still rather confused by the physical meaning of frequency as a function of the wavevector. After discussing the acoustic and optical modes, Hasdeo-san introduced the model of thermal conductivity for a crystal lattice. He taught me how the transfer of energy is related to the phase shift in the two particles. I still do not completely understand this concept and plan on studying it more. After lunch I went with Thomas-san and he introduced the laplacian in polar coordinates. We had some trouble converting the initial definition, but I asked Saito-sensei for helped and he helped me work my way through. With some help from the internet, I understand the process of converting it and will begin to studying solving the differential equation. Hasdeo-san also came to my room again and discussed the thermal conductivity again and how it relates to the velocity of the center of mass. For homework, I am to study how heat transfer relates to the phase of different atoms in the atomic lattice.
** June 11: [#ib956120]
Today Tatsumi-san was absent, but I managed to work on the problems that were assigned to me yesterday. First I worked on plotting the dispersion plot where frequency is a function of wavevector for a two atom periodic lattice. While I attained the right plot, I am still rather lost behind the meaning of the plot. I am still confused as to the strict definition of a phonon and how the modes are determined at various k and w values. After achieving this graph, I began to work on Hasdeo-san's homework which consisted of calculating a constant which describes heat transfer in the lattice. I managed to calculate this lattice,but am still confused as to how the relationship is attained. I managed to plot the relationship of the constant vs frequency and learned the characteristics of this relationship. However, I do have some questions regarding the imaginary part of the solution and what it means for the conductivity to approach 'negative infinity'. Next I want to attempt to solve for this constant with forces at both walls and then to work on achieving this constant for a periodic lattice. Afterwards, I met with Mizuno-san and he taught me some more about Mathematica. I learned about rules and different functions like Flatten, Transpose, Real, and Refine. These different functions can be really useful for manipulating data in Mathematica. He also helped me work on the periodic lattice problem I had been considering. I also met with Saito-sensei today and he forced me to present what I have learned. He was pretty picky, but I think I did rather well for coming up with it on the spot and barely understanding it myself. I wish I could have talked to Saito-sensei more about science, but I understand the importance of presenting what I know. For tomorrow, I am to work on a presentation of a single atom infinite lattice and learn about boltzmann's constant and heat transfer. I also want to ask Hasdeo-san about the meaning of the velocity of the center of mass and how it relates to energy.
**June 12: [#s4419843]
Today, I met with Mizuno-san and he continued to show me different functions of Mathematica. I learned different input methods for basic functions and how to handle matrices and vectors.I also learned how to manipulate solutions of differential equations with NDSolve. For homework, I was assigned to look up Graphics and understand them. I looked up different graphics and have began understanding the various how to use lines, polygons, circles and how to edit their style. I first want to make a solid graphic of a single oscillator and then I want to try to animate it. I hope to ask Mizuno-san for help on animating the object. Afterwards, I met with Saito-sensei and he introduced the relationship of energy to our model of springs. He showed me various laws and equations that relate energy, motion, heat, and conductivity. Today, I will study and attempt to understand these various equations.
I continued to work on diffusion equations and I have a strong understanding of their derivation and physical implications. I continued to work on understand these and began to research the relationships between phonon modes, energy, and temperature. I stumbled upon the Debye and Einstein models, which Saito-sensei later encouraged me to look at. They involve some difficult math and physics I don't quite understand, but I may need some information that it has to offer. I met with Hasdeo-san to ask him various questions about the wave vector, energy equations and heat relationships. I now have a better understanding of the wave vector and the concept of phonons in solid materials. I also met with Saito-sense again and he has advised to attempt to solve the Heat equation with certain boundary conditions and to study the Debye model.
** June 13: [#zbf07e54]
Today I met met with Thomas-san in the morning and we worked on solving the 2D polar laplacian. I understand what we are trying to do and can solve the equation in a general method, but am having trouble with the Fourier series solution. I borrowed Thomas-san's book on Fourier series and plan to study the solution to this differential equation. On my own time, I attempted to study some basic statistical mechanics, but am having a decent amount of trouble with the subject and plan on asking Hasdeo-san and Nugraha-san for help on it tomorrow. I particularly stuck on the derivation of Boltzman's factor and the partition function. I want to understand these topics so that I can understand the Debye Model which provides a relationship between phonon modes, heat, and energy. Hasdeo-san has provided me the assignment of learning to derive the distribution functions, but I am having trouble with this as well. I hope to ask Hasdeo-san for help tomorrow and hopefully I can receive a better understanding of the subject.
** June 14: [#r357f51e]
Today I met with Nugraha-san in he morning and he introduced the expansion of coupled oscillators to containing N oscillators. He showed me how to solve for the frequency and amplitudes of movement for the oscillators. In order to solve them, I was instructed to use Mathematica. I have attempted to solve the matrix, but am having some trouble with obtaining the final solution. The program provides several solutions, that I don't quite understand. I also found the relationship of N coupled oscillators with a damping constant in the springs. This was not very difficult, but I still cannot solve the matrix in mathematica, which I will have to learn to do.
** June 17: [#cf8661e7]
I spent the first part of my day discussing my project with Hasdeo-san. We talked over the basic ideas behind the project and tried developing a simple model describing energy transfer in a system of oscillators. We left the room confused as we could not understand how to relate the temperature of the wall vibrations and force of the wall. After lunch I met with Hasdeo-san again and we may have created a simple way to observe heat transfer. This method consists of comparing the amplitudes of vibration to energy for specific 'points'. These points are atoms in the lattice which may provide a general structure for a plot fit. We plan to use the general solution to the heat equation to fit the data. I will want to use Mathematica to generate the exact solution. I will begin working on this simple model right now and plan to extend this model to multiple atoms later. I also want to consider using phonons in the model in time.
I met with Thomas-san today and we worked on general solutions and fourier series of the heat equation.
** June 18: [#oe7b407b]
Tatsumi-san wasn't here to lecture me, so I worked on my project on my own. My initial idea with Hasdeo-san has hit a road block as the 'spring constant' appears in the relationship of the amplitudes. I do not know if there is a way to solve for this constant, or any assumptions I can make to eliminate the constant. I will have to talk with Hasdeo-san later to see if we can find a way to complete the relationship between the amplitudes of movement. I am also interested in determining the frequency of movement of the wall. We are assuming the wall vibrates at a set frequency, but is there anyway to determine frequency? I also studied the heat equation and using Fourier series to find a solution. I also continued to study phonons and have began looking into the quantum nature of phonons, but this topic seems to be to difficult to allow easy application in our model.
In the afternoon, Mizuno-san taught me more about mathematica. He showed me how to work with graphics and animations and we created a small animation of the 'balls between springs'. This animation is not that great, but its a start! I think I understand the process very well and he has also taught me several functions as well. I also continued to work on my project, and talked to Hasdeo-san about several problems I had. I think we have a running model, but I will have to input values for the 'spring constant' and the 'frequency of vibration'. I also found errors in some of my derivations, so I will have to find these relationships again. I plan to work on finishing a plot tomorrow and to continue expanding the idea to many atoms.
** June 19: [#f92b7927]
This morning I meant with Mizuno-san and we continued to work on the animation of the atoms between springs. We managed to get a realistic animation for the case with several atoms, but are now trying to extend the model to multiple atoms. After showing me how to use Tables, and Block, Mizuno-san assigned me to generate a code that would produce the animation for 'n' atoms. I've been working on this for a while and have hit a problem where I cannot assign the equations of motion to the animated objects. I will need to ask Mizuno-san for help later on.
I spent the rest of my day working on my project. I used mathematica to try to solve for 'lambda' which represents the thermal diffusivity of the solid. I managed to solve for constants representing the general solution of the heat equation for my two atom system, but lambda is both negative and imaginary. I do not know if this makes physical sense and will need to ask Hasdeo-san for his opinion on the matter. Hasdeo-san also showed me some software on flex that may help fitting a plot to the points I obtained.
** June 20: [#keffb9a7]
This morning I began looking into Fourier transforms as they keep appearing in the quantum definition of phonons. I met with Thomas-san and we worked through some differential equations using Fourier series and we derived the fourier transform for the Gaussian distribution. I understand that Fourier transforms are used to make a time dependent function, only dependent on the frequency of a periodic function, but I need to look into the derivation of fourier transforms and their application physics. However, Thomas-san showed me some interesting applications that Fourier Transforms have in imaging.
I spent the afternoon continuing to work on my project. Hasdeo-san and I came to the conclusion that Boltzmann's constant was to small and affecting the calculations on my computer, so we placed Boltzmann's constant=1. This should hopefully keep the values on a larger magnitude which makes it easier to relate. I used this value and obtained a much better list plot of the three points which seems to make physical sense, but I cannot find an acceptable best fit line using the general solution to the heat equation. I keep receiving graphs whose data is way off point. I think I may need may points and I may have to take the time dependence of x along with time to use more points for a better fit. However, I would not know how to fit this into Mathematica and may need to ask for help. I believe that these points are correct, but the general solution of the heat solution is not the correct line that we should be fitting to. Hasdeo-san also began teaching me about Carbon Nanotubes and he reviewed the basic structure of the CNT and began teaching me about energy dispersion and the relationship to the brillioun zone. I hope to learn more about the carbon nanotube and to understand phonons in the CNT so I can later model it.
** June 21: [#t0eb9e69]
Today Nugraha-san was busy with his thesis, so I continued to study various things on my own. I continued to study the brillioun zone and phonons in CNT's. I have begun to understand the process of analyzing phonons in CNT's but am a little lost when the math shows up. I will need to ask Hasdeo-san to explain some details to me. Nugraha-san also brought some sheets for me to read that taught me about modes of vibration in solid materials. In the afternoon, I met with Hasdeo-san and we continued to discuss the project. I told him how I had doubt in our heat solution and we checked it to find out the solution was in fact wrong. We found a new solution that I would use for a fit later. When I worked on my own, the fit worked much better and realistic lambda had been found!
** June 24: [#f62b8cf4]
Today I met with Hasdeo-san and we continued to talk about my project. I showed him the lambda that I had calculated using our new heat solution. We continued to analyze my mathematica program to make sure all details were taken accounted for and we discussed certain points of the model. These points included talking about 'complex temperature' and 'complex energy' which may not be reasonable. We also noticed that 'lambda' (our constant) changed with time. Since this constant may be dynamic, Hasdeo-san has assigned me to find the 'constant's' dependence on time. I will do this by trying to find multiple values at various t's and plotting for a dependence. I hope that we can find a simple relationship between the values. I have also been assigned to work on a presentation and to begin studying phonons in CNT's and graphene.
In the afternoon, I went with Thomas-san to go a library card and had Hasdeo-san connect me to the printer. I then began working on my powerpoint and got several introductory slides done. I then worked on my project some more. I managed to obtain a plot of \lambda vs time. There is a small deviation in the values over time, but most of the values are contained in a small (about 1 unit) area. Hasdeo-san wants me to make an animation where the plot will change with time accordingly, so we can monitor the time dependence. He also gave me a paper to read that discusses thermal properties of carbon nanotubes. I am also continuing to look into phonons in carbon nanotubes. I also would like to begin expanding the simple model to multiple atoms and then to carbon nanotubes in time.
**June 25: [#p83a0ed3]
Today, I spent my morning working on my powerpoint. I managed to finish all the slides, but will need to add graphs, tables, etc from the project I have been working on.I also continued to look up different details about phonons in graphene and carbon nanotubes. However, I am still a little stuck on some math parts that I do not understand. I will need to ask Hasdeo-san or someone to explain these concepts to me.
In the afternoon I worked with Mizuno-san on animations of connected atoms. We managed to create an animation on Mathematica of compounded oscillators. I used these skills to create different animations and graphs for my powerpoint. These graphs include simple pictures of the two atom system along with a '1D' object view that lets me label different point, which should make explaining facts simpler.
**June 26: [#k5e87d6a]
This morning I went with Mizuno-san to continue working on our animation of 'n' atoms connected by springs. The animation is working out pretty well, except that we have to chose our constants right so that the graph works out pretty. He also showed me some functions that will help me later on and began showing me how to analyze the motion of the ith atom in the series. We hope to translate the motion of the ith atom to energy in time. Afterwards, I continued to work on my powerpoint presentation and continued to study different things about the carbon nanotubes.
In the afternoon, I continued working on the powerpoint and went to Hasdeo-san for help in the afternoon. He showed me zone folding in the carbon nanotube and how phonons relate to raman spectroscopy. He then reviewed my powerpoint and I continued to study carbon nanotubes. Before I left I finished up m presentation.
**June 27 [#wbda2962]
This morning I went with Thomas-san and we began studying thermodynamic relations in crystal lattices. We looked at different relationships, and began to analyze a gas model that relates conductivity to mean free distance, etc. I then practiced my powerpoint afterwards and gave the presentation afterwards. I think I did a decent job, but it was very relax because no Professors were there and it was only the students. After my presentation, we discussed my project and what would be the next steps for getting there. I listened to the grad students discuss different topics from types of transport to how to relate heat to vibrational frequencies without quantum mechanics. We could not come to a simple answer and decided to separate to continue studying. I spent my time afterwards studying quantum transport and have a decent understanding, but will need to ask Hasdeo-san for help on specific topics later on. I also continued to study phonons in CNT's and zone folding. I think I'm beginning to understand teh derivation of phonons in CNT's, but I do not quite understand their physical relevance. For the time being, I want to continue studying quantum transport in materials. | 5,868 | 27,078 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2022-49 | latest | en | 0.821644 |
http://mathoverflow.net/questions/17331/extreme-value-theory | 1,462,037,715,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860112228.39/warc/CC-MAIN-20160428161512-00071-ip-10-239-7-51.ec2.internal.warc.gz | 181,770,488 | 15,742 | # Extreme value theory
I have been asked to provide an "approximation at infinity" of an expression that at the end simplifies to $-\frac{b e^{-a t}-a e^{-b t}}{a-b}$, in a course about extreme value theory. In the course, we saw "approximations" such as $2 t^{-\alpha }-t^{-2 \alpha }$ being approximated to $2 t^{-\alpha }$ whatever the vague word approximation means. In words, this "approximation" states that the distribution tail is dominated by the term $2 t^{-\alpha }$ at infinity. I think that there is no polynomial term $t^{-\alpha }$ which dominates at infinity in the given question, since $e^{-k t}$ decreases faster than any term $t^{-\alpha }$. However, is there any sense in which this "approximation at infinity" can be taken? I tried to take the Taylor series at infinity and Mathematica returns the same expression (unevaluated?) and computing manually, the first order approximation of $e^{-k t}$ is 0. Any ideas or references about how to compute these sort of approximations in extreme value theory are appreciated.
-
Do you have information on the relative sizes of $a$ and $b$? If we have, say $a>b$, then we can say that as $t \to \infty$, $-\frac{b e^{-a t}-a e^{-b t}}{a-b}=-e^{-b t}(\frac{b e^{-(a-b) t}-a }{a-b}) \sim \frac{a e^{-b t}}{a-b} =\frac{e^{-b t}}{1-\frac{b}{a}}$. It is rather simple, but I don't think there's much more that you can say about it.
-
If $\alpha > 0$, then you have $$\lim_{t \to \infty} {2t^{-\alpha} - t^{-2\alpha} \over 2t^{-\alpha}} = 1$$ and this is the sense in which such approximations are meant. Of course something stronger can be said -- how quickly do these limits go to 1? -- but usually when these approximations come without any qualifiers, this is what they mean.
-
So, assuming b>a>1, we would have that a an "approximation" of $\frac{1}{\left(a-b\right)}\left(ae^{-bs}-be^{-as}\right)$, when $s\rightarrow\infty$, is $\frac{1}{\left(a-b\right)}\left(-be^{-as}\right)$ in the sense that $\lim_{s\rightarrow\infty}\frac{\frac{1}{\left(a-b\right)}\left(ae^{-bs}-be^{-as}\right)}{e^{-as}}=\lim_{s\rightarrow\infty}\frac{1}{\left(a-b\right)}\left(ae^{\left(a-b\right)s}-b\right)=\frac{-b}{\left(a-b\right)}$, is that correct? Which are first order and second order approximations in this sense? – JOspina Mar 7 '10 at 0:27
Maple says it this way...
f := -(b*exp(-a*t)-a*exp(-b*t))/(a-b):
asympt(f,t) assuming a>b;
$${\frac {a}{ \left( a-b \right) {{\rm e}^{bt}}}}-{\frac {b}{ \left( a-b \right) {{\rm e}^{at}}}}$$
The first term is the most significant one.
- | 799 | 2,541 | {"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} | 3.75 | 4 | CC-MAIN-2016-18 | longest | en | 0.862813 |
http://www.jiskha.com/display.cgi?id=1361539488 | 1,495,797,987,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608659.43/warc/CC-MAIN-20170526105726-20170526125726-00462.warc.gz | 692,722,837 | 3,891 | # Math
posted by on .
2+5+8+11+14+17+20+...........=?
• Math - ,
look for combos or just add
2+8 + 20 + 5 + 11+14 +17 =
10+25+25 + 17 =
60 + 17 = 77
• Math - ,
I think you are asking for the sum of a series.
You are starting with 2 and there is a difference of 3. There is a formula to calculate this total and it won't be 77. | 118 | 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} | 3.65625 | 4 | CC-MAIN-2017-22 | latest | en | 0.927868 |
https://www.studysmarter.co.uk/explanations/math/pure-maths/operator-algebras/ | 1,719,245,354,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865401.1/warc/CC-MAIN-20240624151022-20240624181022-00717.warc.gz | 890,124,417 | 72,029 | # Operator algebras
Operator algebras are a foundational cornerstone in the realm of functional analysis, bridging the gap between abstract algebra and functional analysis through the study of algebraic operations over linear operators on a Hilbert space. This discipline offers profound insights into the structure and behaviour of operators, underpinning much of modern mathematical physics, particularly in quantum mechanics. By exploring the intricate relationships and properties within operator algebras, mathematicians unveil deeper understandings of both mathematical theories and physical phenomena.
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## Understanding Operator Algebras: Definition and Basics
Exploring the concept of Operator Algebras offers a window into a fascinating area of mathematics that intersects with various branches including functional analysis, quantum physics, and differential equations. As you delve into this topic, you'll understand its foundational role and wide-reaching applications.
### What is Operator Algebras? An Introductory Definition
Operator Algebras are mathematical structures that arise from the study of linear operators acting on a Hilbert space. These algebras consist of sets of operators that are closed under addition, scalar multiplication, operator multiplication, and taking adjoints.
Imagine a space where mathematical objects can move and interact in predictable ways. For instance, consider the operation of multiplying matrices or applying a function to a vector. These operations fall under the realm of Operator Algebras when they abide by specific mathematical rules.
Operator Algebras can be viewed as a generalization of more familiar mathematical concepts like matrices, but with a significantly broader and more complex application.
### The Role of Operator Algebras in Pure Maths
In pure mathematics, Operator Algebras play a crucial role in providing a concrete framework for understanding the abstract properties of linear operators. This framework aids in the exploration of fundamental questions in topology, complex analysis, and quantum mechanics. It's a tool that helps mathematicians navigate the complexities of infinite-dimensional spaces.
One fascinating application of Operator Algebras in pure maths involves the use of C*-algebras and Von Neumann algebras in quantum theory. These algebras help in the mathematical formulation of quantum mechanics, offering a rigorous framework for the study of quantum states and operations.
Operator algebras have found profound implications in quantum physics, particularly in understanding the Heisenberg uncertainty principle and the structure of quantum mechanics.
### Different Types of Operator Algebras
• C*-Algebras: These algebras are closed under complex conjugation and possess a norm that satisfies the C*-identity. They are foundational in the study of quantum mechanics and statistical mechanics.
• Von Neumann Algebras: Also known as W*-algebras, they are characterized by a uniquely defined dual space and are critical in quantum field theory and the mathematical formulation of quantum mechanics.
• Banach Algebras: While not exclusively considered operator algebras, they are topological algebras with a norm that allows for the analysis of convergence. They are used in various applications, including solving differential and integral equations.
Comparing these types highlights how Operator Algebras adapt to fit the needs of different mathematical and physical theories. It’s a versatile mathematical tool that underpins many modern scientific advancements.
The C*-identity, which is a condition that all C*-algebras satisfy, is given by the formula $$\|a^*a\| = \|a\|^2$$, where $$a^*$$ denotes the adjoint of $$a$$, and $$\|\cdot\|$$ is the norm.
## Exploring C*-Algebra Examples in Operator Algebras
Operator Algebras, particularly C*-algebras, are at the crest of enriching modern mathematics and theoretical physics. Their structure and the examples they encompass provide intriguing insights into mathematical concepts and real-world phenomena.
### Fundamentals of C*-Algebras
C*-algebras form a cornerstone of operator algebras, offering a framework for understanding various mathematical phenomena. They are algebraic structures that consist of linear operators acting on complex Hilbert spaces, equipped with a norm and an involutive operation known as an adjoint operation.
A C*-algebra is defined as a complex algebra $$A$$ of bounded linear operators on a Hilbert space that is closed under taking adjoints and under the norm topology, and satisfies the C*-identity: $$\|aa^{*}\| = \|a\|^2$$, for all $$a$$ in $$A$$.
An example of a C*-algebra is the set of all bounded linear operators on a Hilbert space $$\mathcal{H}$$, known as $$\mathcal{B}(\mathcal{H})$$. This includes operations such as addition, scalar multiplication, and conjugate transposition, which respect the structure of a C*-algebra.
### C*-Algebra Examples and Their Significance
Exploring examples of C*-algebras illuminates their importance in various areas of mathematics and physics. From the algebra of continuous functions on a compact space to the observables in quantum mechanics, these examples exemplify the versatility and power of C*-algebras.
• Continuous Function Algebras: The set of all continuous complex-valued functions on a compact space forms a C*-algebra. This example is crucial in topology and functional analysis.
• Matrix Algebras: Collections of $$n \times n$$ matrices over the complex numbers with matrix addition and multiplication also constitute C*-algebras, playing a key role in linear algebra and quantum mechanics.
The significance of C*-algebras in quantum mechanics lies in their ability to model observables and states, offering a rigorous mathematical foundation for the theory.
### How C*-Algebras Shape Modern Mathematics
C*-algebras not only enrich mathematical theory but also have profound implications for modern mathematics. Their influence extends to areas such as non-commutative geometry, quantum field theory, and topological dynamical systems.
By enabling mathematicians to model and analyse the structure of various spaces and phenomena, C*-algebras foster deeper understanding and facilitate breakthroughs in these advanced fields. Their algebraic and topological properties have paved the way for innovative concepts, such as K-theory and non-commutative differential geometry, shaping the landscape of contemporary mathematics.
One notable application of C*-algebras is in the area of quantum computing, where they are used to model quantum noise and error correction. This highlights the role of C*-algebras not just in theoretical groundwork but also in practical advancements in technology and science.
## Spectral Theory in Operator Algebras: A Deep Dive
Spectral theory plays a pivotal role in the study of operator algebras, offering deep insights into the behaviour and properties of linear operators. As a mathematical discipline, it stands at the convergence of analysis, algebra, and geometry, providing powerful tools for solving complex problems across various fields.
### Introduction to Spectral Theory within Operator Algebras
Spectral theory investigates the spectrum of operators, including eigenvalues, eigenvectors, and continuous spectra, within the context of operator algebras. This exploration is vital for understanding the structure and dynamics of mathematical systems represented by operators on Hilbert spaces.
The theory’s foundational concept, the spectrum of an operator, provides insight into how operators can be decomposed, setting the stage for their analysis within the frameworks of C*-algebras and Von Neumann algebras.
The spectrum of an operator $$A$$ in a Hilbert space $$\mathcal{H}$$ is the set of all $$\lambda \in \mathbb{C}$$ for which $$A - \lambda I$$ is not invertible, where $$I$$ is the identity operator on $$\mathcal{H}$$.
The spectrum of an operator reveals not just the set of its eigenvalues but also includes other points of interest like the essential spectrum, resonances, and spectral singularities.
### Applying Spectral Theory to Operator Algebras Problems
Applying spectral theory within operator algebras allows for the resolution of various mathematical and physical problems, from understanding quantum systems to solving differential equations. By utilising the spectral decomposition of operators, it is possible to analyse the stability, response, and evolution of systems modelled by these operators.
One powerful application is the spectral theorem, which provides a framework for breaking down self-adjoint, normal, and unitary operators into simpler components, facilitating their study and manipulation in both pure and applied mathematics.
Spectral theory serves as a mathematical microscope, revealing the atomic detail of operator behaviour and properties.
### Key Concepts of Spectral Theory in Operator Algebras
• Spectrum: Comprises eigenvalues and continuous spectra, offering a comprehensive view of an operator’s action.
• Spectral Decomposition: Involves expressing an operator as a direct integral of operators over its spectrum, simplifying its analysis.
• Spectral Theorem: Provides conditions under which an operator can be expressed in terms of its spectral properties, crucial for self-adjoint and normal operators.
• Functional Calculus: Enables functions to be applied to operators, extending the concept of applying functions to numbers or matrices.
Consider a self-adjoint operator $$A$$ with a purely discrete spectrum. The spectral theorem implies $$A$$ can be represented as $$A = \sum_\lambda \lambda P_\lambda$$, where $$\lambda$$ ranges over the spectrum of $$A$$ and $$P_\lambda$$ are the orthogonal projections onto the eigenspaces corresponding to $$\lambda$$.
Functional calculus allows for the elegant treatment of differential operators in quantum mechanics. By applying functions directly to operators, spectral theory constructs a bridge between abstract mathematical operators and observable physical quantities, profoundly influencing the development of modern physics.
## Advanced Topics in Operator Algebras
Delving into the advanced topics of Operator Algebras unveils a treasure trove of techniques and theories that significantly contribute to both mathematics and theoretical physics.
### Von Neumann Algebras Technique: A Closer Look
Von Neumann algebras, named after the renowned mathematician John von Neumann, represent a class of operator algebras that play a critical role in functional analysis and quantum mechanics. They are characterised by unique properties that facilitate the understanding of operator theory and quantum statistical mechanics.
Von Neumann algebras are paramount for the analysis and classification of factors, which are the building blocks of these algebras. This analysis has profound implications on quantum field theory and the mathematical foundations of quantum mechanics.
A Von Neumann algebra is a *-algebra of bounded operators on a Hilbert space that is closed in the weak operator topology and contains the identity operator.
The classification of Von Neumann algebras into types I, II, and III has had a substantial impact on the understanding of quantum theory, particularly in the study of quantum entanglement and the structure of space-time in quantum field theories.
Von Neumann algebras provide a rigorous mathematical framework that supports the algebraic approach to quantum mechanics, known as the Hilbert space formulation.
### Exploring Vertex Operator Algebra and Its Applications
Vertex Operator Algebras (VOA) are intrinsically linked to string theory and conformal field theory. They are algebraic structures that arise naturally in the study of two-dimensional conformal field theory, playing a central role in the enumeration of operator product expansions (OPE) and the classification of conformal field theories.
The development of VOA has provided rich mathematical structures that are utilised in various areas of mathematics, including number theory and geometry. Their link to monstrous moonshine and the theory of modular forms is one of the most celebrated applications.
A Vertex Operator Algebra is an algebraic structure that includes a vacuum vector, a conformal vector, and a collection of vertex operators satisfying certain axioms, reflecting the algebraic properties of correlation functions in conformal field theories.
An example of the utility of VOA is found in the proof of the monstrous moonshine conjecture, which connects the j-function in number theory with the Monster group, showcasing the deep interplay between algebra, geometry, and mathematical physics.
The term 'moonshine’ in the context of VOA, reflects the unexpected and mysterious connections between seemingly unrelated areas of mathematics and physics.
### Operator Algebras in Quantum Physics: Bridging Mathematics and Science
Operator algebras, particularly C*-algebras and Von Neumann algebras, serve as mathematical frameworks for quantum physics, bridging the gap between abstract mathematical concepts and observable physical phenomena. They are pivotal in the formulation of quantum mechanics, providing a rigorous approach to the study of quantum observables, states, and dynamics.
Through the application of operator algebras, quantum theories can be expressed with mathematical precision, allowing for the study of quantum entanglement, decoherence, and the non-commutative geometry of quantum spacetime.
The algebraic approach to quantum field theory, leveraging the structures offered by operator algebras, has led to significant insights into the nature of quantum field interactions, particle statistics, and the fundamental forces of the universe. This approach consolidates the mathematical underpinnings of quantum physics with experimental observations, driving forward both theoretical understanding and technological advances.
Operator algebras facilitate the rigorous treatment of infinities that arise in quantum field theories, providing a coherent structure for renormalisation techniques.
## Operator algebras - Key takeaways
• Operator Algebras Definition: Mathematical structures including sets of linear operators acting on a Hilbert space, closed under addition, scalar multiplication, operator multiplication, and taking adjoints.
• C*-Algebra Examples: Include the set of all bounded linear operators on a Hilbert space and continuous function algebras on a compact space, fundamental in quantum mechanics and topology.
• Spectral Theory in Operator Algebras: Investigates the spectrum of operators, informing the decomposition and analysis of operators within C*-algebras and von Neumann algebras.
• Von Neumann Algebras Technique: *-algebras of bounded operators on a Hilbert space, closed in the weak operator topology, essential in quantum mechanics and quantum statistical mechanics.
• Vertex Operator Algebra: Algebraic structures connected to string theory and conformal field theory, involving a vacuum vector, a conformal vector, and vertex operators satisfying specific axioms.
#### Flashcards in Operator algebras 24
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What is an operator algebra and how does it differ from ordinary algebra?
An operator algebra is a subfield of functional analysis focusing on the study of algebras of bounded linear operators on a Hilbert space. It differs from ordinary algebra in its emphasis on operators and their properties, such as norm and spectrum, rather than on purely algebraic constructs like groups or rings.
What are the fundamental types of operator algebras and their applications in mathematics and physics?
The fundamental types of operator algebras are C*-algebras and von Neumann algebras. They have widespread applications in mathematics and physics, particularly in quantum mechanics for analysing observables and states, and in non-commutative geometry for studying spaces that are not amenable to traditional geometrical approaches.
How are operator algebras used in the study of quantum mechanics and quantum field theory?
Operator algebras, particularly $$C^*$$-algebras and von Neumann algebras, provide a rigorous mathematical framework for quantum mechanics and quantum field theory by enabling the description of observable quantities as operators on Hilbert spaces, encapsulating quantum states and dynamics, and facilitating the analysis of symmetries and interactions in quantum systems.
What are the key differences between C*-algebras and von Neumann algebras, and how do they impact their respective uses in mathematical analysis?
C*-algebras are defined by norm and involution conditions, emphasizing algebraic structure and continuity, ideal for analysis on topological spaces. Von Neumann algebras, subsets of C*-algebras, require a double-commutant in a Hilbert space, focusing on operator theory and quantum mechanics, allowing weak closure and direct integrals. These distinctions influence their applications in functional analysis and quantum physics, respectively.
How does one construct a spectral decomposition in the context of operator algebras, and what is its significance?
In operator algebras, a spectral decomposition involves expressing an operator as an integral or sum over its spectrum, using projection operators. This decomposition signifies the operator's action by breaking it down into simpler components, crucial for understanding spectral properties and operator behaviour.
## Test your knowledge with multiple choice flashcards
What fundamentally composes Operator Algebras?
What is a primary application of operator algebras in pure mathematics?
What are C*-algebras and their significance in the study of operator algebras?
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1. ### Calculus
Derivative of (Sec(z))^2
2. ### calculus
What is the derivative of x^2 -4y -tanxy=0
3. ### Calculus
find f'(x) f(x)= sin(3x)/2x^2-4
4. ### Calculus
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5. ### Calculus
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6. ### calculus
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7. ### calculus
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8. ### Calculus
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9. ### ap calculus ab
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10. ### ap calculus ab
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11. ### calculus
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12. ### calculus
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13. ### calculus
simply the summation of (2/n)(2i/n)
14. ### calculus
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15. ### calculus
differentiate and do not simplify: Y=e^ -sq.rt x + e^3 - 2^(3x-1)
16. ### Calculus
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17. ### calculus
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18. ### Calculus- math
integral of (2x) / [(x+2)(x+1)]
19. ### Calculus
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20. ### CALCULUS
Find the derivative : g(t)=t/(t+1)
21. ### calculus
Let be a function that satisfies
22. ### Calculus
Solve: 20 = 10 (2^x)(ln2)
23. ### calculus
(3x-4)^5 x(sqaure root)9x^2+4
24. ### Calculus
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25. ### calculus
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26. ### Calculus
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27. ### calculus
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28. ### Calculus
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29. ### calculus
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30. ### pre- calculus
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31. ### calculus
lim tan x - x/ x -> 0 x^2
32. ### calculus
lim x-> 0 if tan x - x / x^2
33. ### Calculus
Lim (tan^2(x)/x) x->0
34. ### calculus
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35. ### calculus
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36. ### Calculus
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37. ### calculus
find the derivative of y= (1-x/7)^-7
38. ### calculus
differentiate f(x)=Tan^3(e^2x +1)
39. ### calculus
find h'(2) given that f(2) = -3, g(2) = 4, f'(2)= -2 and g'(2)= 7. A. h(x) = 5f(x) - 4g(x) B. h(x) = f(x)g(x) C. h(x) = f(x)/g(x) D. h(x) = g(x)/ 1+ f(x)
40. ### calculus
lim f(x) sin 3x / x x¨0
41. ### calculus
Suppose F(x)=f(x)g(2x) . If f(1)=3 , f′(1)=3 , g(2)=2 , and g′(2)=7 , find F′(1). F′(1)=
42. ### Calculus
Inflection points of y= x/ x^2 + 25
43. ### calculus
Find the derivative of y=(s^3+12)^(1/6)
44. ### Calculus
Find the derivative of y = (8+e^5x)^3
45. ### Calculus
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46. ### Calculus
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47. ### calculus
Find the derivative of y=t^2(5t-8)^4.
48. ### Calculus
I'm not sure how to do #22. LINK: h t t p : / / g o o . g l / P 0 S e
49. ### calculus
solve tis 5x^+4x+1+3(1-3x)
50. ### calculus
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51. ### calculus
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52. ### Calculus
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53. ### calculus
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54. ### calculus
find the inverse of q(x)=(x-5)^2
55. ### calculus
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56. ### calculus
Calculate the integral from 0 to 4 of |x^2 - 1|. Thanks!
57. ### LIMIT(CALCULUS)
lim(y^2+5y-1) y-->2
58. ### LIMIT(CALCULUS)
lim(x-1)(2x+3) x-->3
59. ### LIMIT(CALCULUS)
lim y^3-y^2-y-2/2y^3-5y^2+5y-6 y-->2
60. ### Calculus
show that y= x + e^x is a solution of dy/dx = y - x +1
61. ### Calculus
Compute the derivatives of : f(x)= x^7 + 7^x+ 7^7
62. ### calculus
DIFERENTIATE THE FUNCTION: y=[(x^2)+1][(1-x^2)]
63. ### calculus
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64. ### calculus
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65. ### calculus
the integral 49dx/x^2(x^2+36)^(1/2)
66. ### Calculus
If f(x)= sqrt(4sin(x)+2), then f'(0)=
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70. ### Calculus
How can I find the derivative of y = [ln(x+4)^3]^2 ?
71. ### Calculus
solve the following: 2^-x=3^2x+2
72. ### Calculus
solve the following: 2^-x=3^2x+2
73. ### Calculus
solve the following: 2^-x=3^2x+2
74. ### Calculus
lnx+ ln (x+3)= 2ln2
75. ### calculus
what is the vertical asymptote of f(x)=x^3+2/x+5?
76. ### calculus
Factor completely x^3 - 5x^2 - x + 5
77. ### calculus
what is the value of y in the equation (y/(2y-x))=0.5
78. ### calculus
How do I integrate [1/(x^2+1)] Thx
79. ### calculus
integrate x+5/(x^2+10x+26)
80. ### calculus
lim (x^3-6)/(x^k+3) x->infinity
81. ### calculus
∫[3√3:(2x^2-3x),]
82. ### calculus
find the derivative of e^8e^x *8e^x
83. ### Calculus
Find the derivative of 4e^x+x^3
84. ### calculus
how did you figure that out
85. ### pre calculus
let f(x)= -x^2 - 1. evaluate f(x)at x=6, -6
86. ### calculus
find the derivative of 3/ln(3z)
87. ### calculus
Sketch the graph of F(x).
88. ### calculus
For each function, find a. f''(x), and b. f''(3). f(x)= 1 --- 6x^2
89. ### calculus
solve x^6<16x^4
90. ### calculus
integrate xsin(2x)dx=
91. ### pre calculus
what is the vertices of 7x^2+5y^2=35
92. ### pre calculus
what is the vertices of (x+1)^2/4+(y+2)^2/25=1 ?
93. ### calculus
what is the derivative of: V(t) = 130sin5t + 18
94. ### Calculus
what is the derivative of: y=2sinxcosx
95. ### calculus
what are the steps to solve for x 2x^2-1/x^2=0
96. ### calculus
how do you separate this variable dy/dx=(x+1)/xy, x>0, y(1)=-4
97. ### calculus
sketch the curve of y=x-1/x^2-x-6
98. ### calculus
sketch the curve of y=x-1/x^2-x-6
99. ### Pre Calculus 1
ln(3x-1) - ln(x+2) = 2ln5
100. ### Pre Calculus 1
lnx + ln(x+3) = 2 | 2,447 | 5,875 | {"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} | 3.390625 | 3 | CC-MAIN-2018-26 | latest | en | 0.44332 |
https://testbook.com/question-answer/a-girl-spends-82-of-her-income-if-her-income-inc--5ea17fe2f60d5d3022ff1866 | 1,627,845,505,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154219.62/warc/CC-MAIN-20210801190212-20210801220212-00346.warc.gz | 589,756,024 | 19,781 | # A girl spends 82% of her income. If her income increases by 15% and the savings decrease by 15%, then what is the percent increase in the expenditure (correct to one decimal place)?
Free Practice With Testbook Mock Tests
## Options:
1. 20
2. 23.5
3. 21.6
4. 17.5
### Correct Answer: Option 3 (Solution Below)
This question was previously asked in
CIL MT Electrical: 2020 Official Paper
## Solution:
Let the income of girl be Rs. 100
Expenditure of girl = Rs. 100 × (82/100) = Rs. 82
Savings of girl = 100 – 82 = 18
New income of girl = 100 × (115/100) = 115
New savings of girl = 18 × 85/100 = 15.3
New expenditure = 115 – 15.3 = 99.7
Her expenditure increased by 99.7 – 82 = 17.7
∴ Her expenditure increased by (in %) = 17.7/82 × 100 = 21.6% | 256 | 764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-31 | latest | en | 0.901065 |
http://math.ucla.edu/~cratsch/151b.2.05s/Homeworks/homework7.html | 1,513,373,224,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948579567.73/warc/CC-MAIN-20171215211734-20171215233734-00588.warc.gz | 170,168,600 | 1,669 | ## Homework Assignment #7
(due: May 27 in class)
### Theory Part
Problem 1 (30%):
Show that the following linear system of equations converges for both, Gauss Jacobi and Gauss Seidel methods:
E1: 4*x1 - x2 -x4 = 0
E2: -x1 +4*x2 -x3 -x5 = 5
E3: -x2 +4*x3 -x6 = 0
E4: -x1 +4*x4 -x5 = 6
E5: -x2 -x4 +4*x5 -x6 = -2
E6: -x3 -x5 + 4*x6 = 6
Bonus Problem(20%):
Proof part ii) of Corollary 7.20.
i.e., proof that for ||T||<1, the following error bound holds for the sequence {x(k)}, defined by x(k) = Tx(k-1) + c:
||x(k) - x|| <= ||T||(k) / (1-||T||) * ||x(1) - x(0)||
Hints
i) Consider ||x(m) - x(k)|| for any m > k > 1. Show that this is less or equal then ||x(m) - x(m-1)|| + ||x(m-1) - x(m-2)|| + ... + ||x(k+1) - x(k)||
ii) Now use that ||x(k+1) - x(k)|| = ||Tx(k) - Tx(k-1)||
iii) Since ||T|| < 1, the series converges, i.e. x(m) --> x as m --> infinity.
iv) The geometrical series sum(i=0,infinity) (xi) = 1/(1-x) for 0 < x < 1.
### Numerics Part
Problem 3 (70%):
a) Solve problem 1 above using the Gauss Jacobi Method.
b) Solve problem 1 above using the Gauss Seidel Method.
c) Solve problem 1 above using the SOR method with omega = 1.1, 1.2, 1.3, 1.4, 1.5
Use a tolerance of 0.001. Set the maximum number of iterations to 100 (you will need fewer !). Use the following stopping criteria:
|| x(k) - x(k-1) ||infinity / || x(k) ||infinity < tolerance.
d) Compare the efficiency of the algorithms in (a), (b), and (c) (i.e., compare the number of iterations)
What you should turn in:
The codes you used in (a), (b), and (c).
Your final results for x for (a), (b), and (c). | 601 | 1,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2017-51 | latest | en | 0.733943 |
https://www.coursehero.com/file/6081958/Nov-07/ | 1,513,186,081,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948529738.38/warc/CC-MAIN-20171213162804-20171213182804-00119.warc.gz | 738,901,092 | 28,810 | # Nov_07 - Enthalpy of Formation Standard States Definitions...
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Enthalpy of Formation
Standard States Definitions of Standard States : • Gases: gas at partial pressure of 1 atm • Solutes: a conc. of 1M at an applied P of 1atm • Pure liquids or solids: the pure liquid or solid • Elements: the elemental form that exists at 1 atm
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Standard Enthalpy of Formation ( H f °) ∆Η f º - enthalpy change for the formation of 1 mole of a compound from its elements all at standard states H f º for an element in its standard state equals zero • Normally reported in kJ per mole of the compound • There is no standard temperature. ∆Η f º depends on temperature and is often reported for 298 K
reactants → products ΔHº reaction ΔHº reaction = Σ ΔH f º (products) – Σ ΔH f º (reactants)
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H Δ f º - example e.g. Calculate Hº Δ reaction at 298 K for : 4 NH 3 (g) + 7 O 2 (g) -----> 4 NO 2 (g) + 6 H 2 O(l) Look up in table of H Δ f º values at 298 K: H Δ f º (NH 3 (g)) = - 46 kJ/mol H Δ f º (O 2 (g)) = 0 kJ/mol <----- Definition H Δ f º (NO 2 (g)) = 34 kJ/mol H Δ f º (H 2 O(l)) = - 286 kJ/mol Δ reaction
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Ask a homework question - tutors are online | 584 | 2,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-51 | latest | en | 0.803512 |
https://aibcapital.com/stocks/quick-answer-how-do-you-determine-number-of-shareholders.html | 1,686,118,268,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224653608.76/warc/CC-MAIN-20230607042751-20230607072751-00141.warc.gz | 119,620,903 | 19,825 | # Quick Answer: How do you determine number of shareholders?
Contents
If you know the market cap of a company and you know its share price, then figuring out the number of outstanding shares is easy. Just take the market capitalization figure and divide it by the share price. The result is the number of shares on which the market capitalization number was based.
## How do you calculate number of shareholders?
Shareholders’ equity may be calculated by subtracting its total liabilities from its total assets—both of which are itemized on a company’s balance sheet.
## What determines how many shares there are?
The number of authorized shares per company is assessed at the company’s creation and can only be increased or decreased through a vote by the shareholders. If at the time of incorporation the documents state that 100 shares are authorized, then only 100 shares can be issued.
## How do I find all shareholders of a company?
All companies in India have to file their financials and details of shareholders with the Ministry of Corporate Affairs (MCA21). You can access these documents through the website Ministry Of Corporate Affairs.
## How do I know how many shares to buy?
How many shares can you buy based on price?
1. Find the current share price of the stock you want. …
2. Divide the amount of money you have available to invest in the stock by its current share price.
3. If your broker allows you to buy fractional shares, the result is the number of shares you can buy.
## How do you find the percentage of shares you own?
Divide the number of issued shares by the number of authorized shares, and then multiply by 100 to convert to a percentage.
## How do you know how many shares to issue?
When the founders have agreed on the ownership percentages (i.e. percentage of common shares issued), they can then determine how many shares in total to issue. This number is usually kept small at the beginning, e.g. 100 or 1000. This number can be “split” (multiplied by 2, 10 or whatever) as required.
## How many shares do you need to own a company?
Many experts suggest starting with 10,000, but companies can authorize as little as one share. While 10,000 may seem conservative, owners can file for more authorized stocks at a later time. Typically, business owners should choose a number that includes the stocks being issued and some for reservation.
## How do you calculate a company’s share price?
To figure out how valuable the shares are for traders, take the last updated value of the company share and multiply it by outstanding shares. Another method to calculate the price of the share is the price to earnings ratio.
IT IS IMPORTANT: How do you record closing entries for dividends?
## Do companies know who their shareholders are?
Do public companies actually know where their issued stock is and who holds it? In a technical sense, the answer is yes, but for most public shareholders, the company really only knows about the broker that is holding the stock for the shareholder.
## Does a company know who owns their stock?
Generally no. They might not pay dividends. But they also have to send shareholder reports, shareholder meeting notices, and proxy forms.
## Can you see who owns shares in a company?
To find out who owns the majority shares of a public company’s stock, use the EDGAR database at SEC.gov [ http://sec.gov ] (there is a link to it on the SEC’s home page) and search for the company’s proxy statements DEF-14A.
## How do I know how many shares I have left to buy?
How to calculate outstanding shares
1. Go to the balance sheet of the company in question and look in the shareholders’ equity section, which is near the bottom of the report.
2. Look in the line item for preferred stock. …
3. Look in the line item for common stock. …
4. Look in the line item for treasury stock.
## Is buying 1 share of stock worth it?
While purchasing a single share isn’t advisable, if an investor would like to purchase one share, they should try to place a limit order for a greater chance of capital gains that offset the brokerage fees. … Buying a small number of shares may limit what stocks you can invest in, leaving you open to more risk.
## What are 100 stock shares called?
A round lot is a standard number of securities to be traded on an exchange. In stocks, a round lot is considered 100 shares or a larger number that can be evenly divided by 100. In bonds, a round lot is usually \$100,000 worth.
IT IS IMPORTANT: Why are some shares cheaper than others? | 962 | 4,546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-23 | latest | en | 0.952739 |
https://socratic.org/questions/how-do-you-rationalize-the-denominator-and-simplify-12-5-2sqrt7 | 1,702,148,651,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100942.92/warc/CC-MAIN-20231209170619-20231209200619-00781.warc.gz | 591,028,113 | 5,662 | # How do you rationalize the denominator and simplify 12/(5-2sqrt7)?
You may try multiplying top and bottom by $5 + \sqrt{7}$ to get: | 39 | 134 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 1, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2023-50 | latest | en | 0.797227 |
https://www.coursehero.com/file/6075451/ipf/ | 1,487,716,355,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170839.31/warc/CC-MAIN-20170219104610-00106-ip-10-171-10-108.ec2.internal.warc.gz | 806,415,777 | 69,055 | # ipf - cup &amp;lt;-colSums(xmed) ; sc &amp;lt;-c /...
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PUBLISH OR PERISH JAN DE LEEUW Suppose A is an n × m matrix with non-negative elements a ij . Also suppose p and p are two vectors with n and m non-negative ele- ments such that n X i = 1 p i = m X j = 1 q j . We now want to find b with n elements and c with m elements such that m X j = 1 b i c j a ij = p i , n X i = 1 b i c j c ij = q j . Thus we rescale rows and columns of A in such a way that the marginals become p and q . Suppose ˜ b and ˜ c are tentative solutions, giving a matrix ˜ A with elements ˜ b i ˜ c j a ij . Update the matrix by a + ij = p i ˜ a i ˜ a ij , which gives a matrix with the correct row sums, and then a ++ ij = q j a + j a + ij , which gives a matrix with the correct column sums. Iterate. Date : May 2, 2007. 1
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2 JAN DE LEEUW 1 ipf < - function (x , r= rep (1 , nrow (x) ) , c = rep (1 , ncol (x) ) ,itmax =100,eps=1e - 6,verbose=FALSE) { 2 xold < - x ; i te l < - 1 3 repeat { 4 rup < - rowSums( xold ) ; sr < - r / rup ; xmed < - sr * xold 5
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Unformatted text preview: cup <-colSums(xmed) ; sc <-c / cup ; xnew <-t ( sc * t (xmed) ) 6 if ( verbose ) cat ( 7 " Iteration : " , formatC ( itel , width=3 , format ="d" ) , 8 " rscal : " , formatC ( range ( sr ) , digits =8,width =12, format =" f " ) , 9 " cscla : " , formatC ( range ( sc ) , digits =8,width =12, format =" f " ) , 10 "\n" ) 11 if ( ( max ( abs (xnew-xold ) ) < eps ) | | ( it el == itmax ) ) return (xnew) 12 xold <-xnew ; it el <-i te l+1 13 } 14 } Department of Statistics, University of California, Los Angeles, CA 90095-1554 E-mail address , Jan de Leeuw: deleeuw@stat.ucla.edu URL , Jan de Leeuw: http://gifi.stat.ucla.edu...
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## This note was uploaded on 01/14/2011 for the course STATS 102A 102A taught by Professor Jandeleeuw during the Fall '10 term at UCLA.
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ipf - cup &amp;lt;-colSums(xmed) ; sc &amp;lt;-c /...
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Ask a homework question - tutors are online | 819 | 2,403 | {"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} | 3.46875 | 3 | CC-MAIN-2017-09 | longest | en | 0.643911 |
https://www.meritnation.com/ask-answer/question/how-to-multiply-or-divide-the-fractions/fractions/540582 | 1,611,206,951,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703522242.73/warc/CC-MAIN-20210121035242-20210121065242-00759.warc.gz | 893,408,276 | 9,449 | # how to multiply or divide the fractions
Dear Student!
While multiplying the fractions, numerators of all given fractions are multiplied together and similarly the denominators of all given fractions are multiplied to each other.
That is,
For mixed fraction, convert mixed into improper fraction and then follow the steps explained above.
Cheers!
• 2
What are you looking for? | 74 | 381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-04 | longest | en | 0.844711 |
https://d2mvzyuse3lwjc.cloudfront.net/doc/en/Tutorials/Fitting-Datasets | 1,726,541,788,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651722.42/warc/CC-MAIN-20240917004428-20240917034428-00422.warc.gz | 183,591,086 | 40,539 | # 4.2.2.8 Fitting One Dataset as a Function of Other Datasets
## Summary
Sometimes, one may want to perform "Dataset Fitting", that is, the output may be composed of one or several datasets, like:
$Output = A_1*Dataset_1+A_2*Dataset_2$
For example, you may want to analyze a composite spectrum to find the contributions/ratio from individual component spectra. This can be accomplished either by defining multiple independent variables or by calculating the "combination" inside the fitting function.
Minimum Origin Version Required: Origin 8.0 SR6
## What you will learn
This tutorial will show you how to:
• Perform "Dataset Fitting"
• Define multiple independent variable fitting function
## Steps
Import the Composite Spectrum.dat file from the \Samples\Curve Fitting\ folder. In this sample data, we can see that column A is the index, columns B and C are the values for the spectrum of components A and B. Column D contains values obtained after reading a composite spectrum of components A and B. By fitting column D to an equation determined by the component spectra of the pure forms of columns B and C, the coefficients for the contributions of B and C (call them c1 and c2 respectively) can be found. (Note: In this case, we supposed the independent and dependent variables have the same size. If not, interpolation is need.)
Bring up the Fitting Function Organizer and define a new fitting function as follow:
Function Name: MultiIndep Function Type: User-Defined Independent Variables: a, b Dependent Variables: ab Parameter Names: C1, C2 Function Form: Origin C Function: ab = C1*a + C2*b;
Initialize both C1 and C2 to 1 in the Parameter Initialization edit box by entering:
C1=1;
C2=1;
Save the fitting function and close Fitting Function Organizer. Highlight ONLY Column D and bring up the NLFit dialog, specify the input datasets in the Data Selection page as follow:
Then you can click the Fit button to generate results.
## Results
You are supposed to get these results:
Value Standard Error C1 0.37169 0.00483 C2 0.66469 0.0047
To verify the fitted results, you can add a new column and Copy + Paste the fitted value, which comes from the fitted Y (col(C)) in the worksheet FitNLCurve1, into it. Then Highlight the Composite and the fitted data and plot a line graph to see how good the fit is: | 537 | 2,334 | {"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": 1, "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-2024-38 | latest | en | 0.814811 |
https://www.flopturnriver.com/casino/blackjack-strategy-21179 | 1,532,001,019,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590866.65/warc/CC-MAIN-20180719105750-20180719125750-00520.warc.gz | 871,829,903 | 18,178 | Select Page
# Blackjack Strategy
Blackjack players face the house, not each other. Psychology and game theory do not come into play with blackjack unlike poker; instead it is all about mathematics. Computer simulations have been run to determine the optimal strategy to use in blackjack, and this is commonly referred to as Basic Strategy. The Basic Strategy chart for a single deck game is slightly different than the Basic Strategy chart for a multi-deck game. A good blackjack player knows the Basic Strategy charts like the back of his hand. It is not enough to casually memorize the Basic Strategy charts. One needs to know the charts backwards and forwards so that the right decision is made even when there are distractions. The best decision should be made automatically. If a player does not follow Basic Strategy, he is hurting his own chances of winning in the long term. The laws of probability cannot be ignored. Blackjack, when played perfectly, only offers a very slight advantage to the house. These house advantages only increase with any slight deviations from Basic Strategy. Play free online blackjack until you are comfortable with the game and ready to move on to real money.
Memorize the Basic Strategy Chart!
#### Good Blackjack Rules
When choosing a casino or table to play blackjack, certain rules are better for the player. Most casinos force the dealer to hit a soft 17 which is bad for the player in the long run. Look for casinos where the dealer is forced to stand on soft 17.
The surrender rule is also very beneficial to the player. The surrender rule allows a player to surrender his hand, prior to any action, and sacrifice half the bet. For example, if the player holds a 10 and a 6 then this 16 should be surrendered when the dealer shows 9, 10 or Ace, allowing the player to take back half his bet.
Being allowed to double down after splitting is probably the most beneficial rule for the player, and fortunately, most casinos allow a player to double down after splitting. For example, if the dealer is showing a 6, and the player has 7, 7 then the right play is to split. If the first 7 then gets hit a 4, then the correct play is to double down on this 11.
Finally, another good rule for the player is the allowance of resplitting Aces. Although this may be used rarely, I have personally resplit aces 4 times. For example, your are dealt A A, and split those aces. You are then hit another ace. With this rule, you can split again. Without this rule, you are stuck with 12 and must move on to the next hand.
Summary of Good Blackjack Rules:
• Dealer Stands on Soft 17
• Surrender
• Double Down Allowed After Split
• Resplitting Aces
#### Bad Blackjack Rules
In a brick and mortar casino, good players, in addition to utilizing Basic Strategy, also count cards and then take advantage of situations late in the deck(s), making intelligent modifications to Basic Strategy based on the situation. Many casinos, including online casinos, have countered the ability to count cards by implementing continuous shufflers. What this means is that as soon as you play a hand, those cards are immediately introduced back into the deck or shoe and reshuffled, so you could get those exact two cards on the very next hand. Obviously, this is very bad for the card counter.
Some casinos do not allow you to double down a soft hand. A hand like Ace, 6 is a soft 17. This hand should be doubled down when the dealer shows a 6, but some casinos do not allow this play. Thankfully, most casinos do allow you to double down a soft hand – which is a very important rule for the player.
Summary of Bad Blackjack Rules
• Continuous Shuffle
• No Soft Double Downs
• No Re-splits | 791 | 3,712 | {"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-2018-30 | latest | en | 0.955758 |
https://hextobinary.com/unit/angle/from/sign/to/minangle/935 | 1,620,418,202,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988802.93/warc/CC-MAIN-20210507181103-20210507211103-00034.warc.gz | 309,550,796 | 13,847 | Sign
Minute
###### How many Minutes are in 935 Signs?
The answer is 935 Signs are equal to 1683000 Minutes. Feel free to use our online unit conversion calculator to convert the unit from Sign to Minute. Just simply enter value 935 in Sign and see the result in Minute. You can also Convert 936 Sign to Minute
###### How to Convert 935 Signs to Minutes (sign to min)
By using our Sign to Minute conversion tool, you know that one Sign is equivalent to 1800 Minute. Hence, to convert Sign to Minute, we just need to multiply the number by 1800. We are going to use very simple Sign to Minute conversion formula for that. Pleas see the calculation example given below.
Convert 935 Sign to Minute935 Sign = 935 × 1800 = 1683000 Minute
###### What is Sign Unit of Measure?
Sign is a unit of angular measurement which is mostly used in astrology related work. It is almost equal to 30 degrees and 1/12 of turn.
###### What is the symbol of Sign?
The symbol of Sign is sign. This means you can also write one Sign as 1 sign.
###### What is Minute Unit of Measure?
Minute of angle is a unit of angular measurement. It is also known as arcmin. A minute of angle is equal to 60 arcsec.
###### What is the symbol of Minute?
The symbol of Minute is min. This means you can also write one Minute as 1 min.
###### Sign to Minute Conversion Table
Sign [sign]Minute [min]
11800
23600
35400
47200
59000
610800
712600
814400
916200
1018000
100180000
10001800000
###### Sign to Other Units Conversion Chart
Sign [sign]Output | 377 | 1,521 | {"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-2021-21 | latest | en | 0.887516 |
https://www.physicsforums.com/threads/calculating-viscous-and-pressure-drag-on-a-sphere.414820/ | 1,521,777,585,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257648177.88/warc/CC-MAIN-20180323024544-20180323044544-00517.warc.gz | 847,832,440 | 16,425 | Calculating Viscous and Pressure Drag On A Sphere
1. Jul 8, 2010
I've been doing some research on sphere aerodynamics, in particular that of soccer balls, and was wondering if there was any way to separately calculate the pressure / form drag, and also the surface / viscous / skin drag.
I know that Stoke's Law of F=6(pi)RnVc, where R is the radius of the sphere, n is the viscosity, and Vc is the velocity through a continuous fluid, can give the viscous drag on a sphere, but was informed that this only applies to very small spheres, so it is no use in my application for the drag on a soccer ball.
I also know of the formula that Force(drag) = 0.5 C.P.A.v^2 where C = coefficient of drag, P = fluid density, A = area of the object, v = velocity of the object, but I didn't know which aspect of the drag force (pressure, viscous or total drag) this formula calculates.
If anyone could help me out here that would be greatly appreciated.
2. Jul 8, 2010
RandomGuy88
You are correct in assuming Stokes drag cannot be used here.
The equation you wrote will give you the total drag on the sphere. You might be able to find some information about the pressure distribution around a sphere and integrate that to determine the pressure drag.
3. Jul 8, 2010
arildno
That would be the total drag coefficient.
An engineer might rewrite C=c_v+c_p, defined in obvious manners, and sought to determine experimentally what goes into c_v and what goes into c_p.
Perhaps, they probably use more refined approaches than my dumb brutish ones..
4. Jul 10, 2010
That actually does sound like quite a good idea, unfortunately I don't know either c_v or c_p. If I knew one I'm sure I could derive the other from the overall formula for C, but without knowing the formula for one of these components of the drag force I'm afraid I'm still stuck... :/
5. Jul 10, 2010
Sybren
the aerodynamics around the ball are turbulent, and this makes it difficult to treat analytically. what you could do to obtain the drag force on the ball is to perform an experiment on the trajectory through air, and compare that to the ideal case of no air resistance (hyperbola). better yet, search scholar.google.com or arxiv.org for articles on this topic.
6. Jul 10, 2010
Thanks, I like this idea and it sounds pretty useful as a good comparison showing the affects of the drag on the ball's flight.
My issue is that using the formula I stated before I can find the overall drag, its just I was wondering if there was any way I could specifically calculate the pressure / form drag, and then the viscous / surface drag, and try to compare them, so that I can show the vast majority of the drag around a sphere is pressure drag, compared to an aerofoil for example, where the majority is viscous drag.
Thanks for all the responses so far, and for any future input it's all been really helpful!
7. Jul 16, 2010
In regards to the adidas jabulani soccer ball that was used at the recent world cup in south africa, it seems that the ball is not any lighter as some may say (it is actually on the heavier side of the strictly regulated limits for ball weight), but just moving faster through the air. I understand that this is due to all the dimples and ridges on the surface of the ball (tripping a turbulent boundary layer, less pressure drag), but would this not be negated by fewer seams (i.e. less seams, less places to trip turbulent boundary layer, more pressure drag) ? | 809 | 3,455 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2018-13 | longest | en | 0.955526 |
https://www.tutorialkart.com/cpp/cpp-not-equal-operator/ | 1,716,790,303,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059037.23/warc/CC-MAIN-20240527052359-20240527082359-00731.warc.gz | 891,375,527 | 17,995 | In this tutorial, you shall learn about Not-Equal Relational Operator in C++ programming language, its syntax, and how to use this operator with the help of examples.
## C++ Not-Equal Operator
In C++, Not Equal Relational Operator is used to check if left operand is not equal to second operand.
The syntax to check if `x` does not equal `y` using Not Equal Operator is
`x != y`
The operator returns a boolean value of `true` if `x` is not equal to `y`, or `false` if not.
## Examples
### 1. Check if two numbers are not equal
In the following example, we take two integer values in `x` and `y`, and check if these two are not equal, using Not Equal Operator.
main.cpp
```#include <iostream>
using namespace std;
int main() {
int x = 1;
int y = 6;
if (x != y) {
cout << "x and y are not equal." << endl;
} else {
cout << "x and y are equal." << endl;
}
}```
Output
```x and y are not equal.
Program ended with exit code: 0```
Since values in `x` and `y` are not equal, `x != y` returned true.
### 2. Check if two strings are not equal
Now, let us take two strings, and check if they are not equal using Not Equal Operator.
main.cpp
```#include <iostream>
using namespace std;
int main() {
string x = "apple";
string y = "apple";
if (x != y) {
cout << "x and y are not equal." << endl;
} else {
cout << "x and y are equal." << endl;
}
}```
Output
```x and y are equal.
Program ended with exit code: 0```
Since values in `x` and `y` are equal, `x != y` returned false.
## Conclusion
In this C++ Tutorial, we learned about Not Equal Operator in C++, with examples. | 437 | 1,587 | {"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-2024-22 | latest | en | 0.776255 |
https://www.scribd.com/presentation/47016867/markov | 1,508,464,301,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823605.33/warc/CC-MAIN-20171020010834-20171020030834-00547.warc.gz | 964,052,782 | 29,872 | # Markov Analysis
To accompany Quantitative Analysis for Management, 8e by Render/Stair/Hanna
Introduction
Markov analysis is a technique that deals with the probabilities of future occurrences by analyzing presently known probabilities. The technique has numerous applications in business, including market share analysis, bad debt prediction, university enrollment predictions, and determining whether a machine will break down in the future.
To accompany Quantitative Analysis for Management, 8e by Render/Stair/Hanna
Introduction Markov analysis makes the assumption that the system starts in an initial state or condition. respectively. Perhaps in two months the market shares for the two companies will change to 45% and 55% of the market. For example. two competing manufacturers might have 40% and 60% of the market sales. as initial states. Predicting these future states involves knowing the system's likelihood or probability of changing from one state to another. 8e by Render/Stair/Hanna . respectively. To accompany Quantitative Analysis for Management.
and it enables us to predict future states or conditions.Introduction For a particular problem. these probabilities can be collected and placed in a matrix or table. This matrix of transition probabilities shows the likelihood that the system will change from one time period to the next. 8e by Render/Stair/Hanna . To accompany Quantitative Analysis for Management. This is the Markov process.
8e by Render/Stair/Hanna .Assumptions of Markov Analysis There are a limited or finite number of possible states. To accompany Quantitative Analysis for Management. The probability of change remains the same over time. Size of system remains the same. Future state predictable from current state.
and we can call the incorrect functioning the second state. To accompany Quantitative Analysis for Management.States and State Probabilities States are used to identify all possible conditions of a process or a system. 8e by Render/Stair/Hanna . it is possible to identify specific states for many processes or systems. It can be either functioning correctly or not functioning correctly. a machine can be in one of two states at any point in time. We can call the proper operation of the machine the first state. Indeed. For example.
States and State Probabilities In Markov analysis we also assume that the states are both collectively exhaustive and mutually exclusive. To accompany Quantitative Analysis for Management. 8e by Render/Stair/Hanna . Collectively exhaustive means that we can list all of the possible states of a system or process. A student can be in only one of the three management specially areas and not in two or more areas at the same time. Our discussion of Markov analysis assumes that there is a finite number of states for any system. It also means that a person can only be a customer of one of the three grocery stores at any point in time. Mutually exclusive means that a system can be in only one state at any point in time.
«. state 2. « Tn) Where n = number of states T1. T3. T2.States and State Probabilities After the states have been identified. To accompany Quantitative Analysis for Management. Such information is then placed into a vector of the probabilities. T3. T2.. T(i) = vector of State probabilities = (T1.. « Tn = probability of being in state 1. state n. 8e by Render/Stair/Hanna . the next step is to determine the probability that the system is in this state.
such as one machine. it is possible to know with complete certainty what state this item is in. 8e by Render/Stair/Hanna . where we are only dealing with one item. if we are investigating only one machine.0) Where T(1)= vector of states for the machine in period 1 T1=1= probability of being in the first state T2=0= probability of being in the second state To accompany Quantitative Analysis for Management. we may know that at this point in time the machine is functioning correctly. Then the vector of states can be represented as follows: T(1)=(1. For example.States and State Probabilities In some cases.
8e by Render/Stair/Hanna . is 0 for the first period. To accompany Quantitative Analysis for Management.States and State Probabilities The last one machine example shows that the probability the machine is functioning correctly. is 1. state 2. In most cases. and the probability that the machine is functioning incorrectly. state 1. however. we are dealing with more than one item.
which will be called state 1. The probability that a person will be shopping at one of these three grocery stores is as follows: State 1 ± American Food Store: 40.30=30% To accompany Quantitative Analysis for Management. which will be called state 3.30=30% State 3 ± Atlas Foods: 30.000/100.000 people that shop at the three grocery stores during any given month. which will be called state2.000=0.Grocery Store Example Let's look at the vector of states for people in the small town with the three grocery stores. 8e by Render/Stair/Hanna .000 people may be shopping at Atlas Foods. Thirty thousand people may be shopping at Food Mart. Forty thousand people may be shopping at American Food Store.000/100.000/100.000=0. and 30.000=0.40=40% State 2± Food Mart: 30. There could be a total of 100.
3.3= probability that a person will shop at Atlas Food.Grocery Store Example These probabilities can be placed in the vector of state probabilities shown as follows: T(1)=(0. 0. 8e by Render/Stair/Hanna . state3 To accompany Quantitative Analysis for Management.3= probability that a person will shop at Food Mart.4. 0.3) Where T(1)= vector of state probabilities for the three grocery stores for period 1 T1= 0. state2 T3= 0. state1 T2= 0.4= probability that a person will shop at American Food.
Grocery Store Example You should also notice that the probabilities in the vector of states for the three grocery stores represent the market shares for these three stores for the first period. and Atlas Foods has 30% of the market in period 1. To accompany Quantitative Analysis for Management. the market shares can be used in place of probability values. the next step is to find the matrix of transition probabilities. This matrix is used along with the state probabilities in predicting the future. 8e by Render/Stair/Hanna . When the initial states and state probabilities have been determined. when we are dealing with market shares. American Food has 40% of the market. Food Mart has 30%. Thus.
The following definition is helpful: Let Pij = conditional probability of being in state j in the future given the current state of i. such as market shares. For example. This is a matrix of conditional probabilities of being in a future state given a current state. to a future state is the matrix of transitions probabilities.Matrix of Transition Probabilities The concept that allows us to get from a current state. To accompany Quantitative Analysis for Management. P12 is the probability of being in state 2 in the future given the event was in state 1 in the period before. 8e by Render/Stair/Hanna .
. 8e by Render/Stair/Hanna ... For example.Pmn ½ Individual Pij values are usually determined empirically......1n » ¬ P 21 P 22 P 23...... if we have observed over time that 10% of the people currently shopping at store 1 (or state 1) will be shopping at store 2 (state 2) next period............ then we know that P12 = 0........1 or 10%...P3n ¼ ¼ ¬ Pm1..Matrix of Transition Probabilities Let P = matrix of transition probabilities « P11 P12 P13..... To accompany Quantitative Analysis for Management.......P 2n ¼ ¼ P!¬ ¬ P31 P32 P33.........
as follows: To accompany Quantitative Analysis for Management.1 .2¼ ¬ ¼ ¬.2 . The meaning of these probabilities can be expressed in terms of the various states.1 .Transition Probabilities for the Three Grocery Stores Let's say we can determine the matrix of transition probabilities for the three grocery stores by using historical data. and Atlas Foods is state 3.6¼ ½ Recall that American Food represents state 1.8 .7 . 8e by Render/Stair/Hanna .1» ¬. Food Mart is state 2. The results of our analysis appear in the following matrix: P «.2 .
1 = P13 = probability of being in state 3 after being in state 1 the preceding period. Row 2 0. 8e by Render/Stair/Hanna . 0.8 = P11 = probability of being in state 1 after being in state 1 the preceding period. To accompany Quantitative Analysis for Management. 0. 0.1 = P21 = probability of being in state 1 after being in state 2 the preceding period.7 = P22 = probability of being in state 2 after being in state 2 the preceding period.1 = P12 = probability of being in state 2 after being in state 1 the preceding period. 0.2 = P23 = probability of being in state 3 after being in state 2 the preceding period.Transition Probabilities for the Three Grocery Stores Row 1 0.
0. it is possible to predict future state probabilities. After the state probabilities have been determined along with the matrix of transition probabilities. 0.Transition Probabilities for the Three Grocery Stores Row 3 0. To accompany Quantitative Analysis for Management.6 = P33 = probability of being in state 3 after being in state 3 the preceding period. Note that the three probabilities in the top row sum to 1.2 = P31 = probability of being in state 1 after being in state 3 the preceding period. 8e by Render/Stair/Hanna .2 = P32 = probability of being in state 2 after being in state 3 the preceding period. The probabilities for any row in a matrix of transition probabilities will also sum to 1.
we are able to compute the probability that a person will be shopping at one of the grocery stores in the future. 8e by Render/Stair/Hanna . Because this probability is equivalent to market share. and Atlas Foods. With this type of analysis. To accompany Quantitative Analysis for Management.Predicting Future Markets Shares Given the vector of state probabilities and the matrix of transition probabilities. Food Mart. it is not very difficult to determine the state probabilities at a future date. it is possible to determine future market shares for American Food.
if we are in any period n.Predicting Future Markets Shares When the current period is 1. The computations are: To accompany Quantitative Analysis for Management. we can compute the state probabilities for period n+1 as follows: T(n+1)= T (n)P «««« (2) Equation (1) can be used to answer the question of next period's market shares for the grocery stores. 8e by Render/Stair/Hanna . calculating the state probabilities for the next period (period 2) can be accomplished as follows: T(2)= T (1)P «««« (1) Furthermore.
3)(0.3 .1) + (0. 0.3)(0.8) + (0. (0.4)(0.6¼ ½ = [(0.2¼ .31.41.8 . To accompany Quantitative Analysis for Management.2) + (0.3)(0.1) + (0.4)(0. 0.3A¬. 8e by Render/Stair/Hanna .28) T(2)= T (1)P As you can see.3)(0.2).7 .6)] = (0.1» T (2) ! ? . the market share for American Food and Food Mart has increased while the market share for Atlas Foods has decreased.2 .2).3)(0.3)(0.1) + (0.4 ¬ ¼ ¬.2 .1 .Predicting Future Markets Shares «. (0.1 .7)+ (0.4)(0. | 2,707 | 11,197 | {"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-2017-43 | latest | en | 0.893256 |
https://www.ifnipedia.es/ball-mill/13480.html | 1,627,613,896,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153931.11/warc/CC-MAIN-20210730025356-20210730055356-00543.warc.gz | 840,857,211 | 5,081 | Welcome to visit us!
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# Ball Size Calculation For Ball Mill
Ball Mill Presentation – 2fishygirl on Scribd Scribd. Design IMPORTANT DATA TO DESIGN A Ball MILL Feed material Ball Mill Group No.3 Calculation Feed capacity → Mill.
### Ball Mill Calculations Ball Mill
Calculation Of Grinding Media Charging Ball Mill. Ball top size bond formula calculation of the top size grinding media balls or cylpebsmodification of the ball charge this calculator analyses the granulometry of the material inside the mill and proposes a modification of the ball charge in order to improve the mill.Ball charges calculators - thecementgrindingoffice - Ball top size (bond formula) calculation of the top size grinding media (balls or cylpebs) -Modification of the Ball Charge This calculator analyses the granulometry of the material inside the mill and proposes a modification of the ball charge in order to improve the mill efficiency - Ball charges composition These calculators give the.Ball mill charge calculation in cement ball mill grinding media calculation the ball charge mill consists handbook for dry process plants pdf 187 technical notes 8 grinding r. p. king. media or charge in the mill and dm is the diameter of e is a function of the ball media size distribution in the mill.Ball Mill Design Power Calculation. Ball Mill Power Calculation Example 1. A wet grinding ball mill in closed circuit is to be fed 100 TPH of a material with a work index of 15 and a size distribution of 80 passing inch (6350 microns). The required product size distribution is to be 80 passing 100 mesh (149 microns).
### Ball Mill Specifications Ball Mill Speed Calculation
Speed And Feed Calculators Ball Mill Finish Calculator Part Spacing Calculator G And M Code Characters Standard End Mill Sizes Standard Drill Sizes Drill And Counterbore Sizes. Contact. End Mill Speed Feed Calculator. Tool Dia. In. Radial (Side) Depth of Cut. This will adjust the feedrate if less than the tool rad. In. Num of Flutes.Calculation Of Torque For Ball Mill Calculation Of Torque For Ball Mill. Batch tests were run in the 20 litre stirred ball mill to achieve efficient grinding It has been shown that the median size of the product can be calculated by the pin spacing, pin diameter and ball diameter significantly affected the mill torque.Ball Mill Motor Power Draw Sizing And Design Formula. The following equation is used to determine the power that wet grinding overflow ball mills should draw for mills larger than 33 meters 10 feet diameter inside liners the top size of the balls used affects the power drawn by the mill this is called the ball size factor s rod and ball mills by ca rowland and dm kjos allischalmers.Ball Mill Working Principle And Calculation. Ball mill working principle and calculationWorking principle operationThe apparent difference in capacities between grinding mills listed as being the same size is due to the fact that there is no uniform method of designating the size of a mill for example a x ball mill has a working diameter of inside the liners and has per cent.
### Ball Mill Design Power Calculation
Calculate and Select Ball Mill Ball Size for Optimum Grinding. In Grinding, selecting (calculate) the correct or optimum ball size that allows for the best and optimum ideal or target grind size to be achieved by your ball mill is an important thing for a Mineral Processing Engineer AKA Metallurgist to do.Jun 26, 2017 Ball Nose Milling Without a Tilt Angle. Ball nose end mills are ideal for machining 3-dimensional contour shapes typically found in the mold and die industry, the manufacturing of turbine blades, and fulfilling general part radius requirements.To properly employ a ball nose end mill (with no tilt angle) and gain the optimal tool life and part finish, follow the 2-step process below (see Figure 1).Involve grinding). With Lloyd's ball milling book having sold over 2000 copies, there are probably over 1000 home built ball mills operating in just America alone. This article borrows from Lloyd's research, which was obtained from the commercial ball milling industry, and explains some of the key design criteria for making your own ball mill.Ball Mill Design Power Calculation. The basic parameters used in ball mill design (power calculations), rod mill or any tumbling mill sizing are material to be ground, characteristics, Bond Work Index, bulk density, specific density, desired mill tonnage capacity DTPH, operating solids or pulp density, feed size as F80 and maximum ‘chunk size’, product size as P80 and maximum and. | 992 | 4,851 | {"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 | 0.748766 |
https://www.coursehero.com/file/1651148/ch05/ | 1,487,619,822,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170600.29/warc/CC-MAIN-20170219104610-00359-ip-10-171-10-108.ec2.internal.warc.gz | 812,469,137 | 706,014 | # ch05 - Chapter Five: Decisions Chapter Goals To be able to...
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Chapter Five: Decisions
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To be able to implement decisions using if statements To understand how to group statements into blocks To learn how to compare integers, floating-point numbers, strings, and objects To recognize the correct ordering of decisions in multiple branches To program conditions using Boolean operators and variables To understand the importance of test coverage Chapter Goals
The if statement lets a program carry out different actions depending on a condition If (amount <= balance) balance = balance – amount; The if Statement
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If (amount <= balance) balance = balance – amount; else balance = balance – OVERDRAFT_PENALTY The if / else Statement
Simple statement balance = balance - amount; Compound statement if (balance >= amount) balance = balance - amount; Also while, for , etc. (loop statements – Chapter 6) Block statement { double newBalance = balance - amount; balance = newBalance; } Statement Types
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if( condition ) statement if ( condition ) statement 1 else Example: if (amount <= balance) balance = balance - amount; if (amount <= balance) balance = balance - amount; else Purpose: To execute a statement when a condition is true or false. Syntax 5.1 The if Statement
{ statement 1 statement 2 . . . } Example: { double newBalance = balance - amount; balance = newBalance; } Purpose: To group several statements together to form a single statement. Syntax 5.2 Block Statement
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Why did we use the condition amount <= balance and not amount < balance in the example for the if/else statement? Answer: If the withdrawal amount equals the balance, the result should be a zero balance and no penalty. Self Check 5.1
What is logically wrong with the statement if (amount <= balance) newBalance = balance - amount; balance = newBalance; and how do you fix it? Answer: Only the first assignment statement is part of the if statement. Use braces to group both assignment statements into a block statement. Self Check 5.2
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Relational operators compare values Java Math Notation Description > > Greater than >= Greater than or equal < < Less than <= Less than or equal == = Equal != Not equal The == denotes equality testing a = 5; // Assign 5 to a if (a == 5) . . . // Test whether a equals 5 Comparing Values: Relational Operators
Consider this code: double r = Math.sqrt(2); double d = r * r -2; if (d == 0) System.out.println("sqrt(2)squared minus 2 is 0"); else System.out.println("sqrt(2)squared minus 2 is not 0 but " + d); It prints: sqrt(2)squared minus 2 is not 0 but 4.440892098500626E-16 Comparing Floating-Point Numbers
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To avoid roundoff errors, don't use == to compare floating-point numbers To compare floating-point numbers test whether they are close enough : | x - y | ≤ ε final double EPSILON = 1E-14; if (Math.abs(x - y) <= EPSILON) // x is approximately equal to y ε is a small number such as 10 -14 Comparing Floating-Point Numbers
Don't use == for strings! if (input == "Y") // WRONG!!!
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This is the end of the preview. Sign up to access the rest of the document. | 871 | 3,864 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-09 | longest | en | 0.80646 |
test.hotcrp.com | 1,638,768,742,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363290.39/warc/CC-MAIN-20211206042636-20211206072636-00479.warc.gz | 610,115,186 | 4,385 | Formulas
Program committee members and administrators can search and display formulas that calculate properties of paper scores—for instance, the standard deviation of papers’ Overall merit scores, or average Overall merit among reviewers with high Reviewer expertise.
To display a formula, use a search term such as “show:var(OveMer)” (show the variance in Overall merit scores, along with statistics for all papers). You can also graph formulas. To search for a formula, use a search term such as “formula:var(OveMer)>0.5” (select papers with variance in Overall merit greater than 0.5). Or save formulas using Search > View options > Edit formulas.
Formulas use a familiar expression language. For example, this computes the sum of the squares of the overall merit scores:
sum(OveMer**2)
This calculates an average of overall merit scores, weighted by expertise (high-expertise reviews are given slightly more weight):
wavg(OveMer, RevExp >= 4 ? 1 : 0.8)
And there are many variations. This version gives more weight to PC reviewers with the “#heavy” tag:
wavg(OveMer, re:#heavy + 1)
(“re:#heavy + 1” equals 2 for #heavy reviews and 1 for others.)
Formulas work better for numeric scores, but you can use them for letter scores too. HotCRP uses alphabetical order for letter scores, so the “min” of scores A, B, and D is A. For instance:
count(confidence=X)
Expressions
Formula expressions are built from the following parts:
Arithmetic
2Numbers
true, falseBooleans
nullThe null value
(e)Parentheses
e + e, e - eAddition, subtraction
e * e, e / e, e % eMultiplication, division, remainder
e ** eExponentiation
e == e, e != e,
e < e, e > e, e <= e, e >= e
Comparisons
!eLogical not
e1 && e2Logical and (returns e1 if e1 is false, otherwise returns e2)
e1 || e2Logical or (returns e1 if e1 is true, otherwise returns e2)
test ? iftrue : iffalseIf-then-else operator
let var = val in eLocal variable definition
greatest(e, e, ...)Maximum
least(e, e, ...)Minimum
coalesce(e, e, ...)Null coalescing: return first of es that is not null
log(e)Natural logarithm
log(e, b)Log to the base b
round(e[, m])Round to the nearest multiple of m
Submission properties
pidPaper ID
auNumber of authors
au:pcNumber of PC authors
au:textNumber of authors matching text
topicsNumber of topics
topics:textNumber of topics matching text
Tags
#tagnameTrue if this paper has tag tagname
tagval:tagnameThe value of tag tagname, or null if this paper doesn’t have that tag
Scores
overall-meritThis review’s Overall merit score
Only completed reviews are considered.
OveMerAbbreviations also accepted
OveMer:externalOverall merit for external reviews, null for other reviews
OveMer:R2Overall merit for round R2 reviews, null for other reviews
Submitted reviews
re:typeReview type
re:roundReview round
re:auwordsReview word count (author-visible fields only)
re:primaryTrue for primary reviews
re:secondaryTrue for secondary reviews
re:externalTrue for external reviews
re:pcTrue for PC reviews
re:sylviaTrue if reviewer matches “sylvia”
Review preferences
prefReview preference
prefexpPredicted expertise
Aggregate functions
Aggregate functions calculate a value based on all of a paper’s reviews and/or review preferences. For instance, “max(OveMer)” would return the maximum Overall merit score assigned to a paper.
An aggregate function’s argument is calculated once per viewable review or preference. For instance, “max(OveMer/RevExp)” calculates the maximum value of “OveMer/RevExp” for any review, whereas “max(OveMer)/max(RevExp)” divides the maximum overall merit by the maximum reviewer expertise.
The top-level value of a formula expression cannot be a raw review score or preference. Use an aggregate function to calculate a property over all review scores.
Aggregates
max(e), min(e)Maximum, minimum
count(e)Number of reviews where e is not null or false
sum(e)Sum
avg(e)Average (mean)
wavg(e, weight)Weighted average; equals “sum(e * weight) / sum(weight)”
median(e)Median
quantile(e, p)Quantile; 0≤p≤1; 0 yields min, 0.5 median, 1 max
stddev(e)Population standard deviation
var(e)Population variance
stddev_samp(e), var_samp(e)Sample standard deviation, sample variance
any(e)True if any of the reviews have e true
all(e)True if all of the reviews have e true
argmin(x, e)Value of x when e is minimized
argmax(x, e)Value of x when e is maximized | 1,109 | 4,364 | {"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-2021-49 | longest | en | 0.731229 |
http://www.studymode.com/essays/Anthony-Case-3-2-And-11-2-480629.html | 1,524,173,086,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937045.30/warc/CC-MAIN-20180419204415-20180419224415-00563.warc.gz | 535,216,204 | 25,658 | Anthony Case 3-2 and 11-2
Pages: 2 (326 words) Published: November 16, 2010
Anthony Cases 3-2, 11-2, and Problem 5-7
Anthony Cases 3-2: Loan Pine Café (B)
1. Below is the income statement for the Lone Pine Café from November 2, 2005-March 30, 2006. [pic]
*Notes regarding two calculations:
a. Sales = \$43,480 (cash) + \$870 (accounts receivable) = \$44, 350 b. Inventory (food & beverage) = \$10,016 (purchase from suppliers) + \$1,583 (accounts payable) + \$370 (which is the change from inventory of \$2800-\$2430) = \$11,969. 2. This income statement communicates to Mrs. Antoine that during this accounting period for operating the business, the business had a net loss. This resulted as the business expenses grossly exceeded the total sales revenues. Since the income statement and the balance sheet are said to articulate with one another, this income statement also validates the loss in cash during the same operating period found on the balance sheet. Therefore, in order for Mrs. Antoine to continue to operate her business, not only must she shore up her dissolved partnership and payoff the bank loan and accounts payable suppliers, but she must also find a way to lesson her expenses and generate more sales – therefore, providing her a positive net income.
Anthony Case 11-2: Amerbran Company (A)
1. Below is the state of cash flows for Amerbran Company for the year ended December 31, 20x1. (Please see in the following examples: the balance sheet with net change and then the cash flow statement). [pic]
[pic]
Anthony Problem 5-7
a. Quick & Current Ratios are found below:
[pic]
Current Ratio = Current assets/Current liabilities = \$125,200/\$71,300 = 1.76 Quick Ratio = Monetary current assets/Current liabilities = (\$23,100 + 32,800)/\$71,300 = 0.78 b.) Number of days’ cash on hand:
[pic]
Days’ cash = Cash/(Total cash expenses/365) = \$23,100/(\$231,000/365) = 36.5... | 501 | 1,899 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2018-17 | latest | en | 0.92039 |
https://xianblog.wordpress.com/2018/06/04/prior-against-truth/?shared=email&msg=fail | 1,603,378,464,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107879673.14/warc/CC-MAIN-20201022141106-20201022171106-00411.warc.gz | 957,092,449 | 17,006 | ## prior against truth!
A question from X validated had interesting ramifications, about what happens when the prior does not cover the true value of the parameter (assuming there ? In fact, not so much in that, from a decision theoretic perspective, the fact that that π(θ⁰)=0, or even that π(θ)=0 in a neighbourhood of θ⁰ does not matter [too much]. Indeed, the formal derivation of a Bayes estimator as minimising the posterior loss means that the resulting estimator may take values that were “impossible” from a prior perspective! Indeed, taking for example the posterior mean, the convex combination of all possible values of θ under π may well escape the support of π when this support is not convex. Of course, one could argue that estimators should further be restricted to be possible values of θ under π but that would reduce their decision theoretic efficiency.
An example is the brilliant minimaxity result by George Casella and Bill Strawderman from 1981: when estimating a Normal mean μ based on a single observation xwith the additional constraint that |μ|<ρ, and when ρ is small enough, ρ1.0567 quite specifically, the minimax estimator for this problem under squared error loss corresponds to a (least favourable) uniform prior on the pair {ρ,ρ}, meaning that π gives equal weight to ρ and ρ (and none to any other value of the mean μ). When ρ increases above this bound, the least favourable prior sees its support growing one point at a time, but remaining a finite set of possible values. However the posterior expectation, 𝔼[μ|x], can take any value on (ρ,ρ).
In an even broader suspension of belief (in the prior), it may be that the prior has such a restricted support that it cannot consistently estimate the (true value of the) parameter, but the associated estimator may remain admissible or minimax.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 418 | 1,912 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2020-45 | latest | en | 0.920536 |
http://westclintech.com/Blog/tabid/132/EntryId/22/Its-Opening-Day.aspx | 1,670,287,811,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711064.71/warc/CC-MAIN-20221205232822-20221206022822-00576.warc.gz | 53,978,128 | 34,487 | ## It's Opening Day!
Apr 8
Written by: Charles Flock
4/8/2009 7:07 PM
Using the XLeratorDB/statistics module to calculate some baseball statistics
Today is opening day for the New York teams, so it’s raining with temperatures in the 50’s. But, since we have two new stadia in New York that are not quite ready to open, the Mets and Yankees are both opening on the road. The Mets are at Cincinnati and the Yanks are at Baltimore. When I go to www.mlb.com they talk about the Orioles needing to have several players have career years. That got me to thinking about baseball statistics and what it actually means to have a career year. And, specifically it got me to thinking about Barry Bonds and the 2001 season, when he set the major league record for most home runs in a season (73).
Of course you cannot really discuss that record without talking about steroids and what has been called the ‘juiced era’. I am not really interested in getting involved in that discussion. I am interested in seeing what the numbers can tell us (or not tell us) about some baseball records.
From my perspective, baseball is a good sport to start doing some statistical analysis because it has records going back over a hundred years. There are a lot of things that need to be taken into consideration, though. First is that the recordkeeping has not been consistent. Second is that the rules and the equipment have changed over time. Third is that playing conditions are not uniform. Fourth is that for a good chunk of its statistical history, many players were excluded on the basis of skin color and ethnic origin. These facts, and dozens of others, confound the statistical analysis to a degree, but they don’t prevent us from doing some basic things, like making some estimation of the statistical significance of the 2001 record. In doing this we will use a variety of XLeratorDB functions to calculate the minimum (MIN), the maximum (MAX), the kurtosis (KURT), the skewness (SKEW), the standard deviation (STDEVP), and mean (AVERAGE) of the data. We will also demonstrate other statistical functions like STANDARDIZE and NORMDIST which we will use to compare the actual data to a theoretically normal distribution, and I think that you will find this much easier to do using SQL Server and XLeratorDB than EXCEL or Access.
In order to do analysis, you need to have data; www.baseball-reference.com and www.baseball1.com are great sources of data. The data can be delivered in a variety of formats. For example, if we want the data for Barry Bonds career, we can get it displayed on the screen, in an Access file, in a SQL Server database, as table formatted data, in EXCEL, as a CSV, and probably even other formats that I don’t know anything about. There are copyright restrictions so I am not going to publish the data here. And www.baseball1.com is free, though it does request donations, which is appropriate as what they provide is really good value. I downloaded their database which contains an astonishing amount of information. The data is not normalized, but I will discuss that in another blog, so it very much looks like an EXCEL spreadsheet. One of the download options is a comma delimited file, but I downloaded the Access version and imported it into SQL Server.
The first thing I did, was extract the maximum number of home runs for each year since 1920 (when the number of games increased from 140 to 154 and the ball changed). This is a straightforward piece of SQL, the results of which I pasted into EXCEL to create a graph.
select yearid,
max(hr)
from batting
where yearid > 1919
group by yearid
order by yearid
While this is pretty interesting, there is some information that I am interested in that is not immediately obvious from the graphical representation. Specifically, I would like to know the lowest number of home runs, the highest number, the average (or mean), and the standard deviation.
In SQL Server, there is a MIN, MAX, AVG, and STDEVP function, but none of them allow expressions that contain aggregates. XLeratorDB allows aggregates to be included, and we can use the QUARTILE function to calculate the MIN and MAX. This saves us a few steps, since we do not need to create a common table expression or a view, and the SQL runs very quickly.
The following SQL will return the data we are looking for:
select wct.QUARTILE_q('select
max(hr)
from batting
where yearid > 1919
group by yearid', 0) as [Min],
wct.QUARTILE_q('select
max(hr)
from batting
where yearid > 1919
group by yearid', 4) as [Max],
wct.AVERAGE_q('select
max(hr)
from batting
where yearid > 1919
group by yearid') as [Average],
wct.STDEVP_Q('select
max(hr)
from batting
where yearid > 1919
group by yearid') as [St. Dev.]
Which returns the following information:
Min Max Average St. Dev.
-------------- -------------- ---------------------- ----------------------
28 73 46.8988764044944 7.91960296334884
(1 row(s) affected)
This means that since 1920, the lowest number of home runs to lead the major leagues was 28 (in 1945) and highest number was 73 (in 2001). The average number of home runs required to lead the league in home runs is almost 47 and the standard deviation is almost 8.
In very general terms, and assuming that the distribution is a ‘normal’ distribution, that means that about 68% (68.2689492%) of the time, we would expect the home run leader for the season to have 47 ± 8 home runs (39–55). A little more than 95% (95.4499736%) of the time, we would expect the home run leader to have approximately 47 ± 16 home runs (31–63).
The STANDARDIZE function will actually show us the number of standard deviations that a particular value represents, which gives us some idea of how the actual performance differences from the hypothetical normal distribution. We can enter the following SQL, which uses a common table expression (just to keep things simple) and the average (mean) and standard deviation from the previous SELECT:
with HR as
(
select yearid,
max(hr) as [Home Runs]
from batting
where yearid > 1919
group by yearid
) select HR.yearid
,wct.STANDARDIZE([Home Runs], 46.8988764044944, 7.91960296334884)
from HR
order by yearid
I then pasted the results into EXCEL and produced the following graph:
We can immediately see that 2001, with 73 home runs, is more than 3 (3.2958) standard deviations away. In a way, that’s a very interesting number, as in a normal distribution, you expect 99.9% of the results would be within 3.2906 standard deviations. It might be possible to conclude that hitting 73 home runs, purely from this analysis, is not really a statistically significant event. In fact, we have a function, the NORMDIST function, that will tell us the exact probability:
select wct.NORMDIST(73
,46.8988764044944
,7.91960296334884
,0)
Which returns the following information:
----------------------
0.000220568560862717
(1 row(s) affected)
This means that there was a .0221% chance that a result of 73 would be contained in this dataset; in other words, a remote, but not improbable, chance.
There are a couple of additional things that we might want to look at. First, the number of games changed in 1961 from 154 to 162. Second, the 1994 season was shortened by about 50 games. While making those kind of adjustments in a spreadsheet can be tricky and time-consuming, it’s a pretty easy change to the WHERE clause in all of our SELECT statements:
SELECT. . .
where yearid > 1960 and
yearid <> 1994
We would then get the following statistics:
Min Max Average St. Dev.
-------------- -------------- ---------------------- ----------------------
31 73 48.063829787234 8.08835321144887
(1 row(s) affected)
From which we can then produce the following graph:
This puts the 2001 season 3.083 standard deviations away, giving it a likelihood of .0426%, slightly less than double the previous calculation.
Earlier in this blog in talking about doing this type of analysis, we assumed that the data represented a normal distribution. There are two measures, SKEW and KURTOSIS that we can use to make that evaluation. And, if you want a good explanation of what a normal distribution is, go here http://en.wikipedia.org/wiki/Normal_distribution.
First, let’s create a different graphical representation of the data by eliminating the time dimension. We are simply going to count the number of times a particular number led MLB in home runs. That graph looks like this:
Visually, we can tell that this isn’t a perfectly normal distribution. The first thing that we will want to measure is the SKEW. We can do this with the following SELECT statement:
select wct.SKEW_q('select
max(hr)
from batting
where yearid > 1919
group by yearid')
Which returns the following results:
----------------------
0.542542382246413
(1 row(s) affected)
A positive value for skewness indicates that a distribution has an asymmetric tail which extends more positive. What this means to me is that normal statistical measures will tend to miss the upside to a greater degree than they miss the downside.
The second measure that we want to look at is KURTOSIS. Again, a simple SELECT statement will give us that information:
select wct.KURT_q('select
max(hr)
from batting
where yearid > 1919
group by yearid')
Which returns the following results:
----------------------
1.14592237200248
(1 row(s) affected)
Kurtosis, like skewness, is a non-dimensional quality. Kurtosis measures the relative peakedness or flatness of a distribution as compared to a normal distribution. Since the kurtosis of this distribution is positive, it is characterized as leptokurtic. One other note, is that the Kurtosis function is identical to the EXCEL KURT function and thus measures excess kurtosis.
Since we know the mean and the standard deviation of the dataset, we can actaully calculate what the normal distriibution would be like. We can create a CTE which gives us every home run value from 28 to 73, and we can then calculate the NORMDIST value for each home run value. We then multilpy that value by 46, which is the range of the values (from 28 to 73). The following SELECT creates that data for us:
with NHR as
(
select 28 as [Home Runs]
UNION ALL
select [Home Runs] + 1
FROM NHR
where [Home Runs] < 73
) select [Home Runs]
,wct.NORMDIST([Home Runs],46.8988764044944,7.91960296334884,0)*46
from NHR
order by 1 option(maxrecursion 46)
Which we can then just paste into EXCEL and graph:
And finally, we can simply plot the actual values against the normalized values and create the following graph:
This provides a visualization of the postive skew and positive kurtosis for the dataset.
In very short order we have been able to perform a pretty sophisticated analysis of this one statistic. I think that this would be quite hard to do in EXCEL and there are almost 100,000 player-seasons that need to be sifted through to get the 89 player-seasons that we are most interested in. That argues very much for putting the data into a database, but SQL Server (and Access), do not have the sophisticated statistical functions which EXCEL provides. XLeratorDB brings those functions into the database and creates a very easy-to-use but powerful environment for doing this kind of analysis.
In this article we have demonstrated a number of statistical functions that can be used to analyze some common statistics. Next week, we will look at a some more baseball statistics and then we will build a simple Monte Carlo simulation, all in SQL Server, for the home run statistic.
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Go | 2,778 | 11,850 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2022-49 | longest | en | 0.970756 |
https://www.delftstack.com/howto/python/nested-for-loop-in-one-line-in-python/ | 1,669,843,060,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710771.39/warc/CC-MAIN-20221130192708-20221130222708-00205.warc.gz | 753,831,946 | 11,372 | # Nested for Loop in One Line in Python
The `for` loop is one of the most commonly used loops to iterate items from a list. In Python, we write the `for` loop in one line, but how can we write it in one line when we have to use another loop inside it?
This tutorial will discuss some different methods that can be used to write a nested `for` loop in just one line.
## Nested `for` Loop in One Line Using List Comprehension in Python
There are three methods in which we can easily write a nested for loop in one line. One of them is list comprehension. It is a compact way of creating lists.
Syntax:
``````# python
[print(a, b) for a in listOne for b in listTwo]
``````
Let’s go through an example and use list comprehension. First of all, we will create two lists, `listOne` and `listTwo`, and then use them to print the result as shown below.
``````# python
listOne = [1,3,5]
listTwo = [2,4,6]
[print(a+b) for a in listOne for b in listTwo]
``````
Output:
From the example, we have added 2 different lists using the nested `for` loop in one line using list comprehension.
## Nested `for` Loop in One Line Using the `exec()` Function in Python
Now, we will discuss another function that can help us achieve a nested `for` loop in one line, the `exec()` function. It is used for the dynamic execution of a Python program that can be a string or an object code.
Let’s go through an example in which we will add two lists and print the sum as shown below.
``````# python
listOne = [1,3,5]
listTwo = [2,4,6]
exec("for a in listOne:\n for b in listTwo:\n print(a+b)")
``````
Output:
As you can see from the example, using the `exec()` function, we added 2 different lists using nested `for` loop in one line.
## Summary
This tutorial taught us two different methods, list comprehension and `exec()`, that can be used to nest `for` loops in one line.
Rana is a computer science graduate passionate about helping people to build and diagnose scalable web application problems and problems developers face across the full-stack. | 516 | 2,052 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-49 | longest | en | 0.856192 |
https://nrich.maths.org/public/leg.php?code=72&cl=3&cldcmpid=379 | 1,569,133,334,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575168.82/warc/CC-MAIN-20190922053242-20190922075242-00047.warc.gz | 610,468,199 | 9,713 | # Search by Topic
#### Resources tagged with Generalising similar to Converse:
Filter by: Content type:
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### Janine's Conjecture
##### Age 14 to 16 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
### Multiplication Square
##### Age 14 to 16 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### Magic Squares II
##### Age 14 to 18
An article which gives an account of some properties of magic squares.
### AMGM
##### Age 14 to 16 Challenge Level:
Can you use the diagram to prove the AM-GM inequality?
### Pair Products
##### Age 14 to 16 Challenge Level:
Choose four consecutive whole numbers. Multiply the first and last numbers together. Multiply the middle pair together. What do you notice?
### Steel Cables
##### Age 14 to 16 Challenge Level:
Some students have been working out the number of strands needed for different sizes of cable. Can you make sense of their solutions?
### What's Possible?
##### Age 14 to 16 Challenge Level:
Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make?
### More Twisting and Turning
##### Age 11 to 16 Challenge Level:
It would be nice to have a strategy for disentangling any tangled ropes...
### One, Three, Five, Seven
##### Age 11 to 16 Challenge Level:
A game for 2 players. Set out 16 counters in rows of 1,3,5 and 7. Players take turns to remove any number of counters from a row. The player left with the last counter looses.
### All Tangled Up
##### Age 14 to 18 Challenge Level:
Can you tangle yourself up and reach any fraction?
### Pentanim
##### Age 7 to 16 Challenge Level:
A game for 2 players with similarities to NIM. Place one counter on each spot on the games board. Players take it is turns to remove 1 or 2 adjacent counters. The winner picks up the last counter.
### Generating Triples
##### Age 14 to 16 Challenge Level:
Sets of integers like 3, 4, 5 are called Pythagorean Triples, because they could be the lengths of the sides of a right-angled triangle. Can you find any more?
### Sums of Pairs
##### Age 11 to 16 Challenge Level:
Jo has three numbers which she adds together in pairs. When she does this she has three different totals: 11, 17 and 22 What are the three numbers Jo had to start with?”
### Lower Bound
##### Age 14 to 16 Challenge Level:
What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 =
### Games Related to Nim
##### Age 5 to 16
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning.
### Square Pizza
##### Age 14 to 16 Challenge Level:
Can you show that you can share a square pizza equally between two people by cutting it four times using vertical, horizontal and diagonal cuts through any point inside the square?
### Winning Lines
##### Age 7 to 16
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games.
### Nim
##### Age 14 to 16 Challenge Level:
Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The loser is the player who takes the last counter.
### Arithmagons
##### Age 14 to 16 Challenge Level:
Can you find the values at the vertices when you know the values on the edges?
### Tourism
##### Age 11 to 14 Challenge Level:
If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.
### For Richer for Poorer
##### Age 14 to 16 Challenge Level:
Charlie has moved between countries and the average income of both has increased. How can this be so?
### Of All the Areas
##### Age 14 to 16 Challenge Level:
Can you find a general rule for finding the areas of equilateral triangles drawn on an isometric grid?
### Loopy
##### Age 14 to 16 Challenge Level:
Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture?
### Regular Hexagon Loops
##### Age 11 to 14 Challenge Level:
Make some loops out of regular hexagons. What rules can you discover?
### Hypotenuse Lattice Points
##### Age 14 to 16 Challenge Level:
The triangle OMN has vertices on the axes with whole number co-ordinates. How many points with whole number coordinates are there on the hypotenuse MN?
### Pick's Theorem
##### Age 14 to 16 Challenge Level:
Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.
### Odd Differences
##### Age 14 to 16 Challenge Level:
The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = n² Use the diagram to show that any odd number is the difference of two squares.
### Konigsberg Plus
##### Age 11 to 14 Challenge Level:
Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.
### In a Spin
##### Age 14 to 16 Challenge Level:
What is the volume of the solid formed by rotating this right angled triangle about the hypotenuse?
### Pareq Calc
##### Age 14 to 16 Challenge Level:
Triangle ABC is an equilateral triangle with three parallel lines going through the vertices. Calculate the length of the sides of the triangle if the perpendicular distances between the parallel. . . .
### Take Three from Five
##### Age 14 to 16 Challenge Level:
Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?
### Converging Means
##### Age 14 to 16 Challenge Level:
Take any two positive numbers. Calculate the arithmetic and geometric means. Repeat the calculations to generate a sequence of arithmetic means and geometric means. Make a note of what happens to the. . . .
### Nim-like Games
##### Age 7 to 16 Challenge Level:
A collection of games on the NIM theme
### Jam
##### Age 14 to 16 Challenge Level:
A game for 2 players
### Consecutive Negative Numbers
##### Age 11 to 14 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
##### Age 7 to 14 Challenge Level:
I added together some of my neighbours house numbers. Can you explain the patterns I noticed?
### Harmonic Triangle
##### Age 14 to 16 Challenge Level:
Can you see how to build a harmonic triangle? Can you work out the next two rows?
### Mystic Rose
##### Age 14 to 16 Challenge Level:
Use the animation to help you work out how many lines are needed to draw mystic roses of different sizes.
### Problem Solving, Using and Applying and Functional Mathematics
##### Age 5 to 18 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### Magic Squares
##### Age 14 to 18
An account of some magic squares and their properties and and how to construct them for yourself.
### Pinned Squares
##### Age 14 to 16 Challenge Level:
What is the total number of squares that can be made on a 5 by 5 geoboard?
### Beelines
##### Age 14 to 16 Challenge Level:
Is there a relationship between the coordinates of the endpoints of a line and the number of grid squares it crosses?
### Building Gnomons
##### Age 14 to 16 Challenge Level:
Build gnomons that are related to the Fibonacci sequence and try to explain why this is possible.
### Gnomon Dimensions
##### Age 14 to 16 Challenge Level:
These gnomons appear to have more than a passing connection with the Fibonacci sequence. This problem ask you to investigate some of these connections.
### Multiplication Arithmagons
##### Age 14 to 16 Challenge Level:
Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons?
##### Age 14 to 16 Challenge Level:
A counter is placed in the bottom right hand corner of a grid. You toss a coin and move the star according to the following rules: ... What is the probability that you end up in the top left-hand. . . .
### Partly Painted Cube
##### Age 14 to 16 Challenge Level:
Jo made a cube from some smaller cubes, painted some of the faces of the large cube, and then took it apart again. 45 small cubes had no paint on them at all. How many small cubes did Jo use?
### Plus Minus
##### Age 14 to 16 Challenge Level:
Can you explain the surprising results Jo found when she calculated the difference between square numbers?
### Equilateral Areas
##### Age 14 to 16 Challenge Level:
ABC and DEF are equilateral triangles of side 3 and 4 respectively. Construct an equilateral triangle whose area is the sum of the area of ABC and DEF.
### Jam
##### Age 14 to 16 Challenge Level:
To avoid losing think of another very well known game where the patterns of play are similar. | 2,180 | 9,476 | {"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} | 4.09375 | 4 | CC-MAIN-2019-39 | latest | en | 0.83881 |
http://www.promegaconnections.com/converting-rpm-to-g-force-rcf-and-vice-versa-2/ | 1,558,329,107,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255562.23/warc/CC-MAIN-20190520041753-20190520063753-00231.warc.gz | 326,210,113 | 15,702 | Converting RPM to g Force (RCF) and Vice Versa
g Force or Relative Centrifugal Force (RCF) is the amount of acceleration to be applied to the sample. It depends on the revolutions per minute (RPM) and radius of the rotor, and is relative to the force of Earth’s gravity.
A good, precise protocol for centrifugation instructs you to use the g force rather than RPMs because the rotor size might differ, and g force will be different while the revolutions per minute stay the same. Unfortunately, many protocols are written in hurry and instructions are given in RPMs. Therefore, you have to convert g force (RCF) into revolutions per minute (rpms) and vice versa.
Modern centrifuges have an automatic converter but older ones do not. There is a simple formula to calculate this, but it takes some time to do the calculation. Meanwhile, your cells might die or the biochemical reaction goes on for three times longer than it should.
There are several ways to make conversion:
1. Use a website where you can select centrifuge model and rotor head.
2. Use online converters.
3. Use a Nomogram (Nomograph).
4. Make the calculation yourself.
1. The most precise way to make conversion is to use converters for a particular centrifuge and a rotor head. Companies provide the data. Here are the links to a resources online from where you can pick up the model and rotor:
Beckman, Sorvall or Thermo Fisher Scientific
Sorvall, Beckman and Eppendorf
Eppendorf
Thermo Fisher Scientific
2. Here are several links to converters online:
Insilico
Broken Symetry
EnCor Biotechnology
Insilicase
3. Print and use Nomogram (a graphical calculator). They are available from the following:
University of Maryland web site
DJB Labcare
Optics Planet
4. The formula is always good to have in case your internet is down or the above links cease to exist.
g Force (RCF) = (rpm)2 × 1.118 × 10-5 × r
RPM = √[RCF/(r × 1.118)] × 1 × 105
g = Relative Centrifuge Force
N = Revolutions Per Minute (RPM)
I hope the above information will make your calculation easy.
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Promega products are used by life scientists who are asking fundamental questions about biological processes and by scientists who are applying scientific knowledge to diagnose and treat diseases, discover new therapeutics, and use genetics and DNA testing for human identification. Originally, founded in 1978 in Madison, Wisconsin, USA, Promega has branches in 16 countries and more than global distributors serving 100 countries. | 583 | 2,548 | {"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-2019-22 | latest | en | 0.856004 |
http://hst-archive.web.cern.ch/archiv/HST2003/publish/ask%20an%20expert/website/FAQs/Higgs.html | 1,656,415,124,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103516990.28/warc/CC-MAIN-20220628111602-20220628141602-00176.warc.gz | 25,707,076 | 5,432 | FAQ HIGGS BOSON
1. When has the theory about Higgs boson appeared ?
In the early 60's theoretical particle physics made big progress with the emergence of a yet more complete description of nature, namely the realization of a theoretical unification of the electromagnetic and the weak force. Despite being very successful in some cases, the formulation suffered initially several serious problems both mathematically and experimentally. In the formulation it had to be assumed that all particles are massless, but this in itself causes other problems. There was a need to find a new way to introduce mass of the particles in the theory. In 1966 Professor Peter W. Higgs proposed a solution. Elaborating on his proposal, today's version suggests that in the very early Universe all particles were massless. Peter HIGGS
2. Why is this theory so important/interesting ?
Shortly after the birth of the Universe in the Big Bang, as the universe expanded the temperature fell below a critical value where a new type of field developed everywhere in the Universe (field, cmp. magnetic field around a magnet. Every point in space has a property: a measurable magnetic force and direction). We call this particular field the Higgs field. Some particles coupled to this field and the property they acquired is what we measure as mass. That is, particles are not solid in themselves but can be seen as a wave on a water surface. Although a wave moves no water from one side of a lake to another, it carries a lot of information: energy, momentum, amplitude, wavelength, etc. For particles mass is just another property acquired by interacting with the ever pervading Higgs field and that property we perceive as mass. Thus as the field developed below the critical temperature some particles acquired mass and some others not. The fact that the mass is "given" to the particles through an external mechanism solved the problems in the theory. You may find it strange that this field suddenly appeared in the Universe but there are several similar physics phenomena around us. A ferromagnet heated to a temperature higher than the "Curie temperature" will lose its magnetization. However, as it is cooled, it will again develop a magnetic field with a specific direction. We call this type of phenomena "spontaneous symmetry breaking". As you might know, interactions between particles is mediated by special types of particles. For instance, the electromagnetic field is carried by the photon. In the same way the Higgs field has its mediator, the Higgs boson.
3. Have we already seen the Higgs boson ?
The fascinating thing is that the theory mentioned above together with the Higgs mechanism is incredibly successful. It has been tested against experiments at extreme conditions for the last 30 years in particle physics experiments and it describes reality with an almost unbelievable precision. However, we have never seen a Higgs boson! You can imagine that confirming the Higgs mechanism is one of the most hot topics! How could we otherwise explain the success of the electroweak theory?
4. Whatís the mass of the Higgs boson ?
One problem is that we can't predict the mass of the Higgs boson itself. This means that we are obliged to scan off bit by bit the possible range for the Higgs boson. Today we have excluded that the Higgs boson is lighter than the total mass of 120 protons! With the new accelerator that we are building at CERN, the Large Hadron Collider (link), we should be able to cover the entire possible range and that's really exciting! If on the other hand we have to really exclude the Higgs mechanism completely, then there must be something entirely unexpected out there!
5. Why a particle should have a mass and what is the mechanism that assigns a mass ?
This is an interesting question for several reasons: first of all there are some particles whose mass is 0. So we want to understand why some particles have mass and others don't. The Higgs boson role in physics is to give a mass to elementary particles. That particles have a mass, is a well known experimental fact. In trying to understand all of these issues, we invented the so-called Higgs mechanism. This mechanism introduces a new particle: the Higgs boson, where "boson" indicates that the intrinsic spin of the particle is an integer; in fact, the spin turns out to be precisely 0, so that often the Higgs is also called the Higgs "scalar", instead of boson. This particle interacts with all particles which are required to have a mass. The interaction is exactly such that the particles start behaving as if they had a mass. You can think of this Higgs as a substance permeating the Universe (we would call this a "field"). When a particle moves through the Higgs field, it interacts with it. This interaction causes a sort of delay in the motion of the particle, as if it were moving through a viscous medium. It is not exactly friction, since friction would bring the particle to a halt, while here the particle does not loose energy. But the Higgs introduces some inertia, namely resistance to the motion. So although the particle itself is massless, interaction with the Higgs field makes it behave as if it had mass (this is, by the way, how the problem with the W bosons mentioned earlier gets solved).
6. How can we detect the Higgs boson ?
We cannot really detect directly the presence of this underlying Higgs field, since it is uniformly distributed throughout the Universe. We can claim that it is there because we see that particles have masses. But we want a more direct proof to confirm the theory. We want to "see and touch" the Higgs. This could be achieved by concentrating enough energy in a point of space. This energy would disturb the continuous Higgs field, and would generate waves (it is a bit like throwing a stone in a lake, to turn the flat surface into a sequence of waves). These waves can then be detected, and the existence of the Higgs field proved. These waves are associated to the Higgs particle. Since the Higgs particle interacts with its underlying field, the Higgs itself has a mass. To produce a Higgs particle, the amount of energy we need to concentrate is at least equal to the Higgs mass. This number is large, and that's why we still have not seen it.
Going back to the example of the lake, the situation is similar to what follows: if we were immersed in an infinite body of still water, we could not know we are in a medium. We could easily think that we are in the perfect vacuum, the only difference being that if we try to move we experience some resistance (let's call it inertia). If we make a sudden movement, however, with energy large enough, we could generate some underwater wave. The wave is detected because it moves, and if it reaches some other observer imbedded in our lake-universe, he will feel the pressure of it. In the case of the real Higgs, we don't feel any pressure, but detect the decay products of the created Higgs particle. This, after having lived as a wave for a very short amount of time, decays to other particles, which will reach the detectors.
7. How is Higgs boson likely to manifest itself within a particle collision?
It will not live very long and decay into energy (photons) and other particles, the total energy of which must of course add up to that of the Higgs. What we will observe is a set of particles coming out of the point where the Higgs decays, and the pattern will be distinctive. Such a distinctive decay pattern is sometimes referred to as the "signature" of the particle. The width of the Higgs particle should be of the order of GeVís so its lifetime should be of the order of 10-22 s.
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https://search.r-project.org/CRAN/refmans/cutpointr/html/tpr.html | 1,653,430,006,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662577259.70/warc/CC-MAIN-20220524203438-20220524233438-00087.warc.gz | 596,402,003 | 2,024 | tpr {cutpointr} R Documentation
## Calculate true / false positive / negative rate
### Description
Calculate the true positive rate (tpr, equal to sensitivity and recall), the false positive rate (fpr, equal to fall-out), the true negative rate (tnr, equal to specificity), or the false negative rate (fnr) from true positives, false positives, true negatives and false negatives. The inputs must be vectors of equal length.
tpr = tp / (tp + fn)
fpr = fp / (fp + tn)
tnr = tn / (tn + fp)
fnr = fn / (fn + tp)
### Usage
tpr(tp, fn, ...)
fpr(fp, tn, ...)
tnr(fp, tn, ...)
fnr(tp, fn, ...)
### Arguments
tp (numeric) number of true positives. fn (numeric) number of false negatives. ... for capturing additional arguments passed by method. fp (numeric) number of false positives. tn (numeric) number of true negatives.
Other metric functions: F1_score(), Jaccard(), abs_d_ppv_npv(), abs_d_sens_spec(), accuracy(), cohens_kappa(), cutpoint(), false_omission_rate(), metric_constrain(), misclassification_cost(), npv(), odds_ratio(), p_chisquared(), plr(), ppv(), precision(), prod_ppv_npv(), prod_sens_spec(), recall(), risk_ratio(), roc01(), sensitivity(), specificity(), sum_ppv_npv(), sum_sens_spec(), total_utility(), tp(), youden()
tpr(10, 5, 20, 10) | 344 | 1,265 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2022-21 | latest | en | 0.397666 |
financialhorse.com | 1,701,310,235,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100164.15/warc/CC-MAIN-20231130000127-20231130030127-00091.warc.gz | 294,879,123 | 86,923 | # How to calculate Price when buying Property in Singapore? How much is your House worth at market value? – FH Property Series: Part II
0
Pricing.
After location, the most important thing you need to look at when buying property is the price.
As the other famous saying goes – “There’s no such thing as bad real estate, only bad pricing”.
You may be buying the worst piece of real estate in the lousiest location ever, but if you’re getting it at an unbelievable price, it may still turn out to be a great investment.
So… how do you determine the right price to pay for a property?
This is Part II of the FH Property Series, where I’ll be sharing everything I know on buying property in Singapore.
Check out Part I on how to pick a property here.
## Theory – How to calculate Price when buying Property in Singapore?
If you go to real estate school, they’ll teach you that the 3 ways to determine price are:
• Sales Comparison
• Replacement Cost Approach
• Income Capitalization
Let’s touch on each of them briefly.
## Sales Comparison
The most basic form of valuation.
You find a similar property of similar size and age, and you see the last transacted price (on a psf basis).
That’s the market value of your property.
Everyone knows how to do sales comparison, and intuitively it’s the first method we will use.
I’ve extracted a screengrab from Property Lim Brothers below that gives you an idea of how sales comparison is done.
## Replacement Cost Approach
Sometimes the property you’re buying is unique, and you just can’t find anything similar in the area.
Maybe it’s a small development, maybe the age is unique, or some other reason that makes it hard to find a similar property to compare.
In such cases you fall back to the cost approach – How much does it cost to build an equivalent property.
You calculate (1) the value to buy the empty land, (2) add on building and development costs, and (3) subtract a bit for depreciation and old age.
This is the main method that I use, so we’ll go in depth into the numbers and method of calculation later.
## Income Capitalization (Cap Rate)
The core concept with income capitalisation is that you calculate the rental income, less off expenses, and apply a multiplier known as the “capitalisation rate” (also known as cap rate).
So let’s say the rental income per year after deducting expenses is \$50,000.
If you apply a 3% capitalisation rate, the value of the real estate is \$50,000 / 0.03 = \$1.67 million.
For stocks investors who are familiar with Price to Earnings, Income Capitalization is basically P/E valuation.
Only difference is that you replace earnings with rental income, and replace the P/E Multiple with the cap rate.
## How do I do it? Calculate Price for a Residential Property in Singapore?
Okay so that’s the theory.
How do I do it in practice, for Singapore residential properties?
Income Capitalization I’ve found to be most useful for commercial properties.
If you’re buying a shopping mall or an office building or an industrial property – absolutely use income capitalization. Use Income Capitalization and cross check against a mix of Sales Comparison / Cost Approach.
But income capitalisation I find less effective for Singapore residential properties.
For Residential Property in Singapore, the way I like to do it is to use the Replacement Cost Approach – it’s to find the replacement price. And crosscheck against sales comparison.
BTW – we share commentary on Singapore Investments every week, so do join our Telegram Channel (or Telegram Group), Facebook and Instagram to stay up to date!
[mc4wp_form id=”173″]
## What Price does the Property Developer need to sell at to break even?
To understand replacement price, you first need to understand how a property developer prices a property when they bid for land.
Let’s say GuocoLand is bidding for the white site in Lentor.
What is the price they need to sell to break even?
The way you do it, is to add:
• \$300 – 400 psf for construction cost (used to be \$200 – \$300 pre-COVID but the price has gone up since due to inflation in labour and raw materials)
• 15% for Sales Expenses (property agent fee, marketing and advertising expenses etc)
• 10% for developer profit
So GuocoLand paid \$1204 psf ppr for the land (per square foot per plot ratio – basically the cost per square foot of buildable area).
The cost of the land + building is \$1204 + \$400 (we’ll take the higher end of the range) = \$1600 psf (round up for the sake of discussion).
The cost of land + building + Sales Expenses is \$1600 X 115% = \$1840 psf.
This means that Guocoland will need to sell at about \$1840 just to break even.
But of course Guocoland needs to make a profit too, so you add in 10% profit.
Realistically, you’re looking at \$2000 average selling prices across the board.
## These are Average PSF numbers
Now of course these are average PSF figures for the entire development. The price you actually pay will depend on the size of the unit you buy.
If you buy a 1200 sf 3 bedroom, you may pay below the average \$2000 psf (PSF goes down the bigger you buy).
If you buy a 400 sf studio, you’ll probably pay above the \$2000 psf.
So you adjust the price / market value accordingly for what you buy.
## How to calculate Replacement Cost (or market value)?
Using the example above, replacement price is basically land value + construction cost.
So back to the lentor example – land value + building cost is about \$1600.
That’s the cost price / replacement cost / market of a brand new building in that area, before you add in things like sales expenses or developer profit.
So if you’re buying a 20 year old freehold, you subtract a bit for the condition of the property.
Say 100 – 150 psf because it’s a bit old and dirty.
If it’s brand new of course (or 2 – 3 years old), then you can pay up to \$1600 and still be close to fair value.
## How to calculate Land Value?
The trickiest part with the replacement cost valuation method is to calculate land price.
There are two main ways to do it, ranked from easiest to hardest:
• Government Land Sale (GLS)
• En Bloc
• Take recent property sales in the area and work backwards
## Easy – Government Land Sale (GLS)
The easiest way to calculate land value is if there is a recent Government Land Sale (GLS).
If there is, you just check the psf ppr the property developer paid for the land tender, and that should give you a very close approximation of the land value.
## Medium – En Bloc
En Blocs are also very easy – you can use the value that the property developer paid for the land.
The only thing to take note of is that if the En Bloc is for leasehold land, then you need to add back the land premium to calculate the value of a full 99 year lease.
So let’s say the developer paid \$1200 psf ppr to the homeowners, for a development with 60 years left on the lease.
The developer then pays another \$200 land premium to URA to “top up” the lease back to a fresh 99 year lease.
The land cost here is \$1400.
Don’t forget this step, otherwise your number will be meaningless.
## Hard – Take recent property sales in the area and work backwards
Sometimes though, there’s just no recent en bloc or GLS.
Which makes your life a lot harder.
In such cases the next best thing is to approximate based on recent land sales in the area. In other words, you use the Sales Comparison approach to calculate land value.
Eg. A 30 year old leasehold property in the area was sold at \$1600 psf last week.
Based on the condition of that property, you think about \$150 – \$200 psf of repair / reno work is required.
So you add this back into the sale price, to get the replacement cost.
That would give you \$1600 + \$200 = a rough valuation of \$1800 psf for the replacement cost of the land.
And then you repeat this for a few properties in the area, varying the age / location / floor each time.
If you do a few of this, that should allow you to triangulate the average land price for the area.
It’s basically using what we learnt about the developer pricing methodology, and working backwards to find the replacement cost of the land.
## But FH, the Pros use “Gut Feel” when it comes to valuation
Okay I know that the 20 year veteran property investors will tell you that they can just “tikam tikam” the valuation.
But their “Gut Feel”, more often than not, is usually a variation of one of these 3 valuation methods above.
So get your basics right first, and understand the theory behind each of these 3 valuations methods, and their strengths and weaknesses.
Once you’re familiar with them, you can very easily apply all 3 of them on the go to give you a gut feel valuation. And then you can laugh at others when they whip out a calculator. 😉
## How to use the Replacement Cost Valuation?
Now that you have your replacement cost, how do you use it to guide your buying decision?
The broad idea is that as long as you pay close to replacement cost, you won’t lose too much money.
But there are some nuances that we’ll discuss below.
## Fair Valuation is a Range
The biggest one to remember, is that valuation is both an art and a science.
There are a lot of numbers and mathematics above, but DON’T fool yourself into thinking that this is the 100% correct valuation of your property.
Just like with stocks – fair value for property is always a range.
So if your fair value is \$1400 – \$1500 psf, that is your fair value range.
Don’t be so stubborn that you can only buy / sell property within that range.
Just like if your fair valuation of DBS is \$25, the share price can still trade at \$30 for years.
## The Property Market can move very quickly
This one is especially relevant right now, when the Singapore residential property market is red hot.
The replacement cost methodology is fundamentally backward looking.
You’re looking at past transacted prices from a few months ago, and trying to calculate price today.
Prices could legitimately have moved 2% – 3% in the time since.
So you need to be aware of market sentiment as well, and you may need to make adjustments to your fair valuation range accordingly.
If the URA Price Index tells you house prices have moved 3% quarter on quarter, and the numbers you’re using are from 6 months back, you probably need to add 3% – 5% to get the fair market value.
## Don’t Forget Subjective Value
A massive factor to consider, especially if you’re buying for own stay, is subjective value.
If you love the property because of its arched pillars, you may be willing to pay slightly more for the property.
Or if you hate it because of the disgusting layout (to you at least), that subjectively the property is worth less to you, and you may not want to pay anything more than the low end of the valuation range.
The important thing to note is that replacement cost just tells you what the fair value of the land is, approximately.
It doesn’t tell you the price that you should actually pay for it.
That is your decision, and yours only.
If you decide that buying real estate in Lentor at \$1450 psf is stupid, then it’s your decision to walk away and buy in another area, or to put all that money into stocks and rent a house instead.
## This isn’t the 1990s – Lowball offers don’t work anymore
What you shouldn’t be doing though, is that if you know fair value is \$1450, but you insist on only buying at \$1350.
That to me – is a fool’s errand.
This isn’t the 1990s anymore.
Today, anyone can go onto the URA website or Property Guru and pull up 10 years worth of transactions with a single click.
Everybody out there, including your competing buyers and the seller himself, knows the fair valuation range of the property.
With the internet, there is no information arbitrage anymore.
In other words – this isn’t your grandfather’s generation where you can lowball an offer and get accepted.
## Can you buy Singapore Property at an undervalue?
That said, there are exceptions to this.
The only reason why a seller would sell at an undervalue today is because (1) he is desperate for cash, or (2) illiquidity.
In this low interest rate environment that is flush with liquidity, almost no investor is desperate for cash. There is always the option to rent it out, and even if they want to sell, there’s a long list of buyers waiting.
During COVID the government even gave homeowners the opportunity to defer mortgage payments – when have we ever had that in Singapore’s history?
Fire sales are much less common today.
Which leave illiquidity – basically a house where nobody wants to buy.
But be objective with this one.
If you’re buying a \$1.5 million condo in a great location, chances are there will be 5 other bidders competing with you in this market.
If you’re buying a \$10 million penthouse in the CBD, or a leasehold landed with 30 years left on the lease in Sembawang, then ok maybe you’re the only bidder and you may have more room to negotiate if you put in a serious offer.
How to negotiate a real estate transaction is a whole other story in itself, which we will cover with a full article.
## Buy Great Real Estate at a Fair Price
My advice is this – Think of it like stocks.
Instead of obsessing over trying to get the perfect price, focus on the bigger picture.
Accept that whatever real estate you want to buy, you have to pay the fair market value for it at the time you are buying.
Spend your time more productively by picking great real estate that will appreciate in value over time.
In other words – buy Great Real Estate, at a Fair Price.
And earn the long term gains.
Love to hear what you think!
If you enjoyed this article, do support Financial Horse as a Patreon to get exclusive early access to the FH Property series, as well as other premium content.
Check out Part I on how to pick a property here.
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All are THE best quality investment courses available to Singapore investors out there! | 3,401 | 15,246 | {"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-2023-50 | longest | en | 0.917074 |
https://smallbusiness.chron.com/npv-investment-appraisal-method-capital-rationing-66781.html | 1,670,027,114,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.70/warc/CC-MAIN-20221202215443-20221203005443-00214.warc.gz | 551,700,167 | 23,536 | # NPV Investment Appraisal Method & Capital Rationing
If you run a small business, you realize early that money can be scarce and you must deploy it wisely. You may want to invest in one or more new projects or expansion ideas but have only limited funds to do so. Capital rationing is the constraint on your spending due to limited funds. The net present value, or NPV, investment appraisal method helps you choose which projects to adopt within your constraints.
## Net Present Value
1. The net present value of an investment is the profit or loss you expect to experience from the investment’s cash flows, expressed in current dollars. The idea behind NPV is that money in your pocket today is worth more than money you might receive in the future, because of the uncertainty factor. In the NPV method, you discount future cash outflows and inflows by an appropriate interest rate to arrive at the current-dollar value. At the very least, you would want to only pursue projects that give a positive NPV.
## Capital Rationing
1. You can raise capital by contributing your own cash to the business, plowing profits back into the company, borrowing money from investors and privately placing common and preferred stock. You will normally have more uses for capital than capital itself, which requires you to ration it. Your ability to raise external capital is the first hurdle in funding projects. You next must budget the capital you have to maximize possible return while minimizing risk of loss.
## Cost of Capital
1. Capital costs money. For debt, the cost is the interest you pay. For stock, it’s the dividend and the price growth that investors demand. You can calculate the weighted average cost of capital to find out the minimum percentage return you need from a project so it pays for the capital you expend on it. By converting NPV to a percentage of the project costs, known as the internal rate of return, you can eliminate projects that return less than the weighted average cost of capital.
## Example
1. Suppose a project requires a \$100,000 expenditure up front but has cash flows over the following four years that bring in \$50,000, \$40,000, \$30,000 and \$20,000, respectively. You are investing \$100,000 today to receive \$140,000 in future dollars. Your weighted average cost of capital is 3.125 percent, so you discount your future cash flows by that percentage per year to calculate NPV, which works out to \$31,136. You calculate the internal rate of return, which is a healthy 17.8 percent. If you have no other projects that return more, you would ration a substantial portion of your investment capital for this project, as it provides a positive NPV and a return above that of other projects and well above your cost of capital. | 577 | 2,770 | {"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-2022-49 | longest | en | 0.94079 |
http://www.brainia.com/essays/Math-533-Dery/364672.html | 1,495,545,059,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607636.66/warc/CC-MAIN-20170523122457-20170523142457-00084.warc.gz | 434,892,958 | 5,971 | # MATH 533 dery new Courses/ uophelp
## MATH 533 dery new Courses/ uophelp
MATH 533 Entire Course
For more course tutorials visit
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MATH 533 Week 1 Homework
MATH 533 Week 1 Quiz
MATH 533 Week 2 DQ 1 Case Let's Make a Deal
MATH 533 Week 2 Homework (2 Sets)
*********************************************************************************************************************************
MATH 533 Final Exam Set 1
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(TCO D) PuttingPeople2Work has a growing business placing out-of-work MBAs. They claim they can place a client in a job in their field in less than 36 weeks. You are given the following data from a sample.
Sample size: 100
Population standard deviation: 5
*********************************************************************************************************************************
MATH 533 Week 1 Homework
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1. Complete the table to the right.
2. In one university, language incorporated a 10-week extensive reading program to improve students’ Japanese reading comprehension. The professors collected 267 books originally written for Japanese children and required their students to read at least 40 of them as part of the grade in the course. The books were categorized into reading levels (color-coded for easy selection) according to length and complexity. Complete parts a through c.
*********************************************************************************************************************************
MATH 533 Week 2 DQ 1 Case Let
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A number of years ago, there was a popular television game show called Let's Make a Deal. The host, Monty Hall, would randomly select contestants from the audience and, as the title suggests, he would make deals for prizes. Contestants would be given relatively modest prizes and then would be offered the opportunity to risk that prize to win better ones.... | 399 | 2,003 | {"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-2017-22 | longest | en | 0.85445 |
https://www.jiskha.com/display.cgi?id=1315495840 | 1,527,219,026,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866938.68/warc/CC-MAIN-20180525024404-20180525044404-00577.warc.gz | 788,526,406 | 3,532 | # math
posted by aishat
using only the number eight,place ten money amounts in the ten blocks of a standard pyramid with 4 blocks on the bottom row
1. tanu
46*3=300
## Similar Questions
1. ### MATHS
USING ONLY THE NUMBER FOUR, PLACE TEN MONEY AMOUNTS IN THE TEN BLOCKS OF THIS PYRAMID TO TOTAL \$10,000
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USING ONLY THE NUMBER FOUR, PLACE TEN MONEY AMOUNTS IN THE TEN BLOCKS OF THIS PYRAMID TO TOTAL \$10,000
3. ### Arithmetic Sequence
A child organizes animal blocks in rows on the ground. There are 64 animal blocks in the 5th row & 92 of those blocks in the 9th and last row. Suppose that the number of animal blocks frm row to row form an arithmetic sequence. The …
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A child organizes animal blocks in rows on the ground. There are 64 animal blocks in the 5th row & 92 of those blocks in the 9th and last row. Suppose that the number of animal blocks frm row to row form an arithmetic sequence. The …
5. ### MATH Word Problem
A child organizes animal blocks in rows on the ground. There are 64 animal blocks in the 5th row & 92 of those blocks in the 9th and last row. Suppose that the number of animal blocks frm row to row form an arithmetic sequence. The …
6. ### MATH
A child organizes animal blocks in rows on the ground. There are 64 animal blocks in the 5th row & 92 of those blocks in the 9th and last row. Suppose that the number of animal blocks frm row to row form an arithmetic sequence. The …
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A child organizes animal blocks in rows on the ground. There are 64 animal blocks in the 5th row & 92 of those blocks in the 9th and last row. Suppose that the number of animal blocks frm row to row form an arithmetic sequence. The …
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You decide to build a pyramid out of wooden blocks. Each block must rest on the corners off 4 other blocks. If there is one block on top and 36 on the bottom, what is the total number of blocks you will need to complete the pyramid?
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A pile of blocks has 60 blocks in the bottom row, 54 blocks in the second row, 48 blocks in the third row, and so on until there are only 6 blocks on the top row. How many blocks are in the 8th row?
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More Similar Questions | 616 | 2,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-22 | latest | en | 0.886317 |
http://www.ck12.org/physics/Sound-in-a-Tube/?by=all&difficulty=all | 1,477,678,275,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988725451.13/warc/CC-MAIN-20161020183845-00213-ip-10-171-6-4.ec2.internal.warc.gz | 366,237,795 | 17,964 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Sound in a Tube
## Sound waves trapped in a tube generate sounds a frequencies dependent on the length of the tube.
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
## Sound in a Tube
Students will learn how to analyze and solve problems where standing waves (and hence sound) is produced in a tube.
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## Unit 11.3 Sound in a Tube
Students will learn how to analyze and solve problems where standing waves (and hence sound) is produced in a tube.
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• Interactive Simulations
## Pan Flute
Learn about wave harmonics in a column of air by looking closely at the sound produced by a pan flute.
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• Video
## Harmonics 2: Standing Waves in a Closed Air Column - Overview
by CK-12 //basic
Overview
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• Video
## Sound in an Open Pipe: Demonstration
An demonstration is used to gather data on the frequency of sound in a tube.
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## Harmonics 3: Standing Waves in an Open Air Column - Overview
by CK-12 //basic
Overview
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• Video
## Sound in an Open Pipe: Explanation
The physics behind sound in a tube and the different harmonic frequencies created by such reverberations are explained.
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• Video
## Harmonics 2: Standing Waves in a Closed Air Column - Example 3
by CK-12 //basic
Determining harmonic (resonant) frequencies using the equation f[n] = nv / 4L or f[n ]= nf[1] for any odd integer n
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• Video
## Harmonics 3: Standing Waves in an Open Air Column - Example 3
by CK-12 //basic
Determining harmonic (resonant) frequencies using the equation f[n] = nv / 2L or f[n ]= nf[1] for any integer n
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• Video
## Harmonics 2: Standing Waves in a Closed Air Column - Example 1
by CK-12 //basic
Determining the length of a tube based on the standing wave frequency or wavelength
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• Video
## Harmonics 2: Standing Waves in a Closed Air Column - Example 2
by CK-12 //basic
Determining harmonic (resonant) wavelengths using the equation λ[n] = 4L/n or λ[n] = λ[1]/n for any odd integer n
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• Video
## Harmonics 3: Standing Waves in an Open Air Column - Example 1
by CK-12 //basic
Determining the length of a tube based on the standing wave frequency or wavelength
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• Video
## Harmonics 3: Standing Waves in an Open Air Column - Example 2
by CK-12 //basic
Determining harmonic (resonant) wavelengths using the equation λ[n] = 2L/n or λ[n] = λ[1]/n for any integer n
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• 0
## Sound in a Tube Quiz - PPB
Calculating wavelength in open and closed pipes.
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• Critical Thinking
## Sound in a Tube Discussion Questions
A list of student-submitted discussion questions for Sound in a Tube.
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• Flashcards
## Sound in a Tube Flashcards
by CK-12 //basic
These flashcards help you study important terms and vocabulary from Sound in a Tube.
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0 | 1,076 | 4,417 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2016-44 | longest | en | 0.877434 |
https://www.vedantu.com/question-answer/write-dfrac1112-as-a-decimal-class-6-maths-cbse-600af645ae059809921768dd | 1,721,764,593,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518115.82/warc/CC-MAIN-20240723194208-20240723224208-00431.warc.gz | 884,801,365 | 25,807 | Courses
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# How do you write $\dfrac{{11}}{{12}}$ as a decimal?
Last updated date: 23rd Jul 2024
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Hint: The line in a fraction that separates the numerator and the denominator can be rewritten using the division symbol. So to convert a fraction to a decimal, divide the numerator by the denominator. If required. This will give us our answer as a decimal.
Complete step-by-step solution:
In the given question we are required $\dfrac{{11}}{{12}}$ into a decimal number for which you have to divide the numerator by the denominator.
$\Rightarrow 11/12 \\ \\$
$= 0.916$
Which is approximately equal to 0.92.
Note:
Decimal is a number that describes the base-10 number system, probably the most commonly used number system. The decimal number system consists of ten single-digit numbers- 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. The number after 9 is 10. The number after 19 is 20 and so on. Additional power of 10 requires the addition of another positional digit.
In algebra, a decimal number can be defined as a number whose whole number part and the fractional part is separated by a decimal point. The dot in the decimal number is called the decimal point. The digit following the decimal point shows a value smaller than one. | 342 | 1,350 | {"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} | 4.375 | 4 | CC-MAIN-2024-30 | latest | en | 0.840736 |
https://www.rdsharmasolutions.in/class-9-chapter-9-triangle-and-its-angles/ | 1,685,432,898,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645417.33/warc/CC-MAIN-20230530063958-20230530093958-00463.warc.gz | 1,053,739,891 | 39,645 | # RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles
In this chapter, we provide RD Sharma Solutions for Chapter 9 Triangle and its Angles for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 9 Triangle and its Angles Maths pdf, free RD Sharma Solutions for Chapter 9 Triangle and its Angles Maths book pdf download. Now you will get step by step solution to each question.
### RD Sharma Solutions for Class 9 Chapter 9 Triangle and its Angles Download PDF
Question 1.
Define the following tenns :
(i) Line segment
(ii) Collinear points
(iii) Parallel lines
(iv) Intersecting lines
(v) Concurrent lines
(vi) Ray
(vii) Half-line.
Solution:
(i) A line segment is a part of a line which lies between two points on it and it is denoted as AB¯¯¯¯¯¯¯¯ or only by AB. It has two end points and is measureable.
(ii) Three or more points which lie on the same straight line, are called collinear points.
(iii) Two lines which do not intersect each other at any point are called parallel lines.
(iv) If two lines have one point in common, are called intersecting lines.
(v) If two or more lines which pass through a common point are called concurrent lines.
(vi) Ray : A part of a line which has one end point.
(vii) Half line : If A, B, C, be the points on a line l, such that A lies between B and C and we delete the point from line l, two parts of l that remain are each called a half-line.
Question 2.
(i) How many lines can pass through a given point?
(ii) In how many points can two distinct lines at the most intersect?
Solution:
(i) Infinitely many lines can pass through a given point.
(ii) Two distinct lines at the most intersect at one point.
Question 3.
(i) Given two points P and Q, find how many line segments do they determine?
(ii) Name the line segments determined by the three collinear points P, Q and R.
Solution:
(i) Only one line segment can be drawn through two given points P and Q.
(ii) Three collinear points P, Q and R, three lines segments determine : PQ¯¯¯¯¯¯¯¯ , QR¯¯¯¯¯¯¯¯ and PR¯¯¯¯¯¯¯¯ .
Question 4.
Write the truth value (T/F) of each of the following statements:
(i) Two lines intersect in a point.
(ii) Two lines may intersect in two points.
(iii) A segment has no length.
(iv) Two distinct points always determine a line.
(v) Every ray has a finite length.
(vi) A ray has one end-point only.
(vii) A segment has one end-point only.
(viii) The ray AB is same as ray BA.
(ix) Only a single line may pass through a given point.
(x) Two lines are coincident if they have only one point in common.
Solution:
(i) False : As two lines do not intersect also any a point.
(ii) False : Two lines intersect at the most one point.
(iii) False : A line segment has definitely length.
(iv) True.
(v) False : Every ray has no definite length.
(vi) True.
(vii) False : A segment has two end point.
(viii)False : Rays AB and BA are two different rays.
(ix) False : Through a given point, infinitely many lines can pass.
(x) False : Two lines are coincident of each and every points coincide each other.
Question 5.
In the figure, name the following:
(i) Five line segments.
(ii) Five rays.
(iii) Four collinear points.
(iv) Two pairs of non-intersecting line segments.
Solution:
From the given figure,
(i) Five line segments are AC, PQ, PR, RS, QS.
(ii) Five rays : Undefined control sequence \xrightarrow , Undefined control sequence \xrightarrow , Undefined control sequence \xrightarrow , Undefined control sequence \xrightarrow , Undefined control sequence \xrightarrow .
(iii) Four collinear points are : CDQS, APR, PQL, PRB.
(iv) Two pairs of non-intersecting line segments an AB and CD, AP and CD, AR and CS, PR and QS.
Question 6.
Fill in the blanks so as to make the following statements true:
(i) Two distinct points in a plane determine a _____ line.
(ii) Two distinct_____ in a plane cannot have more than one point in common.
(iii) Given a line and a point, not on the line, there is one and only _____ line which passes through the given point and is_____ to the given line.
(iv) A line separates a plane into ____ parts namely the____ and the____ itself.
Solution:
(i) Two distinct points in a plane determine a unique line.
(ii) Two distinct lines in a plane cannot have more than one point in common.
(iii) Given a line and a point, not on the line, there is one and only perpendicular line which passes through the given point and is perpendicular to the given line.
(iv) A line separates a plane into three parts namely the two half planes, and the one line itself.
All Chapter RD Sharma Solutions For Class 9 Maths
—————————————————————————–
All Subject NCERT Exemplar Problems Solutions For Class 9
All Subject NCERT Solutions For Class 9
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If these solutions have helped you, you can also share rdsharmasolutions.in to your friends. | 1,237 | 5,207 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2023-23 | latest | en | 0.94333 |
https://helpful.knobs-dials.com/index.php?title=Staples&oldid=97875 | 1,721,722,327,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.81/warc/CC-MAIN-20240723072353-20240723102353-00262.warc.gz | 252,262,693 | 7,573 | # Staples
This article/section is a stub — some half-sorted notes, not necessarily checked, not necessarily correct. Feel free to ignore, or tell me about it.
## Sizing
### If it looks like 23/6 and the first number is 23, 24, 25
...then it's likely that
the first number is the thickness, in american wire gauge
the second number is the leg length, in mm - 6mm leg length
So 23/6 is 0.573 mm thick and 8mm legs.
The fact that it's 13mm wide is left implied.
24 is a little thinner and doesn't really go longer than 8mm.
23 goes further and has more and longer sizes available.
25 and 26 don't seem so common
(outside measurements , so the usable leg length is a little shorter)
### In a size like 13/6 and the first number is something like like 13, 14, 17, 32, 35, 50, 53, 71, 80, 90, 91, 97, 140, or 4000
...this seems to be more arbitrary sizing?(verify)
The second size is still the leg length in mm.
Most common seem to be
• 13/
10.65 wide
• 53/
11.35 wide
### If it's called No. 10, No. 16, No. 18, No. 25, No. 56, No. 123, or similar
This is another system.
It seems that
• No. 16 is 24/6
• No. 18 is 24/8
• No. 10 is 10/4
It's American
### What about width?
Crown width (side to side) is not directly specified, but is implied from the type.
Even so, it is moderately well correlated with the gauge.
For example
• 24/ and 26/ are 12.9mm wide
• 23/ is 13.0 mm wide
• 53 series is 11.3 mm wide
• 13 series is 10.6 mm wide
• No. 8 a.k.a. 6/4 is 7.1mm wide
• No. 10 a.k.a. 10/4 is 9.4mm wide
## What seems to be common?
### Classroom and office staples
This article/section is a stub — some half-sorted notes, not necessarily checked, not necessarily correct. Feel free to ignore, or tell me about it.
Common staples:
• EU:
mostly 24/6 or the longer 24/8
mini: No. 10 (10/5 or 10/4?(verify))
6/4 ?(verify) (No. 8)
• US
home use: 6/6, 24/6, 24/8, 13/6, 13/8, 26/6 (No.56)
mini: No. 10 (10/5 or 10/4?(verify))
heavier duty: 23/8, 23/12, 23/15, 23/20, 23/24, 13/10, and 13/14
and apparently a whole bunch of imperial sizes because why not?
### Staple guns
This article/section is a stub — some half-sorted notes, not necessarily checked, not necessarily correct. Feel free to ignore, or tell me about it. | 751 | 2,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-30 | latest | en | 0.938589 |
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