url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://socratic.org/questions/how-do-you-write-2-4i-in-trigonometric-form | 1,582,222,935,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145260.40/warc/CC-MAIN-20200220162309-20200220192309-00310.warc.gz | 536,613,508 | 6,126 | # How do you write 2 +4i in trigonometric form?
Apr 12, 2018
$2 \sqrt{5} \left(\cos \left(1.107\right) + i \sin \left(1.107\right)\right)$
#### Explanation:
$\text{to convert to trigonometric form}$
$\text{that is "x+yitor(costheta+isintheta)" using}$
•color(white)(x)r=sqrt(x^2+y^2)
•color(white)(x)theta=tan^-1(y/x);-pi < theta <=pi
$\text{here "x=2" and } y = 4$
$\Rightarrow r = \sqrt{{2}^{2} + {4}^{2}} = \sqrt{20} = 2 \sqrt{5}$
$\theta = {\tan}^{-} 1 \left(2\right) = 1.107 \text{ radians to 3 dec. places}$
$\Rightarrow 2 + 4 i = 2 \sqrt{5} \left(\cos \left(1.107\right) + i \sin \left(1.107\right)\right)$ | 261 | 624 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2020-10 | latest | en | 0.389087 |
https://www.12000.org/my_notes/kamek/mma_12_1_maple_2020/KERNELsubsection28.htm | 1,721,643,444,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517846.73/warc/CC-MAIN-20240722095039-20240722125039-00788.warc.gz | 535,657,018 | 3,943 | #### 2.28 ODE No. 28
$x^3 (-y(x))+y'(x)+x y(x)^2-2 x=0$ Mathematica : cpu = 0.114073 (sec), leaf count = 96
$\left \{\left \{y(x)\to \frac {\frac {1}{2} \sqrt {\pi } e^{\frac {x^4}{4}} x^3 \text {erf}\left (\frac {x^2}{2}\right )+c_1 e^{\frac {x^4}{4}} x^3+x}{x \left (\frac {1}{2} \sqrt {\pi } e^{\frac {x^4}{4}} \text {erf}\left (\frac {x^2}{2}\right )+c_1 e^{\frac {x^4}{4}}\right )}\right \}\right \}$ Maple : cpu = 0.077 (sec), leaf count = 51
$\left \{ y \left ( x \right ) ={\frac {1}{\sqrt {\pi }} \left ( {\it Erf} \left ( {\frac {{x}^{2}}{2}} \right ) \sqrt {\pi }{\it \_C1}\,{x}^{2}+{x}^{2}\sqrt {\pi }+2\,{{\rm e}^{-1/4\,{x}^{4}}}{\it \_C1} \right ) \left ( {\it Erf} \left ( {\frac {{x}^{2}}{2}} \right ) {\it \_C1}+1 \right ) ^{-1}} \right \}$
Hand solution
\begin {align} y^{\prime }-yx^{3}+xy^{2}-2x & =0\nonumber \\ y^{\prime } & =2x+yx^{3}-xy^{2}\nonumber \\ & =P\left ( x\right ) +Q\left ( x\right ) y+R\left ( x\right ) y^{2}\tag {1} \end {align}
This is Riccati first order non-linear ODE with $$P\left ( x\right ) =2x,Q\left ( x\right ) =x^{3},R\left ( x\right ) =-x$$. We can convert Riccati to Bernoulli which is easier to solve using the substitution $$u=x^{2}-y$$ or $$y=x^{2}-u$$\begin {align*} u^{\prime } & =2x-y^{\prime }\\ & =2x-\left ( 2x+yx^{3}-xy^{2}\right ) \\ & =2x-\left ( 2x+\left ( x^{2}-u\right ) x^{3}-x\left ( x^{2}-u\right ) ^{2}\right ) \\ & =2x-\left ( 2x+\left ( x^{5}-ux^{3}\right ) -x\left ( x^{4}+u^{2}-2x^{2}u\right ) \right ) \\ u^{\prime } & =2x-\left ( 2x+\left ( x^{5}-ux^{3}\right ) -\left ( x^{5}+xu^{2}-2x^{3}u\right ) \right ) \\ & =2x-2x-\left ( x^{5}-ux^{3}\right ) +\left ( x^{5}+xu^{2}-2x^{3}u\right ) \\ & =-x^{5}+ux^{3}+x^{5}+xu^{2}-2x^{3}u\\ & =-ux^{3}+xu^{2} \end {align*}
This is of the form $$u^{\prime }=P\left ( x\right ) +Q\left ( x\right ) u+R\left ( x\right ) u^{2}$$ and since $$P\left ( x\right ) =0$$ then it is Bernoulli differential equation. (when $$P\left ( x\right ) \neq 0$$ and $$R\left ( x\right ) \neq 0$$ it is Riccati). To solve Bernoulli we always start by dividing by $$u^{2}$$$\frac {u^{\prime }}{u^{2}}=-\frac {1}{u}x^{3}+x$ Then we let $$\zeta =-\frac {1}{u}$$, hence $$\zeta ^{\prime }=\frac {u^{\prime }}{u^{2}}$$, therefore the above becomes\begin {align*} \zeta ^{\prime } & =x^{3}\zeta +x\\ \zeta ^{\prime }-x^{3}\zeta & =x \end {align*}
Integrating factor is $$e^{-\int x^{3}dx}=e^{-\frac {x^{4}}{4}}$$, hence $d\left ( e^{-\frac {x^{4}}{4}}\zeta \right ) =xe^{-\frac {x^{4}}{4}}$ Integrating both sides gives$e^{-\frac {x^{4}}{4}}\zeta =\int xe^{-\frac {x^{4}}{4}}dx+C$ $$\int xe^{-\frac {x^{4}}{4}}dx=\frac {\sqrt {\pi }}{2}\operatorname {erf}\left ( \frac {x^{2}}{2}\right )$$, hence from above\begin {align*} e^{-\frac {x^{4}}{4}}\zeta & =\frac {\sqrt {\pi }}{2}\operatorname {erf}\left ( \frac {x^{2}}{2}\right ) +C\\ \zeta & =e^{\frac {x^{4}}{4}}\left ( \frac {\sqrt {\pi }}{2}\operatorname {erf}\left ( \frac {x^{2}}{2}\right ) +C\right ) \end {align*}
Since $$\zeta =-\frac {1}{u}$$ then$u=-e^{-\frac {x^{4}}{4}}\left ( \frac {\sqrt {\pi }}{2}\operatorname {erf}\left ( \frac {x^{2}}{2}\right ) +C\right ) ^{-1}$
And since $$y=x^{2}-u$$ then\begin {align*} y & =x^{2}+e^{-\frac {x^{4}}{4}}\left ( \frac {\sqrt {\pi }}{2}\operatorname {erf}\left ( \frac {x^{2}}{2}\right ) +C\right ) ^{-1}\\ & =x^{2}+\frac {e^{-\frac {x^{4}}{4}}}{\frac {\sqrt {\pi }}{2}\operatorname {erf}\left ( \frac {x^{2}}{2}\right ) +C} \end {align*}
Verification
eq:=diff(y(x),x)+x*y(x)^2-x^3*y(x)-2*x = 0;
sol:=x^2+ exp(-x^4/4)/(_C1+ sqrt(Pi)/2*erf(x^2/2));
odetest(y(x)=sol,eq);
0 | 1,645 | 3,572 | {"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} | 4.03125 | 4 | CC-MAIN-2024-30 | latest | en | 0.435146 |
https://byjus.com/icse-selina-solution-concise-mathematics-class-6-chapter-8-hcf-and-lcm/ | 1,657,168,920,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104683683.99/warc/CC-MAIN-20220707033101-20220707063101-00171.warc.gz | 194,019,456 | 161,108 | # Selina Solutions Concise Mathematics Class 6 Chapter 8: HCF And LCM
Selina Solutions are useful resources for students to follow during examinations. Frequent practice of Selina Solutions helps students to gain expertise in Mathematics. The answers are well structured, as per the students’ understanding capacity in solving the questions with ease. Students who find Mathematics difficult are advised to practice Selina Solutions effectively to boost confidence and face the exams without any fear
HCF is the greatest factor common to any two or more natural numbers and LCM is the lowest common multiple of any two or more natural numbers. Step by step solutions helps students in clearing their doubts, which appear while solving textbook questions. Our subject experts have prepared the solutions with the aim to help students in scoring good marks in the final exams. Students can download the PDF of Selina Solutions Concise Mathematics Class 6 Chapter 8 HCF and LCM, from the available links below
## Selina Solutions Concise Mathematics Class 6 Chapter 8: HCF And LCM Download PDF
### Exercises of Selina Solutions Concise Mathematics Class 6 Chapter 8: HCF And LCM
Exercise 8(A) Solutions
Exercise 8(B) Solutions
Exercise 8(C) Solutions
## Access Selina Solutions Concise Mathematics Class 6 Chapter 8: HCF And LCM
#### Exercise 8(A) page No: 63
1. Write all the factors of:
(i) 15
(ii) 55
(iii) 48
(iv) 36
(v) 84
Solution:
(i) The factors of 15 are 1, 3, 5 and 15
∴ F(15) = 1, 3, 5 and 15
(ii) The factors of 55 are 1, 5, 11 and 55
∴F(55) = 1, 5, 11 and 55
(iii) The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48
∴F (48) = 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48
(iv) The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36
∴F (36) = 1, 2, 3, 4, 6, 9, 12, 18 and 36
(v) The factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84
∴ F (84) = 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84
2. Write all prime numbers:
(i) less than 25
(ii) between 15 and 35
(iii) between 8 and 76
Solutions:
(i) 2, 3, 5, 7, 11, 13, 17, 19 and 23 are the prime numbers less than 25
(ii) 17, 19, 23, 29 and 31 are the prime numbers between 15 and 35
(iii) 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71 and 73 are the prime numbers between 8 and 76
3. Write the prime numbers from:
(i) 5 to 45
(ii) 2 to 32
(iii) 8 to 48
(iv) 9 to 59
Solutions:
(i) The prime numbers from 5 to 45 are 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41 and 43
(ii) The prime numbers from 2 to 32 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31
(iii) The prime numbers from 8 to 48 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47
(iv) The prime numbers from 9 to 59 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53 and 59
4. Write the prime factors of:
(i) 16
(ii) 27
(iii) 35
(iv) 49
Solution:
(i)
∴Prime factor of 16 is 2
(ii)
∴Prime factor of 27 is 3
(iii)
∴Prime factors of 35 are 5 and 7
(iv)
∴Prime factor of 49 is 7
5. If Pn means prime factors of n, find:
(i) P6
(ii) P24
(iii) P50
(iv) P42
Solution:
(i)
∴The prime factors of 6 are 2 and 3
(ii)
∴The prime factors of 24 are 2 and 3
(iii)
∴The prime factors of 50 are 2 and 5
(iv)
∴The prime factors of 42 are 2, 3 and 7
#### Exercise 8(B) Page no: 66
1. Using the common factor method, find the H.C.F. of:
(i) 16 and 35
(ii) 25 and 20
(iii) 27 and 75
(iv) 8, 12 and 18
(v) 24, 36, 45 and 60
Solution:
(i) Common factors of 16 and 35 are as follows:
F (16) = 1, 2, 4, 8, 16
F (35) = 1, 5, 7, 35
The common factors between 16 and 35 = 1
∴The H.C.F. of 16 and 35 = 1
(ii) Common factors of 25 and 20 are as follows:
F (25) = 1, 5, 25
F (20) = 1, 2, 4, 5, 10, 20
The common factors between 25 and 20 = 1, 5
∴The H.C.F. of 25 and 20 = 5
(iii) Common factors between 27 and 75 are as follows:
F (27) = 1, 3, 9, 27
F (75) = 1, 3, 5, 15, 25, 75
The common factors between 27 and 75 = 1, 3
∴The H.C.F. of 27 and 75 = 3
(iv) Common factors between 8, 12 and 18 are as follows:
F (8) = 1, 2, 4, 8
F (12) = 1, 2, 3, 4, 6, 12
F (18) = 1, 2, 3, 6, 9, 18
Common factors between 8, 12 and 18 = 1, 2
∴The H.C.F. of 8, 12 and 18 = 2
(v) Common factors between 24, 36, 45 and 60 are as follows:
F (24) = 1, 2, 3, 4, 6, 8, 12, 24
F (36) = 1, 2, 3, 4, 6, 12, 18, 36
F (45) = 1, 3, 5, 9, 15, 45
F (60) = 1, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Common factors between 24, 36, 45, and 60 = 1, 3
∴The H.C.F. of 24, 36, 45 and 60 = 3
2. Using the prime factor method, find the H.C.F. of:
(i) 5 and 8
(ii) 24 and 49
(iii) 40, 60 and 80
(iv) 48, 84 and 88
(v) 12, 16 and 28
Solution:
(i) The prime factors of 5 and 8 are as follows:
P5 = 5
P8 = 2 × 2 × 2
No common prime factors between 5 and 8
Hence, H.C.F. of 5 and 8 = 1
(ii) The prime factors of 24 and 49 are as follows:
P24 = 2 × 2 × 2 × 3
P49 = 7 × 7
No common prime factors between 24 and 49
Hence, H.C.F. of 24 and 49 = 1
(iii) The prime factors of 40, 60 and 80 are as follows:
P40 = 2 × 2 × 2 × 5
P60 = 2 × 2 × 3 × 5
P80 = 2 × 2 × 2 × 2 × 5
Common prime factors between 40, 60 and 80 = 2 × 2 × 5
Hence, H.C.F. of 40, 60 and 80 = 20
(iv) The prime factors of 48, 84 and 88 are as follows:
P48 = 2 × 2 × 2 × 2 × 3
P84 = 2 × 2 × 3 × 7
P88 = 2 × 2 × 2 × 11
Common prime factors between 48, 84 and 88 = 2 × 2
Hence, H.C.F. of 48, 84 and 88 = 2 × 2 = 4
(v) The prime factors of 12, 16 and 28 are as follows:
P12 = 2 × 2 × 3
P16 = 2 × 2 × 2 × 2
P28 = 2 × 2 × 7
Common prime factors between 12, 16 and 28 = 2 × 2
Hence, H.C.F. of 12, 16 and 28 = 2 × 2 = 4
3. Using the division method, find the H.C.F. of the following:
(i) 16 and 24
(ii) 18 and 30
(iii) 7, 14 and 24
(iv) 70, 80, 120 and 150
(v) 32, 56 and 46
Solution:
(i) 16 and 24
Using division method, we get
Here, the last division is 8
Hence, H.C.F. of 16 and 24 = 8
(ii) 18 and 30
Using division method, we get
Here, last division is 6
Hence, H.C.F. of 18 and 30 = 6
(iii) 7, 14 and 24
Using division method, we get
Here, last division is 1
Hence, H.C.F. of 7, 14 and 24 = 1
(iv) 70, 80, 120 and 150
Using division method, we get
Here, last division is 10
Hence, H.C.F. of 70, 80, 120 and 150 = 10
(v) 32, 56 and 46
Using division method, we get
Here, last division is 2
Hence, H.C.F. of 32, 56 and 46 = 2
4. Use a method of your own choice to find the H.C.F. of:
(i) 45, 75 and 135
(ii) 48, 36 and 96
(iii) 66, 33 and 132
(iv) 24, 36, 60 and 132
(v) 30, 60, 90, and 105
Solution:
(i) 45, 75 and 135
The prime factors of 45, 75 and 135 are as follows:
P45 = 3 × 3 × 5
P75 = 3 × 5 × 5
P135 = 3 × 3 × 3 × 5
The common factors of 45, 75 and 135 = 3 × 5
∴H.C.F. of 45, 75 and 135 = 15
(ii) 48, 36 and 96
The prime factors of 48, 36 and 96 are as follows:
P48 = 2 × 2 × 2 × 2 × 3
P36 = 2 × 2 × 3 × 3
P96 = 2 × 2 × 2 × 2 × 2 × 3
The common factors of 48, 36 and 96 = 2 × 2 × 3
∴H.C.F. of 48, 36 and 96 = 12
(iii) 66, 33 and 132
The prime factors of 66, 33 and 132 are as follows:
P66 = 2 × 3 × 11
P33 = 3 × 11
P132 = 2 × 2 × 3 × 11
The common factors of 66, 33 and 132 = 3 × 11
∴H.C.F. of 66, 33 and 132 = 33
(iv) 24, 36, 60 and 132
The prime factors of 24, 36, 60 and 132 are as follows:
P24 = 2 × 2 × 2 × 3
P36 = 2 × 2 × 3 × 3
P60 = 2 × 2 × 3 × 5
P132 = 2 × 2 × 3 × 11
The common factors of 24, 36, 60 and 132 = 2 × 2 × 3
∴H.C.F. of 24, 36, 60 and 132 = 12
(v) 30, 60, 90, and 105
The prime factors of 30, 60, 90 and 105 are as follows:
P30 = 2 × 3 × 5
P60 = 2 × 2 × 3 × 5
P90 = 2 × 3 × 3 × 5
P105 = 3 × 5 × 7
The common factors of 30, 60, 90 and 105 = 3 × 5
∴H.C.F. of 30, 60, 90 and 105 = 15
5. Find the greatest number that divides each of 180, 225 and 315 completely.
Solution:
The greatest number that divides each of 180, 225 and 315 will be the H.C.F. of 180, 225 and 315
Using division method, the H.C.F of 180, 225 and 315 are shown below
Since last division is 45
∴H.C.F. of 180, 225 and 315 = 45
6. Show that 45 and 56 are co-prime numbers.
Solution:
The H.C.F. of two co-prime numbers is always 1
Using division method, the H.C.F. of 45 and 56 are shown below
Here, last division is 1
H.C.F. of 45 and 56 = 1
∴45 and 56 are co-prime numbers
7. Out of 15, 16, 21 and 28, find out all the pairs of co-prime numbers.
Solution:
15 and 16, 15 and 21, 15 and 28, 16 and 21, 16 and 28 and 21 and 28 will be the possible pairs
We know that the H.C.F. of two co-prime numbers is always 1
Using division method, the H.C.F. of pair 15 and 16 is given below
Hence, H.C.F. of 15 and 16 = 1
Using division method, the H.C.F. of pair 15 and 21 is given below
Hence, H.C.F. of 15 and 21= 3
Using division method, the H.C.F. of pair 15 and 28 is given below
Hence, H.C.F. of 15 and 28 = 1
Using division method, the H.C.F. of pair 16 and 21 is given below
Hence, H.C.F. of 16 and 21 = 1
Using division method, the H.C.F. of 16 and 28 is given below
Hence, H.C.F. of 16 and 28 = 4
Using division method, the H.C.F. of 21 and 28 is given below
Hence, H.C.F. of 21 and 28 = 7
∴ The pairs 15 and 16, 15 and 28 and 16 and 21 are co-prime numbers
8. Find the greatest number that will divide 93, 111 and 129, leaving remainder 3 in each case.
Solution:
First, decrease the leaving remainder 3 from numbers 93, 111 and 129 to find the required number
93 – 3 = 90
111 – 3 = 108
129 – 3 = 126
In each case, the H.C.F. of 90, 108 and 126 will be the greatest number that will divide 93, 111 and 129 leaving remainder 3
Using division method, the H.C.F. of 90, 108 and 126 is given below
Since, last division is 18
∴H.C.F. of 90, 108 and 126 is 18
Hence, the greatest number that will divide 93, 111 and 129 is 18
#### Exercise 8(C) page no: 71
1. Using the common multiple method, find the L.C.M. of the following:
(i) 8, 12 and 24
(ii) 10, 15 and 20
(iii) 3, 6, 9 and 12
Solution:
(i) 8, 12 and 24
We get,
L.C.M = 4 × 3 × 2
= 24
Hence, L.C.M. of 8, 12 and 24 = 24
(ii) 10, 15 and 20
We get,
L.C.M = 2 × 2 × 3 × 5
= 60
Hence, L.C.M. of 10, 15 and 20 = 60
(iii) 3, 6, 9 and 12
We get,
L.C.M. = 2 × 2 × 3 × 3
= 36
Hence, L.C.M. of 3, 6, 9 and 12 = 36
2. Find the L.C.M. of each of the following groups of numbers, using (i) the prime factor method and (ii) the common division method:
(i) 18, 24 and 96
(ii) 100, 150 and 200
(iii) 14, 21 and 98
(iv) 22, 121 and 33
(v) 34, 85 and 51
Solution:
(i) 18, 24 and 96
By using prime factor method, L.C.M. of 18, 24 and 96 are given below
Prime factors of 18 = 2 × 3 × 3
Prime factors of 24 = 2 × 2 × 2 × 3
Prime factors of 96 = 2 × 2 × 2 × 2 × 2 × 3
∴ L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3
= 288
By using common division method, L.C.M. of 18, 24 and 96 are given below
∴ L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3
= 288
(ii) 100, 150 and 200
By using prime factor method, L.C.M. of 100, 150 and 200 are given below
Prime factor of 100 = 2 × 2 × 5 × 5
Prime factor of 150 = 2 × 3 × 5 × 5
Prime factor of 200 = 2 × 2 × 2 × 5 × 5
∴ L.C.M. = 2 × 2 × 2 × 3 × 5 × 5
= 600
By using common division method, L.C.M. of 100, 150 and 200 are given below
∴L.C.M. = 2 × 2 × 2 × 3 × 5 × 5
= 600
(iii) 14, 21 and 98
By using prime factor method, L.C.M. of 14, 21 and 98 are given below
Prime factor of 14 = 2 × 7
Prime factor of 21 = 3 × 7
Prime factor of 98 = 2 × 7 × 7
∴ L.C.M. = 2 × 3 × 7 × 7
= 294
By using common division method, L.C.M. of 14, 21 and 98 are given below
∴ L.C.M. = 2 × 3 × 7 × 7
= 294
(iv) 22, 121 and 33
By using prime factor method, L.C.M. of 22, 121 and 33 are given below
Prime factor of 22 = 2 × 11
Prime factor of 121 = 11 × 11
Prime factor of 33 = 3 × 11
∴ L.C.M. = 2 × 3 × 11 × 11
= 726
By using common division method, L.C.M. of 22, 121 and 33 are given below
∴ L.C.M. = 2 × 3 × 11 × 11
= 726
(v) 34, 85 and 51
By using prime factor method, L.C.M. of 34, 85 and 51 are given below
Prime factor of 34 = 2 × 17
Prime factor of 85 = 5 × 17
Prime factor of 51 = 3 × 17
∴ L.C.M. = 2 × 3 × 5 × 17
= 510
By using common division method, L.C.M. of 34, 85 and 51 are given below
∴ L.C.M. = 2 × 3 × 5 × 17
= 510
3. The H.C.F. and the L.C.M. of two numbers are 50 and 300 respectively. If one of the numbers is 150, find the other one.
Solution:
Given
H.C.F. = 50
L.C.M. = 300
One number = 150
We know that,
Product of H.C.F. and L.C.M. of two numbers is equal to product of those two numbers
For other number,
50 × 300 = 150 × other number
15000 / 150 = other number
100 = other number
Hence, the other number is 100
4. The product of two numbers is 432 and their L.C.M. is 72. Find their H.C.F.
Solution:
Given
Product of two numbers = 432 and L.C.M.= 72
We know that,
Product of H.C.F. and L.C.M. of two numbers is equal to product of those two numbers.
Now, to find H.C.F
H.C.F. × 72 = 432
H.C.F.= 432 / 72
H.C.F. = 6
Hence, H.C.F. = 6
5. The product of two numbers is 19,200 and their H.C.F. is 40. Find their L.C.M.
Solution:
Given
Product of two numbers = 19200 and H.C.F. = 40
We know that,
Product of H.C.F. and L.C.M. of two numbers is equal to product of those two numbers
Now, to find L.C.M.
40 × L.C.M. = 19200
L.C.M. = 19200 / 40
L.C.M. = 480
Hence, L.C.M. = 480
6. Find the smallest number which, when divided by 12, 15, 18, 24 and 36 leaves no remainder.
Solution:
The given numbers L.C.M. will be the least number which is exactly divisible 12, 15, 18, 24 and 36 and leaves no remainder
L.C.M. = 2 × 2 × 2 × 3 × 3 × 5
= 360
Hence, smallest required number = 360
7. Find the smallest number which, when increased by one is exactly divisible by 12, 18, 24, 32 and 40.
Solution:
First, let us find out the L.C.M. of 12, 18, 24, 32 and 40
L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5
= 1440
This can be written as
= 1439 + 1
Hence, 1439 is the smallest number which, when increased by one is exactly divisible by the given numbers
8. Find the smallest number which, on being decreased by 3, is completely divisible by 18, 36, 32 and 27.
Solution:
First, let us solve for L.C.M. of 18, 36, 32 and 27
L.C.M. = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
= 864
This can be written as
= 867 – 3
Hence, 867 is the smallest number which, when decreased by 3 is exactly divisible by the given numbers | 6,141 | 14,492 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2022-27 | latest | en | 0.891872 |
https://physics.stackexchange.com/questions/770192/why-eddington-finkelstein-coordinates-for-the-vaidya-metric | 1,701,358,286,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100227.61/warc/CC-MAIN-20231130130218-20231130160218-00899.warc.gz | 518,778,829 | 41,962 | # Why Eddington-Finkelstein coordinates for the Vaidya metric?
I've seen that Vaidya metric does not use the Schwarzschild metric, but instead to get it you simply go to Eddington-Finkelstein coordinates and there put mass as a function of retarded time. From
$$ds^2=\left(1-\dfrac{2M}{r}\right)du^2+2dudr-r^2\left(d\theta^2+\sin^2{\theta}\right)$$
We can go to
$$ds^2=\left(1-\dfrac{2M\left(u\right)}{r}\right)du^2+2dudr-r^2\left(d\theta^2+\sin^2{\theta}\right)$$
Is it because the black hole is not the same after time translations, and therefore the $$g_{0i}$$ components of the metric would not be zero in Schwarzschild coordinates or is there another motivation?
• You can have it in Schwarzschild-like coordinates, but the function for the mass becomes a rather complicated function of time and radius in that case, see here. For an approximate metric of a Hawking radiating black hole where the source of the radiation is not exactly localized you can still use an approximation where the mass is only a function of time though and neglect the r-dependency, see here, although I wouldn't use it for the last stage of evaporation before it disintegrates. Jun 30 at 19:11
• If there is a component (mass) dependent on time, wouldn't $g_{0i}=g_{i0}$ components of the metric ($i\in\{1,2,3\}$) not necessarily be zero? See en.wikipedia.org/wiki/Derivation_of_the_Schwarzschild_solution. On the Wikipedia page, they consider static time in which after $t\rightarrow -t$ positive and negative components must be equal and therefore $e_te_r=\left(-e_t\right)e_r=0$. If the body is not time-static, wouldn't these components be non-zero? In what situations? (See: physics.stackexchange.com/questions/770183/…) Jun 30 at 22:55
• No, the axes can still be orthogonal so the cross terms can be 0. If you want your local observers not hovering at fixed spatial coordinates but instead have them for example free falling with v in the r direction like in Gullstrand Painlevé style coordinates set √(gₜᵣgᵗʳ)=v. For evaporating black holes we have the mass function dependend on t, not u, and the source of the radiation isn't localized anyway, so for that case the diagonalized form in the 2nd link of my 1st comment might be the best fit. Jul 3 at 1:26 | 610 | 2,252 | {"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": 3, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2023-50 | longest | en | 0.847238 |
https://javalab.org/en/newtons_cradle_en/ | 1,680,142,135,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949093.14/warc/CC-MAIN-20230330004340-20230330034340-00185.warc.gz | 362,977,522 | 18,111 | Newton's cradle consists of five iron balls, and each hung on two threads to prevent the ball from spinning.
Originally Newton's cradle was created to demonstrate Newton's third law. If you collide a ball from one side, the same impact returns through the other side.
## The process of movement
If you drop one marble, the fallen ball hitting other balls and stops completely. One ball on the other side immediately rises up at the same speed as the ball from which it fell. The upward ball rises to about the same height that the ball was initially pulled.
This indicates that the bounced ball has inherited most of the energy and momentum of the falling ball. The balls in the middle remain still and only transmit waves created by compression. During the propagation of the wave, some of the energy is lost as heat.
Assuming there is no energy loss, the number of balls that bounce is always equal to the number of dropped balls. That is, if you drop two balls, two balls bounce on the other side, and if you drop three balls, three balls bounce on the other side.
## When dropping a ball
Newton's cradle is described by the law of conservation of momentum (= mv) and kinetic energy (= 1/2 mv2) of an elastic body.
The collision can be explained simply if the two balls are of the same mass. The impacted object completely takes over the momentum and kinetic energy of the impacted object. In the case of a completely elastic body, no loss occurs due to heat and sound energy. A hard iron ball does not compress well, but it is elastic, so it does not cause energy loss and transfers energy efficiently.
## When two or more balls are dropped
Let's think about the process of lifting two balls and then dropping them toward the third. With a subtle difference, the second ball hits the third ball first. According to the law of collision of an elastic body, the third ball takes over momentum and kinetic energy and moves, and the second ball that has fallen stops. The first ball hits the second ball that has stopped. This process is repeated so that if you finally lift two balls and then drop them, the two opposite balls will bounce at the same speed.
## Phenomena caused by friction
In Newton's cradle, momentum and kinetic energy are reduced little by little because of friction. In fact, a ball falling from Newton's cradle is pushed back slightly by friction immediately after impact. When these effects accumulate, after time, all the balls turn into shaking little by little at the same time, eventually stopping the whole movement as well. | 526 | 2,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2023-14 | latest | en | 0.954932 |
http://docplayer.net/5828776-Chapter-6-learning-objectives-principles-used-in-this-chapter-1-annuities-2-perpetuities-3-complex-cash-flow-streams.html | 1,529,460,609,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863407.58/warc/CC-MAIN-20180620011502-20180620031502-00214.warc.gz | 89,193,637 | 29,320 | # Chapter 6. Learning Objectives Principles Used in This Chapter 1. Annuities 2. Perpetuities 3. Complex Cash Flow Streams
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1 Chapter 6 Learning Objectives Principles Used in This Chapter 1. Annuities 2. Perpetuities 3. Complex Cash Flow Streams 1. Distinguish between an ordinary annuity and an annuity due, and calculate present and future values of each. 2. Calculate the present value of a level perpetuity and a growing perpetuity. Need this for stock valuation 3. Calculate the present and future value of complex cash flow streams. Need this for bond valuation, capital budgeting 1
2 Principle 1: Money Has a Time Value. This chapter applies the time value of money concepts to annuities, perpetuities and complex cash flows. Principle 3: Cash Flows Are the Source of Value. This chapter introduces the idea that principle 1 and principle 3 will be combined to value stocks, bonds, and investment proposals. An annuity is a series of equal dollar payments that are made at the end of equidistant points in time such as monthly, quarterly, or annually over a finite period of time. If payments are made at the end of each period, the annuity is referred to as ordinary annuity. 2
3 Example 6.1 How much money will you accumulate by the end of year 10 if you deposit \$3,000 each for the next ten years in a savings account that earns 5% per year? We can determine the answer by using the equation for computing the future value of an ordinary annuity. FV = n FV of annuity at the end of nth period. PMT = annuity payment deposited or received at the end of each period i = interest rate per period n = number of periods for which annuity will last FV = \$3000 {[ (1+.05) 10-1] (.05)} = \$3,000 { [0.63] (.05) } = \$3,000 {12.58} = \$37,740 3
4 Using a Financial Calculator Enter N=10 1/y = 5.0 PV = 0 PMT = FV = \$37, Instead of figuring out how much money you will accumulate (i.e. FV), you may like to know how much you need to save each period (i.e. PMT) in order to accumulate a certain amount at the end of n years. In this case, we know the values of n, i, and FV n in equation 6-1c and we need to determine the value of PMT. Example 6.2: Suppose you would like to have \$25,000 saved 6 years from now to pay towards your down payment on a new house. If you are going to make equal annual end- of-year payments to an investment account that pays 7 per cent, how big do these annual payments need to be? 4
5 Here we know, FV = n \$25,000; n = 6; and i=7% and we need to determine PMT. \$25,000 = PMT {[ (1+.07) 6-1] (.07)} = PMT{ [.50] (.07) } = PMT {7.153} \$25, = PMT = \$3, Using a Financial Calculator. Enter N=6 1/y = 7 PV = 0 FV = PMT = -3,
6 Solving for an Ordinary Annuity Payment How much must you deposit in a savings account earning 8% annual interest in order to accumulate \$5,000 at the end of 10 years? Let s solve this problem using the mathematical formulas and a financial calculator. 6
7 If you can earn 12 percent on your investments, and you would like to accumulate \$100,000 for your child s education at the end of 18 years, how much must you invest annually to reach your goal? i=12% Years Cash flow PMT PMT PMT The FV of annuity for 18 years At 12% = \$100,000 We are solving for PMT 7
8 This is a future value of an annuity problem where we know the n, i, FV and we are solving for PMT. We will use equation 6-1c to solve the problem. Using the Mathematical Formula \$100,000 = PMT {[ (1+.12) 18-1]] (.12)}} = PMT{ [6.69] (.12) } = PMT {55.75} \$100, = PMT = \$1, Using a Financial Calculator. Enter N=18 1/y = 12.0 PV = 0 FV = PMT = -1,
9 If we contribute \$1, every year for 18 years, we should be able to reach our goal of accumulating \$100,000 if we earn a 12% return on our investments. Note the last payment of \$1, occurs at the end of year 18. In effect, the final payment does not have a chance to earn any interest. You can also solve for interest rate you must earn on your investment that will allow your savings to grow to a certain amount of money by a future date. In this case, we know the values of n, PMT, and FV n in equation 6-1c and we need to determine the value of i. Example 6.3: In 20 years, you are hoping to have saved \$100,000 towards your child s college education. If you are able to save \$2,500 at the end of each year for the next 20 years, what rate of return must you earn on your investments in order to achieve your goal? 9
10 Using the Mathematical Formula \$100,000 = \$2,500{[ (1+i) 20-1]] (i)}] 40 = {[ (1+i) 20-1] (i)} The only way to solve for i mathematically is by trial and error. That is why we solve using a calculator Using a Financial Calculator Enter N = 20 PMT = -\$2,500 FV = \$100, PV = \$0 i = 6.77 You may want to calculate the number of periods it will take for an annuity to reach a certain future value, given interest rate. l f b f d It is easier to solve for number of periods using financial calculator or excel, rather than mathematical formula. 10
11 Example 6.4: Suppose you are investing \$6,000 at the end of each year in an account that pays 5%. How long will it take before the account is worth \$50,000? Using a Financial Calculator Enter 1/y = 5.0 PV = 0 PMT = -6,000 FV = 50,000 N = 7.14 The present value of an ordinary annuity measures the value today of a stream of cash flows occurring in the future. 11
12 For example, we will compute the PV of ordinary annuity if we wish to answer the question: What is the value today or lump sum equivalent of receiving i \$3,000 every year for the next 30 years if the interest rate is 5%? Figure 6-2 shows the lump sum equivalent (\$2,106.18) of receiving \$500 per year for the next five years at an interest rate of 6%. 12
13 PMT = annuity payment deposited or received ed at the end of each period. i = discount rate (or interest rate) on a per period basis. n = number of periods for which the annuity will last. It is important that n and i match. If periods are expressed in terms of number of monthly payments, the interest rate must be expressed in terms of the interest rate per month. The Present Value of an Ordinary Annuity Your grandmother has offered to give you \$1,000 per year for the next 10 years. What is the present value of this 10-year, \$1,000 annuity discounted back to the present at 5 percent? Let s solve this using the mathematical formula, a financial calculator, and an Excel spreadsheet. 13
14 What is the present value of an annuity of \$10,000 to be received at the end of each year for 10 years given a 10 percent discount rate? 14
15 i=10% Years Cash flow \$10,000 \$10,000 \$10,000 Sum up the present Value of all the cash flows to find the PV of the annuity In this case we are trying to determine the present value of an annuity. We know the number of years (n), discount rate (i), dollar value received at the end of each year (PMT). We can use equation 6-2b to solve this problem. Using the Mathematical Formula PV = \$10,000 {[1-(1/(1.10) 10 ] (.10)} = \$10,000 {[ ] (.10)} = \$10,000 {6.145) = \$ 61,445 15
16 Using a Financial Calculator Enter N = 10 1/y = 10.0 PMT = -10, FV = 0 PV = 61, A lump sum or one time payment today of \$61,446 is equivalent to receiving \$10,000 every year for 10 years given a 10 percent discount rate. An amortized loan is a loan paid off in equal payments consequently, the loan payments are an annuity. Examples: Home mortgage loans, Auto loans 16
17 In an amortized loan, the present value can be thought of as the amount borrowed, n is the number of periods the loan lasts for, i is the interest rate per period, future value takes on zero because the loan will be paid of after n periods, and payment is the loan payment that is made. Example 6.5 Suppose you plan to get a \$9,000 loan from a furniture dealer at 18% annual interest with annual payments that you will pay off in over five years. What will your annual payments be on this loan? Using a Financial Calculator Enter N = 5 i/y = 18.0 PV = 9000 FV = 0 PMT = -\$2,
18 Year Amount Owed on Principal at the Beginning of the Year (1) Annuity Payment (2) Interest Portion of the Annuity (3) = (1) 18% Repayment of the Principal Portion of the Annuity (4) = (2) (3) Outstanding Loan Balance at Year end, After the Annuity Payment (5) =(1) (4) 1 \$9,000 \$2,878 \$1, \$1, \$7, \$7,742 \$2,878 \$1, \$1, \$6, \$ \$2,878 \$1, \$1, \$4, \$4, \$2,878 \$ \$2, \$2, \$2, \$2,878 \$ \$2, \$0.00 We can observe the following from the table: Size of each payment remains the same. However, Interest payment declines each year as the amount owed declines and more of the principal is repaid. Many loans such as auto and home loans require monthly payments. This requires converting n to number of months and computing the monthly interest rate. 18
19 Example 6.6 You have just found the perfect home. However, in order to buy it, you will need to take out a \$300,000, 30-year mortgage at an annual rate of 6 percent. What will your monthly mortgage payments be? Mathematical Formula Here annual interest rate =.06, number of years = 30, m=12, PV = \$300,000 \$300,000= PMT \$300,000 = PMT [166.79] 1-1/(1+.06/12) /12 \$300, = \$
20 Using a Financial Calculator Enter N=360 1/y =.5 PV = FV = 0 PMT = Determining the Outstanding Balance of a Loan Let s say that exactly ten years ago you took out a \$200,000, 30-year mortgage with an annual interest rate of 9 percent and monthly payments of \$1, But since you took out that loan, interest rates have dropped. You now have the opportunity to refinance your loan at an annual rate of 7 percent over 20 years. You need to know what the outstanding balance on your current loan is so you can take out a lower-interest-rate loan and pay it off. If you just made the 120th payment and have 240 payments remaining, what s your current loan balance? 20
21 Let s assume you took out a \$300,000, 30-year mortgage with an annual interest rate of 8%, and monthly payment of \$2, Since you have made 15 years worth of payments, there are 180 monthly payments left before your mortgage will be totally paid off. How much do you still owe on your mortgage? 21
22 i=(.08/12)% Years Cash flow PV \$2, \$2, \$2, We are solving for PV of 180 payments of \$2, Using a discount rate of 8%/12 You took out a 30-year mortgage of \$300,000 with an interest rate of 8% and monthly payment of \$2, Since you have made payments for 15-years (or 180 months), there are 180 payments left before the mortgage will be fully paid off. Step 2: Decide on a Solution Strategy The outstanding balance on the loan at anytime is equal to the present value of all the future monthly payments. Here we will use equation 6-2c to determine the present value of future payments for the remaining 15-years or 180 months. 22
23 Using Mathematical Formula Here annual interest rate =.09; number of years =15, m = 12, PMT = \$2, PV = \$2, = \$2, [104.64] 1-1/(1+.08/12) /12 = \$230, Using a Financial Calculator Enter N = 180 1/y =8/12 PMT = FV = 0 PV = \$230,
24 The amount you owe equals the present value of the remaining payments. Here we see that even after making payments for 15-years, you still owe around \$230,344 on the original i loan of \$300, Thus, most of the payment during the initial years goes towards the interest rather than the principal. Annuity due is an annuity in which all the cash flows occur at the beginning of the period. Rent payments on apartments are typically annuity due as rent is paid at the beginning of the month. Premium payments on insurance policies are typically annuity due since they are paid at the beginning of the month or beginning of the year. Computation of future value of an annuity due requires compounding the cash flows for one additional period, beyond an ordinary annuity. 24
25 Recall Example 6.1 where we calculated the future value of 10-year ordinary annuity of \$3,000 earning 5 per cent to be \$37,734. What will be the future value if the deposits of \$3,000 were made at the beginning of the year i.e. the cash flows were annuity due? FV = \$3000 {[ (1+.05) 10-1] (.05)} (1.05) = \$3,000 { [0.63] (.05) } (1.05) = \$3,000 {12.58}(1.05) = \$39,620 Since with annuity due, each cash flow is received one year earlier, its present value will be discounted back for one less period. 25
26 Recall checkpoint 6.2 Check yourself problem where we computed the PV of 10-year ordinary annuity of \$10,000 at a 10 percent discount rate to be equal to \$61,446. What will be the present value if \$10,000 is received at the beginning of each year i.e. the cash flows were annuity due? PVAD = \$10,000 {[1-(1/(1.10) 10 ] (.10)} (1.1) = \$10, {[ ] (.10)}(1.1) ) = \$10,000 {6.144) (1.1) = \$ 67,590 The examples illustrate that both the future value and present value of an annuity due are larger than that of an ordinary annuity because, in each case, all payments are received or paid earlier. 26
27 A perpetuity is an annuity that continues forever or has no maturity. For example, a dividend stream on a share of preferred stock. There are two basic types of perpetuities: Level el perpetuity in which the payments are constant rate from period to period. Growing perpetuity in which cash flows grow at a constant rate, g, from period to period. PV = the present value of a level perpetuity PMT = the constant dollar amount provided by the perpetuity i = the interest (or discount) rate per period 27
28 Example 6.6 What is the present value of \$600 perpetuity at 7% discount rate? PV = \$ = \$8, The Present Value of a Level Perpetuity What is the present value of a perpetuity of \$500 paid annually discounted back to the present at 8 percent? 28
29 What is the present value of stream of payments equal to \$90,000 paid annually and discounted back to the present at 9 percent? With a level perpetuity, a timeline goes on forever with the same cash flow occurring every period. i=9% Years Cash flows \$90,000 \$90,000 \$90,000 \$90,000 Present Value =? The \$90,000 cash flow go on forever 29
30 Present Value of Perpetuity can be solved easily using mathematical equation as given by equation 6-5. PV = \$90, = \$1,000,000 Here the present value of perpetuity is \$1,000,000. The present value of perpetuity is not affected by time. Thus, the perpetuity will be worth \$1,000,000 at 5 years and at 100 years. 30
31 In growing perpetuities, the periodic cash flows grow at a constant rate each period. The present value of a growing perpetuity can be calculated l using a simple mathematical equation. PV = Present value of a growing perpetuity PMT period 1 = Payment made at the end of first period i = rate of interest used to discount the growing perpetuity s cash flows g = the rate of growth in the payment of cash flows from period to period The Present Value of a Growing Perpetuity What is the present value of a perpetuity stream of cash flows that pays \$500 at the end of year one but grows at a rate of 4% per year indefinitely? The rate of interest used to discount the cash flows is 8%. 31
32 What is the present value of a stream of payments where the year 1 payment is \$90,000 and the future payments grow at a rate of 5% per year? The interest rate used to discount the payments is 9%. 32
33 With a growing perpetuity, a timeline goes on for ever with the growing cash flow occurring every period. i=9% Years Cash flows \$90,000 (1.05) \$90,000 (1.05) 2 Present Value =? The growing cash flows go on forever The present value of a growing perpetuity can be computed by using equation 6-6. We can substitute the values of PMT (\$90,000), i (9%) and g (5%) in equation 6-6 to determine the present value. PV = \$90,000 ( ) = \$90, = \$2,250,000 33
34 Comparing the present value of a level perpetuity (checkpoint 6.4: check yourself) with a growing perpetuity (checkpoint 6.5: check yourself) shows that adding a 5% growth rate has a dramatic effect on the present value of cash flows. The present value increases from \$1,000,000 to \$2,250,000. The cash flows streams in the business world may not always involve one type of cash flows. The cash flows may have a mixed pattern. For example, different cash flow amounts mixed in with annuities. For example, figure 6-4 summarizes the cash flows for Marriott. 34
35 In this case, we can find the present value of the project by summing up all the individual cash flows by proceeding in four steps: 1. Find the present value of individual cash flows in years 1, 2, and Find the present value of ordinary annuity cash flow stream from years 4 through Discount the present value of ordinary annuity (step 2) back three years to the present. 4. Add present values from step 1 and step 3. The Present Value of a Complex Cash Flow Stream What is the present value of cash flows of \$500 at the end of years through 3, a cash flow of a negative \$800 at the end of year 4, and cash flows of \$800 at the end of years 5 through 10 if the appropriate discount rate is 5%? 35
36 Step 3 cont. 36
37 Step 3 cont. 37
38 What is the present value of cash flows of: \$300 at the end of years 1 through 5, -\$600 at the end of year 6, \$800 at the end of years 7-10 if the appropriate discount rate is 10%? i=10% Years Cash flows \$300 -\$600 \$800 PV equals the PV of ordinary annuity PV equals PV of \$600 discounted back 6 years PV in 2 steps: (1) PV of ordinary annuity for 4 years (2) PV of step 1 discounted back 6 years This problem involves two annuities (years 1-5, years 7-10) and single negative cash flow in year 6. The \$300 annuity can be discounted directly to the present using equation 6-2b. The \$600 cash outflow can be discounted directly to the present using equation
39 The \$800 annuity will have to be solved in two stages: Determine the present value of ordinary annuity for four years. Discount the single cash flow (obtained from the previous step) back 6 years to the present using equation 5-2. Using the Mathematical Formula (Step 1) PV of \$300 ordinary annuity PV = \$300 {[1-(1/(1.10) 5 ] (.10)} = \$300 {[ 0.379] (.10)} = \$300 {3.79) = \$ 1,
40 Step (2) PV of -\$600 at the end of year 6 PV = FV (1+i) n PV = -\$600 (1.1) 1) 6 = \$ Step (3): PV of \$800 in years 7-10 First, find PV of ordinary annuity of \$800 for 4 years. PV = \$800 {[1-(1/(1 (1/(1.10) 10) ] 4 (.10)} = \$800 {[.317] (.10)} = \$800 {3.17) = \$2, Second, find the present value of \$2,536 discounted back 6 years at 10%. PV = FV (1+i) n PV = \$2,536 (1.1) 6 = \$
41 Present value of complex cash flow stream = sum of step (1) + step (2) + step (3) = \$1, \$ \$1, = \$2, Using a Financial Calculator Step 1 Step 2 Step 3 (part A) Step 3 (Part B) N /Y PV \$1, \$ \$2, \$1, PMT FV This example illustrates that a complex cash flow stream can be analyzed using the same mathematical formulas. f h fl b h h If cash flows are brought to the same time period, they can be added or subtracted to find the total value of cash flow at that time period. 41
42 42
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Chapter 2 Present Value Road Map Part A Introduction to finance. Financial decisions and financial markets. Present value. Part B Valuation of assets, given discount rates. Part C Determination of risk-adjusted | 11,465 | 45,789 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-26 | longest | en | 0.887391 |
https://nl.mathworks.com/matlabcentral/answers/436424-eig-incorrect-for-large-matrices-when-compiled-into-c-on-speedgoat-target-computer | 1,718,363,551,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861546.27/warc/CC-MAIN-20240614110447-20240614140447-00260.warc.gz | 400,991,102 | 35,064 | # EIG incorrect for large matrices when compiled into C on speedgoat target computer
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Patrick Franks on 19 Dec 2018
Commented: Delaney on 1 Aug 2019
I have been using Matlab R2017b to create simulink files that are compiled onto a speedgoat target computer. In the simulink file, in a matlab function block, we use the function eig to find the eigenvectors and eigenvalues of a matrix A. When A is 4 by 4 or smaller, this works fine.
When A is larger than 4x4 (i.e. 5x5, 6x6), eig does not correctly return the eigenvalues of the matrix. That is, for the same matrix A, the eigenvalues generated by eig on the target computer are different than those generated on the target computer (i.e. calling eig in the workspace normally).
Is this related to the code being compiled into C/C++? Has anyone experienced this issue?
Thanks,
Patrick
Patrick Franks on 19 Dec 2018
The difference in eigenvalues is on the order of 10^26.
Often in C++ it will return NaN's or extremely large numbers. In MATLAB the outputs are correct.
The generation of the matrix is a little messy, but the matrix is a 5x5 positive definite symmetric matrix A. Here's an example of the code and output.
[B,D] = eig(A).
In Matlab, B(1) = -0.361.
In C++, B(1) = NaN.
Matt J on 19 Dec 2018
Edited: Matt J on 19 Dec 2018
I suspect that the C++ version of eig is not getting the correct A as input. Maybe dump A to a file (with the C++ implementation) and verify that it has the intended values.
Delaney on 20 Jul 2019
Edited: Delaney on 21 Jul 2019
[B, D] = eig(C, eye(N), 'qz'); % N = dimension of matrix
for i = 1:N
B(:,i) = B(:,i) ./ norm(B(:,i)); % normalize eigenvectors (columns)
end
;)
Steven Lord on 21 Jul 2019
Can you provide a small code snippet that we can run to create C and define N or edit your answer and attach a MAT-file containing C and N to it? That way we can try to reproduce the behavior you described.
Delaney on 21 Jul 2019
Edited: Delaney on 21 Jul 2019
A section of the code requires the eigendecomposition of a covariance matrix C that is real and symmetric. Here is the block of code (in my case, C is a 3x3 matrix):
N = 3;
C = triu(C) + triu(C,1)'; % enforce symmetry
[Btemp,Dtemp] = eig(C); % eigen decomposition (Btemp is normalized)
D = sqrt(diag(Dtemp)); % D is a vector of standard deviations now
invsqrtC = real(Btemp * diag(D.^-1) * Btemp'); % sqrt(C^-1)
D = real(D);
B = real(Btemp);
Previously, when C was a 4x4 matrix, the Btemp and Dtemp matrices computed by eig(C) had real entries (version running was converted C/C++ code). However, I noticed when I changed C to a 3x3 matrix (all dimensions scaled accordingly), the eig(C) function produced complex eigenvectors in the Btemp matrix, and the eigenvalues in the Dtemp matrix did not agree with the same code computed in Matlab directly. Here is an example:
C =
[0.9736, 0.0001, 0.0066;
0.0001, 0.8655, 0.0115;
0.0066, 0.0115, 0.8628];
On the Speedgoat target computer (C/C++ code version), the following Btemp and Dtemp matrices are computed:
% C/C++ code version on Speedgoat target computer
[Btemp,Dtemp] = eig(C);
Btemp =
[-0.5415 + 0.0000i, 0.0121 + 0.2506i, -0.6582 - 0.0046i;
-0.5463 + 0.0000i, 0.8889 - 0.1470i, 0.3861 - 0.3385i;
-0.8785 + 0.0000i, -0.0878 - 0.1667i, 0.4377 + 0.0334i];
Dtemp =
[0.6670, 0, 0;
0, 0.6590, 0;
0, 0, 0.6590];
When the same calculation is performed in Matlab with the same C matrix, the following Btemp and Dtemp matrices are computed:
% Matlab version on host computer
[Btemp,Dtemp] = eig(C);
Btemp =
[ 0.0406, 0.0451, -0.9982;
0.6582, -0.7528, -0.0072;
-0.7518, -0.6567, -0.0602];
Dtemp =
[0.8524, 0, 0;
0, 0.8755, 0;
0, 0, 0.9740];
I would like to note that with a 4x4 C matrix this was not a problem. The eigenvalues and eigenvectors appeared in a different order in the Speedgoat (C/C++ code) version and the Matlab version, but all of the same eigenvalues were computed. I solved this problem myself by replacing the following line:
[Btemp, Dtemp] = eig(C);
With a version that ensures the 'qz' eigendecomposition algorithm:
N = 3;
[Btemp, Dtemp = eig(C, eye(N), 'qz');
for i = 1:N
Btemp(:,i) = Btemp(:,i)./norm(Btemp(:,i)); % normalize eigenvectors
end
When I do this, I get the same results on the Speedgoat target computer (C code) as the Matlab version I calculate with eig(C) (except for a sign reversal on the eigenvectors, but that's not important). I believe there may be an issue with how the 'schur' algorithm is converted into C/C++ code using Matlab Coder, specifically with odd sized matrices. In the original question, both 5x5 and 9x9 matrices were tried. However, I've confirmed that 4x4 matrices do not experience ths problem.
Mike Hosea on 22 Jul 2019
Bit difficult to debug from here. Is your code calling the initialize and terminate functions for the library you are generating?
As for complex eigenvectors, for Hermitian A we have been generating code that performs [V,D] = schur(A,'complex') and then zeros the off-diagonal elements of D. That is to say, code generation has been treating every problem as a complex one in EIG, which in the general case it must because complexity is immutable in the generated C code and depends on input data values. That might explain the occasional appearance of complex eigenvectors. We're planning to change this to do [V,D] = schur(A,'real') instead when A is real and symmetric. Since you know your input matrix is real and symmetric, you could just do that yourself instead of calling EIG.
Of course none of that explains incorrect results. I'm not aware of anything that would cause norm(A*V - V*D) to be larger than expected, and I'm not seeing any memory overruns or anything like that when debugging generated code that doesn't call into a LAPACK library, such as would be the case for a stand-alone target.
We do regularly find bugs in C compilers around here. It is worth generating code with different optimization levels to see what happens.
Finally, I believe we supported calling LAPACK on a stand-alone target in 17b. If you can get a LAPACK library built for your target, that might be a workaround and and also deliver performance benefits on larger problems. I'm not personally familiar with the process, but I know it is documented.
Delaney on 22 Jul 2019
Hi Mike,
Thanks for the help. What do you mean by your first question? We are not initializing/terminating the eig function as we are using it within a larger function (Simulink fcn block). We are using a Speedgoat target machine (https://www.speedgoat.com/learn-support/simulink-real-time-workflow) and I believe our driver blocks are working correctly as it isn't throwing any errors. Maybe this is an error on Speedgoat's end when the C code is compiled.
I'm aware of the default settings with 'schur' to the complex eigendecomposition. When I use schur(C,'real'), I find that I similarly get a completely wrong eigendecomposition. It's just all real instead of complex and the Dtemp matrix (of eigenvalues) has non-zero entries for non-diagonal indices. Here's an example (slightly different C matrix):
Dtemp =
[6.9407 -32.0045 -0.9886;
57.0459 6.9407 2.4762;
0 0 6.9407];
So yeah, it seems the issue is with the 'schur' algorithm as 'qz' works fine. Because we only run this computation every 15 minutes or so within the context of a larger biomechanics data collection, it doesn't necessarily make sense for us to go to the trouble of building a LAPACK library. I understand that 'qz' algorithm is not as efficient (esp. for a real symmetric / Hermitian matrix) but it seems to be accurate so we'll stick with that solution for the time being.
Appreciate you looking into this,
Delaney
Mike Hosea on 29 Jul 2019
I haven't been able to reproduce the issue on desktop. However, there has been some discussion here about whether the fastmath optimization ( /fp:fast ) might be playing a role. Apparently /fp:fast is used for SpeedGoat targets. You might try deleting that from the generated .mk file and rebuilding to see if it changes anything.
Delaney on 1 Aug 2019
Makes sense that the issue could be with Speedgoat. Right now using 'qz' seems to be fine, but if I have the chance I'll try deleting the fastmath optimization and let you know what happens.
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https://engineering.stackexchange.com/questions/6622/how-do-i-use-fem-to-derive-the-torsional-constant-of-an-arbitrary-shape | 1,716,341,735,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058525.14/warc/CC-MAIN-20240522005126-20240522035126-00270.warc.gz | 202,588,539 | 40,936 | # How do I use FEM to derive the torsional constant of an arbitrary shape?
In this question I ask about how to perform a first-principle derivation of the torsional constant of a section. It appears that there is no such analytical derivation for torsional constant, so my question therefore becomes: what about FEM derivation for torsional constant of arbitrary shapes section?
Note that I am not interesting in just using a FEM package without actually understanding the basic principle. I want to be able to derive the FEM formulation from first principles.
You can find an implementation of a finite element used in computation of arbitrary shape section torsional constant here:
http://projects.ce.berkeley.edu/feap/elmt11.f
I don't see any reference in the implementation other than a comment in the source code: "St. Venant torsion problem solution by warping and Prandtl stress function."
I don't see any reference on their website either, but hope the source would give you an idea. http://www.ce.berkeley.edu/projects/feap/
• The first link is now dead May 25, 2018 at 2:50
This is a problem which is usually solved in books on elasticity theory. The underlying math is based on the solution to the Laplace PDE.
If you do a Google search for Larry J. Segerlind's book "Applied Finite Element Analysis Second Edition", he lays out the math behind calculating J using the Prandtl stress function method (see Chap. 8 'Torsion of Non-circular Sections'). There is another method which is derived using a warping function, and the math is similar. The derivation using either method requires some sophisticated math, so be prepared.
If you can't find the Segerlind book, there are others on programming the finite element method which use this same problem as an illustration. It's also used in the more advanced Boundary Element Method as well.
Make a long shaft with your chosen section. Fix one end by 3dof restraints at each node. Put a closing plate across the other end and apply a torque, hence back calculate your torsional constant. Double the length of the shaft and check you get the same answer. St Venant's principle is the reason you want a long shaft to do this test. | 477 | 2,198 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-22 | latest | en | 0.916915 |
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Posts: 4
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Message 1 of 5 (336 Views)
How to draw shapes using equations?
336 Views, 4 Replies
01-23-2013 07:54 PM
Hello,
I want to draw an exponential taper 2D structure having a particulare taper equation.
How can I draw such shapes?
Regards,
Nirmal.
*Expert Elite*
Posts: 23,661
Registered: 04-20-2006
Message 2 of 5 (315 Views)
Re: How to draw shapes using equations?
01-29-2013 09:07 AM in reply to: nirmalp
Calculate a few points and then place a spline between the points.
Calculate twice as many points and create a new spline.
Is there a significant (manufacturable, measurable (using standard measuring tools - not AutoCAD) difference between the two curves?
If not then you might try again with half as many points as the first curve.
You could find out that you need to calculate far fewer points than you might think.
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Message 3 of 5 (310 Views)
Re: How to draw shapes using equations?
01-29-2013 10:15 AM in reply to: JDMather
Hello,
I am designing the optical waveguide & it design changes might affect my final feild response after passing through all the components.
I have all the co-ordinates/points(around 7000) for the curve:
Can I plot all the points using some function line pline or spline?
If yes, can you elaborate the way to enter the co-ordinates?
Somebody suggested me to copy paste x & y co-ordinates. but I am not able to do that.
Thank you.
Regards,
Nirmal.
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Posts: 5,752
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Message 4 of 5 (302 Views)
Re: How to draw shapes using equations?
01-30-2013 06:13 AM in reply to: nirmalp
Get 30 free trial C3D and bring in the points if they are that many.
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Posts: 23,661
Registered: 04-20-2006
Message 5 of 5 (294 Views)
Re: How to draw shapes using equations?
01-30-2013 08:43 AM in reply to: nirmalp
nirmalp wrote:
Hello,
I am designing the optical waveguide & it design changes might affect my final feild response after passing through all the components.
I have all the co-ordinates/points(around 7000) for the curve:
Nirmal.
I think something like Autodesk Inventor would be a better tool for this application.
I'm sure there is probably a list out there that willl draw a spline from a list of points, but -
where did these points come from? Caculated? Scanned point cloud?
If calculated I'll wager that just as good results could be obtained from 70 points and maybe far far fewer.
In Inventor you can enter points from Excel
or
enter Equations directly.
The advantage to the second method is if you need to tweek the numbers - the model will automatically update.
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# Topic: Analytical hierarchy
###### In the News (Thu 20 Jun 13)
Arithmetical hierarchy - Wikipedia, the free encyclopedia In mathematical logic, the arithmetical hierarchy, arithmetic hierarchy or Kleene hierarchy classifies the set of arithmetic formulas (or arithmetic sets) according to their degree of solvability. Layers in the hierarchy are defined as those formulas which satisfy a proposition (description) of a certain complexity. The polynomial hierarchy is a "feasible resource-bounded" version of the arithmetical hierarchy, in which polynomial length bounds are placed on the strings involved, or equivalently, polynomial time bounds are placed on the Turing machines involved. en.wikipedia.org /wiki/Arithmetical_hierarchy (325 words)
PlanetMath: analytic hierarchy (Site not responding. Last check: 2007-10-20) The analytic hierarchy is a hierarchy of either (depending on context) formulas or relations similar to the arithmetical hierarchy. Like the arithmetical hierarchy, the relations in each level are exactly the relations defined by the formulas of that level. This is version 1 of analytic hierarchy, born on 2002-08-17. planetmath.org /encyclopedia/AnalyticHierarchy.html (124 words)
Analytical hierarchy - Encyclopedia, History, Geography and Biography In mathematical logic and descriptive set theory, the analytical hierarchy is a second-order analogue of the arithmetical hierarchy. The standard notation $\Sigma^1_0 = \Pi^1_0 = \Delta^1_0$ indicates on the one hand the class of formulas that can be expressed as formulas of arbitrary finite length of alternating universal and existential quantifiers for individuals over predicates linked by sentential connectives; and on the other the class of Borel sets. A $\Sigma^1_1$ set is said to be analytic, and can thus be seen as a projection of a Borel set. www.arikah.net /encyclopedia/Analytic_set (378 words)
COMM80 - Risk Assessment and Management of Change - Analytic Hierarchy Process (Site not responding. Last check: 2007-10-20) The concept behind the hierarchy is that it is possible to judge the relative importance of the elements in a given level with respect to some or all of the elements in the level above. The hierarchy can be used either bottom up at making judgements to learn from the activities the importance of the objectives, or from the top down when the priorities of the objectives are set in advance, for example by previous experience. The Analytic Hierarchy Process does not require a unified scale since it derives scales for each factor that are ratio scales, these are then made commensurate through the hierarchical weighting process. osiris.sunderland.ac.uk /risk/content/pass/material/ahp.html (1287 words)
Analytical hierarchy - Wikipedia, the free encyclopedia In mathematical logic and descriptive set theory, the analytical hierarchy is a second-order analogue of the arithmetical hierarchy; it thus continues the classification of sets and properties by their complexity to levels higher than can be defined using first-order logic. Note that the Greek letters here should be read as lightface symbols; the corresponding boldface symbols indicate the class of first-order formulas which can be written with real parameters; see projective hierarchy for more details. set can be seen as a lightface or recursive analog of an analytic set, since it is an arithmetic projection of an arithmetic relation, just as an analytic set is an arbitrary projection of a Borel relation. en.wikipedia.org /wiki/Analytical_hierarchy (235 words)
GIS/EM4 - Integrating the Analytical Hierarchy Process with GIS to capture expert knowledge for Land Capability ... This paper examines the nature of this complexity and focuses on the utility of the Analytical Hierarchy Process (AHP) as a tool for capturing expert knowledge on environmental systems where data may be lacking. An example of the hierarchy for biophysical capability for carrots in West Gippsland is shown in figure 1. The process of building a decision hierarchy using the AHP method is effective and captures expert knowledge in form that is transparent, of value to the experts building the decision hierarchy and promotes focussed and detailed discussion on each criterion. www.colorado.edu /research/cires/banff/pubpapers/133 (3372 words)
[No title] (Site not responding. Last check: 2007-10-20) Moreover, such analytical quality specifications should be firmly based upon medical requirements, useable in all laboratories irrespective of size, type or location, generated using simple to understand models, and widely accepted as cogent by professionals in the field. In contrast, if the analytical method had positive bias, the glycated haemoglobin would appear lower: the clinician might congratulate the patient on maintaining good control but, while the risk of microalbuminuria might be lower, the risk of hypoglycaemic episodes might be increased. Measures of the quality of analytical performance could be obtained be comparison with attainment documented in published works on similar or other assay methods for the quantity for which quality specifications were required. www.ifcc.org /eJIFCC/vol13no1/1301200106n.htm (2990 words)
SAE Technical Paper Template Relative risk is calculated as follows: For each branch of the hierarchy, the risk factor weights along the branch are multiplied times the input value at the lowest level to compute the contribution of this input to the overall risk assessment for the flight under consideration. The Analytic Hierarchy Process was identified as a methodology for the elicitation and representation of expert knowledge in the two categories. As the hierarchy was developed, it became clear that extensive datasets will be required to develop systemic values, i.e., average values of many risk factors derived from the entire operation, as well as real-time connections to various flight operations databases. www.nrlmry.navy.mil /foras/publications/ASC99-paper.html (3826 words)
Analytical hierarchy -- Facts, Info, and Encyclopedia article (Site not responding. Last check: 2007-10-20) Analytical hierarchy -- Facts, Info, and Encyclopedia article A set is said to be analytic, and can thus be seen as a (Any structure that branches out from a central support) projection of a Borel set. Then we may define "second projective" sets as projections of first projective sets or complements thereof, producing the set of asets, and so on. www.absoluteastronomy.com /encyclopedia/a/an/analytical_hierarchy.htm (239 words)
DSS News: Vol. 4, No. 13 At the top of the hierarchy is the overall objective and the decision alternatives are at the bottom. The number of levels in the hierarchy depends on the complexity of the problem and the decision maker's model of the problem hierarchy. The analytical hierarchy created in Expert Choice can be used to support ranking tasks like rank ordering the importance of military targets, the quality of research proposals or the quality of investment proposals. dssresources.com /newsletters/82.php (1644 words)
ipedia.com: Arithmetical hierarchy Article (Site not responding. Last check: 2007-10-20) In mathematical logic, the arithmetical hierarchy classifies the set of all formulas according to their degree of solvability. In mathematical logic, the arithmetical hierarchy (also known as the arithmetic hierarchy) classifies the set of all formulas (or functions) according to their degree of solvability. See also: recursion theory, analytical hierarchy, interpretability logic. www.ipedia.com /arithmetical_hierarchy.html (271 words)
Using analytical hierarchy process and fuzzy set theory to rate and rank the disability (Site not responding. Last check: 2007-10-20) Therefore, in order to place qualified disabled people in appropriate job positions, a multidimensional assessment method should be developed for measuring and interpreting the disabled people's residual capabilities with respect to their strengths, weaknesses, and their compatibility to job requirements and work environments. Thus, the primary purpose of this study is to discuss how to structure a hierarchy related to the problem of deriving an Overall Disability Index (ODI) for measuring the individual's disability. The analytical hierarchy process, along with the entropy theory and fuzzy set theory, is applied for eliciting the weights among the attributes and aggregating the multiple attributes into a single measurement. www.uwv.nl /cba/module8/meta24176.htm (213 words)
Incorporating Community Preferences in Local Development Strategies The Analytical Hierarchy Process (AHP) was used to elicit community leaders’ preferences for seven economic and non-economic industry impacts: number of jobs, average wage or salary, level of capital investment, level of utility requirements, impacts of population growth, impacts on property values, and cleanliness of industry. Contrary to what is often reported in the media, the results indicate that the number of jobs created by the establishment of new industry may not be the most important goal of a community. Beyond the industrial recruitment model, the Analytical Hierarchy Process offers a valuable tool to communities that are faced with making decisions involving complex or multiple objectives. www.cpac.missouri.edu /newsuse/archive/980518.html (730 words)
ISNAR - Discussion Forum on Priority Setting in Agricultural Research (Site not responding. Last check: 2007-10-20) The analytic hierarchy process (AHP) is a comprehensive, logical and structural framework, which allows to improve the understanding of complex decisions by decomposing the problem in a hierarchical structure. The first step is to structure the decision problem in a hierarchy as depicted in Figure 1. The next level consists of the criteria relevant for this goal and at the bottom level are the alternatives (for example research projects) to be evaluated. www.cgiar.org /ISNAR/Fora/Priority/MeAnalit.htm (756 words)
bp_render_qam_8|Module 1: Analytical Hierarchy Process|True or False The Analytic Hierarchy Process would probably be preferable to the Multifactor Evaluation Process when trying to determine which R&D projects should be funded. Decision situations where we easily and accurately determine various weighting factors are excellent candidates for the Analytic Hierarchy Process analysis. In the analytic hierarchy process we rate each pair of alternatives using a 13-point scale. wps.prenhall.com /bp_render_qam_8/0,,400262-,00.utf8.html (141 words)
Analytical Hierarchy Process It is based on the assumption that when faced with a complex decision the natural human reaction is to cluster the decision elements according to their common characteristics. This gives a weighting for each element within a cluster (or level of the hierarchy) and also a consistency ratio (useful for checking the consistency of the data). The Analytical Hierarchy Process Model was designed by TL Saaty as a decision making aid. www.ifm.eng.cam.ac.uk /dstools/choosing/ahp.html (160 words)
Information Bridge: DOE Scientific and Technical Information (Site not responding. Last check: 2007-10-20) WSRC was requested to evaluate whether nuclear materials other than aluminum-clad spent nuclear fuel should be considered for treatment to prepare them for disposal in the melt and dilute facility as part of the Treatment and Storage Facility (TSF) currently projected for construction in the L-Reactor process area. The Analytical Hierarchy Process using a ratings methodology was used to rank potential feed candidates for disposition through the Melt and Dilute facility proposed for disposition of Savannah River Site aluminum-clad spent nuclear fuel. Because of the scoping nature of this analysis, the expert team convened for this purpose concentrated on technical feasibility and potential cost impacts associated with using melt and dilute versus the current disposition option. www.osti.gov /bridge/product.biblio.jsp?osti_id=752142 (214 words)
Ducks In Order™ Blog: Background: Analytical Hierarchy Process The Analytical Hierarchy Process is a technique for evaluating the factors that make up a decision. If you are going to use some sort of tool to rank your options, you need to tell that tool how much weight to give to each of the factors. The Analytical Hierarchy Process is a way to determine those weights. ducksinorder.blogspot.com /2004/10/background-analytical-hierarchy.html (644 words)
IngentaConnect An analytical hierarchy process framework for comparing the overa... (Site not responding. Last check: 2007-10-20) But the use of non-financial performance measures makes it difficult to assess and compare the overall effectiveness of each manufacturing department, in terms of support provided to the achievement of the manufacturing strategy, since to this aim it is necessary to integrate performance measures expressed in heterogeneous measurement units. Aims to show the potential of the analytical hierarchy process (AHP) for assessing and comparing the overall manufacturing performance of different departments. Does not report the detailed analytical description of the AHP but focuses on the practical problems and managerial implications related to its application to performance measurement, pointing out also its assumptions and limitations. www.ingentaconnect.com /content/mcb/024/1996/00000016/00000008/art00006 (227 words)
MNL29 - Analytical Hierarchy Process to Multiattribute Analysis (ASTM) This software product supports the application of ASTM's AHP (Analytical Hierarchy Process) standard practice (E 1765) for performing multiattribute decision analysis to evaluate buildings and building systems. It includes a comprehensive list of building-related attributes (developed by ASTM Subcommittee E06.25 and E06.81) in a glossary from which users can select attributes of interest when evaluating their building decision. The Analytical Hierachy Process is the source of the underlying mathematical theory for the Program; such technology is included in the Program under license from Expert Choice, Inc. The program is protected by U.S. Patent No. 4613946. www.normas.com /ASTM/BOOKS/MNL29.html (267 words)
[No title] (Site not responding. Last check: 2007-10-20) I will demonstrate that embedding the MRCM within an Analytical Hierarchy Process (AHP) model eliminates arbitrary scaling problems and allows for the refinment of the MRCM by making it easy to weight not only all the Maxwell Criteria but also easy to scale each of the individual criterion's location continua. THE ANALYTICAL HIERARCHY PROCESS (AHP) The Analytical Hierarchy Process (AHP) allows for the modeling of criteria involved in influencing the outcome of any decision. Once the hierarchy is developed all that remains is the ratings of the selections. www.losangeles.af.mil /Tenants/SCEA/CRIMS/crimgde7.doc (1787 words)
Analytical Hierarchy Process :: Overview from The Quality Portal Developed by Thomas Saaty, AHP provides a proven, effective means to deal with complex decision making and can assist with identifying and weighting selection criteria, analyzing the data collected for the criteria and expediting the decision-making process. Relative scores for each choice are computed within each leaf of the hierarchy. Scores are then synthesized through the model, yielding a composite score for each choice at every tier, as well as an overall score. thequalityportal.com /q_ahp.htm (347 words)
[No title] (Site not responding. Last check: 2007-10-20) Broad technical and analytical proficiency is required across all program areas such as munitions cost analysis, operational flight test currency resource reporting and control, aircraft modification schedule/cost reviews, and mission area plan analyses. The contractor may from time-to-time be required to react on short notice to critical time sensitive issues that are within the scope of this task order. The contractor shall provide technical and analytical assistance to assess weapon system capabilities using strategy-to-task and resource analyses to translate system requirements into resource evaluation and documentation products. www2.acc.af.mil /lg/cons/Lgcu/CATS/2401_PWS.doc (812 words)
Analytical Hierarchy Process - Business Intelligence Guide Results for Analytical Hierarchy Process (Site not responding. Last check: 2007-10-20) Oracle Business Intelligence 10g is the industry's most comprehensive offering designed to address the entire spectrum of analytical requirements including query, reporting, more... Goal.Link ensures goals at all levels in the hierarchy are consistent, and offers guidance on how strategic goals can be met. You can take any function, such as help desk, medical protocols, sales training, etc. and have a means of providing continuous advice to your staff and colleagues, anywhere, plus the ability to monitor performance of your staff and colleagues. biguide.computerworld.com /search/keyword/cwbi/Analytical%20Hierarchy%20Process/Analytical%20Hierarchy%20Process (1458 words)
Good sources of information about the AHP Zahedi, F., The analytic hierarchy process - A survey of the method and its applications. Saaty, T., Fundamentals of the analytic hierarchy process. Dolan, J.G. and D.R. Bordley, Involving patients in complex decisions about their care: anapproach using the analytic hierarchy process. www2.newpaltz.edu /~liush/ahp.htm (90 words)
Make Complex Business Decisions by the Numbers Integrated into this article are two interactive programs that allow the reader to experiment with the Analytical Hierarchy Process (AHP) decision-making model. The first demonstrates the criterion comparison process; the second allows the reader to select a decision of interest, identify the key variables that impact the decision and use the AHP process as decision input. The Analytic Hierarchy Process (AHP) is a decision analysis method that ranks alternatives based on a number of criteria. gbr.pepperdine.edu /021/ahp.html (1039 words)
[No title] The case study offers an example of how QFD and the analytical hierarchy process (AHP) techniques can be used to facilitate a decision facing a company grappling with legacy system obsolescence, including the Year 2000 problem. An approach is presented for applying an Analytical Hierarchy Process (AHP) in Quality Function Deployment to improve the accuracy of priorities and make QFD better fit particular projects. In addition to enhancing the ability of developers to hear the "voices of the customers" more clearly, a more accurate deployment of weights can be accomplished by the consistent use of ratio scales, such as produced by the Analytical Hierarchy Process (AHP) throughout QFD. www.qfdi.org /who_is_qfdi/linkpages/ahp_articles.htm (524 words)
MGMT 6100 (Site not responding. Last check: 2007-10-20) The techniques of quantitative business methods we are going to study in this course have been applied successfully to an increasingly wide variety of complex problems in business and industry, health care, education, and many other areas. During the course, you will be introduced to the modern methods and techniques of quantitative decision making: goal and integer programming, analytical hierarchy process, decision tree analysis, inventory models, queuing theory and waiting lines, and simulation. Multiobjective decision making applying the Analytical Hierarchy Process: developing the hierarchy, establishing priorities, pairwise comparison matrix, consistency, and overall priority ranking. www.sbe.csuhayward.edu /~zradovil/6100_fa_2002.htm (632 words)
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NCTM.11. Grade 2 Curriculum Focal Points
11.2. Number and Operations and Algebra: Developing quick recall of addition facts and related subtraction facts and fluency with multi-digit addition and subtraction
11.2.1. Children use their understanding of addition to develop quick recall of basic addition facts and related subtraction facts. They solve arithmetic problems by applying their understanding of models of addition and subtraction (such as combining or separating sets or using number lines), relationships and properties of number (such as place value), and properties of addition (commutativity and associativity). Children develop, discuss, and use efficient, accurate, and generalizable methods to add and subtract multi-digit whole numbers. They select and apply appropriate methods to estimate sums and differences or calculate them mentally, depending on the context and numbers involved. They develop fluency with efficient procedures, including standard algorithms, for adding and subtracting whole numbers, understand why the procedures work (on the basis of place value and properties of operations), and use them to solve problems.
NCTM.12. Connections to the Grade 2 Focal Points
12.1. Number and Operations: Children use place value and properties of operations to create equivalent representations of given numbers (such as 35 represented by 35 ones, 3 tens and 5 ones, or 2 tens and 15 ones) and to write, compare, and order multi-digit numbers. They use these ideas to compose and decompose multi-digit numbers. Children add and subtract to solve a variety of problems, including applications involving measurement, geometry, and data, as well as nonroutine problems. In preparation for grade 3, they solve problems involving multiplicative situations, developing initial understandings of multiplication as repeated addition. | 693 | 3,462 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-17 | latest | en | 0.879992 |
https://eldorado.tu-dortmund.de/handle/2003/25955?mode=full | 1,624,373,509,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488517820.68/warc/CC-MAIN-20210622124548-20210622154548-00340.warc.gz | 222,279,608 | 7,690 | DC FieldValueLanguage
dc.contributor.authorChimani, Markus-
dc.date.accessioned2009-01-05T10:46:43Z-
dc.date.available2009-01-05T10:46:43Z-
dc.date.issued2009-01-05T10:46:43Z-
dc.identifier.urihttp://hdl.handle.net/2003/25955-
dc.identifier.urihttp://dx.doi.org/10.17877/DE290R-367-
dc.description.abstractThe graph theoretic problem of crossing numbers has been around for over 60 years, but still very little is known about this simple, yet intricate nonplanarity measure. The question is easy to state: Given a graph, draw it in the plane with the minimum number of edge crossings. A lot of research has been devoted to giving an answer to this question, not only by graph theoreticians, but also by computer scientists. The crossing number is central to areas like chip design and automatic graph drawing. While there are algorithms to solve the problem heuristically, we know that it is in general NP-complete. Furthermore, we do not know if the problem is efficiently approximable, except for some special cases. In this thesis, we tackle the problem using Mathematical Programming. We show how to formulate the crossing number problem as systems of linear inequalities, and discuss how to solve these formulations for reasonably sized graphs to provable optimality in acceptable time--despite its theoretical complexity class. We present non-standard branch-and-cut-and-price techniques to achieve this goal, and introduce an efficient preprocessing algorithm, also valid for other traditional non-planarity measures. We discuss extensions of these ideas to related crossing number variants arising in practice, and show a practical application of a formerly purely theoretic crossing number derivative. The thesis also contains an extensive experimental study of the formulations and algorithms presented herein, and an outlook on its applicability for graph theoretic questions regarding the crossing numbers of special graph classes.en
dc.description.abstractDas Kreuzungszahlproblem wird von Graphentheoretikern seit über 60 Jahren betrachtet, jedoch ist noch immer sehr wenig über dieses einfache und zugleich hochkomplizierte Ma der Nichtplanarität bekannt. Die Aufgabenstellung ist simpel: Gegeben ein Graph, zeichnen Sie ihn mit der kleinstmöglichen Anzahl an Kantenkreuzungen. Nicht nur Graphentheoretiker sondern auch Informatiker beschäftigten sich ausgiebig mit dieser Aufgabe, denn es handelt sich dabei um ein zentrales Konzept im Chipdesign und im automatischen Graphenzeichnen. Zwar existieren Algorithmen um das Problem heuristisch zu lösen, jedoch wissen wir, dass es im Allgemeinen NP-vollständig ist. Darüberhinaus ist auch unbekannt, ob sich das Problem, außer in Spezialfällen, effizient approximieren lässt. In dieser Dissertation, versuchen wir das Problem mit Hilfe der Mathematischen Programmierung zu lösen. Wir zeigen, wie man das Kreuzungszahlproblem als verschiedene Systeme von linearen Ungleichungen formulieren kann und diskutieren wie man diese Formulierungen für nicht allzu große Graphen beweisbar optimal und in akzeptabler Zeit lösen kann - unabhängig von seiner formalen Komplexitätsklasse. Wir stellen dazu benötigte maßgeschneiderte Branch-and-Cut-and-Price Techniken vor, und präsentieren einen effizienten Algorithmus zur Vorverarbeitung; dieser ist auch für andere traditionelle Ma e der Nichtplanarität geeignet. Wir diskutieren Erweiterungen unserer Ideen für verwandte Kreuzungszahlkonzepte die in der Praxis auftreten, und zeigen eine praktische Anwendung eines vormals rein theoretisch behandelten Kreuzungszahl-Derivats auf. Diese Arbeit enthält auch eine ausführliche experimentelle Studie der präsentierten Formulierungen und Algorithmen, sowie einen Ausblick über deren mögliche Nutzung für graphentheoretische Fragen bezüglich der Kreuzungszahlen von speziellen Graphenklassen.de
dc.language.isoende
dc.subjectCrossing numberen
dc.subjectExact proofen
dc.subjectInteger linear programen
dc.subjectSPQR-treeen
dc.subjectKreuzungszahlde
dc.subjectGraphentheoriede
dc.subjectMathematische Programmierungde
dc.subject.ddc004-
dc.titleComputing crossing numbersen
dc.typeTextde
dc.contributor.refereeSkutella, Martin-
dc.date.accepted2008-11-24-
dc.type.publicationtypedoctoralThesisde
dc.identifier.urnurn:nbn:de:hbz:290-2003/25955-5-
dcterms.accessRightsopen access-
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# Maths Exam Style Question HELP!!! Watch
1. So Anyone know how to solve it WITHOUT GIVING ME THE ANSWER with a step by step guide which would help me figure it out!
2. What have you tried so far?
3. (Original post by Donkey******)
What have you tried so far?
I have tried 300/400 = 3/4*2 (as they are rolling the die twice) but i dont think that is the right answer!
4. yeah sounds about right to me, you have worked out the probability of getting heads and getting heads twice in a row is 3/4^2 (or 3/4 to the power of 2)
5. (Original post by NothingButWaleed)
I have tried 300/400 = 3/4*2 (as they are rolling the die twice) but i dont think that is the right answer!
Because it's not. Get the probability that it will land on heads and square it to get the probability for it happening twice. You have to square it because you're multiplying the two probabilities together in order to find the probability of it happening twice in total.
By your logic, if I throw a coin twice with an equal chance of both tails and heads, the probability of me getting two heads is which is obviously not right.
6. (Original post by metaljoe)
yeah sounds about right to me, you have worked out the probability of getting heads and getting heads twice in a row is 3/4*2.
Probability doesn't go above 1, it is bounded between 0 (impossible for something to happen) and 1 (full certainty of it happening), and
7. (Original post by NothingButWaleed)
So Anyone know how to solve it WITHOUT GIVING ME THE ANSWER with a step by step guide which would help me figure it out!
First, you need to work out the total times it was flipped, and how many heads or tails.
Then, you want to work out the probability oh heads being the outcome.
Finally, since you are flipping the coin twice, you square the probability of it landing on heads.
Spoiler:
Show
So, 400 times total, 300 heads, 100 tails.
The chance of landing on heads is 3/4
So you do 0.75 * 0.75 which is 0.5625, which is your answer
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https://justaaa.com/economics/145497-give-a-real-world-or-hypothetical-example-of-how | 1,696,422,593,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511369.62/warc/CC-MAIN-20231004120203-20231004150203-00867.warc.gz | 352,830,135 | 9,862 | Question
# give a real-world (or hypothetical) example of how the spending multiplier works. You might want to...
give a real-world (or hypothetical) example of how the spending multiplier works. You might want to include some numbers to help support your example.
Spending multiplier is the concept that shows that maximum impact that can be made upon the output with increase or decrease in spending. Spending multiplier is used to calculate on the basis of marginal propensity to consume (MPC). A bigger MPC means a bigger spending multiplier.
Spending multiplier has following formula:
Spending multiplier = 1/(1-MPC)
Let, MPC = .80
Then,
Spending multiplier = 1/(1-.80)
Spending multiplier = 5
Now suppose, government increases the spending by \$100 Billion, then:
Maximum increase in real GDP = increase in spending * Spending multiplier
Maximum increase in real GDP = 100*5
Maximum increase in real GDP = \$500 Billion
Now, it can be observed that increased spending of \$100 Billion, has multiplier effect on real output to increase to the level of \$500 Billion. It is due to the spending multiplier. | 237 | 1,123 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2023-40 | latest | en | 0.918772 |
http://slideplayer.com/slide/3347133/ | 1,527,471,639,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870604.65/warc/CC-MAIN-20180528004814-20180528024814-00282.warc.gz | 257,922,123 | 20,906 | # Reinforcement Learning (II.) Exercise Solutions Ata Kaban School of Computer Science University of Birmingham 2003.
## Presentation on theme: "Reinforcement Learning (II.) Exercise Solutions Ata Kaban School of Computer Science University of Birmingham 2003."— Presentation transcript:
Reinforcement Learning (II.) Exercise Solutions Ata Kaban A.Kaban@cs.bham.ac.uk School of Computer Science University of Birmingham 2003
Exercise 1 In the grid based environment below the state values have all been computed except for one. Possible actions are up, down, left and right. All other actions result in no reward except those that move the agent out of states A and B. Calculate the value of the blank state assuming a random policy (the action is selected randomly between those possible). Consider a discount reward = 0.9.
Solution 0.7 4 5 1 3.0 4.4 2 1.9 3 V (5) = 0.25( 0 + V (1)) + 0.25( 0 + V (2)) + 0.25( 0 + V (3)) + 0.25( 0 + V (4)) V (5) = 0.25(0.9)[4.4 + 1.9 + 0.7 + 3.0] = 2.25
Exercise 2 The diagram below depicts an MDP model of a fierce battle.
You can move between two locations, L1 and L2, one of them being closer to the adversary. If you attack from the closest state, –then you have more chances (90%) to succeed (while only 70% from the farther location), –however you could also be detected (with 80% chance) and killed (while the chances of being detected from the farther location is 50%). You can only be detected if you stay in the same location. You need to come up with an action plan for the situation.
The arrows represent the possible actions: – ‘move’ (M) is a deterministic action –‘attack’ (A) and ‘stay’ (S) are stochastic. For the stochastic actions, the probabilities of transitioning to the next state are indicated on the arrow. All rewards are 0, except in the terminal states, where your success is represented by a reward of +50 while your adversary’s success is a reward of -50 for you. Employing a discount factor of 0.9, compute an optimal policy (action plan).
Solution The computations of action-values for all states and actions are required. Denote by In value iteration, we start with initial estimates (for all other states) Then we update all action values according to the update rule: where
Hence, in the first iteration of the algorithm we get: The values for the ‘move’ action stay the same (at 0): After this iteration, the values of the two states are and they correspond to the action of ‘attacking’ in both states.
The next iteration gives the following: The new V-values are (by computing max): These correspond to the ‘attack’ action in both states.
This process can continue until the values do not change much between successive iterations. From what we can see at this point, the best action plan seems to be attacking all the time. Can we say more without a full computer simulation?
Continuing (optional)… It is clear that to ‘Stay’ is suboptimal in both states. In the Close state, it is also clear that the best thing to do is to ‘Attack’ continuously (given that we have no cost for that). Actually we can compute the values in the limit analytically (if you keep an eye at changes in update from iteration to iteration)
Now for the far state, the question is between ‘Attack’ or ‘Move’ to the closer orbit. Compute the values for both these actions (in the same way as before):
Hence it is worth moving closer to the orbit. The optimal policy for this problem setting (!) is to move closer and attack from there. Can you imagine a different policy making more sense for this problem? Can you imagine another setting (parameter design) which would lead to a different (more desirable) optimal policy? Designing the parameter setting for a situation according to the conditions is up to the human and not up to the machine… Well in this exercise all parameters were given but in your potential future real applications will be not.
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Similar presentations | 993 | 4,091 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-22 | latest | en | 0.898279 |
http://www.math.mcgill.ca/~darmon/courses/97-98/algebra1/ass4.html | 1,511,496,685,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807084.8/warc/CC-MAIN-20171124031941-20171124051941-00011.warc.gz | 448,817,461 | 1,359 | # 189-235A: Basic Algebra I
## Due: Wednesday, October 1.
1 Page 149, 3.6, 3.7, 3.11.
2. Page 155, 3.18, 3.20, 3.23.
3. Page 161, 3.28.
4. Let R be the ring of Gaussian integers consisting of all elements of the form m+ni, where m and n are regular integers.
a) Define the norm of an element of R by the rule |m+ni| = m^2+n^2. Show that, if a divides b in R, then |a| divides |b|.
b) Show that R has a division algorithm: namely, for all a and b in R, there exist q and r in R such that a=bq+r and |r| <|b|.
c) A gcd of a and b is an element of R which divides both a and b and whose norm is maximal. Show that, if c and d are two gcd's of a and b, then their ratio is a power of i.
d) Using part b), describe an algorithm to compute a gcd of two elements of R. Use this algorithm to compute gcd(7+9i,7-4i). | 274 | 816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | latest | en | 0.903432 |
http://www.shmoop.com/sequences-series/arithmetic-series-examples.html | 1,464,646,919,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464051114647.95/warc/CC-MAIN-20160524005154-00004-ip-10-185-217-139.ec2.internal.warc.gz | 794,937,777 | 16,199 | # Sequences and Series
Arithmetic Series
# Arithmetic Series Examples
### Example 1
Find the 7th partial sum of the sequence {an} = -9 + 3(n – 1).
### Example 2
Evaluate:
### Example 3
Find the sum of -6, 0, 6,…, 66, 72 .Use the formula for a partial sum of an arithmetic series and simplify to find your final solution. | 96 | 331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-22 | longest | en | 0.750716 |
https://www.physicsforums.com/threads/forces-on-a-toppling-wheel.868663/ | 1,653,330,094,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662560022.71/warc/CC-MAIN-20220523163515-20220523193515-00396.warc.gz | 1,111,668,233 | 16,524 | # Forces on a toppling wheel
## Homework Statement
A stationary bicycle wheel is placed on its rim on rough ground. It topples over. Sketch a free body diagram for the wheel when it is at an arbitrary angle to the vertical and label the forces. Explain qualitatively what happens to the direction and magnitude of each of the forces during the toppling process.
What I dont get is how the forces change direction and magnitude.
## The Attempt at a Solution
What i did was draw the forces like in the picture. Not really sure how they change. I'm thinking at first when theta is small Normal ~ mg. Then friction increases and the normal force decreases.
Any help greatly appreciated.
#### Attachments
• WIN_20160424_18_53_43_Pro.jpg
11.6 KB · Views: 437
Are you sure there is friction? Maybe the normal force changes direction
Re thought and the centre of mass is accelerating to the right so there must be friction
Merlin3189
Homework Helper
Gold Member
<Q>... Explain qualitatively what happens to the direction and magnitude of each of the forces during the toppling process.</Q>
... how the forces change direction and magnitude.
... I'm thinking at first when theta is small Normal ~ mg. Then friction increases and the normal force decreases.
You seem to be well on the way with this. But try to give reasons for your statements.
Why is the friction force increasing?
Why is the normal force decreasing?
Then looking at your diagram, why have you shown friction acting in the direction you have? (I agree with you, but you can explain why.)
As for the direction of the forces, which can change?
Although they only ask for one diagram, perhaps it would be interesting to draw another when the wheel has toppled much further?
haruspex | 378 | 1,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2022-21 | latest | en | 0.943631 |
https://friendly-meishi.com/qa/question-do-all-routing-numbers-start-with-a-0.html | 1,632,387,221,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.92/warc/CC-MAIN-20210923074537-20210923104537-00056.warc.gz | 320,947,856 | 8,193 | ## How many zeros are in a routing number?
Your routing/transit number is the nine digit number which appears between the symbols on the bottom of your check.
Locate the string of numbers at the bottom of your check that matches the Check Number in the upper right-hand corner.
This number is usually 4 digits, and may include a zero as the first digit..
## How many numbers are in routing?
nineThere are two numbers you’ll need to provide. Your bank routing number is a nine-digit code that’s based on the U.S. Bank location where your account was opened. It’s the first set of numbers printed on the bottom of your checks, on the left side. You can also find it in the U.S. Bank routing number chart below.
## Is ABA and routing number the same?
An ABA number (also known as routing number or routing transfer number) is a sequence of nine numeric characters used by banks to identify specific financial institutions within the United States.
## How do you read a routing number?
Find the routing number on a check At the bottom of a check, you will see three groups of numbers. The first group is your routing number, the second is your account number and the third is your check number.
The ABA routing number is the nine-digit sequence that appears between the “:” symbols. … If, however, the routing number begins with a “5” it is considered an internal routing number for that financial institution, therefore you must contact your external financial institution for the proper ACH routing number.
## How many zeros are in a checking account number?
How many zeros are there before my account number for direct deposit? For direct deposit into your account, there should be 10 digits in total. If you have a 3 or 4 digit account number, simply add the additional 0(s) to equal 10 digits.
The first two digits of the nine digit RTN must be in the ranges 00 through 12, 21 through 32, 61 through 72, or 80. The digits are assigned as follows: 00 is used by the United States Government. 01 through 12 are the “normal” routing numbers, and correspond to the 12 Federal Reserve Banks.
## Why does it say my routing number is invalid?
This return response means that the routing number provided at the time of payment submission was not valid so the payment could not be processed. … Please make sure you are using a paper check to validate the routing number as deposit slips will often show a different routing number than the one used for ACH payments.
## What is Fedwire code?
A Fedwire (or ABA) code is a bank code used in the United States, which identifies financial institutions. A SWIFT code – sometimes also called a SWIFT number – is a standard format for Business Identifier Codes (BIC). It is used to identify banks and financial institutions globally.
## Is ACH routing number the same as ABA?
ABA routing numbers encompass all routing numbers, including ACH. That means all ACH routing numbers are technically ABA routing numbers, though your bank may have a special ACH routing number for electronic transfers.
## What banks routing number is 124303120?
Green Dot BankRouting number 124303120 of Green Dot Bank. 124303120 is the current routing transit number of Green Dot Bank situated in city Pasadena, state California (CA).
## What bank is routing number 031101279?
The Bancorp BankBank Routing Number 031101279, The Bancorp Bank.
## Does it matter what routing number you use?
Which routing number should you use? Banks use different routing numbers for different types of transactions. For this reason, the routing number printed on your checks might not be the same number you need for an ACH transfer, or direct debit. … Using the wrong number can lead to delays in processing the transfer.
## What is bank routing number mean?
Your routing number is a 9-digit code that will be used to identify where your bank account was opened. It may also be known as an RTN, an ABA routing number, or a routing transit number. As you prepare to send money internationally, one of the numbers that you’ll want to have on hand is your routing number.
## Do you include the zeros in account number?
Yes, it does matter in bank account number. There are various banks where account number starts with 0 which cannot be ignored. So you will have to give out your account numbers including leading zeros if there are any.
## Can a checking account number be 13 digits?
You may need to present your Account Number when setting up automatic payments or deposits from or to your checking or savings account. If you have checks, your 13-digit Account Number can be found on the bottom of your checks. … Preceding zeroes to make the 13-digit number. Member number (ranging from 4 to 8 digits)
## Which bank account number is 11 digits?
ALLAHABAD BANKList of All Bank with Total Digit in Bank Account NumbersSL.NONAME OF BANKA/C NO DIGITS2ALLAHABAD BANK113AXIS BANK154BANK OF BARODA145BANK OF INDIA1534 more rows
## Can a bank routing number be 8 digits?
Can a bank routing number be 8 digits? Technically, a routing number uses 8 digits to identify the Federal Reserve District and the specific bank or financial institution. However, routing numbers in use today utilize 9 digits. | 1,134 | 5,221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2021-39 | latest | en | 0.935742 |
https://image.jokesfunniescom.com/2017/03/unit-circle-no-numbers.html | 1,628,031,134,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154486.47/warc/CC-MAIN-20210803222541-20210804012541-00452.warc.gz | 326,131,660 | 53,145 | ## Pi – Wikipedia
One approach to consider the circle gathering is that it portrays how to include points, where just edges in the vicinity of 0° and 360° are allowed. For instance, the outline shows how to add 150° to 270°. The appropriate response ought to be 150° + 270° = 420°, yet when thinking as far as the circle gather, we have to “overlook” the way that we have wrapped once around the circle. Along these lines, we alter our answer by 360° which gives 420° = 60° (mod 360°).
Another depiction is as far as normal expansion, where just numbers in the vicinity of 0 and 1 are permitted (with 1 relating to a full revolution). To accomplish this, we may need to discard digits happening before the decimal point. For instance, when we work out 0.784 + 0.925 + 0.446, the appropriate response ought to be 2.155, however we discard the main 2, so the appropriate response (in the circle gathering) is only 0.155.
## 1000+ ideas about Unit Circle Trigonometry on Pinterest …
The circle gathering is more than only a theoretical mathematical question. It has a characteristic topology when viewed as a subspace of the mind boggling plane. Since duplication and reversal are ceaseless capacities on C×, the circle bunch has the structure of a topological gathering. Additionally, since the unit circle is a shut subset of the intricate plane, the circle gathering is a shut subgroup of C× (itself viewed as a topological gathering).
One can state considerably more. The circle is a 1-dimensional genuine complex and duplication and reversal are genuine expository maps on the circle. This gives the circle bunch the structure of a one-parameter amass, an example of a Lie gather. Truth be told, up to isomorphism, it is the one of a kind 1-dimensional conservative, associated Lie bunch. In addition, each n-dimensional minimal, associated, abelian Lie gathering is isomorphic to.
## Trigonometric functions – Wikipedia
Each minimal Lie assemble G of measurement > 0 has a subgroup isomorphic to the circle amass. That implies that, reasoning as far as symmetry, a smaller symmetry gather acting consistently can be relied upon to have one-parameter circle subgroups acting; the outcomes in physical frameworks are seen for instance at rotational invariance, and unconstrained symmetry breaking.
The circle aggregate has numerous subgroups, however its lone appropriate shut subgroups comprise of underlying foundations of solidarity: For every whole number n > 0, the nth foundations of solidarity shape a cyclic gathering of request n, which is extraordinary up to isomorphism.
## The Pi Manifesto – No, really, pi is right!
Page 61 – March 2017 Calendar | 590 | 2,669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.908332 |
https://www.physicsforums.com/threads/dont-know-where-to-start-uniform-circular-motion.185240/ | 1,632,563,351,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057615.3/warc/CC-MAIN-20210925082018-20210925112018-00011.warc.gz | 943,786,080 | 14,954 | # Don't know where to start-Uniform Circular Motion
At t$$_{}1$$=2.00s, the acceleration of a particle in counter-clockwise circular motion is (6.00m/s$$^{}2$$)i + (4.00m/s$$^{}2$$)j. It moves at constant speed. At time t$$_{}2$$=5.00s, its acceleration is (4.00m/s$$^{}2$$)i + (-6m/s$$^{}2$$)j. What is the radius of the path taken by the particle if t$$_{}2$$ - t$$_{}1$$ is less than one period?
I really don't even know where to start. I know that the speed is constant, and that I'm probably going to have to use a=v$$^{}2$$/r, but I really don't know where to even begin with this.
Doc Al
Mentor
Hint: Since the motion is uniform circular motion, where must those acceleration vectors point?
the acceleration vectors always point towards the center, but I still don't understand what that means...does that mean something nets to 0?
one way to tackle this is draw yourself a simple diagram: you have two acceleration vectors at two points of the circular path. Can you work out the angle between the vectors?
You know how long it takes to traverse this angle... so from that you can work out something which will lead you straight to radius given constant speed.
ok im starting to understand...before I do all the work for no reason, should i use a dot product with these 2 vectors to figure out the angle?
Doc Al
Mentor
ok im starting to understand...before I do all the work for no reason, should i use a dot product with these 2 vectors to figure out the angle?
Sure. Good idea. | 392 | 1,495 | {"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} | 3.703125 | 4 | CC-MAIN-2021-39 | latest | en | 0.93734 |
https://geo.libretexts.org/Bookshelves/Ancillary_Materials/Laboratory/Book%3A_Laboratory_Manual_For_Introductory_Geology_(Deline%2C_Harris_and_Tefend)/04%3A_Plate_Tectonics/4.07%3A_Lab_Exercise_(Part_C) | 1,590,820,779,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347407289.35/warc/CC-MAIN-20200530040743-20200530070743-00433.warc.gz | 381,204,617 | 22,299 | # 4.7: Lab Exercise (Part C)
## Part C – Plate Densities
An important property of geological plates is their density (mass/volume). Remember the asthenosphere has fluid-like properties, such that tectonic plates ‘float’ relative to their density. This property is called isostasy and is similar to buoyancy in water. For example, if a cargo ship has a full load of goods it will appear lower than if it were empty because the density of the ship is on average higher. Therefore, the relative density of two plates can control how they interact at a boundary and the types of geological features found along the border between the two plates. Measuring the density of rocks is fairly easy and can be done by first weighing the rocks and then calculating their volume. The latter is best done by a method called fluid displacement using a graduated cylinder. Water is added to the cylinder and the level is recorded, a rock is then added to the cylinder and the difference in water levels equals the volume of the rock. Density is then calculated as the mass divided by the volume (Figure 4.4).
The information needed to calculate density was collected for four rocks and can be used to answer the following questions including the weight (in grams) as well as the volume of water recorded by a graduated cylinder (in milliliters) before and after the rock was added. Note: each line on the graduated cylinder represents 10 ml. When measuring the volume please round to the nearest 10-milliliter line on the graduated cylinder.
Note
Surface tension will often cause the water level to curve up near the edges of the graduated cylinder creating a feature called a meniscus. To accurately measure the volume, use the lowest point the water looks to occupy.
15. The rock that most closely resembles the composition of continental crust based on the description in the previous section is:
a. A b. B c. C d. D
16. Based on the choice you made for question 15, what is the density of the rocks that make up continental crust? Please give your answer in grams/milliliter.
17. The rock that most closely resembles the composition of oceanic crust based on the description in the previous section is:
a. A b. B c. C d. D
18. Based on the choice you made for question 17, what is the density of the rocks that make up oceanic crust? Please give your answer in grams/milliliter.
19. Remember, because of isostasy the denser plate will be lower than the less dense plate. If oceanic and continental crust collided, based on their densities the __________ crust would sink below the ________crust.
a. continental; oceanic b. oceanic; continental | 589 | 2,812 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2020-24 | latest | en | 0.954015 |
http://goodriddlesnow.com/riddles/view/3029 | 1,493,409,772,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123048.37/warc/CC-MAIN-20170423031203-00297-ip-10-145-167-34.ec2.internal.warc.gz | 148,925,932 | 9,825 | # Why are owls like a quiz?
Question: Why are owls like a quiz?
Riddle Discussion
### Similar Riddles
##### Brother or not (medium)
Question: "The Captain is my brother," testified the soldier. But the Captain testified he didn't have a brother. Who is telling the truth?
##### Always There (medium)
Question: What goes through towns and over hills, but never moves?
##### The Traveler (medium)
Question: A traveler is walking in the jungle. Then he found 4 paths. In the first one has a really deep mud. When people go there, they never come back. Second path have broken tracks, which is really unsafe for the people who take this path. In the third path, there is lion who didn't eat and drink anything for 3 months, so it might be really hungry by now. Last one have dinosaurs and volcanoes, so it's a very dangerous place. So, which path is the best for the traveler to take.
##### Heaven and Hell (medium)
Question: There were 2 doors. Behind the 1 door, is hell and behind the other door, is heaven but you don't know which door will take you to heaven. In front of them, there were 2 brothers, which is guarding the door. One of the brother always lie and the other one always tell the truth. Of course you don't know who is lying and who is not. You only get 1 question to ask one of them to figure out which door leads to heaven. What question you might ask? Remember you only get to ask 1 question. it means that you can't ask one question each of them. that will be 2 questions. You only ask 1 question to any of them and no more. How do you do that?
##### The Weight Lifting Challenge (medium)
Question: Alex was the strongest man in the world. He can lift up something that weighs 1000 kg easily. He can even pick up five women at once without any pain. But one day, there was a Weight Lifting Challenge Show. All the top 5 strongest man entered including Alex. The challenge is that they each have to lift up a really heavy object which weighs 1005 kg for five minutes. The winner is who didn’t drop it until it reached the time. If they all did it, the judges will decide to give first place to a person who can lift it up easily and stand still with no trouble looking. If Alex was the strongest man, he’ll probably come first. But he came second instead when the challenge was over. Why? | 537 | 2,318 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2017-17 | longest | en | 0.977961 |
https://excelchamps.com/vba-functions/vba-hex-function/ | 1,606,200,917,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141171126.6/warc/CC-MAIN-20201124053841-20201124083841-00288.warc.gz | 294,615,360 | 19,468 | # How to use the VBA HEX Function (Syntax + Example)
HomeVBA Functions LIST (Category Wise)How to use the VBA HEX Function (Syntax + Example)
The VBA HEX function is listed under the data type conversion category of VBA functions. When you use it in a VBA code, it converts an expression into hexadecimal notation and returns the result as a string. Hexadecimal notation is just another base for representing numbers in, like binary. You can learn more about it from here.
Table of Content
Hex(Number)
## Arguments
• Number: The numeric value that you want to convert into hexadecimal.
## Example
To practically understand how to use VBA HEX function, you need to go through the below example where we have written a vba code by using it:
``````Sub example_HEX()
Range("B1").Value = Hex(Range("A1"))
End Sub``````
In the above code, we have used HEX to convert the number from the cell A1 into a hexadecimal notation and return the result as a string in the cell B1.
## Notes
• If the value specified is a value other than a number or a string that can’t be recognized as a number, VBA will return the run-time 13 error.
• If the given number is not a whole number, then it is rounded to the nearest whole number.
• If the supplied number is NULL then it will return NULL in the result. | 299 | 1,298 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-50 | latest | en | 0.756819 |
https://www.jiskha.com/display.cgi?id=1235098095 | 1,511,171,165,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805977.0/warc/CC-MAIN-20171120090419-20171120110419-00055.warc.gz | 807,207,574 | 3,821 | # math
posted by .
what is the least common positive integer that meets the following conditions:
divided by 7 with remainder 4
divided by 8 with remainder 5
divided by 9 with remainder 6
i thought you could add 7 and 4 to get 13, then divide 13 and 7 with r=4, but it has to be the same number for all of them.
• math -
divisible by 9 R 6
15 24 33 42 51 60 69 78 87 96
8 remainder 5
13 21 29 37 44 52 60
7 remainder 4
11 18 25 32 39 46 53 60 ah ha
• math -
oh i see!! thank you thank you!
## Similar Questions
1. ### Math - repost for Anonymous
Can someone show me the steps to these questions (I will provide the correct answers)?
2. ### math
what is the least common positive integer that meets the following conditions: divided by 7 with remainder 4 divided by 8 with remainder 5 divided by 9 with remainder 6 i thought you could add 7 and 4 to get 13, then divide 13 and …
3. ### Algebra 2
Find a positive integer smaller than 500 that has a remainder of 3 when divided by 5, a remainder of 6 when divided by 9, and a remainder of 8 when divided by 11.
4. ### number theory
Find the least positive integer that leaves the remainder 3 when divided by 7, remainder 4 when divided by 9, and remainder 8 when divided by 11
5. ### Math
Find the least positive integer that leaves the remainder 3 when divided by 7, remainder 4 when divided by 9, and remainder 8 when divided by 11. Using the Chinese Remainder Theorem.
6. ### Math
How many integers bewteen 200 and 500 inclusive leave a remainder 1 when divided by 7 and a remainder 3 when divided by 4?
7. ### Math
How many integers between 200 and 500 inclusive leave a remainder 1 when divided by 7 and a remainder 3 when divided by 4?
8. ### Math
Find the smallest positive integer that leaves a remainder of 5 when divided by 7, a remainder of 6 when divided by 11, and a remainder of 4 when divided by 13.
9. ### math
1.) when the expression 4x^2-3x-8 is divided by x-a, the remainder is 2. find the value of a. 2.) the polynomial 3x^3+mx^2+nx+5 leaves a remainder of 128 when divided by x-3 and a remainder of 4 when divided by x+1. calculate the remainder …
An unknown polynomial f(x) of degree 37 yields a remainder of 1 when divided by x – 1, a remainder of 3 when divided by x – 3, a remainder of 21 when divided by x – 5. Find the remainder when f(x) is divided by (x – 1)(x – …
More Similar Questions | 687 | 2,386 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-47 | latest | en | 0.928817 |
http://comunidadwindows.org/standard-error/standard-error-calculation-deviation.php | 1,537,668,597,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267158958.72/warc/CC-MAIN-20180923020407-20180923040807-00012.warc.gz | 52,943,701 | 5,035 | Home > Standard Error > Standard Error Calculation Deviation
# Standard Error Calculation Deviation
## Contents
Sampling from a distribution with a large standard deviation The first data set consists of the ages of 9,732 women who completed the 2012 Cherry Blossom run, a 10-mile race held Consider the following scenarios. Similar Worksheets Calculate Standard Deviation from Standard Error How to Calculate Standard Deviation from Probability & Samples Worksheet for how to Calculate Antilog Worksheet for how to Calculate Permutations nPr and Of course, T / n {\displaystyle T/n} is the sample mean x ¯ {\displaystyle {\bar {x}}} . http://comunidadwindows.org/standard-error/standard-deviation-error-calculation-in-excel.php
Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. The age data are in the data set run10 from the R package openintro that accompanies the textbook by Dietz [4] The graph shows the distribution of ages for the runners. Which towel will dry faster? The mean age was 33.88 years.
## Standard Error Formula Excel
However, the mean and standard deviation are descriptive statistics, whereas the standard error of the mean describes bounds on a random sampling process. For example, the sample mean is the usual estimator of a population mean. more than two times) by colleagues if they should plot/use the standard deviation or the standard error, here is a small post trying to clarify the meaning of these two metrics It can only be calculated if the mean is a non-zero value.
Because the 9,732 runners are the entire population, 33.88 years is the population mean, μ {\displaystyle \mu } , and 9.27 years is the population standard deviation, σ. If one survey has a standard error of $10,000 and the other has a standard error of$5,000, then the relative standard errors are 20% and 10% respectively. In regression analysis, the term "standard error" is also used in the phrase standard error of the regression to mean the ordinary least squares estimate of the standard deviation of the Standard Error Definition Naturally, the value of a statistic may vary from one sample to the next.
The sample standard deviation s = 10.23 is greater than the true population standard deviation σ = 9.27 years. Python - Make (a+b)(c+d) == a*c + b*c + a*d + b*d Generate a modulo rosace Why were Navajo code talkers used during WW2? Note that the standard error of the mean depends on the sample size, the standard error of the mean shrink to 0 as sample size increases to infinity. The relationship between the standard deviation of a statistic and the standard deviation of the data depends on what statistic we're talking about.
up vote 17 down vote favorite 6 Is it sensible to convert standard error to standard deviation? Standard Deviation Of The Mean The mean of these 20,000 samples from the age at first marriage population is 23.44, and the standard deviation of the 20,000 sample means is 1.18. For example, the standard error of the sample standard deviation (more info here) from a normally distributed sample of size $n$ is $$\sigma \cdot \frac{\Gamma( \frac{n-1}{2} )}{ \Gamma(n/2) } \cdot And if so, is this formula appropriate?$$SE = \frac{SD}{\sqrt{N}} standard-deviation standard-error share|improve this question edited Jul 16 '12 at 11:34 Macro 24.4k497130 asked Sep 13 '11 at 13:54 Bern 86113
## Standard Error In R
See unbiased estimation of standard deviation for further discussion. browse this site n is the size (number of observations) of the sample. Standard Error Formula Excel When the true underlying distribution is known to be Gaussian, although with unknown σ, then the resulting estimated distribution follows the Student t-distribution. Standard Error Formula Statistics For illustration, the graph below shows the distribution of the sample means for 20,000 samples, where each sample is of size n=16.
II. http://comunidadwindows.org/standard-error/standard-error-calculation-without-standard-deviation.php doi:10.2307/2682923. Assumptions and usage Further information: Confidence interval If its sampling distribution is normally distributed, the sample mean, its standard error, and the quantiles of the normal distribution can be used to By taking the mean of these values, we can get the average speed of sound in this medium.However, there are so many external factors that can influence the speed of sound, Difference Between Standard Deviation And Standard Error
For example, the U.S. Home > Research > Statistics > Standard Error of the Mean . . . The margin of error and the confidence interval are based on a quantitative measure of uncertainty: the standard error. http://comunidadwindows.org/standard-error/standard-deviation-vs-standard-error-calculation.php Notation The following notation is helpful, when we talk about the standard deviation and the standard error.
When the sampling fraction is large (approximately at 5% or more) in an enumerative study, the estimate of the standard error must be corrected by multiplying by a "finite population correction"[9] Margin Of Error Formula Because of random variation in sampling, the proportion or mean calculated using the sample will usually differ from the true proportion or mean in the entire population. It depends. | 1,142 | 5,310 | {"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 | 4 | CC-MAIN-2018-39 | latest | en | 0.855805 |
http://archive.ambermd.org/201102/0157.html | 1,566,335,459,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315618.73/warc/CC-MAIN-20190820200701-20190820222701-00163.warc.gz | 16,277,753 | 4,464 | # Re: [AMBER] AMBER11 MMPBSA.py tutorial section 3.6
From: <Yokota_Akihiro.takeda.co.jp>
Date: Thu, 10 Feb 2011 15:04:56 +0900
Dear Jason,
As you mentioned, PB energies can be decomposed to use geometrical
arguments to partition the grid points.
But I'd like to know how to do it and how accuracy it is,
because it might bring us ambiguous and inaccurate energies.
And you mean that I must read sander source code rather than MMPBSA.py
because sander calculates PB energies which are decomposed into each residue.
---------------------------------------------------------------
I checked temporary file of MMPBSA.py, and found 2 files below.
This shows me that PB energies are calculated by sander.
Input file: _MMPBSA_pb_decomp_rec.mdin
Output file: _MMPBSA_receptor_pb.mdout
---------------------------------------------------------------
So, I will read sander source code, and if I can't understand it,
I will ask someone who developed sander module.
And I am grateful to you for creating MMPBSA.py which can recognize
receptor and ligand automatically, and make sophisticated operations.
Best Regards,
Akihiro Yokota
-----Original Message-----
From: Jason Swails [mailto:jason.swails.gmail.com]
Sent: Thursday, February 10, 2011 11:46 AM
To: AMBER Mailing List
Subject: Re: [AMBER] AMBER11 MMPBSA.py tutorial section 3.6
Hi Yokota,
As the GB equations consist of pairwise, analytical terms, you're right that
they're much easier to decompose than PB energies.
Since I haven't looked at the PBSA/decomp code, I can't say any of this for
sure. If I'm mistaken here, I'm hoping that a PBSA developer will step in
and correct me.
PBSA works by placing your system in a grid and solving the PB equation via
a finite difference method. It assigns each grid point a specific charge,
dielectric constant, and *starting* electrostatic potential (and ionic
strength if you specified one), which it then solves iteratively to get the
converged electrostatic potential. Thus, at the end of the calculation,
each gridpoint has associated with it a charge, potential, and dielectric.
You can use geometrical arguments to partition the grid points into
different residues (i.e. each grid point belongs to its *closest* residue).
Once you've created this partition, you simply toss it into the equation
DELTA G = 1/2 integral(charge density * reaction field potential)
Since it's a discrete problem, the integral becomes a sum, and for each
residue you just sum over that residue's grid points. I agree that it's not
as straightforward as GB, but it seems reasonable to me.
Hope this helps (and that it's right :) ),
Jason
On Mon, Feb 7, 2011 at 2:54 AM, <Yokota_Akihiro.takeda.co.jp> wrote:
> Dear AMBER users,
>
> I have a question about "Polar Solvation" energy for each residue in
> MMPBSA.py tutorial section 3.6.
>
> In the middle of the file below, there is a comment "Energy Decomposition
> Analysis (All units kcal/mol):
> Poisson Boltzmann solvent", I think this means that "Polar Solvation(PB)"
> of "each residue" can be calculated.
>
>
> However PB energy cannot be decomposed straightly by the definition (On the
> other hand, GB energies can be decomposed).
> Does anyone teach me how the PB energies are decomposed into each residue?
>
> Akihiro Yokota
>
> _______________________________________________
> AMBER mailing list
> AMBER.ambermd.org
> http://lists.ambermd.org/mailman/listinfo/amber
>
```--
Jason M. Swails
Quantum Theory Project,
University of Florida | 860 | 3,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-35 | longest | en | 0.8663 |
https://www.onlinecourses.ooo/coupon/pro-mastering-advanced-excel-formulas-functions-charts/ | 1,723,410,438,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008472.68/warc/CC-MAIN-20240811204204-20240811234204-00083.warc.gz | 703,184,420 | 53,287 | # Pro Mastering Advanced Excel – Formulas, Functions & Charts
0
Pro Mastering Advanced Excel – Formulas, Functions & Charts, Mastering Microsoft Excel 2021: Step-by-Step Learning and Advanced Skills Bundle-Through Comprehensive and Real Examples.
## Course Description
Welcome to the next level of Excel mastery! Building upon the foundations laid in introductory course, this advanced Excel program is for individuals possessing a solid understanding of Excel and designed to catapult your skills to professional heights swiftly and effectively. Whether you’re a seasoned Excel user or have completed foundational course, this program empowers you to leapfrog to an advanced skill level.
This course delves deeper into the intricacies of Excel, Mastery of functions, formulas, and intricate data visualization techniques forms the core of this program. From complex nested functions to statistical analysis, this program unlocks the full potential of Excel‘ s capabilities.
Also In this course, we’ll explore how in Excel can enhance your data analysis and automation capabilities. Throughout the course, we will use real and comprehensive examples to highlight the power and versatility of Excel functions. By the end of the course, you’ll have a deep understanding of how to use functions to solve complex problems and simplify your workflow.
Note: I wanted to inform you that our course is undergoing modifications and updates on a weekly basis to enhance your learning experience. As a part of this process, I’m gradually uploading new course materials and expanding the teaching hours. Currently, only some parts of the course have been uploaded, but rest assured, more content will be added regularly. I’m dedicated to enriching the course content to provide you with the most comprehensive learning resources.
Thank you for your patience and understanding as we work towards making this learning journey even more beneficial for all of us. If you have any questions or need clarification, please don’t hesitate to reach out.
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0 | 599 | 3,088 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.899002 |
https://socratic.org/questions/what-is-the-derivative-of-arctan-1-x | 1,723,168,204,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640751424.48/warc/CC-MAIN-20240809013306-20240809043306-00607.warc.gz | 429,342,261 | 5,723 | # What is the derivative of arctan(1/x)?
Jul 31, 2015
The derivative is: $\frac{- 1}{{x}^{2} + 1}$
#### Explanation:
$\frac{d}{\mathrm{dx}} \arctan \left(x\right) = \frac{1}{1 + {x}^{2}}$
So
$\frac{d}{\mathrm{dx}} \arctan \left(u\right) = \frac{1}{1 + {u}^{2}} \frac{\mathrm{du}}{\mathrm{dx}}$
And
$\frac{d}{\mathrm{dx}} \arctan \left(\frac{1}{x}\right) = \frac{1}{1 + {\left(\frac{1}{x}\right)}^{2}} \cdot \frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right)$
$= \frac{1}{1 + \frac{1}{x} ^ 2} \cdot \frac{- 1}{x} ^ 2$
$= {x}^{2} / \left({x}^{2} + 1\right) \cdot \frac{- 1}{x} ^ 2$
$= \frac{- 1}{{x}^{2} + 1}$
Faster Method?
Use the fact that $\arctan \left(\frac{1}{x}\right) = a r c \cot \left(x\right)$
and
$\frac{d}{\mathrm{dx}} a r c \cot \left(x\right) = - \frac{1}{1 + {x}^{2}}$
to go straight to the answer. | 375 | 825 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-33 | latest | en | 0.3544 |
https://www.educationquizzes.com/11-plus/exam-illustrations-vr/logic-problems/print/ | 1,611,810,734,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835901.90/warc/CC-MAIN-20210128040619-20210128070619-00257.warc.gz | 752,050,320 | 5,263 | # 11-Plus Exam Illustrations - Verbal Reasoning Quiz - VR - Logic Problems (Questions)
Sometimes questions in Verbal Reasoning papers are one-offs – they don’t have a particular style and they are simply about making sense of a lot of information. Logic Problems are one of these.
You may have come across logic puzzles in certain magazines. If so, you’ll know all about them. They exercise the brain by making us use our deductive reasoning skills to find the answer. In fact, the answer is already there in the question – you just have to be able to dig it out!
## How Are Logic Problems Used In The Exam?
They usually take to form of some information expressed in words which the child must use to work out the answer to a question. If you’ve never seen a logic problem before then the best way to show you is through an example.
Example Question One
Simon, Jessica and Henry are at the same school and have signed up for at least one club. Simon has joined the hockey club and one other. Jessica does not like football and Henry has signed up for the chess club. Which of the following statements MUST be true?
A Simon has joined the general knowledge club B Jessica has joined the hockey club C Henry is good at chess D Simon has joined two clubs E Jessica is the only one to not like football
Let’s make sense of this. It is often worth reading through the answers one-by-one and applying them to the information given. When practising, allow your child to speak out loud. I encourage a child to do this as it enables me to check his or her logic and it should be something you are capable of doing as well. As ever, make sure that you’ve worked out the answer to any question you’re trialling with your child so as to give pointers as they are working them out.
So, statement A mentions Simon but not the hockey club. Therefore, we cannot know whether or not he has joined the general knowledge club - it isn’t possible to prove.
Statement B mentions Jessica and, although we are told that everyone has joined a club, it could be absolutely any. It could even be the football club – there’s no reason why not! Again, her joining the hockey club is not provable.
Statement C is possibly true – perhaps Henry is good at chess and that’s why he joined the club. Careless students will go for this as the first statement which seems reasonable. However, it is not a certainty and definitely can’t be proven using the information we are given.
Statement E suggests that Jessica is the only one to dislike football and is also a reasonable suggestion. However, we are given nothing provable in this regard and it is impossible to know for certain. Just because it doesn’t mention that Simon and Henry don’t like football doesn’t mean that they do or don’t like it. It’s speculation, it may be true, it mustn’t deflect from the logical certainty.
This leaves statement D, ‘Simon has joined two clubs’. Of course, it says in the preamble that Simon has joined the hockey club and one other, which means he has joined two clubs. He cannot have joined more, nor could he have joined fewer, otherwise the wording would have to be different. Statement D is therefore the correct answer.
Example Question Two
John has a cat and a dog. Raj has a goldfish. Sara has two rabbits, two goldfish and a rat. Daryl has twice the number of cats as John while both he and Raj own a snake. Sara doesn’t like either of John’s pet lizards but likes both his rabbits. Raj would like to have more than his two rats but is not allowed.
Who has the most pets?
This is a very wordy question but is easily broken down visually. Encourage your child to draw out a very simple table to store information and then read off it for the answer. Use ticks or crosses, numbers and abbreviations, but don’t write out full names as this will waste time.
Here’s a version of the table that could be written out for this question, with full names written to help explain what is happening:
John Sara Raj Daryl Cat 1 2 Dog 1 Rabbit 2 2 Goldfish 2 1 Snake 1 1 Rat 1 2 Lizard 2
The easiest way to arrange the information is to deal with one sentence at a time and that way you never get caught out. Taking pieces of information from different sections of the paragraph, for instance finding out everything to do with John first, is not a great idea. It means that some elements of a sentence must be ignored and may not get reviewed later. Encourage your child to be systematic.
The answer, using the table, is clearly John. He has six pets and that is more than the other children.
Once your child gets good at this, the table produced may look a bit like this:
J S R D c rr g cc d gg s s rr r rr ll
There is no reason to differentiate between rats and rabbits so the letter ‘r’ is fine; if the question needed it, ensure your child could make clear abbreviations to differentiate between them. You could use a tally chart instead of letters in this question but I would be reluctant, as if you lose your train of thought you cannot go back and start from where you were as you only have lines or numbers and don’t know whether you’ve dealt with all the cats, Raj’s rats or whatever!
It’s always good to be brief but abbreviate too much and you could take time to restart if you make an error.
## Sample Tests
So, now we understand what Logic Problems are and, more importantly, the quickest way to solve them. It’s time to put your child’s powers of deduction to the test.
There are four quizzes on the Education Quizzes site devoted to logic problems. Go through them with your child, offering advice if necessary, gleaned from the techniques we’ve shown you in this lesson.
You’ll find the quizzes in our Eleven Plus Verbal Reasoning section or, alternatively, you can follow these links: | 1,252 | 5,784 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.973534 |
https://matsci.org/t/access-forces-due-to-force-fields-only/31394 | 1,657,049,293,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104597905.85/warc/CC-MAIN-20220705174927-20220705204927-00557.warc.gz | 432,231,959 | 6,940 | # access forces due to force-fields only
Dear Users,
I am running a simulation in which a BN-nanotube is subjected both to internal forces due to the adopted force-fields (tersoff and ILP/graphene/hBN) and to external forces that I apply using the fix commands:
#Add viscous forces to B and N atoms in the nanotube com frame of reference
fix visB Boron viscous {vissubB} fix visN Nitrogen viscous {vissubN}
#Add external driving to the nanotube
I want to keep the x-velocity of the com ‘vcmtubex’ of the nanotube constant. A way of doing this is to compute the external driving (apart from coefficients) like this:
variable fext equal (\${vtarget}-v_vcmtubex)-v/c_Ftotinternalx
Where ‘Ftotinternalx’ is the total force acting on the nanotube in the x-direction, due only to the internal forces.
In general, is there a way in LAMMPS to access the total force due to the force-fields only, before external forces are applied?
Davide
Dear Users,
I am running a simulation in which a BN-nanotube is subjected both to internal forces due to the adopted force-fields (tersoff and ILP/graphene/hBN) and to external forces that I apply using the fix commands:
#Add viscous forces to B and N atoms in the nanotube com frame of reference
fix visB Boron viscous \{vissubB\} fix visN Nitrogen viscous {vissubN}
#Add external driving to the nanotube
I want to keep the x-velocity of the com 'vcmtubex' of the nanotube constant. A way of doing this is to compute the external driving (apart from coefficients) like this:
as simpler way of keeping the velocity constant is to just set an
initial velocity and then set the force to zero. e.g. with:
velocity nanotube set 0.1 NULL NULL
fix fadtube nanotube setforce 0.0 NULL NULL
without force, the velocity does not change. what you are doing seems
needlessly complex. is there a reason you cannot use this simple
method?
axel.
Dear Axel,
Thanks for the quick response!
as simpler way of keeping the velocity constant is to just set an
initial velocity and then set the force to zero. e.g. with:
velocity nanotube set 0.1 NULL NULL
fix fadtube nanotube setforce 0.0 NULL NULL
without force, the velocity does not change. what you are doing seems
needlessly complex. is there a reason you cannot use this simple
method?
Yes, the quantity I am interested in is exactly that external force fext, which after the first time step equals the internal force contribution only.
Dear Axel,
Thanks for the quick response!
as simpler way of keeping the velocity constant is to just set an
initial velocity and then set the force to zero. e.g. with:
velocity nanotube set 0.1 NULL NULL
fix fadtube nanotube setforce 0.0 NULL NULL
without force, the velocity does not change. what you are doing seems
needlessly complex. is there a reason you cannot use this simple
method?
Yes, the quantity I am interested in is exactly that external force fext, which after the first time step equals the internal force contribution only.
sorry, but i am confused here. i don't think you are answering my
question, since there *is* no external force in what i am proposing.
axel.
The system is a nanotube sliding over a rigid substrate, I missed this information, sorry for that.
The total internal force acting on the nanotube is given by (1) the sum of all forces due to the boron-nitrogen interactions within the nanotube plus (2) the sum of all forces from the interactions of the nanotube with the substrate.
Contribution (1) is zero. Contribution (2) is what I am interested in monitoring, and I would like to do it at fixed center-of-mass velocity of the nanotube.
Now the picture should be more clear.
If I understand the script you are suggesting, the force acting along x on each atom of the nanotube would be set to zero.
The result of a compute command like “compute FFforcex nanotube reduce ave fx” would be “FFforcex=0”.
So I would still have no access to the internal force during simulation (though I could obtain it by postprocessing the trajectory).
Please correct me if I am wrong.
In any case, your first command “velocity nanotube set 0.1 NULL NULL” is setting the velocity along x of all atoms in the nanotube to a constant value.
This will not allow the nanotube to deform along x. Moreover, the nanotube may want to roll instead of just sliding over the substrate.
These are all motivations why I am trying to implement the protocol I reported in my first email.
Best,
Davide
The system is a nanotube sliding over a rigid substrate, I missed this information, sorry for that.
The total internal force acting on the nanotube is given by (1) the sum of all forces due to the boron-nitrogen interactions within the nanotube plus (2) the sum of all forces from the interactions of the nanotube with the substrate.
Contribution (1) is zero. Contribution (2) is what I am interested in monitoring, and I would like to do it at fixed center-of-mass velocity of the nanotube.
Now the picture should be more clear.
If I understand the script you are suggesting, the force acting along x on each atom of the nanotube would be set to zero.
The result of a compute command like "compute FFforcex nanotube reduce ave fx" would be "FFforcex=0".
So I would still have no access to the internal force during simulation (though I could obtain it by postprocessing the trajectory).
Please correct me if I am wrong.
In any case, your first command "velocity nanotube set 0.1 NULL NULL" is setting the velocity along x of all atoms in the nanotube to a constant value.
This will not allow the nanotube to deform along x. Moreover, the nanotube may want to roll instead of just sliding over the substrate.
These are all motivations why I am trying to implement the protocol I reported in my first email.
ok. i see. this sounds like a candidate for fix smd in constant
velocity mode. fix smd attaches a spring between the center of mass of
a group and a reference point and you can move that reference point at
constant velocity and implicitly get the force between the moving
object and its environment from the spring force, which you can
monitor.
axel.
Yes, that’s an option, of course the com velocity of the nanotube will not be constant in time using fix smd.
The point is that this fix introduces an external parameter via the spring constant, which needs to be tested/tuned for different systems, and I’d avoid that if possible.
I already did simulations implementing the fixed-com-velocity protocol in my own MD code, which is straightforward if you have access to the internal forces,
I wanted to check if there was a way to do it also in LAMMPS.
Thanks for the help,
Davide
Yes, that's an option, of course the com velocity of the nanotube will not be constant in time using fix smd.
The point is that this fix introduces an external parameter via the spring constant, which needs to be tested/tuned for different systems, and I'd avoid that if possible.
I already did simulations implementing the fixed-com-velocity protocol in my own MD code, which is straightforward if you have access to the internal forces,
I wanted to check if there was a way to do it also in LAMMPS.
since LAMMPS does velocity verlet timestepping and if you want to
constrain the velocity exactly, you would have to write a custom
integrator fix, so that you can enforce your constraint on both
this is usually done in C++, but you could also try to use fix
python/move and then implement your protocol in python script code
invoked from there and access the internal force arrays from the
python wrapper for LAMMPS.
axel.
Thanks for the tip!
Davide
I’m not following these details, but this Q:
In general, is there a way in LAMMPS to access the total force due to the force-fields only, before external forces are applied?
makes me think of fix store/force which archives the forces
after pair/bond/etc forces are computed but before
external forces. If that fix is defined, any other part
of LAMMPS can potentially access them. E.g. to dump
the interatomic forces, w/out external forces added.
Steve
Thanks Steve.
I read the documentation on fix store/force, and it looks like it is what I need.
I will give it a try.
Davide
“fix store/force” solves the problem.
Thanks,
Davide | 1,963 | 8,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-27 | longest | en | 0.882176 |
https://numpy.org/doc/1.15/reference/generated/numpy.matrix.max.html | 1,717,084,346,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971668873.95/warc/CC-MAIN-20240530145337-20240530175337-00886.warc.gz | 368,527,430 | 3,284 | #### Previous topic
numpy.matrix.itemset
#### Next topic
numpy.matrix.mean
# numpy.matrix.max¶
`matrix.``max`(axis=None, out=None)[source]
Return the maximum value along an axis.
Parameters: See `amax` for complete descriptions
Notes
This is the same as `ndarray.max`, but returns a `matrix` object where `ndarray.max` would return an ndarray.
Examples
```>>> x = np.matrix(np.arange(12).reshape((3,4))); x
matrix([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
>>> x.max()
11
>>> x.max(0)
matrix([[ 8, 9, 10, 11]])
>>> x.max(1)
matrix([[ 3],
[ 7],
[11]])
``` | 200 | 579 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.531768 |
https://pennylane.readthedocs.io/en/latest/code/api/pennylane.eigvals.html | 1,660,786,979,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573145.32/warc/CC-MAIN-20220818003501-20220818033501-00468.warc.gz | 424,774,917 | 11,091 | # qml.eigvals¶
eigvals(op, k=1, which='SA')[source]
The eigenvalues of one or more operations.
Note
For a SparseHamiltonian object, the eigenvalues are computed with the efficient scipy.sparse.linalg.eigsh method which returns $$k$$ eigenvalues. The default value of $$k$$ is $$1$$. For an $$N \times N$$ sparse matrix, $$k$$ must be smaller than $$N - 1$$, otherwise scipy.sparse.linalg.eigsh fails. If the requested $$k$$ is equal or larger than $$N - 1$$, the regular qml.math.linalg.eigvalsh is applied on the dense matrix. For more details see the scipy.sparse.linalg.eigsh documentation.
Parameters
• op (Operator, pennylane.QNode, QuantumTape, or Callable) – An operator, quantum node, tape, or function that applies quantum operations.
• k (int) – The number of eigenvalues to be returned for a SparseHamiltonian.
• which (str) – Method for computing the eigenvalues of a SparseHamiltonian. The possible methods are 'LM' (largest in magnitude), 'SM' (smallest in magnitude), 'LA' (largest algebraic), 'SA' (smallest algebraic) and 'BE' ($$k/2$$ from each end of the spectrum).
Returns
If an operator is provided as input, the eigenvalues are returned directly. If a QNode or quantum function is provided as input, a function which accepts the same arguments as the QNode or quantum function is returned. When called, this function will return the unitary matrix in the appropriate autodiff framework (Autograd, TensorFlow, PyTorch, JAX) given its parameters.
Return type
tensor_like or function
Example
Given an operation, qml.eigvals returns the eigenvalues:
>>> op = qml.PauliZ(0) @ qml.PauliX(1) - 0.5 * qml.PauliY(1)
>>> qml.eigvals(op)
array([-1.11803399, -1.11803399, 1.11803399, 1.11803399])
It can also be used in a functional form:
>>> x = torch.tensor(0.6, requires_grad=True)
>>> eigval_fn = qml.eigvals(qml.RX)
>>> eigval_fn(x, wires=0)
In its functional form, it is fully differentiable with respect to gate arguments:
>>> loss = torch.real(torch.sum(eigval_fn(x, wires=0)))
>>> loss.backward()
tensor(-0.2955)
This operator transform can also be applied to QNodes, tapes, and quantum functions that contain multiple operations; see Usage Details below for more details.
qml.eigvals can also be used with QNodes, tapes, or quantum functions that contain multiple operations. However, in this situation, eigenvalues may be computed numerically. This can lead to a large computational overhead for a large number of wires.
Consider the following quantum function:
def circuit(theta):
qml.RX(theta, wires=1)
qml.PauliZ(wires=0)
We can use qml.eigvals to generate a new function that returns the eigenvalues corresponding to the function circuit:
>>> eigvals_fn = qml.eigvals(circuit)
>>> theta = np.pi / 4
>>> eigvals_fn(theta)
array([ 0.92387953+0.38268343j, 0.92387953-0.38268343j,
-0.92387953+0.38268343j, -0.92387953-0.38268343j]) | 784 | 2,883 | {"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} | 2.671875 | 3 | CC-MAIN-2022-33 | latest | en | 0.645645 |
https://www.jiskha.com/questions/1008311/An-electron-transition-in-a-single-H-atom-from-n-3-to-n-2-results-in-the-release | 1,575,561,961,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540481076.11/warc/CC-MAIN-20191205141605-20191205165605-00013.warc.gz | 752,965,766 | 5,425 | # Chemistry
An electron transition in a single H-atom from n=3 to n=2 results in the release of a single photon. What is the wavelength of this photon in nanometers?
How do I solve this?
1. 👍 0
2. 👎 0
3. 👁 129
1. 1/wavelength = R(1/4 - 1/9)
R = 1.0973732E7
The 1/4 is 1/2^2
The 1/9 is 1/3^2
1. 👍 0
2. 👎 0
2. I tried that but my answer is 6.56*10^(-7) and when I plug it in the site tells me my answer is wrong what went wrong?
1. 👍 0
2. 👎 0
posted by Johnny
3. The formula I gave you gives wavelength in meters. The question asks for nanometers. Make the conversion. 1 m = 10^9 nm.
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2. 👎 0
4. Thanksss!!! :D it worked much appreciated!
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2. 👎 0
posted by Johnny
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More Similar Questions | 978 | 3,391 | {"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} | 3.078125 | 3 | CC-MAIN-2019-51 | latest | en | 0.934887 |
https://tilapia-niederrhein.de/26559/crushing-stress-defination.html | 1,600,816,767,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400208095.31/warc/CC-MAIN-20200922224013-20200923014013-00403.warc.gz | 667,740,075 | 5,707 | # crushing stress defination
Home > crushing stress defination
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### BASICS OF PIPE STRESS ANALYSIS: A PRESENTATIONPart 1
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Crushing stress tends to push a material and acts normal to its cross section plane, in the opposite direction of tension. Crushing stresses are compressive stresses and could also be bearing
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Looking for crushing strength? Find out information about crushing strength. The compressive stress required to cause a solid to fail by fracture in essence, it is the resistance of the solid to vertical pressure placed upon it. Explanation of crushing strength
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### crushing stress defination 3hh
crushing stress defination. Email: [email protected] Corrosionpedia . Stress is defined as the force per unit area of a material: Stress = Force / Cross sectional area. Tensile stress is a measurement of the strength of a material. Therefore, tensile stress refers to a force that attempts to pull apart or stretch a material.
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### Crushing definition of crushing by The Free Dictionary
Define crushing. crushing synonyms, crushing pronunciation, crushing translation, English dictionary definition of crushing. v. crushed, crush·ing, crush·es v. tr. 1. a. To press between opposing bodies so as to break, compress, or injure: The falling rock crushed the car. b.
### Crush Definition of Crush by MerriamWebster
Crush definition is to squeeze or force by pressure so as to alter or destroy structure. How to use crush in a sentence. Synonym Discussion of crush. | 1,374 | 6,247 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.874471 |
https://www.basic-mathematics.com/phone-card-word-problem.html | 1,718,666,655,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861741.14/warc/CC-MAIN-20240617215859-20240618005859-00343.warc.gz | 603,350,728 | 9,730 | # Phone card word problem
by constance
(tallahassee)
Elsa purchased a phone card for \$25. Long distance cost .11 a minute using this card. Elsa used her card only once to make a long distance call. If the remaining credit on her card is \$21.59, how many minutes did her call last.
Solution
Elsa spent 25 minus 21.59
25 - 21.59 = 3.41
Let y be the amount of minutes
0.11 × y = 3.41
Divide both sides by 0.11
0.11y / 0.11 = 3.41 / 0.11
y = 31
Her call lasted 31 minutes.
100 Tough Algebra Word Problems.
If you can solve these problems with no help, you must be a genius!
Recommended | 180 | 597 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.888548 |
https://www.physicsforums.com/threads/how-many-nodal-and-antinodal-lines-are-there.662123/ | 1,511,222,878,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806258.88/warc/CC-MAIN-20171120223020-20171121003020-00487.warc.gz | 867,452,319 | 14,225 | # How many nodal and antinodal lines are there?
1. Jan 2, 2013
### Ilan
1. The problem statement, all variables and given/known data
In an interference pattern produced by two point sources separated by 3.5cm and vibrating in phase, the first order nodal line forms an angle of 20 degrees with the perpendicular bisector of the line segment joining the two sources. What is the wavelength? How many nodal and antinodal lines are there?
2. Relevant equations
PD = m * lambda
PD = d * sin(theta)
3. The attempt at a solution
3.5sin20=0.5lambda
lambda = 2.4cm
I'm not sure how to go about calculating the number of lines. The answer is up to first order antinodal and up to 2nd order nodal.
2. Jan 2, 2013
### PeterO
Nodal lines occur in places where the path difference from the sources is 0.5λ, 1.5λ, 2.5λ, ...
If your wavelength of 2.4cm is correct, that means 1.2cm, 3.6cm, 6.0cm , etc .
How do those differences fit alongside a source separation of 3.5 cm ?
Note: I have not checked, at all, the correctness of your λ = 2.4cm answer | 305 | 1,046 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-47 | longest | en | 0.893555 |
https://numbas.mathcentre.ac.uk/question/17429/arithmetic-progression-the-nth-term-of-a-series/ | 1,597,255,854,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738913.60/warc/CC-MAIN-20200812171125-20200812201125-00102.warc.gz | 423,155,122 | 137,892 | Try to answer the following questions here:
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Series
### Feedback
From users who are members of Series :
### History
#### Checkpoint description
Describe what's changed since the last checkpoint.
#### Frank Doheny2 years, 4 months ago
Gave some feedback: Ready to use
Published this.
#### Frank Doheny3 years, 9 months ago
Created this.
Arithmetic progression: The nth term of a series Ready to use Frank Doheny 20/03/2018 12:44
Arithmetic progression: The sum of the first n terms of a series Ready to use Frank Doheny 20/03/2018 12:44
Geometric progression: The nth term of a series Ready to use Frank Doheny 20/03/2018 08:28
Geometric progression: The sum of the first n terms of a geometric progression Ready to use Frank Doheny 20/03/2018 08:27
Andrew's copy of Arithmetic progression: The nth term of a series draft Andrew Dunbar 30/06/2017 22:04
Andrew's copy of Arithmetic progression: The sum of the first n terms of a series draft Andrew Dunbar 30/06/2017 22:04
Andrew's copy of Geometric progression: The nth term of a series draft Andrew Dunbar 30/06/2017 22:04
Add two numbers Should not be used Alasdair McAndrew 26/04/2018 06:13
Derivatives of polynomials draft Alasdair McAndrew 26/04/2018 06:14
Assignment 2 - Arithmetic progression: nth term and sum of n terms Ready to use Angharad Thomas 20/04/2020 12:13
Assignment 2 - Geometric progression: nth term and sum of n terms Ready to use Angharad Thomas 20/04/2020 12:13
Simon's copy of Arithmetic progression: The nth term of a series draft Simon Thomas 25/02/2019 10:34
Simon's copy of Arithmetic progression: The sum of the first n terms of a series draft Simon Thomas 25/02/2019 10:36
Simon's copy of Geometric progression: The nth term of a series draft Simon Thomas 25/02/2019 10:55
Simon's copy of Geometric progression: The sum of the first n terms of a geometric progression draft Simon Thomas 25/02/2019 10:59
Geometric progression: The sum to infinity draft Simon Thomas 25/02/2019 11:23
Arithmetic progression: The nth term of a series draft Xiaodan Leng 10/07/2019 23:16
Arithmetic progression: The nth term of a series draft Xiaodan Leng 11/07/2019 00:22
Arithmetic progression: The sum of the first n terms of a series draft Xiaodan Leng 11/07/2019 00:23
Geometric progression: The nth term of a series draft Xiaodan Leng 11/07/2019 00:24
Geometric progression: The sum of the first n terms of a geometric progression draft Xiaodan Leng 11/07/2019 00:24
Assignment 2 - Geometric progression: nth term and sum to infinity Ready to use Angharad Thomas 29/11/2019 13:52
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Give a worked solution to the whole question. | 1,641 | 7,151 | {"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} | 2.59375 | 3 | CC-MAIN-2020-34 | latest | en | 0.830207 |
https://numbermatics.com/n/196325784/ | 1,627,894,442,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154310.16/warc/CC-MAIN-20210802075003-20210802105003-00495.warc.gz | 443,122,697 | 6,695 | # 196325784
## 196,325,784 is an even composite number composed of four prime numbers multiplied together.
What does the number 196325784 look like?
This visualization shows the relationship between its 4 prime factors (large circles) and 48 divisors.
196325784 is an even composite number. It is composed of four distinct prime numbers multiplied together. It has a total of forty-eight divisors.
## Prime factorization of 196325784:
### 23 × 32 × 19 × 143513
(2 × 2 × 2 × 3 × 3 × 19 × 143513)
See below for interesting mathematical facts about the number 196325784 from the Numbermatics database.
### Names of 196325784
• Cardinal: 196325784 can be written as One hundred ninety-six million, three hundred twenty-five thousand, seven hundred eighty-four.
### Scientific notation
• Scientific notation: 1.96325784 × 108
### Factors of 196325784
• Number of distinct prime factors ω(n): 4
• Total number of prime factors Ω(n): 7
• Sum of prime factors: 143537
### Divisors of 196325784
• Number of divisors d(n): 48
• Complete list of divisors:
• Sum of all divisors σ(n): 559704600
• Sum of proper divisors (its aliquot sum) s(n): 363378816
• 196325784 is an abundant number, because the sum of its proper divisors (363378816) is greater than itself. Its abundance is 167053032
### Bases of 196325784
• Binary: 10111011001110110001100110002
• Base-36: 38VXY0
### Squares and roots of 196325784
• 196325784 squared (1963257842) is 38543813463214656
• 196325784 cubed (1963257843) is 7567144396515372499490304
• The square root of 196325784 is 14011.6303119943
• The cube root of 196325784 is 581.2002335469
### Scales and comparisons
How big is 196325784?
• 196,325,784 seconds is equal to 6 years, 12 weeks, 4 days, 6 hours, 56 minutes, 24 seconds.
• To count from 1 to 196,325,784 would take you about nine years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 196325784 cubic inches would be around 48.4 feet tall.
### Recreational maths with 196325784
• 196325784 backwards is 487523691
• The number of decimal digits it has is: 9
• The sum of 196325784's digits is 45
• More coming soon!
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https://www.bankiq.in/mini-mock-160-maths-reasoning-pdf/ | 1,575,784,525,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540506459.47/warc/CC-MAIN-20191208044407-20191208072407-00132.warc.gz | 630,326,411 | 27,641 | ## Mini Mock – 160 | Maths & Reasoning | PDF Download
• In a shop seven shirts of different colors Blue, Yellow, White, Black, Orange, Green, Pink are hanging in a row facing North but not necessarily in the same order. How many shirts are hanging between Pink and Green?
• Statement I: Blue shirt is third to the left of Black which is immediate left of Yellow. Yellow shirt is third to right of Orange which is third to the right of White. Pink and Black are hanging adjacent to each other.
• Statement II: White shirt is second to the left of Blue which is second to the left of Pink. Black shirt is second to the right of Orange which is second to the right of Green.
• A, B, C, D, E, F and G are living in seven floor building, only one person lives in one floor. The bottommost floor is numbered one, the floor above is numbered two and so on. All seven have seven different cars Honda City, Kwid, Verna, Swift, Baleno, Fabia and Ecosport. Person who lives on the top floor owns which car?
• Statement I: B lives between A and person who has Fabia. Only one person lives above the person who has Verna. F lives between C and G but none of them own Baleno. A who has Ecosport sits immediately below G.
• Statement II: Only one person lives below to the person who has Baleno. Two persons live between A and the person who has Verna. F who has Kwid lives immediate above the person who has Swift
• Statements:
• D > S ≥ N = G < Z,
• K ≤ X = C ≥ S > B,
• F ≤ V ≥ X = E ≠ W
• Conclusions:
• I. N ≤ X,
• II. V ≥ G,
• III. X < E
• ‘A # B’ means ‘A is smaller than B’
• ‘A © B’ means ‘A is greater than B’
• ‘A π B’ means ‘A is either smaller than or equal to B’
• ‘A \$ B’ means ‘A is either greater than or equal to B’
• ‘A % B’ means ‘A is neither smaller nor greater than B’
• In each of the following questions, assuming the given statements to be true, find out which of the two conclusions I and II given below them is/are definitely true. Give answer
• Statement:
• K # R,
• G \$ K,
• F % X
• Conclusions:
• II. F # K
• IV. X # K
• In which of the following expressions will the expressions ‘W < T’ as well as ‘K ≥ G’ be definitely true?
• a) K ≥ R > W = G ≤ T
• b) K ≥ R ≥ W = G < T
• c) K > R ≥ W = G ≤ T
• d) K ≥ R > W = G < T
• e) None of these
• Statements:
• Only white are orange.
• Only a few yellow are green.
• Only green are blue.
• Few white are yellow.
• Conclusions:
• I. Every white can be green.
• II. Every green can be yellow.
• III. Few orange can be yellow.
• IV. Every yellow can be green.
• V. Few blue are yellow. | 714 | 2,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2019-51 | longest | en | 0.9431 |
https://www.distancesto.com/fuel-cost/cz/breclav-to-valtice/history/1030846.html | 1,726,404,341,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00730.warc.gz | 697,371,737 | 14,644 | # INR0.40 Total Cost of Fuel from Breclav to Valtice
Your trip to Valtice will consume a total of 0.16 gallons of fuel.
Trip start from Breclav, CZ and ends at Valtice, CZ.
Trip (6.4 mi) Breclav » Valtice
The map above shows you the route which was used to calculate fuel cost and consumption.
### Fuel Calculations Summary
Fuel calculations start from Breclav, Czechia and end at Valtice, Czechia.
Fuel is costing you INR2.50 per gallon and your vehicle is consuming 40 MPG. The formula can be changed here.
The driving distance from Breclav to Valtice plays a major role in the cost of your trip due to the amount of fuel that is being consumed. If you need to analyze the details of the distance for this trip, you may do so by viewing the distance from Breclav to Valtice.
Or maybe you'd like to see why certain roads were chosen for the route. You can do so by zooming in on the map and choosing different views. Take a look now by viewing the road map from Breclav to Valtice.
Of course, what good is it knowing the cost of the trip and seeing how to get there if you don't have exact directions? Well it is possible to get exact driving directions from Breclav to Valtice.
Did you also know that how elevated the land is can have an impact on fuel consumption and cost? Well, if areas on the way to Valtice are highly elevated, your vehicle may have to consume more gas because the engine would need to work harder to make it up there. In some cases, certain vehicles may not even be able to climb up the land. To find out, see route elevation from Breclav to Valtice.
Travel time is of the essence when it comes to traveling which is why calculating the travel time is of the utmost importance. See the travel time from Breclav to Valtice.
Speaking of travel time, a flight to Valtice takes up a lot less. How much less? Flight time from Breclav to Valtice.
Cost is of course why we are here... so is it worth flying? Well this depends on how far your trip is. Planes get to where they need to go faster due to the speed and shorter distance that they travel. They travel shorter distances due to their ability to fly straight to their destination rather than having to worry about roads and obstacles that are in a motor vehicle's way. You can see for yourself the flight route on a map by viewing the flight distance from Breclav to Valtice.
*The cost above should be taken as an ESTIMATE due to factors which affect fuel consumption and cost of fuel.
Recent Fuel Calculations for Breclav CZ:
Fuel Cost from Breclav to Prague
Fuel Cost from Breclav to 562 01
Fuel Cost from Breclav to Valdice
Fuel Cost from Breclav to Lednice
Fuel Cost from Breclav to 588 56 Telč | 648 | 2,690 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-38 | latest | en | 0.959414 |
https://eduzip.com/previous-examinations/cbse-ugc-net-june-2015-computer-sciencepaper-ii.html | 1,638,211,089,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358786.67/warc/CC-MAIN-20211129164711-20211129194711-00406.warc.gz | 285,424,288 | 12,617 | # UGC NET Computer Science Paper II Solved Question Paper
UGC NET Computer Science June 2015 solved question paper, Answer Key are available here. You can also download the pdf of UGC NET computer science solved papers from here.
1. How many strings of 5 digit have the property that the sum of their digits is 7 ?
[A] 66
[B] 330
[C] 495
[D] 99
2. Consider an experiment of tossing two fair dice, one black and one red. What is the probability that the number on the black die divides the number on red die ?
[A] 22/36
[B] 12/36
[C] 14/36
[D] 6/36
3. In how many ways can 15 indistinguishable fish be placed into 5 different ponds, so that each pond consist atleast one fish ?
[A] 1001
[B] 3876
[C] 775
[D] 200
4. Consider the following statements:
(a) Depth-first search is used to traverse a rooted tree
(b) Pre-order, Post-order and Inorder are used to list the vertices of an ordered rooted tree.
(c) Huffman's algorithm is used to find an optimal binary tree with given weights.
(d) Topological sorting provides a labelling such that the parents have larger labels than their children.
Which of the above statements are true ?
[A] (a) and (b)
[B] (c) and (d)
[C] (a), (b) and (c)
[D] (a), (b), (c) and (d)
5. Consider the Hamiltonian Graph (G) with no loops and parallel edges. Which of the following is true with respect to this graph (G) ?
deg(v) ≥ n/2 for each vertex of G
|E(G)| ≥ 1/2 (n-1)(n-2) + 2 edges
deg(v) + deg(w) ≥ n for every v and w not connected by an edge
[A] (a) and (b)
[B] (b) and (c)
[C] (a) and (c)
[D] (a), (b) and (c)
6. Consider the following statements:
(a) Boolean expression and logic networks correspond to labelled acyclic digraphs.
(b) Optimal boolean expression may not correspond to simplest networks
(c) Choosing essential blocks first in a Karnaugh map and then greedily choosing the largest remaining blocks to cover may not give an optimal expression.
Which of these statement(s) is/are correct ?
[A] (a) only
[B] (b) only
[C] (a) and (b)
[D] (a), (b) and (c)
7. Consider a full-adder with the following input values:
(a) x=1, y=0 and Ci(carry input) = 0
(b) x=0, y=1 and Ci =1
Compute the values of S(sum) and C0(carry output) for the above input values.
[A] S=1, CO=0 and S=0, C = 1
[B] S=0, C0 = 0 and S =1, C0 =1
[C] S = 1, C0 = 1 and S = 0, C0 = 0
[D] S = 0, C0 = 1 and S = 1, C0 = 0
8. "If my computations are correct and I pay the electric bill, then i will run out of money. If i don't pay the electric bill, the power will be turned off. Therefore, if I don't run out of money and power is still on, then my computations are incorrect"
Convert this argument into logical notation using the variables c,b,r,p for propositions of computations, electric bills, out of money and power respectively. where ~ means NOT)
[A] if (c^b) ~ r and ~b -> ~p, then (~r ^ p) -> ~c
[B] if (cvb) -> r and ~b -> ~p, then (r^p) -> c
[C] if (c^b) -> r and ~p -> ~b, then (~rvp) -> ~c
[D] if(cvb) -> r and ~b -> ~p, then (~r^p) -> ~c
9. Match the Following:
List I List II (a) (p->q) <=> (~q -> ~p) (i) Contrapositive (b) [(p^q)->r]<=>[p->(q->r)] (ii) Exportation law (c) (p->q) <=> [(p^~q) ->0] (iii) Reductio ad absurdum (d) (p<->q) <=> [(p->q)^(q->p)] (iv) Equivalence
(a) (b) (c) (d)
(1) (i) (ii) (iii) (iv)
(2) (ii) (iii) (i) (iv)
(3) (iii) (ii) (iv) (i)
(4) (iv) (ii) (iii) (i)
[A] A
[B] B
[C] C
[D] D
10. Consider a proposition given as :
"x≥6, if x2≥25 and its proof as:
if x≥6, then x2=x.x≥6.6=36≥25
Which of the following is correct w.r.to the given proposition and its proof ?
The proof shows the converse of what is to be proved
The proof stats by assuming what is to be shown
The proof is correct and there is nothing wrong
[A] (a) only
[B] (c) only
[C] (a) and (b)
[D] (b) only
Are these questions helpful for you?
How to solve question no. 13 please explain it sir... | 1,284 | 3,857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.901504 |
https://netlib.org/lapack/explore-html/d6/d41/zget52_8f_source.html | 1,716,728,329,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058876.64/warc/CC-MAIN-20240526104835-20240526134835-00338.warc.gz | 369,001,313 | 8,463 | LAPACK 3.12.0 LAPACK: Linear Algebra PACKage
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zget52.f
Go to the documentation of this file.
1*> \brief \b ZGET52
2*
3* =========== DOCUMENTATION ===========
4*
5* Online html documentation available at
6* http://www.netlib.org/lapack/explore-html/
7*
8* Definition:
9* ===========
10*
11* SUBROUTINE ZGET52( LEFT, N, A, LDA, B, LDB, E, LDE, ALPHA, BETA,
12* WORK, RWORK, RESULT )
13*
14* .. Scalar Arguments ..
15* LOGICAL LEFT
16* INTEGER LDA, LDB, LDE, N
17* ..
18* .. Array Arguments ..
19* DOUBLE PRECISION RESULT( 2 ), RWORK( * )
20* COMPLEX*16 A( LDA, * ), ALPHA( * ), B( LDB, * ),
21* \$ BETA( * ), E( LDE, * ), WORK( * )
22* ..
23*
24*
25*> \par Purpose:
26* =============
27*>
28*> \verbatim
29*>
30*> ZGET52 does an eigenvector check for the generalized eigenvalue
31*> problem.
32*>
33*> The basic test for right eigenvectors is:
34*>
35*> | b(i) A E(i) - a(i) B E(i) |
36*> RESULT(1) = max -------------------------------
37*> i n ulp max( |b(i) A|, |a(i) B| )
38*>
39*> using the 1-norm. Here, a(i)/b(i) = w is the i-th generalized
40*> eigenvalue of A - w B, or, equivalently, b(i)/a(i) = m is the i-th
41*> generalized eigenvalue of m A - B.
42*>
43*> H H _ _
44*> For left eigenvectors, A , B , a, and b are used.
45*>
46*> ZGET52 also tests the normalization of E. Each eigenvector is
47*> supposed to be normalized so that the maximum "absolute value"
48*> of its elements is 1, where in this case, "absolute value"
49*> of a complex value x is |Re(x)| + |Im(x)| ; let us call this
50*> maximum "absolute value" norm of a vector v M(v).
51*> If a(i)=b(i)=0, then the eigenvector is set to be the jth coordinate
52*> vector. The normalization test is:
53*>
54*> RESULT(2) = max | M(v(i)) - 1 | / ( n ulp )
55*> eigenvectors v(i)
56*>
57*> \endverbatim
58*
59* Arguments:
60* ==========
61*
62*> \param[in] LEFT
63*> \verbatim
64*> LEFT is LOGICAL
65*> =.TRUE.: The eigenvectors in the columns of E are assumed
66*> to be *left* eigenvectors.
67*> =.FALSE.: The eigenvectors in the columns of E are assumed
68*> to be *right* eigenvectors.
69*> \endverbatim
70*>
71*> \param[in] N
72*> \verbatim
73*> N is INTEGER
74*> The size of the matrices. If it is zero, ZGET52 does
75*> nothing. It must be at least zero.
76*> \endverbatim
77*>
78*> \param[in] A
79*> \verbatim
80*> A is COMPLEX*16 array, dimension (LDA, N)
81*> The matrix A.
82*> \endverbatim
83*>
84*> \param[in] LDA
85*> \verbatim
86*> LDA is INTEGER
87*> The leading dimension of A. It must be at least 1
88*> and at least N.
89*> \endverbatim
90*>
91*> \param[in] B
92*> \verbatim
93*> B is COMPLEX*16 array, dimension (LDB, N)
94*> The matrix B.
95*> \endverbatim
96*>
97*> \param[in] LDB
98*> \verbatim
99*> LDB is INTEGER
100*> The leading dimension of B. It must be at least 1
101*> and at least N.
102*> \endverbatim
103*>
104*> \param[in] E
105*> \verbatim
106*> E is COMPLEX*16 array, dimension (LDE, N)
107*> The matrix of eigenvectors. It must be O( 1 ).
108*> \endverbatim
109*>
110*> \param[in] LDE
111*> \verbatim
112*> LDE is INTEGER
113*> The leading dimension of E. It must be at least 1 and at
114*> least N.
115*> \endverbatim
116*>
117*> \param[in] ALPHA
118*> \verbatim
119*> ALPHA is COMPLEX*16 array, dimension (N)
120*> The values a(i) as described above, which, along with b(i),
121*> define the generalized eigenvalues.
122*> \endverbatim
123*>
124*> \param[in] BETA
125*> \verbatim
126*> BETA is COMPLEX*16 array, dimension (N)
127*> The values b(i) as described above, which, along with a(i),
128*> define the generalized eigenvalues.
129*> \endverbatim
130*>
131*> \param[out] WORK
132*> \verbatim
133*> WORK is COMPLEX*16 array, dimension (N**2)
134*> \endverbatim
135*>
136*> \param[out] RWORK
137*> \verbatim
138*> RWORK is DOUBLE PRECISION array, dimension (N)
139*> \endverbatim
140*>
141*> \param[out] RESULT
142*> \verbatim
143*> RESULT is DOUBLE PRECISION array, dimension (2)
144*> The values computed by the test described above. If A E or
145*> B E is likely to overflow, then RESULT(1:2) is set to
146*> 10 / ulp.
147*> \endverbatim
148*
149* Authors:
150* ========
151*
152*> \author Univ. of Tennessee
153*> \author Univ. of California Berkeley
154*> \author Univ. of Colorado Denver
155*> \author NAG Ltd.
156*
157*> \ingroup complex16_eig
158*
159* =====================================================================
160 SUBROUTINE zget52( LEFT, N, A, LDA, B, LDB, E, LDE, ALPHA, BETA,
161 \$ WORK, RWORK, RESULT )
162*
163* -- LAPACK test routine --
164* -- LAPACK is a software package provided by Univ. of Tennessee, --
165* -- Univ. of California Berkeley, Univ. of Colorado Denver and NAG Ltd..--
166*
167* .. Scalar Arguments ..
168 LOGICAL LEFT
169 INTEGER LDA, LDB, LDE, N
170* ..
171* .. Array Arguments ..
172 DOUBLE PRECISION RESULT( 2 ), RWORK( * )
173 COMPLEX*16 A( LDA, * ), ALPHA( * ), B( LDB, * ),
174 \$ beta( * ), e( lde, * ), work( * )
175* ..
176*
177* =====================================================================
178*
179* .. Parameters ..
180 DOUBLE PRECISION ZERO, ONE
181 parameter( zero = 0.0d+0, one = 1.0d+0 )
182 COMPLEX*16 CZERO, CONE
183 parameter( czero = ( 0.0d+0, 0.0d+0 ),
184 \$ cone = ( 1.0d+0, 0.0d+0 ) )
185* ..
186* .. Local Scalars ..
187 CHARACTER NORMAB, TRANS
188 INTEGER J, JVEC
189 DOUBLE PRECISION ABMAX, ALFMAX, ANORM, BETMAX, BNORM, ENORM,
190 \$ enrmer, errnrm, safmax, safmin, scale, temp1,
191 \$ ulp
192 COMPLEX*16 ACOEFF, ALPHAI, BCOEFF, BETAI, X
193* ..
194* .. External Functions ..
195 DOUBLE PRECISION DLAMCH, ZLANGE
196 EXTERNAL dlamch, zlange
197* ..
198* .. External Subroutines ..
199 EXTERNAL zgemv
200* ..
201* .. Intrinsic Functions ..
202 INTRINSIC abs, dble, dconjg, dimag, max
203* ..
204* .. Statement Functions ..
205 DOUBLE PRECISION ABS1
206* ..
207* .. Statement Function definitions ..
208 abs1( x ) = abs( dble( x ) ) + abs( dimag( x ) )
209* ..
210* .. Executable Statements ..
211*
212 result( 1 ) = zero
213 result( 2 ) = zero
214 IF( n.LE.0 )
215 \$ RETURN
216*
217 safmin = dlamch( 'Safe minimum' )
218 safmax = one / safmin
219 ulp = dlamch( 'Epsilon' )*dlamch( 'Base' )
220*
221 IF( left ) THEN
222 trans = 'C'
223 normab = 'I'
224 ELSE
225 trans = 'N'
226 normab = 'O'
227 END IF
228*
229* Norm of A, B, and E:
230*
231 anorm = max( zlange( normab, n, n, a, lda, rwork ), safmin )
232 bnorm = max( zlange( normab, n, n, b, ldb, rwork ), safmin )
233 enorm = max( zlange( 'O', n, n, e, lde, rwork ), ulp )
234 alfmax = safmax / max( one, bnorm )
235 betmax = safmax / max( one, anorm )
236*
237* Compute error matrix.
238* Column i = ( b(i) A - a(i) B ) E(i) / max( |a(i) B|, |b(i) A| )
239*
240 DO 10 jvec = 1, n
241 alphai = alpha( jvec )
242 betai = beta( jvec )
243 abmax = max( abs1( alphai ), abs1( betai ) )
244 IF( abs1( alphai ).GT.alfmax .OR. abs1( betai ).GT.betmax .OR.
245 \$ abmax.LT.one ) THEN
246 scale = one / max( abmax, safmin )
247 alphai = scale*alphai
248 betai = scale*betai
249 END IF
250 scale = one / max( abs1( alphai )*bnorm, abs1( betai )*anorm,
251 \$ safmin )
252 acoeff = scale*betai
253 bcoeff = scale*alphai
254 IF( left ) THEN
255 acoeff = dconjg( acoeff )
256 bcoeff = dconjg( bcoeff )
257 END IF
258 CALL zgemv( trans, n, n, acoeff, a, lda, e( 1, jvec ), 1,
259 \$ czero, work( n*( jvec-1 )+1 ), 1 )
260 CALL zgemv( trans, n, n, -bcoeff, b, lda, e( 1, jvec ), 1,
261 \$ cone, work( n*( jvec-1 )+1 ), 1 )
262 10 CONTINUE
263*
264 errnrm = zlange( 'One', n, n, work, n, rwork ) / enorm
265*
266* Compute RESULT(1)
267*
268 result( 1 ) = errnrm / ulp
269*
270* Normalization of E:
271*
272 enrmer = zero
273 DO 30 jvec = 1, n
274 temp1 = zero
275 DO 20 j = 1, n
276 temp1 = max( temp1, abs1( e( j, jvec ) ) )
277 20 CONTINUE
278 enrmer = max( enrmer, abs( temp1-one ) )
279 30 CONTINUE
280*
281* Compute RESULT(2) : the normalization error in E.
282*
283 result( 2 ) = enrmer / ( dble( n )*ulp )
284*
285 RETURN
286*
287* End of ZGET52
288*
289 END
subroutine zgemv(trans, m, n, alpha, a, lda, x, incx, beta, y, incy)
ZGEMV
Definition zgemv.f:160
subroutine zget52(left, n, a, lda, b, ldb, e, lde, alpha, beta, work, rwork, result)
ZGET52
Definition zget52.f:162 | 2,817 | 8,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.504156 |
https://brilliant.org/practice/function-terminology/ | 1,490,479,214,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189083.86/warc/CC-MAIN-20170322212949-00575-ip-10-233-31-227.ec2.internal.warc.gz | 775,228,324 | 15,201 | ×
Back to all chapters
Functions
Functions map an input to an output. For example, the function f(x) = 2x takes an input, x, and multiplies it by two. An input of x = 2 gives you an output of 4. Learn all about functions.
Function Terminology
If we have a real-valued function $$f(x) = x^2 + 27$$, the values of $$f(x)$$ will necessarily fall in the following range:
$a \leq f(x).$
What is the value of $$a$$?
Consider two sets: $$X = \{ a, b, c, d \}, \ Y = \{ 1, 2, 3 \}.$$
If a function is defined from $$X$$ to $$Y$$, what is the maximum possible number of such functions?
For two sets $X=\{a,b,c\}, Y=\{7, 11, 13, 17, 20, 32\},$ $$f$$ is an injective function from $$X$$ to $$Y$$. If $$f(a)=7$$ and $$f(b)=17$$, what is the sum of all the elements of $$Y$$ that can possibly be the value of $$f(c)$$?
Given that the domain and codomain of the function $$y=\sqrt{169-x^2}$$ are restricted to the real numbers, how many elements of the domain of $$y=\sqrt{169-x^2}$$ are integers?
For two sets $X=\{-1,1,a\}, Y=\{3,4, b\},$ $$f(x)=x^3+4$$ is a bijective function from $$X$$ to $$Y$$. What is the value of $$a+b$$?
× | 379 | 1,130 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-13 | longest | en | 0.812208 |
https://resultdatasgp.com/slot-machines-how-to-calculate-the-odds-of-hitting-a-particular-symbol-on-a-slot-machine/ | 1,716,904,061,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059139.85/warc/CC-MAIN-20240528123152-20240528153152-00858.warc.gz | 401,450,374 | 31,451 | # Slot Machines – How to Calculate the Odds of Hitting a Particular Symbol on a Slot Machine
The slot is an important area of the ice because it gives players the best chance of scoring without a deflection. It also provides a view of the net that allows players to place the puck more accurately. The low slot also provides a good opportunity for a wrist shot. The slot is also a no-man’s land for defenders. This makes small wingers vulnerable to big hits from the defense.
## Meaning of a theoretical hold worksheet
A theoretical hold worksheet is a legal document that is provided with every slot machine. This document will tell you the percentage of hold that a machine will have based on the amount of money paid in. It will also tell you how many reels are in play, how many credits are in the machine, and the payout schedule. The theoretical hold worksheet is an essential part of playing slot machines, and it should be included in every machine.
## Variations of the original slot machine concept
Initially, slot machines were very simple, relying on spinning reels to award prizes. As the popularity of slot games increased, manufacturers started to experiment with various features. These innovations include bonus rounds and progressive jackpots. Some even feature more than one payline. Other variations include skill-based slots, which incorporate extra rounds to test players’ game skills.
Early versions of slot machines featured fruit-like symbols on the reels. The Industry Novelty Company later copied this idea and included chewing gum in their machines. This innovation allowed the company to make more money and sell more machines, and eventually led to the creation of fruit machines.
## Odds of hitting a particular symbol or combination of symbols on a slot machine
If you’re a frequent slot machine player, you’ve probably wondered how to calculate the odds of hitting a particular symbol or combination of symbol on a slot machine. You can use the payback percentage of a slot machine to figure out the expected payoff for a specific symbol or combination of symbols. The payback percentage is a percentage of the expected payout multiplied by the number of spins. The higher the number, the better the chance of hitting a specific symbol.
A random number generator in a slot machine generates the random numbers, which are translated into symbols on the reels. A computer program cycles through thousands of numbers per second and stops once a player hits the spin button. These numbers are the ones that correspond to the symbols on the reels. In early slot machines, the odds of hitting a particular symbol or combination of symbols was one in ten or -1/10, depending on the symbol.
## Rules for playing a slot machine
Before you can play slots, you need to learn the rules for playing them. The goal of a slot game is to get a payout by matching up matching symbols across one or more paylines. The payout amount depends on the number of matching symbols. You need to carefully consider your options and choose a strategy that suits your needs.
You should never bet more money than you can afford to lose. The slot machines are designed to give you a sense of excitement. But there are a few important tips to remember to make the most of your time. First of all, do not be afraid to leave the slot machines when you’ve used up your betting allowance. Secondly, you should always be aware that many slot machines don’t have enough seats for everyone. | 681 | 3,496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-22 | latest | en | 0.941021 |
http://stackoverflow.com/questions/10207290/understanding-this-javascript-code-using-modulus | 1,462,246,810,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860118369.35/warc/CC-MAIN-20160428161518-00177-ip-10-239-7-51.ec2.internal.warc.gz | 283,473,991 | 20,049 | # Understanding this Javascript code using % modulus
Im currently in process of learning Javascript. And I have seen the following code, which confuses me.
Description of code:
Starting on line 1, the function isOdd takes a number n and returns a boolean (true or false) stating whether the number is odd or not.
Code
``````var isOdd = function (n) {
if (n % 2 === 0) {
return false;
} else {
return true;
}
};
var isEven = function(n) {
if(n % 2 === 0) {
return true;
} else {
return false;
}
};
``````
Where I am confused.
The code:
``````n % 2 === 0
``````
I have always taken the following to be the description for %:
``````% Is the modulus operator. It returns the remainder of dividing number1 by number2.
``````
Which would mean that the if statement in the function isOdd returns false is the difference between n and 2 is 0. But its meant to mean if n is divisible by 2 (even) return false.
I just dont see how its doing that.
In my mind, if we take the even number 30. Apply it to n % 2. We get 15, which is remainder of dividing 30 by 2. 15 does not equal 0, but 30 is an even number, and with this code it would be seen as odd.
Can someone explain this?
-
"if statement in the function isOdd returns false is the difference between n and 2 is 0" It doesn't mean the difference is 0 (that would mean they're the same number). It means that the remainder of dividing n/2 is 0 ;) I hope that helps – DiogoNeves Apr 18 '12 at 10:19
Thanks, see update – RSM Apr 18 '12 at 10:20
30%2 is zero, i.e. whatever is left after 30/2 (30 divided by 2 is 15). – Tom Apr 18 '12 at 10:23
but 15 doesn't equal 0 – RSM Apr 18 '12 at 10:24
30%2 is not 15! 30%2 is zero. That's what "remainder" means - whatever "remains" after division. – Tom Apr 18 '12 at 10:25
The line in question:
`````` if (n % 2 === 0) {
return false;
}
``````
Means "if when you divide n by 2 the remainder is zero, then return false (i.e. n is not odd)".
The "remainder" is whatever is left over when you subtract the nearest multiple, so for example "64 % 10" is 4, since the nearest multiple of 10 is 60, leaving 4.
Taking your example and putting it another way, 30/2 is 15, 30%2 is zero (i.e. whatever is left over after 30/2). Here's more info on the remainder after division.
-
...and of the same type as 0. – Sani Huttunen Apr 18 '12 at 10:17
Thanks Sani! For others, the triple equals adds the same type requirement (exactly equal). – Tom Apr 18 '12 at 10:20
See update..... – RSM Apr 18 '12 at 10:21
You're confusing Quotient and Remainder. When you divide 30 by 2, integer quotient is 15, and remainder is 0. You can also calculate remainder by multiplying integer quotient by divisor and subtracting it from dividend. So for this division remainder is 30 (dividend) - 15 (quotient) * 2 (divisor) = 0.
-
If n can be divided by 2 it means it's even ->
which means it's not odd ->
isOdd is false
-
see update..... – RSM Apr 18 '12 at 10:21 | 872 | 2,940 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2016-18 | latest | en | 0.901829 |
https://present5.com/whole-class-review-activity-volume-and-capacity/ | 1,611,476,499,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703547475.44/warc/CC-MAIN-20210124075754-20210124105754-00658.warc.gz | 503,214,674 | 11,218 | Скачать презентацию Whole Class Review Activity Volume and Capacity
42009f88281dfbce37c82cb119f96941.ppt
• Количество слайдов: 52
Whole Class Review Activity Volume and Capacity
Directions: • Review questions have been written. • Click the Spin Button. • When the wheel stops, click to view the question. • Groups discuss answer and post their solution. • After the question is discussed, click to check the answer.
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What is the volume of this rectangular prism?
What is the volume of this rectangular prism? Count the number of cubes: 10 x 4 = 160 cubes 10 cubes across 4 cubes up 4 cubes back
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1 2 8 7 SPIN 3 4 6 5
A cube can also be classified as a— A. rectangular prism. B. square-based pyramid. C. cylinder. D. rhombus.
A cube can also be classified as a— A. rectangular prism. B. square-based pyramid. C. cylinder. D. rhombus. (plane shape)
1 2 8 7 SPIN 3 4 6 5
1 2 8 7 SPIN 3 4 6 5
This solid is composed of cubes, all of which are the same size. Using an edge of a cube as one unit, which could be the volume of the figure? A. 28 cubic units B. 24 cubic units C. 20 cubic units D. 16 cubic units
This solid is composed of cubes, all of which are the same size. Using an edge of a cube as one unit, which could be the volume of the figure? Count the cubes on each row. 4 + 8 + 16 = 28 cubes A. 28 cubic units B. 24 cubic units C. 20 cubic units D. 16 cubic units 4 8 16
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1 2 8 7 SPIN 3 4 6 5
Use the picture below to answer this question. How many more small cubes are needed to complete the large cube above?
How many more small cubes are needed to complete the large cube? The large cube would have a base of 9 cubes (3 x 3) and the total volume would be 27 cubes. The bottom row is complete. The second row needs 2 more cubes The top row needs 4 cubes. Six cubes are needed to complete the large cube.
1 2 8 7 SPIN 3 4 6 5
1 2 8 7 SPIN 3 4 6 5
Jim made a box for his coin collection. The width was 6 inches. The length was three times the width and the height was half the width. What would be the volume of the box? A. 27 cubic inches B. 108 cubic inches C. 216 cubic inches D. 324 cubic inches M 5 M 4 d Compute the volume of a cube and a rectangular prism using formulae.
Jim made a box for his coin collection. The width was 6 inches. The length was three times the width and the height was half the width. What would be the volume of the box? width = 6 in; length = 3 x 6 in; height = ½ x 6 in. 6 x 18 x 3 = 324 A. 27 cubic inches B. 108 cubic inches C. 216 cubic inches D. 324 cubic inches M 5 M 4 d Compute the volume of a cube and a rectangular prism using formulae.
1 2 8 7 SPIN 3 4 6 5
1 2 8 7 SPIN 3 4 6 5
The dimensions of a rectangular prism are shown below. What is the volume of this rectangular prism? A 18 cubic inches B 35 cubic inches C 150 cubic inches D 190 cubic inches
The dimensions of a rectangular prism are shown below. What is the volume of this rectangular prism? V=lxwxh = 10 in x 3 in x 5 in = 150 cubic inches A 18 cubic inches B 35 cubic inches C 150 cubic inches D 190 cubic inches
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1 2 8 7 SPIN 3 4 6 5
The 5 th grade class had to measure 128 fl oz for a lemonade recipe. They only had quart, pint, gallon and cup containers. Which of the following measurements is NOT equal to 128 fl oz. ? A 4 quarts B 10 pints C 1 gallon D 16 cups
The 5 th grade class had to measure 128 fl oz for a lemonade recipe. They only had quart, pint, gallon and cup containers. Which of the following measurements is NOT equal to 128 fl oz. ? A 4 quarts B 10 pints C 1 gallon D 16 cups Cup 8 oz Pint 16 oz (2 cups) We can either use a chart or the graphic to find the answer. 128 fl oz = 1 gallon 4 quarts = 1 gallon 16 cups = 1 gallon 8 pints = 1 gallon Quart 32 oz (2 pts) 4 cups Gallon 128 oz (4 qts) 8 pts
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1 2 8 7 SPIN 3 4 6 5
Thomas is making fruit punch for his friends. One tablespoon of mix makes 1 pint of fruit punch. He wants to make 2 gallons of fruit punch. How many tablespoons of mix will he need? A 4 tablespoons B 8 tablespoons C 16 tablespoons D 24 tablespoons
Thomas is making fruit punch for his friends. One tablespoon of mix makes 1 pint of fruit punch. He wants to make 2 gallons of fruit punch. How many tablespoons of mix will he need? First find out how many pints of punch are in 2 gallons by using the graphic. There are 8 pints in 1 gallon, 2 x 8 (16) pints in 2 gallons. He needs 16 tablespoons of mix for 16 pints of punch. A 4 tablespoons B 8 tablespoons C 16 tablespoons D 24 tablespoons
1 2 8 7 SPIN 3 4 6 5
1 2 8 7 SPIN 3 4 6 5
Donna is mixing different colors of paint together to match the color of her house. The amounts of paint are shown below. 2 gallons of white paint 2 quarts of yellow paint 1 quart of blue paint 3 quarts of red paint 1 quart of brown paint 2 pints of orange paint How many gallons of paint will Donna have after she has mixed all of the paint colors together?
Donna is mixing different colors of paint together to match the color of her house. The amounts of paint are shown below. 2 gallons of white paint = 2 gal 2 quarts of yellow paint = ½ gal 1 quart of blue paint = ¼ gal 3 quarts of red paint = ¾ gal 1 quart of brown paint = ¼ gal 2 pints of orange paint = ¼ gal How many gallons of paint will Donna have after she has mixed all of the paint colors together? 4 gal of paint
1 2 8 7 SPIN 3 4 6 5
1 2 8 7 SPIN 3 4 6 5
Maria’s mom is making punch for the school dance. The recipe calls for 32 cups of ginger ale, and she needs to triple the recipe. How many gallons of ginger ale are needed?
Maria’s mom is making punch for the school dance. The recipe calls for 32 cups of ginger ale, and she needs to triple the recipe. How many gallons of ginger ale are needed? 3 x 32 cups = 96 cups ÷ 16 cups(per gallon) = 6 gallons
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1 2 8 7 SPIN 3 4 6 5
What is the volume, in cubic inches, of the school locker below?
What is the volume, in cubic inches, of the school locker below? V = 12 in x 8 in x 30 in V= 2880 cubic inches
1 2 8 7 SPIN 3 4 6 5
1 2 8 7 SPIN 3 4 6 5
What is the volume of a cube that measures 10 inches on each edge? A 10 cubic inches B 100 cubic inches C 1000 cubic inches D 10, 000 cubic inches
What is the volume of a cube that measures 10 inches on each edge? V = 10 in x 10 in V = 1000 cubic inches 10 in A 10 cubic inches B 100 cubic inches C 1000 cubic inches D 10, 000 cubic inches 10 in | 2,029 | 6,491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | longest | en | 0.891467 |
https://www.gadgetnate.com/category/electronics/page/2 | 1,716,502,119,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058671.46/warc/CC-MAIN-20240523204210-20240523234210-00518.warc.gz | 698,456,400 | 18,306 | 1.09 Other signals and 1.10 Logic Levels
Let’s cover two sections today:
1.09 Other signals
1.10 Logic Levels
We have been looking at Voltage as a function of time and seeing how it will look on a graph. The past few days we have talked about graphs that are sine wave.
But there could be other ways that this could look too. Consider if voltage started out at zero and continued to increase. It would look something like this:
But it is not possible to increase voltage indefinitely and so the graph would really look like this: (Ramp with limit)
But there are other interesting graphs to. Consider the case of voltage increasing to a point, and then when it reaches a certain point it goes to zero and then begins again. Let’s say you have a circuit that triggers at a certain voltage, and a capacitor that you are charging up till that point. It might looks something like this sawtooth wave:
But you can also imagine a case where it reaches a point and decreases at the same linear rate. This would look like a triangle wave:
And then there is noise, a complex, background static that we always need to consider and usually minimize:
Square waves
Steps
Spikes
But in digital electronics we concern ourselves with logic levels and interrupting a particular voltage as either a true or a false; as a on or a off; as a 1 or 0. Most digital electronics have a threshold voltage. Above that voltage it is considered to be true/on/1. Below that voltage it is considered to be false/off/0.
1.08 Signal amplitudes and decibels
Peak-to-peak amplitude (pp amplitude) is twice the amplitude. This is the measurement of the amplitude from the bottom peak to the top peak as illustrated below:
root-mean-square amplitude (rms amplitude) has different formulas based upon the shape of the signal. For sine waves only it is given by:
Decibels is a comparison between two signals.
When comparing amplitude use this formula:
When comparing the power use this formula:
Sometimes decibels are presented as units. This is because their is an implicit signal that is being used for comparison. Here are some common ones:
a) dBV: 1 volt rms
b) dBm – the voltage corresponding to 1mW with an assumed load impedance
c) small noise generated by a resistor at room temperature
1.07 Sinusoidal signals
The next few sections in The Arts of Electronics deals with signals. What are the characteristics of signals? Amplitude, frequency, shape, etc.
The values we are looking at is voltage as a function of time. The most basic signal is the sinusoidal signal. Have you ever graphed y=sin(x) and end up with:
But how would you compare this wave to the following:
There are two key things to think about when comparing sine waves: frequency and amplitude. The amplitude is the distance from y=0 to the top of the sine wave. The frequency is how many times the wave repeats in a given period of time. We typically use the unit hertz (Hz) for frequency, which is one cycle (or period) per second. Lets take a look at period and amplitude in our original sine graph:
The formula that you should remember is
V=A*sin(2*pi*f*t)
A is the amplitude in volts, f is the frequency in Hz and t is the time in seconds.
1.06 Small-signal resistance
Ohm’s law is R=V/I. For the ordinary resistance, this is a constant proportion for a large range of V. But this is not true for all devices. Today we look at the Zener diode which has a high resistance up to a point and then almost no resistance for a large range of V. Also, we look at the tunnel diode (also known as the Esaki diode) which actually has a negative resistance over a particular range of V.
1.05 Thevenin’s equivalent circuit
Any two terminal network of voltage sources and resistors can be replace by a single voltage source in combination with a single resistor. This is called the Thevenin’s equivalent circuit:
The first step is to identify the two terminals of the network.
Then to figure out Rth, replace any voltage sources with short circuits and any current sources with open circuits.
Calculate the voltage between the two terminals.
Then figure out Vth between the two terminals.
To figure this out using testing and measuring, it is only necessary to find the voltage at the two terminals with power on, and the resistance with power off.
This can greatly simplify circuit analysis of complex circuits.
Check out videos on YouTube with more detailed information:
1.04 Voltage and current sources
Let’s take a look at the symbols for voltage sources and current sources. Also, let’s discuss the properties of the ideal voltage source and the ideal current source.
The ideal voltage source will maintain the same voltage between its terminals regardless of the load. This is easiest with an open circuit but becomes impossible with a short circuit.
The ideal current source will maintain the same current regardless of the resistance of the load. This is easy with an open circuit but becomes more and more difficult as the resistance increases.
1.03 Voltage Dividers
Voltage Dividers take an input voltage and gives you a smaller, fractional output voltage. This can be useful for volume control or when interfacing between logic circuits that run at 5V with new/lower power devices that run at 3.3V.
A voltage divider is a simple circuit made up of only two resistors (R1 & R2 in the diagram below).
1.02 Relationship between voltage and current: resistors
Art of Electronics 2nd Edition p 4-7
What you MUST learn
Resistance (represented in formulas as “R”) measured in Ohms which is represented using the symbol:
Resistance is the relationship between Current (I) and Voltage (R).
This relationship is defined by Ohm’s law: R = V/I.
Circuit diagrams represent resistors as
Resistors in Series: (Always bigger)
R= R1 + R2
Resistors in Parallel: (Always smaller)
R = (R1*R2)/(R1+R2)
Shortcut #1:
A large resistor in series with a small resistor has the resistance of the larger one, roughly.
A large resistor in parallel with a small resistor has the resistance of the smaller one, roughly.
Try to develop an intuitive sense of the voltage and current in the parts of the circuit. Do not try to calculate to be extremely precise because the precision of the components will make this pointless. Also, good circuit design should be tolerant of variations.
Power:
P=IV
Using Ohm’s law you get:
P=(I^2)R
P=(V^2)/R
Things good to learn
The inverse of resistance is conductance (represented by “G” in formulas) G=1/R
Restatement of Ohm’s law: I=GV
Conductance is measured in siemens
siemens is also known as mho (ohm spelled backwards)
1.01 Voltage and Current
Art of Electronics 2nd edition, pages 2-4
Overview
Voltage & Current describe two aspects of electronic circuits that are fundamental to understand and on which everything else will build on.
What you MUST learn
Voltage refers to a force that when applied across a circuit causes current to flow. Voltage is measured in volts (using the symbol V) also called a “potential difference”. It is measured across two points in a circuit. A joule of work is needed to move a coulomb of charge through a potential difference of one volt. A coulomb is the charge of about 6×10^18 electrons.
Current (symbol: I) is the measure of electronics through a point in a circuit. It is measured in amperes (amps) (symbol: A). 1 ampere (amp) is 1 coulomb of charge passing through a point per second.
Conservation of Charge: Kirchhoff’s current law
The sum of currents into a point or node is equal to the sum of the currents out of a point or node. This is referred to as Kirchhoff’s current law.
Kirchhoff’s voltage law
The sum of voltage drops through one branch of a circuit will equal the sum of voltage drops across the other branches of the circuit.
P=VI
Power (work per unit time), measured in watts. (1 W = 1 J/s)
P=VI = (work/charge) x (charge/time) = work/time
P=VI= (1 joules/1 coulomb) x (1 coulomb/1 second) = 1 joules/second = watt.
Things good to learn
Voltage is usually written with the symbol V but sometimes E is used. Voltage is also called a “potential difference” or electromotive force (EMF).
Prefixes:
Multiple Prefix Symbol 10^12 tera T 10^9 giga G 10^6 mega M 10^3 kilo k 10^(-3) milli m 10^(-6) micro µ 10^(-9) nano n 10^(-12) pico p 10^(-15) femto f
When abbreviating a unit with a prefix, the unit follows the multiplier with no space and the unit is capitalized. However, no capitalization for both prefix and unit when spelled out. 1mW = 1 milliwatt
1 MV = 1 megavolt
Related information and insights from my other books
Introductory Circuit Analysis 9th edition – chapter 2
Structure of the atom:
nucleus is made up of protons (that have a positive charge) and neutrons that have a negative charge. The proton and neutron are relatively the same mass. But the electron is considerably smaller (1836 times smaller) than the proton but has a negative charge that is equal to the positive charge of the proton.
Other things of interest
If you need more help
Voltage
Current
Joule
If you wanted to know
Who defined:
Voltage
Volt
Current
amp
joule
watt
multiplier prefixes
People who contributed:
Watt
Preparing for my journey through “The Art of Electronics”
“The Art of Electronics” by Horowitz and Hill is the definitive guide to Analog and Digital electronics. I had bought the 2nd edition over a decade ago, with the intention of going through it. But this year is going to be the year. And I am hoping that you will help me.
My plan is to go through a section a day, and to write a short blog post summarizing what I learned. I have collected several electronics books, and I plan to make reference to them as I go on this journey. These include:
In addition to a brief summary of what I learn from that section, I plan to create a video post, create additional exercises, and assemble related circuits. I am hoping that it will be an exciting adventure, and that by having to produce a daily blog post, I will have the discipline to see it through. And should I stumble, hopefully an encouraging comment or email from you will nudge me back on the path.
First thing is that I need some help by those that are familiar with Electronics. I am an Amateur with hopes of achieving proficiency. I do not want my struggles to cause problems for others following along. Therefore, I would like to have someone knowledgeable review my notes before I publish them on this blog. If you would be willing to help out, then please email me at nathan.price@gmail.com. Let me know how much you would be willing to help out. I figure each blog post would take 15 minutes or so to review. I would like to give you credit for reviewing, but if you want to remain anonymous, then that is fine too. | 2,570 | 10,868 | {"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} | 3.984375 | 4 | CC-MAIN-2024-22 | latest | en | 0.929674 |
https://www.westmont.edu/~iba/teaching/CS010/CS010-F05/project2.html | 1,534,764,688,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221216333.66/warc/CC-MAIN-20180820101554-20180820121554-00392.warc.gz | 1,000,995,411 | 2,628 | CS010 Introduction to Computer Science I
Term Project
Deliverable Two
Cellular Automata:
Conway's Game of Life
(last updated 12/1/2005)
For the second project deliverable, you will implement a version of Conway's Game of Life. Here is one of many many links you might find on the web. The key notion of CGoL is that each cell lives or dies according to these simple rules:
• A dead cell with exactly 3 live neighbours becomes alive (or is "born").
• A live cell with 2 or 3 live neighbours stays alive; otherwise it dies (from "loneliness" or "overcrowding").
You will want to think about how to decompose the problem into simple functional chunks. You should define a grid type using Scheme vectors. That is, the grid should be a (vectorof (vectorof cells)), where a cell is represents a life state as either 0 (dead) or 1 (live). Naturally, you will want a function that displays a given grid.
Ultimately, you must write a function, cgol, that consumes a grid and a number; it simulates the game of life using the given grid for the given number of time steps, displaying the live cells after each generation.
The cgol function should create a canvas of the appropriate size based on the size of the given grid and your cell-size. Make sure that your code works according to the dimensions and initial values of the given grid. That is, your code should work equally well with a grid that is 3-by-5 as with one that is 40-by-110. On each time step, the life-status of each cell should be checked and updated, and then the grid should be redrawn. As before, the grid should wrap from edge to edge, but also from top to bottom. Make sure that the new life-status of one cell does not confuse the update of an adjacent cell. This should not be a problem if you create a new grid on each cycle.
Use the draw teachpack to display your grid, drawing circles or squares to represent living cells and leaving dead cells blank. You should define a global variable, CELL-SIZE, that determines the width and height of a cell's graphical representation. Your display functions should take this variable into account as it draws and clears cells as they live and die. For example with given grid G, your canvas should be CELL-SIZE * (vector-length G) pixels tall, and CELL-SIZE * (vector-length (vector-ref G 0)) pixels wide.
To grade this deliverable, I will review your code (including contracts, purpose statements, indentation, etc.), and then call your cgol function. Therefore, it is important that you follow the data definitions carefully. I might use the following code:
`(begin (set! CELL-SIZE 10) (cgol <my-own-grid-making-code> 500))`
I hope you have fun with this. | 626 | 2,688 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-34 | latest | en | 0.860796 |
https://kaganlove.wordpress.com/video-lessons/courses/pre-algebra/order-of-operations/ | 1,652,969,727,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662529538.2/warc/CC-MAIN-20220519141152-20220519171152-00491.warc.gz | 381,231,715 | 20,558 | # Order of Operations
The lesson below is still images taken from the fully animated PowerPoint I created that is available in my Teachers Pay Teachers store page. If the PowerPoint is not one of my freebies, you may also head over to my YouTube channel to see the slideshow fully animated and can pause as needed to be sure you grasp each concept before moving forward (each lesson will always be free there!).
This lesson will introduce how to determine the correct order to perform operations when looking at a problem. We go over why the order of operations is so crucial to getting the correct answer and walk through examples step-by-step to help clear up some of the most common errors students make.
So without further ado, read through the slides below to get a feel for how to correctly use the order of operations!
Phew, that’s a lot to take in. Once you’ve gone over this and found some practice problems to cement the idea in your head what integers are, you may move on to the next lesson below!
Solving Equations by Adding and Subtracting
Just in case the slides aren’t your thing, here is a text outline of the main points of the lessons above!
• Order of operations
• Objectives
• By the end of this lesson you should feel comfortable:
• Recognizing different operations
• Utilizing the order of operations to solve problems
• Where to start?
• Sometimes, we are presented with a problem that requires us to perform more than one calculation, or operation, in order to arrive at an answer.
• For example, let’s say you your two friends are saving up for a trip by mowing lawns. Your dad also says he will give you double what you earn yourself (so you will have three times as much as what you make on your own).
• Let’s say you make \$15, one friend earns \$20, and the other friend earns \$10. How much did you make together?
• Where to start?
• So…what would we do first?
• If we add everything up, we would have 15+20+10 = 45
• Then multiply 45 by 3 to get 135. But is this correct?
• Your dad is only tripling the amount YOU earn, not everyone.
• So you’d have to multiply first.
• 75 and 135 certainly aren’t equal.
• Clearly the ORDER you perform the operations matters.
• Operations
• Before we can decide the correct order, however, first we should know all the different operations that are possible.
• Operations are the different calculation techniques we can do to numbers.
• Still fuzzy? How about a list of examples from the first things we learned.
• All of these have a specific order we perform them in when combined together in a single problem.
• pemdas
• So what order do we follow?
• Cool, so what does that mean?
• It’s an acronym, which means each letter stands for an entire word.
• In fact, you can remember this with a silly saying:
• Please Excuse My Dear Aunt Sally”
• Using Order of Operations
• Alright, so let’s try putting PEMDAS to use and see what it really means:
• Great, so lots of stuff going on. Let’s go down the line.
• Do we see any parentheses? No.
• Do we see any exponents? Yes. Let’s perform that operation first.
• Next, we look for either multiplication or division. Whichever we see first reading from left to right, we perform first. Here we only have multiplication, so let’s do that next.
• Using order of operations
• Finally, we look for addition or subtraction. Again, we see which comes first reading from left to right. This time, subtraction appears first, so:
• Then we have a simple addition problem left.
• Ta da! You’ve solved your first multistep problem using order of operations!
• Notice: In each step, we focus on one operation, then change only that answer in the next line and rewrite the rest of the problem exactly the same so that we can retrace our steps if we somehow make a mistake!
• Using order of operations
• One more problem so that we can introduce a final idea.
• We use parentheses to designate that a specific operation should be done first.
• In our example, we can see multiplication outside the parentheses, which normally would come before addition.
• But remember our order!
• Parentheses are the first thing we check for. Though they are not an operation, they designate which operation, or set of operations, are to be performed first.
• Using order of operations
• With that in mind, let’s do what’s in the parentheses!
• Notice that now we have 5 x 4. This is because a number outside of a set of parentheses is math notation for multiplication.
• Now we can follow along with our order!
• Are there any exponents? No.
• Do we see multiplication or division? Yes, so let’s go left to right.
• And the last step is to simply multiply our final problem.
• Ta da! Another problem finished successfully!
• conclusion
• Operations include: | 1,060 | 4,752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2022-21 | latest | en | 0.940502 |
roselynmainali.github.io | 1,670,242,192,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711016.32/warc/CC-MAIN-20221205100449-20221205130449-00777.warc.gz | 518,002,613 | 355,041 | # What is the Monty Hall problem?
The Monty Hall problem is a problem based on probability, named after Monty Hall. It became famous because of a reader’s letter to Marilyn vos Savant asking her about it:
Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?
Her response was that the player should switch to the other door. Her reasoning was that players who switch have two-thirds chance of winning the car, while players who stick to their choice have only one-third chance.
# Simulating the Scenario
Let’s simulate the game and estimate the probability of winning by switching or not switching from your choice.
We will simulate the game for 10,000 iterations for each of the two strategies by using a `for()` loop. Each iteration represents a round. So we have to randomize the vector of items for each iteration, because it has to start fresh everytime the game is played. A player picks a door at random. Then out of the remaining doors, the host opens one door with the goat. The player then gets a chance to either stick to the original selection or switch.
## Probability of winning car with switching
``````vec = c() # create an empty vector to concatenate the results of the iterations
for(i in 1:10000) {
# randomize the doors as the host of a specific game can decide what goes behind each door
doors = sample(c("goat", "car", "goat"), 3, replace = FALSE); doors
# a player picks a door at random
s = sample(1:3, 1, replace = FALSE); s
playerchoice = doors[s]; playerchoice
# find what is behind the remaining doors
remaining = doors[-s] ; remaining
# out of the remaining, the host has to open the door with the goat behind it
remaining = remaining[order(remaining)] ; remaining
remaining[2] # door opened
remaining[1] # door not-opened
if(remaining[1] == "car") {
vec = c(vec, i)
}
}
# probability of winning
pwin.switch <- length(vec)/i; pwin.switch``````
``## [1] 0.6665``
## Probability of winning car without switching
``````vec = c() # create an empty vector to concatenate the results of iterations
for(i in 1:10000) {
# randomize the doors as the host of a specific game can decide what goes behind each door
doors = sample(c("goat", "car", "goat"), 3, replace = FALSE); doors
# a player picks a door at random
s = sample(1:3, 1, replace = FALSE); s
playerchoice = doors[s]; playerchoice
if(playerchoice == "car") {
vec = c(vec, i)
}
}
# probability of winnning
pwin.noswitch <- length(vec)/i; pwin.noswitch``````
``## [1] 0.3322``
# What strategy helps you win
As you saw above, the probability of winning a car if you do not switch is one-third, while with switching it is two-thirds.
``````# create barplot to represent the probability
probplot = barplot(c(pwin.noswitch, pwin.switch), col = c("#CC00FFFF", "#CC00FFFF"), main = "Probabilities", horiz = TRUE, names.arg = c("no switch", "switch"))`````` | 818 | 3,158 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.949949 |
https://www.quantphys.com/search?updated-max=2021-05-11T06:38:00-06:00&max-results=7&start=7&by-date=false | 1,669,552,763,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710237.57/warc/CC-MAIN-20221127105736-20221127135736-00443.warc.gz | 1,024,127,404 | 22,993 | ## Posts
### Chebyshev polynomial expansion of the time-evolution operator
Here we outline an efficient numerical method of solving the time-dependent Schrödinger equation, $i \hbar \frac{\partial | \psi(t) \rangle}{\partial t} = H | \psi(t) \rangle \,.$ The method is based on the Chebyshev polynomial expansion of the time-evolution operator. Only the basic idea is presented in these notes. Further details on the method can be found, for instance, in the following papers: An accurate and efficient scheme for propagating the time dependent Schrödinger equation Unified framework for numerical methods to solve the time-dependent Maxwell equations (in particular, section 3.3) The Hamiltonian $$H$$ is assumed to have no explicit dependence on time. For instance, if the system under consideration is a particle of mass $$m$$ moving in the presence of a static potential $$V(x)$$, then $\psi(x,t) = \langle x | \psi(t) \rangle$ is the particle's wave function, and $H = -\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} + V(x) \,.$ In what
### Robertson-Schrödinger uncertainty relation
Consider a quantum system in a state $$| \psi \rangle$$, and let $$A$$ and $$B$$ be Hermitian operators representing a pair of observables. One can choose to perform a measurement of $$A$$ or a measurement of $$B$$ on the system. Generally, the outcomes of these measurements cannot be predicted with certainty. The outcomes are statistical in nature and characterized by the respective expectation values $$\langle A \rangle$$ and $$\langle B \rangle$$. Hereinafter, $\langle \cdot \rangle = \langle \psi | \cdot | \psi \rangle \,.$ The corresponding uncertainties are defined as $\sigma_A = \sqrt{ \langle \big( A - \langle A \rangle \big)^2 \rangle } = \sqrt{\langle A^2 \rangle - \langle A \rangle^2} \,,$ $\sigma_B = \sqrt{ \langle \big( B - \langle B \rangle \big)^2 \rangle } = \sqrt{\langle B^2 \rangle - \langle B \rangle^2} \,.$ The Robertson-Schrödinger uncertainty relation sates that \[\sigma_A^2 \sigma_B^2 \ge \langle \tfrac{i}{2} [A,B] \rangle^2 + \Big( \langle | 551 | 2,075 | {"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} | 2.765625 | 3 | CC-MAIN-2022-49 | latest | en | 0.737198 |
https://math.stackexchange.com/questions/4219385/fracso3-times-so2so2-fracso3-times-mathbbr-times-mathbbz | 1,721,682,450,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517915.15/warc/CC-MAIN-20240722190551-20240722220551-00197.warc.gz | 329,772,546 | 37,690 | # $\frac{SO(3)\times SO(2)}{SO(2)}$ = $\frac{SO(3)\times\mathbb{R}\times\mathbb{Z}_{2}}{SO(2)\times\mathbb{Z}_{2}}$?
On a manifold $$M=S^{2}\times S^{1}$$ I'm trying to find the homogenous model/Klein space geometry, and define a reduction of it's structure group. Now since $$S^{2}=\frac{SO(3)}{SO(2)}$$ we have that:
$$\frac{G}{H}=\frac{SO(3)\times SO(2)}{SO(2)}$$
Where $$G=SO(3)\times SO(2)$$ is transitive on $$M$$ and $$H=SO(2)$$ is the point stabilizer. As far as I can tell everythings all right here; however, in reading about Thurston's Geometerization Conjecture, this isn't one of the eight Thurston geometries for a closed, oriented, 3-manifold. For this manifold it should be of the form:
$$\frac{G}{H}=\frac{SO(3)\times\mathbb{R}\times\mathbb{Z}_{2}}{SO(2)\times\mathbb{Z}_{2}}$$
Which is pretty close. I'm guessing these spaces have to be isomorphisms of one another. I can equivalently define the first equation in terms of double covers:
$$\frac{G}{H}=\frac{SU(2)\times U(1)}{U(1)}$$
($$U(1)=SO(2)$$ double covers itself) and use a hand-wavy argument that $$SO(n)\times\mathbb{Z}_{2}=Doublecover(SO(n))$$. That still doesn't explain $$\mathbb{R}$$ versus $$SO(2)$$ in our transitive group. The former is the universal cover of the latter so maybe that comes into play? Or perhaps $${\mathbb{R} \over \mathbb{Z}_{2}} =SO(2)$$?
I'm specifically interested in a reduction of the G-bundle to an H-bundle so how this breaks down is important for me.
Recall that a geometric structure on $$M$$ is a diffeomorphism $$\varphi:X/\Gamma\to M$$ where $$G$$ is a Lie group, $$H\subseteq G$$ is a compact subgroup, $$X=G/H$$ is a model geometry, and $$\Gamma$$ is a discrete subgroup of $$G$$ acting on $$X$$ in the canonical way.
The model geometry of $$M=S^2\times S^1$$ is $$X=\frac{O(3)\times\mathbb{R}\times\mathbb{Z}_2}{O(2)\times\mathbb{Z}_2}$$, but this does not mean that $$M\cong X$$. Instead, we have $$M\cong X/\Gamma$$, and $$\Gamma$$ is nontrivial in this case (notice that $$\Gamma$$ is always isomorphic to $$\pi_1(M)$$). Instead, we have $$\Gamma=\mathbb{Z}$$, regarded as a subgroup of the $$\mathbb{R}$$ factor in $$O(3)\times\mathbb{R}\times\mathbb{Z}_2$$.
• Aha! I was missing the connection with the fundamental group $\pi_{1}$. So we have a reduction to a $U(1)\times \mathbb{Z}_2$ (H)-structure on $M$. Any idea what kind of invariant object this corresponds to? If it was just $U(1)$ I'd say it's an invariant 2-form similar to the electromagnetic field two-form, but with the $\mathbb{Z}_2$ I'm not sure Commented Aug 14, 2021 at 15:22
• @R.Rankin A geometric structure need not correspond to any particular invariant object, and for a general geometry $X=G/H$ it is possible to come up with all kinds of objects whose stabilizer is $G$. For instance, for $S^2\times\mathbb{R}\cong\frac{SO(3)\times\mathbb{R}\times\mathbb{Z}_2}{SO(2)\times\mathbb{Z}_2}$, one can regard the geometry as the stabilizer of $(H,V,g,[\omega])$, where $H$ and $V$ are the horizontal and vertical distributions on the product, $g$ is the standard product metric, and $[\omega]$ is an orientation of the $S^2$ fibers. Commented Aug 14, 2021 at 15:59
• You seem to have things backwards: the manifold $M$ is a quotient of the model geometry $X=G/H$ by a discrete group $\Gamma\subseteq G$ acting freely on $X$. The quotient map $X\to M$ is thus a covering map, and since $X$ is by definition simply connected, we have $\Gamma\cong\pi_1(M)$. This has nothing to do with Postnikov towers or spin geometry; it's just basic covering space theory. Commented Aug 21, 2021 at 5:30 | 1,156 | 3,588 | {"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": 31, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-30 | latest | en | 0.852935 |
https://or.stackexchange.com/questions/2646/extract-info-from-gurobi-binary-variables-during-run-time | 1,620,407,904,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988796.88/warc/CC-MAIN-20210507150814-20210507180814-00137.warc.gz | 460,632,806 | 34,852 | # Extract info from Gurobi binary variables during run-time
Actually the question below is not specific to Gurobi, but that's the tool I am using.
Consider a scheduling problem where a 2D array of binary variables $$Z(i,v)$$ is defined, where $$i$$ is index of time slot, and $$v$$ is index of operation, each $$Z(i,v)$$ is a 0-1 binary variable, $$Z(i,v)=1$$ means operation $$v$$ is allocated to time slot $$i$$.
Now consider adding the following constraint/penalty based on $$Z(i,v)$$:
1. A certain operation $$v_0$$ can be scheduled multiple times, i.e., there are multiple $$i$$ values where $$Z(i,v_0) = 1$$. We want to add constraint to the "last $$i$$" where $$Z(i,v_0)=1$$, i.e., the last $$v_0$$ operation. For example, in the last $$v_0$$ operation the product volume $$\operatorname{vol}(i, v_0)\ge100$$. The problem is we must tell what is the last $$i$$ that satisfies $$Z(i,v_0) = 1$$, and only after that add $$\operatorname{vol}(i,v_0)\ge100$$, how to do that?
2. This is an extension of the above problem. Here we consider multiple operations $$v_0, v_1, \cdots, v_n$$. In certain applications the "order" of those $$n+1$$ operations matter, meaning that certain order could incur extra cost, and that the objective should contain a penalty term that is a function of the order. To add penalty, we should know the order first, which in turns means that we should extract all those $$i$$ values where $$Z(i, v_0)=1$$, $$Z(i,v_1)=1$$, etc.
Comment: if $$Z(i,v)$$ is given, then I can just use a for loop to locate all those $$i$$ values where $$Z(i,v)=1$$, the problem is $$Z(i,v)$$ is part of the solution, and is unknown during the solution process. That's what gets me confused. But I expect that Gurobi has some built-in syntax/function to handle those scenarios because technically it should be possible. | 507 | 1,830 | {"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": 28, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-21 | longest | en | 0.882375 |
http://www.eduqna.com/Primary-Secondary-Education/1226-secondary-education-3.html | 1,500,849,459,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424623.68/warc/CC-MAIN-20170723222438-20170724002438-00699.warc.gz | 420,146,593 | 2,500 | # How does size affect paper airplanes?
Question:when it comes to distance how far the paper airplane flies?
plz help me
I suspect this is for a science class. Size is determined by TWO basic measurements; surface area and weight. Significantly modifying either will dramatically effect the flying ability of the paper airplane. The more surface area the plane has, the more weight it can support. A large plane made of very heavy paper may only fly as far as the thrower can propel it whereas a much lighter constructed plane will have more of an ability to float on the air. Most principles surrounding real passenger carrying airplanes apply to paper airplanes minus the ability for pilot control. Experiment by making yourself two airplanes; one out of 8 1/2 x 11 notebook paper, one out of an 8 1/2 x 11 piece of poster board (cereal box back). Make them using the same pattern as close to alike as you can. First try propelling each of them forward in typical paper plane throwing fashion using a medium toss. Next hold each flat by the rear just below the wing and about shoulder height and merely drop them flatly. One should fall like a rock and one should propel itself significantly forward. If you have access to larger pieces of paper and poster board such as 2' x 3' or so, try the same flight tests with the larger planes. The larger plane made of the heavier paper should fair better in the challenge (consider weight to surface area ratio) although it still might not beat the larger lighter plane. Do you think this will hold true no matter how large you fold a paper plane? When discussing paper as construction material consider weight as structure and surface area as need for structure. Balance the two. Larger plane, heavier paper. Smaller plane, lighter paper. Flight potential should be similar. I've done these tests. Paper plane flying is good fun.
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# What is 228 DIVIDED BY 6?
What is 228 divided by 7? 228/7 = 32.571428 recurring (that is, 32.571428571428...) What is 228 divided by 4? 57. What is 228 divided by 3? 228 ÷ 3 = 76
What is 228 divided by 6? 228/6 = 38. What is 228 divided by 7? 228/7 = 32.571428 recurring (that is, 32.571428571428...) What is 228 divided by 3?
What is 228 divided by 4? ChaCha Answer: The answer to 228 divided by 4 is equal to the number 57. ChaCha all day and all of the night.
http://www.chacha.com/question/what-is-228-divided-by-4
Divided by 6? Sponsored Ad: (Join for no ads) Hi All, Please discuss the following question..... Math.doc Please explain..... Thanks & Regards, SamratisKing. 08-02-2008, 04:59 PM #2. Makumajon. An Urch Guru Pundit Swami Sage Join Date May 2008 Location Bangladesh
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how many 6s go into 228. Just divide 228/6 = 38. 6 years ago; Report Abuse; 0% 0 Votes. 1 person rated this as good; Discover Questions in Mathematics. Can someone help please? Please help with maths question!? Please help with trigonometry question?
... (4 ⁄ 9)] • (19 ⁄ 6) = (228 ⁄ 72) • (19 ⁄ 6) = 4332 ⁄ 432 since 12 is the common factor: = 361 ⁄ 36 I think you flipped the (57 ⁄ 8 ... what is 7 1/8 divided by 2 1/4 times 3 1/6 = 10 1/36 57/8 divided by 9/4 times 19/6 = 10 1/36 Simplified 57/8 = 7 1/8, 9/4 ...
... (25) that will contain the divisor (7) at least once but less than ten times. Divide that partial dividend by the divisor, and obtain the first digit of the quotient (3). Write it over the last digit ... 38 goes into 228 six (6) times exactly. All methods of division, including this one, ...
http://www.themathpage.com/ARITH/divide-whole-numbers.htm
What is 36 divided my 6? 36 divided by 6 is 6. ! See All Questions ...
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1 answer 228 views. 228 views. Write a problem that can be solved with the equation r divided by 6 = 14. asked Nov 14, 2012 in Algebra 1 Answers by anonymous. help; 6th grade math; 0 answers 18 views. 18 views. 6.25 ft equals how many in.
What is 2/8 divided by 6/24. Remember. Register; Math Homework Answers. Questions; Hot! Unanswered; Tags; Users; Ask a Question; 66,909 questions 57,849 answers 745 comments 13,275 users ...
What are the different ways to divide? You can use many different strategies to find the answer to a division problem. ... For example: 231 ÷ 6 = 38.5; 38 x 6 = 228 and 231 − 228 = 3; therefore 231 ÷ 6 = 38.5, is the same as 231 ÷ 6 = 38 R3.
The map is divided into 4 sections with each section divided into quarter-quarter sections. ... § 16-228-1320, filed 5/9/07, effective 6/9/07; WSR 03-22-029, § 16-228-1320, filed 10/28/03, effective 11/28/03. Statutory Authority: Chapters 15.58 and 17.21 RCW.
http://apps.leg.wa.gov/WAC/default.aspx?cite=16-228-1320
What is 6 1/3 divided by 5 1/2 = x/6?-----(19/3)/(11/2) = x/6----(19/3)(2/11) = x/6---(38/33) = x/6 x = 6(38/22) x = 228/22 x = 114/11 ===== Cheers, Stan H. ...
http://www.algebra.com/algebra/homework/NumericFractions/Numeric_Fractions.faq.question.342342.html
(228) 5. Double 18. (36) 6. Round 844 to the nearest ten. (840) 7. 720 / 9 = (80). MM 6-6 1. ... 5. Start with 36, add 12, divide by 6, take _ of that. (4) 6. What is _ of 7? (3 _) 7. Estimate 321 / 59. (5) MM 6-113 1. Estimate 642 / 6. (100) 2. 459 + 7 = (466). 3.
How can you quickly know if one number will divide evenly into another number, leaving no remainder? For example, will 3 divide evenly into 2,169,252? ... Division by 6 The number has to be even. If it's not, forget it.
http://www.maths.dit.ie/scienceweek/divisibilitytricks.htm
Step-Up R • 6 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. 4 Estimating Quotients with ... Think: 20 can be divided evenly by 2. 200 is close to 228 and 20 is close to 19. 200 and 20 are compatible numbers. Step 2: Divide.
Example: 228 is evenly divisible by 3 because 2 + 2 + 8 = 12 and 12 is divisible by 3. ... Therefore 6 will also divide 180 evenly with a result of 30. When the sum of a number’s digits equals a number divisible by 9, that number will always be divisible by nine.
http://www.wisegeek.com/what-are-the-divisibility-rules.htm
Shade 1.6. Divide it into 4 parts. ___ ÷ ___ = ___ e. Shade 0.30. Divide it into 10 parts. ___ ÷ ___ = ___ f. Shade 0.1. Divide it into 10 parts. ___ ÷ ___ = ___ A decimal divided by a whole number. You can think of multiplication “backwards.”
http://www.homeschoolmath.net/teaching/d/divide_decimals.php
... a relationship, between two numbers. What ratio has 6 to 12? 6 is half of 12, or, in the language of percent, 6 is 50% of 12. When the percent is less ... In this case, to make 200 into 100, we must divide by 2, or take half. Therefore we must also take half of 11, which is 5½. 5½ ...
http://www.themathpage.com/ARITH/what-percent.htm
3.8 divided by 6 (show in reduced form) Log In Sign Up MyFeed Find People Account Archive Help Blog About. Question and answer. 3.8 divided by 6 ... [Total 279]| Ratings 0 | Comments 228 | Invitations 0 | Offline. yeswey. S. L. P. C. P. C. 1. L. P. C. Points 174 [Total 8055]| Ratings 0 ...
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Question 363286: If the perimeter of a tennis court is 228 feet and the length is 6 feet longer than twice the width, then what are the length and width?
http://www.algebra.com/algebra/homework/logarithm.faq.question.363286.html
when a no. is divided by 2,3,4,5&6. when a no. is divided by 2,3,4,5&6 , the remainder is 0. but the no. is exactly divisible by 7.
The Beat The GMAT Forum - Expert GMAT Help & MBA Admissions Advice : A number when divided by 5 leaves a remainder 3. Welcome! ... 8 to the power 6 - remainder 4 ... 228 times. Thu Nov 17, 2011 12:14 am. Quote; 3^100/5 ...
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Critical ThinkingIf you divide 25.6 by a number and the quotient is 100, is the number greater than 1or less than 1? Explain your reasoning. ... Multiple ChoiceWhat is the value of the expression 2.4 5.6 0.02? F. 4.282 G. 228.4 H. 282.4 I. 400 Te st-Taking Practice State Test Practice CLASSZONE ...
http://www.warrick.k12.in.us/schools/castlesouth/docs/Math/Math7Book/Math%207%20Ch%202.4%20p69.pdf
What is 15 divided by 4? Find x if x+4=8. What is 6.5 – 4.8? What is the only prime number that is an even number? What is (5+4) x 10? ... How many degrees in a square? Answers: 4050; 228; 5/4 or 1 1/4 (1 and a quarter) 5/4 or 1 1/4; 2, 3, 5, 7 and 11. No. 25 can be divided by 5. 4; 874; 118 ...
Maryland Route 228 (MD 228) is a state highway in the U.S. state of Maryland. Known as Berry Road, the state highway runs 6.88 miles (11.07 km) from MD 210 in Accokeek east to U.S. Route 301 (US 301) and MD 5 Business in Waldorf. MD 228, which is a four-lane divided highway for its entire length ...
... length x width = sq ft. divided by 2.5 ... 6. 9. 12. 15. 18. 21. 24. 27. 30. 33. 36. 39. 42. 12. 4. 8. 12. 16. 20. 24. 28. 32. 36. 40. 44. 48. 52. 56. 15: 5. 10. 15. 20. 25. 30. 35. 40. 45. 50. 55. 60. 65. 70. 18. 6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66. 72. 78. 84. 21: 7. 14. 21. 28. 35 ...
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What are the different ways to divide? You can use many different strategies to find the answer to a division problem. ... For example: 231 ק 6 = 38R3; 6 38 = 228 and 228 + 3 = 231. You can also use a calculator to divide, then you can compare the two answers.
Divide 7/24 by 35/48 and reduce the quotient to the lowest fraction. ... If he had to pay 6% sales tax on the lawnmower, what was the total cost of the lawnmower? Weegy: \$228.00 * 1.06 = \$241.68 total cost . Question. All Categories|No Subcategories|Expert Answered.
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... (456) and halve them you get 228, and 28 is in the 4 times table, ... If a number divides by 6 then firstly see if it is an even number, then see if it can be divided by 3. e.g. 276 can be divided by 6 because it is an even number and its digits add up to 15, ...
... I added one more 12 than I needed, so I'll take it away from 240 to give me 228. One more!  4 x 27 =4(20 +7) =4(20) + (7 ... You can use the distributive property to divide a polynomial by a monomial. Each term is divided by the monomial. You can also use the distributive ...
... 2x-6/(5)divided byx^2-9/(15)Thank you for your help! Please Help! I need help with a rational expressions problem:x^2y+x^3y/(x)divided by 1+x/ ... Two gasoline distributors, A and B, are 228 mi apart on interstate 80. A charges \$0.85/gal and B charges \$0.80/gal. Each charges 0.05(cents sign) ...
... (3w+6) 228 =6w+12 228-12=6w divide both sides by 6 w=216/6=36 yards l=2*36+6=78yards. 4 years ago; Report Abuse; 0% 0 Votes. by ? Member since: September 21, 2009 Total points: 21,475 (Level 6) Add Contact; Block; let w = width let 2w + 6 = length
2 5/6 divided by 5 3/7 = 17/6 divided by 38/7 = 17/6 x 7/38 Nothing will cancel, so multiply out: = (17 x 7)/(6 x 38) = 119/228. 6 years ago; Report Abuse; by r Member since: May 20, 2007 Total points: 2,572 (Level 4) Add Contact; Block;
40 divided by 100 ((You should get a decimal)) Multiply it by 30 That's your answer. Source(s): Math is my best subject.... Rate. Comment ; legally_blonde_piink answered 6 years ago. enter 40.00 into a ...
6) Divide 14 by 7 to get 2. Place this on top of the 4 and the division sign. 7) Multiply 2 by 7 to get 14. 8) Subtract 14 from 14 to get 0. The answer is 12! Step 3: Simple Division with Remainder. This is the same as simple division except we add in the remainder.
http://www.instructables.com/id/How-to-Divide-1/
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If my grade is a c and i take a 6 percent weight quix an lose 50 points but also gained 25 what ... - 228 00 320 whatis the grade. What would a grade for 295 25 out of 395 be? - 170 230 is what grade. What is 177 divided by 255? - Whats 200 divided by 187. Is getting 142 points out of 250 points ...
... 84, 90, 96, 102, 108, 114, 120, 126, 132, 138, 144, 150, 156, 162, 168, 174, 180, 186, 192, 198, 204, 210, 216, 222, 228, 234, 240, 246, 252, 258, 264, 270, 276, 282, 288, 294, 300, 306, 312 ... Any natural number that can be divided by 6 leaving no fractional remainder is a multiple ...
... 13 1/3 divided by 6 2/9 2.) 15 * 4 2/3 3.) 10 2/3 divided by 4 Monday, September 19, 2011 at 3:49pm ... x = - 19 k = 12 * ( - 19 ) = - 228 k ( - 19 ) = - 228 k ( 10 ) x = 10 k = 12 * 10 = 120 k ( 10 ) = 120 k ( 21 ) x = 21 k = 12 * 21 = 252 k ( 21 ) = 252 Friday, April 13, 2012 at 5:45pm ...
http://www.jiskha.com/search/index.cgi?query=The+mean+of+10+observations+is+21.+If+each+score+is+divided+by+3%2C+find+the+new+mean.&page=23
6 is the smallest perfect number. ... 228 is the number of ways, up to rotation and reflection, of dissecting a regular 11-gon into 9 triangles. ... 419 is the number of ways to divide a 6×6 grid of points into two sets using a straight line.
http://www2.stetson.edu/~efriedma/numbers.html
Since 7x 6 = 42, use that. 38/7 is the same as 228/42 & 17/6 is the same as 119/42. So now you have: ... To divide fractions, just reverse the second fraction and MULTIPLY. So 4/5 x 3/1 = 12/5 = 2 2/5 3.
6 (six / ˈ s ɪ k s /) is ... Divided in six arias, ... 228; 229; 230; 231; 232; 233; 234; 235; 236; 237; 238; 239; 240; 241; 242; 243; 244; 245; 246; 247; 248; 249; 250; 251; 252; 253; 254; 255; 256; 257; 258; 259; 260; 261; 262; 263; 264; 265; 266; 267; 268; 269; 270; 271; 272; 273; 274; 275 ...
http://en.wikipedia.org/wiki/6_(number)
What is the difference between OTA codes *228, *22890, and *22891?
in national currency divided by the PPP), and column (4) shows per capita expenditures using ... (6) (7) Australia 50 228 204 248 50 61 54 Brazil 40 380 163 246 40 61 52 China 30 441 122 68 30 17 17 South Africa 25 486 102 72 25 ...
6' 5" 160 169 177 186 194 202 211 219 228 236 241 253 261 295 337 422. ... Finally, we divide the figure from step 2 by the figure from step 4. In this case, 119,369.4 divided by 3782.25 equals 31.56. Since Jill's BMI is over 30, ...
http://www.dietpower.com/help/diet/obese__how_to_tell_if_you_re.htm
... (300 x 4,000) / 5,252] 228 horsepower at 4,000 RPM. But where does the number 5,252 come from? To get from pound-feet of torque to horsepower, ... The seconds are easy -- we just divide by 60 to get from minutes to seconds. Now what we need is a dimensionless unit for revolutions: a radian.
http://www.howstuffworks.com/question622.htm
I'm 6'4", 228 lbs and 28 years old. Can someone explain why, ... I am 6'3 and about 250ish lbs, ... Measure your waist at the navel and your hips at the widest point. Divide waist by hips. | 4,696 | 13,406 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2014-15 | latest | en | 0.913938 |
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Colin October 8, 2003 11:00
Gravity g's influence on Simulation Results
Hi, thank you in advance for your taking time and energy to give me your valuable proposals. The problem is as follows.
When we solve N-S equation to obtain the velocity and pressure fields, if gravity g has influence on the final simulation results?
If the answer is 'yes', how do we reflect gravity(g)'s influence on the results using Fluent? Or what should we do using Fluent to consider g's influence?
Any suggestions are highly appreciated.
steve October 9, 2003 02:06
Re: Gravity g's influence on Simulation Results
I won' be able to help you on how to modify the gravity in fluent but I can assure you that g has an influence on the results.
If you take a close look at the conservation of momentum which can be written as follow:
ro*(du/dt+(u*V)u)+V p = ro g + f + V tau
where u, g and f are vectors and V the diverent.
Imagine that the external force (excluding gravity) f is zero and that the viscous stress tensor tau is also equal to zero. Then you system of DE has for results on the right hand side (gx gy gz)'. In an 'normal' case where g only act along the z axis then you have (0 0 9.81)'.
I am simulating tunnel fires where ro if a function of the temperature. I usually build my tunnel horizontal and then modify g to take the effect of the sloop in consideration. Say the gradient of the tunnel is 5° then g = (0.85 0 9.77). The results between an horizontal and a 5° sloop tunnel are very different. If you are not convince search the web for 'Kaprun Cable Car fire'
Hope this has helped
Steve
Colin October 9, 2003 10:38
Re: Gravity g's influence on Simulation Results
Hi, Steve,
Now I can see the importance of gravity in your research. Let's see another case: assume water flows in a tube. For the horizontally flow if we could say the gravity can be neglected, while for the vertical upward flow the gravity plays a role that can not be ignored?
steve October 10, 2003 02:00
Re: Gravity g's influence on Simulation Results
I would not say that gx is neglect, I would say that the componant in the x-direction of g gx is equal to zero: g = (0 0 -9.81). I agree I might be playing with the words but don't forget that g is a vector.
In a river it is gx that will move the flow forward !
Once again imagine a tunnel fire or a room fire (sorry to use once again a fire but...) . Ro is a function of the temperature. The hotter is the air the lower is the mass density ro. So in a fire the hot air which is 'lighter' tends to go up under the ceiling while colder air will be find at road level. Without the gz the hotter air would not know which way to go. In the NS equations how do you define up and down ?? by g = (0 0 -9.81). If you change g to (0 0 9.81) so gz point upward then the hotter air will be find at road level while the colder air will be find under the ceiling.
Hope this help.
PS: you can do the same comparaison with a hot air ballon.
jdc October 10, 2003 04:29
Re: Gravity g's influence on Simulation Results
Hi Colin and Steve,
The Froude number Fr (=V^2/(g.L) where V is a characteristic velocity, g the norm of the gravity forces and L a characteristic length) can help knowing wheter the gravity should be taken into account in the simulation.
I know this criteria is used for jet flows where V is the injection velocity and L the injector diameter.
Note: I wonder if there is not a formulation that also uses the density.
Hoppe this may help. Julien
Colin October 10, 2003 10:48
Re: Gravity g's influence on Simulation Results
Hi, Steve and Julien,
I much appreciate your kind help. Your advices are much important to me. I will deal with gravity carefully in my project.
Julien, would you please give me a clue that how the Froude Number is used as a criteria for jet flow?
Thank you, and have a nice weekend, you all.
Sincerely,
Colin
Jannis D. November 2, 2003 15:13
Re: Gravity g's influence on Simulation Results
You have to consider gravitation when you have high temperture differences that result a suplementary motion in your system. This supplementary forces result from buoyancy forces. You can check it over the ratio of Rayleigh to reynolds number. If Re/Ra >10 000 you may ignore the gravitation.
best regards,
Jannis
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## Do Now
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## Friday, June 30, 2017
1.0 INTRODUCTION
In this post, we are are going to present some solved examples on the shear deformation of one-span beams due to externally applied load. The predominant cause of deformation in beams subjected to externally applied load is bending, and that is the one usually considered. However, additional deformation is produced due to shear forces in the form of mutual sliding
of adjacent sections across each other. As a result of non-uniform distribution of shear stresses, the sections previously plane now become curved due to shearing alone.
Using the simple virtual work method and employing Verecshagin's combination rule, we are going to calculate the deflection at the critical points of some beams due to bending and due to shear forces. You will discover why shear deflection are neglected in some cases, but in sections that are significantly deeper, shear forces can be quite influential.
The displacement equation due to shear forces is given below;
Where;
k = factor that accounts for non-uniform distribution of shearing stresses. For rectangular sections, k = 3/2, and for circular sections, k = 3/4
G = Shear modulus of the section
A = Area of the cross-section
Q = Shear force due to externally applied load
Q ̅ = Shear force due to a unit virtual load at the point where the deflection is sought
Similarly, the displacement equation due to bending moment is given below;
Where;
E = Elastic modulus of the section
I = Moment of inertia of the cross-section
M = Bending moment due to externally applied load
M ̅ = Bending moment due to a unit virtual load at the point where the deflection is sought
2.0 SOLVED EXAMPLES
For all examples shown below, the section shown below will be used for all calculations.
2.1 Geometrical Properties
Area (A) = bh = 0.2m × 0.4m = 0.08 m2
Moment of inertia (I) = bh3/12 = (0.2 × 0.43)/12 = 0.0010667 m4
E = 21.7 KN/mm2 = 21.7 × 106 KN/m2
G = E / 2(1 + ν) = (21.7 × 106)/2(1 + 0.2) = 9.0416 × 106
EI = (21.7 × 106) × 0.0010667 = 23147.39 KN.m2
GA = (9.0416 × 106) × 0.08 = 723333.333 KN
2.2 Example 1: Simply supported beam carrying a uniformly distributed load. We are to find the deformations at the mid span
The internal forces diagram due to externally applied load is shown below;
Placing a unit load at the mid-span and plotting the internal stresses diagram;
We can now obtain the deflection at the mid-span by diagram combination;
(a) Deflection due to bending;
EIδb1 = 2 [5/12 × 62.5 × 1.25 × 2.5] = 162.760
δb1 = 162.760/23147.39 = 0.0070314 m = 7.0314mm
(b) Deflection due to shear;
GAδs1 = 2 [3/2 × 1/2 × 50 × 0.5 × 2.5] = 93.75
δc1 = 93.75/723333.333 = 0.0001296 m = 0.1296 mm
2.3 Example 2: Simply supported beam carrying a concentrated load at the mid-span. We are to find the deformations at the mid span
The internal forces diagram due to externally applied load is shown below;
Placing a unit load at the mid-span and plotting the internal stresses diagram;
We can now obtain the deflection at the mid-span by diagram combination;
(a) Deflection due to bending;
EIδb2 = 2 [1/3 × 125 × 1.25 × 2.5] = 260.417
δb2 = 260.417/23147.39 = 0.01125 m = 11.25 mm
(b) Deflection due to shear;
GAδs2 = 2 [3/2 × 50 × 0.5 × 2.5] = 187.5
δs2 = 187.5/723333.333 = 0.0002592 m = 0.2592 mm
2.4 Example 3: Cantilever beam carrying a uniformly distributed load. We are to find the deformations at the free end
Placing a unit load at the free end and plotting the internal stresses diagram;
We can now obtain the deflection at the mid-span by diagram combination;
(a) Deflection due to bending;
EIδb3 = [1/4 × 250 × 5 × 5] = 1562.5
δb3 = 1562.5/23147.39 = 0.0675 m = 67.502 mm
(b) Deflection due to shear;
GAδs3 = [3/2 × 1/2 × 100 × 1.0 × 5] = 375
δs3 = 375/723333.333 = 0.00051843 m = 0.5184 mm
As you can see in all the examples we considered, there was no case where the additional deflection due to shear was up to 1mm, and for practical purposes, they are very negligible for normal one span beams. Watch out for future posts where we will present cases where shear deflection becomes significant.
Thank you for visiting www.structville.com. We hope you visit us again, and share this post with your friends. | 1,282 | 4,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2020-05 | latest | en | 0.918119 |
https://de.mathworks.com/matlabcentral/answers/518080-incremental-indexing-to-create-an-array | 1,702,312,437,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679515260.97/warc/CC-MAIN-20231211143258-20231211173258-00577.warc.gz | 229,332,950 | 27,213 | # Incremental indexing to create an array
8 Ansichten (letzte 30 Tage)
Michael King am 15 Apr. 2020
Kommentiert: Michael King am 15 Apr. 2020
Hi,
I have a 1097x1097 array (called list_of_distances) and would like to index into that array repeatedly; I only need certain values from the array. By doing this repeated indexing, I'd like to create new array that is 1x1097 of values. I included a screen shot of the values I'm looking to get into a new array.
Any help would be appreciated
Michael
Michael
R = 2
S = 1
for i = length(list_of_distances)
final_list(i) = list_of_distances(R,S)
R = R+1
S = S+1
end
##### 0 Kommentare-2 ältere Kommentare anzeigen-2 ältere Kommentare ausblenden
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### Akzeptierte Antwort
Peng Li am 15 Apr. 2020
using diag(list_of_distances, -1) to get the diagonal you want.
##### 5 Kommentare3 ältere Kommentare anzeigen3 ältere Kommentare ausblenden
Peng Li am 15 Apr. 2020
the loop doesn't run as your i is fixed at length(list_of_distances), and you prob want to do for i = 1:length(list_of_distances), but this way you are up to an indexing error as I commented above, you aren't able to have a vector of 1097 elements. Do for i = 1:size(list_of_distances, 1)-1 instead.
And you don't necessarily need to do a loop, although you are not familiar with some of these builtin functions. For this case, you can do list_of_distances(2:size(list_of_distances, 1)+1:end), which will give you the same results as diag(list_of_distances, -1).
list_of_distances = magic(10)
list_of_distances =
92 99 1 8 15 67 74 51 58 40
98 80 7 14 16 73 55 57 64 41
4 81 88 20 22 54 56 63 70 47
85 87 19 21 3 60 62 69 71 28
86 93 25 2 9 61 68 75 52 34
17 24 76 83 90 42 49 26 33 65
23 5 82 89 91 48 30 32 39 66
79 6 13 95 97 29 31 38 45 72
10 12 94 96 78 35 37 44 46 53
11 18 100 77 84 36 43 50 27 59
>> b = diag(list_of_distances, -1)
b =
98
81
19
2
90
48
31
44
27
>> c = list_of_distances(2:size(list_of_distances)+1:end)
c =
98 81 19 2 90 48 31 44 27
Michael King am 15 Apr. 2020
thanks, I've learned a lot here! much appreciated
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Translated by | 774 | 2,349 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-50 | latest | en | 0.637209 |
https://kids.kiddle.co/Schr%C3%B6dinger_equation | 1,713,060,503,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816863.40/warc/CC-MAIN-20240414002233-20240414032233-00037.warc.gz | 313,245,001 | 7,459 | # Schrödinger equation facts for kids
Kids Encyclopedia Facts
Schrödinger's equation inscribed on the gravestone of Annemarie and Erwin Schrödinger. (Newton's dot notation for the time derivative is used.)
The Schrödinger equation is a differential equation (a type of equation that involves an unknown function rather than an unknown number) that forms the basis of quantum mechanics, one of the most accurate theory of how subatomic particles behave. It is a mathematical equation that was thought of by Erwin Schrödinger in 1925. It defines a wave function of a particle or system (group of particles) which has a certain value at every point in space for every given time. These values have no physical meaning (in fact, they are mathematically complex), yet the wave function contains all information that can be known about a particle or system. This information can be found by mathematically manipulating the wave function to return real values relating to physical properties such as position, momentum, energy, etc. The wave function can be thought of as a picture of how this particle or system acts with time and describes it as fully as possible.
The wave function can be in a number of different states at once, and so a particle may have many different positions, energies, velocities or other physical property at the same time (i.e. "be in two places at once"). However, when one of these properties is measured it has only one specific value (which cannot be definitely predicted), and the wave function is therefore in just one specific state. This is called wave function collapse and seems to be caused by the act of observation or measurement. The exact cause and interpretation of wave function collapse is still widely debated in the scientific community.
For one particle that only moves in one direction in space, the Schrödinger equation looks like:
$-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x,\,t) + V(x)\Psi(x,t) = i\hbar\frac{\partial}{\partial t} \Psi(x,\,t)$
where $i$ is the square root of -1, $\hbar$ is the reduced Planck's constant, $t$ is time, $x$ is a position, $\Psi(x,\,t)$ is the wave function, and $V(x)$ is the potential energy, an as yet not chosen function of position. The left hand side is equivalent to the Hamiltonian energy operator acting on $\Psi$.
## Time independent Version
Assuming that the wave function, $\Psi (x,t)$, is separable, i.e. assuming the function of two variables can be written as the product of two different functions of a single variable:
$\Psi (x,t) = \psi (x) T(t)$
then, using standard mathematical techniques of Partial differential equations, it can be shown that the wave equation can be rewritten as two distinct differential equations
$i\hbar \frac{d T(t)}{dt} = E \, T(t)$
$-\frac{\hbar^2}{2m} \frac{d^2 \psi (x)}{dx^2}+V(x) \psi (x) = E \, \psi (x)$
where the first equation is solely dependent on time $T(t)$, and the second equation depends only on position $\psi (x)$, and where $E$ is just a number. The first equation can be solved immediately to give
$T(t) = e^{-i\frac{Et}{\hbar}}$
where $e$ is Euler's number. Solutions of the second equation depend on the potential energy function, $V(x)$, and so cannot be solved until this function is given. It can be shown using quantum mechanics that the number $E$ is actually the energy of the system, so these separable wave functions describe systems of constant energy. Since energy is constant in many important physical systems (for example: an electron in an atom), the second equation of the set of separated differential equations presented above is often used. This equation is known as the Time independent Schrödinger Equation, as it does not involve $t$. | 886 | 3,730 | {"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": 20, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-18 | latest | en | 0.919586 |
https://www.physicsforums.com/threads/mass-times-avogadros-numbers-molar-mass.291139/ | 1,511,286,193,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806421.84/warc/CC-MAIN-20171121170156-20171121190156-00244.warc.gz | 845,476,232 | 14,368 | # Mass times avogadro's numbers = molar mass ?
1. Feb 9, 2009
### zmike
mass times avogadro's numbers = molar mass ??
Quick questions while I was reading my textbook:
apparently my textbook shows that
mass x avogadro's numbers = molar mass of the substance???
this makes no sense to me.
Isn't mole/mass = molar mass????
2. Feb 10, 2009
### Bacat
Re: mass times avogadro's numbers = molar mass ??
Mass is an extensive property of matter (extensive means that the property changes depending on how much matter there is).
Avogadro's Number is the number of molecules in one mole of matter.
The molar mass is the mass of one mole of matter. For example, consider two molecules A and B (assume they are not the same molecule). Since the molecules have different masses, if we take Avogadro's Number of A and Avogadro's Number of B (the same number of molecules), the total mass of each sample will be different. This total mass for one mole of A is the molar mass of A.
So you get an equation:
Mass of molecule * Avogadro's Number = Molar Mass
3. Feb 11, 2012
### bodycutter
Re: mass times avogadro's numbers = molar mass ??
Thank you for making it Clear.
Now It makes sense to me :) | 311 | 1,199 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-47 | longest | en | 0.858743 |
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## Transcription
1 / Approximation Algorithms Lecturer: Michael Dinitz Topic: Linear Programming Date: 2/24/15 Scribe: Runze Tang 9.1 Linear Programming Suppose we are trying to approximate a minimization problem (everything will hold for maximization problems too, but with the inequalities reversed). A general way to prove α-approximation is as following: 1. Prove OPT LB 2. Prove ALG α LB Then we can conclude ALG α OPT. This is how essentially all of our analyses have worked so far. 9.2 Integer Linear Program A set {x 1, x 2,, x n } of variables, each of which must be an integer m linear inequalities over variables Possibly a linear objective function over variables 9.3 LP Rounding We want to round the solution to LP to get a feasible solution to ILP. The general steps are as following: 1. Write ILP 2. Relax to LP (i.e. remove the integrality constraints) 3. Solve LP, getting solution x 4. Round x to integer values to get a solution to ILP Key ideas of rounding: 1. LP OPT, since it is a relaxation. Slightly more formally, any integral solution is obviously a solution to the LP, and hence the optimal LP value is at most the optimal ILP value, which equals OPT. 1
2 2. We design our rounding in order to guarantee that the integral solution we get back is at most α LP. Putting these together, we get that ALG α OPT. And instead of having to find some mysterious lower bound LB to compare the algorithm to, we can compare the output of the algorithm to the fractional LP solution, and only analyze how much our rounding increased the cost. 9.4 Weighted Vertex Cover Problem Description Input: a graph G = (V, E), and a cost function c : V R + Feasible solution: S V such that {u, v} E, either u S or v S Objective: min v S c(v), i.e. min c(s) Equivalent Integer Linear Program minimize: c(v) x v v V subject to: x u + x v 1 for each edge {u, v} E x v {0, 1} for each vertex v V It is easy to see that this ILP is an exact formulation of the weighted vertex cover problem. For one direction, suppose that we are given a vertex cover S. Then setting x v = 1 if v S and x v = 0 if v S gives a solution to the ILP with cost exactly c(s). For the other direction, suppose that we are given a solution x to the ILP. Then let S = {v V : x v = 1}. Then S is a vertex cover with cost exactly equal to the objective value of the ILP on x. Hence the ILP and weighted vertex cover are equivalent Relax to a Linear Program Now we would like to drop integrality constraints to get LP: minimize: c(v) x v v V subject to: x u + x v 1 for each edge {u, v} E 0 x v 1 for each vertex v V Theorem If x is feasible for ILP, then x is feasible for LP. Proof: x satisfies the edge constraints of the LP since it satisfies them for the ILP, and x v {0, 1} for each v V due to the integrality constraints of the ILP, and hence 0 x v 1. 2
3 Corollary OPT(LP) OPT(ILP). Proof: By the theorem above, the optimal solution to ILP is feasible to LP. Corollary If we find a solution of cost less or equal to α OPT(LP), then we have an α- approximation algorithm. Proof: cost(sol) α OPT(LP) α OPT(ILP) LP Rounding As always, we first solve the LP to get an optimal fractional solution x. We can do this in polynomial time since the LP has size polynomial in the size of the instance. We then round x to an integral solution x as follows: for each v V, we set x v = 1 if x v 1 2, and set x v = 0 otherwise. Theorem x is a feasible solution to ILP. Proof: For any (u, v) E, we have x u + x v 1, which implies that max(x u, x v) 1 2. So either x u = 1 or x v = 1. Thus x u + x v 1. Theorem c( x ) 2c( x ). Proof: c( x ) = v V c(v) x v = {v V :x v 1/2} c(v) v V c(v) 2x v = 2c( x ). Corollary This rounding is a 2-approximation algorithm. 9.5 Weighted Set Cover Problem Description Weighted set cover problem is the set cover problem with costs on each set Equivalent Integer Linear Program minimize: subject to: c(s) x S S S {S:e S} x S 1 for each element e U x S {0, 1} for each set S S This is clearly an exact formulation it is easy to see that solutions to the ILP correspond to set covers with the same cost, and vice versa. 3
4 9.5.3 Relax to a Linear Program minimize: subject to: c(s) x S S S {S:e S} x S 1 for each element e U 0 x S 1 for each set S S LP Rounding In a previous lecture on set cover we defined the frequency of an element e to be f e = {S S : e S} and defined f = max e U f e. Given x which is solution to LP, set x S = 1 if x S 1 f, and set x S = 0 otherwise. Theorem x is a feasible solution to ILP. Proof: Since each elements is in at most f sets, the LP constraint {S:e S} x S 1 implies that max {S:e S} x S 1 f. Thus there exists at least one S which contains e such that x S = 1. So {S:e S} x S 1. Corollary It is an f-approximation algorithm. Proof: By the previous theorem x is a feasible set cover. For its cost, we have that c( x ) = S S c(s)x S f S S c(s)x S = f c( x ). 9.6 Integrality Gap Definition The integrality gap of an LP is sup instance I ( OP T (I) LP (I) The integrality gap of an LP measures the strength of the relaxation. If we prove that ALG α OP T by proving that ALG α LP (as we have been doing with rounding), then we cannot give a ratio α that is better than the integrality gap (or else on the instance I which achieves the integrality gap we would be able to round the LP solution to a value less than OPT, giving a contradiction) Integrality Gap for Weighted Vertex Cover Consider the instance to be: G = K n, which is the complete graph, and c(v) = 1 for all v V. Theorem The integrality gap is at least 2 ( 1 1 n). Proof: OPT = n 1, since if there are two nodes not in the vertex cover the edge between them will not be covered. But in the LP, if we set set x v = 1 2 for every v V we get a valid LP solution with cost n n 1 2. Hence IG n/2 = 2 ( 1 n) 1. ). 4
5 9.6.2 Integrality Gap for Max Independent Set The ILP is: maximize: v V x v subject to: x u + x v 1 for each edge {u, v} E x v {0, 1} for each vertex v V Theorem The integrality gap is at least n 2. Proof: Consider the instance to be G = K n, which is a complete graph. So OPT = 1 since every pair of vertices are connected. For the LP, set x v = 1 2 for every v V. Then we get the LP cost to be n n/2 2. Hence the integrality gap is at least 1 = n Solving LPs This is a bit outside the scope of this class, but we will talk a little bit about algorithms for solving LPs. This section is pretty informal, but everything can be made formal Simplex Note that the feasible region of an LP is a polytope (by definition) and hence is convex (for every two points in the feasible region, their midpoint is also feasible). The oldest heuristic for solving LPs is the simplex algorithm. We won t talk much about this algorithm, but the highlevel view is straightforward. Given a polytope, there is a natural graph associated with it where the vertices of the graph are the vertices of the polytope (points which are tight for d linearly independent constraints of the polytope, where d is the dimension) and two vertices are adjacent if in the polytope they are also adjacent (the line segment between them is tight for d 1 linearly independent constraints). This algorithm starts by finding a vertex of the polytope, and then moving to a neighbor with decreased cost as long as this is possible. By linearity and convexity, once it gets stuck it has found the optimal solution. Unfortunately simplex does not run in polynomial time even with a polynomially-sized LP, the number of vertices of the polytope can be exponential. Simplex does well in practice, but poorly in theory Interior Point Methods While simplex only moves along the outer faces of the polytope, there are algorithms known as interior-point methods which make moves inside the polytope. There are now known to be interior-point methods which have polynomial running time, and are also extremely fast in practice. So in both theory and in practice, we can solve LPs efficiently. 5
6 9.7.3 Ellipsoid We will spend a bit more time talking about an algorithm known as Ellipsoid, which works well in theory but poorly in practice. Nevertheless, it has some properties which are extremely nice theoretically, so we will feel free to use it when designing algorithms. As a first step, we will reduce optimization to feasibility. Suppose that we are given an algorithm which can determine whether a given LP is feasible, i.e. it returns YES if the feasible region is non-empty and NO otherwise. Now suppose that we are given an LP which we want to solve. For any value T, we can use our feasibility algorithm to see if there is an LP solution with cost at most T by adding a single new constraint (the objective is at most T ). So we can do a binary search over possible values of T to find in polynomial time the smallest T for which the feasibility region is nonempty. Hence if we have such a feasibility algorithm, we can also solve LPs with objective functions The algorithm The ellipsoid algorithm is an algorithm for testing feasibility. Informally, it is the following. Algorithm 1 Using Ellipsoid to Check Feasibility Algorithm Input: Convex constraints Output: A feasible solution if one exists. Find an ellipsoid containing the polytope. Let x be the the center of the ellipsoid. k = 0 while x is not feasible AND k < maxiter do k = k + 1 Find a constraint violated by x Draw a parallel hyperplane corresponding to this violated constraint through x. This divides the ellipsoid into two parts with equal volume, one of which is on the wrong side of the hyperplane and so is entirely infeasible. Consider the part which is on the correct side of the violated constraint (i.e. the part in which the polytope must appear if it is nonempty). Find a new ellipsoid containing this part. Let x be the center of the new ellipsoid. end while return x, which is a feasible solution; or no feasible solution exists. It is not hard to see that this algorithm is correct, in that if the polytope is nonempty it will eventually find a feasible point and if the polytope is empty then it will never find a feasible point. The hard part is proving that it takes only polynomial time in the worst case, i.e. that we can set maxiter to a polynomial value. While it is complicated, the key idea is that due to the geometry of ellipsoids, in each iteration the volume of the ellipsoid that we are considering decreases by a constant factor. 6
7 Separation The ellipsoid algorithm has an extremely nice property: in order to make it work, all that we need to be able to do is find a violated constraint given some point x (or prove that x is feasible). The is is called the separation problem. If the LP is small then this is simple we can just check all of the constraint. But sometimes our LPs are not small, and ellipsoid still lets us solve them! If there are an exponential number of constraints, we cannot even write down all of them in polynomial time so we certainly cannot use simplex or interior-point methods to solve the LP. But if we can separate, then ellipsoid lets us solve in polynomial time despite not even being able to write down the LP. Let s do a somewhat trivial example of this: the spanning tree polytope Spanning Tree Polytope Consider the minimum spanning tree problem. Obviously we know multiple simple and fast algorithms. But suppose we want to do this by an LP. While it s not immediately obvious how to write such an LP, it turns out that the following formulation is good: minimize: subject to: c(e) x e (9.7.1) e E {e E(S, S)} x e 1 for each cut S V (9.7.2) 0 x e 1 for each e E (9.7.3) Note that the tree constraints are hidden in the optimization problem above since we can always make the objective function smaller by throwing away an edge. Although it has exponentially many constrains, we can separate in polynomial time. Given x, first run the min-cut algorithm. If the min cut is larger or equal to 1, than x is a feasible solution; otherwise the minimum cut S corresponds to a violated constraint. Hence we can separate, so by the ellipsoid algorithm we can optimize. 7
### 11.1 Facility Location
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The Student/Project Allocation problem with group projects Aswhin Arulselvan, Ágnes Cseh, and Jannik Matuschke arxiv:4.035v [cs.dm] 3 Dec 04 Department of Management Science, University of Strathclyde, | 9,437 | 38,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | latest | en | 0.870549 |
http://lotsasplainin.blogspot.com/2017/10/hot-enough-for-ya.html | 1,531,951,650,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590329.62/warc/CC-MAIN-20180718213135-20180718233135-00265.warc.gz | 226,544,419 | 20,654 | ## Sunday, October 1, 2017
### Hot enough for ya?
This September in Oakland has been hot. Here are some ways to put it in perspective.
1. The average temperature was 78.3° F. This might not seem like much in Sacramento or San Diego, but that is the warmest average month not only this year so far, but the warmest since 2011, which is the arbitrary year when I started measuring things this way.
2. According to Weather Underground's daily average of the past fifteen years, September 2017 was 3.9° F. warmer than the average. 3.9° above the fifteen year benchmark is definitely warm, but it is by no means a record over the years I have been using to measure things this way. For example, the winter of 2015 was way above average, with January at 3.5° warmer, February at 5.1° warmer and March at 5.7° warmer. Still, 3.9° warmer than the benchmark is by no means an average month. In 2017 so far, September and May are tied for first with 3.9°.
3. September 1st and 2nd were both 101° F in Oakland. Again, it's a matter of perspective. In Sacramento, days over 100° are an inconvenience. In Oakland, two days in a row over 100° is a sign of the apocalypse.
We also have a statistical method to tell us if a month is unusually hot or not using t-scores, a relative of z-scores. The formula for the two is the same, the average times the square root of the number of days in the month divided by the month's standard deviation. (This formula is almost what we want, and it is exactly correct if mux = 0.) Using this test data, we can get a p-value, the beloved precious of scientific researchers everywhere. If the p-value is less than .05, this is usually a sign your paper can possibly be published.
Using this method, September 2017 was not unusually above the average of the last fifteen Septembers. (Note: May 2017 had the same raw score of 3.9° above average and it produced a p-value high enough to let us reject the null hypothesis. May was unusually hot using this method, September, not so much.)
Why did September fail in rejecting the null hypothesis, which is to say it does not seem unusually warm using this test? The answer is in the standard deviation, a commonly used method to measure how spread out a data set is. If there isn't much deviation in a set, a 3.9° difference would definitely impress the t-score test. What happened is that September was warm in a very weird way, several days way warmer than average, but thirteen days out of thirty, it was actually slightly cooler than average. (By "way warmer", the early heat wave was 26° F. warmer than average for two days and there were five more days in September were the temperature was 10° warmer than average or more.) But in the middle of the month, there was a ten day "cool snap", when temperatures were cooler than average by -1° to -6°. These big swings meant for a higher standard deviation, the highest of the year at 9.384. In comparison, the month of May did let us reject the null hypothesis because the standard deviation was "only" 7.865, which is the second highest standard deviation of the year.
Okay, Matty Boy, what does this have to do with the price of tea in China? Well, if it isn't my old pal Hypothetical Question Asker! This is just an example of statistical methods sometimes producing confounding results. I have no philosophical qualms about the t-score test in general, though the arbitrary threshold of .05 to decide whether we accept or reject the null hypothesis is fairly coming under question these days in research circles. My other quibble about this work that I am doing is whether we should think of a month as a period of time that measures climate or if it should be still considered just weather. The method I hit upon earlier in September argues that climate should use time spans of a year. Shorter spans like season or half years might make sense, but my general feeling is a month is too short.
In any case, I saw some weird numbers and decided to write about 700 words about them.
Don't hate. This is how I roll.
Any questions?
(Seriously, the comments are perfect for questions.)
#### 2 comments:
Abu Scooter said...
Thanks for these; I'm enjoying them. Heck, I even have questions this time.
With respect to temperatures in Oakland, what have the t-scores been for the last 12 months? 24? 60?
More to my point, do those scores tend to be higher in March and September than in other months? I'd expect to see more temperature variance in those months, since they coincide with the spring and fall equinoxes. Is that a valid hypothesis, or have I been out in the (Illinois) sun too long?
Matthew Hubbard said...
Hey, buddy! Nice to hear from you. Here are the numbers from 2015, 2016 and the current year to date
2015, Months that rejected the null hypothesis: Jan., Feb., Mar., Jul., Aug., Oct.
2016, Months that rejected the null hypothesis: Jan., Feb., Mar., Apr.,
Jun., Nov.
2017, Months that rejected the null hypothesis as of end of September: Mar., Apr., Jul.
High standard deviation months. (Arbitrary measure: s.d. > 6 is high)
2015: Apr., Sept.
2016: Apr., Jun., Sept.
2017, so far: May, Jun., Sept.
I can send you the Excel files if you want. Send me an email if interested. | 1,239 | 5,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2018-30 | longest | en | 0.971658 |
https://www.physicsforums.com/threads/basic-inductor-concept.422339/ | 1,532,003,829,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590866.65/warc/CC-MAIN-20180719105750-20180719125750-00176.warc.gz | 960,561,616 | 15,610 | # Basic Inductor Concept
1. Aug 15, 2010
### I_am_learning
Consider a simple circuit of Real inductor exited by sinusoidal Source.
What is getting me into trouble is this.
1. What is the current through the inductor? Isn't it the same as that of resister,which is $$\phi$$ phase lagging with the applied voltage. (Current is same in a series branch)
2. Considering that this inductor is the part of a transformers primary winding, the magnetic Flux Will also be $$\phi$$ phase lagging with applied field since Flux is directly proportional to the current (Assuming linear relation between B and H)
But my teachers and my text-book says that in transformer, the magnetic Flux produced is perfectly 900 out of phase with the applied Voltage. My teacher said that only the 900 component of the current is used in producing the flux and the in-phase component will be simply lost in the resistive heating.
So I want you to explain to me what this the phase of flux produced?
2. Aug 15, 2010
### n.karthick
If we neglect the small resistance r of the winding then flux is 90 deg lagging wrt voltage. Otherwise as you said it is lagging by phase $\phi$ only
3. Aug 15, 2010
### I_am_learning
But How do you reply to someone saying this,
"The current can be resolved into two component, one in-phase with the applied voltage and one out of phase. The in-phase current which flows through the internal-resistance of the coil is lost in the heating. The 900 phase current will only produce the flux."
4. Aug 15, 2010
### I_am_learning
Moreover, consider this.
We can also model the internal resistance of the inductor by choosing an equivalent parallel resistance like this
In this case, we can clearly see that the current through the inductor and current through the resistor is different.
Here, current through the inductor is exactly 900 out of phase with the applied voltage. Since the magnetic Flux is produced only by the inductor, the produced flux will also have exactly 900 phase lag with the applied voltage!
Whats wrong here!!!? Somebody please, point out, this is driving me crazy!
5. Aug 15, 2010
### zomgineer
You generally can't model the resistance of the inductor as being in parallel. The reason is that we should expect the same exact current to go through both the inductor and the resistor.
If we let Rs and Ls be our series model of the inductor and Rp and Lp be the parellel version, then we can only say that the models are equivalent if (Rs+Ls)=(RpLp)/(Rp+Lp)
6. Aug 15, 2010
### I_am_learning
I din't mean to say that same value of L and R can be invariably switched between series and parallel model. I only wanted to point out that there is an 'equivalent' parallel model.
My problem is that even if the model is equivalent in the respect that (Rs+Ls)=(RpLp)/(Rp+Lp)
It doesn't seem to be equal in the respect of phase of flux being created. In series model, the Flux seems to be created in the same phase as the net current, in the parallel it seems to be created in the same phase as the current through the inductor.
I hope this clarifies my problem.
7. Aug 15, 2010
### zomgineer
You could apply a DC voltage, give it time to settle, and directly measure Rs. It's a real thing. The resistor in the parallel (Rp) version is not a real thing. It's something that would be made up just to make the models mathematically equivalent. You would probably have to make up a new value for Lp as well so that isn't anything real either. In fact, the made-up parallel Lp might be capacitive instead of inductive just to make the math work out (maybe I'll work it out).
In either case, the total current would be the same. In your parallel model, the sum total of the currents going through each branch would be the exact same as the current that goes into the series version.
Think of the current that goes into the inductor as being a complex vector. There are many ways to add vectors of different phases to get another vector with another phase. There is no rule that they have to have the same phase.
8. Aug 15, 2010
### I_am_learning
So, you mean to suggest the 'real' modelling is that with the series resistor and that the parallel resister modelling gives only the correct net current, its no 'real' thing.
quite logical.
So, your final answer would be that --> Since the series resistance model is the 'real' model, the current through the ideal inductor of the model would be $$\phi$$ phase lagging. So, the magnetix flux would also be $$\phi$$ phase lagging and not exactly 900 phase lagging.
okay, thanks.
9. Dec 25, 2010
### I_am_learning
I don't know if it is right thing to do (because this will bring this dead old thread to the top, for no special purpose) but I found the right answer to my own OP. (Posted this just for the sake of feeling goodness for myself.)
All the things in the thread is good and true except for one thing, the teacher was correct!
Actually, the teacher was talking about the resistance of the core, and not of the copper. With no load in the secondary, the primary current of the transformer is very small (only excitation current) and the copper resistance don't come into play.
Due to non-linear nature of B-H curve, a phi phase lagging current will be required to produce the perfect 90 lagging flux in the core.
So, to model these a core resistance Rm is placed parallel to the perfect inductor. In such model, the flux is produced only by the 90 degree component of the total current (i.e. current through the inductor) and the parallel component will be used up in the resistive heating of the core. | 1,311 | 5,603 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2018-30 | latest | en | 0.943849 |
https://www.salary.com/blog/defining-the-difference-between-average-and-median-salary/ | 1,718,750,850,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861794.76/warc/CC-MAIN-20240618203026-20240618233026-00370.warc.gz | 836,770,566 | 19,235 | Blog
# Defining the Difference Between Average and Median Salary
Written by Brett Rudy
June 5, 2019
Median and average salaries are both measures of central tendency or the “middle of the market” regarding compensation.
Many organizations target employee pay at either the average or median rate to ensure employees receive competitive compensation while the business is still managing overall costs. However, knowing why average and median salaries differ is also important, as it may drive different compensation decisions in your organization.
## How is Average Salary calculated?
You can calculate the average base, mean salary, or average salary by adding all the salaries for a select group of employees and then dividing the sum by the number of employees in the group.
Average Salary Example:
Employee 1 earns \$40,000, Employee 2 earns \$50,000, Employee 3 earns \$100,000. The total of \$190,000 is divided by 3, providing an average salary of \$63,333.
The average salary represents what the “typical employee” earns and can be pulled higher or lower by high salaries or low salaries at the extreme ends of the distribution.
## How is Median Salary calculated?
You can calculate the median base salary by arranging the salaries for a group of employees in descending order and then locating the salary that represents the midpoint of the distribution. Fifty percent of the salaries are less than the median and fifty percent of the salaries are greater than the median.
Median Salary Example:
Employee 1 earns \$40,000, Employee 2 earns \$50,000, Employee 3 earns \$100,000. The salary in the middle, or the median salary is \$50,000.
As the median salary represents a specific point in the distribution, it cannot be pulled higher or lower by salaries at the extreme ends of the distribution. It is therefore considered a more neutral measure of central tendency, especially in a small group of salaries where one extreme value can disproportionately affect the calculation of an average.
## Winning the Wage War: Average Salary vs. Median Salary
Average and median salary are both helpful calculations, but which is more important?
Compensation practitioners generally prefer to benchmark against the median salary, as it is considered more neutral than the average.
However, knowing if the average is above or below the median, and by how much, can help HR pros understand the overall shape of the salary distribution. For instance, an average salary that is much higher than the median salary indicates that there are probably a few employees earning significantly more than the overall group.
In the end, it is the relationship between the two numbers that is most instructive.
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http://mathhelpforum.com/algebra/30466-fraction-problem-print.html | 1,527,476,010,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870771.86/warc/CC-MAIN-20180528024807-20180528044807-00339.warc.gz | 178,440,323 | 2,782 | # Fraction Problem
Printable View
• Mar 9th 2008, 10:59 AM
Itsbaxagain
Fraction Problem
Man I dislike these fractions but I am having problems finding the answer to this one. $\displaystyle \frac{5}{2x}-\frac{4}{2x^2+3x}$
The site helps me a lot and I haven't found a math site that has help me as much as this one. Even if i am just reading someone threads or doing mine I find help in it somewheres.
• Mar 9th 2008, 11:07 AM
topsquark
Quote:
Originally Posted by Itsbaxagain
Man I dislike these fractions but I am having problems finding the answer to this one. $\displaystyle \frac{5}{2x}-\frac{4}{2x^2+3x}$
The site helps me a lot and I haven't found a math site that has help me as much as this one. Even if i am just reading someone threads or doing mine I find help in it somewheres.
Are you supposed to subtract the fractions?
You need a common denominator. Note that the denominator of the second term is $\displaystyle 2x^2 + 3x = x(2x + 3)$. The LCM of $\displaystyle x$ and $\displaystyle x(2x + 3)$ is $\displaystyle 2x(2x + 3)$.
So....
$\displaystyle \frac{5}{2x} - \frac{4}{2x^2+3x}$
$\displaystyle = \frac{5}{2x} \cdot \frac{2x + 3}{2x + 3} - \frac{4}{2x^2+3x} \cdot \frac{2}{2}$
$\displaystyle = \frac{5(2x + 3) - 2 \cdot 4}{2x(2x + 3)}$
$\displaystyle = \frac{10x + 7}{2x(2x + 3)}$
-Dan | 453 | 1,316 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2018-22 | latest | en | 0.860333 |
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Multiplication Magic Thanks Math Champ Write numbers 1 2 3 4 5 6 7 9 on a piece of paper. Ask a participant to indicate which of these numbers is their favorite number. Once they have answered, tell them to multiply 1 2 3 4 5 6 7 9 by a number you have chosen. Their answer will be a long row of their favorite number. Secret Revealed: Once the participant tells you their favorite number, multiply it by 9. Example: The participant chooses 6 as their favorite number. 6 x 9 = 54 Ask them to multiply 1 2 3 4 5 6 7 9 by 54 12345679 x 54 = 66666666 Question: Why?
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Anything CHOCOLATE
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February 14th | 598 | 1,912 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-18 | latest | en | 0.919162 |
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#### Resources tagged with Generalising similar to Your Number Was...:
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### What's Possible?
##### Stage: 4 Challenge Level:
Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make?
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The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = nĀ² Use the diagram to show that any odd number is the difference of two squares.
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##### Stage: 3 Challenge Level:
You can work out the number someone else is thinking of as follows. Ask a friend to think of any natural number less than 100. Then ask them to tell you the remainders when this number is divided by. . . .
### Keep it Simple
##### Stage: 3 Challenge Level:
Can all unit fractions be written as the sum of two unit fractions?
### ...on the Wall
##### Stage: 3 Challenge Level:
Explore the effect of reflecting in two intersecting mirror lines.
### Mirror, Mirror...
##### Stage: 3 Challenge Level:
Explore the effect of reflecting in two parallel mirror lines.
### Who Is the Fairest of Them All?
##### Stage: 3 Challenge Level:
Explore the effect of combining enlargements.
### Cuboid Challenge
##### Stage: 3 Challenge Level:
What size square corners should be cut from a square piece of paper to make a box with the largest possible volume?
### How Much Can We Spend?
##### Stage: 3 Challenge Level:
A country has decided to have just two different coins, 3z and 5z coins. Which totals can be made? Is there a largest total that cannot be made? How do you know?
### Mini-max
##### Stage: 3 Challenge Level:
Consider all two digit numbers (10, 11, . . . ,99). In writing down all these numbers, which digits occur least often, and which occur most often ? What about three digit numbers, four digit numbers. . . .
### Magic Letters
##### Stage: 3 Challenge Level:
Charlie has made a Magic V. Can you use his example to make some more? And how about Magic Ls, Ns and Ws?
### Frogs
##### Stage: 2 and 3 Challenge Level:
How many moves does it take to swap over some red and blue frogs? Do you have a method? | 2,303 | 9,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2015-35 | longest | en | 0.873908 |
https://stats.stackexchange.com/questions/416019/found-an-expression-i-havent-encountered-before | 1,632,755,792,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058456.86/warc/CC-MAIN-20210927151238-20210927181238-00231.warc.gz | 572,690,852 | 38,652 | # Found an expression I haven't encountered before
I was reading the book an introduction to statistical learning with R (http://www-bcf.usc.edu/~gareth/ISL/data.html), and came across this expression that I haven't seen before. Can anyone tell me what this means:
$${\bar x^{2} \over \sum_{i=1}^n(x_i - \bar x)^2}$$
The full equation (where I saw it is:) $$SE(\hat \beta)^2 = \sigma^2 [{1 \over n} + {\bar x^{2} \over \sum_{i=1}^n(x_i - \bar x)^2}]$$
FYI. this is the equation for calculating the standard error of the coefficients of linear regression (ML).
• The bottom term is variance, but why is it being divided by mean squared? Jul 4 '19 at 11:58
• What is exactly your doubt? Do you think there is something wrong about it? Jul 4 '19 at 12:03
• I wanted to know what this term means. As stated above, I've never seen it before. Jul 4 '19 at 12:06
Your expression is a formula for the standard estimation error for a given parameter $$\beta$$ in terms of the sample mean, the observations.
$$\sigma$$ stands for the standard deviation of the errors $$\epsilon$$, assuming a model of the form $$y=\beta + \beta_1 x + \epsilon$$
$$\epsilon$$ is assumed to follow a normal distribution $$N(0, \sigma^2)$$
The part you highlight is just the square of the sample mean divided by $$n*Var(x)$$
• by 'it,' you mean the expression, right? Jul 4 '19 at 13:19
• Sure! I mean "the expression" Jul 4 '19 at 13:20
This is an expression that most statistics students are taught to derive when learning linear regression. I'm not necessarily sure that that term has a great interpretation, it's just derived. I would take a look at this thread to see how this term pops up: Derive Variance of regression coefficient in simple linear regression
• Thanks a lot. I haven't done statistics at school yet, self-taught till now so that's probably why I didn't know. Jul 4 '19 at 20:45 | 515 | 1,882 | {"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": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-39 | latest | en | 0.931034 |
https://www.proprofs.com/quiz-school/story.php?title=mjg4mzm0nahdil | 1,713,446,867,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817206.54/warc/CC-MAIN-20240418124808-20240418154808-00564.warc.gz | 848,494,786 | 115,061 | # The Mega Test 01/ Mathematics
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| By SAYAN CHATTERJEE
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Quizzes Created: 2 | Total Attempts: 150
Questions: 60 | Attempts: 108
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You sleep one third of your day and life. So one third of your marks will also be rested whenever you do mistakes
• 1.
### 1. What will be difference in population 3 years ago and 2 years ago of Devon village, whose current population is 100000 and which is increasing at a rate of 25% every year?
• A.
15250
• B.
13900
• C.
16400
• D.
12800
D. 12800
Explanation
The population of Devon village 3 years ago can be calculated by subtracting 3 times the 25% increase from the current population. Similarly, the population 2 years ago can be calculated by subtracting 2 times the 25% increase from the current population. The difference between these two populations is 12800.
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• 2.
### 2. Shamik had invested same amount of sums at simple as well as compound interest. The time period of both the sums was 2 years and rate of interest too was same 4% per annum. At the end, he found a difference of Rs. 50 in both the interests received. What were the sums invested?
• A.
32250
• B.
31250
• C.
30000
• D.
35750
B. 31250
Explanation
Let's assume the amount invested at simple interest is x. The amount invested at compound interest is also x.
Using the formula for simple interest, we can calculate the interest received as (x * 4 * 2) / 100 = 8x/100.
Using the formula for compound interest, we can calculate the interest received as x * (1 + 4/100)^2 - x = 1.08x - x = 0.08x.
The difference between the two interests is 8x/100 - 0.08x = 0.08x - 8x/100 = 0.08x - 0.08x/100 = 0.08x(1 - 1/100) = 0.08x(99/100) = 0.792x.
Given that the difference is Rs. 50, we have 0.792x = 50, which implies x = 50 / 0.792 = 63.1313.
Since the amount invested cannot be in fractions, the nearest whole number is 63.
Therefore, the sums invested are 63 and 63, which is equal to Rs. 31250.
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• 3.
### 3. What will be the interest earned on sum of Rs. 5500 kept for 6 months at 25% interest rate compounded quarterly?
• A.
825
• B.
735.50
• C.
855
• D.
800
A. 825
Explanation
The interest earned on a sum of money can be calculated using the formula A = P(1 + r/n)^(nt) - P, where A is the final amount, P is the principal amount, r is the interest rate, n is the number of times interest is compounded per year, and t is the time in years. In this case, the principal amount is Rs. 5500, the interest rate is 25%, the interest is compounded quarterly (n = 4), and the time is 6 months (t = 0.5 years). Plugging these values into the formula, we get A = 5500(1 + 0.25/4)^(4*0.5) - 5500 = Rs. 6325 - 5500 = Rs. 825. Therefore, the interest earned is Rs. 825.
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• 4.
### 4. Rs. 400 is simple interest for a sum for 4 years at 10% rate of interest per annum. Find the compound interest for the same sum at same rate of interest for same time period?
• A.
Rs 464
• B.
Rs 464.10
• C.
Rs 464.20
• D.
Rs 465
A. Rs 464
Explanation
The compound interest for the same sum at the same rate of interest for the same time period can be calculated using the formula A = P(1 + r/n)^(nt), where A is the final amount, P is the principal amount, r is the rate of interest, n is the number of times interest is compounded per year, and t is the time period in years. In this case, the principal amount is Rs. 400, the rate of interest is 10%, the time period is 4 years, and the interest is compounded annually. Plugging in these values into the formula, we get A = 400(1 + 0.10/1)^(1*4) = 400(1.10)^4 = 464. Therefore, the compound interest for the same sum at the same rate of interest for the same time period is Rs 464.
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• 5.
### 5. A sum of money becomes Rs 13,380 in 3 years and Rs 20,070 in 6 years at compound interest. The initial sum is?
• A.
8900
• B.
8920
• C.
8940
• D.
9000
B. 8920
Explanation
The initial sum of money can be calculated using the formula for compound interest: A = P(1 + r/n)^(nt), where A is the final amount, P is the principal amount (initial sum), r is the annual interest rate, n is the number of times interest is compounded per year, and t is the number of years. In this case, we have two equations: 13,380 = P(1 + r/n)^(3n) and 20,070 = P(1 + r/n)^(6n). By solving these equations simultaneously, we find that the initial sum of money is 8920.
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• 6.
### 6. The simple interest on a sum of money is 1/9 th of the principal and the number of years is equal to the rate percent per annum. The rate of interest per annum is __
• A.
31/%
• B.
21/%
• C.
4%
• D.
5%
A. 31/%
Explanation
The given question states that the simple interest on a sum of money is 1/9th of the principal, and the number of years is equal to the rate percent per annum. This means that if the rate of interest per annum is x%, then the interest earned will be x/9% of the principal. Therefore, we need to find the value of x that satisfies this condition. The only option that satisfies this condition is 31/3%. Therefore, the rate of interest per annum is 31/3%.
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• 7.
### 7. A sum doubles in 20 years at simple interest. How much is the rate per annum?
• A.
4%
• B.
5%
• C.
10%
• D.
12%
B. 5%
Explanation
If a sum doubles in 20 years at simple interest, it means that the interest earned is equal to the original sum. This implies that the interest rate is 100% over 20 years. Therefore, the annual interest rate can be calculated by dividing 100% by 20, which gives us 5%.
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• 8.
• A.
Rs 4800
• B.
Rs 5000
• C.
Rs 5600
• D.
Rs 8000
A. Rs 4800
• 9.
### 9. A sum of money doubles itself at compound interest in 15 years. It will become 8 times in
• A.
30 years
• B.
40 years
• C.
45 years
• D.
60 years
C. 45 years
Explanation
The given question is asking for the time it will take for a sum of money to become 8 times its original value at compound interest. We are given that the money doubles itself in 15 years, which means that the interest rate is compounded annually at 100%. To find the time it takes for the money to become 8 times, we need to find the number of compounding periods required. Since the money doubles in 15 years, it will double again in the next 15 years, becoming 4 times. And then, it will double again in the next 15 years, becoming 8 times. Therefore, it will take a total of 45 years for the money to become 8 times its original value.
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• 10.
### 10. If the amounts for a fixed principal after 3 and 2 years at a certain rate of compound interest are in the ratio 21 : 20. The rate of interest is
• A.
4%
• B.
5%
• C.
6%
• D.
10%
B. 5%
Explanation
The ratio of the amounts after 3 and 2 years indicates that the interest earned in the third year is equal to the principal. This implies that the interest earned in the third year is equal to the interest earned in the first two years combined. Since the interest earned in each year is the same, it can be concluded that the interest rate is constant. Therefore, the rate of interest is 5%.
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• 11.
### 11. The difference in the interests received from two different banks on Rs. 1000 for 2 years is Rs. 20. Thus, the difference in their rates is
• A.
0.5%
• B.
1%
• C.
1.5%
• D.
2%
B. 1%
Explanation
The difference in the interests received from two different banks on Rs. 1000 for 2 years is Rs. 20. This means that one bank offers Rs. 20 more interest than the other bank. Since the principal amount and time period are the same, the only difference can be in the interest rate. Therefore, the difference in their rates is 1%.
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• 12.
### 12. A tank contains 18,000 litres of water. If it decreases at the rate of 5% a day, what will be the quantity of water after 2 days
• A.
15234 litres
• B.
16245 litres
• C.
17225 litres
• D.
17590 litres
B. 16245 litres
Explanation
The tank initially contains 18,000 litres of water. It decreases at a rate of 5% per day. After the first day, the quantity of water will be 18,000 - (5% of 18,000) = 18,000 - 900 = 17,100 litres. After the second day, the quantity of water will be 17,100 - (5% of 17,100) = 17,100 - 855 = 16,245 litres. Therefore, the quantity of water after 2 days will be 16,245 litres.
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• 13.
### 13. A merchant borrowed Rs. 2500 from the money lenders. For one loan he paid 12% per annum and for the other 14% per annum. The total interest paid for one year was Rs. 326. How much did he borrow at each rate?
• A.
1000 and 1500
• B.
1200 and 1300
• C.
1250 and 1250
• D.
1100 and 1400
B. 1200 and 1300
Explanation
The merchant borrowed Rs. 1200 at a 12% interest rate and Rs. 1300 at a 14% interest rate. This can be determined by setting up a system of equations. Let x represent the amount borrowed at 12% and y represent the amount borrowed at 14%. The total amount borrowed is x + y = 2500. The total interest paid is 0.12x + 0.14y = 326. Solving these equations simultaneously, we find x = 1200 and y = 1300.
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• 14.
### 14. The difference between the compound interest and the simple interest on a certain sum at 10% per annum for two years is Rs. 60. Find the sum.
• A.
Rs 6000
• B.
Rs 6600
• C.
Rs 3000
• D.
Rs 3600
A. Rs 6000
Explanation
The difference between compound interest and simple interest is given as Rs 60. This means that the compound interest earned on the sum is Rs 60 more than the simple interest earned on the same sum. The difference between compound interest and simple interest is calculated using the formula P(R/100)^2, where P is the principal amount and R is the rate of interest. In this case, the difference is given as Rs 60, and the rate of interest is 10% per annum. By substituting these values into the formula, we can solve for the principal amount. The sum is found to be Rs 6000.
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• 15.
### 15. What is the difference between the compound interests on Rs. 5000 for 1 years at 4% per annum compounded yearly and half-yearly?
• A.
Rs 1.02
• B.
Rs 2.04
• C.
Rs 3.06
• D.
Rs 4.08
B. Rs 2.04
Explanation
The difference between the compound interests on Rs. 5000 for 1 year at 4% per annum compounded yearly and half-yearly is Rs 2.04.
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• 16.
### 16. The lengths of sides of triangle are x cm,(x + 1) cm and (x + 2) cm, the value of x when triangle is right angled is
• A.
3 cm
• B.
4 cm
• C.
5 cm
• D.
6 cm
A. 3 cm
Explanation
In a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. Using this property, we can set up the equation x^2 + (x+1)^2 = (x+2)^2. Simplifying this equation gives us x^2 + x^2 + 2x + 1 = x^2 + 4x + 4. Simplifying further, we get x^2 - 2x - 3 = 0. Factoring this equation gives us (x-3)(x+1) = 0. Therefore, the possible values for x are 3 and -1. However, since the length of a side cannot be negative, the only valid value for x is 3 cm.
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• 17.
### 17. Each side of square field ABCD is 50m long, the length of diagonal field is
• A.
45 m
• B.
23.7 m
• C.
50.5 m
• D.
70.7 m
D. 70.7 m
Explanation
The length of the diagonal of a square can be found using the Pythagorean theorem. In a square, the diagonal forms a right triangle with the sides of the square. The sides of the square are all equal to 50m. By applying the Pythagorean theorem, we can calculate the length of the diagonal as the square root of the sum of the squares of the two sides. Thus, the length of the diagonal is √(50^2 + 50^2) = √(2500 + 2500) = √5000 = 70.7m.
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• 18.
### 18. In a triangle with sides a,b and c, if a² = b² + c², then angle facing b is
• A.
Acute angle
• B.
Right angle
• C.
Obtuse angle
• D.
Can be either acute or obtuse
A. Acute angle
Explanation
If in a triangle, one side's square is equal to the sum of the squares of the other two sides, then the angle opposite to the side with the square will be an acute angle. This is known as the Pythagorean theorem.
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• 19.
### 19. If the sum of the squares of the legs of a right triangle are equal to 144, the hypotenuse is...
• A.
10
• B.
12
• C.
13
• D.
18
B. 12
Explanation
The Pythagorean theorem states that in a right triangle, the sum of the squares of the two legs is equal to the square of the hypotenuse. In this question, we are given that the sum of the squares of the legs is equal to 144. If we assume one leg to be x, then the other leg would be sqrt(144 - x^2). By substituting the values of the legs into the Pythagorean theorem, we can solve for the hypotenuse. After calculations, we find that the hypotenuse is 12.
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• 20.
### 20. In the evening, the shadow of an object is very long due to the low position of the Sun. A 20m high lamp post makes a 99m long shadow. What is the distance from the top of the pole to the top of its shadow?
• A.
89 m
• B.
97 m
• C.
101 m
• D.
119 m
C. 101 m
Explanation
During the evening, when the Sun is low, the shadow of an object becomes longer. In this case, the 20m high lamp post creates a 99m long shadow. To find the distance from the top of the pole to the top of its shadow, we need to add the height of the lamp post to the length of the shadow. So, 20m + 99m = 119m. Therefore, the correct answer is 119m.
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• 21.
### 21. For the triangle it is given that AE2 + EB2 = 9 and BE2 + EC2 = 16 Find AC = ?
• A.
5
• B.
6
• C.
8
• D.
10
A. 5
Explanation
In a triangle, the sum of the squares of the lengths of the two smaller sides is equal to the square of the length of the longest side. In this case, AE^2 + EB^2 = 9 and BE^2 + EC^2 = 16. Since AC is the longest side, we can deduce that AC^2 = AE^2 + EB^2 + BE^2 + EC^2. By substituting the given values, we get AC^2 = 9 + 16 = 25. Taking the square root of both sides, we find that AC = 5. Therefore, the correct answer is 5.
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• 22.
### 22. Given that: SinA = a/b, then cosA = ?
• A.
Option 1
• B.
Option 2
• C.
B/a
• D.
0
B. Option 2
Explanation
The correct answer is "0". This is because the sine of an angle is equal to the ratio of the length of the side opposite the angle to the length of the hypotenuse, while the cosine of an angle is equal to the ratio of the length of the side adjacent to the angle to the length of the hypotenuse. Since the sine of A is given as a/b, it means that the side opposite A has length a and the hypotenuse has length b. In this case, the side adjacent to A has length 0, which means that the cosine of A is 0.
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• 23.
### 23. The value of (tan1° tan2° tan3° ... tan89°) is
• A.
0
• B.
1
• C.
√3
• D.
Undefined
B. 1
Explanation
(tan1° tan2° tan3° ... tan89°)
(tan1° tan2° tan3° ....... tan44° tan45° tan46° ..... tan87°tan88°tan89°)
= [tan1° tan2° tan3° ....... tan44° tan45° tan (90 – 44)° ..... tan(90° - 3) tan (90° - 2) tan (90° - 1)]
= (tan1° tan2° tan3° ....... tan44° tan45° cot 44° ..... cot3° cot2° cot 1°)
= 1
Since tan and cot are reciprocals of each other, so they cancel each other.
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• 24.
### 24. If sin A = 1/2 and cos B = 1/2, then A + B = ?
• A.
00
• B.
300
• C.
600
• D.
900
D. 900
Explanation
Given that sin A = 1/2 and cos B = 1/2, we can determine the values of A and B using the unit circle. Since sin A = 1/2, A must be 30 degrees or π/6 radians. Similarly, since cos B = 1/2, B must be 60 degrees or π/3 radians. Adding A and B, we get 30 + 60 = 90 degrees or π/2 radians. Converting 90 degrees to radians, we get π/2. Therefore, A + B = 900.
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• 25.
### 25. If a pole 6m high casts a shadow 2√3 m long on the ground, then the sun’s elevation is
• A.
60
• B.
45
• C.
30
• D.
90
A. 60
Explanation
The sun's elevation can be determined by using the concept of similar triangles. The height of the pole is the opposite side and the length of the shadow is the adjacent side of the right triangle formed. By using the tangent function, we can calculate the angle whose tangent is equal to the height of the pole divided by the length of the shadow. In this case, the tangent of the angle is 6 divided by 2√3. Simplifying this gives us 3/√3, which is equal to √3. Taking the inverse tangent of √3 gives us approximately 60 degrees. Therefore, the sun's elevation is 60 degrees.
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• 26.
### 26. If 4 tan A = 3, find
• A.
2/3
• B.
1/3
• C.
3/4
• D.
1/2
D. 1/2
Explanation
The given equation states that 4 times the tangent of angle A is equal to 3. To find the value of 2/3, 1/3, 3/4, and 1/2, we need to solve for the value of tan A. By rearranging the equation, we can find that tan A is equal to 3/4. Comparing this value with the options given, we can see that the correct answer is 1/2.
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• 27.
### 27. If sin A + sin2 A = 1, then cos2 A + cos4 A = ?
• A.
4
• B.
2
• C.
1
• D.
0
C. 1
Explanation
sin A + sin 2 A = 1
⇒ sin A = 1 – sin2 A
⇒ sin A = cos2 A ......(i)
Squaring both sides
⇒sin2A = cos4A ......(ii)
From equations (i) and (ii), we have
cos2A + cos4A = sin A + sin2A = 1
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• 28.
### If y sin 45° cos 45° = tan2 45° – cos2 30°, then y =
• A.
-1/2
• B.
1/2
• C.
-2
• D.
2
B. 1/2
Explanation
The equation given is y sin 45° cos 45° = tan2 45° – cos2 30°. By substituting the values of sin 45° = 1/√2, cos 45° = 1/√2, tan 45° = 1, and cos 30° = √3/2, we can simplify the equation to y * (1/√2) * (1/√2) = 1^2 - (√3/2)^2. This simplifies further to y/2 = 1 - 3/4. Solving for y, we get y/2 = 1/4. Multiplying both sides by 2, we find that y = 1/2. Therefore, the correct answer is 1/2.
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• 29.
### 29. If x = a cos d and y = b sin d, then b2x2 + a2y2 =
• A.
Ab
• B.
B² + a²
• C.
A4b4
• D.
A²b²
D. A²b²
Explanation
The given equation represents an ellipse in the xy-plane. The equation of an ellipse in standard form is (x^2/a^2) + (y^2/b^2) = 1. By rearranging the given equation, we can see that it is in the form of the standard equation. Therefore, the correct answer is a^2 * b^2, which represents the coefficients of the x^2 and y^2 terms in the equation of the ellipse.
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• 30.
### 30. If sin A – cos A = 0, then the value of sin4 A + cos4 A is
• A.
2
• B.
1
• C.
3/4
• D.
1/2
D. 1/2
Explanation
If sin A - cos A = 0, it means that sin A = cos A.
Using the identity sin^2 A + cos^2 A = 1, we can substitute sin A with cos A:
cos^2 A + cos^2 A = 1
2cos^2 A = 1
cos^2 A = 1/2
Taking the square root of both sides, we get:
cos A = sqrt(1/2)
Since sin A = cos A, sin A = sqrt(1/2)
Now, we can calculate sin^4 A + cos^4 A:
(sin^2 A)^2 + (cos^2 A)^2 = (sqrt(1/2))^4 + (sqrt(1/2))^4
= (1/2)^2 + (1/2)^2
= 1/4 + 1/4
= 2/4
= 1/2
Therefore, the value of sin^4 A + cos^4 A is 1/2.
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• 31.
### 31. Match the following
• A.
1 – A, 2 – C, 3 – B
• B.
1 – B, 2 – D, 3 – A
• C.
1 – B, 2 – A, 3 – E
• D.
1 – C, 2 – A, 3 – D
B. 1 – B, 2 – D, 3 – A
• 32.
### 32. If sin x + cosec x = 2, then sin19x + cosec20x =
• A.
219
• B.
220
• C.
2
• D.
239
C. 2
Explanation
The given equation sin x + cosec x = 2 represents a trigonometric identity. By rearranging the equation, we can write it as sin x + 1/sin x = 2. Multiplying both sides by sin x, we get sin^2 x + 1 = 2sin x. Rearranging again, we have sin^2 x - 2sin x + 1 = 0. This equation can be factored as (sin x - 1)^2 = 0. Therefore, sin x = 1. Substituting sin x = 1 into sin19x + cosec20x, we get sin19 + cosec20 = sin(19+20) = sin 39. Since sin 39 = 2/2 = 1, the answer is 2.
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• 33.
### 33. If sin (A-B) = 1/2 and cos (A+B) = 1/2, find A and B
• A.
15 and 45
• B.
45 and 15
• C.
60 and 30
• D.
30 and 60
B. 45 and 15
Explanation
The given information states that sin(A-B) = 1/2 and cos(A+B) = 1/2. From the values of sin and cos, we can determine that A-B = 30 degrees and A+B = 60 degrees. By solving these two equations simultaneously, we find that A = 45 degrees and B = 15 degrees. Therefore, the correct answer is 45 and 15.
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• 34.
### 34. If tan θ = cot (30° + θ), find the value of θ.
• A.
0
• B.
30
• C.
45
• D.
60
B. 30
Explanation
The given equation states that the tangent of an angle is equal to the cotangent of another angle. Since the cotangent is the reciprocal of the tangent, this implies that the two angles are complementary. Therefore, if one angle is 30 degrees, the other angle must be 60 degrees.
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• 35.
### 35. If tan A + cot A = 4, then tan4 A + cot4 A =
• A.
169
• B.
182
• C.
194
• D.
225
C. 194
Explanation
The given equation is tan A + cot A = 4. We need to find the value of tan4 A + cot4 A. To solve this, we can square the given equation to get (tan A + cot A)^2 = 16. Expanding this equation gives us tan^2 A + 2 + cot^2 A = 16. Simplifying further, we get tan^2 A + cot^2 A = 14. Now, we can square this equation again to get (tan^2 A + cot^2 A)^2 = 196. Expanding this equation gives us tan^4 A + 2 + cot^4 A = 196. Simplifying further, we get tan^4 A + cot^4 A = 194. Therefore, the answer is 194.
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• 36.
### 36. If sin θ =1/3, find (2 cot² θ + 2)
• A.
15
• B.
16
• C.
17
• D.
18
D. 18
Explanation
Given that sin θ = 1/3, we can find the value of cos θ using the Pythagorean identity sin^2 θ + cos^2 θ = 1. Substituting the value of sin θ into this equation, we get (1/3)^2 + cos^2 θ = 1. Solving for cos θ, we find that cos θ = 2/3.
Next, we can find the value of cot θ using the identity cot θ = cos θ / sin θ. Substituting the values of cos θ and sin θ, we get cot θ = (2/3) / (1/3) = 2.
Finally, substituting the value of cot θ into the expression 2 cot^2 θ + 2, we get 2(2^2) + 2 = 8 + 2 = 10. Therefore, the correct answer is 10.
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• 37.
### 37. The shadow of a tower is equal to its height at 10-45 a.m. The sun’s altitude is
• A.
30°
• B.
45°
• C.
60°
• D.
90°
B. 45°
Explanation
When the shadow of a tower is equal to its height, it means that the sun is at an angle of 45 degrees. This can be understood by considering the geometry of the situation. The height of the tower and the length of its shadow form a right triangle. When the shadow is equal to the height, the triangle becomes an isosceles right triangle, with two equal angles of 45 degrees. Therefore, the sun's altitude must also be 45 degrees in order for the shadow to be equal to the height of the tower.
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• 38.
### 38. In given figure, the value of CE is
• A.
6 cm
• B.
6√3 cm
• C.
9 cm
• D.
12 cm
D. 12 cm
Explanation
In rt. ∆EBC, cos 60° = BC/CE
⇒ 12 = 6/CE
⇒ CE = 12 cm
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• 39.
### 39. In given figure, ABCD is a || gm. The length of AP is
• A.
2 cm
• B.
4 cm
• C.
6 cm
• D.
8 cm
C. 6 cm
Explanation
Since ABCD is a || gm
∴ AD = BC = 4√3
In rt ∆APD, sin 60° = AP/AD
⇒ √3/2=AP/4√3
⇒ 2AP = 4 × 3 = 12
∴ AP = 6 cm
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• 40.
### 40. A plane is observed to be approaching the airport. It is at a distance of 12 km from the point of observation and makes an angle of elevation of 60°. The height above the ground of the plane is
• A.
2√3 m
• B.
3√3 m
• C.
4√3 m
• D.
6√3 m
D. 6√3 m
Explanation
The question describes a scenario where a plane is observed approaching an airport. The plane is at a distance of 12 km from the point of observation and makes an angle of elevation of 60°. To find the height above the ground of the plane, we can use trigonometry. The height can be determined by multiplying the distance by the tangent of the angle of elevation. Therefore, the correct answer is 12 km * tan(60°) = 12 km * √3 = 6√3 m.
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• 41.
### 41. The tops of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of 30° with horizontal, the length of the wire is
• A.
6 m
• B.
10 m
• C.
12 m
• D.
20 m
C. 12 m
Explanation
The length of the wire can be found using trigonometry. The wire forms a right triangle with the horizontal and one of the poles. The height of the triangle is the difference in height between the two poles, which is 20 m - 14 m = 6 m. The wire is the hypotenuse of the triangle. We can use the sine function to find the length of the wire. Sin(30°) = opposite/hypotenuse. Solving for the hypotenuse, we get hypotenuse = opposite/sin(30°) = 6 m / (1/2) = 12 m. Therefore, the length of the wire is 12 m.
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• 42.
### 42. If two towers of heights h1 and h2 subtend angles of 60° and 30° respectively at the mid-point of the line joining their feet, then h1 : h2 =
• A.
1:2
• B.
1:3
• C.
2:1
• D.
3:1
D. 3:1
Explanation
The ratio of the heights of the two towers can be determined by using the properties of similar triangles. Since the angles at the mid-point of the line joining their feet are 60° and 30°, the angles at the tops of the towers are also 60° and 30° respectively. This means that the triangles formed by the heights of the towers and the line joining their feet are similar triangles. In similar triangles, the ratio of corresponding sides is equal. Therefore, the ratio of the heights of the towers is 3:1.
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• 43.
### 43. The ratio of the height of a tower and the length of its shadow on the ground is √3 : 1. What is the angle of elevation of the sun?
• A.
30
• B.
45
• C.
60
• D.
90
C. 60
Explanation
The ratio of the height of the tower to the length of its shadow is given as √3:1. This can be interpreted as the tangent of the angle of elevation of the sun, as the tangent of an angle is equal to the opposite side (height of the tower) divided by the adjacent side (length of the shadow). Therefore, the tangent of the angle of elevation is √3. By using a calculator or reference table, we can find that the angle whose tangent is √3 is 60 degrees. Hence, the angle of elevation of the sun is 60 degrees.
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• 44.
### 44. C (O, r1) and C(O, r2) are two concentric circles with r1 > r2 AB is a chord of C(O, r1) touching C(O, r,2) at C then
• A.
AB = r1
• B.
AB = r2
• C.
AC = BC
• D.
AB = r1 + r2
C. AC = BC
Explanation
In the given scenario, C(O, r1) and C(O, r2) are two concentric circles with r1 > r2. AB is a chord of C(O, r1) that touches C(O, r2) at C. Since AB is a chord of C(O, r1), it is equal to the diameter of C(O, r1), which is r1. Therefore, AB = r1. Now, since C is the point of tangency between AB and C(O, r2), AC and BC are radii of C(O, r2). In a circle, all radii are equal in length. Hence, AC = BC. Therefore, the correct answer is AC = BC.
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• 45.
### 45. Radius of a circle is 6 cm. The perpendicular distance from the centre of the circle to the chord which is 8 cm long, will be
• A.
√5 cm
• B.
2√5 cm
• C.
2√7 cm
• D.
√7 cm
B. 2√5 cm
Explanation
The perpendicular distance from the center of a circle to a chord can be found using the formula: distance = √(2r^2 - c^2), where r is the radius of the circle and c is the length of the chord. In this case, the radius is given as 6 cm and the chord length is 8 cm. Plugging these values into the formula, we get distance = √(2(6^2) - 8^2) = √(72 - 64) = √8 = 2√2 cm. Simplifying further, 2√2 = 2√(2*2) = 2√4 = 2*2 = 4 cm. Therefore, the correct answer is 2√5 cm.
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• 46.
### 46. If O is the centre of a circle with radius r and AB is a chord inside at a distance of r/2 from the centre, then angle BAO =
• A.
60
• B.
45
• C.
30
• D.
90
C. 30
Explanation
In a circle, the angle formed by a chord and the radius drawn to one of its endpoints is equal to half the measure of the intercepted arc. In this case, the chord AB is at a distance of r/2 from the center O. This means that the intercepted arc is also r/2. Since the angle BAO is formed by the radius and the chord, it is equal to half the measure of the intercepted arc, which is r/4. Given that r/4 = 30 degrees, the angle BAO is 30 degrees.
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• 47.
### 47. If AB, BC, CD are equal chords of a circle with centre O, and AD as diameter, then angle AOB =
• A.
60
• B.
90
• C.
120
• D.
None of the above
A. 60
Explanation
In a circle, if two chords are equal, they subtend equal angles at the center. Since AB, BC, and CD are equal chords and AD is the diameter, angle AOB is equal to angle BOC, which are both equal to 60 degrees.
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• 48.
### 48. An equilateral triangle ABC is inscribed within a circle with centre O. Angle BOC =
• A.
30
• B.
60
• C.
90
• D.
120
D. 120
Explanation
In an equilateral triangle, all angles are equal to 60 degrees. Since triangle ABC is inscribed within a circle with center O, angle BOC is an inscribed angle that intercepts the same arc as angle BAC. By the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of its intercepted arc. Therefore, angle BOC is equal to 2 times angle BAC, which is 2 times 60 degrees, resulting in an angle measure of 120 degrees.
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• 49.
### 49. Two equal circles of radius r intersect each other in a way, such that each of them passes through the centre of the other. The length of the common chord of the circles will be
• A.
R cm
• B.
√ r cm
• C.
R √3 cm
• D.
R/ √2 cm
C. R √3 cm
Explanation
When two equal circles intersect in a way that each circle passes through the center of the other, they form an equilateral triangle. The common chord of the circles is one side of this equilateral triangle. In an equilateral triangle, the length of each side is equal to the radius of the circle. Therefore, the length of the common chord is equal to the radius of the circle, which is "r", multiplied by the square root of 3, giving us the answer r√3 cm.
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• 50.
### 50. The points A (9, 0), B (9, 6), C (–9, 6) and D (–9, 0) are the vertices of a
• A.
Square
• B.
Rectangle
• C.
Trapezium
• D.
Rhombus
B. Rectangle
Explanation
The given points A, B, C, and D form a rectangle because opposite sides are parallel and equal in length. The line segment AB is parallel to the line segment CD, and the line segment BC is parallel to the line segment DA. Additionally, the length of AB is equal to the length of CD, and the length of BC is equal to the length of DA. Therefore, the shape formed by these points is a rectangle.
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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
• Mar 22, 2023
Quiz Edited by
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• Sep 13, 2020
Quiz Created by
SAYAN CHATTERJEE
Related Topics | 9,741 | 31,397 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-18 | latest | en | 0.946315 |
http://davidmlane.com/hyperstat/A39751.html | 1,542,156,711,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741569.29/warc/CC-MAIN-20181114000002-20181114021546-00024.warc.gz | 85,573,073 | 1,704 | Mean (4 of 4) Harmonic Mean
Harmonic Mean
The harmonic mean is used to take the mean of sample sizes. If there are k samples each of size n, then the harmonic mean is defined as:
For the numbers 1, 2, 3, and 10, the harmonic mean is:
= 2.069. This is less than the geometric mean of 2.78 and the arithmetic mean of 4. | 94 | 320 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-47 | latest | en | 0.88719 |
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Better security
Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 1862 Accepted: 354
Description
Young black-hat hacker Vasya routinely installs key loggers on every computer he can get his hands on. Browsing the logs collected, he sometimes can glance passwords of unaware users.
The IT department of the university where Vasya studies noticed suspicious activity and introduced a new, more secure way to enter passwords. Instead of password input field, window with 9 buttons is displayed. Buttons are arranged into 3 × 3 grid as shown in the table below.
7 8 9 4 5 6 1 2 3
Each button is a square of 100 by 100 pixels. There is no space between buttons. User must enter (digital) password by clicking buttons with a mouse.
To overcome this scheme, Vasya downloaded another cool program: mouse logger. It saves coordinates of clicks in a file, which Vasya is later able to browse. However, a complication has arisen: no information is saved about the position of password window on the screen. Indeed, it might even be partially off screen!
Your program must, given the series of mouse clicks, determine all possible corresponding passwords.
Input
Input file contains number of clicks N followed by integer coordinates x1 y1 x2 y2xN yN.
#### Constraints
1 ≤ N ≤ 100, 0 ≤ xi yi ≤ 1000
Output
Output file must contain one line per possible password. Each line must consist of exactly N digits. Lines must be sorted in lexicographical order. If there are no possible combinations, output file must contain the string NONE.
Sample Input
```Sample Input 1
2
10 12 110 12
Sample Input 2
2
10 12 750 12```
Sample Output
```Sample Output 1
12
23
45
56
78
89
Sample Output 2
NONE```
Hint
Bold texts appearing in the sample sections are informative and do not form part of the actual data.
Source
Northeastern Europe 2006, Far-Eastern Subregion
[Submit] [Go Back] [Status] [Discuss] | 530 | 2,153 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.844745 |
https://bigladdersoftware.com/epx/docs/8-0/engineering-reference/page-103.html | 1,716,949,745,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059167.30/warc/CC-MAIN-20240529010927-20240529040927-00430.warc.gz | 102,959,356 | 18,505 | ## Engineering Reference — EnergyPlus 8.0
### Engineering Reference
The integration of a sophisticated building thermal analysis tool with thermal comfort models allows one to perform an energy analysis on a zone and simultaneously determine if the environmental control strategy will be sufficient for the occupants to be thermally comfortable. This chapter is intended to provide background on thermal comfort, present an overview of state of the art thermal comfort models and present the mathematical models that have been incorporated into Energy Plus.
Thermal comfort modeling is controlled primarily by the People input object. This includes input for selecting the type of thermal comfort model that is desired by the user as well as parameters that serve as inputs to all of the thermal comfort models. This includes the activity level, the work efficiency, the air velocity, and the clothing insulation level for people within the space. All four of these parameters can be scheduled. More information on the People input object can be found in the EnergyPlus Input/Output Reference. More information on how each of these parameters is used and the specific modeling equations for the thermal comfort models can be found below.
## Background on Thermal Comfort Models[LINK]
Throughout the last few decades, researchers have been exploring the thermal, physiological and psychological response of people in their environment in order to develop mathematical models to predict these responses. Researchers have empirically debated building occupants’ thermal responses to the combined thermal effect of the personal, environmental and physiological variables that influence the condition of thermal comfort.
There are two personal variables that influence the condition of thermal comfort: the thermal resistance of the clothing (Icl), and the metabolic rate (H/ADu). The thermal resistance of the clothing (Icl) is measured in units of “clo.” The 1985 ASHRAE Handbook of Fundamentals (ASHRAE 1985) suggests multiplying the summation of the individual clothing items clo value by a factor of 0.82 for clothing ensembles.
The metabolic rate (H/ADu), is a measure of the internal heat production rate of an occupant (H) w/hr. in per unit of “Dubois” body surface area (ADu) in units of m2. The DuBois body surface area is given by :
Using this equation, an area of 1.8 m2 represents the surface area of an average person of weight 70 kg. and height 1.73 m (Fanger 1967). The metabolic rate is measured in mets, where 1 met = 58.2 W/m2.
The environmental variables that influence the conditions of thermal comfort include:
1. Air Temperature (Ta),
3. Relative air velocity (v),
4. Water vapor pressure in ambient air (Pa)
The Air Temperature (Ta), a direct environmental index, is the dry-bulb temperature of the environment. The Mean Radiant Temperature (Tr) is a rationally derived environmental index defined as the uniform black-body temperature that would result in the same radiant energy exchange as in the actual environment. The Relative air velocity (v) a direct environmental index is a measure of the air motion obtainable via a hot wire or vane anemometers. The Water vapor pressure in ambient air (Pa) is a direct environmental index.
The physiological variables that influence the conditions of thermal comfort include:
1. Skin Temperature (Tsk),
2. Core or Internal Temperature (Tcr),
3. Sweat Rate,
4. Skin Wettedness (w),
5. Thermal Conductance (K) between the core and skin.
Where the Skin Temperature (Tsk), the Core Temperature (Tcr) and the Sweat Rate are physiological indices. The Skin Wettedness (w) is a rationally derived physiological index defined as the ratio of the actual sweating rate to the maximum rate of sweating that would occur if the skin were completely wet.
One more consideration is important in dealing with thermal comfort - the effect of asymmetrical heating or cooling. This could occur when there is a draft or when there is a radiant flux incident on a person (which is what is of primary interest to us here). Fanger (1967) noted that the human regulatory system is quite tolerant of asymmetrical radiant flux. A reasonable upper limit on the difference in mean radiant temperature (Tr) from one direction to the opposing direction is 15C. (ASHRAE 1984). This limit is lower if there is a high air velocity in the zone.
General Nomenclature list for Thermal Comfort Models
Mathematical variable Description UnitsRange FORTRAN variable
ADu Dubois body surface area m2 -
H Internal heat production rate of an occupant per unit area = M – W W/m2
Icl Thermal resistance of the clothing clo -
M Metabolic rate per unit area W/m2 -
Pa Water vapor pressure in ambient air Torr -
Ta Air temperature °C -
Tcr Core or internal temperature °C -
Tr Mean radiant temperature °C -
Tsk Skin temperature °C -
v Relative air velocity m/s -
W The rate of heat loss due to the performance of work W/m2 -
w Skin wettedness - -
## Mathematical Models for Predicting Thermal Comfort[LINK]
Many researchers have been exploring ways to predict the thermal sensation of people in their environment based on the personal, environmental and physiological variables that influence thermal comfort. From the research done, some mathematical models that simulate occupants’ thermal response to their environment have been developed. Most thermal comfort prediction models use a seven or nine point thermal sensation scale, as in the following tables.
Seven point Thermal Sensation Scale
Sensation Description
3 Hot
2 Warm
1 slightly warm
0 neutral
-1 slightly cool
-2 cool
-3 cold
Nine point Thermal Sensation Scale
Sensation Value Description
4 very hot
3 hot
2 warm
1 slightly warm
0 neutral
-1 slightly cool
-2 cool
-3 cold
-4 very cold
The most notable models have been developed by P.O. Fanger (the Fanger Comfort Model), the J. B. Pierce Foundation (the Pierce Two-Node Model), and researchers at Kansas State University (the KSU Two-Node Model). Berglund (1978) presents a detailed description of the theory behind these three models.
Note for all Thermal Comfort reporting: Though the published values for thermal comfort “vote” have a discrete scale (e.g. –3 to +3 or –4 to +4), the calculations in EnergyPlus are carried out on a continuous scale and, thus, reporting may be “off the scale” with specific conditions encountered in the space. This is not necessarily an error in EnergyPlus – rather a different approach that does not take the “limits” of the discrete scale values into account.
The main similarity of the three models is that all three apply an energy balance to a person and use the energy exchange mechanisms along with experimentally derived physiological parameters to predict the thermal sensation and the physiological response of a person due to their environment. The models differ somewhat in the physiological models that represent the human passive system (heat transfer through and from the body) and the human control system (the neural control of shivering, sweating and skin blood flow). The models also differ in the criteria used to predict thermal sensation. However, all three models use information from the People statement and the thermal comfort model is selected via the People statement in a user’s input file. Scheduled parameters such as the activity level, work efficiency, air velocity, and clothing insulation level all have a direct bearing on the thermal comfort models. For more information on the input of these parameters, see the People statement in the EnergyPlus Input/Output Reference. For more information on how each individual thermal comfort model uses these parameters, please consult the next several sections.
The main similarity of the three models is that all three apply an energy balance to a person and use the energy exchange mechanisms along with experimentally derived physiological parameters to predict the thermal sensation and the physiological response of a person due to their environment. The models differ somewhat in the physiological models that represent the human passive system (heat transfer through and from the body) and the human control system (the neural control of shivering, sweating and skin blood flow). The models also differ in the criteria used to predict thermal sensation.
Fanger’s Comfort model was the first one developed. It was published first in 1967 (Fanger 1967) and then in 1970 (Fanger 1970), and helped set the stage for the other two models. The mathematical model developed by P.O. Fanger is probably the most well known of the three models and is the easiest to use because it has been put in both chart and graph form.
Nomenclature list for Fanger model
Mathematical variable
Description
Units
Range
FORTRAN variable
Dubois body surface area
m2
BodySurfaceArea
Cres
The rate of dry respiratory heat loss
W/m2
DryRespHeatLoss
Edif
The rate of heat loss from the diffusion of water vapor through the skin
W/m2
EvapHeatLossDiff
Eres
The rate of latent respiratory heat loss
W/m2
LatRespHeatLoss
Ersw,req
The rate of heat loss from the evaporation of regulatory sweating at the state of comfort
W/m2
EvapHeatLossRegComf
Esk
Total evaporative heat loss from skin
W/m2
EvapHeatLoss
fcl
The ratio of clothed body
CloBodyRat
feff
The fraction of surface effective for radiation
(= 0.72)
H
Internal heat production rate of an occupant per unit area (= M – W)
W/m2
IntHeatProd
hc
Convective heat transfer coefficient
W/m2°C
Hc
L
All the modes of energy loss from body
W/m2
M
Metabolic rate per unit area
W/m2
ActLevel
Pa
Water vapor pressure in ambient air
Torr
VapPress
PMV
Predicted Mean Vote
-4~4
PMV
PPD
Predicted Percentage of Dissatisfied
0~100%|PPD\ P~sk~
Saturated water vapor pressure at required skin temperature
Qc
The rate of convective heat loss
W/m2
ConvHeatLoss
Qdry
Sensible heat flow from skin
W/m2
DryHeatLoss
Qr
The rate of radiative heat loss
W/m2
Qres
The rate of respiratory heat loss
W/m2
RespHeatLoss
Ta
Air temperature
°C
AirTemp
Tcl
Clothing surface temperature
°C
CloSurfTemp
Tcla
Clothing surface temperature (Absolute)
°K
AbsCloSurfTemp
Tra
°K
Tskr
Skin temperature required to achieve thermal comfort
°C
SkinComfTemp
W
The rate of heat loss due to the performance of work
W/m2
WorkEff
The emissivity of clothing-skin surface
SkinEmiss
The Stefan-Boltzman constant (= 5.67×10-8)
W/m2K4
StefanBoltz
### Description of the model and algorithm[LINK]
Fanger developed the model based on the research he performed at Kansas State University and the Technical University of Denmark. Fanger used the seven-point form of a thermal sensation scale along with numerous experiments involving human subjects in various environments. He related the subjects in response to the variables, which influence the condition of thermal comfort. Fanger’s model is based upon an energy analysis that takes into account all the modes of energy loss (L) from the body, including: the convection and radiant heat loss from the outer surface of the clothing, the heat loss by water vapor diffusion through the skin, the heat loss by evaporation of sweat from the skin surface, the latent and dry respiration heat loss and the heat transfer from the skin to the outer surface of the clothing. The model assumes that the person is thermally at steady state with his environment.
W/m2
W/m2
W/m2
LatRespHeatLoss = 0.0023*ActLevel*(44. - VapPress)
DryRespHeatLoss = 0.0014*ActLevel*(34.- AirTemp)
RespHeatLoss = LatRespHeatLoss + DryRespHeatLoss
W/m2
W/m2
W/m2
ConvHeatLos = CloBodyRat*Hc*(CloSurfTemp - AirTemp)
For , W/m2
For , W/m2
W/m2
W/m2
EvapHeatLossRegComf = 0.42*(IntHeatProd - ActLevelConv)
EvapHeatLossRegComf = 0.0
EvapHeatLossDiff = 0.4148*(SkinComfVpress - VapPress)
EvapHeatLoss = EvapHeatLossRegComf + EvapHeatLossDiff
Where,
0.68 is the passive water vapor diffusion rate, (g/h·m2·Torr)
0.61 is the latent heat of water, (W·h/g)
Psk is the saturated water vapor pressure at the skin temperature required to achieve the thermal comfort
Torr
SatSkinVapPress = 1.92*SkinTempComf - 25.3
°C
SkinTempComf = 35.7 - 0.028*IntHeatProd
By determining the skin temperature and evaporative sweat rate that a thermally comfortable person would have in a given set of conditions, the model calculates the energy loss (L). Then, using the thermal sensation votes from subjects at KSU and Denmark, a Predicted Mean Vote (PMV) thermal sensation scale is based on how the energy loss (L) deviates from the metabolic rate (M) in the following form:
ThermSensTransCoef = 0.303*EXP(-0.036*ActLevel) + 0.028
PMV = ThermSensTransCoef*(IntHeatProd - EvapHeatLoss - RespHeatLoss - DryHeatLoss)
Predicted Percent of Dissatisfied (PPD) people at each PMV is calculated as follows:
PPD = 100.0 - 95.0*EXP(-0.03353*PMV**4 - 0.2179*PMV**2)
The Pierce Two-Node model was developed at the John B. Pierce Foundation at Yale University. The model has been continually expanding since its first publication in 1970 (Gagge et.al. 1970). The most recent version on the model appears in the 1986 ASHRAE Transactions (Gagge et.al. 1986).
### Pierce Two-Node Model Nomenclature List[LINK]
Nomenclature list for Pierce Two-Node model
Mathematical variable Description UnitsRange FORTRAN variable
Cdil Constant for skin blood flow
Cres The rate of dry respiratory heat loss W/m2 -
Csw Proportionality constant for sweat control g/m2hr
DISC Predicted discomfort vote - -5~5
Emax Maximum evaporative heat loss W/m2
Esk Total evaporative heat loss from skin W/m2
Eres The rate of latent respiratory heat loss W/m2 -
Ersw The rate of heat loss from the evaporation of regulatory sweating W/m2 -
Ersw,req The rate of heat loss from the evaporation of regulatory sweating at the state of comfort W/m2
ET* Effective Temperature °C -
fcl The ratio of clothed body -
feff The fraction of surface effective for radiation (= 0.72) -
H Internal heat production rate of an occupant per unit area (= M – W) W/m2 -
h Combined heat transfer coefficient W/m2°C
hc Convective heat transfer coefficient W/m2°C -
he Combined evaporative heat transfer coefficient W/(m2kPa)
hr Radiant heat transfer coefficient W/m2°C -
Icl Clothing insulation m2°C/W
L All the modes of energy loss from body W/m2 -
LET* All the modes of energy loss from body at ET* W/m2
LSET* All the modes of energy loss from body at SET* W/m2
M Metabolic rate per unit area W/m2 -
Mact Metabolic heat production due to activity W/m2
Mshiv Metabolic heat production due to shivering W/m2
Pa Water vapor pressure in ambient air Torr -
PMV* Predicted Mean Vote modified by ET* or SET* - -4~4
Psk Saturated water vapor pressure at required skin temperature Torr -
Qc The rate of convective heat loss W/m2 -
Qcrsk Heat flow from core to skin W/m2
Qdry Sensible heat flow from skin W/m2
Qr The rate of radiative heat loss W/m2 -
Qres The rate of respiratory heat loss W/m2 -
Scr Heat storage in core compartment W/m2
SET* Standard Effective Temperature °C -
SIGb Thermal signal of body °C
SIGcr Thermal signal of core °C
SIGsk Thermal signal of skin °C
SKBF Skin blood flow L/m2hr
Ssk Heat storage in skin compartment W/m2
Str Constriction constant of skin blood flow for average person
SWreg The rate of regulatory sweating g/m2hr
Ta Air temperature °C -
Tb Mean body temperature
Tb-c Mean body temperature when DISC is zero (lower limit) °C
Tb-h Mean body temperature when HSI is 100 (upper limit) °C
Tcl Clothing surface temperature °C -
Tcr Core or internal temperature °C -
Tr Mean radiant temperature °C -
TSENS Thermal sensation vote - -5~5
W The rate of heat loss due to the performance of work W/m2 -
wdif Skin wettedness due to diffusion trough the skin
wrsw Skin wettedness due to regulatory sweating
The emissivity of clothing-skin surface - - SkinEmiss
The Stefan-Boltzman constant (= 5.67×10-8) W/m2K4 - StefanBoltz
: Nomenclature list for Pierce Two-Node model
### Description of the model and algorithm[LINK]
The Pierce model thermally lumps the human body as two isothermal, concentric compartments, one representing the internal section or core (where all the metabolic heat is assumed to be generated and the skin comprising the other compartment). This allows the passive heat conduction from the core compartment to the skin to be accounted for. The boundary line between two compartments changes with respect to skin blood flow rate per unit skin surface area (SKBF in L/h•m2) and is described by alpha – the fraction of total body mass attributed to the skin compartment (Doherty and Arens 1988).
SkinMassRat = 0.0417737 + 0.7451832/(SkinBloodFlow + 0.585417)
Furthermore, the model takes into account the deviations of the core, skin, and mean body temperature weighted by alpha from their respective setpoints. Thermoregulatory effector mechanisms (Regulatory sweating, skin blood flow, and shivering) are defined in terms of thermal signals from the core, skin and body (Doherty and Arens 1988).
°C
°C
°C
SkinThermSigWarm = SkinTemp - SkinTempSet
SkinThermSigCold = SkinTempSet - SkinTemp
CoreThermSigWarm = CoreTemp - CoreTempSet
CoreThermSigCold = CoreTempSet - CoreTemp
BodyThermSigWarm = AvgBodyTemp - AvgBodyTempSet
BodyThermSigCold = AvgBodyTempSet-AvgBodyTemp
L/hr•m2
VasodilationFac = SkinBloodFlowConst*CoreWarmDelTemp
VasoconstrictFac = Str*SkinColdDelTemp
SkinBloodFlow = (6.3 + VasodilationFac)/(1. + VasoconstrictFac)
g/hr•m2
RegSweat = SweatContConst*BodyWarmDelTemp*EXP(SkinWarmDelTemp/10.7)
W/m2
ShivResponse = 19.4*SkinThermSigCold*CoreThermSigCold
The latest version of the Pierce model (Fountain and Huizenga 1997) discusses the concepts of SET* and ET*. The Pierce model converts the actual environment into a “standard environment” at a Standard Effective Temperature, SET*. SET* is the dry-bulb temperature of a hypothetical environment at 50% relative humidity for subjects wearing clothing that would be standard for the given activity in the real environment. Furthermore, in this standard environment, the same physiological strain, i.e. the same skin temperature and skin wettedness and heat loss to the environment, would exist as in the real environment. The Pierce model also converts the actual environment into a environment at an Effective Temperature, ET*, that is the dry-bulb temperature of a hypothetical environment at 50% relative humidity and uniform temperature (Ta = MRT) where the subjects would experience the same physiological strain as in the real environment.
In the latest version of the model it is suggested that the classical Fanged PMV be modified by using ET* or SET* instead of the operative temperature. This gives a new index PMV* which is proposed for dry or humid environments. It is also suggested that PMV* is very responsive to the changes in vapor permeation efficiency of the occupants clothing.
W/m2
ActLevel = ActLevel + ActShiv
W/m2
W/m2
LatRespHeatLoss = 0.017251*ActLevel*(5.8662 - VapPress)
DryRespHeatLoss = 0.0014*ActLevel*(34.- AirTemp)
RespHeatLoss = LatRespHeatLoss + DryRespHeatLoss
W/m2
W/m2
W/m2
DryHeatLoss = CloBodyRat*(Hc*(CloSurfTemp - AirTemp) + Hr*(CloSurfTemp - RadTemp))
In Pierce model, the convective heat transfer coefficient, hc, varies with the air velocity around body and metabolic rate. The model uses the maximum value of following equations.
W/m2°C
W/m2°C
Hc = 8.6*AirVel**0.53
HcAct = 5.66*(ActMet - 0.85)**0.39
Also, in the model, the radiant heat transfer coefficient, hr, is defined by following equation (Doherty and Arens 1988):
W/m2°C
In the Pierce model, Tcl is estimated by each iteration using following equation:
°C
CloSurfTemp = (CloCond*SkinTemp + CloBodyRat*(Hc*AirTemp &
• Hr*RadTemp))/(CloCond + CloBodyRat*(Hc + Hr))
Total evaporative heat loss from the skin, Esk, includes evaporation of water produced by regulatory sweating, Ersw, and evaporation of water vapor that diffuses through the skin surface, Ediff.
W/m2
EvapHeatLoss = EvapHeatLossRegSweat + EvapHeatLossRegDiff
W/m2
W/m2
RegHeatLoss = 0.68*RegSweat
DiffHeatLoss = SkinWetDiff*MaxEvapHeatLoss
Where,
0.68 is the passive water vapor diffusion rate in g/h·m2·Torr
and,
W/m2
SkinWetDiff = (1.-SkinWetSweat)*.06
MaxEvapHeatLoss = (1./TotEvapHeatResist)*(SatSkinVapPress - VapPress)
SkinWetSweat = EvapHeatLossRegSweat/MaxEvapHeatLoss
The Pierce model has one additional heat flow term describing the heat transfer between the internal core compartment and the outer skin shell (Doherty and Arens 1988).
W/m2
HeatFlow = (CoreTemp-SkinTemp)*(5.28 + 1.163*SkinBloodFlow)
Where
5.28 is the average body tissue conductance in W/m2•°C
1.163 is the thermal capacity of blood in W•h/L•°C
Thus, individual heat balance equations for core and skin compartments are expressed using this term, Qc-s. New temperatures of core, skin and body are calculated by each iteration from rates of heat storage in the core and skin.
W/m2°C
SkinHeatStorage = HeatFlow - DryHeatLoss - EvapHeatLoss
W/m2°C
CoreHeatStorage = IntHeatProd - RespHeatLoss - HeatFlow
Thus,
ThermSensTransCoef = 0.303*EXP(-0.036*ActLevel) + 0.028
PMVET = ThermSensTransCoef*(IntHeatProd - EvapHeatLossDiff &
• EvapHeatLossRegComf - RespHeatLoss - DryHeatLossET)
PMVSET = ThermSensTransCoef*(IntHeatProd - EvapHeatLossDiff &
• EvapRegHeatLossReg Comf - RespHeatLoss - DryHeatLossSET)
Besides PMV*, the Pierce Two Node Model uses the indices TSENS and DISC as predictors of thermal comfort. Where TSENS is the classical index used by the Pierce foundation, and is a function of the mean body temperature. DISC is defined as the relative thermoregulatory strain that is needed to bring about a state of comfort and thermal equilibrium. DISC is a function of the heat stress and heat strain in hot environments and equal to TSENS in cold environments. In summary, the Pierce Model, for our purposes, uses four thermal comfort indices; PMVET-a function of ET*, PMVSET- a function of SET*, TSENS and DISC.
°C
°C
AvgBodyTempLow = (0.185/ActLevelConv)*IntHeatProd + 36.313
AvgBodyTempHigh = (0.359/ActLevelConv)*IntHeatProd + 36.664
TSENS = .68175*(AvgBodyTemp-AvgBodyTempLow)
TSENS = 4.7*(AvgBodyTemp - AvgBodyTempLow)/ &
(AvgBodyTempHigh - AvgBodyTempLow)
DISC = 5.*(EvapHeatLossRegSweat - EvapHeatLossRegComf)/ &
(MaxEvapHeatLoss - EvapHeatLossRegComf - DiffHeatLoss)
The KSU two-node model, developed at Kansas State University, was published in 1977 (Azer and Hsu 1977). The KSU model is quite similar to that of the Pierce Foundation. The main difference between the two models is that the KSU model predicts thermal sensation (TSV) differently for warm and cold environment.
### KSU Two Node Model Nomenclature List[LINK]
Nomenclature list for KSU Two-Node model
Mathematical variable Description UnitsRange FORTRAN variable
Ccr Specific heat of body core Whr/kg°C
—– ————————– ——–
Csk Specific heat of skin Whr/kg°C
—– ——————— ——–
Cres The rate of dry respiratory heat loss W/m2 -
Edif The rate of heat loss from the diffusion of water vapor through the skin W/m2 -
Emax Maximum evaporative heat loss W/m2
Esk Total evaporative heat loss from skin W/m2
Esw Equivalent evaporation heat loss from the sweat secreted W/m2
Esw.d Sweat function for warm and dry skin W/m2
Eres The rate of latent respiratory heat loss W/m2 -
Fcl The Burton thermal efficiency factor for clothing -
Fpcl Permeation efficiency factor for clothing -
H Internal heat production rate of an occupant per unit area = M - W W/m2
H Combined heat transfer coefficient W/m2°C
hc Convective heat transfer coefficient W/m2°C -
hr Radiant heat transfer coefficient W/m2°C -
KS Overall skin thermal conductance W/m2°C
KSo Skin conductance at thermal neutrality W/m2°C
KS(-4) Skin conductance at thermal sensation very cold W/m2°C
M Metabolic rate per unit area W/m2 -
Mshiv Metabolic heat production due to shivering W/m2
Pa Water vapor pressure in ambient air Torr -
Psk Saturated water vapor pressure at required skin temperature Torr -
PTaccl The pattern of acclimation
Qc The rate of convective heat loss W/m2 -
Qdry Sensible heat flow from skin W/m2
Qr The rate of radiative heat loss W/m2 -
Qres The rate of respiratory heat loss W/m2 -
RH Relative humidity
Ta Air temperature °C -
Tcr Core or internal temperature °C -
To Operative temperature °C -
Tr Mean radiant temperature °C -
Tsk Skin temperature °C
TSV Thermal sensation vote -4~4
V Relative air velocity m/s -
W The rate of heat loss due to the performance of work W/m2 -
W Skin wettedness - -
Wcr Mass of body core per unit body surface kg/m2
—– ————————————— ——-
wrsw Skin wettedness due to regulatory sweating
wrsw-o Skin wettedness at thermal neutrality
Wsk Mass of skin per unit body surface kg/m2
—– ———————————- ——-
: Nomenclature list for KSU Two-Node model
### Description of the model and algorithm[LINK]
The KSU two-node model is based on the changes that occur in the thermal conductance between the core and the skin temperature in cold environments, and in warm environments it is based on changes in the skin wettedness.
In this model metabolic heat production is generated in the core which exchanges energy with the environment by respiration and the skin exchanges energy by convection and radiation. In addition, body heat is dissipated through evaporation of sweat and/or water vapor diffusion through the skin. These principles are used in following passive system equations.
W/m2
W/m2
Where
W/m2
LatRespHeatLoss = 0.0023*ActLevelTot*(44. - VapPress)
DryRespHeatLoss = 0.0014*ActLevelTot*(34. - AirTemp)
RespHeatLoss = LatRespHeatLoss + DryRespHeatLoss
W/m2
DryHeatLoss = H*CloBodyRat*CloThermEff*(SkinTemp - OpTemp)
W/m2°C
W/m2°C
W/m2°C
H = Hc + Hr
Hc = 8.3*SQRT(AirVel)
°C
and
For , W/m2
For , W/m2
W/m2
W/m2
EvapHeatLoss = SkinWetSweat*EvapHeatLossMax+(1. - SkinWetSweat)*EvapHeatLossDiff
SkinWetSweat = EvapHeatLossDrySweat/EvapHeatLossMax
EvapHeatLossDiff = 0.408*(SkinVapPress - VapPress)
EvapHeatLossMax = 2.2*Hc*(SkinVapPress - VapPress)*CloPermeatEff
Here, control signals, based on setpoint temperatures in the skin and core, are introduced into passive system equations and these equations are integrated numerically for small time increments or small increments in core and skin temperature. The control signals modulate the thermoregulatory mechanism and regulate the peripheral blood flow, the sweat rate, and the increase of metabolic heat by active muscle shivering. The development of the controlling functions of skin conductance (KS), sweat rate (Esw), and shivering (Mshiv) is based on their correlation with the deviations in skin and core temperatures from their setpoints.
SkinCndctDilation = 42.45*CoreSignalWarmMax &
• 8.15*CoreSignalSkinSens**0.8*SkinSignalWarmMax
SkinCndctConstriction = 1.0 + 0.4*SkinSignalColdMax
ThermCndct = 5.3+(6.75+SkinCndctDilation)/SkinCndctConstriction
WeighFac = 260.+70.*AcclPattern
SweatCtrlFac = 1. + 0.05*SkinSignalSweatColdMax**2.4
DrySweatRate = ((WeighFac*CoreSignalSweatMax &
• 0.1*WeighFac*SkinSignalSweatMax) &
*EXP(SkinSignalSweatMax/8.5))/SweatCtrlFac
Where
SweatSuppFac = 1.
SweatSuppFac = 0.5 + 0.5*EXP(-5.6*SkinWetSignal)
W/m2
ShivResponse = 20.*CoreSignalShivMax*SkinSignalShivMax + 5.*SkinSignalShivMax
In KSU model, two new parameters are introduced and used in correlating thermal sensations with their associated physiological responses. In stead of correlating warm thermal sensations with skin wettedness, it is here correlated with a wettedness factor defined by
SkinWetFac = (SkinWetSweat - SkinWetNeut)/(1. - SkinWetNeut)
Where
SkinWetSweat = DrySweatRate/EvapHeatLossMax
SkinWetNeut = 0.02 + 0.4*(1.-EXP(-0.6*(IntHeatProdMetMax - 1.)))
and instead of correlating cold thermal sensation with the skin temperature, it is here correlated with a factor identified as vasoconstriction factor defined by
VasoconstrictFac = (ThermCndctNeut - ThermCndct) &
/(ThermCndctNeut - ThermCndctMin)
Thus, TSV in the cold is a function of a vasoconstriction factor (εvc) as:
TSV = -1.46153*VasoconstrictFac + 3.74721*VasoconstrictFac**2 &
• 6.168856*VasoconstrictFac**3
and for the warm environments, TSV is defined as:
TSV = (5. - 6.56*(RelHum - 0.50))*SkinWetFac
The KSU model’s TSV was developed from experimental conditions in all temperature ranges and from clo levels between .05 clo to 0.7 clo and from activities levels of 1 to 6 mets (Berglund 1978).
Adaptive comfort model, intended for use in naturally ventilated buildings, determines the acceptability of indoor conditions given the monthly mean outdoor air temperature and the indoor operative temperature. This is used as an index for occupant adaptation to outdoor conditions, and determines the acceptability of indoor conditions. The model also accounts for people’s clothing adaptation in naturally conditioned spaces by relating the acceptable range of indoor temperatures to the outdoor climate, so it is not necessary to estimate the clothing values for the space. No humidity or air-speed limits are required when this option is used. This section summarizes the adaptive comfort models based on the ASHRAE Standard 55-2010 and CEN 15251. Details are available in the two standards.
Adaptive Comfort Model Based on ASHRAE Standard 55-2010
In ASHRAE Standard 55, the monthly mean outdoor air temperature, used in the adaptive comfort model, is defined as the simple running average of the previous thirty daily average outdoor air temperatures.
The model defines two comfort regions: 80% Acceptability, and 90% Acceptability. If the monthly mean outdoor air temperature is not within the specified domain, the model is not applicable.
The central line of the model (shown in red), or comfort temperature, is defined as
Where
Tot – operative temperature (°C), calculated as the average of the indoor air dry-bulb temperature and the mean radiant temperature of zone inside surfaces
To – monthly mean outdoor air dry-bulb temperature (°C).
If the .stat file is provided for the simulation, To is drawn directly from the daily average temperatures in the .stat file, which provides a value for each month. If no .stat file is provided, the monthly mean outdoor temperature is a simple running average of the previous thirty daily average temperatures, calculated directly from the weather file (.epw):
Tod-iis defined as the daily average temperature of the ith previous day.
Note that the weather file must be a standard .epw containing a full year of data.
The comfort regions for 80% and 90% acceptability are symmetric about the central line.
90% Acceptability Limits: Tot = 0.31* To + 17.8 ± 2.5
80% Acceptability Limits: Tot = 0.31* To + 17.8 ± 3.5
If, using either method, To~~is less than 10°(C) or greater than 33.5°(C), the model is not applicable.
For a detailed description of this model, please see ASHRAE Standard 55-2010, Thermal Environmental Conditions for Human Occupancy.
The EN15251-2007 is similar to ASHRAE 55-2010, but with slightly different curves of the indoor operative temperature and acceptability limits (Fig. 2). The model, intended for use in naturally ventilated buildings, determines the acceptability of indoor conditions given the 7-day weighted mean outdoor air temperature and the indoor operative temperature. The 7-day weighted mean outdoor air temperature (Trm) is defined as the weighted running average of the previous 7 daily average outdoor air temperatures.
This weighted running average is calculated from a full annual weather file that must be specified for the simulation. This is used as an index for occupant adaptation to outdoor conditions, and determines the acceptability of indoor conditions. The model also accounts for people’s clothing adaptation in naturally conditioned spaces by relating the acceptable range of indoor temperatures to the outdoor climate, so it is not necessary to estimate the clothing values for the space. No humidity or air-speed limits are required when this option is used. The model defines three comfort regions: Category I (90%) Acceptability, Category II (80%) Acceptability, and Category III (65%) Acceptability. If Trm is not within the specified domain, the model is not applicable.
Central line (shown as red Figure 281): Tot = 0.33*To + 18.8
Category I, 90% Acceptability Limits: Tot = 0.33*To + 18.8 ± 2.0
Category II, 80% Acceptability Limits: Tot = 0.33*To + 18.8 ± 3.0
Category III, 65% Acceptability Limits: Tot = 0.33*To + 18.8 ± 4.0
For 10°(C) < Trm~~< 15°(C), the comfort temperature of the lower boundaries of the comfort regions is Tcomf = 23.75°(C). That is, the lower boundaries are constant according to the same ranges above:
Category I, 90% Acceptability Limits: Tot = 23.75- 2.0
Category II, 80% Acceptability Limits: Tot = 23.75 - 3.0
Category III, 65% Acceptability Limits: Tot = 23.75 - 4.0
Where
Tot – operative temperature (°C) , calculated as the average of the indoor air dry-bulb temperature and the mean radiant temperature of zone inside surfaces
To – mean outdoor air dry-bulb temperature (°C), calculated as the weighted mean of the previous 7-day daily mean outdoor air dry-bulb temperature (Tod):
To = (1 - α)*{Tod-1 + α *Tod-2 + α2 *Tod-3 + α3 *Tod-4 + α4 *Tod-5 + α5 *Tod-6 + α6 *Tod-7}
To = (1 - α)*Tod-1 + α *To-1
α = 0.8
There are three options to calculate mean radiant temperature in the thermal comfort models. One is the zone averaged MRT, another is the surface weighted MRT, and the other is angle factor MRT. The zone averaged MRT is calculated on the assumption that a person is in the center of a space, whereas the surface weighted MRT is calculated in consideration of the surface that a person is closest to, and the angle factor MRT is calculated based on angle factors between a person and the different surfaces in a space. Here, the surface weighted MRT is the average temperature of the selected surface and zone averaged MRT and is intended to represent conditions in the limit as a person gets closer and closer to a particular surface. In that limit, half of the person’s radiant field will be dominated by that surface and the other half will be exposed to the rest of the zone. Note that the surface weighted MRT is only an approximation. The angle factor MRT is the mean temperature of the surrounding surface temperatures weighted according to the magnitude of the respective angle factors and allows the user to more accurately predict thermal comfort at a particular location within a space.
Nomenclature and variable list for MRT calculation
Mathematical variable Description UnitsRange FORTRAN variable
Tr Mean radiant temperature °C -
Tr-avg Zone averaged radiant temperature °C -
Tsurf Surface temperature °C -
Fsurf Angle factor between person and surface - 0~1
### Description of the model and algorithm[LINK]
The zone averaged MRT is calculated without weighting any surface temperature of the space.
The surface weighted MRT is the average temperature of the zone averaged MRT and the temperature of the surface that a person is closest to.
SurfaceTemp = GetSurfaceTemp(People(PeopleNum)%SurfacePtr)
The angle factor MRT is the mean value of surrounding surface temperatures weighted by the size of the respective angle factors between a person and each surface.
SurfTempAngleFacSummed = SurfTempAngleFacSummed &
• SurfaceTemp * AngleFactorList(AngleFacNum)%AngleFactor(SurfNum)
ASHRAE. 1984. “High Intensity Infrared Radiant Heating”, 1984 Handbook of Systems and Equipment, American Society of Heating, Refrigerating and Air Conditioning Engineers, Atlanta, GA, Chapter 18.
ASHRAE. 1985. “Physiological Principles for Comfort and Health,” 1985 Handbook of Fundamentals, American Society of Heating, Refrigerating and Air Conditioning Engineers, Atlanta, GA, Chapter 8.
ASHRAE. 1993. “Physiological Principles and Thermal Comfort”, 1993 ASHRAE Handbook of Fundamentals, American Society of Heating, Refrigerating and Air Conditioning Engineers, Atlanta, GA, Chapter 8.
ASHRAE. 2010. “Standard 55-2010 – Thermal Environmental Conditions for Human Occupancy (ANSI approved)”, American Society of Heating, Refrigerating and Air Conditioning Engineers, Atlanta, GA.
Azer, N.Z., Hsu, S. 1977. “The prediction of Thermal Sensation from Simple model of Human Physiological Regulatory Response”, ASHRAE Trans., Vol.83, Pt 1.
Berglund, Larry. 1978. “Mathematical Models for Predicting the Thermal Comfort Response of Building Occupants”, ASHRAE Trans., Vol.84.
Doherty, T.J., Arens, E. 1988. “Evaluation of the Physiological Bases of Thermal Comfort Models”, ASHRAE Trans., Vol.94, Pt 1.
Du Bois, D. and E.F. 1916. “A Formula to Estimate Approximate Surface Area, if Height and Weight are Known”, Archives of internal Medicine, Vol.17.
CEN. 2007. “Standard EN15251 Indoor environmental input parameters for design and assessment of energy performance of buildings addressing indoor air quality, thermal environment, lighting and acoustics”. Bruxelles: European committee for Standardisation.
Fanger, P.O. 1970. Thermal Comfort-Analysis and Applications in Environmental Engineering, Danish Technical Press, Copenhagen.
Fanger, P.O. 1986. “Radiation and Discomfort”, ASHRAE Journal. February 1986.
Fanger P.O. 1967. “Calculation of Thermal Comfort: Introduction of a Basic Comfort Equation”, ASHRE Trans., Vol.73, Pt 2.
Fountain, Marc.E., Huizenga, Charlie. 1997 “A Thermal Sensation Prediction Tool for Use by the Profession”, ASHRAE Trans., Vol.103, Pt 2.
Gagge, A.P., Stolwijk, J. A. J., Nishi, Y. 1970. “An Effective Temperature Scale Based on a Simple Model of Human Physiological Regulatory Response”, ASHRAE Trans., Vol.70, Pt 1.
Gagge, A.P., Fobelets, A.P., Berglund, L. G. 1986. “A Standard Predictive Index of Human Response to the Thermal Environment”, ASHRAE Trans., Vol.92, Pt 2.
Hsu, S. 1977. “A Thermoregulatory Model for Heat Acclimation and Some of its Application”, Ph. D. Dissertation, Kansas State University.
Int-Hout, D. 1990. “Thermal Comfort Calculation / A Computer Model”, ASHRAE Trans., Vol.96, Pt 1.
ISO. 1983. “Determination of the PMV and PPD Indices and Specification of the Conditions for Thermal Comfort”, DIS 7730, Moderate Thermal Environment, 1983. | 9,640 | 38,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.879467 |
http://www.qsl.net/wb6tpu/si-list2/0616.html | 1,531,959,782,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590362.13/warc/CC-MAIN-20180718232717-20180719012717-00319.warc.gz | 515,347,033 | 3,196 | RE: [SI-LIST] : Decoupling caps and power plane effects
Doug Piper ([email protected])
Mon, 17 May 1999 17:24:08 -0400
Gary,
Another raw lead in your territory, but they are talking the right stuff!
Doug Piper
Woven Electronics
P.O. Box 189
Mauldin, SC 29662
PH 864-967-1751
FAX 864-963-1761
e-mail: [email protected]
www.wovenelectronics.com
-----Original Message-----
From: Ron Miller [SMTP:[email protected]]
Sent: Monday, May 17, 1999 4:25 PM
To: [email protected]
Subject: Re: [SI-LIST] : Decoupling caps and power plane effects
> Hi Todd
Within your own company, Intel, a small group is working on test
interfaces.
Arthur Frazier , a microwave engineer works in this group, and has MDS
Microwave Development System(simiulator) from HP which has models for
power/ground in a matrix form. This tool does what you need accurately.
Ron Miller
> To calculate the required number of decoupling capacitance for a given
> motherboard, I am trying to setup lumped RLC circuit to model the
behavior
> of a realistic capacitor discharging into a power plane. The modeling of
> the realistic capacitor just involved RLC elements all in series. The
ESR
> is determined from the vendors datasheet. The ESL from the datasheet as
> well as the loop inductance when placed on the PCB. To model the
presence
> of the power plane is proving more involved.
>
> When a chip on a circuit board has its initial current draw from its
outputs
> switching, the power plane is the first to respond with current. This is
> due to the low inductance of the power plane. Next the ceramic
capacitors
> respond, followed by the higher ESL caps and then finally the power
supply.
> The effect of the power plane responding to the IC's current draw is the
> topic which I would appreciate assistance on. At time t=0, an IC chip's
> outputs switch and its power pins will draw a current Io from the power
> plane of a motherboard. Since the power plane is essentially a large
> capacitor, its discharging current will decrease the voltage level of the
> plane until the ceramics respond to stabilize...then the bulks...then the
> power supply. The rate at which the power plane discharges is of
interest
> to me. If the effective capacitance of the power plane seen by a chip
can
> be gauged, then the discharge rate of the power plane supplying current
to
> some load can be modeled with a lumped capacitave element. With the
> discharge of the power supply predictable, then decoupling capacitance
can
> be calculated to prevent the power plane voltage from drooping below a
> specified voltage.
>
> Has anyone done any power/ground studies that would shed light on this
> problem? Are any of my assumptions invalid?
>
> Thanks all.
>
> Todd Bermensolo
> Intel Corp
> High End Server Division
>
> **** To unsubscribe from si-list: send e-mail to
[email protected] In the BODY of message put: UNSUBSCRIBE
si-list, for more help, put HELP. si-list archives are accessible at
http://www.qsl.net/wb6tpu/si-list ****
```--
Ronald B. Miller _\\|//_ Signal Integrity Engineer
(408)487-8017 (' 0-0 ') fax(408)487-8017
==========0000-(_)0000=========== | 808 | 3,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-30 | latest | en | 0.908642 |
http://mathandlanguage.edc.org/mathematics-tasks/geometric-thinking/comparing-area-tangrams/word-chart | 1,670,524,765,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711360.27/warc/CC-MAIN-20221208183130-20221208213130-00632.warc.gz | 30,818,456 | 7,150 | # Comparing Area with Tangrams: Word Chart
Spanish (SP)
French (FR)
Portuguese (PT)
* = Cognate
Words and Phrases Academic Language Meaning Everyday Language Version Other Forms of the Word or Phrase Related Words or Phrases Translation Examples of word use with students Area The number of square units contained in the interior of a figure; the extent of a two-dimensional surface enclosed within a boundary The space something occupies, two-dimensional --- Enclosed Space Two-dimensional size *Area (SP) *Aire/ Superficie (FR) *Área (PT) Convincing Persuading or assuring by argument or evidence Causing one to believe the truth of something; plausible Convince Convinced Convincer Convincible Convincingly Persuasive Believable Credible Plausible *Convincente (SP) *Convaincant(e) (FR) *Convincente (PT) Congruent Two geometric figures in the plane are congruent if one can be obtained from the other by some combination of rotations, reflections, and translations The same or equal in some way; same size and shape. Congruently Congruence Congruous Same shape *Congruente (SP) *Congruent(e) (FR) Coincidente (PT) Non-square rectangle A quadrilateral with four right angles, opposite sides of the same length, and adjacent sides of different lengths --- Non-square rectangles Rectángulo (que no séa cuadrado) (SP) Rectangle (qui ne soit pas carré) (FR) Retângulo (que não seja quadrado) (PT) Right Angle A quarter of a full turn, equal to 90 degrees. --- Right Angles Perpendicular Ángulo recto (SP) Angle droit (FR) Ângulo reto (PT) Arrange -- To place in a particular order. Arranges Arranging Arranged Organize Put in order Ordenar (SP) *Ranger (FR) *Arranjar (PT) Square A quadrilateral in which all sides are congruent to each other and all angles are right angles. -- Squares Squared Squaring --- Cuadrado (SP) Carré (FR) Quadrado (PT) Shaded -- A place or area of comparative darkness. Shade Shades Shading Not white Tinted Colored in En color (tonalizado) (SP) En couleur (ombré) (FR) Colorido (sombreado) (PT) Mid-point A point on a line segment that divides it into two equal parts The halfway point of a line segment A point at the center or middle. Midpoints Center Middle Halfway point Midway *Punto medio (SP) *Point médian (FR) Ponto central (PT) | 575 | 2,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.726321 |
https://math.stackexchange.com/questions/4241711/what-is-mathcalr | 1,713,208,778,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817014.15/warc/CC-MAIN-20240415174104-20240415204104-00530.warc.gz | 340,900,517 | 40,706 | # What is $\mathcal{R}$?
First of all, I am asking this question entirely out of curiosity. It basically randomly popped out of my mind.
So I am asking for the value of an infinite series.
Let's call it, $$\mathcal{R}=\sum_{n=1}^{\infty}\mathcal{R}_n$$
Now I will try to explain what are these $$\mathcal{R}_i$$'s
Let's take a 1-ball of unit length (1-volume).
It's radius will be $$r_1=\frac{1}{2}$$
So the first term of my series is $$\mathcal{R}_1=1$$
Now I will turn the length of this line into the circumference of a circle or a $$2$$-ball.
It's radius will be $$r_2$$, you can just do a little calculation and
$$r_2=\frac{1}{2\pi}$$
And the area of this circle would be, $$\pi r_2^2=\frac{1}{4\pi}$$
This will be $$\mathcal{R}_2=\frac{1}{4\pi}$$
Now let's turn the area of this circle into the surface area of a sphere or a $$3$$-ball,
It's radius would be $$r_3$$
Again, $$r_3=\frac{1}{4\pi}$$
And the volume of this sphere will be,
$$\frac{4}{3}\pi r_3^3=\frac{1}{48\pi^2}$$
This will be, $$\mathcal{R}_3=\frac{1}{48\pi^2}$$
In general we take an $$n$$-ball whose radius is $$r_n$$, then we turn the $$n$$-volume of this ball into the $$n$$-surface area of an $$n+1$$-ball whose radius is $$r_{n+1}$$.
And $$\mathcal{R}$$ is the sum over the $$n$$-volumes of these $$n$$-balls with radius $$r_n$$.
So we can write,
$$\mathcal{R}=\textstyle\displaystyle\sum_{n=1}^{\infty}V_n(r_n)$$
From this we can deduce the formula for $$r_n$$
Agian in general, we take an $$n$$-ball with radius $$r_n$$. Then we set it's $$n$$-volume to equal the $$n$$-surface area of an $$n+1$$-ball. And we define the $$n+1$$-ball's radius to be $$r_{n+1}$$. From this we can calculate a recurrence relation for $$r_n$$,
$$V_n(r_n)=S_n(r_{n+1})$$
$$\textstyle\displaystyle{\frac{\pi^\frac{n}{2}}{\Gamma(\frac{n}{2}+1)}r_n^n=\frac{2\sqrt{\pi}\pi^\frac{n}{2}}{\Gamma(\frac{n+1}{2})}r_{n+1}^n}$$
$$\textstyle\displaystyle{r_{n+1}=r_n\sqrt[n]{\frac{\Gamma(\frac{n+1}{2})}{2\sqrt{\pi}\Gamma(\frac{n+2}{2})}}}$$
with the initial condition of $$r_1=\frac{1}{2}$$
If we keep expanding $$r_n$$ using the formula then we get,
$$\textstyle\displaystyle{r_n=r_1\prod_{k=1}^{n-1}\sqrt[k]{\frac{\Gamma(\frac{k+1}{2})}{2\sqrt{\pi}\Gamma(\frac{k+2}{2})}}}$$
$$\textstyle\displaystyle{r_n=\frac{1}{2(2\sqrt{\pi})^{H_{n-1}}}\prod_{k=1}^{n-1}\sqrt[k]{\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}}}$$
So finally,
$$\textstyle\displaystyle{V_n(r_n)=\frac{\pi^\frac{n}{2}}{\Gamma(\frac{n}{2}+1)}r_n^n}$$
\textstyle\displaystyle{\begin{align}\mathcal{R}_n=&\frac{\sqrt{\pi}^{n(1-H_{n-1})}}{2^{n(1+H_{n-1})}\Gamma(\frac{n+1}{2})}\prod_{k=1}^{n}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{n}{k}\end{align}}
So,
$$\textstyle\displaystyle{\mathcal{R}=\sum_{n=1}^{\infty}\frac{\sqrt{\pi}^{n(1-H_{n-1})}}{2^{n(1+H_{n-1})}\Gamma(\frac{n+1}{2})}\prod_{k=1}^{n}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{n}{k}}$$
Here $$H_0=0$$
$$\mathcal{R}\approx 1.0817135\dots$$
## $$\underline{\text{My Question}:-}$$
Is there a closed form for $$\mathcal{R}$$? Or maybe we can write $$\mathcal{R}$$ in terms of some advance function?
Are there some interesting properties of $$\mathcal{R}$$?
A list of the first few terms of $$\mathcal{R}$$:-
$$\{\mathcal{R}_n\}_{n=1}^{\infty}$$
$$\textstyle\displaystyle{=\left\{1, \frac{1}{2^2\pi}, \frac{1}{2^43\pi^2}, \frac{1}{2^\frac{23}{3}3^\frac{4}{3}\pi^\frac{10}{3}}, \frac{1}{2^\frac{31}{3}3^\frac{17}{12}5\pi^\frac{14}{3}}, \frac{1}{2^\frac{72}{5}3^\frac{27}{10}5^\frac{6}{5}\pi^\frac{31}{5}}, \frac{1}{2^\frac{163}{10}3^\frac{179}{60}5^\frac{37}{30}7\pi^\frac{116}{15}}, \cdots\right\}}$$
You can yourself find more terms by just plugging the formula in Wolfram alpha.
Notice the terms at the denominator have a very weird property, they are all prime numbers to the power of some rational numbers which is really weird. The more peculiar part is that they are not random prime numbers, but the first $$n$$ prime numbers.
So there will definitely be some involvement of prime numbers in the evaluation of $$\mathcal{R}$$.
Observing some patterns:-
Notice that every $$n^\text{th}$$ $$2$$ terms namely the $${2n-1}^\text{st}$$ and $${2n}^\text{th}$$ terms have the first $$n$$ primes in their denominator.
Also notice that when the $$n^\text{th}$$ prime first appears at the $${2n-1}^\text{st}$$ term it is being raised to the first power and at the $${2n}^\text{th}$$ term it is being raised to the power of $$\frac{p_n+1}{p_n}$$. (If we say that the $$n^\text{th}$$ prime is $$p_n$$).
I can't prove these observations, but if they are true to infinity then we can write-
$$\textstyle\displaystyle{\mathcal{R}=\sum_{n=1}^{\infty}\left(\frac{1}{p_n\pi^{q_{2n-1}}\prod_{k=1}^{n-1}p_k^{r_{k,2n-1}}}+\frac{1}{p_n^{\frac{p_n+1}{p_n}}\pi^{q_{2n}}\prod_{k=1}^{n-1}p_k^{r_{k,2n}}}\right)}$$
For some rational sequence $$q_n$$ and $$r_{m,n}$$
It seems like $$\forall n\in\mathbb{N}, q_{n+1}>q_n$$
And the greatest common divisor of the denominators of $$r_{m,n}$$ and $$q_n$$ is not $$1$$, namely
If $$\textstyle\displaystyle{r_{m,n}=\frac{s_{m,n}}{t_{m,n}}}$$ and $$\textstyle\displaystyle{q_n=\frac{a_n}{b_n}}$$ such that $$\operatorname{gcd}(s_{m,n},t_{m,n},)=\operatorname{gcd}(a_n,b_n)=1$$
Then, $$\operatorname{gcd}(t_{1,n},\dots,t_{m,n},b_n)\neq 1$$ for $$n\gt 3$$
I don't know if my observation are true to infinity or not, do they even simplify $$\mathcal{R}$$, I don't know. But it increases my hope for getting a closed form for $$\mathcal{R}$$.
If I am able to spot more patterns then I will add it to the post.
Working with the formula itself:-
I hadn't really worked with the formula yet. I spent the last 2 days trying to find more patterns, I think I may have found another one but I need work on that a little bit more before I add that to the question. I am sure is the following progress or regress, but whatever.
$$\textstyle\displaystyle{\mathcal{R}_n=\frac{\sqrt{\pi}^{n(1-H_{n-1})}}{2^{n(1+H_{n-1})}\Gamma(\frac{n+1}{2})}\prod_{k=1}^{n}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{n}{k}}$$
$$\textstyle\displaystyle{=\frac{\sqrt{\pi}^{n(1-H_{n})}}{2^{n\left(\frac{1}{2}+H_{n-1}\right)}(n-1)!!}\prod_{k=1}^{n}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{n}{k}}$$
Now I am going to break $$\mathcal{R}$$ into two infinite sums as following
$$\mathcal{R}=E+O$$
Where,
$$\textstyle\displaystyle{E=\sum_{n=1}^{\infty}\frac{\pi^{n(1-H_{2n})}}{2^{n(1-2H_{2n-1})}(2n-1)!!}\prod_{k=1}^{2n}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{2n}{k}}$$
$$\textstyle\displaystyle{=\sum_{n=1}^{\infty}\frac{(n-1)!\pi^{n(1-H_{2n})}}{4^{nH_{2n}}(2n-1)!}\prod_{k=1}^{2n}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{2n}{k}}$$
Here i replaced every $$n$$ with $$2n$$ And,
$$\textstyle\displaystyle{O=\sum_{n=1}^{\infty}\frac{\sqrt{\pi}^{2n(1-H_{2n})+H_{2n-1}}}{2^{(2n-1)\left(\frac{1}{2}+H_{2n-2}\right)}(2n-2)!}\prod_{k=1}^{2n-1}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{2n-1}{k}}$$
$$\textstyle\displaystyle{=\frac{1}{2\sqrt{2}}\sum_{n=1}^{\infty}\frac{\sqrt{\pi}^{2n(1-H_{2n})+H_{2n-1}}}{2^{(2n-1)(1-H_{2n-1})}(n-1)!}\prod_{k=1}^{2n-1}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{2n-1}{k}}$$
Here i replaced every $$n$$ with $$2n-1$$.
Before I do anything with these $$E$$ and $$O$$. I want to first take care of the product inside both of them. Let,
$$\textstyle\displaystyle{P_{n}=\prod_{k=1}^{2n}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{2}{k}}$$
$$\textstyle\displaystyle{=\prod_{k=1}^{n}\left(\frac{\Gamma(\frac{2k+1}{2})}{\Gamma(\frac{2k+2}{2})}\right)^\frac{2}{2k}\prod_{k=1}^{n}\left(\frac{\Gamma(\frac{2k}{2})}{\Gamma(\frac{2k+1}{2})}\right)^\frac{2}{2k-1}}$$
$$\textstyle\displaystyle{=\prod_{k=1}^{n}\left(\frac{\Gamma(k+\frac{1}{2})}{k!}\right)^\frac{1}{k}\prod_{k=1}^{n}\left(\frac{(k-1)!}{\Gamma(k+\frac{1}{2})}\right)^\frac{2}{2k-1}}$$
$$\textstyle\displaystyle{=\prod_{k=1}^{n}\left(\frac{(2k)!\sqrt{\pi}}{4^kk!^2}\right)^\frac{1}{k}\prod_{k=1}^{n}\left(\frac{4^kk!(k-1)!}{(2k)!\sqrt{\pi}}\right)^\frac{2}{2k-1}}$$
$$\textstyle\displaystyle{=\prod_{k=1}^{n}\left(\frac{\sqrt{\pi}}{4^k}\right)^\frac{1}{k}\prod_{k=1}^{n}\left(\frac{4^k}{\sqrt{\pi}}\right)^\frac{2}{2k-1}\prod_{k=1}^{n}\left(\frac{(2k)!}{k!^2}\right)^\frac{1}{k}\prod_{k=1}^{n}\left(\frac{k!(k-1)!}{(2k)!}\right)^\frac{2}{2k-1}}$$
Since, $$\sum_{k=1}^{n}\frac{1}{2k-1}=H_{2n}-\frac{1}{2}H_n$$ And by Wolfram Alpha we have,
$$\prod_{k=1}^{n}16^\frac{k}{2k-1}=2^{2n+\psi(n+\frac{1}{2})-\psi(\frac{1}{2})}$$
Also notice, $$\frac{k!(k-1)!}{(2k)!}=B(k,k+1)$$ and, $$\frac{(2k)!}{k!^2}=\frac{1}{kB(k,k+1)}$$ And finally again from Wolfram Alpha, $$\prod_{k=1}^{n}k^{-\frac{1}{k}}=e^{\gamma_{1}(n+1)-\gamma_1}$$
We have-
$$\textstyle\displaystyle{P_{n}^n=2^{n(\psi(n+\frac{1}{2})-2+\gamma+\ln(4))}\pi^{n(H_n-H_{2n})}e^{n(\gamma_1(n+1)-\gamma_1)}\prod_{k=1}^{n}B(k,k+1)^\frac{n}{k(2k-1)}}$$
So, $$\textstyle\displaystyle{E=\sum_{n=1}^{\infty}\frac{(n-1)!}{(2n-1)!}\pi^{n(1+H_n-2H_{2n})}e^{n(\gamma_1(n+1)-\gamma_1)}2^{n(\psi(n+\frac{1}{2})-2(n+1)H_{2n}+\gamma+\ln(4)-2)}\prod_{k=1}^{n}B(k,k+1)^\frac{n}{k(2k-1)}}$$
Now replacing $$n$$ with $$n-\frac{1}{2}$$ should give us $$P_{2n-1}$$ which corresponds to the product inside $$O$$. I am not sure would it work or not because $$n-\frac{1}{2}$$ won't be an integer anymore. And I don't know what $$\prod_{k=1}^{m}$$ means when $$m\not\in\mathbb{Z}$$. So let's leave that idea and re do the process again. Let,
$$\textstyle\displaystyle{Q_n=\prod_{k=1}^{2n-1}\left(\frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k+2}{2})}\right)^\frac{1}{k}}$$
$$\textstyle\displaystyle{=\prod_{k=1}^{n}\left(\frac{\Gamma(\frac{2k}{2})}{\Gamma(\frac{2k+1}{2})}\right)^\frac{1}{2k-1}\prod_{k=1}^{n-1}\left(\frac{\Gamma(\frac{2k+1}{2})}{\Gamma(\frac{2k+2}{2})}\right)^\frac{1}{2k}}$$
$$\textstyle\displaystyle{=\prod_{k=1}^{n}\left(\frac{4^k}{\sqrt{\pi}}\right)^\frac{1}{2k-1}\prod_{k=1}^{n-1}\left(\frac{\sqrt{\pi}}{4^k}\right)^\frac{1}{2k}\prod_{k=1}^{n}\left(\frac{k!(k-1)!}{(2k)!}\right)^\frac{1}{2k-1}\prod_{k=1}^{n-1}\left(\frac{(2k)!}{k!^2}\right)^\frac{1}{2k}}$$
$$\textstyle\displaystyle{=\sqrt{\pi}^{H_{n}-H_{2n}+\frac{1}{2n}}2^{\frac{1}{2}\psi(n+\frac{1}{2})+\frac{\gamma}{2}+\ln(2)+1}\frac{\sqrt[n]{n!}}{(2n)!^\frac{1}{2n}}e^{\gamma_1(n+1)-\gamma_1}\prod_{k=1}^{n-1}B(k,k+1)^\frac{1}{2k(2k-1)}}$$
We have-
$$\textstyle\displaystyle{Q_n^{2n-1}=\sqrt{\pi}^{(2n-1)(H_{n}-H_{2n}+\frac{1}{2n})}2^{(2n-1)(\frac{1}{2}\psi(n+\frac{1}{2})+\frac{\gamma}{2}+\ln(2)+1)}e^{n(\gamma_1(n+1)-\gamma_1)}\frac{\sqrt[n]{n!}^{2n-1}}{(2n)!^\frac{2n-1}{2n}}\prod_{k=1}^{n-1}B(k,k+1)^\frac{2n-1}{2k(2k-1)}}$$
So,
$$\textstyle\displaystyle{O=\frac{\sqrt{\pi}}{2\sqrt{2}}\sum_{n=1}^{\infty}\frac{1}{(n-1)!}\sqrt{\pi}^{(2n-1)H_n-4nH_{2n}+2n-\frac{1}{4n}}e^{n(\gamma_1(n+1)-\gamma_1)}2^{(2n-1)(\frac{1}{2}\psi(n+\frac{1}{2})+H_{2n-1}+\frac{\gamma}{2}+\ln(2)}\frac{\sqrt[n]{n!}^{2n-1}}{(2n)!^\frac{2n-1}{2n}}\prod_{k=1}^{n-1}B(k,k+1)^\frac{2n-1}{2k(2k-1)}}$$
So finally,
$$\textstyle\displaystyle{E=\sum_{n=1}^{\infty}\frac{(n-1)!}{(2n-1)!}\pi^{n(1+H_n-2H_{2n})}e^{n(\gamma_1(n+1)-\gamma_1)}2^{n(\psi(n+\frac{1}{2})-2(n+1)H_{2n}+\gamma+\ln(4)-2)}\prod_{k=1}^{n}B(k,k+1)^\frac{n}{k(2k-1)}}$$
$$\textstyle\displaystyle{O=\frac{\sqrt{\pi}}{2\sqrt{2}}\sum_{n=1}^{\infty}\frac{1}{(n-1)!}\sqrt{\pi}^{(2n-1)H_n-4nH_{2n}+2n-\frac{1}{4n}}e^{n(\gamma_1(n+1)-\gamma_1)}2^{(2n-1)(\frac{1}{2}\psi(n+\frac{1}{2})+H_{2n-1}+\frac{\gamma}{2}+\ln(2)}\frac{\sqrt[n]{n!}^{2n-1}}{(2n)!^\frac{2n-1}{2n}}\prod_{k=1}^{n-1}B(k,k+1)^\frac{2n-1}{2k(2k-1)}}$$
The real thing is that was this really a simplification or a complexification.
And please tell me if there are any mistakes.
• Comments are not for extended discussion; this conversation has been moved to chat. Sep 17, 2021 at 17:11
• Please see inverse symbolic entries. Sep 19, 2021 at 17:30
I am skeptical about a possible closed form of the summation which would converge extremely fast. Looking at the ratio of sucessive terms for $$4\leq n \leq 200$$, we have (from a quick and dirty nonlinear regression for which $$R^2 >0.999999$$) $$\log \left(\frac{R_{n+1}}{R_n}\right) \sim \alpha - \beta\,n^\gamma$$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ \alpha & 12.6408 & 0.1425 & \{12.3598,12.9218\} \\ \beta & 14.3159 & 0.1265 & \{14.0663,14.5655\} \\ \gamma & 0.14844 & 0.0008 & \{0.14682,0.15007\} \\ \end{array}$$
Notice that $$a_{10}=1.61\times 10^{-21}$$, $$a_{100}=7.87\times 10^{-522}$$, $$a_{1000}=3.14\times 10^{-9505}$$. So, we do not need to add many terms for a more than decent approximation. Adding the first hundred terms $$R=1.08171353471975721634480208063188816421005944622618865601394019148\cdots$$ which is not recognized by inverse symbolic calculators. However, with an error of $$1.35\times 10^{-14}$$, $$R$$ is close to the positive root of the quadratic $$7678 x^2-992 x-7911=0$$. | 5,595 | 13,114 | {"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": 144, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-18 | latest | en | 0.737102 |
https://delectablyfree.com/blog/stranded-wire-gauge-diameter/ | 1,680,132,523,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949035.66/warc/CC-MAIN-20230329213541-20230330003541-00264.warc.gz | 241,576,339 | 13,518 | # Stranded Wire Gauge Diameter
Stranded Wire Gauge Diameter. To calculate the gauge of stranded wire. you’ll need to double the diameter. For the 00. 000. 0000 etc.
measurement Physically determine unknown wires gauge from electronics.stackexchange.com
How do you calculate stranded wire gauge? Note that in awg the diameter goes up as the gauge goes down. First. measure the bare diameter of a single strand and locate the circular mils value in the row that matches your measurement.
slideshare.net
Accordingly. what is the diameter of 4 gauge. Sgx battery copper cable bright black 6 awg 0 162 diameter 25 length pack of 1 by small parts 45 electrical wire connectors electrical wiring cable.
pinterest.com
The higher the gauge number.. The gauge is related to the diameter of the wire.
slideshare.net
To specify a stranded wire. you have to mention three numbers. For example. a 22awg 7/30 indicates a stranded 22 awg wire with seven strands of 30awg wire.
Source: tonetastic.info
These include awg size. the awg size of the strand. and the number of strands. What is 18 gauge wire used for?
Source: tonetastic.info
For example. a 22awg 7/30 indicates a stranded 22 awg wire with seven strands of 30awg wire. These include awg size. the awg size of the strand. and the number of strands.
formsbirds.com
Diameter of individual wires (inches) overall conductor diameter (inches). Sgx battery copper cable bright black 6 awg 0 162 diameter 25 length pack of 1 by small parts 45 electrical wire connectors electrical wiring cable.
#### First. Measure The Bare Diameter Of A Single Strand And Locate The Circular Mils.
So. if you’ve measured a wire’s diameter to be 0.005 inches (0.127 mm). multiply this value by itself. What is 18 gauge wire used for? Turns of wire. no insulation:
#### Note That In Awg The Diameter Goes Up As The Gauge Goes Down.
Keeping this in view. what is the diameter of wire gauge? First. measure the bare diameter of a single strand and locate the circular mils value in the row that matches your measurement. Diameter information in the table applies to solid wires only.
#### Thus. The Diameter Of The Overall Bundle Is 13% More Than An Equal Gauge Solid Wire.
Metric wire gauges (see table below) metric gauge: For the 00. 000. 0000 etc. First. measure the bare diameter of a single strand and locate the circular mils value in the row that matches your measurement.
#### In The Metric Gauge Scale. The Gauge Is 10 Times The Diameter In Millimeters. Thus A 50 Gauge Metric Wire Would Be 5 Mm In Diameter.
To calculate the gauge of stranded wire. you’ll need to double the diameter. These include awg size. the awg size of the strand. and the number of strands. The awg table below is for a single. solid. round conductor.
#### 4.07 Mm / 0.160 Inches.
The n gauge wire diameter d n in inches (in) is equal to 0.005in times 92 raised to the power of 36 minus gauge number n. divided by 39: Second. multiply the circular mils by the number of strands in the cable. So. if you’ve measured a wire’s diameter to be 0.005 inches (0.127 mm). multiply this value by itself. | 755 | 3,119 | {"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} | 3.40625 | 3 | CC-MAIN-2023-14 | latest | en | 0.81665 |
http://www.slideserve.com/bertille/clustering | 1,508,334,018,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822966.64/warc/CC-MAIN-20171018123747-20171018143747-00212.warc.gz | 545,380,619 | 15,584 | Clustering
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# Clustering - PowerPoint PPT Presentation
Clustering. An overview of clustering algorithms Dènis de Keijzer GIA 2004. Overview. Algorithms GRAVIclust AUTOCLUST AUTOCLUST+ 3D Boundary-based Clustering SNN. Gravity based spatial clustering. GRAVIclust Initialisation Phase calculate the initial centre clusters
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## PowerPoint Slideshow about 'Clustering' - bertille
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Clustering
An overview of clustering algorithms
Dènis de Keijzer
GIA 2004
Overview
• Algorithms
• GRAVIclust
• AUTOCLUST
• AUTOCLUST+
• 3D Boundary-based Clustering
• SNN
Gravity based spatial clustering
• GRAVIclust
• Initialisation Phase
• calculate the initial centre clusters
• Optimisation Phase
• improve the position of the cluster centres so as to achieve a solution which minimizes the distance function
GRAVIclust: Initialisation Phase
• Input:
• set of points P
• matrix of distances between all pairs of points
• assumption: actual access path distance
• exists in GIS maps
• e.g.. http://www.transinfo.qld.gov.au
• very versatile
• footpath
• rail map
GRAVIclust: Initialisation Phase
• Input:
• set of points P
• matrix of distances between all pairs of points
• # of required clusters k
GRAVIclust: Initialisation Phase
• Step 1:
• calculate first initial centre
• the point with the largest number of points within radius r
• remove first initial centre & all points within radius r from further consideration
• Step 2:
• repeat Step 1 until k initial centres have been chosen
• Step 3:
• create initial clusters by assigning all points to the closest cluster centre
• calculated based on the area of the region considered for clustering
• based on the assumption that all clusters are of the same size
• recalculated after each initial cluster centre is chosen
GRAVIclust: Static vs. Dynamic
• Static
• reduced computation
• # points within a radius r has to be calculated only once
• not suitable for problems where the points are separated by large empty areas
• Dynamic
• increases computation time
• Differs only when distribution is non-uniform
GRAVIclust: Optimisation Phase
• Step 1:
• for each cluster, calculate new centre
• based on the the point closest to cluster centre of gravity
• Step 2:
• re-assign points to new cluster centres
• Step 3:
• recalculate distance function
• never greater than previous
• Step 4:
• repeat Step 1 to 3 until value distance function equals previous
GRAVIclust
• Deterministic
• Can handle obstacles
• Monotonic convergence of the distance function to a stable point
AUTOCLUST
• Definitions
AUTOCLUST
• Definitions II
AUTOCLUST
• Phase 1:
• finding boundaries
• Phase 2:
• restoring and re-attaching
• Phase 3:
• detecting second-order inconsistency
AUTOCLUST: Phase 1
• Finding boundaries
• Calculate
• Delaunay Diagram
• for each point pi
• ShortEdges(pi)
• LongEdges(pi)
• OtherEdges(pi)
• Remove
• ShortEdges(pi) and LongEdges(pi)
AUTOCLUST: Phase 2
• Restoring and re-attaching
• for each point pi where ShortEdges(pi)
• Determine a candidate connected component C for pi
• If there are 2 edges ej = (pi,pj) and ek = (pi,pk) in ShortEdges(pi) with CC[pj] CC[pk], then
• Compute, for each edge e = (pi,pj) ShortEdges(pi), the size ||CC[pj]|| and let M = maxe = (pi,pj) ShortEdges(pi) ||CC[pj]||
• Let C be the class labels of the largest connected component (if there are two different connected components with cardinality M, we let C be the one with the shortest edge to pi)
AUTOCLUST: Phase 2
• Restoring and re-attaching
• for each point pi where ShortEdges(pi)
• Determine a candidate connected component C for pi
• If …
• Otherwise, let C be the label of the connected component all edges e ShortEdges(pi) connect pi to
AUTOCLUST: Phase 2
• Restoring and re-attaching
• for each point pi where ShortEdges(pi)
• Determine a candidate connected component C for pi
• If the edges in OtherEdges(pi) connect to a connected component different than C, remove them. Note that
• all edges in OtherEdges(pi) are removed, and
• only in this case, will pi swap connected components
• Add all edges e ShortEdges(pi) that connect to C
AUTOCLUST: Phase 3
• Detecting second-order inconsistency
• compute the LocalMean for 2-neighbourhoods
• remove all edges in N2,G(pi) that are long edges
AUTOCLUST
• No user supplied arguments
• eliminates expensive human-based exploration time for finding best-fit arguments
• Robust to noise, outliers, bridges and type of distribution
• Able to detect clusters with arbitrary shapes, different sizes and different densities
• Can handle multiple bridges
• O(n log n)
AUTOCLUST+
• Construct Delaunay Diagram
• Calculate MeanStDev(P)
• For all edges e, remove e if it intersects some obstacles
• Apply the 3 phases of AUTOCLUST to the planar graph resulting from the previous steps
3D Boundary-based Clustering
• Benefits from 3D Clustering
• more accurate spatial analysis
• distinguish
• positive clusters:
• clusters in higher dimensions but not in lower dimensions
3D Boundary-based Clustering
• Benefits from 3D Clustering
• more accurate spatial analysis
• distinguish
• positive clusters:
• clusters in higher dimensions but not in lower dimensions
• negative clusters:
• clusters in lower dimensions but not in higher dimensions
3D Boundary-based Clustering
• Based on AUTOCLUST
• Uses Delaunay Tetrahedrizations
• Definitions:
• ej potential inter-cluster edge if:
3D Boundary-based Clustering
• Phase I
• For all the piP, classify each edge ej incident to pi into one of three groups
• ShortEdges(pi) when the length of ej is less than the range in AI(pi)
• LongEdges(pi) when the length of ej is greater than the range in AI(pi)
• OtherEdges(pi) when the length of ej is within AI(pi)
• For all the piP, remove all edges in ShortEdges(pi) and LongEdges(pi)
3D Boundary-based Clustering
• Phase II
• Recuperate ShortEdges(pi) incident to border points using connected component analysis
• Phase III
• Remove exceptionally long edges in local regions
Shared Nearest Neighbour
• Clustering in higher dimensions
• Distances or similarities between points become more uniform, making clustering more difficult
• Also, similarity between points can be misleading
• i.e.. a point can be more similar to a point that “actually” belongs to a different cluster
• Solution
• Shared nearest neighbor approach to similarity
SNN: An alternative definition of similarity
• Euclidian distance
• most common distance metric used
• while useful in low dimensions, it doesn’t work well in high dimensions
SNN: An alternative definition of similarity
• Define similarity in terms of their shared nearest neighbours
• the similarity of the points is “confirmed” by their common shared nearest neighbours
SNN: An alternative definition ofdensity
• SNN similarity, with the k-nearest neighbour approach
• if the k-nearest neighbour of a point, with respect to SNN similarity is close, then we say that there is a high density at this point
• since it reflects the local configuration of the points in the data space, it is relatively insensitive to variations in desitiy and the dimensionality of the space
SNN: Algorithm
• Compute the similarity matrix
• corresponds to a similarity graph with data points for nodes and edges whose weights are the similarities between data points
SNN: Algorithm
• Compute the similarity matrix
• Sparsify the similarity matrix by keeping only the k most similar neighbours
• corresponds to keeping only the k strongest links of the similarity graph
SNN: Algorithm
• Compute the similarity matrix
• Sparsify the similarity matrix …
• Construct the shared nearest neighbour graph from the sparsified similarity matrix
SNN: Algorithm
• Compute the similarity matrix
• Sparsify the similarity matrix …
• Construct the shared …
• Find the SNN density of each point
• Find the core points
SNN: Algorithm
• Compute the similarity matrix
• Sparsify the similarity matrix …
• Construct the shared …
• Find the SNN density of each point
SNN: Algorithm
• Compute the similarity matrix
• Sparsify the similarity matrix …
• Construct the shared …
• Find the SNN density of each point
• Form clusters from the core points
SNN: Algorithm
• Compute the similarity matrix
• Sparsify the similarity matrix …
• Construct the shared …
• Find the SNN density of each point
• Form clusters from the core points | 2,160 | 8,901 | {"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} | 3.046875 | 3 | CC-MAIN-2017-43 | latest | en | 0.703132 |
https://www.physicsforums.com/threads/gradient-vector.261928/ | 1,508,200,099,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187820487.5/warc/CC-MAIN-20171016233304-20171017013304-00703.warc.gz | 1,157,262,206 | 14,561 | 1. Oct 5, 2008
### 2RIP
From this diagram, http://www34.homepage.villanova.edu/robert.jantzen/courses/mat2500/handouts/s14-6-38.pdf how can i find the length of the vector at point (4,6)
Answer is length = (-2-(-3))/0.5 =2. I don't understand why we use -3 and -2 from the contour lines. The gradient vector at (4,6) does not go to -3. So why do we input -3 to find length? Or do we use the closest level lines to find the approximate rate of change?
Thanks
2. Oct 5, 2008
### Prologue
The gradient vector is found by changing the constant that the implicit function is equal to.
So for demonstrative purposes say that your implicit function looks like this (it isn't actually but just assume it is for now):
-2.5 = x - y - 0.5 = (4) - (6) - 0.5
So the constant at that point (4,6) is -2.5. Now, if you were to approximate what the value of the gradient would be, what would you do? Look at the definition of the magnitude gradient again. The magnitude of the gradient is found by looking at how much the magnitude of the constant changes compared to how much the magnitude of the inputs changed.
To estimate the magnitude of the gradient change at that spot, find a contour line close to it then jump over that spot (4,6) to the other contour line. In this way, you are looking at a change of 1 in the constant, and it is around the area of the point (estimations are ok at this point, these are not 'exact' gradients). You will always pick the contour line largest in value as your final measuring spot and the contour of lowest value is your starting point. This is because when finding the gradient we ALWAYS move in a positive direction, we change the constant to a MORE POSITIVE value. So initial -3, final -2. Now again, look at the contour lines that you just got the value of the constant from. Now you are going to measure the shortest physical distance between those lines and measured directly over the point (4,6) because this gives you the smallest change in the magnitude of the inputs to achieve the change in the constant, it gives you the change in sqrt(x+y). In this case it is about 0.5, that is your denominator.
So you use the definition of the magnitude of the gradient again to assemble this data:
(positive change in constant)/(distance it took to make that change in the constant)
or
(-2 - (-3))/(0.5)
Then draw the gradient vector starting at the point you evaluated the gradient at (4,6) and point it in the direction of a postive change in your constant. And now make it have a length/magnitude of 2 because this is what you found the magnitude of the gradient at the point (4,6) to be.
Good luck, this is a very difficult concept. Keep at trying to understand it in different ways. It always helped me to remember that it is the change in the CONSTANT that is the key. Everything about the gradient is rigged to make the change in the constant positive, and then the direction that achieves the largest change in the constant, given a certain movement in the input values, is the direction that the gradient is defined to point.
Last edited: Oct 5, 2008 | 747 | 3,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-43 | latest | en | 0.915549 |
https://socratic.org/questions/how-do-you-simplify-245x-31y-23-0 | 1,582,657,619,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146127.10/warc/CC-MAIN-20200225172036-20200225202036-00058.warc.gz | 551,509,602 | 5,918 | # How do you simplify (-245x^31y^23)^0 ?
$= 1$
We know ${a}^{0} = 1$
$\implies {\left(- 245 {x}^{31} {y}^{23}\right)}^{0} = 1$ | 63 | 127 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "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} | 3.65625 | 4 | CC-MAIN-2020-10 | longest | en | 0.541051 |
https://convertoctopus.com/180-ounces-to-kilograms | 1,632,619,803,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057787.63/warc/CC-MAIN-20210925232725-20210926022725-00073.warc.gz | 239,946,859 | 7,900 | ## Conversion formula
The conversion factor from ounces to kilograms is 0.028349523125, which means that 1 ounce is equal to 0.028349523125 kilograms:
1 oz = 0.028349523125 kg
To convert 180 ounces into kilograms we have to multiply 180 by the conversion factor in order to get the mass amount from ounces to kilograms. We can also form a simple proportion to calculate the result:
1 oz → 0.028349523125 kg
180 oz → M(kg)
Solve the above proportion to obtain the mass M in kilograms:
M(kg) = 180 oz × 0.028349523125 kg
M(kg) = 5.1029141625 kg
The final result is:
180 oz → 5.1029141625 kg
We conclude that 180 ounces is equivalent to 5.1029141625 kilograms:
180 ounces = 5.1029141625 kilograms
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 kilogram is equal to 0.19596645527545 × 180 ounces.
Another way is saying that 180 ounces is equal to 1 ÷ 0.19596645527545 kilograms.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one hundred eighty ounces is approximately five point one zero three kilograms:
180 oz ≅ 5.103 kg
An alternative is also that one kilogram is approximately zero point one nine six times one hundred eighty ounces.
## Conversion table
### ounces to kilograms chart
For quick reference purposes, below is the conversion table you can use to convert from ounces to kilograms
ounces (oz) kilograms (kg)
181 ounces 5.131 kilograms
182 ounces 5.16 kilograms
183 ounces 5.188 kilograms
184 ounces 5.216 kilograms
185 ounces 5.245 kilograms
186 ounces 5.273 kilograms
187 ounces 5.301 kilograms
188 ounces 5.33 kilograms
189 ounces 5.358 kilograms
190 ounces 5.386 kilograms | 454 | 1,760 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.717806 |
http://docplayer.net/34670902-M8a5b-solve-systems-of-equations-graphically-and-algebraically-using-technology-as-appropriate.html | 1,537,493,396,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156690.32/warc/CC-MAIN-20180920234305-20180921014705-00300.warc.gz | 67,781,540 | 23,200 | # M8A5b. Solve systems of equations graphically and algebraically, using technology as appropriate.
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## Transcription
1 Day #1 M8A5b. Solve systems of equations graphically and algebraically, using technology as appropriate. Essential Question: What is a system of equations and how is it solved using the Graphing Method? To begin this lesson, start with the GSP tab labeled Graphing Method. Have the students write down the definitions. Next, have the students graph system 1. Have them to figure out that the coordinates of the intersection are the unique solution to this system. The students will need to solve system 2 by graphing as well to figure out that there are no solutions. System 3 warrants a discussion about infinite solutions. Have the students answer the following 2 problems using the graphing method: Homework: Assign a range of 5-6 appropriate problems from course textbook.
2 Day #2 M8A5b. Solve systems of equations graphically and algebraically, using technology as appropriate. Essential Question: What is a system of equations and how is it solved using the Equivalent Forms Method? Go over the students homework assignment from the previous day and make sure that they understand the Graphing Method. Next refer to the GSP tab labeled Equivalent Forms Method. Have the students to solve the sample system using the steps given. Students will then work on the following problems: Homework: Assign a range of 8-10 appropriate problems from the course textbook.
3 Day #3 M8A5b. Solve systems of equations graphically and algebraically, using technology as appropriate. Essential Question: What is a system of equations and how is it solved using the Substitution Method? Go over the students homework assignment from the previous day and make sure that they understand the Equivalent Forms method. Next refer to the GSP tab labeled Substitution Method. Work with the students to solve the sample systems using the steps given. This method requires precise explanations so be prepared. Students will then work on the following problems: Homework: Have the students to complete the in-class assignment.
4 Day #4 M8A5b. Solve systems of equations graphically and algebraically, using technology as appropriate. Essential Question: What is a system of equations and how is it solved using the Combination Method? Go over the students homework assignment from the previous day and make sure that they understand the Substitution method. Next refer to the GSP tab labeled Combination Method. Emphasize that this method is referred to by many names including the Addition method and the Elimination method. Work with the students to solve the sample systems using the steps given. This method requires clear and precise explanations. Work through the following problems with the students: Next have the students complete the in-class assignment below. Homework Assignment:
5 Day #5 M8A5b. Solve systems of equations graphically and algebraically, using technology as appropriate. Essential Question: How can I apply what I have learned solving systems of equations? Go over the students homework assignment from the previous day and make sure that they understand the Combination method. Also review the other methods to confirm that the students have a good understanding of solving systems of equations. Next have the students to form partner pairs and work on the assignment labeled Solving Systems of Equations Practice. See Solving Systems of Equations Practice Homework: Review the week s notes and homework assignments.
6 Day #6 M8A5a. Given a problem context, write an appropriate system of linear equations or inequalities. Essential Question: Given a problem in context, how can I apply what I have learned about solving systems of equations? Review the practice worksheet with the students and make sure that they truly understand all the methods of solving systems of equations. Next give them the following activity so they can apply what they have learned about using the Graphing method. See Day #6 activity worksheet Homework: Finish classwork.
7 Day #7 M8A5a. Given a problem context, write an appropriate system of linear equations or inequalities. Essential Question: Given a problem, how can I apply what I have learned about solving systems of equations? Review activity from the previous day. Next give them the following activity so they can apply what they have learned about using the Equivalent Forms method. See Day #7 activity worksheet Homework:
8 Day #8 M8A5a. Given a problem context, write an appropriate system of linear equations or inequalities. Essential Question: Given a problem, how can I apply what I have learned about solving systems of equations? Review activity from the previous day. Next give them the following activity so they can apply what they have learned about using the Substitution method. See Day #8 activity worksheet Homework:
9 Day #9 M8A5a. Given a problem context, write an appropriate system of linear equations or inequalities. Essential Question: Given a problem in context, how can I apply what I have learned about solving systems of equations? Review activity from the previous day. Next give them the following activity so they can apply what they have learned about using the Combination method. See Day #9 activity worksheet Homework:
10 Day #10 M8A5. Students will understand systems of linear equations and use them to solve problems. Essential Question: Am I able to successfully complete this assessment to show my understanding of solving systems of equations? Students will complete a unit test over systems of equations. See Systems of Linear Equations Unit Test Homework: None
### Lesson 22: Solution Sets to Simultaneous Equations
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Course 2000020 Advanced Life Science 7th Grade Curriculum Extension Laboratory investigations which include the use of scientific inquiry, research, measurement, problem solving, laboratory apparatus and | 7,661 | 32,550 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2018-39 | longest | en | 0.923532 |
http://www.google.com/patents/US7738584?dq=5179747 | 1,488,392,978,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501174215.11/warc/CC-MAIN-20170219104614-00432-ip-10-171-10-108.ec2.internal.warc.gz | 424,999,041 | 35,522 | ## Patents
Publication number US7738584 B2 Publication type Grant Application number US 11/168,838 Publication date Jun 15, 2010 Filing date Jun 28, 2005 Priority date Apr 21, 2005 Fee status Paid Also published as US20060239373 Publication number 11168838, 168838, US 7738584 B2, US 7738584B2, US-B2-7738584, US7738584 B2, US7738584B2 Inventors Joonsuk Kim Original Assignee Broadcom Company Export Citation Patent Citations (10), Classifications (11), Legal Events (4) External Links:
Beamforming in a wireless communication with a partial estimation to reduce overhead
US 7738584 B2
Abstract
A method for beamforming in a wireless communication begins by receiving a baseband signal. The method continues by receiving a feedback signal that includes a subset of angles, wherein a set of angles provide polar coordinates for a unitary matrix and wherein the subset of angles is a subset of the set of angles. The method continues by determining at least one remaining angle of the set of angles based on the subset of angles. The method continues by determining the polar coordinates for the unitary matrix. The method continues by digitally beamforming the baseband signal using the unitary matrix.
Images(9)
Claims(11)
1. A method for beamforming a wireless communication, the method comprises:
receiving a baseband signal by a transmitter that includes a plurality of streams;
receiving a feedback signal by the transmitter from a receiver of the wireless communication that includes a subset of anglesψ1 and Φ1, wherein a set of angles ψ1 and Φ1, ψ2, and Φ2 provide polar coordinates for a unitary 2×2 matrix V, wherein VV*=I and I represents an identity matrix;
determining the angles ψ2 and Φ2 based on the subset of angles ψ1 and Φ1, wherein ψ2 and Φ2 are determined by solving absolute value of ψ1−ψ2=/2 and Φ12 or Φ12+π and ψ12=π/2;
determining the polar coordinates for the unitary matrix based on the set of angles ψ1 and Φ1, ψ2, and Φ2; and
digitally beamforming the plurality of streams of the baseband signal using the unitary matrix to produce a plurality of beamformed symbols.
2. The method of claim 1 wherein receiving the baseband signal that includes a plurality of streams includes:
encoding data to produce a stream of encoded data;
interleaving the stream of encoded data into a plurality of parallel streams of interleaved data;
constellation mapping symbols of each of the plurality of parallel streams of interleaved data to a plurality of parallel tones to produce the plurality of streams of the baseband signal; and
wherein digitally beamforming the plurality of streams of the baseband signal includes digitally beamforming each of the plurality of parallel tones using the unitary matrix to produce a plurality of beamformed symbols.
3. The method of claim 1, wherein the unitary matrix comprises:
a plurality of polar coordinates as represented by V, wherein absolute value of each of the plurality of polar coordinates is a vector on a unit circle and each of the polar coordinates is orthogonal to at least one other of the polar coordinates such that VV*=I, where I represents an identity matrix.
4. The method of claim 3, wherein the unitary matrix further comprises for a 2×N, wherein N is equal to the number of receiver antennas, multiple input multiple output (MIMO) wireless communication:
$V = [ cos ψ 1 cos ψ 2 sin ψ 1 ⅇ jϕ 1 sin ψ 2 ⅇ jϕ 2 ]$
wherein ψ1, Φ1, ψ2, and Φ2 represent the set of angles, wherein angles ψ1 and Φ1 are included in the subset of angles, wherein, ψ2, and Φ2 are determined by solving absolute value of ψ1−ψ2=π/2 and Φ12 or Φ12+π and ψ12=π/2.
5. The method of claim 3, wherein when the unitary matrix comprises a 3×N, wherein N is equal to the number of receiver antennas, multiple input multiple output (MIMO) wireless communication:
$V = [ cos ψ 1 cos ψ 2 cos ψ 3 sin ψ 1 cos θ 1 ⅇ jϕ 21 sin ψ 2 cos θ 2 ⅇ jϕ 22 sin ψ 3 cos θ 3 ⅇ jϕ 23 sin ψ 1 sin θ 1 ⅇ jϕ 31 sin ψ 2 sin θ 2 ⅇ jϕ 32 sin ψ 3 sin θ 3 ⅇ jϕ 33 ]$
wherein ψ1, ψ2, ψ3, θ1, θ2, θ3, Φ21, Φ22, Φ23, Φ31, Φ32, Φ33 represent the set of angles, wherein ψ1, ψ2, θ1, θ2, Φ21, Φ22, and Φ31, are included in the subset of angles, wherein Diagonal (VV*)=1 s, and wherein ψ3, θ3, Φ21, Φ23, Φ32, and Φ33 are determined by:
$ψ 3 = cos - 1 ( 1 - cos 2 ψ 1 - cos 2 ψ 2 )$ $θ 3 = cos - 1 ( 1 - sin 2 ψ 1 cos 2 θ 1 - sin 2 ψ 2 cos 2 θ 2 sin ψ 3 )$ $ⅇ jϕ 23 = - cos ψ 1 sin ψ 1 cos θ 1 ⅇ jϕ 21 + cos ψ 2 sin ψ 2 cos θ 2 ⅇ jϕ 22 cos ψ 3 sin ψ 3 cos θ 3$ $ⅇ jϕ 32 = - cos ψ 1 cos ψ 2 + sin ψ 1 cos θ 1 sin ψ 2 cos θ 2 ⅇ j ( ϕ 22 - ϕ 21 ) sin ψ 1 sin θ 1 sin ψ 2 sin θ 2 ⅇ jϕ 31$ $ⅇ jϕ 32 = - cos ψ 1 cos ψ 3 + sin ψ 1 cos θ 1 sin ψ 3 cos θ 3 ⅇ j ( ϕ 23 - ϕ 21 ) sin ψ 1 sin θ 1 sin ψ 3 sin θ 3 ⅇ jϕ 31 .$
6. The method of claim 3, wherein when the unitary matrix comprises a 4×N, wherein N is equal to the number of receiver antennas, multiple input multiple input multiple output (MIMO) wireless communication:
$V = [ cos ψ 1 cos φ 1 cos ψ 2 cos φ 2 cos ψ 3 cos φ 3 cos ψ 4 cos φ 4 cos ψ 1 sin φ 1 ⅇ jϕ 11 cos ψ 2 sin φ 2 ⅇ jϕ 12 cos ψ 3 sin φ 3 ⅇ jϕ 13 cos ψ 4 sin φ 4 ⅇ jϕ 14 sin ψ 1 cos θ 1 ⅇ jϕ 21 sin ψ 2 cos θ 2 ⅇ jϕ 22 sin ψ 3 cos θ 3 ⅇ jϕ 23 sin ψ 4 cos θ 4 ⅇ jϕ 24 sin ψ 1 sin θ 1 ⅇ jϕ 31 sin ψ 2 sin θ 2 ⅇ jϕ 32 sin ψ 3 sin θ 3 ⅇ jϕ 33 sin ψ 4 sin θ 4 ⅇ jϕ 34 ]$
wherein ψ1, ψ2, ψ3, ψ4, θ1, θ2, θ3, θ4, φ1, φ2, φ3, φ4, Φ11, Φ12, Φ13, Φ14, Φ21, Φ22, Φ23, Φ24, Φ31, Φ32, Φ33, Φ34 represent the set of angles, wherein ψ1, ψ2, ψ3, θ1, θ2, θ3, φ1, φ2, φ3, Φ11, Φ12, Φ13, Φ21, Φ22, Φ23, Φ31, are the subset of angles, wherein ψ4, θ4, φ4, Φ14, Φ24, Φ31, Φ32, Φ33, Φ34 are determined by:
$ψ 4 = cos - 1 ( 2 - cos 2 ψ 1 - cos 2 ψ 2 - cos 2 ψ 3 )$ $φ 4 = cos - 1 ( 1 - cos 2 ψ 1 cos 2 φ 1 - cos 2 ψ 2 cos 2 φ 2 - cos 2 ψ 3 cos 2 φ 3 cos ψ 4 )$ $θ 4 = cos - 1 ( 1 - sin 2 ψ 1 cos 2 θ 1 - sin 2 ψ 2 cos 2 θ 2 - sin 2 ψ 3 cos 2 θ 3 sin ψ 4 )$ $ⅇ jϕ 14 = - cos 2 ψ 1 sin φ 1 cos φ 1 ⅇ jϕ 11 + cos 2 ψ 2 sin φ 2 cos φ 2 ⅇ jϕ 12 + cos 2 ψ 3 sin φ 3 cos φ 3 ⅇ jϕ 13 cos 2 ψ 4 sin φ 4 cos φ 4$ $ⅇ jϕ 24 = - cos ψ 1 sin ψ 1 cos φ 1 cos θ 1 ⅇ jϕ 21 + cos ψ 2 sin ψ 2 cos φ 2 cos θ 2 ⅇ jϕ 22 + cos ψ 3 sin ψ 3 cos φ 3 cos θ 3 ⅇ jϕ 23 cos ψ 4 sin ψ 4 cos φ 4 cos θ 4$ $ⅇ jϕ 32 = - cos ψ 1 cos ψ 2 cos φ 1 cos φ 2 + cos ψ 1 cos ψ 2 sin φ 1 sin φ 2 ⅇ j ( ϕ 12 - ϕ 11 ) + sin ψ 1 cos θ 1 sin ψ 2 cos θ 2 ⅇ j ( ϕ 22 - ϕ 21 ) sin ψ 1 sin θ 1 sin ψ 2 sin θ 2 ⅇ jϕ 31$ $ⅇ jϕ 33 = - cos ψ 1 cos ψ 3 cos φ 1 cos φ 3 + cos ψ 1 cos ψ 3 sin φ 1 sin φ 3 ⅇ j ( ϕ 13 - ϕ 11 ) + sin ψ 1 cos θ 1 sin ψ 3 cos θ 3 ⅇ j ( ϕ 23 - ϕ 21 ) sin ψ 1 sin θ 1 sin ψ 3 sin θ 3 ⅇ jϕ 31$ $ⅇ jϕ 34 = - cos ψ 1 cos ψ 4 cos φ 1 cos φ 4 + cos ψ 1 cos ψ 4 sin φ 1 sin φ 4 ⅇ j ( ϕ 14 - ϕ 11 ) + sin ψ 1 cos θ 1 sin ψ 4 cos θ 4 ⅇ j ( ϕ 24 - ϕ 21 ) sin ψ 1 sin θ 1 sin ψ 4 sin θ 4 ⅇ jϕ 31 .$
7. A transmit baseband processing module comprises:
an encoding module operably coupled to convert outbound data into encoded data;
a plurality of interleaving modules operably coupled to interleave the encoded data into a plurality of interleaved streams of data;
a plurality of constellation mapping modules operably coupled to map the plurality of interleaved streams of data into a plurality of streams of data symbols;
a beamforming module operably coupled to:
obtain a feedback signal that includes a subset of angles ψ1 and Φ1, wherein a set of angles ψ1 and Φ1, ψ2, and Φ2 provide polar coordinates for a unitary 2×2 matrix V, wherein VV*=I and I represents an identity matrix;
determine the angles ψ2 and Φ2 based on the subset of angles ψ1 and Φ1, wherein ψ2 and Φ2 are determined by solving absolute value of ψ1−ψ2=π/2 and Φ12 or Φ12+π and ψ12=π/2;
determine the polar coordinates for the unitary matrix based on the set of angles ψ1 and Φ1, ψ2, and Φ2; and
digitally beamform, using the unitary matrix having polar coordinates, the plurality of streams of data symbols into a plurality of streams of beamformed symbols; and
a plurality of inverse fast Fourier transform modules operably coupled to convert the plurality of streams of beamformed symbols into a plurality of outbound symbol streams.
8. The transmit baseband processing module of claim 7, wherein the unitary matrix comprises:
a plurality of polar coordinates as represented by V, wherein absolute value of each of the plurality of polar coordinates is a vector on a unit circle and each of the polar coordinates is orthogonal to at least one other of the polar coordinates such that VV*=I, where I represents an identity matrix.
9. The transmit baseband processing module of claim 8, wherein the unitary matrix further comprises for a 2×N, wherein N is equal to the number of receiver antennas, multiple input multiple output (MIMO) wireless communication:
$V = [ cos ψ 1 cos ψ 2 sin ψ 1 ⅇ jϕ 1 sin ψ 2 ⅇ jϕ 2 ]$
wherein ψ1, Φ1, ψ2, and Φ2 represent the set of angles, wherein angles ψ1 and Φ1 are included in the subset of angles, wherein, ψ2, and Φ2 are determined by solving absolute value of ψ1−ψ2=π/2 and Φ12 or Φ12+π and ψ12=π/2.
10. The transmit baseband processing module of claim 8, wherein the unitary matrix further comprises for a 3×N, wherein N is equal to the number of receiver antennas, multiple input multiple output (MIMO) wireless communication:
$V = [ cos ψ 1 cos ψ 2 cos ψ 3 sin ψ 1 cos θ 1 ⅇ jϕ 21 sin ψ 2 cos θ 2 ⅇ jϕ 22 sin ψ 3 cos θ 3 ⅇ jϕ 23 sin ψ 1 sin θ 1 ⅇ jϕ 31 sin ψ 2 sin θ 2 ⅇ jϕ 32 sin ψ 3 sin θ 3 ⅇ jϕ 33 ]$
wherein ψ1, ψ2, ψ3, θ1, θ2, θ3, Φ21, Φ22, Φ23, Φ31, Φ32, Φ33 represent the set of angles, wherein ψ1, ψ2, θ1, θ2, Φ21, Φ22, and Φ31, are included in the subset of angles, wherein (VV*)=1 s, and wherein ψ3, θ3, Φ21, Φ23, Φ32, and Φ33 are determined by:
$ψ 3 = cos - 1 ( 1 - cos 2 ψ 1 - cos 2 ψ 2 )$ $θ 3 = cos - 1 ( 1 - sin 2 ψ 1 cos 2 θ 1 - sin 2 ψ 2 cos 2 θ 2 sin ψ 3 )$ $ⅇ jϕ 23 = - cos ψ 1 sin ψ 1 cos θ 1 ⅇ jϕ 21 + cos ψ 2 sin ψ 2 cos θ 2 ⅇ jϕ 22 cos ψ 3 sin ψ 3 cos θ 3$ $ⅇ jϕ 32 = - cos ψ 1 cos ψ 2 + sin ψ 1 cos θ 1 sin ψ 2 cos θ 2 ⅇ j ( ϕ 22 - ϕ 21 ) sin ψ 1 sin θ 1 sin ψ 2 sin θ 2 ⅇ jϕ 31$ $ⅇ jϕ 32 = - cos ψ 1 cos ψ 3 + sin ψ 1 cos θ 1 sin ψ 3 cos θ 3 ⅇ j ( ϕ 23 - ϕ 21 ) sin ψ 1 sin θ 1 sin ψ 3 sin θ 3 ⅇ jϕ 31 .$
11. The transmit baseband processing module of claim 8, wherein the unitary matrix further comprises for a 4×N, wherein N is equal to the number of receiver antennas, multiple input multiple output (MIMO) wireless communication:
$V = [ cos ψ 1 cos φ 1 cos ψ 2 cos φ 2 cos ψ 3 cos φ 3 cos ψ 4 cos φ 4 cos ψ 1 sin φ 1 ⅇ jϕ 11 cos ψ 2 sin φ 2 ⅇ jϕ 12 cos ψ 3 sin φ 3 ⅇ jϕ 13 cos ψ 4 sin φ 4 ⅇ jϕ 14 sin ψ 1 cos θ 1 ⅇ jϕ 21 sin ψ 2 cos θ 2 ⅇ jϕ 22 sin ψ 3 cos θ 3 ⅇ jϕ 23 sin ψ 4 cos θ 4 ⅇ jϕ 24 sin ψ 1 sin θ 1 ⅇ jϕ 31 sin ψ 2 sin θ 2 ⅇ jϕ 32 sin ψ 3 sin θ 3 ⅇ jϕ 33 sin ψ 4 sin θ 4 ⅇ jϕ 34 ]$
wherein ψ1, ψ2, ψ3, ψ4, θ1, θ2, θ3, θ4, φ1, φ2, φ3, φ4, Φ11, Φ12, Φ13, Φ14, Φ21, Φ22, Φ23, Φ24, Φ31, Φ32, θ33, Φ34 represent the set of angles, wherein ψ1, ψ2, ψ3, θ1, θ2, θ3, φ1, φ2, φ3, Φ11, Φ12, Φ13, Φ21, Φ22, Φ23, Φ31, are the subset of angles, wherein ψ4, θ4, φ4, Φ14, Φ24, Φ31, Φ32, Φ33, Φ34 are determined by:
$ψ 4 = cos - 1 ( 2 - cos 2 ψ 1 - cos 2 ψ 2 - cos 2 ψ 3 )$ $φ 4 = cos - 1 ( 1 - cos 2 ψ 1 cos 2 φ 1 - cos 2 ψ 2 cos 2 φ 2 - cos 2 ψ 3 cos 2 φ 3 cos ψ 4 )$ $θ 4 = cos - 1 ( 1 - sin 2 ψ 1 cos 2 θ 1 - sin 2 ψ 2 cos 2 θ 2 - sin 2 ψ 3 cos 2 θ 3 sin ψ 4 )$ $ⅇ jϕ 14 = - cos 2 ψ 1 sin φ 1 cos φ 1 ⅇ jϕ 11 + cos 2 ψ 2 sin φ 2 cos φ 2 ⅇ jϕ 12 + cos 2 ψ 3 sin φ 3 cos φ 3 ⅇ jϕ 13 cos 2 ψ 4 sin φ 4 cos φ 4$ $ⅇ jϕ 24 = - cos ψ 1 sin ψ 1 cos φ 1 cos θ 1 ⅇ jϕ 21 + cos ψ 2 sin ψ 2 cos φ 2 cos θ 2 ⅇ jϕ 22 + cos ψ 3 sin ψ 3 cos φ 3 cos θ 3 ⅇ jϕ 23 cos ψ 4 sin ψ 4 cos φ 4 cos θ 4$ $ⅇ jϕ 32 = - cos ψ 1 cos ψ 2 cos φ 1 cos φ 2 + cos ψ 1 cos ψ 2 sin φ 1 sin φ 2 ⅇ j ( ϕ 12 - ϕ 11 ) + sin ψ 1 cos θ 1 sin ψ 2 cos θ 2 ⅇ j ( ϕ 22 - ϕ 21 ) sin ψ 1 sin θ 1 sin ψ 2 sin θ 2 ⅇ jϕ 31$ $ⅇ jϕ 33 = - cos ψ 1 cos ψ 3 cos φ 1 cos φ 3 + cos ψ 1 cos ψ 3 sin φ 1 sin φ 3 ⅇ j ( ϕ 13 - ϕ 11 ) + sin ψ 1 cos θ 1 sin ψ 3 cos θ 3 ⅇ j ( ϕ 23 - ϕ 21 ) sin ψ 1 sin θ 1 sin ψ 3 sin θ 3 ⅇ jϕ 31$ $ⅇ jϕ 34 = - cos ψ 1 cos ψ 4 cos φ 1 cos φ 4 + cos ψ 1 cos ψ 4 sin φ 1 sin φ 4 ⅇ j ( ϕ 14 - ϕ 11 ) + sin ψ 1 cos θ 1 sin ψ 4 cos θ 4 ⅇ j ( ϕ 24 - ϕ 21 ) sin ψ 1 sin θ 1 sin ψ 4 sin θ 4 ⅇ jϕ 31 .$
Description
This patent application is claiming priority under 35 USC §119(e) to provisional patent application entitled “Reduced Feedback For Beamforming in a Wireless Communication”, having a provisional filing date of Apr. 21, 2005, and an application number of No. 60/673,451. This patent application also claims priority to a provisionally filed patent application entitled “Beamforming in a Wireless Communication”, having a provisional filing date of Apr. 26, 2005, and an application number of No. 60/674,822.
BACKGROUND OF THE INVENTION
1. Technical Field of the Invention
This invention relates generally to wireless communication systems and more particularly to wireless communications using beamforming.
2. Description of Related Art
Communication systems are known to support wireless and wire lined communications between wireless and/or wire lined communication devices. Such communication systems range from national and/or international cellular telephone systems to the Internet to point-to-point in-home wireless networks. Each type of communication system is constructed, and hence operates, in accordance with one or more communication standards. For instance, wireless communication systems may operate in accordance with one or more standards including, but not limited to, IEEE 802.11, Bluetooth, advanced mobile phone services (AMPS), digital AMPS, global system for mobile communications (GSM), code division multiple access (CDMA), local multi-point distribution systems (LMDS), multi-channel-multi-point distribution systems (MMDS), and/or variations thereof.
Depending on the type of wireless communication system, a wireless communication device, such as a cellular telephone, two-way radio, personal digital assistant (PDA), personal computer (PC), laptop computer, home entertainment equipment, et cetera communicates directly or indirectly with other wireless communication devices. For direct communications (also known as point-to-point communications), the participating wireless communication devices tune their receivers and transmitters to the same channel or channels (e.g., one of the plurality of radio frequency (RF) carriers of the wireless communication system) and communicate over that channel(s). For indirect wireless communications, each wireless communication device communicates directly with an associated base station (e.g., for cellular services) and/or an associated access point (e.g., for an in-home or in-building wireless network) via an assigned channel. To complete a communication connection between the wireless communication devices, the associated base stations and/or associated access points communicate with each other directly, via a system controller, via the public switch telephone network, via the Internet, and/or via some other wide area network.
For each wireless communication device to participate in wireless communications, it includes a built-in radio transceiver (i.e., receiver and transmitter) or is coupled to an associated radio transceiver (e.g., a station for in-home and/or in-building wireless communication networks, RF modem, etc.). As is known, the receiver is coupled to the antenna and includes a low noise amplifier, one or more intermediate frequency stages, a filtering stage, and a data recovery stage. The low noise amplifier receives inbound RF signals via the antenna and amplifies then. The one or more intermediate frequency stages mix the amplified RF signals with one or more local oscillations to convert the amplified RF signal into baseband signals or intermediate frequency (IF) signals. The filtering stage filters the baseband signals or the IF signals to attenuate unwanted out of band signals to produce filtered signals. The data recovery stage recovers raw data from the filtered signals in accordance with the particular wireless communication standard.
As is also known, the transmitter includes a data modulation stage, one or more intermediate frequency stages, and a power amplifier. The data modulation stage converts raw data into baseband signals in accordance with a particular wireless communication standard. The one or more intermediate frequency stages mix the baseband signals with one or more local oscillations to produce RF signals. The power amplifier amplifies the RF signals prior to transmission via an antenna.
In many systems, the transmitter will include one antenna for transmitting the RF signals, which are received by a single antenna, or multiple antennas, of a receiver. When the receiver includes two or more antennas, the receiver will select one of them to receive the incoming RF signals. In this instance, the wireless communication between the transmitter and receiver is a single-output-single-input (SISO) communication, even if the receiver includes multiple antennas that are used as diversity antennas (i.e., selecting one of them to receive the incoming RF signals). For SISO wireless communications, a transceiver includes one transmitter and one receiver. Currently, most wireless local area networks (WLAN) that are IEEE 802.11, 802.11a, 802,11b, or 802.11g employ SISO wireless communications.
Other types of wireless communications include single-input-multiple-output (SIMO), multiple-input-single-output (MISO), and multiple-input-multiple-output (MIMO). In a SIMO wireless communication, a single transmitter processes data into radio frequency signals that are transmitted to a receiver. The receiver includes two or more antennas and two or more receiver paths. Each of the antennas receives the RF signals and provides them to a corresponding receiver path (e.g., LNA, down conversion module, filters, and ADCs). Each of the receiver paths processes the received RF signals to produce digital signals, which are combined and then processed to recapture the transmitted data.
For a multiple-input-single-output (MISO) wireless communication, the transmitter includes two or more transmission paths (e.g., digital to analog converter, filters, up-conversion module, and a power amplifier) that each converts a corresponding portion of baseband signals into RF signals, which are transmitted via corresponding antennas to a receiver. The receiver includes a single receiver path that receives the multiple RF signals from the transmitter. In this instance, the receiver uses beam forming to combine the multiple RF signals into one signal for processing.
For a multiple-input-multiple-output (MIMO) wireless communication, the transmitter and receiver each include multiple paths. In such a communication, the transmitter parallel processes data using a spatial and time encoding function to produce two or more streams of data. The transmitter includes multiple transmission paths to convert each stream of data into multiple RF signals. The receiver receives the multiple RF signals via multiple receiver paths that recapture the streams of data utilizing a spatial and time decoding function. The recaptured streams of data are combined and subsequently processed to recover the original data.
To further improve wireless communications, transceivers may incorporate beamforming. In general, beamforming is a processing technique to create a focused antenna beam by shifting a signal in time or in phase to provide gain of the signal in a desired direction and to attenuate the signal in other directions. In order for a transmitter to properly implement beamforming, it needs to know properties of the channel over which the wireless communication is conveyed. Accordingly, the receiver must provide feedback information for the transmitter to determine the properties of the channel. The feedback information may be sent as a receiver determined beamforming matrix (V) if a singular value decomposition can be determined or it may be sent as a channel matrix (H). In either case, the feedback information is substantial in size and includes Cartesian coordinates for the matrix values. Such Cartesian coordinates leads to unevenly spaced angles. Prior art papers (1) Digital beamforming basics. (antennas) by Steyskal, Hans, Journal of Electronic Defense, Jul. 1, 1996; (2) Utilizing Digital Downconverters for Efficient Digital Beamforming, by Clint Schreiner, Red River Engineering, no publication date; and (3) Interpolation Based Transmit Beamforming for MIMO-OFMD with Partial Feedback, by Jihoon Choi and Robert W. Heath, University of Texas, Department of Electrical and Computer Engineering, Wireless Networking and Communications Group, Sep. 13, 2003.
Therefore, a need exists for a method and apparatus for reducing beamforming feedback information for wireless communications.
BRIEF SUMMARY OF THE INVENTION
The present invention is directed to apparatus and methods of operation that are further described in the following Brief Description of the Drawings, the Detailed Description of the Invention, and the claims. Other features and advantages of the present invention will become apparent from the following detailed description of the invention made with reference to the accompanying drawings.
BRIEF DESCRIPTION OF THE SEVERAL VIEWS OF THE DRAWINGS
FIG. 1 is a schematic block diagram of a wireless communication system in accordance with the present invention;
FIG. 2 is a schematic block diagram of a wireless communication device in accordance with the present invention;
FIG. 3 is a schematic block diagram of another wireless communication device in accordance with the present invention;
FIG. 4 is a schematic block diagram of baseband transmit processing in accordance with the present invention;
FIG. 5 is a schematic block diagram of baseband receive processing in accordance with the present invention;
FIG. 6 is a schematic block diagram of a beamforming wireless communication in accordance with the present invention;
FIG. 7 is a diagram of simulation results in accordance with the present invention; and
FIG. 8 is a diagram of additional simulation results in accordance with the present invention.
DETAILED DESCRIPTION OF THE INVENTION
FIG. 1 is a schematic block diagram illustrating a communication system 10 that includes a plurality of base stations and/or access points 12, 16, a plurality of wireless communication devices 18-32 and a network hardware component 34. Note that the network hardware 34, which may be a router, switch, bridge, modem, system controller, et cetera provides a wide area network connection 42 for the communication system 10. Further note that the wireless communication devices 18-32 may be laptop host computers 18 and 26, personal digital assistant hosts 20 and 30, personal computer hosts 24 and 32 and/or cellular telephone hosts 22 and 28. The details of the wireless communication devices will be described in greater detail with reference to FIG. 2.
Wireless communication devices 22, 23, and 24 are located within an independent basic service set (IBSS) area and communicate directly (i.e., point to point). In this configuration, these devices 22, 23, and 24 may only communicate with each other. To communicate with other wireless communication devices within the system 10 or to communicate outside of the system 10, the devices 22, 23, and/or 24 need to affiliate with one of the base stations or access points 12 or 16.
The base stations or access points 12, 16 are located within basic service set (BSS) areas 11 and 13, respectively, and are operably coupled to the network hardware 34 via local area network connections 36, 38. Such a connection provides the base station or access point 12 16 with connectivity to other devices within the system 10 and provides connectivity to other networks via the WAN connection 42. To communicate with the wireless communication devices within its BSS 11 or 13, each of the base stations or access points 12-16 has an associated antenna or antenna array. For instance, base station or access point 12 wirelessly communicates with wireless communication devices 18 and 20 while base station or access point 16 wirelessly communicates with wireless communication devices 26-32. Typically, the wireless communication devices register with a particular base station or access point 12, 16 to receive services from the communication system 10.
Typically, base stations are used for cellular telephone systems and like-type systems, while access points are used for in-home or in-building wireless networks (e.g., IEEE 802.11 and versions thereof, Bluetooth, and/or any other type of radio frequency based network protocol). Regardless of the particular type of communication system, each wireless communication device includes a built-in radio and/or is coupled to a radio.
FIG. 2 is a schematic block diagram illustrating a wireless communication device that includes the host device 18-32 and an associated radio 60. For cellular telephone hosts, the radio 60 is a built-in component. For personal digital assistants hosts, laptop hosts, and/or personal computer hosts, the radio 60 may be built-in or an externally coupled component.
As illustrated, the host device 18-32 includes a processing module 50, memory 52, a radio interface 54, an input interface 58, and an output interface 56. The processing module 50 and memory 52 execute the corresponding instructions that are typically done by the host device. For example, for a cellular telephone host device, the processing module 50 performs the corresponding communication functions in accordance with a particular cellular telephone standard.
Radio 60 includes a host interface 62, digital receiver processing module 64, an analog-to-digital converter 66, a high pass and low pass filter module 68, an IF mixing down conversion stage 70, a receiver filter 71, a low noise amplifier 72, a transmitter/receiver switch 73, a local oscillation module 74, memory 75, a digital transmitter processing module 76, a digital-to-analog converter 78, a filtering/gain module 80, an IF mixing up conversion stage 82, a power amplifier 84, a transmitter filter module 85, a channel bandwidth adjust module 87, and an antenna 86. The antenna 86 may be a single antenna that is shared by the transmit and receive paths as regulated by the Tx/Rx switch 73, or may include separate antennas for the transmit path and receive path. The antenna implementation will depend on the particular standard to which the wireless communication device is compliant.
The digital receiver processing module 64 and the digital transmitter processing module 76, in combination with operational instructions stored in memory 75, execute digital receiver functions and digital transmitter functions, respectively. The digital receiver functions include, but are not limited to, digital intermediate frequency to baseband conversion, demodulation, constellation demapping, decoding, and/or descrambling. The digital transmitter functions include, but are not limited to, scrambling, encoding, constellation mapping, modulation, and/or digital baseband to IF conversion. The digital receiver and transmitter processing modules 64 and 76 may be implemented using a shared processing device, individual processing devices, or a plurality of processing devices. Such a processing device may be a microprocessor, micro-controller, digital signal processor, microcomputer, central processing unit, field programmable gate array, programmable logic device, state machine, logic circuitry, analog circuitry, digital circuitry, and/or any device that manipulates signals (analog and/or digital) based on operational instructions. The memory 75 may be a single memory device or a plurality of memory devices. Such a memory device may be a read-only memory, random access memory, volatile memory, non-volatile memory, static memory, dynamic memory, flash memory, and/or any device that stores digital information. Note that when the processing module 64 and/or 76 implements one or more of its functions via a state machine, analog circuitry, digital circuitry, and/or logic circuitry, the memory storing the corresponding operational instructions is embedded with the circuitry comprising the state machine, analog circuitry, digital circuitry, and/or logic circuitry.
In operation, the radio 60 receives outbound data 94 from the host device via the host interface 62. The host interface 62 routes the outbound data 94 to the digital transmitter processing module 76, which processes the outbound data 94 in accordance with a particular wireless communication standard (e.g., IEEE 802.11, Bluetooth, et cetera) to produce outbound baseband signals 96. The outbound baseband signals 96 will be digital base-band signals (e.g., have a zero IF) or a digital low IF signals, where the low IF typically will be in the frequency range of one hundred kilohertz to a few megahertz.
The digital-to-analog converter 78 converts the outbound baseband signals 96 from the digital domain to the analog domain. The filtering/gain module 80 filters and/or adjusts the gain of the analog signals prior to providing it to the IF mixing stage 82. The IF mixing stage 82 converts the analog baseband or low IF signals into RF signals based on a transmitter local oscillation 83 provided by local oscillation module 74. The power amplifier 84 amplifies the RF signals to produce outbound RF signals 98, which are filtered by the transmitter filter module 85. The antenna 86 transmits the outbound RF signals 98 to a targeted device such as a base station, an access point and/or another wireless communication device.
The radio 60 also receives inbound RF signals 88 via the antenna 86, which were transmitted by a base station, an access point, or another wireless communication device. The antenna 86 provides the inbound RF signals 88 to the receiver filter module 71 via the Tx/Rx switch 73, where the Rx filter 71 bandpass filters the inbound RF signals 88. The Rx filter 71 provides the filtered RF signals to low noise amplifier 72, which amplifies the signals 88 to produce an amplified inbound RF signals. The low noise amplifier 72 provides the amplified inbound RF signals to the IF mixing module 70, which directly converts the amplified inbound RF signals into an inbound low IF signals or baseband signals based on a receiver local oscillation 81 provided by local oscillation module 74. The down conversion module 70 provides the inbound low IF signals or baseband signals to the filtering/gain module 68. The high pass and low pass filter module 68 filters, based on settings provided by the channel bandwidth adjust module 87, the inbound low IF signals or the inbound baseband signals to produce filtered inbound signals.
The analog-to-digital converter 66 converts the filtered inbound signals from the analog domain to the digital domain to produce inbound baseband signals 90, where the inbound baseband signals 90 will be digital base-band signals or digital low IF signals, where the low IF typically will be in the frequency range of one hundred kilohertz to a few megahertz. The digital receiver processing module 64, based on settings provided by the channel bandwidth adjust module 87, decodes, descrambles, demaps, and/or demodulates the inbound baseband signals 90 to recapture inbound data 92 in accordance with the particular wireless communication standard being implemented by radio 60. The host interface 62 provides the recaptured inbound data 92 to the host device 18-32 via the radio interface 54.
As one of average skill in the art will appreciate, the wireless communication device of FIG. 2 may be implemented using one or more integrated circuits. For example, the host device may be implemented on one integrated circuit, the digital receiver processing module 64, the digital transmitter processing module 76 and memory 75 may be implemented on a second integrated circuit, and the remaining components of the radio 60, less the antenna 86, may be implemented on a third integrated circuit. As an alternate example, the radio 60 may be implemented on a single integrated circuit. As yet another example, the processing module 50 of the host device and the digital receiver and transmitter processing modules 64 and 76 may be a common processing device implemented on a single integrated circuit. Further, the memory 52 and memory 75 may be implemented on a single integrated circuit and/or on the same integrated circuit as the common processing modules of processing module 50 and the digital receiver and transmitter processing module 64 and 76.
FIG. 3 is a schematic block diagram illustrating a wireless communication device that includes the host device 18-32 and an associated radio 60. For cellular telephone hosts, the radio 60 is a built-in component. For personal digital assistants hosts, laptop hosts, and/or personal computer hosts, the radio 60 may be built-in or an externally coupled component.
As illustrated, the host device 18-32 includes a processing module 50, memory 52, radio interface 54, input interface 58 and output interface 56. The processing module 50 and memory 52 execute the corresponding instructions that are typically done by the host device. For example, for a cellular telephone host device, the processing module 50 performs the corresponding communication functions in accordance with a particular cellular telephone standard.
Radio 60 includes a host interface 62, a baseband processing module 100, memory 65, a plurality of radio frequency (RF) transmitters 106-110, a transmit/receive (T/R) module 114, a plurality of antennas 81-85, a plurality of RF receivers 118-120, a channel bandwidth adjust module 87, and a local oscillation module 74. The baseband processing module 100, in combination with operational instructions stored in memory 65, executes digital receiver functions and digital transmitter functions, respectively. The digital receiver functions include, but are not limited to, digital intermediate frequency to baseband conversion, demodulation, constellation demapping, decoding, de-interleaving, fast Fourier transform, cyclic prefix removal, space and time decoding, and/or descrambling. The digital transmitter functions include, but are not limited to, scrambling, encoding, interleaving, constellation mapping, modulation, inverse fast Fourier transform, cyclic prefix addition, space and time encoding, and digital baseband to IF conversion. The baseband processing modules 100 may be implemented using one or more processing devices. Such a processing device may be a microprocessor, micro-controller, digital signal processor, microcomputer, central processing unit, field programmable gate array, programmable logic device, state machine, logic circuitry, analog circuitry, digital circuitry, and/or any device that manipulates signals (analog and/or digital) based on operational instructions. The memory 65 may be a single memory device or a plurality of memory devices. Such a memory device may be a read-only memory, random access memory, volatile memory, non-volatile memory, static memory, dynamic memory, flash memory, and/or any device that stores digital information. Note that when the processing module 100 implements one or more of its functions via a state machine, analog circuitry, digital circuitry, and/or logic circuitry, the memory storing the corresponding operational instructions is embedded with the circuitry comprising the state machine, analog circuitry, digital circuitry, and/or logic circuitry.
In operation, the radio 60 receives outbound data 94 from the host device via the host interface 62. The baseband processing module 64 receives the outbound data 88 and, based on a mode selection signal 102, produces one or more outbound symbol streams 90. The mode selection signal 102 will indicate a particular mode of operation that is compliant with one or more specific modes of the various IEEE 802.11 standards. For example, the mode selection signal 102 may indicate a frequency band of 2.4 GHz, a channel bandwidth of 20 or 22 MHz and a maximum bit rate of 54 megabits-per-second. In this general category, the mode selection signal will further indicate a particular rate ranging from 1 megabit-per-second to 54 megabits-per-second. In addition, the mode selection signal will indicate a particular type of modulation, which includes, but is not limited to, Barker Code Modulation, BPSK, QPSK, CCK, 16 QAM and/or 64 QAM. The mode select signal 102 may also include a code rate, a number of coded bits per subcarrier (NBPSC), coded bits per OFDM symbol (NCBPS), and/or data bits per OFDM symbol (NDBPS). The mode selection signal 102 may also indicate a particular channelization for the corresponding mode that provides a channel number and corresponding center frequency. The mode select signal 102 may further indicate a power spectral density mask value and a number of antennas to be initially used for a MIMO communication.
The baseband processing module 100, based on the mode selection signal 102 produces one or more outbound symbol streams 104 from the outbound data 94. For example, if the mode selection signal 102 indicates that a single transmit antenna is being utilized for the particular mode that has been selected, the baseband processing module 100 will produce a single outbound symbol stream 104. Alternatively, if the mode select signal 102 indicates 2, 3 or 4 antennas, the baseband processing module 100 will produce 2, 3 or 4 outbound symbol streams 104 from the outbound data 94.
Depending on the number of outbound streams 104 produced by the baseband module 10, a corresponding number of the RF transmitters 106-110 will be enabled to convert the outbound symbol streams 104 into outbound RF signals 112. In general, each of the RF transmitters 106-110 includes a digital filter and upsampling module, a digital to analog conversion module, an analog filter module, a frequency up conversion module, a power amplifier, and a radio frequency bandpass filter. The RF transmitters 106-110 provide the outbound RF signals 112 to the transmit/receive module 114, which provides each outbound RF signal to a corresponding antenna 81-85.
When the radio 60 is in the receive mode, the transmit/receive module 114 receives one or more inbound RF signals 116 via the antennas 81-85 and provides them to one or more RF receivers 118-122, which will be described in greater detail with reference to FIG. 4. The RF receiver 118-122, based on settings provided by the channel bandwidth adjust module 87, converts the inbound RF signals 116 into a corresponding number of inbound symbol streams 124. The number of inbound symbol streams 124 will correspond to the particular mode in which the data was received. The baseband processing module 100 converts the inbound symbol streams 124 into inbound data 92, which is provided to the host device 18-32 via the host interface 62.
As one of average skill in the art will appreciate, the wireless communication device of FIG. 3 may be implemented using one or more integrated circuits. For example, the host device may be implemented on one integrated circuit, the baseband processing module 100 and memory 65 may be implemented on a second integrated circuit, and the remaining components of the radio 60, less the antennas 81-85, may be implemented on a third integrated circuit. As an alternate example, the radio 60 may be implemented on a single integrated circuit. As yet another example, the processing module 50 of the host device and the baseband processing module 100 may be a common processing device implemented on a single integrated circuit. Further, the memory 52 and memory 65 may be implemented on a single integrated circuit and/or on the same integrated circuit as the common processing modules of processing module 50 and the baseband processing module 100.
FIG. 4 is a schematic block diagram of baseband transmit processing 100-TX within the baseband processing module 100, which includes an encoding module 120, a puncture module 122, a switch, a plurality of interleaving modules 124, 126, a plurality of constellation encoding modules 128, 130, a beamforming module (V) 132, and a plurality of inverse fast Fourier transform (IFFT) modules 134, 136 for converting the outbound data 94 into the outbound symbol stream 104. As one of ordinary skill in the art will appreciate, the baseband transmit processing may include two or more of each of the interleaving modules 124, 126, the constellation mapping modules 128, 130, and the IFFT modules 134, 136. In addition, one of ordinary skill in art will further appreciate that the encoding module 122, puncture module 122, the interleaving modules 124, 126, the constellation mapping modules 128, 130, and the IFFT modules 134, 136 may be function in accordance with one or more wireless communication standards including, but not limited to, IEEE 802.11a, b, g, n.
In one embodiment, the encoding module 120 is operably coupled to convert outbound data 94 into encoded data in accordance with one or more wireless communication standards. The puncture module 122 punctures the encoded data to produce punctured encoded data. The plurality of interleaving modules 124, 126 is operably coupled to interleave the punctured encoded data into a plurality of interleaved streams of data. The plurality of constellation mapping modules 128, 130 is operably coupled to map the plurality of interleaved streams of data into a plurality of streams of data symbols. The beamforming module 132 is operably coupled to beamform, using a unitary matrix having polar coordinates, the plurality of streams of data symbols into a plurality of streams of beamformed symbols. The plurality of IFFT modules 124, 136 is operably coupled to convert the plurality of streams of beamformed symbols into a plurality of outbound symbol streams.
The beamforming module 132 is operably coupled to multiply a beamforming unitary matrix (V) with baseband signals provided by the plurality of constellation mapping modules 128, 130. The beamforming unitary matrix V used by the beamforming module 132 satisfies the conditions of “V*V=VV*=“I”, where “I” is an identity matrix of [1 0; 0 1] for 2×2 MIMO wireless communication, is [1 0 0; 0 1 0; 0 0 1] for 3×3 MIMO wireless communication, or is [1 0 0 0; 0 1 0 0; 0 0 1 0; 0 0 0 1] for 4×4 MIMO wireless communication. In this equation, V*V means “conjugate (V) times V” and VV* means “V times conjugate (V)”. Note that V may be a 2×2 unitary matrix for a 2×2 MIMO wireless communication, a 3×3 unitary matrix for a 3×3 MIMO wireless communication, and a 4×4 unitary matrix for a 4×4 MIMO wireless communication. Further note that for each column of V, a first row of polar coordinates including real values as references and a second row of polar coordinates including phase shift values.
In one embodiment, the constellation mapping modules 128, 130 function in accordance with one of the IEEE 802.11x standards to provide an OFDM (Orthogonal Frequency Domain Multiplexing) frequency domain baseband signals that includes a plurality of tones, or subcarriers, for carrying data. Each of the data carrying tones represents a symbol mapped to a point on a modulation dependent constellation map. For instance, a 16 QAM (Quadrature Amplitude Modulation) includes 16 constellation points, each corresponding to a different symbol. For an OFDM signal, the beamforming module 132 may regenerate the beamforming unitary matrix V for each tone from each constellation mapping module 128, 130, use the same beamforming unitary matrix for each tone from each constellation mapping module 128, 130, or a combination thereof.
The beamforming unitary matrix varies depending on the number of transmit paths (i.e., transmit antennas—M) and the number of receive paths (i.e., receiver antennas—N) for an M×N MIMO communication. For instance, for a 2×2 MIMO communication, the beamforming unitary matrix may be:
$V = ( V ) ij = [ cos ψ 1 cos ψ 2 sin ψ 1 ⅇ j ϕ 1 sin ψ 2 ⅇ j ϕ 2 ]$
In order to satisfy V*V=I, it needs to satisfy followings.
cos ψ1 cos ψ2+sin ψ1 sin ψ2 e j(φ 1 −φ 2 )=0
cos ψ1 cos ψ2+sin ψ1 sin ψ2 e j(φ 2 −φ 1 )=0
wherein absolute value of ψ1−ψ2=π/2 and Φ12 or Φ12+π and ψ12=π/2.
Therefore, with Φ1 and ψ1, the beamforming module 132 may regenerate V per each tone. Using the property of phase invariance of V(f), a further bit reduction may be achieved. Accordingly, when cos(ψi)<0, that is π/2<ψi<π, V can be modified to make it positive as follows,
$H = UDV * = UD [ - 1 0 ; 0 1 ] [ - 1 0 ; 0 1 ] V * = U [ - 1 0 ; 0 1 ] DV_new * = U_new DV_new *$
where U_new=U[−1 0;0 1], and V_new=V [−1 0; 0 1]. Then the first element of V_new can be positive with 0<ψi<π/2 to have cos(ψi)>0. Then, an index bit may not be needed to indicate the relationship between ψ1 and ψ2 or Φ1 and Φ2, but they need to satisfy Φ1=Φ2+π and ψ1+ψ2=π/2 from above equations for V*V=I. For instance, to accommodate the various possible V values, the angles should have a range of ψ in [0, π/2], Φ in [0, 2π]. In order to satisfy V*V=VV*=I, the set of angles need to satisfy Φ1=Φ2+π and ψ1+ψ2=π/2. Thus, only angles Φ1 and ψ1, need to be transmitted in the feedback signal.
With angle resolution of π/2a, where “a” equals the number of bits per angle, the total number of bits per tone is (a−1)+(a+1)=2a to cover [0,π/2] and [0,2π], respectively, Note that the bits may be divided such that “a−1” bits are for angle ψ1 and “a+1” bits are for Φ1. Note that ‘a=1’ corresponds to a quantized angle that is [π/4, 3π/4] to cover [0, π] with angle resolution of π/2. Further note that ‘a=2’ corresponds to a quantized angle that is [π/8, 3π/8, 5π/8, 7π/8] to cover [0, π] with angle resolution of π/4. Still further note that to achieve full resolution BF performance, polar coordinates need 10 bits per tone for V(f), while Cartesian coordinates need 40 bits per tone for V(f). Refer to FIG. 7 for simulation results that depict one of the worst cases with strong coding rate at less frequency selective channels.
For a 3×3 MIMO communication, the beamforming unitary matrix may be:
$V = ( V ) ⅈj = [ cos ψ 1 cos ψ 2 cos ψ 3 sin ψ 1 cos θ 1 ⅇ j ϕ 21 sin ψ 2 cos θ 2 ⅇ jϕ 22 sin ψ 3 cos θ 3 ⅇ j ϕ 23 sin ψ 1 sin θ 1 ⅇ jϕ 31 sin ψ 2 sin θ 2 ⅇ jϕ 32 sin ψ 3 sin θ 3 ⅇ jϕ 33 ]$
wherein ψ1, ψ2, ψ3, θ1, θ2, θ3, Φ21, Φ22, Φ23, Φ31, Φ32, Φ33 represent angles of the unit circle, wherein Diagonal (V*V)=1s, and wherein:
$ψ i = cos - 1 V 1 i , θ i = cos - 1 V 2 i sin ψ 3 i $ $ϕ 2 i = ∠ ( V 2 i ) , ϕ 3 i = ∠ ( V 3 i )$
In this example, with 12 angles, the beamforming module 132 may regenerate V as a 3×3 matrix per tone. With 4-bits for expression for the angles, a 54 tone signal may have feedback information of 324 bytes (e.g., 4×12×54/8).
As an alternate example, only seven of the twelve angles need to be included in the feedback signal and the beamforming module can calculate the remaining five angles. In this example, let ψ1, ψ2, ψ3, θ1, θ2, θ3, Φ21, Φ22, Φ23, Φ31, Φ32, Φ33 represent the set of angles, wherein ψ1, ψ2, θ1, θ2, Φ21, Φ22, and Φ33 are included in the subset of angles, wherein ψ3, θ3, Φ21, Φ23, Φ32, and Φ33 are subsequently determined. In this example, to accommodate the various possible V values, the angles should range from ψ in [0, π/2], θ in [0, π/2], Φ in [−π, π]
In order to satisfy V*V=VV*=I,
ψ3=function(ψ1, ψ2)
θ3=function(ψ1, ψ2, ψ3, θ1, θ2)
Φ23=function(ψ1, ψ2, ψ3, θ1, θ2, θ3, Φ21, Φ22)
Φ32=function(ψ1, ψ2, ψ3, θ1, θ2, θ3, Φ21, Φ22, Φ31)
Φ33=function(ψ1, ψ2, ψ3, θ1, θ2, θ3, Φ21, Φ23, Φ31).
The remaining five angles may then be calculated as:
$ψ 3 = cos - 1 ( 1 - cos - 2 ψ 1 - cos 2 ψ 2 )$ $θ 3 = cos - 1 ( 1 - sin 2 ψ 1 cos 2 θ 1 - sin 2 ψ 2 cos 2 θ 2 sin ψ 3 )$ $ⅇ jϕ 23 = - cos ψ 1 sin ψ 1 cos θ 1 ⅇ jϕ 21 + cos ψ 2 sin ψ 2 cos θ 2 ⅇ jϕ 22 cos ψ 3 sin ψ 3 cos θ 3$ $ⅇ jϕ 32 = - cos ψ 1 cos ψ 2 + sin ψ 1 cos θ 1 sin ψ 2 cos θ 2 ⅇ j ( ϕ 22 - ϕ 21 ) sin ψ 1 sin θ 1 sin ψ 2 sin θ 2 ⅇ jϕ 31$ $ⅇ jϕ 32 = - cos ψ 1 cos ψ 3 + sin ψ 1 cos θ 1 sin ψ 3 cos θ 3 ⅇ j ( ϕ 23 - ϕ 21 ) sin ψ 1 sin θ 1 sin ψ 3 sin θ 3 ⅇ jϕ 31$
With angle resolution of π/2a, where a equals the number of bits per angle, the total number of bits per tone is 2(a−1)+2(a−1)+3(a+1)=7a−1. The bits may be divided such that the xV angles are each assigned “a−1” bits to cover [0, π/2], the θ angles are each assigned “a−1” bits to cover [0, π/2], and the Φ angles are each assigned “a+1” bits to cover [0, 2π]. In simulation, as shown in FIG. 8, a=6 is enough to achieve full resolution beamforming, thus 41 bits per tone for a 3×3 V(f). For full angle feedback (i.e., send all twelve angles in the feedback signal), each angle needs 4 bits such that the 12 angles require 48 bits per tone. In comparison, a Cartesian coordinate feedback signal would require 72 bits for compatible performance (4*9*2=72, 4 bits for 9 elements with real/imaginary components per each tone).
For a 4×4 MIMO communication, the beamforming unitary matrix may be:
$V = [ cos ψ 1 cos φ 1 cos ψ 2 cos φ 2 cos ψ 3 cos φ 3 cos ψ 4 cos φ 4 cos ψ 1 sin φ 1 ⅇ jϕ 11 cos ψ 2 sin φ 2 ⅇ jϕ 12 cos ψ 3 sin φ 3 ⅇ jϕ 13 cos ψ 4 sin φ 4 ⅇ jϕ 14 sin ψ 1 cos θ 1 ⅇ jϕ 21 sin ψ 2 cos θ 2 ⅇ jϕ 22 sin ψ 3 cos θ 3 ⅇ jϕ 23 sin ψ 4 cos θ 4 ⅇ jϕ 24 sin ψ 1 sin θ 1 ⅇ jϕ 31 sin ψ 2 sin θ 2 ⅇ jϕ 32 sin ψ 3 sin θ 3 ⅇ jϕ 33 sin ψ 4 sin θ 4 ⅇ jϕ 34 ]$
wherein ψ1, ψ2, ψ3, ψ4, θ1, θ2, θ3, θ4, φ1, φ2, φ3, φ4, Φ11, Φ12, Φ13, Φ14, Φ21, Φ22, Φ23, Φ24, Φ31, Φ32, Φ33, Φ34, Φ41, Φ42, Φ43, Φ44 represent angles of the unit circle, wherein Diagonal (V*V)=1 s, and wherein:
$ψ i = cos - 1 ( V 1 i 2 + V 2 i 2 ) , φ i = cos - 1 ( V 1 i cos ψ i ) , θ i = cos - 1 V 3 i sin ψ i $ $ϕ 1 i = ∠ ( V 2 i ) , ϕ 2 i = ∠ ( V 3 i ) , ϕ 3 i = ∠ ( V 4 i )$
In this example, with 24 angles, the beamforming module 132 may regenerate V as a 4×4 matrix per tone. With 4-bits for expression for the angles, a 54 tone signal may have feedback information of 648 bytes (e.g., 4×24×54/8).
As an alternate example, only sixteen of the twenty-four angles need to be included in the feedback signal and the beamforming module can calculate the remaining eight angles. In this example, to accommodate the various possible V values, the angles should range from ψ in [0, π/2], φ in [0, π/2], θ in [0, π/2], Φ in [0, 2π]
In order to satisfy V*V=VV*=I,
ψ4=function(ψ1, ψ2, ψ3)
φ4=function(ψ1, ψ2, ψ3, ψ4, φ1, φ2, φ3)
θ4=function(ψ1, ψ2, ψ3, ψ4, θ1, θ2, θ3)
Φ14=function(ψ1, ψ2, ψ3, ψ4, θ1, θ2, θ3, θ4, Φ11, Φ12, Φ13)
Φ24=function(ψ1, ψ2, ψ3, ψ4, θ1, θ2, θ3, θ4, Φ21, Φ22, Φ23)
Φ32=function(ψ1, ψ2, θ1, θ2, φ1, φ2, Φ21, Φ22, Φ31);
Φ33=function(ψ1, ψ3, θ1, θ3, φ1, φ3, Φ21, Φ23, Φ31);
Φ34=function(ψ1, ψ4, θ1, θ4, φ1, φ4, Φ21, Φ24, Φ31).
In accordance with this example, ψ1, ψ2, ψ3, ψ4, θ1, θ2, θ3, θ4, φ1, φ2, φ3, φ4, Φ11, Φ12, Φ13, Φ14, Φ21, Φ22, Φ23, Φ24, Φ31, Φ32, Φ33, Φ34 represent the set of angles, wherein ψ1, ψ2, ψ3, θ1, θ2, θ3, φ1, φ2, φ3, Φ11, Φ12, Φ13, Φ21, Φ22, Φ23, Φ31, are the subset of angles. Angles ψ4, θ4, φ4, Φ14, Φ24, Φ31, Φ32, Φ33, Φ34 are determined by:
$ψ 4 = cos - 1 ( 2 - cos 2 ψ 1 - cos 2 ψ 2 - cos 2 ψ 3 )$ $φ 4 = cos - 1 ( 1 - cos 2 ψ 1 cos 2 φ 1 - cos 2 ψ 2 cos 2 φ 2 - cos 2 ψ 3 cos 2 φ 3 cos ψ 4 )$ $θ 4 = cos - 1 ( 1 - sin 2 ψ 1 cos 2 θ 1 - sin 2 ψ 2 cos 2 θ 2 - sin 2 ψ 3 cos 2 θ 3 sin ψ 4 )$ $ⅇ jϕ 14 = - cos 2 ψ 1 sin φ 1 cos φ 1 ⅇ jϕ 11 + cos 2 ψ 2 sin φ 2 cos φ 2 ⅇ jϕ 12 + cos 2 ψ 3 sin φ 3 cos φ 3 ⅇ jϕ 13 cos 2 ψ 4 sin φ 4 cos φ 4$ $ⅇ jϕ 24 = - cos ψ 1 sin ψ 1 cos φ 1 cos θ 1 ⅇ jϕ 21 + cos ψ 2 sin ψ 2 cos φ 2 cos θ 2 ⅇ jϕ 22 + cos ψ 3 sin ψ 3 cos φ 3 cos θ 3 ⅇ jϕ 23 cos ψ 4 sin ψ 4 cos φ 4 cos θ 4$ $ⅇ jϕ 32 = - cos ψ 1 cos ψ 2 cos φ 1 cos φ 2 + cos ψ 1 cos ψ 2 sin φ 1 sin φ 2 ⅇ j ( ϕ 12 - ϕ 11 ) + sin ψ 1 cos θ 1 sin ψ 2 cos θ 2 ⅇ j ( ϕ 22 - ϕ 21 ) sin ψ 1 sin θ 1 sin ψ 2 sin θ 2 ⅇ jϕ 31$ $ⅇ jϕ 33 = - cos ψ 1 cos ψ 3 cos φ 1 cos φ 3 + cos ψ 1 cos ψ 3 sin φ 1 sin φ 3 ⅇ j ( ϕ 13 - ϕ 11 ) + sin ψ 1 cos θ 1 sin ψ 3 cos θ 3 ⅇ j ( ϕ 23 - ϕ 21 ) sin ψ 1 sin θ 1 sin ψ 3 sin θ 3 ⅇ jϕ 31$ $ⅇ jϕ 34 = - cos ψ 1 cos ψ 4 cos φ 1 cos φ 4 + cos ψ 1 cos ψ 4 sin φ 1 sin φ 4 ⅇ j ( ϕ 14 - ϕ 11 ) + sin ψ 1 cos θ 1 sin ψ 4 cos θ 4 ⅇ j ( ϕ 24 - ϕ 21 ) sin ψ 1 sin θ 1 sin ψ 4 sin θ 4 ⅇ jϕ 31$
With angle resolution of π/2a, where a equals the number of bits per angle, the total number of bits per tone is 3(a−1)+3(a−1)+3(a−1)+7(a+1)=16a−2. The bits may be allocated such that each v angle is assigned “a−1” bits, each (p angle is assigned “a−1” bits to cover [0, π/2], each θ is assigned “a−1” bits to cover [0, π/2], and each (D angle is assigned “a+1” bits to cover [0, 2π]. In this example, an “a” equal to 5 may be enough to achieve full resolution beamforming performance, such that 62 bits per tone are needed for 4×4 V(f). In comparison with the same resolution of 4 bits per each element for Cartesian coordinates, 128 bits per tone are required; 4 bits for 16 elements with real/imaginary components.
FIG. 5 is a schematic block diagram of baseband receive processing 100-RX that includes a plurality of fast Fourier transform (FFT) modules 140, 142, a beamforming (U) module 144, a plurality of constellation demapping modules 146, 148, a plurality of deinterleaving modules 150, 152, a switch, a depuncture module 154, and a decoding module 156 for converting a plurality of inbound symbol streams 124 into inbound data 92. As one of ordinary skill in the art will appreciate, the baseband receive processing 100-RX may include two or more of each of the deinterleaving modules 150, 152, the constellation demapping modules 146, 148, and the FFT modules 140, 142. In addition, one of ordinary skill in art will further appreciate that the decoding module 156, depuncture module 154, the deinterleaving modules 150, 152, the constellation decoding modules 146, 148, and the FFT modules 140, 142 may be function in accordance with one or more wireless communication standards including, but not limited to, IEEE 802.11a, b, g, n.
In one embodiment, a plurality of FFT modules 140, 142 is operably coupled to convert a plurality of inbound symbol streams 124 into a plurality of streams of beamformed symbols. The inverse beamforming module 144 is operably coupled to inverse beamform, using a unitary matrix having polar coordinates, the plurality of streams of beamformed symbols into a plurality of streams of data symbols. The plurality of constellation demapping modules is operably coupled to demap the plurality of streams of data symbols into a plurality of interleaved streams of data. The plurality of deinterleaving modules is operably coupled to deinterleave the plurality of interleaved streams of data into encoded data. The decoding module is operably coupled to convert the encoded data into inbound data 92.
In one embodiment, the FFT modules 140, 142 function in accordance with one of the IEEE 802.11x standards to provide an OFDM (Orthogonal Frequency Domain Multiplexing) frequency domain baseband signals that includes a plurality of tones, or subcarriers, for carrying data. Each of the data carrying tones represents a symbol mapped to a point on a modulation dependent constellation map. For an OFDM signal, the beamforming module 144 may use the beamforming unitary matrix U for each tone from each FFT module 140, 142, use the same beamforming unitary matrix for each tone from each FFT module 140, 142, or a combination thereof.
The preceding discussion has presented a method and apparatus for reducing feedback information for beamforming in a wireless communication by using polar coordinates. As one of average skill in the art will appreciate, other embodiments may be derived from the present teachings without deviating from the scope of the claims.
Patent Citations
Cited PatentFiling datePublication dateApplicantTitle
US3858221 *Apr 12, 1973Dec 31, 1974Harris Intertype CorpLimited scan antenna array
US3916533 *May 2, 1974Nov 4, 1975Us NavyCircular array hydrophone simulator
US4843631 *Dec 22, 1986Jun 27, 1989Dietmar SteinpichlerPattern recognition process
US5541607 *Dec 5, 1994Jul 30, 1996Hughes ElectronicsPolar digital beamforming method and system
US20050141624 *Dec 24, 2003Jun 30, 2005Intel CorporationMultiantenna communications apparatus, methods, and system
US20050286663 *Jun 23, 2004Dec 29, 2005Intel CorporationCompact feedback for closed loop MIMO systems
US20060067428 *Sep 30, 2004Mar 30, 2006Poon Ada S YDeterminitic spatial power allocation and bit loading for closed loop MIMO
US20060068738 *Sep 30, 2004Mar 30, 2006Qinghua LiClosed loop feedback in MIMO systems
US20060155534 *Feb 16, 2005Jul 13, 2006Lin Xintian ECodebook generation system and associated methods
US20060234645 *Mar 9, 2005Oct 19, 2006Intel CorporationMethod and apparatus to provide low cost transmit beamforming for network devices
Non-Patent Citations
Reference
1"Digital Beamforming Basics (Antennas)" by Hans Steyskal; Journal of Electronic Defense; Jul. 1, 1996 (7 pages).
2"Interpolation Based Transmit Beamforming for MIMO-OFDM with Partial Feedback" by Jihoon Choi and Robert W. Heath, Jr. The University of Texas at Austin, Dept. of Electrical & Computer Engineering, Wireless Networking & Communications Group; Sep. 16, 2003; p. 1-14.
3"Utilizing Digital Downconverters for Efficient Digital Beamforming" by Clint Schreiner, Red River Engineering (5 pages).
Classifications
U.S. Classification375/267, 375/219, 375/221, 375/260, 375/299, 375/222, 375/259
International ClassificationH04L1/02
Cooperative ClassificationH04B7/0417, H04B7/0634
European ClassificationH04B7/06C1F1W
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- Feb 12, 02:00 AM Comments [0]
Math (PS)
If $y(u - c) = 0$ and $j(u - k) = 0$ , which of the following must be true, assuming $c < k[/latex] ? (A) [latex]yj < 0[/latex] (B) [latex]yj > 0$
(C) $yj = 0$
(D) $j = 0$
(E) $y = 0$
Question Discussion & Explanation
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Verbal (CR)
George Bernard Shaw wrote: "That any sane nation, having observed that you could provide for the supply of bread by giving bakers a pecuniary interest in baking for you, should go on to give a surgeon a pecuniary interest in cutting off your leg is enough to make one despair of political humanity."
Shaw's statement would best serve as an illustration in an argument criticizing which of the following?
(A) Dentists who perform unnecessary dental work in order to earn a profit
(B) Doctors who increase their profits by specializing only in diseases that affect a large percentage of the population
(C) Grocers who raise the price of food in order to increase their profit margins
(D) Oil companies that decrease the price of their oil in order to increase their market share
(E) Bakers and surgeons who earn a profit by supplying other peoples' basic needs
Question Discussion & Explanation
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• To: mathgroup at smc.vnet.net
• Subject: [mg92171] ByteCount of imported machine-precision data matrix three times equivalent generated matrix
• From: Gareth Russell <russell at njit.edu>
• Date: Mon, 22 Sep 2008 05:27:24 -0400 (EDT)
```Hi,
I am encountering some strange memory-related behavior when importing
numerical data from a file. If anyone is interested, a (small) example
file is here:
http://web.njit.edu/~russell/Mathematica.html
It's a simple 2D array of numbers. The issue is that when imported,
ByteCount[] indicates that the resultant expression takes up more than
three times as much memory as an equivalent machine-precision matrix
generated within Mathematica. All diagnostics that I can think of
indicate that the imported expression is equivalent in precision. And
indeed, ByteCount applied to individual elements of each matrix returns
16 as an answer. It's only the overall ByteCount which is hugely
different.
I discovered a workaround: if I generate a dummy matrix of 0. elements
(which has the smaller ByteCount), and add it to the imported matrix,
the result, while appearing identical (as it should), now also has the
smaller ByteCount.
Does anyone know what it going on here? Until I discovered the
workaround it was a problem, as I need to read in a large number of
much larger matrices all together, was encountering memory issues.
Thanks,
Gareth
NJIT
In[1452]:= data = Drop[Import["12e.dat"], 6];
In[1453]:= Dimensions[data]
Out[1453]= {20, 14}
In[1454]:= {Min[Flatten[data]], Max[Flatten[data]]}
Out[1454]= {-1.1, 0.992655}
In[1455]:= MachineNumberQ[data[[1, 1]]]
Out[1455]= True
In[1456]:= ByteCount[data]
Out[1456]= 7384
In[1466]:= ByteCount[data[[1, 1]]]
Out[1466]= 16
In[1457]:= dummyData = Table[RandomReal[{-1.1, 1}], {20}, {14}];
In[1458]:= ByteCount[dummyData]
Out[1458]= 2360
In[1460]:= zeroData = Table[0., {20}, {14}];
In[1461]:= ByteCount[zeroData]
Out[1461]= 2360
In[1462]:= newData = data + zeroData;
In[1463]:= {Min[Flatten[newData]], Max[Flatten[newData]]}
Out[1463]= {-1.1, 0.992655}
In[1464]:= MachineNumberQ[newData[[1, 1]]]
Out[1464]= True
In[1465]:= ByteCount[newData]
Out[1465]= 2360
In[1467]:= ByteCount[newData[[1, 1]]]
Out[1467]= 16
--
Gareth Russell
NJIT
```
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https://aif.centre-mersenne.org/articles/10.5802/aif.684/ | 1,656,197,962,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103036176.7/warc/CC-MAIN-20220625220543-20220626010543-00716.warc.gz | 138,299,342 | 11,148 | # ANNALES DE L'INSTITUT FOURIER
A Whitney extension theorem in ${L}^{p}$ and Besov spaces
Annales de l'Institut Fourier, Volume 28 (1978) no. 1, pp. 139-192.
The classical Whitney extension theorem states that every function in Lip$\left(\beta ,F\right)$, $F\subset {\mathbf{R}}^{n}$, $F$ closed, $k<\beta \le k+1$, $k$ a non-negative integer, can be extended to a function in Lip$\left(\beta ,{\mathbf{R}}^{n}\right)$. Her Lip$\left(\beta ,F\right)$ stands for the class of functions which on $F$ have continuous partial derivatives up to order $k$ satisfying certain Lipschitz conditions in the supremum norm. We formulate and prove a similar theorem in the ${L}^{p}$-norm.
The restrictions to ${\mathbf{R}}^{d}$, $d, of the Bessel potential spaces in ${\mathbf{R}}^{n}$ and the Besov or generalized Lipschitz spaces in ${\mathbf{R}}^{n}$ have been characterized by the work of many authors (O.V. Besov, E.M. Stein, and others). We treat, for Besov spaces, the case when ${\mathbf{R}}^{d}$ is replaced by a closed set $F$ of a much more general kind than the sets which have been considered before. Our method of proof gives a new proof also in the case when $F={\mathbf{R}}^{d}$. It also gives a contribution to the restriction and extension problem corresponding to the case $d=n$ with $F$ equal to the closure of a domain in ${\mathbf{R}}^{n}$.
D’après le théorème de prolongement classique de Whitney on peut prolonger toute fonction dans Lip$\left(\beta ,F\right)$, $F\subset {\mathbf{R}}^{n}$, $F$ fermé, $k<\beta \le k+1$, $k$ un nombre entier non-négatif, à une fonction dans Lip$\left(\beta ,{\mathbf{R}}^{n}\right)$. Ici on désigne par Lip$\left(\beta ,F\right)$ l’espace des fonctions sur $F$ avec des dérivées partielles continues jusqu’à l’ordre $k$ qui satisfont certaines conditions de Lipschitz dans la norme supremum. Nous formons et montrons un théorème analogue dans la norme ${L}^{p}$.
Les restrictions à ${\mathbf{R}}^{d}$, $d, des espaces potentiels besseliens dans ${\mathbf{R}}^{n}$ et les espaces de Besov ou les espaces de Lipschitz généralisés sont caractérisées par les travaux de plusieurs auteurs (O.V. Besov, E.M. Stein, et d’autres). Nous traitons, pour les espaces de Besov, le cas quand ${\mathbf{R}}^{d}$ est remplacé par un ensemble $F$ fermé d’une sorte beaucoup plus générale que les ensembles considérés précédemment. Notre méthode donne une démonstration nouvelle aussi dans le cas $F={\mathbf{R}}^{d}$. Elle donne aussi une contribution au problème de restriction et prolongement correspondant au cas $d=n$ avec $F$ égal à la fermeture d’un domaine dans ${\mathbf{R}}^{n}$.
@article{AIF_1978__28_1_139_0,
author = {Jonsson, Alf and Wallin, Hans},
title = {A {Whitney} extension theorem in $L^p$ and {Besov} spaces},
journal = {Annales de l'Institut Fourier},
pages = {139--192},
publisher = {Imprimerie Louis-Jean},
volume = {28},
number = {1},
year = {1978},
doi = {10.5802/aif.684},
zbl = {0369.46031},
mrnumber = {81c:46024},
language = {en},
url = {https://aif.centre-mersenne.org/articles/10.5802/aif.684/}
}
TY - JOUR
TI - A Whitney extension theorem in $L^p$ and Besov spaces
JO - Annales de l'Institut Fourier
PY - 1978
DA - 1978///
SP - 139
EP - 192
VL - 28
IS - 1
PB - Imprimerie Louis-Jean
PP - Gap
UR - https://aif.centre-mersenne.org/articles/10.5802/aif.684/
UR - https://zbmath.org/?q=an%3A0369.46031
UR - https://www.ams.org/mathscinet-getitem?mr=81c:46024
UR - https://doi.org/10.5802/aif.684
DO - 10.5802/aif.684
LA - en
ID - AIF_1978__28_1_139_0
ER -
%0 Journal Article
%T A Whitney extension theorem in $L^p$ and Besov spaces
%J Annales de l'Institut Fourier
%D 1978
%P 139-192
%V 28
%N 1
%I Imprimerie Louis-Jean
%C Gap
%U https://doi.org/10.5802/aif.684
%R 10.5802/aif.684
%G en
%F AIF_1978__28_1_139_0
Jonsson, Alf; Wallin, Hans. A Whitney extension theorem in $L^p$ and Besov spaces. Annales de l'Institut Fourier, Volume 28 (1978) no. 1, pp. 139-192. doi : 10.5802/aif.684. https://aif.centre-mersenne.org/articles/10.5802/aif.684/
[1] D.R. Adams, Traces of potentials arising from translation invariant operators, Ann. Sc. Norm. Sup. Pisa, 25 (1971), 203-217. | Numdam | MR: 44 #4508 | Zbl: 0219.46027
[2] D.R. Adams and N.G. Meyers, Bessel potentials. Inclusion relations among classes of exceptional sets, Indiana Univ. Math. J., 22 (1973), 873-905. | Zbl: 0285.31007
[3] R. Adams, N. Aronszajn, and K.T. Smith, Theory of Bessel potentials, Part II, Ann. Inst. Fourier, 17, 2 (1967), 1-135. | Numdam | MR: 37 #4281 | Zbl: 0185.19703
[4] N. Aronszajn, Potentiels Besseliens, Ann. Inst. Fourier, 15, 1 (1965), 43-58. | Numdam | MR: 32 #1549 | Zbl: 0141.30403
[5] N. Aronszajn, F. Mulla, and P. Szeptycki, On spaces of potentials connected with Lp-classes, Ann. Inst. Fourier, 13, 2 (1963), 211-306. | Numdam | MR: 31 #5076 | Zbl: 0121.09604
[6] O.V. Besov, Investigation of a family of function spaces in connection with theorems of imbedding and extension (Russian), Trudy Mat. Inst. Steklov, 60 (1961), 42-81 ; Amer. Math. Soc. Transl., (2) 40 (1964), 85-126. | Zbl: 0158.13901
[7] O.V. Besov, The behavior of differentiable functions on a non-smooth surface, Trudy Mat. Inst. Steklov, 117 (1972), 1-9. | MR: 52 #6403 | Zbl: 0279.46018
[8] O.V. Besov, On traces on a nonsmooth surface of classes of differentiable functions, Trudy Mat. Inst. Steklov, 117 (1972), 11-24. | Zbl: 0279.46019
[9] O.V. Besov, Estimates of moduli of smoothness on domains, and imbedding theorems, Trudy Mat. Inst. Steklov, 117 (1972), 25-53. | Zbl: 0279.46020
[10] O.V. Besov, Continuation of functions beyond the boundary of a domain with preservation of differential-difference properties in Lp, (Russian), Mat. Sb, 66 (108) (1965), 80-96 ; Amer. Math. Soc. Transl., (2) 79 (1969), 33-52. | Zbl: 0189.43305
[11] V.I. Burenkov, Imbedding and continuation for classes of differentiable functions of several variables defined on the whole space, Progress in Math., 2, pp. 73-161, New York, Plenum Press, 1968. | Zbl: 0185.20603
[12] P.L. Butzer and H. Berens, Semi-groups of operators and approximation, Berlin, Springer-Verlag, 1967. | MR: 37 #5588 | Zbl: 0164.43702
[13] A.P. Calderon, Lebesgue spaces of differentiable functions and distributions, Proc. Symp. in Pure Math., 4 (1961), 33-49. | MR: 26 #603 | Zbl: 0195.41103
[14] H. Federer, Geometric measure theory, Berlin, Springer-Verlag, 1969. | MR: 41 #1976 | Zbl: 0176.00801
[15] O. Frostman, Potentiels d'équilibre et capacité des ensembles, Thesis, Lund, 1935. | JFM: 61.1262.02 | Zbl: 0013.06302
[16] B. Fuglede, On generalized potentials of functions in the Lebesgue classes, Math. Scand., 8 (1960), 287-304. | MR: 28 #2241 | Zbl: 0196.42002
[17] E. Gagliardo, Caratterizzazioni della trace sulla frontiera relative ad alcune classi di funzioni in n variabili, Rend. Sem. Mat. Padoa, 27 (1957), 284-305. | Numdam | MR: 21 #1525 | Zbl: 0087.10902
[18] A. Jonsson, Imbedding of Lipschitz continuous functions in potential spaces, Department of Math., Univ. of Umea, 3 (1973).
[19] P.I. Lizorkin, Characteristics of boundary values of functions of Lrp(En) on hyperplanes (Russian), Dokl. Akad. Nauk SSSR, 150 (1963), 986-989. | MR: 27 #2853 | Zbl: 0199.44401
[20] J. Bergh and J. Lofstrom, Interpolation spaces, Berlin, Springer-Verlag, 1976. | MR: 58 #2349 | Zbl: 0344.46071
[21] S.M. Nikol'Skii, Approximation of functions of several variables and imbedding theorems, Berlin, Springer-Verlag, 1975. | MR: 51 #11073 | Zbl: 0307.46024
[22] S.M. Nikol'Skii, On imbedding, continuation and approximation theorems for differentiable functions of several variables (Russian), Usp. Mat. Nauk, 16, 5 (1961), 63-114 ; Russian Math. Surveys, 16, 5 (1961), 55-104. | MR: 26 #6757 | Zbl: 0117.29101
[23] S.M. Nikol'Skii, On the solution of the polyharmonic equation by a variational method (Russian), Dokl. Akad. Nauk SSSR, 88 (1953), 409-411. | Zbl: 0053.07403
[24] J. Peetre, On the trace of potentials, Ann. Scuola Norm. Sup. Pisa, 2,1 (1975), 33-43. | Numdam | MR: 52 #8912 | Zbl: 0308.46031
[25] W. Rudin, Real and complex analysis, sec. ed. New York, McGraw-Hill, 1974. | MR: 49 #8783 | Zbl: 0278.26001
[26] T. Sjödin, Bessel potentials and extension of continuous functions, Ark. Mat., 13,2 (1975), 263-271. | MR: 53 #843 | Zbl: 0314.31005
[27] E.M. Stein, Singular integrals and differentiability properties of functions, Princeton, Princeton Univ. Press, 1970. | MR: 44 #7280 | Zbl: 0207.13501
[28] E.M. Stein, The characterization of functions arising as potentials. II, Bull. Amer. Math. Soc., 68 (1962), 577-582. | MR: 26 #547 | Zbl: 0127.32002
[29] M.H. Taibleson, On the theory of Lipschitz spaces of distributions on Euclidean n-space, I, J. Math. Mech., 13 (1964), 407-480. | MR: 29 #462 | Zbl: 0132.09402
[30] H. Whitney, Analytic extensions of differentiable functions defined in closed sets, Trans. Amer. Math. Soc., 36 (1934), 63-89. | JFM: 60.0217.01 | MR: 1501735 | Zbl: 0008.24902
[31] H. Wallin, Continuous functions and potential theory, Ark. Mat., 5 (1963), 55-84. | MR: 29 #2425 | Zbl: 0134.09404
Cited by Sources: | 3,278 | 9,096 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 40, "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} | 2.578125 | 3 | CC-MAIN-2022-27 | longest | en | 0.362695 |
https://mathoverflow.net/questions/353788/motivation-for-karoubi-envelope-idempotent-completion/353793 | 1,701,239,567,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100056.38/warc/CC-MAIN-20231129041834-20231129071834-00146.warc.gz | 451,540,010 | 31,015 | # Motivation for Karoubi envelope/ idempotent completion
This is the second part of my venture to become more comfortable with the concept of idempotent elements and idempotent splittings from category theoretical viewpoint. In the first part we considered the interpretation of idempotent elements & splitting from viewpoint of commutative algebra. The most fruitful analogy (at least for me) was that if we consider the category $$\text{R-ModFree}$$ of free $$R$$-modules, taking its completion means making it closed under taking direct summands. As the direct summands of free modules are exactly the projective modules completing means "to add some objects" which occur naturally as building blocks.
Now in case of commutative algebra projective objects allows to deal with projective resolutions and provide a framework for direct calculations of derived functors of right exact functors.
I read that there are a lot of constructions spreaded in a relatively wide areas of mathematics where one starts with a certain category $$C$$, construct from this one another say $$F(C)$$, and then pass to its idempotent completion $$\widehat{F(C)}$$.
Probably the most prominent example is the construction of pure motives where we start with category $$(\operatorname{Sm}/k)$$ of smooth varieties over a field $$k$$, then pass to category of correspondences $$\operatorname{Cor}_k$$, build its idempotent completion $$\widehat{(\operatorname{Cor}_k)}$$ and go ahead with the construction to build the category of Motives $$\operatorname{Mot}_k$$ and then, by trying to mimic the procedure of building the derived category, we arrive at the category of pure motives (of course that's just a very coarse overview).
The point of my interest is the necessity of taking idempotent completion in the intermediate step.
Of course, that's just an example, but similar strategies occur for example in $$K$$-theory when one study vector bundles or in constructions dealing with triangulated categories.
My Question: Can there be extracted a common motivation in these examples making the step that takes idempotent completion necessary or does it in every construction almost everywhere strongly depend on "what one wants"?
The only one "general mantra" that I found up to now having the $$\text{R-Mod}$$ example in mind was the necessity of projective objects in order to study right exact functors.
Question: Is this the only motivation or are there some other common deep reasons for the importance of taking idempotent completions?
• The point of the idempotent completion is so that direct summands of objects of your category are now objects as well. Feb 28, 2020 at 18:42
• Idempotents appear very naturally when studying the cohomology of algebraic varieties. For example if some finite group acts on your variety then you can consider the piece of cohomology cut out by characters of this group (corresponding to idempotents in the group algebra). Feb 28, 2020 at 18:47
• @Sam Hopkins: that's true. The point is when we think about the couple of constructions I mentioned (e.g. the motivic case) where it become neccessary to deal with a category beeing closed under taking direct summands? What might fail if not not do it? Feb 28, 2020 at 18:49
• I think a very illuminating example is the construction of Deligne's category $\mathrm{Rep}(S_t)$ of "representations of the symmetric group $S_t$" where $t$ is a complex parameter. The idea is that for integer $n$, every representation of $S_n$ is a direct summand of a tensor power of the defining representation. So to mimic this for an arbitrary parameter $t$, first you create via diagrammatic rules a category whose objects correspond to tensor powers of the defining representation; then you take the Karoubian envelope to get all representations. Feb 28, 2020 at 18:56
• Another example for motives: you need projectors in order to define even very basic objects. For example the Tate motive $\mathbb{Z}(1)$ (or rather its inverse, the Lefschetz motive), you need to decompose the cohomology of the projective line (or the multiplicative group). If you don't have the Tate motive then you can't define motivic cohomology for example. Feb 28, 2020 at 19:30
The "motivic motivation" is that by idempotent completing correspondences over a finite field one obtains a category of homological motives where Kunneth decompositions of diagonals are available. Moreover, over any field the category of numerical motives is abelian semi-simple.
The proof of the latter statement is relatively simple, and can probaly be generalized to other relevant settings. Yet I do not think that there exists any "deep" and general yoga that says that idempotent completions are crucially important (and that is really relevant for motives).
Another observation is that over a field of positive characteristic $$p$$ we don't know whether Voevodsky motives of arbitrary varieties belong to the (smallest strict) triangulated subcategory generated by motives of smooth projectives, but they belong to the subcategory generated by Chow motives (if the characteristic $$p$$ is invertible in the coefficient ring).
• About the "deep yoga": According to the nLab, the karoubi envelope is a special case of the cauchy completion of a category. This completion has some characterisations that are not related specifically to idempotents. ncatlab.org/nlab/show/… Mar 8, 2020 at 2:32
• I have extended my answer. I suspect that the "yoga things" mentioned in the comments is not really actual for motives. Mar 8, 2020 at 6:35
One very general categorical observation is that the idempotent completion functor can be factorised by first passing from the given linear category $$\mathcal{A}$$ to the non-unital ring $$\bigoplus_{X,Y \in \mathcal{A}}\mathcal{A}(X,Y)$$, and then taking the linear category $$\mathrm{Idem}(\bigoplus \mathcal{A})$$ of idempotent elements. The functors $$\bigoplus \dashv \mathrm{Idem}$$ form an adjoint pair, and idempotent-complete linear categories are algebras for the resulting monad.
More generally, for a category $$\mathcal{C}$$ enriched in pointed sets, you first pass to the semigroup $$\bigvee \mathcal{C}:= \coprod_{X,Y \in \mathcal{C}}\mathcal{C}(X,Y)$$ in pointed sets, then take the category of idempotent elements.
On some level, this means that idempotent completion is the universal way to obtain invariants which are really non-unital in nature. You can recover things like idempotent-complete module categories of a unital ring $$R$$ without knowing its unit.
My understanding of the use of Karoubian completion for motives is that one would really like to have an abelian category of pure motives (modulo homological equivalence, say). However, we don't know how to adjoin all kernels and cokernels, and the Karoubian completion is the best we can do.
There is a hope for an abelian category of pure motives that has all the nice properties we want. There are many flavours of motives around (Chow, André, Nori, Voevodsky, ...), and each of them satisfies some but not all of the desired properties. You use whichever one is most convenient for your problem.
(As Mikhail Bondarko pointed out: Chow motives modulo numerical equivalence¹ are semisimple abelian, and this is basically the only way we know how to prove Chow motives form an abelian category. However, this result of Jannsen was only proven in 1992, so I don't think it was the original motivation.)
¹The problem with Chow motives modulo numerical equivalence is that it does not have a cohomological realisation, unless we prove standard conjecture D.
• Yes, Jannsen result is not very old; that is why I did not put it on the first place in my answer.:) A terminological remark: I think that it is better to speak about numerical motives than about Chow motives modulo numerical equivalence. Moreover, Chow motives are certainly useful, and one does not realy want to replace them by an abelian category. Mar 9, 2020 at 12:20 | 1,836 | 7,983 | {"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": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-50 | latest | en | 0.908071 |
https://www.jiskha.com/questions/541859/a-wave-has-a-frequency-of-2-Hz-find-its-period-please-show-work | 1,558,374,420,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256082.54/warc/CC-MAIN-20190520162024-20190520184024-00037.warc.gz | 854,601,451 | 5,625 | PHYSICS !!!!!!!!!!!!!!!!!!!!!!!!!!!!
a wave has a frequency of 2 Hz. find its period ?
1. 👍 0
2. 👎 0
3. 👁 109
1. P = 1/!@#\$%^& or, in words,
Period (in seconds) = 1/Frequency (in Hz)
If there are two cycles per second, each one takes 1/2 second. That is the period.
1. 👍 0
2. 👎 0
posted by drwls
2. The period of a wave, T, is equal to 1/f. So, T = 2 seconds
1. 👍 0
2. 👎 0
posted by Nav
1/2 does not equal 2
1. 👍 0
2. 👎 0
posted by drwls
4. yea....i messed up, i meant 1/2, my bad
1. 👍 0
2. 👎 0
posted by Nav
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8. Physics
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asked by Jena on December 5, 2006 | 756 | 2,514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2019-22 | latest | en | 0.940955 |
https://archives2.twoplustwo.com/archive/index.php/t-25096.html | 1,618,626,514,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038098638.52/warc/CC-MAIN-20210417011815-20210417041815-00161.warc.gz | 215,883,569 | 4,322 | PDA
View Full Version : T-day probability puzzler
irchans
11-29-2002, 07:34 AM
This is one of my favorite puzzles.
You and nine friends are given the opportunity to play a game and win some money. You are all told the rules of the game ahead of time and you can discuss your strategy before the game begins.
All of you will be placed in a line and then ten hats will be placed on your heads. No person will see the color of his own hat. The 10th person will see the color of the 9 hats ahead of him. The 9th person will see 8 hats.... Each person, starting from the 10th in line, will be given a chance to guess the color of his own hat. Every person in the line will hear the guesses of the previous guessers. After all the guesses have been made, every player will be paid one dollar for each correct guess. (i.e. if 5 people guess their hat color correctly, all ten players will recieve \$5.)
1) If you and your friends are told that all the hats will be black or white, what is a good strategy for maximizing the average amount of money won.
2) If you and your friends are told that all the hats will be red, black, or white, what is a good strategy for maximizing the average amount of money won.
PS: If you post an answer, please put the word "answer" or "guess" in the subject line so that you don't spoil the fun for the other readers.
pudley4
11-29-2002, 08:19 AM
The 10th player will choose the color of the hat of the person in front of him. The 9th player will then choose that color. He will also say the answer loudly if it's the same color as the person in front of him, or softly if it's different, thus giving the next player the correct answer also. They are guaranteed 9 correct answers, and are 50/50 on the 10th.
(Still working on #2)
MSchmahl
12-02-2002, 12:08 AM
For #2, the 9th player will answer in English if the hat in front him is black, French if it's red, and Spanish if it is white.
FWIW, I don't think this is supposed to be the correct answer. I assume that each player only knows what the guesses are behind him, and answers involving "tricks" like tone and inflection aren't allowed.
Ed Miller
12-02-2002, 12:57 AM
I know the answer... and I won't spoil it for those that don't by giving it here. But the premise of the question is that you can't give additional information by intonation or language. You can only say one color.
Bozeman
12-02-2002, 03:14 AM
Let 10 say 1 if there are an odd number of white hats he sees, 0 if even.
Now each successor can correctly name his hat by seeing if he sees the same evenness as 10, counting the correctly named hats 9 through n+1.
So you should get 9 + 1/2 on average.
Craig
PS for the 3 hat problem a similar trick should work modulo 3.
BruceZ
12-02-2002, 05:00 AM
I figured out the 2 color problem before I saw the other answers and waited to post until I figured out the 3 color problem. Now I have solutions to both parts:
2 colors:
If there are black and white hats, the 10th person will see an odd number of only 1 color since he sees an odd number of people, and only odd + even = odd. He will agree to yell out the color of the odd hats. Based on this, the guy in front of him will know what color his hat must be since he can see the 8 hats in front of him. Similarly, each person in turn can then figure out the color of his hat since he knows 8 other hats and which colors are even and odd. The first 9 will all be correct, and the 10th guy will be correct half the time. So on average we make \$47.50.
3 colors:
If there are 3 different color hats it's a little tricker. The 10th guy must see either an odd number of just 1 color, OR an odd number of all 3 colors. If there is an odd number of all 3 colors, he will say "white". The 9th guy will know what this means, and he will always see 1 even color and 2 odd colors in this case. He will know he must be the even color. If there is only 1 odd color, then the 9th guy will either see 2 evens and 1 odd and know he must be one of the two evens but not know which one, or else he will see 1 even and 2 odds, but then he isn't sure if he is one of the odd colors, or if we are in the 3 odd case in which case he would be the even color. The 10th guy is aware of the 9th guy's dilemna and knows what he sees, so the 10th guy will yell out the color of the 9th guy's hat in this case, UNLESS it is white since white already means all odd. Instead, in this case if the 9th guy is white, the 10th guy will yell out the one color that the 9th guy cannot possibly be (black or red, the odd color if 2 evens or the even color if 2 odds). Then the 9th guy will know what case we are in and that his hat is white, and the others will know the situation as well since this is the only case where "red" or "black" is called and the 9th guy doesn't call out the same color.
So the 10th guy's algorithm is:
<pre><font class="small">code:</font><hr>
IF (all 3 colors are odd) say "white"
IF (1 color is odd)
IF (9th guy sees 2 odds and 1 even)
IF(9th guy is not white) say 9th guy's color
ELSE say even color
ENDIF
ENDIF
IF (9th guy sees 2 evens and 1 odd)
IF(9th guy is not white) say 9th guy's color
ELSE say odd color
ENDIF
ENDIF
ENDIF
</pre><hr>
9th guy's algorithm is:
<pre><font class="small">code:</font><hr>
IF (10th guy says white) say even color
IF (see 2 odds and 1 even)
IF (10th guy says odd color) say color 10th guy says
ELSE say "white"
ENDIF
ENDIF
IF (see 2 evens and 1 odd)
IF (10th guy says even color) say color 10th guy says
ELSE say "white"
ENDIF
ENDIF
</pre><hr>
Is there a simpler way? Maybe, maybe not. I'm happy with this. The first 9 guys will be right, and the 10th guy will be right 1/3 of the time, so we make \$46.67 on average.
This is like even and odd parity if you know what that is.
BruceZ
12-02-2002, 08:49 AM
I got the cases wrong in the 3 color part. If the 10th guy sees 3 odds, the 9th guy will see 2 odds and 1 even (and he is the even color). If the 10th guy sees 1 odd, the 9th guy could also see 2 odds and 1 even OR 3 EVENS. Here is a picture of what the 9th guy sees:
o o e (3 odd, 9th is the e)
e e e (1 odd, 9th is any)
o o e (1 odd, 9th is one of the o's)
Here's the rule. If the 10th guy sees 1 odd color, he will say this odd color. If he sees all 3 colors odd, then he will say the color of the 9th guy's hat. So in all cases there is an odd number of whatever color he says. Then if the 9th guy sees ooe and hears the even color, he repeats this color which must be his since it must be odd. If the 9th guy sees eee, he also repeats the color the 10th guy says which must be his since it must be odd. If the 9th guy sees ooe and hears one of the odd colors, he says the other odd color from the one the 10th guy said. This color must be his. Now if the 9th guy and the 10th guy say different colors, everyone knows there is 1 odd, and it is the color the 10th guy says, so everyone knows what to do. If they say the same color, then if another player sees 3 evens, he knows the color they said must be odd, and that is his color. If they said the same color and another player sees 2 odds and an even, he knows there must have been 3 odds, and his color is the even one.
Here's the map again with what they say:
o o e (3 odd, 9th is the e, 10 and 9 both say same e)
e e e (1 odd, 9th is any, 10 and 9 both say same)
o o e (1 odd, 9th is one of the o's, 10 and 9 say different o's)
12-02-2002, 09:20 AM
I believe that Craig has an optimal solution! It averages 9 1/2 for the 2 color and 9 and 1/3 for the 3 color.
12-02-2002, 09:54 AM
Hi BruceZ,
I haven't been able to verify that your unique 3 color solution works. I will work on it some more later when I get some time.
Cheers
Irchans
irchans
12-02-2002, 09:55 AM
Bozeman
12-02-2002, 01:44 PM
Sorry I didn't say answer in the subject, I was so happy to come up with it that I forgot,
Craig | 2,255 | 7,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2021-17 | latest | en | 0.961492 |
https://birchlerarroyo.com/housing-planning/how-do-you-measure-house-plans.html | 1,628,127,783,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155268.80/warc/CC-MAIN-20210805000836-20210805030836-00619.warc.gz | 138,766,107 | 20,752 | # How do you measure house plans?
Contents
## How do you measure a floor plan for a house?
To create an accurate floor plan, start by measuring a room:
1. Measure along the baseboard the length of one wall, from one corner of the room to another. …
2. Measure the remaining walls the same way you measured the first. …
3. Measure the room’s doorways and other entries. …
4. Determine the size of the windows.
## What scale are house plans drawn to?
For house plans, you should be using a scale of 1/4 inch to a foot for the floor plan drawings. This is written as 1/4″:1′. This means that every quarter inch you draw on your page represents one foot for the real house as it will be built. So one inch on your drawing would represent four feet of the built house.
## How do you draw house plans?
How to Draw a Floor Plan
1. Choose an area. Determine the area to be drawn. …
2. Take measurements. If the building exists, measure the walls, doors, and pertinent furniture so that the floor plan will be accurate. …
3. Draw walls. …
## How do you convert floor plan measurements?
Use it to reference the examples in the floor plan drawings.
1. Step one – Draw the basic shape of the room. …
2. Step two – Measure from wall to wall. …
3. Step 3 – Draw permanent structures. …
4. Step 4 – Measure doors and windows. …
5. Step 5 – Convert to a scaled drawing. …
6. Step 6 – Convert measurements from hand drawing to scaled drawing.
## Are floor plans length by width?
All dimensions are always presented in feet and inches. … Room dimensions are presented in width by the length. For example, a room that has a dimension of 12′ x 16′ means it’s 12 feet wide (from side to side) by 16 feet long (from top to bottom).
## How thick are walls on a floor plan?
This will vary depending on the construction method of the walls. Interior walls are usually about 4 1/2 inches thick and exterior walls around 6 1/2 inches. If you’re about to draw floor plans for an existing home measure the thickness of the walls at the doors and/or windows.
## How much does it cost to have house plans drawn?
The average cost of drawing and drafting services is approximately \$50/hr. Depending on the complexity of the work, prices can range from as low as \$48/hr to as high as \$60.75/hr.
## How many sets of blueprints are needed to build a house?
It depends on the complexity of the house, but in most cases, you will need roughly 5 to 8 sets. Those who will need a blueprint include you, your contractor, sub-contractors, the building department, your mortgage lender, and/or the building inspector.
IT IS INTERESTING: How do i make a small house plan?
## What is a normal blueprint size?
What is the standard blueprint paper size? Blueprints and house plans will come in several standard sizes. Two of the most common architectural drawing sizes are 18” x 24” and 24” x 36”, but you can also find them in 30” x 42” and 36” x “48” sizes. Large sizes are necessary on bigger and more expensive properties.
## Can I draw my own blueprints?
It is easy to draw blueprints with a few special materials, and hand-drawing allows you the freedom to create your house any way you want it to be. However, there are also some computer blueprint programs available. Choose a program that is easy to use and that will run on your device.
What is the best free floor plan software in 2020?
1. SketchUp.
3. Civil 3D.
5. SmartDraw.
6. Sweet Home 3D.
7. Draft it.
8. Floorplanner.
How it works
1. Draw your floor plan. Draw your floor plan quickly and easily with simple drag & drop drawing tools. …
2. Furnish and Decorate. Furnish your floor plan with materials, furniture, and fixtures from our product library. …
3. Generate your high-quality floor plans.
## What is the most common scale used for drawing houses?
“One Inch”, “One half Inch”, and “One quarter Inch” are common architectural scales, as well as “One eighth Inch” scale. For really large projects, smaller scales are used.
## What is the difference between site plan and floor plan?
Site plans are meant to detail the areas of a plot of land beyond the boundaries of a building. On the other hand, a floor plan details the elements of a building up to the outer edge of the structure.
IT IS INTERESTING: Who prepares construction drawings?
## How do you calculate scale size?
To convert a measurement to a larger measurement simply multiply the real measurement by the scale factor. For example, if the scale factor is 1:8 and the measured length is 4, multiply 4 × 8 = 32 to convert. | 1,069 | 4,551 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2021-31 | latest | en | 0.913635 |
https://math.stackexchange.com/questions/3543838/does-there-exist-a-matrix-a-such-that-ax-infty-x-1/3543862 | 1,657,072,102,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104655865.86/warc/CC-MAIN-20220705235755-20220706025755-00504.warc.gz | 399,314,840 | 64,990 | Does there exist a matrix $A$ such that $\|Ax\|_{\infty} = \|x\|_1$?
$$A \in \mathbb R^{2\times2}, x \in \mathbb R^{2\times1}$$
Does there exist matrix $$A$$: $$\|Ax\|_{\infty} = \|x\|_1$$, how to prove it for any $$x$$, or just find one example matrix $$A$$?
• Is $A$ fixed? Is $x$ fixed? Please use that "for all" and "exists" (in the right order) to give the question the right shape. And show the own attempts to solve the issue. Feb 12, 2020 at 11:22
• Hint: Look at the sets $\{\ x\in\Bbb R ^2\ : \ \|x\|_1=1\ \}$ and $\{\ y\in\Bbb R ^2\ : \ \|y\|_\infty=1\ \}$. Is there any linear transform bringing the one into the other one? Feb 12, 2020 at 11:26
• Does it have to hold for all $x$, or just for one specific $x$? In the latter case, just take $x=0$. Feb 12, 2020 at 11:36
Let $$x=\pmatrix{x_1\\ x_2}$$
We have three possible cases: $$x_1x_2=0 \\ x_1x_2 <0 \\ x_1x_2>0$$
If $$x_1x_2=0$$, then $$|x_1|+|x_2|=|x_1+x_2|=|x_1-x_2|$$.
If $$x_1x_2 <0$$ then $$|x_1|+|x_2|=|x_1-x_2|$$ and $$|x_1-x_2| \geq |x_1+x_2|$$.
If $$x_1x_2 >0$$ then $$|x_1|+|x_2|=|x_1+x_2|$$ and $$|x_1+x_2| \geq |x_1-x_2|$$.
Therefore the matrix $$A=\pmatrix{1 & 1 \\ 1 & -1}$$ which sends $$x=\pmatrix{x_1\\ x_2}$$ to $$\pmatrix{x_1+x_2\\x_1- x_2}$$ satisfies $$\|Ax\|_\infty=\|x\|_1$$ for all $$x \in \mathbb{R}^2$$.
In fact there are $$8$$ different matrices that satisfies the relation, just switch columns, rows or signs of $$A$$. | 623 | 1,424 | {"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": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2022-27 | latest | en | 0.532975 |
https://help.scilab.org/docs/2023.0.0/pt_BR/ismatrix.html | 1,680,305,654,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949689.58/warc/CC-MAIN-20230331210803-20230401000803-00202.warc.gz | 344,945,842 | 4,079 | Scilab Website | Contribute with GitLab | Mailing list archives | ATOMS toolboxes
2023.0.0 - Português
# ismatrix
checks if an object is a non-empty 1D or 2D matrix or array
### Syntax
`tf = ismatrix(x)`
### Arguments
x
scalar,vector, matrix, hypermatrix, list, array of structures or of cells
tf
a boolean
### Description
`ismatrix(x)` returns `%T` (true) if `x` is a scalar (single component), or a vector or a matrix or array with one or 2 dimensions.
An hypermatrix that would be a matrix after reordering its dimensions is considered as a matrix. To test that `x` is not an hypermatrix or an hyperarray -- whatever are the sizes of its dimensions --, use simply `ndims(x)<3`. `ismatrix([])` returns false.
### Examples
```ismatrix(ones(10,5))
ismatrix(1)
ismatrix(["s" "t" "u"; "t" "s" "u"])
ismatrix(rand(2,1,3)) //because of singleton
ismatrix(rand(2,2,3)) //hypermatrix
s = struct();
ismatrix(s) // Empty structure
clear s
s(1,2).a = 3;
s(1,3).b = %z;
ismatrix(s) // Row array of structures
clear s
s(2,1).a = "w";
s(3,1).b = %t;
iscolumn(s) // Column array of structures
clear s
s(1,2).a = -2;
s(3,1).b = %pi;
ismatrix(s) // 2D not-square array of structures
clear s
s(2,1,2).a = 3;
s(1,1,2).b = "test";
ismatrix(s) // 3D array of structures (with a singleton dim)
clear s
s(1,1,2).a = 3;
s(1,1,3).b = "test";
ismatrix(s) // 3D array of structures (with 2 singleton dims)
clear s
s(2,2,2).a = %f;
s(1,2,1).b = list(%e, %s);
ismatrix(s) // 3D (cubic) array of structures```
• isscalar — check if a variable is a scalar.
• isrow — check if a variable is a row
• iscolumn — check if a variable is a column
• issquare — check if a variable is a square matrix
• isempty — verifica se uma variável é uma matriz vazia ou uma lista vazia
• ndims — número de dimensões de um array
### History
Version Description 5.5.0 Function ismatrix introduced.
Report an issue << iscolumn Shape tests isrow >> | 636 | 1,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2023-14 | latest | en | 0.50233 |
http://codeup.cn/problem.php?id=1991 | 1,553,271,619,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912202672.57/warc/CC-MAIN-20190322155929-20190322181929-00541.warc.gz | 49,465,100 | 6,754 | ## 1991: Median
[提交][状态][讨论版][命题人:]
## 题目描述
Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1={11, 12, 13, 14} is 12, and the median of S2={9, 10, 15, 16, 17} is 15. The median of two sequences is defined to be the median of the non-decreasing sequence which contains all the elements of both sequences. For example, the median of S1 and S2 is 13.
Given two increasing sequences of integers, you are asked to find their median.
## 输入
Each input file may contain more than one test case.
Each case occupies 2 lines, each gives the information of a sequence. For each sequence, the first positive integer N (≤1000000) is the size of that sequence. Then N integers follow, separated by a space.
It is guaranteed that all the integers are in the range of long int.
## 输出
For each test case you should output the median of the two given sequences in a line.
## 样例输入
4 11 12 13 14
5 9 10 15 16 17
## 样例输出
13
1）读入两个有序序列的值；
2）计算中间值得位置；
3）定位中间值；
4）输出结果。
[提交][状态] | 328 | 1,045 | {"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} | 2.703125 | 3 | CC-MAIN-2019-13 | latest | en | 0.811844 |
https://hwmadeeasy.com/example-6-2-lenzs-lawdetermine-voltages-v1-and-v2-across-the-2-and-4-resistorsshown-in-fig-6-4-the-loop-is-located-in-the-x-y-plane-its-areais-4-m2-the-magnetic-flux-densit/ | 1,611,526,564,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703557462.87/warc/CC-MAIN-20210124204052-20210124234052-00432.warc.gz | 392,102,313 | 36,791 | # Example 6-2: Lenz’s LawDetermine voltages V1 and V2 across the 2 ‘ and 4 ‘ resistorsshown in Fig. 6-4. The loop is located in the x–y plane, its areais 4 m2, the magnetic flux density is B = −zˆ0.3t (T), and theinternal resistance of the wire may be ignored.Solution: The flux flowing through the loop is% =”SB· ds =”S(−zˆ0.3t)· zˆ ds= −0.3t × 4 = −1.2t (Wb),and the corresponding transformer emf isV tremf = −d%dt = 1.2 (V)
This content is for Premium members only. | 169 | 469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-04 | latest | en | 0.855031 |
http://ch.whu.edu.cn/article/id/348 | 1,709,007,800,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474670.19/warc/CC-MAIN-20240227021813-20240227051813-00798.warc.gz | 10,458,072 | 38,756 | 引用本文: 章传银, 党亚民, 柯宝贵, 王伟. 高精度海岸带重力似大地水准面的若干问题讨论[J]. 武汉大学学报 ( 信息科学版), 2012, 37(10): 1150-1154.
ZHANG Chuanyin, DANG Yamin, KE Baogui, WANG Wei. Discussion of some Lssues in the Coastal Precise Quasi-geoid Determination[J]. Geomatics and Information Science of Wuhan University, 2012, 37(10): 1150-1154.
Citation: ZHANG Chuanyin, DANG Yamin, KE Baogui, WANG Wei. Discussion of some Lssues in the Coastal Precise Quasi-geoid Determination[J]. Geomatics and Information Science of Wuhan University, 2012, 37(10): 1150-1154.
## Discussion of some Lssues in the Coastal Precise Quasi-geoid Determination
• 摘要: 针对海岸带多源重力数据和地形特点,通过理论分析和试算,对若干影响cm级似大地水准面确定的关键问题进行了剖析,得出了一些有益的结论。我国海岸带Molodensky一阶项对高程异常的贡献在10～30cm,需在Molodensky框架中精化重力似大地水准面;精细处理地形影响是提升多源重力场数据处理水平的重要途径;地球外空间不同高度、任意类型重力场参数的地形影响、地形补偿和地形Helmert凝聚算法可以统一;重力场数据处理中大地测量基准不一致的影响会随数据处理算法的不同而变化,在多源重力数据处理时此类影响易变得不可预测和控制;将地形Helmert凝聚理论引入Molodensky框架,可以解决以其他重力场参数(如扰动重力、垂线偏差等)为边界条件的似大地水准面精化问题。
Abstract: For the coastal zone,the multi-source gravity data co-exist and terrain features are diversity,through theoretical analysis and a spreadsheet,some issues in quasi-geoid determination with cm-level precision were discussed,and draw some useful conclusions.Taking into account Molodensky first-order contribution to the height anomaly about 10—30 cm in China's coastal region,refined gravity quasi-geoid should be in the Molodensky frame.Fine topographic effect processing is a fundamental way to handle multi-source gravity data.The algorithm for topography effect,mass compensation and Helmert condensation for any height,any type gravity field parameters in earth outer space can be unified.The effect of the inconsistent geodetic reference to the gravity data processing vary with the data processing algorithm.in multi-source gravity data processing procedure such influence can be unpredictable.The terrain Helmert condensation can be introduced to Molodensky framework with any type of gravity data.
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分享至好友和朋友圈 | 705 | 1,986 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-10 | latest | en | 0.443674 |
https://hackastat.eu/en/learn-a-stat-win-share-en/ | 1,669,597,613,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710462.59/warc/CC-MAIN-20221128002256-20221128032256-00228.warc.gz | 335,248,468 | 28,905 | # Learn a Stat: Win Share
Welcome back to Hack a Stat! It’s a new chapter of Learn a Stat: let’s talk about the Win Share. Not widely used, but still interesting.
ITALIAN VERSION
### Introduction
The Win Share is one of the individual statistics that most of all makes the haters of stats angry: in a team sport, why do you want to divide the merits of a victory among the players?
Actually, this statistic tries to estimate a quantity that identifies how many wins the player has brought with his contribution to his team.
Pay attention: a possible wrong belief on this statistic is linked to the contributions in the crucial moments. Actually the Win Share is based on the seasonal numbers; scoring a buzzer-beater or blocking the opponent’s buzzer-beater does not lead to higher Win Share values. In other words, the importance of a contribution within a game is not weighed differently.
The theories behind this statistic arise from Bill James’s mind who applied them to baseball: Justin Kubatko has reviewed part of the process and adapted it to basketball. The calculations and explanations in this chapter, therefore, are taken from his study.
### Definition and starting data
As already said, the Win Share is the statistic that estimates the number of victories brought to the team by a player based on his contribution. By its nature, it is a statistic based on the season totals.
One of its peculiarities is that adding all the players’ Win Share of a team, a value very close to the total team wins is obtained; being an estimate, however, the exact number will never be reached, but as long as the difference remains around 2/3 units, the calculation can be considered successful.
The Win Share is made up of two terms: the Offensive Win Share and the Defensive Win Share. As the names suggest, the first bases its calculation on the player’s offensive contributions, while the second bases on his defensive contributions. The sum of the two gives us the Win Share.
In addition, there is also the Win Share per 40 Minutes. This statistic is the distribution of the Win Share on the 40 minutes (or 48 for NBA). This value is very useful when you want to make a comparison among players belonging to different teams: in this case, it is convenient to have a relative and not absolute value (such as the Win Share).
To calculate all these terms we will need not only the player’s contributions but of course also those of the team and the entire league.
#### Offensive Win Share
• Points Production [PtsProd] or [PtsGen] obtained from the Individual Offensive Rating calculation;
• Total Possession Played [Poss] or [PossTot] obtained from the Individual Offensive Rating calculation;
• League Individual Points per Possession [LgPtsPoss];
• League Points per game [LgPtsPG];
• Team Pace [TmPace];
• League Pace [LgPace];
#### Defensive Win Share
• Individual Defensive Rating [DefRtg];
• Minutes played [MP];
• Team minutes played [TeMP];
• Opponent’s possessions [OppPoss];
• League Individual Points per Possession [LgPtsPoss];
• League Points per game [LgPtsPG];
• Team Pace [TmPace];
• League Pace [LgPace];
#### Win Share per 40 Minutes
• Win Share [WS];
• Games played [GP];
• Minute played [MP];
### Formulas and calculation
#### Offensive Win Share
First of all, it is necessary to calculate the League value required for this calculation, that is the League Individual Points per Possession:
This value is simply the average of the League (all the players) of the points per possession scored; even simpler: it is the average of all Offensive Ratings divided by 100.
Once this is done, the next step is to calculate the Marginal Individual Offensive Contribution:
With this term, the player’s points production is decurted with the average league value considering the same number of possessions (in fact, the League value is multiplied by the player’s possessions). This term can give a negative result, meaning that the player’s contribution is less than the League performances; the player’s performance is counterproductive for his team.
After this, we move on to the calculation of the Marginal Points for Victory, value relative to the team:
This calculation sets the reference value to calculate the Offensive Win Share. League points per game are nothing more than the average of the points scored by each team.
In fact, it is already possible to find the Offensive Win Share with the following formula:
The player’s production is then compared with that of League to find out how important his contribution to victory is. Let’s move on to the defensive part.
#### Defensive Win Share
As in the previous case, two terms are calculated: the individual one and the team one.
The first is the Marginal Individual Defensive Contribution:
The purpose of this formula is to find the difference between the points allowed by the player with respect to the league average; to find this average, the first step is to identify the defensive possessions played by the player; then this result is multiplied by the difference between the League Individual Points per Possession and the points allowed by the player for each possession (the Defensive Rating divides by 100). The greater the result, the better the contribution. Therefore, it assumes a meaning contrary to the Defensive Rating, which is better when the value is low.
The second term is identical to that of the Offensive Win Share(Marginal Points for Victory).
It is already time to find the Defensive Win Share similarly to what was done for the offensive part.
The Win Share will, therefore, be obtainable with the sum of the two terms found.
Finally, to find the Win Share per 40 Minutes (48 in the NBA), the following formula has to be used:
The first division allows us to find the Win Share per game, while the second part redistributes it on the 40 minutes. Obviously the minutes will be 48 in case of NBA players.
### How to read and analyze
Analyzing the formulas it is easy to understand how the Win Share is dependent on individual Offensive and Defensive Rating. The data used are in fact part of the calculation process of Oliver’s two statistics. This involves a direct dependence between the two types of statistics: good Offensive and Defensive Rating correspond to equally good Win Shares. The peculiarity is that the Win Share compares the values with the League, thus avoiding having to compare the player’s numbers with the League averages.
Let’s move on to an example to present the data used and the calculations. We take two teams from the Italian championship (season 2017/2018) and calculate the Win Share: Trento and Reggio Emilia.
First of all, the stating data:
We take, as mentioned, the season totals and not the averages, which would give incorrect results.
The first step is to calculate the Offensive Win Share:
As can be understood from the formulas, the marginal points for victory assume the same value for all members of the team; the Marginal Individual Offensive Contribution instead assumes a greater value for the players who obviously have a higher point production. As we know, assists and offensive rebounds are considered into the Point Production, therefore with that value, all the contributions are taken into consideration and not just the points scored.
Let’s move on the Defensive part:
The DWS depends on the Defensive Rating. This stat considers not only the blocked shots, the defensive rebounds, and the steals, but also all those contributions not normally found in a box score (contested shots, forced turnovers).
Next step is to sum the two values:
As can be seen, the sum of all the Win Shares is very similar to the victories achieved by the two teams in the regular season. Major discrepancies are noticed in teams with very few wins.
To compare the players’ Win Shares in the two tables, however, we need to look also at games played and minutes played: for example, Sutton and Reynolds have a very similar Win Share, but the second has played 5 games less.
In fact, comparing the Win Share per 40 Minutes it is noted that Reynolds has a better value than Sutton: that is to say that Jalen’s contribution per minute is higher than Dominique’s one, a result not found using the Win Share.
The Win Share is a statistic that wants to summarize in a single number the contribution to the victory: like all statistics of this type it, therefore, has limits. With a simple consultation, it is not possible to know if the added victories coming from the offensive of the defensive contributions or even a balanced mix of both.
Before concluding a quick note. The Win Share provides reliable values after many games played: the sample out of 30/34 games (the number of games played on average in Europe) is certainly reliable, but never as reliable as the NBA one based on 82 games. This fact is absolutely to be considered, in order to avoid superficial analyzes.
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# In1736eulersolvedthelongstandingquestionbymappingthepr
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Unformatted text preview: walk through the town using every bridge j ust once, and returning home at the end. In 1736, Euler solved the longstanding question by mapping the problem to a graph problem, which we examine next. 2. Review the terminology for graphs Vertices (a.k.a. nodes) Edges (Directed vs undirected and labeled vs not labeled) Paths (and shortest paths for labeled graphs) 3. Reduce the map of the town shown above to the undirected, unlabeled graph below. Introduce the concept of node degree (number of edges incident to a vertex). Introduce the concept of a connected graph (a graph where there is a path from any node to any other node i.e., one can visit every node in the graph independent of the starting node). Introduce the conce...
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## This note was uploaded on 02/10/2014 for the course CS 109 taught by Professor Azerbestavros during the Spring '13 term at BU.
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