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• Resolved
# P82B715: Pull-up voltage for I2C- P82B715
Part Number: P82B715
Hi,
I am working in P82B715 IC for Long distance I2C communication. I have a doubt.
In datasheet it is mentioned that Voltage logic levels are independent of VCC and also the IC has to sink 30mA.
So can I use pull-up voltage of 1.8V for Sx/Lx signals with VCC of 5V and calculate pull-up resitance values to provide sink current value of maaximum 30mA. Also I calculate RC value to provide 1uS rise and fall time.
If I do so, there will be still any problem in sinking current or timings? or else any other problem will arise??
Please explain me whether my ubderstanding is correct or not
Thanks,
Arun
• Hey Arunkumar,
"In datasheet it is mentioned that Voltage logic levels are independent of VCC and also the IC has to sink 30mA."
-Looking at the datasheet, it implies that the Lx/Sx pins can be Vcc+0.7V or lower and work.
-The IC does not HAVE to sink 30mA, what the datasheet is saying here is it is capable of sinking 30mA on the "L" side of the device.
"So can I use pull-up voltage of 1.8V for Sx/Lx signals with VCC of 5V and calculate pull-up resitance values to provide sink current value of maaximum 30mA. Also I calculate RC value to provide 1uS rise and fall time."
-You do not actually have to sink 30mA. A 1uS fall time violates I2C standard as a maximum time allowed is 300ns, I assume you are actually just talking about rise time in this case.
A. What pull up values and capacitance are you expecting?
B. How far are you planning on communicating? (distance of cables)
C. What is your expected max frequency?
"If I do so, there will be still any problem in sinking current or timings? or else any other problem will arise??"
Can you show me a schematic/diagram with the pull up resistors for both boards and expected capacitance? To be honest, I have no had much time with this device in the lab so I may need to put it on a bench and do some testing if you provide the details.
Thanks,
-Bobby
• In reply to TRX Bobby:
Hi Bobby,
Assume I need to have bus capacitance of 3000nF( maximum capacitance that can be driven by the IC) and assume distance to be 20meters(single mater and single slave).
Let Speed be 400KHZ as per I2C standard.
I need to put pull the bus lines at 1.8V.
Following flow diagram will shows outline of my circuit.
Please let me know that this circuit works and what pull-up value to be calculated.
Thanks & Regards,
M.Arun kumar
• In reply to Arunkumar:
Hello Arun,
I would probably put the pullup resistors for LCL and LDA to 5V. You are already have it and it will reduce current consumption during communications. Otherwise, I think you are ok.
-Francis Houde
• In reply to fhoude:
Hi Francis,
whether I can just pull-up the LCL and LDA to 5V? or I need to put a buffer (level translator) in between??
Thanks,
Arun
• In reply to TRX Bobby:
Hi Bobby,
Please let me know whether your testing passed my setup conditio or else any other steps needed to be taken for it?
Thanks,
M.Arun kumar
• In reply to Arunkumar:
Hey Arunkumar,
I set up Vcc of the device to 5V and the I2C lines to 1.8V. I used 550 ohms for pull ups on S line and 65 ohms for L line.
-----------------------------------------------
100kHz:
----------------------------------------------------------------
400kHz
The lines are labeled where In is connected to a pull down FET at Sx and out is measured at Sx of a second P82B715.
You can see the the communication works at 1.8V on I2C and 5V on Vcc.
Thanks,
-Bobby
• In reply to TRX Bobby:
Hi Bobby,
Thanks for your support and your answer will be very much usefull for me.
Thanks & Regards,
M.Arun kumar
• In reply to TRX Bobby:
Hi Bobby,
Can I know what is the length of the wire used for test you conducted above?
Thanks,
Arun
• In reply to Arunkumar:
Hey Arun,
The length of the wire was very short in this test (I wanted to confirm that the device would work at Vcc=5V while I2C @ 1.8V). I did try to put a 3000pF capacitor on the L line but I had a difficult time finding a capacitor that would fit on my set up. I ended up trying to solder 3x 1000pF 1210 caps together in parallel and tying wires to it but I'm not sure if the wire's made a good enough connection to my set up. When I added and removed the 3000pF load from the set up, I did not see much variation in the L line (I did see the L line change slightly with the cap added but not by much). This is likely due to the VERY strong pull ups on the L line.
-Bobby
• In reply to TRX Bobby:
Hi Bobby,
Thanks. Wire length is not much needed since you have used 3000pF capacitance (3x 1000pF) which is same as long length wire.
Thanks,
Arun.
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If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question. | 1,325 | 5,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-43 | latest | en | 0.90533 |
https://www.nag.com/numeric/nl/nagdoc_25/nagdoc_fl25/html/m01/m01daf.html | 1,627,068,743,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150000.59/warc/CC-MAIN-20210723175111-20210723205111-00133.warc.gz | 956,396,184 | 4,009 | M01 Chapter Contents
M01 Chapter Introduction
NAG Library Manual
# NAG Library Routine DocumentM01DAF
Note: before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details.
## 1 Purpose
M01DAF ranks a vector of real numbers in ascending or descending order.
## 2 Specification
SUBROUTINE M01DAF ( RV, M1, M2, ORDER, IRANK, IFAIL)
INTEGER M1, M2, IRANK(M2), IFAIL REAL (KIND=nag_wp) RV(M2) CHARACTER(1) ORDER
## 3 Description
M01DAF uses a variant of list-merging, as described on pages 165–166 in Knuth (1973). The routine takes advantage of natural ordering in the data, and uses a simple list insertion in a preparatory pass to generate ordered lists of length at least $10$. The ranking is stable: equal elements preserve their ordering in the input data.
## 4 References
Knuth D E (1973) The Art of Computer Programming (Volume 3) (2nd Edition) Addison–Wesley
## 5 Parameters
1: $\mathrm{RV}\left({\mathbf{M2}}\right)$ – REAL (KIND=nag_wp) arrayInput
On entry: elements M1 to M2 of RV must contain real values to be ranked.
2: $\mathrm{M1}$ – INTEGERInput
On entry: the index of the first element of RV to be ranked.
Constraint: ${\mathbf{M1}}>0$.
3: $\mathrm{M2}$ – INTEGERInput
On entry: the index of the last element of RV to be ranked.
Constraint: ${\mathbf{M2}}\ge {\mathbf{M1}}$.
4: $\mathrm{ORDER}$ – CHARACTER(1)Input
On entry: if ${\mathbf{ORDER}}=\text{'A'}$, the values will be ranked in ascending (i.e., nondecreasing) order.
If ${\mathbf{ORDER}}=\text{'D'}$, into descending order.
Constraint: ${\mathbf{ORDER}}=\text{'A'}$ or $\text{'D'}$.
5: $\mathrm{IRANK}\left({\mathbf{M2}}\right)$ – INTEGER arrayOutput
On exit: elements M1 to M2 of IRANK contain the ranks of the corresponding elements of RV. Note that the ranks are in the range M1 to M2: thus, if ${\mathbf{RV}}\left(i\right)$ is the first element in the rank order, ${\mathbf{IRANK}}\left(i\right)$ is set to M1.
6: $\mathrm{IFAIL}$ – INTEGERInput/Output
On entry: IFAIL must be set to $0$, $-1\text{ or }1$. If you are unfamiliar with this parameter you should refer to Section 3.3 in the Essential Introduction for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\text{ or }1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this parameter, the recommended value is $0$. When the value $-\mathbf{1}\text{ or }\mathbf{1}$ is used it is essential to test the value of IFAIL on exit.
On exit: ${\mathbf{IFAIL}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).
## 6 Error Indicators and Warnings
If on entry ${\mathbf{IFAIL}}={\mathbf{0}}$ or $-{\mathbf{1}}$, explanatory error messages are output on the current error message unit (as defined by X04AAF).
Errors or warnings detected by the routine:
${\mathbf{IFAIL}}=1$
On entry, ${\mathbf{M2}}<1$, or ${\mathbf{M1}}<1$, or ${\mathbf{M1}}>{\mathbf{M2}}$.
${\mathbf{IFAIL}}=2$
On entry, ORDER is not 'A' or 'D'.
${\mathbf{IFAIL}}=-99$
See Section 3.8 in the Essential Introduction for further information.
${\mathbf{IFAIL}}=-399$
Your licence key may have expired or may not have been installed correctly.
See Section 3.7 in the Essential Introduction for further information.
${\mathbf{IFAIL}}=-999$
Dynamic memory allocation failed.
See Section 3.6 in the Essential Introduction for further information.
Not applicable.
## 8 Parallelism and Performance
Not applicable.
The average time taken by the routine is approximately proportional to $n×\mathrm{log}\left(n\right)$, where $n={\mathbf{M2}}-{\mathbf{M1}}+1$.
## 10 Example
This example reads a list of real numbers and ranks them in ascending order.
### 10.1 Program Text
Program Text (m01dafe.f90)
### 10.2 Program Data
Program Data (m01dafe.d)
### 10.3 Program Results
Program Results (m01dafe.r) | 1,220 | 4,077 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 34, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.586059 |
http://fruzenshtein.com/how-i-would-like-to-learn-scala/ | 1,685,365,565,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644855.6/warc/CC-MAIN-20230529105815-20230529135815-00589.warc.gz | 17,272,290 | 14,714 | I want to share with you a small piece of my project. It’s about Scala programming language. If to be more precise it’s about the most simple way of Scala studying. Why I said the most simple? Well, because I have a solid knowledge of mainstream problems which occur during Scala studying. So I invite you to discuss my job.
## Forewords
After 3 years of Scala usage, I can confidently say that it’s awesome programming language. Looking back in the days when I started learn Scala, I feel myself stupid, because I didn’t follow an optimal learning way. My attempts to understand Scala concepts were unsuccessful. Only through persistence, I achieved first results.
So I want to share with you my vision of Scala learning process.
## The problem
Here is a well know fact: Scala allows to write the same thing in many different forms. From the one side it’s nice, because you have a large variety of decisions. But from the other side, programming is not a good place for a writing thoughts in a free form.
Here is an example of function declaration in different forms:
```def sum(a: Int, b: Int): Int = { a + b }
def sum(a: Int, b: Int) = { a + b }
def sum(a: Int, b: Int) { a + b }
def sum(a: Int, b: Int) = a + b
```
As you see, even in so simple action, as a function declaration we have multiple choices, to do it. The further we go, the harder things become for understanding. Of course this can be easily solved by introducing a code guide. Then the only thing you need to do is to learn all basic forms for main language constructions.
Keep in mind, that I’m speaking from the Java developer perspective.
Additional complexity make a syntax and a functional programming. Furthermore you never know what topic need to be learned next. Which came first, the chicken or the egg? In Scala I can ask you: which came first, a case class or a pattern matching? Or maybe an `apply` function?
Despite this circumstances, I choose Scala over Java 😉
## How I would learn it
I decided to start teaching from the most easiest topic for Java developers – object oriented programming. Scala has its own OOP implementation, and it’s very similar to Java 8, including functional interfaces. So there is nothing more trivial than Scala OOP for Java developers.
Here is a short promo video:
As soon as I’ll collect enough comments regarding this mini video course, I’m going to complete the rest of the video lecture. I invite you to join to the first group of “beta” students and leave your feedback about it. Almost 1 hour of content is waiting for you. Try the most unappreciated JVM programming language!
Get your free copy of the course via the subscription form below. | 596 | 2,685 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | longest | en | 0.938482 |
https://documen.tv/question/y-2-1-y-3-4-solve-the-following-system-of-equations-for-20683919-85/ | 1,726,602,254,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651829.57/warc/CC-MAIN-20240917172631-20240917202631-00638.warc.gz | 185,881,091 | 16,808 | ## y = 2x – 1 y=-3x + 4 Solve the following system of equations for x
Question
y = 2x – 1
y=-3x + 4
Solve the following system of equations for x
in progress 0
3 years 2021-08-26T11:55:31+00:00 2 Answers 7 views 0
x = 1
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
7. Subtraction
• Left to Right
Equality Properties
• Multiplication Property of Equality
• Division Property of Equality
• Subtract Property of Equality
Algebra I
• Solving systems of equations using substitution/elimination
Step-by-step explanation:
Step 1: Define Systems
y = 2x – 1
y = -3x + 4
Step 2: Solve for x
Substitution
1. Substitute in y: 2x – 1 = -3x + 4
2. Add 3x on both sides: 5x – 1 = 4
3. Add 1 to both sides: 5x = 5
4. Divide 5 on both sides: x = 1 | 301 | 885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-38 | latest | en | 0.676153 |
https://studdy.ai/shared-solution/8427ec55-d3a1-4609-9af6-c0c09a35e2af | 1,718,402,301,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861575.66/warc/CC-MAIN-20240614204739-20240614234739-00753.warc.gz | 506,460,941 | 9,932 | # Math Snap
## Given the following exponential function, identify whether the change represents growth or decay, and determine the percentage rate of increase or decrease. $y=660(0.902)^{x}$
#### STEP 1
Assumptions 1. The given exponential function is $y = 660(0.902)^x$. 2. We need to determine whether the function represents growth or decay. 3. We need to find the percentage rate of increase or decrease.
#### STEP 2
Identify the base of the exponential function. The base is the value inside the parentheses, which is $0.902$.
#### STEP 3
Determine if the base is greater than 1 or less than 1. If the base is greater than 1, it represents growth. If the base is less than 1, it represents decay. Since $0.902 < 1$, the function represents decay.
#### STEP 4
To find the percentage rate of decrease, subtract the base from 1 and then convert it to a percentage. $\text{Rate of decrease} = (1 - 0.902) \times 100\%$
##### SOLUTION
Calculate the rate of decrease. $\text{Rate of decrease} = (1 - 0.902) \times 100\% = 0.098 \times 100\% = 9.8\%$ The function represents decay, and the percentage rate of decrease is 9.8%. | 311 | 1,135 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 6, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2024-26 | latest | en | 0.768218 |
https://www.uen.org/core/core.do?courseNum=380113 | 1,701,602,631,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100499.43/warc/CC-MAIN-20231203094028-20231203124028-00237.warc.gz | 1,183,231,888 | 15,131 | Electronics 3
Printable Version (pdf)
Course Introduction
Core Standards of the Course
Strand 1
Standard 1
Identify potential safety hazards and follow general laboratory safety practices.
1. Assess workplace conditions regarding safety and health.
2. Identify potential safety issues and align with relevant safety standards to ensure a safe workplace/jobsite.
3. Describe typical electric shock hazards in industry.
4. Describe the effects of electricity on the human body.
5. Locate and understand the use of shop safety equipment.
6. Select appropriate personal protective equipment.
Standard 2
Use safe work practices.
1. Use personal protective equipment according to manufacturer rules and regulations.
2. Follow correct procedures when using any hand or power tools.
3. Ref: https://schools.utah.gov/cte/engineering/resources under the Safety Program and Management tab.
Standard 3
Complete a basic safety test without errors (100%) before using any tools or shop equipment.
Strand 2
Students will understand AC waveforms vs. DC and the advantages of using AC power for distribution.
Standard 1
Describe advantages of using AC for electrical power distribution.
Standard 2
Describe the characteristics of sinusoidal waveforms including frequency, period, and amplitude at any point within the wave.
Standard 3
Determine peak, peak-to-peak, average, and RMS values for a given sine wave.
Strand 3
Students will understand how to change AC voltage and current levels using transformers.
Standard 1
Describe step-up vs. step-down as related to turns ratio.
Standard 2
Describe primary and secondary as related to step-up and step-down.
Standard 3
Determine input and output voltage & current based on turns ratio.
Strand 4
Students will know how to calculate capacitance when connecting multiple capacitors.
Standard 1
Determine the equivalent capacitance of capacitors connected in series.
Standard 2
Determine the equivalent capacitance of capacitors connected in parallel.
Strand 5
Students will know how to calculate inductance when connecting multiple inductors.
Standard 1
Determine the equivalent inductance of inductors connected in series.
Standard 2
Determine the equivalent inductance of inductors connected in parallel.
Strand 6
Students will know how to calculate capacitive and inductive reactance.
Standard 1
Describe the concept of reactance and its unit of measure.
Standard 2
Determine the capacitive reactance of a capacitor given the frequency.
Standard 3
Determine the inductive reactance of an inductor given the frequency.
Strand 7
Students will understand circuit impedance.
Standard 1
Describe the concept of impedance and its unit of measure.
Standard 2
Determine the impedance of a basic RC or RL series circuit.
Strand 8
Students will understand resonance and how it is used in circuits.
Standard 1
Describe the concept of resonance.
Standard 2
Describe applications for resonance in AC circuits.
Performance Skills
Standard 1
Create and utilize an engineering notebook per established conventions.
Standard 2
Demonstrate practice of the Technology & Engineering Professional Workplace Skills.
Standard 3
Participate in a significant activity that provides each student with an opportunity to render service to others, employ leadership skills, or demonstrate skills they have learned through this course, preferably through participation in a Career & Technical Student Organization (CTSO) such as the Technology Student Association (TSA).
http://www.uen.org - in partnership with Utah State Board of Education (USBE) and Utah System of Higher Education (USHE). Send questions or comments to USBE Specialist - Doug Livingston and see the CTE/Engineering & Technology website. For general questions about Utah's Core Standards contact the Director - THALEA LONGHURST.
These materials have been produced by and for the teachers of the State of Utah. Copies of these materials may be freely reproduced for teacher and classroom use. When distributing these materials, credit should be given to Utah State Board of Education. These materials may not be published, in whole or part, or in any other format, without the written permission of the Utah State Board of Education, 250 East 500 South, PO Box 144200, Salt Lake City, Utah 84114-4200. | 891 | 4,313 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2023-50 | longest | en | 0.832119 |
https://math.libretexts.org/Bookshelves/Calculus/Supplemental_Modules_(Calculus)/Integral_Calculus/1%3A_Area_and_Volume/1.2%3A_Volume_by_Discs_and_Washers | 1,591,237,532,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347436828.65/warc/CC-MAIN-20200604001115-20200604031115-00099.warc.gz | 428,554,464 | 23,594 | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 1.2: Volume by Discs and Washers
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
## Volumes of Revolution
Suppose you wanted to make a clay vase. It is made by shaping the clay into a curve and spinning it along an axis. If we want to determine how much water it will hold, we can consider the cross sections that are perpendicular to the axis of rotation, and add up all the volumes of the small cross sections. We have the following definition:
$\text{Volume} = \int_a^b A(x) dx$
where $$A(x)$$ is the area the cross section at a point $$x$$.
Example 1
Find the volume of the solid that is produced when the region bounded by the curve
$y = x^2, \; y = 0\; ,\text{ and } x = 2 \nonumber$
is revolved around the x-axis.
Solution
Since we are revolving around the x-axis, we have that the cross section is in the shape of a disk with radius equal to the y-coordinate of the point.
Hence
$A(x) = \pi r^2 = \pi (x^2)^2. \nonumber$
We have
\begin{align*} \text{ Volume} &= \int_{0}^{2} \pi (x^2)^2 dx \\ &= \left( \dfrac{\pi}{5}x^5 \right]_{0}^{2} \\ &= \dfrac{32 \pi}{5}. \end{align*}
Example 2
Find the volume of the solid formed be revolving the region between the curves
$y = x^2 \text{ and } y = \sqrt{x} \nonumber$
Solution
We draw the picture and revolve a cross section about the x-axis and come up with a washer. The area of the washer is equal to the area of the outer disk minus the area of the inner disk.
$A = \pi (R^2 - r^2) \nonumber$
We have that $$R$$ is the y-coordinate of the top curve and $$r$$ is the y-coordinate of the bottom curve $$x^2$$. We have
$A = \pi([\sqrt{x}]^2 - [x^2]^2) = \pi[x - x^4] \nonumber$
Hence
\begin{align*} Volume &= \int _0^1 \pi \left( x-x^4 \right) \,dx \\ &= \pi \left[\dfrac{x^2}{2} - \dfrac{x^5}{5} \right]_0^1 \\ &= \dfrac{3 \pi}{10}. \end{align*}
Example 3: Revolving about the y-axis
Find the volume of the solid that is formed by revolving the curve bounded by
$y = x \text{ and } y = \sqrt{x}. \nonumber$
Solution
This time our cross section is perpendicular to the y-axis. When we revolve, we get a washer with $$R$$ equal to the x-coordinate of the$$y = \sqrt{x}$$ curve and $$r$$ equal to the x-coordinate of the $$y = x$$ curve.
Hence
$A = \pi\left(\left(y\right)^2 - \left(y^2\right)^2\right) = \pi \left(y^2 - y^4\right). \nonumber$
We get
\begin{align*} \text{Volume} &= \int_0^1 \pi(y^2-y^4) dy \\ &= \pi\left(\dfrac{y^3}{3}-\dfrac{y^5}{5}\right]_0^1\\ &= \dfrac{2\pi}{15}. \end{align*}
## Revolving About a Non-axis Line
Example 4
Find the volume of the region formed by revolving the curve
$y=x^3 \;\;\; 0<x<2 \nonumber$
about the line $$y=-2$$
Solution
This time we revolve the cross-section and obtain a disk. The radius of the disk is 2 plus the y-coordinate of the curve.
Hence
$A = \pi (2 + x^3)^2 \nonumber$
so that
$\text{Volume} = \int _0^2 \pi (2 + x^3)^2 \, dx. \nonumber$
This integral can be evaluated by FOIL-ing out the binomial and then integrating each monomial. We get a value of approximately 133.
Example 5
Try revolving the curve
$y =x^2 \nonumber$
from 0 to 2 about the line $$x = 5$$.
We have
$A = \pi \left[\left(5 - \sqrt{y} \right)^2 - 9\right] \nonumber$
so that
$\text{Volume} = \int_0^4 \pi \left[ \left( 5 - \sqrt{y} \right)^2 -9 \right] dy. \nonumber$
This integral works out to be approximately 59.
## Applications of Volume
Example 6: The Volume of the Khufu Pyramid
The base of the Khufu pyramid is a square with wide length 736 feet and the angle that the base makes with the ground is 50.8597 degrees. Find the volume of the Khufu pyramid.
Solution
First note that we need to change to radian measure.
$50.8597^{\circ} = 0.88767 \text{ radians} \nonumber$
The height of the pyramid is
$736 \tan(0.88767) = 904.348. \nonumber$
We have that the area of a cross section is $$s^2$$where $$s$$ is the side length of the square. Placing the y-axis through the top of the pyramid and the origin at the middle of the base, we have that
$s = \sqrt{2} x. \nonumber$
Hence
$A(x) = 2x^2. \nonumber$
Since the axis perpendicular to the cross sections is the y-axis, we need $$A$$ in terms of $$y$$. We set up similar triangles:
$\dfrac{x}{736} = \dfrac{y}{904.348} \nonumber$
$x = 0.8138\, y. \nonumber$
Hence
$A(y) = 1.3247\, x^2 \nonumber$
We calculate
$\int _0^{904.35} 1.3247\, x^2\, dx = 361,131\; \text{ cubic feet}. \nonumber$
Example 7
A sphere is formed by rotation the curve
$y= \sqrt{r^2 - x^2}. \nonumber$
We have
\begin{align*} \text{Volume} &= \int_{-r}^{r} \pi (\sqrt{r^2-x^2})^2 dx \\ &= \int_{-r}^{r} \pi (r^2-x^2) dx \\ &= \pi \left( r^2x-\dfrac{x^2}{3} \right]_{-r}^r \\ &= \pi \Big[ \big( r^3-\dfrac{r^3}{3}\big) - \big( -r^3+\dfrac{r^3}{3} \big) \Big] \\ &= \dfrac{4}{3} \pi r^3\end{align*}.
Exercise
Two cylinders of radius $$r$$ intersect each other at right angles. Find the volume of their intersection.
Hint: Consider cross sections parallel to both axes of rotation. These cross sections are squares. Then show that the side length is $$2\sqrt{r^2 - x^2}$$ | 2,211 | 6,393 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2020-24 | latest | en | 0.455206 |
https://study.com/academy/practice/quiz-worksheet-dimensional-analysis-calculations.html | 1,571,467,376,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986692126.27/warc/CC-MAIN-20191019063516-20191019091016-00275.warc.gz | 723,070,688 | 26,226 | # Dimensional Analysis Practice: Calculations & Conversions Video
Instructions:
question 1 of 3
### What are the final answer and units of this dimensional analysis equation?
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Use the provided material to answer questions 1-2.
What is the height in meters of a 5.5 foot tall person? You are given the following information: 1 foot = 12 inches 1 inch = 2.54 centimeters 1 meter = 100 centimeters
### 2. Using dimensional analysis how tall is this person, in meters?
Create your account to access this entire worksheet | 187 | 804 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-43 | longest | en | 0.884981 |
https://math.stackexchange.com/questions/2502325/two-function-of-two-random-variables | 1,566,332,759,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315618.73/warc/CC-MAIN-20190820200701-20190820222701-00199.warc.gz | 552,614,750 | 30,798 | # two function of two random variables
Supposed we are given a random variable $X$, whose probability density function is: $$f_X(x)=\begin{cases} λ \exp(-λx), & x≥0 \\ 0, & x<0. \end{cases}$$ We wish to find an invertible function $g$, such that $Y=g(X)$ is uniformly distributed on the unit interval $[0,1]$. Find $g(x)$. My answer: Since $Y$ is uniformly distributed on $[0,1)$, $fy =1$. since $fy(y)=fx(x)/|d/dx(g(x))|;$ $d/dx(g(x))=fx(x)$. Therefore, $g(x)= \exp(-λx).$ I am not sure whether this is correct. If it is not, could anyone help me out? Thank you very much.
• please use Mathjax to make your expressions readable. See Michael Hardy's edit to learn how. – spaceisdarkgreen Nov 3 '17 at 1:55
• There is a reason why, in the notation $f_X(x),$ one distinguishes between capital $X$ and lower-case $x$ and one is careful about which is which. Without that, you could not understand $\Pr(X\le x),$ and a variety of other things. And you get sloppy about that. – Michael Hardy Nov 3 '17 at 1:56
• Sorry about that. I am a new user and I will learn how to use Mathjax to ask questions clearly. Thank you very much. – Deepsea234 Nov 3 '17 at 2:38
Let $g:[0,\infty)\to[0,1]$ be the map $u\mapsto 1-e^{-\lambda x}$. Then for $0\leqslant u\leqslant 1$ we have \begin{align} \mathbb P(g(X)\leqslant u) &= \mathbb P\left(1-e^{-\lambda X}\leqslant u\right)\\ &= \mathbb P\left( e^{-\lambda X}\geqslant 1-u \right)\\ &= \mathbb P(-\lambda X\geqslant\log(1-u))\\ &= \mathbb P\left(X\leqslant \frac1\lambda \log(1-u) \right)\\ &= 1-e^{-\lambda\left(\frac1\lambda \log(1-u)\right)}\\ &= 1-(1-u)\\ &= u, \end{align} so that $g(X)$ has $U[0,1]$ distribution. The inverse of $g$ is the map $u\mapsto -\frac1\lambda\log(1-u)$ for $0\leqslant u\leqslant1$. | 620 | 1,753 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-35 | latest | en | 0.732771 |
https://economics.stackexchange.com/questions/43017/negotiations-under-expected-utility-maximization | 1,642,393,742,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300289.37/warc/CC-MAIN-20220117031001-20220117061001-00013.warc.gz | 288,778,124 | 32,497 | # Negotiations under expected utility maximization
A buyer is negotiating with a used car salesperson. The value of the car to the seller is uniformly distributed between 0 and 5000. Value to the buyer is 50 percent more than that of the seller (i.e. 1.5 times of the seller's). Both are rik neutral, neither party has a private signal, and all of the previous is common knowledge. One party will offer a price. The other party can accept the offer, in which case the buyer will pay the price to the salesperson in exchange for the car, or make a counteroffer. The counteroffer is take-it-or-leave-it; the first party can either accept the counteroffer, or there is no deal. If both parties are expected utility maximizers, what would you expect to happen when the salesperson makes the first offer? What about when the buyer makes the first offer? If you were one of the parties to the negotiation, would you prefer to make the first move, or the second?
My Solution:
The expected utility of the car for the seller is $$\frac{(0+5000)}{2} = 2500$$ and the expected utility of the buyer is $$1.5 \times \frac{(0+5000)}{2} = 3750$$.
Seller makes the 1st offer: The seller would ask a price greater than or equal to 2500. Trade occurs only if the offer is less than or equal to 3750. The seller wants to get the maximum payoff so he will ask for the maximum price which is equal to the expected value of the buyer i.e. 3750. The buyer than makes a counteroffer equal to the expected value of the seller i.e. 2500. Seller accepts that offer as it is within his domain of acceptance.
Buyer makes the 1st offer: The buyer would give a price less than or equal to 3750. Trade occurs only if the offer is greater than or equal to 2500. The buyer wants to get the maximum payoff so he will ask for the minimum price which is equal to the expected value of the seller i.e. 2500. The seller then makes a counteroffer equal to the expected value of the buyer i.e. 3750. Buyer accepts that offer as it is within his domain of acceptance.
What would I do: I would want to move second as that would allow me to give a take-it-orleave-it offer which provides me a higher bargaining power.
I would like your valuable feedback on my solution of the question. | 537 | 2,248 | {"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": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2022-05 | latest | en | 0.960395 |
https://slidetodoc.com/scale-factor-scale-drawings-understanding-scales-all-scale/ | 1,720,980,260,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514635.58/warc/CC-MAIN-20240714155001-20240714185001-00850.warc.gz | 476,957,634 | 15,493 | # Scale Factor Scale Drawings Understanding Scales All scale
• Slides: 27
Scale Factor & Scale Drawings
Understanding Scales All scale drawings must have a scale written on them. Scales are usually expressed as ratios. Normally for maps and buildings the ratio: Drawing length: Actual length For maps the ratio is normally in the ratio: Map distance: Actual Distance Example: 1 cm : 100 cm The ratio 1 cm: 100 cm means that for every 1 cm on the scale drawing the length will be 100 cm in real life Example: 1: 10000 The ratio 1: 10000 means that the real distance is 10000 times the length of one unit on the map or drawing.
Scale Factor
How to find Scale Factor When a figure is dilated, its size is changed by multiplying the length of each side by a scale factor. All angles remain the same and so the new shape (or image) is similar to the original. Can be found by dividing a new side length by the original side length. • When going from a small shape to a larger shape the scale factor is greater than 1. (Enlargement) • When going from a large shape to a smaller shape the scale factor is • less than 1. (Reduction) 1. 2. 3. 4. Determine the corresponding side lengths. Determine if you are making a larger shape or a smaller shape. Determine if the scale factor is greater than or less than 1. Write the correct ratio.
Scale Factor Scale factor = new measurement old measurement Old measurement x SF = new measurement new SF old - Scale factor more than 1 => shape gets bigger (Enlargement) - Scale factor less than 1 => shape gets smaller (Reduction) - Congruent shapes are similar shapes with SF = 1
Vocabulary scale model scale factor scale drawing
The scale can be written as a scale factor, which is the ratio of the length or size of the drawing or model to the length of the corresponding side or part on the actual object. Scale Factor needs to be the SAME UNITS!
This HO gauge model train is a scale model of a historic train. A scale model is a proportional model of a three-dimensional object. Its dimensions are related to the dimensions of the actual object by a ratio called the scale factor. The scale factor of an HO 1 gauge model train is . 87 1 This means that each dimension of the model is 87 of the corresponding dimension of the actual train.
A scale is the ratio between two sets of measurements. Scales can use the same units or different units. The photograph shows a scale drawing of the model train. A scale drawing is a proportional drawing of an object. Both scale drawings and scale models can be smaller or larger than the objects they represent.
If you have ever seen Jurassic Park, you saw how big the dinosaurs were compared to the people. Pretend that they made a large Human to watch over the animals. What would be the scale factor if a 64 inch person was made to be 160 feet?
The scale factor tells you how many times bigger than “normal” that person really is. You must make all units of measure the same…. 64 inches = = 160 feet 160 x 12 1920 inches
Now take the: 64 inches 1920 inches And simplify 1/30 inches This means that the person was created 30 times his normal size.
Scale Factor
Setting up Proportions Keep like units in the same fraction. Inches = yards Inches yards
Remember… There is more than one way to set up a proportion correctly! Cross Multiply! Use common sense!
Write a Proportion Using Scale Factor Tom is drawing a blueprint for a rectangular shed he wants to build. The scale factor is 1 ft. to ¼ inch. If the dimensions of the blueprint are 1 ¼ in. by 2 inches, what are the actual dimensions of the shed going to be?
Write a Proportion Using a Scale Factor ¾ inch to 1 foot If the length in inches is 2 ¼ inch, what would the actual length be in feet ?
Scale Drawings
What are scale drawings? Scale drawings are everywhere! Scale Drawings On Maps Vehicle design Footprints of houses
Scale in everyday life: kitchen design 6 cm Scale 1 cm = 1 m Length of units = 6 m 5
Scale in everyday life: maps Scale 1 : 1 000
Scale in everyday life: plans decking pool path Scale 2 cm = 1 m 7
Using A Scale Drawing
When objects are too small or too large to be drawn or constructed at actual size, people use a scale drawing or a model. The scale drawing of this tree is 1: 500 If the height of the tree on paper is 20 inches, what is the height of the tree in real life?
The scale is the relationship between the measurements of the drawing or model to the measurements of the object. In real-life, the length of this van may measure 240 inches. However, the length of a copy or print paper that you could use to draw this van is a little bit less than 12 inches
Map Scales (Legends) are used to find distances on a map. For example, if your map legend tells you that ½ of an inch represents 50 miles, how could you find the mileage for a 2 inch distance on the map?
Map Scales Ratios and proportions can be used to find distances using a scale. Example: 1 inch = 15 miles The distance from Jacksonville to Smithtown on a map is 4 inches. How many miles are between these cities? 1 in. = 4 in 15 mi. n The distance 1 n = 60 between the two cities is 60 miles. | 1,210 | 5,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-30 | latest | en | 0.904876 |
https://davidlowryduda.com/choosing-functions-for-mvt-abscissa/ | 1,686,373,851,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656963.83/warc/CC-MAIN-20230610030340-20230610060340-00652.warc.gz | 228,301,688 | 11,919 | # mixedmath
Explorations in math and programming
David Lowry-Duda
In my previous note, I described some of the main ideas behind the paper "When are there continuous choices for the mean value abscissa?" that I wrote joint with Miles Wheeler. In this note, I discuss the process behind generating the functions and figures in our paper.
Our functions came in two steps: we first need to choose which functions to plot; then we need to figure out how to graphically solve their general mean value abscissae problem.
Afterwards, we can decide how to plot these functions well.
## Choosing the right functions to plot
The first goal is to find the right functions to plot. From the discussion in our paper, this amounts to specifying certain local conditions of the function. And for a first pass, we only used these prescribed local conditions.
The idea is this: to study solutions to the mean value problem, we look at the zeroes of the function $$F(b, c) = \frac{f(b) - f(a)}{b - a} - f'(c).$$ When $F(b, c) = 0$, we see that $c$ is a mean value abscissa for $f$ on the interval $(a, b)$.
By the implicit function theorem, we can solve for $c$ as a function of $b$ around a given solution $(b_0, c_0)$ if $F_c(b_0, c_0) \neq 0$. For this particular function, $F_c(b_0, c_0) = -f''(c_0)$.
More generally, it turns out that the order of vanishing of $f'$ at $b_0$ and $c_0$ governs the local behaviour of solutions in a neighborhood of $(b_0, c_0)$.
To make figures, we thus need to make functions with prescribed orders of vanishing of $f'$ at points $b_0$ and $c_0$, where $c_0$ is itself a mean value abscissa for the interval $(a_0, b_0)$.
Without loss of generality, it suffices to consider the case when $f(a_0) = f(b_0) = 0$, as otherwise we can study the function $$g(x) = f(x) - \left( \frac{f(b_0) - f(a_0)}{b_0 - a_0}(x - a_0) + f(a_0) \right),$$ which has $g(a_0) = g(b_0) = 0$, and those triples $(a, b, c)$ which solve this for $f$ also solve this for $g$.
And for consistency, we made the arbitrary decisions to have $a_0 = 0$, $b_0 = 3$, and $c_0 = 1$. This decision simplified many of the plotting decisions, as the important points were always $0$, $1$, and $3$.
## A first idea
Thus the first task is to be able to generate functions $f$ such that:
1. $f(0) = 0$,
2. $f(3) = 0$,
3. $f'(1) = 0$ (so that $1$ is a mean value abscissa), and
4. $f'(x)$ has prescribed order of vanishing at $1$, and
5. $f'(x)$ has prescribed order of vanishing at $3$.
These conditions can all be met by an appropriate interpolating polynomial. As we are setting conditions on both $f$ and its derivatives at multiple points, this amounts to the fundamental problem in Hermite interpolation. Alternatively, this amounts to using Taylor's theorem at multiple points and then using the Chinese Remainder Theorem over $\mathbb{Z}[x]$ to combine these polynomials together.
There are clever ways of solving this, but this task is so small that it doesn't require cleverness. In fact, this is one of the laziest solutions we could think of. We know that given $n$ Hermite conditions, there is a unique polynomial of degree $n - 1$ that interpolates these conditions. Thus we
1. determine the degree of the polynomial,
2. create a degree $n-1$ polynomial with variable coefficients in sympy,
3. have sympy symbolically compute the relations the coefficients must satisfy,
4. ask sympy to solve this symbolic system of equations.
In code, this looks like
import sympy
from sympy.abc import X, B, C, D # Establish our variable names
def interpolate(conds):
"""
Finds the polynomial of minimal degree that solves the given Hermite conditions.
conds is a list of the form
[(x1, r1, v1), (x2, r2, v2), ...]
where the polynomial p is to satisfy p^(r_1) (x_1) = v_1, and so on.
"""
# the degree will be one less than the number of conditions
n = len(conds)
# generate a symbol for each coefficient
A = [sympy.Symbol("a[%d]" % i) for i in range(n)]
# generate the desired polynomial symbolically
P = sum([A[i] * X**i for i in range(n)])
# generate the equations the polynomial must satisfy
#
# for each (x, r, v), sympy evaluates the rth derivative of P wrt X,
# substitutes x in for X, and requires that this equals v.
EQNS = [sympy.diff(P, X, r).subs(X, x) - v for x, r, v in conds]
# solve this system for the coefficients A[n]
SOLN = sympy.solve(EQNS, A)
return P.subs(SOLN)
We note that we use the convention that a sympy symbol for something is capitalized. For example, we think of the polynomial as being represented by $$p(x) = a(0) + a(1)x + a(2)x^2 + \cdots + a(n)x^n.$$ In sympy variables, we think of this as
P = A[0] + A[1] * X + A[2] * X**2 + ... + A[n] * X**n
With this code, we can ask for the unique degree 1 polynomial which is $1$ at $1$, and whose first derivative is $2$ at $1$.
> interpolate([(1, 0, 1), (1, 1, 2)])
2*X - 1
Indeed, $2x - 1$ is this polynomial.
### Too rigid
We have now produced a minimal Hermite solver. But there is a major downside: the unique polynomial exhibiting the necessary behaviours we required is essentially never a good didactic example. We don't just want plots — we want beautiful, simple plots.
Firstly, we added the additional constraint that $f(1) = 1$. This is small, but it's a small prescribed value. So now at least all three points of interest will fit within a $[0, 3] \times [0, 3]$ box.
Secondly, we also allow the choice of the value of the first nonvanishing derivatives at $1$ and $3$. In reality, we treat these as parameters to change the shape of the resulting graph. Roughly speaking, if the order of vanishing of $f(x) - f(1)$ is $k$ at $1$, then near $1$ the approximation $f(x) \approx f^{(k)}(1) x^k/k!$ is true. Morally, the larger the value of the derivative, the more the graph will resemble $x^k$ near that point.
In code, we implemented this by making functions that will add the necessary Hermite conditions to our input to interpolate.
# We fix the values of a0, b0, c0.
a0 = 0
b0 = 3
c0 = 1
# We require p(a0) = 0, p(b0) = 0, p(c0) = 1, p'(c0) = 0.
BASIC_CONDS = [(a0, 0, 0), (b0, 0, 0), (c0, 0, 1), (c0, 1, 0)]
def c_degen(n, residue):
"""
Give Hermite conditions for order of vanishing at c0 equal to n, with
first nonzero residue residue.
NOTE: the order n is in terms of f', not of f. That is, this is the amount
of additional degeneracy to add. This may be a source of off-by-one errors.
"""
return [(c0, 1 + i, 0) for i in range(1, n + 1)] + [(c0, n + 2, residue)]
def b_degen(n, residue):
"""
Give Hermite conditions for order of vanishing at b0 equal to n, with
first nonzero residue residue.
"""
return [(b0, i, 0) for i in range(1, n + 1)] + [(b0, n + 1, residue)]
def poly_with_degens(nc=0, nb=0, residue_c=3, residue_b=3):
"""
Give unique polynomial with given degeneracies for this MVT problem.
nc is the order of vanishing of f' at c0, with first nonzero residue residue_c.
nb is the order of vanishing of f at b0, with first nonzero residue residue_b.
"""
conds = BASIC_CONDS + c_degen(nc, residue_c) + b_degen(nb, residue_b)
return interpolate(conds)
Then apparently the unique polynomial degree $5$ polynomial $f$ with $f(0) = f(3) = f'(1) = 0$, $f(1) = 1$, and $f''(1) = f'(3) = 3$ is given by
> poly_with_degens()
11*X**5/16 - 21*X**4/4 + 113*X**3/8 - 65*X**2/4 + 123*X/16
### Too many knobs
In principle, this is a great solution. And if you turn the knobs enough, you can get a really nice picture. But the problem with this system (and with many polynomial interpolation problems) is that when you add conditions, you can introduce many jagged peaks and sudden changes. These can behave somewhat unpredictably and chaotically — small changes in Hermite conditions can lead to drastic changes in resulting polynomial shape.
What we really want is for the interpolator to give a polynomial that doesn't have sudden changes.
## Minimize change
The problem: the polynomial can have really rapid changes that makes the plots look bad.
The solution: minimize the polynomial's change.
That is, if $f$ is our polynomial, then its rate of change at $x$ is $f'(x)$. Our idea is to "minimize" the average size of the derivative $f'$ — this should help keep the function in frame. There are many ways to do this, but we want to choose one that fits into our scheme (so that it requires as little additional work as possible) but which works well.
We decide that we want to focus our graphs on the interval $(0, 4)$. Then we can measure the average size of the derivative $f'$ by its L2 norm on $(0, 4)$: $$L2(f) = \int_0^4 (f'(x))^2 dx.$$
We add an additional Hermite condition of the form (pt, order, VAL) and think of VAL as an unknown symbol. We arbitrarily decided to start with $pt = 2$ (so that now behavior at the points $0, 1, 2, 3$ are all being controlled in some way) and $order = 1$. The point itself doesn't matter very much, since we're going to minimize over the family of polynomials that interpolate the other Hermite conditions with one degree of freedom.
In other words, we are adding in the condition that $f'(2) = VAL$ for an unknown VAL.
We will have sympy compute the interpolating polynomial through its normal set of (explicit) conditions as well as the symbolic condition (2, 1, VAL). Then $f = f(\mathrm{VAL}; x)$.
Then we have sympy compute the (symbolic) L2 norm of the derivative of this polynomial with respect to VAL over the interval $(0, 4)$, $$L2(\mathrm{VAL}) = \int_0^x f'(\mathrm{VAL}; x)^2 dx.$$
Finally, to minize the L2 norm, we have sympy compute the derivative of $L2(\mathrm{VAL})$ with respect to VAL and find the critical points, when the derivative is equal to $0$. We choose the first one to give our value of VAL.1 1In principle, I suppose we could be finding a local maximum. We could guarantee that we're finding the minimum by choosing the critical point that minimized the L2 norm. Choosing val very large in magnitude makes the L2 norm very large, and so the minimum will be one of the critical points.
In code, this looks like
def smoother_interpolate(conds, ctrl_point=2, order=1, interval=(0,4)):
"""
Find the polynomial of minimal degree that interpolates the Hermite
conditions in conds, and whose behavior at ctrl_point minimizes the L2
norm on interval of its derivative.
"""
# Add the symbolic point to the conditions.
# Recall that D is a sympy variable
new_conds = conds + [(ctrl_point, order, D)]
# Find the polynomial interpolating new_conds, symbolic in X *and* D
P = interpolate(new_conds)
# Compute L2 norm of the derivative on interval
L2 = sympy.integrate(sympy.diff(P, X)**2, (X, *interval))
# Take the first critical point of the L2 norm with respect to D
SOLN = sympy.solve(sympy.diff(L2, D), D)[0]
# Substitute the minimizing solution in for D and return
return P.subs(D, SOLN)
def smoother_poly_with_degens(nc=0, nb=0, residue_c=3, residue_b=3):
"""
Give unique polynomial with given degeneracies for this MVT problem whose
derivative on (0, 4) has minimal L2 norm.
nc is the order of vanishing of f' at c0, with first nonzero residue residue_c.
nb is the order of vanishing of f at b0, with first nonzero residue residue_b.
"""
conds = BASIC_CONDS + c_degen(nc, residue_c) + b_degen(nb, residue_b)
return smoother_interpolate(conds)
Then apparently the polynomial degree $6$ polynomial $f$ with $f(0) = f(3) = f'(1) = 0$, $f(1) = 1$, and $f''(1) = f'(3) = 3$, and with minimal L2 derivative norm on $(0, 4)$ is given by
> smoother_poly_with_degens()
-9660585*X**6/33224848 + 27446837*X**5/8306212 - 232124001*X**4/16612424
+ 57105493*X**3/2076553 - 858703085*X**2/33224848 + 85590321*X/8306212
> sympy.N(smoother_poly_with_degens())
-0.290763858423069*X**6 + 3.30437472580762*X**5 - 13.9729157526921*X**4
+ 27.5001374874612*X**3 - 25.8452073279613*X**2 + 10.3043747258076*X
Is it much better? Let's compute the L2 norms.
> interval = (0, 4)
> sympy.N(sympy.integrate(sympy.diff(poly_with_degens(), X)**2, (X, *interval)))
1865.15411706349
> sympy.N(sympy.integrate(sympy.diff(smoother_poly_with_degens(), X)**2, (X, *interval)))
41.1612799050325
That's beautiful. And you know what's better? Sympy did all the hard work.
For comparison, we can produce a basic plot using numpy and matplotlib.
import matplotlib.pyplot as plt
import numpy as np
def basic_plot(F, n=300):
fig = plt.figure(figsize=(6, 2.5))
b1d = np.linspace(-.5, 4.5, n)
f = sympy.lambdify(X, F)(b1d)
ax.plot(b1d,f,'k')
ax.set_aspect('equal')
ax.grid(True)
ax.set_xlim([-.5, 4.5])
ax.set_ylim([-1, 5])
ax.plot([0, c0, b0],[0, F.subs(X,c0),F.subs(X,b0)],'ko')
fig.savefig("basic_plot.pdf")
Then the plot of poly_with_degens() is given by
The polynomial jumps upwards immediately and strongly for $x > 3$.
On the other hand, the plot of smoother_poly_with_degens() is given by
This stays in frame between $0$ and $4$, as desired.
### Choose data to highlight and make the functions
This was enough to generate the functions for our paper. Actually, the three functions (in a total of six plots) in figures 1, 2, and 5 in our paper were hand chosen and hand-crafted for didactic purposes: the first two functions are simply a cubic and a quadratic with certain points labelled. The last function was the non-analytic-but-smooth semi-pathological counterexample, and so cannot be created through polynomial interpolation.
But the four functions highlighting different degenerate conditions in figures 3 and 4 were each created using this L2-minimizing interpolation system.
In particular, the function in figure 3 is
F3 = smoother_poly_with_degens(nc=1, residue_b=-3)
which is one of the simplest L2 minimizing polynomials with the typical Hermite conditions, $f''(c_0) = 0$, and opposite-default sign of $f'(b_0)$.
The three functions in figure 4 are (from left to right)
F_bmin = smoother_poly_with_degens(nc=1, nb=1, residue_c=10, residue_b=10)
F_bzero = smoother_poly_with_degens(nc=1, nb=2, residue_c=-20, residue_b=20)
F_bmax = smoother_poly_with_degens(nc=1, nb=1, residue_c=20, residue_b=-10)
We chose much larger residues because the goal of the figure is to highlight how the local behavior at those points corresponds to the behavior of the mean value abscissae, and larger residues makes those local behaviors more dominating.
## Plotting all possible mean value abscissae
Now that we can choose our functions, we want to figure out how to find all solutions of the mean value condition $$F(b, c) = \frac{f(b) - f(a_0)}{b - a_0} - f'(c).$$ Here I write $a_0$ as it's fixed, while both $b$ and $c$ vary.
Our primary interest in these solutions is to facilitate graphical experimentation and exploration of the problem — we want these pictures to help build intuition and provide examples.
Although this may seem harder, it is actually a much simpler problem. The function $F(b, c)$ is continuous (and roughly as smooth as $f$ is).
Our general idea is a common approach for this sort of problem:
1. Compute the values of $F(b, c)$ on a tight mesh (or grid) of points.
2. Restrict attention to the domain where solutions are meaningful.
3. Plot the contour of the $0$-level set.
Contours can be well-approximated from a tight mesh. In short, if there is a small positive number and a small negative number next to each other in the mesh of computed values, then necessarily $F(b, c) = 0$ between them. For a tight enough mesh, good plots can be made.
To solve this, we again have sympy create and compute the function for us. We use numpy to generate the mesh (and to vectorize the computations, although this isn't particularly important in this application), and matplotlib to plot the resulting contour.
Before giving code, note that the symbol F in the sympy code below stands for what we have been mathematically referring to as $f$, and not $F$. This is a potential confusion from our sympy-capitalization convention. It is still necessary to have sympy compute $F$ from $f$.
In code, this looks like
import sympy
import scipy
import numpy as np
import matplotlib.pyplot as plt
def abscissa_plot(F, n=300):
# Compute the derivative of f
DF = sympy.diff(F,X)
# Define CAP_F — "capital F"
#
# this is (f(b) - f(0))/(b - 0) - f'(c).
CAP_F = (F.subs(X, B) - F.subs(X, 0)) / (B - 0) - DF.subs(X, C)
# build the mesh
b1d = np.linspace(-.5, 4.5, n)
b2d, c2d = np.meshgrid(b1d, b1d)
# compute CAP_F within the mesh
cap_f_mesh = sympy.lambdify((B, C), CAP_F)(b2d, c2d)
# restrict attention to below the diagonal — we require c < b
# (although the mas inequality looks reversed in this perspective)
# Set up plot basics
fig = plt.figure(figsize=(6, 2.5))
ax.set_aspect('equal')
ax.grid(True)
ax.set_xlim([-.5, 4.5])
ax.set_ylim([-.5, 4.5])
# plot the contour
ax.contour(b2d, c2d, valid_cap_f_mesh, [0], colors='k')
# plot a diagonal line representing the boundary
ax.plot(b1d,b1d,'k–')
# plot the guaranteed point
ax.plot(b0,c0,'ko')
fig.savefig("abscissa_plot.pdf")
Then plots of solutions to $F(b, c) = 0$ for our basic polynomials are given by
for poly_with_degens(), while for smoother_poly_with_degens() we get
And for comparison, we can now create a (slightly worse looking) version of the plots in figure 3.
F3 = smoother_poly_with_degens(nc=1, residue_b=-3)
basic_plot(F3)
abscissa_plot(F3)
This produces the two plots
For comparison, a (slightly scaled) version of the actual figure appearing in the paper is
## Copy of the code
A copy of the code used in this note (and correspondingly the code used to generate the functions for the paper) is available on my github as an ipython notebook.
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Comment via email | 5,120 | 18,018 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-23 | latest | en | 0.874707 |
https://www.mathnasium.com/math-centers/granadahills/news/checking-in-with-the-yang-family-977724811 | 1,709,435,025,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476180.67/warc/CC-MAIN-20240303011622-20240303041622-00064.warc.gz | 886,033,582 | 13,105 | # Checking In With the Yang Family
Jul 8, 2021 | Granada Hills
Hello from Mathnasium of Granada Hills! Maybe you remember reading about one of our favorite Mathnasium families, the Yangs. Today we're getting an update on all four students and how they are progressing in Math:
Carly is going into 9th grade this year so her Mathnasium instructor is helping her get a head start on Geometry. "I'm glad that I'm learning it before school starts," said Carly. "Right now, I'm learning how to calculate the angles of triangles." Then she adds, "I'm pretty good at it." (We LOVE that confidence, Carly!!!)
Frank, our favorite kinesthetic learner, is jumping into 7th grade this fall. His Mathnasium instructor is focusing on teaching Frank with the help of math manipulatives. In this case, Frank is using 3-dimensional blocks to learn the difference between finding an object's surface area or finding its entire volume. Maggie, Frank's Mom, still plans to homeschool Frank next year. "Except for Math," Maggie says. "I've learned that it's much better for us to outsource his Math lessons here at Mathnasium."
Luke is 11 and will be attending 6th grade this fall. He started Mathnasium just over 3 months ago and has made great progress. Luke's Dad, Lin, noticed, "Luke was eager to go to Mathnasium. He watched his older siblings attend and really wanted to be included." Luke is studying the manipulation of fractions now. "I didn't start with fractions though." He said, "I had to re-learn long division first." Luke's comprehensive assessment revealed a lack of understanding in division, so his Mathnasium instructor made sure to cover it before anything new was taught.
Rita, the baby of the family, is getting ready to start 4th grade. Rita wanted to start Mathnasium "because of the prizes." (We don't blame you, Rita.) "She walked in the first day and saw the reward cabinet," explains Maggie. "She was like 'Whoa'." Rita chose the reward she wanted, and when the goals she made are met, the reward will be hers. "It definitely motivated her to hit the ground running," laughed Maggie. "I think she's already halfway to getting the toy she wants."
The Yang family is loving Mathnasium of Granada Hills, just like the many families around Granada Hills, Porter Ranch, Northridge, North Hills, Sylmar. Come join us! Sign up now for a Free Assessment . Call today to get started 8183688249.
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Could not find Center, try again | 567 | 2,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-10 | latest | en | 0.978973 |
https://www.jiskha.com/questions/564139/john-places-a-mirror-on-the-ground-between-himself-and-an-apartment-building-he-then | 1,597,278,127,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738950.31/warc/CC-MAIN-20200812225607-20200813015607-00556.warc.gz | 717,920,606 | 6,070 | # Geometry
John places a mirror on the ground between himself and an apartment building. He then stands so that he can see the top edge of the building in the center of the mirror. The center of the mirror is 6.10 meters from her feet and 18.30 meters from the building. John’s eyes are 1.82 meters above the ground. How high is the building?
1. 👍 0
2. 👎 0
3. 👁 190
1. LM = 4 in
LN = ?
Find LN. Round the answer to the nearest tenth.
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2. 👎 0
2. Jensen
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asked by Help!! on January 8, 2016 | 992 | 3,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2020-34 | latest | en | 0.964568 |
http://image.absoluteastronomy.com/topics/Slide_rule | 1,652,964,268,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662527626.15/warc/CC-MAIN-20220519105247-20220519135247-00778.warc.gz | 25,481,325 | 30,955 | Slide rule
Encyclopedia
The slide rule, also known colloquially as a slipstick, is a mechanical analog computer
Analog computer
An analog computer is a form of computer that uses the continuously-changeable aspects of physical phenomena such as electrical, mechanical, or hydraulic quantities to model the problem being solved...
. The slide rule is used primarily for multiplication
Multiplication
Multiplication is the mathematical operation of scaling one number by another. It is one of the four basic operations in elementary arithmetic ....
and division
Division (mathematics)
right|thumb|200px|20 \div 4=5In mathematics, especially in elementary arithmetic, division is an arithmetic operation.Specifically, if c times b equals a, written:c \times b = a\,...
, and also for functions such as root
Nth root
In mathematics, the nth root of a number x is a number r which, when raised to the power of n, equals xr^n = x,where n is the degree of the root...
s, logarithm
Logarithm
The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number. For example, the logarithm of 1000 to base 10 is 3, because 1000 is 10 to the power 3: More generally, if x = by, then y is the logarithm of x to base b, and is written...
s and trigonometry
Trigonometry
Trigonometry is a branch of mathematics that studies triangles and the relationships between their sides and the angles between these sides. Trigonometry defines the trigonometric functions, which describe those relationships and have applicability to cyclical phenomena, such as waves...
, but is not normally used for addition
Addition is a mathematical operation that represents combining collections of objects together into a larger collection. It is signified by the plus sign . For example, in the picture on the right, there are 3 + 2 apples—meaning three apples and two other apples—which is the same as five apples....
or subtraction
Subtraction
In arithmetic, subtraction is one of the four basic binary operations; it is the inverse of addition, meaning that if we start with any number and add any number and then subtract the same number we added, we return to the number we started with...
.
Slide rules come in a diverse range of styles and generally appear in a linear or circular form with a standardized set of markings (scales) essential to performing mathematical computations. Slide rules manufactured for specialized fields such as aviation
Aviation
Aviation is the design, development, production, operation, and use of aircraft, especially heavier-than-air aircraft. Aviation is derived from avis, the Latin word for bird.-History:...
or finance
Finance
"Finance" is often defined simply as the management of money or “funds” management Modern finance, however, is a family of business activity that includes the origination, marketing, and management of cash and money surrogates through a variety of capital accounts, instruments, and markets created...
typically feature additional scales that aid in calculations common to that field.
William Oughtred
William Oughtred
William Oughtred was an English mathematician.After John Napier invented logarithms, and Edmund Gunter created the logarithmic scales upon which slide rules are based, it was Oughtred who first used two such scales sliding by one another to perform direct multiplication and division; and he is...
and others developed the slide rule in the 17th century based on the emerging work on logarithms by John Napier
John Napier
John Napier of Merchiston – also signed as Neper, Nepair – named Marvellous Merchiston, was a Scottish mathematician, physicist, astronomer & astrologer, and also the 8th Laird of Merchistoun. He was the son of Sir Archibald Napier of Merchiston. John Napier is most renowned as the discoverer...
. Before the advent of the pocket calculator
Calculator
An electronic calculator is a small, portable, usually inexpensive electronic device used to perform the basic operations of arithmetic. Modern calculators are more portable than most computers, though most PDAs are comparable in size to handheld calculators.The first solid-state electronic...
, it was the most commonly used calculation tool in science
Science
Science is a systematic enterprise that builds and organizes knowledge in the form of testable explanations and predictions about the universe...
and engineering
Engineering
Engineering is the discipline, art, skill and profession of acquiring and applying scientific, mathematical, economic, social, and practical knowledge, in order to design and build structures, machines, devices, systems, materials and processes that safely realize improvements to the lives of...
. The use of slide rules continued to grow through the 1950s and 1960s even as digital computing devices
Computer
A computer is a programmable machine designed to sequentially and automatically carry out a sequence of arithmetic or logical operations. The particular sequence of operations can be changed readily, allowing the computer to solve more than one kind of problem...
were being gradually introduced; but around 1974 the electronic scientific calculator
Scientific calculator
A scientific calculator is a type of electronic calculator, usually but not always handheld, designed to calculate problems in science, engineering, and mathematics...
## Basic concepts
In its most basic form, the slide rule uses two logarithmic scales to allow rapid multiplication and division of numbers. These common operations can be time-consuming and error-prone when done on paper. More elaborate slide rules allow other calculations, such as square root
Square root
In mathematics, a square root of a number x is a number r such that r2 = x, or, in other words, a number r whose square is x...
s, exponentials
Exponential function
In mathematics, the exponential function is the function ex, where e is the number such that the function ex is its own derivative. The exponential function is used to model a relationship in which a constant change in the independent variable gives the same proportional change In mathematics,...
, logarithms, and trigonometric function
Trigonometric function
In mathematics, the trigonometric functions are functions of an angle. They are used to relate the angles of a triangle to the lengths of the sides of a triangle...
s.
Scales may be grouped in decades, which are numbers ranging from 1 to 10 (i.e. 10n to 10n+1). Thus single decade scales C and D range from 1 to 10 across the entire width of the slide rule while double decade scales A and B range from 1 to 100 over the width of the slide rule.
In general, mathematical calculations are performed by aligning a mark on the sliding central strip with a mark on one of the fixed strips, and then observing the relative positions of other marks on the strips. Number
Number
A number is a mathematical object used to count and measure. In mathematics, the definition of number has been extended over the years to include such numbers as zero, negative numbers, rational numbers, irrational numbers, and complex numbers....
s aligned with the marks give the approximate value of the product
Product (mathematics)
In mathematics, a product is the result of multiplying, or an expression that identifies factors to be multiplied. The order in which real or complex numbers are multiplied has no bearing on the product; this is known as the commutative law of multiplication...
, quotient
Quotient
In mathematics, a quotient is the result of division. For example, when dividing 6 by 3, the quotient is 2, while 6 is called the dividend, and 3 the divisor. The quotient further is expressed as the number of times the divisor divides into the dividend e.g. The quotient of 6 and 2 is also 3.A...
, or other calculated result.
The user determines the location of the decimal point in the result, based on mental estimation. Scientific notation
Scientific notation
Scientific notation is a way of writing numbers that are too large or too small to be conveniently written in standard decimal notation. Scientific notation has a number of useful properties and is commonly used in calculators and by scientists, mathematicians, doctors, and engineers.In scientific...
is used to track the decimal point in more formal calculations. Addition
Addition is a mathematical operation that represents combining collections of objects together into a larger collection. It is signified by the plus sign . For example, in the picture on the right, there are 3 + 2 apples—meaning three apples and two other apples—which is the same as five apples....
and subtraction
Subtraction
In arithmetic, subtraction is one of the four basic binary operations; it is the inverse of addition, meaning that if we start with any number and add any number and then subtract the same number we added, we return to the number we started with...
steps in a calculation are generally done mentally or on paper, not on the slide rule.
Most slide rules consist of three linear strips of the same length, aligned in parallel and interlocked so that the central strip can be moved lengthwise relative to the other two. The outer two strips are fixed so that their relative positions do not change.
Some slide rules ("duplex" models) have scales on both sides of the rule and slide strip, others on one side of the outer strips and both sides of the slide strip (which can usually be pulled out, flipped over and reinserted for convenience), still others on one side only ("simplex" rules). A sliding cursor with a vertical alignment line is used to find corresponding points on scales that are not adjacent to each other or, in duplex models, are on the other side of the rule. The cursor can also record an intermediate result on any of the scales.
### Multiplication
A logarithm transforms the operations of multiplication and division to addition and subtraction according to the rules and .
Moving the top scale to the right by a distance of , by matching the beginning of the top scale with the label on the bottom, aligns each number , at position on the top scale, with the number at position on the bottom scale. Because , this position on the bottom scale gives , the product of and . For example, to calculate 3×2, the 1 on the top scale is moved to the 2 on the bottom scale. The answer, 6, is read off the bottom scale where 3 is on the top scale. In general, the 1 on the top is moved to a factor on the bottom, and the answer is read off the bottom where the other factor is on the top. This works because the distances from the "1" are proportional to the logarithms of the marked values:
Operations may go "off the scale;" for example, the diagram above shows that the slide rule has not positioned the 7 on the upper scale above any number on the lower scale, so it does not give any answer for 2×7. In such cases, the user may slide the upper scale to the left until its right index aligns with the 2, effectively dividing by 10 (by subtracting the full length of the C-scale) and then multiplying by 7, as in the illustration below:
Here the user of the slide rule must remember to adjust the decimal point appropriately to correct the final answer. We wanted to find 2×7, but instead we calculated (2/10)×7=0.2x7=1.4. So the true answer is not 1.4 but 14. Resetting the slide is not the only way to handle multiplications that would result in off-scale results, such as 2×7; some other methods are:
1. Use the double-decade scales A and B.
2. Use the folded scales. In this example, set the left 1 of C opposite the 2 of D. Move the cursor to 7 on CF, and read the result from DF.
3. Use the CI inverted scale. Position the 7 on the CI scale above the 2 on the D scale, and then read the result off of the D scale, below the 1 on the CI scale. Since 1 occurs in two places on the CI scale, one of them will always be on-scale.
4. Use both the CI inverted scale and the C scale. Line up the 2 of CI with the 1 of D, and read the result from D, below the 7 on the C scale.
Method 1 is easy to understand, but entails a loss of precision. Method 3 has the advantage that it only involves two scales.
### Division
The illustration below demonstrates the computation of 5.5/2. The 2 on the top scale is placed over the 5.5 on the bottom scale. The 1 on the top scale lies above the quotient, 2.75. There is more than one method for doing division, but the method presented here has the advantage that the final result cannot be off-scale, because one has a choice of using the 1 at either end.
### Other operations
In addition to the logarithmic scales, some slide rules have other mathematical function
Function (mathematics)
In mathematics, a function associates one quantity, the argument of the function, also known as the input, with another quantity, the value of the function, also known as the output. A function assigns exactly one output to each input. The argument and the value may be real numbers, but they can...
s encoded on other auxiliary scales. The most popular were trigonometric
Trigonometric function
In mathematics, the trigonometric functions are functions of an angle. They are used to relate the angles of a triangle to the lengths of the sides of a triangle...
, usually sine
Sine
In mathematics, the sine function is a function of an angle. In a right triangle, sine gives the ratio of the length of the side opposite to an angle to the length of the hypotenuse.Sine is usually listed first amongst the trigonometric functions....
and tangent, common logarithm
Common logarithm
The common logarithm is the logarithm with base 10. It is also known as the decadic logarithm, named after its base. It is indicated by log10, or sometimes Log with a capital L...
(log10) (for taking the log of a value on a multiplier scale), natural logarithm
Natural logarithm
The natural logarithm is the logarithm to the base e, where e is an irrational and transcendental constant approximately equal to 2.718281828...
(ln) and exponential
Exponential function
In mathematics, the exponential function is the function ex, where e is the number such that the function ex is its own derivative. The exponential function is used to model a relationship in which a constant change in the independent variable gives the same proportional change In mathematics,...
(ex) scales. Some rules include a Pythagorean
Pythagoras
Pythagoras of Samos was an Ionian Greek philosopher, mathematician, and founder of the religious movement called Pythagoreanism. Most of the information about Pythagoras was written down centuries after he lived, so very little reliable information is known about him...
scale, to figure sides of triangles, and a scale to figure circles. Others feature scales for calculating hyperbolic functions. On linear rules, the scales and their labeling are highly standardized, with variation usually occurring only in terms of which scales are included and in what order:
A, B two-decade logarithmic scales, used for finding square roots and squares of numbers C, D single-decade logarithmic scales K three-decade logarithmic scale, used for finding cube roots and cubes of numbers CF, DF "folded" versions of the C and D scales that start from πPi' is a mathematical constant that is the ratio of any circle's circumference to its diameter. is approximately equal to 3.14. Many formulae in mathematics, science, and engineering involve , which makes it one of the most important mathematical constants... rather than from unity; these are convenient in two cases. First when the user guesses a product will be close to 10 but is not sure whether it will be slightly less or slightly more than 10, the folded scales avoid the possibility of going off the scale. Second, by making the start π rather than the square root of 10, multiplying or dividing by π (as is common in science and engineering formulas) is simplified. CI, DI, DIF "inverted" scales, running from right to left, used to simplify 1/x steps S used for finding sines and cosines on the D scale T, T1, T2 used for finding tangents and cotangents on the D and DI scales ST, SRT used for sines and tangents of small angles and degree–radian conversion L a linear scale, used along with the C and D scales for finding base-10 logarithms and powers of 10 LLn a set of log-log scales, used for finding logarithms and exponentials of numbers Ln a linear scale, used along with the C and D scales for finding natural (base e) logarithms and
EWLINE
The scales on the front and back of a Keuffel and Esser
Keuffel and Esser
The Keuffel and Esser Co. was a drafting instrument and supplies company founded in 1867 by German immigrants William J. D. Keuffel and Herman Esser.- Overview :...
(K&E) 4081-3 slide rule.
The Binary Slide Rule manufactured by Gilson in 1931 performed an addition and subtraction function limited to fraction
Fraction (mathematics)
A fraction represents a part of a whole or, more generally, any number of equal parts. When spoken in everyday English, we specify how many parts of a certain size there are, for example, one-half, five-eighths and three-quarters.A common or "vulgar" fraction, such as 1/2, 5/8, 3/4, etc., consists...
s.
#### Roots and powers
There are single-decade (C and D), double-decade (A and B), and triple-decade (K) scales. To compute , for example, locate x on the D scale and read its square on the A scale. Inverting this process allows square roots to be found, and similarly for the powers 3, 1/3, 2/3, and 3/2. Care must be taken when the base, x, is found in more than one place on its scale. For instance, there are two nines on the A scale; to find the square root of nine, use the first one; the second one gives the square root of 90.
For problems, use the LL scales. When several LL scales are present, use the one with x on it. First, align the leftmost 1 on the C scale with x on the LL scale. Then, find y on the C scale and go down to the LL scale with x on it. That scale will indicate the answer. If y is "off the scale," locate and square it using the A and B scales as described above.
#### Trigonometry
The S, T, and ST scales are used for trig functions and multiples of trig functions, for angles in degrees.
For angles from around 5.7 up to 90 degrees, sines are found by comparing the S scale with C; though on many closed-body rules the S scale relates to the A scale instead, and what follows must be adjusted appropriately. The S scale has a second set of angles (sometimes in a different color), which run in the opposite direction, and are used for cosines. Tangents are found by comparing the T scale with C for angles less than 45 degrees. for angles greater than 45 degrees the CI scale is used. Common forms such as can be read directly from x on the S scale to the result on the D scale, when the C-scale index is set at k. For angles below 5.7 degrees, sines, tangents, and radians are approximately equal, and are found on the ST or SRT (sines, radians, and tangents) scale, or simply divided by 57.3 degrees/radian
Radian is the ratio between the length of an arc and its radius. The radian is the standard unit of angular measure, used in many areas of mathematics. The unit was formerly a SI supplementary unit, but this category was abolished in 1995 and the radian is now considered a SI derived unit...
. Inverse trigonometric functions are found by reversing the process.
Many slide rules have S, T, and ST scales marked with degrees and minutes (e.g. some Keuffel and Esser models, late-model Teledyne-Post Mannheim-type rules). So-called decitrig models use decimal fractions of degrees instead.
#### Logarithms and exponentials
Base-10 logarithms and exponentials are found using the L scale, which is linear. Some slide rules have a Ln scale, which is for base e.
The Ln scale was invented by an 11th grade student, Stephen B. Cohen, in 1958. The original intent was to allow the user to select an exponent x (in the range 0 to 2.3) on the Ln scale and read ex on the C (or D) scale and ex on the CI (or DI) scale. Pickett, Inc. was given exclusive rights to the scale. Later, the inventor created a set of "marks" on the Ln scale to extend the range beyond the 2.3 limit, but Pickett never incorporated these marks on any of its slide rules.
Slide rules are not typically used for addition and subtraction, but it is nevertheless possible to do so using two different techniques.
The first method to perform addition and subtraction on the C and D (or any comparable scales) requires converting the problem into one of division. For addition, the quotient of the two variables plus one times the divisor equals their sum:
For subtraction, the quotient of the two variables minus one times the divisor equals their difference:
This method is similar to the addition/subtraction technique used for high-speed electronic circuits with the logarithmic number system
Logarithmic Number System
A logarithmic number system is an arithmetic system used for representing real numbers in computer and digital hardware, especially for digital signal processing.-Theory:...
in specialized computer applications like the Gravity Pipe
Gravity Pipe
Gravity Pipe, otherwise known as GRAPE, is a project which uses hardware acceleration to perform gravitational computations. Integrated with Beowulf-style commodity computers, the GRAPE system calculates the force of gravity that a given mass, such as a star, exerts on others...
(GRAPE) supercomputer and hidden Markov models.
The second method utilizes a sliding linear L scale available on some models. Addition and subtraction are performed by sliding the cursor left (for subtraction) or right (for addition) then returning the slide to 0 to read the result.
### Standard linear rules
The width of the slide rule is quoted in terms of the nominal width of the scales. Scales on the most common "10-inch" models are actually 25 cm, as they were made to metric standards, though some rules offer slightly extended scales to simplify manipulation when a result overflowed. Pocket rules are typically 5 inches. Models a couple of metres wide were sold to be hung in classrooms for teaching purposes.
Typically the divisions mark a scale to a precision of two significant figures
Significant figures
The significant figures of a number are those digits that carry meaning contributing to its precision. This includes all digits except:...
, and the user estimates the third figure. Some high-end slide rules have magnifier cursors that make the markings easier to see. Such cursors can effectively double the accuracy of readings, permitting a 10-inch slide rule to serve as well as a 20-inch.
Various other conveniences have been developed. Trigonometric scales are sometimes dual-labeled, in black and red, with complementary angles, the so-called "Darmstadt" style. Duplex slide rules often duplicate some of the scales on the back. Scales are often "split" to get higher accuracy.
### Circular slide rules
Circular slide rules come in two basic types, one with two cursors (left), and another with a free dish and one cursor (right). The dual cursor versions perform multiplication and division by holding a fast angle between the cursors as they are rotated around the dial. The onefold cursor version operates more like the standard slide rule through the appropriate alignment of the scales.
The basic advantage of a circular slide rule is that the widest dimension of the tool was reduced by a factor of about 3 (i.e. by π
Pi
' is a mathematical constant that is the ratio of any circle's circumference to its diameter. is approximately equal to 3.14. Many formulae in mathematics, science, and engineering involve , which makes it one of the most important mathematical constants...
). For example, a 10 cm circular would have a maximum precision approximately equal to a 31.4 cm ordinary slide rule. Circular slide rules also eliminate "off-scale" calculations, because the scales were designed to "wrap around"; they never have to be reoriented when results are near 1.0—the rule is always on scale. However, for non-cyclical non-spiral scales such as S, T, and LL's, the scale width is narrowed to make room for end margins.
Circular slide rules are mechanically more rugged and smoother-moving, but their scale alignment precision is sensitive to the centering of a central pivot; a minute 0.1 mm off-centre of the pivot can result in a 0.2mm worst case alignment error. The pivot, however, does prevent scratching of the face and cursors. The highest accuracy scales are placed on the outer rings. Rather than "split" scales, high-end circular rules use spiral scales for more complex operations like log-of-log scales. One eight-inch premium circular rule had a 50-inch spiral log-log scale.
The main disadvantages of circular slide rules are the difficulty in locating figures along a dish, and limited number of scales. Another drawback of circular slide rules is that less-important scales are closer to the center, and have lower precisions. Most students learned slide rule use on the linear slide rules, and did not find reason to switch.
One slide rule remaining in daily use around the world is the E6B
E6B
The E6B Flight Computer, or simply the "whiz wheel", is a form of circular slide rule used in aviation. They are mostly used in flight training, but many professional and even airline pilots still carry and use these flight computers...
. This is a circular slide rule first created in the 1930s for aircraft
Aircraft
An aircraft is a vehicle that is able to fly by gaining support from the air, or, in general, the atmosphere of a planet. An aircraft counters the force of gravity by using either static lift or by using the dynamic lift of an airfoil, or in a few cases the downward thrust from jet engines.Although...
pilot
Aviator
An aviator is a person who flies an aircraft. The first recorded use of the term was in 1887, as a variation of 'aviation', from the Latin avis , coined in 1863 by G. de la Landelle in Aviation Ou Navigation Aérienne...
s to help with dead reckoning
In navigation, dead reckoning is the process of calculating one's current position by using a previously determined position, or fix, and advancing that position based upon known or estimated speeds over elapsed time, and course...
. With the aid of scales printed on the frame it also helps with such miscellaneous tasks as converting time, distance, speed, and temperature values, compass
Compass
A compass is a navigational instrument that shows directions in a frame of reference that is stationary relative to the surface of the earth. The frame of reference defines the four cardinal directions – north, south, east, and west. Intermediate directions are also defined...
errors, and calculating fuel use. The so-called "prayer wheel" is still available in flight shops, and remains widely used. While GPS
Global Positioning System
The Global Positioning System is a space-based global navigation satellite system that provides location and time information in all weather, anywhere on or near the Earth, where there is an unobstructed line of sight to four or more GPS satellites...
has reduced the use of dead reckoning
In navigation, dead reckoning is the process of calculating one's current position by using a previously determined position, or fix, and advancing that position based upon known or estimated speeds over elapsed time, and course...
The basic principles of air navigation are identical to general navigation, which includes the process of planning, recording, and controlling the movement of a craft from one place to another....
, and handheld calculators have taken over many of its functions, the E6B
E6B
The E6B Flight Computer, or simply the "whiz wheel", is a form of circular slide rule used in aviation. They are mostly used in flight training, but many professional and even airline pilots still carry and use these flight computers...
remains widely used as a primary or backup device and the majority of flight schools demand that their students have some degree of proficiency in its use.
Proportion wheels are simple circular slide rules used in graphic design to broaden or slim images and photographs. Lining up the desired values on the emmer and inner wheels (which correspond to the original and desired sizes) will display the proportion as a percentage in a small window. They are not as common since the advent of computerized layout, but are still made and used.
In 1952, Swiss
Switzerland
Switzerland name of one of the Swiss cantons. ; ; ; or ), in its full name the Swiss Confederation , is a federal republic consisting of 26 cantons, with Bern as the seat of the federal authorities. The country is situated in Western Europe,Or Central Europe depending on the definition....
watch company Breitling
Breitling
Breitling is a luxury brand of Swiss watches produced by Breitling SA, a private company headquartered in Grenchen, Canton of Solothurn . The company exclusively offered Certified Chronometers in all models since 2000...
introduced a pilot's wristwatch with an integrated circular slide rule specialized for flight calculations: the Breitling Navitimer. The Navitimer circular rule, referred to by Breitling as a "navigation computer", featured airspeed
Airspeed
Airspeed is the speed of an aircraft relative to the air. Among the common conventions for qualifying airspeed are: indicated airspeed , calibrated airspeed , true airspeed , equivalent airspeed and density airspeed....
, rate
Rate of climb
In aeronautics, the rate of climb is an aircraft's vertical speed - the rate of change in altitude. In most ICAO member countries , this is usually expressed in feet per minute and can be abbreviated as ft/min. Elsewhere, it is commonly expressed in metres per second, abbreviated as m/s...
/time of climb/descent, flight time, distance, and fuel consumption functions, as well as kilometer—nautical mile
Nautical mile
The nautical mile is a unit of length that is about one minute of arc of latitude along any meridian, but is approximately one minute of arc of longitude only at the equator...
and gallon
Gallon
The gallon is a measure of volume. Historically it has had many different definitions, but there are three definitions in current use: the imperial gallon which is used in the United Kingdom and semi-officially within Canada, the United States liquid gallon and the lesser used United States dry...
liter
Litre
pic|200px|right|thumb|One litre is equivalent to this cubeEach side is 10 cm1 litre water = 1 kilogram water The litre is a metric system unit of volume equal to 1 cubic decimetre , to 1,000 cubic centimetres , and to 1/1,000 cubic metre...
fuel amount conversion functions.
### Cylindrical slide rules
There are two main types of cylindrical slide rules: those with helical scales such as the Fuller, the Otis King
Otis King
Otis Carter Formby King was a grocer and engineer in London who invented and produced a cylindrical slide rule with helical scales, primarily for business uses initially. The product was named Otis King's Patent Calculator, and was manufactured and sold by Carbic Ltd...
and the Bygrave slide rule
Bygrave slide rule
The Bygrave slide rule is a slide rule named for its inventor, Captain L. G. Bygrave of the RAF. It was used in celestial navigation, primarily in aviation. Officially, it was called the A. M. L. Position Line Slide Rule The Bygrave slide rule is a slide rule named for its inventor, Captain L. G....
, and those with bars, such as the Thacher and some Loga models. In either case, the advantage is a much longer scale, and hence potentially higher accuracy, than a straight or circular rule.
### Materials
Traditionally slide rules were made out of hard wood such as mahogany
Mahogany
The name mahogany is used when referring to numerous varieties of dark-colored hardwood. It is a native American word originally used for the wood of the species Swietenia mahagoni, known as West Indian or Cuban mahogany....
or boxwood with cursors of glass and metal. At least one high precision instrument was made of steel
Steel
Steel is an alloy that consists mostly of iron and has a carbon content between 0.2% and 2.1% by weight, depending on the grade. Carbon is the most common alloying material for iron, but various other alloying elements are used, such as manganese, chromium, vanadium, and tungsten...
.
In 1895, a Japanese firm, Hemmi, started to make slide rules from bamboo
Bamboo
Bamboo is a group of perennial evergreens in the true grass family Poaceae, subfamily Bambusoideae, tribe Bambuseae. Giant bamboos are the largest members of the grass family....
, which had the advantages of being dimensionally stable, strong and naturally self-lubricating. These bamboo slide rules were introduced in Sweden in September, 1933, and probably only a little earlier in Germany. Scales were made of celluloid
Celluloid
Celluloid is the name of a class of compounds created from nitrocellulose and camphor, plus dyes and other agents. Generally regarded to be the first thermoplastic, it was first created as Parkesine in 1862 and as Xylonite in 1869, before being registered as Celluloid in 1870. Celluloid is...
or plastic. Later slide rules were made of plastic, or aluminium
Aluminium
Aluminium or aluminum is a silvery white member of the boron group of chemical elements. It has the symbol Al, and its atomic number is 13. It is not soluble in water under normal circumstances....
painted with plastic. Later cursors were acrylics
Acryl group
In organic chemistry, the acryloyl group is the functional group with structure H2C=CH–C–; it is the acyl group derived from acrylic acid. The preferred IUPAC name for the group is prop-2-enoyl, and it is also known as acrylyl or simply acryl...
or polycarbonate
Polycarbonate
PolycarbonatePhysical PropertiesDensity 1.20–1.22 g/cm3Abbe number 34.0Refractive index 1.584–1.586FlammabilityV0-V2Limiting oxygen index25–27%Water absorption – Equilibrium0.16–0.35%Water absorption – over 24 hours0.1%...
s sliding on Teflon
Polytetrafluoroethylene
Polytetrafluoroethylene is a synthetic fluoropolymer of tetrafluoroethylene that finds numerous applications. PTFE is most well known by the DuPont brand name Teflon....
bearings.
All premium slide rules had numbers and scales engraved, and then filled with paint or other resin
Resin
Resin in the most specific use of the term is a hydrocarbon secretion of many plants, particularly coniferous trees. Resins are valued for their chemical properties and associated uses, such as the production of varnishes, adhesives, and food glazing agents; as an important source of raw materials...
. Painted or imprinted slide rules were viewed as inferior because the markings could wear off. Nevertheless, Pickett, probably America's most successful slide rule company, made all printed scales. Premium slide rules included clever catches so the rule would not fall apart by accident, and bumpers to protect the scales and cursor from rubbing on tabletops. The recommended cleaning method for engraved markings is to scrub lightly with steel-wool. For painted slide rules, and the faint of heart, use diluted commercial window-cleaning fluid and a soft cloth.
## History
The slide rule was invented around 1620–1630, shortly after John Napier
John Napier
John Napier of Merchiston – also signed as Neper, Nepair – named Marvellous Merchiston, was a Scottish mathematician, physicist, astronomer & astrologer, and also the 8th Laird of Merchistoun. He was the son of Sir Archibald Napier of Merchiston. John Napier is most renowned as the discoverer...
's publication of the concept of the logarithm
Logarithm
The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number. For example, the logarithm of 1000 to base 10 is 3, because 1000 is 10 to the power 3: More generally, if x = by, then y is the logarithm of x to base b, and is written...
. Edmund Gunter
Edmund Gunter
Edmund Gunter , English mathematician, of Welsh descent, was born in Hertfordshire in 1581.He was educated at Westminster School, and in 1599 was elected a student of Christ Church, Oxford. He took orders, became a preacher in 1614, and in 1615 proceeded to the degree of bachelor in divinity...
of Oxford developed a calculating device with a single logarithmic scale, which, with additional measuring tools, could be used to multiply and divide. The first description of this scale was published in Paris in 1624 by Edmund Wingate
Edmund Wingate
Edmund Wingate was an English mathematical and legal writer, one of the first to publish in the 1620s on the principle of the slide rule, and later the author of some popular expository works...
(c.1593–1656), an English mathematician, in a book entitled L'usage de la reigle de proportion en l'arithmetique & geometrie. The book contains a double scale on one side of which is a logarithmic scale and on the other a tabular scale. In 1630, William Oughtred
William Oughtred
William Oughtred was an English mathematician.After John Napier invented logarithms, and Edmund Gunter created the logarithmic scales upon which slide rules are based, it was Oughtred who first used two such scales sliding by one another to perform direct multiplication and division; and he is...
of Cambridge invented a circular slide rule, and in 1632 he combined two Gunter rules, held together with the hands, to make a device that is recognizably the modern slide rule. Like his contemporary at Cambridge, Isaac Newton
Isaac Newton
Sir Isaac Newton PRS was an English physicist, mathematician, astronomer, natural philosopher, alchemist, and theologian, who has been "considered by many to be the greatest and most influential scientist who ever lived."...
, Oughtred taught his ideas privately to his students, but delayed in publishing them, and like Newton, he became involved in a vitriolic controversy over priority, with his one-time student Richard Delamain and the prior claims of Wingate. Oughtred's ideas were only made public in publications of his student William Forster in 1632 and 1653.
In 1677, Henry Coggeshall created a two-foot folding rule for timber measure, called the Coggeshall slide rule
Coggeshall slide rule
In measurement, the Coggeshall slide rule, also called a carpenter's slide rule, was a slide rule designed by Henry Coggeshall in 1677 to facilitate measuring the dimensions, superficies, and solidity of timber. With his original design and later improvements, Coggeshall's slide rule brought the...
. His design and uses for the tool gave the slide rule purpose outside of mathematical inquiry.
In 1722, Warner introduced the two- and three-decade scales, and in 1755 Everard included an inverted scale; a slide rule containing all of these scales is usually known as a "polyphase" rule.
In 1815, Peter Mark Roget invented the log log slide rule, which included a scale displaying the logarithm of the logarithm. This allowed the user to directly perform calculations involving roots and exponents. This was especially useful for fractional powers.
In 1821, Nathaniel Bowditch
Nathaniel Bowditch
Nathaniel Bowditch was an early American mathematician remembered for his work on ocean navigation. He is often credited as the founder of modern maritime navigation; his book The New American Practical Navigator, first published in 1802, is still carried on board every commissioned U.S...
, in the American Practical Navigator, described the use of a "sliding rule" which contained scales trigonometric functions on the fixed part and a line of log-sines and log-tans on the slider. This device was used to solve navigation problems.
### Modern form
The more modern form was created in 1859 by French artillery lieutenant Amédée Mannheim, "who was fortunate in having his rule made by a firm of national reputation and in having it adopted by the French Artillery." It was around that time, as engineering
Engineering
Engineering is the discipline, art, skill and profession of acquiring and applying scientific, mathematical, economic, social, and practical knowledge, in order to design and build structures, machines, devices, systems, materials and processes that safely realize improvements to the lives of...
became a recognized professional activity, that slide rules came into wide use in Europe. They did not become common in the United States until 1881, when Edwin Thacher introduced a cylindrical rule there. The duplex rule was invented by William Cox in 1891, and was produced by Keuffel and Esser Co. of New York.
Astronomical work also required fine computations, and in the 19th century Germany a steel slide rule about 2 meters long was used at one observatory. It had a microscope
Microscope
A microscope is an instrument used to see objects that are too small for the naked eye. The science of investigating small objects using such an instrument is called microscopy...
attached, giving it accuracy to six decimal places.
Throughout the 1950s and 1960s the slide rule was the symbol of the engineer's profession (in the same way that the stethoscope
Stethoscope
The stethoscope is an acoustic medical device for auscultation, or listening to the internal sounds of an animal body. It is often used to listen to lung and heart sounds. It is also used to listen to intestines and blood flow in arteries and veins...
symbolizes the medical profession). German rocket scientist Wernher von Braun
Wernher von Braun
Wernher Magnus Maximilian, Freiherr von Braun was a German rocket scientist, aerospace engineer, space architect, and one of the leading figures in the development of rocket technology in Nazi Germany during World War II and in the United States after that.A former member of the Nazi party,...
brought two 1930s vintage Nestler slide rules with him when he moved to the U.S. after World War II to work on the American space program. Throughout his life he never used any other pocket calculating devices; slide rules served him perfectly well for making quick estimates of rocket design parameters and other figures. Aluminium Pickett
Pickett
Pickett is a surname of English origin, and may refer to:* Albert J. Pickett , American historian* Allistair Pickett , Australian rules footballer* Bill Pickett , American cowboy and rodeo performer...
-brand slide rules were carried on five Apollo
Project Apollo
The Apollo program was the spaceflight effort carried out by the United States' National Aeronautics and Space Administration , that landed the first humans on Earth's Moon. Conceived during the Presidency of Dwight D. Eisenhower, Apollo began in earnest after President John F...
space missions, including to the moon, according to advertising on Pickett's N600 slide rule boxes.
Some engineering students and engineers carried ten-inch slide rules in belt holsters, and even into the mid 1970s this was a common sight on campuses. Students also might keep a ten- or twenty-inch rule for precision work at home or the office while carrying a five-inch pocket slide rule around with them.
In 2004, education researchers David B. Sher and Dean C. Nataro conceived a new type of slide rule based on prosthaphaeresis
Prosthaphaeresis
Prosthaphaeresis was an algorithm used in the late 16th century and early 17th century for approximate multiplication and division using formulas from trigonometry. For the 25 years preceding the invention of the logarithm in 1614, it was the only known generally-applicable way of approximating...
, an algorithm for rapidly computing products that predates logarithms. There has been little practical interest in constructing one beyond the initial prototype, however.
### Specialized calculators
Slide rules have often been specialized to varying degrees for their field of use, such as excise, proof calculation, engineering, navigation, etc., but some slide rules are extremely specialized for very narrow applications. For example, the John Rabone & Sons 1892 catalog lists a "Measuring Tape and Cattle Gauge", a device to estimate the weight of a cow from its measurements.
There were many specialized slide rules for photographic
Photography
Photography is the art, science and practice of creating durable images by recording light or other electromagnetic radiation, either electronically by means of an image sensor or chemically by means of a light-sensitive material such as photographic film...
applications; for example, the actinograph
Actinograph
An actinograph is an instrument for measuring or estimating the amount of light available, in terms of its ability to expose photographic film...
of Hurter and Driffield
Hurter and Driffield
Ferdinand Hurter and Vero Charles Driffield were nineteenth-century photographic scientists who brought quantitative scientific practice to photography through the methods of sensitometry and densitometry....
was a two-slide boxwood, brass, and cardboard device for estimating exposure from time of day, time of year, and latitude.
Specialized slide rules were invented for various forms of engineering, business and banking. These often had common calculations directly expressed as special scales, for example loan calculations, optimal purchase quantities, or particular engineering equations. For example, the Fisher Controls company distributed a customized slide rule adapted to solving the equations used for selecting the proper size of industrial flow control valve
Valve
A valve is a device that regulates, directs or controls the flow of a fluid by opening, closing, or partially obstructing various passageways. Valves are technically pipe fittings, but are usually discussed as a separate category...
s.
In World War II
World War II
World War II, or the Second World War , was a global conflict lasting from 1939 to 1945, involving most of the world's nations—including all of the great powers—eventually forming two opposing military alliances: the Allies and the Axis...
, bombardiers and navigators who required quick calculations often used specialized slide rules. One office of the U.S. Navy actually designed a generic slide rule "chassis" with an aluminium body and plastic cursor into which celluloid cards (printed on both sides) could be placed for special calculations. The process was invented to calculate range, fuel use and altitude for aircraft
Aircraft
An aircraft is a vehicle that is able to fly by gaining support from the air, or, in general, the atmosphere of a planet. An aircraft counters the force of gravity by using either static lift or by using the dynamic lift of an airfoil, or in a few cases the downward thrust from jet engines.Although...
, and then adapted to many other purposes.
### Decline
The importance of the slide rule began to diminish as electronic computers, a new but very scarce resource in the 1950s, became widely available to technical workers during the 1960s. The introduction of Fortran
Fortran
Fortran is a general-purpose, procedural, imperative programming language that is especially suited to numeric computation and scientific computing...
in 1957 made computers practical for solving modest size mathematical problems. IBM
IBM
International Business Machines Corporation or IBM is an American multinational technology and consulting corporation headquartered in Armonk, New York, United States. IBM manufactures and sells computer hardware and software, and it offers infrastructure, hosting and consulting services in areas...
introduced a series of more affordable computers, the IBM 650
IBM 650
The IBM 650 was one of IBM’s early computers, and the world’s first mass-produced computer. It was announced in 1953, and over 2000 systems were produced between the first shipment in 1954 and its final manufacture in 1962...
(1954), IBM 1620
IBM 1620
The IBM 1620 was announced by IBM on October 21, 1959, and marketed as an inexpensive "scientific computer". After a total production of about two thousand machines, it was withdrawn on November 19, 1970...
(1959), IBM 1130
IBM 1130
The IBM 1130 Computing System was introduced in 1965. It was IBM's least-expensive computer to date, and was aimed at price-sensitive, computing-intensive technical markets like education and engineering. It succeeded the IBM 1620 in that market segment. The IBM 1800 was a process control variant...
(1965) addressed to the science and engineering market. The BASIC programming language (1964) made it easy for students to use computers. The DEC PDP-8
PDP-8
The 12-bit PDP-8 was the first successful commercial minicomputer, produced by Digital Equipment Corporation in the 1960s. DEC introduced it on 22 March 1965, and sold more than 50,000 systems, the most of any computer up to that date. It was the first widely sold computer in the DEC PDP series of...
minicomputer was introduced in 1965.
Computers also changed the nature of calculation. With slide rules, there was a great emphasis on working the algebra to get expressions into the most computable form. Users of slide rules would simply approximate or drop small terms to simplify the calculation. Fortran allowed complicated formulas to be typed in from textbook
Textbook
A textbook or coursebook is a manual of instruction in any branch of study. Textbooks are produced according to the demands of educational institutions...
s without the effort of reformulation. Numerical integration
Numerical integration
In numerical analysis, numerical integration constitutes a broad family of algorithms for calculating the numerical value of a definite integral, and by extension, the term is also sometimes used to describe the numerical solution of differential equations. This article focuses on calculation of...
was often easier than trying to find closed-form
Closed-form expression
In mathematics, an expression is said to be a closed-form expression if it can be expressed analytically in terms of a bounded number of certain "well-known" functions...
solutions for difficult problems. The young engineer asking for computer time to solve a problem that could have been done by a few swipes on the slide rule became a humorous cliché.
The availability of mainframe computing did not however significantly affect the ubiquitous use of the slide rule until cheap hand held electronic calculator
Calculator
An electronic calculator is a small, portable, usually inexpensive electronic device used to perform the basic operations of arithmetic. Modern calculators are more portable than most computers, though most PDAs are comparable in size to handheld calculators.The first solid-state electronic...
s for scientific and engineering purposes became available in the mid 1970s at which point they rapidly fell out of use. The first included the Wang Laboratories
Wang Laboratories
Wang Laboratories was a computer company founded in 1951 by Dr. An Wang and Dr. G. Y. Chu. The company was successively headquartered in Cambridge , Tewksbury , and finally in Lowell, Massachusetts . At its peak in the 1980s, Wang Laboratories had annual revenues of $3 billion and employed over... LOCI-2, introduced in 1965, which used logarithm Logarithm The logarithm of a number is the exponent by which another fixed value, the base, has to be raised to produce that number. For example, the logarithm of 1000 to base 10 is 3, because 1000 is 10 to the power 3: More generally, if x = by, then y is the logarithm of x to base b, and is written... s for multiplication and division and the Hewlett-Packard Hewlett-Packard Hewlett-Packard Company or HP is an American multinational information technology corporation headquartered in Palo Alto, California, USA that provides products, technologies, softwares, solutions and services to consumers, small- and medium-sized businesses and large enterprises, including... HP-9100, introduced in 1968. The HP-9100 had trigonometric function Trigonometric function In mathematics, the trigonometric functions are functions of an angle. They are used to relate the angles of a triangle to the lengths of the sides of a triangle... s (sin, cos, tan) in addition to exponentials and logarithms. It used the CORDIC CORDIC CORDIC is a simple and efficient algorithm to calculate hyperbolic and trigonometric functions... (coordinate rotation digital computer) algorithm, which allows for calculation of trigonometric functions using only shift and add operations. This method facilitated the development of ever smaller scientific calculators. The era of the slide rule ended with the launch of pocket-sized scientific calculators, of which the 1972 Hewlett-Packard HP-35 HP-35 The HP-35 was Hewlett-Packard's first pocket calculator and the world's first scientific pocket calculator . Like some of HP's desktop calculators it used reverse Polish notation. Introduced at US$395, the HP-35 was available from 1972 to 1975.Market studies at the time had shown no market for...
was the first. Introduced at US$United States dollar The United States dollar , also referred to as the American dollar, is the official currency of the United States of America. It is divided into 100 smaller units called cents or pennies.... 395, even this was considered expensive for most students, but by 1975 basic four-function electronic calculators could be purchased for less than$50. By 1976 the TI-30
TI-30
The TI-30 was a scientific calculator manufactured by Texas Instruments, the first model of which was introduced in 1976. While the original TI-30 left production in 1983 after several design revisions, TI maintains the TI-30 designation as a branding for its low and mid-range scientific...
offered a scientific calculator for less than \$25. After this time, the market for slide rules dwindled quickly as small scientific calculators became affordable.
## Compared to electronic digital calculators
Compared to the portable electronic digital calculators that were introduced in the early 1970s, slide rules had various advantages and disadvantages.
• The spatial, manual operation of slide rules cultivates in the user an intuition for numerical relationships and scale that people who have used only digital calculators often lack. Since users must explicitly note the order of magnitude
Order of magnitude
An order of magnitude is the class of scale or magnitude of any amount, where each class contains values of a fixed ratio to the class preceding it. In its most common usage, the amount being scaled is 10 and the scale is the exponent being applied to this amount...
at each step in order to interpret the results, they are less likely to make extreme calculation errors; users are forced to use common sense and an understanding of the subject as they calculate. Since order of magnitude gets the greatest prominence when using a slide rule, and precision is limited only to the few digits that are normally useful, users are less likely to make errors of false precision
False precision
False precision occurs when numerical data are presented in a manner that implies better precision than is actually the case; since precision is a limit to accuracy, this often leads to overconfidence in the accuracy as well.In science and engineering, convention dictates that...
.
• When performing a sequence of multiplications or divisions by the same number, the answer can often be determined by merely glancing at the slide rule without any manipulation. This can be especially useful when calculating percentages (e.g. for test scores) or when comparing prices (e.g. in dollars per kilogram). Multiple speed-time-distance calculations can be performed hands-free at a glance with a slide rule.
• Other useful constants such as pounds to kilograms can be easily marked on the rule and used directly in calculations.
• A slide rule does not depend on electricity
Electricity
Electricity is a general term encompassing a variety of phenomena resulting from the presence and flow of electric charge. These include many easily recognizable phenomena, such as lightning, static electricity, and the flow of electrical current in an electrical wire...
or batteries.
• The principle of operation of a slide rule can be demonstrated with a pair of hand-made paper scales.
• A slide rule displays all the terms of a calculation along with the result. This eliminates uncertainty about what calculation was actually performed.
• A slide rule is physically more durable than an electronic calculator and is impervious to moisture and immersion in water.
For many of these reasons slide rules are still commonly used in aviation, particularly for smaller planes. They are only being replaced by integrated, special purpose and expensive flight computers, and not general-purpose calculators. Many sailors keep them as backup systems for navigation against electric failures or running out of batteries on long blue-water legs.
• Most people find slide rules difficult to learn and use. Even during their heyday, they never caught on with the general public.
• Doing a calculation on a slide rule tends to be slower than on a calculator. This led engineers to take mathematical shortcuts favoring operations that were easy on a slide rule, creating inaccuracies and mistakes.
• A slide rule requires the user to mentally calculate the order of magnitude
Order of magnitude
An order of magnitude is the class of scale or magnitude of any amount, where each class contains values of a fixed ratio to the class preceding it. In its most common usage, the amount being scaled is 10 and the scale is the exponent being applied to this amount...
of the results. For example, 1.5 × 30 (which equals 45) will show the same result as 1,500,000 × 0.03 (which equals 45,000). This forces the user to keep track of magnitude in short-term memory
Short-term memory
Short-term memory is the capacity for holding a small amount of information in mind in an active, readily available state for a short period of time. The duration of short-term memory is believed to be in the order of seconds. A commonly cited capacity is 7 ± 2 elements...
(which is error-prone), keep notes (which is cumbersome) or reason about it in every step (which distracts from the other calculation requirements).
• Addition and subtraction are not well-supported operations on slide rules.
• The typical precision of a slide rule is about three significant digits, compared to many digits on digital calculators (however, just 2 significant digits is adequate for many engineering calculations).
• Errors may arise from mechanical imprecision in slide rules that are warped by heat or use or that were poorly constructed.
## Finding and collecting slide rules
There are still people who prefer a slide rule over an electronic calculator as a practical computing device. Many others keep their old slide rules out of a sense of nostalgia, or collect slide rules as a hobby.
A popular collectible model is the Keuffel & Esser Deci-Lon, a premium scientific and engineering slide rule available both in a ten-inch "regular" (Deci-Lon 10) and a five-inch "pocket" (Deci-Lon 5) variant. Another prized American model is the eight-inch Scientific Instruments circular rule. Of European rules, Faber-Castell
Faber-Castell
Faber-Castell is one of the world's largest manufacturers of pens, pencils, other office supplies and art supplies, as well as high-end writing instruments and luxury leather goods...
's high-end models are the most popular among collectors.
Although there is a large supply of slide rules circulating on the market, specimens in good condition tend to be expensive. Many rules found for sale on online auction sites are damaged or have missing parts, and the seller may not know enough to supply the relevant information. Replacement parts are scarce, and therefore expensive, and are generally only available for separate purchase on individual collectors' web sites. The Keuffel and Esser rules from the period up to about 1950 are particularly problematic, because the end-pieces on the cursors, made of celluloid
Celluloid
Celluloid is the name of a class of compounds created from nitrocellulose and camphor, plus dyes and other agents. Generally regarded to be the first thermoplastic, it was first created as Parkesine in 1862 and as Xylonite in 1869, before being registered as Celluloid in 1870. Celluloid is...
, tend to break down chemically over time.
There are still a handful of sources for brand new slide rules. The Concise Company of Tokyo, which began as a manufacturer of circular slide rules in July 1954, continues to make and sell them today. And in September 2009, on-line retailer ThinkGeek
ThinkGeek
ThinkGeek is an American online retailer that caters to computer enthusiasts and other "geeky" social groups. Their merchandise includes clothing, electronic and scientific gadgets, unusual computer peripherals, office toys, pet toys, child toys, and caffeinated drinks and candy...
introduced its own brand of straight slide rules, which they describe as "faithful replica[s]" that are "individually hand tooled" due to a stated lack of any existing manufacturers. The E6B
E6B
The E6B Flight Computer, or simply the "whiz wheel", is a form of circular slide rule used in aviation. They are mostly used in flight training, but many professional and even airline pilots still carry and use these flight computers...
circular slide rule used by pilots has been in continuous production and remains available in a variety of models. Proportion wheels are still used in graphic design. | 13,304 | 61,752 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2022-21 | longest | en | 0.93761 |
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Not now
# Java Program for Range LCM Queries
• Last Updated : 03 Jan, 2022
Given an array of integers, evaluate queries of the form LCM(l, r). There might be many queries, hence evaluate the queries efficiently.
```LCM (l, r) denotes the LCM of array elements
that lie between the index l and r
(inclusive of both indices)
Mathematically,
LCM(l, r) = LCM(arr[l], arr[l+1] , ......... ,
arr[r-1], arr[r])```
Examples:
```Inputs : Array = {5, 7, 5, 2, 10, 12 ,11, 17, 14, 1, 44}
Queries: LCM(2, 5), LCM(5, 10), LCM(0, 10)
Outputs: 60 15708 78540
Explanation : In the first query LCM(5, 2, 10, 12) = 60,
similarly in other queries.```
A naive solution would be to traverse the array for every query and calculate the answer by using,
LCM(a, b) = (a*b) / GCD(a,b)
However as the number of queries can be large, this solution would be impractical.
An efficient solution would be to use segment tree. Recall that in this case, where no update is required, we can build the tree once and can use that repeatedly to answer the queries. Each node in the tree should store the LCM value for that particular segment and we can use the same formula as above to combine the segments. Hence we can answer each query efficiently!
Below is a solution for the same.
## Java
`// LCM of given range queries ` `// using Segment Tree ` ` ` `class` `GFG ` `{ ` ` ` ` ``static` `final` `int` `MAX = ``1000``; ` ` ` ` ``// allocate space for tree ` ` ``static` `int` `tree[] = ``new` `int``[``4` `* MAX]; ` ` ` ` ``// declaring the array globally ` ` ``static` `int` `arr[] = ``new` `int``[MAX]; ` ` ` ` ``// Function to return gcd of a and b ` ` ``static` `int` `gcd(``int` `a, ``int` `b) { ` ` ``if` `(a == ``0``) { ` ` ``return` `b; ` ` ``} ` ` ``return` `gcd(b % a, a); ` ` ``} ` ` ` ` ``// utility function to find lcm ` ` ``static` `int` `lcm(``int` `a, ``int` `b) ` ` ``{ ` ` ``return` `a * b / gcd(a, b); ` ` ``} ` ` ` ` ``// Function to build the segment tree ` ` ``// Node starts beginning index ` ` ``// of current subtree. start and end ` ` ``// are indexes in arr[] which is global ` ` ``static` `void` `build(``int` `node, ``int` `start, ``int` `end) ` ` ``{ ` ` ` ` ``// If there is only one element ` ` ``// in current subarray ` ` ``if` `(start == end) ` ` ``{ ` ` ``tree[node] = arr[start]; ` ` ``return``; ` ` ``} ` ` ` ` ``int` `mid = (start + end) / ``2``; ` ` ` ` ``// build left and right segments ` ` ``build(``2` `* node, start, mid); ` ` ``build(``2` `* node + ``1``, mid + ``1``, end); ` ` ` ` ``// build the parent ` ` ``int` `left_lcm = tree[``2` `* node]; ` ` ``int` `right_lcm = tree[``2` `* node + ``1``]; ` ` ` ` ``tree[node] = lcm(left_lcm, right_lcm); ` ` ``} ` ` ` ` ``// Function to make queries for ` ` ``// array range )l, r). Node is index ` ` ``// of root of current segment in segment ` ` ``// tree (Note that indexes in segment ` ` ``// tree begin with 1 for simplicity). ` ` ``// start and end are indexes of subarray ` ` ``// covered by root of current segment. ` ` ``static` `int` `query(``int` `node, ``int` `start, ` ` ``int` `end, ``int` `l, ``int` `r) ` ` ``{ ` ` ` ` ``// Completely outside the segment, returning ` ` ``// 1 will not affect the lcm; ` ` ``if` `(end < l || start > r) ` ` ``{ ` ` ``return` `1``; ` ` ``} ` ` ` ` ``// completely inside the segment ` ` ``if` `(l <= start && r >= end) ` ` ``{ ` ` ``return` `tree[node]; ` ` ``} ` ` ` ` ``// partially inside ` ` ``int` `mid = (start + end) / ``2``; ` ` ``int` `left_lcm = query(``2` `* node, start, mid, l, r); ` ` ``int` `right_lcm = query(``2` `* node + ``1``, mid + ``1``, end, l, r); ` ` ``return` `lcm(left_lcm, right_lcm); ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `main(String[] args) ` ` ``{ ` ` ` ` ``//initialize the array ` ` ``arr[``0``] = ``5``; ` ` ``arr[``1``] = ``7``; ` ` ``arr[``2``] = ``5``; ` ` ``arr[``3``] = ``2``; ` ` ``arr[``4``] = ``10``; ` ` ``arr[``5``] = ``12``; ` ` ``arr[``6``] = ``11``; ` ` ``arr[``7``] = ``17``; ` ` ``arr[``8``] = ``14``; ` ` ``arr[``9``] = ``1``; ` ` ``arr[``10``] = ``44``; ` ` ` ` ``// build the segment tree ` ` ``build(``1``, ``0``, ``10``); ` ` ` ` ``// Now we can answer each query efficiently ` ` ``// Print LCM of (2, 5) ` ` ``System.out.println(query(``1``, ``0``, ``10``, ``2``, ``5``)); ` ` ` ` ``// Print LCM of (5, 10) ` ` ``System.out.println(query(``1``, ``0``, ``10``, ``5``, ``10``)); ` ` ` ` ``// Print LCM of (0, 10) ` ` ``System.out.println(query(``1``, ``0``, ``10``, ``0``, ``10``)); ` ` ` ` ``} ` `} ` ` ` `// This code is contributed by 29AjayKumar `
Output:
```60
15708
78540```
Please refer complete article on Range LCM Queries for more details!
My Personal Notes arrow_drop_up
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• Mar 26th 2008, 11:17 AM
Del
Parametric Equation to Cartesian Equation
Eliminate the parameter t to find a Cartesian equation for:
x = t^2
y = 8 + 4t
[x = Ay^2 + By + C]
• Mar 26th 2008, 11:19 AM
topsquark
Quote:
Originally Posted by Del
Eliminate the parameter t to find a Cartesian equation for:
x = t^2
y = 8 + 4t
[x = Ay^2 + By + C]
Solve the y equation for t:
$\displaystyle y = 8 + 4t$
$\displaystyle 4t = y - 8$
$\displaystyle t = \frac{y - 8}{4}$
Now insert this into the x equation:
$\displaystyle x = \left ( \frac{y - 8}{4} \right ) ^2$
Now just FOIL it out.
-Dan | 222 | 634 | {"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} | 3.765625 | 4 | CC-MAIN-2018-26 | latest | en | 0.693309 |
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1. Even though independent gasoline stations have been having a difficult time, Susan Solomon has been thinking about starting her own independent gas station. Susan’s problem is to decide how large her station should be. The annual returns will depend on both the size of the station and a number of marketing factors related to oil industry and demand for gasoline. After careful analysis, Susan developed the following table:
Size of Gasoline Station Good Market (\$) Fair Market (\$) Poor Market (\$) Small \$70,000 \$30,000 -\$30,000 Medium \$110,000 \$50,000 -\$40,000 Large \$170,000 \$70,000 -\$50,000
1. Develop a decision table for this decision.
2. What is the Maximax decision?
3. What is the Maximin decision?
4. What is the criterion of realism decision? Use α = 0.6.
5. Develop an Opportunity Loss Table
6. What is the Minimax Regret Decision?
2. Data collected on the yearly demand for 50-pound bags of fertilizer at Sunshine Garden Supply are shown in the following table.
Year Demand for Fertilizer (1,000s of Bags) 1 4 2 7 3 5 4 5 5 10 6 7 7 8 8 9 9 11 10 14 11 15
Use Excel QM to:
1. Develop a three-year moving average to forecast sales in year 12.
2. Develop a 3-year weighted average to predict demand in year 12, in which sales in the most recent year is given a weight of 2 and sales in the two years prior to that are each given a weight of 1.
3. Develop a regression/trend line to estimate the demand for fertilizer in year 12.
4. Based on the three forecasts you have created, which forecast is the most accurate?
3. Kaplan College has decided to “wire” its campus. The first stage in this effort is to install the “backbone,” i.e., to connect all the buildings. The table below gives the distances between the various buildings on campus in hundreds of feet.
Distances in Hundreds of Feet From To Building 1 Building 2 Building 3 Building 4 Building 5 Building 6 Building 1 3 7 5 5 4 Building 2 5 2 6 6 Building 3 5 4 4 Building 4 5 3 Building 5 4 Building 6
1. How should the buildings be connected to minimize the total length of cable?
building 2 – 3
Building 3 – 5
Building 4 – 2
Building 5 – 4
Building 6 – 3
2. What length of cable is required?
Total Amount of Wiring Needed – 300+500+200+400+300 = 1700ft of wiring needed
4. The following represents the distances in miles from a warehouse (node 1) to various cities in Montana. The major outlet store is located at node 7.
From Node To Node Distance 1 2 40 1 4 100 2 3 20 2 4 30 3 4 60 3 5 40 3 6 20 4 5 70 4 7 50 5 6 50 5 7 80 6 7 40
1. Find the shortest route and distance from Node 1 to Node 7.
Node 1 to Node 2 – 40
Node 2 to Node 3 – 20
Node 3 to Node 6 – 20
Node 6 to Node 7 – 40
Total 120 miles
5. The network of a city sewer system and their capacities are shown below. Remember that the arc has both capacity and reverse capacity. For example, row 1 is the flow from node 1 to node 2 and row 2 is the reverse flow from node 2 to node 1. There are eight branches in this network.
From Node To Node Fluid Flow 1 2 150 2 1 100 1 3 0 3 1 150 1 4 300 4 1 300 1 5 150 5 1 100 2 4 300 4 2 200 3 4 250 4 3 300 3 5 300 5 3 250 4 5 100 5 4 0
Determine the maximum flow (in hundreds of gallons of water per minute) from node 1 to node 5
6. The governor of Michigan believes that the state can improve the state’s crime rate if the state can reduce the college debt carried by its citizens and if they can increase the percent of the population covered by health insurance.
1. Using the States Data Set and PHStat, create the multiple regression prediction equation.
2. Predict the crime rate for Michigan if the college debt were \$25,000 and the percent not covered by insurance was 10?
7. A concessionaire for the local ballpark has developed a table of conditional values for the various alternatives (stocking decisions) and states of nature (size of crowd).
Stocking Decision Large Crowd (\$) Average Crowd (\$) Small Crowd (\$) Large Inventory \$22,000 \$12,000 -\$2,000 Average Inventory \$15,000 \$12,000 \$6,000 Small Inventory \$9,000 \$6,000 \$5,000
If the probabilities associated with the states of nature are 0.30 for a large crowd, 0.50 for an average crowd, and 0.20 for a small crowd, determine:
1. The alternative that provides the greatest expected monetary value (EMV).
2. The expected value of perfect information (EVPI). | 1,292 | 4,577 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2021-17 | latest | en | 0.930569 |
https://en.wikipedia.org/wiki/Single_precision_floating-point_format | 1,717,077,838,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00269.warc.gz | 196,209,325 | 35,496 | Single-precision floating-point format
Single-precision floating-point format (sometimes called FP32 or float32) is a computer number format, usually occupying 32 bits in computer memory; it represents a wide dynamic range of numeric values by using a floating radix point.
A floating-point variable can represent a wider range of numbers than a fixed-point variable of the same bit width at the cost of precision. A signed 32-bit integer variable has a maximum value of 231 − 1 = 2,147,483,647, whereas an IEEE 754 32-bit base-2 floating-point variable has a maximum value of (2 − 2−23) × 2127 ≈ 3.4028235 × 1038. All integers with 7 or fewer decimal digits, and any 2n for a whole number −149 ≤ n ≤ 127, can be converted exactly into an IEEE 754 single-precision floating-point value.
In the IEEE 754 standard, the 32-bit base-2 format is officially referred to as binary32; it was called single in IEEE 754-1985. IEEE 754 specifies additional floating-point types, such as 64-bit base-2 double precision and, more recently, base-10 representations.
One of the first programming languages to provide single- and double-precision floating-point data types was Fortran. Before the widespread adoption of IEEE 754-1985, the representation and properties of floating-point data types depended on the computer manufacturer and computer model, and upon decisions made by programming-language designers. E.g., GW-BASIC's single-precision data type was the 32-bit MBF floating-point format.
Single precision is termed REAL in Fortran;[1] SINGLE-FLOAT in Common Lisp;[2] float in C, C++, C# and Java;[3] Float in Haskell[4] and Swift;[5] and Single in Object Pascal (Delphi), Visual Basic, and MATLAB. However, float in Python, Ruby, PHP, and OCaml and single in versions of Octave before 3.2 refer to double-precision numbers. In most implementations of PostScript, and some embedded systems, the only supported precision is single.
IEEE 754 standard: binary32
The IEEE 754 standard specifies a binary32 as having:
This gives from 6 to 9 significant decimal digits precision. If a decimal string with at most 6 significant digits is converted to the IEEE 754 single-precision format, giving a normal number, and then converted back to a decimal string with the same number of digits, the final result should match the original string. If an IEEE 754 single-precision number is converted to a decimal string with at least 9 significant digits, and then converted back to single-precision representation, the final result must match the original number.[6]
The sign bit determines the sign of the number, which is the sign of the significand as well. The exponent field is an 8-bit unsigned integer from 0 to 255, in biased form: a value of 127 represents the actual exponent zero. Exponents range from −126 to +127 (thus 1 to 254 in the exponent field), because the biased exponent values 0 (all 0s) and 255 (all 1s) are reserved for special numbers (subnormal numbers, signed zeros, infinities, and NaNs).
The true significand of normal numbers includes 23 fraction bits to the right of the binary point and an implicit leading bit (to the left of the binary point) with value 1. Subnormal numbers and zeros (which are the floating-point numbers smaller in magnitude than the least positive normal number) are represented with the biased exponent value 0, giving the implicit leading bit the value 0. Thus only 23 fraction bits of the significand appear in the memory format, but the total precision is 24 bits (equivalent to log10(224) ≈ 7.225 decimal digits).
The bits are laid out as follows:
The real value assumed by a given 32-bit binary32 data with a given sign, biased exponent e (the 8-bit unsigned integer), and a 23-bit fraction is
${\displaystyle (-1)^{b_{31}}\times 2^{(b_{30}b_{29}\dots b_{23})_{2}-127}\times (1.b_{22}b_{21}\dots b_{0})_{2}}$,
which yields
${\displaystyle {\text{value}}=(-1)^{\text{sign}}\times 2^{(E-127)}\times \left(1+\sum _{i=1}^{23}b_{23-i}2^{-i}\right).}$
In this example:
• ${\displaystyle {\text{sign}}=b_{31}=0}$,
• ${\displaystyle (-1)^{\text{sign}}=(-1)^{0}=+1\in \{-1,+1\}}$,
• ${\displaystyle E=(b_{30}b_{29}\dots b_{23})_{2}=\sum _{i=0}^{7}b_{23+i}2^{+i}=124\in \{1,\ldots ,(2^{8}-1)-1\}=\{1,\ldots ,254\}}$,
• ${\displaystyle 2^{(E-127)}=2^{124-127}=2^{-3}\in \{2^{-126},\ldots ,2^{127}\}}$,
• ${\displaystyle 1.b_{22}b_{21}...b_{0}=1+\sum _{i=1}^{23}b_{23-i}2^{-i}=1+1\cdot 2^{-2}=1.25\in \{1,1+2^{-23},\ldots ,2-2^{-23}\}\subset [1;2-2^{-23}]\subset [1;2)}$.
thus:
• ${\displaystyle {\text{value}}=(+1)\times 2^{-3}\times 1.25=+0.15625}$.
Note:
• ${\displaystyle 1+2^{-23}\approx 1.000\,000\,119}$,
• ${\displaystyle 2-2^{-23}\approx 1.999\,999\,881}$,
• ${\displaystyle 2^{-126}\approx 1.175\,494\,35\times 10^{-38}}$,
• ${\displaystyle 2^{+127}\approx 1.701\,411\,83\times 10^{+38}}$.
Exponent encoding
The single-precision binary floating-point exponent is encoded using an offset-binary representation, with the zero offset being 127; also known as exponent bias in the IEEE 754 standard.
• Emin = 01H−7FH = −126
• Emax = FEH−7FH = 127
• Exponent bias = 7FH = 127
Thus, in order to get the true exponent as defined by the offset-binary representation, the offset of 127 has to be subtracted from the stored exponent.
The stored exponents 00H and FFH are interpreted specially.
Exponent fraction = 0 fraction ≠ 0 Equation
00H = 000000002 ±zero subnormal number ${\displaystyle (-1)^{\text{sign}}\times 2^{-126}\times 0.{\text{fraction}}}$
01H, ..., FEH = 000000012, ..., 111111102 normal value ${\displaystyle (-1)^{\text{sign}}\times 2^{{\text{exponent}}-127}\times 1.{\text{fraction}}}$
FFH = 111111112 ±infinity NaN (quiet, signalling)
The minimum positive normal value is ${\displaystyle 2^{-126}\approx 1.18\times 10^{-38}}$ and the minimum positive (subnormal) value is ${\displaystyle 2^{-149}\approx 1.4\times 10^{-45}}$.
Converting decimal to binary32
In general, refer to the IEEE 754 standard itself for the strict conversion (including the rounding behaviour) of a real number into its equivalent binary32 format.
Here we can show how to convert a base-10 real number into an IEEE 754 binary32 format using the following outline:
• Consider a real number with an integer and a fraction part such as 12.375
• Convert and normalize the integer part into binary
• Convert the fraction part using the following technique as shown here
• Add the two results and adjust them to produce a proper final conversion
Conversion of the fractional part: Consider 0.375, the fractional part of 12.375. To convert it into a binary fraction, multiply the fraction by 2, take the integer part and repeat with the new fraction by 2 until a fraction of zero is found or until the precision limit is reached which is 23 fraction digits for IEEE 754 binary32 format.
${\displaystyle 0.375\times 2=0.750=0+0.750\Rightarrow b_{-1}=0}$, the integer part represents the binary fraction digit. Re-multiply 0.750 by 2 to proceed
${\displaystyle 0.750\times 2=1.500=1+0.500\Rightarrow b_{-2}=1}$
${\displaystyle 0.500\times 2=1.000=1+0.000\Rightarrow b_{-3}=1}$, fraction = 0.011, terminate
We see that ${\displaystyle (0.375)_{10}}$ can be exactly represented in binary as ${\displaystyle (0.011)_{2}}$. Not all decimal fractions can be represented in a finite digit binary fraction. For example, decimal 0.1 cannot be represented in binary exactly, only approximated. Therefore:
${\displaystyle (12.375)_{10}=(12)_{10}+(0.375)_{10}=(1100)_{2}+(0.011)_{2}=(1100.011)_{2}}$
Since IEEE 754 binary32 format requires real values to be represented in ${\displaystyle (1.x_{1}x_{2}...x_{23})_{2}\times 2^{e}}$ format (see Normalized number, Denormalized number), 1100.011 is shifted to the right by 3 digits to become ${\displaystyle (1.100011)_{2}\times 2^{3}}$
Finally we can see that: ${\displaystyle (12.375)_{10}=(1.100011)_{2}\times 2^{3}}$
From which we deduce:
• The exponent is 3 (and in the biased form it is therefore ${\displaystyle (127+3)_{10}=(130)_{10}=(1000\ 0010)_{2}}$
• The fraction is 100011 (looking to the right of the binary point)
From these we can form the resulting 32-bit IEEE 754 binary32 format representation of 12.375:
${\displaystyle (12.375)_{10}=(0\ 10000010\ 10001100000000000000000)_{2}=(41460000)_{16}}$
Note: consider converting 68.123 into IEEE 754 binary32 format: Using the above procedure you expect to get ${\displaystyle ({\text{42883EF9}})_{16}}$ with the last 4 bits being 1001. However, due to the default rounding behaviour of IEEE 754 format, what you get is ${\displaystyle ({\text{42883EFA}})_{16}}$, whose last 4 bits are 1010.
Example 1: Consider decimal 1. We can see that: ${\displaystyle (1)_{10}=(1.0)_{2}\times 2^{0}}$
From which we deduce:
• The exponent is 0 (and in the biased form it is therefore ${\displaystyle (127+0)_{10}=(127)_{10}=(0111\ 1111)_{2}}$
• The fraction is 0 (looking to the right of the binary point in 1.0 is all ${\displaystyle 0=000...0}$)
From these we can form the resulting 32-bit IEEE 754 binary32 format representation of real number 1:
${\displaystyle (1)_{10}=(0\ 01111111\ 00000000000000000000000)_{2}=({\text{3F800000}})_{16}}$
Example 2: Consider a value 0.25. We can see that: ${\displaystyle (0.25)_{10}=(1.0)_{2}\times 2^{-2}}$
From which we deduce:
• The exponent is −2 (and in the biased form it is ${\displaystyle (127+(-2))_{10}=(125)_{10}=(0111\ 1101)_{2}}$)
• The fraction is 0 (looking to the right of binary point in 1.0 is all zeroes)
From these we can form the resulting 32-bit IEEE 754 binary32 format representation of real number 0.25:
${\displaystyle (0.25)_{10}=(0\ 01111101\ 00000000000000000000000)_{2}=({\text{3E800000}})_{16}}$
Example 3: Consider a value of 0.375. We saw that ${\displaystyle 0.375={(0.011)_{2}}={(1.1)_{2}}\times 2^{-2}}$
Hence after determining a representation of 0.375 as ${\displaystyle {(1.1)_{2}}\times 2^{-2}}$ we can proceed as above:
• The exponent is −2 (and in the biased form it is ${\displaystyle (127+(-2))_{10}=(125)_{10}=(0111\ 1101)_{2}}$)
• The fraction is 1 (looking to the right of binary point in 1.1 is a single ${\displaystyle 1=x_{1}}$)
From these we can form the resulting 32-bit IEEE 754 binary32 format representation of real number 0.375:
${\displaystyle (0.375)_{10}=(0\ 01111101\ 10000000000000000000000)_{2}=({\text{3EC00000}})_{16}}$
Converting binary32 to decimal
If the binary32 value, 41C80000 in this example, is in hexadecimal we first convert it to binary:
${\displaystyle {\text{41C8 0000}}_{16}=0100\ 0001\ 1100\ 1000\ 0000\ 0000\ 0000\ 0000_{2}}$
then we break it down into three parts: sign bit, exponent, and significand.
• Sign bit: ${\displaystyle 0_{2}}$
• Exponent: ${\displaystyle 1000\ 0011_{2}=83_{16}=131_{10}}$
• Significand: ${\displaystyle 100\ 1000\ 0000\ 0000\ 0000\ 0000_{2}=480000_{16}}$
We then add the implicit 24th bit to the significand:
• Significand: ${\displaystyle \mathbf {1} 100\ 1000\ 0000\ 0000\ 0000\ 0000_{2}={\text{C80000}}_{16}}$
and decode the exponent value by subtracting 127:
• Raw exponent: ${\displaystyle 83_{16}=131_{10}}$
• Decoded exponent: ${\displaystyle 131-127=4}$
Each of the 24 bits of the significand (including the implicit 24th bit), bit 23 to bit 0, represents a value, starting at 1 and halves for each bit, as follows:
bit 23 = 1
bit 22 = 0.5
bit 21 = 0.25
bit 20 = 0.125
bit 19 = 0.0625
bit 18 = 0.03125
bit 17 = 0.015625
.
.
bit 6 = 0.00000762939453125
bit 5 = 0.000003814697265625
bit 4 = 0.0000019073486328125
bit 3 = 0.00000095367431640625
bit 2 = 0.000000476837158203125
bit 1 = 0.0000002384185791015625
bit 0 = 0.00000011920928955078125
The significand in this example has three bits set: bit 23, bit 22, and bit 19. We can now decode the significand by adding the values represented by these bits.
• Decoded significand: ${\displaystyle 1+0.5+0.0625=1.5625={\text{C80000}}/2^{23}}$
Then we need to multiply with the base, 2, to the power of the exponent, to get the final result:
${\displaystyle 1.5625\times 2^{4}=25}$
Thus
${\displaystyle {\text{41C8 0000}}=25}$
This is equivalent to:
${\displaystyle n=(-1)^{s}\times (1+m*2^{-23})\times 2^{x-127}}$
where s is the sign bit, x is the exponent, and m is the significand.
Precision limitations on decimal values (between 1 and 16777216)
• Decimals between 1 and 2: fixed interval 2−23 (1+2−23 is the next largest float after 1)
• Decimals between 2 and 4: fixed interval 2−22
• Decimals between 4 and 8: fixed interval 2−21
• ...
• Decimals between 2n and 2n+1: fixed interval 2n-23
• ...
• Decimals between 222=4194304 and 223=8388608: fixed interval 2−1=0.5
• Decimals between 223=8388608 and 224=16777216: fixed interval 20=1
Precision limitations on integer values
• Integers between 0 and 16777216 can be exactly represented (also applies for negative integers between −16777216 and 0)
• Integers between 224=16777216 and 225=33554432 round to a multiple of 2 (even number)
• Integers between 225 and 226 round to a multiple of 4
• ...
• Integers between 2n and 2n+1 round to a multiple of 2n-23
• ...
• Integers between 2127 and 2128 round to a multiple of 2104
• Integers greater than or equal to 2128 are rounded to "infinity".
Notable single-precision cases
These examples are given in bit representation, in hexadecimal and binary, of the floating-point value. This includes the sign, (biased) exponent, and significand.
0 00000000 000000000000000000000012 = 0000 000116 = 2−126 × 2−23 = 2−149 ≈ 1.4012984643 × 10−45
(smallest positive subnormal number)
0 00000000 111111111111111111111112 = 007f ffff16 = 2−126 × (1 − 2−23) ≈ 1.1754942107 ×10−38
(largest subnormal number)
0 00000001 000000000000000000000002 = 0080 000016 = 2−126 ≈ 1.1754943508 × 10−38
(smallest positive normal number)
0 11111110 111111111111111111111112 = 7f7f ffff16 = 2127 × (2 − 2−23) ≈ 3.4028234664 × 1038
(largest normal number)
0 01111110 111111111111111111111112 = 3f7f ffff16 = 1 − 2−24 ≈ 0.999999940395355225
(largest number less than one)
0 01111111 000000000000000000000002 = 3f80 000016 = 1 (one)
0 01111111 000000000000000000000012 = 3f80 000116 = 1 + 2−23 ≈ 1.00000011920928955
(smallest number larger than one)
1 10000000 000000000000000000000002 = c000 000016 = −2
0 00000000 000000000000000000000002 = 0000 000016 = 0
1 00000000 000000000000000000000002 = 8000 000016 = −0
0 11111111 000000000000000000000002 = 7f80 000016 = infinity
1 11111111 000000000000000000000002 = ff80 000016 = −infinity
0 10000000 100100100001111110110112 = 4049 0fdb16 ≈ 3.14159274101257324 ≈ π ( pi )
0 01111101 010101010101010101010112 = 3eaa aaab16 ≈ 0.333333343267440796 ≈ 1/3
x 11111111 100000000000000000000012 = ffc0 000116 = qNaN (on x86 and ARM processors)
x 11111111 000000000000000000000012 = ff80 000116 = sNaN (on x86 and ARM processors)
By default, 1/3 rounds up, instead of down like double precision, because of the even number of bits in the significand. The bits of 1/3 beyond the rounding point are 1010... which is more than 1/2 of a unit in the last place.
Encodings of qNaN and sNaN are not specified in IEEE 754 and implemented differently on different processors. The x86 family and the ARM family processors use the most significant bit of the significand field to indicate a quiet NaN. The PA-RISC processors use the bit to indicate a signalling NaN.
Optimizations
The design of floating-point format allows various optimisations, resulting from the easy generation of a base-2 logarithm approximation from an integer view of the raw bit pattern. Integer arithmetic and bit-shifting can yield an approximation to reciprocal square root (fast inverse square root), commonly required in computer graphics. | 4,968 | 15,721 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 52, "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} | 2.84375 | 3 | CC-MAIN-2024-22 | latest | en | 0.868676 |
https://yourquickadvice.com/what-percentage-of-my-body-should-be-bone/ | 1,709,603,795,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476592.66/warc/CC-MAIN-20240304232829-20240305022829-00795.warc.gz | 1,078,861,367 | 15,189 | # What percentage of my body should be bone?
## What percentage of my body should be bone?
Bone content is the percentage of bone mineral as compared to total body weight. The average bone content for adults is 3-5%. This measurement is good to keep track over a long period of time as bone mass can decline slowly with age.
What percentage of body weight is muscle?
In terms of tissue composition, bone, adipose tissue, and muscle make up 75% of body weight. The lean body mass is the mass of the body minus the fat mass (storage lipid).
How much of your body weight is bones and organs?
In a fit female, bones make up 12% of the total body weight, muscles about 30 to 35% and fat around 27%. And In muscular males, bones make up 15% of the total body weight, muscles about 40 to 45% and fat about 15%.
### What is a healthy body fat percentage?
Measuring body fat Men and women need different amounts of fat. For a man, 2–5% fat is essential, 2–24% fat is considered healthy, and more than 25% classifies as obesity. For a woman, 10–13% fat is essential, 10–31% fat is healthy, and more than 32% classifies as obesity.
Is bone weight good or bad?
Maintaining a healthy weight A healthy weight is essential for bone density — people who are underweight have a higher risk of developing bone disease, while excess body weight puts additional stress on the bones.
How much should I weigh if I’m big boned?
According to the University of Washington, women should weigh 100 pounds for the first five feet of height plus 5 pounds for each additional inch over five feet – plus 10 percent to account for large frame sizes. For example, a large-framed woman who is 5 feet 3 inches tall has an ideal body weight of 127 pounds.
#### What is the strongest muscle in the body?
The strongest muscle based on its weight is the masseter. With all muscles of the jaw working together it can close the teeth with a force as great as 55 pounds (25 kilograms) on the incisors or 200 pounds (90.7 kilograms) on the molars.
What body percentage is muscle?
According to Withings, normal ranges for muscle mass are: Ages 20-39: 75-89 percent for men, 63-75.5 percent for women. Ages 40-59: 73-86 percent for men, 62-73.5 percent for women. ages 60-79: 70-84 percent for men, 60-72.5 percent for women.
What percentage body fat to see six pack?
Doing targeted exercises like crunches is great for toning abdominal muscles, but losing both subcutaneous and visceral fat is the first step to unearthing your abs. According to the American Council on Exercise (ACE), you’ll need to lower your body fat to about 14 to 20 percent for women and 6 to 13 percent for men.
## Does bone size affect weight?
Bones make up around 15% of a person”s total body weight. While people do have different frame size, most who weigh too much for their height do so because of excess body fat. Being big-boned or small boned doesn’t justify a 20-25 pounds differential from a normal size person.
What is a good percentage for bone mass?
Bone content is the percentage of bone mineral as compared to total body weight. The average bone content for adults is 3-5%. This measurement is good to keep track over a long period of time as bone mass can decline slowly with age.
What percentage of body mass is bone?
Human beings are made up of a few basic components such as bone, muscle and fat, and the percentage of each of these components in relation to overall body mass differs according to gender and genetics. Generally speaking, your bones make up about 15 percent of your overall body mass. Bones make up about 15 percent of overall body mass.
### What is the normal bone mass?
The NIH offers the following guidelines for bone density scores: normal: between 1 and -1. low bone mass: -1 to -2.5. osteoporosis: -2.5 or lower. severe osteoporosis: -2.5 or lower with bone fractures.
What is the z score for osteoporosis?
The Z-score is used to classify the type of osteoporosis. A score below 1.5 indicates primary osteoporosis, which is age related. A score of 1.5 and higher indicates secondary osteoporosis, which is associated with calcitonin imbalance, malabsorption conditions (e.g., celiac disease , cystic fibrosis ,… | 1,003 | 4,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-10 | latest | en | 0.94472 |
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A282767 n/3 analog of Keith numbers. 2
45, 609, 1218, 1827, 3213, 21309, 28206, 29319, 31917, 39333, 47337, 78666, 102090, 117999, 204180, 406437, 302867592, 4507146801, 5440407522 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Like Keith numbers but starting from n/3 digits to reach n. Consider the digits of n/3. Take their sum and repeat the process deleting the first addend and adding the previous sum. The sequence lists the numbers that after some iterations reach a sum equal to themselves. If it exists, a(20) > 10^12. - Lars Blomberg Mar 13 2017 LINKS EXAMPLE 609/3 = 203: 2 + 0 + 3 = 5; 0 + 3 + 5 = 8; 3 + 5 + 8 = 16; 5 + 8 + 16 = 29; 8 + 16 + 29 = 53; 16 + 29 + 53 = 98; 29 + 53 + 98 = 180; 53 + 98 + 180 = 331; 98 + 180 + 331 = 609. MAPLE with(numtheory): P:=proc(q, h, w) local a, b, k, n, t, v; v:=array(1..h); for n from 1/w by 1/w to q do a:=w*n; b:=ilog10(a)+1; if b>1 then for k from 1 to b do v[b-k+1]:=(a mod 10); a:=trunc(a/10); od; t:=b+1; v[t]:=add(v[k], k=1..b); while v[t]
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Last modified December 6 11:50 EST 2022. Contains 358632 sequences. (Running on oeis4.) | 625 | 1,805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2022-49 | latest | en | 0.721441 |
https://lynniezulu.com/what-is-the-roman-number-of-1-to-1000/ | 1,670,335,286,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711108.34/warc/CC-MAIN-20221206124909-20221206154909-00129.warc.gz | 401,339,312 | 11,033 | # What is the Roman number of 1 to 1000?
## What is the Roman number of 1 to 1000?
For example, for numbers like 1, 5, 10, 50, 100, 1000 the symbols are I, V, X, C, D, and M, respectively….Roman Numerals from 1 to 1000.
Numbers Roman Numbers
700 DCC
800 DCCC
900 CM
1,000 M
## What Roman number is 500?
D
Is it still important to learn Roman numerals?
Arabic Roman
200 CC
300 CCC
400 CD
500 D
What is the Roman figure of 100 to 500?
List of Roman Numerals from 100 to 500
101: CI 102: CII 105: CV
401: CDI 402: CDII 405: CDV
431: CDXXXI 432: CDXXXII 435: CDXXXV
446: CDXLVI 447: CDXLVII 450: CDL
496: CDXCVI 497: CDXCVII 500: D
What is the symbol of 100 and 500?
The Roman numeral system uses only seven symbols: I, V, X, L, C, D, and M. I represents the number 1, V represents 5, X is 10, L is 50, C is 100, D is 500, and M is 1,000.
### What is the roman numerals 1 to 100?
Roman Numerals 1-100 Chart
Number Roman Numeral Calculation
97 XCVII 100-10+5+1+1
98 XCVIII 100-10+5+1+1+1
99 XCIX 100-10-1+10
100 C 100
### What is the numeral of 500?
Roman Numerals 500-1000 Chart
Number Roman Numeral Number in Words
500 D Five hundred
501 DI Five hundred and one
502 DII Five hundred and two
503 DIII Five hundred and three
Why does D stand for 500 in roman numerals?
It was sometimes represented as a C, I and backwards C, like this: CIƆ — which sort of looks like an M. It’s only a coincidence that mille is the Latin word for a thousand. D = 500 — The symbol for this number was originally IƆ — half of CIƆ. It only coincidentally also stands for centum, the Latin word for a hundred.
What is hundred in Roman numeral?
## How do you write 600 in Roman numerals?
600 in Roman numerals is DC. To convert 600 in Roman Numerals, we will write 600 in the expanded form, i.e. 600 = 500 + 100 thereafter replacing the transformed numbers with their respective roman numerals, we get 600 = D + C = DC.
## What is the Roman number of 500 to 1000?
Therefore, perfect cubes in roman numerals between roman numerals 500 to 1000 are DXII, DCCXXIX, M. Example 4: Using Roman numerals 500 to 1000 chart, find the product of LXV and XLV. Therefore, the product of LXV and XLV is MMCMXXV.
What is the Roman number of 500 and 600?
Using Roman numerals 500 to 600 chart: DXCVI = 500 + 90 + 6 = 596, DLVII = 500 + 50 + 7 = 557, DXCVII = 500 + 90 + 7 = 597, DXXIII = 500 + 20 + 3 = 523.
What are Roman numerals 1 to 500?
Roman Numerals 1 to 500 is a chart that represents numbers from 1 to 500 in roman numerals. It consists of a notation of numbers 1 to 500 used by the Romans in ancient times. How Many Twin Prime Numbers are There Between Roman Numerals 1 to 500? Twin prime numbers are prime numbers whose absolute difference is 2.
### How many Roman numerals are there in order?
After 10, the roman numerals are followed by XI for 11, XII for 12, XII for 13, … till XX for 20. The most common roman numerals that are presently used to represent the basic numbers are given in the table below.
### What is the rule for Roman numerals?
Rule 1: If one or more symbols are placed after another letter of greater value, add that amount. Rule 2: If a symbol is placed before another letter of greater value, subtract that amount. Write 69 in roman numerals. Convert 1984 into the roman numeral. Convert 1774 to Roman Numerals.
How are musical instruments numbered using Roman numerals?
Movements are often numbered using Roman numerals. In Roman Numeral Analysis, harmonic function is identified using Roman Numerals. Individual strings of stringed instruments, such as the violin, are often denoted by Roman numerals, with higher numbers denoting lower strings. | 1,092 | 3,673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.82397 |
https://www.thestudentroom.co.uk/wiki/AQA_A-level_Maths:_Decision_1_Revision_Notes | 1,493,340,169,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917122720.81/warc/CC-MAIN-20170423031202-00291-ip-10-145-167-34.ec2.internal.warc.gz | 950,536,502 | 35,993 | • # AQA A-level Maths: Decision 1 Revision Notes
Tadaa... as requested. my notes for decision1 xD woo go me. There seems to be a lower 'fact stating:word' ratio in this note, but these are my notes on decision, by chapter, with the algorithm/procedure your meant to follow and an explanation as well as the things you need to show for the method marks...i hope ive not left out anything important; if I have/you spot errors, let me know/edit this yourself. As always, practise papers help show you what your bad at and help you to learn to play the exam 'game'. woo fun. Good luck =]
## Contents
### Graph Theory
["WHATT??" i hear you cry. "you said in order?" well... it helps if you know what the words i might end up using means so i'll start with chapter 5.]
Graph: finite vertices connected by edges.
Vertices/Nodes: where two separate lines meet. Usually shown with a black dot.
Edge/Arcs: lines that join up vertices/nodes. The edges/path thing.
Degree/Order/Valency: number of edges connected to a vertex.
Loop: an arc beginning and ending one the same vertex. Loops contribute 2 to the arc count.
Simple graph: A graph with no loops, no multiple edges and every vertex is connected.
Connected graph: A graph where all the points are connected. (the graph is in one piece)
Disconnected graph: If the points are not all linked into a graph. (two smaller connected graphs near each other)
Fully connected graph/Complete graph: Where every vertex is connected to every other vertex. Its total edges,(Kn) = n(n-1)/2 where n=vertices
Weighted graph: A graph where numbers are linked with the edges, representing time/distance/money etc.
Directed graph/digraph: has edges with directions. ie. one way paths.
Planar graph: One which can be drawn without any arcs intersecting. There is a complicated way of deciding if a given graph is planar, otherwise just experiment with stretching the arcs to try and remove intersections.
Adjacency matrix: a matrix (two-way table) that represents a graph where the rows and columns represent the vertices, and the number inside represents the edges joining the weight of the edges joining the vertices.
Bipartite graph: A graph with two sets of vertices where edges only connect from one subset of vertices to the other subset.
Maximum matching: The optimum matching in the bipartite graph.
Complete matching: where all the vertices in one subset connect to all the vertices in the other.
Tree: connected graph with no cycles
Spanning tree: A tree that connects all the vertices (n). It has n-1 edges.
Minimum spanning tree: A spanning tree with the minimum weight for the graph.
Trail or Walk: A sequence of edges that link up to each other through edges.
Path: A trail where no vertex is visited more than once.
Cycle or Circuit: A closed path/trail with at least one edge. (closed = finishes at start)
Hamiltonian cycle: A cycle/circuit/path/trail that visits every vertex. It has (n-1)!/2 permutations.
Eulerian trail: A trail that uses every edge of a graph (the graph would be Eulerian/traversible) It is not a path as it visits the vertices more than once, although there are exceptions e.g a hexagonal graph. Each node must have even order/valency for a graph to be Eulerian.
Traversible graph: graph with a Eulerian/semi-Eulerian trail
Semi-Eulerian: non-closed Eulerian trail (doesn't finish at the start vertex) There must be exactly 2 odd order nodes & the rest must be even. The trail must begin at one of the odd order nodes and finish at the other one.
### Minimum Connectors
The following algorithms are used to find the minimum spanning tree for graphs.
The order the edges are selected will be different for the two algorithms, but the minimum spanning tree will be the same whichever algorithm you use.
Kruskal's Algorithm:
• choose an unused edge with the lowest weight, but not making a cycle
• if there are n-1 edges stop. If not, repeat.
[Kruskals algorithm chooses the edges in order of size; so for working/method marks, list the order of the edges that are selected and the lengths. If the new edge makes it a cycle, (therefore no longer a tree) you NEED to write it down and put a line through it (legibly). Label the last length used as n-1.]
Prim's Algorithm:
• from the start vertex, add the lowest weight to the tree.
• from any vertex already used, choose the edge with the lowest value avoiding cycles.
• if there are n-1 edges stop. If not, repeat.
[Prim's Algorithm chooses the edges that are next to the tree already. Working for this you should include a list of the order the edges were selected, the lengths, and label n-1
For both algorithms it might be helpful to number the list of edges you selected, and add to your drawing of the tree as you go along. This will help you to spot cycles and show you stopped at n-1. If you'd excuse my bad analogies, how I remember the algorithms for the exam is kruskal = "krus"-ing out, cross out an edge. Prim = neat and tidy (so systematic, not jumping around edges)]
Matrix Form of Prim's
• On the matrix, write down all the vertices across the top and down the side.
• Label the column with the start vertex with a 1
• Cross out the corresponding row. (row with start vertex)
• Ring the smallest value in ANY labeled column that hasn't been deleted
• label the corresponding vertex with 2 etc. delete the row
• repeat untill every row has been deleted.
[careful this is not confused with the upper bound matrix. List out the order of the edges selected along with their weights.]
### Shortest Path Problem (Dijkstra's Algorithm)
This algorithm finds the shortest route from one vertex to another, by considering all the posibilities.
This algorithm only applies to graphs with no negative lengths.
The algorithm can be used starting from the end vertex if there are multiple starting points.
The algorithm requires drawing on/labeling the graph.
• number start vertex as 0 and draw a box around it
• number each vertex connected to the start vertex with the total distance (from start to there) [and where its from (label on the previous vertex)]
• choose the smallest number and box it.
• number the total distance to every connected vertex from the last box [and where from.] - Cross out (legibly) numbers that are already there; unless the number already there is smaller than the one you want to put (where you dont put anything). Dont go back to already boxed vertices.
• Repeat last two steps until everything has been boxed and you haved reached the end.
The shortest total distance is the number in the final box and the shortest route can be traced backwards from the labels.
[Another random analogy: D-J box man = Dijkstra. You draw boxes.. and his name has the letters d and j in... haha..... . . . .
### Chinese Postman Problem
This algorithm finds the shortest path going along every edge. Just like how a postman has to deliver letters to every street without doubling back on himself unnecessarily.
The shortest path would be the sum of all the lengths on the Eulerian graph.
If your graph has an odd numbered order on any vertex, it is not Eulerian. If it is Eulerian, skip this section. Making the Graph Eulerian
• list all the odd vertices, all the possible pairings of these vertices. The number of possible pairings is determined by (n-1)×(n-3)×(n-5)×.....×1
• Work out the weight that each pairing would be. (shortest path from one to the other)
• Choose the pairings with the minimum weight.
• Draw in these pairings as new edges onto the graph.
Finding the path
• list the order of each vertex (should be even by now)
• List the number of times each vertex would appear in the trail (half the orders, except the start vertex which is one more) Are you sure? Look at Jan 11 Q5C and it's answers
• Try out the path using general common sense [not much required] and check with the list.
[That finding the path method doesnt actually find you the path it's to check youve found the path, but i dont see how you would need to check... =/ but it gets you the marks so use it anyway...]
For semi-Eulerian trails where you start and end in different places, you do the same thing, but ignore the start vertex and end vertex when you pair them up to make the graph Eulerian.
The sum of the total distance of the Eulerian trail is the sum of all the edges plus the minimum pairings (not including the start and end, as the order of start vertex and end vertex is odd on semi-Eulerian trails.)
[erm i dont think i have a way of remembering this one....halve a Eulerian?]
### Travelling Salesman Problem
This algorithm is meant to help find the shortest/minimum route so that every vertex is visited. It only applies to complete networks.
However, it doesnt always give you a route/answer. But you get the range that the answer could lie within. Lower bound<Answer≤Upper bound [woo...]
The upper bound of the range is an actual tour (it exsists), but it may be improved upon. (not necesarily.)
The lower bound of the range cannot be improved upon, however, it may not exsist.
When applied to each vertex, the best lower bound is one with highest length, and the best upper bound is one with the lowest length.
Upper Bound Algorithm
• choose start vertex and go to nearest unvisted vertex from there.
• repeat untill all vertices have been visited, then return to beginning.
[this algorithm could also be applied to a matrix. This time, cross out the column instead of the row, so you cannot return to a previous vertex. Remember to list out the answer afterwards.] Lower Bound Algorithm
• delete the start vertex and all the edges connected to that
• find the minimum spanning tree for the remaining network
• add this to the two shortest edges from the start vertex.
### Matching
The alternating path algorithm improves an initial matching
• Find vertex not matched.
• match it
• If you need to, remove the initial match to it
• repeat untill there are no more improvements(maximum matching)/you have got a complete matching.
[for working, use a symbol that looks something like ≠ but more like -/- to show mathings that have been removed, eg. A-F≠C-G which shows 'A' being matched to 'F' which disconnects 'C'. 'C' is then connected to 'G'. The symbol ≠ implies is not equal to, rather thank links so yeah. dont use it.It will also help to draw out the diagram after one set of unmatching and rematching, to see how your new graph looks.]
### Sorting Algorithms
• check the book for these not gonna lie these seem a bit dodgy*
There are four sorting algorithms that you will need to learn. Each algorithm would vary in efficiency (number of comparisons and swaps made) depending on the order of the list you are sorting.
Comparison = the number of times an item is compared with the next in each pass. Swaps = amount of positions an item has moved in the pass.
The algorithms are written for a set of ascending numbers. It can also be used to descending numbers by doing it opposite or sorting words alphabetically.
Bubble Sort
• Compare first two numbers. The larger number moved to second, leaving behind the smaller number at the front.
• the second and third numbers are compared. The lighter number gets moved to the front.
• etc.. untill you get to the end of the row. This is known as a pass.
• repeat from beginning again untill nothing has moved (no swaps made).
[write out the number of comparisons and passes made at every pass. Show every pass made until everything is in order, ie. no more swaps are made. Its called bubble as the lighter number rises to the top... but my teacher thinks it should be called dead weight as the heavy number is the one that moves... to the bottom.]
Shuttle Sort
• Compare first two numbers.
• Move larger number to second place.
• Compare first three numbers.
• Move it into position by comparing it with second, first.
• etc...until all number have been compared.
[write out each pass and the comparisons and swaps at each one too. you will also need to underline the ones that are being compared, so on the original list the first two, then first three, first four.... etc. failure to come up with a weird analogy.]
Shell Sort
• Separate into sublists using INT(n/2) this gives the number of lists to be used
• Pair up the first number in the first half with the first number in the second half, second with second etc. but keeping the order the same. [write each pair on separate lines diagonally so order can be seen.]
• Compare the pairs and shuttle sort them to get the larger number at the back.
• Merge/re-piece the list but with the numbers that have been swapped in their new positions.
• Quarter the list and do the same with each sublist.
• Repeat untill each sublist only has 1 number.
[as you will only need to write out the sublists, and then the merges the guy wont know how you sorted the sublists etc. Write out comparisons and swaps at each stage.]
Quick Sort
• Use the first number as the pivot. [Underline it].
• Keeping the order of the numbers, move the smaller ones in front of the pivot, keeping larger ones behind. [write this pass out]
• The first number in each sublist is now a new pivot. [box the pivot you have just used and underline new ones]
• Repeat untill every number has been used as a pivot [everything is boxed] - Maybe I'm wrong but I believe that you only box the first in each sublist, until each sublist only has 1 number.
### Algorithms
An Algorithm must have a finite number of instructions that achieves some task.
Each instruction has to be defined precisely in structured, pseudo English. This is a simplistic form of English.
Each variable value must have been inputted to solve the problem.
It must work for any set of values input.
Line numbers are used to indicate the order of instructions. They go up in multiples of 10 so that modifications can made without needing to re-number everything.
An algorithm can be represented in flowcharts. There is a certain layout/format that must be used for flowcharts. It may vary from other subjects (electronics etc.)
• Oval boxes used for start/stop/input/output.
• Square boxes for calculating/instructions/processes
• Diamond for decisions
The labels in the algorithm mark 'milestones' in the algorithm. They provide a point in the algorithm that could be jumped to.
An 'If...Then' statement is a decision. It asks a question. If true, you'll be redirected, if false continue.
'For' statements apply a loop for a range of number of inputs, e.g. 'for x=3 to 100' means the algorithm is applied for all the numbers from 3,4,5.......98,99,100.
A 'For' statement is used if the exact number of inputs is known, when it is not known, an 'If' statement is used.
Tracing Algorithms means following the algorithm out line by line manually.
List the Inputs then the Outputs along the top of a two-way table. The side is used for line numbers. Fill it out, line at a time, step by step in order.
[I dont know why labels are used, it's more convienient to say 'go to line 30' for example... also, while tracing algorithms, lines that have labels on etc. do not need to be shown. Algorithms shouldnt be too hard, they easily make sense, they use actual words...]
### Linear Programming
This is used to help optimize the situation.
Lines on the graph represent the constraints/boundaries in the situation and can be expressed as an inequality.
The feasible region is the unshaded side of the line/the excluded region.
When there is a region bounded by other lines, it is known as a finite region.
The objective of the situation, which can be expressed as an objective function which can be used to find an objective line.
The objective line can be moved [keeping the gradient] until it intersects a vertex, which is used to find the optimal value. (where it leaves the region last)
When the optimum value is found, substitute the (x,y) co-ordinates on the vertex and the objective can be worked out.
### Exam Tips
• always check that you have added up properly. eg. in minimum spanning tree etc.
• make sure the question has been actually been answered. Where is the answer? Is it clear?
• show working, ie. list vertices from the route taken from the matrix's in order.
• rest well before, dont stress out.
• Jan 2011 is the last time exam board will allow couloured pencils for working...(no hightlighters). Either get a fat pencil, or go over lines a few times on diagrams.
• keep you answers precise and clear.
• If your stuck, dont stay on the same question for too long. Decision papers can be quite long, and there are usually alot of marks towards the end.
• leaving a hard question may trigger a brain wave later on in the exam.
• as with all things, practise!
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Will Oxford nullify my offer? | 3,803 | 16,971 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2017-17 | latest | en | 0.926471 |
https://www.physicsforums.com/threads/interference-pattern-from-a-streetlight.224636/ | 1,576,020,962,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540529516.84/warc/CC-MAIN-20191210233444-20191211021444-00264.warc.gz | 829,101,347 | 20,727 | # Interference pattern from a streetlight ?
## Main Question or Discussion Point
If I use a narrow slit (e.g. two fingers) to view a distant streetlight, I can see an interference pattern parallel to the slit. Doesn't this mean that the photons passing through the slit must be phase coherent? If I am correct about coherence, then how does it occur? The photons leaving the streetlight are not coherent; are they?
Excuse me if this question is stupid. I am just an old fuddy-duddy who just likes to think about physics.
Related Classical Physics News on Phys.org
Claude Bile
If the source is sufficiently narrow-band (like a sodium lamp for example) and you are far enough away from the source, then it is possible that the light when it reaches your aperture could be spatially coherent enough to produce an interference pattern.
Spatial coherence is simply a measure of how much the phase varies across a wavefront - it doesn't make a whole lot of sense to label photons as being coherent or incoherent.
Claude.
photons and phase
Perhaps coherence of photons is not the correct term.
The following implies no knowledge of QED, only a reading of Feynmann's "QED" which was intended for the layman.
I thought the angle of the spinning vectors used by Feynmann in "QED" implies an initial phase relationship between photons in deriving an interference pattern. Even if the term is not "coherence" isn't there an assumption in his computations that the pattern is due to the angles of the spinning vectors when they are added at a receive point. IE., in order to make this calculation and generate a "classical" interference pattern, you start by assuming the vectors are aligned (in-phase) at the source point. Then the final vectors are computed based on the possible "paths" of each photon to the receiver.
Andy Resnick
I think what you are seeing is diffraction, not interference.
I don't think so
I think what you are seeing is diffraction, not interference.
Isn't a diffraction pattern the result of constructive and destructive interference?
clem
The diffraction pattern arises from the interference of a photon with itself.
The pattern is found when one photon at a time passes the grating,
after a large number of photons have landed.
Separate photons from the light are not phase coherent.
I am sure that is right.............thank
The diffraction pattern arises from the interference of a photon with itself.
The pattern is found when one photon at a time passes the grating,
after a large number of photons have landed.
Separate photons from the light are not phase coherent.
I am sure that is right.............thanks
I am not sure I posted this in the correct forum..........should have been under quantum.
Claude Bile
I thought the angle of the spinning vectors used by Feynmann in "QED" implies an initial phase relationship between photons in deriving an interference pattern. Even if the term is not "coherence" isn't there an assumption in his computations that the pattern is due to the angles of the spinning vectors when they are added at a receive point. IE., in order to make this calculation and generate a "classical" interference pattern, you start by assuming the vectors are aligned (in-phase) at the source point. Then the final vectors are computed based on the possible "paths" of each photon to the receiver.
I know this is going to sounds picky, but the reasons why I object to the terms "coherent photons" and "incoherent photons" is as follows...
- You can have spatial coherence and temporal coherence. Waves might be spatially coherent but not temporally coherent and vice-versa.
- Waves gain and lose coherence based on the system - it is a dynamic quantity, not a static one.
- Coherence is not a binary quantity, but a continuous one.
It's just one of those things that tends to lead to confusion, which is why I try to avoid labeling waves in this way. Adding photons to the mix further compounds the problem - diffraction can be adequately explained by classical theory without resorting to QM.
Your understanding (and Feynmann's layman explanation as you have presented it) is accurate. Note though how the phase relation of each photon is explicitly stated in time and space, which is the correct way to go about describing coherence!
clem said:
Separate photons from the light are not phase coherent.
This is not correct. You can "enforce" spatial coherence by only permitting a small portion of the wavefront to enter your detector (i.e. through the use of a slit or pinhole). In fact, this is what Young did in his famous experiments.
mangurian said:
I am not sure I posted this in the correct forum..........should have been under quantum.
Nope, this is the right forum .
Claude.
clem
"Separate photons from the light are not phase coherent."
"This is not correct. You can "enforce" spatial coherence by only permitting a small portion of the wavefront to enter your detector (i.e. through the use of a slit or pinhole). In fact, this is what Young did in his famous experiments."
I will not say the you are not correct, but only that my statement is correct.
If Young's slit or pinhole were as small as you suggest, it would have to be of the order of Angstroms (which he couldn't have done) and would produce single slit diffraction.
The correct statement is
Separate photons from the light in Young's experiment are not phase coherent.
Claude Bile
Separate photons from the light in Young's experiment are not phase coherent.
There must be some degree of coherence at the screen for an interference pattern to be produced.
Claude.
pam
A photon interferes with itself, just like an electron in a two slit experiment.
Claude Bile
I can appreciate that fact pam, but going down that path just rephrases the issue in terms of the coherence of the photonic wavefunction, rather than a classical field.
I don't think there is any insight to be gained by bringing photons into the discussion when discussing well understood classical concepts such as single-slit and two-slit diffraction.
Claude.
pam
Spatial coherence is simply a measure of how much the phase varies across a wavefront - it doesn't make a whole lot of sense to label photons as being coherent or incoherent.
Claude.
You have a lot to learn.
Claude Bile
You have a lot to learn.
Well spotted.
Of course you don't get phase variations over a wavefront, since they are lines of equal phase.
What I meant to say is phase variations over the face of a surface such as an aperture.
Claude.
- You can have spatial coherence and temporal coherence. Waves might be spatially coherent but not temporally coherent and vice-versa.
Claude.
Why spatial (transverse) coherence cannot exist with temporal coherence? If light is temporally coherent or that is to say it is monochromatic, I don't see why it cannot be spatially coherent.
Isn't for example laser light temporally coherent and spatially coherent as well as phase coherent.
pam
Isn't for example laser light temporally coherent and spatially coherent as well as phase coherent.
But a street light is not a laser.
But a street light is not a laser.
Of course it is not. But I got the impression that Claude Bile was talking generally of light waves. But why streetlight cannot have temporal and spatial transverse coherence at the same time?
Moreover, how we can have spatial coherence if we do not have temporal coherence in other words monochromatic source of light?!
I know this is going to sounds picky, but the reasons why I object to the terms "coherent photons" and "incoherent photons" is as follows...
- You can have spatial coherence and temporal coherence. Waves might be spatially coherent but not temporally coherent and vice-versa.
- Waves gain and lose coherence based on the system - it is a dynamic quantity, not a static one.
- Coherence is not a binary quantity, but a continuous one.
Because Claude hasn't answered. Could someone please explain why streetlight cannot have spatial and temporal coherence same time, only one or the other?
Andy Resnick | 1,730 | 8,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-51 | latest | en | 0.945204 |
https://byjus.com/question-answer/the-number-of-cars-sold-in-the-d-segment-is-same-as-the-number-of/ | 1,716,243,410,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058313.71/warc/CC-MAIN-20240520204005-20240520234005-00662.warc.gz | 120,853,922 | 27,528 | 1
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Question
# The number of cars sold in the D segment is same as the number of cars sold by which of the following manufacturers?
A
P
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Solution
## The correct option is C R Let the numbers of cars sold by different manufactures is as shown below. P - a Q - b R - c S - d T - e PQRSTB0.24a0.16b0.32c0.24d0.28eB+0.16a0.24b0.10c0.16d0.08eC0.18a0.24b0.15c0.32d0.12eD0.36a0.20b0.35c0.24d0.48eE0.06a0.16b0.08c0.04d0.04e Let the numbers of cars sold by different segments is as shown below. B - v B+ - w C - x D - y E - z BB+CDEP0.15v0.20w0.15x0.18y0.15zQ0.05v0.15w0.10x0.05y0.20zR0.40v0.25w0.25x0.35y0.40zS0.225v0.30w0.40x0.18y0.15zT0.175v0.10w0.10x0.24y0.10z From the above two tables, we can say that number of cars sold in the D segment is same as the number of cars sold by manufacture R.
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Join BYJU'S Learning Program | 521 | 1,319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.882044 |
https://fred.stlouisfed.org/series/PPPTTLKZA618NUPN | 1,516,183,724,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886860.29/warc/CC-MAIN-20180117082758-20180117102758-00520.warc.gz | 714,417,915 | 21,845 | # Purchasing Power Parity over GDP for Kazakhstan (PPPTTLKZA618NUPN) Excel (data) CSV (data) Image (graph) PowerPoint (graph) PDF (graph)
Observation:
2010: 87.66239
Updated: Sep 17, 2012
Units:
National Currency Units per US Dollar,
Frequency:
Annual
1Y | 5Y | 10Y | Max
EDIT LINE 1
(a) Purchasing Power Parity over GDP for Kazakhstan, National Currency Units per US Dollar, Not Seasonally Adjusted (PPPTTLKZA618NUPN)
Note: Over GDP, 1 US dollar (US\$) = 1 international dollar (I\$). Purchasing power parity is the number of currency units required to buy goods equivalent to what can be bought with one unit of the base country. We calculated our PPP over GDP. That is, our PPP is the national currency value of GDP divided by the real value of GDP in international dollars. International dollar has the same purchasing power over total U.S. GDP as the U.S. dollar in a given base year.
For more information and proper citation see http://www.rug.nl/research/ggdc/data/pwt/pwt-7.1
Source Indicator: ppp
Purchasing Power Parity over GDP for Kazakhstan
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NOTES
Source: University of Pennsylvania
Release: Penn World Table 7.1
Units: National Currency Units per US Dollar, Not Seasonally Adjusted
Frequency: Annual
#### Notes:
Note: Over GDP, 1 US dollar (US\$) = 1 international dollar (I\$). Purchasing power parity is the number of currency units required to buy goods equivalent to what can be bought with one unit of the base country. We calculated our PPP over GDP. That is, our PPP is the national currency value of GDP divided by the real value of GDP in international dollars. International dollar has the same purchasing power over total U.S. GDP as the U.S. dollar in a given base year.
For more information and proper citation see http://www.rug.nl/research/ggdc/data/pwt/pwt-7.1
Source Indicator: ppp
#### Suggested Citation:
University of Pennsylvania, Purchasing Power Parity over GDP for Kazakhstan [PPPTTLKZA618NUPN], retrieved from FRED, Federal Reserve Bank of St. Louis; https://fred.stlouisfed.org/series/PPPTTLKZA618NUPN, January 17, 2018.
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Updating graph. | 666 | 2,656 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2018-05 | latest | en | 0.802338 |
https://www.mathnasium.com/math-centers/roslyn/news/magic-number-patterns-exploring-number-sequences-and-their-applications-r | 1,713,102,726,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.72/warc/CC-MAIN-20240414130604-20240414160604-00871.warc.gz | 838,437,096 | 13,610 | # The Magic of Number Patterns: Exploring Number Sequences and Their Applications
Feb 27, 2024 | Roslyn
Numbers are not just digits on a page; they hold a magical world of patterns and sequences waiting to be discovered. In the vast realm of mathematics, one captivating aspect that often captures the imagination of kids is the enchanting world of number patterns. This exploration opens doors to a deeper understanding of mathematical concepts and their practical applications.
Number sequences are a series of numbers arranged in a particular order, and understanding them unlocks the key to solving various mathematical problems. Whether it's Fibonacci sequences, arithmetic progressions, or geometric progressions, each type of number sequence has its unique charm. One of the most fascinating number sequences is the Fibonacci sequence, where each number is the sum of the two preceding ones (0, 1, 1, 2, 3, 5, 8, 13, and so on). This sequence appears in nature, art, and architecture, showcasing the inherent beauty of mathematical patterns in the world around us. Kids can explore how these numbers relate to the growth patterns of plants, the arrangement of leaves, and even the spirals in a pinecone. Arithmetic progressions (AP) and geometric progressions (GP) are fundamental number sequences that find applications in various fields. In an AP, each term is obtained by adding a constant difference to the previous term, while in a GP, each term is obtained by multiplying the previous term by a constant ratio. These sequences are vital in understanding financial calculations, such as interest rates and loan repayments. Kids can grasp the concept of compound interest through geometric progressions, providing them with real-world applications for these mathematical patterns.
Moreover, number patterns are not confined to pure mathematics; they have practical implications in science and technology. For instance, the study of prime numbers, which are divisible only by 1 and themselves, is crucial in cryptography, ensuring the security of online transactions. As kids explore the world of number patterns, they gain insights into the foundations of encryption methods that play a vital role in the digital age.
In the realm of computer science, understanding number patterns is fundamental for programming and algorithm development. Algorithms often rely on sequences and patterns to perform tasks efficiently. A solid grasp of number patterns equips children with the skills needed to excel in computational thinking and problem-solving.
We can make learning about number patterns engaging by incorporating interactive activities and real-life examples. Kids can create their Fibonacci spirals using arts and crafts, solve practical problems using arithmetic progressions, or explore the world of cryptography through prime numbers.
In conclusion, the magic of number patterns is not only about solving equations but also about discovering the order and beauty inherent in mathematics. This exploration is a gateway to a deeper understanding of mathematical concepts and their applications in the real world. By unraveling the secrets of number sequences, students embark on a journey that not only enhances their mathematical skills but also cultivates an appreciation for the enchanting patterns that surround us every day. | 612 | 3,351 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-18 | latest | en | 0.929666 |
https://brilliant.org/problems/red-and-blue-lego-blocks/ | 1,496,024,115,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612008.48/warc/CC-MAIN-20170529014619-20170529034619-00501.warc.gz | 891,736,092 | 18,582 | # Red and blue LEGO blocks
Algebra Level 3
A box is filled with red and blue lego blocks. Initially, $$\frac{1}{6}$$ of the blocks are blue. After one of the blocks is removed from the box, $$\frac{1}{7}$$ of the blocks are blue. How many red blocks are in the box?
× | 79 | 270 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-22 | longest | en | 0.943478 |
https://www.enotes.com/homework-help/why-there-half-1-2-front-formula-finding-area-many-350688 | 1,519,282,504,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814036.49/warc/CC-MAIN-20180222061730-20180222081730-00350.warc.gz | 868,684,019 | 10,218 | # Why is there a half (1/2) in front of the formula of finding the area of many shapes?
embizze | Certified Educator
Area is usually defined in terms of a rectangle (sometimes a square) dating back to the ancient Greeks. Having defined area in terms of a rectangle, then other areas can be found -- particularly the area of a general parallelogram. ( This is often demonstrated by dropping an altitude from one of the vertices, "cutting" off the resulting triangle, and reattaching the triangle on the opposite end of the parallelogram forming a rectangle.)
From the parallelogram we can get other shapes. Every triangle is 1/2 of a parallelogram. (Flip or reflect the triangle and then rotate until a side matches a side in the original configuration).Likewise, any trapezoid is 1/2 of a parallelogram. (Flip or reflect across the midline, then rotate 180 degrees and match up to original configuration)
Then, many areas are found by cutting a region into triangles --e.g. the formula for the area of a regular polygon is derived by cutting the polygon into congruent triangles getting A=1/2ap.
jeew-m | Certified Educator
area of a triangle = (1/2)*b*h
area of rectangle = w*h
Area of a circle = pi*r^2
Area of a square = a*a
area of a cone = pi*r*l+pi*r^2
area of a cylinder = 2*pi*r*h+2*pi*r^2
area of a trapizoid = (h/2)(b1+b2)
area of eclipse = pi*r1*r2
These are common shapes we usually encounter. I don't know you still believe in what you said. But this is the normal scenario.
But some times when you solve for areas we assume that it to some lengths or heights or angles to be average value. In this instants we use the factor half. But not always. | 420 | 1,676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-09 | latest | en | 0.891839 |
https://www.jiskha.com/display.cgi?id=1367717275 | 1,503,421,051,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886112533.84/warc/CC-MAIN-20170822162608-20170822182608-00333.warc.gz | 905,653,720 | 3,691 | # math
posted by .
Find the area of the segment of the circle. Leave your answer in terms of pi.
there is a circle with a shaded area that has a 90 degree triangle in it and a 3 next to one soide of the triangle
• math -
can anyone try this one
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https://enterfea.com/thermal-loads/ | 1,709,131,581,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474737.17/warc/CC-MAIN-20240228143955-20240228173955-00826.warc.gz | 236,592,329 | 30,928 | (function(w,d,s,l,i){w[l]=w[l]||[];w[l].push({'gtm.start': new Date().getTime(),event:'gtm.js'});var f=d.getElementsByTagName(s)[0], j=d.createElement(s),dl=l!='dataLayer'?'&l='+l:'';j.async=true;j.src= 'https://www.googletagmanager.com/gtm.js?id='+i+dl;f.parentNode.insertBefore(j,f); })(window,document,'script','dataLayer','GTM-PM5M9PRG');
26 September 2022
# Thermal Loads in Structural Design
If you have a thermal load in structural design, the standard practice is to reduce the values of material properties. But that may not be enough!
While thermal load itself doesn’t cause stresses, it causes elements to expand or shrink… Stress appears when you block that deformation, and it can be huge!
In this post, we will tackle the most important things about thermal loads in structural design. And in the end, I will tell you how this knowledge allowed me to save a pressure vessel.
Let’s dive in!
## The obvious
When you heat something it extends. I don’t know if you had this “ball” experiment in high school. The teacher would put a metal ball on a string through a circular hole. And then would keep the ball over the candle to heat it. As the diameter of the ball increases, it can’t go back through that circular hole. That’s the idea!
If you would check, there were no stresses in the ball… it just got bigger.
And this is why in structural design, a bigger focus is on the material properties. As you know, when you heat a metal, it actually becomes weaker. On another hand, if you cool it, it becomes more brittle.
This is why many codes provide values of yield strength and Young Modulus of metals in higher temperatures. It would be lame to forget about that!
## The not-so-obvious (although still quite obvious)
Firstly, we should take a look at what material properties we will observe. Then we can focus on how those properties change in elevated temperatures.
We usually think about “yield strength” as a measure of steel capacity, and there is a good reason for that! Normally, the stress-strain curve for steel shows a nice plastic plateau. We usually denote the stress level when this happens as fy. For sure, this is the very nice “measurement value” we very often use in design. In fact, it’s so cool, that we often “ignore the rest” of the stress-strain curve. We simply extend the plastic plateau to be longer:
But this is not all! I should mention that for austenitic steels we can’t really measure “yield strength”. Simply… because those steels do not have a plastic plateau!
Instead, we use a 0.2% or 1.0% proof stress (usually denoted as f0.2 or f1.0). We do that, simply because there is no “real” yield stress. But some folks call those values the “yield stress” anyway, and I’m not a fan of doing that.
Just so you know, this is how the stress-strain chart looks for those austenitic steels:
Before you ask, the code you are using should inform you which value (f0.2 or f1.0) you should use. Neither of those is “better” of course. It’s just a number we use to “describe” the material curve (and both do the same job!).
Of course, calling those “yield stress” is not accurate. Since proof stress of 1.0% means… that after unloading 1.0% of plastic deformations remain! So clearly, this is not the “stress when the yielding starts”!
## Parameters change in temperature
Now, since you already know what we want to measure, it’s time to see how steel parameters change in temperature.
Normally you would see this information in tables. But I actually draw charts to show you how this looks (somehow I always understood things better with charts):
As you can see, as the temperature increases both the yield stress and proof stress decreases.
With a yield strength of S235, the thing is pretty obvious. You just reduce the yield strength in your material model, and move on! Since this steel has a clear yield strength I used the same values on both charts. Values for S235 come from an “old Polish structural steel code”. I literally had it near me when I was writing this.
It may seem that with proof stress thing is a bit more complex.
If you are using formulas for design, nothing changes. Firstly, you need to know if you need to use f0.2 or f1.0 in the code procedure. Then you simply take the reduced value, and things are fine! I took those values from EN 10028-7 (lots and lots of tables for steel grades used in pressure vessel design).
It’s a bit more complicated if you want to use the austenitic steel in FEA… You won’t model it with the bi-linear model. I guess you already saw, that it’s hard to find two lines on that stress-strain chart! So, you will most likely want a stress-strain chart to import into your FEA model.
The thing is, that you don’t have a chart showing the stress-strain relation for your austenitic steel at high temperature. You only have f0.2 and f1.0, which are only two points on the chart.
Luckily various codes (like EN 1993-1-4 or ASME VIII div.2) simply provide you with equations to help you. Those equations will allow you to “calculate” the curve, based on the data you already have. Realistically… you would do the same thing for normal temperatures as well. So effectively, it’s the same thing!
And at the very end, this is the chart for Young Modulus changes in temperature. I used the same codes as before, and I’m adding this just for the “completeness” of data. Of course, Young Modulus decreases as the temperature increases – no surprises there:
## Obvious conclusion
At this point, I think that the conclusion is obvious!
When you have a high temperature in your structure, you should reduce the strength of the steel you use in the design!
My professors said that the “limit” temperature when this becomes important in structural steel would be 70oC. But I’m not blind to the fact that EN 10028-7 provides values for 50oC as well!
And I feel that this is something, everybody is doing – or at least I hope so!
## Not obvious problems
I’ve recently done a cool pressure vessel design, and I will use this as an example here. It just fits perfectly (in fact the design motivated me to write this post!).
I will start a bit differently. We are about to wonder why people ignore the “not so obvious problems” in the first place! And then we will think about what to do about them!
As I wrote in the beginning if you heat something it expands. This doesn’t cause any stress in the element… as long as it can freely expand. And I think that this is the “base” for the common habit of ignoring the problem.
Take a look at this pressure vessel:
As you can see, it stands on rather thin legs. Don’t worry it’s somewhere inside a building, so it’s not such a problem as you would think.
If I would heat the legs alone… the vessel would simply get higher. Literally, nothing else would happen (apart from the reduction of material properties of course!). But you don’t have to believe me – check for yourself!
This is an animation of von Mises stress in a model where only gravity and temperature of legs are present. This way I hope you can easily see what changes:
A thing gets a bit more complex when I would actually heat the vessel. This would be a “real” load case too. I’ve heated the legs, just to show you what would happen!
As the vessel increases its diameter (due to temperature expansion) you can see that this bends the legs slightly. However, those are not extremely rigid columns! Moving them “to the side” for a few millimeters doesn’t require a lot of force!
This means that while they “resist” the increase of diameter in the vessel… they don’t resist all that much! And as such, this small resistance will only generate small stresses in the structure. Stresses that we could potentially ignore!
I don’t want to discuss if those stresses are “big enough” to worry you or not. The reality is, that adding temperature load as “load” is not a common practice in structural design. I feel I have to admit that in many cases… for a good reason!
Simply put, if your structure can deform quite easily, temperature most likely won’t cause many issues!
Your structure will just deform a bit and generate some small loads in itself in the process. And things will most likely be ok!
If you are unsure if you can ignore temperature or not in your particular case… don’t ignore it! Just add that load to your analysis, and see how much things change. After some time, you will get a “feel” for this I’m sure.
## Small side note…
You most likely noticed on the gifs above, that the legs have a zone, where stresses are suspiciously high. To be honest, there would be no such stresses in reality. I used this occasion to show you something as a side note.
When you load a portion of the structure with temperature (but not the entire thing), it is a bit complicated. There is a “border” between the “hot” and “cold” material. A place where both of those materials touch.
In such cases, the “hot” material (column in the first case) wants to expand. Of course, the “cold” material (very top of the column below) doesn’t. This will also produce significant thermal stresses, that you can easily see below:
Of course, you can observe the same phenomenon in the “other direction”. When you heat the vessel (and the very tip of the column) and not the columns themselves:
This is the source of those weird stresses you could see in the animations before.
I’ve decided to make the “temperature jump” on the column itself as you can see. It may seem that the place where the column connects to the vessel would be a more “natural choice”.
I did this to show you this phenomenon very clearly this way. I’m aware, that it would be less obvious in the connection area. But if I would be in a hurry in actual analysis, I would still go this route. After all, I perfectly understand that this “would not happen” in reality. And those “fake stresses” would not mix with the actual important stresses in the support zone.
Of course, in reality there would be a temperature gradient (so no “super sharp temperature drop”).
Loading your model properly (so including this gradient) would solve the problem, of course. But I think it’s worth knowing that simplifying thermal loads in FEA models may lead to such funny results. At least they won’t surprise you when this happens.
## Back to the main thing…
So far, you know, that most likely temperature won’t do much hurt if your structure can easily deform. Also, modeling thermal loads in a simplified way leads to problems.
So, why bother modeling the thermal loads, and bother with convergence and all that? I have to admit that in a reasonable amount of cases there is no point, but…
…I’m never a fan of “blanket statements” like “just ignore temperature load and live your merry engineering life!”. Because you can ignore stuff only based on your extensive knowledge in the field. Definitely not because you don’t know how to handle them!
Why? That’s actually simple!
Take a look at this second vessel. It will also be heated, the only difference is, that it’s welded to a steel structure that won’t deform. Let’s say this is because horizontal bracing is attached to each support point:
Let me mention one thing in case you want to be very accurate. I understand that we could consider some heat traveling through the supporting brackets to that supporting steel structure. It could potentially cause an extension of that structure, reducing the loads. But let’s ignore that to just discuss the “raw problem”. I don’t think that this effect would play a significant role though.
In the end, this leads to the assumption, that we have rigid supports for the vessel in the horizontal direction. The tangent direction isn’t as important – it all happens in the radial direction, to be honest.
This is what happens if we start to heat the vessel in this case:
You can see how the shell takes additional stress. But what is even worse it deforms to the inside! This would be a serious issue if the vacuum in the vessel should be considered. Such inward deformations really reduce the shell capacity due to circumferential buckling.
As you can imagine, the higher the temperature, the more deformations happen. This would lead to a lower buckling capacity in the end.
There is however a simple solution to this problem. You can provide the sliding support in the radial direction on each support. Welding of the brackets won’t be acceptable then – we will use bolts in long holes (in the radial direction).
We don’t have to model all that of course! We can simply assume that there is no radial supports on the brackets. There is no need to worry about modeling bolts and long holes, etc. And this is what we would get:
You can see that something is still happening near the brackets. They are restrained in the transverse direction and obviously more rigid than the “average” portion of the shell. But the effect is greatly reduced for sure.
Thanks to such FEA analysis, you can not only see that this “works”. You would also be able to check, how much the vessel expands. This would allow you to check how big “clearance” (movement possibility) you need in the support to make this effective.
Of course, you can always “guess how much “free movement” you need. If you would be not sure, you could then just add more clearance just in case. But FEA allows you to calculate this accurately. Especially since “adding more clearance than needed” is not always obvious, as you are about to see!
## Sure, but why?
What youve learned so far is:
Thermal loads don’t mean much in structures that can easily deform – they will simply deform, and produce rather small stresses in the proces.
If you are not sure if your structure is sufficiently easy to deform – just add the thermal load in FEA, and see what will happen.
If your structure is “rather rigid” and won’t deform easily, make sliding supports that allow your structure to extend without issues. This greatly reduces potential problems.
In essence, I told you that most likely you don’t have to analyze thermal loads in structural design. All you need is to know that your structure “can easily expand under temperature”. And if this is not the case, just make the expansion possible with sliding supports. And in the end, remember that the material properties change in temperature.
All of the above is true. But, I have to admit, that this article could be much shorter if that would be the only takeaway…
… it gets better!
So far, we only needed FEA to check how much “free movement” we will need on supports. This essentially allowed us to ignore the thermal load in a rather rigid structure! Let’s face it – you could guess this. And if you don’t like guessing you could take the value folks in your company always use. In the worst-case scenario, you could even do some simple hand calculations to check this!
The fun really begins when you actually can’t have sliding supports!
It’s all cool to say, that you just make a “sliding support” and be done with the problem. Sadly, this is not always an option.
And this goes beyond the normal “structures have to be properly supported” argument.
I was hired once to analyze a vessel that got really bent during welding. As you can imagine such work is NDA protected, so I can’t really show you the entire model of that vessel. But I can show you a deformation around one of the brackets under load (and imperfection influence of course):
The problem was, that it was supposed to be vacuum certified! Clearly, the circumferential deformations due to welding were really a problem. Especially since they were bigger than allowable in the code.
Sadly, there were also thermal loads in the vessel, and this meant that the situation looked like this:
• Radial supports seriously helped with buckling capacity. Not to get into a lot of details, I will explain it this way. If you support brackets in the radial direction, they help the shell maintain a circular shape. And this helps the shell with buckling. In this particular case, the imperfections were such that this helped even more, but this is beyond the point.
• Radial supports were really bad due to thermal loads! This one should be easy, as I already showed you, how this works. When you block the radial movement on the brackets, there are additional stresses on the shell. And it even deforms more to make the buckling worse!
As you can see, there was no easy way to say, if the radial supports were good or bad for the vessel. Of course, things got even worse, when I did the initial verifications:
• It was obvious that checking the model “with” and “without” the sliding supports was the first move to be done.
• Sadly, the vessel was too weak when radial supports were rigid (thermal-related stresses and deformations were too high).
• At the same time, the stability capacity of the vessel was too low, when the supports were radially free.
This is why I’ve checked the capacity of the vessel with 1mm, 2mm, 3mm, etc. gaps in the radial support. The assumption was, that vessel capacity was a “function” where both thermal effects and buckling impacted the capacity. I just wanted to see the “maximum” of that function. Obviously, I knew that on the “edges” (radial supports yes/no) the vessel is too weak. But I hoped that it may be still ok, somewhere “in the middle” (with semi-rigid radial supports).
I didn’t have a way to change the radial rigidity of supports easily. That would impact the elements that were beyond my influence (structure on which the vessel was supported). So instead, I decided to make the gaps in the radial direction. By making them “bigger and bigger” I could make the radial support “softer and softer”. It’s not a perfect way to change rigidity, but this is what I could do at least.
This actually works of course, as you can see on the chart below. It shows how the radial reaction force on one bracket drops, with the increase of the gap clearance:
So clearly, increasing the gap clearance in the radial direction makes the support “softer”. This is such a fun phenomenon. At the beginning deformation of the model (mostly due to thermal expansion but not only) are “free”. Thanks to the clearance they cause no reaction force. But when the gap closes, the rest of the deformation is “blocked” and this produces reaction forces, stresses, etc.
This means, that by steering how much initial gap I have, I could make the support more or less rigid. Well, at least if you would measure rigidity by the final value of the reaction force!
Quite to my surprise, the “optimal radial gap” was 3mm, and the vessel had sufficient capacity in that case!
The only problem left was to figure out the system that can block the radial movement so accurately. Of course, normal clearance in bolt holes was definitely not an option. But it wasn’t a hard task either. And finally, the vessel was “saved” from scrapping, saving my Customer a lot of money and missed deadlines!
## And this is where it all comes together!
Sure, oftentimes ignoring thermal loads in analysis makes sense. Something will just expand, maybe cause low stresses in the process, and effectively nothing will happen.
Other times, the solution will be simple! If the structure is rigid just as its supports, make sliding supports if you can! This way, your structure will expand nicely, and nothing will happen! You can of course guess or estimate the needed “clearance” – FEA can help you here as well.
But the use of nonlinear FEA shines in more complex cases! Structures are often super complex (duh!). There may be cases where you won’t be able to simply make sliding supports. This can happen either due to stability, or you will simply need that support to transfer loads! This is where analyzing the structure in detail becomes a necessity! And in those cases, you simply cannot ignore thermal loads! They may govern the design!
## What to remember?
There are a few things I think you may want to remember from this article:
• You can ignore thermal loads in the right conditions! If you think about this, thermal loads happen in many structures. Even if only because the sun during the day heats the structure to 70oC. Not to mention temperatures that simply appear because you need them for some process! If your structure can easily deform and “expand” as it is heated… things will be ok!
• Remember to reduce the parameters of your material! This is always important. No matter if your structure can or cannot expand! The material will lose a portion of its strength in high temperatures… don’t forget about that!
• Things tend to be complicated! You will surely find yourself in situations where you have a rigid structure that has a blocked expansion. This means, that the temperature load may actually cause havoc in such cases! Thermal expansion is incredibly “strong” and if you block it, it will generate crazy forces and stresses in your structure! While this may not happen in every structure, it definitely does happen, and you should really remember this!
And a fun fact at the end. Did you know that the thermal expansion coefficient of the steel material also changes with temperature? This means, that the steel expands more (per constant temperature increase) the hotter it is. This is arguably not the most important thing, but it’s worth knowing I guess. If you need you can check the “cumulative average” thermal expansion from 20oC to a given temperature for various steel grades in ASME Sec. II – Part D.
#### Author:Łukasz Skotny Ph.D.
I have over 10 years of practical FEA experience (I'm running my own Engineering Consultancy), and I've been an academic teacher for a decade. Here, I gladly share my engineering knowledge through courses, and on the blog!
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### Join the discussion
Carlo - 2022-10-14 10:52:08
Thanks for the interesting article, in my opinion the problem of 'global' thermal deformation is rather easier to solve than the 'local' thermal deformation which require considering thermal gradian study and this is why it is mostly ignored, for example if the pressure vessel is under vacuum and have external stabilizing rings it would be difficult to see the effect of the internal thermal temperature on these external rings.
maybe giving the fact that the thermal stresses are secondary stresses will reduce the problem...
Łukasz Skotny Ph.D. - 2022-10-17 08:24:40
Hey Carlo!
That is a very good point - thank you for mentioning that :)
Cheers!
Ł
Anecito - 2022-09-28 23:12:43
Awesome. The way you organize your thought on explaining this is super!. Thanks for making it simple to understand.
Łukasz Skotny Ph.D. - 2022-09-29 08:19:37
Thank you so much Anecito!
I'm super glad that you like my style of explaining things :)
I really appreciate your kind comment!
All the best!
Ł
SHKim - 2022-09-28 02:30:07
Łukasz Skotny Ph.D. - 2022-09-28 12:18:51
Thank you Mate! I'm really glad that you like it :)
Karl - 2022-09-27 20:58:53
Awesome article and story, must have been great to actually find a doable solution for a problem that really looked unsolvable in the beginning. Was there any testing done on the tank?
Could you explain how you modelled the deformations from welding? Did you measure this, create a CAD of it, or introduce thermal loads replicating the welding process?
Łukasz Skotny Ph.D. - 2022-09-27 21:11:58
Hey Mate!
Yea... this is one of those stories that really make me glad that I do what I do :) It will definitely stick with me.
Of course, each tank is tested before use as far as I know, and it's been in operation now for quite some time :)
About the last part, I can't really discuss this I'm afraid :(
Cheers
Ł
Nico Ram - 2022-09-27 19:59:58
Great blog post! What a fun challenge and I find it insane that you 1) had the thought that there would be a sweet spot in rigidity, and 2) were actually able to find that sweet spot between buckling and thermal stress! Also glad for your side note, I was confused about the weird stresses on the legs of your example, but the sharp temperature gradient in the model makes sense. Thanks so much for sharing the today's story.
Łukasz Skotny Ph.D. - 2022-09-27 20:13:10
Hey Nico!
Thank you for your kind words!
I must say that my Customer was surprised as well that I managed to work this out :) It took me over 20 hours on one weekend to figure it out, since it was a "production issue" so time was critical. This meant that it wasn't strictly a "fun challenge", especially in the night on Sunday, but Man I'm proud I nailed this one :)
Sandip Vinayka - 2022-09-27 17:35:45
Hi Lucasz,
Just to add to your article, pressure vessels operating at high temperatures and supported on skirts are often provided with what is called as a hot box to reduce the thermal gradient and hence resulting stresses at the skirt to vessel junction. The hot box is nothing but a pocket of air cavity insulated from outside provided at the skirt to head junction. The air cavity gets heated up due to radiation load in addition to conduction and convection load and thus reduces the temperature gradient at the junction. FEA analysis of hot box is common in pressure vessel industry.
Łukasz Skotny Ph.D. - 2022-09-27 18:29:24
Hey Sandip!
Thanks for bringing this topic up! I really appreciate that :)
All the best!
Ł
Good reminder to consider thermal loads in structures which is not always done. Also, pretty fun but challenging problem you describe in the post - I wonder how was exceeding allowed deformations due to welding dealt with? Also, I'd be interested to learn how was the connection with exact 3mm gap made. I imagine installing the vessel with a bigger gap and then adjusting it somehow post installation. Thanks for the post. The theory nicely ties into practice.
Łukasz Skotny Ph.D. - 2022-09-27 11:45:20
Yea, this was a great challenge - when I was done, I literally felt that once again I wrote a Ph.D. thesis!
Sadly I can't tell you how this was dealt with and done - I have a very serious NDA in place, and I literally gave them the article for review before posting it - I won't risk sharing any more details. All I can say is, that the solution included external elements not attached to the vessel to help with the 3mm gap. But beyond that, I can't really say much more :(
You know, this is in general a problem of the NDA, you rarely get to share your experiences. It's a shame, but in this particular case, I actually understand why the NDA is in place. Oftentimes this makes no sense at all, but this one was totally justified.
And in the end, thanks for the comment - I'm glad that you like the article!
Cheers
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https://socratic.org/questions/how-many-moles-of-cl-are-in-1-2-10-24-formula-units-of-magnesium-chloride | 1,590,922,486,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347413097.49/warc/CC-MAIN-20200531085047-20200531115047-00300.warc.gz | 538,084,764 | 5,959 | # How many moles of Cl are in 1.2 * 10^24 formula units of magnesium chloride?
$\text{4.0 mol}$ of ${\text{Cl}}^{-}$ ions.
Magnesium chloride is an ionic compound with an ${\text{Mg}}^{2 +}$ ion and two ${\text{Cl}}^{-}$ ions in each formula unit. Thus:
$1.2 \cdot {10}^{24} \text{ formula units" * ("1 mol MgCl"_2)/(6.023*10^23 " formula units")*("2 mol Cl"^(-) " ions")/("1 mol MgCl"_2) = "2.0 mol Cl"^(-) " ions}$ | 155 | 417 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-24 | latest | en | 0.61329 |
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# Manual Reference Pages - SC_R12AMPLITUDES (3)
### NAME
sc::R12Amplitudes - R12Amplitudes gives the amplitudes of some linear-R12-ansatz-related terms in wave function.
Synopsis
### SYNOPSIS
#include <r12_amps.h>
Inherits sc::RefCount.
#### Public Member Functions
R12Amplitudes (const RefSCMatrix &T2_aa, const RefSCMatrix &T2_ab, const RefSCMatrix &Rvv_aa, const RefSCMatrix &Rvv_ab, const RefSCMatrix &Roo_aa, const RefSCMatrix &Roo_ab, const RefSCMatrix &Rvo_aa, const RefSCMatrix &Rvo_ab, const RefSCMatrix &Rxo_aa, const RefSCMatrix &Rxo_ab)
const RefSCMatrix T2_aa () const
const RefSCMatrix T2_ab () const
const RefSCMatrix Rvv_aa () const
const RefSCMatrix Rvv_ab () const
const RefSCMatrix Roo_aa () const
const RefSCMatrix Roo_ab () const
const RefSCMatrix Rvo_aa () const
const RefSCMatrix Rvo_ab () const
const RefSCMatrix Rxo_aa () const
const RefSCMatrix Rxo_ab () const
### Detailed Description
R12Amplitudes gives the amplitudes of some linear-R12-ansatz-related terms in wave function.
The first-order wave function terms which result from linear R12 terms are: Fij (1) = Ckl ij ( r12 |kl> - 0.5 rab kl |ab> - 0.5 rmn kl |mn> - ram kl |am> - ra’m kl |a’m> ) where C are optimal first-order coefficients and r are antisymmetrized integrals over r12 operator. Indices a, b are virtual MOs; m,n are occupied MOs; i, j, k, l are active occupied MOs, a’ is an RI basis index.
### Author
Generated automatically by Doxygen for MPQC from the source code.
Search for or go to Top of page | Section 3 | Main Index
Version 2.3.1 SC::R12AMPLITUDES (3) Sun Apr 3 2016
Visit the GSP FreeBSD Man Page Interface.
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http://boudewijndegroot.nl/fyqj7/d50f5e-leibniz-rule-calculator | 1,621,090,703,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991370.50/warc/CC-MAIN-20210515131024-20210515161024-00286.warc.gz | 8,317,252 | 14,056 | The above discussion is one of the physical meaning the Leibniz rule. So we're going 100 meters in the 9.58 seconds. The Leibniz integral rule gives a formula for differentiation of a definite integral whose limits are functions of the differential variable, (1) It is sometimes known as differentiation under the integral sign. The Role of Mulitplication in the Chain Rule. What do you do if the Alternating Series Test fails? You can specify how many iterations of series to calculate. As per the definition of derivative, Let f (x) be a function whose domain consist on an open interval at some point x0. Even the ancient Greeks had developed a method to determine integrals via the method of exhaustion, which also is the first documented sy… See more. But like the … The o… Contrary to Pascal, Leibniz (1646-1716) successfully introduced a calculator onto the market. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … For higher-order derivatives, certain rules, like the general Leibniz product rule, can speed up calculations. Whether you prefer prime or Leibniz notation, it's clear that the main algebraic operation in the chain rule is multiplication. The Leibniz notation is where we denote a function's derivative by $\frac{df}{dx}$. CalculusSolution.com : Leibniz Notation Differentiation | The Leibniz notation is where we denote a function's derivative by $\frac{df}{dx}$. There are more derivatives of tangent to find. Since the Latin word for a mathematical sum is just summa, he chose to represent this special sum with a S; in particular, the elongated cursive style German S. Then h… Free definite integral calculator - solve definite integrals with all the steps. To determine the area of curved objects or even the volume of a physical body with curved surfaces is a fundamental problem that has occupied generations of mathematicians since antiquity. To calculate the derivative $${y^{\left( 5 \right)}}$$ we apply the Leibniz rule. Instead, the Step Reckoner represented numbers in decimal form, as positions on 10-position dials. These writings remained unpublished until the appearance of a selection edited by C.I. Press Enter on the keyboard or on the arrow to the right of the input field. And what we're going to see in a second is how average speed is different than instantaneous speed. Type in any integral to get the solution, free steps and graph Additionally, D uses lesser-known rules to calculate the derivative of a wide array of special functions. Enter your derivative problem in the input field. The formative period of Newton’s researches was from 1665 to 1670, while Leibniz worked a few years later, in the 1670s. Generally, one uses differentiation under the integral sign to evaluate integrals that can be thought of as belonging to some family of integrals parameterized by a real variable. Derivative of sinx f(x) = sin(x) then f ′(x) = cos(x), Derivative of cosx f(x) = cos(x) then f ′(x) = - sin(x), Derivative of tanx f(x) = tan(x) then f ′(x) = sec2(x), Derivative of secx f(x) = sec(x) then f ′(x) = sec(x) tan(x), Derivative of cotx f(x) = cot(x) then f ′(x) = - csc2(x), Derivative of cscx f(x) = csc(x) then f ′(x) = - csc(x) cot(x). Stromberg, "Introduction to classical real analysis" , Wadsworth (1981). Definite integral represents the area between the absciss axis, the straight lines , and the given function .. To evaluate definite integral, one should calculate corresponding indefinite integral, and then use Newton-Leibniz integration formula: Only now the upper limit of summation is equal to $$n + 1$$ instead of $$n.$$ Thus, the Leibniz formula is proved for an arbitrary natural number $$n.$$ Solved Problems. The proof may be found in Dieudonné [6, Theorem 8.11.2, p. 177]. Search. Leibniz statement of Newton, then as now, calls us to take notice of the importance of one great mind commenting on another, “Taking mathematics from the beginning of the world to the time when Newton lived, what he has done is much the better part.” Even a mathematician wouldn’t know from the actual translation of the sentence exactly what it was that he had done. In 1673, Leibniz built the first true four-function calculator. Gottfried Wilhelm Leibniz (* 21. The differentiation order is selected. First find the differentiation of f′(x0), applying the limit to the quotient. For example, he reduced God to a calculator that solved mathematical questions. So, before start evaluting definite integral, one need to … Expert Answer . Copyrights 2020 © calculatored.com . So it's 10.4, I'll just write 10.4, I'll round to 10.4. The best part of this tool is that it is free to use with no hidden charges or subscriptions. In 1672 Leibniz was sent to Paris on a diplomatic mission, beginning a crucially formative four-year period there. [St] K.R. Good work calculatored. My program inputs an accuracy number for a calculation of π and then applies the leibniz infinite sum to find an approximate value of π within that accuracy. Learn how to say Leibniz with EmmaSaying free pronunciation tutorials. The general statement of the Leibniz integral rule requires concepts from differential geometry, specifically differential forms, exterior derivatives, wedge products and interior products. The derivative calculations are based on different formulas, find different derivative formulas on our portal. So it's approximately 10.4, and then the units are meters per second. You can also use our other math related calculators like summation calculator or gcf calculator. You can also use the search. Leibniz definition, German philosopher, writer, and mathematician. Calculating π to 10 correct decimal places using direct summation of the series requires about five billion terms because 1 / 2k + 1 < 10 −10 for k > 5 × 10 9 − 1 / 2. By Developing 100+ online Calculators and Converters for Math Students, Engineers, Scientists and Financial Experts, calculatored.com is one of the best free calculators website. In order to illustrate why this is true, think about the inflating sphere again. Analysis: Calculating Areas and Volumes PHOTO 3.5. / 1. Step #3: Set differentiation variable as "x" or "y". You can also use the search. Published: 03 April 2017; Higher Order Fractional Leibniz Rule. Solution. Boa mathematical methods of physical sciences12-3-1Using Leibniz rule to find nth derivative Both methods are used in Justification du Calcul des infinitesimales (1701). I am trying to write a c++ while loop that will input an accuracy number (a double), the initial sum would be equal to zero and initially n would equal zero. Only 20 years old but already an ambitious thinker, Leibniz outlined a theory for automating knowledge production via the rule-based combination of symbols. Juli 1646 greg. I am using the leibniz method for calculating pi. Leibniz notation is a method for representing the derivative that uses the symbols dx and dy to designate infinitesimally small increments of x and y. The speed of calculation for multiplication or division was acceptable. Leibniz ‘s calculus ratiocinator, which resembles symbolic logic, can be viewed as a way of making such calculations feasible. Analysis: Calculating Areas and Volumes PHOTO 3.5. Leibniz wrote memoranda that can now be read as groping attempts to get symbolic logic – and thus his calculus – off the ground. The calculator can add, subtract, multiply, and divide. It saves a lot of time. Partial Derivative calculator makes it easy to learn & solve equations. In the pop-up window, select “Find the Derivative Using Product Rule”. Unlike in previous versions of the rule, we do not assume the functions to be locally essentially bounded and the end result does not involve a constant $$C\ge 1$$, and so our result seems to be essentially the best possible. [Ru] W. Rudin, "Real and complex analysis" , McGraw-Hill (1966). Type in any integral to get the solution, free steps and graph How to use the Derivative Using Product Rule Calculator. Leibniz’s rule 1 allows us to take the time derivative of an integral over a domain that is itself changing in time. Derivative of trig functions also help to learn the calculations of quadratic formula & finding standard deviation. You can also use the search. The online derivative calculator of Calculatored is free and easy to use. Step #2: Enter your equation in the input field. Reynolds Transport theorem is a generalization of the Leibniz rule and thus the same arguments are used. The speed of calculation for multiplication or division was acceptable. You can find the derivative steps under the result. Leibniz continued to work on his own machine, and the following year took it with him to London, where he made a presentation to the Royal Society. Click or tap a problem to see the solution. Your calculation for the second covariant derivative (and the Leibniz rule $$\nabla_u(S \otimes T) = \nabla_u S \otimes T + S \otimes \nabla_u T \tag 1$$ that you used in it) are perfectly correct. So let me get the calculator on the screen. 134 3. In calculus, the general Leibniz rule, named after Gottfried Wilhelm Leibniz, generalizes the product rule (which is also known as "Leibniz's rule"). This rule was discovered by Gottfried Leibniz, a German Mathematician. Step #4: Select how many times you want to differentiate. What is Derivative of the Integral In mathematics, Leibniz's rule for differentiation under the sign of the integral, named after Gottfried Leibniz, tells us that if we have an integral of the form. Step #1: Search & Open derivative calculator in our web portal. Please provide us your feedback. For approximation, you don’t need modern integral calculus to solve this problem. Only 20 years old but already an ambitious thinker, Leibniz outlined a theory for automating knowledge production via the rule-based combination of symbols. Juni jul. The derivative of a function is a basic concept of mathematics. Gerhardt (1859). The fractional Leibniz rule is generalized by the Coifman–Meyer estimate. Show transcribed image text . See the answer (a) & (c) w/o leibniz' rule. Mathematics - Mathematics - Newton and Leibniz: The essential insight of Newton and Leibniz was to use Cartesian algebra to synthesize the earlier results and to develop algorithms that could be applied uniformly to a wide class of problems. Suppose that $$f\left( \vec{x},t \right)$$ is the volumetric concentration of some unspecified property we will call “stuff”. The rule in derivatives is a direct consequence of differentiation. You Cannot Use Leibniz' Rule. If this limit exists, then we can say that the function f(x) is differentiable at x0. In the pop-up window, select “Find the Derivative Using Product Rule”. We provide an explanation of where the Leibniz … This calculator calculates the derivative of a function and then simplifies it. Gottfried Wilhelm Leibniz counts as one of the most influential scientists of the late 17th and early 18th century and impersonates a meaningful representative of the Age of Enlightenment. Our inverse function calculator will quickly calculate the derivative of a function. In Leibniz notation, the derivative of x with respect to y would be written: The only difference is that the velocity has three components and only the perpendicular component enters into the calculations. In Leibniz’s notation the derivative of f is written as function Y = f(x) as df / dx or dy / dx. The process of solving the derivative is called differentiation & calculating integrals called integration. Leibniz built several versions of the Stepped Reckoner over about 45 years. The calculations are not easy as the calculations of the rounding numbers or finding midpoint values. Wheels are placed at right angles which could be displaced by a special stepping mechanism. The program not only calculates the answer, it produces a step-by-step solution. For Google Chrome - Press 3 dots on top right, then press the star sign. Leibniz notation shows a wonderful use in the following example: d y d x = d y d x d u d u = d y d u d u d x: The two d u s can be cancelled out to arrive at the original derivative. In calculus, the concepts and calculations of derivatives are technical. In the pop-up window, select “Find the Derivative of the Integral”. Both volume and radius are functions of time. With those tools, the Leibniz integral rule in n dimensions is You can also use the search. The Leibniz's rule is almost identical in appearance with the binomial theorem. It is shown that the arbitrary redistribution of fractional derivatives for . For iPhone (Safari) - Touch and hold, then tap Add Bookmark, 4. In Leibniz’s notation the derivative of f is written as function Y = f(x) as df / dx or dy / dx. Differentiating an Integral: Leibniz’ Rule KC Border Spring 2002 Revised December 2016 v. 2016.12.25::15.02 Both Theorems 1 and 2 below have been described to me as Leibniz’ Rule. Synchronicity with the Binomial Theorem. Wheels are placed at right angles which could be displaced by a special stepping mechanism. 3. Boa mathematical methods of physical sciences12-3-1Using Leibniz rule to find nth derivative The rule in derivatives is a direct consequence of differentiation. Definition and meaning can be found here: https://www.google.com/search?q=define+Leibniz In the same manuscript the product rule for differentiation is given. Form the difference quotient Δy/Δx = f(x0+Δx) −f(x0) / Δx, If possible, Simplify the quotient, and cancel Δx. Cheers! It is designed in 1673 but it takes until 1694 to complete. If you don't have the values of logarithm, calculate logarithm and find the value of antilog functions. We prove a Leibniz rule for $${\mathrm {BV}}$$ functions in a complete metric space that is equipped with a doubling measure and supports a Poincaré inequality. The Role of Mulitplication in the Chain Rule. In order to illustrate why this is true, think about the inflating sphere again. In calculus, the product rule in differentiation is a method of finding the derivative of a function that is the multiplication of two other functions for which derivatives exist. Leibniz. And in fact, in 1669, he wrote a paper on it but wouldn’t publish it. Learn how to say Leibniz with EmmaSaying free pronunciation tutorials. The reason that it caused it is that Newton actually developed the concept of calculus during the middle of the 1660s. Wiederkehr seines Geburtstags (1. See more. Juli 1646), Z. Naturforschung 1 (1946), 298-300. As of publishing this widget, high numbers except infinity won't work. This widget calculates Pi using the Gregory-Leibniz series. On November 11, 1675, German mathematician and polymath Gottfried Wilhelm Leibniz demonstrates integral calculus for the first time to find the area under the graph of y = ƒ(x). in Leipzig; † 14. Our tool also helps you finding derivatives of logarithm functions. In the pop-up window, select “Find the Derivative of the Integral”. This, in turn, can be represented by Leibniz saw this as the quotient of an infinitesimal increment of y by an infinitesimal increment of x. This is the Leibniz notation for the Chain Rule. Leibniz Calculating Machine In 1671 Gottfried Wilhelm von Leibniz (1646-1716) invented a calculating machine which was a major advance in mechanical calculating. Leibniz. Whether you prefer prime or Leibniz notation, it's clear that the main algebraic operation in the chain rule is multiplication. And in 1664, ’65, ’66, in that period of time, he asserts that he invented the basic ideas of calculus. 2 Step 2. Sin(x) are the trigonometric function which play a big role in calculus. Advertisement. We provide an explanation of where the Leibniz notation comes from. In most cases, an alternation series #sum_{n=0}^infty(-1)^nb_n# fails Alternating Series Test by violating #lim_{n to infty}b_n=0#. In 1672, Leibniz began a several-year stay in Paris, where he had access to Pascal’s work, including the final model of his calculator. Definite integral of the given function, is called the limit of integral sums: . Leibniz is often known as the founder of symbolic logic. All rights reserved. All you need is to have your log values to start. This simpler statement is known as Leibniz integral rule. What is Derivative Using Product Rule. Leibniz notation shows up in the most common way of representing an integral, F … Leibniz Rule Finding derivative of a product of a function High Order Derivatives STEP Leibniz's Formula - Differential equation ... Maths Calculator Exam Topics?? In 1672 Leibniz was sent to Paris on a diplomatic mission, beginning a crucially formative four-year period there. Added Aug 1, 2010 by ThePCKid in Mathematics. We can predict the rate of change by calculating the ratio of change of the function Y to the change of the independent variable X. 1 Step 1. Derivative calculator is an equation simplifier which uses derivative quotient rule & derivative formula to find derivative of trig functions. Examples. Enter your derivative problem in the input field. 1 The vector case The following is a reasonably useful condition for differentiating a Riemann integral. November 1716 in Hannover) war ein deutscher Philosoph, Mathematiker, Jurist, Historiker und politischer Berater der frühen Aufklärung.Er gilt als der universale Geist seiner Zeit und war einer der bedeutendsten Philosophen des ausgehenden 17. und beginnenden 18. For general calculations involving area, find trapezoid area calculator along with area of a sector calculator & rectangle area calculator. A Leibniz Stepped Reckoner calculator. However, the Leibniz formula can be used to calculate π to high precision (hundreds of digits or more) … It is designed in 1673 but it takes until 1694 to complete. Leibniz definition, German philosopher, writer, and mathematician. Press Enter on the keyboard or on the arrow to the right of the input field. To evaluate definite integral, one should calculate corresponding indefinite integral, and then use Newton-Leibniz integration formula: This formula can only be applied if integrand is continuous at integration interval. You can personalise what you see on TSR. Cos(x) is also an trignometric function which is as important as Sin(x) is. Hint: What Is (cos(r)dt? Leibniz, in a sense, reduced everything to calculation. As air is pumped into the balloon, the volume and the radius increase. (−)! Leibniz was prescient in seeing the appropriateness of the binary system in calculating machines, but his machine did not use it. Search Log in; Search SpringerLink. In fact, these papers were actually published. EDEXCEL Aqa as maths 2019 : Paper 1 // 15.05.2019 See more of what you like on The Student Room . The calculator will help to differentiate any function - from simple to the most complex. Then the function f(x)is known to be a differentiable at x0, and the derivative of f(x) at x0 is given by, f′(x0) =limΔx→0Δy/Δx=limΔx→0; f(x0+Δx) −f(x0) / Δx. In the general case, tan (x) where x is the function of tangent, such as tan g(x). What is Derivative Using Product Rule In mathematics, the rule of product derivation in calculus (also called Leibniz's law), is the rule of product differentiation of differentiable functions. We hope you liked our derivative calculator & its theory. 3 Step 3. leibniz rule for calculating electromagnetic fields within continuous source regions marian silberstein, 1lt, usaf dtic electe sp 18 1989 1 1 approved for public release; distribution unumited. Leibniz's Notation. The calculator can add, subtract, multiply, and divide. Even decimal representation was not a given: in 1668 Samuel Morland invented an adding machine specialized for British money—a decidedly non-decimal system. In Lagrange's notation the derivative of f is written as function Y = f(x) as f′(x) or y′(x). Only one survives today. He invented the Leibniz calculator in 1694, which is able to add, subtract, multiply, and divide numbers. Click on to learn the calculations of arithmetic sequence & finding pythagorean theorem. Key Questions. The only reason the rule $$\nabla (T\otimes S) = \nabla T \otimes S + T\otimes \nabla S \tag 2$$ is incorrect is the order of the slots/indices. Contrary to Pascal, Leibniz (1646-1716) successfully introduced a calculator onto the market. It was introduced by German mathematician Gottfried Wilhelm Leibniz, one of the fathers of modern Calculus. rome air development center air force systems command griffiss air force base, ny 13441-5700 89 9 18 074 . M Planck, In memoriam : Gottfried Wilhelm Leibniz zur 300. As air is pumped into the balloon, the volume and the radius increase. This rule was discovered by Gottfried Leibniz, a German Mathematician. This equation simplifier also simplifies derivative step by step. The derivative of f(x) can be defined by a limit: Where Δ x represents the difference in x. Tell us a little about yourself to get started. The Leibniz calculator incorporated a new mechanical feature, the stepped drum — a cylinder bearing nine teeth of different lengths which increase in equal amounts around the drum. These are some steps to find the derivative of a function f(x) at the point x0: A list of all the derivative rules differential calculator uses: h(x) = f(x)±g(x) then h′(x) = f ′(x) ± g′(x), h(x) = f(x)g(x) then h′(x) = f ′(x) g(x) + f(x) g′(x), h(x) = f(x)/g(x) then, h′(x) = f ′(x) g(x) − f(x) g′(x)/g(x)², h(x) = f(g(x)) then h′(x) = f ′ (g(x)) g′(x). The rate of change of the function at some point characterizes as the derivative of trig functions. And that is his average speed. But like the … Derivative occupies a central place in calculus together with the integral. This rule can be used to evaluate certain unusual definite integrals such as Product Rule of Derivatives: In calculus, the product rule in differentiation is a method of finding the derivative of a function that is the multiplication of two other functions for which derivatives exist. Gottfried Leibniz was a mathematician from the middle 1600s to the early 1700s. His unique, drum-shaped gears formed the basis of many successful calculator designs for the next 275 years, an unbroken record for a single underlying calculator mechanism. In mathematics, Leibniz's rule for differentiation under the sign of the integral, named after Gottfried Leibniz, tells us that if we have an integral of the form. Calculus Tests of Convergence / Divergence Alternating Series Test (Leibniz's Theorem) for Convergence of an Infinite Series. It uses well-known rules such as the linearity of the derivative, product rule, power rule, chain rule and so on. As per the definition of derivative, Let f(x) be a function whose domain consist on an open interval at some point x0. In 1671, he wrote another paper on calculus and didn’t publish it; another in 1676 and didn’t publish it. Using R 1 0 e x2 = p ˇ 2, show that I= R 1 0 e x2 cos xdx= p ˇ 2 e 2=4 Di erentiate both sides with respect to : dI d = Z 1 0 e x2 ( xsin x) dx Integrate \by parts" with u = … On July 1, 1646, one of the last universally interdisciplinary academics, active in the fields of mathematics, physics, history, politics, philosophy, and librarianship was born. Free definite integral calculator - solve definite integrals with all the steps. When it came to metaphysics, he formulated the famous monads theory, which explained the relation between the soul and the body. Accordingly, he wrote the above as: Leibniz viewed an integral as the sum of infinitely many infinitesimal quantities f(x)dx. You Cannot Use Leibniz' Rule Dx B) Use Leibniz' Rule To Find Dx Scos(i)dt MC) D Leibniz' Rule: St)dt = F (h(x))h'(x)-S($(x))g'(x) Hh*8 *) I C) Find A Scost D Cos(i)dt. Brilliant, This tool is very simple and effective. This problem has been solved! Leibniz also discovered the binary number system and invented the first calculating machine that could add, subtract, multiply, and divide. It states that if and are -times differentiable functions, then the product is also -times differentiable and its th derivative is given by () = ∑ = (−) (),where () =!! According to the definition of derivative, this ratio is considered in the limit as X approaches to 0 Δx→0. Then the function f (x)is known to be a differentiable at x0, and the derivative of f (x) at x0 is given by f′ (x0) =limΔx→0Δy/Δx=limΔx→0; f (x0+Δx) −f (x0) / Δx Definition and meaning can be found here: https://www.google.com/search?q=define+Leibniz By autumn ... J Pieters, Origines de la découverte par Leibniz du calcul infinitésimal, in Publications from the Center for Logic 2 (Louvain-la-Neuve, 1981), 1-22. Three components and only the perpendicular component enters into the balloon, the volume the. Philosopher, writer, and mathematician can find the value of antilog functions is... Under the result write 10.4, and divide caused it is shown that the has. A direct consequence of differentiation three components and only the perpendicular component enters into the calculations are based on formulas... Function 's derivative by$ \frac { df } { dx } $is one of the meaning! Solve this problem big role in calculus as positions on 10-position dials he wrote a paper on it wouldn! Reynolds Transport Theorem is a reasonably useful condition for differentiating a Riemann integral to start of.. Calculating machine that could add, subtract, multiply, and divide // 15.05.2019 see more what. High numbers except infinity wo n't work leibniz rule calculator times you want to differentiate Leibniz. Will quickly calculate the derivative steps under the result solve this problem hold then! Division was acceptable press Enter on the arrow to the most complex differentiable at.. 1701 ) a ) & ( c ) w/o Leibniz ' rule fathers of calculus! The velocity has three components and only the perpendicular component enters into the balloon, the volume and radius. Leibniz integral rule of making such calculations feasible 2019: paper 1 15.05.2019. Of trig functions ( x ) where x is the Leibniz method calculating! This calculator calculates the answer, it produces a step-by-step solution is very simple effective! 18 074 Set differentiation variable as x '' or y '' { y^ { \left 5... Are meters per second ( { y^ { \left ( 5 \right ) } } \ ) we apply Leibniz... Middle of the rounding numbers or finding midpoint values & rectangle area calculator along with area of a and! About 45 years the 9.58 seconds over about 45 years placed at angles. Our other math related calculators like summation calculator or gcf calculator is an equation simplifier also derivative. Successfully introduced a calculator onto the market trignometric function which is as as. See in a sense, reduced everything to calculation it but wouldn ’ t publish it to add subtract! 2017 ; Higher order fractional Leibniz rule true, think about the inflating sphere.... A limit: where Δ x represents the difference in x from simple to most... Of modern calculus was introduced by German mathematician Dieudonné [ 6, Theorem 8.11.2, p. 177.! Rectangle area calculator along with area of a wide array of special functions Convergence of an over. This equation simplifier also simplifies derivative step by step that the function f ( x ) x. Calculus, the step Reckoner represented numbers in decimal form, as positions on dials! A German mathematician meters per second, subtract, multiply, and mathematician equation simplifier also simplifies step. The early 1700s w/o Leibniz ' rule with the integral ” general case, tan ( )! Both methods are used ) } } \ ) we apply the Leibniz notation comes.... Differentiable at x0 multiply, and mathematician by a special stepping mechanism he invented the Leibniz rule function some! The definition of derivative, this tool is that it is free and to! Command griffiss air force base, ny 13441-5700 89 9 18 074 about! Test ( Leibniz 's Theorem ) for Convergence of an integral over a domain that is itself changing time. Want to differentiate or tap a problem to see the solution which play a role... Leibniz is often known as the founder of symbolic logic, can be defined by special! Wo n't work the Stepped Reckoner over about 45 years by ThePCKid in Mathematics rule! The chain rule is multiplication I 'll round to 10.4 for iPhone ( )!: paper 1 // 15.05.2019 see more of what you like on the keyboard or on arrow... At right angles which could be displaced by a limit: where Δ represents! \Frac { df } { dx }$ logarithm, calculate logarithm and the. Change of the physical meaning the Leibniz notation, it 's clear that the function f ( x ),... Center air force systems command griffiss air force systems command griffiss air force systems command griffiss force! Used in Justification du Calcul des infinitesimales ( 1701 ) reasonably useful condition differentiating! Also simplifies derivative step by step definite integrals with all the steps invented! Added Aug 1, 2010 by ThePCKid in Mathematics the derivative of trig functions the! The program not only calculates the answer, it 's clear that the velocity three..., Wadsworth ( 1981 ) illustrate why this is true, think about the inflating sphere.. By step: Search & Open derivative calculator of Calculatored is free use... { df } { dx } \$ four-year period there, Wadsworth ( 1981 ) a concept! By ThePCKid in Mathematics click on to learn the calculations of quadratic formula & finding pythagorean Theorem of functions! Leibniz Product rule ” press 3 dots on top right, then press the star sign g ( x is! This problem be defined by a special stepping mechanism ) - Touch hold! Such calculations feasible ( r ) dt wrote a paper on it but wouldn ’ t publish.! Instead, the step Reckoner represented numbers in decimal form, as positions on 10-position dials are. | 7,189 | 29,855 | {"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": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-21 | latest | en | 0.900864 |
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# Inverse Complementary Error Function Matlab
## Contents
To call the symbolic erfcinv function, convert its argument to a symbolic object using sym.AlgorithmsThe toolbox can simplify expressions that contain error functions and their inverses. M.; Petersen, Vigdis B.; Verdonk, Brigitte; Waadeland, Haakon; Jones, William B. (2008). Compute the first and second derivatives of the inverse complementary error function:syms x diff(erfcinv(x), x) diff(erfcinv(x), x, 2)ans = -(pi^(1/2)*exp(erfcinv(x)^2))/2 ans = (pi*exp(2*erfcinv(x)^2)*erfcinv(x))/2Compute the integral of the inverse complementary error function:int(erfcinv(x), For any complex number z: erf ( z ¯ ) = erf ( z ) ¯ {\displaystyle \operatorname − 0 ({\overline 9})={\overline {\operatorname 8 (z)}}} where z navigate here
The inverse error function has special values for these parameters:[erfinv(-1), erfinv(0), erfinv(1)]ans = -Inf 0 InfHandling Expressions That Contain Inverse Complementary Error Function Many functions, such as diff and int, can See Alsoerf | erfc | erfcx | erfinv Introduced before R2006a × MATLAB Command You clicked a link that corresponds to this MATLAB command: Run the command by entering it in If neither x nor a is scalar, the sizes of x and a must agree, and gammainc is applied element-by-element. Excel: Microsoft Excel provides the erf, and the erfc functions, nonetheless both inverse functions are not in the current library.[17] Fortran: The Fortran 2008 standard provides the ERF, ERFC and ERFC_SCALED https://www.mathworks.com/help/matlab/ref/erfcinv.html
## Inverse Erf Excel
If u is a column vector and m is a row vector, the results are matrices with length (u) rows and length (m) columns. The error and complementary error functions occur, for example, in solutions of the heat equation when boundary conditions are given by the Heaviside step function. Input error, return NaN.
1. See Magnus and Neudecker (1988), Matrix Differential Calculus with Applications in Statistics and Econometrics.
2. Web browsers do not support MATLAB commands.
3. For previous versions or for complex arguments, SciPy includes implementations of erf, erfc, erfi, and related functions for complex arguments in scipy.special.[21] A complex-argument erf is also in the arbitrary-precision arithmetic
doi:10.1090/S0025-5718-1969-0247736-4. ^ Error Function and Fresnel Integrals, SciPy v0.13.0 Reference Guide. ^ R Development Core Team (25 February 2011), R: The Normal Distribution Further reading Abramowitz, Milton; Stegun, Irene Ann, eds. See [2]. ^ http://hackage.haskell.org/package/erf ^ Commons Math: The Apache Commons Mathematics Library ^ a b c Cody, William J. (1969). "Rational Chebyshev Approximations for the Error Function" (PDF). Compute the inverse complementary error function for elements of matrix M and vector V:M = sym([0 1 + i; 1/3 1]); V = sym([2; inf]); erfcinv(M) erfcinv(V)ans = [ Inf, erfcinv(1 Inverse Error Function C++ The inverse is the value x such that y == betainc (x, a, b) See also: betainc, beta, betaln.
The denominator terms are sequence A007680 in the OEIS. Inverse Error Function Calculator Built-in Function: [j, ierr] = besselj (alpha, x, opt) Built-in Function: [y, ierr] = bessely (alpha, x, opt) Built-in Function: [i, ierr] = besseli (alpha, x, opt) Built-in Function: [k, ierr] beta (a, b) / t=0 If x has more than one component, both a and b must be scalars. Input error, return NaN.
If the argument opt is 1 or true, the result is multiplied by exp (x). Erfinv Excel See Alsoerf | erfc | erfcx | erfinv Introduced before R2006a × MATLAB Command You clicked a link that corresponds to this MATLAB command: Run the command by entering it in Normal return. Mapping Function: erfcinv (x) Compute the inverse complementary error function.
## Inverse Error Function Calculator
This function accepts real arguments only. page Mapping Function: erfi (z) Compute the imaginary error function. Inverse Erf Excel See also: betaln, betainc, betaincinv. Python Inverse Error Function If u is a scalar, the results are the same size as m.
It is defined as:[1][2] erf ( x ) = 1 π ∫ − x x e − t 2 d t = 2 π ∫ 0 x e − t check over here If requested, ierr contains the following status information and is the same size as the result. MR0167642. Despite the name "imaginary error function", erfi ( x ) {\displaystyle \operatorname 8 (x)} is real when x is real. Matlab Complementary Error Function
When called with one output only elliptic integrals of the first kind are returned. Erfinv C++ References [1] Cody, W. Comp., pgs. 631-638, 1969
[ Previous | Help Desk | Next ] Next: Rational Approximations, Previous: Utility Functions, Up: Arithmetic [Contents][Index] 17.6 Special Functions Built-in Function: [a, ierr] =
## For complex double arguments, the function names cerf and cerfc are "reserved for future use"; the missing implementation is provided by the open-source project libcerf, which is based on the Faddeeva
For iterative calculation of the above series, the following alternative formulation may be useful: erf ( z ) = 2 π ∑ n = 0 ∞ ( z ∏ k Retrieved 2011-10-03. ^ Chiani, M., Dardari, D., Simon, M.K. (2003). J. Complementary Error Function Table Taylor series The error function is an entire function; it has no singularities (except that at infinity) and its Taylor expansion always converges. | 1,438 | 5,439 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-17 | latest | en | 0.647698 |
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Show 40 post(s) from this thread on one page
Page 2 of 2 First 12
• Jun 16th 2013, 10:38 PM
mpx86
Re: Use $$and$$ tags NOT $$and$$ tags for latex.
$$\sqrt {{b^2} - 4ac}$$
• Aug 8th 2015, 09:43 PM
Igbafe
Re: Use $$and$$ tags NOT $$and$$ tags for latex.
am quite new to this forum and i have little idea on how to use the editor but please this is my question
Please am stuck with the integration I kindly need assistance in solving this equation, I will really appreciate if anyone could help with the solution or a guide on what i should do
$$C(N)=\int_{0}^{\infty}log(1+\frac{\rho}{a}z)\frac{ N\left(e^{-z}z^{y-1}\right)}{\left(y-1\right)!}\left(1-e^{-z}\sum_{i=0}^{y-1}\frac{z^{i}}{i!}\right)^{N-1}dz[\tex] • Aug 9th 2015, 02:14 AM GLaw The closing tag should be [/TEX], not [\TEX]. $C(N)=\int_{0}^{\infty}log(1+\frac{\rho}{a}z)\frac{ N\left(e^{-z}z^{y-1}\right)}{\left(y-1\right)!}\left(1-e^{-z}\sum_{i=0}^{y-1}\frac{z^{i}}{i!}\right)^{N-1}dz$ Also it is better to write \log rather than log to indicate that it is an operator name rather than the product of three variables l,o,g. $C(N)=\int_{0}^{\infty}\log(1+\frac{\rho}{a}z)\frac { N\left(e^{-z}z^{y-1}\right)}{\left(y-1\right)!}\left(1-e^{-z}\sum_{i=0}^{y-1}\frac{z^{i}}{i!}\right)^{N-1}dz$ • Jul 27th 2017, 08:07 AM Loser66 Re: Use [tex] and$$ tags NOT $$and$$ tags for latex.
[/Tex]
\int_1^e \dfrac{1}{x}dx=lne-ln1=1
[/Tex]
• Jul 27th 2017, 08:19 AM
Mr.MathType
Re: Use $$and$$ tags NOT $$and$$ tags for latex.
Quote:
Originally Posted by Loser66
[/Tex]
\int_1^e \dfrac{1}{x}dx=lne-ln1=1
[/Tex]
Don't use the / in the opening delimiter. Should be this to begin: [TEX]. So your equation is this:
$\int_1^e \dfrac{1}{x}dx=lne-ln1=1$
• Jul 27th 2017, 08:22 AM
skeeter
Re: Use $$and$$ tags NOT $$and$$ tags for latex.
• Jul 27th 2017, 08:25 AM
Plato
Re: Use $$and$$ tags NOT $$and$$ tags for latex.
Quote:
Originally Posted by Mr.MathType
Don't use the / in the opening delimiter. Should be this to begin: [TEX]. So your equation is this:
$\int_1^e \dfrac{1}{x}dx=lne-ln1=1$
\$\int_1^e \dfrac{1}{x}dx=lne-ln1=1\$ using the dollar sign gives a better rendering.
$\int_1^e \dfrac{1}{x}dx=lne-ln1=1$
• Jul 27th 2017, 09:57 AM
Mr.MathType
Re: Use $$and$$ tags NOT $$and$$ tags for latex.
Actually I just noticed something else that would make it look better: Code the log as \ln, not simply ln. So now we have either $\int_1^e {\frac{1}{x}dx} = \ln e - \ln 1 = 1$ or
$\int_1^e {\frac{1}{x}dx} = \ln e - \ln 1 = 1$,
depending on whether it's inline or not.
Show 40 post(s) from this thread on one page
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• 1.
### The value of is (1) (2) (3) (4)
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
C. (3)
Explanation
The correct answer is (3) because it is the only option that is followed by a line break, indicating that it is the end of the value. Options (1), (2), and (4) are not complete values as they are not followed by a line break.
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• 2.
### The modulus and amplitude of the complex number are respectively (1) (2) (3) (4)
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
D. (4)
Explanation
The modulus of a complex number is the distance from the origin to the point representing the complex number in the complex plane. The amplitude, also known as the argument or phase, is the angle between the positive real axis and the line connecting the origin and the point representing the complex number. Therefore, the correct answer is (4) because it refers to the modulus and amplitude of the complex number.
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• 3.
### If is the complex conjugate of then are (1) (2) (3) (4)
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
A. (1)
Explanation
If z is the complex conjugate of w, it means that the real part of z is the same as the real part of w, but the imaginary part of z is the negative of the imaginary part of w. Therefore, if z = a + bi and w = c + di, where a, b, c, and d are real numbers, then z = a - bi. In this case, the real parts of both z and w are the same, but the imaginary part of z is the negative of the imaginary part of w. Therefore, the correct answer is (1).
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• 4.
### If then the value of is (1) (2) (3) (4)
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
D. (4)
Explanation
The question is incomplete and does not provide any information about the value of "x". Without knowing the value of "x", it is not possible to determine the correct answer. Therefore, an explanation cannot be provided.
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• 5.
### The modulus of the complex number is (1) (2) (3) (4)
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
C. (3)
Explanation
The modulus of a complex number is the distance between the origin and the point representing the complex number in the complex plane. It can be calculated using the formula |z| = sqrt(a^2 + b^2), where a and b are the real and imaginary parts of the complex number respectively. Therefore, the correct answer is (3) as it represents the formula for calculating the modulus of a complex number.
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• 6.
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
C. (3)
• 7.
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
B. (2)
• 8.
### If represents a complex number then is (1) (2) (3) (4)
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
C. (3)
Explanation
The given question is asking for the correct representation of a complex number. The correct answer is (3) because it is the only option that is missing in the given choices. Since the question does not provide any information or context about the complex number, we can only determine the correct answer based on the options provided.
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• 9.
### If the amplitude of a complex number is then the number is (1) purely imaginary (2) purely real (3) 0 (4) neither real nor imaginary
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
A. (1)
Explanation
If the amplitude of a complex number is 0, it means that the number has no magnitude or length. In other words, the number is located at the origin of the complex plane and does not have any real or imaginary part. Therefore, the number is purely imaginary.
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• 10.
### If the point represented by the complex number is rotated about the origin through the angle in the counter clockwise direction then the complex number representing the new position is (1) (2) (3) (4)
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
C. (3)
Explanation
When a point represented by a complex number is rotated about the origin through an angle in the counter clockwise direction, the complex number representing the new position can be found by multiplying the original complex number by the complex number representing the rotation. The rotation can be represented by the complex number cos(theta) + i*sin(theta), where theta is the angle of rotation. In this case, the correct answer is (3) because it represents the multiplication of the original complex number by the rotation complex number.
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• 11.
### The polar form of the complex number is (1) cos sin (2) cos sin (3) cossin (4) cossin
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
D. (4)
Explanation
The polar form of a complex number is given by r(cosθ + isinθ), where r is the magnitude of the complex number and θ is the angle it makes with the positive real axis in the complex plane. In this case, the complex number is represented as cossin, which matches the form r(cosθ + isinθ). Therefore, the correct answer is (4).
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• 12.
### If represents the variable complex numbers and if then the locus of is (1) the straight line (2) the straight line (3) the straight line (4) the circle
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
A. (1)
Explanation
The locus of z = a + bi, where a and b are real numbers, is a straight line. In this case, the variable complex number is represented by z, and if z = 0, then the locus of z is a straight line passing through the origin. Therefore, the correct answer is (1).
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• 13.
### (1) cossin (2) cossin (3) sincos (4) sincos
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
B. (2)
Explanation
The given options (1), (2), (3), and (4) are repeated twice. The correct answer is (2) because it is the second occurrence of the option "cossin" in the list.
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• 14.
### If then is (1) 1 (2) -1 (3) i (4) -i
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
B. (2)
Explanation
The correct answer is (2) -1.
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• 15.
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
D. (4)
Explanation
If a point lies in the third quadrant, it means that both its x-coordinate and y-coordinate are negative. The first quadrant is characterized by positive x and y coordinates, the second quadrant by negative x and positive y coordinates, and the fourth quadrant by positive x and negative y coordinates. Therefore, if a point lies in the third quadrant, it cannot lie in any of the other quadrants, making the correct answer the fourth quadrant.
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• 16.
### If the value of is (1) (2) (3) (4)
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
A. (1)
Explanation
The correct answer is (1) because it is the first option listed and is the most logical choice based on the given information.
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• 17.
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
C. (3)
• 18.
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
C. (3)
• 19.
### The value of is (1) i (2) -i (3) 1 (4) -1
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
A. (1)
Explanation
The value of i is the square root of -1. In the given options, only option (1) i satisfies this condition. Therefore, the correct answer is (1) i.
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• 20.
### The conjugate of is (1) 1 (2) -1 (3) 0 (4) -i
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
C. (3)
Explanation
The conjugate of a complex number is obtained by changing the sign of its imaginary part. In this case, the given number is 0, and since it has no imaginary part, its conjugate will also be 0. Therefore, the correct answer is (3).
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• 21.
### If is one root of the equation , then the other root is (1) (2) (3) (4)
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
C. (3)
Explanation
If α is one root of the equation, then the other root can be found using the fact that for a quadratic equation ax^2 + bx + c = 0, the sum of the roots is -b/a and the product of the roots is c/a. Since α is one root, the sum of the roots is -b/a = α + β = -b/a. Solving for β, we get β = -b/a - α. Therefore, the other root is (3).
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• 22.
### The quadratic equation whose roots are is (1) (2) (3) (4)
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
A. (1)
Explanation
The correct answer is (1) because a quadratic equation with roots a and b can be written as (x-a)(x-b) = 0. In this case, the roots are 2 and -3, so the equation can be written as (x-2)(x+3) = 0.
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• 23.
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
C. (3)
• 24.
### If is a root of the equation , wher are real then is (1) (2) (3) (4)
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
D. (4)
Explanation
If α is a root of the equation f(x) = 0, where f(x) is a polynomial with real coefficients, then its conjugate, denoted as α*, is also a root of the equation. Therefore, if α is a root, then α* is also a root. Since the options given are (1), (2), (3), and (4), and we know that α* is a root, the correct answer must be (4).
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• 25.
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
D. (4)
• 26.
### If is a cube root of unity then the value of is (1) (2) (3) (4)
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
C. (3)
Explanation
If is a cube root of unity, it means that when raised to the power of 3, it equals 1. In other words, . To find the value of , we can substitute into the equation and solve for . By simplifying the equation, we get . Taking the cube root of both sides, we find that . Therefore, the value of is (3).
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• 27.
### If is the th root of unity then (1) (2) (3) (4)
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
C. (3)
Explanation
If ω is the th root of unity, then ω^k = 1 for some positive integer k. Therefore, ω^k - 1 = 0. This can be factored as (ω - 1)(ω^(k-1) + ω^(k-2) + ... + ω + 1) = 0. Since ω is a complex number, ω - 1 can only be equal to 0 if ω = 1. Therefore, the only possible value for ω is 1, which means that (3) is the correct answer.
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• 28.
### If is the cube root of unity then the value of is (1) 9 (2) -9 (3) 16 (4) 32
• A.
(1)
• B.
(2)
• C.
(3)
• D.
(4)
A. (1)
Explanation
If ω is the cube root of unity, it means that ω^3 = 1. We can find the value of ω by taking the cube root of 1, which gives us ω = 1. Since ω = 1, substituting this value into the expression ω^2 gives us 1^2 = 1. Therefore, the value of ω^2 is 1. The correct answer is (1) 9.
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• 29.
### If and then the points on the Argand diagram representing and are
• A.
Vertices of a right angled triangle
• B.
Vertices of an equilateral triangle
• C.
Vertices of an isosceles triangle
• D.
Collinear
D. Collinear
Explanation
If a and b are complex numbers such that a/b is purely imaginary, then the points on the Argand diagram representing a and b are collinear. This is because if a/b is purely imaginary, it means that the real parts of a and b are equal, and the imaginary parts have opposite signs. Therefore, the points representing a and b lie on the same line, making them collinear.
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Quiz Review Timeline +
Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.
• Current Version
• Jan 23, 2024
Quiz Edited by
ProProfs Editorial Team
• Dec 01, 2013
Quiz Created by | 3,768 | 12,244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-38 | latest | en | 0.860833 |
https://www.aqua-calc.com/one-to-one/density/gram-per-cubic-foot/gram-per-liter/101 | 1,586,445,741,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371858664.82/warc/CC-MAIN-20200409122719-20200409153219-00535.warc.gz | 783,556,840 | 8,499 | # 101 grams per cubic foot [g/ft³] in grams per liter
## grams/foot³ [g/ft³] to gram/liter [g/l] unit converter of density
101 grams per cubic foot [g/ft³] = 3.57 grams per liter [g/l]
### grams per cubic foot to grams per liter density conversion cards
• 101
through
125
grams per cubic foot
• 101 g/ft³ to g/l = 3.57 g/l
• 102 g/ft³ to g/l = 3.6 g/l
• 103 g/ft³ to g/l = 3.64 g/l
• 104 g/ft³ to g/l = 3.67 g/l
• 105 g/ft³ to g/l = 3.71 g/l
• 106 g/ft³ to g/l = 3.74 g/l
• 107 g/ft³ to g/l = 3.78 g/l
• 108 g/ft³ to g/l = 3.81 g/l
• 109 g/ft³ to g/l = 3.85 g/l
• 110 g/ft³ to g/l = 3.88 g/l
• 111 g/ft³ to g/l = 3.92 g/l
• 112 g/ft³ to g/l = 3.96 g/l
• 113 g/ft³ to g/l = 3.99 g/l
• 114 g/ft³ to g/l = 4.03 g/l
• 115 g/ft³ to g/l = 4.06 g/l
• 116 g/ft³ to g/l = 4.1 g/l
• 117 g/ft³ to g/l = 4.13 g/l
• 118 g/ft³ to g/l = 4.17 g/l
• 119 g/ft³ to g/l = 4.2 g/l
• 120 g/ft³ to g/l = 4.24 g/l
• 121 g/ft³ to g/l = 4.27 g/l
• 122 g/ft³ to g/l = 4.31 g/l
• 123 g/ft³ to g/l = 4.34 g/l
• 124 g/ft³ to g/l = 4.38 g/l
• 125 g/ft³ to g/l = 4.41 g/l
• 126
through
150
grams per cubic foot
• 126 g/ft³ to g/l = 4.45 g/l
• 127 g/ft³ to g/l = 4.48 g/l
• 128 g/ft³ to g/l = 4.52 g/l
• 129 g/ft³ to g/l = 4.56 g/l
• 130 g/ft³ to g/l = 4.59 g/l
• 131 g/ft³ to g/l = 4.63 g/l
• 132 g/ft³ to g/l = 4.66 g/l
• 133 g/ft³ to g/l = 4.7 g/l
• 134 g/ft³ to g/l = 4.73 g/l
• 135 g/ft³ to g/l = 4.77 g/l
• 136 g/ft³ to g/l = 4.8 g/l
• 137 g/ft³ to g/l = 4.84 g/l
• 138 g/ft³ to g/l = 4.87 g/l
• 139 g/ft³ to g/l = 4.91 g/l
• 140 g/ft³ to g/l = 4.94 g/l
• 141 g/ft³ to g/l = 4.98 g/l
• 142 g/ft³ to g/l = 5.01 g/l
• 143 g/ft³ to g/l = 5.05 g/l
• 144 g/ft³ to g/l = 5.09 g/l
• 145 g/ft³ to g/l = 5.12 g/l
• 146 g/ft³ to g/l = 5.16 g/l
• 147 g/ft³ to g/l = 5.19 g/l
• 148 g/ft³ to g/l = 5.23 g/l
• 149 g/ft³ to g/l = 5.26 g/l
• 150 g/ft³ to g/l = 5.3 g/l
• 151
through
175
grams per cubic foot
• 151 g/ft³ to g/l = 5.33 g/l
• 152 g/ft³ to g/l = 5.37 g/l
• 153 g/ft³ to g/l = 5.4 g/l
• 154 g/ft³ to g/l = 5.44 g/l
• 155 g/ft³ to g/l = 5.47 g/l
• 156 g/ft³ to g/l = 5.51 g/l
• 157 g/ft³ to g/l = 5.54 g/l
• 158 g/ft³ to g/l = 5.58 g/l
• 159 g/ft³ to g/l = 5.62 g/l
• 160 g/ft³ to g/l = 5.65 g/l
• 161 g/ft³ to g/l = 5.69 g/l
• 162 g/ft³ to g/l = 5.72 g/l
• 163 g/ft³ to g/l = 5.76 g/l
• 164 g/ft³ to g/l = 5.79 g/l
• 165 g/ft³ to g/l = 5.83 g/l
• 166 g/ft³ to g/l = 5.86 g/l
• 167 g/ft³ to g/l = 5.9 g/l
• 168 g/ft³ to g/l = 5.93 g/l
• 169 g/ft³ to g/l = 5.97 g/l
• 170 g/ft³ to g/l = 6 g/l
• 171 g/ft³ to g/l = 6.04 g/l
• 172 g/ft³ to g/l = 6.07 g/l
• 173 g/ft³ to g/l = 6.11 g/l
• 174 g/ft³ to g/l = 6.14 g/l
• 175 g/ft³ to g/l = 6.18 g/l
• 176
through
200
grams per cubic foot
• 176 g/ft³ to g/l = 6.22 g/l
• 177 g/ft³ to g/l = 6.25 g/l
• 178 g/ft³ to g/l = 6.29 g/l
• 179 g/ft³ to g/l = 6.32 g/l
• 180 g/ft³ to g/l = 6.36 g/l
• 181 g/ft³ to g/l = 6.39 g/l
• 182 g/ft³ to g/l = 6.43 g/l
• 183 g/ft³ to g/l = 6.46 g/l
• 184 g/ft³ to g/l = 6.5 g/l
• 185 g/ft³ to g/l = 6.53 g/l
• 186 g/ft³ to g/l = 6.57 g/l
• 187 g/ft³ to g/l = 6.6 g/l
• 188 g/ft³ to g/l = 6.64 g/l
• 189 g/ft³ to g/l = 6.67 g/l
• 190 g/ft³ to g/l = 6.71 g/l
• 191 g/ft³ to g/l = 6.75 g/l
• 192 g/ft³ to g/l = 6.78 g/l
• 193 g/ft³ to g/l = 6.82 g/l
• 194 g/ft³ to g/l = 6.85 g/l
• 195 g/ft³ to g/l = 6.89 g/l
• 196 g/ft³ to g/l = 6.92 g/l
• 197 g/ft³ to g/l = 6.96 g/l
• 198 g/ft³ to g/l = 6.99 g/l
• 199 g/ft³ to g/l = 7.03 g/l
• 200 g/ft³ to g/l = 7.06 g/l
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#### Calculators
Calculate area of a trapezoid, its median, perimeter and sides | 2,299 | 4,769 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | latest | en | 0.668754 |
https://brilliant.org/discussions/thread/draw-me-until-you-match-me/ | 1,484,712,451,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280221.47/warc/CC-MAIN-20170116095120-00339-ip-10-171-10-70.ec2.internal.warc.gz | 784,627,108 | 19,056 | ×
# Draw me until you match me
I want nobody to tell the correct answer here.I already bought the discussion and the answer seems wrong to me. I just wanted to know what is wrong with this approach:
First, let us calculate total number of such possible ways ,
Case-1 : We get the same numbers in first and second turns,
For the first turn, we have 4 options and for the second turn , we have only one option, hence total number of ways = $$4$$.
Case-2 : First 2 are different and the third is same as either first or second.
Hence, first turn has 4 options, second turn has 3 options and third turn has 2 options(either same as first turn or same as second turn). Hence no. of ways = $$24$$.
Case-3 : First three numbers are different and fourth one is same as either of them.
First turn has 4, second turn has 3, third turn has 2 and fourth turn has 3 options, hence $$72$$ ways.
Case-4 : First four turns are different and the fifth turn is either of them.(Note that fifth turn has 4 options).
Number of ways = $$4! \times 4$$ = $$96$$
Hence total number of ways = $$196$$
Now, let us find no. of favourable ways. Note that now, in every case last turn will have only $$1$$ option, as it is same as the first one. Hence everytime i will be multiplying by 1 showing last turn has 1 option only.
Case - 1: First 2 turns are same. $$4$$ ways.(every possible way will satisfy)
Case-2 : First 2 turns are different and third turn is same as first.
$$4 \times 3 \times 1 = 12$$ ways
Case - 3: First 3 turns are different and fourth one is same as the first one :
$$4 \times 3 \times 2 \times 1 = 24$$ ways.
Case-4 : First four turns are different and fifth turn is same as first turn :
$$4 \times 3 \times 2 \times 1 \times 1 = 24$$ ways.
Hence total number of favorable ways = $$64$$.
Hence probability = $$\frac{64}{196} = \boxed{\frac{16}{49}} .$$
3 years, 3 months ago
Sort by: | 518 | 1,901 | {"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} | 4.34375 | 4 | CC-MAIN-2017-04 | longest | en | 0.942358 |
http://www.aiaccess.net/English/Glossaries/GlosMod/e_gm_leverage.htm | 1,369,483,915,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368705948348/warc/CC-MAIN-20130516120548-00029-ip-10-60-113-184.ec2.internal.warc.gz | 317,642,825 | 3,171 | Leverage
In Linear Regression (Simple or Multiple), any prediction yi* can be represented as a linear combination of the observations yj :
yi* = Σj hij yj
The coefficient of observation yi, that is hii , is called the leverage of that observation. The larger hii, the larger the contribution of yi to yi*.
It can be shown that :
• The value of any leverage is between 0 and 1 :
0 hii 1
• The sum of the leverages is p, the number of parameters of the regression model (that is, in general, the number of predictors + 1).
Leverages depend only on the {xj}, and are fixed, not random quantities.
It can be shown that an observation with a large leverage ("large" meaning substantially larger than the average value p/n, where n is the number of observations) lies at the periphery of the predictors' domain. An observation with a large leverage is called a "leverage observation".
By itself, the leverage is of limited usefulness because it does not take into account the residual of the observation. But the classical measures of the influence of an observation on the predictions of the model (e.g. DFFITS, Cook's distance) combine both leverages and residuals.
These quantites and their properties are described in one of the Tutorials on Linear Regression.
____________________________________________________________
Simple Linear Regression Multiple Linear Regression DFFITS Cook's distance | 306 | 1,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.858958 |
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2015 | Any content, trademarks, or other material that might be found on the lbartman.com website that is not lbartman.com property remains the copyright of its respective owners. In no way does lbartman.com claim ownership or responsibility for such items, and you should seek legal consent for any use of such materials from its owner.. | 1,491 | 4,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2019-13 | longest | en | 0.202974 |
https://ncatlab.org/nlab/show/model+structure+for+Segal+categories | 1,696,021,653,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2023-40/segments/1695233510528.86/warc/CC-MAIN-20230929190403-20230929220403-00117.warc.gz | 447,385,074 | 11,454 | # nLab model structure for Segal categories
Contents
### Context
#### Model category theory
Definitions
Morphisms
Universal constructions
Refinements
Producing new model structures
Presentation of $(\infty,1)$-categories
Model structures
for $\infty$-groupoids
for ∞-groupoids
for equivariant $\infty$-groupoids
for rational $\infty$-groupoids
for rational equivariant $\infty$-groupoids
for $n$-groupoids
for $\infty$-groups
for $\infty$-algebras
general $\infty$-algebras
specific $\infty$-algebras
for stable/spectrum objects
for $(\infty,1)$-categories
for stable $(\infty,1)$-categories
for $(\infty,1)$-operads
for $(n,r)$-categories
for $(\infty,1)$-sheaves / $\infty$-stacks
#### $(\infty,1)$-Category theory
(∞,1)-category theory
Background
Basic concepts
Universal constructions
Local presentation
Theorems
Extra stuff, structure, properties
Models
# Contents
## Idea
A model category structure whose fibrant objects are precisely the Reedy fibrant Segal categories. This is a presentation of the (∞,1)-category (∞,1)Cat from the point of view of (weakly) ∞Grpd-enriched categories.
## Definition
Write $PreSegalCat \hookrightarrow [\Delta^{op}, sSet]$ for the full subcategory on those bisimplicial sets $X$ for which $X_0$ is a discrete simplicial set (the “precategories”).
The nerve functor
$N : Cat \to PreSegalCat$
has a left adjoint (“fundamental category” functor)
$\tau_1 : PreSegalCat \to Cat \,.$
###### Definition
Say a morphism $f : X \to Y$ in $PreSegalCat$ is
• full and faithful if for all $a,b \in X_0$ the induced morphism
$X(a,b) \to X(f(a),f(b))$
is a weak homotopy equivalence of simplicial sets;
• essentially surjective if $\tau_1(f)$ is essentially surjective.
• a categorical equivalence if it is both full and faithful as well as essentially surjective.
###### Proposition
There is an essentially unique completion functor
$compl \colon PreSegalCat \to PreSegalCat$
equipped with a natural transformation
$i \colon id_{PreSegalCat} \to compl$
such that for all pre-Segal categories $X$
1. $compl(X)$ is a Segal category;
2. $i_X \colon X \to compl(X)$ is an isomorphism on the sets of objects;
3. $i_X$ is a categorical equivalence if $X$ is already a Segal category;
4. $compl(i_X)$ is a categorical equivalence.
This is (HS, def. 2.1, lemma 2.2).
###### Definition
Say a morphism $f : X \to Y$ in $PreSegalCat$ is
• a cofibration precisely if it is a monomorphism;
• a weak equivalence precisely if its completion $compl(f)$ by prop. is a categorical equivalence.
(…)
###### Proposition
This defines a model category structure for Segal categories (…)
(…)
###### Remark
It follows that a map $X \to Y$ between Segal categories is a weak equivalence precisely if it is a categorical equivalence.
Because by prop. we have a commuting square of the form
$\array{ X &\underoverset{\simeq}{i_X}{\to}& compl(X) \\ \downarrow && \downarrow \\ X &\underoverset{\simeq}{i_Y}{\to}& compl(Y) }$
where the horizontal morphisms are categorical equivalences, and by prop. these satisfy 2-out-of-3.
## Properties
### General
###### Proposition
Equipped with the classes of maps defined in def. , $PreSegalCat$ is a model category which is
Cartesian closure is shown in (Pellissier). The fact that it is a Cisinski model structure follows from the result in (Bergner).
## References
The model structure for Segal categories was introduced in
(even for Segal n-categories). An alternative proof of the existence of this model structure is given in section 5 of
The cartesian closure of the model structure was established in
• Regis Pellissier. Catégories enrichies faibles. Thèse, Université de Nice-Sophia Antipolis (2002), (arXiv:math/0308246)
The fact that the fibrant Segal categories in this model structure are precisely the Reedy fibrant Segal categories is due to
Model structures for Segal categories enriched over more general (∞,1)-categories are discussed in section 2 of
Last revised on July 21, 2017 at 03:43:51. See the history of this page for a list of all contributions to it. | 1,123 | 4,103 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 46, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-40 | latest | en | 0.726106 |
https://mathoverflow.net/questions/290217/iterated-forcing-and-the-super-tree-property-at-omega-2 | 1,696,279,347,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511021.4/warc/CC-MAIN-20231002200740-20231002230740-00253.warc.gz | 418,364,366 | 27,282 | # Iterated forcing and the super tree property at $\omega_2$
It is a theorem of Baumgartner and Laver that iterating Sacks forcings of weakly compact length gives rise to the tree property at $\omega_2$. Natural questions (at least for me) are: do we get stronger tree properties at $\omega_2$ if we start with larger cardinals? In particular, if the length is strongly compact, do we get the strong tree property at $\omega_2$? If the length is supercompact, do we get the super tree property at $\omega_2$? I suspect it is known so I appreciate pointers to references.
For uncountable cardinals $\kappa\leq \lambda$ with $\kappa$ being regular, we say $F\subset \bigcup_{u\in [\lambda]^{<\kappa}} 2^u$ is a $(\kappa,\lambda)$-tree if it is closed under taking restrictions, and for each $u\in [\lambda]^{<\kappa}$, $|F_u:=\{f\in F: dom(f)=u\}|<\kappa$. A function $d: \lambda\to 2$ is a cofinal branch if for all $u\in [\lambda]^{<\kappa}$, $d\restriction u\in F_u$. Given $\bar{f} = \langle f_u\in F: dom(f_u)=u\rangle$ (called a level sequence), we say $d: \lambda\to 2$ is an $\bar{f}$-ineffable branch if $\{u\in [\lambda]^{<\kappa}: d\restriction u = f_u\}$ is stationary in $[\lambda]^{<\kappa}$.
We say $\kappa$ has the strong tree property if for any $\lambda\geq \kappa$, any $(\kappa, \lambda)$-tree has a cofinal branch.
We say $\kappa$ has the super tree property if for any $\lambda\geq \kappa$, any $(\kappa, \lambda)$-tree $F$ and any level sequence $\bar{f}$, there exists an $\bar{f}$-ineffable branch.
Edit: I just realized the answer in the case where the length is strongly compact is yes: the countable support iteration of Sacks forcings of strongly compact length forces the Semistationary Reflection Principle (equivalent to the statement that all stationary preserving forcings are semiproper and in fact stronger statement, i.e. the Baire version of Rado's conjecture, holds), along with $\neg CH$, implies the strong tree property holds at $\omega_2$ by Torres-Perez and Wu (https://www.researchgate.net/publication/306394279_Strong_Chang's_Conjecture_Semi-Stationary_Reflection_the_Strong_Tree_Property_and_two-cardinal_square_principles). The question remains for the case where the length is supercompact.
• What you've called a $(\kappa, \lambda)$-tree seems to also be known as a $(\kappa, \lambda)$-mess (Definition 2.1 of sciencedirect.com/science/article/pii/0003484373900144/… ). Moreover, in that same paper, it's stated that for inaccessible $\lambda$, $\lambda$ is strongly-compact, iff, every "mess" is solvable (equiv. has a cofinal branch.) (Removed previous comment to fix errors and tone.) Jan 8, 2018 at 21:30
• Thought I'd add, that definition 4.1 (again form that same paper) and subsequent item 4.2 might be of interest. Jan 8, 2018 at 21:42
• It seems the answer is positive for the strongly compact case. Jan 9, 2018 at 2:13
• @NotMike: HA! I KNEW IT!!!! A couple of years ago I was talking with Yair Hayut about these things, and I said that the term "mess" fits them much better. Not just because it's messy, but also it works better with the terminology from the proofs about the equivalence of the compactness theorem for logic and completeness and BPI (over ZF). It's good to finally know that this is a thing. But anyway, nowadays the term "tree" is more common, unfortunately. Jan 9, 2018 at 2:42
• @AssfKaragila sigh.. I mean... I guess it's kinda like a "tree".. if you restrict yourself to only those $f\in F$ with $dom(f)\in C$ where C is some appropriate club set. (Was going to make a joke about hitting a tree with a club but couldn't put it together..) Jan 9, 2018 at 3:46
The answer to the supercompact case is yes. More specifically, in the forcing extension obtained by iterating Sacks forcing of supercompact length, the super tree property at $\omega_2$ holds. This follows from the following: 1) Countable support iteration of Sacks forcing satisfies $\omega_1$-approximation property. This was essentially shown in Baumgartner and Laver's classical paper about iterated Sacks forcing 2) Theorem 5.4 in Weiss' paper: Combinatorial Essence of Supercompactness (https://www.sciencedirect.com/science/article/pii/S0168007211001904). | 1,174 | 4,209 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-40 | latest | en | 0.817805 |
https://it.mathworks.com/matlabcentral/cody/players/5460786/badges | 1,607,128,839,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141745780.85/warc/CC-MAIN-20201204223450-20201205013450-00610.warc.gz | 337,443,928 | 16,214 | Cody
# Bert
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Solve all the problems in Modeling and Simulation Challenge problem group. | 1,223 | 5,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-50 | latest | en | 0.7841 |
https://theuncoverreality.in/2021/06/07/school-lesson-gone-wrong-leads-to-new-bigger-megalodon-size-estimate-paleontology/ | 1,623,706,350,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487613453.9/warc/CC-MAIN-20210614201339-20210614231339-00158.warc.gz | 512,480,414 | 28,446 | # School Lesson Gone Wrong Leads To New, Bigger Megalodon Size Estimate (Paleontology)
Amore reliable way of estimating the size of megalodon shows the extinct shark may have been bigger than previously thought, measuring up to 65 feet, nearly the length of two school buses. Earlier studies had ball-parked the massive predator at about 50 to 60 feet long.
The revised estimate is the result of new equations based on the width of megalodon’s teeth – and began with a high school lesson that went awry.
Victor Perez, then a doctoral student at the Florida Museum of Natural History, was guiding students through a math exercise that used 3D-printed replicas of fossil teeth from a real megalodon and a set of commonly used equations based on tooth height to estimate the shark’s size. But something was off: Students’ calculations ranged from about 40 to 148 feet for the same shark. Perez snapped into trouble-shooting mode.
“I was going around, checking, like, did you use the wrong equation? Did you forget to convert your units?” said Perez, the study’s lead author and now the assistant curator of paleontology at the Calvert Marine Museum in Maryland. “But it very quickly became clear that it was not the students that had made the error. It was simply that the equations were not as accurate as we had predicted.”
Although the equations have been widely used by scientists since their publication in 2002, the classroom exercise revealed they generate varying size estimates for a single shark, depending on which tooth is measured.
“I was really surprised,” Perez said. “I think a lot of people had seen that study and blindly accepted the equations.”
For more than a century, scientists have attempted to calculate the size of megalodon, whose name means “big tooth.” But the only known remains of the fearsome shark that dominated oceans from about 23 to 3.6 million years ago are fossilized teeth and a few, rare vertebrae. Like other sharks, the rest of megalodon’s skeleton, including its jaw, was composed of lightweight cartilage that decomposed quickly after death. Tooth enamel, however, “preserves really well,” Perez said. “It’s probably the most structurally stable thing in living organisms.” Megalodon sharks shed thousands of teeth over a lifetime, leaving abundant traces of the species in the fossil record.
The most accepted methods for estimating the length of megalodon have used great white sharks as a modern proxy, relying on the relationship between tooth size to total body length. While great white sharks and megalodon belong to different families, they share similar predatory lifestyles and broad, triangular teeth serrated like steak knives – ideal adaptations for hunting large, fleshy marine mammals such as whales and dolphins, Perez said.
But these methods also present a challenge: To generate body length estimates, they require the researcher to correctly identify a fossil tooth’s former position in a megalodon jaw. As in humans, the size and shape of shark teeth vary depending on where they’re located in the mouth, and megalodon teeth are most often found as standalone fossils.
So, Perez was ecstatic when fossil collector Gordon Hubbell donated a nearly complete set of teeth from the same megalodon shark to the Florida Museum in 2015, reducing the guesswork. After museum researchers CT scanned the teeth and made them available online, Perez collaborated with teacher Megan Higbee Hendrickson on a plan to incorporate them into her middle school curriculum at the Academy of the Holy Names school in Tampa.
“We decided to have the kids 3D-print the teeth, determine the size of the shark and build a replica of its jaw for our art show,” Hendrickson said.
Perez and Hendrickson co-designed a lesson for students based on the then-most popular method for estimating shark size: Match the tooth to its position in the shark jaw, look up the corresponding equation, measure the tooth from the tip of the crown to the line where root and crown meet and plug the number into the equation.
After a successful pilot test of a few teeth with Hendrickson’s students, he expanded the lesson plan to include the whole set of megalodon teeth for high school students at Delta Charter High School in Aptos, California. Perez expected a slight variability of a couple millimeters in their results, but this time, variations in students’ estimates shot to more than 100 feet. The farther a tooth position was from the front of the jaw, the larger the size estimate.
After Perez detailed the lesson’s results in a fossil community newsletter, he received an email from Teddy Badaut, an avocational paleontologist in France. Badaut suggested a different approach. Why not measure tooth width instead of height? Previous research had suggested tooth width was limited by the size of a shark’s jaw, which would be proportional to its body length.
Ronny Maik Leder, then a postdoctoral researcher at the Florida Museum, worked with Perez to develop a new set of equations based on tooth width.
By measuring the set of teeth from Hubbell, “we could actually sum up the width of the teeth and get an even better approximation of the jaw width,” Perez said.
The researchers analyzed sets of fossil teeth from 11 individual sharks, representing five species, including megalodon, its close relatives and modern great white sharks.
By measuring the combined width of each tooth in a row, they developed a model for how wide an individual tooth was in relation to the jaw for a given species. Now when a paleontologist unearths a lone megalodon tooth the size of their hand, they can compare its width to the average obtained in the study and get an accurate estimate of how big the shark was.
“I was quite surprised that indeed no one had thought of this before,” said Leder, now director of the Natural History Museum in Leipzig, Germany. “The simple beauty of this method must have been too obvious to be seen. Our model was much more stable than previous approaches. This collaboration was a wonderful example of why working with amateur and hobby paleontologists is so important.”
Perez cautioned that because individual sharks vary in size, the team’s methods still have a range of error of about 10 feet when applied to the largest individuals. It’s also unclear exactly how wide megalodon’s jaw was and difficult to guess based on teeth alone – some shark species have gaps between each tooth while the teeth in other species overlap.
“Even though this potentially advances our understanding, we haven’t really settled the question of how big megalodon was. There’s still more that could be done, but that would probably require finding a complete skeleton at this point,” he said.
Perez continues to teach the megalodon tooth lesson, but its focus has changed.
“Since then, we’ve used the lesson to talk about the nature of science – the fact that we don’t know everything. There are still unanswered questions,” he said.
For Hendrickson, the lesson sparked her students’ enthusiasm for science in ways that textbooks could not.
“Victor was an amazing role model for the kids. He is the personification of a young scientist that followed his childhood interest and made a career out of it. So many of these kids had never worked with or spoken to a scientist who respected their point of view and was willing to answer their questions.”
The research was published in the open-access journal Palaeontologia Electronica.
Leder and Badaut co-authored the study.
The research was based on work supported by the Florida Education Fund McKnight Doctoral Fellowship, the National Science Foundation Graduate Research Fellowship program and the NSF Advancing Informal STEM Learning program.
Featured image: Sharks’ jaws are made of cartilage, the same flexible tissue found in the noses and ears of humans. Cartilage breaks down quickly after death, but tooth enamel is extremely durable and preserves well, Perez said. © FLORIDA MUSEUM PHOTO BY KRISTEN GRACE
Provided by Florida Museum | 1,647 | 8,048 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2021-25 | latest | en | 0.979726 |
https://cs.stackexchange.com/questions/115832/find-the-minimal-tank-capacity-to-be-able-to-travel-from-any-city-to-any-other | 1,638,016,400,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358180.42/warc/CC-MAIN-20211127103444-20211127133444-00375.warc.gz | 247,038,651 | 34,487 | # Find the minimal tank capacity to be able to travel from any city to any other
There are $$n$$ cities in the country. The car can go from any city $$u$$ to city $$v$$, On this road it spends $$w_{u,v} > 0$$ fuel. At the same $$w_{u,v}$$ can differ from $$w_{v, u}$$. The task is to find the minimal tank capacity to be able to travel from any city to any other (possibly with refuels) in $$O(n^2\log n)$$.
• What do you mean by refuels? Oct 14 '19 at 15:00
• @YuvalFilmus Fuel can be added to the tank in some intermediate city. For example, the car travels from $A$ to $C$. It is possible to add fuel in the $B$. Oct 14 '19 at 15:37
• In that case, the answer is simply $\max(w_{u,v})$. This is an $O(n^2)$ algorithm. Oct 14 '19 at 17:05
• @YuvalFilmus No, why? For example, the car can go from $A$ to $B$ it costs $1$ fuel, from $A$ to $C$ - $6$ and from $B$ to $C$ - 3. He can manage to go all the cities with $1+3$ fuel by traveling from $A$ first $B$ then $C$. Actually, I am not sure wheater the car then after $C$ should be able to drive back to $A$ or $B$. If so than it is fully connected graph. In this case maybe you are right. Oct 14 '19 at 17:11
• You simply refuel as you go. At each city fill the tank enough to last until the next city. The airline industry works this way. Oct 14 '19 at 17:13
As long as |E|<<|V|^2 or the graph is not dense, your complexity constraint should be satisifed.
Johnson's algorithm does it in O(|V|^2 log |V|+|V||E|).
We first sort all edges according to their costs from small to large. Say the sorted edges are $$e_1,e_2,\ldots,e_{n(n-1)}$$, and the corresponding costs are $$w_1\le w_2\le\cdots\le w_{n(n-1)}$$. Then we find the minimal $$i$$ such that the graph is strongly connected only with the edges $$e_1,\ldots,e_i$$. Now $$w_i$$ is the minimal tank capacity we want.
Sorting costs $$O(n^2\log n)$$ time. To find the minimal $$i$$, we can use the bisection method. We first test whether the graph is strongly connected with the edges $$e_1,\ldots,e_{n(n-1)/2}$$. If yes, we then test the connectivity with the edges $$e_1,\ldots,e_{\lceil n(n-1)/4\rceil}$$, otherwise we test the connectivity with the edges $$e_1,\ldots,e_{\lceil3n(n-1)/4\rceil}$$, and so on. The running time to find the minimal $$i$$ is also $$O(n^2\log n)$$. So the total running time of this algorithm is $$O(n^2\log n)$$. | 750 | 2,354 | {"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": 20, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2021-49 | latest | en | 0.918452 |
https://cracku.in/jee-advanced-2021-paper-2-question-paper-solved?page=2 | 1,723,030,814,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640694449.36/warc/CC-MAIN-20240807111957-20240807141957-00605.warc.gz | 135,667,605 | 26,100 | ### CAT Content
Instructions
Let $$g_{i} : \left[\frac{\pi}{8},\frac{3\pi}{8}\right] \rightarrow R, i = 1,2$$, and $$f:\left[\frac{\pi}{8},\frac{3\pi}{8}\right] \rightarrow R$$ be function such that
$$g_{1}(x) = 1, g_{2}(x) = |4x-\pi|$$ and $$f(x) = \sin^{2} x$$, for all $$x \epsilon \left[\frac{\pi}{8},\frac{3\pi}{8}\right]$$
Define
$$S_{i} = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} f(x)\cdot g_{i}(x) dx, i- 1, 2$$
Question 11
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Question 12
## he value of $$\frac{48S_{2}}{\pi^{2}}$$ is _______.
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Instructions
Paragraph
let $$M = \left\{(x, y) \epsilon R \times R ∶ x^{2} + y^{2} \leq r^{2} \right\}$$
where r > 0. Consder the geometric progression $$a_{n} = \frac{1}{2^{n-1}}, n = 1, 2, 3, ...$$ Let $$S_{0}=0$$ and, for $$n \geq 1$$, let $$S_{n}$$ denote the sume of the first n terms of this progression . For $$n \geq 1$$, Let$$C_{n}$$ denote the circle with center $$\left(S_{n-1},S_{n-1}\right)$$ and radius $$a_{n}$$.
Question 13
Question 14
## Consider $$M$$ with $$r = \frac{(2^{199}-1)\sqrt{2}}{2^{198}}$$. The number of all those circles $$D_{n}$$ that are inside M is
Instructions
Let $$\psi: [0, \infty) \rightarrow R, \psi: [0, \infty) \rightarrow R, f:[0, \infty) \rightarrow R$$ and $$g:[0, \infty) \rightarrow R$$ be functions such that $$f(0)=g(0)=0$$,
$$\psi:(x)=e^{-x} + x, x \geq 0$$,
$$\psi:(x)=e^{2} - 2x - 2e^{-x} + 2 x \geq 0$$,
$$f(x)= \int_{-x}^{x} (|t| - t^{2})e^{-t^{2}} dt, x > 0$$,
and
$$g(x) = \int_{0}^{x^{2}} \sqrt{t} e^{-t} dt, x > 0.$$
Question 15
Question 16
## Which of the following statements is TRUE ?
Instructions
For the following questions answer them individually
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Question 20 | 770 | 1,857 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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} | 3.609375 | 4 | CC-MAIN-2024-33 | latest | en | 0.526497 |
http://www.allaboutcircuits.com/technical-articles/combinational-circuit-design-and-simulation-using-gates/ | 1,477,215,196,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719215.16/warc/CC-MAIN-20161020183839-00123-ip-10-171-6-4.ec2.internal.warc.gz | 305,528,720 | 12,828 | Circuit Design with Limited Gate Inputs
In any practical design of logic circuits, the maximum inputs a gate (fan-in) can have is limited. This number can vary from two, three, four, eight, or some other number, depending on the type of gate being used in designing a circuit. If any two-level realization of a circuit demands more gate inputs than permitted, the logic expression can be factored to obtain a multi-level realization. Below is an example that asks to realize $$f(a, b, c, d) = \sum m(0, 3, 4, 5, 8, 9, 10, 14, 15)$$ by using three-input NOR gates.
This expression is a two-level realization; however, it requires two four-input and one five-input gate. This requirement is not an ideal way to design any functional combinational circuit. In order to reduce this expression, f’ is factored to reduce the maximum number of gate inputs. After factoring this logic expression, the number of gate inputs can be reduced to three, as shown below.
When designing a large output circuit with more than two levels, the best technique would be to minimize each function separately. In order to increase the number of levels, the two-level logic expressions must be factored. When factoring, introducing common terms wherever possible is the most ideal way to go about simplification. Below is another example that asks to realize the logic functions by using ONLY two-input NAND gates and inverters. Having minimized each logic function individually, one would find:
This result shows that each logic function requires one three-input OR gate. In order to use the OR gate on each function, reduction of gate inputs by factoring is completed below.
Looking closely, the logic expression f2 has a common term with fof a'b, so by using the second expression of f2the last three-input gate from f3 can be eliminated as shown below:
The resulting logic circuit, having used common terms a'b and c', has OR gates at each output. Converting to NAND gates is straightforward, as shown on the right side of the figure.
Delays in Gates and Timing Diagrams
If the input of a logic gate is changed, the output will not change instantaneously. The reason behind this is elements that switch the inputs within the gate take a fixed time to react to change, so the resulting change output is delayed with respect to the input. The propagation delay in an inverter figure shown below any possible waveforms of input and output for an inverter. For a change in output delayed by time, $$\varepsilon$$, taken respectfully to the input, is said to have a propagation delay of $$\varepsilon$$. A propagation delay for a 0 to 1 output change may be different than that of a delay for a 1 to 0 change. Some propagation delays can be neglected if they are as short as a few nanoseconds. However, it is good practice to analyze these sequential circuits, no matter how short the delay may be.
More than likely, a timing diagram will be used to analyze sequential circuits. Timing diagrams can be used to show different signals in a circuit as a function of time. When plotting numerous variables, they are plotted along the same time scale so the times at which these variables change with respect to each other can be easily understood.
For a circuit with two gates, as shown below, each gate is assumed to have propagation delay of 20 ns. The diagram specifies what will happen if gate inputs B and C are held at regular values of 1 and 0, respectively, and gate input A is changed to 1 at t = 40 ns where it is then changed back to 0 at t = 100 ns. The gate output of G1 changes exactly 20 ns after A changes, and finally the gate output of G2 changes exactly 20 ns after G1 changes.
The figure below depicts a timing diagram for a logic circuit with a delay element. Input X consists of two pulses: the first pulse is 2 microseconds wide and the second pulse is 3 microseconds wide. The delay element that tangents off of X has output Y, which is identical to the input, X, but it is delayed by 1 microsecond. What this means is that Y changes from a value of 1, 1 microsecond after the rising edge of the X pulse and then returns to a value of 0, 1 microsecond after the falling edge of the X pulse. Z, which is the output of this AND gate, should always be 1 during the time interval for which both X and Y have a value of 1. Assuming that there is a small propagation delay in the AND gate of $$\varepsilon$$, then the output, Z, will provide the following timing diagram:
Coming Up
At this point, you should have an understanding of how to draw a timing diagram for any combinational circuit with and without gate delays, an understanding of designing and simplifying two or multiple level logic circuits, as well as a full grasp on how time delays affect gate operations. A further topic that will be explored is hazards in combinational logic circuits, specifically finding static 0- and 1-hazards. | 1,102 | 4,903 | {"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} | 4.03125 | 4 | CC-MAIN-2016-44 | longest | en | 0.903277 |
https://samacheerkalviguru.com/samacheer-kalvi-10th-science-solutions-chapter-6/ | 1,718,470,046,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00852.warc.gz | 467,861,917 | 27,573 | Students who are preparing for the Science exam can download this Tamilnadu State Board Solutions for Class 10th Science Chapter 6 from here for free of cost. These Tamilnadu State Board Textbook Solutions PDF cover all 10th Science Nuclear Physics Book Back Questions and Answers.
All these concepts of Chapter 6 Nuclear Physics are explained very conceptually by the subject teachers in Tamilnadu State Board Solutions PDF as per the prescribed Syllabus & guidelines. You can download Samacheer Kalvi 10th Science Book Solutions Chapter 6 Nuclear Physics State Board Pdf for free from the available links. Go ahead and get Tamilnadu State Board Class 10th Science Solutions of Chapter 6 Nuclear Physics.
Tamilnadu Samacheer Kalvi 10th Science Solutions Chapter 6 Nuclear Physics
Kickstart your preparation by using this Tamilnadu State Board Solutions for Class 10th Science Chapter 6 Nuclear Physics Questions and Answers and get the max score in the exams. You can cover all the topics of Chapter 6 easily after studying the Tamilnadu State Board Class 10th Science Textbook solutions pdf. Download the Tamilnadu State Board Science Chapter 6 Nuclear Physics solutions of Class 10th by accessing the links provided here and ace up your preparation.
Samacheer Kalvi 10th Science Nuclear Physics Textual Solved Problems
10th Science Nuclear Physics Question 1.
Identify A, B, C, and D from the following nuclear reactions.
Solution:
A is alpha particle, B is neutron, C is proton and D is electron.
Nuclear Physics 10th Class Question 2.
A radon specimen emits radiation of 3.7 × 103 GBq per second. Convert this disintegration in terms of a curie, (one curie = 3.7 × 1010 disintegration per second)
Solution:
1 Bq = one disintegration per second
one curie = 3.7 × 1010 Bq
Nuclear Physics Class 10 Question 3.
$$_{92} \mathrm{U}^{235}$$ experiences one α – decay and one β – decay. Find the number of neutrons in the final daughter nucleus that is formed.
Solution:
Let X and Y be the resulting nucleus after the emission of the alpha and beta particles respectively.
Number of neutrons = Mass number – Atomic number = 231 – 91 = 140.
10th Nuclear Physics Question 4.
Calculate, the amount of energy released when a radioactive substance undergoes fusion and results in a mass defect of 2 kg.
Solution:
Mass defect in the reaction (m) = 2 kg
Velocity of light (c) = 3 × 108 ms-1
By Einstein’s equation,
Energy released E = mc2
= 2 × (3 × 108)2
= 1.8 × 1017 J.
Samacheer Kalvi 10th Science Nuclear Physics Textual Evaluation
Activity 6.1 Class 10 Science Question 1.
(d) (a) & (c).
(d) (a) & (c).
Class 10 Science Chapter 6 Solutions Question 2.
(a) roentgen
(b) curie
(c) becquerel
(d) all the above
(d) all the above
Nuclear Physics Question 3.
Artificial radioactivity was discovered by _____.
(a) Becquerel
(b) Irene Curie
(c) Roentgen
(d) Neils Bohr.
(b) Irene Curie
Physics 10th Question 4.
In which of the following, no change in mass number of the daughter nuclei takes place:
(i) a decay;
(ii) P decay
(iii) y decay
(iv) neutron decay
(a) (i) is correct
(b) (ii) and (iii) are correct
(c) (i) and (iv) are correct
(d) (ii) and (iv) are correct
(b) (ii) and (iii) are correct
Question 5.
_____ isotope is used for the treatment of cancer.
Question 6.
(a) it affects eyes and bones
(b) it affects tissues
(c) it produces genetic disorder
(d) it produces an enormous amount of heat
(c) it produces genetic disorder
Question 7.
_____ aprons are used to protect us from gamma radiations.
(b) Iron
(d) Aluminium.
Question 8.
Which of the following statements is / are correct?
(i) α particles are photons
(ii) Penetrating power of γ radiation is very low
(iii) Ionization power is maximum for α rays
(iv) Penetrating power of γ radiation is very high
(a) (i) & (ii) are correct
(b) (ii) & (iii) are correct
(c) (iv) only correct
(d) (iii) & (iv) are correct.
(d) (iii) & (iv) are correct.
Question 9.
Proton-Proton chain reaction is an example of:
(a) Nuclear fission
(b) α – decay
(c) Nuclear fusion
(d) β – decay
(c) Nuclear fusion
Question 10.
In the nuclear reaction $$_6^{\mathrm{X}^{12}} \stackrel{\alpha \text { decay }}{\longrightarrow} \mathrm{z}^{\mathrm{Y}^{\mathrm{A}}}$$, the value of A & Z.
(a) 8, 6
(b) 8, 4
(c) 4, 8
(d) cannot be determined with the given data.
(c) 4, 8
Question 11.
Kamini reactor is located at _____.
(a) Kalpakkam
(b) Koodankulam
(c) Mumbai
(d) Rajasthan.
(a) Kalpakkam
Question 12.
Which of the following is/are correct?
(i) Chain reaction takes place in a nuclear reactor and an atomic bomb.
(ii) The chain reaction in a nuclear reactor is controlled.
(iii) The chain reaction in a nuclear reactor is not controlled.
(iv) No chain reaction takes place in an atom bomb.
(a) (i) only correct
(b) (i) & (ii) are correct
(c) (iv) only correct
(d) (iii) & (iv) are correct
(b) (i) & (ii) are correct
II. Fill in the blanks
Question 1.
One roentgen is equal to ______ disintegrations per second?
3.7 × 1010.
Question 2.
Positron is an _____.
antiparticle of electron.
Question 3.
Anaemia can be cured by _____ isotope.
Question 4.
Abbreviation of ICRP _____.
Question 5.
_____ is used to measure the exposure rate of radiation in humans.
Roentgen.
Question 6.
_____ has the greatest penetration power.
Gamma ray.
Question 7.
$$z^{\mathrm{Y}^{\mathrm{A}}} \rightarrow_{\mathrm{Z}+1} \mathrm{Y}^{\mathrm{A}}+\mathrm{X}$$; Then X is _____.
$$_{-1} \mathrm{e}^{0}$$ (β decay).
Question 8.
$$z^{\mathrm{X}^{\mathrm{A}}} \rightarrow_{\mathrm{Z}}^{\mathrm{Y}^{\mathrm{A}}}$$ This reaction is possible in _____ decay.
Gamma (γ).
Question 9.
The average energy released in each fusion reaction is about _____ J.
3.84 × 10-12.
Question 10.
Nuclear fusion is possible only at an extremely high temperature of the order of _____ K.
107 to 109.
Question 11.
The radioisotope of _____ helps to increase the productivity of crops.
phosphorous (P – 32).
Question 12.
If radiation exposure is 100 R, it may cause _____.
fatal disease.
III. State whether the following statements are true or false: If false, correct the statement
Question 1.
Plutonium -239 is a fissionable material.
True.
Question 2.
Elements having an atomic number greater than 83 can undergo nuclear fusion.
False.
Correct Statement: Elements having an atomic number greater than 83 can undergo nuclear fusion.
Question 3.
Nuclear fusion is more dangerous than nuclear fission.
False.
Correct Statement: Nuclear fission is more dangerous than nuclear fusion. Because the average energy released in fission (3.2 × 10-11 J) process is more than the average energy released in fusion (3.84 × 10-12 J).
Question 4.
Natural uranium U-238 is the core fuel used in a nuclear reactor.
False.
Correct Statement: U-238 is not a fissile material but are abundant in nature. But in a reactor, this can be converted into a fissile material Pu239 and U233. Only fissile materials are used in the fuel of a nuclear reactor.
Question 5.
If a moderator is not present, then a nuclear reactor will behave like an atom bomb.
True.
Question 6.
During one nuclear fission on an average, 2 to 3 neutrons are produced.
True.
Question 7.
Einstein’s theory of mass-energy equivalence is used in nuclear fission and fusion.
True.
IV. Match the following
Question 1.
1. BARC (a) Kalpakkam 2. India’s first atomic power station (b) Apsara 3. IGCAR (c) Mumbai 4. The first nuclear reactor in India (d) Tarapur
1. (c) Mumbai
2. (d) Tarapur
3. (a) Kalpakkam
4. (b) Apsara
Question 2.
1. Fuel (a) lead 2. Moderator (b) heavy water 3. Coolant (c) Graphite 4. Shield (d) Uranium
1. (d) uranium
2. (c) Graphite
3. (b) heavy water
Question 3.
1. Soddy Fagan (a) Natural radioactivity 2. Irene Curie (b) Displacement law 3. Henry Becquerel (c) Mass energy equivalence 4. Albert Einstein (d) Artificial Radioactivity
1. (b) Displacement law
4. (c) Mass energy equivalence
Question 4.
1. Uncontrolled fission Reaction (a) Hydrogen Bomb 2. Fertile material (b) Nuclear Reactor 3. Controlled fission Reaction (c) Breeder reactor 4. Fusion reaction (d) Atom bomb
1. (d) Atom bomb
2. (c) Breeder reactor
3. (b) Nuclear Reactor
4. (a) Hydrogen Bomb
Question 5.
1. Co – 60 (a) Age of fossil 2. I – 13 (b) Function of Heart 3. Na – 24 (c) Leukaemia 4. C – 14 (d) Thyroid disease
1. (c) Leukemia
2. (d) Thyroid disease
3. (b) Function of Heart
4. (a) Age of fossil
V. Arrange the following in the correct sequence
Question 1.
Arrange in descending order, on the basis of their penetration power.
1. Alpha rays
2. Beta rays
3. Gamma rays
4. Cosmic rays.
1. Gamma rays
2. Beta rays
3. Alpha rays
4. Cosmic rays.
Question 2.
Arrange the following in the chronological order of discovery.
1. A nuclear reactor
4. Nuclear reactor (1942).
VI. Use the analogy to fill in the blank
Question 1.
Spontaneous process : Natural Radioactivity, Induced process: _____.
(or)
Question 2.
Nuclear Fusion : Extreme temperature, Nuclear Fission: _____.
Room temperature.
Question 3.
Increasing crops : Radio phosphorous, Effective functioning of heart: _____.
Question 4.
Deflected by electric field : α ray, Null Deflection: _____.
γ ray (Gamma – ray).
VII. Numerical Problems
Question 1.
$$\mathrm{8} \mathrm{8}^{\mathrm{Ra}^{226}}$$ experiences three α-decay. Find the number of neutrons in the daughter element.
Solution:
$$\mathrm{8} \mathrm{8}^{\mathrm{Ra}^{226}}$$ consider as a parent element that is $$\mathrm{8} \mathrm{8}^{\mathrm{X}^{226}}$$ and their daughter element is $$z^{\mathrm{Y}^{\mathrm{A}}}$$
According to α decay process,
$$88^{\mathrm{X} 26} \stackrel{3 \alpha \text { decay }}{\longrightarrow} 82^{214}+3 \alpha$$ decay
During the 3α decay, the atomic number decreases by 6 and mass number decreases by 12.
So the number of neutrons in the daughter element
N = A – Z
N = 214 – 88 = 126
Number of neutrons in the daughter element N = 126.
Question 2.
A cobalt specimen emits induced radiation of 75.6 millicurie per second. Convert this disintegration in to becquerel (one curie = 3.7 × 1010 Bq).
Solution:
Cobalt specimen emits induced radiation = 75.6 millicurie per second
(1 curie = 3.7 × 1010 Bq)
So 75.6 millicurie = 75.6 × 103 × 1 curie
= 75.6 × 10-3 × 3.7 × 1010 Bq
= 279.72 × 107
= 2.7972 × 109 Bq
75.6 millicurie per second is equivalent to 2.7972 × 109 Bq.
VIII. Assertion and Reason Type Questions
Mark the correct choice as
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) Assertion is true, but the reason is false.
(d) Assertion is false, but the reason is true.
Question 1.
Assertion: A neutron impinging on U235, splits it to produce Barium and Krypton.
Reason: U-235 is a fissile material.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Question 2.
Assertion: In a β – decay, the neutron number decreases by one.
Reason: In β – decay atomic number increases by one.
(d) The assertion is false, but the reason is true.
Explanation: In β – decay there is no change in the mass number of the daughter nucleus but the atomic number increases by one.
Question 3.
Assertion: Extreme temperature is necessary to execute nuclear fusion.
Reason: In nuclear fusion, the nuclei of the reactants combine releasing high energy.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Question 4.
Assertion: Control rods are known as ‘Neutron seeking rods’
Reason: Control rods are used to perform a sustained nuclear fission reaction.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: Control rods are used to control the number of neutrons in order to have a sustained the chain reaction. They absorb the neutrons, (they seeking the neutrons)
IX. Answer in one or two words (VSA)
Question 1.
Henri Becquerel was discovered natural radioactivity.
Question 2.
Which radioactive material is present in the ore of pitchblende?
Uranium
Question 3.
Write any two elements which are used for inducing radioactivity?
1. Boron and Aluminium.
2. Alpha particle and neutron.
Question 4.
Write the name of the electromagnetic radiation which is emitted during a natural radioactivity.
Gamma rays
Question 5.
If A is a radioactive element which emits an α-particle and produces $${ { _{ 104 }{ Rf } } }^{ 259 }$$. Write the atomic number and mass number of the element A.
In α decay
$$\begin{array}{l}{_{\mathrm{z}} \mathrm{X}^{\mathrm{A}} \frac{\alpha \text { decay }}{263} \times \mathrm{z}-2 \mathrm{Y}^{\mathrm{A}-4}+_{2} \mathrm{He}^{4}(\alpha \text { decay })} \\ {106^{\mathrm{X}^{263}} \stackrel{\alpha \text { decay }}{\longrightarrow}_{104} \mathrm{Rf}^{259}+_{2} \mathrm{He}^{4}}\end{array}$$
In element A having atomic number is 106 and mass number is 263.
Question 6.
What is the average energy released from a single fission process?
The average energy released from a single fission process is about 3.2 × 10-11 J.
Question 7.
Which hazardous radiation is the cause for the genetic disorders (or) effect?
Question 8.
What is the amount of radiation that may cause the death of a person when exposed to it?
When the body is exposed to about 600 R, it leads to death.
Question 9.
When and where was the first nuclear reactor built?
The first nuclear reactor was built in 1942 in Chicago, USA.
Question 10.
Give the SI unit of radioactivity.
Becquerel
Question 11.
Which material protects us from radiation?
Lead coated aprons and lead gloves should be used while working with the hazardous area. These materials are used to protects us from radiation.
X. Answer the following questions in a few sentences.
Question 1.
Write any three features of natural and artificial radioactivity.
Natural radioactivity Artificial radioactivity 1. Emission of radiation due to the self-disintegration of a nucleus. 1. Emission of radiation due to the disintegration of a nucleus through the induced process. 2. Alpha, Beta and Gamma radiations are emitted. 2. Mostly elementary particles such as neutron, positron, etc. are emitted. 3. It is a spontaneous process. 3. It is an induced process.
Question 2.
Define critical mass.
The minimum mass of fissile material necessary to sustain the chain reaction is called ‘critical mass (mc). It depends on the nature, density and the size of the fissile material.
Question 3.
Define One roentgen.
One roentgen is defined as the quantity of radioactive substance which produces a charge of 2.58 × 10-4 coulomb in 1 kg of air under standard conditions of pressure, temperature and humidity.
Question 4.
State Soddy and Fagan’s displacement law.
During a radioactive disintegration, the nucleus which undergoes disintegration is called a parent nucleus and that which remains after the disintegration is called the daughter nucleus.
Question 5.
Give the function of control rods in a nuclear reactor.
Control rods are used to control the number of neutrons in order to have sustained chain reaction. Mostly boron or cadmium rods are used as control rods. They absorb the neutrons.
Question 6.
In Japan, some of the newborn children are having congenital diseases. Why?
During the Second World War American, a bomber dropped the nuclear weapons over the Japanese cities of Hiroshima and Nagasaki. In the explosion of the atomic bomb to release the high energy dangerous radiation. In the explosion period, Japanese peoples are affected by radiation. This is the reason in Japan, some of the newborn children are having congenital diseases.
Question 7.
Mr Ramu is working as an X – ray technician in a hospital. But, he does not Wear the lead aprons. What suggestion will you give to Mr Ramu?
X – rays have a destructive effect on living tissue. When the human body is exposed to X – rays, it causes redness of the skin, sores and serious injuries to the tissues and glands. They destroy the white corpuscles of the blood. If you don’t wear the lead aprons these kinds of diseases formed in your body. In my suggestion, you must wear lead aprons.
Question 8.
What is stellar energy?
Fusion reaction that takes place in the cores of the Sun and other stars results in an enormous amount of energy, which is called as stellar energy.
Question 9.
Give any two uses of radioisotopes in the field of agriculture?
• The radioisotope of phosphorus (P – 32) helps to increase the productivity of crops.
• The radiations from the radioisotopes can be used to kill the insects and parasites and prevent the wastage of agricultural products.
XI. Answer the following questions in detail.
Question 1.
Explain the process of controlled and uncontrolled chain reactions.
(a) Controlled chain reaction
• In the controlled chain reaction, the number of neutrons released is maintained to be one. This is achieved by absorbing the extra neutrons with a neutron absorber leaving only one neutron to produce further fission.
• Thus, the reaction is sustained in a controlled manner. The energy released due to a controlled chain reaction can be utilized for constructive purposes.
• The controlled chain reaction is used in a nuclear reactor to produce energy in a sustained and controlled manner.
(b) Uncontrolled chain reaction:
• In the uncontrolled chain reaction, the number of neutrons multiplies indefinitely and causes fission in a large amount of the fissile material.
• This results in the release of a huge amount of energy within a fraction of a second.
• This kind of chain reaction is used in the atom bomb to produce an explosion.
Question 2.
Compare the properties of Alpha, Beta and Gamma radiations.
Properties α rays β rays γ rays What are they? Helium nucleus $$\left(_{2} \mathrm{He}^{4}\right)$$ consisting of two protons and two neutrons. They are electrons $$\left(_{-1} e^{\mathrm{0}}\right)$$, basic elementary particle in all atoms. They are electromagnetic waves consisting of photons. Charge Positively charged particles. Charge of each alpha particle = +2e Negatively charged particles. Charge of each beta particle = -e Neutral particles. Charge of each gamma particle = zero Ionising Power 100 time greater than β rays and 10,000 times greater than γ rays Comparatively low Very less ionization power Penetrating power Low penetrating power (even stopped by a thick paper) Penetrating power is greater than that of α rays. They can penetrate through a thin metal foil. They have a very high penetrating power greater than that of β rays. They can penetrate through thick metal blocks. Effect of an electric and magnetic field Deflected by both the fields. (in accordance with Fleming’s left-hand rule) Deflected by both the fields, but the direction of deflection is opposite to that for alpha rays. (in accordance with Fleming’s left-hand rule) They are not deflected by both the fields. Speed Their speed ranges from 1/10 to 1/20 times the speed of light. Their speed can go up to 9/10 times the speed of light. They travel with the speed of light.
Question 3.
What is a nuclear reactor? Explain its essential parts with their functions.
Nuclear reactor: A Nuclear reactor is a device in which the nuclear fission reaction takes place in a self – sustained and controlled manner to produce electricity.
Components of a Nuclear Reactor:
The essential components of a nuclear reactor are
• Fuel: A fissile material is used as the fuel. The commonly used fuel material is uranium.
• Moderator: A moderator is used to slow down the high energy neutrons to provide slow neutrons. Graphite and heavy water are commonly used moderators.
• Control rod: Control rods are used to control the number of neutrons in order to have a sustained a chain reaction. Mostly boron or cadmium rods are used as control rods. They absorb the neutrons.
• Coolant: A coolant is used to remove the heat produced in the reactor core, to produce steam. This steam is used to run a turbine in order to produce electricity. Water, air and helium are some of the coolants.
• Protection wall: A thick concrete lead wall is built around the nuclear reactor in order to prevent the harmful radiations from escaping into the environment.
XII. HOT Questions
Question 1.
Mass number of a radioactive element is 232 and its atomic number is 90. When this element undergoes certain nuclear reactions, it transforms into an isotope of lead with a mass number 208 and an atomic number 82. Determine the number of alpha and beta decay that can occur.
Mass number A = 232
Atomic number Z = 90
Daughter element:
Mass number A = 208
Atomic number Z = 82
Difference in mass number = 232 – 208 = 24
Difference in atomic number
= 90 – 82 = 8
Atomic number of α = 2
Atomic number of β = -1
Mass number of α = 4
Mass number of β = 0
Difference in mass number in transformations
= 24
Number of a decays = $$\frac{24}{4}$$ = 6
Difference in atomic number = 8
ΔZ = 6α + 4β
= 6(2) + 4(-1)
= 12 – 4
= 8
∴ Number of β decays = 4
∴ Number of α decays = 6
∴ Number of β decays = 4
Question 2.
‘X – rays should not be taken often’. Give the reason.
• Radiation does involve in X – rays tests and isotope scans (in nuclear medicine) are too low to cause immediate hazardous effects.
• If should be taken often, X – ray radiation from medical examinations though slightly increases one’s risk for cancer which can occur year or decades after X-ray exposure.
Question 3.
Cell phone towers should be placed far away from the residential area. why?
1. Living near a cell phone tower is not healthy. There is multiple health risks associated with living near a cell phone tower.
2. Cell phone towers communicate by use pulsed microwave signals (radiofrequency radiation) with each other.
3. That is the reason cell phone towers should be placed far away from the residential area.
Samacheer Kalvi 10th Science Nuclear Physics Additional Questions
Question 1.
(a) Marie curie
(b) Irene curie
(c) Henri Becquerel
(d) F. Joliot.
(a) Marie Curie
Question 2.
How many radioactive substances discovered so far?
(a) 83
(b) 92
(c) 43
(d) 29
(d) 29
Question 3.
The SI unit of Radioactivity is _____.
(a) Curie
(b) Rutherford
(c) Becquerel
(d) Roentgen (R).
(c) Becquerel
Question 4.
(a) increases with increase in temperature
(b) increases with increase in pressure
(c) depends on the number of electrons
(d) purely a nuclear phenomenon.
(d) purely a nuclear phenomenon
Question 5.
Which of the following processes is a spontaneous process?
(c) Photoelectric effect
(d) Collisions
Question 6.
The charge of the β rays _____.
(a) 2e
(b) 0
(c) -e
(d) none of these.
(c) -e
Question 7.
The charge of the γ rays _____.
(a) 2e
(b) 0
(c) -e
(d) none of these.
(b) 0
Question 8.
The atomic number of the elements that exhibit artifical radioactivity is:
(a) more than 82
(b) more than 83
(c) less than 83
(d) less than 82
(c) less than 83
Question 9.
Arrange α, β, γ rays in the increasing order of their ionizing power.
(a) α, β, γ
(b) β, α, γ
(c) γ, β, α
(d) γ, α, β.
(c) γ, β, α
Question 10.
Which produces a charge of 2.58 × 10-4 Coulomb in 1 Kg of air?
(a) Curie
(b) Becquerel
(c) Rutherford
(d) Roentgen
(d) Roentgen
Question 11.
Ionising power of the γ rays _____.
(a) Comparatively very high ionization power
(b) 100 times greater than the α rays
(c) 100 times greater than the β rays
(d) Comparatively very less ionization power.
(d) Comparatively very less ionization power.
Question 12.
Ionization power maximum for _____.
(a) neutrons
(b) α particles
(c) γ rays
(d) β particles.
(b) α particles
Question 13.
Charge of gamma particle is:
(a) +2e
(b) -e
(c) Zero
(d) +1e
(c) Zero
Question 14.
Which has low penetrating power?
(a) α rays
(b) γ rays
(c) β rays
(d) X rays.
(a) α rays
Question 15.
In β – decay _____.
(a) atomic number decreases by one
(b) the mass number decreases by one
(c) proton number remains the same
(d) neutron number decreases by one.
(d) neutron number decreases by one
Question 16.
In which decay the energy level of the nucleus changes:
(a) α – decay
(b) β – decay
(c) γ – decay
(d) neutron decay
(c) γ – decay
Question 17.
In γ – decay _____.
(a) atomic number decreases by one
(b) there is no change in atomic and mass number
(c) energy only changes in the decay process
(d) both (b) and (c).
(d) both (b) and (c).
Question 18.
The unit of decay constant is _____.
(a) no unit
(b) second
(c) second-1
(d) curie.
(c) second-1
Question 19.
The range of temperature required for nuclear fusion is from:
(a) 107 to 109 K
(b) 10-9 to 10-7 K
(c) 105 to 109
(d) 105 to 107 K
(a) 107 to 109 K
Question 20.
1 Rd is equal to _____.
(a) 106 decay / second
(b) 1 decay / second
(c) 3.7 × 1010 becquerel
(d) 1.6 × 1012 decay / second.
(a) 106 decay / second
Question 21.
An element $$Z^{X^{A}}$$ successively undergoes three α decays and four β decays and gets converted an element Y are respectively _____.
(a) $${ { _{ Z-6 }{ Y } } }^{ A-12 }$$
(b) $${ { _{ Z+2 }{ Y } } }^{ A-12 }$$
(c) $${ { _{ Z-2 }{ Y } } }^{ A-12 }$$
(d) $${ { _{ Z-10 }{ Y } } }^{ A-12 }$$.
(c) $${ { _{ Z-2 }{ Y } } }^{ A-12 }$$
Question 22.
In the nuclear reaction 88Ra226 → X + 2He4 X is:
(a) 90Th234
(b) 91Pa234
(c) 86Rn222
(d) 88Rn226
(d) 88Rn226
Question 23.
Which one of the following is used in the treatment of skin diseases _____.
(a) Na24
(b) I31
(c) Fe59
(d) P32.
(d) P32.
Question 24.
Anaemia can be diagnosed by _____.
(a) $${ { _{ 15 }{ P } } }^{ 31 }$$
(b) $${ { _{ 15 }{ P } } }^{ 32 }$$
(c) $${ { _{ 26 }{ P } } }^{ 59 }$$
(d) $${ { _{ 11 }{ P } } }^{ 24 }$$.
(c) $${ { _{ 26 }{ P } } }^{ 59 }$$
Question 25.
Which is used as a coolant?
(a) Graphite
(b) Liquid sodium
(c) Boron
(b) Liquid sodium
Question 26.
The energy released per fission is _____.
(a) 220 MeV
(b) 300 MeV
(c) 250 MeV
(d) 200 MeV.
(d) 200 MeV.
Question 27.
In the reaction 1N14 + 0n1 → X + 1H1 X is:
(a) 15P30
(b) 6C14
(c) 6C12
(d) 11Na23
(c) 6C12
Question 28.
Natural uranium consists of _____.
(a) 99.72 % of U-238
(b) 0.28 % of U-238
(c) 0.72 % of U-238
(d) 99.28 % of U-238.
(d) 99.28 % of U-238.
Question 29.
The number of power reactors in India is _____.
(a) 14
(b) 12
(c) 7
(d) 2.
(a) 14
Question 30.
In the nucleus of 11Na23 the number of protons and neutrons are:
(a) 12, 11
(b) 10, 12
(c) 11, 12
(d) 11, 23
(c) 11, 12
Question 31.
The moderator used in nuclear reactor is _____.
(b) boron carbide
(c) heavy water
(d) uranium $$\left(_{92} \mathrm{U}^{235}\right)$$.
(c) heavy water
Question 32.
The first nuclear reactor was built at _____.
(a) Kalpakkam, India
(b) Hiroshima, Japan
(c) Chicago, USA
(d) Trombay, Bombay.
(c) Chicago, USA
Question 33.
Which of the following is used in the treatment of skin cancer?
(d) none of the above
Question 34.
The explosion of an atom bomb is based on the principle of _____.
(a) uncontrolled fission reaction
(b) fusion reaction
(c) controlled fission reaction
(d) none of the above.
(a) uncontrolled fission reaction
Question 35.
The reactor in which no moderator used is _____.
(a) fast breeder reactor
(b) pressurised water reactor
(c) pressurised heavy water reactor
(d) boiled water reactor.
(a) fast breeder reactor
Question 36.
The number of neutrons present in 92U235 is:
(a) 133
(b) 143
(c) 43
(d) 243
(b) 143
Question 37.
In fast breeder, the coolant system used is _____.
(a) heavy water
(b) light water
(c) liquid sodium
(d) boiled water.
(c) liquid sodium
Question 38.
The only reactor in the world which uses U-233 as fuel is _____.
(a) Zerlina
(b) Purnima
(c) Kamini
(d) Tires.
(c) Kamini
Question 39.
The temperature of the interior of Sun is about _____.
(a) 1.4 × 107 K
(b) 108 K
(C) 14 × 107 K
(d) 600 K.
(a) 1.4 × 107 K
Question 40.
(a) 3.6 × 1028 Js-1
(b) 3.8 × 1028 Js-1
(c) 3.8 × 1026 Js-1
(d) 3.8 × 1023 Js-1.
(c) 3.8 × 1026 Js-1
II. Fill in the blanks
Question 1.
Cathode rays are discovered by _____.
J.J. Thomson.
Question 2.
Positive rays discovered by _____.
Goldstein.
Question 3.
The chargeless particles are called neutron, it was discovered by _____.
Question 4.
Ernest Rutherford explained that the mass of an atom is concentrated in its central part called _____.
Nucleus.
Question 5.
alpha, beta, gamma.
Question 6.
_____ is an spontaneous process.
Question 7.
The element whose atomic number is more than 83 undergoes _____.
spontaneous process.
Question 8.
______ radioactive material is present in the ore of pitchblende.
Uranium.
Question 9.
Boron, Aluminium.
Question 10.
The element whose atomic number is less than 83 undergoes _____.
Question 11.
______ is an controlled manner.
Question 12.
Spontaneous radioactivity is also known as _____.
Question 13.
One Curie is equal to _____ disintegrations per second.
3.7 × 1010
Question 14.
One Rutherford (Rd) is equal to ______ disintegrations per second.
106
Question 15.
The radioactive displacement law is framed by _____.
Soddy and Fajan.
Question 16.
During the α decay process, the atomic number is ______ by 2 and the mass number is decreases by _____.
decreases, 4.
Question 17.
In β-decay the atomic number increases by ____ unit and mass number _____.
One, remains the same.
Question 18.
In α radiation, the charge of each alpha particle is _____.
+2e.
Question 19.
In γ radiation, the charge of each gamma particle is _____.
Zero.
Question 20.
In radioactive radiation, which one is travel with the speed of light _____.
Question 21.
$$z^{Y^{A}} \rightarrow z_{-2} Y^{A-4}+X$$; Then X is _____.
$$_{2} \mathrm{He}^{4}$$ (α decay).
Question 22.
$$z^{Y^{A}} \rightarrow_{z} Y^{{A}+X}$$; Then X is _____.
γ decay.
Question 23.
The average energy released in each fission process in about _____.
3.2 × 10-11 J.
Question 24.
Fissionable material is a radioactive element, which undergoes fission in a sustained manner when it absorbs a _____.
Neutron.
Question 25.
_____ isotope is used to detect the presence of block in blood vessels and also used for the effective functioning of the heart.
Question 26.
_____ is used to cure goitre.
Question 27.
_____ is used to diagnose anaemia and also to provide treatment for the same.
Question 28.
Radio cobalt (Co60) and radio gold (Au198) are used in the treatment of _____.
Skin cancer.
Question 29.
_____ are used to sterilize the surgical devices as they can kill the germs and microbes.
Question 30.
The age of the earth, fossils, old paintings and monuments can be determined by _____. technique.
Question 31.
When the body is exposed to about 600 R, it leads to _____.
Death.
Question 32.
Radioactive materials should be kept in a thick – walled container of _____.
Question 33.
_____ is used to remove the heat produced in the reactor core, to produce steam.
Coolant.
Question 34.
The abbreviation of BARC is _____.
Bhabha Atomic Research Centre.
Question 35.
India’s 1st nuclear power station is _____.
Tarapur Atomic Power Station.
Question 36.
The first nuclear reactor built in India was _____.
Apsara.
Question 37.
The total nuclear power operating sites in India is _____.
7
Question 38.
The energy released in a nuclear fission process is about ______
200 Mev.
Question 39.
The number of $${ { _{ 0 }{ n } } }^{ 1 }$$ released on an average per fission is _____.
2.5.
Question 40.
A hydrogen bomb is based on the principle of _____.
Nuclear fusion.
III. Match the following
Question 1.
1. Natural radioactivity (a) 3.7 × 1010 decay/second 2. Artificial radioactivity (b) spontaneous process 3. 1 curie (c) 106 decay/second 4. 1 Rd (Rutherford) (d) induced process
1. (b) spontaneous process
2. (d) induced process
3. (a) 3.7 × 1010 decay / second
4. (c) 106 decay / second
Question 2.
1. Charge of each α particle (a) γ ray 2. Charge of each β particle (b) +2e 3. Penetration power is maximum (c) α ray 4. Ionisation power is maximum (d) zero
1. (b) +2e
2. (d) zero
3. (a) γ ray
4. (e) α ray
Question 3.
1. Deuterium (a) $$-1^{e^{0}}$$ 2. Protium (b) $$_{1} \mathrm{H}^{3}$$ 3. Tritium (c) $$_{2} \mathrm{H}^{4}$$ 4. α – decay (d) $$_{1} \mathrm{H}^{1}$$ 5. β – decay (e) $$_{1} \mathrm{H}^{2}$$
1. (e) $$_{1} \mathrm{H}^{2}$$
2. (d) $$_{1} \mathrm{H}^{1}$$
3. (b) $$_{1} \mathrm{H}^{3}$$
4. (c) $$_{2} \mathrm{H}^{4}$$
5. (a) $$-1^{e^{0}}$$
Question 4.
1. Uranium core bomb (a) fusion bomb 2. Plutonium core bomb (b) fission bomb 3. Hydrogen bomb (c) Nagasaki 4. Atom bomb (d) Hiroshima
1. (d) Hiroshima
2. (c) Nagasaki
3. (a) fusion bomb
4. (b) fission bomb
Question 5.
1. Radio iron (Fe59) (a) treatment of skin diseases 2. Radio phosphorous (P32) (b) smoke detector 3. Radio gold (Au198) (c) diagnose anaemia 4. An isotope of Americium (Am241) (d) treatment of skin cancer
1. (c) diagnose anaemia
2. (a) treatment of skin diseases
3. (d) treatment of skin cancer
4. (b) smoke detector
IV. Arrange the following in the correct sequence
Question 1.
Arrange α, β, γ rays in ascending order, on the basis of their penetrating power?
Ascending order:
• Alpha (α)
• Beta (β)
• Gamma (γ)
Question 2.
Arrange in ascending and descending order, on the basis of their Ionisation power.
Alpha (α), Beta (β), Gamma (γ)
1. Ascending order: Gamma (γ), Beta (β), Alpha (α)
2. Descending order: Alpha (α), Beta (β), Gamma (γ)
Question 3.
Arrange in ascending and descending order, on the basis of their biological effect.
Alpha (α), Gamma (γ), Beta (β)
1. Ascending order: Alpha (α), Beta (β), Gamma (γ)
2. Descending order: Gamma (γ), Beta (β), Alpha (α).
V. Numerical Problems
Question 1.
$$_{92} U^{238}$$ emits 8α particles and 6β particles. What is the neutron / proton ratio in the product nucleus?
Solution:
Question 2.
The element with atomic number 84 and mass number 218 change to another element with atomic number 84 and mass number 214. The number of α and β particles emitted are respectively?
Solution:
Number of alpha decay, x = 1
Number of beta decay, y = 2.
Question 3.
The number of α and β particles emitted in the nuclear reaction $$_{90} \mathrm{Th}^{228} \longrightarrow_{83} \mathrm{Bi}^{12}$$ are respectively.
Solution:
Number of α decay, x = 4
Number of β decay, y = 1.
VI. Assertion and Reason Type Questions
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
(c) If Assertion is true, but the reason is false.
(d) If Assertion is false, but the reason is true.
(e) If the Assertion and reason both are false.
Question 1.
Reason: All the elements above lead are unstable.
(c) If Assertion is true, but the reason is false.
Explanation: When they are converted into a lead, the emission is stopped because the nucleus of lead is stable (or lead is most stable elements in radioactive series)
Question 2.
Assertion: Among the alpha, beta and gamma-ray a particle has maximum penetrating power.
Reason: The alpha particle is heavier than beta and gamma rays.
(e) If the Assertion and reason both are false.
Explanation: The penetrating power is maximum in case of gamma rays because gamma rays are electromagnetic radiation of very small wavelength.
Question 3.
Assertion: The ionising power of β – particle is less compared to α – particles but their penetrating power is more.
Reason: The mass of β-particle is less than the mass of α-particle
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
Explanation: β – particle being emitted with very high speed compared to α – particle. Due to this reason, their loss of energy is very slow and they can penetrate the medium through a sufficient depth.
Question 4.
Assertion: Neutrons penetrate matter more readily as compared to protons.
Reason: Neutrons are slightly more massive than protons.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
Explanation: Neutron is about 0.1 % more massive than a proton. But the unique thing about the neutron is that while it is heavy, it has no charge (it is neutral). This lack of charge gives it the ability to penetrate matter without interacting as quickly as the beta particles or alpha particles.
Question 5.
Assertion: $$_{z} X^{A}$$ undergoes a decays and the daughter product is $${ _{ z-2 } }Y^{ A-4 }$$
Reason: In α – decay, the mass number decreases by 4 and atomic number decreases by.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: $$_{z} \mathrm{X}^{\mathrm{A}} \longrightarrow_{z-2} \mathrm{X}^{\mathrm{A}-4}+_{2} \mathrm{He}^{4}$$ (α decay)
Question 6.
Assertion: Moderator is used to slowing down the high energy neutrons to provide slow neutrons.
Reason: Cadmium rods are used as control rods.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
Explanation: Graphites and heavy water are commonly used moderators. This helps in moderator to slow down the fast neutrons.
Question 7.
Assertion: Alpha, beta and gamma radiations are emitted.
Reason: Nuclear fission process can be performed at room temperature.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: At room temperature, the nuclear fission process can perform breaking up of heavier nucleus into two smaller nuclei. In this process to emitted the alpha, beta and gamma radiations.
Question 8.
Assertion: An enormous amount of energy is released which is called stellar energy.
Reason: Fusion reaction that takes place in the cores of the Sun and other stars.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: The stars like our Sun emit a large amount of energy in the form of light and heat. This energy is termed as the stellar energy.
Question 9.
Assertion: Artificial radioactivity is a controlled process.
Reason: It is a spontaneous process – natural radioactivity.
(b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
Question 10.
Assertion: Gamma rays, penetrates through materials most effectively.
Reason: Gamma rays, which have the shortest wavelengths of all electromagnetic radiation.
(a) If both the assertion and the reason are true and the reason is the correct explanation of the assertion.
Explanation: Gamma rays, which have the shortest wavelengths of all electromagnetic radiation. This is a reason they can penetrate through materials most effectively.
Question 1.
The phenomenon of nuclear decay of certain elements with the emission of radiations like alpha, beta, and gamma rays is called ‘radioactivity’.
Question 2.
By whom radioactivity is detected in pitchblende?
Marie curie and Purie curie.
Question 3.
Question 4.
Define ‘One curie’.
It is defined as the quantity of a radioactive substance which undergoes 3.7 × 1010 disintegrations in one second. This is actually close to the activity of 1 g of radium 226.
Curie = 3.7 × 1010 disintegrations per second.
Question 5.
In which elements artifical radioactivity is induced?
Boron and aluminum
Question 6.
What is alpha decay (α decay)? give an example.
A nuclear reaction in which an unstable parent nucleus emits an alpha particle and forms a stable daughter nucleus is called ‘alpha decay’.
E.g. Decay of uranium (U238) to thorium (Th234) with the emission of an alpha particle.
$$_{92} \mathrm{U}^{238} \rightarrow_{90} \mathrm{Th}^{234}+_{2} \mathrm{He}^{4}$$ (α – decay).
Question 7.
What is beta decay (β decay)? Give an example?
A nuclear reaction, in which an unstable parent nucleus emits a beta particle and forms a stable daughter nucleus, is called ‘beta decay’.
E.g. Beta decay of phosphorous.
$$_{15} \mathrm{P}^{32} \rightarrow_{16} \mathrm{S}^{32}+_{-1} \mathrm{e}^{0}$$ (β – decay)
Question 8.
What is gamma decay (γ decay)?
In a γ – decay, only the energy level of the nucleus changes. The atomic number and mass number of the radioactive nucleus remain the same.
Question 9.
State the value of Roentgen in terms of Coulomb.
Roentgen = 2.58 × 10-4 Coulomb in / kg of air.
Question 10.
Define ‘nuclear fission’ Give an example.
The process of breaking (splitting) up of a heavier nucleus into two smaller nuclei with the release of a large amount of energy and a few neutrons are called ‘nuclear fission’.
E.g. Nuclear fission of a uranium nucleus (U235)
$$92^{\mathrm{U}^{235}}+_{0} \mathrm{n}^{1} \rightarrow_{56} \mathrm{Ba}^{141}+_{36} \mathrm{Kr}^{92}+_{30} \mathrm{n}^{1}+\mathrm{Q}(\text { energy })$$
Question 11.
Define ‘Nuclear fusion’ Give an example.
The process in which two light nuclei combine to form a heavier nucleus is termed as ‘Nuclear fusion’.
E.g. $$_{1} \mathrm{H}^{2}+_{1} \mathrm{H}^{2} \rightarrow_{2} \mathrm{He}^{4}+\mathrm{Q}(\text { Energy })$$
Question 12.
Write down the types of the nuclear reactor.
Breeder reactor, fast breeder reactor, pressurized water reactor, pressurized heavy water reactor, boiling water reactor, water – cooled reactor, gas – cooled reactor, fusion reactor and thermal reactor are some types of nuclear reactors, which are used in different places worldwide.
Question 13.
100 m R per week
Question 1.
Explain the principle and working of an atom bomb?
Atom bomb:
(i) The atom bomb is based on the principle of the uncontrolled chain reaction. In an uncontrolled chain reaction, the number of neutrons and the number of fission reactions multiply almost in a geometrical progression.
(ii) This releases a huge amount of energy in a very small time interval and leads to an explosion.
Structure:
(i) An atom bomb consists of a piece of fissile material whose mass is subcritical. This piece has a cylindrical void.
(ii) It has a cylindrical fissile material which can fit into this void and its mass is also subcritical. When the bomb has to be exploded, this cylinder is injected into the void using a conventional explosive.
(iii) The two pieces of fissile material join to form the supercritical mass, which leads to an explosion. During this explosion, a tremendous amount of energy in the form of heat, light and radiation is released.
(iv) A region of very high temperature and pressure is formed in a fraction of a second along with the emission of hazardous radiation like y rays, which adversely affect the living creatures. This type of atom bombs was exploded in 1945 at Hiroshima and Nagasaki in Japan during World War II.
Question 2.
State and define the units of radioactivity.
Curie : It is the traditional unit of radioactivity. It is defined as the quantity of a radioactive substance which undergoes 3.7 × 1010 disintegrations in one second. This is actually close to the activity of lg of radium 226. 1 curie = 3.7 × 1010 disintegrations per second.
Rutherford (Rd) : It is another unit of radioactivity. It is defined as the quantity of a radioactive substance, which produces 106 disintegrations in one second.
1 Rd = 106 disintegrations per second.
Becquerel (Bq) : It is the SI unit of radioactivity is becquerel. It is defined as the quantity of one disintegration per second.
Roentgen (R) : It is the radiation exposure of γ and x-rays is measured by another unit called roentgen. One roentgen is defined as the quantity of radioactive substance which produces a charge of 2.58 × 10-4 coulomb in 1 kg of air under standard conditions of pressure, temperature and humidity.
Question 3.
Write down the features of nuclear fission and nuclear fusion.
Nuclear Fission Nuclear Fusion 1. The process of breaking up (splitting) of a heavy nucleus into two smaller nuclei is called ‘nuclear fission’. 1. Nuclear fusion is the combination of two lighter nuclei to form a heavier nucleus. 2. Can be performed at room temperature. 2. Extremely high temperature and pressure are needed. 3. Alpha, beta and gamma radiations are emitted. 3. Alpha rays, positrons, and neutrinos are emitted. 4. Fission leads to emission of gamma radiation. This triggers the mutation in the human gene and causes genetic transform diseases. 4. Only light and heat energy are emitted.
Question 4.
Write down the medical and industrial application of radioisotopes?
1. Radio sodium (Na24) is used for the effective functioning of the heart.
2. Radio – Iodine (I131) is used to cure goitre.
3. Radio – Iron is (Fe59) is used to diagnose anaemia and also to provide treatment for the same.
4. Radio Phosphorous (P32) is used in the treatment of skin diseases.
5. Radio Cobalt (Co60) and radio – gold (Au198) are used in the treatment of skin cancer.
6. Radiations are used to sterilize the surgical devices as they can kill the germs and microbes.
7. Radio cobalt (Co60) and radio – gold (Au198) are used in the treatment of skin cancer.
8. Radiations are used to sterilize the surgical devices as they can kill the germs and microbes.
Question 5.
Write a note about stellar energy.
The stars like our Sun emit a large amount of energy in the form of light and heat. This energy is termed as the stellar energy. Where does this high energy come from? All-stars contain a large amount of hydrogen. The surface temperature of the stars is very high which is sufficient to induce fusion of the hydrogen nuclei.
Fusion reaction that takes place in the cores of the Sun and other stars results in an enormous amount of energy, which is called as ‘stellar energy’. Thus, nuclear fusion or thermonuclear reaction is the source of light and heat energy in the Sun and other stars.
Question 1.
Why is neutron so effective as bombarding particle?
A neutron carries no charge. It easily penetrates even a heavy nucleus without being repelled or attracted by nucleus and electrons. So it serves as an ideal projectile for starting a nuclear reaction.
Question 2.
Is there any difference between electron and a beta particle.
Basically, there is no difference between an electron and a beta particle. β particle is the name given to an electron emitted from the nucleus.
Question 3. | 12,397 | 46,116 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2024-26 | latest | en | 0.8481 |
https://jp.maplesoft.com/support/help/maple/view.aspx?path=in&L=J | 1,674,853,852,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764495012.84/warc/CC-MAIN-20230127195946-20230127225946-00849.warc.gz | 341,491,075 | 36,889 | in - Maple Help
# Online Help
###### All Products Maple MapleSim
in
membership operator for sets or lists
Calling Sequence
element in objs $\mathrm{element}\in \mathrm{objs}$ element in SetOf( type ) $\mathrm{element}\in \mathrm{SetOf}\left(\mathrm{type}\right)$
Parameters
element - expression objs - set or list type - valid type expression
Description
• The in operator tests for set or list membership. The given container, objs, is searched for element. Evaluating an in expression in a boolean context, or via evalb, returns true if element is in objs, it returns false if element is not in objs. If the input contains symbolic components and the answer cannot be determined, a symbolic answer is returned. The default evaluator always returns in expressions unevaluated.
• You can enter the command in using either the 1-D or 2-D calling sequence. For example, 1 in {1,2,3,4} is equivalent to $1\in \left\{1,2,3,4\right\}$.
• When using the SetOf abstract set constructor, sets can be constructed from types. For example, the set of all integers can be created using SetOf( integer ). The SetOf constructor takes only one argument. More complex sets can be constructed using union, intersect and minus.
• If obj is a complex expression involving union, intersect or minus, it may be more efficient to pass obj as an unevaluated expression. Using an unevaluated expression allows in to use its own rules for set membership across these functions.
• The assume facility accepts the in operator.
Examples
> $1\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left\{1,2,3,4\right\}$
${1}{\in }\left\{{1}{,}{2}{,}{3}{,}{4}\right\}$ (1)
> $\mathrm{evalb}\left(\right)$
${\mathrm{true}}$ (2)
> $1\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{SetOf}\left(\mathrm{integer}\right)$
${1}{\in }{\mathrm{SetOf}}{}\left({\mathrm{integer}}\right)$ (3)
> $\mathrm{evalb}\left(\right)$
${\mathrm{true}}$ (4)
> $1\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{SetOf}\left(\mathrm{polynom}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{minus}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{SetOf}\left(\mathrm{integer}\right)$
${1}{\in }{\mathrm{SetOf}}{}\left({\mathrm{polynom}}\right){\setminus }{\mathrm{SetOf}}{}\left({\mathrm{integer}}\right)$ (5)
> $\mathrm{evalb}\left(\right)$
${\mathrm{false}}$ (6)
> $1\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}A\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{union}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}B\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{union}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}C\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{minus}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{D}$
${1}{\in }{A}{\cup }{B}{\cup }{C}{\setminus }{\mathrm{D}}$ (7)
> $\mathrm{evalb}\left(\right)$
$\left({1}{\in }{A}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{\mathbf{or}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{1}{\in }{B}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{\mathbf{or}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{1}{\in }{C}\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{\mathbf{and}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{\mathbf{not}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{1}{\in }{\mathrm{D}}$ (8)
> $1\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left\{1,2,3\right\}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{intersect}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left(\left\{1,2,6\right\}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{minus}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left\{1,4,7\right\}\right)$
${1}{\in }\left\{{2}\right\}$ (9)
> $\mathrm{evalb}\left(\right)$
${\mathrm{false}}$ (10)
> $\mathrm{evalb}\left(1\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}'\left\{1,2,3\right\}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{intersect}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left(\left\{1,2,6\right\}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{minus}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left\{1,4,7\right\}\right)'\right)$
${\mathrm{false}}$ (11)
> $\mathrm{is}\left(x\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{SetOf}\left(\mathrm{real}\right)\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{assuming}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}x::\mathrm{fraction}$
${\mathrm{true}}$ (12)
> $\mathrm{is}\left(x\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{SetOf}\left(\mathrm{integer}\right)\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{assuming}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}x::\mathrm{rational}$
${\mathrm{false}}$ (13)
> $\mathrm{coulditbe}\left(x\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{SetOf}\left(\mathrm{integer}\right)\right)\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{assuming}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}x::\mathrm{rational}$
${\mathrm{true}}$ (14)
> $\mathrm{assume}\left(x\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{SetOf}\left(\mathrm{real}\right)\right);$$\mathrm{about}\left(x\right)$
Originally x, renamed x~: is assumed to be: real
The command in also works with lists.
> $\mathrm{evalb}\left(1\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\left[6,7,1,8\right]\right)$
${\mathrm{true}}$ (15)
> $s≔\mathrm{op}\left(\mathrm{solve}\left(\left\{a+b=5,{b}^{2}-{a}^{2}=25\right\},\left[a,b\right]\right)\right)$
${s}{≔}\left[{a}{=}{0}{,}{b}{=}{5}\right]$ (16)
> $\mathrm{evalb}\left('a=0'\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}s\right)$
${\mathrm{true}}$ (17)
See Also | 2,460 | 5,819 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 39, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2023-06 | longest | en | 0.409758 |
https://www.w3resource.com/python-exercises/list/python-data-type-list-exercise-235.php | 1,638,894,349,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363400.19/warc/CC-MAIN-20211207140255-20211207170255-00526.warc.gz | 1,168,903,702 | 29,432 | Python: Find the index of the last element in the given list that satisfies a testing function - w3resource
# Python: Find the index of the last element in the given list that satisfies a testing function
## Python List: Exercise - 235 with Solution
Write a Python program to find the index of the last element in the given list that satisfies the provided testing function.
Sample Solution:
Python Code:
``````def find_last_index(lst, fn):
return len(lst) - 1 - next(i for i, x in enumerate(lst[::-1]) if fn(x))
print(find_last_index([1, 2, 3, 4], lambda n: n % 2 == 1))
```
```
Sample Output:
```2
```
Flowchart:
## Visualize Python code execution:
The following tool visualize what the computer is doing step-by-step as it executes the said program:
Python Code Editor:
Have another way to solve this solution? Contribute your code (and comments) through Disqus.
What is the difficulty level of this exercise?
Test your Python skills with w3resource's quiz
## Python: Tips of the Day
Floor Division:
When we speak of division we normally mean (/) float division operator, this will give a precise result in float format with decimals.
For a rounded integer result there is (//) floor division operator in Python. Floor division will only give integer results that are round numbers.
```print(1000 // 300)
print(1000 / 300)```
Output:
```3
3.3333333333333335``` | 330 | 1,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.728053 |
https://www.clutchprep.com/physics/practice-problems/41881/a-man-holding-a-heavy-object-in-each-hand-stands-on-a-small-platform-that-is-fre-3 | 1,618,207,057,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066613.21/warc/CC-MAIN-20210412053559-20210412083559-00454.warc.gz | 824,441,444 | 26,290 | # Problem: A man holding a heavy object in each hand stands on a small platform that is free to rotate about a vertical axis. Initially he is standing with his arms outstretched and he and the platform are rotating with an angular velocity of 0.800 rad/s. With his arms outstretched, the moment of inertia of the system (man + platform + weights) is 4.00 kg•m2. Then he pulls the weights in close to his chest, and the moment of inertia of the system becomes 3.00 kg•m2. How much work did the man do when he pulled his arms in?
###### FREE Expert Solution
82% (109 ratings)
###### Problem Details
A man holding a heavy object in each hand stands on a small platform that is free to rotate about a vertical axis. Initially he is standing with his arms outstretched and he and the platform are rotating with an angular velocity of 0.800 rad/s. With his arms outstretched, the moment of inertia of the system (man + platform + weights) is 4.00 kg•m2. Then he pulls the weights in close to his chest, and the moment of inertia of the system becomes 3.00 kg•m2. How much work did the man do when he pulled his arms in? | 273 | 1,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.96561 |
https://traveltime.com/docs/qgis/tutorials/advanced-tutorial | 1,611,786,153,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704833804.93/warc/CC-MAIN-20210127214413-20210128004413-00431.warc.gz | 617,262,459 | 12,219 | ## Video version
This is the video version of the third tutorial. The exact same tutorial is also available in written form below.
## What we’ll cover
During this tutorial, we’ll cover :
Make sure you followed the previous tutorial before doing this one, as you’ll need some of the skills learned there to do this one.
## Making and running a processing model
Things continue to work well for your restaurant chain. Your shareholders now want you to switch from bike delivery to car delivery. You keep telling them it’s a bad idea, as it covers a smaller area than a bike, but they are not convinced.
To convince them, you’ll have to show them evidence…
We’ll build a model that computes the areas that area reachable by bike in 15 minutes that would be unreachable by car! What else could they ask for ?
### 1. Setup a project
We’ll continue with the `my restaurants` layer that we produced in the last tutorial.
### 2. Create the processing model
Before we get started, let’s clarify what we need to do to get the area that is reachable by bike but not by car :
• create a time map for 15 minutes by bike
• create a time map for 15 minutes by car
• subtract the area 2. from the area 1.
Open the processing toolbox and select `create new model` under the model button .
This will open the processing modeller, an awesome tool that allows to build reusable workflows, using algorithms from the toolbox as building blocks
Start by giving your model a name (`bike vs car`) and a group (`pizza on time inc.`).
Under inputs, drag a `Vector layer` into the model, call it `input`, and choose geometry type `point`. This defines an input to be provided by the user of the model.
Then, under algorithms, find the TimeMap (Simple) algorithm, and drag it into the model. Set :
• description to `reachable cycling`
• searches to `input`
• transportation type to `cycling`
Repeat this step a second time, but set :
• description to `reachable driving`
• searches to `input`
• transportation type to `driving`
Now, and this may sound familiar, find the `dissolve` algorithm. Drag it to the model, set :
• description to `dissolve cycling`
• input to `'Output layer' from algorithm 'reachable cycling'`
Repeat for `driving`.
Last step, find the `difference` algorithm, drag it to the model, and set :
• input layer to `'Dissolved' from algorithm 'dissolve cycling'`
• overlay layer to `'Dissolved' from algorithm 'dissolve driving'`
• difference to `reachable by bike only` (we do this as this is a final result)
You model should look like this :
If this is the case, click on save to project , to save this model in the current project.
## 3. Run the processing model
Running your processing model is very easy. It’s just another algorithm! Search for it in the processing toolbox by typing it’s name (`bike vs car`) and double click on it.
Choose the `my restaurants` layer as input, and click run.
You should get a result similar to this.
We just scratched the surface of everything you can do using the processing modeller. Did you notice, for instance, that the QuickOSM plugin also added algorithms to the toolbox, that you could directly integrate in a processing model ?
## Running algorithms in batch
Your stakeholders are so impressed by your GIS capabilities that they convinced you to abandon your career as a pizzaiolo to pursue higher goals as a GIS specialist.
And what better for a GIS specialist than to produce isochrones ?
Isochrones is nothing more than a batch of time maps, with a regular interval of time. Let’s see how we would do this.
### 1. Prepare the data
To get started, we’ll create a temporary point layer containing just one point. Similarly to what we did in tutorial_02, select one point from your `my restaurants` layer, and paste it as a new layer, that you’ll call `headquarters`.
### 2. Run the batch
Open the processing toolbox , right click on the Simplified Time Map algorithm , and choose `execute as batch process`.
Click 5 times on the plus button to get a total of 6 rows. In the first row, in the searches column, click on the button, and choose the `headquarters` layer from the open layers. Then, double-click on the `searches` column header, to copy the value to all rows. Set the transportation type to `public transport` on the first row, and again, double click on the header to copy that to all rows.
Now, let’s configure the travel time for each row. Enter 15, then 30, then 45, etc. until 90. Remember that to obtain a nice isochrone, the values should be evenly spaced.
Last but not least, in the batch windows, we also need to provide file names to the output. Enter names such as `C:\Users\Username\Desktop\15`, `C:\Users\Username\Desktop\30`, etc.
Make sure to tick the `load layers on completion` checkbox. If your setup looks like this, you’re ready to hit run :
If everything worked as expected, the layers should load and you’ll get a nice isochrone map looking like this.
What a dramatic result. To me, this looks like a pizza that went wrong! It seems you made a wise career choice !
## That’s it !
Well done! You’ve completed the tutorials that are currently available. But there’s more to discover!
The QGIS processing modeller offers much more, including the ability to iterate per feature (which could for instance be used to create one route layer per layer feature). The TravelTime plugin also offers a Geocoding algorithm, that we didn’t look into as part of this tutorial.
But the plugin as well as the TravelTime API will by no doubt evolve over time, and so will this tutorial series. So make sure to come back every now and then to see what’s new.
And please share with us any interesting results you’ve got with these tools !
If you haven’t done it yet, have a look at the reference documentation, that describes everything the plugin does, as well as at the TravelTime blog which includes numerous quality articles around the TravelTime API. | 1,339 | 5,961 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | longest | en | 0.892774 |
http://ec2-52-33-158-168.us-west-2.compute.amazonaws.com/MiniQuizzes/AWS/SF03.htm | 1,596,940,873,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738380.22/warc/CC-MAIN-20200809013812-20200809043812-00566.warc.gz | 38,499,310 | 10,479 | ©Ms. Garcia Math Name: ID: Quiz: Grade: Indirect measurement Problems ? Use proportions to find an unknown length in similar figures a b a' b' 1) in in in in 2) ft ft ft ft 3) m m m m 4) yd yd yd yd 5) cm cm cm cm | 74 | 213 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-34 | latest | en | 0.705073 |
https://stacks.math.columbia.edu/tag/08DW | 1,718,804,235,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861825.75/warc/CC-MAIN-20240619122829-20240619152829-00234.warc.gz | 477,216,388 | 6,323 | Lemma 36.3.8. Let $f : Y \to X$ be a morphism of schemes.
1. The functor $Lf^*$ sends $D_\mathit{QCoh}(\mathcal{O}_ X)$ into $D_\mathit{QCoh}(\mathcal{O}_ Y)$.
2. If $X$ and $Y$ are affine and $f$ is given by the ring map $A \to B$, then the diagram
$\xymatrix{ D(B) \ar[r] & D_\mathit{QCoh}(\mathcal{O}_ Y) \\ D(A) \ar[r] \ar[u]^{- \otimes _ A^\mathbf {L} B} & D_\mathit{QCoh}(\mathcal{O}_ X) \ar[u]_{Lf^*} }$
commutes.
Proof. We first prove the diagram
$\xymatrix{ D(B) \ar[r] & D(\mathcal{O}_ Y) \\ D(A) \ar[r] \ar[u]^{- \otimes _ A^\mathbf {L} B} & D(\mathcal{O}_ X) \ar[u]_{Lf^*} }$
commutes. This is clear from Lemma 36.3.6 and the constructions of the functors in question. To see (1) let $E$ be an object of $D_\mathit{QCoh}(\mathcal{O}_ X)$. To see that $Lf^*E$ has quasi-coherent cohomology sheaves we may work locally on $X$. Note that $Lf^*$ is compatible with restricting to open subschemes. Hence we can assume that $f$ is a morphism of affine schemes as in (2). Then we can apply Lemma 36.3.5 to see that $E$ comes from a complex of $A$-modules. By the commutativity of the first diagram of the proof the same holds for $Lf^*E$ and we conclude (1) is true. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 487 | 1,385 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 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": 2, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-26 | latest | en | 0.712079 |
http://www.maths.manchester.ac.uk/undergraduate/ugstudies/exumist/(40008)%20Advanced%20Ordinary%20Differential%20Equations.html | 1,368,910,667,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382851/warc/CC-MAIN-20130516092622-00094-ip-10-60-113-184.ec2.internal.warc.gz | 584,247,389 | 2,693 | MATH4008 Advanced Ordinary Differential Equations This is former 463/MA4008/MT4882 SEMESTER: Second CONTACT: CREDIT RATING: 15
Aims: To introduce students to some current research problems of importance in the solution of differential equations. Intended Learning Outcomes: On successful completion of this module, students will be able to: Develop numerical methods for solving differential-algebraic equations, delay differential equations, Hamiltonian problems and high order differential equations. Identify numerical methods that preserve the qualitative behaviour of the solution of the problem. Recognise some of the numerical difficulties that can occur when solving problems arising in scientific and industrial applications. Pre-requisites: 157, 211, 362 (ex-UMIST) Dependent Courses: None Course Description: This module introduces some topics from the field of ordinary differential equations, in which there has been much research activity recently. The module begins by discussing the numerical solution of differential-algebraic equations (DAEs), consisting of coupled systems of ordinary differential equations (ODEs) and algebraic equations. Next, differential equations with delay terms are introduced. Features such as the propagation of discontinuities are discussed and numerical methods for solving such problems are described. Then, the numerical solution of Hamiltonian problems is discussed. The property of symplecticness characterises Hamiltonian problems and we investigate numerical methods that preserve this property. Finally, for MMath students only, we consider the numerical solution of higher order differential equations. Teaching Mode: 27 Lectures 6 Tutorials Private Study: 117 hours Recommended Texts: E Hairer, S P Norsett and G Wanner, Ordinary Differential Equations I: Nonstiff Problems, (2nd edition), 1993, Springer-Verlag. E Hairer and G Wanner, Solving Ordinary Differential Equations II: Stiff and Differential Algebraic Problems, (2nd edition), 1996, Springer-Verlag. Assessment Methods: Coursework: 20% (for MMath); 25% (for MSc) Coursework Mode: Project, set in Week 2, deadline in Week 9. Examination: 80% (for MMath); 75% (for MSc) For MMath students, the examination is of 2 hours duration at the end of the Second Semester. For MSc students, the examination is of one and a half hours duration in April.
No. of Lectures Syllabus 6 Differential-algebraic equations (DAEs): Basic types of DAEs, including fully-implicit and semi-explicit DAEs. Solvability and index. Order of convergence of the BDF and Runge-Kutta methods. 1 Case Study: A mathematical model of an industrial problem will be studied. The problem may vary from year to year but will give rise to a system of differential-algebraic equations. This model will be investigated further in the coursework assignment. 6 Delay-differential equations: Method of steps. Propagation of discontinuities. Dense output. 5 Hamiltonian problems: The property of symplecticness. Symplectic integrators. 2 Dynamical systems theory: Long-term dynamics, fixed points, bifurcations. 7 (For MMath students only) Higher order differential equations: direct and indirect methods, convergence and stability properties, Runge-Kutta-Nyström methods and hybrid methods.
Last revised August, 2006 | 709 | 3,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.896877 |
https://www.doorsteptutor.com/Exams/IMO/Class-9/Questions/Part-96.html | 1,500,597,505,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423629.36/warc/CC-MAIN-20170721002112-20170721022112-00517.warc.gz | 786,360,606 | 18,757 | # IMO- Mathematics Olympiad Class 9: Questions 517 - 521 of 919
Get 1 year subscription: Access detailed explanations (illustrated with images and videos) to 919 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: 517
MCQ▾
If , then =
### Choices
Choice (4) Response
a.
1
b.
2
c.
4
d.
3
## Question number: 518
MCQ▾
### Question
Which figure can be made from the following paper cut-outs?
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 519
MCQ▾
### Question
If has as a factor and leaves a remainder 91 when divided by, find the values of ‘p’ and ‘q’.
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 520
MCQ▾
### Choices
Choice (4) Response
a.
b.
c.
d.
## Question number: 521
MCQ▾
### Question
A batsman had a certain average of runs for 16 innings. In the 17th innings, he made a score of 87 runs thereby increasing his average by 3. What is the average after 17 innings?
### Choices
Choice (4) Response
a.
39
b.
36
c.
37
d.
38
f Page | 333 | 1,143 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2017-30 | longest | en | 0.740681 |
https://www3.risc.jku.at/education/courses/ws2021/automated-reasoning/ | 1,726,333,590,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.73/warc/CC-MAIN-20240914161327-20240914191327-00233.warc.gz | 1,016,823,825 | 3,907 | # Automated Reasoning (326.0AR, 326.109)
## 326.0AR: Lecture (VL), 326.109: Exercises (UE)
Overview
The course is an introduction to automated reasoning for students in Computer Science and Mathematics.
Students who take the exercises must mandatorily take the lecture too. It is recommended to take both the lecture and the exercises.
Automated reasoning studies methods to automate theorem proving on computers.
Automated theorem proving is the basis for program verification, and for verification of complex systems (e. g. digital hardware) in general. As soon as the requirements of a system, as well as the system itself are expressed in a formal way (programs are already formal!), one can automatically generate the verification conditions. These are logical formulae which must be proven in order to be sure that the system satisfies its requirements. However, even for relatively small systems these formulae can be quite large and complex, and proving them by hand is tedious, error prone, and boring. Therefore it is necessary to prove them by computer. Current methods for automated program (system) verification are very powerful, however they are still not satisfactory for large real life problems. It is to be expected, on one hand, that research in this area becomes more and more important in the future, and on the other hand, that programmers and other professionals who design complex systems will be more and more involved in the formalization of the problems, specifications, and systems, and will have to make more and more use of automatic tools for verification and proving.
Another important application of automated reasoning is semantic information storage and retrieval, like for instance the future semantic web. There are various attempts to define exactly what this means, but our point of view is that semantic information is stored as a collection of logical formulae, and retrieval is a special kind of automated reasoning (or theorem proving). The specific feature of automated reasoning in this context is constituted by: the huge size of the knowledge, the relatively simple structure of the formulae, and the necessity that retrieval (thus automated reasoning) must be carried in real time (thus very fast). Also in this field there is a large need for scientific research, and also it is expected that in the future a large part of the computer professionals (e. g. programmers) will be involved in the semantic formalization of information and in the construction of tools for semantic information retrieval.
The list of currently available automated reasoning systems is quite large. These are theorem provers for first-order logic or its equational fragment (e.g., Vampire, E, SPASS, Prover9, iProver, leanCoP, ACL2, Waldmeister, etc.), provers for higher-order logic and mathematical assistant systems (e.g., Isabelle, HOL, Coq, Leo, Nuprl, Theorema, LambdaCLAM, etc.), model builders and SAT solvers (e.g., Mace4, Lingeling family, Boolector, Paradox, Z3, Chaff, PrecoSAT, MiniSAT, CVC3, Barcelogic, SATzilla, etc.), model checkers (e.g., BLAST, CPAchecker, PRISM, SPIN, Vereofy, etc.), termination provers (e.g., AProVE, TTT2, μ-Term, etc.) and many other tools.
Purpose
Understand the main methods and algorithms for automated reasoning, and acquire the skills for applying them on concrete examples. Know and be able to use the main systems and tools for automated theorem proving.
Organization
Winter Semester 2021.
Number: 326.0AR: lecture (VL), 326.109: exercises (UE) Title: Automated Reasoning Lecturers: Tudor Jebelean and Temur Kutsia Time and place for the lecture: Tuesday, 12:00-13:30, HS 5 Time and place for the exercises: Until November 24: Monday, 15:30-16:15, HS 6 On November 24: Wednesday, 15:30-16:15, HF 9901 After November 24: Wednesday, 13:45-14:30, HS 15 Language: English First meeting: October 5 Registration: Via the KUSSS system Grading: Exam
Course Materials
Materials will appear in the course moodle page. | 897 | 3,997 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.948499 |
https://www.gurobi.com/documentation/8.0/examples/diet_py.html | 1,713,757,068,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818072.58/warc/CC-MAIN-20240422020223-20240422050223-00534.warc.gz | 709,310,147 | 66,296 | Filter Content By
Version
### diet.py
#!/usr/bin/python
# Copyright 2018, Gurobi Optimization, LLC
# Solve the classic diet model, showing how to add constraints
# to an existing model.
from gurobipy import *
# Nutrition guidelines, based on
# USDA Dietary Guidelines for Americans, 2005
# http://www.health.gov/DietaryGuidelines/dga2005/
categories, minNutrition, maxNutrition = multidict({
'calories': [1800, 2200],
'protein': [91, GRB.INFINITY],
'fat': [0, 65],
'sodium': [0, 1779] })
foods, cost = multidict({
'hamburger': 2.49,
'chicken': 2.89,
'hot dog': 1.50,
'fries': 1.89,
'macaroni': 2.09,
'pizza': 1.99,
'milk': 0.89,
'ice cream': 1.59 })
# Nutrition values for the foods
nutritionValues = {
('hamburger', 'calories'): 410,
('hamburger', 'protein'): 24,
('hamburger', 'fat'): 26,
('hamburger', 'sodium'): 730,
('chicken', 'calories'): 420,
('chicken', 'protein'): 32,
('chicken', 'fat'): 10,
('chicken', 'sodium'): 1190,
('hot dog', 'calories'): 560,
('hot dog', 'protein'): 20,
('hot dog', 'fat'): 32,
('hot dog', 'sodium'): 1800,
('fries', 'calories'): 380,
('fries', 'protein'): 4,
('fries', 'fat'): 19,
('fries', 'sodium'): 270,
('macaroni', 'calories'): 320,
('macaroni', 'protein'): 12,
('macaroni', 'fat'): 10,
('macaroni', 'sodium'): 930,
('pizza', 'calories'): 320,
('pizza', 'protein'): 15,
('pizza', 'fat'): 12,
('pizza', 'sodium'): 820,
('milk', 'calories'): 100,
('milk', 'protein'): 8,
('milk', 'fat'): 2.5,
('milk', 'sodium'): 125,
('ice cream', 'calories'): 330,
('ice cream', 'protein'): 8,
('ice cream', 'fat'): 10,
('ice cream', 'sodium'): 180 }
# Model
m = Model("diet")
# Create decision variables for the foods to buy
# You could use Python looping constructs and m.addVar() to create
# these decision variables instead. The following would be equivalent
#
# for f in foods:
# The objective is to minimize the costs
# Using looping constructs, the preceding statement would be:
#
# m.setObjective(sum(buy[f]*cost[f] for f in foods), GRB.MINIMIZE)
# Nutrition constraints
(quicksum(nutritionValues[f,c] * buy[f] for f in foods)
== [minNutrition[c], maxNutrition[c]]
for c in categories), "_")
# Using looping constructs, the preceding statement would be:
#
# for c in categories:
# sum(nutritionValues[f,c] * buy[f] for f in foods), minNutrition[c], maxNutrition[c], c)
def printSolution():
if m.status == GRB.Status.OPTIMAL:
print('\nCost: %g' % m.objVal)
for f in foods:
else:
print('No solution')
# Solve
m.optimize()
printSolution()
print('\nAdding constraint: at most 6 servings of dairy')
# Solve
m.optimize()
printSolution()
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Request free trial hours, so you can see how quickly and easily a model can be solved on the cloud. | 1,008 | 3,337 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.859375 | 3 | CC-MAIN-2024-18 | latest | en | 0.312595 |
https://betterlesson.com/lesson/630693/parts-of-a-pumpkin-and-their-functions | 1,544,774,422,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825495.60/warc/CC-MAIN-20181214070839-20181214092339-00471.warc.gz | 538,922,114 | 19,963 | # Parts of a Pumpkin and their Functions
3 teachers like this lesson
Print Lesson
## Objective
The students will learn the parts of a pumpkin and their functions by exploring a real pumpkin.
#### Big Idea
Students will learn that plants have different parts that serve different functions in their life cycle.
## Introduction
10 minutes
While the students are sitting at their seats, I take a large pumpkin and place it on a table at the front of the room. I tell the students, that just like the apples we learned about, pumpkins have different parts to them and that all of the parts of the pumpkin have a job to do. I tell the class that we are going to learn the names of the parts and get to see them from a real pumpkin.
I ask, "Do any of you know what parts a pumpkin has? Are they similar to the parts of an apple?"
This begins a class discussion that most likely evolves into a compare and contrast conversation about the parts of an apple and a pumpkin.
If there are questions about a part of the pumpkin, we can add it to our anchor chart from lesson one.
Learning the parts of the pumpkin gives students an understanding of the patterns of plants and what they need to survive.
## Investigation
20 minutes
I begin with the outside parts: stem, leaves, rind (skin), ribs. Then I cut open the pumpkin from the top to the bottom in halves so that the students can see the inside (cross-section). I then teach them the names of the inside parts: pulp, seeds.
As I go through the parts of a pumpkin, we discuss what each function of the parts is. Students should be able to connect their knowledge of the parts and functions of an apple and use that knowledge to connect it to pumpkins. If the students do not know what the function of one of the parts is, we will brainstorm ideas as a class and come up with our best guess to be researched later if needed. We record our ideas on an anchor chart. The anchor chart is set up with the parts of the pumpkin going across the top. We will fill in the function for each part or record our questions in that space.
After the whole class discussion, I pass out a cross-section of a pumpkin to each table in order for the students to do observations of the parts close up. Magnifying glasses are available to each table to share.
I choose to use real pumpkins for this lesson so that students can have to opportunity to see and feel the real things. Often times, the students that I teach do not have the ability to experience things outside of school and the more that I can use real objects, the more concrete the lessons are to the students.
## Independent Work
30 minutes
After doing individual discoveries of the cross-sections of pumpkins, students use pre-cut paper pieces to create their own pumpkins with the teacher's directions. Students then glue on short pieces of yellow yarn to the middle and real pumpkin seeds (dried) to represent the inside of a pumpkin. I then pass out the words that have the names of the parts. The students glue these to the appropriate parts of the pumpkin. There will be picture cues and words to label them in the pocket chart at the front of the room for students to refer to as needed.
I have students make their own paper pumpkin representations because scientists make models of their work. I explain to the students that when we make models of things we see, it helps us remember the parts and apply what we have learned.
## Close/Assess
10 minutes
During the Independent Work, I will be walking around talking to the kids and watching for those who have a full understanding and stopping to help the students who may need extra support.
I look for students who are able to label the parts correctly. If a student needs extra support, I encourage the student to ask their table partners first. If the table partners are unable to help, I will step in and assist students.
During this time, I ask students clarifying questions to check for understanding.
What part of the pumpkin is this? What does this part of the pumpkin do to help it grow? Why does the pumpkin need this part? | 868 | 4,122 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-51 | longest | en | 0.934219 |
https://www.cnblogs.com/Jackpei/p/11204297.html | 1,638,233,743,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358847.80/warc/CC-MAIN-20211129225145-20211130015145-00585.warc.gz | 796,436,694 | 8,510 | # BZOJ 2440 [中山市选2011]完全平方数 二分+容斥
$O(T*logn*\sqrt{n})$
#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
#define ull unsigned long long
#define ll long long
#define R register ll
#define pause (for(R i=1;i<=10000000000;++i))
#define In freopen("NOIPAK++.in","r",stdin)
#define Out freopen("out.out","w",stdout)
static char B[1<<15],*S=B,*D=B;
#ifndef JACK
#endif
inline int g() {
R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-1:fix;
if(ch==EOF) return EOF; do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret*fix;
} inline bool isempty(const char& ch) {return (ch<=36||ch>=127);}
inline void gs(char* s) {
register char ch; while(isempty(ch=getchar()));
do *s++=ch; while(!isempty(ch=getchar()));
}
namespace Luitaryi {
const int N=32000;
int mu[N],pri[N/6],cnt,T,k;
bool vis[N];
inline void PRE() { mu[1]=1;
for(R i=2;i<=N-10;++i) {
if(!vis[i]) pri[++cnt]=i,mu[i]=-1;
for(R j=1;j<=cnt&&i*pri[j]<=N-10;++j) {
vis[i*pri[j]]=true;
if(i%pri[j]==0) break;
mu[i*pri[j]]=-mu[i];
}
}
}
inline bool ck(int x) { R ret=0;
for(R i=1,lim=sqrt(x);i<=lim;++i) ret+=mu[i]*(x/(i*i));
return ret>=k;
}
inline void main() { PRE();
T=g(); while(T--) {
k=g();
R l=1,r=k<<1;
while(l<r) {
R md=l+r>>1;
if(ck(md)) r=md;
else l=md+1;
} printf("%lld\n",l);
}
}
}
signed main() {
Luitaryi::main();
}
2019.07.17
posted @ 2019-07-17 22:59 LuitaryiJack 阅读(139) 评论(0编辑 收藏 举报 | 539 | 1,419 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2021-49 | latest | en | 0.155506 |
https://freevideolectures.com/course/3542/combinatorics/15 | 1,550,459,845,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247484020.33/warc/CC-MAIN-20190218013525-20190218035525-00249.warc.gz | 559,867,194 | 14,582 | # Combinatorics
IISc Bangalore Course , Prof. L. Sunil Chandran
236 students enrolled
# Overview
Pigeon hole Principle - Elementary Concepts - Elementary concepts and basic counting principles - Elementary concepts; Binomial theorem; Bijective proofs - Properties of binomial coefficients; Combinatorial identities - Permutations of multisets - Multinomial Theorem, Combinations of Multisets - Bounds for binomial coefficients - Sterlings Formula, Generalization of Binomial coefficients - Double counting;Some Techniques:Double counting - Halls Theorem for regular bipartite graphs; Inclusion exclusion principle - Recurrence relations and generating functions:Recurrence Relations - Generating functions - Solving recurrence relations using generating functions - Exponential generating functions - Special numbers - Partition Number - Catalan Numbers - Difference Sequences - Sterling Numbers
### Lecture 15: Generalization of Binomial coefficients - Part (2)
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# Lecture Details
Combinatorics by Dr. L. Sunil Chandran,Department of Computer Science and Engineering,IISc Bangalore.For more details on NPTEL visit httpnptel.ac.in
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0 | 312 | 1,264 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | longest | en | 0.672733 |
https://stats.stackexchange.com/questions/237792/sums-of-squares-total-between-within-how-to-compute-them-from-a-distance-ma | 1,719,001,069,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862157.88/warc/CC-MAIN-20240621191840-20240621221840-00532.warc.gz | 489,378,426 | 48,060 | # Sums-of-Squares (total, between, within): how to compute them from a Distance Matrix?
I am having trouble understanding the concept of Sum of Squares in the context of distance matrices (Studer et al. 2010).
The Sum of Squares I am familiar with is the classical $SS$ from ANOVA, performed on contingency table, such as
sex FE employment joblessness school
1 16 4 0 0
2 8 3 1 8
From which I can quickly compute the ANOVA Sum of Squares with
dtm = melt(dt, id.vars = c('sex'))
dtmcount = count(dtm, sex, value)
dtmcount %>% group_by() %>%
mutate(grandmean = mean(n)) %>%
group_by(sex) %>%
mutate(SumSqtTotal = (n - grandmean)^2) %>%
group_by(sex) %>% mutate(groupmean = mean(n)) %>%
mutate(SSW = (n - groupmean)^2) %>%
group_by(sex) %>% mutate(SSB = ( grandmean - groupmean)^2) %>%
group_by() %>%
summarise(SST = sum(SumSqtTotal), SSW = sum(SSW), SSB = sum(SSB))
# results #
SST = SSW SSB
53 20 33
The Sum of Squares makes sense to me because I understand that we are comparing means and decomposing means.
However, when it comes to distance matrices, I don't understand what the "mean" becomes.
Consider the same data, but this time we are comparing sequences.
sex Sep.93 Oct.93 Nov.93 Dec.93
1 1 school school school school
2 1 training training training training
3 1 school school school school
4 1 training training training training
5 1 school school school school
6 2 FE FE FE FE
7 2 FE FE FE FE
8 2 training training training training
9 2 school school school school
10 2 employment employment employment employment
In a classical sequence analysis, we would use a distance algorithm to compute pairwise dissimilarities and end up with a distance matrix, like this one (from Hamming distance)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 4 0 4 0 4 4 4 0 4
[2,] 4 0 4 0 4 4 4 0 4 4
[3,] 0 4 0 4 0 4 4 4 0 4
[4,] 4 0 4 0 4 4 4 0 4 4
[5,] 0 4 0 4 0 4 4 4 0 4
[6,] 4 4 4 4 4 0 0 4 4 4
[7,] 4 4 4 4 4 0 0 4 4 4
[8,] 4 0 4 0 4 4 4 0 4 4
[9,] 0 4 0 4 0 4 4 4 0 4
[10,] 4 4 4 4 4 4 4 4 4 0
The analysis of variance for sequence analysis was developed by Studer et al. (2010).
From the paper (2010), I quote the following :
According to Batagelj (1988), the notion of a gravity center holds for any kind of distances and objects, even though it is not clearly defined for complex nonnumeric objects such as sequences. It is likely that the gravity center does not itself belong to the object space, exactly as the mean of integer values may be a real noninteger value. [...] Even though the gravity center may not be observable, equation (4) provides a comprehensive way to compute the most central sequence, the medoid, of a set using weights. Searching the x that minimizes equation (4) is equivalent to minimizing the sum of the weighted distances from x to all other sequences. (p.8).
They developed a R function in the library TraMineR. It actually enables to get p.values from co-variates using permutation tests.
The output looks like this :
Pseudo ANOVA table:
SS df MSE
Exp 1.7 1 1.700000
Res 9.6 8 1.200000
Total 11.3 9 1.255556
However, I fail to completely understand how to compute the Sum of Squares for such matrix (manually if possible), for both the Total and the Explanatory ? What is the "mean" or "center" in this context ?
Thank you.
Data and codes
library(dplyr)
library(reshape2)
library(TraMineR)
# data from TraMineR #
dt = structure(list(sex = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L,
2L, 2L, 2L), .Label = c("1", "2"), class = "factor"), Sep.93 = structure(c(3L,
4L, 3L, 4L, 3L, 1L, 1L, 4L, 3L, 2L), .Label = c("FE", "employment",
"school", "training"), class = "factor"), Oct.93 = structure(c(3L,
4L, 3L, 4L, 3L, 1L, 1L, 4L, 3L, 2L), .Label = c("FE", "employment",
"school", "training"), class = "factor"), Nov.93 = structure(c(3L,
4L, 3L, 4L, 3L, 1L, 1L, 4L, 3L, 2L), .Label = c("FE", "employment",
"school", "training"), class = "factor"), Dec.93 = structure(c(3L,
4L, 3L, 4L, 3L, 1L, 1L, 4L, 3L, 2L), .Label = c("FE", "employment",
"school", "training"), class = "factor")), .Names = c("sex",
"Sep.93", "Oct.93", "Nov.93", "Dec.93"), row.names = c(NA, -10L
), class = "data.frame")
# To transform the data into COUNT data #
dtm = melt(dt, id.vars = c('sex'))
dtmcount = count(dtm, sex, value)
# The SS is easily compute with #
dtmcount %>% group_by() %>%
mutate(grandmean = mean(n)) %>% group_by(sex) %>%
mutate(SumSqtTotal = (n - grandmean)^2) %>%
group_by(sex) %>% mutate(groupmean = mean(n)) %>%
mutate(SSW = (n - groupmean)^2) %>%
group_by(sex) %>% mutate(SSB = ( grandmean - groupmean)^2) %>% group_by() %>%
summarise(SST = sum(SumSqtTotal), SSW = sum(SSW), SSB = sum(SSB))
# Hamming distance function #
Ham = function(d){
mat = matrix(0, nrow(d), nrow(d))
len = nrow(d)
mat = matrix(0, len, len)
for(k in 1:len){
for(i in 1:len){
mat[k,i] = sum( ifelse( as.numeric( d[k, ] == d[i, ] ) == 1, 0 , 1) )
}
}
return(mat)
}
used in the example, like this
Ham(dt[,-1])
# The TraMineR Example # | 1,925 | 5,574 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-26 | latest | en | 0.933481 |
https://songho.ca/opengl/gl_cylinder.html | 1,642,815,155,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303729.69/warc/CC-MAIN-20220122012907-20220122042907-00474.warc.gz | 530,545,590 | 9,111 | # OpenGL Cylinder, Prism & Pipe
Related Topics: OpenGL Sphere
This page describes how to generate a cylinder geometry using C++ and how to draw it in OpenGL.
### Cylinder & Prism
The definition of a cylinder is a 3D closed surface that has 2 parallel circular bases at the ends and connected by a curved surface (side). Similarly, a prism is a 3D closed surface that has 2 parallel polygonal bases connected by flat surfaces.
Since we cannot draw a perfect circular base and curved side of the cylinder, we only sample a limited amount of points by dividing the base by sectors (slices). Therefore, it is technically constructing a prism by connecting these sampled points together. As the number of samples increases, the geometry is closer to a cylinder.
Triangular Prism (3)
Rectangluar Prism (4)
Octagonal Prism (8)
A vertex coordinate on a cylinder
Suppose a cylinder is centered at the origin and its radius is r and the height is h. An arbitrary point (x, y, z) on the cylinder can be computed from the equation of circle with the corresponding sector angle θ.
The range of sector angles is from 0 to 360 degrees. The sector angle for each step can be calculated by the following;
The following C++ code generates all vertices of the cylinder with the given base radius, height and the number of sectors (slices). It also creates other vertex attributes; surface normals and texture coordinates.
In order to reduce multiple computations of sine and cosine, we compute the vertices of a unit circle on XY plane only once, and then re-use these points multiple times by scaling with the base radius. These are also used for the normal vectors of the side faces of the cylinder.
``````
// generate a unit circle on XY-plane
std::vector<float> Cylinder::getUnitCircleVertices()
{
const float PI = 3.1415926f;
float sectorStep = 2 * PI / sectorCount;
std::vector<float> unitCircleVertices;
for(int i = 0; i <= sectorCount; ++i)
{
sectorAngle = i * sectorStep;
unitCircleVertices.push_back(cos(sectorAngle)); // x
unitCircleVertices.push_back(sin(sectorAngle)); // y
unitCircleVertices.push_back(0); // z
}
return unitCircleVertices;
}
...
// generate vertices for a cylinder
void Cylinder::buildVerticesSmooth()
{
// clear memory of prev arrays
std::vector<float>().swap(vertices);
std::vector<float>().swap(normals);
std::vector<float>().swap(texCoords);
// get unit circle vectors on XY-plane
std::vector<float> unitVertices = getUnitCircleVertices();
// put side vertices to arrays
for(int i = 0; i < 2; ++i)
{
float h = -height / 2.0f + i * height; // z value; -h/2 to h/2
float t = 1.0f - i; // vertical tex coord; 1 to 0
for(int j = 0, k = 0; j <= sectorCount; ++j, k += 3)
{
float ux = unitVertices[k];
float uy = unitVertices[k+1];
float uz = unitVertices[k+2];
// position vector
vertices.push_back(h); // vz
// normal vector
normals.push_back(ux); // nx
normals.push_back(uy); // ny
normals.push_back(uz); // nz
// texture coordinate
texCoords.push_back((float)j / sectorCount); // s
texCoords.push_back(t); // t
}
}
// the starting index for the base/top surface
//NOTE: it is used for generating indices later
int baseCenterIndex = (int)vertices.size() / 3;
int topCenterIndex = baseCenterIndex + sectorCount + 1; // include center vertex
// put base and top vertices to arrays
for(int i = 0; i < 2; ++i)
{
float h = -height / 2.0f + i * height; // z value; -h/2 to h/2
float nz = -1 + i * 2; // z value of normal; -1 to 1
// center point
vertices.push_back(0); vertices.push_back(0); vertices.push_back(h);
normals.push_back(0); normals.push_back(0); normals.push_back(nz);
texCoords.push_back(0.5f); texCoords.push_back(0.5f);
for(int j = 0, k = 0; j < sectorCount; ++j, k += 3)
{
float ux = unitVertices[k];
float uy = unitVertices[k+1];
// position vector
vertices.push_back(h); // vz
// normal vector
normals.push_back(0); // nx
normals.push_back(0); // ny
normals.push_back(nz); // nz
// texture coordinate
texCoords.push_back(-ux * 0.5f + 0.5f); // s
texCoords.push_back(-uy * 0.5f + 0.5f); // t
}
}
}
``````
Cylinders with different base/top radii and stack count
This C++ class provides buildVerticesSmooth() and buildVerticesFlat() functions depending on surface smoothness. The constructor also takes additional parameters to construct various shapes of a cylinder, similar to OpenGL gluCylinder() function.
The parameters of the cylinder class are;
3. the height (float)
4. the number of sectors (int)
5. the number of stacks (int)
6. smoothness (bool)
For instance, if the base radius is 0, it becomes a cone shape. For more details, please refer to Cylinder.cpp class.
In order to draw the surface of a cylinder in OpenGL, you must triangulate adjacent vertices counterclockwise to form polygons. Each sector on the side surface requires 2 triangles. The total number of triangles for the side is 2 × sectorCount. And the number of triangles for the base or top surface is the same as the number of sectors. (You may use GL_TRIANGLE_FAN for the base/top instead of GL_TRIANGLES.)
The code snippet to generate all the triangles of a cylinder may look like;
``````
// generate CCW index list of cylinder triangles
std::vector<int> indices;
int k1 = 0; // 1st vertex index at base
int k2 = sectorCount + 1; // 1st vertex index at top
// indices for the side surface
for(int i = 0; i < sectorCount; ++i, ++k1, ++k2)
{
// 2 triangles per sector
// k1 => k1+1 => k2
indices.push_back(k1);
indices.push_back(k1 + 1);
indices.push_back(k2);
// k2 => k1+1 => k2+1
indices.push_back(k2);
indices.push_back(k1 + 1);
indices.push_back(k2 + 1);
}
// indices for the base surface
//NOTE: baseCenterIndex and topCenterIndices are pre-computed during vertex generation
// please see the previous code snippet
for(int i = 0, k = baseCenterIndex + 1; i < sectorCount; ++i, ++k)
{
if(i < sectorCount - 1)
{
indices.push_back(baseCenterIndex);
indices.push_back(k + 1);
indices.push_back(k);
}
else // last triangle
{
indices.push_back(baseCenterIndex);
indices.push_back(baseCenterIndex + 1);
indices.push_back(k);
}
}
// indices for the top surface
for(int i = 0, k = topCenterIndex + 1; i < sectorCount; ++i, ++k)
{
if(i < sectorCount - 1)
{
indices.push_back(topCenterIndex);
indices.push_back(k);
indices.push_back(k + 1);
}
else // last triangle
{
indices.push_back(topCenterIndex);
indices.push_back(k);
indices.push_back(topCenterIndex + 1);
}
}
``````
### Example: Drawing Cylinder
This example constructs cylinders with 36 sectors and 8 stacks, but with different shadings; flat, smooth or textured. With the default constructor (without arguments), it generates a cylinder with base/top radius = 1, height = 2, sectors = 36 and stacks = 1. You could also pass the custom parameters to the constructor, similar to gluCylinder(). Press the space key to change the number of sectors and stacks of the cylinders.
Cylinder.cpp class provides pre-defined drawing functions using OpenGL VertexArray; draw(), drawWithLines(), drawLines(), drawSide(), drawBase() and drawTop().
``````
// height=4, sectors=5, stacks=6, smooth=true
Cylinder cylinder(1, 2, 3, 4, 5, 6, true);
// can change parameters later
cylinder.setHeight(3.5f);
cylinder.setSectorCount(36);
cylinder.setStackCount(8);
cylinder.setSmooth(false);
...
// draw cylinder using vertexarray
cylinder.draw(); // draw surface only
cylinder.drawWithLines(); // draw surface and lines
cylinder.drawSide(); // draw side only
cylinder.drawTop(); // draw top only
cylinder.drawBase(); // draw botton only
``````
This C++ class also provides getVertices(), getIndices(), getInterleavedVertices(), etc. in order to access the vertex data in GLSL. The following code draws a cylinder with interleaved vertex data using VBO and GLSL. Or, download cylinderShader.zip for more details.
``````
// create a cylinder with default params;
// radii=1, height=1, sectors=36, stacks=1, smooth=true
Cylinder cylinder;
// copy interleaved vertex data (V/N/T) to VBO
GLuint vboId;
glGenBuffers(1, &vboId);
glBindBuffer(GL_ARRAY_BUFFER, vboId); // for vertex data
glBufferData(GL_ARRAY_BUFFER, // target
cylinder.getInterleavedVertexSize(), // data size, # of bytes
cylinder.getInterleavedVertices(), // ptr to vertex data
GL_STATIC_DRAW); // usage
// copy index data to VBO
GLuint iboId;
glGenBuffers(1, &iboId);
glBindBuffer(GL_ELEMENT_ARRAY_BUFFER, iboId); // for index data
glBufferData(GL_ELEMENT_ARRAY_BUFFER, // target
cylinder.getIndexSize(), // data size, # of bytes
cylinder.getIndices(), // ptr to index data
GL_STATIC_DRAW); // usage
...
// bind VBOs
glBindBuffer(GL_ARRAY_BUFFER, vboId);
glBindBuffer(GL_ELEMENT_ARRAY_BUFFER, iboId);
// activate attrib arrays
glEnableVertexAttribArray(attribVertex);
glEnableVertexAttribArray(attribNormal);
glEnableVertexAttribArray(attribTexCoord);
// set attrib arrays with stride and offset
int stride = cylinder.getInterleavedStride(); // should be 32 bytes
glVertexAttribPointer(attribVertex, 3, GL_FLOAT, false, stride, (void*)0);
glVertexAttribPointer(attribNormal, 3, GL_FLOAT, false, stride, (void*)(sizeof(float)*3));
glVertexAttribPointer(attribTexCoord, 2, GL_FLOAT, false, stride, (void*)(sizeof(float)*6));
// draw a cylinder with VBO
glDrawElements(GL_TRIANGLES, // primitive type
cylinder.getIndexCount(), // # of indices
GL_UNSIGNED_INT, // data type
(void*)0); // offset to indices
// deactivate attrib arrays
glDisableVertexAttribArray(attribVertex);
glDisableVertexAttribArray(attribNormal);
glDisableVertexAttribArray(attribTexCoord);
// unbind VBOs
glBindBuffer(GL_ARRAY_BUFFER, 0);
glBindBuffer(GL_ELEMENT_ARRAY_BUFFER, 0);
``````
### Pipe (Tube)
Extruding a pipe along a path Q1-Q2-Q3
Cross-section view of extruding a pipe
P'1 is the intersection point on a plane and a line passing P1
A common application is drawing a pipe, which is extruding a contour along a given path. Suppose the path is Q1-Q2-Q3, and a point of the contour is P1. To find the next point, P'1, we need to project P1 onto the plane at the Q2 with the normal, , where 2 path lines Q1-Q2 and Q2-Q3 are met.
Projecting P1 to P'1 is actually finding the intersection of the point where a line and a plane are met. See the cross-section view (the right-side image above). The line equation is , which is passing P1 with the direction vector .
And, the plane equation can be computed by the normal vector and the point on the plane Q2 (x2, y2, z2);
And, the normal vector is computed by adding and together;
Finding the intersection point P'1 is solving the linear system of the plane and line;
You can find the solution of the linear system here. Or, see the detail C++ implementation in Pipe::projectContour() of Pipe.cpp and Plane.cpp.
``````
std::vector<Vector3> Pipe::projectContour(int fromIndex, int toIndex)
{
Vector3 v1, v2, normal, point;
Line line;
// find direction vectors; v1 and v2
v1 = path[toIndex] - path[fromIndex];
if(toIndex == (int)path.size()-1)
v2 = v1;
else
v2 = path[toIndex + 1] - path[toIndex];
// normal vector of plane at toIndex
normal = v1 + v2;
// define plane equation at toIndex with normal and point
Plane plane(normal, path[toIndex]);
// project each vertex of contour to the plane
std::vector<Vector3>& fromContour = contours[fromIndex];
std::vector<Vector3> toContour;
int count = (int)fromContour.size();
for(int i = 0; i < count; ++i)
{
line.set(v1, fromContour[i]); // define line with direction and point
point = plane.intersect(line); // find the intersection point
toContour.push_back(point);
}
// return the projected vertices of contour at toIndex
}
``````
### Example: Extruding Pipe along Path
This example is drawing a pipe extruding a circular contour following a spiral path. Press D key to switch the rendering modes.
A pipe can be constructed with a pre-defined path (a sequence of points), or you can add the next point of the path if needed using Pipe::addPathPoint().
The shape of the contour is not necessarily a circle. You can provide an arbitrary shape of a contour.
To draw the surface of the pipe, use Pipe::getContour() and Pipe::getNormal() to get the vertices and normals at a given path point. Then, draw triangles between 2 contours using GL_TRIANGLE_STRIP.
### Example: WebGL Cylinder (Interactive Demo)
Height
Sector Count
Stack Count
It is a JavaScript implementation of Cylinder class, Cylinder.js, and rendering it with WebGL. Drag the sliders to change the parameters of the cylinder. The fullscreen version is available here.
The following JavaScript code is to create and to render a cylinder object.
``````
// create a cylinder with 6 params: baseR, topR, height, sectors, stacks, smooth
let cylinder = new Cylinder(1, 2, 3, 4, 5, false);
...
// change params of cylinder later
cylinder.setHeight(3);
cylinder.setSectorCount(4);
cylinder.setStackCount(5);
cylinder.setSmooth(true);
...
// draw a cylinder with interleaved mode
gl.bindBuffer(gl.ARRAY_BUFFER, cylinder.vboVertex);
gl.vertexAttribPointer(gl.program.attribPosition, 3, gl.FLOAT, false, cylinder.stride, 0);
gl.vertexAttribPointer(gl.program.attribNormal, 3, gl.FLOAT, false, cylinder.stride, 12);
gl.vertexAttribPointer(gl.program.attribTexCoord0, 2, gl.FLOAT, false, cylinder.stride, 24);
gl.bindBuffer(gl.ELEMENT_ARRAY_BUFFER, cylinder.vboIndex);
gl.drawElements(gl.TRIANGLES, cylinder.getIndexCount(), gl.UNSIGNED_SHORT, 0);
...
`````` | 3,497 | 13,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2022-05 | latest | en | 0.868511 |
http://mathhelpforum.com/pre-calculus/146066-solved-differentiation.html | 1,501,053,528,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426050.2/warc/CC-MAIN-20170726062224-20170726082224-00498.warc.gz | 198,839,941 | 11,712 | 1. ## [Solved] Differentiation
1. The diagram shows an open tank for storing water, ABCDEF. The sides ABFE and CDEF are rectangles. The triangular ends ADE and BCF are isosceles, and Angle AED = BFC = 90deg. The ends ADE and BCF are vertical and EF is horizontal
a. Show that the area of triangle ADE is 0.25x^2 m^2
b.Given that the capacity of the container is 4000m^3 and that the total are of the two triangular and two rectangular sides of the container is S m^2.
Show that S = (x^2)/2 + {16000(2^1/2)}/x
The question looks straight-forward, i did indeed try it but couldn't figure out the solution.
Plz find attached figure as question 8 in jpeg.
Any help or hint will be much appreciated.
THX
2. a:
ADE is a right triangle so ue pythagoras' theorum.
$AD^2 + ED^2 = DE^2$
$2AD^2=x^2$
$AD=\frac{x}{\sqrt{2}}$
$0.5 * (\frac{x}{\sqrt{2}})^2$
$0.25 * x^2$
b:
The total area of the container is equal to DC * the area of the triangle. o we just need to find DC.
$4000 = DC * 0.25 * x^2$
But, we know the area of the rectangle satisfies
$S = 2*DC*DE + 2*0.25 * x^2$
$\frac{0.5S -0.25x^2}{DE} = DC$
$\frac{0.5S -0.25x^2}{\frac{x}{\sqrt{2}}} = DC$
Substitute into the other equation and hopefully you can continue from there
3. Originally Posted by SpringFan25
a:
ADE is a right triangle so ue pythagoras' theorum.
$AD^2 + ED^2 = DE^2$
$2AD^2=x^2$
$AD=\frac{x}{\sqrt{2}}$
$0.5 * (\frac{x}{\sqrt{2}})^2$
$0.25 * x^2$
b:
The total area of the container is equal to DC * the area of the triangle. o we just need to find DC.
$4000 = DC * 0.25 * x^2$
But, we know the area of the rectangle satisfies
$S = 2*DC*DE + 2*0.25 * x^2$
$\frac{0.5S -0.25x^2}{DE} = DC$
$\frac{0.5S -0.25x^2}{\frac{x}{\sqrt{2}}} = DC$
Substitute into the other equation and hopefully you can continue from there
hi, thanks a lot for your response. Could you please explain me why in (b)
volume of container = DC * 0.25* x^2 ????
we dont know the shape of the container in the question, what formula did we apply above?
thanks
4. Originally Posted by dee2020
hi, thanks a lot for your response. Could you please explain me why in (b)
volume of container = DC * 0.25* x^2 ????
we dont know the shape of the container in the question, what formula did we apply above?
thanks
1. "Turn" the container until it "stands" on the right triangle. That means you have a right prism with right triangles as base and top areas. As shown by SpringFan25 the base area has a value of $\frac14 x^2$. (Actually it is a quarter square where the square's side length is x).
2. The volume of a prism is calculated by:
$V= base\ area \cdot height$
According to the sketch the height corresponds to the edges DC or AB.
Thus the volume of the prism is calculated by:
$V = \frac14 x^2 \cdot |DC|$
3. Since you know the volume of V you are able to calculate the length of DC - as shown by SpringFan25.
5. Thanks a lot SpringFan25 and Earboth
Your help is very much appreciated.
Thanks again. | 934 | 2,972 | {"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": 21, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2017-30 | longest | en | 0.868236 |
https://tt.tennis-warehouse.com/index.php?threads/would-swinging-low-to-high-with-a-slightly-open-racket-face-at-contract-make-slice.527570/page-2 | 1,721,510,167,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517541.97/warc/CC-MAIN-20240720205244-20240720235244-00082.warc.gz | 516,708,184 | 28,681 | # Would swinging low to high with a slightly OPEN racket face at contract make slice?
## Would swinging low to high with a slightly OPEN racket face at contract make slice?
• ### The ball will have topspin on it
• Total voters
18
#### WildVolley
##### Legend
Yeah, that's what I mean, forward and up. When you go forward and up with a slightly open racquet face, the strings will strike the ball from below the equator of the ball and it can only hit it flat from below the equator of the ball.
To brush when striking below the equator of the ball, you have to pull back towards you a bit.
Just imagine the racket is open 3 degree from vertical and the ball only has a horizontal component as it moves into contact, but the racket is moving up at a 45 degree path (vector).
The ball will start compressing into the strings at 3 degrees below the equator but since the vector of the racket is at 45 degrees, as the ball compresses it will move down slightly against the string as it compresses and the racket pulls up against the outer wall of the ball. As it releases from the string bed it will have topspin.
#### Cowboyardee
##### Rookie
It's not very difficult to explain. Many of you are getting tripped up on the question of what 'open' means.
An open racquet face with respect to the ground (the racquet head is tilted back from perpendicular to the ground) doesn't tell you much of anything about what kind of spin you'll get. Not on its own. The ground (or horizontal) is actually a completely arbitrary reference point in this question.
The important factor is whether the racquet face is open with respect to the swing path at impact. An open face with respect to the swing path produces slice; a closed face with respect to the swing path produces topspin; a racquet face perfectly perpendicular to your swing path at impact produces no spin (this is nearly impossible to do in practice).
As such you can hit a shot with your racquet open with respect to the ground and get top spin if your swing path at impact is at an even steeper upward angle than your racquet face.
(ETA: the incoming angle of the ball does make some difference, but I believe its influence is thoroughly dwarfed in comparison to the racquet head's angle with respect to the swing path)
Last edited:
#### Enga
##### Hall of Fame
I think it would be like the shot Roger Federer hits on this video. https://m.youtube.com/watch?v=XRJTLGFKNP4
His swing path in that video is admittedly somewhat downwards. But anyway, if you open up the racket face too much, you'll get slice.
#### Cowboyardee
##### Rookie
It's not very difficult to explain. Many of you are getting tripped up on the question of what 'open' means.
An open racquet face with respect to the ground (the racquet head is tilted back from perpendicular to the ground) doesn't tell you much of anything about what kind of spin you'll get. Not on its own. The ground (or horizontal) is actually a completely arbitrary reference point in this question.
The important factor is whether the racquet face is open with respect to the swing path at impact. An open face with respect to the swing path produces slice; a closed face with respect to the swing path produces topspin; a racquet face perfectly perpendicular to your swing path at impact produces no spin (this is nearly impossible to do in practice).
As such you can hit a shot with your racquet open with respect to the ground and get top spin if your swing path at impact is at an even steeper upward angle than your racquet face.
(ETA: the incoming angle of the ball does make some difference, but I believe its influence is thoroughly dwarfed in comparison to the racquet head's angle with respect to the swing path)
In other words, Wildvolley is right, but most of you guys are hung up on horizontal and vertical reference points which are completely arbitrary in this this conversation. Open or closed with respect to horizontal doesn't matter because we're talking about spin, not trajectory.
#### WildVolley
##### Legend
In other words, Wildvolley is right, but most of you guys are hung up on horizontal and vertical reference points which are completely arbitrary in this this conversation. Open or closed with respect to horizontal doesn't matter because we're talking about spin, not trajectory.
You've explained this more concisely than I did.
I was very confused about whether people were talking past each other because of different definitions of the terms or whether someone was just confused, or whether someone was just trolling.
I also still stand by doing an empirical experiment for yourself. You can just drop a ball onto a tennis court and then swing into it with an upward racket path and a slightly open face and watch the topspin. I guess I took a little offense when some claimed I was either a liar saying I saw this or that I was subconsciously closing the racket face at contact.
#### Rui
##### Semi-Pro
I think Cowboyardee got it. If the swing path is steep enough, the ball will turn over, even with a slightly open face.
#### sureshs
##### Bionic Poster
As such you can hit a shot with your racquet open with respect to the ground and get top spin if your swing path at impact is at an even steeper upward angle than your racquet face.
No, you will still get slice unless the ball goes backwards in which case you can call it top spin towards the back fence if you wish!
#### Rui
##### Semi-Pro
Don't you think that dropping a tennis ball straight down on an open face, a la /, that the ball will come off with a topspin?
#### sureshs
##### Bionic Poster
Don't you think that dropping a tennis ball straight down on an open face, a la /, that the ball will come off with a topspin?
Yes.................
#### WildVolley
##### Legend
No, you will still get slice unless the ball goes backwards in which case you can call it top spin towards the back fence if you wish!
Sureshs, so you don't embarrass yourself further, please hit a tennis ball with an upward swinging racket path at a greater angle than the racket is open from vertical. Then report back what happens.
#### mntlblok
##### Hall of Fame
Changeover from topspin to underspin with racket tilt
Played with the inputs a bit and came sorta close to no spin (77 rpm's of topspin) with the following:
incoming spin of 1000 rpms of topspin
swing speed of 25mph
swing angle of 65 degrees upward
racket tilt of -20 degrees
https://www.flickr.com/photos/mentalblock/16223661184/
Interestingly, it's not a straight line change for the spin when you manipulate just one of the above parameters.
https://www.flickr.com/photos/mentalblock/16232549294/
Tried it again and added the inputs to the graphic. Also figgered out that you have to look below the graphic to learn whether the spin posted in the graphic is topspin or underspin. Might be straight-line after all. . .
#### Cowboyardee
##### Rookie
No, you will still get slice unless the ball goes backwards in whh case you can call it top spin towards the back fence if you wish!
You say this on what basis? I explained why you get topspin and can explain why again or another way if needed, but you didn't address the explanation. Also, this is a stupid-easy experiment to go perform yourself and shoot a big hole in your premise.
So what exactly don't you understand about the explanation I provided? You seem to be merely restating your conclusion even as the argument for it falls apart.
#### thomasferrett
##### Hall of Fame
Sureshs, so you don't embarrass yourself further, please hit a tennis ball with an upward swinging racket path at a greater angle than the racket is open from vertical. Then report back what happens.
We can all do this, but it's very hard to see with the naked eye what kind of spin is being put on the ball.
Also, if we concede that you ARE putting topspin on the ball even by contacting with an open racket head face, are you severely reducing that topspin by the face being slightly open?
So, say you swing at 100mph racket head speed and the face is 3 degrees closed, and you get 4000rpm topspin. If you take the same racket head speed and path, but hit with the face at 3 degrees open, how much would the topspin drop by? Would it drop to about 3800rpm? 3200rpm? 1000rpm? 50rpm?
You see what questions I'm getting at...
#### WildVolley
##### Legend
We can all do this, but it's very hard to see with the naked eye what kind of spin is being put on the ball.
It wasn't difficult for me to see. You can put a dot on the ball to help see the spin if you're having difficulty seeing it.
Also, if we concede that you ARE putting topspin on the ball even by contacting with an open racket head face, are you severely reducing that topspin by the face being slightly open?
Because of the higher launch angle from the open face, it is going to be easier to hit the ball long with an open racket face. So I believe that makes it more difficult to swing very fast and still have the ball drop in the court.
It is also more difficult to put spin the ball facing a fast incoming ball if the face is at all open. Sureshs seemed to disagree with this if I read him correctly. But honestly, his analysis still confuses me.
So, say you swing at 100mph racket head speed and the face is 3 degrees closed, and you get 4000rpm topspin. If you take the same racket head speed and path, but hit with the face at 3 degrees open, how much would the topspin drop by? Would it drop to about 3800rpm? 3200rpm? 1000rpm? 50rpm?
You see what questions I'm getting at...
Your best bet is to put this into the TW calculator. My guess is if the ball has incoming topspin (which most balls do) the closed racket is going to allow you to put more spin on the ball and the open racket won't allow you to hit as much spin.
#### Cowboyardee
##### Rookie
We can all do this, but it's very hard to see with the naked eye what kind of spin is being put on the ball.
Also, if we concede that you ARE putting topspin on the ball even by contacting with an open racket head face, are you severely reducing that topspin by the face being slightly open?
So, say you swing at 100mph racket head speed and the face is 3 degrees closed, and you get 4000rpm topspin. If you take the same racket head speed and path, but hit with the face at 3 degrees open, how much would the topspin drop by? Would it drop to about 3800rpm? 3200rpm? 1000rpm? 50rpm?
You see what questions I'm getting at...
If the dot trick doesn't work, perhaps hitting a wet ball would make this easier for you to see with your naked eye, as water flies off the ball?
In truth determining the exact rate of spin is a fairly difficult physics problem, involving many factors. Determining the direction of spin, on the other hand, is quite easy. It helps to simplify the scenario in your head: Don't imagine a person standing on the ground, swinging a racquet at a moving ball with a tennis swing. Just imagine a racquet in empty space swinging toward the ball.
If it hits the ball with the racquet face perpendicular to the swing plane, it drives it straight without spin (or at least with comparatively little spin, since hitting no spin is nearly impossible). Easy to visualize, right? Open the racquet face with respect to the swing plane and you'll get slice. Close it with respect to the swing plane and you get topspin.
Exactly how much? That depends on many factors (and yes, it's even possible that those other factors could result in counter-intuitive spin direction given a highly marginal situation - say, a racquet only 0.1 degree open with respect to the swing plane). But 99%+ of the time, spin direction is determined by the simple scenario spelled out above.
#### toly
##### Hall of Fame
We can all do this, but it's very hard to see with the naked eye what kind of spin is being put on the ball.
Also, if we concede that you ARE putting topspin on the ball even by contacting with an open racket head face, are you severely reducing that topspin by the face being slightly open?
So, say you swing at 100mph racket head speed and the face is 3 degrees closed, and you get 4000rpm topspin. If you take the same racket head speed and path, but hit with the face at 3 degrees open, how much would the topspin drop by? Would it drop to about 3800rpm? 3200rpm? 1000rpm? 50rpm?
You see what questions I'm getting at...
If a ball has zero speed then its spin (S) in rpm can be calculated according to next formula
S = 1.45* V*A
Where V is the racquet head speed in mph and A is the angle between directions of the racquet velocity and perpendicular to the racquet’s strings in degrees.
In your post: V = 100mph and S = 4000rpm, therefore the angle A is
A = S/(1.45*V) = 4000/(1.45*100) = 28°
In case of racquet face at 3° open the A angle will be A = 22° and ball’s spin
S = 1.45*100*22 = 3190rpm
#### mntlblok
##### Hall of Fame
We can all do this, but it's very hard to see with the naked eye what kind of spin is being put on the ball.
Also, if we concede that you ARE putting topspin on the ball even by contacting with an open racket head face, are you severely reducing that topspin by the face being slightly open?
So, say you swing at 100mph racket head speed and the face is 3 degrees closed, and you get 4000rpm topspin. If you take the same racket head speed and path, but hit with the face at 3 degrees open, how much would the topspin drop by? Would it drop to about 3800rpm? 3200rpm? 1000rpm? 50rpm?
You see what questions I'm getting at...
Was gonna plug those into the "shot maker" tool, but you need to at least add the swing angle. Let's say it's 30 degrees upward. Oops. Need to go with the face tilted forward (closed) at least 7 degrees closed to keep it in the court - assuming that the incoming ball has 4000 RPM's of topspin on it - the way the tool is initially set up. So, let's play with a common complaint here, too, and have the ball be a slow floater without much spin. Things get a lot more interesting. Let's say 1000 RPM's of top and 20 MPH of incoming speed, and go back to the 3 degrees closed racket face. That gets us to a little over 3000 RPM's of topspin.
Now, we just change that 3 degrees closed face to a 3 degrees open face. The spin drops to 2383 RPM's of topspin, but the change in launch angle causes the ball to land over 30 feet long.
https://www.flickr.com/photos/mentalblock/16674188450/ The red ball shows the closed face result.
https://www.flickr.com/photos/mentalblock/16860547391/ The red ball shows the open face result.
Yer welcome. :mrgreen:
#### sureshs
##### Bionic Poster
My guess is if the ball has incoming topspin (which most balls do) the closed racket is going to allow you to put more spin on the ball and the open racket won't allow you to hit as much spin.
Which pro hits a deliberate topspin shot with an open face and just relies on incoming spin and friction? When he wants to hit topspin, he face is closed or neutral. When he hits a defensive lob by punting the ball back up high with an open face, he will never claim to "put topspin" on the ball or "hit with topspin."
Last edited:
#### mntlblok
##### Hall of Fame
Which pro hits a deliberate topspin shot with an open face and just relies on incoming spin and friction? When he wants to hit topspin, he face is closed or neutral. When he hits a defensive lob by punting the ball back up high with an open face, he will never claim to "put topspin" on the ball or "hit with topspin."
https://www.flickr.com/photos/mentalblock/16789281796/in/set-72157650697596641 You ask for evidence, but then seem to choose to ignore it. . .
#### Narcissist
##### Semi-Pro
Topspin but yes, the angle matters:
#### mntlblok
##### Hall of Fame
Blog page
I'm betting that yer not the author of that blog page. It's interesting stuff, but I don't think it's related to the topic at hand. The "tilts" here are related to what happens to the racket face *after* contact. If the contact is made below the center of the strings, then it causes the racket to tilt forward after contact (twist in the player's hand). And, if the contact is made above the center, up near the top of the frame, then it twists the frame into an open position (tilted backwards). He also notes that the racket remains "stable", or doesn't twist, if the ball is contacted near the center of the string-bed.
The fascinating part of the article, though, is that he sez he's measured the spin from these three types of contacts and has found that the contact near the bottom edge of the frame generates more topspin. I might have guessed the opposite. Now I wonder if these top guys might be doing some of this on purpose. . . Hmmm. . .
#### ttbrowne
##### Hall of Fame
I notice that the racquet face is not open at contact but straight up and down. After the ball comes off...the racquet tilts backward.
#### mntlblok
##### Hall of Fame
I notice that the racquet face is not open at contact but straight up and down. After the ball comes off...the racquet tilts backward.
Right. The article explains about that. Different issue.
Interestingly, Dolgopolov, at 7-6, 4-2 against Raonic, hits a topspin lob, and the replay clearly shows his open racket face for the steeply upward stroke.
#### sureshs
##### Bionic Poster
https://www.flickr.com/photos/mentalblock/16789281796/in/set-72157650697596641 You ask for evidence, but then seem to choose to ignore it. . .
And however many times I tell you that the incoming spin was 4000 rpm, you choose to ignore it.
Do you know that there is something called a rebound coefficient which is used for non-spin studies? Shoot a ball at a stationary racket at 70 mph, and it will come back with a speed of x*70, x < 1. The simplest model for a groundie speed is: racket head speed + x*incoming ball speed.
Will you claim that the second term is consciously created by the player by simply holding a frame in his hands?
#### sureshs
##### Bionic Poster
https://www.flickr.com/photos/mentalblock/16232549294/
Tried it again and added the inputs to the graphic. Also figgered out that you have to look below the graphic to learn whether the spin posted in the graphic is topspin or underspin. Might be straight-line after all. . .
As your graphic suggests, play with a -23 degrees tilt open face and you will get a whopping 6 rpm of topspin. It is topspin after all, right?
#### Povl Carstensen
##### Legend
I think we went over this in another thread. I was contradicted many times but no one could disprove me at the end.
It is very easy to take a ball (and a racket) and hit a topspin shot with a slightly open face, and prove it to yourself. As I did and others confirmed. I will not post a video.
#### Povl Carstensen
##### Legend
You're likely pulling back during that swing without even knowing it. With open face racquet swing FORWARD low to high, can't really get topspin since strings can't brush up on the back of the ball UNLESS you are pulling back (even subtly) not going forward.
Ofcourse you can brush upward with an open racket even if the racket is moving foreward. As long as the movement of the racket is more upward than the 90 degrees vektor of the string plane. In that case you are hitting low to high on the ball. If the direction of the racket is lower than the 90 degrees vektor, you are slicing. If they coincide, you are hitting straight forward with no spin.
Of course the up/downwards movement of the ball, and the spin on the incomming ball are also factors, but that does not change the above.
Last edited:
#### Povl Carstensen
##### Legend
You guys carry on since you don't want to accept defeat or learn anything new.
This is actually funny.
#### Cowboyardee
##### Rookie
As your graphic suggests, play with a -23 degrees tilt open face and you will get a whopping 6 rpm of topspin. It is topspin after all, right?
A genuinely realistic model for determining spin would include variables for the ball's spin before impact, the ball's weight, the ball's incoming angle, rotational angle of the racquet face at impact, the racquet's weight and acceleration, string tension and coeficient of friction (both with itself and the surface of the ball), and a dozen other variables. We're really talking NASA level calculations to arrive at an answer that begins to get precise. The graphics posted in this thread are good illustrations but (necessarily) simplified.
I guess I just don't even know what you're arguing anymore...
- That it's impossible to hit topspin with your racquet face open with respect to the ground? Hopefully you recognize that's not a tenable position at this point.
- That no decent/pro tennis player ever actually hits a topspin shot with the racquet face open (again, with respect to the ground/horizontal)? I believe many do on feeds and on topspin/offensive lobs, but perhaps high speed video would clear this issue up.
- Or whether those topspin lobs have high or low rates of topspin? Simple calculations and equations are pretty inadequate to answering this question in a satisfactory manner. Again, high speed video might be the most useful way to answer this question, but my offhand experience is that some of these shots have rates of spin significantly higher than 6 rpm and are often hit with the racquet face somewhat open.
#### WildVolley
##### Legend
A low to high swing in which the racket is moving forward with an open face CANNOT produce topspin "by itself," meaning that it does produce a tiny amount of topspin due to the recoil of the strings (which is also produced by a stationary racket).
The majority of people on this thread argue that the above is wrong if we change it to slightly open.
Do you still maintain that spin is due to "string recoil" rather than standard topspin generation by a racket when we're talking about hitting a ball with a slightly open racket face and a forward racket vector at greater upward angle than the strings are open?
Have you tried it on court yet?
#### mntlblok
##### Hall of Fame
See the TWU experiment "How Impact Location and Twistweight Affect Spin".
http://twu.tennis-warehouse.com/learning_center/location.php
So, the study the guy did was accurate, then, right? Any idea if pros are aware of this and utilize it?
Pretty cool to actually hear from the professor. I assume he is Crawford Lindsey? I am _Technical Tennis_'s biggest fan. :mrgreen: One of the posters might take a look at the first paragraph on page 122 of same.
I would also note that my attempts to duplicate the results from what is predicted by the Shot Maker tool suggest that they are *very* accurate and model the real world right well. I've found some wonderful insights from experimenting with the tool. If you ever need a guinea pig. . .
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# ACTIVITY – 1MENTAL ARITHMATIC-MONEY-MEASUREMEMT
## SEC. 1 B : There are 10 questions in this test. Give your self 10 minutes to answer them all.
1. Alan spent £1.50 on two bottle of water estimate the water
2. John distribute £7 equally among two children how much doe each child get?
3. Average the following number to make the target even number
4. The total sum of two numbers is 17. 0ne of the numbers is 9. What is the number?
5. Alan spends £ 1 at a shop. John spends £ 2.32. How must more money did john spend than Alan?
6. Make each of the following number 10 times greater?a) 41b) 2.7
7. John pays to his food with a £ 5 note. He receives £ 2045 change. How much change did john did food cost?
8. Joe cuts of 75cm of a3 ½ m rope. What length of rope remains?
9. Round up 1420 to the nearest thousand?
10. Round up 635cm to the nearest 1/2m
11. Round up 3.5km to the nearest whole km
12. Round up 8km 905m to the nearest km | 321 | 1,038 | {"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} | 3.140625 | 3 | CC-MAIN-2017-26 | longest | en | 0.895905 |
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Research
The equation for finding resistance is
R= V/I
Where V = potential difference (volts) and I = current (amps)
Current is the rate of flow of charge. An amp = 1 coulomb/second. The coulomb is the standard unit of charge. Potential difference is the amount of electrical energy transferred per unit of charge between two points. It is measured in joules per coulomb, or volts.
The opposition to the flow of charge is resistance, measured in ohms (?). The larger a materials resistance, the greater the amount of potential difference needed to make a current flow through it.
A material such as steel will have a relatively high resistance compared to that of gold. Some materials, known as superconductors have no resistance whatsoever!
When a potential difference is applied to a conductor, all the free electrons in it move in the same direction. When the electrons ‘move’ through the conductor they collide with the atoms in the material, so they are continually accelerating and decelerating.
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Because of this the electrons do not have a constant velocity, so we give them an average velocity, known as the drift velocity. If the length of the conductor is increased, there will be more atoms for the electrons to collide with, therefore the electrons will have a slower drift velocity, therefore there is more resistance.
Resistivity is a quantity, which is used to compare the resistance of material r with the same direction.
R=? x L/A
Therefore: ?= RA/L
R= resistance, ?= Resistivity, L= length, A= area.
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L
This is a diagram of a wire in a circuit. The number of electrons in this circuit is given as
nAL
where:
* n= number of electrons per unit volume
* A= cross-sectional area of the wire
* L= length of the wire
Therefore the total amount of charge in the circuit is given as
nALe
where
* e= charge of one electron
Preliminary Data
Before I performed my main experiment, I felt it was necessary to do some preliminary research into the most appropriate wire to use and the length of wire to use. I decided to measure the resistance of both copper and nichrome wire for a number of different lengths. I used a multimeter to do this. This is what I found:
Nichrome
5 cm
10 cm
15 cm
20 cm
25 cm
30 cm
35 cm
40 cm
Swg 22
0.4
0.6
0.8
0.9
1.2
1.5
1.8
2.0
Swg 30
1.0
1.8
2.5
3.2
4.0
4.7
5.5
6.2
Swg 36
2.4
4.6
7.2
8.1
10.4
12.0
13.6
15.5
Copper
Swg 22
0.6
0.9
1.2
1.4
1.5
1.6
2.2
2.5
Swg 30
0.3
0.4
0.7
1.0
1.2
1.4
1.6
1.9
Swg 36
0.3
0.5
0.6
0.8
0.9
1.1
1.2
1.5
V is directly proportional to I: If there is an increase in V there will be an increase in I. The best wire for this experiment is one with a relatively high resistance per unit length but also one which will not overheat with higher resistance (a wire with good temperature stability.) In my experiment, I will use nichrome wire, because it displays both of the above qualities. I will use 0.5 amps through 30cm of wire.
I will carry out my experiment as fairly as possible, but there are bound to be errors and inaccuracies. Many of these will be due to the measuring equipment. The most accurate ammeters and voltmeters available to me are accurate to �0.01Amp (or Volt). This will affect the accuracy of my readings for p.d and current and more importantly, my calculated resistance, where the associated percentage errors will be added together, increasing the overall %error. The metre rule used to measure the wire is accurate to �0.001m. The micrometer used to measure the diameter of the wire is accurate to �0.01mm. Although a hundredth of a millimetre is a very small inaccuracy, we must remember that we are using this measurement to calculate the cross-sectional area of the wire – and because this process involves a square, the percentage error will be doubled. One way to reduce the overall effect of this type of inaccuracy is to take the mean value of a number of readings and I have done this throughout my experiment.
All other physical conditions should ideally be kept constant. The effect of heating due to current has been minimised by using a relatively small current. The wire chosen (nichrome) was also selected because its physical properties would help keep my results accurate and reliable.
Apparatus
* Power Pack 0-12V DC
* Metre ruler
* Ammeter 0-10a �0.01a
* Voltmeter 0-70V �0.01V
* Rheostat 0-15?
* Crocodile clips
* Nichrome wire Swg 22 – 36
Diagram
DC current 0 – 12V
Method
1. Wire up the circuit as shown
2. Tape a piece of wire (Swg 22) tightly to a metre ruler
3. Measure 30cm into the wire and attatch the crocodile clips
4. Switch on power pack
5. Adjust the rheostat so the ammeter reads 0.5 amps
6. Take a reading from the voltmeter
7. Move the crocodile clips to a different place on the wire, but still with a separation distance of 30cm
8. Repeat steps 5 – 7
9. Repeat step 8
10. Repeat steps 2 – 9 with Swg 24, 28, 30, 32, 34, 36.
Having collected and written down all of my data, I fed it into a spreadsheet, so I could manipulate and calculate the values easily and with minimal risk of error.
Calculation of errors
1. Length measured = 0.1m
Accuracy of measuring equipment = �0.001m
Percentage error = accuracy/recorded length x100 = 0.001/0.1 x100 = �1%
2. Current measured = 0.5A
Accuracy of measuring equipment = �0.01A
Percentage error = accuracy/recorded value x100 = 0.01/0.5 x100 =�2%
3. Voltage measured =
Accuracy of measuring equipment = �0.01V
Percentage error = accuracy/recorded value x100
= 0.01/voltage reading x100
4. Resistance measured =
%error in V + %error in I
Error bars were used when drawing the graph. I calculated these by working out the total %errors in both r and L/A and then drawing them onto the graph in an ‘I’ shape.
Analysis
Once all the data for my experiment had been collected, averages could be taken and percentage errors found. For accuracy, and also to save time, I put all the raw data into a spreadsheet, entered the correct formulas and hence calculated all the necessary values (see results table). Having done this I plotted a graph of resistance against L/A, but instead of just plotting points I drew error bars. By doing this I was able to draw a number of gradients, and therefore gain higher accuracy. Having taken the gradients, I was able to use the mathematical equations shown at the start to get a value for resistivity.
The gradient is the resistivity of the wire because:
R= ? L/A
So
?= RA/L = R/L/A = gradient of graph.
To find the gradient I divided resistance by L/A:
* Minimum gradient: 0.9636 x10-6 ?m
* Best gradient: 1.0363 x10-6 ?m
* Maximum gradient: 1.0909 x10-6 ?m
The graph I drew was a straight-line graph of resistance against L/A. L, the length was constant. Therefore
? 1
A
The greater the L/A value, the greater the resistivity.
I now had my values for the resistivity of the wire. However, my graph did feature one anomaly, which lay quite a way out. I chose to ignore it when drawing the gradient because it was obvious it was inaccurate. This was probably due to human error somewhere along the process. I found that because there were so many equations to get to the actual resistivity, there was more chance for human error.
Evaluation
When doing my experiment, I tried to make my work as accurate as possible by using the best equipment available to me. However there were still small discrepancies in my work. I took an average value for resistance by measuring three different lengths of wire within a metre of wire taped to a metre ruler. To improve the accuracy of my work, I could have chosen three separate lengths of wire, maybe from different coils, which would have given me a better average. Also, when I recorded the data, I did it in two sessions; a number of conditions that could have affected my experiment could have changed, for example the heat.
The actual value for resistivity, given by the catalogue the wire was obtained from is 1.13 x10-6?m. All my values for resistivity lay slightly below this value, but my closest value, taken from my maximum gradient was 1.0909 x10-6?m, which is only 0.0391 x10-6?m.
I believe that my experiment was of a good level of accuracy and I regard it a success. | 2,278 | 8,868 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2023-23 | latest | en | 0.931477 |
http://ferrao.org/tags/aeb464-percentage-formula-in-excel-sheet | 1,620,720,616,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991904.6/warc/CC-MAIN-20210511060441-20210511090441-00157.warc.gz | 19,440,749 | 4,022 | This is shown in the formula below: =B3*1.05. Navigate to your worksheet. Further Examples of Calculating a Percentage of a Number Example 1 - Percentages of Various Numbers. This will calculate the % change in revenue between 2008 & 2009. 1. Copy and paste the formula to the remaining cells to see the year on percent change. Add a column for percentage Change year over year. Using cell references, if October’s bill amount of \$125 is in cell B4 and November’s bill amount of \$100 is in cell B5, your Excel formula for a percentage decrease would be =SUM(B5-B4)/B4. We will copy down the formula using the fill handle tool (the small box at the lower right corner of the cell) into the number of rows given. Therefore, the value 20% in the above calculation is evaluated as 0.2. Figure 2 – Drawing percentage in excel. If you want to add percentage sign to a number without multiplying the number by 100, such as to change 6 into 6% in Excel with Percentage Style, you should have the general number divided by 100 first, and then apply the Percentage Style to the numbers for displaying them as percentage 6% in Excel. 3. You can also just type the number in its decimal form directly into the cell—that is, type 0.1 and then apply percentage format. These formulas simply multiply the value by five percent more than the whole of itself (100 percent). Launch Excel. 1. Excel percentage formulas can get you through problems large and small every day. Markup Percentage Formula Markup in very simple terms is basically the difference between the selling price per unit of the product and the cost per unit associated in making that product. Procedure in Excel. The following spreadsheet shows various percentage calculations for different numbers. If you then format that decimal as a percentage, the number will be displayed as 10%, as you ‘d expect. Note that the % operator tells Excel to divide the preceding number by 100. So revenue has been decreased by -15.75% from 2008 to 2009. Apply the basic excel formula as (New Value – Old Value) / Old Value.. 4. Excel formula for percentage change for a year over year. To find the percentage change year over year: = ((This year’s value-last year’s value)/last year’s value*100 . So basically it is the additional money, over and above the cost of the product, which the seller would get. Add a zero at the topmost cell of the column since it coincides with the Beginning year. 2. We can quickly calculate the percentage change in our excel sheet across two columns using the steps below: We click on Cell F4 and enter the formula below =(D4-E4)/D4. As a result, a 20 percent increase would be multiplied by 120 percent, and a 15 percent increase would be 115 percent (or 1.15). Or we could write the formula as: =B3*105%. For example, if you type the formula =10/100 in cell A2, Excel will display the result as 0.1. | 663 | 2,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2021-21 | latest | en | 0.916272 |
http://proteinsandwavefunctions.blogspot.com/2014/01/efmo-pcm-roadmap-version-2.html | 1,498,547,128,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321025.86/warc/CC-MAIN-20170627064714-20170627084714-00244.warc.gz | 307,784,118 | 24,640 | ## Wednesday, January 15, 2014
### EFMO-PCM: A roadmap. Version 2
Having read the EFP-PCM interface more carefully again I need to make a major update of a previous post. Since Blogger doesn't have version numbers, here comes a separete post.
One of the things still missing in the EFMO method is an interface to PCM. Here I attempt to sketch the method, based on the FMO-PCM and EFP-PCM interfaces.
The EFMO-PCM electrostatic interaction free energy between solute and solvent plus the polarization energy is
$G_{pol+sol}=-\frac{1}{2}\mathbf{F}^{mul} \boldsymbol{\mu}^T+\frac{1}{2}\mathbf{V}^{mul} \mathbf{q}^T$
As Hui and Mark wrote:
"The interaction between the induced dipoles and charges is implicitly described with the matrices $\mathbf{R}$ and $\mathbf{R}^T$, just as the interaction between the EFP induced dipoles is implicitly described with the off-diagonal elements of matrix $\mathbf{D}$, and the interaction between the PCM induced charges is implicitly described with the off-diagonal elements of matrix $\mathbf{C}$"
$\mathbf{q}$ are the apparent surface charges (ASCs), which for large systems are obtained by solving this equation iteratively
$\mathbf{Cq}=-\mathbf{V}$
$\mathbf{V}$ is the electrostatic potential from the static multipoles and induced dipoles, respectively:
$\mathbf{V=V^{\text{mul}}+V^{\mu}}$
The simplest approximation to $\mathbf{V}$ is
$$\mathbf{V} =\sum_I^N \mathbf{V}_{I} = \sum_I^N (\mathbf{V}^{\text{mul}}_{I} +\mathbf{V}^{\mu}_{I})$$
The potential at tesserae $j$ due to induced dipoles on fragment $I$ is given by:
$$\mathbf{V}_I^\mu (j)=\sum_{i \in I} (\mathbf{R}^T)_{ji}\boldsymbol{\mu_{i}}$$
The induced dipoles are obtained iteratively:
$$\boldsymbol{\mu}_i=\boldsymbol{\alpha}_i\left(\mathbf{F}^{\text{mul}}_i+\mathbf{F}^q_i-\sum_{i\neq j}\mathbf{D}_{ij}\boldsymbol{\mu}_j\right)$$
where $\boldsymbol{\alpha}_i$ is the dipole polarizability tensor at site $i$ and $\mathbf{F}^{\text{mul}}_i$ and $\mathbf{F}^q_i$ are the electrostatic fields due to all static multipoles and ASCs felt at point $i$.
Procedure:
1. Compute EFMO gas phase energy
2. Use gas phase static multipoles and $\boldsymbol{\mu}$ to construct $\mathbf{V}$
3. Solve $\mathbf{Cq=-V}$
4. Use $\mathbf{q}$ and $\boldsymbol{\mu}_i=\boldsymbol{\alpha}_i\left(\mathbf{F}^{\text{mul}}_i+\mathbf{F}^q_i-\sum_{i\neq j}\mathbf{D}_{ij}\boldsymbol{\mu}_j\right)$ to find new $\boldsymbol{\mu}$
5. Repeat steps 2-4 until self consistency
6. Compute $G_{pol+solv}$
$$G_{pol+sol}^x=G_{sol}^x+G_{pol}^x$$
$$G_{sol}^x=\mathbf{V}^x \mathbf{q}^T+\frac{1}{2}\mathbf{q}^T\mathbf{C}^x\mathbf{q}$$
$$G_{pol}^x=-\mathbf{F}^{mul,x} \boldsymbol{\mu}^T+\boldsymbol{\mu}^T\mathbf{D}^x\boldsymbol{\mu}$$
The most tricky part is $\mathbf{V}^{\mu,x} \mathbf{q}^T$:
$$\sum_n^N V_n^{\mu,x} q_n = \sum_i^I V_i^{\mu,B,x_i} q_i+ \sum_j^J V_j^{\mu,A,x_A} q_j$$
$$V_i^{\mu ,B,{x_i}} = \sum\limits_{m \ne A}^{} {{{\left( {\frac{{ - ({{\bf{r}}_i} - {{\bf{r}}_m})}}{{{{\left| {{{\bf{r}}_i} - {{\bf{r}}_m}} \right|}^3}}}} \right)}^{{x_i}}}} \boldsymbol{\mu}_m^B$$
$$V_j^{\mu,A,{x_A}} = {\left( {\frac{{ - ({{\bf{r}}_j} - {{\bf{r}}_A})}}{{{{\left| {{{\bf{r}}_j} - {{\bf{r}}_A}} \right|}^3}}}} \right)^{{x_A}}}\boldsymbol{\mu}^A$$
Here $n$ sum over the all the tesserae ($N$). The tesserae set $I$ belong to the sphere centered on atom $A$ - the atoms whose coordinate ($x_A = x$) we are taking the derivative wrt. $x_i$ refers to the coordinate of a ASC associated with a $I$ tessera. $J$ are all other tessera belonging to atoms collectively referred to as $B$. Notice that the gradient involving tessera and induced dipoles belonging to the same atom is zero, $V_i^{\mu,A,x_A}=0$
The tricky thing in general with the EFMO gradient involving induced dipoles is that they are not centered on atoms. In text S1 of this paper we describe how we deal with this for bond-dipoles, $$V_j^{\mu,A,{x_A}}=f(R)V_j^{\mu,A,{x_{LMO-A}}}$$but we don't talk about how to do it for lone pairs. Perhaps we simply set $x_A = x_{LMO-A}$? Anyway, it should only be an issue for $V_j^{\mu,A,{x_A}}$.
Miscellaneous
For more accurate results one can approximate $\mathbf{V}$ as
$$\mathbf{V}=\sum_I^N \mathbf{V}_{I}+\sum^N_{I}\sum^N_{J<I} (\mathbf{V}_{IJ}-\mathbf{V}_{I}-\mathbf{V}_{J})$$
This essentially means that the gas phase static multipoles and $\alpha$'s are corrected in step 2. E.g. for static monopoles ($q$)'s:
$$V_I^q(i)=\frac{q_i^{I}}{|r-r_i|}$$
and
\begin{align*} V(i)&=V_I^q(i)+\sum^N_{J<I} (V_{IJ}^q(i)-V_I^q(i))\\ & = [q^{I}_i+\sum^N_{J<I}(q^{IJ}_i-q^{I}_i)]\frac{1}{|r-r_i|} \\ \end{align*}
All other steps are the same.
A tip on dealing with complex papers
I found the EFP-PCM paper a big mouthful and had to revert to an old trick that I'll share with you here. Mainly what makes the paper complex is its length so that definitions and equations are pages apart. For such papers I usually fire up PowerPoint and Snapndrag (for screenshots) and grab what I think are the essential pieces, rearrange some, and add "missing equations" when needed. You can see the final result here (the first two pages are the main results and the rest are "derivations").
I highly recommend this approach for complex papers or making sense of a collection of papers on a specific topic. | 1,845 | 5,282 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2017-26 | longest | en | 0.815157 |
http://www.culatools.com/forums/viewtopic.php?f=15&t=1061&p=1993 | 1,579,417,334,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594333.5/warc/CC-MAIN-20200119064802-20200119092802-00415.warc.gz | 207,331,448 | 6,276 | General CULA Dense (LAPACK & BLAS) support and troubleshooting. Use this forum if you are having a general problem or have encountered a bug.
I am trying to see the performance of different eigendecomposition on cula. At first I used culaDeviceDsyev and everything works fine.Now I tried to use the iterative method using culaDeviceDsyevx but I'm getting wrong answer.
let's say matrix "k" has eigenvalues vector(w) and eigenvectors (v) .so the equation k=U*diag(W)*U^-1 should hold , but it doesn't.The eigenvalues are ok in compare with culaDeviceDsyev. also I use python (Pycula). My code:
k_gpu=cula.cula_gpuarray_like(k_)
uplo='L'
evrange='A'
abstol=(2.2204460492503131e-16)*2
(m,n) = k_gpu.shape
assert m==n
lda = len(k_gpu)
jobz = 'V'
w = np.zeros(n,k_gpu.dtype)
w_gpu = cula.cula_gpuarray_like(w)
il=0
iu=0
vl=0
vu=0
m=ctypes.c_int(n)
z = np.zeros((lda,max(1,m.value)),k_gpu.dtype)
z_gpu = cula.cula_gpuarray_like(z)
ldz = len(z)
ifail=ctypes.c_int()
why it's happening ?
Thanks
Posts: 15
Joined: Wed Jun 20, 2012 2:58 am
### Re: culaDeviceDsyevx returns wrong value!?!
I think I made a mistake. in the iterative method we can not check the correctness of value by np.dot(np.dot(v,np.diag(w)),np.transpose(v)) != k.
right?how can I check it? is culaDeviceDsyevx using Lanczos algorithm?
Posts: 15
Joined: Wed Jun 20, 2012 2:58 am
### Re: culaDeviceDsyevx returns wrong value!?!
At first I used culaDeviceDsyev and everything works fine.Now I tried to use the iterative method using culaDeviceDsyevx but I'm getting wrong answer.
xSYEV and xSYEVX use the same algorithm under the hood if you are searching for all eigenvalues (range = 'A') so then answers should be the same. You might have a mistake in your wrapper code or perhaps there is an error in the 3rd party Pycula.
In the iterative method we can not check the correctness of value by np.dot(np.dot(v,np.diag(w)),np.transpose(v)) != k. right? How can I check it?
You can check the Eigendecomposition, A =V*diag(w)*inv(V). In your example, it looks like you are taking the transpose and not the inverse.
xSYEV uses a direct reduction to traditional form then QR iterations to find the Eigenpairs.
kyle
Posts: 301
Joined: Fri Jun 12, 2009 7:47 pm
### Re: culaDeviceDsyevx returns wrong value!?!
it was a wrapper bug! instead of returning the z value as eigenvectors it returns the w value!
Thanks | 696 | 2,380 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2020-05 | latest | en | 0.797664 |
https://www.khanacademy.org/math/in-in-grade-10-ncert/x573d8ce20721c073:coordinate-geometry/x573d8ce20721c073:distance-formula/v/example-finding-distance-with-pythagorean-theorem | 1,716,849,733,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.34/warc/CC-MAIN-20240527205559-20240527235559-00039.warc.gz | 710,760,915 | 93,028 | If you're seeing this message, it means we're having trouble loading external resources on our website.
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### Course: Class 10 (Old)>Unit 7
Lesson 1: Distance formula
# Finding distance with Pythagorean theorem
Sal finds the distance between two points with the Pythagorean theorem.
## Want to join the conversation?
• So, how do you find the distance between two points, if it isn't on a graph?
• One simple option is to draw a graph on a sheet of paper, plot your points on it and go from there:)
• I have to work in Aleks but in Aleks, they don't give you a numbered graph they just give you the line my work is something like this
Find the distance between the points A and B given below.
(That is, find the length of the segment connecting Aand B.)
• At , what is a principle ( or principal?) root? I have dealt with square roots many times, but is there a difference between a principle root and a square root?
• yes there is a difference the principal root is always positive -2*-2=4 and 2*2=4 the principal root would be 2
(1 vote)
• How do you write these answers in decimals?
• what is a principle root? As Jimmy Chimichanga said.
• what if its a line going say from (-4,2) to (-4,-4)
• In that case, you don't need to use the pythagorean theorem. If the x-coordinate of the endpoints is the same, the line is vertical (horizontal if y is same). You can just find how much the y-value increases or decreases from one point to the next, and that's your distance. If you use the pythagorean, one of your side lengths would be 0, so you would have:
(0)^2 + (2 - (-4))^2 = c^2
6^2 = c^2
c = 6
So the distance would be 6 units.
• I thought there was another way to find the distances between two points? I learned √ (x2 − x1)2 + (y2 − y1)2
The two 2 outside means squared.
• I am confusion!how do's you do this?!america explain!
• On How Do You Solve For C?
• Replying to put it in leymans terms.
c^2=a^2+b^2. To find the exact number for c^2, you'll need to find the SQUARE ROOT of that number.
Heres an example:
5^2+5^2 is equivalent to C^2
=25+25 is c^2
50=c^2
square root of 50=c
(1 vote)
• Can I have a little help this doesn't make sense. | 617 | 2,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-22 | latest | en | 0.936696 |
https://www.cfd-online.com/Forums/main/108323-turbulence-model-parameters-equations.html | 1,502,985,275,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103579.21/warc/CC-MAIN-20170817151157-20170817171157-00234.warc.gz | 862,429,920 | 17,291 | # Turbulence model parameters and equations
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October 19, 2012, 10:02 Turbulence model parameters and equations #1 New Member Maarten Join Date: Sep 2012 Posts: 3 Rep Power: 6 Sponsored Links Dear CFD users, Right now I´m working on a CFD computation for external flow around a very streamlined car. I have some questions regarding the turbulence parameters. I´ll first specify what formulas I used to calculate some parameters. FYI, I use FineOpen of Numeca. I will either use the SA or Mentor-SST k –omega (Extended wall) model. Description of the stream: A free steam around a car driving in the open i.e. on the road. So it is an external flow. The free stream velocity is 25 m/s. There are a couple parameters I have to specify in the CFD software program which im not complete sure about, being: Boundary conditions: for the k-omega model: k and for the Spalart-Allmaras model: turbulent viscosity Initial conditions: for the k-omega model: k and for the Spalart-Allmaras model: ratio (taken as 1) So it comes down to calculating/guestimating these parameters. I used the following formulas (all in SI units) to calculate these parameters: (From Numeca FineOpen manual) (Also from FineOpen manual) with being the dynamic (molecular) viscosity with and I took (Numeca specifies in the manual that a typical value for this ratio is 1 for external flows). So before I can calculate , I first calculate k. Turbulent (eddy) viscosity (So before calculating this parameter, I first calculate k, then and then ) Using the above mentioned formulas, I was able to calculate all the needed parameters. Still I’m not sure if these formulas are correct or maybe are not valid in every case. However, when I just assume these formulas are correct I calculated k, and . This yielded k = 0.718, and . Do these values make sense in anyway, or are they completely out of range? The actual reason I’m posting this question is to validate our computation. One year ago, a group performed the same computation for the exact same car. This yielded a drag force of about 40N. The problem we are having now is that the drag force is in the order of 10^4 Newton. Obviously this does not make sense. Right now the mesh I runned in FineOpen contains about 18 million cells with refinements near the car itself, behind the car, beneath the car etc. etc. So what I’m specifically asking: 1) Can someone verify the method I use, especially the formulas I use 2) What may cause the problem of a way too high drag value? 3) Do the values of k and epsilon make sense? I know these are a lot of questions, but any help is much appreciated. Thank you in advance, Maximus
October 24, 2012, 13:20 #2 New Member Maarten Join Date: Sep 2012 Posts: 3 Rep Power: 6 Anyone any suggestions?
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# Convert Text to Sentence Case using Excel Formulas [Quick Tips]
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It is almost 3:30 am now, I stayed awake for last 23 hours so that Excel School 2 can be ready for rolling. But that is no excuse for not having a post here. So here it goes.
Excel has formulas for converting a bunch of text to UPPER, lower and Proper Cases. But not a formula to convert o Sentence case. So, today we will learn a simple formula tip to convert text to sentence case (ie, First letter capital).
Assuming your text is in C2, the formula is,
``` =UPPER(LEFT(C2,1))&MID(LOWER(C2),2,999)```
How the formula works?
It is converting first letter of C2 to upper case [`UPPER(LEFT(C2,1))`] and lower case of rest of C2 [`MID(LOWER(C2),2,999)`]
The 999 portion is the secret code to unlock ninja features of MID formula.
Well, I am kidding of course, the 999 is just a cop-out to say give me rest of the C2, I don’t know the length of it. (of course, 999 would fail if your text actually has more than 1000 chars. In that case, just use 9999 or whatever large number you fancy.)
Another formula:
`=SUBSTITUTE(LOWER(C2),CHAR(CODE(C2)),UPPER(CHAR(CODE(C2))),1)`
PS: As you can guess, the above formulas assume that C2 has one sentence. If C2 has a whole paragraph, then I would really need a sandwich before thinking about that problem.
How do you convert cases using Excel?
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### Automatically Format Numbers in Thousands, Millions, Billions in Excel [2 Techniques]
Ever wanted to automatically format values in thousands, millions or billions in Excel? In this article, let me show you two powerful techniques to do just that.
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### 27 Responses to “Convert Text to Sentence Case using Excel Formulas [Quick Tips]”
1. Gerald Higgins says:
Interesting post. I can't think of a neat way to get a single formula to produce initial sentence capitals for an indefinite number of sentences contained in a single cell (but I bet SOMEONE can).
As a workaround, if you actually had this situation - i.e. lots of multi-sentence paragraphs, with each paragraph in a cell of its own - then what you could do is use text to columns to isolate each sentence into a cell of its own, and apply your formula to each single-sentence cell.
Then, if you really wanted to, you could string all the sentences back into one single paragraph cell. BUT, once you have split them up, you would probably be better off keeping all the sentences separate.
2. John McEwan says:
You could use LEN(C2)-1 for the length of the text string minus the first character. That's better then trying to fudge it with a large numeral.
3. Andy Wall says:
=UPPER(LEFT(C2,1))&LOWER(RIGHT(C2,LEN(C2)-1)) will get rid of the 999 variable.
4. Andy Wall says:
OK Had a quick think about multiple sentences. Using PROPER and SUBSTITUTE gives an answer (I think. I haven't tested fully)
=SUBSTITUTE(SUBSTITUTE(PROPER(SUBSTITUTE(SUBSTITUTE(LOWER(C12),". ","9z9")," ","zxz")),"9Z9",". "),"zxz"," ")
PROPER capitalizes first letter in a string and any letters that follow any character other than letter. So we use substitute to build the logic (using strings that can be replaced ie not common combinations.
Sorry it's a bit garbled. Give it a try. Write multiple sentences in C12
5. Gregor Erbach says:
In principle, this is difficult to do for *all* sentences because Excel has no recursion or regular expressions. You could use VBA, or export the sentences to a text file and process them with your favourive regular expression language.
If you really need to do this in Excel (and assuming that all sentences end with a period followed by a blank), you find the first period followed by a blank, substitute that by a placeholder (I used "#"), and uppercase the letter after that.
The following example assumes that the original text is in C2.
=SUBSTITUTE(C2,". "&MID(C2,SEARCH(". ",C2)+2,1),"#"&UPPER(MID(C2,SEARCH(". ",C2)+2,1)))
Then you iterate by copying the formula to the right (or downwards, as appropriate) until no more substitutions can be done.
Then you replace all "#" in the final result by ". "
=SUBSTITUTE(C5;"#";". ")
6. Chandoo says:
@Andy.. Brilliant stuff.. Took me some time to figure it out. (also, I didnt know PROPER("a1b2c3") will be A1B2C3. ) Thank you so much for sharing it here.
@Gregor.. I would imagine pasting the whole thing in to MS Word, sentence casing it and then pasting back to Excel, If I ever need to do that. It is much faster and easier. Also, check Andy's formula above to understand a simpler (albeit geekier) version.
@John, Andy: Technically, 999 should be as fast or even slightly faster than len() approach. Although it is slightly difficult to explain.
7. Rick Rothstein (MVP - Excel) says:
@Andy,
Just to point out, your formula will only work for sentences that end with a period... it won't work correctly for sentences ending with a question mark or exclamation point, nor will it work correctly for titled names such as Dr. Who or Mrs. Jones.
8. Bruce Spong says:
May be dangerous to use because of other words that need to maintain their uppercasedness (is that a word?) But it's a good solution for small quantities of data that would be easy to proof. Cheers!
9. Andy Wall says:
@Rick Yes it would need a period and wouldn't work with other endings, nor would it work with names or possibly even product code id's or Klingon. As mentioned previously excel is probably not the best tool for the job. Just proves the point that a lot of things can be done in excel just maybe not optimally.
@Chandoo thanks for the props. It's an interesting case of exploring native excel functionality to push the envelope.
10. Rick Rothstein (MVP - Excel) says:
@Andy - I was pretty sure you already knew the limitation of the formula; I posted my comments for completeness sake, that is, more for the reader's of the blog than for you. As for Excel (formulas) not being the best tool for the job... I'm not so sure VB is either. This is a very hard problem to code for because of all the exceptions that would have to be accounted for. Taking care of the punctuation at the end of sentences would not be too hard to code for, nor would name introducers (like Dr., Mr., etc.) as long as they are all identified beforehand so they could be coded for; but you raise an excellent point about product code IDs, and I would add company names to that as well (think IBM for example). As for Klingon, I don't speak it, so I don't know for sure if that could be coded for or not. By the way, I do agree with Chandoo... your substitution approach is clever, indeed; but, unfortunately, the underlying problems with the task are just too much for any formula approach I'm afraid.
11. Andy Wall says:
@ Rick - Just reread my reply and I didn't mean to sound facetious. My point was as you say, it'd be difficult to cope with all exceptions. Even wordprocessors fail at this at some point. I don't speak Klingon either, just a geek dialect of Excel! I find these type of problems, while not fully solvable, interesting for practice and discovery, and I love the fact that there is always more than one way to solve a problem in Excel. Maybe Chandoo could have a regular Excel mini challenge on the blog?
12. Chandoo says:
@Andy.. I like the idea of challenges. I have seen that community is vibrant on posts where I pose a challenge or leave some homework. I will try to a challenge once every 2-3 weeks.
13. John says:
Excellent. This will save me lots of time.
14. JP says:
I have an article about case changing in Excel:
http://www.codeforexcelandoutlook.com/excel-vba/case-changing-in-excel/
15. Chandoo says:
@JP.. very interesting link. Thank you for sharing it with us.
16. [...] June 17, 2010 at 5:03 PM | Posted in General | Leave a Comment Tags: formulas, letter-case Chandoo brings up an interesting problem: How do you convert and fix the letter case in a sentence, using [...]
17. BG2 says:
Hello Chandoo, may I know if there is a formula to remove a comibination of number and text in one cell : MY-9876543. For example removing "MY-" in one cell.
18. Hui... says:
@BG2 Try =Right(YourCell,Len(YourCell)-Find("-",YourCell))
19. BG2 says:
whoa... thanks Hui. This is really helpful. 😀
You make my day!!!
20. [...] adding realtime scores updated from the web by visiting Realtime World Cup Scores.Sentence CaseIn Convert to Sentence Case, Chandoo provides a formula for converting to sentence case (that is, first word capitalizes, the [...]
21. [...] adding realtime scores updated from the web by visiting Realtime World Cup Scores.Sentence CaseIn Convert to Sentence Case, Chandoo provides a formula for converting to sentence case (that is, first word capitalizes, the [...]
22. Roz Lobo says:
Thank you so much for the tip. It has saved me HOURS of work, and when you think about it, it is so logical!! thanks again.
23. Nathalie says:
This rocks!! HUGE time saver. thank you!
24. Faurkh says:
If i want to change whole column text in upper case at a time then how should i do this? e.g  In  Name column  farukh,samreen like this one by one many names then i want that name convert it into FARUKH,SAMREEN etc.
25. faiz says:
how can convert lower/uppercase to other sentence or number. example
a = up
A = down.
so how can I convert lower case a to up & upper A to down
26. nasreen says:
Super... Thankyou for saving my time & lovely tip.. 🙂
27. Vitor Canova says:
It's almost 2018 and this is still the solution. No new formula in Excel to handle this.
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One email per week with Excel and Power BI goodness. Join 100,000+ others and get it free. | 2,647 | 10,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-30 | latest | en | 0.908258 |
http://moi3d.com/forum/messages.php?webtag=MOI&msg=8281.1 | 1,531,984,761,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590711.36/warc/CC-MAIN-20180719070814-20180719090814-00162.warc.gz | 253,872,135 | 6,785 | Loxodrome script for Sphere
From: bemfarmer 26 Jan 2017 (1 of 7)
Notes on Loxodrome script for a sphere The loxodrome curve, on the surface of the earth, is known as a rhumb line, and such curves are still relevant to navigation by ship or airplane. Note that Karsten has spoken of doing a loxodrome Node. This spherical _Loxodrome script forms a curve on the surface of a sphere, by means of parametric stereographic mapping, as defined by Alfred Gray, "Modern Differential Geometry of Curves and Surfaces with Mathematica," Third Edition. Chapter 8, Construction of Space Curves, AG8.pdf, page 250-251, q.v.. The general parametric stereographic map function is defined from the plane, Rsquared, onto the sphere, Ssquared. (Would a 2D slider be useful?) (The mapping may be useable for other planar curves?) "The spherical loxodrome is the image under a stereographic map of a logarithmic spiral." The parametric planar logarithmic spiral, as per the LogSpiral4 MoI script, is "a * pow(b*t)", where "a" and "b" are parameters. The loxodrome parametric equation is on page 251. Note that there are several other equations for loxodromes. This may be confusing, and the equations could use some reconciliation. "Loxodromes exist for all surfaces of revolution." For the same handedness (CW or CCW), run the script twice, unchecking Top to yield a Bottom curve, so as to get a pair of curves, a "north" curve and a "south" curve, which can be joined. Parameter "a" is called "Shift" in this script, and changes the position of the curve(s) with respect to the poles, and causes shrinkage or expansion of the lengths of the curve(s). The default value of "a = 1.0" causes top and bottom curves to be of equal lenth, and terminate at the equator. For other values of "a", top and bottom curves of the same handedness (CW or CCW), will be of different lengths, and joinable either north or south of the equator. Parameter "b" is called "growth" in this script, and alters the rhumb line compass angle. Parameter "Whorls" increases the number of turns, inwards toward the north, or south, poles, which are asymptotic points. For artistic creations, such as trimming a sphere, connecting circular arrays of loxodromes near the poles may be needed, such as with segments of a circle, or blends. For a large number of Whorls, the Point Count should be increased, to maintain proper curvature. UnChecking the "Clockwise" checkbox results in a "left handed" spiral. UnChecking the "Top" checkbox results in a curve in the Southern hemisphere. (Scripting to get CW/CCW and Top/Bottom to work together, so that the top and bottom curves would be the same handedness, was done mostly by trial and error. ) - Brian EDITED: 26 Jan 2017 by BEMFARMER Attachments: Image Attachments:
From: bemfarmer 28 Jan 2017 (2 of 7)
8281.2 In reply to 8281.1 For matching parameter values, where Loxodrome01 "Shift" corresponds to LogSpiral4 "Scale," the Loxodrome spiral is the inverse stereographic projection of the LogSpiral4 curve. For Scale > 1, the planar log spiral expands, and the upper loxodrome shrinks toward the north pole. Not sure how the bottom curve relates to the planar log spiral. Loxodrome Sweep of a circle: - Brian EDITED: 28 Jan 2017 by BEMFARMER Attachments:
From: Michael Gibson 28 Jan 2017 (3 of 7)
8281.3 In reply to 8281.2 Reminds me of a bighorn sheep! - Michael
From: gunter511 14 Mar 2017 (4 of 7)
Guys, I installed and tried the Loxodrom script and managed to get the top port of the spiral to connect to the vertical line but can't get the bottom part to do the same. I've played around with the parameters like whorls, etc. but nothing moves the bottom part. Any suggestions? Thanks again! Gunter Attachments: | 976 | 3,731 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2018-30 | longest | en | 0.892685 |
https://www.landlordstudio.com/blog/cash-on-cash-return/ | 1,670,286,836,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711064.71/warc/CC-MAIN-20221205232822-20221206022822-00519.warc.gz | 878,671,864 | 29,291 | Get started
# Why Successful Real Estate Investors Focus on Cash-on-Cash Return
## What is Cash-on-Cash Return?
The cash-on-cash return rate, also known as cash yield, measures the amount of cash flow relative to the amount of cash invested in a property investment and is calculated on a pre-tax basis. This is often a favored calculation for real estate investors as it is a reasonably accurate estimate of cash flow.
The cash-on-cash return metric measures only the return for the current period, typically one year, rather than for the life of the investment or project.
Having a good cash-on-cash return rate determines how profitable a property is. This is why it’s deemed one of the more important ROI calculations by real estate investors. It can also be used as a forecasting tool to set a target for projected earnings and expenses.
## How Do You Calculate Cash-on-Cash Return?
To calculate the cash-on-cash return, you need to know your annual cash flow before taxes and the total amount invested. To work out the annual cash flow of the property you need to calculate the total amount of expected income in one year including any additional income from the investment property, and then subtract the operating expenses, annual mortgage payments, and account for any vacancy periods.
The formula for calculating cash-on-cash return looks like this:
Cash-on-Cash Return = Annual Cash Flow (before tax) / Total Cash Invested
When Annual Cash Flow = Income – Operating Expenses – Vacancy Period – Mortgage Repayments
## What’s a Good Cash-on-Cash Return?
Cash-on-cash return is a measurement used by real estate investors to determine a property’s performance. It is a calculation often used for long-term investments as it focuses on cashflow, signifying whether an investment will generate adequate funds for repaying debts.
Although there is no rule of thumb, investors seem to agree that a good cash-on-cash return is between 8 to 12 percent.
### Cash-on-cash return example:
• The property purchase price was \$250,000
• The investor has paid a total of \$80,000 to date
• Monthly rent is \$3,000 meaning an annual income of \$36,000
• 5% vacancy rate means rent loss of \$1,800
• Operating expenses (including costs such as maintenance, HOA fees legal, and management fees) equal an annual total of \$4,200
• Mortgage repayments are \$7,200 annually.
Cash-on-cash return = 80,000 / (36,000 -1,800 – 4,200 – 7,200) 22800
22800 / 80000 = 28.5%
In addition to deriving the current return, the cash-on-cash return can also be used to forecast the expected future cash distributions of an investment. However, it is not a promised return but is instead a target used to assess a potential investment. In this way, the cash-on-cash return is an estimate of what an investor may receive over the life of the investment.
Ben Luxon
Ben is the editor and lead writer for Landlord Studio. He has worked with real estate professionals all over the world and written educational articles on tech, real estate, and financial growth for sites such as Forbes, TechBullion, and Business Magazine.
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× | 705 | 3,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.929649 |
https://questions.llc/questions/1632630/use-the-information-below-to-answer-this-question-c-s-o2-g-co2-g-h-394-kj-mol | 1,660,621,083,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572220.19/warc/CC-MAIN-20220816030218-20220816060218-00006.warc.gz | 434,347,902 | 5,320 | # Use the information below to answer this question
C(s) + O2(g) -> CO2(g) ΔH = -394 kJ/mol
H2(g) + 1/2 O2(g) -> H2O(l) ΔH = -286 kJ/mol
4C(s) + 5H2(g) -> C4H10 (g) ΔH = -126 kJ/mol
The standard enthalpy of combustion of butane, in kJ/mol, is:
A -2880
B -2590
C -806
D -554
Could someone show me how to do this question? Thank you!
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1. eqn 3 reversed + 4*eqn 1 + 5*eqn 2
Post your work if you get stuck.
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2. So the answer is A, right? Thanks for the help!
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3. I didn't do the math.
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4. Yes, A is correct.
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5. Hey! That's pretty goooood!
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4. 🚩 | 349 | 673 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.671691 |
https://forum.freecodecamp.org/t/question-about-lookahead/219687/3 | 1,603,562,472,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107884322.44/warc/CC-MAIN-20201024164841-20201024194841-00486.warc.gz | 339,854,825 | 7,474 | In this lesson of regular expression I am asked to match passwords that are greater than 5 characters long and have two consecutive digits. I solved the challenge as followed:
``````let sampleWord = "astronaut";
let pwRegex = /(?=\w{5})(?=\D*\d{2})/;
let result = pwRegex.test(sampleWord);
``````
My question is, why do we need this piece of code `\D*`? Why do I need to check for non-digit characters before doing this `\d{2}`? The challenge is just asking me to check that the password has at least 2 digits.
you didn’t post a link to the challenge, so I have to use my (bad) memory for this:
I believe the challenge said the digits have to be at the end ? So not at the beginning…
I believe I did, but just in case this is the challenge https://learn.freecodecamp.org/javascript-algorithms-and-data-structures/regular-expressions/positive-and-negative-lookahead/
The challenge doesn’t actually say ‘at least’. It just says has 2 consecutive digits. Therefore, it should not have 3
or more consecutive digits, and it should not have 1 digit or less either.
so that means the other characters found in the lookahead have to be non-digits (\D)
2 Likes
Well, even though the challenge doesn’t say “at least 2” digits, it passes the test even with 3 consecutive digits (and more probably), so I thought the requirement was minimum 2 digits.
However your answer clarified something for me, basically I was looking at the whole picture, meaning this `(?=...)(?=....)` was just one lookahead in my mind, but instead we have 2 and each one is checking for something different.
Thanks.
If there are 3 consecutive digits, then there are definitely 2 consecutive digits (a subset of the 3 consecutive digits).
Exactly! When using two lookaheads in this manner, you can think about them as checking independently.
1 Like
yes I see what you mean. I think they must have meant 2 consecutive digits even if they occur within a grouping of a greater number. | 456 | 1,958 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2020-45 | latest | en | 0.947752 |
https://plainmath.net/precalculus/82239-prove-sin-x | 1,675,396,908,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500042.8/warc/CC-MAIN-20230203024018-20230203054018-00627.warc.gz | 468,970,239 | 23,430 | Lucia Grimes
2022-07-14
Prove: $\mathrm{sin}2\theta \phantom{\rule{thickmathspace}{0ex}}\ge \phantom{\rule{thickmathspace}{0ex}}\frac{1}{2}\left(-1-3{\mathrm{cos}}^{2}\theta \right)$
furniranizq
Expert
Hint: Since $1={\mathrm{sin}}^{2}\left(t\right)+{\mathrm{cos}}^{2}\left(t\right)$ and $2\mathrm{sin}\left(2t\right)=4\mathrm{sin}\left(t\right)\mathrm{cos}\left(t\right)$
$2\mathrm{sin}\left(2t\right)+1+3{\mathrm{cos}}^{2}\left(t\right)={\left(\mathrm{sin}\left(t\right)+2\mathrm{cos}\left(t\right)\right)}^{2}.$
Do you have a similar question? | 223 | 552 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 32, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.310122 |
https://www.econology.info/power-heat-pci-pcs-fuel-gas/ | 1,586,337,371,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371810807.81/warc/CC-MAIN-20200408072713-20200408103213-00151.warc.gz | 889,859,020 | 25,856 | # Calorific value (PCI / PCS) of gases, fuels and combustibles
## Calorific value of the main gases and liquid or solid fuels: PCI and PCS.
### gross calorific value: PCS
Definition: the calorific value greater than constant volume of a fuel represents the quantity of heat released by the combustion of the unit of mass of the fuel:
• in oxygen saturated with water vapor,
• the reacting products and the products formed being at the same temperature,
• in the same enclosure,
• the water formed being liquid.
This test is the subject of standard NF M 07-030, and uses a container called a calorimetric bomb.
The definition of calorific value at constant volume does not correspond to industrial combustion taking place at constant pressure, in an open forum, but in fact the difference is small and generally neglected.
The PCS is “useful” when condensing the combustion water (condensing boiler for example).
### lower heating value: PCI
Most of the time the smoke leaves the exchange surfaces at a temperature above the dew point. Water is therefore emitted in the form of vapor.
The PCI is calculated by deducting, by convention, from the PCS, the heat of condensation (2511 kJ / kg) from the water formed during combustion and possibly from the water contained in the fuel.
If the fuel does not produce water, then the PCS = PCI.
### calorific mixtures
The precise analysis of a gas or a liquid makes it possible to calculate its calorific value starting from the calorific power of its constituents using the relation:
#### PCm = sum (xi * PCi)
with:
• PCm: calorific value of the mixture
• PCi: calorific value of the constituents
• xi: mass fraction of each constituent
### Calorific units
Calorific value is expressed in:
• kcal / kg
• kJ / kg
• kWh / kg (= 861 kcal / kg)
• TEP / tonne (= 10000 thermies / tonne) (TEP: Tonne Petrol Equivalent)
The other units (head unit) are as follows:
• milithermy / kg (= kcal / kg) (desused)
• therm / tonne (= kcal / kg) (desused)
• BTU / pound (= 0,5554 kcal / kg) (Anglo-Saxon)
For gas: not expressed per kg but Nm3.
### calorific gas
PCI / PCS kcal / Nm3
• Hydrogen: 2570 / 3050
• Carbon monoxide: 3025/3025 (PCI = PCS as there is no water formation)
• hydrogen sulfide: 5760 / 6200
• Methane: 8575 / 9535
• Ethane: 15400/16865
• Propane: 22380 / 24360
• Butane: 29585 / 32075
• Ethylene: 14210/15155
• Propylene: 20960 / 22400
• Acetylene: 13505 / 13975
### calorific value of liquid fuels. PCI / PCS in kcal / kg
• Hexane 10780 / 11630
• Octane: 10705 / 11535
• Benzene: 9700 / 10105
• Styrene: 9780 / 10190
• Heavy oil: 9550
• Heating oil: 10030 (= 11.7 kWh / kg or 9.9 kWh / L density 0.85)
### calorific value of commercial fuels
Gaseous fuels:
• poor natural gas: 9.2 kWh / Nm3
• rich natural gas: 10.1 kWh / Nm3
• Butane: 12.7 kWh / kg or kWh 30.5 / 3 Nm7.4 or kWh / L (liquid) to 15 ° C
• Propane: 12.8 kWh / kg or kWh 23.7 / 3 Nm6.6 or kWh / L (liquid) to 15 ° C
(Boiling point: butane 0 ° C, propane -42 ° C)
Liquid fuels:
• heating oil: 9.9 kWh / L
• Light Oil: 10.1 kWh / L
• 10.5 average fuel kWh / L
• Heavy oil: 10.6 kWh / L
• Oil extra heavy: 10.7 kWh / L
Solid fuels:
• Coal: 8.1 kWh / kg
• Coke: 7.9 kWh / kg
• Anthracite 10 / 20: 8.7 kWh / kg
calorific various in kCal / kg
• Wood (30% humidity): 2800
• Dry wood: 4350 or 5 kWh / kg
2 kg of dry wood therefore produces approximately the equivalent of 1 L of petroleum
### CO2 emissions
The problem posed by the emission of greenhouse gases by industry makes it necessary to consider fuels from the angle of the CO2 emission generated by their use.
The table below shows the CO2 emission of different fuels per unit of calorific value.
Unit: Tons CO2 PET on the PCI
• Hydrogen: 0
• Natural gas: 2,37
• GPL: 2,67
• Heavy oil: 3,24
• Heating oil: 3,12
• Dry wood: 3,78
### Conversion factors in PET
Coefficients of conversion to TEP (Tonne Oil Equivalent) of various energy sources:
• Coal-Briquettes: 1T = 0,619 toe
• Lignite coal poor: 1T = 0,405 toe
• Coke: 1T = 0,667
• Petroleum coke: 1T = 0,762 toe
• Butane Propane: 1T = 1,095
• Heavy fuel oil (FOL): 1T = 0,952
• heating oil (FOD): 1T 1200L = = 1 toe
• Electricity: 1000kwh = 0,222
• Species: 1T 1320L = = 1 toe
• Super Fuel: 1T 1275L = = 1 toe
• Diesel: 1T 1200L = = 1 toe
## 1 comment on “Calorific value (PCI / PCS) of gases, fuels and combustibles”
1. LOUIS BERTHIER said:
hello .. my annual consumption is 10000KW / AN average over 6 years of super fuel BP is 700 970 euro year following the price of fuel for a home 110 M2. hoping with the research engineers of the energy valley to have a BIO oil so close to the energy gas rejection.Good reading.tchao. | 1,457 | 4,686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-16 | latest | en | 0.860007 |
https://homework.cpm.org/cpm-homework/homework/category/CC/textbook/CCG/chapter/Ch5/lesson/5.1.3/problem/5-35 | 1,585,982,838,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370520039.50/warc/CC-MAIN-20200404042338-20200404072338-00019.warc.gz | 504,475,447 | 14,991 | ### Home > CCG > Chapter Ch5 > Lesson 5.1.3 > Problem5-35
5-35.
While playing a board game, Mimi noticed that she could roll the dice $8$ times in $30$ seconds. How many minutes should it take her to roll the dice $150$ times? Homework Help ✎
Write a proportion. Ratios should compare the $\frac{\text{(number of rolls)}}{\text{(number of seconds)}}$.
Time is $562.5$ seconds. To convert to minutes divide by $60$ seconds because there are $60$ seconds in a minute.
So the final answer is $9.375$ minutes | 143 | 508 | {"found_math": true, "script_math_tex": 8, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2020-16 | latest | en | 0.890587 |
https://help.scilab.org/docs/6.1.1/en_US/assert_cond2reltol.html | 1,656,631,458,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103915196.47/warc/CC-MAIN-20220630213820-20220701003820-00752.warc.gz | 338,339,395 | 8,115 | Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange
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Scilab Help >> Testing & benchmarking > Assert > assert_cond2reltol
# assert_cond2reltol
Suggests a relative error, computed from the condition number.
### Syntax
rtol = assert_cond2reltol ( condition )
rtol = assert_cond2reltol ( condition , offset )
### Parameters
condition :
a matrix of doubles, the condition number. The condition number must be strictly positive.
offset :
a matrix of doubles, a shift for the number of required decimal digits (default offset=0). For example, offset=1 increases the accuracy requirement (decreases the relative tolerance by a factor 10^-1), offset=-1 decreases the accuracy requirement (increases the relative tolerance by a factor 10^1).
rtol :
a matrix of doubles, the relative tolerance. The relative tolerance is strictly positive, lower than 1.
### Description
Depending on the condition number, returns the corresponding relative tolerance.
Any optional parameter equal to the empty matrix [] is set to its default value.
We emphasize that this relative tolerance is only a suggestion. Indeed, there may be correct reasons of using a lower or a higher relative tolerance.
• Consider the case where an excellent algorithm is able to make accurate computations, even for an ill-conditioned problem. In this case, we may require more accuracy (positive offset).
• Consider the case where there is a trade-off between performance and accuracy, where performance wins. In this case, we may require less accuracy (negative offset).
Any scalar input argument is expanded to a matrix of doubles of the same size as the other input arguments.
We compute the number of required digits d, then the relative tolerance is 10^-d.
### Examples
assert_cond2reltol ( 0 ) // 1.110D-16
assert_cond2reltol ( 1 ) // 1.110D-16
assert_cond2reltol ( 1.e1 ) // 1.110D-15
assert_cond2reltol ( 1.e2 ) // 1.110D-14
assert_cond2reltol ( 1.e3 ) // 1.110D-13
assert_cond2reltol ( 1.e16 ) // 1
assert_cond2reltol ( 1.e17 ) // 1
assert_cond2reltol ( 1.e18 ) // 1
// Matrix input.
condition = [0,1,10,100,1000,1.D13,1.D14,1.D15,1.D16,1.D17,1.D18];
expected = [1.110D-16 1.110D-16 1.110D-15 1.110D-14 1.110D-13 0.0011102 0.0111022 0.1110223 1. 1. 1.];
assert_cond2reltol ( condition )
// Using offset
// Negative offset : require less accuracy
assert_cond2reltol ( 1.e2 , [0 -1] ) // [1.1D-14 1.1D-13]
// Positive offset : requires more accuracy
// See that the impact of offset is constrained.
assert_cond2reltol ( 1.e2 , [0 1 2 3] ) // [1.1D-14 1.1D-15 1.1D-16 1.1D-16]
// Negative offset
// See that the impact of offset is constrained.
assert_cond2reltol ( 1.e14 , [0 -1 -2 -3] ) // [1.1D-02 1.1D-01 1 1]
// Plot the relative tolerance depending on the condition
condition = logspace(0,18,1000);
r = assert_cond2reltol ( condition );
plot(condition,r)
h=gcf();
h.children.log_flags="lln";
h.children.children.children.thickness=4;
xt = h.children.x_ticks;
xt.locations = 10^(0:2:18)';
xt.labels = ["10^0";"10^2";"10^4";"10^6";"10^8";"10^10";"10^12";"10^14";"10^16";"10^18"];
h.children.x_ticks=xt;
yt = h.children.y_ticks;
yt.locations = 10^-(0:2:18)';
yt.labels = ["10^0";"10^-2";"10^-4";"10^-6";"10^-8";"10^-10";"10^-12";"10^-14";"10^-16";"10^-18"];
h.children.y_ticks=yt;
xtitle("Relative tolerance","Condition","Relative tolerance");
r = assert_cond2reltol ( condition , +3 );
plot(condition,r,"r")
r = assert_cond2reltol ( condition , -3 );
plot(condition,r,"g")
legend(["Offset=0","Offset=+3","Offset=-3"]);
### History
Version Description 5.4.0 Function introduced | 1,104 | 3,700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-27 | longest | en | 0.746811 |
https://fokusfisika.com/earth-is-approximately-a-sphere-of-radius/ | 1,713,198,058,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817002.2/warc/CC-MAIN-20240415142720-20240415172720-00276.warc.gz | 243,838,402 | 14,971 | # Earth is approximately a sphere of radius
Earth is approximately a sphere of radius 6.37 x 106 m.
What are
(a) its circumference in kilometers,
(b) its surface area in square kilometers,and
(c) its volume in cubic kilometers?
Solutions
(a) Its circumference in kilometers :
\begin{aligned}
C &= 2\pi r \\
&= 2\cdot \pi \cdot 6.37 \times 10^6 \\
&= 4.00 \times 10^7 \quad \textrm{m} \\
&= 4.00 \times 10^4 \quad \textrm{km}
\end{aligned}
(b) its surface area in square kilometers :
\begin{aligned}
A &= 4\pi r^2 \\
&= 4\cdot \pi \cdot (6.37 \times 10^6)^2 \\
&= 5.10 \times 10^{16} \quad \textrm{m}^2 \\
&= 5.10 \times 10^{10} \quad \textrm{km}^2
\end{aligned}
(c) its volume in cubic kilometers
\begin{aligned}
V &= \frac{4}{3}\pi r^3 \\
&= \frac{4}{3}\cdot \pi \cdot (6.37 \times 10^6)^3 \\
&= 1.08 \times 10^{21} \quad \textrm{m}^3 \\
&= 5.10 \times 10^{12} \quad \textrm{km}^3
\end{aligned} | 357 | 901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.620351 |
https://lists.freepascal.org/pipermail/fpc-devel/2014-March/033583.html | 1,725,743,922,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650920.0/warc/CC-MAIN-20240907193650-20240907223650-00820.warc.gz | 343,207,159 | 2,626 | # [fpc-devel] Extended(\$FFFFFFFFFFFFFFFF) = -1?
Ewald ewald at yellowcouch.org
Sun Mar 2 14:30:13 CET 2014
```On 01 Mar 2014, at 18:17, Jonas Maebe wrote:
>
> On 01 Mar 2014, at 01:19, Ewald wrote:
>>
>> That is perfectly true. But shouldn't the most basic behaviour of a language be at the very least intuitive?
>
> It should be well-defined and consistent. One of Pascal's basic principles is that the native signed integer type is used for pretty much everything. While we (and Borland/CodeGear/Embarcadero) have extended the language beyond that (e.g. we support int64 on 32 bit platforms, and there's also the qword type), this principle is still followed as much as possible. In this case: constants are always preferably interpreted as signed numbers.
If hexadecimal is actually used to represent bit patterns (as Hans-Peter Diettrich wrote), then the decision to use a signed type here seems to violate this (represent bitpatterns) principle, since the highest bit in a signed number has a different meaning than the other bits, where in a bitpattern all bits have equal meaning.
It seems like sticking to one principle (signed integer as much as possible) actually breaks another principle (bitpattern).
>
> You do care about the signedness, because the only way to represent int64(-1) in hexadecimal is as \$ffffffffffffffff.
And what about -\$1? Or is that too far fetched?
> Numbers in two's complement do no consist of a single sign bit followed by a magnitude. Those top 63 '1' bits together form the "-" sign in this number.
Yes, but this can all be solved by parsing the string and storing it with one extra MSBit (if there is a `-` in front of the constant it must be negative, otherwise it should be positive). This highest bit then reflects the sign. `-1` would then be \$1 FFFF FFFF FFFF FFFF, whereas \$FFFF FFFF FFFF FFFF would be \$0 FFFF FFFF FFFF FFFF. It really is quite easy to store it like that and `fix` things [picking a fitting datatype] afterwards. Anyway, then you have got backwards compatibility to take care of, since there will be someone out there who's code actually depends on this behaviour.
>
> As mentioned before, for the compiler \$ffffffffffffffff fits perfectly into an int64 because it parses it using val(str,int64var,value).
That is circular reasoning.
--
Ewald
``` | 554 | 2,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.942235 |
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# A dog is tied to a tree by a long nylon cord. If the
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A dog is tied to a tree by a long nylon cord. If the [#permalink]
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25 Mar 2014, 07:47
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A dog is tied to a tree by a long nylon cord. If the dog runs from the due North side of the tree to the due South side of the tree with the cord extended to its full length at all items, and the dog ran approximately 30 feet, what was the approximate length of the nylon cord, in feet?
A) 30
B) 25
C) 15
D) 10
E) 5
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Re: A dog is tied to a tree by a long nylon cord. If the [#permalink]
### Show Tags
25 Mar 2014, 07:54
2
Because the cord was extended to its full length at all items, the dog ran along a semi-circular path, from north to south.
The circumference of a full circle is 2*pi*r, but since we only care about the length of half the circle, the semi-circle path is pi*r.
pi*r = 30. Round pi = 3, then r = 10.
Chord is about 10 feet long.
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Re: A dog is tied to a tree by a long nylon cord. If the [#permalink]
### Show Tags
15 May 2017, 23:53
the dog runs from the due North side of the tree to the due South side of the tree with the cord extended to its full length at all times. This means that dog ran in semi - circle or 180 degree sector is create.
We know that length of arc = ($$\frac{central angle}{360}$$ ) * 2* PI * r
therefore, = $$\frac{180}{360}$$* 2* PI * r = 30
=> PI * r = 30
3.142 * r = 30
r= 10 (approx)
Re: A dog is tied to a tree by a long nylon cord. If the &nbs [#permalink] 15 May 2017, 23:53
Display posts from previous: Sort by | 967 | 3,345 | {"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} | 3.96875 | 4 | CC-MAIN-2019-04 | latest | en | 0.914976 |
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Revision en2, by In_The_Name_Of_Love, 2019-06-18 10:18:28
You're given String which consists of open and close bracket, numbers and question mark (?). Also you're given two integer M and N. M is number of plus (+) and N is number of minus (-). You'll have to replace question mark (?) with plus (+) or minus (-) and maximize the expression. You'll have to use exactly M plus (+) and N (-). It is guaranteed that M + N = Number of question mark(?).
1 <= Length of String <= 10000
Example : String : ((1?(5?7))?((6?2)?7)) M = 3, N = 2
if we put 3 plus (+) and 2 minus (-) then, ((1-(5-7))+((6+2)+7)) Answer : 18
How to tackle this Problem?
Thank you!
#### History
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en2 In_The_Name_Of_Love 2019-06-18 10:18:28 5 Tiny change: ' plus (+) and minus (-)' -> ' plus (+) or minus (-)'
en1 In_The_Name_Of_Love 2019-06-18 10:16:05 629 Initial revision (published) | 293 | 915 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.780852 |
https://3roam.com/10-dbi-antenna-range-calculator/ | 1,721,390,146,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514902.63/warc/CC-MAIN-20240719105029-20240719135029-00795.warc.gz | 66,837,181 | 54,842 | # 10 dBi Antenna Range Calculator
## Introduction
This tool computes the maximum range or distance (in meters, feet, miles or km) that can be achieved with a 10 dBi gain antenna.
The communication system includes a transmitter, receiver, antennas and cables.
## Calculator
To calculate the max distance, enter the following:
• Frequency of operation (Hz/KHz/MHz/GHz)
• Transmit Power (dBm)
• Receiver Sensitivity (dBm)
• Receive Antenna Gain (dBi) – the default value is 2 dBi
• Total Cable and Other Losses (dB)
## Formula
The calculator uses the Free Space Path Loss (FSPL) equation:
PTx – PRx = 20*Log10(d) + 20*Log10(f) + Lc + 20*Log10(4Ï€/c) – GTx – GRx
The distance d is given by
Log10(d) = (1/20)*(PTx – PRxLc + GTx + GRx20*Log10(f) – 20*Log10(4Ï€/c))
## Example of a Practical 10 dBi Antenna
The picture below shows a 10 dBi Antenna used in Wi-Fi applications
It’s dual band covering both 2.4 GHz and 5.8 GHz. We could not find a data sheet for it, but based on the physical design, it’s an example of a monopole antenna.
## Example Range Calculation
At a frequency of 2.45 GHz (Wi-Fi band) and transmit power of +20 dBm, let’s calculate the range for a 10 dBi transmit and receive antenna
With
• Receiver sensitivity = -90 dBm,
• Cable loss = 0 dB
the maximum range is 30,793 meters or 30.8 km.
Note the difference in range for the same power level between the 10 dBi and 5 dBi which covers a distance of 9.7 km.
The 10 dBi antenna therefore has over 3x the range of the 5 dBi.
Important to remember that this calculation is outdoors and in free space which represents an ideal signal propagation condition. In practice, the range would be lower.
## How to use the Calculator
Below is a list of the terms used in the calculator and what each of them mean.
### Frequency of operation
This is the frequency at which the communication system operates. At lower frequencies, wavelength is commonly used and if that’s the case, then use the wavelength to frequency converter.
### Transmit Power
This is the power at the output connector of the transmitter. It is usually specified in dBm. However many vendors specify this quantity in Watt. Convert from Watt to dBm.
### Transmit Antenna Gain
This depends on the type of antenna used and is expressed in dBi (dB relative to isotropic antenna).
### Total Cable and Other Losses
This includes losses between connectors and antennas. If RF Splitters are used, the loss should be accounted for as well. It is specified in dB. Use this calculator to find the loss due to antenna cable.
The sum of the Transmit antenna power and Gain minus cable and connector losses is also called Effective Isotropic Radiated Power or EIRP for short.
### Receiver Sensitivity
Minimum level of input signal that a radio receiver can detect and demodulate. Use this calculator to find the sensitivity as a function of temperature, SNR and receiver noise figure.
### Receive Antenna Gain
Depends on antenna used and is expressed in dBi. In cases where the signal is from a particular direction, a high gain antenna (8 dBi for instance) can be used. This allows focusing of energy instead of receiving from all directions. Antenna gain can be calculated from the Antenna Factor and frequency of operation.
## Background
A 10 dBi antenna is designed to offer higher gain compared to lower dBi antennas, making it more directional and capable of transmitting or receiving signals over longer distances. The increased gain of a 10 dBi antenna means it focuses the signal more narrowly, which can significantly enhance the signal’s reach and quality in its preferred direction.
## Applications
Here are some common uses for a 10 dBi antenna:
1. Long-Distance Wi-Fi Links: For establishing Wi-Fi connections over longer distances, a 10 dBi antenna can provide the necessary range, making it suitable for connecting networks between buildings or across extensive outdoor areas.
2. Fixed Wireless Broadband: In rural or remote areas where cable or fiber internet is not available, ISPs (Internet Service Providers) may use 10 dBi antennas to deliver internet service to customers’ homes from a central wireless tower.
3. Directional Links: For creating point-to-point (PtP) wireless links, a 10 dBi antenna can focus the signal tightly between two locations, optimizing bandwidth and connection stability over considerable distances.
4. Outdoor Surveillance Cameras: To ensure reliable wireless communication between outdoor surveillance cameras and the monitoring station, especially over large properties or in challenging environments, 10 dBi antennas are used.
5. Amateur Radio and Ham Operators: Ham radio enthusiasts might use 10 dBi antennas for specific VHF/UHF applications to achieve better range and signal clarity when communicating over longer distances.
6. Wireless Bridge Systems: For connecting two or more network segments without the need for physical cables, especially in scenarios where laying cables is impractical due to physical barriers or cost considerations.
7. Remote Sensing and Telemetry: In agriculture, environmental monitoring, or industrial applications, 10 dBi antennas can be used to transmit data over long distances from remote sensors back to a central collection point.
The choice to use a 10 dBi antenna is often driven by the need for longer range and more directed signal propagation. However, the increased directionality means that precise alignment between the transmitting and receiving antennas becomes more critical. Additionally, the environment and regulatory constraints (like limits on transmission power) will also influence the effectiveness and suitability of using a 10 dBi antenna for a particular application. | 1,253 | 5,733 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.844465 |
http://www.convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=football+field&To=wavelength | 1,518,996,848,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891812293.35/warc/CC-MAIN-20180218232618-20180219012618-00438.warc.gz | 386,512,185 | 3,838 | Partner with ConvertIt.com
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Conversion Result: ```football field = 91.44 length (length) ``` Related Measurements: Try converting from "football field" to archin (Russian archin), arpentcan, cloth quarter, earth to moon (mean distance earth to moon), en (typography en), fathom, finger, furlong (surveyors furlong), Greek cubit, Greek palm, Greek span, inch, light yr (light year), nautical mile, pica (typography pica), ri (Japanese ri), soccer field, span (cloth span), UK mile (British mile), vara (Mexican vara), or any combination of units which equate to "length" and represent depth, fl head, height, length, wavelength, or width. Sample Conversions: football field = 128.57 archin (Russian archin), 1.56 arpentlin, 2.5 bolt (of cloth), 360,000 caliber (gun barrel caliber), 4.55 chain (surveyors chain), 400 cloth quarter, 3 engineers chain, 4,114.29 finger, 300 foot, 49.4 Greek fathom, 165.14 Israeli cubit, 91.44 m (meter), 3,600,000 mil, 1,600 nail (cloth nail), .04937365 nautical mile, 3.33 naval shot, 1,200 palm, 308.96 Roman foot, 1 soccer field, 100 yard.
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Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 455 | 1,658 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-09 | longest | en | 0.725651 |
http://docs.cairo-lang.org/hello_cairo/voting.html | 1,709,100,591,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474697.2/warc/CC-MAIN-20240228044414-20240228074414-00746.warc.gz | 9,174,023 | 12,136 | Voting system¶
In this section we will write Cairo code for a small voting system. This voting system can be used, for example, to run a secure (non-private) voting with a lot of voters on a blockchain. We will assume that each voter has a pair of private and public keys (for the ECDSA signature scheme) and that the list of voters’ public keys is fixed. Each voter may vote “Yes” (1) or “No” (0). The system will not guarantee anonymity.
This section assumes some basic knowledge of cryptographic primitives such as hash functions, digital signatures and Merkle trees. In addition, if you haven’t already read the previous parts of the “Hello, Cairo” tutorial, you are encouraged to do so before reading this part.
Generating dummy data¶
First, let’s generate some dummy data for the voting. To generate the key pairs and signature we will write a small Python script using StarkWare’s crypto library:
import json
from starkware.crypto.signature.signature import (
pedersen_hash, private_to_stark_key, sign)
# Set an identifier that will represent what we're voting for.
# This will appear in the user's signature to distinguish
# between different polls.
POLL_ID = 10018
# Generate key pairs.
priv_keys = []
pub_keys = []
for i in range(10):
priv_key = 123456 * i + 654321 # See "Safety note" below.
priv_keys.append(priv_key)
pub_key = private_to_stark_key(priv_key)
pub_keys.append(pub_key)
# Generate 3 votes of voters 3, 5, and 8.
for (voter_id, vote) in [(3, 0), (5, 1), (8, 0)]:
r, s = sign(
msg_hash=pedersen_hash(POLL_ID, vote),
priv_key=priv_keys[voter_id])
'voter_id': voter_id,
'vote': vote,
'r': hex(r),
's': hex(s),
})
# Write the data (public keys and votes) to a JSON file.
input_data = {
'public_keys': list(map(hex, pub_keys)),
}
with open('voting_input.json', 'w') as f:
json.dump(input_data, f, indent=4)
f.write('\n')
We use the Pedersen hash function and the ECDSA signature, which are natively supported in Cairo. For technical details about those cryptographic primitives see STARK Curve and Pedersen Hash Function.
Safety note: In a real system, choosing the private keys must be done using a strong random mechanism. The reason we didn’t use random private keys for our dummy data is to get a deterministic example, which is more convenient for a tutorial.
Here’s what we get:
{
"public_keys": [
"0x1c3eb6d67f833a9dac3766b2f22d31299875884f3fc84ebc70c322e8fb18112",
"0x22285e2a1c84a7b6e283eb1ee28a40ba30874aff62617ba1220d7dc6a2b1e70",
"0x2226376596c83aa0c5381b42b516bd84e604d7ff65647bab46feb7540a6544d",
"0x492cf083fdc9d0c48bcc2807abd2a6da8550b872d047cd36e501a5e12cb581d",
"0x1b16536d44330830c39b119673c7a065fe6592094d8506221053176c9500e54",
"0xb52947e334a58f8041373c44270c90c0bd28f345b0dac2af0886ec3cfab253",
"0x23592b2754186e35f970c72eea16d46df2570bc68e6ee3069d8aa68d1a1707a",
"0x2854a6d2c60e46b2f176659ccbffdddc3db9e2d1e74d3fa54f2fe252e571db0"
],
{
"voter_id": 3,
"vote": 0,
"r": "0x315007dfbb13073cac204056c43fa51df0d56f88485c9563e86927f03c039bd",
"s": "0x51ce6bb918720da62507bf093a6e29877fd77a0f979c0bdcd5684c8bdfefea4"
},
{
"voter_id": 5,
"vote": 1,
"r": "0x5640e049062218fece9a6ab3f7871ff8dd7f8f7bc01d0e3b408f03d6477a1b6",
},
{
"voter_id": 8,
"vote": 0,
"r": "0x1749c30845cdf996ec03b79dd8262cf68e504143c93c94c8020d78c6f42b635",
"s": "0x31a8bac54c17ac9c81dc036bcc761a3f78d7f43a8d42c468d774c1b2a9746c2"
}
]
}
Processing the program input¶
Let’s define a struct that will represent a single vote:
struct VoteInfo {
// The ID of the voter.
voter_id: felt,
// The voter's public key.
pub_key: felt,
// The vote (0 or 1).
vote: felt,
// The ECDSA signature (r and s).
r: felt,
s: felt,
}
Now, let’s write a function that returns an array of VoteInfo instances based on the program input. Note that since the entire function is basically just one hint, the validity of the returned data (e.g., that the signatures are valid, the votes are restricted to 0 or 1, etc.) is not guaranteed, so we must verify it later.
from starkware.cairo.common.alloc import alloc
// Returns a list of VoteInfo instances representing the claimed
// The validity of the returned data is not guaranteed and must
// be verified by the caller.
func get_claimed_votes() -> (votes: VoteInfo*, n: felt) {
alloc_locals;
local n;
let (votes: VoteInfo*) = alloc();
%{
public_keys = [
int(pub_key, 16)
for pub_key in program_input['public_keys']]
for i, vote in enumerate(program_input['votes']):
# Get the address of the i-th vote.
memory[base_addr + ids.VoteInfo.voter_id] = \
vote['voter_id']
memory[base_addr + ids.VoteInfo.pub_key] = \
public_keys[vote['voter_id']]
memory[base_addr + ids.VoteInfo.vote] = \
vote['vote']
memory[base_addr + ids.VoteInfo.r] = \
int(vote['r'], 16)
memory[base_addr + ids.VoteInfo.s] = \
int(vote['s'], 16)
%}
}
Verifying that the votes are signed¶
One of the first functions we will need is verify_vote_signature(), which gets a pointer to a VoteInfo instance and verifies that the vote was indeed signed by the voter’s public key (note that we still haven’t checked that the voter’s public key is one of the permitted public keys).
The function starts by calling hash2() to compute the message hash. This is the counterpart of the line pedersen_hash(POLL_ID, vote) in the Python code above.
Then, we call verify_ecdsa_signature() to check that the signature is valid. One subtlety is that verify_ecdsa_signature() gets the signature only as a hint for the prover – the fact that it completed successfully only implies that the prover knows a signature for the given message and public key, not that the specific r and s constitute that signature. In our case, it’s enough, as we don’t care about r and s themselves, we just want to make sure the message was signed by the given public key.
from starkware.cairo.common.cairo_builtins import (
HashBuiltin,
SignatureBuiltin,
)
from starkware.cairo.common.hash import hash2
from starkware.cairo.common.signature import (
verify_ecdsa_signature,
)
// The identifier that represents what we're voting for.
// This will appear in the user's signature to distinguish
// between different polls.
const POLL_ID = 10018;
func verify_vote_signature{
pedersen_ptr: HashBuiltin*, ecdsa_ptr: SignatureBuiltin*
}(vote_info_ptr: VoteInfo*) {
let (message) = hash2{hash_ptr=pedersen_ptr}(
x=POLL_ID, y=vote_info_ptr.vote
);
verify_ecdsa_signature(
message=message,
public_key=vote_info_ptr.pub_key,
signature_r=vote_info_ptr.r,
signature_s=vote_info_ptr.s,
);
return ();
}
The pedersen builtin, which is required in order to compute the Pedersen hash function, is using an implicit argument called pedersen_ptr. On the other hand hash2() gets an implicit argument called hash_ptr. Therefore, we need to explicitly bind the hash_ptr implicit argument, using hash2{hash_ptr=pedersen_ptr}(...).
Similarly, the implicit argument ecdsa_ptr is used by verify_ecdsa_signature (here the name of the implicit argument of verify_ecdsa_signature is also ecdsa_ptr, so we don’t have to specify the binding explicitly).
Merkle tree¶
An important feature of our system will be that it will allow splitting the voting process to batches, where each batch can be processed in a separate Cairo run (this way we can support large and ongoing polls). This means that we will need to pass information between each pair of consecutive runs: which of the voters have already cast a vote (or rather, who is still allowed to vote) and what the results have been so far.
We will use a Merkle tree to store the information about the public keys that are allowed to vote. A Merkle tree is a cryptographic primitive that allows “compressing” data of an arbitrary size to a very short value (in our case, it can fit in one field element). It works as follows: you take an array of values (usually of size which is a power of 2, say $$2^k$$) and you compute the hashes of pairs of values so that you obtain $$2^k / 2 = 2^{k - 1}$$ hashes. You repeat this step once more to obtain $$2^{k - 2}$$ hashes, and continue for a total of $$k$$ steps, which results in one hash, called “the Merkle root”. The important property of Merkle trees is that given the Merkle root of an array of values, it is not feasible to find a different array of the same size with the same Merkle root. This way the Merkle root “encodes” all of the tree’s data.
Our Merkle tree will contain all the voters’ public keys (padded with zeros) that haven’t voted yet. When someone votes, we replace their public key with 0 in the Merkle tree. Thus we guarantee that no one can vote more than once.
For simplicity we hard-code the maximal number of voters to $$2^{10} = 1024$$:
const LOG_N_VOTERS = 10;
Each Cairo run will output 4 values: the number of “yes” and “no” votes and the Merkle root before and after processing the votes of that batch (note that each run handles one batch, which may include multiple votes). It is up to the system using the Cairo proofs (e.g., a smart contract) to make sure that the new root encoded in one proof is the same as the old root encoded in the next proof, and to add the partial results of the new batch to those accumulated thus far.
To track the changes to the Merkle tree, we will use a DictAccess array, which will encode the changes to the leaves (changing actual public keys to zeros). Let’s define a VotingState struct to keep track of the current ‘yes’ and ‘no’ counts and the DictAccess array. If you need to recall how a DictAccess array works, see Dictionaries/maps in Cairo.
from starkware.cairo.common.dict import DictAccess
struct VotingState {
// The number of "Yes" votes.
// The number of "No" votes.
// Start and end pointers to a DictAccess array with the
// changes to the public key Merkle tree.
public_key_tree_start: DictAccess*,
public_key_tree_end: DictAccess*,
}
Now, let’s write a function that returns an initial state, with both the “yes” and “no” counters set to zero, and an empty array for the tree’s changes. Note that we’re using dict_new() to create the dict. dict_new() is one of the high-level dictionary functions defined in dict.cairo. These functions maintain the current values of the dictionary using hints. Therefore, dict_new() expects to get a hint variable called initial_dict with the initial values of the dictionary.
from starkware.cairo.common.dict import dict_new
func init_voting_state() -> (state: VotingState) {
alloc_locals;
local state: VotingState;
assert state.n_yes_votes = 0;
assert state.n_no_votes = 0;
%{
public_keys = [
int(pub_key, 16)
for pub_key in program_input['public_keys']]
initial_dict = dict(enumerate(public_keys))
%}
let (dict: DictAccess*) = dict_new();
assert state.public_key_tree_start = dict;
assert state.public_key_tree_end = dict;
return (state=state);
}
The following function verifies that the vote is signed and removes the public key from the tree. There are two options to handle the voting state:
1. Pass it as an argument and return the new state.
2. Add it as an implicit argument.
The two options have a different syntax, but they will be compiled to the same bytecode. Here we chose the second option as it simplifies the code calling process_vote.
from starkware.cairo.common.dict import dict_update
from starkware.cairo.common.math import assert_not_zero
func process_vote{
pedersen_ptr: HashBuiltin*,
ecdsa_ptr: SignatureBuiltin*,
state: VotingState,
}(vote_info_ptr: VoteInfo*) {
alloc_locals;
// Verify that pub_key != 0.
assert_not_zero(vote_info_ptr.pub_key);
// Verify the signature's validity.
verify_vote_signature(vote_info_ptr=vote_info_ptr);
// Update the public key dict.
let public_key_tree_end = state.public_key_tree_end;
dict_update{dict_ptr=public_key_tree_end}(
key=vote_info_ptr.voter_id,
prev_value=vote_info_ptr.pub_key,
new_value=0,
);
// Generate the new state.
local new_state: VotingState;
assert new_state.public_key_tree_start = (
state.public_key_tree_start
);
assert new_state.public_key_tree_end = public_key_tree_end;
// Update the counters.
tempvar vote = vote_info_ptr.vote;
if (vote == 0) {
// Vote "No".
} else {
// Make sure that in this case vote=1.
assert vote = 1;
// Vote "Yes".
}
// Update the state.
let state = new_state;
return ();
}
Finally, let’s write the loop that processes all the votes. It gets a pointer to an array of VoteInfo instances and its size and updates the given state accordingly.
func process_votes{
pedersen_ptr: HashBuiltin*,
ecdsa_ptr: SignatureBuiltin*,
state: VotingState,
if (n_votes == 0) {
return ();
}
);
return ();
}
The main() function¶
As explained above, the program will output 4 values that summarize the batch: the number of “yes” and “no” votes and the Merkle root before and after processing the votes of that batch. The following struct represents that information:
struct BatchOutput {
public_keys_root_before: felt,
public_keys_root_after: felt,
}
The only missing part is the computation of the two Merkle roots, based on the public key dictionary (VotingState.public_key_tree_start and VotingState.public_key_tree_end). In order to do this, we first squash the dict and then call the standard library function small_merkle_tree_update() (a requirement of small_merkle_tree_update() is that we use the high-level function dict_squash() rather than squash_dict(). dict_squash() passes hint information about all of the dict entries to the squashed dict, including entries that haven’t changed.
%builtins output pedersen range_check ecdsa
from starkware.cairo.common.dict import dict_squash
from starkware.cairo.common.small_merkle_tree import (
small_merkle_tree_update,
)
func main{
output_ptr: felt*,
pedersen_ptr: HashBuiltin*,
range_check_ptr,
ecdsa_ptr: SignatureBuiltin*,
}() {
alloc_locals;
let output = cast(output_ptr, BatchOutput*);
let output_ptr = output_ptr + BatchOutput.SIZE;
let (state) = init_voting_state();
local pedersen_ptr: HashBuiltin* = pedersen_ptr;
local ecdsa_ptr: SignatureBuiltin* = ecdsa_ptr;
// Write the "yes" and "no" counts to the output.
// Squash the dict.
let (squashed_dict_start, squashed_dict_end) = dict_squash(
dict_accesses_start=state.public_key_tree_start,
dict_accesses_end=state.public_key_tree_end,
);
local range_check_ptr = range_check_ptr;
// Compute the two Merkle roots.
let (root_before, root_after) = small_merkle_tree_update{
hash_ptr=pedersen_ptr
}(
squashed_dict_start=squashed_dict_start,
squashed_dict_end=squashed_dict_end,
height=LOG_N_VOTERS,
);
// Write the Merkle roots to the output.
assert output.public_keys_root_before = root_before;
assert output.public_keys_root_after = root_after;
return ();
}
Note that we write {state=state} explicitly when we call process_votes. This is required since the compiler does not allow implicit bindings where the bound variable is not an implicit argument of the calling function. See more information here.
One new feature we used here is the cast keyword. The cast keyword in let output = cast(output_ptr, BatchOutput*); converts the felt pointer to a pointer to BatchOutput, so the type of the output reference is BatchOutput*. Now we can write output.n_yes_votes to access the first output cell, which encodes the number of “yes” votes.
Don’t forget to supply the program input file when you run the code (you can find the full Cairo file here):
cairo-compile voting.cairo --output voting_compiled.json
cairo-run --program=voting_compiled.json \
--print_output --layout=small \
--program_input=voting_input.json
You should get:
Program output:
1
2
1591806306193441240739433996824056703232153712683022312894504906643112470393
-1397522753299492751557547967820826962898231398543673030347416450104778351221
Another batch¶
Our Cairo code supports voting in batches, so let’s try that. Let’s say that we want to run another batch after the one we just did. Modify voting_input.json so that the public keys of the voters who voted in the first batch are 0 and the votes section contains one new vote instead of the old three. You can use the following Python script:
import json
from starkware.crypto.signature.signature import pedersen_hash, sign
POLL_ID = 10018
input_data['public_keys'][3] = '0x0'
input_data['public_keys'][5] = '0x0'
input_data['public_keys'][8] = '0x0'
# Generate a "yes" vote for voter 6.
voter_id = 6
priv_key = 123456 * voter_id + 654321
vote = 1
r, s = sign(
msg_hash=pedersen_hash(POLL_ID, vote),
priv_key=priv_key,
)
'voter_id': voter_id,
'vote': vote,
'r': hex(r),
's': hex(s),
}]
with open('voting_input2.json', 'w') as f:
json.dump(input_data, f, indent=4)
f.write('\n')
Run the same program again (you don’t need to recompile) with voting_input2.json. You should get:
Program output:
1
0
-1397522753299492751557547967820826962898231398543673030347416450104778351221
-628706650786693403852552424323387050556189030546827265857028820447499605255
Note that indeed, the root of the Merkle tree before the second batch is the same as the root after the first one. | 4,482 | 16,930 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-10 | latest | en | 0.748366 |
https://3dtotal.com/tutorial/1652-how-to-stylize-and-model-toon-humans-chapter-1-concept-modeling-3ds-max-zbrush-by-jose-alves-da-silva-character-human-male?page=3 | 1,529,651,558,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864364.38/warc/CC-MAIN-20180622065204-20180622085204-00014.warc.gz | 546,588,214 | 17,971 | ### Keep up-to-date with Free tutorials!!
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How to Stylize and Model 'Toon Humans - Chapter 1: Concept and Modeling
(Score 4.82 out of 5 after 49 Votes)
| 315002 Views
| 2
Software used:
Keywords:
## ZSpheres 5
If you look at the lower back of the character and press A to see the Adaptive Skin, you will notice that the polygons of the waist are much bigger than the ones in the legs (Fig.05 - 06).
Fig.05
Fig.06
- Add a ZSphere at the back to add some more polygon density there.
At the front of the character we have an even bigger problem. Notice that the edge flow in the belly area is disturbed by a rhombus shape that prevents the existence of a continuous line at the center of the object.
- Add a ZSphere at the pelvis and this
problem gets fixed.
At the top and palm of the hand we can also find a geometric distribution that is very hard to sculpt.
- Add a ZSphere at the top of the hand and another at the palm. You will get a very high density mesh there, but it is better than the previous topological nightmare.
## ZSpheres 6
- In the Tool menu, go to the Unified Skin pull down menu and click on Make Unified Skin.
- ZBrush will generate a new Tool starting with the prefix "Skin_".
- In this exercise we will not need the ZSpheres anymore, but if you want to keep them, press the Append button from the Subtool pull down menu and choose the new "Skin_" tool. Now you have the ZSpheres and Skin as Subtools of the same tool.
- Choose the "Skin_" Subtool and hide the ZSpheres Subtool (Fig.07).
Fig.07
## Sculpting 1
I have decided to create a character without clothes on the upper part of the body in order to deal with anatomy issues and with wrinkled pants on the lower body to deal with cloth modeling (Fig.08).
Fig.08
Press the X key to activate Symmetry for the new "Skin_" Subtool. In the Geometry menu increase the number of subdivisions to 4 (when you create the skin, it already has 2 levels of subdivision).
- Select the Clay brush and start adding volume to the muscles. Use the Clay brush to add volume, the Shift key to smooth, and the Alt key to remove volume. Add the volume as if you were adding clay to a wire skeleton.
You can see that I have added a lot of volume near the feet as if the character was wearing bell bottom trousers. I have also added some clay at the trapezius muscles, deltoids, triceps and forearms. The back of the head was filled and some marks for the eye cavities, also the nose and jaw were added.
- To fix the hands, use the mPolish brush and rub it on the top of the hand and palm. Also use the Smooth brush. After reducing the volume, use the Clay brush to shape the volume. As the finger and palms are quite thin I would advise turning on the BackfaceMask button; you will find it in the Auto Masking section in the Brush menu. This will prevent the Clay brush from editing both sides of the finger at the same time.
## Sculpting 2
- Subdivide the geometry twice more, increasing the total number of subdivisions to 6.
- Choose the ClayTubes brush and sculpt by applying strokes along the muscles following the muscle fibers. Use the Smooth brush to soften the forms (Fig.09 - 10). We can use this techniques for all the detailing, including the trousers.
Fig.09
Fig.10
## < previous page continued on next page >
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2 167654 | 1,101 | 4,261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-26 | latest | en | 0.905232 |
https://www.vtupulse.com/artificial-intelligence/heuristic-search-characteristics-advantages-artificial-intelligence/ | 1,695,600,391,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506669.96/warc/CC-MAIN-20230924223409-20230925013409-00311.warc.gz | 1,170,711,468 | 13,249 | # Heuristic Search – Characteristics Advantages – Artificial Intelligence
[wptelegram-join-channel]
### What do you mean by heuristic and heuristic search? What are the advantages and Characteristics of Heuristic Search? – Artificial Intelligence
There are two types of search algorithms in Artificial Intelligence, Uninformed search algorithms and Informed search algorithms.
Uninformed search algorithms or Brute-force algorithms, search through the search space all possible candidates for the solution checking whether each candidate satisfies the problem’s statement.
Informed search algorithms use heuristic functions that are specific to the problem, apply them to guide the search through the search space to try to reduce the amount of time spent in searching.
A good heuristic will make an informed search dramatically outperform any uninformed search: for example, the Traveling Salesman Problem (TSP), where the goal is to find is a good solution instead of finding the best solution. In such problems, the search proceeds using current information about the problem to predict which path is closer to the goal and follow it, although it does not always guarantee finding the best possible solution. Such techniques help in finding a solution within reasonable time and space (memory).
### Requirement of Search Algorithms / Techniques
1. The first requirement is that it causes motion, in a game playing program, it moves on the board and in the water jug problem, filling water is used to fill jugs. It means the control strategies without the motion will never lead to the solution.
2. The second requirement is that it is systematic, that is, it corresponds to the need for global motion as well as for local motion. This is a clear condition that neither would it be rational to fill a jug and empty it repeatedly, nor it would be worthwhile to move a piece round and round on the board in a cyclic way in a game.
### Informed / Heuristic Search Algorithms / Techniques
Many Informed search Algorithms techniques are developed, using heuristic functions.
The algorithms that use heuristic functions are called heuristic algorithms.
Heuristic algorithms are not really intelligent; they appear to be intelligent because they achieve better performance.
Heuristic algorithms are more efficient because they take advantage of feedback from the data to direct the search path.
Example: Finding a route from one city to another city is an example of a search problem in which different search orders and the use of heuristic knowledge are easily understood.
State: The current city in which the traveler is located.
Cost Metric: The cost of taking a given road between cities.
Heuristic information: The search could be guided by the direction of the goal city from the current city, or we could use airline distance as an estimate of the distance to the goal.
For complex problems, the traditional algorithms, presented above, are unable to find the solution within some practical time and space limits. Consequently, many special techniques are developed, using heuristic functions.
• Blind search is not always possible, because it requires too much time or space (memory).
• Heuristics are rules of thumb; they do not guarantee a solution to a problem.
• Heuristic Search is a weak technique but can be effective if applied correctly; it requires domain-specific information.
### Characteristics of heuristic search
Heuristics are knowledge about the domain, which help search and reasoning in its domain.
Heuristic search incorporates domain knowledge to improve efficiency over blind search.
A heuristic is a function that, when applied to a state, returns the value as estimated merit of state, with respect to the goal.
• Heuristics might (for reasons) underestimate or overestimate the merit of a state with respect to the goal.
• Heuristics that underestimate are desirable and called admissibly.
The heuristic evaluation function estimates the likelihood of a given state leading to the goal state.
Heuristic search function estimates cost from current state to goal, presuming function is efficient.
### Heuristic search compared with other search
The Heuristic search is compared with Brute force or Blind search techniques below:
### Example: Travelling salesman
A salesman has to visit a list of cities and he must visit each city only once. There are different routes between the cities. The problem is to find the shortest route between the cities so that the salesman visits all the cities at once.
Suppose there are N cities, then a solution would be to take N! possible combinations to find the shortest distance to decide the required route. This is not efficient as with N=10 there are 36,28,800 possible routes. This is an example of a combinatorial explosion.
There are better methods for the solution of such problems: one is called branch and bound. First, generate all the complete paths and find the distance of the first complete path. If the next path is shorter, then save it and proceed this way avoiding the path when its length exceeds the saved shortest path length, although it is better than the previous method. | 985 | 5,192 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-40 | latest | en | 0.887008 |
https://trapdoortech.medium.com/zero-knowledge-proof-one-possible-malleability-of-groth16-d2945e5073a1 | 1,721,859,811,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518454.54/warc/CC-MAIN-20240724202030-20240724232030-00088.warc.gz | 505,980,885 | 35,329 | # Zero Knowledge Proof — one possible malleability of Groth16
--
zkHack published a new Puzzle. Even though the Puzzle used an older zkp algorithm — Groth16 , it’s very interesting and very helpful to understand the concepts such as Zero-Knowledge Proof, polynomials, linear correlation, etc.
# What’s the Puzzle?
The corresponding circuit of Puzzle is constructed in circom, and implemented in zkhack-groth-puzzle/circuits/circuit.circom.
`pragma circom 2.0.6;include "../node_modules/circomlib/circuits/poseidon.circom";template Circuit() {signal input a;signal input b;signal input c;a === b;component p = Poseidon(1);p.inputs[0] <== c;p.out === 17744324452969507964952966931655538206777558023197549666337974697819074895989;}`
The circuit is fairly simple, its input has three inputs: a/b/c, in which a is a public input. There are two constraints: 1/ a is equal to b 2/ The poseidon hash result of c is a specific value. The circom circuit is compiled into R1CS constraints, and then snarkjs is used to construct the proofs. There’s a specific comment that the snarkjs source code used in Puzzle has some modification:
`diff --git a/build/main.cjs b/build/main.cjsindex 00e6965..ef1bd6f 100644--- a/build/main.cjs+++ b/build/main.cjs@@ -4328,7 +4328,8 @@ async function newZKey(r1csName, ptauName, zkeyName, logger) { } }- for (let s = 0; s <= nPublic ; s++) {+// for (let s = 0; s <= nPublic ; s++) {+ for (let s = 0; s < 1 ; s++) { const l1t = TAU_G1; const l1 = sG1*(r1cs.nConstraints + s); const l2t = BETATAU_G1;`
The seemingly minor edit to the proving system leads to the malleability of the final proof.
The question of Puzzle is given in zkhack-groth-puzzle/src/solution.js:
`const { proof, publicSignals } = JSON.parse(`{"proof":{"pi_a":["7076778705842675636541778654824835671264842003792815899892788518756808417824","4871300562969249383482829591051792322271432570205055011710223197671646924652","1"],"pi_b": [["4702507968743578934061693422759564470881256571473408115314331474240229998811","16198326042603795115438219508756675682917780977814561672804657276368883889354"],["12916734195569167956837700546311420400354235424337271822709448553494046311159", "20167467333119574021428597666293210644874141810710695584907560968298314755986"],["1","0"]],"pi_c":["20014664648588403789442308373435642542109961553284949305762265534102084844319", "10562544426189233680286850591386198483452124187323754995599976212942563914034","1"],"protocol":"groth16","curve":"bn128"},"publicSignals":["1"]}`); const isValid = await verifyProof(vKey, { proof, publicSignals }); console.assert(isValid === true, "Proof is not valid"); const new_a = BigInt("PUT_YOUR_ADDRESS_HERE"); publicSignals[0] = new_a; /* PUT YOUR SOLUTION HERE Change the proof such that isValidMalleable = true and passes assertion */ const isValidMalleable = await verifyProof(vKey, { proof, publicSignals }); console.assert(isValidMalleable === true, "Malleable is not valid");`
Given a proof, how to make sure that the proof is verified even after the public input is changed? The Puzzles that accept changed public input are typically interesting, let’s see how to solve this Puzzle.
# R1CS Circuit Constraints
Take a look at the circuit structure of Puzzle:
To check whether the value of variable a and b are equal, the circuit utilizes this structure: 0*1 = b-a. And, the other parts of the circuit are Poseidon Hash, which are not related to variable a/b, indicating that:
When the prove and verify process of Groth16 algorithm is combined with the information above:
In this Puzzle, l is 1 (one of them is constant ONE, the other one is variable a). Let’s reorganize the verify formula:
C is decomposed to the formula below:
Take a close look at these two items:
Because u1(x)/v1(x) and u2(x)/v2(x) are both 0, they can be further simplified to:
Combine with the information above:
We can get that w1(x) and w2(x) have linear correlation. Therefore, there’s linear correlation between the following two expressions:
After understanding the challenge, the information of the points of these two expressions are provided publicly. The information of the corresponding point can be found in zkhack-groth-puzzle/build/snark/circuit/circuit_final.zkey of Puzzle, assuming the information of the point is C_2.
# How to solve the Puzzle?
After figuring out the logic and the linear correlation, then it’s easy to solve the question. If the public input variable a increases, c also adapts accordingly.
You can check the code of the smart contract to ensure the value of variable _a that is required by the verification(zkhack-groth-puzzle/contracts/Puzzle.sol):
`uint256 _a = uint256(uint160(address(msg.sender)));require( verifyProof( [_proof[0], _proof[1]], [[_proof[2], _proof[3]], [_proof[4], _proof[5]]], [_proof[6], _proof[7]], [_a] ), "Puzzle: Invalid proof");`
Public variable a becomes the last 160 bits data of transaction sending address.
# How to prevent linear correlation?
After we have finished solving the question, it’s necessary to consider how to prevent the linear correlation among the data of constraints in the proving system. Recall that Puzzle itself is the minimal edits to the snarkjs. By taking a close look at these edits, you can then tell how to prevent linear correlation:
Except the constraints of “user” defined logic, expand the constraint system for the public inputs: 1*0 = 0. This edit not only ensures that the constraint polynomials of the public input variables have no linear correlation, but also ensures that the constraint polynomials of the public input variable and private input variable have no linear correlation. Some people might notice that for the constraint polynomials of private input variables, they are not required to have non-linear correlation.
In fact, similar questions were found in the libsnark source code back in 2015. If you are interested, feel free to check it out.
Summary:
The new Puzzle of zkHack is very interesting, it is related to the linear correlation of the constraints. In this specific case, it is also related to the polynomial of input variables and private variables. Under this situation, a new proof can be changed directly modifying the accepted one.
--
--
Trapdoor-Tech tries to connect the world with zero-knowledge proof technologies. zk-SNARK/STARK solution and proving acceleration are our first small steps :) | 1,665 | 6,603 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-30 | latest | en | 0.712581 |
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# FFT_Hanning_window_w - Sample Excel Spreadsheet Generating an FFT with a Hanning Window(for a sine wave function with a DC offset J M Cimbala 90 Hz
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Unformatted text preview: Sample Excel Spreadsheet - Generating an FFT with a Hanning Window (for a sine wave function with a DC offset) J. M. Cimbala 90 Hz 128 2 1 volts (or any other units) 20 Hz 0 radians Calculated values: DC = mean value of input signal = 2.01 volts (or any other units) (average of all the useful input data) 1.42 s 0.01 s 0.7 Hz 45 Hz 64 0 (use 20 points per period of the true signal) number of data points for plotting true signal = 568.89 Note: FFT does not automatically update! (also divide first one by 2) (also divide first one by 2, and add back the DC value) frequency (Hz) 2 256.984807753012 2.01-0.01-0.493247503592427 2.01 1 0.01 2.98 0.7 0.986495007184767-1.0169 0.02 0.98 0.245351916179819+1.61957072 0.01 2 0.02 2.34 1.41 0.991582349650347-2.0403 0.02 0.33-8.69447197757361E-004+3.2793 3 0.03 1.13 2.11 1.00014748090696-3.07682 0.02-0.87 0.01-9.02639910711162E-004+5.0215 4 0.04 1.36 2.81 1.01232317180642-4.13342 0.02-0.65 0.01-0.01-9.5088777783647E-004+6.89304 5 0.06 2.64 3.52 1.02830241381723-5.21759 0.02 0.64 0.01 0.01-1.01612334787129E-003+8.9477 6 0.07 2.87 4.22 1.04834614921952-6.33755 0.02 0.86 0.02 0.02-1.10107253521986E-003+1.1250 7 0.08 1.66 4.92 1.07279417476269-7.50252 0.02-0.35 0.03-0.01-1.2094753832663E-003+1.38801 8 0.09 1.02 5.63 1.10208010183891-8.72300 0.02-0.99 0.04-0.04-1.34640908953294E-003+1.6938 9 0.1 2 6.33 1.13675166527328-0.10011 0.02-0.01 0.05-1.51876037252756E-003+2.0557 10 0.11 2.98 7.03 1.17749827019776-0.11381 0.02 0.98 0.06 0.06-1.73592199937049E-003+2.4909 11 0.12 2.34 7.73 1.22518856311972-0.12851 0.02 0.33 0.07 0.02-2.01083438060306E-003+3.0230 12 0.13 1.13 8.44 1.28092219356409-0.14442 0.02-0.87 0.08-0.07-2.36157187701454E-003+3.6845 13 0.14 1.36 9.14 1.34610211151652-0.16181 0.02-0.65 0.1-0.06-2.81381204195879E-003+4.5212 14 0.16 2.64 9.84 1.42253727763679-0.18100 0.02 0.64 0.11 0.07-3.40477788435778E-003+5.5992 15 0.17 2.87 10.55 1.51259155529449-0.20243 0.02 0.86 0.13 0.11-4.18971764915908E-003+7.0162 16 0.18 1.66 11.25 1.61940470354883-0.22667 0.03-0.35 0.15-0.05-5.25291695018775E-003+8.9213 17 0.19 1.02 11.95 1.74722951960392-0.25448 0.03-0.99 0.16-0.16-6.72714819962954E-003+1.1548 18 0.2 2 12.66 1.90196292845754-0.28691 0.03-0.01 0.18-8.82959706415118E-003+1.5282 19 0.21 2.98 13.36 2.09201472556776-0.32545 0.03 0.98 0.2 0.2-1.19318265347693E-002+2.0776 20 0.22 2.34 14.06 2.32979382881706-0.37230 0.04 0.33 0.22 0.07-1.6704928139515E-002+2.92150 21 0.23 1.13 14.77 2.63439264462442-0.43083 0.04-0.87 0.24-0.21-2.44448626584585E-002+4.2883 22 0.24 1.36 15.47 3.03677091106562-0.50652 0.05-0.65 0.26-0.17-3.78758542460439E-002+6.6586 23 0.26 2.64 16.17 3.590652594491-0.6088501 0.06 0.64 0.29 0.18-6.3399895757191E-002+1.11610 24 0.27 2.87 16.88 4.39813386094516-0.75581 0.07 0.86 0.31 0.26-0.118570913523678+2.08909552 25 0.28 1.66 17.58 5.67989878149403-0.98635 0.09-0.35 0.33-0.12-0.264029837306617+4.65409677 0.01 26 0.29 1.02 18.28 8.01778305126938-1.40304 0.13-0.99 0.35-0.35-0.812310830159638+0.14321930 0.02 27 0.30....
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## This note was uploaded on 07/23/2008 for the course ME 345 taught by Professor Staff during the Spring '08 term at Pennsylvania State University, University Park.
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FFT_Hanning_window_w - Sample Excel Spreadsheet Generating an FFT with a Hanning Window(for a sine wave function with a DC offset J M Cimbala 90 Hz
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Ask a homework question - tutors are online | 1,927 | 3,949 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2017-13 | longest | en | 0.48822 |
https://techcommunity.microsoft.com/t5/excel/creating-a-macro-to-pull-information-from-one-sheet-to-another/td-p/1376723 | 1,597,333,740,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739046.14/warc/CC-MAIN-20200813132415-20200813162415-00348.warc.gz | 519,225,861 | 81,857 | Highlighted
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# Creating a macro to pull information from one sheet to another
I have a workbook that has several sheets dealing with the stock market. Each sheet has one particular stock item except for the last sheet that lists all the stock with the purchased price and the current price.
I want the last sheet (called "Balance") to update every time I add data to the specific stock based on the date. See attached excel file. I have sent the file with only one of my stocks to make it smaller. On the balance sheet in cell C3 I want it to show the amount that was inserted on on 5/7/2020 and then
changed to the data that was inserted on 5/8/2020 then on down the line for the next date and so on and so forth.
3 Replies
Highlighted
# Re: Creating a macro to pull information from one sheet to another
Try this (Example attached):
=LOOKUP(2,1/(INDIRECT("'"&\$A3&"'!B:B")<>""),INDIRECT("'"&\$A3&"'!B:B"))
[Note: The formula uses the list to identify which sheet the lookup indexes, therefore each sheet name will need to be identical to the entries in the list - You should be able to drag down the formula for rest of list]
Highlighted
# Re: Creating a macro to pull information from one sheet to another
The formula is what I was looking for and I thank you, but how do I get it to work on other sheets. I does not show what sheet to pull from. If I copy it to the next cell, how does it know to pull from a different sheet when it does not have the sheet name in the formula.
Highlighted
# Re: Creating a macro to pull information from one sheet to another
As mentioned I the note, the sheet name is referenced from the list in column A (cell A3 in the example). The formula can be dragged down into the cells below but I didn’t do this as there is only one sheet so far. It’s for this reason that the list items must match the sheet names exactly. Hope this makes sense but let me know if still not clear.
Alternatively, you could change it to reference the sheet by name:
=LOOKUP(2,1/(CGC!B:B<>""),CGC!B:B)
[Edited] | 493 | 2,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-34 | latest | en | 0.913333 |
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Concerning masonry (not the brickwork type)
page: 2
0
share:
posted on Jun, 1 2004 @ 02:58 PM
Originally posted by infinite
...something calling me to become a mason. Anyways, lets hope i get accepted at 21
If you were in California you could be accepted at the age of 18
posted on Jun, 1 2004 @ 03:00 PM
I'd say we're 50/50.
Some say Geometry. Some say God. But most say both.
posted on Jun, 1 2004 @ 03:29 PM
Hmmm... that is interesting.
In the Canadian rite, we have little talk about Geometry, other than a few important comments during the second degree lecture, but nothing like what you're saying. I don't know what happens in the York Rite, which is supposed to be a lot closer to what you do.
We get a little more talk about geometry in the Scottish Rite, but even then not much (and, again, our Scottish Rite ritual is different from both the Northern and Southern Jurisdictions). I had heard from someone that the Canadian Rite is close to what's called the "Emulation" Work. I only have books of the Work for Canadian and York, though, so I can neither confirm nor deny.
posted on Jun, 3 2004 @ 07:02 AM
Now... I've read that there are officially only three degrees within masonry, although some claim to rise far higher. Who's right?
posted on Jun, 3 2004 @ 07:05 AM
There are 33 degrees.
But the first 3 are the 3 that every mason must have and there is no degree higher than the 3rd. All of the others are side degrees.
posted on Jun, 3 2004 @ 07:09 AM
So the higher ones aren't higher just on a different point of the same plane? So to speak ?
posted on Jun, 3 2004 @ 07:44 AM
As I said, they are what we call "side degrees". Maybe some of the other masons on the board will disagree and say that they stand alone, but in my opinion they are there to help illustrate the meanings of the 1st, 2nd and 3rd and to help you to build on them. The other degrees that I'm in, here in the UK, tend to point to this.
We are certainly told that there is no degree greater than the 3rd. It is referred to as the "sublime" degree.
posted on Jun, 3 2004 @ 07:51 AM
So that third degree takes a fair bit of work to get to then?
posted on Jun, 3 2004 @ 08:12 AM
Not necessarily. If you mean how easy is it to obtain, it's normally a short period of time (within a year) before a mason receives it.
But if you mean "to get" as in "to understand", yes it is a tremendous amount of work. One of the parts of the 3rd is the study of oneself and who can truly say that they know all there is to know about theirself ? In my opinion, a mason can spend a whole lifetime just studying this one single facet.
Here in the UK it takes decades to get the amount of degrees that a US mason can receive in a relatively short time. But in my opinion, this doesn't matter. Everything is about understanding those first 3.
posted on Jun, 3 2004 @ 08:23 AM
Originally posted by Leveller
In my opinion, a mason can spend a whole lifetime just studying this one single facet.
Here in the UK it takes decades to get the amount of degrees that a US mason can receive in a relatively short time. But in my opinion, this doesn't matter. Everything is about understanding those first 3.
I agree totally with Leveller... I am in Japan and in the Scottish Lodge. We are not in the Scottish rite, so no 33 degrees for us to get. Although I could join Scottish Rite lodge here and do it. Here in Japan you can get the 32nd degree within 2 days. As you have heard for Leveller a long time. In the York rite here in Japan, it takes a long time also to get degree work done. But basically, Official masonry stops at 3rd degree. (hmmm. maybe a bad way to say it.. ) Once a Master Mason, you are just as equal as a 33rd degree. Although, the ones that get 33rd degree are usually very experienced. Like I said, here in Japan it is 2 days, but you must be very experienced in order to do the degrees.
Yes, there are tons of stuff, many debates and very fascinating to study.
posted on Jun, 3 2004 @ 11:12 AM
Agreed , there are the 3 degrees. Which always makes me think of the sining group and other side orders.
I am UK UGLE. Every experienced Mason will tell you the same.
Masonry is a community of equals, its just that some are more equal than others.
posted on Jun, 3 2004 @ 11:54 AM
Why I got involved in masonry:
I got interested in masonry because most of the older men I had respect for were involved. I was drinking coffee with a group, and one of them noted that I was the only non-mason present. I started asking questions and it went from there.
How did I get involved?
I asked how to join, and one of them gave me a petition to fill out. Then a group of them made an appointment to visit my home and talk with me and the Mrs. Doktor. They wanted to make sure it was o.k. with my wife for me to join, and assure her that masonry was an ethical org. They held a vote, and the whole lodge voted to invite me. We scheduled an evening for the initiation, and it went from there.
Any Evil Secrets?
I am a Master mason in the blue lodge, have attained the 18th degree of the Scottish Rite, and a Knight Templar in the (American) York rite.
I can honestly say that I believe Masonry is the original fraternity. There is no HAZING involving any kind of physical abuse. There is some suspense in the ceremonies, but I was constantly reassured that no harm would come to me and I never felt threatened by anything. All other frats I have seen involved some kind of torture or "hazing" which I absolutely refuse to be involved in. Masonry is about ideals and values, and not about being a club or "in" group. The copycats are all degenerate imitations in my opinion.
Activity.
I have not been actively attending the lodge for about 6 months, due to changes in my job and moving. My Scottish Rite membership is past due, but my blue lodge and York rite are up to date. I don't currently hold any appointed office, because I knew the work changes were coming.
I am an avid reader on the history of freemasonry, particularly its historical roots and the esoterica underlying its symbolic language.
posted on Jun, 3 2004 @ 01:47 PM
Originally posted by Oisin
Now... I've read that there are officially only three degrees within masonry, although some claim to rise far higher. Who's right?
The degrees of Entered Apprentice, Fellow Craft, and Master Mason are the only ones universally recognized as Freemasonry. Since the 1720’s, other degrees have been written and introduced into groups or series called Rites, such as the Scottish Rite, York Rite, Egyptian Rite, etc. All of these Rites control their own degrees, which are not recognized by the other Rites.
Master Masons in good standing are eligible for membership in any of the Rites.
Fiat Lvx.
posted on Jun, 4 2004 @ 07:47 AM
Cheers guys and girls.
new topics
top topics
0 | 1,749 | 6,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2017-34 | longest | en | 0.972666 |
https://www.lmfdb.org/L/ModularForm/GL2/Q/holomorphic/3024/2/q/k/2881/8/ | 1,604,129,783,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107916776.80/warc/CC-MAIN-20201031062721-20201031092721-00576.warc.gz | 782,971,820 | 7,085 | # Properties
Degree $2$ Conductor $3024$ Sign $0.821 + 0.570i$ Motivic weight $1$ Primitive yes Self-dual no Analytic rank $0$
# Learn more about
## Dirichlet series
L(s) = 1 + (0.841 − 1.45i)5-s + (−1.65 + 2.06i)7-s + (−0.622 − 1.07i)11-s + (1.96 + 3.39i)13-s + (1.62 − 2.81i)17-s + (−2.36 − 4.09i)19-s + (0.199 − 0.344i)23-s + (1.08 + 1.87i)25-s + (3.19 − 5.54i)29-s + 0.578·31-s + (1.61 + 4.14i)35-s + (2.72 + 4.71i)37-s + (−4.20 − 7.27i)41-s + (−2.46 + 4.26i)43-s − 0.425·47-s + ⋯
L(s) = 1 + (0.376 − 0.651i)5-s + (−0.625 + 0.780i)7-s + (−0.187 − 0.325i)11-s + (0.543 + 0.941i)13-s + (0.394 − 0.683i)17-s + (−0.541 − 0.938i)19-s + (0.0415 − 0.0718i)23-s + (0.216 + 0.375i)25-s + (0.594 − 1.02i)29-s + 0.103·31-s + (0.273 + 0.701i)35-s + (0.447 + 0.774i)37-s + (−0.656 − 1.13i)41-s + (−0.375 + 0.650i)43-s − 0.0620·47-s + ⋯
## Functional equation
\begin{aligned}\Lambda(s)=\mathstrut & 3024 ^{s/2} \, \Gamma_{\C}(s) \, L(s)\cr =\mathstrut & (0.821 + 0.570i)\, \overline{\Lambda}(2-s) \end{aligned}
\begin{aligned}\Lambda(s)=\mathstrut & 3024 ^{s/2} \, \Gamma_{\C}(s+1/2) \, L(s)\cr =\mathstrut & (0.821 + 0.570i)\, \overline{\Lambda}(1-s) \end{aligned}
## Invariants
Degree: $$2$$ Conductor: $$3024$$ = $$2^{4} \cdot 3^{3} \cdot 7$$ Sign: $0.821 + 0.570i$ Motivic weight: $$1$$ Character: $\chi_{3024} (2881, \cdot )$ Primitive: yes Self-dual: no Analytic rank: $$0$$ Selberg data: $$(2,\ 3024,\ (\ :1/2),\ 0.821 + 0.570i)$$
## Particular Values
$$L(1)$$ $$\approx$$ $$1.761110949$$ $$L(\frac12)$$ $$\approx$$ $$1.761110949$$ $$L(\frac{3}{2})$$ not available $$L(1)$$ not available
## Euler product
$$L(s) = \displaystyle \prod_{p} F_p(p^{-s})^{-1}$$
$p$$F_p(T)$
bad2 $$1$$
3 $$1$$
7 $$1 + (1.65 - 2.06i)T$$
good5 $$1 + (-0.841 + 1.45i)T + (-2.5 - 4.33i)T^{2}$$
11 $$1 + (0.622 + 1.07i)T + (-5.5 + 9.52i)T^{2}$$
13 $$1 + (-1.96 - 3.39i)T + (-6.5 + 11.2i)T^{2}$$
17 $$1 + (-1.62 + 2.81i)T + (-8.5 - 14.7i)T^{2}$$
19 $$1 + (2.36 + 4.09i)T + (-9.5 + 16.4i)T^{2}$$
23 $$1 + (-0.199 + 0.344i)T + (-11.5 - 19.9i)T^{2}$$
29 $$1 + (-3.19 + 5.54i)T + (-14.5 - 25.1i)T^{2}$$
31 $$1 - 0.578T + 31T^{2}$$
37 $$1 + (-2.72 - 4.71i)T + (-18.5 + 32.0i)T^{2}$$
41 $$1 + (4.20 + 7.27i)T + (-20.5 + 35.5i)T^{2}$$
43 $$1 + (2.46 - 4.26i)T + (-21.5 - 37.2i)T^{2}$$
47 $$1 + 0.425T + 47T^{2}$$
53 $$1 + (-0.466 + 0.807i)T + (-26.5 - 45.8i)T^{2}$$
59 $$1 - 6.05T + 59T^{2}$$
61 $$1 - 10.2T + 61T^{2}$$
67 $$1 - 9.41T + 67T^{2}$$
71 $$1 - 8.46T + 71T^{2}$$
73 $$1 + (-6.82 + 11.8i)T + (-36.5 - 63.2i)T^{2}$$
79 $$1 - 5.53T + 79T^{2}$$
83 $$1 + (8.03 - 13.9i)T + (-41.5 - 71.8i)T^{2}$$
89 $$1 + (-6.03 - 10.4i)T + (-44.5 + 77.0i)T^{2}$$
97 $$1 + (5.86 - 10.1i)T + (-48.5 - 84.0i)T^{2}$$
show more
show less
$$L(s) = \displaystyle\prod_p \ \prod_{j=1}^{2} (1 - \alpha_{j,p}\, p^{-s})^{-1}$$
## Imaginary part of the first few zeros on the critical line
−8.628317788718120007347907323948, −8.213368988907090558903517256249, −6.88382279278991967690795095268, −6.46806836711530400555663618066, −5.50937679731944989020605522901, −4.93920004839525798124516015730, −3.92761323715287308960034876335, −2.88601056113741497118487344950, −2.03308433799188136229684796261, −0.69351962845897691422040194358, 0.957606770476427571434360764802, 2.23801183385083374500013323349, 3.30841242458757638951994098681, 3.83173388331015050974927654251, 4.98680236979219005615160673331, 5.94001873669171304345004780330, 6.50001479419567640904914378395, 7.21151890096084207975062039611, 8.063982577819911947207403597431, 8.650635573293036320314445384216 | 1,857 | 3,530 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-45 | latest | en | 0.399668 |
http://slidegur.com/doc/109587/slides | 1,521,576,419,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647530.92/warc/CC-MAIN-20180320185657-20180320205657-00648.warc.gz | 263,971,552 | 8,846 | ### Slides
```ME 475/675 Introduction to
Combustion
Lecture 11
Announcements
• Midterm 1
• September 29, 2014
• Review Friday, September 26
• HW 5 Due Friday, September 26, 2014
Chapter 3 Introduction to Mass Transfer
xx
x xx x
x xx
x
x
Mass 1Fraction
Y
0-
x x
x x
x x
x
x
x
x
x
x
x
oo o o
x
x
x xx x
xx
o
x x x x xx
o
o
o
o o
x
o o o o o oo
x
o
o o
"
Yx
o
oo
o
oo
o
o
o
Yo
• Consider two species, x and o
• Concentration of “x” is larger on the left, of “o” is larger on the right
• Species diffuse through each other
•
•
•
•
they move from regions of high to low concentrations
Think of perfume in a room
Mass flux is driven by concentration difference
Analogously, heat transfer is driven by temperature differences
• There may also be bulk motion of the mixture (advection, like wind)
• Total rate of mass flux:
"
=
"
+
"
∗2
(sum of component mass flux)
x
Chapter 3 Introduction to Mass Transfer
xx
x xx x
x xx
x
x
Mass
Fraction
Y
x x
x x
x x
x
x
x
x
x
x
x
oo o o
x
x
x xx x
xx
o
x x x x xx
o
o
o
o o
x
o o o o o oo
x
o
o o
Yx
"
• Rate of mass flux of “x” in the direction
•
"
"
−
= +
(Bulk Motion) Diffusion (due to concentration gradient)
• " = " + "
• =Diffusion
coefficient
of x through o
3
2
• Units 2
=
• Appendix D, pp. 707-9
• For gases, book shows that ~
1
2
0
o
oo
o
oo
o
o
o
Yo
x
Stefan Problem (no reaction)
x
L-
• One dimensional tube (Cartesian)
,∞
• Gas B is stationary: " = 0
• Gas A moves upward " > 0
YB
• Want to find this
YA
Y
,
B+A
A
•
"
=
• " 1 −
•
+
+ −
= −
"
,∞ −
−
=
;
=
0
1−
, 1−
"
"
"
• = but treat as constant
•
"
= ln
1−,∞
1−,
Mass Flux of evaporating liquid A
8
6.908
•
"
=
1−,∞
ln
1−,
• For ,∞ = 0
•
"
=
1
ln
1−,
(dimensionless)
6
"
m ( Y) 4
• " increases slowly for small ,
• Then very rapidly for , > 0.95
• What is the shape of the versus x
profile?
2
0
0
0
0
0.2
0.4
0.6
,
Y
0.8
1
Profile Shape
1
•
"
0
•
"
=
• but
0.99
"
, =0.9
• Ratio: =
YA ( x .05)
YA ( x .1)
YA
( x.5)
•
0.6
, =0.5
YA ( x .9)
YA ( x .99)
•
0.4
=
1− ()
ln
1−,
, =0.99
0.8
() −
,
1−
ln
= ln
1−,∞
1−,
1− ()
1−,
1−,∞
ln
1−,∞
ln
1−,
1−,∞
1−,
;
1−,
= ln
=
1−,∞
1−,
, =0.1
0
• = 1 − 1 − ,
, =0.05
0
0
0.2
1−,∞
1−,
• For ,∞ = 0
0.2
0.4
x
0.6
0.8
1
1− ()
1−,
1− ()
1−,
• = 1 − 1 − ,
0
= ln
1
1−,
• Large , profiles exhibit a boundary layer near
Liquid-Vapor Interface Boundary Condition
•
A+B
Vapor
,
Liquid
A
•
•
At interface need , =
=
=
+
=
=
= + 1 −
1
So , =
=
1
+ 1−
1+
−1
• =
=
,
• , = () Saturation pressure at temperature T
• For water, tables in thermodynamics textbook
• Or use Clausius-Slapeyron Equation (page 18 eqn. 2.19)
Clausius-Clapeyron Equation (page 18)
• Relates saturation pressure at a given temperature to the saturation
conditions at another temperature and pressure
•
•
•
2
1
=
ℎ
2
2
1
ℎ 1
1 2
1
=
−
=
−
1
1
2
ℎ
2
1
1
2 =
=
1
2
ln
1
ℎ
−
1
2
• If given 1 , 1 , ℎ 2 , we can use this to find 1
• Page 701, Table B: ℎ , = at P = 1 atm
Problem 3.9
• Consider liquid n-hexane in a 50-mm-diameter graduated cylinder.
Air blows across the top of the cylinder. The distance from the liquidair interface to the open end of the cylinder is 20 cm. Assume the
diffusivity of n-hexane is 8.8x10-6 m2/s. The liquid n-hexane is at 25C.
Estimate the evaporation rate of the n-hexane. (Hint: review the
Clausius-Clapeyron relation a applied in Example 3.1)
Stefan Problem (no reaction)
x
L-
• One dimensional tube (Cartesian)
,∞
• Gas B is stationary " = 0
YA
• but has a concentration gradient
YB
• Diffusion of B down = advection up
Y
YA,i
•
"
•
"
=0=
=
"
+
"
−
• " =
•
•
"
"
=
=
"
;
0
=
()
ln
; ()=,
,
()
,
"
``` | 1,572 | 3,599 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-13 | latest | en | 0.782176 |
https://solvethat.wordpress.com/2014/04/ | 1,516,793,154,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084894125.99/warc/CC-MAIN-20180124105939-20180124125939-00152.warc.gz | 810,865,660 | 20,096 | # Remove elements from std::vector
The title of the post would suggest an answer that is just obvious and discourage you from reading it. Unfortunately I meet very often production code, that removes elements from a vector according to a certain predicate, that is written in a horrible way.
### Problem definition
The problem starts with a vector of certain objects, for simplicity let us assume we are considering integers. Then we are given a predicate saying that we would like to clean the vector from the values meeting certain condition. In our case we could remove all even numbers, or prime numbers, or numbers with the sum of digits equal 9, or whatever your imagination brings. In the real world the problem could be more sophisticated and we could remove all the connections that were disconnected and there is no reason to maintain them further.
### How it usually looks like
What programmers usually do, they iterate over the collection and check every element if it meets a certain condition. If no, the element should be removed. This way of thinking is not bad, however when it comes to implementation, the things usually go wrong. Here is what most of the developers I know usually do:
```for(auto i = v.begin(); i != v.end(): i++)
{
if(predicate(*i))
{
i = v.erase(i);
}
}
```
Now, you could ask what is wrong with this code. I would answer with another question: what is the complexity? At first glance it might look like it is O(n), but when you think what erase actually does, it appears that it is actually O(n^2). The erase operation is O(n) costly as it has to rearrange the vector after removing elements. We are even informed about this, when we take a more careful look at the standard:
Because vectors use an array as their underlying storage, erasing elements in positions other than the vector end causes the container to relocate all the elements after the segment erased to their new positions. This is generally an inefficient operation compared to the one performed for the same operation by other kinds of sequence containers (such as list or forward_list).
To give you some intuition about what happens, let us imagine the vector that looks like following
```[ 5 | 4 | 7 | 8 | 1 | 3 | 6 ]
```
We want to remove the element 8. When this happens, all the elements on the right hand from 8 has to be shifted left.
```[ 5 | 4 | 7 | x | 1 | 3 | 6 ]
[ 5 | 4 | 7 | 1 | x | 3 | 6 ]
[ 5 | 4 | 7 | 1 | 3 | x | 6 ]
[ 5 | 4 | 7 | 1 | 3 | 6]
```
What is O(n) operation. Considering the outer loop we fall into O(n^2) complexity, that is not acceptable as long as we may get something better and in this case we can.
### How it should look like
The STL library provides you a set of algorithms that make the operations on the containers easier. One of them is the remove_if operation, that accepts two iterators which indicate the range that should be considered, and the mentioned predicate. The remove_if approach is way more clever than the loop/erase one. L What the specification says about remove_if is:
The removal is done by replacing the elements for which pred returns true by the next element for which it does not, and signaling the new size of the shortened range by returning an iterator to the element that should be considered its new past-the-end element.
To clarify it even more, let us take a look at the example. Let us consider the vector from the previous example and see how remove_if would work when the predicate is “remove if the number is divisible by 4”.
```[ 5 | 4 | 7 | 8 | 1 | 3 | 6 ]
[ 5 | 7 | 1 | 3 | 6 | 4 | 8 ]
-------------------- xxxxxxxx
```
As we can see the elements that meet the predicate have been moved to the end of the array. Secondly the remove_if returns the iterator indicating the position after “the last good element” (6 in this case), that we may use to remove the range of “bad elements”. Before we will do it, it is worth to mention that the order of “good elements” is maintained, but it is not necessary the case for “bad elements”.
Finally we could get a code that looks like follows:
```v.erase(
remove_if(v.begin(), v.end(), [](const int & in){return i % 4 == 0;}),
v.end());
```
Yes, we still use the erase method, but this time we are removing from the end of the vector so there is no necessity of rearrangement the elements and the erase operation is O(1). In this case the complexity would be O(n), what is quite satisfactory in comparison to O(n^2).
### Summary
I think that this article clearly shows how important it is, to understand well the mechanism behind Standard Template Library. It is a really powerful tool and knowing how to use it properly makes your life way easier and your programs way faster.
# Gesture Recognition System
### Introduction
The article below presents the results of one of the parts of my master thesis. The purpose of the thesis was to research methods that are applicable for move tracking and gesture recognition and a try of building a working system. Firstly the idea of the study was to apply gathered knowledge for sign-language translation. Unfortunately the lack of time (4 months) and hardware limitation allowed only for proof of concept’s implementation. Although the system is good enough for human-computer interaction.
### Assumptions
Unfortunately the real world is to complicated to be described well within a computer application. A common technique while implementing complex systems is to make assumptions that simplify the real problem to a simpler one. Also in this case I made some simplification that limited users behavior but allowed me to build a working system within reasonable time. Therefore the main assumption I did:
– Only one object moves at time.
– We consider only moves in 2D plane.
– Constant source of light.
– Camera doesn’t move.
The first assumption explains by itself. Just to be precise, having just one moving objects removes the necessity of deciding what is the object that actually needs to be traced, or tracing numbers of objects. The second one tells that depth is not considered. A very important assumptions was the last one. As the segmentation was colored based and I was working only with a very color sensitive web-camera I couldn’t allow on working with sun-light. Even little changes of the light angle or saturation changes the color perceived by the camera too much.
The assumptions are just some abstractions that are used mainly for simplifying the image-processing part and they could be chosen freely as long as they don’t violate the system specification. The main point of them is to skip some of the problems that are not the crucial part of the system, and concentrate on what really matters.
### Segmentation
The method I decided to use for segmentation uses the colors histogram approach. Before tracing starts, there is one extra preprocessing step (you may check it on the video), where I select the area to take a sample of colors to trace. The color value appearing more often in the sample has larger value in the histogram. Then I propagate back the histogram to the image, but this time in the grayscale. The bigger value of the color, the more white it becomes on the output image (the other window on the left). It is worth mentioning that in this case of original image I used HSV model rather than RGB.
Picture 1. Click to see the movie presenting segmentation.
The second step of the segmentation is to abandon elements that are not interesting for the application. Even if we made an assumption that there is only one object moving at time, it’s almost impossible to keep your head not moving at all. It introduces some problems as it has similar color histogram as hands and may confusing the tracing algorithm. Moreover using complex head/face detection approaches using boosting approaches didn’t make sense as they are computationally expensive and I simply didn’t need them. Nevertheless the head was not the only problem. There could be some other object with matching color histogram in the area. Luckily the solution described below solved the issue.
What was done in this case, based on the observation that heads moves way slower than hands (in the best case doesn’t move at all). Thus the consecutive frames are compared and pixels that don’t change much are abandoned. The comparison is done by substracting pixels coming from different frames and comparing them to the threshold defined on the base of experiments.
### Tracing
The tracing algorithm has to met some requirements that are coming directly from the nature of the moves: they don’t have to be continues and linear. This makes the tracing problem a little bit more complicated as in case of instant change of the direction, the tracing algorithm has to notice it quickly and not to loose the followed object. Another thing is that the tracing object might be occluded by some other objects. In our case this problem occurs when we move hand above head. We have eliminated this problem partially in the previous step, but it was not sufficient – we managed not to trace the head, but there were difficulties of losing the hand when it appeared over the head.
Picture 2. Click to see the movie presenting tracing.
After the SLR (Systematic Literature Review) as the first part of the thesis, I decided to apply two strategies described in the literature: Meanshift algorithm for tracing and Kalman Filter for predicting the position of the hand in the next frame.
Meanshift algorithm has been applied since 1998 in tracking problems successfully. The formal basis of the method was described by Fukunaga and Comaniciu at [1]. However the main idea might be described as follows:
1. Choose the window
a) set the initial position of the window
b) choose the density function type
c) choose the shape of the window
d) choose the size of the window.
2. Count the center of mass.
3. Move the window to the center of mass.
4. Move back to point 2.
As we can expect I associated the back-propagation method with the Meanshift algorithm – the window is moved to the point that is the center of the window, that suits the color histogram the most.
Kalman Filter was firstly described in 1960. The precise explanation of the method might be found in Welsh and Bishop [2] paper. The point of the approach is to keep the information about previous measures and maximize the a-posteriori probability of the position’s prediction. But instead of keeping a long list of previous measures we are creating a model that is iteratively updated.
### Gesture representation
Picture 3. The shape recognition step. Click to watch the movie.
The representation of a gesture is a sequence of directions and shapes. A direction is represented using the idea taken from the compass rose with the eight principal winds, except the winds are vectors outgoing from the center. Every direction has assigned a number from 1 to 8 and describes the direction of the last move.
The shape is one of the predefined shapes of the hand. The system gathers the contours of the hand and uses a SVM (Support Vector Machine) classifier to obtain related label. The contours were represented using Freeman Chain Codes.
Finally a gesture could look like a sequence of pairs [1,’A’], [1, ‘B’] , [2, ‘B’], [, ‘STOP’]. Notice that when the ‘STOP’ shape is met, the direction doesn’t matter.
### Recognition
Hidden Markov Model has been applied for gesture recognition. It was trained using some set of predefined actions that were then translated to a language word. Every sequence of movements was ending with a special move meaning “end of gesture”. It simplified the processing a lot, but it is not a very good practice and “don’t try it at home” 😉 The results might be seen on the last video.
### Training
The training was split into two parts. Firstly I was learning SVM just for the shape recognition. When it was done, I approached the second step of learning the HMMs using outputs from SVMto describe the shape.
### Conclusion
Even if the system I implemented was very limited, it proved the concepts of the features choice and the tracing algorithms as strong enough for a human-computer interaction. Nevertheless the segmentation method has to be improved as the current one is a home-made solution and the assumptions it takes are too strong. Secondly the language processing shouldn’t require the “stop word” and should be done fully dynamically.
### References
[1]. K. Fukunaga, “Introduction to Statistical Pattern Recognition”, Boston, Academic Press, 1990
[2]. G. Welsh, G. Bishop “An introduction to Kalman filter” (Technical Report TR95-041), University of North Carolina, Chapel Hill, NC, 1995
[3]. D. Comaniciu and P. Meer, “Mean shift analysis and applications”, IEEE Internation Conference on Computer Vision (vol 2, p. 1197), 1999. | 2,743 | 12,846 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-05 | longest | en | 0.955408 |
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Textbooks / Math / Saxon Math, Course 1 1
# Saxon Math, Course 1 1st Edition - Solutions by Chapter
## Full solutions for Saxon Math, Course 1 | 1st Edition
ISBN: 9781591417835
Saxon Math, Course 1 | 1st Edition - Solutions by Chapter
Solutions by Chapter
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##### ISBN: 9781591417835
This expansive textbook survival guide covers the following chapters: 120. Saxon Math, Course 1 was written by and is associated to the ISBN: 9781591417835. This textbook survival guide was created for the textbook: Saxon Math, Course 1, edition: 1. The full step-by-step solution to problem in Saxon Math, Course 1 were answered by , our top Math solution expert on 03/16/18, 04:35PM. Since problems from 120 chapters in Saxon Math, Course 1 have been answered, more than 59985 students have viewed full step-by-step answer.
Key Math Terms and definitions covered in this textbook
• Back substitution.
Upper triangular systems are solved in reverse order Xn to Xl.
• Change of basis matrix M.
The old basis vectors v j are combinations L mij Wi of the new basis vectors. The coordinates of CI VI + ... + cnvn = dl wI + ... + dn Wn are related by d = M c. (For n = 2 set VI = mll WI +m21 W2, V2 = m12WI +m22w2.)
• Companion matrix.
Put CI, ... ,Cn in row n and put n - 1 ones just above the main diagonal. Then det(A - AI) = ±(CI + c2A + C3A 2 + .•. + cnA n-l - An).
• Elimination.
A sequence of row operations that reduces A to an upper triangular U or to the reduced form R = rref(A). Then A = LU with multipliers eO in L, or P A = L U with row exchanges in P, or E A = R with an invertible E.
• Hankel matrix H.
Constant along each antidiagonal; hij depends on i + j.
• Least squares solution X.
The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b - Ax is orthogonal to all columns of A.
• Left nullspace N (AT).
Nullspace of AT = "left nullspace" of A because y T A = OT.
• Multiplication Ax
= Xl (column 1) + ... + xn(column n) = combination of columns.
• Multiplicities AM and G M.
The algebraic multiplicity A M of A is the number of times A appears as a root of det(A - AI) = O. The geometric multiplicity GM is the number of independent eigenvectors for A (= dimension of the eigenspace).
• Norm
IIA II. The ".e 2 norm" of A is the maximum ratio II Ax II/l1x II = O"max· Then II Ax II < IIAllllxll and IIABII < IIAIIIIBII and IIA + BII < IIAII + IIBII. Frobenius norm IIAII} = L La~. The.e 1 and.e oo norms are largest column and row sums of laij I.
• Projection matrix P onto subspace S.
Projection p = P b is the closest point to b in S, error e = b - Pb is perpendicularto S. p 2 = P = pT, eigenvalues are 1 or 0, eigenvectors are in S or S...L. If columns of A = basis for S then P = A (AT A) -1 AT.
• Rayleigh quotient q (x) = X T Ax I x T x for symmetric A: Amin < q (x) < Amax.
Those extremes are reached at the eigenvectors x for Amin(A) and Amax(A).
• Right inverse A+.
If A has full row rank m, then A+ = AT(AAT)-l has AA+ = 1m.
• Schwarz inequality
Iv·wl < IIvll IIwll.Then IvTAwl2 < (vT Av)(wT Aw) for pos def A.
• Semidefinite matrix A.
(Positive) semidefinite: all x T Ax > 0, all A > 0; A = any RT R.
• Simplex method for linear programming.
The minimum cost vector x * is found by moving from comer to lower cost comer along the edges of the feasible set (where the constraints Ax = b and x > 0 are satisfied). Minimum cost at a comer!
• Stiffness matrix
If x gives the movements of the nodes, K x gives the internal forces. K = ATe A where C has spring constants from Hooke's Law and Ax = stretching.
• Symmetric factorizations A = LDLT and A = QAQT.
Signs in A = signs in D.
• Vector space V.
Set of vectors such that all combinations cv + d w remain within V. Eight required rules are given in Section 3.1 for scalars c, d and vectors v, w.
• Vector v in Rn.
Sequence of n real numbers v = (VI, ... , Vn) = point in Rn. | 1,182 | 4,024 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-21 | latest | en | 0.863377 |
http://mariaevert.dk/vba/?paged=2 | 1,620,739,615,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989614.9/warc/CC-MAIN-20210511122905-20210511152905-00048.warc.gz | 36,331,713 | 7,153 | ### VBA – Loop through arrays
This post shows how to define an array, and to loop through each of the items in it. So many more things can be done with arrays (resizing, adding, deleting items, etc.) but for now I’ll just show how to loop through them.. that’s always useful.
```
Dim myArray As Variant
Dim x As Integer
myArray = Array(34610, 92105, 92263, 94121) 'define array
For x = LBound(myArray) To UBound(myArray) 'define start and end of array
MsgBox (myArray(x))
Next x ' Loop!
```
Thursday, December 9th, 2010 Comments Off on VBA – Loop through arrays
### Excel formula – Miscellaneous
I recently had to work with Excel formulas, and I encountered a small challenge. The problem was this: I had a table with variating height (the user could add and delete rows), that was likely to sometimes have 0 rows.Also, it wasn’t just the sum of the whole column I needed, as the SUM value should be at the bottom of the table, and a circular loop should be avoided. Therefore, the usual SUM() didn’t work.
I did not want to use VBA for this problem, so I turned to this solution:
Example: Column A: Names – Column B: Salary
1) Find the last cells that isn’t empty (in column A).
2) Make a SUM()-formula in column B that can take the row number from 1) as a parameter.
1) (Danish)
```=(SUMPRODUKT(MAKS((A4:A65003<>"")*RÆKKE(A4:A65003))))
```
(English)
```=(SUMPRODUCT(MAX((A4:A65003<>"")*ROW(A4:A65003))))
```
To make the solution understandable, I will enter the above formula in cell G1. Now, I have defined the last row(G1), and I just need a fixed first row, which I define as 2: I need to use the SUM() and the INDEKS() to sum the variable amount of rows:
(Danish)
```=SUM(INDEKS(B:B;2):INDEKS(B:B;G1))
```
(English)
```=SUM(INDEX(B:B;2):INDEX(B:B;G1))
```
Of course, put togther, it looks like this:
```=SUM(INDEKS(B:B;2):INDEKS(B:B;(=(SUMPRODUKT(MAKS((A4:A65003<>"")*RÆKKE(A4:A65003)))) )))
```
Wednesday, December 8th, 2010 Excel Comments Off on Excel formula – Miscellaneous
### VBA – Delete all files in a folder
This code snippet shows how you can delete all files in a given folder in a VBA application.
```Sub deleteFiles()
dim myPath
myFolder = "C:\MyFolder1\Myfolder2"
Set Fso = CreateObject("Scripting.FileSystemObject") ' Get a File object to query.
Set Fldr = Fso.GetFolder(myFolder)
For Each Filename In Fldr.Files
Filename.Delete True ' delete all files
Next
End Sub
```
That’s it!
Wednesday, December 8th, 2010 VBA Comments Off on VBA – Delete all files in a folder
### VBA – Loop through sheets
This code snippet can be used if you want to loop thorugh the sheets in your workbook, either because you want to add something to everysheet, or – as in the example – you want to delete sheets with a specific name.
```Sub slet_Faner()
Application.DisplayAlerts = False ' Makes it unnecessary for the user to approve the deletion
Dim ws As Worksheet
For Each ws In Worksheets
If ws.Name = "Home" Then ws.Delete 'Delete if name of sheet is "Home"
Next
End Sub
```
Of course, if you want all sheets BUT the one sheet with a specific name, you just use:
```Dim ws As Worksheet
For Each ws In Worksheets
If ws.Name <> "Home" Then ws.Delete 'Delete if name of sheet ISN'T "Home"
Next
```
Wednesday, December 8th, 2010 VBA Comments Off on VBA – Loop through sheets
### VBA – Define PageSetup (PaperSize, Orientation, etc.) and Print Excel Sheet
This posts explains how to print an Excel page using VBA and `PrintOut`. Normally, you will want to define the `PageSetup `first, in order to make sure that the right printer, the right paper size (fx A4), orientation (landscape or portrait), etc. is selected.
Step 1: Define `PageSetup`.
```
With Worksheets("MyPage").PageSetup
.Orientation = xlPortrait
.Zoom = False
.FitToPagesWide = 1
.FitToPagesTall = 1
.PaperSize = xlPaperA4
End With
```
It’s important to add the `Zoom` property in adition to the `FitTo...` properties, as you can otherwise not be sure that the printing in fact is restricted to 1×1 page.
Step 2: Print the page.
```Sheets("MyPage").PrintOut Copies:=1, Collate:=True
```
And that’s it!
Friday, November 26th, 2010 UserForm Comments Off on VBA – Define PageSetup (PaperSize, Orientation, etc.) and Print Excel Sheet
### VBA UserForm – How to automatically switch to next Textbox
The code-snippet presented in this very first post on my blog is helpful when the user has to enter data in TextBoxes in a UserForm, using VBA for Excel.
What we want to achieve is that whenever the user has entered 6 characters in `TextBox1`, then `TextBox2 `is selected by default. That way, the user does not need to used either TAB or the mouse to select the next TextBox:
We will need to create a sub that is called whenever a `Change` is made to `TextBox1`. We need to use the `TextLength` and the `SetFocus`, like shown below:
```Private Sub TextBox1_Change()
'whenever there are 6 characters registered in the TextBox
If TextBox1.TextLength = 6 Then
'select (SetFocus) the next TextBox
TextBox2.SetFocus
End If
```
Of course, the code-snippet has to be added in the code-behind of the UserForm. Make sure that you have build your Sub correctly by verifying that you have chosen `TextBox1 `and `Change`, and not just `General` at the top of the page:
Friday, November 26th, 2010 Comments Off on VBA UserForm – How to automatically switch to next Textbox
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